CHAPTER 1
d1
PROBLEM 1.1
d2
125 kN B
C
A
Two solid cylindrical rods AB and BC are
welded together at B and loaded as shown.
Knowing that d1 30 mm and d 2 50 mm,
find the average normal stress at the
midsection of (a) rod AB, (b) rod BC.
60 kN
125 kN
0.9 m
1.2 m
SOLUTION
(a)
Rod AB:
Force:
P 60 103 N tension
Area:
A
Normal stress:
(b)
AB
4
d12
4
(30 103 ) 2 706.86 106 m 2
P
60 103
84.882 106 Pa
A 706.86 106
AB 84.9 MPa
Rod BC:
Force:
P 60 103 (2)(125 103 ) 190 103 N
Area:
A
Normal stress:
BC
4
d 22
4
(50 103 )2 1.96350 103 m 2
P
190 103
96.766 106 Pa
A 1.96350 103
BC 96.8 MPa
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3
d1
PROBLEM 1.2
d2
125 kN B
C
A
Two solid cylindrical rods AB and BC are
welded together at B and loaded as shown.
Knowing that the average normal stress must
not exceed 150 MPa in either rod, determine
the smallest allowable values of the diameters
d1 and d2.
60 kN
125 kN
0.9 m
1.2 m
SOLUTION
(a)
Rod AB:
Force:
P 60 103 N
Stress:
AB 150 106 Pa
2
A
Area:
AB
4
d1
4
P
P
A
A
AB
d12
d12
P
AB
4P
AB
(4)(60 103 )
509.30 106 m 2
(150 106 )
d1 22.568 103 m
(b)
d1 22.6 mm
Rod BC:
Force:
Stress:
Area:
P 60 103 (2)(125 103 ) 190 103 N
BC 150 106 Pa
2
A
BC
d2
4
P
4P
A d 22
d 22
4P
BC
(4)(190 103 )
1.61277 103 m 2
(150 106 )
d 2 40.159 103 m
d 2 40.2 mm
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4
PROBLEM 1.3
A
Two solid cylindrical rods AB and BC are welded together at B and loaded as
shown. Knowing that P = 10 kips, find the average normal stress at the
midsection of (a) rod AB, (b) rod BC.
30 in.
1.25 in.
B
12 kips
25 in.
0.75 in.
C
P
SOLUTION
(a)
Rod AB:
P 12 10 22 kips
A
AB
(b)
d12
(1.25) 2 1.22718 in 2
4
4
P
22
17.927 ksi
A 1.22718
AB 17.93 ksi
Rod BC:
P 10 kips
AB
d 22
(0.75)2 0.44179 in 2
4
4
P
10
22.635 ksi
A 0.44179
A
AB 22.6 ksi
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5
PROBLEM 1.4
A
Two solid cylindrical rods AB and BC are welded together at B and loaded as
shown. Determine the magnitude of the force P for which the tensile stresses in
rods AB and BC are equal.
30 in.
1.25 in.
B
12 kips
25 in.
0.75 in.
C
P
SOLUTION
(a)
Rod AB:
P P 12 kips
A
d2
4
4
(1.25 in.)2
A 1.22718 in 2
AB
(b)
P 12 kips
1.22718 in 2
Rod BC:
P P
A
4
d2
4
(0.75 in.)2
A 0.44179 in 2
BC
P
0.44179 in 2
AB BC
P 12 kips
P
2
1.22718 in
0.44179 in 2
5.3015 0.78539 P
P 6.75 kips
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6
1200 N
PROBLEM 1.5
A strain gage located at C on the surface of bone AB indicates that the average normal stress
in the bone is 3.80 MPa when the bone is subjected to two 1200-N forces as shown.
Assuming the cross section of the bone at C to be annular and knowing that its outer diameter
is 25 mm, determine the inner diameter of the bone’s cross section at C.
A
C
B
1200 N
SOLUTION
Geometry:
A
4
P
P
A
A
(d12 d 22 )
d 22 d12
4A
d12
d 22 (25 103 )2
4P
(4)(1200)
(3.80 106 )
222.92 106 m 2
d 2 14.93 103 m
d 2 14.93 mm
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7
PROBLEM 1.6
A
a
15 mm
B
100 m
b
Two brass rods AB and BC, each of uniform diameter, will be brazed together
at B to form a nonuniform rod of total length 100 m, which will be suspended
from a support at A as shown. Knowing that the density of brass is 8470 kg/m3,
determine (a) the length of rod AB for which the maximum normal stress in
ABC is minimum, (b) the corresponding value of the maximum normal stress.
10 mm
C
SOLUTION
Areas:
AAB
ABC
4
4
(15 mm) 2 176.715 mm 2 176.715 106 m 2
(10 mm)2 78.54 mm 2 78.54 106 m 2
b 100 a
From geometry,
Weights:
WAB g AAB AB (8470)(9.81)(176.715 106 ) a 14.683 a
WBC g ABC BC (8470)(9.81)(78.54 106 )(100 a) 652.59 6.526 a
Normal stresses:
At A,
PA WAB WBC 652.59 8.157a
A
At B,
(a)
PA
3.6930 106 46.160 103a
AAB
PB WBC 652.59 6.526a
B
(1)
(2)
PB
8.3090 106 83.090 103a
ABC
Length of rod AB. The maximum stress in ABC is minimum when A B or
4.6160 106 129.25 103a 0
a 35.71 m
(b)
AB a 35.7 m
Maximum normal stress.
A 3.6930 106 (46.160 103 )(35.71)
B 8.3090 106 (83.090 103 )(35.71)
A B 5.34 106 Pa
5.34 MPa
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8
PROBLEM 1.7
0.4 m
C
0.25 m
0.2 m
B
Each of the four vertical links has an 8 36-mm uniform rectangular
cross section and each of the four pins has a 16-mm diameter. Determine
the maximum value of the average normal stress in the links connecting
(a) points B and D, (b) points C and E.
E
20 kN
D
A
SOLUTION
Use bar ABC as a free body.
M C 0 :
(0.040) FBD (0.025 0.040)(20 103 ) 0
FBD 32.5 103 N
Link BD is in tension.
3
M B 0 : (0.040) FCE (0.025)(20 10 ) 0
FCE 12.5 103 N
Link CE is in compression.
Net area of one link for tension (0.008)(0.036 0.016) 160 106 m 2
For two parallel links,
(a)
BD
A net 320 106 m 2
FBD
32.5 103
101.563 106
6
Anet
320 10
BD 101.6 MPa
Area for one link in compression (0.008)(0.036) 288 106 m 2
For two parallel links,
(b)
CE
A 576 106 m 2
FCE
12.5 103
21.701 106
6
A
576 10
CE 21.7 MPa
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9
PROBLEM 1.8
B
2 in.
Link AC has a uniform rectangular cross section
12 in.
in. thick and 1 in. wide.
Determine the normal stress in the central portion of the link.
120 lb
4 in.
30⬚
1
8
120 lb
A
C
10 in.
8 in.
SOLUTION
Use the plate together with two pulleys as a free body. Note that the cable tension causes at 1200 lb-in.
clockwise couple to act on the body.
M B 0: (12 4)( FAC cos 30) (10)( FAC sin 30) 1200 lb 0
FAC
1200 lb
135.500 lb
16 cos 30 10 sin 30
Area of link AC:
Stress in link AC:
1
in. 0.125 in 2
8
F
135.50
AC
1084 psi 1.084 ksi
A
0.125
A 1 in.
AC
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10
PROBLEM 1.9
0.100 m
E
P
P
P
D
A
0.150 m
B
Three forces, each of magnitude P 4 kN, are applied to the mechanism
shown. Determine the cross-sectional area of the uniform portion of rod
BE for which the normal stress in that portion is 100 MPa.
C
0.300 m
0.250 m
SOLUTION
Draw free body diagrams of AC and CD.
Free Body CD:
M D 0: 0.150P 0.250C 0
C 0.6 P
Free Body AC:
Required area of BE:
M A 0: 0.150 FBE 0.350 P 0.450 P 0.450C 0
FBE
1.07
P 7.1333 P (7.133)(4 kN) 28.533 kN
0.150
BE
FBE
ABE
ABE
FBE
BE
28.533 103
285.33 106 m 2
100 106
ABE 285 mm 2
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11
4 kips
6
C
in.
u
B
1
308
Link BD consists of a single bar 1 in. wide and
1
in. thick. Knowing that each pin has a 83 -in.
2
diameter, determine the maximum value of the
average normal stress in link BD if (a) = 0,
(b) = 90.
.
2 in
A
PROBLEM 1.10
D
SOLUTION
Use bar ABC as a free body.
(a)
0.
M A 0: (18 sin 30)(4) (12 cos30) FBD 0
FBD 3.4641 kips (tension)
Area for tension loading:
Stress:
(b)
3 1
A (b d )t 1 0.31250 in 2
8 2
F
3.4641 kips
BD
A
0.31250 in 2
11.09 ksi
90.
M A 0: (18 cos30)(4) (12 cos 30) FBD 0
FBD 6 kips i.e. compression.
Area for compression loading:
Stress:
1
A bt (1) 0.5 in 2
2
F
6 kips
BD
A
0.5 in 2
12.00 ksi
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12
B
D
PROBLEM 1.11
F
12 ft
H
A
C
9 ft
E
9 ft
80 kips
For the Pratt bridge truss and loading shown, determine the
average normal stress in member BE, knowing that the crosssectional area of that member is 5.87 in2.
G
9 ft
80 kips
9 ft
80 kips
SOLUTION
Use entire truss as free body.
M H 0: (9)(80) (18)(80) (27)(80) 36 Ay 0
Ay 120 kips
Use portion of truss to the left of a section cutting members
BD, BE, and CE.
Fy 0: 120 80
BE
12
FBE 0
15
FBE 50 kips
FBE
50 kips
A
5.87 in 2
BE 8.52 ksi
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13
45 in.
A
B
PROBLEM 1.12
30 in.
C
480 lb
4 in.
4 in.
40 in.
D
15 in.
E
30 in.
The frame shown consists of four wooden members, ABC,
DEF, BE, and CF. Knowing that each member has a 2 4-in.
rectangular cross section and that each pin has a 12 -in.
diameter, determine the maximum value of the average
normal stress (a) in member BE, (b) in member CF.
F
SOLUTION
Add support reactions to figure as shown.
Using entire frame as free body,
M A 0: 40Dx (45 30)(480) 0
Dx 900 lb
Use member DEF as free body.
Reaction at D must be parallel to FBE and FCF .
Dy
4
Dx 1200 lb
3
4
M F 0: (30) FBE (30 15) DY 0
5
FBE 2250 lb
4
M E 0: (30) FCE (15) DY 0
5
FCE 750 lb
Stress in compression member BE:
A 2 in. 4 in. 8 in 2
Area:
(a)
BE
FBE
2250
A
8
BE 281 psi
Minimum section area occurs at pin.
Amin (2)(4.0 0.5) 7.0 in 2
Stress in tension member CF:
(b)
CF
FCF
750
Amin
7.0
CF 107.1 psi
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14
PROBLEM 1.13
Dimensions in mm
1150
D
100
C
G
A
F
850
B
250
E
500
450
675
825
An aircraft tow bar is positioned by means of a single
hydraulic cylinder connected by a 25-mm-diameter steel
rod to two identical arm-and-wheel units DEF. The mass
of the entire tow bar is 200 kg, and its center of gravity
is located at G. For the position shown, determine the
normal stress in the rod.
SOLUTION
FREE BODY – ENTIRE TOW BAR:
W (200 kg)(9.81 m/s 2 ) 1962.00 N
M A 0: 850R 1150(1962.00 N) 0
R 2654.5 N
FREE BODY – BOTH ARM & WHEEL UNITS:
tan
100
675
8.4270
M E 0: ( FCD cos )(550) R(500) 0
FCD
500
(2654.5 N)
550 cos 8.4270
2439.5 N (comp.)
CD
2439.5 N
FCD
ACD
(0.0125 m)2
4.9697 106 Pa
CD 4.97 MPa
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15
150 mm
300 mm
A
D
F
150 mm
PROBLEM 1.14
Two hydraulic cylinders are used to control the position
of the robotic arm ABC. Knowing that the control rods
attached at A and D each have a 20-mm diameter and
happen to be parallel in the position shown, determine the
average normal stress in (a) member AE, (b) member DG.
C
B
400 mm
E
800 N
600 mm
G
200 mm
SOLUTION
Use member ABC as free body.
M B 0: (0.150)
4
FAE (0.600)(800) 0
5
FAE 4 103 N
Area of rod in member AE is
Stress in rod AE:
A
AE
4
d2
4
(20 103 ) 2 314.16 106 m 2
FAE
4 103
12.7324 106 Pa
6
A
314.16 10
(a)
AE 12.73 MPa
Use combined members ABC and BFD as free body.
4
4
M F 0: (0.150) FAE (0.200) FDG (1.050 0.350)(800) 0
5
5
FDG 1500 N
Area of rod DG:
Stress in rod DG:
A
4
d2
DG
4
(20 103 ) 2 314.16 106 m 2
FDG
1500
4.7746 106 Pa
A
3.1416 106
(b)
DG 4.77 MPa
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16
PROBLEM 1.15
Determine the diameter of the largest circular hole that can be punched into a sheet of polystyrene 6 mm
thick, knowing that the force exerted by the punch is 45 kN and that a 55-MPa average shearing stress is
required to cause the material to fail.
SOLUTION
For cylindrical failure surface:
A dt
Shearing stress:
Therefore,
Finally,
P
P
or
A
A
P
dt
d
P
t
45 103 N
(0.006 m)(55 106 Pa)
43.406 103 m
d 43.4 mm
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17
5
8
P'
1 in.
2 in.
PROBLEM 1.16
in.
5
8
in.
2 in.
1 in.
9 in.
Two wooden planks, each 12 in. thick and 9 in.
wide, are joined by the dry mortise joint shown.
Knowing that the wood used shears off along its
grain when the average shearing stress reaches
1.20 ksi, determine the magnitude P of the axial
load that will cause the joint to fail.
P
SOLUTION
Six areas must be sheared off when the joint fails. Each of these areas has dimensions
5
8
in.
1
2
in., its area
being
A
5 1
5 2
in 0.3125 in 2
8 2 16
At failure, the force carried by each area is
F A (1.20 ksi)(0.3125 in 2 ) 0.375 kips
Since there are six failure areas,
P 6 F (6)(0.375)
P 2.25 kips
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18
PROBLEM 1.17
0.6 in.
P
P'
Steel
3 in.
Wood
When the force P reached 1600 lb, the wooden specimen shown failed
in shear along the surface indicated by the dashed line. Determine the
average shearing stress along that surface at the time of failure.
SOLUTION
Area being sheared:
A 3 in. 0.6 in. 1.8 in 2
Force:
P 1600 lb
Shearing stress:
P 1600 lb
8.8889 102 psi
2
A 1.8 in
889 psi
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19
PROBLEM 1.18
40 mm
10 mm
8 mm
12 mm
A load P is applied to a steel rod supported as shown by an aluminum
plate into which a 12-mm-diameter hole has been drilled. Knowing that
the shearing stress must not exceed 180 MPa in the steel rod and 70 MPa
in the aluminum plate, determine the largest load P that can be applied to
the rod.
P
SOLUTION
A1 dt (0.012 m)(0.010 m)
For steel:
376.99 106 m 2
1
P
P A11 (376.99 106 m 2 )(180 106 Pa)
A
67.858 103 N
A2 dt (0.040 m)(0.008 m) 1.00531 103 m 2
For aluminum:
2
P
P A2 2 (1.00531 103 m 2 )(70 106 Pa) 70.372 103 N
A2
P 67.9 kN
Limiting value of P is the smaller value, so
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20
PROBLEM 1.19
The axial force in the column supporting the timber beam shown is
P 20 kips. Determine the smallest allowable length L of the bearing
plate if the bearing stress in the timber is not to exceed 400 psi.
L
6 in.
P
SOLUTION
Bearing area: Ab Lw
b
L
P
P
Ab
Lw
20 103 lb
P
8.33 in.
b w (400 psi)(6 in.)
L 8.33 in.
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21
d
PROBLEM 1.20
Three wooden planks are fastened together by a series of bolts to form
a column. The diameter of each bolt is 12 mm and the inner diameter
of each washer is 16 mm, which is slightly larger than the diameter of
the holes in the planks. Determine the smallest allowable outer
diameter d of the washers, knowing that the average normal stress in
the bolts is 36 MPa and that the bearing stress between the washers
and the planks must not exceed 8.5 MPa.
12 mm
SOLUTION
Bolt:
ABolt
Tensile force in bolt:
d2
4
(0.012 m)2
4
1.13097 104 m 2
P
P A
A
(36 106 Pa)(1.13097 104 m 2 )
4.0715 103 N
Bearing area for washer:
Aw
and
Aw
4
d
2
o
di2
P
BRG
Therefore, equating the two expressions for Aw gives
4
d
2
o
di2
d o2
d o2
P
BRG
4P
BRG
di2
4 (4.0715 103 N)
(0.016 m) 2
(8.5 106 Pa)
d o2 8.6588 104 m 2
d o 29.426 103 m
d o 29.4 mm
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22
PROBLEM 1.21
P 5 40 kN
120 mm
b
A 40-kN axial load is applied to a short wooden post that is
supported by a concrete footing resting on undisturbed soil.
Determine (a) the maximum bearing stress on the concrete
footing, (b) the size of the footing for which the average bearing
stress in the soil is 145 kPa.
100 mm
b
SOLUTION
(a)
Bearing stress on concrete footing.
P 40 kN 40 103 N
A (100)(120) 12 103 mm 2 12 103 m 2
(b)
P
40 103
3.3333 106 Pa
A 12 103
Footing area. P 40 103 N
P
A
3.33 MPa
145 kPa 45 103 Pa
A
P
40 103
0.27586 m 2
3
145 10
Since the area is square, A b 2
b
A
0.27586 0.525 m
b 525 mm
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23
PROBLEM 1.22
a
P
a
An axial load P is supported by a short W8 40 column of crosssectional area A 11.7 in 2 and is distributed to a concrete foundation
by a square plate as shown. Knowing that the average normal stress in
the column must not exceed 30 ksi and that the bearing stress on the
concrete foundation must not exceed 3.0 ksi, determine the side a of
the plate that will provide the most economical and safe design.
SOLUTION
For the column,
P
or
A
P A (30)(11.7) 351 kips
For the a a plate, 3.0 ksi
A
P
351
117 in 2
3.0
Since the plate is square, A a 2
a
A 117
a 10.82 in.
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24
PROBLEM 1.23
Link AB, of width b = 2 in. and thickness t = 14 in., is used to support the end of a
horizontal beam. Knowing that the average normal stress in the link is 20 ksi and
that the average shearing stress in each of the two pins is 12 ksi, determine (a) the
diameter d of the pins, (b) the average bearing stress in the link.
A
d
b
t
B
d
SOLUTION
Rod AB is in compression.
A bt
1
in.
4
1
P A (20)(2) 10 kips
4
P
P
AP
Pin:
AP
and
(a)
where b 2 in. and t
d
4 AP
4P
P
4
d2
(4)(10)
1.03006 in.
(12)
d 1.030 in.
(b)
b
P
10
38.833 ksi
(1.03006)(0.25)
dt
b 38.8 ksi
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25
PROBLEM 1.24
P
Determine the largest load P which may be applied at A when
60°, knowing that the average shearing stress in the 10-mmdiameter pin at B must not exceed 120 MPa and that the average
bearing stress in member AB and in the bracket at B must not
exceed 90 MPa.
A
16 mm
750 mm
750 mm
50 mm
B
C
12 mm
SOLUTION
Geometry: Triangle ABC is an isoseles triangle with angles shown here.
Use joint A as a free body.
Law of sines applied to force triangle:
P
FAB
FAC
sin 30 sin 120 sin 30
P
FAB sin 30
0.57735FAB
sin 120
P
FAC sin 30
FAC
sin 30
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26
PROBLEM 1.24 (Continued)
If shearing stress in pin at B is critical,
A
4
d2
4
(0.010) 2 78.54 106 m 2
FAB 2 A (2)(78.54 106 )(120 106 ) 18.850 103 N
If bearing stress in member AB at bracket at A is critical,
Ab td (0.016)(0.010) 160 106 m 2
FAB Ab b (160 106 )(90 106 ) 14.40 103 N
If bearing stress in the bracket at B is critical,
Ab 2td (2)(0.012)(0.010) 240 106 m 2
FAB Ab b (240 106 )(90 106 ) 21.6 103 N
Allowable FAB is the smallest, i.e., 14.40 103 N
Then from statics,
Pallow (0.57735)(14.40 103 )
8.31 103 N
8.31 kN
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27
PROBLEM 1.25
P
A
16 mm
750 mm
750 mm
50 mm
B
Knowing that 40° and P 9 kN, determine (a) the smallest
allowable diameter of the pin at B if the average shearing stress in
the pin is not to exceed 120 MPa, (b) the corresponding average
bearing stress in member AB at B, (c) the corresponding average
bearing stress in each of the support brackets at B.
C
12 mm
SOLUTION
Geometry: Triangle ABC is an isoseles triangle with angles shown here.
Use joint A as a free body.
Law of sines applied to force triangle:
P
FAB
FAC
sin 20 sin110 sin 50
P sin110
FAB
sin 20
(9)sin110
24.727 kN
sin 20
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28
PROBLEM 1.25 (Continued)
(a)
Allowable pin diameter.
d2
FAB
F
2FAB
AB 2
where FAB 24.727 103 N
2
2 AP
24d
d
2 FAB
(2)(24.727 103 )
131.181 106 m 2
(120 106 )
d 11.4534 103 m
(b)
11.45 mm
Bearing stress in AB at A.
Ab td (0.016)(11.4534 103 ) 183.254 106 m 2
b
(c)
FAB
24.727 103
134.933 106 Pa
Ab
183.254 106
134.9 MPa
Bearing stress in support brackets at B.
A td (0.012)(11.4534 103 ) 137.441 106 m 2
b
1
2
FAB
A
(0.5)(24.727 103 )
89.955 106 Pa
137.441 106
90.0 MPa
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29
PROBLEM 1.26
175 mm
100 mm
D
B
208
C
u
E
200 mm
P
A
The hydraulic cylinder CF, which partially controls the position of rod
DE, has been locked in the position shown. Member BD is 15 mm
thick and is connected at C to the vertical rod by a 9-mm-diameter
bolt. Knowing that P 2 kN and 75, determine (a) the average
shearing stress in the bolt, (b) the bearing stress at C in member BD.
F
45 mm
SOLUTION
Free Body: Member BD.
40
9
FAB (100 cos 20) FAB (100 sin 20)
41
4
(2 kN) cos 75(175sin 20) (2 kN)sin 75(175cos 20) 0
M c 0:
100
FAB (40 cos 20 9sin 20) (2 kN)(175)sin(75 20)
41
FAB 4.1424 kN
Fx 0: C x
Fy 0: C y
9
(4.1424 kN) (2 kN) cos 75 0
41
C x 0.39167 kN
40
(4.1424 kN) (2 kN)sin 75 0
41
C y 5.9732 kN
C 5.9860 kN
(a)
ave
(b)
b
86.2°
C
5.9860 103 N
94.1 106 Pa 94.1 MPa
2
A
(0.0045 m)
C
5.9860 103 N
44.3 106 Pa 44.3 MPa
(0.015 m)(0.009 m)
td
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30
PROBLEM 1.27
0.4 m
For the assembly and loading of Prob. 1.7, determine (a) the average
shearing stress in the pin at B, (b) the average bearing stress at B in
member BD, (c) the average bearing stress at B in member ABC,
knowing that this member has a 10 50-mm uniform rectangular cross
section.
C
0.25 m
0.2 m
B
E
20 kN
PROBLEM 1.7 Each of the four vertical links has an 8 36-mm
uniform rectangular cross section and each of the four pins has a 16-mm
diameter. Determine the maximum value of the average normal stress in
the links connecting (a) points B and D, (b) points C and E.
D
A
SOLUTION
Use bar ABC as a free body.
M C 0 : (0.040) FBD (0.025 0.040)(20 103 ) 0
FBD 32.5 103 N
(a)
Shear pin at B.
where
A
(b)
Bearing: link BD.
Bearing in ABC at B.
4
d2
4
(0.016) 2 201.06 10 6 m 2
32.5 103
80.822 106 Pa
(2)(201.06 106 )
80.8 MPa
A dt (0.016)(0.008) 128 106 m 2
b
(c)
FBD
for double shear
2A
1
2
FBD
A
(0.5)(32.5 103 )
126.95 106 Pa
6
128 10
b 127.0 MPa
A dt (0.016)(0.010) 160 106 m 2
b
FBD
32.5 103
203.12 106 Pa
6
A
160 10
b 203 MPa
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31
PROBLEM 1.28
A
B
12 in.
C
D
12 in.
1500 lb
15 in.
E
16 in.
Two identical linkage-and-hydraulic-cylinder systems control the
position of the forks of a fork-lift truck. The load supported by the one
system shown is 1500 lb. Knowing that the thickness of member BD is
5
in., determine (a) the average shearing stress in the 12 -in.-diameter
8
pin at B, (b) the bearing stress at B in member BD.
16 in.
20 in.
SOLUTION
Use one fork as a free body.
M B 0: 24 E (20)(1500) 0
E 1250 lb
Fx 0: E Bx 0
Bx E
Bx 1250 lb
Fy 0: By 1500 0
B
(a)
Bx2 By2 12502 15002 1952.56 lb
Shearing stress in pin at B.
Apin
(b)
By 1500 lb
4
2
d pin
1
2
2
0.196350 in
4 2
B
1952.56
9.94 103 psi
Apin
0.196350
9.94 ksi
Bearing stress at B.
B 1952.56
6.25 103 psi
dt
12 85
6.25 ksi
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32
P'
150 mm
P
PROBLEM 1.29
Two wooden members of uniform rectangular cross section are joined
by the simple glued scarf splice shown. Knowing that P 11 kN,
determine the normal and shearing stresses in the glued splice.
45⬚⬚
45
75 mm
SOLUTION
90 45 45
P 11 kN 11 103 N
A0 (150)(75) 11.25 103 mm 2 11.25 103 m 2
P cos 2
(11 103 ) cos 2 45
489 103 Pa
A0
11.25 103
489 kPa
P sin 2
(11 103 )(sin 90)
489 103 Pa
2 A0
(2)(11.25 103 )
489 kPa
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33
P'
150 mm
45⬚⬚
45
P
75 mm
PROBLEM 1.30
Two wooden members of uniform rectangular cross section are joined
by the simple glued scarf splice shown. Knowing that the maximum
allowable shearing stress in the glued splice is 620 kPa, determine
(a) the largest load P that can be safely applied, (b) the corresponding
tensile stress in the splice.
SOLUTION
90 45 45
A0 (150)(75) 11.25 103 mm 2 11.25 103 m 2
620 kPa 620 103 Pa
P sin 2
2 A0
(a)
P
2 A0
(2)(11.25 103 )(620 103 )
sin2
sin 90
13.95 103 N
(b)
P 13.95 kN
P cos 2
(13.95 103 )(cos 45) 2
A0
11.25 103
620 kPa
620 103 Pa
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34
PROBLEM 1.31
P
5.0 in.
The 1.4-kip load P is supported by two wooden members of uniform cross section
that are joined by the simple glued scarf splice shown. Determine the normal and
shearing stresses in the glued splice.
3.0 in.
608
P'
SOLUTION
P 1400 lb
90 60 30
A0 (5.0)(3.0) 15 in 2
P cos 2
(1400)(cos30) 2
A0
15
70.0 psi
P sin 2
(1400)sin 60
2 A0
(2)(15)
40.4 psi
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35
PROBLEM 1.32
P
5.0
Two wooden members of uniform cross section are joined by the simple scarf splice
shown. Knowing that the maximum allowable tensile stress in the glued splice is 75 psi,
determine (a) the largest load P that can be safely supported, (b) the corresponding
shearing stress in the splice.
3.0 in.
608
P'
SOLUTION
A0 (5.0)(3.0) 15 in 2
90 60 30
P cos 2
A0
(a)
P
A0
(75)(15)
1500 lb
2
cos
cos 2 30
(b)
P sin 2
(1500)sin 60
2 A0
(2)(15)
P 1.500 kips
43.3 psi
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36
PROBLEM 1.33
P
A centric load P is applied to the granite block shown. Knowing that the
resulting maximum value of the shearing stress in the block is 2.5 ksi, determine
(a) the magnitude of P, (b) the orientation of the surface on which the maximum
shearing stress occurs, (c) the normal stress exerted on that surface, (d ) the
maximum value of the normal stress in the block.
6 in.
6 in.
SOLUTION
A0 (6)(6) 36 in 2
max 2.5 ksi
45 for plane of max
| P|
| P | 2 A0 max (2)(36)(2.5)
2 A0
(a)
max
(b)
sin 2 1 2 90
(c)
45
(d )
max
P 180.0 kips
45.0
P
P
180
cos 2 45
A0
2 A0
(2)(36)
P
180
36
A0
45 2.50 ksi
max 5.00 ksi
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37
PROBLEM 1.34
P
A 240-kip load P is applied to the granite block shown. Determine the resulting
maximum value of (a) the normal stress, (b) the shearing stress. Specify the
orientation of the plane on which each of these maximum values occurs.
6 in.
6 in.
SOLUTION
A0 (6)(6) 36 in 2
(a)
(b)
240
P
cos 2
cos 2 6.67 cos 2
36
A0
max tensile stress 0 at 90.0
max. compressive stress 6.67 ksi at 0
P
240
max
2 A0
(2)(36)
max 3.33 ksi
at 45
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38
PROBLEM 1.35
P
10 mm
A steel pipe of 400-mm outer diameter is fabricated from 10-mm thick
plate by welding along a helix that forms an angle of 20 with a plane
perpendicular to the axis of the pipe. Knowing that a 300-kN axial
force P is applied to the pipe, determine the normal and shearing
stresses in directions respectively normal and tangential to the weld.
Weld
208
SOLUTION
d o 0.400 m
1
d o 0.200 m
2
ri ro t 0.200 0.010 0.190 m
ro
Ao (ro2 ri2 ) (0.2002 0.1902 )
12.2522 103 m 2
20
P
300 103 cos 2 20
cos 2
21.621 106 Pa
Ao
12.2522 103
21.6 MPa
P
300 103 sin 40
sin 2
7.8695 106 Pa
2 A0
(2)(12.2522 103 )
7.87 MPa
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39
PROBLEM 1.36
P
A steel pipe of 400-mm outer diameter is fabricated from 10-mm
thick plate by welding along a helix that forms an angle of 20° with a
plane perpendicular to the axis of the pipe. Knowing that the
maximum allowable normal and shearing stresses in the directions
respectively normal and tangential to the weld are 60 MPa and
36 MPa, determine the magnitude P of the largest axial force that
can be applied to the pipe.
10 mm
Weld
208
SOLUTION
d o 0.400 m
1
d o 0.200 m
2
ri ro t 0.200 0.010 0.190 m
ro
Ao (ro2 ri2 ) (0.2002 0.1902 )
12.2522 103 m 2
20
Based on
| | 60 MPa:
P
cos 2
A0
Ao
(12.2522 103 )(60 106 )
832.52 103 N
cos 2
cos 2 20
P
| | 30 MPa:
sin 2
2 Ao
P
Based on
P
2 Ao
(2)(12.2522 103 )(36 106 )
1372.39 103 N
sin 2
sin 40
P 833 kN
Smaller value is the allowable value of P.
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40
PROBLEM 1.37
Q
12 in.
12 in.
E
B
9 in.
1 in.
C
A
9 in.
3
8
A steel loop ABCD of length 5 ft and of 83 -in. diameter is placed as
shown around a 1-in.-diameter aluminum rod AC. Cables BE and DF,
each of 12 -in. diameter, are used to apply the load Q. Knowing that the
ultimate strength of the steel used for the loop and the cables is 70 ksi,
and that the ultimate strength of the aluminum used for the rod is 38 ksi,
determine the largest load Q that can be applied if an overall factor of
safety of 3 is desired.
in.
D
1
2
F
in.
Q'
SOLUTION
Using joint B as a free body and considering symmetry,
2
3
6
FAB Q 0 Q FAB
5
5
Using joint A as a free body and considering symmetry,
4
FAB FAC 0
5
8 5
3
Q FAC 0 Q FAC
5 6
4
2
Based on strength of cable BE,
QU U A U
4
d 2 (70)
1
2
13.7445 kips
4 2
Based on strength of steel loop,
QU
6
6
6
FAB, U U A U d 2
5
5
5
4
2
3
6
(70) 9.2775 kips
5
4 8
Based on strength of rod AC,
QU
3
3
3
3
FAC , U U A U d 2 (38) (1.0)2 22.384 kips
4
4
4
4
4
4
Actual ultimate load QU is the smallest, QU 9.2775 kips
Allowable load:
Q
QU
9.2775
3.0925 kips
3
F.S .
Q 3.09 kips
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41
A
w
908
B
PROBLEM 1.38
Link BC is 6 mm thick, has a width w 25 mm, and is made of a steel with
a 480-MPa ultimate strength in tension. What was the safety factor used if the
structure shown was designed to support a 16-kN load P?
480 mm
C
D
P
SOLUTION
Use bar ACD as a free body and note that member BC is a two-force member.
M A 0:
(480) FBC (600) P 0
FBC
Ultimate load for member BC:
600
(600)(16 103 )
20 103 N
P
480
480
FU U A
FU (480 106 )(0.006)(0.025) 72 103 N
Factor of safety:
F.S.
FU
72 103
FBC
20 103
F.S. 3.60
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42
A
w
908
B
PROBLEM 1.39
Link BC is 6 mm thick and is made of a steel with a 450-MPa ultimate strength in
tension. What should be its width w if the structure shown is being designed to
support a 20-kN load P with a factor of safety of 3?
480 mm
C
D
P
SOLUTION
Use bar ACD as a free body and note that member BC is a two-force member.
M A 0:
(480) FBC 600P 0
FBC
600P (600)(20 103 )
25 103 N
480
480
For a factor of safety F.S. 3, the ultimate load of member BC is
FU (F.S.)( FBC ) (3)(25 103 ) 75 103 N
But FU U A A
FU
U
75 103
166.667 106 m 2
6
450 10
For a rectangular section, A wt or w
A 166.667 106
27.778 103 m
t
0.006
w 27.8 mm
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43
PROBLEM 1.40
0.75 m
A
0.4 m
B
1.4 m
Members AB and BC of the truss shown are made of the same alloy. It is known
that a 20-mm-square bar of the same alloy was tested to failure and that an
ultimate load of 120 kN was recorded. If a factor of safety of 3.2 is to be
achieved for both bars, determine the required cross-sectional area of (a) bar
AB, (b) bar AC.
C
SOLUTION
Length of member AB:
AB
0.752 0.42 0.85 m
Use entire truss as a free body.
M c 0: 1.4 Ax (0.75)(28) 0
Ax 15 kN
Fy 0: Ay 28 0
Ay 28 kN
Use Joint A as free body.
0.75
FAB Ax 0
0.85
(0.85)(15)
FAB
17 kN
0.75
0.4
Fy 0: Ay FAC
FAB 0
0.85
(0.4)(17)
FAC 28
20 kN
0.85
Fx 0:
For the test bar,
For the material,
A (0.020)2 400 106 m 2
U
PU 120 103 N
PU
120 103
300 106 Pa
A
400 106
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44
PROBLEM 1.40 (Continued)
(a)
For member AB:
F.S.
AAB
(b)
For member AC:
F.S.
AAC
PU
A
U AB
FAB
FAB
(F.S.) FAB
U
(3.2)(17 103 )
181.333 106 m 2
300 106
AAB 181.3 mm 2
PU
A
U AC
FAC
FAC
(F.S.) FAC
U
(3.2)(20 103 )
213.33 106 m 2
300 106
AAC 213 mm 2
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45
PROBLEM 1.41
0.75 m
A
0.4 m
B
1.4 m
Members AB and BC of the truss shown are made of the same alloy. It is known
that a 20-mm-square bar of the same alloy was tested to failure and that an
ultimate load of 120 kN was recorded. If bar AB has a cross-sectional area of
225 mm2, determine (a) the factor of safety for bar AB and (b) the crosssectional area of bar AC if it is to have the same factor of safety as bar AB.
C
SOLUTION
Length of member AB:
AB
0.752 0.42 0.85 m
Use entire truss as a free body.
M c 0: 1.4 Ax (0.75)(28) 0
Ax 15 kN
Fy 0: Ay 28 0
Ay 28 kN
Use Joint A as free body.
Fx 0:
0.75
FAB Ax 0
0.85
(0.85)(15)
FAB
17 kN
0.75
0.4
FAB 0
0.85
(0.4)(17)
28
20 kN
0.85
Fy 0: Ay FAC
FAC
For the test bar,
For the material,
A (0.020)2 400 106 m 2
U
PU 120 103 N
PU
120 103
300 106 Pa
A
400 106
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46
PROBLEM 1.41 (Continued)
(a)
For bar AB:
F.S.
FU
A
(300 106 )(225 106 )
U AB
FAB
FAB
17 103
F.S. 3.97
(b)
For bar AC:
F.S.
AAC
FU
A
U AC
FAC
FAC
(F.S.) FAC
U
(3.97)(20 103 )
264.67 106 m 2
300 106
AAC 265 mm 2
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47
A
PROBLEM 1.42
600 lb/ft
35⬚
B
C
D
5 kips
1.4 ft
1.4 ft
E
Link AB is to be made of a steel for which the ultimate normal stress is
65 ksi. Determine the cross-sectional area of AB for which the factor
of safety will be 3.20. Assume that the link will be adequately
reinforced around the pins at A and B.
1.4 ft
SOLUTION
P (4.2)(0.6) 2.52 kips
M D 0 :
(2.8)( FAB sin 35)
(0.7)(2.52) (1.4)(5) 0
FAB 5.4570 kips
AB
AAB
FAB
ult
AAB
F. S .
( F. S .) FAB
ult
(3.20)(5.4570 kips)
65 ksi
0.26854 in 2
AAB 0.268 in 2
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48
PROBLEM 1.43
16 kN
L
Two wooden members are joined by plywood splice plates that are fully glued on
the contact surfaces. Knowing that the clearance between the ends of the members
is 6 mm and that the ultimate shearing stress in the glued joint is 2.5 MPa,
determine the length L for which the factor of safety is 2.75 for the loading shown.
6 mm
125 mm
16 kN
SOLUTION
all
2.5 MPa
0.90909 MPa
2.75
On one face of the upper contact surface,
A
L 0.006 m
(0.125 m)
2
Since there are 2 contact surfaces,
all
0.90909 106
P
2A
16 103
( L 0.006)(0.125)
L 0.14680 m
146.8 mm
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49
PROBLEM 1.44
16 kN
For the joint and loading of Prob. 1.43, determine the factor of safety when
L = 180 mm.
L
6 mm
PROBLEM 1.43 Two wooden members are joined by plywood splice plates that
are fully glued on the contact surfaces. Knowing that the clearance between the
ends of the members is 6 mm and that the ultimate shearing stress in the glued
joint is 2.5 MPa, determine the length L for which the factor of safety is 2.75 for
the loading shown.
125 mm
16 kN
SOLUTION
Area of one face of upper contact surface:
A
0.180 m 0.006 m
(0.125 m)
2
A 10.8750 103 m 2
Since there are two surfaces,
all
P
16 103 N
2 A 2(10.8750 103 m 2 )
all 0.73563 MPa
F.S.
u
2.5 MPa
3.40
all 0.73563 MPa
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50
PROBLEM 1.45
Three 34 -in.-diameter steel bolts are to be used to attach the steel plate shown to
a wooden beam. Knowing that the plate will support a load P = 24 kips and that
the ultimate shearing stress for the steel used is 52 ksi, determine the factor of
safety for this design.
P
SOLUTION
For each bolt,
A
4
d2
3
2
2
0.44179 in
44
PU A U (0.44179)(52)
22.973 kips
For the three bolts,
PU (3)(22.973) 68.919 kips
Factor of safety:
F. S .
PU
68.919
24
P
F. S . 2.87
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51
PROBLEM 1.46
Three steel bolts are to be used to attach the steel plate shown to a wooden beam.
Knowing that the plate will support a load P = 28 kips, that the ultimate shearing
stress for the steel used is 52 ksi, and that a factor of safety of 3.25 is desired,
determine the required diameter of the bolts.
P
SOLUTION
For each bolt,
Required:
P
24
8 kips
3
PU ( F. S.) P (3.25)(8.0) 26.0 kips
U
d
PU
P
4P
U 2 U2
A
d
d
4
4 PU
U
(4)(26.0)
0.79789 in.
(52)
d 0.798 in.
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52
PROBLEM 1.47
1
2
d
A load P is supported as shown by a steel pin that has been inserted in a
short wooden member hanging from the ceiling. The ultimate strength of
the wood used is 60 MPa in tension and 7.5 MPa in shear, while the
ultimate strength of the steel is 145 MPa in shear. Knowing that
b 40 mm, c 55 mm, and d 12 mm, determine the load P if an
overall factor of safety of 3.2 is desired.
P
1
2
c
40 mm
P
b
SOLUTION
Based on double shear in pin,
PU 2 A U 2
4
4
d 2 U
(2)(0.012) 2 (145 106 ) 32.80 103 N
Based on tension in wood,
PU A U w (b d ) U
(0.040)(0.040 0.012)(60 106 )
67.2 103 N
Based on double shear in the wood,
PU 2 AU 2wc U (2)(0.040)(0.055)(7.5 106 )
33.0 103 N
Use smallest
PU 32.8 103 N
Allowable:
P
PU
32.8 103
10.25 103 N
F .S.
3.2
10.25 kN
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53
PROBLEM 1.48
For the support of Prob. 1.47, knowing that the diameter of the pin is
d 16 mm and that the magnitude of the load is P 20 kN, determine
(a) the factor of safety for the pin, (b) the required values of b and c if the
factor of safety for the wooden members is the same as that found in part a
for the pin.
1
2
d
P
1
2
c
b
40 mm
P
PROBLEM 1.47 A load P is supported as shown by a steel pin that has
been inserted in a short wooden member hanging from the ceiling. The
ultimate strength of the wood used is 60 MPa in tension and 7.5 MPa in
shear, while the ultimate strength of the steel is 145 MPa in shear.
Knowing that b 40 mm, c 55 mm, and d 12 mm, determine the
load P if an overall factor of safety of 3.2 is desired.
SOLUTION
P 20 kN 20 103 N
(a)
Pin:
A
Double shear:
4
d2
4
(0.016) 2 2.01.06 106 m 2
P
P
U U
2A
2A
PU 2 A U (2)(201.16 106 )(145 106 ) 58.336 103 N
F .S.
(b)
Tension in wood:
where w 40 mm 0.040 m
b 40.3 mm
PU 58.336 103 N for same F.S.
Double shear: each area is A wc
c
PU
PU
A
w(b d )
PU
58.336 103
0.016
40.3 103 m
w U
(0.040)(60 106 )
Shear in wood:
F .S. 2.92
PU 58.336 103 N for same F.S.
U
bd
PU
58.336 103
P
20 103
U
PU
P
U
2 A 2wc
PU
58.336 103
97.2 103 m
2w U
(2)(0.040)(7.5 106 )
c 97.2 mm
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54
PROBLEM 1.49
a
3
4
in.
1
4
in.
b
P
A steel plate 14 in. thick is embedded in a
concrete wall to anchor a high-strength cable
as shown. The diameter of the hole in the plate
is 34 in., the ultimate strength of the steel used is
36 ksi, and the ultimate bonding stress between
plate and concrete is 300 psi. Knowing that a
factor of safety of 3.60 is desired when
P = 2.5 kips, determine (a) the required width a
of the plate, (b) the minimum depth b to which a
plate of that width should be embedded in the
concrete slab. (Neglect the normal stresses
between the concrete and the end of the plate.)
SOLUTION
Based on tension in plate,
A (a d )t
PU U A
F .S.
PU
(a d )t
U
P
P
Solving for a,
ad
( F .S .) P
3 (3.60)(2.5)
U t
4
(36) 14
(a) a 1.750 in.
Based on shear between plate and concrete slab,
U 0.300 ksi
A perimeter depth 2(a t )b
PU U A 2 U (a t )b
Solving for b,
b
F .S.
PU
P
( F .S .) P
(3.6)(2.5)
2(a t ) U
(2) 1.75 14 (0.300)
(b) b 7.50 in.
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55
PROBLEM 1.50
Determine the factor of safety for the cable
anchor in Prob. 1.49 when P 2.5 kips, knowing
that a 2 in. and b 6 in.
PROBLEM 1.49 A steel plate 14 in. thick is
embedded in a concrete wall to anchor a highstrength cable as shown. The diameter of the hole
in the plate is 34 in., the ultimate strength of the
steel used is 36 ksi, and the ultimate bonding
stress between plate and concrete is 300 psi.
Knowing that a factor of safety of 3.60 is desired
when P = 2.5 kips, determine (a) the required
width a of the plate, (b) the minimum depth b to
which a plate of that width should be embedded
in the concrete slab. (Neglect the normal stresses
between the concrete and the end of the plate.)
a
3
4
in.
1
4
in.
b
P
SOLUTION
Based on tension in plate,
A (a d )t
3 1
2 0.31250 in 2
4 4
PU U A
(36)(0.31250) 11.2500 kips
F .S .
PU
11.2500
4.50
P
3.5
Based on shear between plate and concrete slab,
1
A perimeter depth 2(a t )b 2 2 (6.0)
4
A 27.0 in 2
U 0.300 ksi
PU U A (0.300)(27.0) 8.10 kips
F .S .
PU
8.10
3.240
P
2.5
F .S . 3.24
Actual factor of safety is the smaller value.
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56
PROBLEM 1.51
A
1
2
in.
8 in.
B
C
6 in.
D
4 in.
P
Link AC is made of a steel with a 65-ksi ultimate normal stress and has
a 14 12 -in. uniform rectangular cross section. It is connected to a
support at A and to member BCD at C by 34 -in.-diameter pins, while
member BCD is connected to its support at B by a 165 -in.-diameter pin.
All of the pins are made of a steel with a 25-ksi ultimate shearing stress
and are in single shear. Knowing that a factor of safety of 3.25 is
desired, determine the largest load P that can be applied at D. Note that
link AC is not reinforced around the pin holes.
SOLUTION
Use free body BCD.
8
M B 0 : (6) FAC 10 P 0
10
P 0.48 FAC
Fx 0 : Bx
Bx
(1)
6
FAC 0
10
6
FAC 1.25P
10
M C 0 : 6By 4P 0
2
By P
3
i.e. By
2
P
3
2
B
2
Bx2 By2 1.252 P 1.41667 P
3
P 0.70588B
(2)
Shear in pins at A and C.
FAC Apin
2
U
25 3
d2
0.84959 kips
F. S . 4
3.25 4 8
Tension on net section of A and C.
FAC Anet
U
F. S .
65 1 1 3
Anet
0.625 kips
3.25 4 2 8
Smaller value of FAC is 0.625 kips.
From (1),
P (0.48)(0.625) 0.300 kips
Shear in pin at B.
B Apin
From (2),
P (0.70588)(0.58999) 0.416 kips
2
U
25 5
d2
0.58999 kips
F. S. 4
3.25 4 16
Allowable value of P is the smaller value.
P 0.300 kips
or
P 300 lb
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57
PROBLEM 1.52
Solve Prob. 1.51, assuming that the structure has been redesigned to use
5
-in-diameter pins at A and C as well as at B and that no other changes
16
have been made.
A
1
2
in.
8 in.
B
C
6 in.
PROBLEM 1.51 Link AC is made of a steel with a 65-ksi ultimate
normal stress and has a 14 12 -in. uniform rectangular cross section. It is
connected to a support at A and to member BCD at C by 34 -in.-diameter
pins, while member BCD is connected to its support at B by a 165 -in.diameter pin. All of the pins are made of a steel with a 25-ksi ultimate
shearing stress and are in single shear. Knowing that a factor of safety of
3.25 is desired, determine the largest load P that can be applied at D.
Note that link AC is not reinforced around the pin holes.
D
4 in.
P
SOLUTION
Use free body BCD.
8
M B 0 : (6) FAC 10 P 0
10
P 0.48 FAC
6
Fy 0 : Bx
FAC 0
10
6
Bx
FAC 1.25P
10
M C 0 : 6By 4 P 0
2
By P
3
i.e. By
(1)
2
P
3
2
B
Bx2
By2
2
1.25 P 1.41667 P
3
2
P 0.70583 B
(2)
Shear in pins at A and C.
FAC Apin
2
U
25 5
d2
0.58999 kips
F. S . 4
3.25 4 16
Tension on net section of A and C.
FAC Anet
U
5
65 1 1
Anet
0.9375 kips
F. S .
3.25 4 2 16
Smaller value of FAC is 0.58999 kips.
From (1),
P (0.48)(0.58999) 0.283 kips
Shear in pin at B.
B Apin
From (2),
P (0.70588)(0.58999) 0.416 kips
2
U
25 5
d2
0.58999 kips
F. S. 4
3.25 4 16
Allowable value of P is the smaller value.
P 0.283 kips
or
P 283 lb
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58
PROBLEM 1.53
250 mm
400 mm
A
Each of the two vertical links CF connecting the two horizontal
members AD and EG has a 10 40-mm uniform rectangular cross
section and is made of a steel with an ultimate strength in tension of
400 MPa, while each of the pins at C and F has a 20-mm diameter and
are made of a steel with an ultimate strength in shear of 150 MPa.
Determine the overall factor of safety for the links CF and the pins
connecting them to the horizontal members.
250 mm
B
C
D
E
F
G
24 kN
SOLUTION
M E 0 : 0.40 FCF (0.65)(24 103 ) 0
FCF 39 103 N
Based on tension in links CF,
A (b d ) t (0.040 0.02)(0.010) 200 106 m 2
6
(one link)
FU 2 U A (2)(400 10 )(200 10 ) 160.0 10 N
6
3
Based on double shear in pins,
A
4
d2
4
(0.020) 2 314.16 106 m 2
FU 2 U A (2)(150 106 )(314.16 106 ) 94.248 103 N
Actual FU is smaller value, i.e. FU 94.248 103 N
Factor of safety:
F. S .
FU
94.248 103
FCF
39 103
F. S . 2.42
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59
PROBLEM 1.54
250 mm
400 mm
A
Solve Prob. 1.53, assuming that the pins at C and F have been replaced
by pins with a 30-mm diameter.
250 mm
B
PROBLEM 1.53 Each of the two vertical links CF connecting the two
horizontal members AD and EG has a 10 40-mm uniform rectangular
cross section and is made of a steel with an ultimate strength in tension of
400 MPa, while each of the pins at C and F has a 20-mm diameter and
are made of a steel with an ultimate strength in shear of 150 MPa.
Determine the overall factor of safety for the links CF and the pins
connecting them to the horizontal members.
C
D
E
F
G
24 kN
SOLUTION
Use member EFG as free body.
M E 0 : 0.40FCF (0.65)(24 103 ) 0
FCF 39 103 N
Based on tension in links CF,
A (b d ) t (0.040 0.030)(0.010) 100 106 m 2
6
6
(one link)
3
FU 2 U A (2)(400 10 )(100 10 ) 80.0 10 N
Based on double shear in pins,
A
4
d2
4
(0.030) 2 706.86 106 m 2
FU 2 U A (2)(150 106 )(706.86 106 ) 212.06 103 N
Actual FU is smaller value, i.e. FU 80.0 103 N
Factor of safety:
F. S .
FU
80.0 103
FCF
39 103
F. S . 2.05
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60
PROBLEM 1.55
Top view
200 mm
180 mm
12 mm
In the structure shown, an 8-mm-diameter pin
is used at A, and 12-mm-diameter pins are
used at B and D. Knowing that the ultimate
shearing stress is 100 MPa at all connections
and that the ultimate normal stress is 250 MPa
in each of the two links joining B and D,
determine the allowable load P if an overall
factor of safety of 3.0 is desired.
8 mm
A
B
C
B
A
C
B
20 mm
P
8 mm
8 mm
D
D
12 mm
Front view
Side view
SOLUTION
Statics: Use ABC as free body.
M B 0 : 0.20 FA 0.18P 0
M A 0 : 0.20FBD 0.38P 0
Based on double shear in pin A, A
FA
2 U A
4
10
FA
9
10
P
FBD
19
P
d2
4
(0.008)2 50.266 106 m 2
(2)(100 106 )(50.266 106 )
3.351 103 N
3.0
F .S .
10
P
FA 3.72 103 N
9
Based on double shear in pins at B and D, A
FBD
2 U A
4
d2
4
(0.012) 2 113.10 106 m 2
(2)(100 106 )(113.10 106 )
7.54 103 N
3.0
F .S.
10
P
FBD 3.97 103 N
19
Based on compression in links BD, for one link, A (0.020)(0.008) 160 106 m 2
2 U A (2)(250 106 )(160 106 )
26.7 103 N
3.0
F .S .
10
P
FBD 14.04 103 N
19
Allowable value of P is smallest, P 3.72 103 N
FBD
P 3.72 kN
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61
PROBLEM 1.56
Top view
200 mm
180 mm
In an alternative design for the structure of
Prob. 1.55, a pin of 10-mm-diameter is to be
used at A. Assuming that all other
specifications remain unchanged, determine
the allowable load P if an overall factor of
safety of 3.0 is desired.
12 mm
8 mm
A
B
C
B
A
C
B
20 mm
P
8 mm
8 mm
D
D
12 mm
Front view
Side view
PROBLEM 1.55 In the structure shown, an 8mm-diameter pin is used at A, and 12-mmdiameter pins are used at B and D. Knowing
that the ultimate shearing stress is 100 MPa at
all connections and that the ultimate normal
stress is 250 MPa in each of the two links
joining B and D, determine the allowable load
P if an overall factor of safety of 3.0 is
desired.
SOLUTION
Statics: Use ABC as free body.
10
FA
9
10
P
FBD
19
M B 0: 0.20 FA 0.18P 0
P
M A 0: 0.20 FBD 0.38P 0
Based on double shear in pin A, A
4
d2
4
(0.010) 2 78.54 106 m 2
2 U A (2)(100 106 )(78.54 106 )
5.236 103 N
3.0
F .S .
10
P
FA 5.82 103 N
9
FA
Based on double shear in pins at B and D, A
4
d2
4
(0.012) 2 113.10 106 m 2
2 U A (2)(100 106 )(113.10 106 )
7.54 103 N
3.0
F .S .
10
P
FBD 3.97 103 N
19
FBD
Based on compression in links BD, for one link, A (0.020)(0.008) 160 106 m 2
2 U A (2)(250 106 )(160 106 )
26.7 103 N
3.0
F .S.
10
P
FBD 14.04 103 N
19
FBD
Allowable value of P is smallest, P 3.97 103 N
P 3.97 kN
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62
PROBLEM 1.57
C
A 40-kg platform is attached to the end B of a 50-kg wooden beam AB,
which is supported as shown by a pin at A and by a slender steel rod BC
with a 12-kN ultimate load. (a) Using the Load and Resistance Factor
Design method with a resistance factor 0.90 and load factors
D 1.25 and L 1.6, determine the largest load that can be safely
placed on the platform. (b) What is the corresponding conventional
factor of safety for rod BC?
1.8 m
A
B
2.4 m
SOLUTION
3
M A 0 : (2.4) P 2.4W1 1.2W2
5
5
5
P W1 W2
3
6
For dead loading,
W1 (40)(9.81) 392.4 N, W2 (50)(9.81) 490.5 N
5
5
PD (392.4) (490.5) 1.0628 103 N
3
6
For live loading,
W1 mg W2 0
From which
m
Design criterion:
PL
5
mg
3
3 PL
5 g
D PD L PL PU
PL
PU D PD (0.90)(12 103 ) (1.25)(1.0628 103 )
1.6
L
5.920 103 N
(a)
m
Allowable load.
3 5.92 103
5 9.81
m 362 kg
Conventional factor of safety:
P PD PL 1.0628 103 5.920 103 6.983 103 N
(b)
F. S.
PU
12 103
P
6.983 103
F. S. 1.718
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63
P
P
PROBLEM 1.58
The Load and Resistance Factor Design method is to be used to select
the two cables that will raise and lower a platform supporting two
window washers. The platform weighs 160 lb and each of the window
washers is assumed to weigh 195 lb with equipment. Since these
workers are free to move on the platform, 75% of their total weight and
the weight of their equipment will be used as the design live load of each
cable. (a) Assuming a resistance factor 0.85 and load factors
D 1.2 and L 1.5, determine the required minimum ultimate load of
one cable. (b) What is the corresponding conventional factor of safety
for the selected cables?
SOLUTION
D PD L PL PU
(a)
PU
D PD L PL
1
3
(1.2) 160 (1.5) 2 195
2
4
0.85
PU 629 lb
Conventional factor of safety:
P PD PL
(b)
F. S.
1
160 0.75 2 195 372.5 lb
2
PU
629
P
372.5
F. S. 1.689
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64
15 m
25 m
PROBLEM 1.59
3m
B
In the marine crane shown, link CD is known to have a
uniform cross section of 50 150 mm. For the loading
shown, determine`` the normal stress in the central portion
of that link.
35 m
80 Mg
C
15 m
D
A
SOLUTION
W (80 Mg)(9.81 m/s 2 ) 784.8 kN
Weight of loading:
Free Body: Portion ABC.
M A 0: FCD (15 m) W (28 m) 0
28
28
W
(784.8 kN)
15
15
1465 kN
FCD
FCD
CD
FCD
1465 103 N
195.3 106 Pa
A
(0.050 m)(0.150 m)
CD 195.3 MPa
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65
PROBLEM 1.60
0.5 in.
Two horizontal 5-kip forces are applied to pin B of the assembly shown.
Knowing that a pin of 0.8-in. diameter is used at each connection,
determine the maximum value of the average normal stress (a) in link
AB, (b) in link BC.
B
1.8 in.
A
5 kips
5 kips
60⬚
0.5 in.
45⬚
1.8 in.
C
SOLUTION
Use joint B as free body.
Law of Sines:
FAB
FBC
10
sin 45 sin 60 sin 95
FAB 7.3205 kips
FBC 8.9658 kips
Link AB is a tension member.
Minimum section at pin: Anet (1.8 0.8)(0.5) 0.5 in 2
(a)
Stress in AB :
AB
FAB
7.3205
Anet
0.5
AB 14.64 ksi
Link BC is a compression member.
Cross sectional area is A (1.8)(0.5) 0.9 in 2
(b)
Stress in BC:
BC
FBC
8.9658
A
0.9
BC 9.96 ksi
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66
PROBLEM 1.61
0.5 in.
For the assembly and loading of Prob. 1.60, determine (a) the average
shearing stress in the pin at C, (b) the average bearing stress at C in
member BC, (c) the average bearing stress at B in member BC.
B
1.8 in.
A
5 kips
5 kips
PROBLEM 1.60 Two horizontal 5-kip forces are applied to pin B of the
assembly shown. Knowing that a pin of 0.8-in. diameter is used at each
connection, determine the maximum value of the average normal stress
(a) in link AB, (b) in link BC.
0.5 in.
60⬚
1.8 in.
45⬚
C
SOLUTION
Use joint B as free body.
Law of Sines:
FAB
FBC
10
sin 45 sin 60 sin 95
(a)
Shearing stress in pin at C.
FBC 8.9658 kips
FBC
2 AP
4
d2
(0.8)2 0.5026 in 2
4
8.9658
8.92
(2)(0.5026)
AP
8.92 ksi
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67
PROBLEM 1.61 (Continued)
(b)
Bearing stress at C in member BC. b
FBC
A
A td (0.5)(0.8) 0.4 in 2
b
(c)
Bearing stress at B in member BC. b
8.9658
22.4
0.4
b 22.4 ksi
FBC
A
A 2td 2(0.5)(0.8) 0.8 in 2
b
8.9658
11.21
0.8
b 11.21 ksi
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68
PROBLEM 1.62
Two steel plates are to be held together by means of 16-mmdiameter high-strength steel bolts fitting snugly inside cylindrical
brass spacers. Knowing that the average normal stress must not
exceed 200 MPa in the bolts and 130 MPa in the spacers,
determine the outer diameter of the spacers that yields the most
economical and safe design.
SOLUTION
At each bolt location the upper plate is pulled down by the tensile force Pb of the bolt. At the same time, the
spacer pushes that plate upward with a compressive force Ps in order to maintain equilibrium.
Pb Ps
For the bolt,
b
Fb
4 Pb
Ab
db2
or
Pb
For the spacer,
s
Ps
4 Ps
2
As
(d s db2 )
or
Ps
4
4
b db2
s (d s2 db2 )
Equating Pb and Ps ,
4
b db2
ds
4
s (d s2 db2 )
b
1
d
s b
200
1
(16)
130
d s 25.2 mm
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69
PROBLEM 1.63
P
A couple M of magnitude 1500 N m is applied to the crank of an engine. For the
position shown, determine (a) the force P required to hold the engine system in
equilibrium, (b) the average normal stress in the connecting rod BC, which has a
450-mm2 uniform cross section.
C
200 mm
B
M
80 mm
A
60 mm
SOLUTION
Use piston, rod, and crank together as free body. Add wall reaction H
and bearing reactions Ax and Ay.
M A 0 : (0.280 m) H 1500 N m 0
H 5.3571 103 N
Use piston alone as free body. Note that rod is a two-force member;
hence the direction of force FBC is known. Draw the force triangle
and solve for P and FBE by proportions.
l
2002 602 208.81 mm
P
200
H
60
P 17.86 103 N
(a)
P 17.86 kN
FBC
208.81
FBC 18.6436 103 N
H
60
Rod BC is a compression member. Its area is
450 mm 2 450 106 m 2
Stress:
BC
FBC
18.6436 103
41.430 106 Pa
A
450 106
(b)
BC 41.4 MPa
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70
4 in.
PROBLEM 1.64
4 in.
12 in.
E
2 in.
B
Knowing that the link DE is 18 in. thick and 1 in. wide, determine
the normal stress in the central portion of that link when
(a) 0, (b) 90.
D
C
J
6 in.
D
8 in.
A
F
60 lb
SOLUTION
Use member CEF as a free body.
M C 0 : 12 FDE (8)(60 sin ) (16)(60 cos ) 0
FDE 40 sin 80 cos lb
1
ADE (1) 0.125 in 2
8
F
DE DE
ADE
(a)
0: FDE 80 lb
DE
(b)
80
0.125
DE 640 psi
90: FDE 40 lb
DE
40
0.125
DE 320 psi
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71
1500 lb
1 in.
750 lb
A
4 in.
D
750 lb
B
C
PROBLEM 1.65
A 85 -in.-diameter steel rod AB is fitted to a round hole near end C of
the wooden member CD. For the loading shown, determine (a) the
maximum average normal stress in the wood, (b) the distance b for
which the average shearing stress is 100 psi on the surfaces indicated
by the dashed lines, (c) the average bearing stress on the wood.
b
SOLUTION
(a)
Maximum normal stress in the wood.
Anet (1) 4
P
Anet
(b)
5
2
3.375 in
8
1500
444 psi
3.375
444 psi
Distance b for = 100 psi.
For sheared area see dotted lines.
P
P
A 2bt
1500
P
7.50 in.
b
2t
(2)(1)(100)
(c)
b 7.50 in.
Average bearing stress on the wood.
b
P
P
1500
2400 psi
5
Ab
dt
(1)
8
b 2400 psi
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72
PROBLEM 1.66
D
Front view
D
6 mm
18 mm
B
A
B
160 mm
120 mm
C
Side view
P
A
B
Top view
In the steel structure shown, a 6-mmdiameter pin is used at C and 10-mmdiameter pins are used at B and D. The
ultimate shearing stress is 150 MPa at all
connections, and the ultimate normal
stress is 400 MPa in link BD. Knowing
that a factor of safety of 3.0 is desired,
determine the largest load P that can be
applied at A. Note that link BD is not
reinforced around the pin holes.
C
SOLUTION
Use free body ABC.
M C 0 : 0.280 P 0.120 FBD 0
P
3
FBD
7
(1)
M B 0 : 0.160 P 0.120 C 0
P
3
C
4
(2)
Tension on net section of link BD:
400 106
3
3
3
Anet
(6 10 )(18 10)(10 ) 6.40 10 N
F. S.
3
U
FBD Anet
Shear in pins at B and D:
FBD Apin
150 106
3 2
3
d 2
(10 10 ) 3.9270 10 N
F. S. 4
3
4
U
Smaller value of FBD is 3.9270 103 N.
3
P (3.9270 103 ) 1.683 103 N
7
From (1),
150 106
3 2
3
d 2 (2)
(6 10 ) 2.8274 10 N
F. S . 4
3
4
U
Shear in pin at C:
C 2 Apin 2
From (2),
3
P (2.8274 103 ) 2.12 103 N
4
Smaller value of P is allowable value.
P 1.683 103 N
P 1.683 kN
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73
40⬚
PROBLEM 1.67
D
P
A
Member ABC, which is supported by a pin and bracket at C and a cable
BD, was designed to support the 16-kN load P as shown. Knowing that
the ultimate load for cable BD is 100 kN, determine the factor of safety
with respect to cable failure.
30⬚
B
0.6 m
C
0.8 m
0.4 m
SOLUTION
Use member ABC as a free body, and note that member BD is a two-force member.
M c 0 : ( P cos 40)(1.2) ( P sin 40)(0.6)
( FBD cos 30)(0.6)
( FBD sin 30)(0.4) 0
1.30493P 0.71962FBD 0
FBD 1.81335P (1.81335)(16 103 ) 29.014 103 N
FU 100 103 N
F. S .
FU
100 103
FBD
29.014 103
F. S. 3.45
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74
PROBLEM 1.68
d
L
P
A force P is applied as shown to a steel reinforcing bar that has
been embedded in a block of concrete. Determine the smallest
length L for which the full allowable normal stress in the bar can be
developed. Express the result in terms of the diameter d of the bar,
the allowable normal stress all in the steel, and the average
allowable bond stress all between the concrete and the cylindrical
surface of the bar. (Neglect the normal stresses between the
concrete and the end of the bar.)
SOLUTION
A dL
For shear,
P all A all dL
A
For tension,
4
d2
P all A all d 2
4
Equating,
all dL all
4
d2
Lmin alld/4 all
Solving for L,
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75
PROBLEM 1.69
2.4 kips
The two portions of member AB are glued together along a plane
forming an angle with the horizontal. Knowing that the ultimate stress
for the glued joint is 2.5 ksi in tension and 1.3 ksi in shear, determine
(a) the value of for which the factor of safety of the member is
maximum, (b) the corresponding value of the factor of safety. (Hint:
Equate the expressions obtained for the factors of safety with respect to the
normal and shearing stresses.)
A
B
1.25 in.
2.0 in.
SOLUTION
A0 (2.0)(1.25) 2.50 in 2
At the optimum angle,
( F. S.) ( F. S.)
U A0
P
cos 2 PU ,
A0
cos 2
Normal stress:
( F. S .)
Shearing stress:
U A0
P cos
Solving,
(b)
PU
2
P
U A0
P cos 2
U A0
P
sin cos PU ,
sin cos
A0
( F. S.)
Equating,
PU ,
PU ,
P
U A0
P sin cos
U A0
P sin cos
sin
1.3
tan U
0.520
cos
U
2.5
(a)
opt 27.5
U A0
(12.5)(2.50)
7.94 kips
2
cos
cos 2 27.5
F. S.
PU
7.94
P
2.4
F. S. 3.31
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76
PROBLEM 1.70
2.4 kips
The two portions of member AB are glued together along a plane
forming an angle with the horizontal. Knowing that the ultimate stress
for the glued joint is 2.5 ksi in tension and 1.3 ksi in shear, determine
the range of values of for which the factor of safety of the members is
at least 3.0.
A
B
1.25 in.
2.0 in.
SOLUTION
A0 (2.0)(1.25) 2.50 in.2
P 2.4 kips
PU ( F. S.) P 7.2 kips
Based on tensile stress,
U
cos 2
PU
cos 2
A0
U A0
PU
cos 0.93169
Based on shearing stress,
U
sin 2
2 64.52
(2.5)(2.50)
0.86806
7.2
21.3
21.3
PU
P
sin cos U sin 2
A0
2 A0
2 A0U
PU
(2)(2.50)(1.3)
0.90278
7.2
32.3
32.3
21.3 32.3
Hence,
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77
PROBLEM 1.C1
Element n
Pn
A solid steel rod consisting of n cylindrical elements welded together is subjected to the
loading shown. The diameter of element i is denoted by di and the load applied to its
lower end by Pi with the magnitude Pi of this load being assumed positive if Pi is
directed downward as shown and negative otherwise. (a) Write a computer program
that can be used with either SI or U.S. customary units to determine the average stress
in each element of the rod. (b) Use this program to solve Problems 1.1 and 1.3.
Element 1
P1
SOLUTION
Force in element i:
It is the sum of the forces applied to that element and all lower ones:
Fi
i
P
k
k 1
Average stress in element i:
1
Area Ai di2
4
Fi
Ave. stress
Ai
Program outputs:
Problem 1.1
Problem 1.3
Element
Stress (MPa)
Element
Stress (ksi)
1
84.883
1
22.635
2
96.766
2
17.927
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78
PROBLEM 1.C2
A 20-kN load is applied as shown to the horizontal member ABC.
Member ABC has a 10 50-mm uniform rectangular cross section
and is supported by four vertical links, each of 8 36-mm uniform
rectangular cross section. Each of the four pins at A, B, C, and D has the
same diameter d and is in double shear. (a) Write a computer program
to calculate for values of d from 10 to 30 mm, using 1-mm increments,
(i) the maximum value of the average normal stress in the links
connecting pins B and D, (ii) the average normal stress in the links
connecting pins C and E, (iii) the average shearing stress in pin B,
(iv) the average shearing stress in pin C, (v) the average bearing stress
at B in member ABC, and (vi) the average bearing stress at C in member
ABC. (b) Check your program by comparing the values obtained for
d 16 mm with the answers given for Probs. 1.7 and 1.27. (c) Use this
program to find the permissible values of the diameter d of the pins,
knowing that the allowable values of the normal, shearing, and bearing
stresses for the steel used are, respectively, 150 MPa, 90 MPa, and
230 MPa. (d) Solve Part c, assuming that the thickness of member ABC
has been reduced from 10 to 8 mm.
0.4 m
C
0.25 m
0.2 m
B
E
20 kN
D
A
SOLUTION
P = 20 kN
Forces in links.
F.B. diagram of ABC:
M C 0: 2FBD ( BC ) P( AC ) 0
FBD P( AC )/2( BC ) (tension)
M B 0: 2FCE ( BC ) P( AB) 0
(i)
Link BD.
Thickness t L
FCE P( AB)/2( BC ) (comp.)
(ii)
Link CE.
Thickness t L
ACE tL wL
ABD t L ( wL d )
CE FCE / ACE
BD FBD / ABD
(iv)
(iii) Pin B.
C FCE /( d 2 /4)
B FBD /( d 2 /4)
(v)
Pin C.
Shearing stress in ABC under Pin B.
FB AC t AC ( wAC /2)
Bearing stress at B.
Thickness of member AC t AC
Fy 0: 2FB 2 FBD
Sig Bear B FBD /(dt AC )
(vi) Bearing stress at C.
Sig Bear C FCE /( dt AC )
AC
2 FBD
AC wAC
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79
PROBLEM 1.C2 (Continued)
Program Outputs
Input data for Parts (a), (b), (c):
P 20 kN, AB 0.25 m, BC 0.40 m, AC 0.65 m,
TL 8 mm, WL 36 mm, TAC 10 mm, WAC 50 mm
(c) Answer: 16 mm d 22 mm (c)
Check: For d 22 mm, Tau AC = 65 MPa < 90 MPa O.K.
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80
PROBLEM 1.C2 (Continued)
Input data for Part (d): P 20 kN,
AB = 0.25 m, BC = 0.40 m,
AC = 0.65 m, TL = 8 mm, WL = 36 mm,
TAC 8 mm, WAC 50 mm
(d) Answer: 18 mm d 22 mm (d)
Check: For d = 22 mm, Tau AC = 81.25 MPa < 90 MPa O.K.
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81
PROBLEM 1.C3
0.5 in.
B
1.8 in.
5 kips
5 kips
60⬚
A
0.5 in.
45⬚
1.8 in.
C
Two horizontal 5-kip forces are applied to Pin B of the assembly
shown. Each of the three pins at A, B, and C has the same diameter d
and is double shear. (a) Write a computer program to calculate for
values of d from 0.50 to 1.50 in., using 0.05-in. increments, (i) the
maximum value of the average normal stress in member AB, (ii) the
average normal stress in member BC, (iii) the average shearing stress
in pin A, (iv) the average shearing stress in pin C, (v) the average
bearing stress at A in member AB, (vi) the average bearing stress at C
in member BC, and (vii) the average bearing stress at B in member BC.
(b) Check your program by comparing the values obtained for
d 0.8 in. with the answers given for Problems 1.60 and 1.61. (c) Use
this program to find the permissible values of the diameter d of the
pins, knowing that the allowable values of the normal, shearing, and
bearing stresses for the steel used are, respectively, 22 ksi, 13 ksi, and
36 ksi. (d) Solve Part c, assuming that a new design is being
investigated in which the thickness and width of the two members are
changed, respectively, from 0.5 to 0.3 in. and from 1.8 to 2.4 in.
SOLUTION
Forces in members AB and BC.
Free body: Pin B.
From force triangle:
F
FAB
2P
BC
sin 45 sin 60 sin 75
FAB 2 P (sin 45/sin 75)
FBC 2 P (sin 60/sin 75)
(i)
Max. ave. stress in AB.
(ii)
Ave. stress in BC.
Width w
ABC wt
Thickness t
BC FBC / ABC
AAB ( w d ) t
AB FAB / AAB
(iv)
(iii) Pin A.
C ( FBC /2) /( d 2 /4)
A ( FAB /2)/( d 2 /4)
(v)
Pin C.
(vi)
Bearing stress at A.
Bearing stress at C.
Sig Bear C FBC /dt
Sig Bear A FAB /dt
(vii) Bearing stress at B in member BC.
Sig Bear B FBC /2dt
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82
PROBLEM 1.C3 (Continued)
Program Outputs
Input data for Parts (a), (b), (c):
P = 5 kips, w = 1.8 in., t = 0.5 in.
(c) Answer: 0.70 in. d 1.10 in.
(c)
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83
PROBLEM 1.C3 (Continued)
Input data for Part (d),
P = 5 kips, w 2.4 in., t 0.3 in.
(d) Answer: 0.85 in. d 1.25 in.
(d)
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84
PROBLEM 1.C4
a
D
P
b
A
B
15 in.
C
18 in.
12 in.
A 4-kip force P forming an angle with the vertical is applied as
shown to member ABC, which is supported by a pin and bracket at C
and by a cable BD forming an angle with the horizontal. (a) Knowing
that the ultimate load of the cable is 25 kips, write a computer program
to construct a table of the values of the factor of safety of the cable for
values of and from 0 to 45, using increments in and
corresponding to 0.1 increments in tan and tan . (b) Check that
for any given value of , the maximum value of the factor of safety is
obtained for 38.66 and explain why. (c) Determine the smallest
possible value of the factor of safety for 38.66, as well as the
corresponding value of , and explain the result obtained.
SOLUTION
(a)
Draw F.B. diagram of ABC:
M C 0 : (P sin )(1.5 in.) ( P cos )(30 in.)
( F cos )(15 in.) ( F sin )(12 in.) 0
15 sin 30 cos
15 cos 12 sin
F .S . Fult /F
FP
Output for P 4 kips and Fult 20 kips:
(b)
When 38.66°, tan 0.8 and cable BD is perpendicular to the lever arm BC.
(c)
F .S . 3.579 for 26.6; P is perpendicular to the lever arm AC.
Note: The value F . S . 3.579 is the smallest of the values of F.S. corresponding to 38.66 and the
largest of those corresponding to 26.6. The point 26.6, 38.66 is a “saddle point,” or
“minimax” of the function F .S . ( , ).
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85
P
a
b
PROBLEM 1.C5
A load P is supported as shown by two wooden members of uniform rectangular cross
section that are joined by a simple glued scarf splice. (a) Denoting by U and U ,
respectively, the ultimate strength of the joint in tension and in shear, write a
computer program which, for given values of a, b, P, U and U , expressed in either
SI or U.S. customary units, and for values of from 5 to 85 at 5 intervals, can be
used to calculate (i) the normal stress in the joint, (ii) the shearing stress in the joint,
(iii) the factor of safety relative to failure in tension, (iv) the factor of safety relative
to failure in shear, and (v) the overall factor of safety for the glued joint. (b) Apply
this program, using the dimensions and loading of the members of Probs. 1.29 and
1.31, knowing that U 150 psi and U 214 psi for the glue used in Prob. 1.29, and
that U 1.26 MPa and U 1.50 MPa for the glue used in Prob. 1.31. (c) Verify in
each of these two cases that the shearing stress is maximum for a 45.
a
P'
SOLUTION
(i) and (ii) Draw the F.B. diagram of lower member:
Fx 0: V P cos 0
Fy 0:
F P sin 0
V P cos
F P sin
Area ab/sin
Normal stress:
F
( P/ab) sin 2
Area
Shearing stress:
V
( P/ab) sin cos
Area
(iii)
F.S. for tension (normal stresses):
FSN U /
(iv)
F.S. for shear:
FSS U /
(v)
Overall F.S.:
F.S. The smaller of FSN and FSS.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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86
PROBLEM 1.C5 (Continued)
Program Outputs
Problem 1.29
a 150 mm
b 75 mm
P 11 kN
U 1.26 MPa
U 1.50 MPa
ALPHA
SIG (MPa)
TAU (MPa)
FSN
FSS
FS
5
0.007
0.085
169.644
17.669
17.669
10
0.029
0.167
42.736
8.971
8.971
15
0.065
0.244
19.237
6.136
6.136
20
0.114
0.314
11.016
4.773
4.773
25
0.175
0.375
7.215
4.005
4.005
30
0.244
0.423
5.155
3.543
3.543
35
0.322
0.459
3.917
3.265
3.265
40
0.404
0.481
3.119
3.116
3.116
45
0.489
0.489
2.577
3.068
2.577
(b), (c)
50
0.574
0.481
2.196
3.116
2.196
55
0.656
0.459
1.920
3.265
1.920
60
0.733
0.423
1.718
3.543
1.718
65
0.803
0.375
1.569
4.005
1.569
70
0.863
0.314
1.459
4.773
1.459
75
0.912
0.244
1.381
6.136
1.381
80
0.948
0.167
1.329
8.971
1.329
85
0.970
0.085
1.298
17.669
1.298
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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87
PROBLEM 1.C5 (Continued)
Problem 1.31
a 5 in.
b 3 in.
P 1400 lb
U 150 psi
U 214 psi
ALPHA
SIG (psi)
TAU (psi)
FSN
FSS
FS
5
0.709
8.104
211.574
26.408
26.408
10
2.814
15.961
53.298
13.408
13.408
15
6.252
23.333
23.992
9.171
9.171
20
10.918
29.997
13.739
7.134
7.134
25
16.670
35.749
8.998
5.986
5.986
30
23.333
40.415
6.429
5.295
5.295
35
30.706
43.852
4.885
4.880
4.880
40
38.563
45.958
3.890
4.656
3.890
45
46.667
46.667
3.214
4.586
3.214
(c)
50
54.770
45.958
2.739
4.656
2.739
55
62.628
43.852
2.395
4.880
2.395
60
70.000
40.415
2.143
5.295
2.143
65
76.663
35.749
1.957
5.986
1.957
70
82.415
29.997
1.820
7.134
1.820
75
87.081
23.333
1.723
9.171
1.723
80
90.519
15.961
1.657
13.408
1.657
85
92.624
8.104
1.619
26.408
1.619
(b)
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88
PROBLEM 1.C6
Top view
200 mm
180 mm
12 mm
8 mm
A
B
C
B
A
C
B
20 mm
P
8 mm
8 mm
D
D
12 mm
Front view
Side view
Member ABC is supported by a pin and
bracket at A and by two links, which are pinconnected to the member at B and to a fixed
support at D. (a) Write a computer program to
calculate the allowable load Pall for any given
values of (i) the diameter d1 of the pin at A,
(ii) the common diameter d2 of the pins at B
and D, (iii) the ultimate normal stress U in
each of the two links, (iv) the ultimate shearing
stress U in each of the three pins, and (v) the
desired overall factor of safety F.S. (b) Your
program should also indicate which of the
following three stresses is critical: the normal
stress in the links, the shearing stress in the pin
at A, or the shearing stress in the pins at B and
D. (c) Check your program by using the data
of Probs. 1.55 and 1.56, respectively, and
comparing the answers obtained for Pall with
those given in the text. (d) Use your program to
determine the allowable load Pall, as well as
which of the stresses is critical, when d1
d 2 15 mm, U 110 MPa for aluminum links,
U 100 MPa for steel pins, and F.S. 3.2.
SOLUTION
(a)
F.B. diagram of ABC:
200
FBD
380
200
M B 0: P
FA
180
M A 0: P
(i)
For given d1 of Pin A:
FA 2( U /FS )( d12 /4),
P1
200
FA
180
(ii)
For given d 2 of Pins B and D :
FBD 2( U /FS )( d 22 /4),
P2
200
FBD
380
(iii)
For ultimate stress in links BD:
(iv)
For ultimate shearing stress in pins: P4 is the smaller of P1 andP2 .
(v)
For desired overall F.S.:
FBD 2 ( U /FS )(0.02)(0.008), P3
200
FBD
380
P5 is the smaller of P3 and P4 .
If P3 < P4 , stress is critical in links.
If P4 < P3 and P1 < P2 , stress is critical in Pin A.
If P4 P3 and P2 P1 , stress is critical in Pins B and D.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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89
PROBLEM 1.C6 (Continued)
Program Outputs
(b)
Problem 1.55.
Data: d1 8 mm, d 2 12 mm, U , 250 MPa, U 100 MPa,
F .S . 3.0
Pall 3.72 kN. Stress in Pin A is critical.
(c)
Problem 1.56.
Data: d1 10 mm, d 2 12 mm, U 250 MPa, U 100 MPa, F .S . 3.0
Pall 3.97 kN. Stress in Pins B and D is critical.
(d)
Data:
d1 d 2 15 mm, U 110 MPa, U 100 MPa, F .S . 3.2
Pall 5.79 kN. Stress in links is critical.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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90
CHAPTER 7
PR
ROBLEM 7.1
7
4 ksi
3 ksi
708
8 ksi
Foor the given sttate of stress, determine thee normal and shearing stressses exerted
onn the oblique face of the shhaded trianguular element shown.
s
Use a method of
anaalysis based on
o the equilibrrium of that ellement, as waas done in the derivations
of Sec. 7.1A.
SOLUTION
F
0:
A
8 A cos 20
2 cos 20
8cos 2 20
3cos 20 sin 20
3 A cos 20 sin 20
2
3 sin 20 cos 20
3 A sin 200 cos 20
4sin 2 20
4 A sin 20 sin 20
2
0
0
9.46 ksi
F
0:
A
8 A cos 200 sin 20
8coos 20 sin 20
3(cos2 20
3A
A cos 20 cos 20
sin 2 20 )
3 A sin 200 sin 20
4A
A sin 20 cos 200
0
4 20 cos 200
4sin
1.013 ksi
PRO
OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
Not authorized
a
for salee or distribution inn any manner. Thiis document may not
n be copied, scannned, duplicated, forwarded, distribbuted, or posted
on a website,
w
in whole or part.
1027
PRO
OBLEM 7.2
2
60 MPa
For th
he given statee of stress, dettermine the noormal and sheearing stressess exerted on
the oblique face off the shaded triangular
t
elem
ment shown. Use
U a methodd of analysis
based
d on the equilibbrium of that element,
e
as waas done in the derivations off Sec. 7.1A.
608
90 MPa
M
SO
OLUTION
F
0:
90
9 A sin 30 coss 30
A
180sin 30 cos
c 30
90 A cos 30 sin 30
60 A cos 30 ccos 30
0
60 coos 2 30
3
32.9
M Pa
F
0:
A
90
0 A sin 30 sin 30
3
90(cos 2 30
sin 2 30 )
90 A cos 30 cos 30
60 A cos 30 sinn 30
0
60 cos 30 sin 30
7
71.0
M Pa
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use.
Not authorized for salle or distribution in
i any manner. Thhis document may not be copied, scaanned, duplicated,, forwarded, distributed, or posted
on a website, in wholee or part.
1028
PROBLEM
M 7.3
10 ksi
For the giveen state of sttress, determiine the normaal and sheariing stresses
exerted on thhe oblique faace of the shaaded triangulaar element shoown. Use a
method of annalysis based on the equilibbrium of that element, as was
w done in
the derivationns of Sec. 7.1A
A.
6 ksi
758
4 ksi
SOLUTION
F
0:
A
4 A cos15 sin15
4 co
os15 sin15
10 cos 2 15
100 A cos15 cos115
6sin 2 15
6 A sin15 sin15
4 A sin15 cos155
0
4
4sin15
cos155
1
10.93
ksi
F
0:
A
4(ccos2 15
4 A cos15 cos15
10 A cos15 sin 15
sin 2 15 )
6) cos15 sin15
(10
6 A sin15 cos15
4 A sin15 sin155
0
0
0.536
ksi
PRO
OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
Not authorized
a
for salee or distribution inn any manner. Thiis document may not
n be copied, scannned, duplicated, forwarded, distribbuted, or posted
on a website,
w
in whole or part.
1029
80 MPa
M
PROBLEM
P
7.4
For
F the given state
s
of stress, determine thhe normal andd shearing streesses exerted
on
o the obliquee face of the shaded trianggular element shown. Use a method of
analysis
a
based on the equilibbrium of that element,
e
as was
w done in thee derivations
of
o Sec. 7.1A.
40 MPa
558
SO
OLUTION
Streesses
F
0
0:
A
80 A cos 55 cos555
80 cos 2 55
F
0
0:
A
Areas
Forces
40 A sin 55 sin 55
40sin 2 55
80 A cos 55 sin 55
5
0
0.5521 MPa
40 A sin 55 cos 55
5 MPa
56.4
1 cos 55 sin 55
120
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use.
Not authorized for salle or distribution in
i any manner. Thhis document may not be copied, scaanned, duplicated,, forwarded, distributed, or posted
on a website, in wholee or part.
1030
PROBLEM 7.5
40 MPa
35 MPa
60 MPa
For the given state of stress, determine (a) the principal planes, (b) the principal
stresses.
SOLUTION
60 MPa
x
(a)
tan 2
2
xy
p
x
2
p
y
(2)(35)
60 40
y
40 MPa
xy
35 MPa
3.50
74.05
37.0 , 53.0
p
2
(b)
x
max, min
y
x
2
60 40
2
y
2
xy
2
60 40
2
2
(35)2
50 36.4 MPa
max
min
13.60 MPa
86.4 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1031
PROBLEM 7.6
10 ksi
For the given state of stress, determine (a) the principal planes, (b) the principal
stresses.
2 ksi
3 ksi
SOLUTION
x
(a)
tan 2
2 ksi
2
p
p
y
3 ksi
xy
(2)( 3)
2 10
xy
x
2
10 ksi
y
0.750
36.87
p
18.4 , 108.4 ◄
2
(b)
x
max,min
x
y
2
2
10
2
6
y
2
2
10
2
2
xy
2
( 3)2
5 ksi
max
11.00 ksi ◄
min
1.000 ksi ◄
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1032
PROBLEM 7.7
30 MPa
For the given state of stress, determine (a) the principal planes, (b) the principal
stresses.
150 MPa
80 MPa
SOLUTION
x
(a)
tan 2
150 MPa,
2
p
2
p
y
53.130
80 MPa
xy
2( 80 MPa)
( 150 MPa 30 MPa)
xy
x
30 MPa,
y
1.33333 MPa
and 126.870
p
(b)
max,min
x
y
x
2
y
2
150 MPa
30 MPa
2
90 MPa
26.6 and 63.4 ◄
2
xy
150 MPa
30 MPa
2
2
( 80 MPa)2
100 MPa
max
min
190.0 MPa ◄
10.00 MPa ◄
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
1033
PROBLEM 7.8
12 ksi
8 ksi
For the given state of stress, determine (a) the principal planes, (b) the principal
stresses.
18 ksi
SOLUTION
x
(a)
tan 2
18 ksi
2
p
2
p
(2)(8)
18 12
xy
x
12 ksi
y
y
xy
8 ksi
0.5333
28.07
14.0 , 104.0 ◄
p
2
(b)
max,min
x
y
x
2
18
12
2
y
2
18
12
2
2
xy
2
(8)2
3 17 ksi
max
min
20.0 ksi ◄
14.00 ksi ◄
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1034
PROBLEM 7.9
40 MPa
35 MPa
For the given state of stress, determine (a) the orientation of the planes of
maximum in-plane shearing stress, (b) the maximum in-plane shearing stress,
(c) the corresponding normal stress.
60 MPa
SOLUTION
x
(a)
tan 2
2
x
s
s
60 40
(2)(35)
y
2
xy
60 MPa
y
40 MPa
xy
35 MPa
0.2857
15.95
s
8.0 , 98.0
2
(b)
(c)
x
max
y
2
xy
2
60 40
2
2
x
y
ave
2
(35) 2
max
60 40
2
36.4 MPa
50.0 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1035
PROBLEM 7.10
10 ksi
For the given state of stress, determine (a) the orientation of the planes of
maximum in-plane shearing stress, (b) the maximum in-plane shearing stress,
(c) the corresponding normal stress.
2 ksi
3 ksi
SOLUTION
2 ksi
x
(a)
tan 2
2
y
x
s
2 10
(2)( 3)
y
2
10 ksi
xy
xy
3 ksi
1.33333
53.13
s
s
26.6 , 63.4
2
(b)
x
max
y
2
xy
2
2
10
2
( 3)2
2
max
(c)
ave
x
y
2
2
5.00 ksi
10
2
6.00 ksi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1036
PROBLEM 7.11
30 MPa
For the given state of stress, determine (a) the orientation of the planes of
maximum in-plane shearing stress, (b) the maximum in-plane shearing stress,
(c) the corresponding normal stress.
150 MPa
80 MPa
SOLUTION
150 MPa,
x
(a)
tan 2
2
x
s
30 MPa,
150 30
2( 80)
y
2
y
xy
xy
80 MPa
0.750
36.87 and 216.87
s
s
18.4 and 108.4
2
(b)
x
max
y
2
xy
2
150
30
2
( 80)2
2
max
(c)
ave
x
100.0 MPa
y
2
150 30
2
90.0 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1037
PROBLEM 7.12
12 ksi
8 ksi
For the given state of stress, determine (a) the orientation of the planes of
maximum in-plane shearing stress, (b) the maximum in-plane shearing stress,
(c) the corresponding normal stress.
18 ksi
SOLUTION
18 ksi
x
(a)
tan 2
2
x
s
18 12
(2)(8)
y
2
12 ksi
y
xy
xy
8 ksi
1.875
61.93
s
31.0 , 59.0
s
2
(b)
x
max
y
2
xy
2
18
12
2
(8)2
2
max
(c)
ave
x
y
2
17.00 ksi
18 12
2
3.00 ksi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1038
PROBLEM 7.13
8 ksi
5 ksi
For the given state of stress, determine the normal and shearing stresses after the
element shown has been rotated through (a) 25 clockwise, (b) 10 counterclockwise.
SOLUTION
x
x
y
2
0
8 ksi
y
x
4 ksi
y
25
2
x
xy
y
y
2
x
y
2
4 sin ( 50 ) 5 cos ( 50 )
xy
xy
sin 2
cos 2
cos 2
xy
sin 2
2.40 ksi
x
0.1498 ksi
xy
4 4 cos ( 50 ) 5 sin ( 50)
y
2
cos 2 +
sin 2 +
4 4 cos ( 50°) + 5 sin ( 50°)
xy
10
y
50
x
(b)
y
2
x
4 ksi
2
x
(a)
x
2
xy
y
y
2
x
x
5 ksi
xy
10.40 ksi
y
20
4 4 cos (20°) + 5 sin (20°)
4 sin (20°) + 5 cos (20°)
4 4 cos (20°)
5 cos (20°)
x
1.951 ksi
xy
6.07 ksi
y
6.05 ksi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1039
PROBLEM 7.14
90 MPa
30 MPa
For the given state of stress, determine the normal and shearing stresses after
the element shown has been rotated through (a) 25 clockwise, (b) 10
counterclockwise.
60 MPa
SOLUTION
x
x
y
2
60 MPa
x
y
(a)
25
2
x
2
sin 2 +
y
2
x
y
2
xy
xy
sin 2
cos 2
cos 2
xy
sin 2
x
56.2 MPa
xy
38.2 MPa
15 75 cos ( 50 ) 30 sin ( 50 )
y
86.2 MPa
20
15 75 cos (20°) + 30 sin (20°)
75 sin (20°) + 30 cos (20°)
xy
y
y
75 sin ( 50 ) 30 cos ( 50 )
y
2
cos 2 +
15 75 cos ( 50 ) 30 sin ( 50 )
xy
10
y
50
x
(b)
x
30 MPa
xy
75 MPa
2
x
x
y
2
2
xy
y
x
15 MPa
x
90 MPa
y
15 75 cos (20°) 30 sin (20°)
x
45.2 MPa
xy
53.8 MPa
y
75.2 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1040
PROBLEM 7.15
12 ksi
For the given state of stress, determine the normal and shearing stresses after
the element shown has been rotated through (a) 25 clockwise, (b) 10
counterclockwise.
8 ksi
6 ksi
SOLUTION
x
x
y
2
8 ksi
x
2 ksi
x
x
y
2
x
2
y
2
xy
xy
sin 2
cos 2
cos 2
xy
sin 2
2 10 cos ( 50 ) 6 sin ( 50 )
9.02 ksi
x
10 sin ( 50 ) 6 cos ( 50 )
y
2
y
sin 2 +
cos 2 +
2 10 cos ( 50 ) 6 sin ( 50 )
xy
10
y
6 ksi
50
x
(b)
y
2
2
x
25
x
xy
10 ksi
2
x
(a)
y
2
xy
y
12 ksi
y
xy
3.80 ksi
13.02 ksi
y
20
x
xy
y
2 10 cos (20°) 6 sin (20°)
x
10 sin (20°) 6 cos (20°)
2 10 cos (20°) + 6 sin (20°)
5.34 ksi
xy
9.06 ksi
y
9.34 ksi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1041
PROBLEM 7.16
80 MPa
For the given state of stress, determine the normal and shearing stresses after the
element shown has been rotated through (a) 25 clockwise, (b) 10 counterclockwise.
50 MPa
SOLUTION
x
x
y
2
0
80 MPa
y
x
x
y
(a)
25
2
2
sin 2 +
y
2
x
y
2
xy
xy
sin 2
xy
sin 2
cos 2
cos 2
40 sin ( 50°) 50 cos ( 50 )
40 40 cos ( 50 ) 50 sin ( 50 )
y
2
y
cos 2
40 40 cos ( 50 ) 50 sin ( 50°)
xy
10
y
40 MPa
50
x
(b)
x
2
x
x
y
2
2
xy
y
x
40 MPa
50 MPa
xy
x
24.0 MPa
1.498 MPa
xy
y
104.0 MPa
x
19.51 MPa
xy
60.7 MPa
y
60.5 MPa
20
x
xy
y
40 40 cos (20°) 50 sin (20°)
40 sin (20°) 50 cos (20°)
40 40 cos (20°) + 50 sin (20°)
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1042
PROBLEM 7.17
250 psi
The grain of a wooden member forms an angle of 15° with the vertical. For the state
of stress shown, determine (a) the in-plane shearing stress parallel to the grain, (b) the
normal stress perpendicular to the grain.
158
SOLUTION
x
(a)
0
y
x
xy
0
250 psi
xy
y
sin 2
2
250cos( 30 )
15
xy cos 2
xy
(b)
x
x
0
y
2
0
x
y
cos 2
2
250sin( 30 )
217 psi
xy sin 2
x
125.0 psi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1043
PROBLEM 7.18
1.8 MPa
The grain of a wooden member forms an angle of 15° with the vertical. For the
state of stress shown, determine (a) the in-plane shearing stress parallel to the
grain, (b) the normal stress perpendicular to the grain.
3 MPa
158
SOLUTION
x
3 MPa
15
(a)
2
x
xy
1.8 MPa
y
y
2
3
1.8
2
xy
0
30
sin 2
xy sin 2
sin( 30 )
0
0.300 MPa
xy
(b)
x
x
y
x
2
3
y
2
1.8
2
3
1.8
2
cos 2
cos( 30 )
xy sin 2
0
x
2.92 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1044
P'
80 mm
m
P
PROBLEM
7
7.19
1200 mm
Tw
wo wooden members
m
of 800 120-mm uniform
u
rectanngular cross
seection are joined by the simpple glued scarrf splice shownn. Knowing
22 and
thhat
a that the maximum
m
allow
wable stresses in the joint
arre, respectivelly, 400 kPa inn tension (perrpendicular to the splice)
annd 600 kPa inn shear (parallel to the splicce), determinee the largest
ceentric load P thhat can be appplied.
b
P
SOLUTION
Forces
Areeas
A (80) (120) 9.6 103 mm 2
N all
a
all
Fy
0: N
Sall
Fx
aall
A/sin
P sin
A/sin
0:: S
P cos
9.6 10 3 m 2
(4400 103 )(9.6 10 3 )
10.2251 103 N
sin 22
0
P
N
sinn
10.251 103
sin 222
27.4 1103 N
(6600 103 )(9.6 10 3 )
15.3376 103 N
sin 22
0
P
S
coos
Thee smaller valuee for P governns.
15.376 103
cos 22
2
16.58 103 N
P 16.58 kN
PRO
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TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
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a
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in whole or part.
1045
P'
PROBLEM 7.20
7
1220 mm
80 mm
Tw
wo wooden members
m
of 800 120-mm uniform
u
rectanngular cross
section are joineed by the simp
mple glued scarrf splice show
wn. Knowing
25 and
a that centriic loads of magnitude
m
P 10 kN are
that
t
in-plane
appplied to the members ass shown, dettermine (a) the
shhearing stresss parallel to the splice, (b) the noormal stress
peerpendicular too the splice.
b
P
SO
OLUTION
Forcess
A
Areas
A (80)(1220) 9.6 103 mm
m 2
(a)
Fx
0: S
N
A/sin
(b)
Fy
0: N
N
A/sin
P cos
0
S
P cos
(9.063 103 )sinn 25
9.6 10 3
P sin
n
0
N
P sin
(4.226 103 )sin 25
9.6 103
9.6 100 3 m 2
(10 103 ) cos 25
9.063 103 N
399 1003 Pa
(10 103 )sin 25
186.0 103 Pa
399 kPa
4.226 103 N
1
186.0
kPa
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use.
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1046
P
PROBL
LEM 7.21
The centrric force P is applied to a short post as shown. Know
wing that the stresses on
5 ksi, determinne (a) the anggle that planne a-a forms
15 ksi and
plane a-a are
with the horizontal,
h
(b) the maximum
m compressivee stress in the post.
p
a
!
a
SOLUTION
x
0
xy
0
y
(a)
P/ A
From the Mohr’s
M
circle,
tan
5
15
P
2A
(b)
P
A
2( )
1 co
os 2
0.33333
18.4
P
cos 2
2A
(2)(115)
1 coss 2
P
16.67
1
ksi
A
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1047
PROBLEM 7.2
22
a
a
25"
50 mm
m
Two
o members of uniform crosss section 50 80 mm are glued
g
togetherr along plane
a-a that forms ann angle of 255 with the horizontal.
h
Knnowing that thhe allowable
800 kPa and
a
600 kPa,
k
determinne the largest
stressses for the gluued joint are
centtric load P thatt can be applieed.
P
SO
OLUTION
Forr plane a-a,
65 .
x
0,
0
x
P
0,
cos 2
x
y
y
P
A
sin 2
2
xy
sin
cos
0
(50 10 3 )(80 10 3 )(800 103 )
sin 2 65
6
A
sin
s 2 65
(
P
xy
y )sin
A
sin
s 65 cos 65
P 2
sin 655
A
0
3.90 103 N
P
sin 65 cos 65 0
A
(50 10 3 )((80 10 3 )(600 103 )
6.277 103 N
sinn 65 cos 65
cos
(
xy (cos
Alllowable value of P is the sm
maller one.
2
sin 2 )
P
3.90 kN
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use.
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1048
PROBLEM 7.23
0.2 m
0.15 m
3 kN
The axle of an automobile is acted upon by the forces and couple
shown. Knowing that the diameter of the solid axle is 32 mm,
determine (a) the principal planes and principal stresses at point H
located on top of the axle, (b) the maximum shearing stress at the
same point.
H
350 N · m
3 kN
SOLUTION
1
d
2
c
Tc
J
Torsion:
I
Bending:
4
2T
c3
c4
4
1
(32)
2
2(350 N m)
(16 10 3 m)3
(16 10 3 )4
(0.15m)(3 103 N)
M
16 mm 16 10 3 m
54.399 MPa
51.472 10 9 m 4
450 N m
(450)(16 10 3 )
51.472 10 9
My
I
54.399 106 Pa
Top view:
139.882 106 Pa
139.882 MPa
Stresses:
x
139.882 MPa
1
(
2
x
y)
R
x
y
ave
y
0
1
( 139.882
2
xy
0)
54.399 MPa
69.941 MPa
2
(a)
2
max
ave
R
69.941
88.606
min
ave
R
69.941
88.606
2
xy
( 69.941)2
( 54.399) 2
88.606 MPa
max
min
18.67 MPa
158.5 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1049
PROBLEM 7.23 (Continued)
tan 2
2
p
x
xy
y
(2)( 54.399)
139.882
0.77778
2
p
37.88
p
(b)
max
R
88.6 MPa
18.9
max
and 108.9°
88.6 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1050
6 in.
PROBLEM 7.24
C
H
A 400-lb vertical force is applied at D to a gear attached to the solid l-in.
diameter shaft AB. Determine the principal stresses and the maximum
shearing stress at point H located as shown on top of the shaft.
B
A
D
2 in.
400 lb
SOLUTION
Equivalent force-couple system at center of shaft in section at point H:
Shaft cross section:
V
400 lb
T
(400)(2)
d
1 in. c
J
c4
2
1
d
2
(400)(6)
2400 lb in.
800 lb in.
0.5 in.
0.098175 in 4
Torsion:
Tc
J
(800)(0.5)
0.098175
Bending:
Mc
I
(2400)(0.5)
0.049087
Transverse shear:
M
1
J
2
I
0.049087 in 4
4.074 103 psi
4.074 ksi
24.446 103 psi
24.446 ksi
Stress at point H is zero.
x
ave
24.446 ksi,
1
(
2
x
y)
x
y
y
0,
xy
4.074 ksi
12.223 ksi
2
R
2
2
xy
(12.223) 2
(4.074) 2
12.884 ksi
a
ave
R
b
ave
R
max
R
a
25.1 ksi
b
0.661 ksi
max
12.88 ksi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1051
PROBLEM 7.25
H
E
A mechanic uses a crowfoot wrench to loosen a bolt at E. Knowing that
the mechanic applies a vertical 24-lb force at A, determine the principal
stresses and the maximum shearing stress at point H located as shown
as on top of the 34 -in. diameter shaft.
6 in.
B
24 lb
A
10 in.
SOLUTION
Equivalent force-couple system at center of shaft in section at point H:
Shaft cross section:
d
V
24 lb M
T
(24)(10)
1
d
2
0.75 in., c
J
2
c4
Tc
J
(240)(0.375)
0.031063
Bending:
Mc
I
(144)(0.375)
0.015532
Transverse shear:
Resultant stresses:
144 lb in.
240 lb in.
0.375 in.
1
J
2
0.031063 in 4 I
Torsion:
(24)(6)
0.015532 in 4
2.897 103 psi
2.897 ksi
3.477 103 psi 3.477 ksi
At point H, the stress due to transverse shear is zero.
x
ave
3.477 ksi,
1
(
2
y
x
y)
x
y
0,
2.897 ksi
xy
1.738 ksi
2
R
2
a
ave
R
b
ave
R
max
2
xy
1.7382
2.897 2
3.378 ksi
a
1.640 ksi
b
R
5.12 ksi
max
3.38 ksi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
1052
P
PROBLEM
7
7.26
y
m
6 mm
A
200 mm
Thhe steel pipe AB
A has a 1022-mm outer diameter and a 6-mm wall
thhickness. Knowing that arm
m CD is rigiddly attached to
t the pipe,
deetermine the principal
p
stressses and the maximum
m
shearing stress
att point K.
51 mm
A
T
D
10 kN
N
C
1 mm
150
H
K
B
x
z
SOLUTION
ro
J
I
do
2
1102
2
51 mm
ri
ro
t
45 mm
ro4 ri4
4.18555 106 mm 4
2
4.18555 10 6 m 4
1
J
2
2.0927 10 6 m 4
Forcce-couple systtem at center of
o tube in the plane
p
containiing points H and
a K:
Fx
10 kN
10 1003 N
My
(10 103 )(200 10 3 )
2000 N m
Mz
(10 103 )(150 10 3 )
15000 N m
Torsion:
At po
oint K, place local
l
x-axis in negative globbal z-directionn.
T
My
c
ro
xy
2000 N m
511 10 3 m
((2000)(51 100 3 )
4.1855 106
24.37 106 Pa
24.37 MPa
Tc
J
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1053
PROBLEM 7.26 (Continued)
Transverse shear:
Stress due to transverse shear V
Fx is zero at point K.
Bending:
|
y|
(1500)(51 10 3 )
2.0927 10 6
|M z |c
I
36.56 106 Pa
36.56 MPa
Point K lies on compression side of neutral axis.
36.56 MPa
y
Total stresses at point K:
x
ave
0,
36.56 MPa,
y
1
(
2
x
y)
x
y
xy
24.37 MPa
18.28 MPa
2
R
2
2
xy
30.46 MPa
max
ave
R
18.28 30.46
max
12.18 MPa
min
ave
R
18.28 30.46
min
48.7 MPa
max
R
max
30.5 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1054
PROBLEM 7.27
#y
20 MPa
60 MPa
For the state of plane stress shown, determine the largest value of y for which
the maximum in-plane shearing stress is equal to or less than 75 MPa.
SOLUTION
x
60 MPa,
y
?,
xy
20 MPa
Let
u
x
y
y
x
R
u2
2
.
Then
Largest value of
y
2
xy
R2
u
y
2u
x
2u
75 MPa
2
xy
752
202
60 (2)(72.284)
is required.
72.284 MPa
84.6 MPa or 205 MPa
y
205 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1055
PROBLEM 7.28
8 ksi
$xy
10 ksi
For the state of plane stress shown, determine (a) the largest value of xy for which
the maximum in-plane shearing stress is equal to or less than 12 ksi, (b) the
corresponding principal stresses.
SOLUTION
x
10 ksi,
8 ksi,
y
xy
2
max
x
R
z
92
122
(a)
xy
(b)
ave
1
(
2
y
2
xy
2
xy
?
10 ( 8)
z
2
2
xy
12 ksi
92
xy
y)
x
7.94 ksi
1 ksi
a
ave
R 1 12 13 ksi
b
ave
R 1 12
a
11 ksi
b
13.00 ksi
11.00 ksi
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1056
P
PROBLEM
7.29
2 MPaa
$xy
12 MPa
75"
For the state of plane stress shown, determ
F
mine (a) the vaalue of xy foor which the
inn-plane shearring stress paarallel to the weld is zeroo, (b) the corrresponding
p
principal
stressses.
SOLUTION
x
Sincce
xy
12 MPa,
y
tan 2
?
15
2
1
(
2
x
y)
tan 2
x
y
2
xyy
y)
7 MPa
M
p
xy
p
x
xy
xy
0, x -direction is a principal direection.
p
(a)
2 MPaa,
y
1
(12 2)) tan( 30 )
2
xy
2.89 MPa
2
R
ave
(b)
2
1
(
2
x
52
2.8992
5.7735 MPa
M
a
ave
R
7 5.77735
a
12..77 MPa
b
ave
R
7 5.77735
b
1.2
226 MPa
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OPRIETARY MAT
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1057
PROBLEM 7.30
15 ksi
8 ksi
Determine the range of values of
is equal to or less than 10 ksi.
x
for which the maximum in-plane shearing stress
#x
SOLUTION
x
Let u
R
x
x
2
u2
2
xy
R2
u
x
y
y
max
2
xy
y
?,
y
15 ksi,
xy
8 ksi
2u
10 ksi
102
2u 15 (2)(6)
82 z
6 ksi
27 ksi or 3 ksi
3 ksi
Allowable range:
x
27 ksi
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1058
PR
ROBLEM 7.31
7
40 MPaa
Soolve Probs. 7.55 and 7.9, usinng Mohr’s circcle.
355 MPa
PR
ROBLEM 7.55 through 7..8 For the givven state of stress,
s
determ
mine (a) the
priincipal planes, (b) the principal stresses.
60 MPa
PR
ROBLEM 7.99 through 7.12 For the giiven state of stress, determ
mine (a) the
oriientation of thhe planes of maximum
m
in-pplane shearing stress, (b) thee maximum
in--plane shearinng stress, (c) thhe correspondiing normal strress.
SOLUTION
x
6 MPa,
60
y
4 MPa,
40
355 MPa
xy
x
ave
y
2
50 MPa
Plottted points forr Mohr’s circlee:
(a)
a
R
(a )
(b )
(c )
x,
xy )
Y:(
y,
xy )
( 40 MPa, 35 MPa)
C:(
ave ,
0)
( 50 MPa, 0)
X 35
GX
CG
G 10
74
4.05
tan
b
(b)
X :(
1
2
180
1
2
( 60 MPa, 35 MPa)
3.5000
37.03
105.995
52.97
CG
C
2
GX
2
10 2
ave
a
R
50 36.4
max
a
ave
R
50 36.4
d
B
45
7.97
e
A
45
97.977
352
36.4 MPa
min
max
86
6.4 MPa
13
3.60 MPa
d
e
R 36.4 MPa
ave
a
53.0
a
min
max
37.0
b
max
50 MPaa
8.0
98.0
36
6.4 MPa
50
0.0 MPa
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1059
PROBLEM 7.32
30 MPa
Solve Probs. 7.7 and 7.11, using Mohr’s circle.
150 MPa
80 MPa
PROBLEM 7.5 through 7.8 For the given state of stress, determine (a) the
principal planes, (b) the principal stresses.
PROBLEM 7.9 through 7.12 For the given state of stress, determine (a) the
orientation of the planes of maximum in-plane shearing stress, (b) the maximum
in-plane shearing stress, (c) the corresponding normal stress.
SOLUTION
x
150 MPa
y
30 MPa
80 MPa
xy
x
ave
y
2
90 MPa
Plotted points for Mohr’s circle:
x
tan 2
p
2
p
y , xy )
C:(
ave ,
30)
(60)2
xy )
0)
(80)2
100
p
90
100
min
ave
R
90
100
max
(90 MPa, 0)
53.130
R
(b′)
(30 MPa, 80 MPa)
80
60
ave
s
(150 MPa, 80 MPa)
60
max
(a′)
(c )
Y:(
2
R
(b)
x,
(150
y
2
(a)
X :(
p
26.6
max
min
45
s
R
18.4
max
and 63.4
190.0 MPa
10.00 MPa
and 108.4
100.0 MPa
90.0 MPa
ave
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1060
PROBLEM 7.33
10 ksi
Solve Prob. 7.10, using Mohr’s circle.
2 ksi
PROBLEM 7.9 through 7.12 For the given state of stress, determine (a) the
orientation of the planes of maximum in-plane shearing stress, (b) the maximum
in-plane shearing stress, (c) the corresponding normal stress.
3 ksi
SOLUTION
x
2 ksi
y
x
ave
10 ksi
y
2
2
10
2
xy
3 ksi
6 ksi
Plotted points for Mohr’s circle:
FX
FC
tan
X: (
x,
Y: (
y , xy )
(10 ksi, 3 ksi)
C: (
ave ,
(6 ksi, 0)
3
4
xy )
0)
(2 ksi, 3 ksi)
0.75
36.87
B
(a)
(b)
(c)
1
2
18.43
D
B
45
E
B
45
R
CF
max
R
5.00 ksi
ave
6.00 ksi
2
26.6
63.4
FX
2
26.6
D
E
42
32
63.4
5 ksi
max
5.00 ksi
6.00 ksi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1061
PROBLEM 7.34
12 ksi
8 ksi
Solve Prob. 7.12, using Mohr’s circle.
18 ksi
PROBLEM 7.9 through 7.12 For the given state of stress, determine (a) the
orientation of the planes of maximum in-plane shearing stress, (b) the
maximum in-plane shearing stress, (c) the corresponding normal stress.
SOLUTION
18 ksi
x
x
ave
12 ksi
y
y
xy
8 ksi
3 ksi
2
Plotted points for Mohr’s circle:
FX
CF
tan
X: (
x,
Y: (
y,
xy )
C: (
ave ,
0)
8
15
xy )
(18 ksi, 8 ksi)
( 12 ksi, 8 ksi)
(3 ksi, 0)
0.5333
28.07
A
(a)
(b)
(c)
1
2
14.04
D
A
45
E
A
45
R
CF
max
R
17.00 ksi
ave
3.00 ksi
2
59.0
D
30.1
FX
2
E
152
82
59.0
30.1
17 ksi
max
17.00 ksi
3.00 ksi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1062
PROBLEM 7.35
8 ksi
5 ksi
Solve Prrob. 7.13, usinng Mohr’s circcle.
PROBL
LEM 7.13 through 7.16 For the given staate of stress, determine
d
the normal and
shearingg stresses afterr the element shown has beeen rotated thrrough (a) 25 clockwise,
(b) 10 counterclockw
c
wise.
SOLUTION
x
0
0,
y
8 ksi,
xy
5 ksi
x
ave
y
4 kssi
2
Plottted points forr Mohr’s circlee:
X : (0, 5 ksi)
k 5 ksi)
Y : (8 ksi,
k 0)
C : (4 ksi,
tan 2
p
2
p
FX 5
1
1.25
FC 4
51.34
R
(a)
25
FC
.
2
X
FX
2
xy
y
10
.
42
52
6.4031 ksi
50
x
(b)
2
2
51.34
50
ave
R cos
1.34
R sin
ave
2.40 ksi
x
0.1497 ksi
xy
R cos
y
10.40 ksi
20
51.34
x
xy
y
ave
20
R cos
R sin
ave
R cos
71.34
x
1.951
1
ksi
xy
6.07 ksi
y
6
6.05
ksi
PRO
OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
Not authorized
a
for salee or distribution inn any manner. Thiis document may not
n be copied, scannned, duplicated, forwarded, distribbuted, or posted
on a website,
w
in whole or part.
1063
PROBLEM
M 7.36
90 MP
Pa
3 MPa
30
Solve Prob. 7.14, using Mohr’s
M
circle.
60 MPa
PROBLEM
M 7.13 througgh 7.16 For thhe given statee of stress, deetermine the
normal and shearing
s
stresses after the element
e
shownn has been rotaated through
(a) 25 clockkwise, (b) 10 counterclockkwise.
SO
OLUTION
60 MP
Pa,
x
y
90 MPa,,
xy
30 MPa
x
ave
y
15 MPa
2
Plootted points for Mohr’s circlle:
X : ( 60 MPa, 30 MPa)
Y : (90 MPa, 300 MPa)
C : (15 MPa, 0)
tan 2
p
2
p
FX
FC
21.80
R
(a)
25
.
30
75
FC
2
0
0.4
10.90
P
FX
X
2
752
300 2
80.78 MP
Pa
50
5
2
2
x
xy
y
2
ave
P
50
21.80
288.20
R cos
R sin
ave
R cos
x
5
56.2
MPa
xy
3
38.2
MPa
y
8 MPa
86.2
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use.
Not authorized for salle or distribution in
i any manner. Thhis document may not be copied, scaanned, duplicated,, forwarded, distributed, or posted
on a website, in wholee or part.
1064
PROB
BLEM 7.36 (Continued)
(
d)
(b)
10
.
2
200
2
x
p
ave
xy
R sin
y
ave
2
21.880
20
41.80
R cos
R cos
x
45
5.2 MPa
xy
53
3.8 MPa
y
755.2 MPa
PRO
OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
Not authorized
a
for salee or distribution inn any manner. Thiis document may not
n be copied, scannned, duplicated, forwarded, distribbuted, or posted
on a website,
w
in whole or part.
1065
PR
ROBLEM 7..37
12 ksi
8 ksi
6 ksi
Solv
ve Prob. 7.15, using Mohr’ss circle.
PRO
OBLEM 7.133 through 7.16 For the giveen state of stress, determinee the normal
and shearing streesses after thee element shoown has beenn rotated through (a) 25
clocckwise, (b) 10 counterclockkwise.
SO
OLUTION
8 ksi,
x
12 ksi,
y
6 ksi
xy
x
y
ave
2 ksi
k
2
Plootted points for Mohr’s circlle:
X : (8 ksi, 6 ksi)
Y : ( 12 ksi, 6 ksi))
C : ( 2 ksi, 0)
(a)
tan 2
p
2
p
25
FX
6
CF 100
30.96
R
CF
.
2
2
0.6
FX
2
x
xy
y
10
.
62
11.66 ksi
k
5
50
5
50
(b)
102
2
ave
30.96
19.04
R cos
R sin
ave
R cos
x
9.02 ksi
xy
3.80 ksi
13.02 ksi
y
2
20
3
30.96
x
xy
y
20
R cos
ave
R sin
ave
50.96
R cos
x
5.34 ksi
xy
9.06 ksi
y
9.34 ksi
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use.
Not authorized for salle or distribution in
i any manner. Thhis document may not be copied, scaanned, duplicated,, forwarded, distributed, or posted
on a website, in wholee or part.
1066
80 MPa
PROB
BLEM 7.38
Solve Prob.
P
7.16, usiing Mohr’s cirrcle.
50 MPa
PROBL
LEM 7.13 thrrough 7.16 Foor the given sttate of stress, determine
d
the normal and
shearinng stresses afteer the elementt shown has been rotated thhrough (a) 25 clockwise,
(b) 10 counterclockkwise.
SOLUTION
x
0,
y
M
80 MPa,
xy
50 MPa
M
x
ave
y
40 MPa
2
Plotted points for Moohr’s circle:
X : (0, 50 MPa)
M
MPa, 50 MPa))
Y : ( 80 M
M 0)
C : ( 40 MPa,
tann 2
p
2
p
R
(a)
25
.
2
FX 50
1.25
CF 40
51.34
2
FX
CF
64.031 MPa
x
xy
y
10
.
402
502
50
51.34
(b)
2
2
x
xy
y
50
1.34
R cos
ave
x
R sinn
1.4497 MPa
xy
R cos
ave
244.0 MPa
y
1044.0 MPa
x
199.51 MPa
xy
600.7 MPa
y
0.5 MPa
60
20
51.34
20
ave
R cos
R sinn
ave
R cos
71.34
PRO
OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
Not authorized
a
for salee or distribution inn any manner. Thiis document may not
n be copied, scannned, duplicated, forwarded, distribbuted, or posted
on a website,
w
in whole or part.
1067
250 psi
PROBLEM 7.39
Solve Prob. 7.17, using Mohr’s circle.
158
PROBLEM 7.17 The grain of a wooden member forms an angle of 15° with the
vertical. For the state of stress shown, determine (a) the in-plane shearing stress
parallel to the grain, (b) the normal stress perpendicular to the grain.
SOLUTION
x
xy
y
0
250 psi
Plotted points for Mohr’s circle:
(a)
xy
X
(0, 250 psi)
Y
(0, 250 psi)
C
(0, 0)
R cos 2
(250 psi)cos30
217 psi
xy
(b)
x
217 psi
R sin 2
(250 psi) sin 30
125.0 psi
x
125.0 psi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
1068
PROBLEM 7.40
1.8 MPa
Solve Prob. 7.18, using Mohr’s circle.
3 MPa
PROBLEM 7.18 The grain of a wooden member forms an angle of 15° with the
vertical. For the state of stress shown, determine (a) the in-plane shearing stress
parallel to the grain, (b) the normal stress perpendicular to the grain.
158
SOLUTION
x
ave
3 MPa
x
1.8 MPa
y
y
xy
0
2.4 MPa
2
Points.
X: (
x,
xy )
( 3 MPa, 0)
Y: (
y,
xy )
( 1.8 MPa, 0)
C: (
ave ,
0)
( 2.4 MPa, 0)
15
CX
(a)
xy
CX sin 30
(b)
x
ave
CX cos 30
2
0.6 MPa
R sin 30
2.4
30
R
0.6 MPa
0.6sin 30
0.6 cos 30
0.300 MPa
2.92 MPa
0.300 MPa
xy
x
2.92 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1069
PROBLEM 7.41
P'
80 mm
Solve Prob. 7.19, using Mohr’s circle.
120 mm
b
P
PROBLEM 7.19 Two wooden members of 80 120-mm uniform
rectangular cross section are joined by the simple glued scarf
22 and that the maximum
splice shown. Knowing that
allowable stresses in the joint are, respectively, 400 kPa in tension
(perpendicular to the splice) and 600 kPa in shear (parallel to the
splice), determine the largest centric load P that can be applied.
SOLUTION
P
,
A
x
0
y
xy
0
Plotted points for Mohr’s circle:
X:
P
,0 ,
A
C:
P
,0
2
R
CX
Y : (0, 0)
P
2A
Coordinates of point Y :
P
(1 cos 2 )
2A
P
sin 2
2A
Data:
A
If
(80)(120)
400 kPa
P
9.6 103 mm 2
400 103 Pa,
(2)(9.6 10 3 )(400 103 )
(1 cos 44 )
2A
1 cos 2
27.4 103 N
If
600 kPa
P
2A
sin 2
9.6 10 3 m 2
27.4 kN
600 103 Pa,
(2)(9.6 10 3 )(600 103 )
(sin 44 )
16.58 103 N
16.58 kN
The smaller value of P governs.
P
16.58 kN
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1070
PROBLEM 7.42
P'
80 mm
Solve Prob. 7.20, using Mohr’s circle.
120 mm
b
P
PROBLEM 7.20 Two wooden members of 80 120-mm uniform
rectangular cross section are joined by the simple glued scarf
25 and that centric loads of
splice shown. Knowing that
magnitude P 10 kN are applied to the members as shown,
determine (a) the in-plane shearing stress parallel to the splice,
(b) the normal stress perpendicular to the splice.
SOLUTION
x
P
A
0
y
xy
0
Plotted points for Mohr’s circle:
X:
P
,0
A
Y : (0, 0)
C:
P
,0
2A
R
CX
P
2A
Coordinates of point Y:
P
(1 cos 2 )
2A
P
sin 2
2A
Data:
A
(80)(120)
9.6 103 mm 2
(a)
(10 103 )sin 50
(2)(9.6 10 3 )
(b)
(10 103 )(1 cos 50 )
(2)(9.6 10 3 )
9.6 10 3 m 2
399 103 Pa
399 kPa
186.0 103 Pa
186.0 kPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1071
PROBLEM 7.43
P
Solve Prob. 7.21, using Mohr’s circle.
PROBLEM 7.21 The centric force P is applied to a short post as shown. Knowing that
5 ksi, determine (a) the angle that
15 ksi and
the stresses on plane a-a are
plane a-a forms with the horizontal, (b) the maximum compressive stress in the post.
a
!
a
SOLUTION
x
0
xy
0
y
P
A
(a)
From the Mohr’s circle,
tan
(b)
P
A
5
0.3333
15
P
P
cos 2
2A 2A
1
2( )
cos 2
1
18.4
(2)(15)
cos 2
16.67 ksi
16.67 ksi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1072
PROBLEM 7.44
Solve Prob. 7.22, using Mohr’s circle.
a
a
25"
50 mm
PROBLEM 7.22 Two members of uniform cross section 50 80 mm are glued
together along plane a-a that forms an angle of 25 with the horizontal. Knowing
that the allowable stresses for the glued joint are
800 kPa and
600 kPa,
determine the largest centric load P that can be applied.
P
SOLUTION
x
0
xy
0
y
P/A
A (50 10 3 )(80 10 3 )
4 10 3 m 2
P
P
(1 cos50 )
2A
2A
1 cos 50
(2)(4 10 3 )(800 103 )
1 cos 50
P 3.90 103 N
P
P
sin 50
2A
P
2A
sin 50
Choosing the smaller value,
(2)(4 10 3 )(600 103 )
sin 50
6.27 103 N
P
3.90 kN
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1073
PROBLEM 7.45
0.2 m
0.15 m
3 kN
Solve Prob. 7.23, using Mohr’s circle.
H
PROBLEM 7.23 The axle of an automobile is acted upon by
the forces and couple shown. Knowing that the diameter of the
solid axle is 32 mm, determine (a) the principal planes and
principal stresses at point H located on top of the axle, (b) the
maximum shearing stress at the same point.
350 N · m
3 kN
SOLUTION
c
Torsion:
1
d
2
1
(32)
2
Tc
J
2T
c3
16 mm 16 10 3 m
2(350 N m)
(16 10 3 m)3
Bending:
I
M
4
c4
4
54.399 106 Pa
(16 10 3 )4
(0.15m)(3 103 N)
My
I
54.399 MPa
51.472 10 9 m 4
450 N m
(450)(16 10 3 )
51.472 10 9
139.882 106 Pa
Top view
Stresses
x
Plotted points:
139.882 MPa
139.882 MPa,
y
X : ( 139.882, 54.399);
ave
1
(
2
0,
xy
54.399 MPa
Y: (0, 54.399); C: ( 69.941, 0)
x
y)
x
y
69.941 MPa
2
R
2
xy
2
139.882
2
2
(54.399)2
88.606 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1074
PROBLEM 7.45 (Continued)
tan 2
2
p
xy
x
y
(2)( 54.399)
139.882
0.77778
(a)
(b)
a
a
ave
R
69.941
88.606
b
ave
R
69.941
88.606
max
R
18.9
,
b
108.9
158.5 MPa
a
b
max
18.67 MPa
88.6 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1075
6 in.
PROBLEM 7.46
C
H
Solve Prob. 7.24, using Mohr’s circle.
B
PROBLEM 7.24 A 400-lb vertical force is applied at D to a gear
attached to the solid 1-in.-diameter shaft AB. Determine the principal
stresses and the maximum shearing stress at point H located as shown
on top of the shaft.
A
D
2 in.
400 lb
SOLUTION
Equivalent force-couple system at center of shaft in section at point H:
V
400 lb
M
T
(400)(2)
800 lb in.
Shaft cross section:
(400)(6)
2400 lb in.
d
1 in.
J
c4
Resultant stresses:
(800)(0.5)
0.098175
Mc
I
Bending:
0.5 in.
0.098175 in 4
Tc
J
Torsion:
Transverse shear:
2
1
d
2
c
I
1
J
2
0.049087 in 4
4.074 103 psi
(2400)(0.5)
0.049087
4.074 ksi
24.446 103 psi
24.446 ksi
Stress at point H is zero.
x
ave
24.446 ksi,
1
(
2
x
y)
x
y
y
0,
xy
4.074 ksi
12.223 ksi
2
R
2
(12.223) 2
a
ave
R
b
ave
R
max
R
2
xy
(4.074) 2
12.884 ksi
a
25.1 ksi
b
0.661 ksi
max
12.88 ksi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1076
PROBLEM 7.47
H
Solve Prob. 7.25, using Mohr’s circle.
E
PROBLEM 7.25 A mechanic uses a crowfoot wrench to loosen a
bolt at E. Knowing that the mechanic applies a vertical 24-lb force at A,
determine the principal stresses and the maximum shearing stress at
point H located as shown as on top of the 34 -in.-diameter shaft.
6 in.
B
24 lb
10 in.
A
SOLUTION
Equivalent force-couple system at center of shaft in section at point H:
V
24 lb
M
T
(24)(10)
240 lb in.
Shaft cross section:
(24)(6)
144 lb in.
d
0.75 in.
J
c4
Resultant stresses:
1
J
2
I
(240)(0.375)
0.031063
Mc
I
Bending:
0.375 in.
0.031063 in 4
Tc
J
Torsion:
Transverse shear:
2
1
d
2
c
0.015532 in 4
2.897 103 psi
(144)(0.375)
0.015532
3.477 103 psi
2.897 ksi
3.477 ksi
At point H, stress due to transverse shear is zero.
x
ave
3.477 ksi,
1
(
2
0,
y
x
y)
x
y
xy
2.897 ksi
1.738 ksi
2
R
2
1.7382
a
ave
R
b
ave
R
max
R
2.8972
xy
2
3.378 ksi
a
5.12 ksi
1.640 ksi
b
max
3.38 ksi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1077
PROBLEM 7.48
y
6 mm
A
200 mm
Solve Prob. 7.26, using Mohr’s circle.
51 mm
A
T
PROBLEM 7.26 The steel pipe AB has a 102-mm outer
diameter and a 6-mm wall thickness. Knowing that arm CD is
rigidly attached to the pipe, determine the principal stresses and
the maximum shearing stress at point K.
D
10 kN
C
150 mm
H
K
B
z
x
SOLUTION
ro
J
I
do
2
102
2
ro4
2
51 mm
ri4
1
J
2
ri
ro
t
4.1855 106 mm 4
45 mm
4.1855 10
6
m4
2.0927 10 6 m 4
Force-couple system at center of tube in the plane containing points H and K:
Fx
My
Mz
Torsion:
10 103 N
(10 103 )(200 10 3 )
(10 103 )(150 10 3 )
T
My
c
ro
xy
2000 N m
1500 N m
2000 N m
51 10 3 m
(2000)(51 10 3 )
4.1855 10 6
Tc
J
24.37 MPa
Note that the local x-axis is taken along a negative global z direction.
Transverse shear:
Bending:
Stress due to V
y
Fx is zero at point K.
(1500)(51 10 3 )
2.0927 10 6
Mz c
I
Point K lies on compression side of neutral axis.
y
36.56 MPa
36.56 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1078
PROBLEM 7.48 (Continued)
Total stresses at point K:
x
ave
0,
1
(
2
36.56 MPa,
y
x
y)
x
y
xy
24.37 MPa
18.28 MPa
2
R
max
min
max
2
ave
ave
R
R
R
2
xy
18.28
18.28
30.46 MPa
30.46
max
12.18 MPa
min
48.7 MPa
30.46
max
30.5 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1079
PROBLEM 7.49
#y
20 MPa
60 MPa
Solve Prob. 7.27, using Mohr’s circle.
PROBLEM 7.27 For the state of plane stress shown, determine the largest
value of y for which the maximum in-plane shearing stress is equal to or
less than 75 MPa.
SOLUTION
x
60 MPa,
y
?,
xy
20 MPa
Given:
max
R
XY
2 R 150 MPa
DY
(2)(
XD
y
75 MPa
XY
x
xy )
2
40 MPa
DY
2
1502
XD 60 144.6
402
144.6 MPa
y
205 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1080
PROBLEM 7.50
8 ksi
$xy
10 ksi
Solve Prob. 7.28, using Mohr’s circle.
PROBLEM 7.28 For the state of plane stress shown, determine (a) the largest
value of xy for which the maximum in-plane shearing stress is equal to or less
than 12 ksi, (b) the corresponding principal stresses.
SOLUTION
The center of the Mohr’s circle lies at point C with coordinates
x
y
2
The radius of the circle is
max (in-plane)
10 8 , 0
2
,0
(1 ksi, 0).
12 ksi.
The stress point ( x , xy ) lies along the line X1 X 2 of the Mohr circle diagram. The extreme points with
R 12 ksi are X 1 and X 2 .
(a)
The largest allowable value of
xy
2
(b)
The principal stresses are
is obtained from triangle CDX.
2
DX 1
DX 2
xy
122
a
1 12
b
1 12
2
CX 1
CD
2
92
xy
a
b
7.94 ksi
13.00 ksi
11.00 ksi
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1081
2 MPa
PROBLEM 7.51
$xy
Solve Prob. 7.29, using Mohr’s circle.
12 MPa
75"
PROBLEM 7.29 For the state of plane stress shown, determine (a) the value
of xy for which the in-plane shearing stress parallel to the weld is zero,
(b) the corresponding principal stresses.
SOLUTION
Point X of Mohr’s circle must lie on X X
that y 2 MPa. The coordinates of C are
so that
2 12 , 0
2
x
12 MPa. Likewise, point Y lies on line Y Y
(7 MPa, 0).
Counterclockwise rotation through 150° brings line CX to CB, where
R
(a)
x
xy
y
2
x
y
2
sec 30
12 2
sec 30
2
0.
5.7735 MPa
tan 30
12 2
tan 30
2
(b)
so
xy
2.89 MPa
a
ave
R
7 5.7735
a
12.77 MPa
b
ave
R
7 5.7735
b
1.226 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1082
PROBLEM 7.52
15 ksi
8 ksi
Solve Prob. 7.30, using Mohr’s circle.
#x
PROBLEM 7.30 Determine the range of values of
in-plane shearing stress is equal to or less than 10 ksi.
x
for which the maximum
SOLUTION
For the Mohr’s circle, point Y lies at (15 ksi, 8 ksi). The radius of limiting circles is R 10 ksi.
Let C1 be the location of the leftmost limiting circle and C2 be that of the rightmost one.
C1Y
10 ksi
C2Y
10 ksi
Noting right triangles C1 DY and C2 DY ,
C1D
2
DY
2
C1Y
2
C1D
2
82
102
C1D
6 ksi
Coordinates of point C1 are (0, 15 6) (0, 9 ksi).
Likewise, coordinates of point C2 are (0, 15 6) (0, 21 ksi).
Coordinates of point X1: (9 6, 8) (3 ksi, 8 ksi)
Coordinates of point X2: (21 6, 8) (27 ksi, 8 ksi)
The point (
x,
xy )
must lie on the line X1 X2.
3 ksi
Thus,
x
27 ksi
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1083
2 MPa
PROBLEM 7.53
$xy
75"
Solve Problem 7.29, using Mohr’s circle and assuming that the weld forms
an angle of 60 with the horizontal.
12 MPa
PROBLEM 7.29 For the state of plane stress shown, determine (a) the value
of xy for which the in-plane shearing stress parallel to the weld is zero,
(b) the corresponding principal stresses.
SOLUTION
12
Locate point C at
Angle XCB
x
y
2
2
7 MPa with
2
0.
120
12
2
2
5 MPa
R
5sec 60
10 MPa
5 tan 60
xy
8.66 MPa
xy
ave
a
7
10
ave
b
7
R
10
a
17.00 MPa
b
3.00 MPa
R
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1084
3 ksi
6 ksi
5 ksi
+
458
PROBLEM 7.54
Determine the principal planes and the principal
stresses for the state of plane stress resulting
from the superposition of the two states of stress
shown.
2 ksi
4 ksi
SOLUTION
Consider state of stress on the right. We shall express it in terms of horizontal and vertical components.
We now can add the two stress elements by superposition.
Principal planes and principal stresses:
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1085
PROBLEM 7.54 (Continued)
ave
x
y
2
1
(6
2
2)
1
(6
2
2)
(4)2
R
tan 2
p
2
p
2
4
(3)2
5
3
4
36.87
p
max
ave
2
R
18.4 , 108.4
5
max
min
ave
R
2
7.00 ksi
5
min
3.00 ksi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1086
PROBLEM 7.55
100 MPa
50 MPa
50 MPa
+
308
75 MPa
Determine the principal planes and the principal
stresses for the state of plane stress resulting
from the superposition of the two states of
stress shown.
SOLUTION
Consider the state of stress on the left. We shall express it in terms of horizontal and vertical components.
x
50 cos 30
43.30
y
43.30
xy
50sin 30
25.0
Principal axes and principal stress:
ave
y
x
2
R
tan 2
p
1
(118.3
2
56.7)
1
(118.3
2
56.7)
(30.8)2
(75)2
75
30.8
2
p
87.5
30.8
81.08
67.67
max
ave
R
87.5
81.08
min
ave
R
87.5
81.08
p
33.8
, and 123.8
max
min
168.6 MPa
6.42 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1087
#0
#0
PROBLEM
M 7.56
#0
#0
Determine thhe principal planes
p
and thhe principal
stresses for the
t state of pllane stress ressulting from
the superposiition of the twoo states of streess shown.
30"
30"
SO
OLUTION
Exppress each state of stress in terms of horizzontal and verrtical componeents.
s
of stresss,
Addding the two states
p
0 and
a 90°
max
m
min
0
0
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use.
Not authorized for salle or distribution in
i any manner. Thhis document may not be copied, scaanned, duplicated,, forwarded, distributed, or posted
on a website, in wholee or part.
1088
PROBLEM 7.57
$0
$0
30"
+
Determine the principal planes and the principal
stresses for the state of plane stress resulting from the
superposition of the two states of stress shown.
SOLUTION
Mohr’s circle for 2nd state of stress:
x
0
y
0
0
xy
xy
3
2
0 sin 60
x
1
2
0 cos 60
0
y
0
y
3
2
0 sin 60
0
0
Resultant stresses:
x
3
2
1
2
0
xy
0
ave
1
(
2
3
2
0
3
2
0
x
y)
x
y
tan 2
2
x
2
2
xy
(2)
xy
y
p
60
a
ave
R
b
ave
R
0
0
2
p
3
2
0
0
2
R
3
2
0
3
2
3
3
2
2
0
3
2
2
0
3
0
3
b
30
a
a
b
60
3
0
3
0
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1089
PROBLEM 7.58
120 MPa
$xy
For the element shown, determine the range of values of
maximum tensile stress is equal to or less than 60 MPa.
xy
for which the
20 MPa
SOLUTION
x
ave
Set
max
R
20 MPa
1
(
2
y)
x
60 MPa
max
120 MPa
y
70 MPa
R
ave
130 MPa
ave
But
2
x
R
2
xy
x
2
2
xy
R
2
x
x
2
1302 502
120.0 MPa
Range of
xy :
120.0 MPa
xy
120.0 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1090
PROBLEM 7.59
120 MPa
$xy
For the element shown, determine the range of values of xy for which the
maximum in-plane shearing stress is equal to or less than 150 MPa.
20 MPa
SOLUTION
20 MPa
x
1(
2
y)
x
Set
max (in-plane)
But
R
120 MPa
y
50 MPa
R 150 MPa
2
x
y
2
xy
2
2
xy
R
2
1502
x
y
2
502
141.4 MPa
Range of
xy :
141.4 MPa
xy
141.4 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1091
!y'
6 ksi
PROBLEM 7.60
"x'y'
!x'
#
16 ksi
For the state of stress shown, determine the
range of values of for which the magnitude
of the shearing stress x y is equal to or less
than 8 ksi.
SOLUTION
x
xy
ave
16 ksi,
0
y
6 ksi
1
(
2
x
y)
x
y
8 ksi
2
R
( 8)2
tan 2
2
p
x
2
2
xy
2
(6) 2
xy
y
p
36.870
b
18.435
10 ksi
(2)(6)
16
0.75
8 ksi for states of stress corresponding to arcs HBK and UAV of Mohr’s circle. The angle
xy
is
calculated from
R sin 2
2
8
8
10
sin 2
53.130
0.8
26.565
k
b
18.435
26.565
45
k
b
18.435
26.565
8.13
u
h
90
45
v
k
90
98.13
Permissible range of :
Also,
h
k
u
v
135
45
45
188.13
and 225
8.13
98.13
278.13
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1092
PROBLEM 7.61
#y'
#x'
For the state of stress shown, determine the
range of values of
for which the normal
stress x is equal to or less than 50 MPa.
%
90 MPa
$x'y'
60 MPa
SOLUTION
x
90 MPa,
0
60 MPa
xy
ave
y
1
(
2
x
y)
x
y
45 MPa
2
R
452
2
tan 2
p
602
xy
x
2
x
2
xy
2
y
p
53.13
a
26.565
75 MPa
(2)( 60)
90
4
3
50 MPa for states of stress corresponding to the arc HBK of Mohr’s circle. From the circle,
R cos 2
50
cos 2
5
75
2
5 MPa
0.066667
86.177
h
2
45
43.089
26.565
a
k
2
k
110.085
h
360
Permissible range of
:
4
43.089
16.524
32.524
360
h
172.355
220.169
k
16.5
Also,
196.5
110.1
290.1
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1093
PROBLEM 7.62
#y'
#x'
For the state of stress shown, determine the
range of values of
for which the normal
stress x is equal to or less than 100 MPa.
%
90 MPa
$x'y'
60 MPa
SOLUTION
x
xy
ave
90 MPa,
y
0
60 MPa
1
(
2
x
y)
x
y
45 MPa
2
R
2
452
2
tan 2
p
xy
x
2
x
602
y
p
53.13
a
26.565
2
xy
75 MPa
(2)( 60)
90
4
3
100 MPa for states of stress corresponding to arc HBK of Mohr’s circle. From the circle,
R cos 2
100
45
cos 2
55
75
0.73333
2
42.833
h
2
55 MPa
21.417
26.565
a
k
2
k
132.02
h
Permissible range of
Also,
360
4
21.417
10.297
is
h
5.15
360
85.666
264.037
k
5.1
132.0
174.8
312.0
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1094
PROBLEM 7.63
#y
$xy
For the state of stress shown, it is known that the normal and shearing stresses are
directed as shown and that x 14 ksi, y 9 ksi, and min 5 ksi. Determine
(a) the orientation of the principal planes, (b) the principal stress max, (c) the
maximum in-plane shearing stress.
#x
SOLUTION
14 ksi,
x
min
9 ksi,
y
R
ave
1
(
2
ave
R
ave
x
y)
11.5 ksi
min
11.5 5 6.5 ksi
2
x
R
y
2
xy
2
2
xy
But it is given that
(a)
tan 2
x
R2
2
xy
2
is positive, thus
6.52
xy
2.52
6 ksi
6 ksi.
xy
p
x
2
y
p
y
(2)(6)
5
67.38
2.4
a
b
(b)
max
ave
max (in-plane)
123.7
R
max
(c)
33.7
18.00 ksi
R
max (in-plane)
6.50 ksi
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1095
!
PROBLEM 7.64
"y
"y'
The Mohr’s circle shown corresponds to the state of stress given
in Fig. 7.5a and b. Noting that x OC (CX )cos (2 p 2 ) and
that x y (CX )sin (2 p 2 ), derive the expressions for x and
given in Eqs. (7.5) and (7.6), respectively. [Hint: Use
xy
sin( A B) sin A cos B cos A sin B and cos ( A B) cos A cos B
sin A sin B.]
Y
Y'
C
O
2#p
2#
!x'y'
X'
"
!xy
X
"x
"x'
SOLUTION
OC
1
(
2
y)
x
CX
CX
x
y
CX cos 2
p
CX cos 2
p
CX sin 2
p
CX sin 2
p
x
OC CX cos (2
p
2 )
OC CX (cos 2
p
cos 2
2
xy
OC CX cos 2
x
y
x
2
xy
p
CX sin 2
p
cos 2
cos 2
y
2
CX sin (2
xy
p
cos 2
sin 2
p
sin 2 )
CX sin 2
xy
p
sin 2
2 )
CX (sin 2
p
cos 2
cos 2
CX cos 2
p
sin 2
x
y
2
sin 2
cos 2
p
sin 2 )
sin 2
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1096
PROBLEM 7.65
2
(a) Prove that the expression x y
x y , where
x , y , and x y are components of the stress along the
rectangular axes x and y , is independent of the orientation of these axes. Also, show that the given
expression represents the square of the tangent drawn from the origin of the coordinates to Mohr’s circle.
(b) Using the invariance property established in part a, express the shearing stress xy in terms of x , y , and
the principal stresses max and min .
SOLUTION
(a)
From Mohr’s circle,
R sin 2
xy
x
y
x
p
ave
R cos 2
p
ave
R cos 2
p
2
xy
y
2
ave
R 2 cos2 2
R 2 sin 2 2
2
ave
R 2 ; independent of
p
p
p.
Draw line OK from origin tangent to the circle at K. Triangle OCK is a right triangle.
OC
OK
2
2
OK
OC
2
2
CK
2
ave
x
(b)
2
CK
2
R2
2
xy
y
x,
Applying above to
x
y
But
x
2
xy
ab
y
a
max
2
xy
x
xy
xy
2
ave
R2
max ,
a
2
xy
and
2
ab
b
0,
y,
b
, and to
a,
b,
min
min
max
y
x
y
min
max
min
The sign cannot be determined from above equation.
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1097
PROBLEM 7.66
y
For the state of plane stress shown, determine the maximum shearing
stress when (a) x 14 ksi and y 4 ksi, (b) x 21 ksi and y 14 ksi.
(Hint: Consider both in-plane and out-of-plane shearing stresses.)
σy
12 ksi
σx
z
x
SOLUTION
(a)
ave
1
(
2
x
y)
1
(14
2
1
(14
2
4)
9
4)
5
(5)2
R
(12)2
13
max
ave
R
9
13
22
min
ave
R
9
13
4
Since max and min have opposite signs, the maximum shearing stress is equal to the maximum inplane shearing stress.
max
R
13.00 ksi
max
13.00 ksi
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1098
PROBLEM 7.66 (Continued)
(b)
ave
1
(
2
x
y)
1
(21 14)
2
1
(21 14)
2
(3.5)2
R
17.5
3.5
(12)2
12.5
max
ave
R
17.5
12.5
30
min
ave
R
17.5
12.5
5
Since max and min have the same sign,
O and A, we have
max
1
2
max
max
is out of the plane of stress. Using Mohr’s circle through
1
(30 ksi)
2
max
15.00 ksi
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1099
PROBLEM 7.67
y
For the state of plane stress shown, determine the maximum shearing
stress when (a) x 20 ksi and y 10 ksi, (b) x 12 ksi and y 5 ksi.
(Hint: Consider both in-plane and out-of-plane shearing stresses.)
σy
12 ksi
σx
z
x
SOLUTION
(a)
ave
1
(20
2
1
( x
2
10)
15
y)
1
(20
2
R
(5)2
10)
5
(12)2
13
max
ave
R
15
13
28
min
ave
R
15
13
2
Since max and min have the same sign,
O and A, we have
max
max
1
2
max
is out of the plane of stress. Using Mohr’s circle through
1
(28 ksi)
2
max
14.00 ksi
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1100
PROBLEM 7.67 (Continued)
(b)
ave
1
(12
2
1
(
2
x
5)
8.5
y)
1
(12
2
R
5)
(3.5)2
3.5
(12)2
12.5
max
ave
R
8.5
12.5
21
min
ave
R
8.5
12.5
4
Since max and min have opposite signs, the maximum shearing stress is equal to the maximum
in-plane shearing stress.
max
R
12.50 ksi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1101
y
PROBLEM 7.68
σy
For the state of stress shown, determine the maximum shearing stress
when (a) y 40 MPa, (b) y 120 MPa. (Hint: Consider both in-plane
and out-of-plane shearing stresses.)
80 MPa
140 MPa
z
x
SOLUTION
(a)
x
ave
140 MPa,
1
(
2
y
x
y)
x
y
40 MPa,
80 MPa
xy
90 MPa
2
R
502 802
a
ave
R 184.34 MPa (max)
b
ave
R
c
4.34 MPa
x
ave
94.34 MPa
(min)
0
1
(
2
1
(
max
2
140 MPa,
max (in-plane)
(b)
2
xy
2
1
(
2
b)
a
y
y)
x
y
1
(
2
120 MPa,
min )
max
x
94.34 MPa
R
b)
a
xy
94.3 MPa
max
94.3 MPa
80 MPa
130 MPa
2
R
2
xy
2
102 802
a
ave
R
210.62 MPa (max)
b
ave
R
49.38 MPa
c
max
80.62 MPa
0 (min)
a
max (in-plane)
max
210.62 MPa
min
c
0
R 86.62 MPa
1
(
2
max
min )
105.3 MPa
max
105.3 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1102
y
PROBLEM 7.69
σy
For the state of stress shown, determine the maximum shearing stress
when (a) y 20 MPa, (b) y 140 MPa. (Hint: Consider both in-plane
and out-of-plane shearing stresses.)
80 MPa
140 MPa
z
x
SOLUTION
(a)
x
ave
140 MPa,
1
(
2
y
x
y)
x
y
20 MPa,
xy
80 MPa
80 MPa
2
R
602
802
100 MPa
a
ave
R 80 100 180 MPa (max)
b
ave
R 80 100
c
x
ave
20 MPa (min)
0
1
(
2
1
(
max
2
140 MPa,
max (in-plane)
(b)
2
xy
2
1
(
2
b)
a
min )
max
y
x
y)
x
y
100 MPa
100 MPa
140 MPa,
xy
max
100.0 MPa
max
110.0 MPa
80 MPa
140 MPa
2
R
2
xy
2
0 802
a
ave
R
220 MPa (max)
b
ave
R
60 MPa
c
80 MPa
0 (min)
max (in-plane)
max
1
(
2
1
(
2
a
max
b)
80 MPa
min )
110 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1103
PROBLEM 7.70
y
For the state of stress shown, determine the maximum shearing stress when
60 MPa, (c) z
60 MPa.
(a) z 0, (b) z
100 MPa
84 MPa
σz
30 MPa
x
z
SOLUTION
The z axis is a principal axis. We determine the other two principal axes by drawing Mohr’s circle for a
rotation in the x y plane.
ave
1
(
2
x
y)
1
(30
2
1
(30
2
R
(35)2
A
ave
B
ave
100)
100)
(84)2
R
R
65
65
65
35
91
91
91
156 MPa
26 MPa
(a)
0. Point Z corresponding to the z axis is located at O between A and B. Therefore, the largest of
the 3 Mohr’s circles is the circle we drew through A and B. We have
R 91.0 MPa
max
(b)
60 MPa. Point Z is located between A and B. The largest of the 3 circles is still the circle
z
through A and B, and we still have
R 91.0 MPa
max
(c)
60 MPa. Point Z is now outside the circle through A and B. The largest circle is the circle
through Z and A.
1
1
108.0 MPa
( ZH )
(60 156)
max
max
2
2
z
z
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1104
PROBLEM 7.71
y
For the state of stress shown, determine the maximum shearing stress
when (a) z 0, (b) z
60 MPa, (c) z
60 MPa.
100 MPa
84 MPa
170 MPa
z
x
z
SOLUTION
ave
1
(
2
x
y)
1
(170
2
1
(170
2
(35)2
R
A
B
(a)
135
100)
35
(84)2
91
91
91
226 MPa
44 MPa
0. Point Z corresponding to the z axis is located at O, outside the circle drawn through A and B.
The largest of the 3 Mohr’s circles is the circle through O and A. We have
z
max
(b)
135
135
100)
1
(OA)
2
1
2
A
1
(226)
2
max
60 MPa. Point Z is located between B and A. The largest of the 3 circles is the one drawn
z
through A and B.
max
(c)
113.0 MPa
R
91.0 MPa
60 MPa. Point Z is located outside the circle drawn through A and B. The largest of the 8
Mohr’s circles is the circle through Z and A. We have
z
max
1
( ZA)
2
1
(60
2
226)
max
143.0 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1105
PRO
OBLEM 7.72
y
For thhe state of strress shown, determine the maximum sheearing stress
when (a) yz 17.5 ksi, (b) yz 8 ksi, (c) yz 0.
τyz
12 ksi
3 ksi
x
z
SO
OLUTION
(a)
(b)
17.55 ksi
yz
R
(6) 2
A
6 18.5
B
6 18.5
A
min
B
max
1
(
2
24.5
12.5
min )
max
(6))2
(8) 2
6 10
1 16
B
6 10
1
18.50 ksi
max
10.00 ksi
10
4
max
A
16 ksi
min
B
4 ksi
max
1
(
2
0
max
3 kssi
x
A
yz
18.5
12.5 ksi
8 kssi
R
(c)
(17.5) 2
24.5 ksi
max
yz
3 ksi
x
min )
max
3 ksi
x
max
z
12 ksi
min
x
3 ksi
max
1
(
2
max
min )
max
7.50 ksi
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use.
Not authorized for salle or distribution in
i any manner. Thhis document may not be copied, scaanned, duplicated,, forwarded, distributed, or posted
on a website, in wholee or part.
1106
PROBLEM 7.73
y
For the state of stress shown, determine the maximum shearing stress
when (a) yz 17.5 ksi, (b) yz 8 ksi, (c) yz 0.
τyz
12 ksi
10 ksi
x
z
SOLUTION
(a)
17.5 ksi
yz
(6)2
R
A
B
max
6
6
A
min
B
max
1
(
2
(17.5)2
18.5
18.5 24.5
18.5
12.5
24.5 ksi
12.5 ksi
max
min )
max
(b)
yz
18.50 ksi
8 ksi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1107
PROBLEM 7.73 (Continued)
(6)2
R
(c)
yz
(8)2
A
6
10
16
B
6
10
4
max
A
min
x
max
1
(
2
10
16 ksi
10 ksi
min )
max
max
13.00 ksi
max
11.00 ksi
0
max
z
min
x
max
1
(
2
12 ksi
10 ksi
max
min )
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1108
PROBLEM 7.74
y
For the state of stress shown, determine the value of
maximum shearing stress is (a) 9 ksi, (b) 12 ksi.
6 ksi
xy
for which the
τ xy
15 ksi
z
x
SOLUTION
15 ksi
1
( x
2
x
(ksi)
ave
y)
x
u
(a)
y
y
R
xy
a
u
18
ave
2
2
xy
R2
u2
R2
c
max
xy
6.00 ksi
u2
11.24 ksi
b
7.5 ksi
For max 12 ksi,
center of Mohr’s circle lies at point C.
R 12 ksi
xy
10.5 12
10.5 12
0
1
( max
2
10.5
4.52
6.00 ksi
a
4.5 ksi
2
7.52
Checking,
10.5 ksi
For max 9 ksi,
center of Mohr’s circle lies at point C.
Lines marked (a) show the limits on max .
Limit on max is max 2 max 18 ksi .
The Mohr’s circle a
max corresponds
to point Aa.
R
(b)
6 ksi
xy
11.24 ksi
22.5 ksi
1.5 ksi
min )
12 ksi
(o.k.)
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1109
PROBLEM 7.75
y
For the state of stress shown, determine the value of
maximum shearing stress is 80 MPa.
70 MPa
xy
for which the
τ xy
120 MPa
z
x
SOLUTION
120 MPa
x
1
(
2
ave
x
2
Assume
min
0
max
2
y)
x
120
y
y
70 MPa
95 MPa
70
25 MPa
2
160 MPa
max
a
max
ave
R
R
max
ave
160
95
65 MPa
652
252
2
x
R2
y
2
xy
2
2
2
xy
R
x
2
y
2
602
xy
b
a
2R
160
130
30 MPa
0
60.0 MPa
(o.k.)
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1110
y
PROBLEM 7.76
σy
For the state of stress shown, determine two values of
maximum shearing stress is 73 MPa.
y
for which the
48 MPa
50 MPa
z
x
SOLUTION
50 MPa,
x
y
u
Let
1
(
2
(1a)
u
ave
55 MPa
1
(
2
a
a
(1b)
u
ave
b
min
R
ave
0
max
ave
78 MPa,
min
2u
y
R
178 MPa,
2
xy
u
R2
u
732
u
x
482
2
xy
55 MPa
5 MPa
b
y)
x
y)
x
60 MPa
x
78 MPa,
55 MPa
1
(
2
2u
y
y)
x
73 MPa,
R
max
2u
y
x
u2
R
Case (1)
x
2
ave
48 MPa
xy
ave
1
(
2
max
73 MPa
160 MPa (reject)
105 MPa,
max
68 MPa
68 MPa,
x
178 MPa,
R
a
c
max
R
ave
0,
0
max
min )
32 MPa
89 MPa
73 MPa
y
60.0 MPa
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1111
PROBLEM 7.76 (Continued)
Assume
Case (2)
0.
x
u
u2
2
xy
2(
x
b )u
u2
146 MPa
min
b
ave
u2
2
xy
x
u2
2
xy
2u
max
(
2
xy
min )
73 MPa
b
b)
x
max
b
R
u
1
(
2
max
(
2
b)
x
x
2
(48)2
b
u
36 MPa
R
u2
a
b
2
xy
2R
y
2u
( 50 146)2
50 146
x
72 MPa
122 MPa
60 MPa
146
120
26 MPa
(o.k.)
y
122.0 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1112
y
PROBLEM 7.77
σy
For the state of stress shown, determine two values of
maximum shearing stress is 10 ksi.
y
for which the
y
2.00 ksi
8 ksi
14 ksi
z
x
SOLUTION
14 ksi,
x
y
u
Let
1
(
2
u
ave
(1b)
6 ksi
1
(
2
x
max
30 ksi,
u
6 ksi
ave
max
1
(
2
x
18 ksi,
y
y)
min
y
y)
min
2u
20 ksi,
2u
a
max
2 ksi,
u
x
R2
u
2
xy
6 ksi
u
1
(
2
30 ksi,
R
ave
min )
max
b
15 ksi
R
ave
10 ksi
7.5 ksi
2 ksi
x
8 ksi,
x
26 ksi (reject)
x
0,
2
xy
10 ksi
max
2u
y)
10 ksi,
R
max
8 ksi,
y
x
u2
R
(1a)
x
2
ave
Case (1)
xy
a
ave
max
R
1
(
2
18 ksi,
max
min )
b
ave
R
2 ksi
10 ksi (o.k.)
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1113
PROBLEM 7.77 (Continued)
Assume
Case (2)
a
ave
a
x
x
(
a
x)
2u
u2
u)2
a
2
x)
u
ave
a
max
y)
x
ave
R
20 ksi,
2
xy
u2
u2
(20
x
2.3333 ksi
1
(
2
x )u
2
xy
a
u2
y
20 ksi =
a
2
xy
14)2 82
20 14
2u
x
0,
b
max
4.6667 ksi
9.3333 ksi
u2
11.6667 ksi R
20 ksi
min
max
2
xy
a
2
2
max
2
xy
u2
2(
a
u
x
u
(
(
R
0.
min
ave
2
xy
R
8.3333 ksi
3.3334 ksi
10 ksi
y
9.33 ksi
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1114
PROBLEM 7.78
y
σ y & 100 MPa
For the state of stress shown, determine the range of values of xz for
which the maximum shearing stress is equal to or less than 60 MPa.
60 MPa
z
x
τ xz
SOLUTION
60 MPa,
x
0,
z
y
100 MPa
For Mohr’s circle of stresses in zx plane,
Assume
max
y
min
b
30
(2)(60)
( 20)
30
xz
u
x
z)
x
z
2
30 MPa
30 MPa
max
20 MPa
b
50 MPa
R
ave
R
2
max
ave
a
1
(
2
100 MPa
100
R
ave
50
u2
R2
502
80 MPa <
y
2
xz
u2
302
40 MPa
40.0 MPa
xz
40.0 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1115
PROBLEM 7.79
y
For the state of stress shown, determine two values of
maximum shearing stress is 80 MPa.
σy
y
for which the
90 MPa
z
x
60 MPa
SOLUTION
x
90 MPa,
0,
z
xz
60 MPa
Mohr’s circle of stresses in zx plane:
ave
1
(
2
x
z)
x
y
45 MPa
2
R
a
Assume
ave
a
max
y
min
120
Assume
min
y
b
max
30
R
120 MPa,
b
ave
2
R
2
zx
452
602
75 MPa
30 MPa
120 MPa.
max
2
max
(2)(80)
40.0 MPa
y
30 MPa.
min
2
max
(2)(80)
y
130.0 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1116
y
PROBLEM 7.80*
σy
For the state of stress of Prob. 7.69, determine (a) the value of y
for which the maximum shearing stress is as small as possible,
(b) the corresponding value of the shearing stress.
80 MPa
140 MPa
z
x
SOLUTION
x
u
Let
1
(
2
ave
Assume
Then
max
y
y
2
y)
x
u
x
u2
a
ave
R
x
u
u2
2
xy
b
ave
R
x
u
u2
2
xy
is minimum if u
2
xy
R
max
0.
2u
140 MPa,
y
x
R
xy
80 MPa
a
ave
R 140 80
b
ave
R 140 80 60 MPa
max
2u
R
is the in-plane shearing stress.
max (in-plane)
x
x
220 MPa,
x
u 140 MPa
220 MPa
0,
min
ave
max
1
(
2
max
min )
110 MPa
Assumption is incorrect.
Assume
max
min
d a
du
a
R
ave
0
max
1
u
u
2
2
xy
x
1
(
2
u
max
0
u2
min )
2
xy
1
2
a
(no minimum)
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1117
PROBLEM 7.80* (Continued)
Optimum value for u occurs when
1
(
2
(
a
x
R)
u )2
2u
max (out-of-plane)
R or
2
x
2u
2
x
2
xy
x
(a)
(b)
y
R
x
u2
a
x
max (in-plane)
R or
u
2
u
u
x
2
140 2 80 2
140
u2
2
xy
2
xy
94.286 MPa
u
2u 140 94.286
2
xy
max
47.143 MPa
y
92.857 MPa
max
45.7 MPa
92.857 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1118
σ0
PROBLEM 7.81
100 MPa
σ0
The state of plane stress shown occurs in a machine component made of a steel with
325 MPa. Using the maximum-distortion-energy criterion, determine whether
Y
yield will occur when (a) 0 200 MPa, (b) 0 240 MPa, (c) 0 280 MPa. If
yield does not occur, determine the corresponding factor of safety.
SOLUTION
2
ave
(a)
0
200 MPa
ave
a
2
a
F . S.
(b)
0
240 MPa
ave
a
2
a
F . S.
(c)
0
280 MPa
ave
a
2
a
2
b
a
b
R
0
x
y
2
xy
2
100 MPa
200 MPa
ave
2
b
R
a
100 MPa,
b
b
ave
R
300 MPa
264.56 MPa < 325 MPa
(No yielding)
325
264.56
F . S . 1.228
240 MPa
ave
2
b
R
a
140 MPa,
b
b
ave
R
340 MPa
295.97 MPa < 325 MPa
(No yielding)
325
295.97
F . S . 1.098
280 MPa
ave
R
180 MPa,
329.24 MPa > 325 MPa
b
ave
R
380 MPa
(Yielding occurs)
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1119
PROBLEM 7.82
σ0
100 MPa
σ0
Solve Prob. 7.81, using the maximum-shearing-stress criterion.
PROBLEM 7.81 The state of plane stress shown occurs in a machine component
made of a steel with Y 325 MPa. Using the maximum-distortion-energy criterion,
determine whether yield will occur when (a) 0 200 MPa, (b) 0 240 MPa,
(c) 0 280 MPa. If yield does not occur, determine the corresponding factor of
safety.
SOLUTION
2
ave
(a)
0
200 MPa:
ave
a
max
2
max
F . S.
(b)
0
240 MPa:
ave
a
max
2
(c)
0
280 MPa:
max
ave
a
max
2
max
R
0
x
y
2
xy
2
100 MPa
200 MPa
ave
0,
R
100 MPa
ave
R
300 MPa
300 MPa
min
max
b
min
300 MPa
325 MPa
(No yielding)
325
300
F . S . 1.083
240 MPa
ave
0,
R
140 MPa,
min
max
min
b
ave
R
340 MPa
340 MPa
340 MPa > 325 MPa
(Yielding occurs)
280 MPa
ave
0,
max
R
180 MPa,
min
min
b
ave
R
380 MPa
380 MPa
380 MPa
325 MPa
(Yielding occurs)
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1120
PROBLEM 7.83
21 ksi
τ xy
The state of plane stress shown occurs in a machine component made of a steel with
45 ksi. Using the maximum-distortion-energy criterion, determine whether
Y
yield will occur when (a) xy 9 ksi, (b) xy 18 ksi, (c) xy 20 ksi. If yield does
not occur, determine the corresponding factor of safety.
36 ksi
SOLUTION
36 ksi,
x
For stresses in xy plane,
1
(
2
ave
x
y
21 ksi,
y
y)
x
z
0
28.5 ksi
7.5 ksi
2
2
(a)
xy
9 ksi
a
ave
2
a
F .S .
2
b
x
R
R
a
y
2
xy
2
40.215 ksi,
ave
b
34.977 ksi
b
(7.5) 2
R
(9) 2
11.715 ksi
16.875 ksi
45 ksi
(No yielding)
45
39.977
F .S .
1.287
2
(b)
x
18 ksi R
xy
ave
a
2
a
F .S .
2
b
R
a
y
2
xy
2
48 ksi,
ave
b
44.193 ksi
b
(7.5)2
R
(18)2
19.5 ksi
9 ksi
45 ksi
(No yielding)
45
44.193
F .S .
1.018
2
(c)
xy
20 ksi
a
ave
2
a
2
b
R
R
a
x
2
xy
2
49.86 ksi,
b
y
b
46.732 ksi
ave
(7.5) 2
R
(20) 2
21.36 ksi
7.14 ksi
45 ksi
(Yielding occurs)
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1121
PROBLEM 7.84
21 ksi
Solve Prob. 7.83, using the maximum-shearing-stress criterion.
τ xy
PROBLEM 7.83 The state of plane stress shown occurs in a machine component
made of a steel with Y 45 ksi. Using the maximum-distortion-energy criterion,
determine whether yield will occur when (a) xy 9 ksi, (b) xy 18 ksi,
(c) xy 20 ksi. If yield does not occur, determine the corresponding factor of
safety.
36 ksi
SOLUTION
36 ksi,
x
21 ksi,
y
0
z
For stress in xy plane,
1
(
2
ave
y)
x
28.5 ksi
x
y
2
7.5 ksi
2
(a)
xy
9 ksi
a
ave
max
2
max
F .S.
x
R
R
2
40.215 ksi,
34.977 ksi,
max
y
b
11.715 ksi
ave
R
16.875 ksi
0
min
40.215 ksi
min
2
xy
(No yielding)
45 ksi
45
40.215
F .S .
1.119
2
(b)
xy
a
2
x
18 ksi R
ave
max
48 ksi
max
max
y
2
xy
2
R
48 ksi,
9 ksi
0
min
min
R
ave
b
19.5 ksi
48 ksi
(Yielding occurs)
45 ksi
2
(c)
xy
a
max
2
max
x
20 ksi R
ave
R
min
min
2
xy
2
49.86 ksi
49.86 ksi
max
y
b
ave
21.36 ksi
R
7.14 ksi
0
49.86 ksi
(Yielding occurs)
45 ksi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1122
PROBLEM 7.85
The 38-mm-diameter shaft AB is made of a grade of steel for which the yield
strength is Y 250 MPa. Using the maximum-shearing-stress criterion,
determine the magnitude of the torque T for which yield occurs when
P 240 kN.
B
T
P
A
d = 38 mm
SOLUTION
P
A
x
y
ave
240 103 N
d2
4
4
1.13411 103 mm 2
240 103
1.13411 10
P
A
0
1
(
2
(38) 2
x
y)
x
y
3
1
2
211.62 106 Pa
2
2R
2
xy
2
Y
4
xy
From torsion:
xy
J
c
T
1
2
1
4
2
xy
2
max
2
x
2
xy
4
2
x
2
x
1
250 2
2
2
x
2
Y
211.62 2
66.553 106 Pa
Tc
J
J
c4
1
d
2
2
xy
y
66.553 MPa
2
211.62 MPa
x
2
R
1.13411 10 3 m 2
T
2
38
2
xy
c
4
204.71 103 mm 4
204.71 10 9 m 4
19 10 3 m
(204.71 10 9 )(66.553 106 )
19 10 3
717 N m
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1123
PROBLEM 7.86
Solve Prob. 7.85, using the maximum-distortion-energy criterion.
B
PROBLEM 7.85 The 38-mm-diameter shaft AB is made of a grade of steel
for which the yield strength is Y 250 MPa. Using the maximum-shearingstress criterion, determine the magnitude of the torque T for which yield
occurs when P 240 kN.
T
P
A
d = 38 mm
SOLUTION
P
A
x
y
ave
240 103 N
4
d2
(38) 2
4
1.13411 103 mm 2
240 103
1.13411 10
P
A
0
1
(
2
x
y)
x
y
1
2
211.62 106 Pa
3
2
a
2
b
1
4
2
xy
2
2
xy
2
x
a
ave
R
1
2
x
1
4
2
x
2
xy
b
ave
R
1
2
x
1
4
2
x
2
xy
a b
1
4
2
x
1
4
2
x
2
xy
xy
1
4
x
2
x
3
x
2
xy
1
4
2
xy
2
x
1
4
211.62 MPa
x
2
R
1.13411 10 3 m 2
2
x
2
xy
1
4
2
x
2
xy
2
x
2
xy
1
4
2
x
1
4
2
x
2
xy
2
Y
1 2
2
Y
x
3
1
2502 211.622
3
76.848 MPa
76.848 106 Pa
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1124
PROBLEM 7.86 (Continued)
From torsion,
xy
J
c
T
Tc
J
2
c4
1
d
2
T
38
2 2
J
xy
c
4
204.71 103 mm 4
204.71 10 9 m 4
19 10 3 m
(204.71 10 9 )(76.848 106 )
19 10 3
828 N m
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1125
PROBLEM 7.87
P
T
A
1.5 in.
The 1.5-in.-diameter shaft AB is made of a grade of steel with a 42-ksi tensile yield
stress. Using the maximum-shearing-stress criterion, determine the magnitude of the
torque T for which yield occurs when P 60 kips.
B
SOLUTION
P
60 kips
A
d2
4
y
ave
4
1.76715 in 2
60
1.76715
P
A
x
(1.5) 2
33.953 ksi
0
1
(
2
x
y)
x
y
1
2
x
2
R
2
2
max
2R
2
xy
2
Y
4
xy
1
2
2
x
1
4
2
xy
4
2
xy
2
x
2
xy
Y
2
x
2
Y
2
x
1
422
2
33.9532
12.3612 ksi
From torsion,
xy
c
J
T
Tc
J
1
d
2
2
c4
T
J
xy
c
0.75 in.
0.49701 in 4
(0.49701)(12.3612)
0.75
8.19 kip in.
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1126
PROBLEM 7.88
P
T
A
Solve Prob. 7.87, using the maximum-distortion-energy criterion.
PROBLEM 7.87 The 1.5-in.-diameter shaft AB is made of a grade of steel with a
42-ksi tensile yield stress. Using the maximum-shearing-stress criterion, determine
the magnitude of the torque T for which yield occurs when P 60 kips.
1.5 in.
B
SOLUTION
P
60 kips
A
d2
4
ave
1.76715 in 2
60
1.76715
p
A
x
y
4
(1.5) 2
33.953 ksi
0
1
(
2
x
y)
x
y
1
2
x
2
R
2
a
ave
b
2
a
2
b
a b
ave
(
ave
2
ave
2
ave
1
4
3
2
xy
2
Y
xy
1
3
1
4
2
xy
2
x
2
xy
R
R
R) 2
2
3R
2
x
3
(
ave R
2
1
4
R) 2
ave
2
R
2
x
(
2
ave
2
xy
2
2
x
R)(
ave
ave
2
ave R
R
2
xy
2
Y
3
R)
2
ave
R2
2
x
2
Y
2
x
1
422
3
33.9532
14.2734 ksi
From torsion,
xy
c
J
T
Tc
J
1
d
2
T
J
xy
c
0.75 in.
c4
(0.75) 4 0.49701 in 4
2
2
(0.49701)(14.2734)
9.46 kip in.
0.75
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1127
PRO
OBLEM 7.89
100 MPa
The state
s
of plane stress shownn is expected to occur in an
a aluminum
8 MPa and
castinng. Knowing that for the aluuminum alloy used UT 80
nd using Mohrr’s criterion, determine
d
wheether rupture
200 MPa an
UC
of thee casting will occur.
o
60 MPa
M
10 MPa
SO
OLUTION
x
y
xy
10 MPa,
M
10
00 MPa,
60 MPa
x
ave
10 1000
2
y
2
45 MPaa
2
x
R
y
2
xy
2
(55) 2
(60)2
a
avee
R
45 81.39 36.39 MPa
b
avee
R
45 81.39
8
81.399 MPa
126.39 MPa
Equuation of 4th quadrant
q
of bo
oundary:
36.39
80
a
b
UT
UC
1
( 1226.39)
1.087 1
2000
Rupture will
w occur.
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OPRIETARY MATERIAL. Copyriight © 2015 McG
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1128
PROBLEM
M 7.90
75 MPa
The state of plane stress shown
s
is expeected to occur in an aluminuum casting.
Knowing thatt for the alumiinum alloy useed UT 80 MPa
M and UC 200 MPa
and using Mohr’s
M
criterioon, determinee whether ruppture of the casting
c
will
occur.
32 MPa
SOLUTION
x
y
xy
ave
32 MPa,
M
0,
M
75 MPa
1
(
2
x
y)
x
y
M
16 MPa
2
R
(16) 2
2
xy
2
(775) 2
a
ave
R
16 766.69 60.69 MPa
M
b
ave
R
16 766.69
76.69 MPa
M
92.69 MPa
Equuation of 4th quadrant
q
of bouundary:
60.69
80
a
b
UT
UC
1
( 92..69)
1.222 1
200
Rupture will
w occur.
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OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
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a
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n be copied, scannned, duplicated, forwarded, distribbuted, or posted
on a website,
w
in whole or part.
1129
PROBLEM
M 7.91
7 ksi
The state off plane stress shown
s
is expeected to occurr in an aluminnum casting.
1 ksi and UC 30 ksi
Knowing thaat for the alum
minum alloy used UT 10
and using Mohr’s
M
criterionn, determine whether
w
rupturee of the castingg will occur.
8 ksi
SO
OLUTION
x
y
xy
ave
8 ksi,
0,
7 ksi
1
(
2
x
y)
x
y
4 ksi
2
R
2
xy
2
42
a
ave
a
R
4 8.062
b
ave
a
R
4 8.062
72
8.062 ksi
4.0622 ksi
12.0062 ksi
Equuation of 4th quadrant
q
of bo
oundary:
4.062
10
a
b
UT
UC
( 122.062)
330
1
0.8088 1
No rupture.
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OPRIETARY MATERIAL. Copyriight © 2015 McG
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on a website, in wholee or part.
1130
PROBLEM
M 7.92
15 ksi
k
The state of plane
p
stress shhown is expeccted to occur in an aluminuum casting.
30 ksi
Knowing thatt for the alum
minum alloy used
u
10 ksi and UC
U
UT
and using Mohr’s criterion, determine whhether rupture of the casting will occur.
9 ksi
2 ksi
SOLUTION
x
2 ksi,
15 ksi,
y
xy
ave
a
9 ksi
1
(
2
x
y)
x
y
6.5 ksii
2
R
2
xy
2
a
ave
R 5.879 ksi
b
ave
R
8.52
92
1
12.379
ksi
18.879 ksii
q
of bouundary:
Equuation of 4th quadrant
5.879
10
a
b
UT
UC
1
( 18.879)
1.217 1
30
Rupture will occur.
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w
in whole or part.
1131
PROBLEM
M 7.93
The state of plane stress shown
s
will occcur at a critical point in an
a aluminum
25 ksi.
casting that is made of ann alloy for whhich UT 10 ksi and UC
U
Using Mohrr’s criterion, determine
d
thee shearing stress 0 for which
w
failure
should be exppected.
8 ksi
t0
SO
OLUTION
x
8 ksi,
y
0,
xyy
0
avee
1
(
2
x
y)
x
y
4 ksi
2
R
R2
0
Sinnce
ave
42
2
xy
2
2
0
42
a
ave
R
(4 R) ksi
b
ave
R
(4 R) ksi
< R, stress point lies in 4th quaddrant. Equatioon of 4th quaddrant boundaryy is
a
b
UT
UC
4 R
10
1
10
4 R
25
1
1
1
4
R 1
25
10
R
4
25
5.429 ksi
0
5.42992
42
0
3.67 ksi
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OPRIETARY MATERIAL. Copyriight © 2015 McG
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on a website, in wholee or part.
1132
PROBLEM
M 7.94
80 MPa
!0
The state off plane stress shown
s
will occcur at a criticaal point in a piipe made of
an aluminum
m alloy for which
w
Using
75 MPa and UC 150 MPa.
M
UT
Mohr’s criteerion, determinne the shearinng stress 0 foor which failurre should be
expected.
SOLUTION
x
y
80 MPa,
0,
xy
0
ave
1
(
2
x
y)
x
y
40 MPa
2
R
a
ave
R
b
ave
R
0
Sincce
ave
2
xy
2
R2
402
2
0
MPa
40 2
< R, stress point lies in 4th quaddrant. Equationn of 4th quadrrant boundary is
a
b
UT
UC
40 R
75
R
75
R
150
1
R 63.33 MP
Pa,
40 R
150
40
75
40
150
0
1
1
1.2667
63.332
402
0
8.49 MPa
PRO
OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
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a
for salee or distribution inn any manner. Thiis document may not
n be copied, scannned, duplicated, forwarded, distribbuted, or posted
on a website,
w
in whole or part.
1133
PR
ROBLEM 7.95
7
T'
Thee cast-aluminnum rod shoown is made of an alloyy for which
a
wing that the magnitude
m
T
70 MPa and
1775 MPa. Know
UT
T
UC
of the
t applied torrques is slowlly increased annd using Mohr’s criterion,
dettermine the shearing stress 0 that shouldd be expected at
a rupture.
t0
T
SO
OLUTION
x
0
y
0
xy
0
ave
1
(
2
x
y)
x
y
0
2
R
Sinnce
ave
2
xy
2
a
ave
a
R
b
ave
a
R
2
xyy
0
xy
R
R
< R, stress point lies in 4th quaddrant. Equatioon of boundaryy of 4th quadrrant is
a
b
UT
UC
R
700
1
70
1
R
1
175
1
R 1
1175
R 50 M
MPa
0
R
0
5
50.0
MPa
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on a website, in wholee or part.
1134
P
PROBLEM
7.96
T cast-alum
The
minum rod shhown is madee of an alloyy for which
U
Mohr’s criterion,
60 MPa and UC 120 MPa. Using
UT
d
determine
the magnitude off the torque T for which faiilure should
b expected.
be
32 mm
B
T
A
26 kN
SOLUTION
P
26 103 N
y
2
Sincce
(32) 2
804.25 mm 2
32.3288 106 Pa
6
804.25 10 6 m 2
322.328 MPa
1
1
( x
(32.328 0)
0 16.164 MP
Pa
y)
2
2
1
(32.328 0)) 16.164 MP
Pa
2
ave
x
4
26 1003
804.25 10
P
A
x
A
a
ave
R 16..164 R MPa
b
ave
R 16.164 R MPa
< R, stress point lies in the 4th quadrant.
q
Equaation of the 4thh quadrant is
ave
a
b
UT
UC
C
1
60
1
16.1644
600
R
1
16.1664
R 1
60
120
16.1644 R
1200
1
16.164
120
R
34
4.612 MPa
2
y
x
R
2
xy
2
R
xy
2
x
y
34.6122 116.1642
2
30.6606 MPa
30.606
6 106 Pa
For torsion,
xy
T
Tc
J
2
c3
2T
c3
x
xy
wherre c
2
1
d
2
166 mm 16 100 3 m
(16 100 3 )3 (30.606 106 )
T
196
6.9 N m
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OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
Not authorized
a
for salee or distribution inn any manner. Thiis document may not
n be copied, scannned, duplicated, forwarded, distribbuted, or posted
on a website,
w
in whole or part.
1135
1!
2 0
1!
2 0
!0
!0
(a)
PROBLEM
M 7.97
1!
2 0
(b)
!0
A machine component
c
is made of a grade
g
of cast
k and UC 20 ksi. For
iron for whicch UT 8 ksi
each of the states
s
of stress shown, and using
u
Mohr’s
criterion, dettermine the normal stress 0 at which
rupture of thee component should
s
be expeected.
0
UT
(c)
SO
OLUTION
(a)
a
b
0
1
2
0
Stress poinnt lies in 1st quadrant.
q
a
(b)
a
0
b
1
2
0
8.00 ksi
0
Stress poinnt lies in 4th quadrant.
q
Equaation of 4th quuadrant bounddary is
a
b
UT
C
UC
1
2
0
8
(c)
a
1
2
0,
b
0,
1
0
1
0
6.67 ksi
0
1
0
8.89 ksi
20
4th quadrannt
1
2
0
8
20
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use.
Not authorized for salle or distribution in
i any manner. Thhis document may not be copied, scaanned, duplicated,, forwarded, distributed, or posted
on a website, in wholee or part.
1136
PROBLEM 7.98
A spherical pressure vessel has an outer diameter of 3 m and a wall thickness of 12 mm. Knowing that for
the steel used all 80 MPa, E 200 GPa, and
0.29, determine (a) the allowable gage pressure, (b) the
corresponding increase in the diameter of the vessel.
SOLUTION
(a)
(b)
r
1
d
2
1
2
all
1
2
pr
2t
p
2t
r
p
1.290 106 Pa
1
d
1
(
E
1
E
d
1
1
(3)
2
t
12 10
3
1.488 m
80 106 Pa
(2)(12 10 3 )(80 106 )
1.488
1
p
1.290 MPa
d
0.852 mm
2)
1
1
1 0.29
(80 106 )
9
200 10
(3)(284 10 6 )
284 10
6
852 10 6 m
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1137
PROBLEM 7.99
A spherical gas container having an inner diameter of 5 m and a wall thickness of 24 mm is made of steel
0.29. Knowing that the gage pressure in the container is increased from zero
for which E 200 GPa and
to 1.8 MPa, determine (a) the maximum normal stress in the container, (b) the corresponding increase in the
diameter of the container.
SOLUTION
(a)
p
1.8 MPa
r
1
d
2
1
2
1
(5)
2
t
pr
2t
24 10
3
(1.8)(2.476)
(2)(24 10 3 )
2.476 m
92.850 MPa
92.9 MPa
1
(b)
d
1
(
E
d
2)
1
1
1
E
1
(5)(329.6 10 6 )
1 0.29
(92.85 106 )
200 109
1.648 10 3 m
329.6
d
1.648 mm
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1138
PROBLEM 7.100
The maximum gage pressure is known to be 1150 psi in a spherical steel pressure vessel having a 10-in. outer
diameter and a 0.25-in. wall thickness. Knowing that the ultimate stress in the steel used is U 60 ksi,
determine the factor of safety with respect to tensile failure.
SOLUTION
r
1
d
t
2
10 in.
0.25 in.
2
4.75 in.
pr
2t
(1150 psi)(4.75 in.)
2(0.25 in.)
2
10.925 ksi
F.S.
U
max
60 ksi
10.925 ksi
F.S.
5.49
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1139
PROBLEM 7.101
A spherical pressure vessel of 750-mm outer diameter is to be fabricated from a steel having an ultimate stress
400 MPa. Knowing that a factor of safety of 4.0 is desired and that the gage pressure can reach 4.2 MPa,
U
determine the smallest wall thickness that should be used.
SOLUTION
r
We have
and
max
F.S.
1
d t
2
1
(0.750 m)
2
0.375t (m)
1
t
pr
2t
2
U
max
Combining these two equations gives
F.S.
or
2
Ut
2t U
pr
(F.S.) pr
Substituting for r gives
2(400 106 Pa)t
6
816.80 10 t
t
(4)(4.2 106 Pa)(0.375
6.30 10
t)
6
7.71 10 3 m
t
7.71 mm
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1140
PROBLEM 7.102
A spherical gas container made of steel has a 20-ft outer diameter and a wall thickness of 167 in. Knowing that
the internal pressure is 75 psi, determine the maximum normal stress and the maximum shearing stress in the
container.
SOLUTION
d
t
r
20 ft
240 in.
7
in. 0.4375 in.
16
1
d t 119.56 in.
2
(75)(119.56)
pr
10.25 103 psi
2t
(2)(0.4375)
max
10.25 ksi
min
0
(Neglecting small radial stress)
1
(
2
max
max
min )
10.25 ksi
max
5.12 ksi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1141
PROBLEM 7.103
A basketball has a 300-mm outer diameter and a 3-mm wall thickness. Determine the normal stress in the wall
when the basketball is inflated to a 120-kPa gage pressure.
SOLUTION
r
1
d t
2
1
(300 mm)
2
147 mm
1
2
3
or
147 10 3 m
pr
2t
(120 103 Pa)(147 10 3 m)
2(3 10 3 m)
2.9400 106 Pa
2.94 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1142
PROBLEM 7.104
8m
14.5 m
h
The unpressurized cylindrical storage tank shown has a 5-mm wall thickness
and is made of steel having a 400-MPa ultimate strength in tension.
Determine the maximum height h to which it can be filled with water if a
factor of safety of 4.0 is desired. (Density of water 1000 kg/m3.)
SOLUTION
d0
t
5 mm
r
1
d
2
all
all
p
but
8m
U
F.S.
pr
t
t
all
r
p
gh,
h
p
g
0.005 m
t
4
0.005
400 MPa
4.0
3.995 m
100 MPa
(0.005 m)(100 106 Pa)
3.995 m
125.156 103 Pa
(1000 kg/m3 )(9.81 m/s 2 )
100 106 Pa
125.156 103 Pa
12.7580 m
h
12.76 m
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1143
PROBLEM 7.105
8m
14.5 m
For the storage tank of Prob. 7.104, determine the maximum normal stress
and the maximum shearing stress in the cylindrical wall when the tank is
filled to capacity (h 14.5 m).
h
PROBLEM 7.104 The unpressurized cylindrical storage tank shown has a
5-mm wall thickness and is made of steel having a 400-MPa ultimate strength
in tension. Determine the maximum height h to which it can be filled with
water if a factor of safety of 4.0 is desired. (Density of water 1000 kg/m3.)
SOLUTION
d0
t
r
p
8m
5 mm 0.005 m
1
d t 4 0.005
2
gh
3.995 m
(1000 kg/m3 )(9.81 m/s2 )(14.5 m)
142.245 103 Pa
1
pr
t
(142.245 103 Pa)(3.995 m)
0.005 m
113.654 106 Pa
max
min
max
1
max
113.7 MPa
0
1
(
2
max
min )
max
56.8 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1144
PROBLEM 7.106
The bulk storage tank shown in Photo 7.3 has an outer diameter of 3.3 m and a wall thickness of 18 mm. At a
time when the internal pressure of the tank is 1.5 MPa, determine the maximum normal stress and the
maximum shearing stress in the tank.
SOLUTION
r
d
2
1
pr
t
max
1
min
p
0
1
(
2
max
max
3.3
2
t
18 10
3
1.632 m,
(1.5 106 Pa)(1.632 m)
18 10 3 m
136 106 Pa
min )
t
18 10 3 m
136 106 Pa
max
68 106 Pa
max
136.0 MPa
68.0 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1145
PROBLEM 7.107
A standard-weight steel pipe of 12-in. nominal diameter carries water under a pressure of 400 psi.
(a) Knowing that the outside diameter is 12.75 in. and the wall thickness is 0.375 in., determine the maximum
tensile stress in the pipe. (b) Solve part a, assuming an extra-strong pipe is used of 12.75-in. outside diameter
and 0.5-in. wall thickness.
SOLUTION
(a)
d0
12.75 in. t
pr
t
(b)
d0
(400)(6.00)
0.375
12.75 in. t
pr
t
0.375 in. r
t
6.00 in.
6400 psi
0.500 in. r
(400)(5.875)
0.500
1
d0
2
6.40 ksi
1
d0
2
t
5.875 in.
4700 psi
4.70 ksi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1146
PROBLEM 7.108
A cylindrical storage tank contains liquefied propane under a pressure of 1.5 MPa at a temperature of 38 C.
Knowing that the tank has an outer diameter of 320 mm and a wall thickness of 3 mm, determine the
maximum normal stress and the maximum shearing stress in the tank.
SOLUTION
r
d
2
t
3 10 3 m
320
2
t
157 mm
157 10 3 m
(1.5 106 Pa)(157 10 3 m)
3 10 3 m
1
pr
t
max
1
min
p
0
1
(
2
max
max
3
78.5 106 Pa
min )
78.5 106 Pa
max
78.5 MPa
max
39.3 MPa
39.25 106 Pa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1147
PROBLEM 7.109
Determine the largest internal pressure that can be applied to a cylindrical tank of 5.5-ft outer diameter and
5
-in. wall thickness if the ultimate normal stress of the steel used is 65 ksi and a factor of safety of 5.0
8
is desired.
SOLUTION
1
r
1
65 ksi
13 ksi 13 103 psi
F .S.
5.0
d
(5.5)(12)
t
0.625 32.375 in.
2
2
U
pr
t
p
t
1
r
(0.625)(13 103 )
32.375
p
251 psi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1148
PROBLEM 7.110
A
A steel penstock has a 36-in. outer diameter, a 0.5-in. wall
thickness, and connects a reservoir at A with a generating station
at B. Knowing that the specific weight of water is 62.4 lb/ft3,
determine the maximum normal stress and the maximum shearing
stress in the penstock under static conditions.
500 ft
B
36 in.
SOLUTION
r
1
d
2
t
1
(36)
2
p
rh
(62.4 lb/ft 3 )(500 ft)
0.5
17.5 in.
31.2 103 lb/ft 2
216.67 psi
1
pr
t
max
1
min
p
max
1
(
2
(216.67)(17.5)
0.5
7583 psi
7583 psi
max
7.58 ksi
max
3.90 ksi
217 psi
max
min )
3900 psi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1149
PROBLEM 7.111
A
A steel penstock has a 36-in. outer diameter and connects a reservoir
at A with a generating station at B. Knowing that the specific weight
of water is 62.4 lb/ft3 and that the allowable normal stress in the
steel is 12.5 ksi, determine the smallest thickness that can be used
for the penstock.
500 ft
B
36 in.
SOLUTION
p
h
(62.4 lb/ft 3 )(500 ft)
31.2 103 lb/ft 2
216.67 psi
1
r
1
18
t
t
18
t
12.5 ksi
12.5 103 psi
1
d t 18 t
2
pr
r
1
,
t
t
p
12.5 103
216.67
57.692
t
58.692
0.307 in.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1150
PROBLEM 7.112
600 mm
b
The cylindrical portion of the compressed-air tank shown is fabricated of 8-mm-thick
plate welded along a helix forming an angle
30° with the horizontal. Knowing
that the allowable stress normal to the weld is 75 MPa, determine the largest gage
pressure that can be used in the tank.
1.8 m
SOLUTION
r
1
2
ave
R
w
p
1
d t
2
pr
t
1 pr
2 t
1
( 1
2
1
(600)
2
6
292 mm
3 pr
4 t
1 pr
1
2
2
4 t
R cos 60
ave
2)
5 pr
8 t
8 wt
5 r
p
8 (75)(8)
5 292
3.29 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1151
PROBLEM 7.113
600 mm
For the compressed-air tank of Prob. 7.112, determine the gage pressure that will
cause a shearing stress parallel to the weld of 30 MPa.
b
1.8 m
PROBLEM 7.112 The cylindrical portion of the compressed-air tank shown is
fabricated of 8-mm-thick plate welded along a helix forming an angle
30° with
the horizontal. Knowing that the allowable stress normal to the weld is 75 MPa,
determine the largest gage pressure that can be used in the tank.
SOLUTION
r
1
2
R
w
p
1
d t
2
pr
t
1 pr
2 t
1
2
2
R sin 60
1
(600)
2
8
292 mm
1 pr
4 t
3 pr
8 t
8 wt
3 R
p
8 (30)(8)
3 292
3.80 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1152
PROBLEM 7.114
The steel pressure tank shown has a 750-mm inner diameter and a 9-mm
wall thickness. Knowing that the butt-welded seams form an angle
50
with the longitudinal axis of the tank and that the gage pressure in the tank is
1.5 MPa, determine (a) the normal stress perpendicular to the weld, (b) the
shearing stress parallel to the weld.
!
SOLUTION
r
d
2
1
pr
t
2
1
2
375 mm
0.375 m
(1.5 106 Pa 0.375 m)
0.009 m
1
31.25 MPa
62.5 106 Pa
2
100
ave
1
(
2
R
1
(a)
2)
1
2
2
w
62.5 MPa
46.875 MPa
15.625 MPa
ave
R cos100
w
(b)
w
44.2 MPa
R sin100
w
15.39 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1153
PROBLEM 7.115
The pressurized tank shown was fabricated by welding strips of plate along
a helix forming an angle
with a transverse plane. Determine the largest
value of
that can be used if the normal stress perpendicular to the weld is
not to be larger than 85 percent of the maximum stress in the tank.
!
SOLUTION
1
pr
t
ave
1
(
2
R
w
0.85
pr
t
cos 2
2
2
pr
2t
3 pr
4 t
1
pr
1
2
2
4 t
R
cos
2
ave
2)
1
3
4
1
cos 2
4
4 0.85
3
4
pr
t
0.4
113.6
56.8
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1154
12 ft
PROBLEM 7.116
12 ft
Square plates, each of 0.5-in. thickness, can be bent and
welded together in either of the two ways shown to form the
cylindrical portion of a compressed-air tank. Knowing that the
allowable normal stress perpendicular to the weld is 12 ksi,
determine the largest allowable gage pressure in each case.
45"
20 ft
(a)
(b)
SOLUTION
d
12ft
1
pr
t
(a)
1
144 in. r
2
1
d
2
pr
2t
t
71.5 in.
12 ksi
1t
p
(12)(0.5)
71.5
r
0.0839 ksi
p
(b)
ave
1
(
2
R
1
w
ave
2)
1
2
45
2
83.9 psi
3 pr
4 t
1 pr
4 t
R cos
3 pr
4 t
p
4 wt
3 r
4 (12)(0.5)
3
71.5
0.1119 ksi
p
111.9 psi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1155
PROBLEM 7.117
3m
The pressure tank shown has a 0.375-in. wall thickness and butt-welded
20° with a transverse plane. For a gage
seams forming an angle
pressure of 85 psi, determine (a) the normal stress perpendicular to the
weld, (b) the shearing stress parallel to the weld.
1.6 m
!
SOLUTION
d
r
1
2
ave
R
(a)
w
(b)
w
ave
R cos 40
R sin 40
5 ft
60 in.
1
d t 30 0.375 29.625 in.
2
(85)(29.625)
pr
6715 psi
0.375
t
1
3357.5 psi
1
2
1
( 1
5036.2 psi
2)
2
1
2
2
1678.75 psi
3750 psi
1079 psi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1156
PROBLEM 7.118
3m
1.6 m
For the tank of Prob. 7.117, determine the largest allowable gage pressure,
knowing that the allowable normal stress perpendicular to the weld is
18 ksi and the allowable shearing stress parallel to the weld is 10 ksi.
PROBLEM 7.117 The pressure tank shown has a 0.375-in. wall thickness
20° with a transverse plane.
and butt-welded seams forming an angle
For a gage pressure of 85 psi, determine (a) the normal stress
perpendicular to the weld, (b) the shearing stress parallel to the weld.
!
SOLUTION
d
5 ft
60 in.
1
d
2
pr
t
pr
2t
1
(
2
r
1
2
ave
t
30
1
w
ave
2
R cos 50
1
cos 50
4
0.58930
p
w
p
wt
0.5893r
(18)(0.375)
(0.58930)(29.625)
R sin 50
0.191511
wt
0.191511r
0.38664 ksi
pr
t
(10)(0.375)
(0.191511)(29.625)
3 pr
4 t
1 pr
4 t
2
3
4
29.625 in.
2)
1
R
0.375
pr
t
pr
t
387 psi
0.66097 ksi
661 psi
p
Allowable gage pressure is the smaller value.
387 psi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1157
PROBLEM 7.119
3m
1.6 m
For the tank of Prob. 7.117, determine the range of values of that can be
used if the shearing stress parallel to the weld is not to exceed 1350 psi
when the gage pressure is 85 psi.
PROBLEM 7.117 The pressure tank shown has a 0.375-in. wall thickness
20° with a transverse plane.
and butt-welded seams forming an angle
For a gage pressure of 85 psi, determine (a) the normal stress perpendicular
to the weld, (b) the shearing stress parallel to the weld.
!
SOLUTION
d
r
1
2
R
w
sin 2
2
a
53.53
a
2
b
53.53
b
26.8
2
c
53.53
c
63.2
2
d
d
116.8
53.53
180
180
126.47
233.53
26.8
a
5 ft
60 in.
1
3
29.625 in.
d t 30
2
8
(85)(29.625)
pr
6715 psi
0.375
t
1
3357.5 psi
1
2
1
2
2
R sin 2
w
R
1678.75
all
1350
1678.75
0.80417
26.8
63.2
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1158
PRO
OBLEM 7.120
4 ft
P'
A
A preessure vessel of
o 10-in. inner diameter annd 0.25-in. waall thickness
is fabbricated from a 4-ft sectioon of spirally--welded pipe AB and is
equippped with two rigid end plattes. The gage pressure
p
insidde the vessel
is 3000 psi and 10-kkip centric axxial forces P and
a P are appplied to the
end plates.
p
Determ
mine (a) the noormal stress peerpendicular to
t the weld,
(b) the shearing streess parallel to the weld.
P
35"
B
SOLUTION
r0
1
d
2
pr
t
pr
2t
r t
A
r02
r
1
2
P
A
Totaal stresses.
1
t 0.25 in.
(10) 5 in.
2
(3000)(5)
6000 psi
p
6 ksi
0.225
(3000)(5)
3000 psi
p
3 ksi
(2)(00.25)
5 0.25 5.25 inn.
(5.252
r2
100 103
8.0803
5.002 ) 8.05003 in 2
12442 psi
Longitudinal:
x
3 1.242 1.7588 ksi
Circumferential:
y
6 ksi
k
Shear:
xy
1.242 ksi
0
Plottted points forr Mohr’s circlee:
X : (1.758, 0)
Y : (6, 0)
C : (3.879)
ave
1
(
2
x
y)
x
y
3.8879 ksi
2
R
2
xy
2
((1.758 6)
2
(a)
(b)
x
|
xy |
avee
R cos 70
R siin 70
2
0
2.121 kssi
3.879 2.1221 cos 70
2.1211 sin 70
x
|
xy
3.15 ksi
| 1.993
1
ksi
PRO
OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
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a
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in whole or part.
1159
PROB
BLEM 7.12
21
Solve Prob. 7.120, assuming thatt the magnituude P of the tw
wo forces is
increassed to 30 kips.
4 ft
P'
A
PROB
BLEM 7.120 A pressure vesssel of 10-in. inner diameterr and 0.25-in.
wall thhickness is faabricated from
m a 4-ft sectioon of spirally--welded pipe
AB andd is equipped with two rigiid end plates. The gage preessure inside
the vesssel is 300 psii and 10-kip centric
c
axial foorces P and P are applied
to the end plates. Determine
D
(a) the normal stress perpenddicular to the
weld, (b)
( the shearinng stress parallel to the weldd.
P
35""
B
SO
OLUTION
r0
1
d
2
pr
t
pr
2t
r t
A
r02
r
1
2
Tottal stresses.
1
(
(10)
5 in.
t 0.25 in.
2
(3000)(5)
6000 psi 6 ksi
0
0.25
(300)(5)
30000 psi 3 ksi
(2))(0.25)
5 0.25 5.25 in.
r2
(5.252
52 ) 8.05033 in 2
P
A
30 103
8.0503
37727 psi
Longitudinall:
x
3 3.727
0.7727 ksi
Circumferen
ntial:
y
6 ksi
Shear:
3.7727 ksi
0
xy
Plootted points for Mohr’s circlle:
X : ( 0.727, 0)
0
Y : (6, 0)
C : (2.66365, 0)
av
ve
1
(
2
x
y)
x
y
2.6365 kssi
2
R
0.7277 6
2
(a)
(b)
x
|
xy
ave
|
R cos 70
R sin 70
2
xy
2
2
0
3
3.3635
ksi
2.6365 3.3635 cos 70
3.36
635 sin 70
x
|
xy
1.486 ksi
| 3.16 ksi
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use.
Not authorized for salle or distribution in
i any manner. Thhis document may not be copied, scaanned, duplicated,, forwarded, distributed, or posted
on a website, in wholee or part.
1160
T
PROBLEM 7.122
2
A torquue of magnituude T 12 kN
nd of a tank containing
N m is applied to the en
compresssed air under a pressure off 8 MPa. Know
wing that the tank
t
has a 1800-mm inner
diameterr and a 12-mm
m wall thicknness, determinne the maximuum normal strress and the
maximuum shearing strress in the tank.
SOLUTION
d
1
d
2
180 mm
m
r
90 mm t
12 mm
Torssion:
c1
90 mm
m c2
J
c24
2
90
c14
12
102 mm
m
66.9668 106 mm 4
(12 103 )(1022 10 3 )
66.968 10 6
Tc
J
66.968 10 6 m 4
188.277 MPa
Presssure:
pr
t
1
(8)(90)
12
600 MPa
2
pr
2t
30 MP
Pa
Sum
mmary of stresses:
x
ave
60 MPa,
M
1
(
2
30 MPa,
y
x
y)
x
y
xy
18.277 MPa
45 MPa
2
R
2
xxy
2
a
avee
R
M
68.64 MPa
b
avee
R
21.36 MPa
M
c
min
max
Pa
23.64 MP
0
max
688.6 MPa
max
344.3 MPa
0
1
(
2
max
min )
PRO
OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
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in whole or part.
1161
T
PROBLEM 7.123
The tank shown has a 180-mm inner diameter and a 12-mm wall thickness. Knowing
that the tank contains compressed air under a pressure of 8 MPa, determine the
magnitude T of the applied torque for which the maximum normal stress is 75 MPa.
SOLUTION
r
1
2
ave
max
R
1
1
d
(180) 90 mm
2
2
pr (8)(90)
60 MPa
t
12
pr
30 MPa
2t
1
( 1
45 MPa
y)
2
75 MPa
max
t 12 mm
30 MPa
ave
2
R
xy
1
2
2
R 2 152
2
xy
152
302 152
2
xy
25.98 MPa
6
25.98 10 Pa
Torsion:
c1
90 mm
c2
90 12 102 mm
J
xy
T 4
c2 c14 66.968 106 mm 4 66.968 10 6 m 4
2
J xy (66.968 10 6 )(25.98 106 )
Tc
T
17.06 103 N m
3
J
c
102 10
T
17.06 kN m
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1162
PROBLEM 7.124
y
150 mm
The compressed-air tank AB has a 250-mm outside diameter and an 8-mm
wall thickness. It is fitted with a collar by which a 40-kN force P is
applied at B in the horizontal direction. Knowing that the gage pressure
inside the tank is 5 MPa, determine the maximum normal stress and the
maximum shearing stress at point K.
B
P
600 mm
K
L
A
z
150 mm
x
SOLUTION
Consider element at point K.
Stresses due to internal pressure:
p
r
x
y
Stress due to bending moment:
5 MPa 5 106 Pa
1
250
d t
8 117 mm
2
2
pr (5 106 )(117 10 3 )
73.125 MPa
t
(8 10 3 )
(5 106 )(117 10 3 )
(2)(8 10 3 )
pr
2t
Point K is on the neutral axis.
0
y
Stress due to transverse shear:
36.563 MPa
V
c2
c1
Q
I
xy
P 40 103 N
1
d 125 mm
2
c2 t 117 mm
2 3 3
2
c2 c1
(1253 1173 )
3
3
234.34 103 mm3 234.34 10 6 m3
(1254 117 4 )
c24 c14
4
4
44.573 106 mm 4 44.573 10 6 m 4
VQ
It
PQ
I (2t )
(40 103 )(234.34 10 6 )
(44.573 10 6 )(16 10 3 )
13.1436 106 Pa 13.1436 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1163
PROBLEM 7.124 (Continued)
Total stresses:
x
Mohr’s circle:
ave
73.125 MPa,
1
(
2
x
y)
36.563 MPa,
y
xy
13.1436 MPa
54.844 MPa
2
x
R
y
2
xy
2
(18.281) 2
(13.1436) 2
a
ave
R
77.360 MPa
b
ave
R
32.328 MPa
22.516 MPa
Principal stresses:
a
77.4 MPa,
b
The 3rd principal stress is the radial stress.
z
max
Maximum shearing stress:
32.3 MPa
max
77.4 MPa,
1
(
2
max
min
min )
0
max
max
0
77.4 MPa
38.7 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
1164
PROB
BLEM 7.12
25
y
150
0 mm
In Probb. 7.124, deterrmine the maxximum normall stress and thee maximum
shearinng stress at poiint L.
PROB
BLEM 7.124 The
T compresssed-air tank AB
B has a 250-m
mm outside
diametter and an 8-m
mm wall thicknness. It is fitteed with a collaar by which
a 40-kN
N force P is appplied at B in the horizontal direction. Knnowing that
the gagge pressure innside the tankk is 5 MPa, determine
d
thee maximum
normall stress and thee maximum shhearing stress at point K.
B
P
600 mm
K
L
A
z
m
150 mm
x
SOLUTION
Connsider elementt at point L.
Streesses due to in
nternal pressurre:
p
r
x
y
ding moment:
Streess due to bend
5 MPa 5 106 Pa
1
250
d t
8 1177 mm
2
2
pr (5 106 )(117 100 3 )
73.125 MPa
t
8 10 3
pr (5 103 )(117 100 3 )
36.563 MPa
2t
(2)(8 10 3 )
M
c2
c1
I
y
(40 kN)(600
k
mm) 24,000 N m
1
d 125 mm
2
c2 t 125 8 117 mm
c24 c14
(1254 117 4 )
4
4
44.573 106 mm 4 44.573 10 6 m 4
Mc
I
(24, 000)(125 10 3 )
44.573 10 6
Pa
67.305 MP
PRO
OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
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a
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n be copied, scannned, duplicated, forwarded, distribbuted, or posted
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in whole or part.
1165
PROBLEM 7.125 (Continued)
Stress due to transverse shear:
Point L lies in a plane of symmetry.
xy
Total stresses:
x
Principal stresses:
max
73.125 MPa,
0
30.742 MPa,
y
xy
0
Since xy 0, x and y are principal stresses. The 3rd principal stress is in the
radial direction, z 0.
73.125 MPa,
min
0,
a
73.1 MPa,
b
Maximum stress:
Maximum shearing stress:
max
1
(
2
max
min )
30.7 MPa,
z
0
max
73.1 MPa
max
51.9 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
1166
PROBLEM 7.126
1.5 in.
A brass ring of 5-in. outer diameter and 0.25-in. thickness fits exactly
inside a steel ring of 5-in. inner diameter and 0.125-in. thickness
when the temperature of both rings is 50 F. Knowing that the
temperature of both rings is then raised to 125 F, determine (a) the
tensile stress in the steel ring, (b) the corresponding pressure exerted
by the brass ring on the steel ring.
STEEL
ts # 81 in.
Es # 29 $ 106 psi
%ss # 6.5 $ 10–6/"F
5 in.
BRASS
tb # 14 in.
Eb # 15 $ 106 psi
%bs # 11.6 $ 10–6/"F
SOLUTION
Let p be the contact pressure between the rings. Subscript s refers to the steel ring. Subscript b refers to the
brass ring.
Steel ring.
Internal pressure p:
s
pr
ts
pr
Es t s
s
Corresponding strain:
sp
Es
Strain due to temperature change:
sT
s
Total strain:
(1)
T
s
pr
Es t s
Ls
2 r
b
pr
tb
s
T
Change in length of circumference:
Brass ring.
External pressure p:
Corresponding strains:
s
2 r
pr
Es ts
bT
b
pr
,
Eb tb
bp
s
T
T
Change in length of circumference:
Lb
Equating
Ls to Lb ,
pr
Es t s
r
Es t s
s
2 r
(
b
pr
Eb tb
2 r
pr
Eb tb
T
r
p
Eb tb
b
b
s)
T
b
T
T
(2)
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1167
PROBLEM 7.126 (Continued)
T 125 F 50 F 75 F
Data:
r
From Equation (2),
2.5
(29 106 )(0.125)
1
d
2
1
(5)
2
2.5 in.
2.5
p
(15 106 )(0.25)
(11.6 6.5)(10 6 )(75)
1.35632 10 6 p 382.5 10
p
From Equation (1),
s
pr
ts
(282.0)(2.5)
0.125
6
282.0 psi
5.64 103 psi
(a)
s
(b) p
5.64 ksi
282 psi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
1168
PROBLEM 7.127
1.5 in.
STEEL
ts # 81 in.
Es # 29 $ 106 psi
%ss # 6.5 $ 10–6/"F
5 in.
BRASS
tb # 14 in.
Eb # 15 $ 106 psi
%bs # 11.6 $ 10–6/"F
Solve Prob. 7.126, assuming that the brass ring is 0.125 in. thick and
the steel ring is 0.25 in. thick.
PROBLEM 7.126 A brass ring of 5-in. outer diameter and 0.25-in.
thickness fits exactly inside a steel ring of 5-in. inner diameter and
0.125-in. thickness when the temperature of both rings is 50 F.
Knowing that the temperature of both rings is then raised to 125 F,
determine (a) the tensile stress in the steel ring, (b) the corresponding
pressure exerted by the brass ring on the steel ring.
SOLUTION
Let p be the contact pressure between the rings. Subscript s refers to the steel ring. Subscript b refers to the
brass ring.
Steel ring.
Internal pressure p:
s
pr
ts
pr
Es t s
s
Corresponding strain:
sp
Es
Strain due to temperature change:
sT
s
Total strain:
(1)
T
s
pr
Es t s
Ls
2 r
s
T
Change in length of circumference:
Brass ring.
External pressure p:
2 r
pr
Es t s
bT
b
s
T
pr
tb
b
Corresponding strains:
s
pr
,
Eb tb
bp
T
Change in length of circumference:
Lb
Equating
Ls to Lb ,
pr
Es t s
r
Es ts
s
2 r
b
pr
Eb tb
T
r
p (
Eb tb
b
pr
Eb tb
2 r
b
s)
T
b
T
T
(2)
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1169
PROBLEM 7.127 (Continued)
Data:
T
125 F 50 F 75 F
1
1
(5) 2.5 in.
r
d
2
2
From Equation (2),
2.5
(29 106 )(0.25)
2.5
p
(15 106 )(0.125)
(11.6 6.5)(10 6 )(75)
1.67816 10 6 p 382.5 10
p
From Equation (1),
s
pr
ts
(227.93)(2.5)
0.25
6
227.93 psi
2279 psi
(a)
s
(b) p
2.28 ksi
228 psi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1170
PROBLEM 7.128
y
y'
x'
&
x
For the given state of plane strain, use the method of Sec. 7.7A to
determine the state of plane strain associated with axes x and y
rotated through the given angle .
800 ,
x
450 ,
y
xy
200 ,
25
SOLUTION
25
x
y
2
x
x
xy
y
y
2
175
(
y
x
( 800
x
xy
625
2
2
175
x
y
x
175
y
2
100
xy
cos 2
sin 2
2
2
( 625 ) cos ( 50 ) (100 )sin ( 50 )
x
y
xy
cos 2
sin 2
2
2
( 625 ) cos ( 50 ) (100 )sin ( 50 )
y )sin
2
653
x
xy
y
303
cos 2
450 )sin ( 50 ) ( 200 ) cos ( 50 )
xy
829
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1171
PROBLEM 7.129
y
y'
For the given state of plane strain, use the method of Sec. 7.7A to
determine the state of plane strain associated with axes x and y
rotated through the given angle .
x'
&
x
240 ,
x
160 ,
y
xy
150 ,
60
SOLUTION
60
x
y
2
x
x
xy
y
2
y
x
y
xy
40
y
x
y
cos 2
2
75
xy
sin 2
2
2
2
200 40 cos ( 120 ) 75 sin ( 120 )
x
y
x
200
xy
sin 2
2
2
2
200 40cos ( 120 ) 75sin ( 120 )
(
x
y )sin
cos 2
2
xy
115.0
x
y
285
cos 2
(240 160)sin ( 120 ) 150 cos ( 120 )
xy
5.72
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1172
PROBLEM 7.130
y
y'
For the given state of plane strain, use the method of Sec. 7.7A to
determine the state of plane strain associated with axes x and y
rotated through the given angle .
x'
!
x
500 ,
x
250 ,
y
xy
0,
15
SOLUTION
15
x
y
2
x
x
xy
2
125
x
y
y
x
( 500
x
y
cos 2
cos 2
2
( 375 ) cos 30
y )sin
2
xy
375
2
( 375 ) cos 30
2
125
(
y
2
y
x
y
x
125
xy
2
xy
2
0
xy
2
0
0
sin 2
x
450
y
199.8
sin 2
cos 2
250 )sin 30
0
xy
375
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1173
PROBLEM 7.131
y
y'
x'
!
x
For the given state of plane strain, use the method of Sec 7.7A to
determine the state of plane strain associated with axes x and y
rotated through the given angle .
0,
x
y
320 ,
xy
100 ,
30
SOLUTION
30
x
y
2
x
160
2
x
x
y
y
x
2
y
2
x
y
2
x
y
2
160 160cos 60
xy
(
x
y )sin
cos 2
xy
2
sin 2
100
sin 60
2
160 160 cos 60
y
160
2
(0 320)sin 60
cos 2
xy
2
100
sin 60
2
xy
x
36.7
sin 2
y
283
xy
227
cos 2
100 cos 60
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1174
PROBLEM 7.132
y
y'
For the given state of plane strain, use Mohr’s circle to determine the
state of plane strain associated with axes x and y rotated through
the given angle .
x'
x
!
800 ,
x
450 ,
y
xy
200 ,
25
SOLUTION
Plotted points:
X : ( 800 , 100 )
Y : ( 450 , 100 )
C : ( 175 , 0)
100
625
tan
(625 ) 2
R
x
ave
9.09
(100 ) 2
2
50
R cos
175
9.09
632.95
40.91
632.95 cos 40.91
653
x
y
ave
R cos
175
632.95 cos 40.91
y
1
2
xy
R sin
632.95 sin 40.91
xy
303
829
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on a website, in whole or part.
1175
PR
ROBLEM 7.133
7
y
y'
x'
!
x
For the given staate of plane sttrain, use Mohhr’s circle to determine
d
the
staate of plane sttrain associateed with axes x and y rotaated through
thee given angle .
240 ,
x
y
160 ,
xy
150 ,
60
SO
OLUTION
Plootted points for Mohr’s circlle:
X : ( 240
2
, 75 )
Y : ( 1160 , 75 )
C : ( 200
2
, 0)
75
1.875
40
tan
n
(40 )2
R
2
1
2
61.933
(75 )2
8
85
x
ave
181.93
120 61.93
R cos
c
200
(85 ) cos ( 181.93 )
y
ave
R cos
c
xy
R sin
200
(85 ) cos ( 181.93 )
85 sin ( 181.93 )
2
2.86
115.0
x
y
xy
285
5.72
PRO
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1176
PROBLEM 7.134
y
y'
x'
x
!
For the given state of plane strain, use Mohr’s circle to determine the
state of plane strain associated with axes x and y rotated through
the given angle .
500 ,
x
y
250 ,
xy
0,
15
SOLUTION
Plotted points:
X : ( 500 ,0)
Y : ( 250 , 0)
C : ( 125 , 0)
1
2
R
375
x
ave
R cos 2
125 375cos 30
y
ave
R cos 2
125 375cos 30
xy
R sin 2
375sin 30
x
y
xy
450
199.8
375
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1177
PR
ROBLEM 7.135
7
y
y'
x'
!
x
For the given staate of plane sttrain, use Mohhr’s circle to determine
d
the
staate of plane strrain associated with axes x and y rotaated through
thee given angle .
0,
x
y
3200 ,
xy
100 ,
30
SO
OLUTION
Plootted points for Mohr’s circlle:
X : (0, 50 )
Y : (320 , 50 )
C : (160 , 0)
50
160
tan
n
17.35
(160 ) 2
R
2
1
2
(50 ) 2
60
167.63
17.35
42.65
x
ave
R cos
c
160
(167.63 ) coos 42.65
y
ave
R cos
c
160
(167.63 ) coos 42.65
xy
R sin
(167.63 )ssin 42.65
x
36.7
y
283
xy
227
PRO
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1178
PR
ROBLEM 7.136
Thee following staate of strain haas been measuured on the surrface of a thinn plate. Knowiing that the suurface of the
platte is unstresseed, determine (a) the direction and magnnitude of the principal
p
strains, (b) the maaximum inplanne shearing strrain, (c) the maximum
m
shearring strain. (U
Use v 13 . )
2600 ,
x
600 ,
y
xy
4880
SOLUTION
For Mohr’s circlee of strain, plot points:
X : ( 2660 , 240 )
Y : ( 600 , 240 )
C : ( 1660 , 0)
tan 2
x
2
ave
R
160
260
b
ave
R
160
260
max (in-plaane)
R
max
m
(c)
max
max
x
1 v
(
160
min
a
min
160
2
(240 )
a
56.3
a
100
2
420
b
2R
max ((in-plane)
v
c
33.7
260
a
1
2
2.4
b
(100 )
R
(b)
y
67.38
p
R
(a)
480
260 60
xy
p
max (in-plane))
v
b)
1 v
(
x
y)
1/3
( 260 60)
2/3
160
420
c
420
520
maxx
160
580
PRO
OPRIETARY MAT
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in whole or part.
1179
PR
ROBLEM 7.137
7
Thee following sttate of strain has
h been measuured on the suurface of a thinn plate. Know
wing that the suurface of the
plaate is unstresseed, determine (a) the direction and magnnitude of the principal straains, (b) the maximum
m
inplaane shearing sttrain, (c) the maximum
m
sheaaring strain. (U
Use v 13 . )
6000 ,
x
y
4000 ,
xy
3
350
SO
OLUTION
Plootted points for Mohr’s circlle:
X : ( 600 , 175 )
Y : ( 400 , 175 )
C : ( 500 , 0)
0
tan 2
p
2
p
175
100
60.26
30.1
b
a
(100 ) 2
R
59.9
(175 ) 2
201.6
(a)
(b)
a
avee
R
500
201.6
a
298
b
avee
R
500
201.6
b
702
max (in-pllane)
2R
v
(
1 v
c
(c)
max
max
500
max
min
a
min
500
5
v
(
1 v
b)
x
y)
1/3
( 600
2/3
max (in-planee)
403
c
500
400 )
702
702
max
1202
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
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on a website, in wholee or part.
1180
PR
ROBLEM 7.138
Thee following staate of strain haas been measuured on the surrface of a thinn plate. Knowiing that the suurface of the
platte is unstresseed, determine (a) the direction and magnnitude of the principal
p
strains, (b) the maaximum inplanne shearing strrain, (c) the maximum
m
shearring strain. (U
Use v 13 . )
160 ,
x
4800 ,
y
xy
6
600
SOLUTION
(a)
For Mohr’ss circle of straain, plot pointss:
X : (160 , 300 )
Y : ( 480 , 300 )
C : ( 160 , 0)
0
(a)
tan 2
x
2
3000
3200
xy
p
p
y
43.15
0.9375
21.58
p
andd
21.58 900
68.42
21.6
a
b
R
(b)
(c)
1
2
c
(320 ) 2
(3000 ) 2
68.4
438.66
a
ave
a
R
1600
438.6
a
279
b
ave
a
R
1600
438.6
b
599
R
(max, in-plaane)
v
(
1 v
a
2R
(maxx, in-plane)
b)
v
(
1 v
max
2778.6
max
m
max
1/3
(160
2/3
y)
x
min
min
(max, in-plane))
480 )
c
877
160.0
598.6
2778.6
598.6
max
877
PRO
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in whole or part.
1181
PR
ROBLEM 7.139
7
Thee following sttate of strain has
h been measuured on the suurface of a thinn plate. Know
wing that the suurface of the
plaate is unstresseed, determine (a) the direction and magnnitude of the principal straains, (b) the maximum
m
inplaane shearing sttrain, (c) the maximum
m
sheaaring strain. (U
Use v 13 .)
x
30 ,
y
5700 ,
xy
7
720
SO
OLUTION
Plootted points for Mohr’s circlle:
X : (30 , 360 )
Y : (570 , 360 )
C : (300 , 0)
tan 2
p
2
p
360
270
53.13
1.3333
(a)
26.6
b
a
(b)
R
(2770 )2
a
ave
R
300
450
b
ave
R
300
450
max (in-planee)
(360 ) 2
(c)
max
450
a
max (in-planee)
1 v
maax
a
maax
min
(
a
750 ,
750
0
b)
min
1/3
(750
2/33
c
150 )
750
150.0
b
2R
v
c
64.4
900
c
300
max
1050
300
( 300 )
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1182
PR
ROBLEM 7.140
For the given staate of plane sttrain, use Mohhr’s circle to determine (a)) the orientatioon and magniitude of the
prinncipal strains, (b) the maxim
mum in-plane strain,
s
(c) the maximum
m
sheearing strain.
60 ,
x
2400 ,
y
xy
5
50
SOLUTION
Plottted points:
X : (60 , 25
2 )
Y : (240 , 25 )
C : (150 , 0)
tan 2
xy
p
x
2
y
50
60 240
0
0.277778
15.52
p
97.8
a
7.8
b
(90 ) 2
R
(a)
(b)
(c)
a
ave
R 150
933.4
b
ave
R 150
933.4
max (in-plane))
c
0,
max
m
(25 ) 2
933.4
a
b
2R
max (in-plane)
243.4 ,
max
m
max
min
0
243
56.6
186.8
c
max
x
m
min
0
243
PRO
OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
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n be copied, scannned, duplicated, forwarded, distribbuted, or posted
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w
in whole or part.
1183
PR
ROBLEM 7.141
7
Forr the given staate of plane strain,
s
use Moohr’s circle to determine (aa) the orientation and magnnitude of the
prinncipal strains,, (b) the maxim
mum in-plane strain, (c) the maximum shearing strain.
4000 ,
x
200 ,
y
xy
3
375
SO
OLUTION
Plootted points for Mohr’s circlle:
X : ( 400 , 187.5 )
Y : ( 200 , 187.5 )
C : ( 300 , 0)
0
tan 2
xy
p
x
2
y
375
400 200
1.875
61.93
p
a
121.0
b
(100 )2
R
(a)
a
ave
R
300
212.5
2
b
ave
R
300
212.5
2
(b)
max (in-planee)
(c)
c
0
(187.5 ) 2
212.5
a
b
2R
max (in-planee)
max
512.5
max
max
min
31.0
0
513
87.5
425
c
m
max
min
0
513
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1184
PR
ROBLEM 7.142
For the given staate of plane sttrain, use Mohhr’s circle to determine (a)) the orientatioon and magniitude of the
prinncipal strains, (b) the maxim
mum in-plane strain,
s
(c) the maximum
m
sheearing strain.
3000 ,
x
600 ,
y
xy
1000
SOLUTION
X : (300 , 500 )
Y : (60 , 50 )
C : (180 , 0)
tan 2
xy
p
x
2
p
y
100
300 60
22.62
a
b
R
(a)
(120 )2
(50 ) 2
11.3
101.3
1
130
a
ave
R 180
1330
a
310
b
ave
R 180
1330
b
50.0
max (in-plane))
260
(b)
max (in-plane))
(c)
c
0,
max
m
2R
310 ,
max
m
max
min
0
c
m
min
maxx
0
310
PRO
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w
in whole or part.
1185
PR
ROBLEM 7.143
7
Forr the given staate of plane strain,
s
use Moohr’s circle to determine (aa) the orientation and magnnitude of the
prinncipal strains,, (b) the maxim
mum in-plane strain, (c) the maximum shearing strain.
1800 ,
x
y
2660 ,
xy
3
315
SO
OLUTION
Plootted points for Mohr’s circlle:
X : ( 180 , 157.5 )
Y : ( 260 , 157.5 )
C : ( 220 , 0)
0
(a)
tan 2
x
2
315
5
80
xy
p
p
y
3.9375
7
75.75
a
b
(40 )2
R
(15
57.5 )2
ave
R
22
20
162.5
b
ave
R
22
20
162.5
max (in-planee)
(c)
c
0,
max
2R
57.5
a
b
383
325
0,
max
127.9
162..5
a
(b)
37.9
max
min
n
min
382.5
c
0 382.5
m
max
0
383
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on a website, in wholee or part.
1186
3
PRO
OBLEM 7.1
144
45"
2
Deterrmine the strain x , knowingg that the folloowing strains have been dettermined by
use of
o the rosette shown:
s
30"
15"
x
1
1
4800
1
15
2
1220
3
800
SOLUTION
c
x cos
2
n
y sin
1
0.9330
os
x co
2
0.75
os
x co
2
3
2
x
n
y sin
0.06699
30
3
75
1
n 1 cos 1
xy sin
0.06699
x
n
y sin
2
2
2
x
2
0.25
y
xy
2
xy sin 2 cos 2
0.25
y
3
0..4330
xy
xy sin 3 cos 3
0.9330
y
0.25
xy
1
4880
(1)
2
(2)
1
120
3
(3)
800
Solvving (1), (2), and
a (3) simultaaneously,
x
253 ,
y
307 ,
xy
8893
x
253
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1187
PROB
BLEM 7.145
y
The strrains determinned by the usee of the rosettte shown durring the test of
o a machine
elemen
nt are
30"
3
2
1
30"
600
1
x
4500
2
755
3
Determ
mine (a) the in--plane principal strains, (b) the in-plane maximum
m
sheaaring strain.
SO
OLUTION
os
x co
2
n
y sin
1
0.75
x
co
os2
2
y
0.75
x
cos2
y
3
2
sin 2
x
2
2
1500
3
90
xy sin 1 cos 1
1
0.433301
xy
6000
cos
2
2
0.433301
xy
4500
y
xy
0.25
sinn 2
30
1
0.25
x
1
3
sin
y
xy
2
sin
0
3 cos 3
y
(1)
(2)
3
0
755
(3)
Sollving (1), (2), and (3) simulttaneously,
x
ave
725
7
,
1
(
2
75 ,
y
x
y)
x
y
(a)
(b)
a
ave
R
b
ave
R
max (in-plane))
2
173.21
325
2
R
xy
2
725 75
2
xy
2
2
173.21
2
2
4
409.3
734
a
84.3
b
2 R 819
max (in-planee)
734
84.3
819
PRO
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on a website, in wholee or part.
1188
PROBLEM 7.146
4
3
45"
45"
The rosette shown has been used to determine the following strains at a point
on the surface of a crane hook:
2
45"
1
420 10 6 in./in.
1
x
45 10 6 in./in.
2
165 10 6 in./in.
4
(a) What should be the reading of gage 3? (b) Determine the principal strains
and the maximum in-plane shearing strain.
SOLUTION
(a) Gages 2 and 4 are 90 apart.
1
( 2
4)
2
1
( 45 10
2
ave
ave
6
165 10 6 )
60 10 6 in./in.
Gages 1 and 3 are also 90 apart.
1
( 1
2
2 ave
ave
3
3)
1
(2)(60 10 6 ) 420 10
6
300 10 6 in./in.
3
(b)
x
xy
420 10 6 in./in.
1
2
2
1
3
y
300 10 6 in./in.
3
(2)( 45 10 6 ) 420 10
6
( 300 10 6 )
210 10 6 in./in.
2
x
R
y
2
xy
2
420 10
2
6
( 300 10 6 )
2
2
210 10
2
6
2
375 10 6 in./in.
a
ave
R
60 10
6
375 10
6
b
ave
R
60 10
6
375 10
6
max (in-plane)
a
b
2R
max (in-plane)
435 10 6 in./in.
315 10 6 in./in.
750 10 6 in./in.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1189
"
2
PROBLEM 7.147
!2
3
45#
Using a 45 rosette, the strains 1, 2 , and 3 have
been determined at a given point. Using Mohr’s
circle, show that the principal strains are:
!3
2
B
O
45#
A
!
C
1
1
2
( 1
[( 1
(
3)
2)
2
2
(Hint: The shaded triangles are congruent.)
max,min
! min
1
!1
! max
2
3)
SOLUTION
Since gage directions 1 and 3 are 90 apart,
1
(
2
ave
Let
u
1
ave
1
(
2
v
2
ave
2
R2
1
(
2
u2
1
4
R
max, min
ave
3)
1
1
(
4
3)
1
3)
1
v2
3)
1
1
2
2
1
1 2
1
2
1
( 1
2
1
[(
2
2
2
2
1
1
4
1 3
2
3
2
2
2 1
2)
2( 1
2)
2
2
2
(
2
2
2 1
1
2
2 3
1
(
2
2
1
(
4
3)
3)
1
2 3
3)
2
1
4
2
1
1
2
1 3
1
4
2
3
2
3
2
3)
2 1/2
]
R gives the required formula.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1190
1
2 2
]
PROBLEM 7.148
2
3
Show that the sum of the three strain measurements made with a 60
rosette is independent of the orientation of the rosette and equal to
60"
60"
1
1
&
where
circle.
x
2
3
3
avg
is the abscissa of the center of the corresponding Mohr’s
avg
SOLUTION
x
1
ave
2
ave
y
2
x
y
2
x
ave
y
2
xy
2
x
cos (2
2
y
2
x
ave
y
2
xy
cos (2
2
xy
2
120 )
sin 120 sin 2 )
3
sin 2
2
(2)
xy
240 )
2
(cos 240 cos 2
y
sin (2
3
cos 2
2
(cos 240 sin 2
x
ave
xy
2
sin 120 cos 2 )
1
cos 2
2
y
x
ave
(1)
120 )
1
sin 2
2
xy
sin 2
(cos 120 cos 2
2
2
2
(cos 120 sin 2
ave
3
xy
cos 2
240 )
sin 240 sin 2 )
sin 240 cos 2 )
1
cos 2
2
1
sin 2
2
sin (2
3
sin 2
2
3
cos 2
2
(3)
Adding (1), (2), and (3),
1
2
3
3
ave
3
0 0
ave
1
2
3
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1191
PROBLEM 7.149
The strains determined by the use of the rosette attached as shown during the test of a
machine element are
3
75"
2
x
75"
1
1
93.1 10 6 in./in.
2
385 10 6 in./in.
3
210 10 6 in./in.
Determine (a) the orientation and magnitude of the principal strains in the plane of the
rosette, (b) the maximum in-plane shearing strain.
SOLUTION
Use
x
1
(
2
x
y)
1
(
2
where
sin 2
for gage 2,
0
and
2
for gage 1,
75
for gage 3.
75
From Eq. (2),
xy
y ) cos 2
x
1
1
(
2
x
y)
1
(
2
x
y ) cos (
2
1
(
2
x
y)
1
(
2
x
y ) cos
0
3
1
(
2
x
y)
1
(
2
x
y ) cos
(150 )
x
z
xy
150 )
xy
2
2
sin ( 150 )
sin 0
xy
2
sin (150 )
(1)
(2)
(3)
385 10 6 in./in.
Adding Eqs. (1) and (3),
1
3
(
x (1
y
y)
x
1
(
y ) cos 150
x
cos 150 )
y (1
cos 150 )
x (1 cos 150 )
(1 cos 150 )
3
93.1 10
6
210 10 6 385 10 6 (1 cos 150 )
1 cos 150
35.0 10 6 in./in.
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1192
PROBLEM 7.149 (Continued)
Subtracting Eq. (1) from Eq. (3),
3
1
sin 150
xy
3
xy
210 10
1
sin 150
6
( 93.1 10 6 )
sin 150
606.2 10 6 in./in.
tan 2
606.2 10 6
385 10 6 35.0 10
xy
p
x
y
1
1
( x
(385 10
y)
2
2
210 10 6 in./in.
ave
2
x
R
6
2
30.0 ,
b
120.0
xy
2
2
6
35.0 10
6
2
606.2
2
a
ave
R
210 10
6
350.0 10
6
b
ave
R
210 10
6
350.0 10
6
R
a
35.0 10 6 )
2
max (in-plane)
(a)
2
y
385 10
(b)
1.732
6
350.0 10 6 in./in.
2
350.0 10
6
a
b
max (in-plane)
560 10 6 in./in.
140.0 10 6 in./in.
700 10 6 in./in.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1193
PROBLE
EM 7.150
y
1 in
n.
P
A centric axial
a
force P and
a a horizonttal force Qx arre both applieed at point C
of the rectaangular bar shown. A 45 sttrain rosette on the surface of the bar at
point A indicates the folloowing strains::
Qx
C
x
12 in.
i
3
A
3 in.
45!
60 10 6 in./in.
2
240 10 6 in./in.
3
200 10 6 in./in.
29 106 psi and v
Knowing thhat E
and Qx.
2
1
0.30, determ
mine the magnitudes of P
1
3 in..
SO
OLUTION
x
1
60 10
6
y
3
200 10
6
xy
x
y
P
A
2
2
1
E
(
1 v2
E
(
1 v2
P
y
3400 10
3
x
v y)
y
v x)
A
y
6
29
[ 60
6 (0.3)(200)] 0
1 (0.3)2
29
[2000 (0.3)( 60)] 5.8 103 psi
p
2
1 (0.3)
(22)(6)(5.8 103 )
69.6 103 lb
G
xy
I
Qˆ
xy
V
E
2(1 v)
G
xy
69.6 kips
Q
30.3 kips
29 1006
11.1538 106 psi
(2)(1.300)
(11.1538)((340) 3.79233 103 psi
1 3 1
(2)(6))3 36 in 4
bbh
12
12
A y (2)(3)(1.5) 9 in 3
t
ˆ
VQ
It
It xy
Qˆ
P
(36)(2)(3..7923 103 )
9
Q V
2 in.
30.338 103 lb
l
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1194
PROBLEM 7.151
y
1 in.
Solve Prob. 7.150, assuming that the rosette at point A indicates the
following strains:
P
Qx
C
x
12 in.
1
30 10 6 in./in.
2
250 10 6 in./in.
3
100 10 6 in./in.
PROBLEM 7.150 A centric axial force P and a horizontal force Qx are both
applied at point C of the rectangular bar shown. A 45 strain rosette on the
surface of the bar at point A indicates the following strains:
3
A
3 in.
2
45"
1
3 in.
Knowing that E
P and Qx.
1
60 10 6 in./in.
2
240 10 6 in./in.
3
200 10 6 in./in.
29 106 psi and v
0.30, determine the magnitudes of
SOLUTION
6
x
1
30 10
y
3
100 10
xy
x
y
2
2
1
E
(
1 v2
0
E
(
1 v2
6
430 10
3
6
x
v y)
29
[ 30 (0.3)(100)]
1 (0.3)2
y
v x)
29
[100 (0.3)( 30)]
1 (0.3)2
2.9 103 psi
P
A
y
P
A
y
(2)(6)(2.9 103 )
34.8 103 lb
P
34.8 kips
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1195
PROB
BLEM 7.151 (Continue
ed)
G
xyy
I
Qˆ
t
xyy
V
E
2(1 v)
G
xy
29 106
(2)(1.30)
111.1538 106 pssi
(11.1538)(430)
4.7962 103 psi
1 3 1
bh
(2)(6)3 36 in 4
12
12
A y (2)(33)(1.5) 9 in 3
2 in.
VQˆ
It
It xy
Qˆ
(366)(2)(4.7962 103 )
9
Q V
38.37 103 lb
Q
3
38.4
kips
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1196
PROBL
LEM 7.152
T'
A single strain gage iss cemented to a solid 4-in.-diameter steell shaft at an
25 with a line parallel to the axis off the shaft. Knnowing that
angle
G 11.5 106 psi, deetermine the toorque T indicaated by a gagee reading of
300 100 6 in./in.
!
T
2 in.
SOLUTION
For torsion,
x
0,
y
1
(
E
1
(
E
x
y
x
v
y)
0
y
v
x)
0
1
2
0
xy
0
G
xy
0
2G
Draaw the Mohr’s circle for straain.
R
x
But
0
T
0
2G
R sin 2
Tc
J
c 3G
sin 2
2T
c3
0
2
2G
sin 2
2G x
ssin 2
x
(2)3(11.5 106 )(300 10 6 )
sinn 50
113.2 103 lbb in.
T 113.2
2 kip in.
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1197
PROB
BLEM 7.153
3
T'
Solve Prob.
P
7.152, asssuming that thhe gage formss an angle
line parrallel to the axxis of the shaftt.
!
35 with a
PROBL
LEM 7.152 A single gagee is cementedd to a solid 4--in.-diameter
steel shhaft at an anglee
25 with a line paralllel to the axis of the shaft.
Knowinng that G 11.5 106 psi, determine thee torque T inddicated by a
gage reaading of 300 10 6 in./in.
T
2 in.
SO
OLUTION
Forr torsion,
0
0,
x
1
(
E
1
(
E
x
y
G
xy
0
x
v
y)
0
y
v
x)
0
1
2
0
xy
0,
y
xy
0
2G
Draaw Mohr’s cirrcle for strain.
R
x
0
2G
R sin 2
0
2
2G
sin 2
But
0
T
Tc
J
c 3G
sin 2
2T
c3
x
2G x
ssin 2
(2)3(11.5 106 )(300 10 6 )
7
sin 70
92.3 103 lb in.
T
92.3 kip in.
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1198
PROBLEM 7.15
54
A singgle strain gage forming an angle
18 with a hoorizontal planee is used to
determ
mine the gage pressure in thhe cylindrical steel tank shoown. The cylinndrical wall
of the tank is 6 mm thick, has a 600-mm
6
insidee diameter, annd is made of a steel with
E 200
2 GPa and v 0.30. Dettermine the prressure in the tank
t
indicatedd by a strain
gage reading of 280 .
!
SOLUTION
x
pr
t
1
1
x,
2
1
( x
E
y
x
0.85
v
0.20
xy
xy
z)
v
y
v
2
1
x
E
x
E
1
( v
E
y
0
z
x
v
y
z)
1
2
v
x
E
x
E
0
G
Draaw Mohr’s circcle for strain.
ave
a
R
1
(
2
1
(
2
x
p
Data:
ave
t
x
r
x
y)
0.525
x
y)
0.325
R cos 2
x
E
(0.5225 0.325cos 2 )
x
E
tE x
r (00.525 0.325ccos 2 )
r
1
d
2
t
6 10 3 m
mm E
1
(6600)
2
3
p
x
E
300 mm
m
0.300 m
200 109 Pa,
9
x
280 10
6
18
6
(6 10 )(200
)
10 )(2880 10 )
1..421 106 Pa
(0.300)((0.525 0.325 cos 36 )
p 1.4
421 MPa
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1199
PRO
OBLEM 7.155
Solvee Prob. 7.154, assuming thaat the gage form
ms an angle
planee.
!
35 with a horizontal
PROBLEM 7.1544 A single straain gage formiing an angle
18 with a horizontal
planee is used to deetermine the gaage pressure in
i the cylindriical steel tank shown. The
cylind
drical wall off the tank is 6 mm thick, has a 600-mm
m inside diam
meter, and is
madee of a steel witth E 200 GP
Pa and v 0.30.
Determinne the pressurre in the tank
0
indicaated by a straiin gage reading of 280 .
SO
OLUTION
x
y
x
y
pr
t
1
1
x,
2
1
( x
E
v
1
( v
E
xy
x
xy
0
z
x
v
y
y
z)
v
v
2
1
1
2
z)
x
0.85
E
x
v
E
x
E
0.20
x
E
0
G
Draaw Mohr’s cirrcle for strain.
ave
R
1
(
2
1
(
2
x
ave
a
x
y)
0.525
x
y)
0.325
x
E
x
E
R cos 2
0.525 0.325 cos
c 2 )
(0
p
Data:
t
x
r
tE x
r (0.525 0.325 cos 2 )
r
1
d
2
t
6 10 3 m E
1
(600)
2
3
p
x
E
300 mm
0.300 m
200 109 Pa,
x
280 10
1
6
35
6
9
(6
6 10 )(200 10 )(280 100 )
1.761 106 Pa
(0.300)(0.525
(
0.325 cos 70 )
p 1.761 MPa
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1200
PROBL
LEM 7.156
150 MPa
The given state of planne stress is knoown to exist on the surface of
o a machine component.
c
G , determ
mine the direection and
Knowingg that E 200 GPa andd G 77.2 GPa
magnitudde of the threee principal strrains (a) by determining
d
thhe correspondiing state of
strain [usse Eq. (2.43) and Eq. (2.38)] and then using Mohr’ss circle for strrain, (b) by
using Moohr’s circle foor stress to deetermine the principal
p
plannes and princippal stresses
and then determining the
t correspondding strains.
75 MP
Pa
SOLUTION
(a)
x
E
G
x
y
xy
xy
2
ave
x
y
taan 2
0,
4877.0
1
( x
2
974
y)
2633
974
974
xyy
y
1.000
45.0
x
R
y
a
ave
R
b
ave
R
x
2
xy
2
v
(
E
22.5
a
2
c
75 106 Pa
E
E
v
1 0.2987
2(1 v)
2G
1
1
( x v y)
[0 (00.2987)(150 1106 )]
E
2000 109
224
1
1
( y v x)
[( 1500 106 ) 0]
E
2000 109
7500
75 106
xy
974
G
77 109
a
a
xy
77 109 Pa
200 109 Pa G
x
2
150 1006 Pa,
y
2
y)
689
(0.2987)(0 150
1
106 )
200 10
1 9
b
67.5
a
426
b
952
c
224
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1201
PROB
BLEM 7.156
6 (Continued
d)
(b)
ave
1
(
2
x
y)
x
y
75 MPa
2
R
0 1150
2
2
xy
2
2
752
1006.07 MPa
a
ave
R
b
ave
R
a
1
(
E
31.07 MPa
181
1.07 MPa
v
a
1
200 109
4226 10
tan 2
2
xy
[31.07 106
(0.29987)( 181.07 106 )]
6
000
1.0
a
x
b)
2
a
426
a
45
y
a
22.5
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1202
PR
ROBLEM 7.157
Thee following staate of strain haas been determ
mined on the suurface of a casst-iron machinne part:
720
x
y
400
xy
6
660
Knoowing that E 69 GPa annd G 28 GP
Pa, determinee the principaal planes and principal streesses (a) by
deteermining the corresponding
c
g state of planee stress [use Eq.
E (2.36), Eqq. (2.43), and the
t first two equations
e
of
Probb. 2.73] and then
t
using Mohr’s
M
circle for
fo stress, (b) by using Moohr’s circle forr strain to dettermine the
orieentation and magnitude
m
of thhe principal strrains and thenn determining the corresponding stresses.
SOLUTION
Thee 3rd principall stress is
z
0.
E
69
E
1
1 0.2321
v
2(1 v)
2G
56
6
69
72.933 GPa
1 (0..232) 2
G
E
1 v2
(a)
x
y
E
( x v y)
1 v2
(72..93 109 )[ 720 10
59
9.28 MPa
E
( y v x)
1 v2
(72..93 109 )[ 4000 10
6
(0.2232)( 400 100 6 )]
6
(0.22321)( 720 10
1 6 )]
41.36 MPa
xy
G
(28 109 )(660 10 6 )
xy
x
18.4
48 MPa
ave
tan 2
1
(
2
2
xy
2.06225
b
y
x
2
b
500.32 MPa
y)
x
64
4.1 ,
b
32.1 ,
a
57.9
2
x
R
y
2
2
xy
20.54 MP
Pa
a
avee
R
a
29.8 MPa
b
avee
R
b
700.9 MPa
PRO
OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
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a
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1203
PROB
BLEM 7.157 (Continue
ed)
(b)
ave
tan 2
1
(
2
xy
b
560
5
2.0625
b
x
2
y)
x
y
6
64.1
,
32.1 ,
b
2
x
R
y
a
5
57.9
2
xy
2
2
a
avve
R
193..26
b
avve
R
926..74
366..74
a
E
(
1 v2
a
v b)
a
2
29.8
MPa
b
E
(
1 v2
b
v a)
b
7
70.9
MPa
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1204
P
PROBLEM
M 7.158
T
1
4
A steel pipe of 12-in. outter diameter iss fabricated frrom 14 -in. -thiick plate by
w a plane peerpendicular
welding alonng a helix thatt forms an anggle of 22.5 with
to the axis off the pipe. Knoowing that a 40-kip
4
axial foorce P and an 80-kip in.
torque T, eaach directed as shown, arre applied to the pipe, dettermine the
normal and in-plane
i
shearring stresses in
i directions, respectively, normal and
tangential to the weld.
in.
Weld
22.5
SOLUTION
1
d2
2
5.75 in.
d2
12 in., c2
c1
c2
t
A
c22
c12
J
c24
c14
2
(62
2
6 in., t
0.25 in.
5.752 )
9
9.2284
in 2
(664
5.754 ) 318.67
3
in 4
Streesses:
P
A
40
4.33444 ksi
9.22284
Tc2
J
(80))(6)
1.5063 ksi
k
318.67
0,
4.33444 ksi,
y
x
xy
1..5063 ksi
Chooose the x an
nd y axes, resspectively, tanngential and noormal to the weld.
w
Theen
w
y
y
and
w
xy
x
y
x
y
2
( 4.3344)
2
4
4.76
ksi
x
xy
22.5
cos 2
x sin 2
xy
2
[ ( 4.3344)]
cos 45 1.5063 sin
s 45°
2
w
4.76 ksi
y
sin 2
xy cos 2
2
[ ( 4.3344)]
s 45 1.50663 cos 45
sin
2
0
0.467
ksi
w
0.467
0
ksi
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1205
100 kN
!
80 mm
100 kN
PROBLEM 7.159
Two steel plates of uniform cross section 10 80 mm are welded
together as shown. Knowing that centric 100-kN forces are applied to
the welded plates and that
25 , determine (a) the in-plane shearing
stress parallel to the weld, (b) the normal stress perpendicular to the
weld.
SOLUTION
Area of weld:
Aw
(10 10 3 )(80 10 3 )
cos 25
882.7 10 6 m 2
(a)
Fs
w
(b)
Fn
w
0: Fs
Fs
Aw
100sin 25
42.26 103
882.7 10 6
0: Fn
Fn
Aw
100 cos 25
90.63 103
882.7 10 6
0
Fs
42.26 kN
47.9 106 Pa
0
Fn
w
47.9 MPa
90.63 kN
102.7 106 Pa
w
102.7 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
1206
100 kN
PROBLEM 7.160
!
Two steel plates of uniform cross section 10 80 mm are welded
together as shown. Knowing that centric 100-kN forces are applied to
the welded plates and that the in-plane shearing stress parallel to the
weld is 30 MPa, determine (a) the angle , (b) the corresponding normal
stress perpendicular to the weld.
80 mm
100 kN
SOLUTION
Area of weld:
Aw
(10 10 3 )(80 10 3 )
cos
800 10
cos
(a)
Fs
w
0: Fs
Fs
Aw
sin cos
(b)
100sin
Aw
800 10 6
cos14.34
Fn
Aw
100sin
100 10 sin
800 10 6 / cos
30 106
125 106
1
sin 2
2
0: Fn
Fs
3
30 106
Fn
0
100 cos
0
kN
100 103 sin
m2
N
125 106 sin cos
0.240
Fn
6
14.34
100cos14.34
96.88 kN
825.74 10 6 m 2
96.88 103
825.74 10 6
117.3 106 Pa
117.3 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
1207
PROBLEM 7.161
'0
&
+
'0
Determine the principal planes and the principal
stresses for the state of plane stress resulting from the
superposition of the two states of stress shown.
SOLUTION
Mohr’s circle for 2nd stress state:
x
y
xy
1
2
1
2
1
2
1
2
1
2
0
0
0
cos 2
0
cos 2
sin 2
0
Resultant stresses:
x
y
0
xy
0
1
2
1
2
1
(
2
2
ave
tan 2
1
2
0
1
2
0
1
2
0
y)
x
xy
1
2
0
y
1
2
0
1
2
0
1
2
0
0
0
cos 2
cos 2
sin 2
0
p
0
sin 2
1 cos 2
3
2
cos 2
cos 2
0
sin 2
0
x
0
sin 2
0 cos 2
tan
p
2
x
R
y
2
1
2
0
2
xy
1
2
1 2 cos 2 + cos 2 2
0
1
2
2
0
sin 2 2
cos 2
2
2
0
1
2
2
(
0 sin 2
1 cos 2
1
2
p
0
)
2
|cos |
a
ave
R
a
0
0
cos
b
ave
R
b
0
0
cos
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1208
y
PROBLEM 7.162
2 ksi
For the state of stress shown, determine the maximum shearing stress
when (a) z
4 ksi, (b) z
4 ksi, (c) z 0.
6 ksi
σz
7 ksi
z
x
SOLUTION
7 ksi,
x
ave
1
(
2
x
y)
x
y
y
2 ksi,
xy
6 ksi
4.5 ksi
2
R
2.52
(a)
(b)
( 6) 2
6.5 ksi
a
ave
R 11 ksi
b
ave
R
z
4 ksi,
max
11 ksi,
z
max
2 ksi
a
a
11 ksi,
min
11 ksi,
11 ksi,
min
2 ksi
b
2 ksi,
min
11 ksi,
0,
11 ksi,
a
4 ksi,
z
max
(c)
2
xy
2
4 ksi,
b
2 ksi,
1
(
2
max
min )
max
max
6.50 ksi
2 ksi
b
max
1
(
2
max
min )
max
7.50 ksi
1
(
2
max
min )
max
6.50 ksi
2 ksi
max
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1209
y
PROBLEM 7.163
40 MPa
For the state of stress shown, determine the value of
xy
for which the
maximum shearing stress is (a) 60 MPa, (b) 78 MPa.
τ xy
100 MPa
z
x
SOLUTION
x
100 MPa,
1
(
2
ave
(a)
y)
x
40 MPa,
0
z
70 MPa
60 MPa.
max
If
y
z
is
min ,
then
max
max
max
2
min
max .
0 (2)(60) 120 MPa
R
ave
R
max
b
max
ave
2R
120 70 50 MPa
20 MPa > 0
2
x
R
xy
(b)
2
xy
2
502
302
2
xy
50 MPa
302
xy
40.0 MPa
xy
72.0 MPa
78 MPa.
max
If
y
z
is
min ,
then
max
min
2
max
ave
R
R
Set
max
0 (2)(78) 156 MPa.
max
156 70 86 MPa >
ave
R
max
78 MPa.
min
ave
R
max
78 MPa
8 MPa < 0
2
R
xy
x
y
2
782
2
xy
302
302
2
xy
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
1210
PROBLEM 7.164
14 ksi
(xy
24 ksi
The state of plane stress shown occurs in a machine component made of a
steel with Y 30 ksi. Using the maximum-distortion-energy criterion,
determine whether yield will occur when (a) xy 6 ksi, (b) xy 12 ksi,
(c) xy 14 ksi. If yield does not occur, determine the corresponding factor
of safety.
SOLUTION
24 ksi
x
For stresses in xy-plane,
(a)
xy
ave
1
(
2
x
y
14 ksi
y
y)
x
0
z
x
19 ksi
y
2
5 ksi
6 ksi
2
R
a
2
a
2
b
a
b
F .S .
(b)
xy
2
xy
2
R
ave
(5)2
26.810 ksi,
(6)2
ave
b
7.810 ksi
R 11.190 ksi
23.324 ksi < 30 ksi
(No yielding)
30
23.324
F .S . 1.286
12 ksi
2
x
R
a
2
a
2
b
a
b
F .S .
(c)
xy
y
2
xy
2
R
ave
32 ksi,
(5)2
b
ave
(12)2
R
13 ksi
6 ksi
29.462 ksi < 30 ksi
(No yielding)
30
29.462
F .S . 1.018
14 ksi
2
x
R
a
2
a
2
b
a
b
y
2
xy
2
ave
R
33.866,
32.00 ksi > 30 ksi
(5)2
b
ave
(14)2
R
14.866 ksi
4.134 ksi
(Yielding occurs)
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1211
PROBLEM 7.165
750 mm
750 mm
The compressed-air tank AB has an inner diameter of 450 mm and a
uniform wall thickness of 6 mm. Knowing that the gage pressure
inside the tank is 1.2 MPa, determine the maximum normal stress
and the maximum in-plane shearing stress at point a on the top of the
tank.
b
a
B
D
A
5 kN
500 mm
SOLUTION
Internal pressure:
r
1
2
Torsion:
c1
J
T
1
d
2
pr
t
pr
2t
225 mm t
6 mm
(1.2)(225)
6
45 MPa
22.5 MPa
225 mm, c2
2
c24
c14
225
6
446.9 106 mm 4
(5 103 )(500 10 3 )
Tc
J
231 mm
2500 N m
3
(2500)(231 10 )
446.9 10 6
1.29224 106 Pa
At point a,
1.29224 MPa
0 at point a.
Transverse shear:
Bending:
446.9 10 6 m 4
I
M
1
J
2
223.45 10 6 m 4 , c
(5 103 )(750 10 3 )
231 10 3 m
3750 N m
Mc
I
(3750)(231 10 3 )
223.45 10 6
3.8767
3.8767 MPa
Total stresses (MPa).
Longitudinal:
x
22.5
Circumferential:
y
45 MPa
Shear:
xy
26.377 MPa
1.29224 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1212
PROBLEM 7.165 (Continued)
ave
1
(
2
x
y)
x
y
35.688 MPa
2
R
max
max(in-plane)
2
xy
2
ave
R
R
45.1 MPa
9.40 MPa
9.4007 MPa
max
45.1 MPa
max (in-plane)
9.40 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1213
PROBLEM 7.166
750 mm
750 mm
b
a
D
A
5 kN
500 mm
B
For the compressed-air tank and loading of Prob. 7.165, determine
the maximum normal stress and the maximum in-plane shearing
stress at point b on the top of the tank.
PROBLEM 7.165 The compressed-air tank AB has an inner
diameter of 450 mm and a uniform wall thickness of 6 mm. Knowing
that the gage pressure inside the tank is 1.2 MPa, determine the
maximum normal stress and the maximum in-plane shearing stress at
point a on the top of the tank.
SOLUTION
Internal pressure:
r
1
2
Torsion:
c1
J
T
1
d
2
pr
t
pr
2t
Bending:
At point b,
(1.2)(225)
6
2
6 mm
45 MPa
22.5 MPa
225 mm, c2
c24
225
c14
6
231 mm
446.9 106 mm 4
(5 103 )(500 10 3 )
Tc
J
Transverse shear:
225 mm t
446.9 10 6 m 4
2500 N m
3
(2500)(231 10 )
446.9 10 6
1.29224 106 Pa
1.29224 MPa
0 at point b.
I
M
1
J
2
223.45 10
(5 103 )(2
6
m4 , c
750 10 3 )
Mc
I
(7500)(231 10 3 )
223.45 10 6
7.7534
231 10 3 m
7500 N m
7.7534 MPa
Total stresses (MPa).
Longitudinal:
x
22.5
Circumferential:
y
45 MPa
Shear:
xy
30.253 MPa
1.29224 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1214
PROBLEM 7.166 (Continued)
ave
1
( x
2
y)
37.626 MPa
2
x
R
max
max (in-plane)
y
2
xy
2
ave
R
R
45.1 MPa
7.49 MPa
7.4859 MPa
max
45.1 MPa
max (in-plane)
7.49 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1215
0.12 in.
A
PROBLEM 7.167
The brass pipe AD is fitted with a jacket used to apply a hydrostatic pressure of 500 psi
to portion BC of the pipe. Knowing that the pressure inside the pipe is 100 psi,
determine the maximum normal stress in the pipe.
B
0.15 in.
C
D
2 in.
4 in.
SOLUTION
The only stress to be considered is the hoop stress. This stress can be obtained by applying
1
pr
t
Using successively the inside and outside pressures (the latter of which causes a compressive stress),
pi
100 psi, ri
(
po
max )i
1
pi ri
t
500 psi, ro
(
max )o
max
0.12
0.88 in., t
(100)(0.88)
0.12
1 in.,
po ro
t
733.33
t
0.12 in.
733.33 psi
0.12 in.
(500)(1)
0.12
4166.7
4166.7 psi
3433.4 psi
max
3.43 ksi (compression)
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1216
0.12 in.
A
PROB
BLEM 7.168
8
For the assembly of Prob. 7.167, determine the normal streess in the jackket (a) in a
o the jacket, (b) in a directtion parallel
directionn perpendicular to the longitudinal axis of
to that axis.
a
B
PROBL
LEM 7.167 Thhe brass pipe AD
A is fitted with
w a jacket ussed to apply a hydrostatic
pressuree of 500 psi too portion BC of the pipe. Knnowing that thee pressure inside the pipe
is 100 psi, determine the
t maximum
m normal stresss in the pipe.
0.15 in.
C
D
2 in.
4 in.
SOLUTION
(a)
Hoop stress.
p
( 1)
500 psii, t
pr
t
0.15 inn., r
((500)(1.85)
0.15
2
0
0.15
1.85 in.
6166.7 psi
1
(b)
6.17 ksi
Longitudin
nal stress.
Free body of portion of jacket
j
above a horizontal seection, consideering vertical forces
f
only:
Fy
Af
0:
Af
p dA
Aj
pA f
Areas :
Af
r22
r12
[(1.85) 2
Aj
r32
r22
[(2) 2
(11)2 ]
(1.855)2 ]
2 dA j
0
2 Aj
0
2
p
Af
(1)
Aj
7.6105 in
i 2
1.814277 in 2
Recalling Eq.
E (1),
2
p
Af
Aj
(500)
7.6105
1
1.81427
20097.4 psi
2
2
2.10
ksi
PRO
OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
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a
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n be copied, scannned, duplicated, forwarded, distribbuted, or posted
on a website,
w
in whole or part.
1217
PROBLEM
P
7.169
1
2
Determine
D
the largest in-planne normal straain, knowing that the follow
wing strains
haave been obtaiined by the usse of the rosettte shown:
3
1
50 100 6 in./in.
2
360 10 6 in./in.
3
1 6 in./in.
315 10
45!
x
45!
SO
OLUTION
455 ,
1
x
cos 2
1
sin 2
y
1
0.55
cos 2
x
y
2
sin 2
x
cos 2
3
y
sin 2
sin
xy
0.5
x
2
0.55
45 ,
2
xy
y
sin
0.5
x
3
xy
2
y
sin
cos
1
3
Eq. (1)
Eq. (2):
Eq. (1)
Eq. (2):
x
x
315
3
10
6
xy
50 10
6
y
1
2
y
1
2
ave
1
(
2
xy
cos
2
0.5
xy
cos
3
360 10
y)
50 10
6
(1)
6
(2)
6
(3)
2
360 10
3
0 0
315 10
4
410
10 6 in./iin.
6
3
360
10
6
3115 10
6
5 10 6 in./in.
1
155
10 6 in./inn.
2
y
xy
2
315 10
6
50 10
2
x
R
1
in.//in.
x
x
1
0
0.5
x
Froom (3),
0
3
2
6
5 10
6
2
410 10
2
2
6
2
260 10 6 in../in.
max
ave
R 1555 10
6
260 10
6
max
415 110 6 in./in.
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use.
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i any manner. Thhis document may not be copied, scaanned, duplicated,, forwarded, distributed, or posted
on a website, in wholee or part.
1218
y'
y
"y
PROBLEM 7.C1
1
y
#
"y'
!x'y'
!xy
x
Q
z
#
"x'
Q
x
"x
x'
x
z
(a)
(
(b)
A statee of plane streess is defined by
b the stress
compoonents x , y , and xy assoociated with
the eleement shown inn Fig. P7.C1a. (a) Write a
compuuter program that can be
b used to
calculaate the stress components x y , and
ssociated withh the element after it has
x y as
rotatedd through an angle
a
abouut the z axis
(Fig. P.7C1
P
b). (b) Use
U this prograam to solve
Probs. 7.13 through 7.16.
7
SOLUTION
Proggram followin
ng equations:
x
Equuation (7.5), Paage 427:
x
Equuation (7.7), Paage 427:
y
x,
y,
xyy
x
2
y
x
2
y
2
x
xy
y
2
x
Equuation (7.6), Paage. 427:
Enteer
y
y
2
sin 2
c 2
cos
xy
sinn 2
c 2
cos
xy
sinn 2
x
xy
cos 2
and
Prinnt values obtaiined for
x
,
y
and
xy
Prooblem Outputts
Probblem 7.13
x
x
xy
0 ksi
8 ksi
5 ksi
Rootation of elem
ment
(+ counterclockw
c
wise)
25
Rotation off element
(+ countercllockwise)
10
x
2.40 ksi
x
1.995 ksi
y
10.40 ksi
y
6.05 ksi
xy
6..07 ksi
xy
0.15 ksi
PRO
OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
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a
for salee or distribution inn any manner. Thiis document may not
n be copied, scannned, duplicated, forwarded, distribbuted, or posted
on a website,
w
in whole or part.
1219
PROB
BLEM 7.C1 (Continued
d)
Prooblem 7.14
y
60 MPa
M
90 MPa
xy
Pa
30 MP
x
Ro
otation of Elem
ment
( counterclockw
wise)
25
Rotation off Element
( countercclockwise)
1
10
x
56.19 MP
Pa
x
455.22 MPa
y
86.19 MP
Pa
y
755.22 MPa
xy
38.17 MP
Pa
xy
53.84 MPa
Prooblem 7.15
x
8 ksii
y
12 ksi
xy
6 ksi
Rotation off Element
( countercclockwise)
1
10
Ro
otation of Elem
ment
( counterclockw
wise)
25
x
9.02 kssi
x
5.344 ksi MPa
y
13.02 kssi
y
9.344 ksi MPa
xy
9.066 ksi MPa
xy
3.80 kssi
Prooblem 7.16
x
0 MPa
M
y
80 MPa
M
xy
50 MPa
M
Ro
otation of Elem
ment
( counterclockw
wise)
25
Rotation off Element
( countercllockwise)
10
x
24.01 MPa
M
x
19.51 MPa
y
104.01 MPa
M
y
60..49 MPa
xy
60.67 MPa
xy
1.50 MPa
M
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
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on a website, in wholee or part.
1220
PROBLE
EM 7.C2
y
"y
A state of plane
p
stress iss defined by thhe stress compponents x , y , and xy
associated with the elem
ment shown inn Fig. P7.C1aa. (a) Write a computer
program thhat can be used to calcullate the princcipal axes, thhe principal
stresses, thee maximum inn-plane sheariing stress, andd the maximuum shearing
stress. (b) Use
U this prograam to solve Prrobs. 7.5, 7.9, 7.68, and 7.699.
!xy
x
Q
x
"x
z
SOLUTION
Proggram followin
ng equations:
2
x
y
Equuation (7.10)
ave
Equuation (7.14)
max
ave
R
J min
ave
R
Equuation (7.12)
p
2
taan
: R
2
1
s
Sheearing stress:
Theen
If
m
max
2
xy
y
x
2
y
xy
0 and
m
min
0 and
m
min
max(out-of-planne)
m
max
R;
max(in-plane)
m
1
2
max(out-of-planne)
R;
max(in-plane)
m
If
Theen
m
max
R;
maax(in-plane)
If
Theen
taan
y
xy
x
Equuation (7.15)
x
0 and
m
min
max(out-of-plaane)
0:
R
0:
1
2
0:
1
|
2
max
min |
Proogram Outputts
Probblems 7.5 and
d 7.9
x
60.00 MPa
y
40.00 MPa
xy
35.00 MPa
PRO
OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
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a
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n be copied, scannned, duplicated, forwarded, distribbuted, or posted
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in whole or part.
1221
PROB
BLEM 7.C2 (Continued
d)
Proogram Outpu
uts (Continued
d)
Anngle between xy
x axes and priincipal axes (+
+ counterclockkwise):
p
37.003
and 522.97°
max
13.660 MPa
min
86.440 MPa
x axis and plaanes of maxim
mum in-plane shearing
s
stresss (+ countercloockwise):
Anngle between xy
s
7.97
and 97.977°
max (in-plane)
36.400 MPa
max
43.200 MPa
Prooblem 7.68
x
140.000 MPa
y
40.000 MPa
xy
80.000 MPa
Anngle between xy
x axes and priincipal axes ( counterclockkwise):
p
29.000
and 1199°
max
1844.34 MPa
min
4.344
MPa
Anngle between xy
x axis and plaanes of maxim
mum in-plane in-plane sheariing stress ( counterclockw
c
wise):
s
74.000
and 1644.00°
max (in-plane)
(
94.334 MPa
max (out--of-plane)
94.334 MPa
x
140.000 MPa
y
120.000 MPa
xy
80.000 MPa
x axes and priincipal axes (+
+ counterclockkwise):
Anngle between xy
p
41.444
and 1311.44°
max
210.62 MPa
min
49.338 MPa
x axis and plaanes of maxim
mum in-plane in-plane sheariing stress (+ counterclockw
c
wise):
Anngle between xy
s
86.444
and 1766.44°
max (in-plane)
(
80.662 MPa
max (out--of-plane)
105.331 MPa
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use.
Not authorized for salle or distribution in
i any manner. Thhis document may not be copied, scaanned, duplicated,, forwarded, distributed, or posted
on a website, in wholee or part.
1222
PROBLEM 7.C2 (Continued)
Program Outputs (Continued)
Problem 7.69
x
140.00 MPa
y
20.00 MPa
xy
80.00 MPa
Angle between xy axes and principal axes (+ counterclockwise):
p
26.57
and 116.57
max
180.00 MPa
min
20.00 MPa
Angle between xy axis and planes of maximum in-plane in-plane shearing stress (+ counterclockwise):
s
71.57
and 161.57
max (in-plane)
100.00 MPa
max (out-of-plane)
100.00 MPa
x
140.00 MPa
y
140.00 MPa
xy
80.00 MPa
Angle between xy axes and principal axes (+ counterclockwise):
p
45
and 135.00
max
220.00 MPa
min
60.00 MPa
Angle between xy axis and planes of maximum in-plane in-plane shearing stress ( counterclockwise):
s
90.00
and 180.00°
max (in-plane)
80.00 MPa
max (out-of-plane)
110.00 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1223
PR
ROBLEM 7.C3
7
(a) Write a com
mputer program
m that, for a given
g
state off plane stress and a given yield
y
strength of a ductile
mine whether the
t material will
w yield. The program shouuld use both thhe maximum
maaterial, can be used to determ
sheearing-strengthh criterion and
d the maximum
m-distortion-ennergy criterionn. It should allso print the values
v
of the
prinncipal stressess and, if the material
m
does noot yield, calculate the factor of safety. (b) Use this proggram to solve
Proobs. 7.81, 7.82, and 7.164.
SO
OLUTION
Principal stressess.
2
x
avve
Maaximum-shearring-stress criterion.
If
a
and
b
havve same sign,
y
2
a
ave
R
b
ave
R
y
1
2
y
maax
1
2
a
If
max
y,
yieelding occurs.
If
max
y,
no yielding occu
urs, and factor of safety
x
; R
y
2
2
xy
y
m
max
Maaximum-distorrtion-energy criterion.
2
a
Compute radicall
a
b
2
b
r
If radical
y,
urs.
yielding occu
If radical
r
y,
no yielding occcurs, and facttor of safety
y
Radical
Proogram Outpu
uts
Prooblems 7.81a and
a 7.82a
Yield strengthh
325 MPa
x
200.00 MPa
M
y
200.00 MPa
M
xy
100.00 MPa
M
maxx
100.00 MPa
M
minn
300.00 MPa
M
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use.
Not authorized for salle or distribution in
i any manner. Thhis document may not be copied, scaanned, duplicated,, forwarded, distributed, or posted
on a website, in wholee or part.
1224
PROB
BLEM 7.C3 (Continued
d)
Proogram Outputts (Continuedd)
Usinng the maximu
um-shearing-sstress criterionn,
matterial will not yield.
y
F .S . 1.0083
Usinng the maximu
um-distortion-energy criteriion,
matterial will not yield.
y
F .S . 1.2228
a 7.82b
Probblems 7.81b and
Yield strenngth
325 MP
Pa
y
240.000 MPa
240.000 MPa
xy
100.00 MPa
x
max
min
140.000 MPa
340.000 MPa
um-shearing-sstress criterionn,
Usinng the maximu
matterial will yield
d.
Usinng the maximu
um-distortion-energy criteriion,
matterial will not yield.
y
F .S . 1.098
Probblems 7.81c and 7.82c
Yield strenngth
325 MP
Pa
y
280.000 MPa
280.000 MPa
xy
100.000 MPa
x
max
min
180.000 MPa
380.000 MPa
um-shearing-sstress criterionn,
Usinng the maximu
matterial will yield
d.
Usinng the maximu
um-distortion-energy criteriion,
matterial will yield
d.
Probblem 7.164a
Yield strenngth
30 ksi
y
24.00 ksi
k
14.00 ksi
k
xy
6.00 ksi
k
x
max
min
26.81 ksi
k
11.19 ksi
k
PRO
OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
Not authorized
a
for salee or distribution inn any manner. Thiis document may not
n be copied, scannned, duplicated, forwarded, distribbuted, or posted
on a website,
w
in whole or part.
1225
PROB
BLEM 7.C3 (Continued
d)
Proogram Outpu
uts (Continued
d)
(a)
Using the maximum-sheearing-stress criterion,
c
material will
w not yield.
F .S . 1.119
(b)
Using the maximum-disstortion-energgy criterion,
material will
w not yield.
F .S . 1.286
Prooblem 7.164b
(a)
Yield strenngth
30 ksi
x
24.00 ksi
k
y
14.00 kssi
xy
12.00 ksi
k
max
32.00 ksi
k
min
6.00 kssi
Using the maximum-sheearing-stress criterion,
c
material will
w yield.
(b)
Using the maximum-disstortion-energyy criterion,
material will
w not yield.
F .S . 1.018
Prooblem 7.164c
(a)
Yield strength
30 ksi
x
24.00 kssi
y
14.00 kssi
xy
14.00 kssi
max
33.87 kssi
min
4.13 kssi
Using the maximum-sheearing-stress criterion,
c
material will
w yield.
(b)
Using the maximum-disstortion-energyy criterion,
material will
w yield.
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use.
Not authorized for salle or distribution in
i any manner. Thhis document may not be copied, scaanned, duplicated,, forwarded, distributed, or posted
on a website, in wholee or part.
1226
PR
ROBLEM 7.C4
(a) Write
W
a computer program based on Mohhr’s fracture criterion
c
for brrittle materialss that, for a givven state of
planne stress and given
g
values of
o the ultimatee strength of thhe material inn tension and compression,
c
c be used
can
to determine
d
wheether rupture will
w occur. Thhe program shhould also print the values of the princippal stresses.
(b) Use
U this progrram to solve Probs.
P
7.89 andd 7.90 and to check
c
the answ
wers to Probs. 7.93 and 7.944.
SOLUTION
Prinncipal stresses.
2
x
avee
y
2
a
ave
R
b
ave
R
R
x
y
2
2
xy
c
Mohhr’s fracture criterion.
If
If
and
b
a
UT
and
a
UT
or
a
a
0 and
hav
ve same sign, and
a
UC ,
b
UC ,
b
b
no faailure;
failuree.
0:
Connsider fourth quadrant
q
of Figgure 7.47.
For no rupture to occur, point (
If
b
a,
b)
mustt lie within Moohr’s envelope (Figure 7.477).
Criterion
n,
thenn rupture occu
urs.
If
b
Criterion
n,
thenn no rupture occcurs.
PRO
OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
Not authorized
a
for salee or distribution inn any manner. Thiis document may not
n be copied, scannned, duplicated, forwarded, distribbuted, or posted
on a website,
w
in whole or part.
1227
PROB
BLEM 7.C4 (Continued
d)
Proogram Outpu
uts
Prooblem 7.89
10.00 MPa
x
100.00 MPa
y
60 MPa
xy
Ulttimate strengthh in tension
80 MPaa
Ulttimate strengthh in compressiion
200 MP
Pa
max
a
336.39 MPa
min
b
1226.39 MPa
Rupture will occcur
Prooblem 7.90
32.000 MPa
x
y
xy
0.00 MPa
M
75.00 MPa
M
Ulttimate strengthh in tension
80 MP
P
Ulttimate strengthh in compressiion
200 M
MP
max
a
60.69 MPa
min
b
92.69 MPa
Rupture will not occur.
To check answerrs to the follow
wing problemss, we check foor rupture usinng given answeers and an adjacent value.
Annswer:
Ruppture occurs for
f
0
3.67 ksi.
k
Prooblem 7.93
x
8.00 ksii
y
0.00 ksii
xy
3.67 ksii
Ulttimate strengthh in tension
10 ksi
Ulttimate strengthh in compressiion
25 ksi
max
a
9.443 ksi
min
b
1.443 ksi
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use.
Not authorized for salle or distribution in
i any manner. Thhis document may not be copied, scaanned, duplicated,, forwarded, distributed, or posted
on a website, in wholee or part.
1228
PROB
BLEM 7.C4 (Continued
d)
Proogram Outputts (Continuedd)
Ruppture will not occur.
o
x
8.00 ksi
y
0.00 ksi
xy
3.68 ksi
Ultiimate strength
h in tension
10 ksi
Ultiimate strength
h in compressioon
25 ksi
m
max
a
9.444 ksi
m
min
b
1.444 ksi
Ruppture will occu
ur.
Ansswer:
Ruppture occurs fo
or
0
49.1 MPa.
M
Probblem 7.94
x
y
xy
80.00 MPa
Pa
0.00 MP
49.10 MPa
M
Ultiimate strength
h in tension
75 MPaa
Ultiimate strength
h in compressioon
150 MP
Pa
max
a
23.33 MPa
min
b
1103.33 MPa
Ruppture will not occur.
o
x
y
xy
80.00 MPa
Pa
0.00 MP
49.20 MPa
M
Ultiimate strength
h in tension
75 MPaa
Ultiimate strength
h in compressioon
150 MP
Pa
max
a
min
b
233.41 MPa
1
103.41
MPa
Ruppture will occu
ur.
PRO
OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
Not authorized
a
for salee or distribution inn any manner. Thiis document may not
n be copied, scannned, duplicated, forwarded, distribbuted, or posted
on a website,
w
in whole or part.
1229
PROBLEM 7.C5
y
A state of plane strain is defined by the strain components x , y ,
and xy associated with the x and y axes. (a) Write a computer
program that can be used to calculate the strain components x , y ,
and x y associated with the frame of reference x y obtained by
rotating the x and y axes through an angle . (b) Use this program to
solve Probs 7.129 and 7.131.
y'
x'
&
x
SOLUTION
Program following equations:
x
Equation (7.44):
x
Equation (7.45):
y
Equation (7.46):
xy
Enter
x,
y,
y
2
y
x
2
(
y
2
x
xy ,
x
y
2
cos 2
1
2
xy
sin 2
sin 2
1
2
xy
cos 2
y )sin 2
x
xy
cos 2
and .
Print values obtained for
x
,
y
, and
xy
.
Program Outputs
Problem 7.129
x
240 micro meters
y
160 micro meters
xy
150 micro radians
Rotation of element, in degrees (+ counterclockwise):
Problem 7.131
x
60
115.05 micro meters
y
284.95 micro meters
xy
5.72 micro radians
x
0 micro meters
y
320 micro meters
xy
100 micro radians
Rotation of element, in degrees (+ counterclockwise):
x
30
36.70 micro meters
y
283.30 micro meters
xy
227.13 micro radians
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1230
PROBLEM 7.C6
A state of strain is defined by the strain components x , y , and xy associated with the x and y axes.
(a) Write a computer program that can be used to determine the orientation and magnitude of the principal
strains, the maximum in-plane shearing strain, and the maximum shearing strain. (b) Use this program to
solve Probs 7.136 through 7.139.
SOLUTION
Program following equations:
2
x
y
Equation (7.50):
ave
Equation (7.51):
max
ave
Equation (7.52):
p
tan
R
2
R
min
y
2
ave
2
xy
2
R
xy
1
x
Shearing strains:
x
y
Maximum in-plane shearing strain
2R
max (in-plane)
Calculate out-of-plane shearing strain and check whether it is the maximum shearing strain.
Let
a
max
b
min
v
Calculate
c
1 v
If
a
b
c,
out-of-plane
a
c
If
a
c
b,
out-of-plane
a
b
If
c
a
c
b
b,
out-of-plane
(
a
b)
2R
Program Printout
Problem 7.136
x
260 micro meters
y
60 micro meters
xy
480 micro radians
0.333
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1231
PROBLEM 7.C6 (Continued)
Program Printout (Continued)
Angle between xy axes and principal axes (
p
a
b
c
33.69
100.00 micro meters
420.00 micro meters
159.98 micro meters
max (in-plane)
520.00 microradians
max
579.98 microradians
Problem 7.137
counterclockwise):
x
600 micrometers
y
400 micrometers
xy
350 microradians
0.333
Angle between xy axes and principal axes (+ = counterclockwise):
p
30.13
a
298.44 micrometers
b
701.56 micrometers
c
500.00 micrometers
max(in-plane)
403.11 microradians
max
Problem 7.138
x
y
xy
1201.56 microradians
160 micrometers
480 micrometers
600.00 microradians
0.333
Angle between xy axes and principal axes (
p
a
b
c
counterclockwise):
21.58
278.63 micrometers
598.63 micrometers
159.98 micrometers
max(in-plane)
877.27 microradians
max
877.27 microradians
Problem 7.139
x
30 micrometers
y
570 micrometers
xy
720 microradians
0.333
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1232
PROBLEM 7.C6 (Continued)
Angle between xy axes and principal axes (
p
a
counterclockwise):
26.57
750.00 micrometers
b
150.00 micrometers
c
300.00 micrometers
max(in-plane)
max
900.00 microradians
1050.00 microradians
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1233
PROBLEM 7.C7
A state of plane strain is defined by the strain components x , y , and xy measured at a point. (a) Write a
computer program that can be used to determine the orientation and magnitude of the principal strains, the
maximum in-plane shearing strain, and the magnitude of the shearing strain. (b) Use this program to solve
Probs 7.140 through 7.143.
SOLUTION
Program following equations:
2
x
y
Equation (7.50)
ave
Equation (7.51)
max
ave
Equation (7.52)
p
tan
2
R
y
2
min
ave
2
xy
2
R
xy
1
x
Shearing strains:
x
R
y
Maximum in-plane shearing strain
2R
xy (in-plane)
Calculate out-of-plane-shearing strain and check whether it is the maximum shearing strain.
Let
a
b
c
If
a
max
min
(Plain strain)
0
b
c,
out-of-plane
a
c
out-of-plane
a
b
out-of-plane
c
b
If
a
c
b,
If
c
a
b,
2R
Program Printout
Problem 7.140
x
60 micrometers
y
240 micrometers
xy
50 microradians
0.000
Angle between xy axes and principal axes (+ = counterclockwise):
7.76 and 82.24
p
a
243.41 micrometers
b
56.59 micrometers
max(in-plane)
0.00 micrometers
186.82 microradians
max
243.41 microradians
c
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1234
PROBLEM 7.C7 (Continued)
Program Printout (Continued)
Problem 7.141
y
400 micrometers
200 micrometers
xy
375 microradians
x
0.000
Angle between xy axes and principal axes (
counterclockwise):
30.96 and 59.04
p
max(in-plane)
512.50 micrometers
87.50 micrometers
0.00 micrometers
425.00 microradians
max
512.50 microradians
a
b
c
Problem 7.142
x
300 micrometers
y
60 micrometers
100 microradians
xy
0.000
Angle between xy axes and principal axes (+ = counterclockwise):
p
11.31 and
78.69
a
310.00 micrometers
b
50.00 micrometers
max(in-plane)
0.00 micrometers
260.00 microradians
max
310.00 microradians
c
Problem 7.143
x
180 micrometers
y
260 micrometers
xy
315 microradians
0.000
Angle between xy axes and principal axes (+ = counterclockwise):
p
37.87 and
52.13
a
57.50 micrometers
b
382.50 micrometers
c
0.00 micrometers
max(in-plane)
325.00 microradians
max
382.50 microradians
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1235
PROBLEM 7.C8
A rosette consisting of three gages forming, respectively, angles of 1 , 2 , and 3 with the x axis is attached
to the free surface of a machine component made of a material with a given Poisson’s ratio v. (a) Write a
computer program that, for given readings 1 , 2 , and 3 of the gages, can be used to calculate the strain
components associated with the x and y axes and to determine the orientation and magnitude of the three
principal strains, the maximum in-plane shearing strain, and the maximum shearing strain. (b) Use this
program to solve Probs 7.144, 7.145, 7.146, and 7.169.
SOLUTION
For n 1 to 3, enter
Enter: NU
n
and
n.
V
Solve Equation (7.60) for
x,
y,
and
xy
using method of determinates or any other method.
2
x
Enter
y
ave
2
a
max
ave
R
b
max
avg
R
V
c
p
Shearing strains:
x
; R
1 V
1
tan
2
(
y
2
xy
2
b)
a
xy
1
x
y
Maximum in-plane shearing strain
max (in plane)
2R
Calculate out-of-plane shearing strain, and check whether it is the maximum shearing strain.
If
c
b,
out-of-plane
a
c
If
c
a,
out-of-plane
c
b
Otherwise,
out-of-plane
2R
Problem Outputs
Problem 7.144
Gage
Theta Degrees
Epsilon Micro Meters
1
–15
480
2
30
–120
3
75
80
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1236
PROBLEM 7.C8 (Continued)
Program Outputs (Continued)
x
253.21 micrometers
y
306.79 micrometers
892.82 microradians
xy
727.21 micrometers
a
167.21 micrometers
b
894.43 microradians
max (in-plane)
Problem 7.145
Gage
Theta Degrees
Epsilon Micro Meters
1
2
3
30
–30
90
600
450
–75
725.00 micrometers
x
y
75.000 micrometers
xy
173.205 microradians
a
734.268 micrometers
b
84.268 micrometers
max (in-plane)
818.535 microradians
Problem 7.146
Observe that Gage 3 is orientated along the y axis. Therefore,
enter
4
and
4
as
3
and
3,
the value of
Gage
y
that is obtained is also the expected reading of Gage 3.
Theta Degrees
Epsilon in./in.
1
0
420
2
45
–45
4
135
165
x
y
xy
a
b
max (in-plane)
420.00 in./in.
300.00 in./in.
210.00 microradians
435.00 in./in.
315.00 in./in.
750.00 microradians
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1237
PROBLEM 7.C8 (Continued)
Program Outputs (Continued)
Problem 7.169
Gage
Theta Degrees
Epsilon in./in.
1
2
3
45
–45
0
–50
360
315
x
315.000 in./in.
y
5.000 in./in.
xy
a
b
max (in-plane)
410.000 microradians
415.048 in./in.
105.048 in./in.
520.096 microradians
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1238
CHAPTER 2
PROBLEM 2.1
A nylon thread is subjected to a 8.5-N tension force. Knowing that E 3.3 GPa and that the length of the
thread increases by 1.1%, determine (a) the diameter of the thread, (b) the stress in the thread.
SOLUTION
(a)
1.1
0.011
100
Stress:
E (3.3 109 )(0.011) 36.3 106 Pa
(b)
Strain:
Area:
A
Diameter:
d
L
P
A
P
4A
8.5
234.16 109 m 2
36.3 106
(4)(234.16 109 )
546 106 m
d 0.546 mm
36.3 MPa
Stress:
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
93
PROBLEM 2.2
A 4.8-ft-long steel wire of 14 -in.-diameter is subjected to a 750-lb tensile load. Knowing that E = 29 × 106 psi,
determine (a) the elongation of the wire, (b) the corresponding normal stress.
SOLUTION
(a) Deformation:
Area:
A
PL
;
AE
A
(0.25 in.)2
4
d2
4
4.9087 10 2 in 2
(750 lb)(4.8 ft 12 in./ft)
(4.9087 102 in 2 )(29 106 psi)
3.0347 10 2 in.
(b) Stress:
Area:
P
A
(750lb)
(4.9087 102 in 2 )
0.0303 in.
1.52790 10 4 psi
15.28 ksi
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94
PROBLEM 2.3
An 18-m-long steel wire of 5-mm diameter is to be used in the manufacture of a prestressed concrete beam.
It is observed that the wire stretches 45 mm when a tensile force P is applied. Knowing that E 200 GPa,
determine (a) the magnitude of the force P, (b) the corresponding normal stress in the wire.
SOLUTION
(a)
PL
, or
AE
P
AE
L
1
1
with A d 2 (0.005)2 19.6350 106 m 2
4
4
P
(0.045 m)(19.6350 106 m 2 )(200 109 N/m 2 )
9817.5 N
18 m
P 9.82 kN
(b)
P
A
9817.5 N
19.6350 10
6
6
m
2
500 MPa
500 10 Pa
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95
PROBLEM 2.4
Two gage marks are placed exactly 250 mm apart on a 12-mm-diameter aluminum rod with E = 73 GPa and
an ultimate strength of 140 MPa. Knowing that the distance between the gage marks is 250.28 mm after a load
is applied, determine (a) the stress in the rod, (b) the factor of safety.
SOLUTION
(a)
L L0
250.28 mm 250 mm
0.28 mm
L0
0.28 mm
250 mm
1.11643 10 4
E
(73 109 Pa)(1.11643 10 4 )
8.1760 107 Pa
81.8 MPa
(b)
F.S.
u
140 MPa
81.760 MPa
1.71233
F.S. 1.712
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96
PROBLEM 2.5
An aluminum pipe must not stretch more than 0.05 in. when it is subjected to a tensile load. Knowing that
E 10.1 106 psi and that the maximum allowable normal stress is 14 ksi, determine (a) the maximum
allowable length of the pipe, (b) the required area of the pipe if the tensile load is 127.5 kips.
SOLUTION
(a)
PL
AE
Thus,
L
EA
E
(10.1 106 ) (0.05)
P
14 103
L 36.1 in.
(b)
P
A
Thus,
A
P
127.5 103
14 103
A 9.11 in 2
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97
PROBLEM 2.6
A control rod made of yellow brass must not stretch more than 3 mm when the tension in the wire is 4 kN.
Knowing that E = 105 GPa and that the maximum allowable normal stress is 180 MPa, determine (a) the
smallest diameter rod that should be used, (b) the corresponding maximum length of the rod.
SOLUTION
(a)
P
;
A
A
d2
4
Substituting, we have
P
d2
4
d
d
4P
4(4 103 N)
(180 106 Pa)
d 5.3192 103 m
d 5.32 mm
(b)
E ;
L
Substituting, we have
E
L
L
L
E
(105 109 Pa) (3 103 m)
(180 106 Pa)
L 1.750 m
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98
PROBLEM 2.7
A steel control rod is 5.5 ft long and must not stretch more than 0.04 in. when a 2-kip tensile load is applied to it.
Knowing that E 29 106 psi, determine (a) the smallest diameter rod that should be used, (b) the
corresponding normal stress caused by the load.
SOLUTION
(a)
PL
AE
: 0.04 in.
(2000 lb) (5.5 12 in.)
6
A (29 10 psi)
A
1 2
d 0.113793 in 2
4
d 0.38063 in.
(b)
P
A
2000 lb
0.113793 in
2
d 0.381 in.
17.58 ksi
17575.8 psi
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99
PROBLEM 2.8
A cast-iron tube is used to support a compressive load. Knowing that E 10 106 psi and that the maximum
allowable change in length is 0.025%, determine (a) the maximum normal stress in the tube, (b) the minimum
wall thickness for a load of 1600 lb if the outside diameter of the tube is 2.0 in.
SOLUTION
(a)
L
100
0.00025
E ;
E
L
L
(10 106 psi)(0.00025)
2.50 103 psi
2.50 ksi
(b)
A
P
;
A
4
d
2
o
di2 d o2
di2
A
1600 lb
0.64 in 2
2.50 103 psi
4A
di2 (2.0 in.) 2
4(0.64 in 2 )
t
P
3.1851 in 2
di 1.78469 in.
1
1
(d o di ) (2.0 in. 1.78469 in.)
2
2
t 0.107655 in.
t 0.1077
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100
PROBLEM 2.9
A 4-m-long steel rod must not stretch more than 3 mm and the normal stress must not exceed 150 MPa when
the rod is subjected to a 10-kN axial load. Knowing that E 200 GPa, determine the required diameter of
the rod.
SOLUTION
L 4m
3 103 m,
150 106 Pa
E 200 109 Pa, P 10 103 N
Stress:
A
Deformation:
P
A
P
10 103 N
66.667 106 m 2 66.667 mm 2
6
150 10 Pa
PL
AE
A
PL
(10 103 )(4)
66.667 106 m 2 66.667 mm 2
E
(200 109 )(3 103 )
The larger value of A governs:
A 66.667 mm 2
A
4
d2
d
4A
4(66.667)
d 9.21 mm
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101
PROBLEM 2.10
A nylon thread is to be subjected to a 10-N tension. Knowing that E 3.2 GPa, that the maximum allowable
normal stress is 40 MPa, and that the length of the thread must not increase by more than 1%, determine the
required diameter of the thread.
SOLUTION
Stress criterion:
40 MPa 40 106 Pa P 10 N
A
P
P
10 N
: A
250 109 m 2
A
40 106 Pa
4
d 2: d 2
A
250 109
2
564.19 106 m
d 0.564 mm
Elongation criterion:
L
1% 0.01
PL
:
AE
A
P /E 10 N/3.2 109 Pa
312.5 109 m 2
/L
0.01
d 2
A
2
312.5 109
630.78 106 m 2
d 0.631 mm
d 0.631 mm
The required diameter is the larger value:
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102
PROBLEM 2.11
A block of 10-in. length and 1.8 × 1.6-in. cross section is to support a centric compressive load P. The
material to be used is a bronze for which E 14 × 106 psi. Determine the largest load that can be applied,
knowing that the normal stress must not exceed 18 ksi and that the decrease in length of the block should be at
most 0.12% of its original length.
SOLUTION
Considering allowable stress,
18 ksi
Cross-sectional area:
A (1.8 in.)(1.6 in.) 2.880 in 2
P
A
or
18 103 psi
P A
(18 103 psi)(2.880 in 2 )
5.1840 104 lb
or 51.840 kips
Considering allowable deformation,
PL
AE
L
0.12%
or
0.0012 in.
P AE
L
P (2.880 in 2 )(14 106 psi)(0.0012 in.)
4.8384 104 lb
or 48.384 kips
The smaller value for P resulting from the required deformation criteria governs.
48.4 kips
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103
PROBLEM 2.12
A square yellow-brass bar must not stretch more than 2.5 mm when it is subjected to a tensile load. Knowing
that E 105 GPa and that the allowable tensile strength is 180 MPa, determine (a) the maximum allowable
length of the bar, (b) the required dimensions of the cross section if the tensile load is 40 kN.
SOLUTION
180 106 Pa P 40 103 N
E 105 109 Pa 2.5 103 m
(a)
PL L
AE
E
E (105 109 )(2.5 103 )
1.45833 m
L
180 106
L 1.458 m
(b)
A
P
A
P
A a2
40 103
222.22 106 m 2 222.22 mm 2
180 106
a
A
a 14.91 mm
222.22
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104
P ⫽ 130 kips
PROBLEM 2.13
Rod BD is made of steel ( E 29 106 psi) and is used to brace the axially
compressed member ABC. The maximum force that can be developed in member
BD is 0.02P. If the stress must not exceed 18 ksi and the maximum change in
length of BD must not exceed 0.001 times the length of ABC, determine the
smallest-diameter rod that can be used for member BD.
A
72 in.
D
B
72 in.
C
54 in.
SOLUTION
FBD 0.02 P (0.02)(130) 2.6 kips 2.6 103 lb
Considering stress, 18 ksi 18 103 psi
FBD
A
A
FBD
2.6
0.14444 in 2
18
Considering deformation, (0.001)(144) 0.144 in.
FBD LBD
AE
A
FBD LBD
(2.6 103 )(54)
0.03362 in 2
6
E
(29 10 )(0.144)
Larger area governs. A 0.14444 in 2
A
4
d2
d
4A
(4)(0.14444)
d 0.429 in.
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105
PROBLEM 2.14
B
2.5 m
The 4-mm-diameter cable BC is made of a steel with E 200 GPa. Knowing
that the maximum stress in the cable must not exceed 190 MPa and that the
elongation of the cable must not exceed 6 mm, find the maximum load P that
can be applied as shown.
P
3.5 m
A
C
4.0 m
SOLUTION
LBC 62 42 7.2111 m
Use bar AB as a free body.
4
FBC 0
3.5P (6)
7.2111
P 0.9509 FBC
M A 0:
Considering allowable stress, 190 106 Pa
A
d2
4
FBC
A
4
(0.004) 2 12.566 106 m 2
FBC A (190 106 )(12.566 106 ) 2.388 103 N
Considering allowable elongation, 6 103 m
FBC LBC
AE
FBC
AE (12.566 106 )(200 109 )(6 103 )
2.091 103 N
7.2111
LBC
Smaller value governs. FBC 2.091 103 N
P 0.9509 FBC (0.9509)(2.091 103 ) 1.988 103 N
P 1.988 kN
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106
PROBLEM 2.15
1.25-in. diameter
d
A single axial load of magnitude P = 15 kips is applied at end C of the steel
rod ABC. Knowing that E = 30 × 106 psi, determine the diameter d of portion
BC for which the deflection of point C will be 0.05 in.
A
4 ft
B
3 ft
C
P
SOLUTION
C
PLi
PL
PL
Ai Ei AE AB AE BC
LAB 4 ft 48 in.;
AAB
d2
4
LBC 3 ft 36 in.
(1.25 in.)2
4
1.22718 in 2
Substituting, we have
15 103 lb 48 in.
36 in.
0.05 in.
6
2
ABC
30 10 psi 1.22718 in
ABC 0.59127 in 2
ABC
or
d
d
d2
4
4 ABC
4(0.59127 in 2 )
d 0.86766 in.
d 0.868 in.
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107
PROBLEM 2.16
36 mm
A 250-mm-long aluminum tube ( E 70 GPa) of 36-mm outer
diameter and 28-mm inner diameter can be closed at both ends by
means of single-threaded screw-on covers of 1.5-mm pitch. With
one cover screwed on tight, a solid brass rod ( E 105 GPa) of
25-mm diameter is placed inside the tube and the second cover is
screwed on. Since the rod is slightly longer than the tube, it is
observed that the cover must be forced against the rod by rotating
it one-quarter of a turn before it can be tightly closed. Determine
(a) the average normal stress in the tube and in the rod, (b) the
deformations of the tube and of the rod.
28 mm
25 mm
250 mm
SOLUTION
Atube
A rod
4
4
(d o2 di2 )
d2
4
4
(362 282 ) 402.12 mm 2 402.12 106 m 2
(25)2 490.87 mm 2 490.87 106 m 2
PL
P(0.250)
8.8815 109 P
9
Etube Atube (70 10 )(402.12 106 )
PL
P(0.250)
4.8505 109 P
Erod Arod (105 106 )(490.87 106 )
tube
rod
1
turn 1.5 mm 0.375 mm 375 106 m
4
*
tube * rod
or tube rod *
8.8815 109 P 4.8505 109 P 375 106
P
(a)
tube
P
27.308 103
67.9 106 Pa
6
Atube 402.12 10
rod
(b)
0.375 103
27.308 103 N
(8.8815 4.8505)(109 )
tube 67.9 MPa
P
27.308 103
55.6 106 Pa
6
Arod
490.87 10
tube (8.8815 109 )(27.308 103 ) 242.5 106 m
rod (4.8505 109 )(27.308 103 ) 132.5 106 m
rod 55.6 MPa
tube 0.243 mm
rod 0.1325 mm
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108
P 5 350 lb
B 0.4 in. C
A
1 in.
1.6 in.
D
PROBLEM 2.17
P 5 350 lb
1 in.
2 in.
1.6 in.
The specimen shown has been cut from a 14 -in.-thick sheet
of vinyl (E = 0.45 × 106 psi) and is subjected to a 350-lb
tensile load. Determine (a) the total deformation of the
specimen, (b) the deformation of its central portion BC.
SOLUTION
AB
PLAB
(350 lb)(1.6 in.)
4.9778 103 in.
EAAB (0.45 106 psi)(1 in.)(0.25 in.)
BC
PLBC
(350 lb)(2 in.)
15.5556 103 in.
EABC (0.45 106 psi)(0.4 in.)(0.25 in.)
CD AB 4.9778 103 in.
(a)
Total deformation:
AB BC CD
25.511 103 in.
25.5 103 in.
(b)
Deformation of portion BC :
BC 15.56 10 3 in.
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109
PROBLEM 2.18
P
D
1 mm
A
The brass tube AB ( E 105 GPa) has a cross-sectional area of
140 mm2 and is fitted with a plug at A. The tube is attached at B to a
rigid plate that is itself attached at C to the bottom of an aluminum
cylinder ( E 72 GPa) with a cross-sectional area of 250 mm2. The
cylinder is then hung from a support at D. In order to close the
cylinder, the plug must move down through 1 mm. Determine the
force P that must be applied to the cylinder.
375 mm
B
C
SOLUTION
Shortening of brass tube AB:
LAB 375 1 376 mm 0.376 m
AAB 140 mm 2 140 106 m 2
E AB 105 109 Pa
AB
PLAB
P(0.376)
25.578 109P
6
9
E AB AAB
(105 10 )(140 10 )
Lengthening of aluminum cylinder CD:
LCD 0.375 m
CD
ACD 250 mm 2 250 106 m 2
ECD 72 109 Pa
PLCD
P(0.375)
20.833 109 P
ECD ACD (72 109 )(250 106 )
A AB CD where A 0.001 m
Total deflection:
0.001 (25.578 109 20.833 109 ) P
P 21.547 103 N
P 21.5 kN
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110
PROBLEM 2.19
P
Both portions of the rod ABC are made of an aluminum for which E 70 GPa.
Knowing that the magnitude of P is 4 kN, determine (a) the value of Q so that
the deflection at A is zero, (b) the corresponding deflection of B.
A
20-mm diameter
0.4 m
B
Q
0.5 m
60-mm diameter
C
SOLUTION
(a)
AAB
ABC
4
4
2
d AB
2
d BC
4
4
(0.020) 2 314.16 106 m 2
(0.060)2 2.8274 103 m 2
Force in member AB is P tension.
Elongation:
AB
PLAB
(4 103 )(0.4)
72.756 106 m
EAAB (70 109 )(314.16 106 )
Force in member BC is Q P compression.
Shortening:
BC
(Q P) LBC
(Q P)(0.5)
2.5263 109(Q P)
9
3
EABC
(70 10 )(2.8274 10 )
For zero deflection at A, BC AB
2.5263 109(Q P) 72.756 106 Q P 28.8 103 N
Q 28.3 103 4 103 32.8 103 N
(b)
AB BC B 72.756 106 m
Q 32.8 kN
AB 0.0728 mm
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111
PROBLEM 2.20
P
The rod ABC is made of an aluminum for which E 70 GPa. Knowing that
P 6 kN and Q 42 kN, determine the deflection of (a) point A, (b) point B.
A
20-mm diameter
0.4 m
B
Q
0.5 m
60-mm diameter
C
SOLUTION
AAB
ABC
4
4
2
d AB
2
d BC
4
4
(0.020)2 314.16 106 m 2
(0.060)2 2.8274 103 m 2
PAB P 6 103 N
PBC P Q 6 103 42 103 36 103 N
LAB 0.4 m LBC 0.5 m
AB
PAB LAB
(6 103 )(0.4)
109.135 106 m
6
9
AAB E A (314.16 10 )(70 10 )
BC
PBC LBC
(36 103 )(0.5)
90.947 106 m
ABC E
(2.8274 103 )(70 109 )
(a)
A AB BC 109.135 106 90.947 106 m 18.19 106 m
(b)
B BC 90.9 106 m 0.0909 mm
or
A 0.01819 mm
B 0.0909 mm
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112
228 kN
PROBLEM 2.21
D
For the steel truss ( E 200 GPa) and loading shown, determine
the deformations of the members AB and AD, knowing that their
cross-sectional areas are 2400 mm2 and 1800 mm2, respectively.
B
A
4.0 m
2.5 m
C
4.0 m
SOLUTION
Statics: Reactions are 114 kN upward at A and C.
Member BD is a zero force member.
LAB 4.02 2.52 4.717 m
Use joint A as a free body.
Fy 0 : 114
2.5
FAB 0
4.717
FAB 215.10 kN
Fx 0 : FAD
FAD
4
FAB 0
4.717
(4)(215.10)
182.4 kN
4.717
Member AB:
AB
FAB LAB
(215.10 103 )(4.717)
EAAB
(200 109 )(2400 106 )
2.11 103 m
Member AD:
AD
AB 2.11 mm
FAD LAD
(182.4 103 )(4.0)
EAAD
(200 109 )(1800 106 )
2.03 103 m
AD 2.03 mm
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113
30 kips
A
PROBLEM 2.22
30 kips
B
For the steel truss ( E 29 106 psi) and loading shown, determine the
deformations of the members BD and DE, knowing that their crosssectional areas are 2 in2 and 3 in2, respectively.
8 ft
C
8 ft
30 kips
D
E
8 ft
F
G
15 ft
SOLUTION
Free body: Portion ABC of truss
M E 0 : FBD (15 ft) (30 kips)(8 ft) (30 kips)(16 ft) 0
FBD 48.0 kips
Free body: Portion ABEC of truss
Fx 0 : 30 kips 30 kips FDE 0
FDE 60.0 kips
BD
PL (48.0 103 lb)(8 12 in.)
AE
(2 in 2 )(29 106 psi)
DE
PL (60.0 103 lb)(15 12 in.)
AE
(3 in 2 )(29 106 psi)
BD 79.4 103 in.
DE 124.1 103 in.
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114
6 ft
PROBLEM 2.23
6 ft
C
B
5 ft
A
D
28 kips
Members AB and BC are made of steel ( E 29 106 psi) with crosssectional areas of 0.80 in2 and 0.64 in2, respectively. For the loading
shown, determine the elongation of (a) member AB, (b) member BC.
E
54 kips
SOLUTION
(a)
LAB
62 52 7.810 ft 93.72 in.
Use joint A as a free body.
5
FAB 28 0
7.810
43.74 kip 43.74 103 lb
Fy 0:
FAB
AB
(b)
FAB LAB
(43.74 103 )(93.72)
EAAB
(29 106 )(0.80)
AB 0.1767 in.
Use joint B as a free body.
Fx 0: FBC
FBC
BC
6
FAB 0
7.810
(6)(43.74)
33.60 kip 33.60 103 lb
7.810
FBC LBC
(33.60 103 )(72)
EABC
(29 106 )(0.64)
BC 0.1304 in.
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115
P
B
C
A
D
PROBLEM 2.24
The steel frame ( E 200 GPa) shown has a diagonal brace BD with
an area of 1920 mm2. Determine the largest allowable load P if the
change in length of member BD is not to exceed 1.6 mm.
6m
5m
SOLUTION
BD 1.6 103 m, ABD 1920 mm 2 1920 106 m 2
52 62 7.810 m, EBD 200 109 Pa
LBD
BD
FBD LBD
EBD ABD
FBD
(200 109 )(1920 106 )(1.6 103 )
EBD ABD BD
7.81
LBD
78.67 103 N
Use joint B as a free body.
Fx 0:
5
FBD P 0
7.810
P
5
(5)(78.67 103 )
FBD
7.810
7.810
50.4 103 N
P 50.4 kN
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116
PROBLEM 2.25
D
225 mm
C
A
B
150 mm
P
Link BD is made of brass ( E 105 GPa) and has a cross-sectional area of
240 mm2. Link CE is made of aluminum ( E 72 GPa) and has a crosssectional area of 300 mm2. Knowing that they support rigid member ABC,
determine the maximum force P that can be applied vertically at point A if
the deflection of A is not to exceed 0.35 mm.
E
225 mm
125 mm
SOLUTION
Free body member AC:
M C 0: 0.350 P 0.225FBD 0
FBD 1.55556 P
M B 0: 0.125P 0.225 FCE 0
FCE 0.55556 P
FBD LBD
(1.55556 P)(0.225)
13.8889 109 P
EBD ABD (105 109 )(240 106 )
F L
(0.55556 P)(0.150)
CE CE
3.8581 109 P
ECE ACE (72 109 )(300 106 )
B BD
C CE
Deformation Diagram:
From the deformation diagram,
Slope:
B C
LBC
A B LAB
17.7470 109 P
78.876 109 P
0.225
13.8889 109 P (0.125)(78.876 109 P)
23.748 109 P
Apply displacement limit. A 0.35 103 m 23.748 109P
P 14.7381 103 N
P 14.74 kN
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117
PROBLEM 2.26
F
C
180 mm
B
E
A
D
Members ABC and DEF are joined with steel links (E 200 GPa).
Each of the links is made of a pair of 25 × 35-mm plates. Determine
the change in length of (a) member BE, (b) member CF.
260 mm
18 kN
240 mm
18 kN
SOLUTION
Free body diagram of Member ABC:
M B 0:
(0.26 m)(18 kN) (0.18 m) FCF 0
FCF 26.0 kN
Fx 0:
18 kN FBE 26.0 kN 0
FBE 44.0 kN
Area for link made of two plates:
A 2(0.025 m)(0.035 m) 1.750 103 m 2
(a)
BE
FBE L
(44.0 103 N)(0.240 m)
EA
(200 109 Pa)(1.75 103 m 2 )
30.171 106 m
BE 0.0302 mm
(b)
CF
FBF L
(26.0 103 N)(0.240 m)
EA
(200 109 Pa)(1.75 103 m 2 )
17.8286 106 m
CF 0.01783 mm
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118
A
D
P = 1 kip
Each of the links AB and CD is made of aluminum ( E 10.9 106 psi)
and has a cross-sectional area of 0.2 in2. Knowing that they support the
rigid member BC, determine the deflection of point E.
18 in.
E
B
22 in.
PROBLEM 2.27
C
10 in.
SOLUTION
Free body BC:
M C 0: (32) FAB (22) (1 103 ) 0
FAB 687.5 lb
Fy 0: 687.5 1 103 FCD 0
FCD 312.5 lb
FAB LAB
(687.5)(18)
5.6766 103 in. B
EA
(10.9 106 )(0.2)
F L
(312.5)(18)
CD CD
2.5803 103 in. C
EA
(10.9 106 )(0.2)
AB
CD
Deformation diagram:
Slope
B C
LBC
3.0963 103
32
96.759 106 rad
E C LEC
2.5803 103 (22)(96.759 106 )
4.7090 103 in.
E 4.71 10 3 in.
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119
PROBLEM 2.28
D
3
The length of the 32
-in.-diameter steel wire CD has been adjusted so that
with no load applied, a gap of 161 in. exists between the end B of the rigid
beam ACB and a contact point E. Knowing that E 29 106 psi,
determine where a 50-lb block should be placed on the beam in order to
cause contact between B and E.
12.5 in.
x
C
50 lb
B
A
E
16 in.
1
16
in.
4 in.
SOLUTION
Rigid beam ACB rotates through angle to close gap.
1/16
3.125 103 rad
20
Point C moves downward.
C 4 4(3.125 103 ) 12.5 103 in.
CD C 12.5 103 in.
ACD
CD
2
d
3
d
F L
CD CD
EACD
FCD
2
6.9029 103 in 2
4 32
EACD CD (29 106 )(6.9029 103 )(12.5 103 )
LCD
12.5
200.18 lb
Free body ACB:
M A 0: 4 FCD (50)(20 x) 0
(4)(200.18)
16.0144
50
x 3.9856 in.
20 x
x 3.99 in.
For contact,
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120
PROBLEM 2.29
A homogeneous cable of length L and uniform cross section is suspended from one end. (a) Denoting by the
density (mass per unit volume) of the cable and by E its modulus of elasticity, determine the elongation of the
cable due to its own weight. (b) Show that the same elongation would be obtained if the cable were horizontal
and if a force equal to half of its weight were applied at each end.
SOLUTION
(a)
For element at point identified by coordinate y,
P weight of portion below the point
gA(L y )
d
Pdy gA( L y )dy g ( L y )
dy
EA
EA
E
(b)
Total weight:
L
g ( L y)
E
0
g
L
1
dy
Ly y 2
2 0
E
g
L2
2
L
2
E
1 gL2
2 E
W gAL
F
EA EA 1 gL2 1
gAL
2
L
L 2 E
1
F W
2
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121
PROBLEM 2.30
P
A a
A vertical load P is applied at the center A of the upper section of a homogeneous
frustum of a circular cone of height h, minimum radius a, and maximum radius b.
Denoting by E the modulus of elasticity of the material and neglecting the effect
of its weight, determine the deflection of point A.
h
b
SOLUTION
Extend the slant sides of the cone to meet at a point O and place the origin of the coordinate system there.
tan
From geometry,
a1
ba
h
a
b
, b1
,
tan
tan
r y tan
At coordinate point y, A r 2
Deformation of element of height dy:
d
Pdy
AE
d
P dy
P
dy
E r 2 E tan 2 y 2
Total deformation:
P
A
E tan 2
b1
a1
1
dy
P
2
2
E tan y
y
b1 a1 P(b1 a1 )
P
2
Eab
E tan a1b1
b1
a1
P
E tan 2
1 1
a1 b1
A
Ph
Eab
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122
PROBLEM 2.31
Denoting by the “engineering strain” in a tensile specimen, show that the true strain is t ln (1 ).
SOLUTION
t ln
L
L
ln 0
ln 1 ln (1 )
L0
L0
L0
t ln (1 )
Thus,
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123
PROBLEM 2.32
The volume of a tensile specimen is essentially constant while plastic deformation occurs. If the initial
diameter of the specimen is d1, show that when the diameter is d, the true strain is t 2 ln(d1 /d ).
SOLUTION
If the volume is constant,
4
d 2L
4
d12 L0
L d12 d1
L0 d 2 d
t ln
2
L
d
ln 1
L0
d
2
t 2ln
d1
d
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124
Brass core
(E = 105 GPa)
Aluminum plates
(E = 70 GPa)
P
PROBLEM 2.33
Rigid
end plate
An axial centric force of magnitude P 450 kN is applied
to the composite block shown by means of a rigid end
plate. Knowing that h 10 mm, determine the normal
stress in (a) the brass core, (b) the aluminum plates.
300 mm
60 mm
h
40 mm
h
SOLUTION
A B ;
PA L
E A AA
P PA PB
and
PB L
EB AB
Therefore,
PA ( E A AA ) ;
L
Substituting,
PA E A AA E B AB
L
L
P
E A AA EB AB
(450 103 N)
(70 109 Pa)(2)(0.06 m)(0.01 m) (105 109 Pa)(0.06 m)(0.04 m)
1.33929 103
E
Now,
(a)
PB ( EB AB )
L
Brass-core:
B (105 109 Pa)(1.33929 103 )
1.40625 108 Pa
B 140.6 MPa
(b)
Aluminum:
A (70 109 Pa)(1.33929 103 )
9.3750 107 Pa
A 93.8 MPa
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125
Brass core
(E = 105 GPa)
Aluminum plates
(E = 70 GPa)
P
Rigid
end plate
PROBLEM 2.34
For the composite block shown in Prob. 2.33, determine
(a) the value of h if the portion of the load carried by the
aluminum plates is half the portion of the load carried by
the brass core, (b) the total load if the stress in the brass is
80 MPa.
PROBLEM 2.33. An axial centric force of magnitude
P 450 kN is applied to the composite block shown by
means of a rigid end plate. Knowing that h 10 mm,
determine the normal stress in (a) the brass core, (b) the
aluminum plates.
300 mm
60 mm
h
40 mm
h
SOLUTION
a b;
Pa L
Ea Aa
P Pa Pb
and
Pb L
Eb Ab
Therefore,
Pa ( Ea Aa ) ;
L
Pa
(a)
Pb ( Eb Ab )
L
1
Pb
2
1
( Ea Aa ) ( Eb Ab )
L 2
L
Aa
1 Eb
Ab
2 E a
Aa
1 105 GPa
(40 mm)(60 mm)
2 70 GPa
Aa 1800 mm 2
1800 mm 2 2(60 mm)(h)
h 15.00 mm
(b)
b
Pb
1
Pb b Ab and Pa Pb
2
Ab
P Pa Pb
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126
PROBLEM 2.34 (Continued)
P
1
( b Ab ) b Ab
2
P ( b Ab )1.5
P (80 106 Pa)(0.04 m)(0.06 m)(1.5)
P 2.880 105 N
P 288 kN
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127
PROBLEM 2.35
P
The 4.5-ft concrete post is reinforced with six steel bars, each with a 1 18 -in. diameter.
Knowing that Es 29 × 106 psi and Ec = 4.2 × 106 psi, determine the normal stresses
in the steel and in the concrete when a 350-kip axial centric force P is applied to the
post.
18 in.
4.5 ft
SOLUTION
Let Pc portion of axial force carried by concrete.
Ps portion carried by the six steel rods.
Pc L
Ec Ac
Pc
Ec Ac
L
Ps L
Es As
Ps
Es As
L
P Pc Ps ( Ec Ac Es As )
L
As 6
Ac
4
L
P
Ec Ac Es As
d s2
6
(1.125 in.) 2 5.9641 in 2
4
d c2 As
4
248.51 in 2
4
(18 in.)2 5.9641 in 2
L 4.5 ft 54 in.
350 103 lb
2.8767 104
6
2
6
2
(4.2 10 psi)(248.51 in ) (29 10 psi)(5.9641 in )
s Es (29 106 psi)( 2.8767 10 4 ) 8.3424 10 psi
s 8.34 ksi
c Ec (4.2 106 psi)(2.8767 104 ) 1.20821 103 psi
c 1.208 ksi
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128
PROBLEM 2.36
P
18 in.
For the post of Prob. 2.35, determine the maximum centric force that can be applied if the
allowable normal stress is 20 ksi in the steel and 2.4 ksi in the concrete.
PROBLEM 2.35 The 4.5-ft concrete post is reinforced with six steel bars, each with
a 1 18 -in. diameter. Knowing that Es 29 × 106 psi and Ec 4.2 × 106 psi, determine
the normal stresses in the steel and in the concrete when a 350-kip axial centric force P
is applied to the post.
4.5 ft
SOLUTION
Allowable strain in each material:
Steel:
s
Concrete:
c
Smaller value governs.
s
Es
c
Ec
L
20 103 psi
6.8966 104
6
29 10 psi
2.4 103 psi
5.7143 104
4.2 106 psi
5.7143 104
Let Pc = Portion of load carried by concrete.
Ps = Portion of load carried by 6 steel rods.
Pc L
E c Ac
Pc Ec Ac Ec Ac
L
Ps L
E s As
Ps Es As Es As
L
6
As 6 d s2
(1.125 in.) 2 5.9641 in 2
4
4
Ac d c2 As (18 in.) 2 5.9641 in 2 2.4851 102 in 2
4
4
P Pc Ps Ec Ac Es As
P [(4.2 106 psi)(2.4851 102 in 2 ) (29 106 psi)(5.9641 in 2 )](5.7143 104 )
P 6.9526 105 lb
P 695 kips
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129
PROBLEM 2.37
25 mm
Brass core
E 105 GPa
An axial force of 200 kN is applied to the assembly shown by means of
rigid end plates. Determine (a) the normal stress in the aluminum shell,
(b) the corresponding deformation of the assembly.
300 mm
Aluminium shell
E 70 GPa
60 mm
SOLUTION
Let Pa = Portion of axial force carried by shell.
Pb = Portion of axial force carried by core.
Pa L
, or
Ea Aa
Pa
Ea Aa
L
Pb L
, or
Eb Ab
Pb
Eb Ab
L
P Pa Pb ( Ea Aa Eb Ab )
Thus,
Aa
with
Ab
4
4
L
[(0.060) 2 (0.025)2 ] 2.3366 103 m 2
(0.025)2 0.49087 103 m 2
P [(70 109 )(2.3366 103 ) (105 109 )(0.49087 103 )]
P 215.10 10
Strain:
L
6
L
L
P
200 103
0.92980 103
6
6
215.10 10
215.10 10
(a)
a Ea (70 109 ) (0.92980 103 ) 65.1 106 Pa
(b)
L (0.92980 103 ) (300 mm)
a 65.1 MPa
0.279 mm
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130
25 mm
Brass core
E 105 GPa
PROBLEM 2.38
The length of the assembly shown decreases by 0.40 mm when an axial
force is applied by means of rigid end plates. Determine (a) the magnitude
of the applied force, (b) the corresponding stress in the brass core.
300 mm
Aluminium shell
E 70 GPa
60 mm
SOLUTION
Let Pa = Portion of axial force carried by shell and Pb = Portion of axial force carried by core.
Thus,
with
Pa L
, or
Ea Aa
Pa
Ea Aa
L
Pb L
, or
Eb Ab
Pb
Eb Ab
L
P Pa Pb ( Ea Aa Eb Ab )
Aa
Ab
4
4
L
[(0.060) 2 (0.025)2 ] 2.3366 103 m 2
(0.025)2 0.49087 103 m 2
P [(70 109 )(2.3366 103 ) (105 109 )(0.49087 103 )]
with
L
215.10 106
L
0.40 mm, L 300 mm
(a)
P (215.10 106 )
(b)
b
0.40
286.8 103 N
300
P 287 kN
Pb
E
(105 109 )(0.40 103 )
b
140 106 Pa
Ab
L
300 103
b 140.0 MPa
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131
PROBLEM 2.39
A
25 in.
1.25 in.
6 kips
6 kips
B
A polystyrene rod consisting of two cylindrical portions AB and BC is restrained at
both ends and supports two 6-kip loads as shown. Knowing that E 0.45 106 psi,
determine (a) the reactions at A and C, (b) the normal stress in each portion of
the rod.
2 in.
15 in.
C
SOLUTION
(a)
We express that the elongation of the rod is zero.
But
PAB LAB
d2 E
4 AB
PAB RA
PBC LBC
4
2
d BC
E
0
PBC RC
Substituting and simplifying,
RA LAB RC LBC
2 0
2
d AB
d BC
L
RC AB
LBC
2
2
d BC
25 2
RA
RA
15 1.25
d AB
RC 4.2667 RA
From the free body diagram,
Substituting (1) into (2),
RA RC 12 kips
(2)
5.2667 RA 12
RA 2.2785 kips
From (1),
(1)
RA 2.28 kips
RC 4.2667 (2.2785) 9.7217 kips
RC 9.72 kips
(b)
AB
PAB RA
2.2785
AAB AAB
(1.25) 2
4
AB 1.857 ksi
BC
PBC RC 9.7217
(2) 2
ABC
ABC
4
BC 3.09 ksi
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132
PROBLEM 2.40
C
A
P
Three steel rods (E = 29 × 106 psi) support an 8.5-kip load P. Each of the
rods AB and CD has a 0.32-in2 cross-sectional area and rod EF has a 1-in2
cross-sectional area. Neglecting the deformation of bar BED, determine
(a) the change in length of rod EF, (b) the stress in each rod.
20 in.
B
D
E
16 in.
F
SOLUTION
Use member BED as a free body.
By symmetry, or by M E 0 :
PCD PAB
Fy 0: PAB PCD PEF P 0
P 2 PAB PEF
AB
PAB LAB
EAAB
CD
PCD LCD
EACD
EF
PEF LEF
EAEF
LAB LCD and AAB ACD , AB CD
Since
Since points A, C, and F are fixed, B AB , D CD , E EF
Since member BED is rigid, E B C
PAB LAB PEF LEF
EAAB
EAEF
PAB
AAB LEF
0.32 16
PEF
PEF 0.256 PEF
1 20
AEF LAB
P 2 PAB PEF 2(0.256 PEF ) PEF 1.512 PEF
P
8.5
5.6217 kips
1.512 1.512
PCD 0.256(5.6217) 1.43916 kips
PEF
PAB
(a)
PEF LEF (5.6217)(16)
0.0031016 in.
EAEF
(29 103 )(1)
P
1.43916
CD AB
4.4974 ksi
0.32
AAB
EF
(b) AB
EF
PEF 5.6217
5.6217 ksi
1
AEF
EF 0.00310 in.
AB CD 4.50 ksi
EF 5.62 ksi
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133
PROBLEM 2.41
Dimensions in mm
180
100
120
A
C
Steel B
D
Brass
60 kN
40-mm diam.
100
E
40 kN
Two cylindrical rods, one of steel and the other of brass, are joined at
C and restrained by rigid supports at A and E. For the loading shown
and knowing that Es 200 GPa and Eb 105 GPa, determine
(a) the reactions at A and E, (b) the deflection of point C.
30-mm diam.
SOLUTION
A to C:
E 200 109 Pa
A
(40) 2 1.25664 103 mm 2 1.25664 103 m 2
4
EA 251.327 106 N
C to E:
E 105 109 Pa
A
(30)2 706.86 mm 2 706.86 106 m 2
4
EA 74.220 106 N
A to B:
P RA
L 180 mm 0.180 m
RA (0.180)
PL
EA 251.327 106
716.20 1012 RA
AB
B to C:
P RA 60 103
L 120 mm 0.120 m
BC
PL ( RA 60 103 )(0.120)
EA
251.327 106
447.47 1012 RA 26.848 106
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134
PROBLEM 2.41 (Continued)
C to D:
P RA 60 103
L 100 mm 0.100 m
PL ( RA 60 103 )(0.100)
EA
74.220 106
1.34735 109 RA 80.841 106
BC
D to E:
P RA 100 103
L 100 mm 0.100 m
DE
PL ( RA 100 103 )(0.100)
EA
74.220 106
1.34735 109 RA 134.735 106
A to E:
AE AB BC CD DE
3.85837 109 RA 242.424 106
Since point E cannot move relative to A,
(a)
(b)
AE 0
3.85837 109 RA 242.424 106 0 RA 62.831 103 N
RA 62.8 kN
RE RA 100 103 62.8 103 100 103 37.2 103 N
RE 37.2 kN
C AB BC 1.16367 109 RA 26.848 106
(1.16369 109 )(62.831 103 ) 26.848 106
46.3 106 m
C 46.3 m
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135
PROBLEM 2.42
Dimensions in mm
180
100
120
A
C
Steel B
D
Brass
60 kN
40-mm diam.
100
E
40 kN
30-mm diam.
Solve Prob. 2.41, assuming that rod AC is made of brass and
rod CE is made of steel.
PROBLEM 2.41 Two cylindrical rods, one of steel and the
other of brass, are joined at C and restrained by rigid supports at A
and E. For the loading shown and knowing that Es 200 GPa
and Eb 105 GPa, determine (a) the reactions at A and E, (b) the
deflection of point C.
SOLUTION
A to C:
E 105 109 Pa
(40) 2 1.25664 103 mm 2 1.25664 103 m 2
4
EA 131.947 106 N
A
C to E:
E 200 109 Pa
(30)2 706.86 mm 2 706.86 106 m 2
4
EA 141.372 106 N
A
A to B:
P RA
L 180 mm 0.180 m
AB
RA (0.180)
PL
EA 131.947 106
1.36418 109 RA
B to C:
P RA 60 103
L 120 mm 0.120 m
BC
PL ( RA 60 103 )(0.120)
EA
131.947 106
909.456 1012 RA 54.567 106
C to D:
P RA 60 103
L 100 mm 0.100 m
CD
PL ( RA 60 103 )(0.100)
EA
141.372 106
707.354 1012 RA 42.441 106
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136
PROBLEM 2.42 (Continued)
D to E:
P RA 100 103
L 100 mm 0.100 m
DE
PL ( RA 100 103 )(0.100)
EA
141.372 106
707.354 1012 RA 70.735 106
A to E:
AE AB BC CD DE
3.68834 109 RA 167.743 106
Since point E cannot move relative to A,
(a)
AE 0
3.68834 109 RA 167.743 106 0 RA 45.479 103 N
RE RA 100 103 45.479 103 100 103 54.521 103
(b)
R A 45.5 kN
RE 54.5 kN
C AB BC 2.27364 109 RA 54.567 106
(2.27364 109 )(45.479 103 ) 54.567 106
48.8 106 m
C 48.8 m
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137
D
PROBLEM 2.43
E
2 kN
225 mm
B
A
F
C
Each of the rods BD and CE is made of brass (E 105 GPa)
and has a cross-sectional area of 200 mm2. Determine the
deflection of end A of the rigid member ABC caused by the
2-kN load.
550 mm
75 mm
100 mm
SOLUTION
Let be the rotation of member ABC as shown.
Then A 0.6251
But
B
PBD
B 0.0751
C 0.1
PBD LBD
AE
EA B (105 109 )(200 10 6 )(0.075 )
0.225
LBD
7 106
Free body ABC:
C
PCE LCE
AE
PCE
EA C (105 109 )(200 10 6 )(0.1 )
0.225
LCE
9.3333 106
From free body of member ABC:
M F 0 : (0.625)(2000) 0.075 PBD 0.1PCE 0
or
(0.625)(2000) 0.075(7 106 ) 0.1(9.3333 106 ) 0
0.85714 103 rad
and
A 0.625 0.625(0.85714 103 ) 0.53571 103 m
A 0.536 mm
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138
PROBLEM 2.44
F
8 in.
E
10 in.
A
B
C
P
D
12 in.
12 in.
The rigid bar AD is supported by two steel wires of 161 -in. diameter
(E 29 × 106 psi) and a pin and bracket at A. Knowing that the
wires were initially taut, determine (a) the additional tension in each
wire when a 220-lb load P is applied at D, (b) the corresponding
deflection of point D.
12 in.
SOLUTION
Let be the notation of bar ABCD.
Then B 12
C 24
B
PBE LBE
AE
EA BE
LBE
PBE
(29 106 )
1
2
(12 )
4 6
10
106.765 103
C
PCF
PCF LCF
EA
EA CE
LCF
(29 106 )
1
2
(24 )
4 16
18
118.628 103
Using free body ABCD,
MA 0 :
12 PBE 24 PCF 36 P 0
(12)(106.765 103 ) (24)(118.628 106 ) (36)(220) 0
4.1283 106 (36)(220)
1.91847 103 rad
(a)
(b)
PBE (106.765 103 )(1.91847 103 ) 204.83 lb
PBE 205 lb
PCF (118.628 103 )(1.91847 10 3 ) 227.58 lb
PCF 228 lb
D 36 (36)(1.91847 103 ) 69.1 103 in.
0.0691 in.
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139
L
B
D
A
3
4
L
PROBLEM 2.45
L
C
The rigid bar ABC is suspended from three wires of the same material. The crosssectional area of the wire at B is equal to half of the cross-sectional area of the
wires at A and C. Determine the tension in each wire caused by the load P shown.
P
SOLUTION
M A 0:
PC
3
LP 0
4
3
1
P PB
8
2
M C 0:
PA
2 LPC LPB
2 LPA LPB
5
LP 0
4
5
1
P PB
8
2
Let l be the length of the wires.
A
PAl
l 5
1
P PB
2
EA EA 8
B
PB l
2l
PB
E ( A/2) EA
C
PC l
1
l 3
P PB
2
EA EA 8
From the deformation diagram,
A B B C
or
1
2
B ( A c )
l
1 l 5
1
3
1
PB
P PB P PB
2 EA 8
2
8
2
E ( A / 2)
5
1
PB P;
2
2
PB
1
P
5
PB 0.200 P
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140
PROBLEM 2.45 (Continued)
PA
5
1 P 21
P
P
8
2 5 40
PA 0.525 P
PC
3
1 P 11
P
P
8
2 5 40
PC 0.275P
Check:
PA PB PC 1.000 P
Ok
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141
PROBLEM 2.46
E
The rigid bar AD is supported by two steel wires of 161 -in. diameter
( E 29 106 psi) and a pin and bracket at D. Knowing that the wires
were initially taut, determine (a) the additional tension in each wire when
a 120-lb load P is applied at B, (b) the corresponding deflection of point B.
F
15 in.
8 in.
A
B
C
8 in.
8 in.
D
8 in.
P
SOLUTION
Let be the rotation of bar ABCD.
Then A 24
C 8
A
PAE
PAE LAE
AE
6
2
EA A (29 10 ) 4 ( 161 ) (24 )
LAE
15
142.353 103
C
PCF LCF
AE
6 1
EA C (29 10 ) 4 16 (8 )
8
LCF
2
PCF
88.971 103
Using free body ABCD,
M D 0 :
24PAE 16 P 8PCF 0
24(142.353 103 ) 16(120) 8(88.971 103 ) 0
0.46510 103 rad哷
(a)
(b)
PAE (142.353 103 )(0.46510 103 )
PAE 66.2 lb
PCF (88.971 103 )(0.46510 103 )
PCF 41.4 lb
B 7.44 103 in.
B 16 16(0.46510 103 )
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142
PROBLEM 2.47
25 mm
The aluminum shell is fully bonded to the brass core and the
assembly is unstressed at a temperature of 15C. Considering only
axial deformations, determine the stress in the aluminum when the
temperature reaches 195C.
Brass core
E 105 GPa
20.9 10–6/C
Aluminum shell
E 70 GPa
23.6 10–6/C
60 mm
SOLUTION
Brass core:
E 105 GPa
20.9 106/ C
Aluminum shell:
E 70 GPa
23.6 106 / C
Let L be the length of the assembly.
Free thermal expansion:
T 195 15 180 C
Brass core:
Aluminum shell:
(T )b L b (T )
(T ) a L a (T )
Net expansion of shell with respect to the core:
L( a b )(T )
Let P be the tensile force in the core and the compressive force in the shell.
Brass core:
Eb 105 109 Pa
(25)2 490.87 mm 2
4
490.87 106 m 2
Ab
( P )b
PL
Eb Ab
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143
PROBLEM 2.47 (Continued)
Aluminum shell:
( p )a
PL
Ea Aa
Ea 70 109 Pa
(602 252 )
4
2.3366 103 mm 2
Aa
2.3366 103 m 2
( P )b ( P ) a
L( b a )(T )
PL
PL
KPL
Eb Ab Ea Aa
where
K
1
1
Eb Ab Ea Aa
1
1
6
9
(105 10 )(490.87 10 ) (70 10 )(2.3366 103 )
9
25.516 109 N 1
Then
( b a )(T )
K
(23.6 106 20.9 106 )(180)
25.516 109
19.047 103 N
P
Stress in aluminum:
a
P
19.047 103
8.15 106 Pa
3
Aa
2.3366 10
a 8.15 MPa
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144
PROBLEM 2.48
25 mm
Solve Prob. 2.47, assuming that the core is made of steel (Es 200 GPa,
s 11.7 106 / C) instead of brass.
Brass core
E 105 GPa
20.9 10–6/C
PROBLEM 2.47 The aluminum shell is fully bonded to the brass core
and the assembly is unstressed at a temperature of 15C. Considering
only axial deformations, determine the stress in the aluminum when the
temperature reaches 195C.
Aluminum shell
E 70 GPa
23.6 10–6/C
60 mm
SOLUTION
E 70 GPa 23.6 106 / C
Aluminum shell:
Let L be the length of the assembly.
T 195 15 180 C
Free thermal expansion:
Steel core:
(T ) s L s (T )
Aluminum shell:
(T )a L a (T )
L( a s )(T )
Net expansion of shell with respect to the core:
Let P be the tensile force in the core and the compressive force in the shell.
Es 200 109 Pa, As
Steel core:
4
(25) 2 490.87 mm 2 490.87 106 m 2
PL
( P ) s
Es As
Ea 70 109 Pa
Aluminum shell:
( P )a
PL
Ea Aa
(602 25)2 2.3366 103 mm 2 2.3366 103 m 2
4
( P ) s ( P ) a
Aa
L( a s )(T )
PL
PL
KPL
Es As Ea Aa
where
K
1
1
Es As Ea Aa
1
1
6
9
(200 10 )(490.87 10 ) (70 10 )(2.3366 103 )
9
16.2999 109 N 1
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145
PROBLEM 2.48 (Continued)
Then
P
( a s )( T ) (23.6 10 6 11.7 10 6 )(180)
131.412 103 N
K
16.2999 10 9
Stress in aluminum: a
P
131.412 103
56.241 106 Pa
3
Aa
2.3366 10
a 56.2 MPa
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146
1
4
1
4
in.
1 in.
in.
1
4
in.
PROBLEM 2.49
1 in.
1
4
in.
Steel core
E 29 106 psi
Brass shell
E 15 106 psi
The brass shell ( b 11.6 106 /F) is fully bonded to the steel core
( s 6.5 106 /F). Determine the largest allowable increase in
temperature if the stress in the steel core is not to exceed 8 ksi.
12 in.
SOLUTION
Let Ps axial force developed in the steel core.
For equilibrium with zero total force, the compressive force in the brass shell is Ps .
s
Strains:
Ps
s (T )
Es As
b
Ps
b (T )
Eb Ab
s b
Matching:
Ps
P
s (T ) s b (T )
Es As
Eb Ab
1
1
Ps ( b s )(T )
Es As Eb Ab
(1)
Ab (1.5)(1.5) (1.0)(1.0) 1.25 in 2
As (1.0)(1.0) 1.0 in 2
b s 5.1 106 /F
Ps s As (8 103 )(1.0) 8 103 lb
1
1
1
1
87.816 109 lb 1
6
Es As Eb Ab (29 10 )(1.0) (15 106 )(1.25)
From (1),
(87.816 109 )(8 103 ) (5.1 106 )(T )
T 137.8 F
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147
PROBLEM 2.50
The concrete post ( Ec 3.6 106 psi and c 5.5 106 / F) is reinforced with six steel
bars, each of 78 -in. diameter ( Es 29 106 psi and s 6.5 106 / F). Determine the
normal stresses induced in the steel and in the concrete by a temperature rise of 65°F.
6 ft
10 in.
10 in.
SOLUTION
As 6
4
2
d 6
7
2
3.6079 in 2
48
Ac 102 As 102 3.6079 96.392 in 2
Let Pc tensile force developed in the concrete.
For equilibrium with zero total force, the compressive force in the six steel rods equals Pc .
Strains:
s
Pc
s (T )
Es As
c
Pc
c (T )
Ec Ac
Pc
P
c (T ) c s (T )
Ec Ac
Es As
Matching: c s
1
1
Pc ( s c )(T )
Ec Ac Es As
1
1
6
Pc (1.0 10 )(65)
6
6
(3.6 10 )(96.392) (29 10 )(3.6079)
Pc 5.2254 103 lb
c
Pc 5.2254 103
54.210 psi
96.392
Ac
s
c 54.2 psi
Pc
5.2254 103
1448.32 psi
3.6079
As
s 1.448 ksi
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148
PROBLEM 2.51
A
A rod consisting of two cylindrical portions AB and BC is restrained at both
ends. Portion AB is made of steel ( Es 200 GPa, s 11.7 106 / C) and
portion BC is made of brass ( Eb 105 GPa, b 20.9 106 / C). Knowing
that the rod is initially unstressed, determine the compressive force induced in
ABC when there is a temperature rise of 50 C.
30-mm diameter
250 mm
B
50-mm diameter
300 mm
C
SOLUTION
AAB
ABC
4
4
2
d AB
2
d BC
4
4
(30) 2 706.86 mm 2 706.86 106 m 2
(50)2 1.9635 103 mm 2 1.9635 103 m 2
Free thermal expansion:
T LAB s (T ) LBC b (T )
(0.250)(11.7 106 )(50) (0.300)(20.9 10 6 )(50)
459.75 106 m
Shortening due to induced compressive force P:
P
PL
PL
Es AAB Eb ABC
0.250 P
0.300 P
6
9
(200 10 )(706.86 10 ) (105 10 )(1.9635 103 )
9
3.2235 109 P
For zero net deflection, P T
3.2235 109 P 459.75 106
P 142.624 103 N
P 142.6 kN
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149
24 in.
A
PROBLEM 2.52
32 in.
B
C
1
2 14 -in. diameter 1 2 -in. diameter
A rod consisting of two cylindrical portions AB and BC is restrained at both
ends. Portion AB is made of steel ( Es 29 106 psi, s 6.5 106 / F) and
portion BC is made of aluminum ( Ea 10.4 106 psi, a 13.3 106 /°F).
Knowing that the rod is initially unstressed, determine (a) the normal stresses
induced in portions AB and BC by a temperature rise of 70°F, (b) the
corresponding deflection of point B.
SOLUTION
AAB
Free thermal expansion.
4
(2.25) 2 3.9761 in 2 ABC
4
(1.5)2 1.76715 in 2
T 70F
(T ) AB LAB s (T ) (24)(6.5 106 )(70) 10.92 103 in.
(T ) BC LBC a (T ) (32)(13.3 106 )(70) 29.792 103 in.
T (T ) AB (T ) BC 40.712 103 in.
Total:
Shortening due to induced compressive force P.
PLAB
24 P
208.14 109 P
Es AAB (29 106 )(3.9761)
PLBC
32 P
1741.18 109 P
6
Ea ABC (10.4 10 )(1.76715)
( P ) AB
( P ) BC
P ( P ) AB ( P ) BC 1949.32 109 P
Total:
For zero net deflection, P T
(a)
(b)
1949.32 109 P 40.712 103
AB
P
20.885 103
5.25 103 psi
AAB
3.9761
BC
P
20.885 103
11.82 103 psi
ABC
1.76715
P 20.885 103 lb
AB 5.25 ksi
BC 11.82 ksi
( P ) AB (208.14 109 )(20.885 103 ) 4.3470 103 in.
B (T ) AB ( P ) AB 10.92 103 4.3470 103
B 6.57 103 in.
or
( P ) BC (1741.18 109 )(20.885 103 ) 36.365 103 in.
B (T ) BC ( P ) BC 29.792 103 36.365 103 6.57 103 in.
(checks)
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150
PROBLEM 2.53
24 in.
A
Solve Prob. 2.52, assuming that portion AB of the composite rod is made of
aluminum and portion BC is made of steel.
32 in.
B
C
1
2 14 -in. diameter 1 2 -in. diameter
PROBLEM 2.52 A rod consisting of two cylindrical portions AB and BC is
restrained at both ends. Portion AB is made of steel ( Es 29 106 psi,
s 6.5 106 /F) and portion BC is made of aluminum ( Ea 10.4 106 psi,
a 13.3 106 /°F). Knowing that the rod is initially unstressed, determine
(a) the normal stresses induced in portions AB and BC by a temperature rise of
70°F, (b) the corresponding deflection of point B.
SOLUTION
AAB
4
(2.25) 2 3.9761 in 2
Free thermal expansion.
ABC
4
(1.5)2 1.76715 in 2
T 70F
(T ) AB LAB a (T ) (24)(13.3 106 )(70) 22.344 103 in.
( T ) BC LBC s (T ) (32)(6.5 106 )(70) 14.56 103 in.
T (T ) AB (T ) BC 36.904 103 in.
Total:
Shortening due to induced compressive force P.
PLAB
24 P
580.39 109 P
Ea AAB (10.4 106 )(3.9761)
PLBC
32 P
624.42 109 P
Es ABC (29 106 )(1.76715)
( P ) AB
( P ) BC
P ( P ) AB ( P ) BC 1204.81 109 P
Total:
For zero net deflection, P T
(a)
1204.81 109 P 36.904 103
P 30.631 103 lb
AB
P
30.631 103
7.70 103 psi
AAB
3.9761
AB 7.70 ksi
BC
P
30.631 103
17.33 103 psi
ABC
1.76715
BC 17.33 ksi
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151
PROBLEM 2.53 (Continued)
(b)
( P ) AB (580.39 109 )(30.631 103 ) 17.7779 103 in.
B (T ) AB ( P ) AB 22.344 103 17.7779 103
or
B 4.57 103 in.
( P ) BC (624.42 109 )(30.631 103 ) 19.1266 103 in.
B (T ) BC ( P ) BC 14.56 103 19.1266 103 4.57 103 in.
(checks)
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152
PROBLEM 2.54
The steel rails of a railroad track (Es 200 GPa, αs 11.7 × 102–6/C) were laid at a temperature of 6C.
Determine the normal stress in the rails when the temperature reaches 48C, assuming that the rails (a) are
welded to form a continuous track, (b) are 10 m long with 3-mm gaps between them.
SOLUTION
(a)
T (T ) L (11.7 106 )(48 6)(10) 4.914 103 m
P
PL L
(10)
50 1012
9
AE
E
200 10
T P 4.914 103 50 1012 0
98.3 106 Pa
(b)
98.3 MPa
T P 4.914 103 50 1012 3 103
3 103 4.914 103
50 1012
38.3 106 Pa
38.3 MPa
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153
PROBLEM 2.55
P⬘
2m
15 mm
Steel
5 mm
Brass
P
Steel
40 mm
Two steel bars ( Es 200 GPa and s 11.7 106/ C) are used to
reinforce a brass bar ( Eb 105 GPa, b 20.9 106/ C) that is subjected
to a load P 25 kN. When the steel bars were fabricated, the distance
between the centers of the holes that were to fit on the pins was made
0.5 mm smaller than the 2 m needed. The steel bars were then placed in
an oven to increase their length so that they would just fit on the pins.
Following fabrication, the temperature in the steel bars dropped back to
room temperature. Determine (a) the increase in temperature that was
required to fit the steel bars on the pins, (b) the stress in the brass bar
after the load is applied to it.
SOLUTION
(a)
Required temperature change for fabrication:
T 0.5 mm 0.5 103 m
Temperature change required to expand steel bar by this amount:
T L s T , 0.5 103 (2.00)(11.7 106 )(T ),
T 0.5 103 (2)(11.7 106 )(T )
T 21.368 C
(b)
21.4 C
*
*
Once assembled, a tensile force P develops in the steel, and a compressive force P develops in the
brass, in order to elongate the steel and contract the brass.
Elongation of steel:
As (2)(5)(40) 400 mm 2 400 106 m 2
( P ) s
F *L
P* (2.00)
25 109 P*
As Es (400 106 )(200 109 )
Contraction of brass: Ab (40)(15) 600 mm 2 600 106 m 2
( P )b
P* L
P* (2.00)
31.746 109 P*
Ab Eb (600 106 )(105 109 )
But ( P ) s ( P )b is equal to the initial amount of misfit:
( P ) s ( P )b 0.5 103 , 56.746 109 P* 0.5 103
P* 8.8112 103 N
Stresses due to fabrication:
Steel:
*s
P * 8.8112 103
22.028 106 Pa 22.028 MPa
6
As
400 10
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154
PROBLEM 2.55 (Continued)
Brass:
b*
P*
8.8112 103
14.6853 106 Pa 14.685 MPa
6
Ab
600 10
To these stresses must be added the stresses due to the 25-kN load.
For the added load, the additional deformation is the same for both the steel and the brass. Let be the
additional displacement. Also, let Ps and Pb be the additional forces developed in the steel and brass,
respectively.
Ps L
PL
b
As Es Ab Eb
As Es
(400 106 )(200 109 )
40 106
L
2.00
AE
(600 106 )(105 109 )
Pb b b
31.5 106
L
2.00
Ps
P Ps Pb 25 103 N
Total:
40 106 31.5 106 25 103
349.65 106 m
Ps (40 106 )(349.65 106 ) 13.9860 103 N
Pb (31.5 106 )(349.65 10 6 ) 11.0140 103 N
s
Ps 13.9860 103
34.965 106 Pa
6
As
400 10
b
Pb 11.0140 103
18.3566 106 Pa
Ab
600 10 6
Add stress due to fabrication.
Total stresses:
s 34.965 106 22.028 106 56.991 106 Pa
s 57.0 MPa
b 18.3566 106 14.6853 106 3.6713 106 Pa
b 3.67 MPa
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155
PROBLEM 2.56
P⬘
2m
15 mm
Steel
5 mm
Brass
P
Steel
40 mm
Determine the maximum load P that may be applied to the brass bar of
Prob. 2.55 if the allowable stress in the steel bars is 30 MPa and the
allowable stress in the brass bar is 25 MPa.
PROBLEM 2.55 Two steel bars ( Es 200 GPa and s 11.7 10–6/C)
are used to reinforce a brass bar ( Eb 105 GPa, b 20.9 10–6/C)
that is subjected to a load P 25 kN. When the steel bars were fabricated,
the distance between the centers of the holes that were to fit on the pins
was made 0.5 mm smaller than the 2 m needed. The steel bars were then
placed in an oven to increase their length so that they would just fit on the
pins. Following fabrication, the temperature in the steel bars dropped back
to room temperature. Determine (a) the increase in temperature that was
required to fit the steel bars on the pins, (b) the stress in the brass bar after
the load is applied to it.
SOLUTION
See solution to Problem 2.55 to obtain the fabrication stresses.
*s 22.028 MPa
b* 14.6853 MPa
Allowable stresses:
s ,all 30 MPa, b,all 25 MPa
Available stress increase from load.
s 30 22.028 7.9720 MPa
b 25 14.6853 39.685 MPa
Corresponding available strains.
s
b
s
Es
b
Eb
7.9720 106
39.860 10 6
200 109
39.685 106
377.95 10 6
105 109
Smaller value governs 39.860 10 6
Areas: As (2)(5)(40) 400 mm 2 400 106 m 2
Ab (15)(40) 600 mm 2 600 106 m 2
Forces Ps Es As (200 109 )(400 10 6 )(39.860 10 6 ) 3.1888 103 N
Pb Eb Ab (105 109 )(600 106 )(39.860 10 6 ) 2.5112 10 3 N
Total allowable additional force:
P Ps Pb 3.1888 103 2.5112 103 5.70 103 N
P 5.70 kN
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156
PROBLEM 2.57
Dimensions in mm
0.15
20
20
200
A
An aluminum rod (Ea 70 GPa, αa 23.6 × 106/C) and a steel link
(Es × 200 GPa, αa 11.7 × 106/C) have the dimensions shown at a
temperature of 20C. The steel link is heated until the aluminum rod
can be fitted freely into the link. The temperature of the whole
assembly is then raised to 150C. Determine the final normal stress
(a) in the rod, (b) in the link.
30
A
20
Section A-A
SOLUTION
T T f Ti 150C 20C 130C
Unrestrained thermal expansion of each part:
Aluminum rod:
( T )a L a ( T )
( T ) a (0.200 m)(23.6 106 / C)(130C)
6.1360 10 4 m
Steel link:
( T ) s L s ( T )
( T ) s (0.200 m)(11.7 106 /C)(130C)
3.0420 10 4 m
Let P be the compressive force developed in the aluminum rod. It is also the tensile force in the steel link.
Aluminum rod:
( P )a
PL
Ea Aa
P (0.200 m)
(70 10 Pa)( /4)(0.03 m)2
9
4.0420 109 P
Steel link:
( P ) s
PL
Es As
P (0.200)
(200 109 Pa)(2)(0.02 m)2
1.250 109 P
Setting the total deformed lengths in the link and rod equal gives
(0.200) (T ) s ( P ) s (0.200) (0.15 103 ) (T )a ( P )a
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157
PROBLEM 2.57 (Continued)
( P ) s ( P )a 0.15 103 (T )a (T ) s
1.25 109 P 4.0420 109 P 0.15 103 6.1360 104 3.0420 104
P 8.6810 104 N
(a)
Stress in rod:
R
P
A
8.6810 104 N
1.22811 108 Pa
2
( /4)(0.030 m)
R 122.8 MPa
(b)
Stress in link:
L
8.6810 104 N
1.08513 108 Pa
(2)(0.020 m)2
L 108.5 MPa
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158
PROBLEM 2.58
0.02 in.
14 in.
Bronze
A 2.4 in2
E 15 106 psi
12 10 –6/F
18 in.
Knowing that a 0.02-in. gap exists when the temperature is 75 F,
determine (a) the temperature at which the normal stress in the
aluminum bar will be equal to 11 ksi, (b) the corresponding exact
length of the aluminum bar.
Aluminum
A 2.8 in2
E 10.6 106 psi
12.9 10 –6/F
SOLUTION
a 11 ksi 11 103 psi
P a Aa (11 103 )(2.8) 30.8 103 lb
Shortening due to P:
P
PLb
PLa
Eb Ab Ea Aa
(30.8 103 )(14) (30.8 103 )(18)
(15 106 )(2.4) (10.6 106 )(2.8)
30.657 103 in.
Available elongation for thermal expansion:
T 0.02 30.657 103 50.657 103 in.
But T Lb b (T ) La a (T )
(14)(12 106 )(T ) (18)(12.9 106 )(T ) (400.2 106 )T
Equating, (400.2 106 )T 50.657 103
(a)
(b)
T 126.6F
Thot Tcold T 75 126.6 201.6F
a La a (T )
Thot 201.6F
PLa
Ea Aa
(18)(12.9 106 )(26.6)
(30.8 103 )(18)
10.712 103 in.
(10.6 106 )(2.8)
Lexact 18 10.712 103 18.0107 in.
L 18.0107 in.
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159
PROBLEM 2.59
0.02 in.
14 in.
18 in.
Determine (a) the compressive force in the bars shown after a
temperature rise of 180F, (b) the corresponding change in length of
the bronze bar.
Bronze
A 2.4 in2
E 15 106 psi
12 10 –6/F
Aluminum
A 2.8 in2
E 10.6 106 psi
12.9 10 –6/F
SOLUTION
Thermal expansion if free of constraint:
T Lb b (T ) La a (T )
(14)(12 106 )(180) (18)(12.9 106 )(180)
72.036 103 in.
Constrained expansion: 0.02 in.
Shortening due to induced compressive force P:
P 72.036 103 0.02 52.036 103 in.
P
But
PLb
PLa Lb
L
a
Eb Ab Ea Aa Eb Ab Ea Aa
P
14
18
9
P 995.36 10 P
6
6
(15
10
)(2.4)
(10.6
10
)(2.8)
995.36 109 P 52.036 103
Equating,
P 52.279 103 lb
P 52.3 kips
(a)
(b)
b Lb b (T )
PLb
Eb Ab
(14)(12 106 )(180)
(52.279 103 )(14)
9.91 103 in.
6
(15 10 )(2.4)
b 9.91 103 in.
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160
PROBLEM 2.60
0.5 mm
300 mm
At room temperature (20C) a 0.5-mm gap exists between the ends of
the rods shown. At a later time when the temperature has reached
140C, determine (a) the normal stress in the aluminum rod, (b) the
change in length of the aluminum rod.
250 mm
A
B
Aluminum
A 5 2000 mm2
E 5 75 GPa
a 5 23 3 10–6/8C
Stainless steel
A 5 800 mm2
E 5 190 GPa
a 5 17.3 3 10–6/8C
SOLUTION
T 140 20 120C
Free thermal expansion:
T La a (T ) Ls s (T )
(0.300)(23 106 )(120) (0.250)(17.3 106 )(120)
1.347 103 m
Shortening due to P to meet constraint:
P 1.347 103 0.5 103 0.847 103 m
PLa
PLs La
L
s P
Ea Aa Es As Ea Aa Es As
0.300
0.250
P
9
6
9
6
(75 10 )(2000 10 ) (190 10 )(800 10 )
3.6447 109 P
P
3.6447 109 P 0.847 103
Equating,
P 232.39 103 N
P
232.39 103
116.2 106 Pa
Aa
2000 106
(a)
a
(b)
a La a (T )
a 116.2 MPa
PLa
Ea Aa
(0.300)(23 106 )(120)
(232.39 103 )(0.300)
363 106 m
(75 109 )(2000 106 )
a 0.363 mm
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161
PROBLEM 2.61
P
5
8
5.0 in.
in. diameter
A standard tension test is used to determine the properties of an experimental plastic.
The test specimen is a 85 -in.-diameter rod and it is subjected to an 800-lb tensile
force. Knowing that an elongation of 0.45 in. and a decrease in diameter of 0.025 in.
are observed in a 5-in. gage length, determine the modulus of elasticity, the modulus
of rigidity, and Poisson’s ratio for the material.
P'
SOLUTION
A
4
d2
5
2
2
0.306796 in
4 8
P 800 lb
y
y
x
E
y
L
x
d
0.45
0.090
5.0
0.025
0.040
0.625
y 2.6076 103
28.973 103 psi
0.090
y
v
P
800
2.6076 103 psi
A 0.306796
E 29.0 103 psi
x 0.040
0.44444
0.090
y
v 0.444
E
28.973 103
10.0291 103 psi
2(1 v) (2)(1 0.44444)
10.03 103 psi
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162
PROBLEM 2.62
640 kN
A 2-m length of an aluminum pipe of 240-mm outer diameter and 10-mm
wall thickness is used as a short column to carry a 640-kN centric axial load.
Knowing that E 73 GPa and v 0.33, determine (a) the change in length of
the pipe, (b) the change in its outer diameter, (c) the change in its wall
thickness.
2m
SOLUTION
d o 0.240
t 0.010
L 2.0
di d o 2t 0.240 2(0.010) 0.220 m P 640 103 N
A
(a)
4
do2 di2 4 (0.240 0.220) 7.2257 103 m2
PL
(640 103 )(2.0)
EA
(73 109 )(7.2257 103 )
2.4267 103 m
L
2.43 mm
2.4267
1.21335 103
2.0
LAT v (0.33)(1.21335 103 )
4.0041 104
(b)
d o d o LAT (240 mm)(4.0041 104 ) 9.6098 102 mm
d o 0.0961 mm
t t LAT (10 mm)(4.0041 104 ) 4.0041 103 mm
t 0.00400 mm
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163
PROBLEM 2.63
200 kN
4
200 kN
10
200 mm
150 mm
A line of slope 4:10 has been scribed on a cold-rolled yellow-brass
plate, 150 mm wide and 6 mm thick. Knowing that E 105 GPa and
v 0.34, determine the slope of the line when the plate is subjected to
a 200-kN centric axial load as shown.
SOLUTION
A (0.150)(0.006) 0.9 103 m 2
x
x
P 200 103
222.22 106 Pa
A 0.9 103
x
E
222.22 106
2.1164 103
9
105 10
y x (0.34)(2.1164 103 )
0.71958 103
tan
4(1 y )
10(1 x )
4(1 0.71958 103 )
10(1 2.1164 103 )
0.39887
tan 0.399
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164
PROBLEM 2.64
50 mm
2.75 kN
2.75 kN
A
B
12 mm
A 2.75-kN tensile load is
applied to a test coupon
made from 1.6-mm flat
steel plate (E 200 GPa,
v 0.30). Determine the
resulting change (a) in the
50-mm gage length, (b) in
the width of portion AB
of the test coupon, (c) in
the thickness of portion
AB, (d) in the crosssectional area of portion
AB.
SOLUTION
A (1.6)(12) 19.20 mm 2
19.20 106 m 2
P 2.75 103 N
x
P
2.75 103
A 19.20 106
143.229 106 Pa
x
x
E
143.229 106
716.15 106
9
200 10
y z x (0.30)(716.15 106 ) 214.84 106
(a)
L 0.050 m
x L x (0.50)(716.15 106 ) 35.808 106 m
0.0358 mm
(b)
w 0.012 m
y w y (0.012)(214.84 106 ) 2.5781 106 m
0.00258 mm
(c)
t 0.0016 m
z t z (0.0016)(214.84 106 ) 343.74 109 m
0.000344 mm
(d)
A w0 (1 y )t0 (1 z ) w0t0 (1 y z y z )
A0 w0 t0
A A A0 w0t0 ( y z negligible term) 2w0 t0 y
(2)(0.012)(0.0016)(214.84 106 ) 8.25 109 m 2
0.00825 mm 2
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165
PROBLEM 2.65
22-mm diameter
75 kN
75 kN
200 mm
In a standard tensile test, a steel rod of 22-mm diameter is subjected to a
tension force of 75 kN. Knowing that v 0.3 and E 200 GPa,
determine (a) the elongation of the rod in a 200-mm gage length, (b) the
change in diameter of the rod.
SOLUTION
P 75 kN 75 103 N
A
x
4
d2
4
(0.022)2 380.13 106 m 2
P
75 103
197.301 106 Pa
A 380.13 106
E
197.301 106
986.51 106
200 109
x L x (200 mm)(986.51 106 )
(a)
x 0.1973 mm
y v x (0.3)(986.51 106 ) 295.95 106
y d y (22 mm)(295.95 106 )
(b) y 0.00651 mm
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166
2.5 in.
PROBLEM 2.66
The change in diameter of a large steel bolt is carefully measured as the
nut is tightened. Knowing that E 29 106 psi and v 0.30, determine
the internal force in the bolt if the diameter is observed to decrease by
0.5 103 in.
SOLUTION
y 0.5 103 in.
y
y
d
v
d 2.5 in.
0.5 103
0.2 103
2.5
y
:
x
x
y
v
0.2 103
0.66667 103
0.3
x E x (29 106 )(0.66667 103 ) 19.3334 103 psi
A
4
d2
4
(2.5) 2 4.9087 in 2
F x A (19.3334 103 )(4.9087) 94.902 103 lb
F 94.9 kips
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167
PROBLEM 2.67
A
The brass rod AD is fitted with a jacket that is used to apply a
hydrostatic pressure of 48 MPa to the 240-mm portion BC of the rod.
Knowing that E 105 GPa and v 0.33, determine (a) the change in
the total length AD, (b) the change in diameter at the middle of the rod.
B
240 mm
600 mm
C
D
50 mm
SOLUTION
x z p 48 106 Pa,
y 0
1
( x y z )
E
1
48 106 (0.33)(0) (0.33)(48 106 )
105 109
306.29 106
x
1
( x y z )
E
1
(0.33)(48 106 ) 0 (0.33)(48 106 )
105 109
301.71 106
y
(a)
Change in length: only portion BC is strained. L 240 mm
y L y (240)(301.71 106 ) 0.0724 mm
(b)
Change in diameter: d 50 mm
x z d x (50)(306.29 106 ) 0.01531 mm
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168
PROBLEM 2.68
y
4 in.
3 in.
A
B
D
z
C
z
x
x
A fabric used in air-inflated structures is subjected to a biaxial
loading that results in normal stresses x 18 ksi and z 24 ksi .
Knowing that the properties of the fabric can be approximated as
E 12.6 × 106 psi and v 0.34, determine the change in length of
(a) side AB, (b) side BC, (c) diagonal AC.
SOLUTION
x 18 ksi
y 0
z 24 ksi
1
1
( x y z )
18,000 (0.34)(24, 000) 780.95 106
E
12.6 106
1
1
z ( x y z )
(0.34)(18,000) 24,000 1.41905 103
6
E
12.6 10
x
(a)
AB ( AB) x (4 in.)(780.95 106 ) 0.0031238 in.
0.00312 in.
(b)
BC ( BC ) z (3 in.)(1.41905 103 ) 0.0042572 in.
0.00426 in.
Label sides of right triangle ABC as a, b, c.
Then
c2 a 2 b2
Obtain differentials by calculus.
2cdc 2ada 2bdb
dc
But a 4 in.
b 3 in.
da AB 0.0031238 in.
(c)
4
5
a
b
da db
c
c
c 42 32 5 in.
db BC 0.0042572 in.
3
5
AC dc (0.0031238) (0.0042572)
0.00505 in.
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169
PROBLEM 2.69
y 6 ksi
A
B
x 12 ksi
1 in.
A 1-in. square was scribed on the side of a large steel pressure vessel.
After pressurization the biaxial stress condition at the square is as
shown. Knowing that E 29 × 106 psi and v 0.30, determine the
change in length of (a) side AB, (b) side BC, (c) diagonal AC.
C
D
1 in.
SOLUTION
1
1
( x y )
E
29 106
6
351.72 10
1
1
y ( y x )
E
29 106
82.759 106
x
12 103 (0.30)(6 103 )
6 103 (0.30)(12 103 )
(a)
AB ( AB)0 x (1.00)(351.72 106 ) 352 106 in.
(b)
BC ( BC )0 y (1.00)(82.759 106 ) 82.8 106 in.
(c)
( AC ) ( AB) 2 ( BC ) 2 ( AB0 x )2 ( BC0 y )2
(1 351.72 106 )2 (1 82.759 106 ) 2
1.41452
AC ( AC )0 307 106
( AC )0 2
or use calculus as follows:
Label sides using a, b, and c as shown.
c2 a 2 b2
Obtain differentials.
dc
from which
2cdc 2ada 2bdc
a
b
da dc
c
c
But a 100 in., b 1.00 in., c 2 in.
da AB 351.72 106 in., db BC 82.8 106 in.
AC dc
1.00
2
(351.7 106 )
1.00
2
(82.8 106 )
307 106 in.
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170
PROBLEM 2.70
The block shown is made of a magnesium alloy, for which
E 45 GPa and v 0.35. Knowing that x 180 MPa,
determine (a) the magnitude of y for which the change in
the height of the block will be zero, (b) the corresponding
change in the area of the face ABCD, (c) the corresponding
change in the volume of the block.
SOLUTION
(a)
y 0
y 0
z 0
1
( x v y v z )
E
y v x (0.35)(180 106 )
y
y 63.0 MPa
63 106 Pa
1
v
(0.35)(243 106 )
( z v x v y ) ( x y )
1.890 103
E
E
45 109
x v y
1
157.95 106
x ( x v y v Z )
3.510 103
E
E
45 109
z
(b)
A0 Lx Lz
A Lx (1 x ) Lz (1 z ) Lx Lz (1 x z x z )
A A A0 Lx Lz ( x z x z ) Lx Lz ( x z )
A (100 mm)(25 mm)(3.510 103 1.890 103 )
(c)
A 4.05 mm 2
V0 Lx Ly Lz
V Lx (1 x ) Ly (1 y ) Lz (1 z )
Lx Ly Lz (1 x y z x y y z z x x y z )
V V V0 Lx Ly Lz ( x y z small terms)
V (100)(40)(25)(3.510 103 0 1.890 103 )
V 162.0 mm3
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171
PROBLEM 2.71
y
A
B
D
z
C
z
x
x
The homogeneous plate ABCD is subjected to a biaxial loading as
shown. It is known that z 0 and that the change in length of
the plate in the x direction must be zero, that is, x 0. Denoting
by E the modulus of elasticity and by v Poisson’s ratio, determine
(a) the required magnitude of x , (b) the ratio 0 / z .
SOLUTION
z 0 , y 0, x 0
x
1
1
( x v y v z ) ( x v 0 )
E
E
x v 0
(a)
(b)
z
1
1
1 v2
(v x v y z ) (v 2 0 0 0 )
0
E
E
E
0
E
z 1 v2
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172
y
P'
PROBLEM 2.72
P
x
x ⫽ P
A
z
(a)
P'
'
'
45⬚
m ⫽ P
x
For a member under axial loading, express the normal strain in a
direction forming an angle of 45 with the axis of the load in terms of the
axial strain x by (a) comparing the hypotenuses of the triangles shown in
Fig. 2.43, which represent, respectively, an element before and after
deformation, (b) using the values of the corresponding stresses of and
x shown in Fig. 1.38, and the generalized Hooke’s law.
P
m
2A
'
' ⫽ P
2A
(b)
SOLUTION
Figure 2.49
(a)
[ 2(1 )]2 (1 x )2 (1 v x )2
2(1 2 2 ) 1 2 x x2 1 2v x v 2 x2
4 2 2 2 x x2 2v x v 2 x2
4 2 x 2v x
Neglect squares as small.
(A)
1 v
x
2
(B)
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173
PROBLEM 2.72 (Continued)
(b)
v
E
E
1 v P
E 2A
1 v
x
2E
1 v
x
2
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174
y
PROBLEM 2.73
x
In many situations, it is known that the normal stress in a given direction is zero.
For example, z 0 in the case of the thin plate shown. For this case, which is
known as plane stress, show that if the strains x and y have been determined
experimentally, we can express x , y , and z as follows:
x E
x v y
1 v
2
y E
y v x
1 v
2
z
v
( x y )
1 v
SOLUTION
z 0
x
1
( x v y )
E
(1)
y
1
(v x y )
E
(2)
Multiplying (2) by v and adding to (1),
x v y
1 v2
x
E
or
x
E
( x v y )
1 v2
or
y
E
( y v x )
1 v2
Multiplying (1) by v and adding to (2),
y v x
1 v2
y
E
1
v
E
(v x v y )
( x v y y v x )
E
E 1 v2
v(1 v)
v
( x y )
( x y )
2
1 v
1 v
z
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175
PROBLEM 2.74
y
y
x
x
z
(a)
z
In many situations, physical constraints
prevent strain from occurring in a given
direction. For example, z 0 in the case
shown, where longitudinal movement of
the long prism is prevented at every point.
Plane sections perpendicular to the
longitudinal axis remain plane and the
same distance apart. Show that for this
situation, which is known as plane strain,
we can express z , x , and y as follows:
z v( x y )
(b)
1
[(1 v 2 ) x v(1 v) y ]
E
1
y [(1 v 2 ) y v(1 v) x ]
E
x
SOLUTION
z 0
1
(v x v y z ) or z v( x y )
E
1
( x v y v z )
E
1
[ x v y v 2 ( x y )]
E
1
[(1 v 2 ) x v(1 v) y ]
E
x
1
(v x y v z )
E
1
[v x y v 2 ( x y )]
E
1
[(1 v 2 ) y v(1 v) x ]
E
y
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176
PROBLEM 2.75
3.2 in.
The plastic block shown is bonded to a rigid support and to a vertical plate to
which a 55-kip load P is applied. Knowing that for the plastic used G 150 ksi,
determine the deflection of the plate.
4.8 in.
P
2 in.
SOLUTION
A (3.2)(4.8) 15.36 in 2
P 55 103 lb
P 55 103
3580.7 psi
A
15.36
G 150 103 psi
G
h 2 in.
3580.7
23.871 103
150 103
h (2)(23.871 103 )
47.7 103 in.
0.0477 in.
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177
PROBLEM 2.76
3.2 in.
What load P should be applied to the plate of Prob. 2.75 to produce a
deflection?
1
-in.
16
PROBLEM 2.75 The plastic block shown is bonded to a rigid support and to a
vertical plate to which a 55-kip load P is applied. Knowing that for the plastic
used G 150 ksi, determine the deflection of the plate.
4.8 in.
2 in.
P
SOLUTION
1
in. 0.0625 in.
16
h 2 in.
h
0.0625
0.03125
2
G 150 103 psi
G (150 103 )(0.03125)
4687.5 psi
A (3.2)(4.8) 15.36 in 2
P A (4687.5)(15.36)
72.0 103 lb
72.0 kips
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178
b
PROBLEM 2.77
a
a
Two blocks of rubber with a modulus of rigidity G 12 MPa are
bonded to rigid supports and to a plate AB. Knowing that c 100 mm
and P 45 kN, determine the smallest allowable dimensions a and b of
the blocks if the shearing stress in the rubber is not to exceed 1.4 MPa
and the deflection of the plate is to be at least 5 mm.
B
A
P
c
SOLUTION
Shearing strain:
a
Shearing stress:
b
a
G
1
2
P
A
G
(12 106 Pa)(0.005 m)
0.0429 m
1.4 106 Pa
P
2bc
45 103 N
P
0.1607 m
2c 2(0.1 m)(1.4 106 Pa)
a 42.9 mm
b 160.7 mm
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179
b
PROBLEM 2.78
a
a
Two blocks of rubber with a modulus of rigidity G 10 MPa are bonded
to rigid supports and to a plate AB. Knowing that b 200 mm and
c 125 mm, determine the largest allowable load P and the smallest
allowable thickness a of the blocks if the shearing stress in the rubber
is not to exceed 1.5 MPa and the deflection of the plate is to be at least
6 mm.
B
A
P
c
SOLUTION
Shearing stress:
1
2
P
A
P
2bc
P 2bc 2(0.2 m)(0.125 m)(1.5 103 kPa)
Shearing strain:
a
a
G
P 75.0 kN
G
(10 106 Pa)(0.006 m)
0.04 m
1.5 106 Pa
a 40.0 mm
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180
PROBLEM 2.79
An elastomeric bearing (G 130 psi) is used to support a bridge girder
as shown to provide flexibility during earthquakes. The beam must not
displace more than 83 in. when a 5-kip lateral load is applied as shown.
Knowing that the maximum allowable shearing stress is 60 psi,
determine (a) the smallest allowable dimension b, (b) the smallest
required thickness a.
P
a
b
8 in.
SOLUTION
Shearing force:
P 5 kips 5000 lb
Shearing stress:
60 psi
and
(a)
b
(b)
P
,
A
or
A
P
5000
83.333 in 2
60
A (8 in.)(b)
A 83.333
10.4166 in.
8
8
b 10.42 in.
60
461.54 103 rad
130
0.375 in.
But , or a
a
461.54 103
a 0.813 in.
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181
PROBLEM 2.80
For the elastomeric bearing in Prob. 2.79 with b 10 in. and a 1 in.,
determine the shearing modulus G and the shear stress for a
maximum lateral load P 5 kips and a maximum displacement
0.4 in.
P
a
b
8 in.
SOLUTION
Shearing force:
P 5 kips 5000 lb
Area:
A (8 in.)(10 in.) 80 in 2
Shearing stress:
Shearing strain:
Shearing modulus:
G
a
P 5000
80
A
62.5 psi
0.4 in.
0.400 rad
1 in.
62.5
0.400
G 156.3 psi
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182
PROBLEM 2.81
P
A
150 mm
100 mm
A vibration isolation unit consists of two blocks of hard rubber bonded
to a plate AB and to rigid supports as shown. Knowing that a force of
magnitude P 25 kN causes a deflection 1.5 mm of plate AB,
determine the modulus of rigidity of the rubber used.
B
30 mm
30 mm
SOLUTION
F
1
1
P (25 103 N) 12.5 103 N
2
2
F
(12.5 103 N)
833.33 103 Pa
A (0.15 m)(0.1 m)
1.5 103 m
h 0.03 m
1.5 103
0.05
h
0.03
833.33 103
G
16.67 106 Pa
0.05
G 16.67 MPa
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183
PROBLEM 2.82
P
A
150 mm
100 mm
A vibration isolation unit consists of two blocks of hard rubber with a
modulus of rigidity G 19 MPa bonded to a plate AB and to rigid
supports as shown. Denoting by P the magnitude of the force applied
to the plate and by the corresponding deflection, determine the
effective spring constant, k P/ , of the system.
B
30 mm
30 mm
SOLUTION
Shearing strain:
Shearing stress:
G
h
G
h
GA
1
P A
h
2
Force:
P
P
k
with
A (0.15)(0.1) 0.015 m 2
k
2GA
h
2GA
h
Effective spring constant:
or
h 0.03 m
2(19 106 Pa)(0.015 m 2 )
19.00 106 N/m
0.03 m
k 19.00 103 kN/m
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184
PROBLEM 2.83
A 6-in.-diameter solid steel sphere is lowered into the ocean to a point where the pressure is 7.1 ksi (about
3 miles below the surface). Knowing that E 29 106 psi and v 0.30, determine (a) the decrease in
diameter of the sphere, (b) the decrease in volume of the sphere, (c) the percent increase in the density of the
sphere.
SOLUTION
For a solid sphere,
V0
6
d 03
(6.00)3
6
113.097 in 3
x y z p
7.1 103 psi
1
( x v y v z )
E
(1 2v) p
(0.4)(7.1 103 )
E
29 106
x
97.93 106
Likewise,
y z 97.93 106
e x y z 293.79 106
(a)
d d 0 x (6.00)(97.93 106 ) 588 106 in.
(b)
V V0 e (113.097)(293.79 106 ) 33.2 103 in 3
(c)
Let m mass of sphere.
d 588 106 in.
V 33.2 103 in 3
m constant.
m 0V0 V V0 (1 e)
0
V
1
m
1
0 1
1
1 e
0
0
V0 (1 e) m
(1 e e 2 e3 ) 1 e e 2 e3
e 293.79 106
0
100% (293.79 106 )(100%)
0
0.0294%
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185
PROBLEM 2.84
85 mm
(a) For the axial loading shown, determine the change in height and the
change in volume of the brass cylinder shown. (b) Solve part a,
assuming that the loading is hydrostatic with x y z 70 MPa.
sy 5 258 MPa
E 5 105 GPa
n 5 0.33
135 mm
SOLUTION
h0 135 mm 0.135 m
A 0
d 02
(85) 2 5.6745 103 mm 2 5.6745 103 m 2
4
4
V0 A0 h0 766.06 103 mm3 766.06 106 m3
(a)
x 0, y 58 106 Pa, z 0
y
y
1
58 106
(v x y v z )
552.38 106
9
E
E
105 10
h h 0 y (135 mm)( 552.38 106 )
e
h 0.0746 mm
(1 2v) y (0.34)(58 106 )
1 2v
( x y z )
187.81 106
9
E
E
105 10
V V0 e (766.06 103 mm3 )(187.81 106 )
(b)
x y z 70 106 Pa
y
V 143.9 mm3
x y z 210 106 Pa
1
1 2v
(0.34)(70 106 )
(v x y v z )
y
226.67 106
E
E
105 109
h h 0 y (135 mm)( 226.67 106 )
e
h 0.0306 mm
1 2v
(0.34)(210 106 )
( x y z )
680 106
E
105 109
V V0 e (766.06 103 mm3 )(680 106 )
V 521 mm3
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186
PROBLEM 2.85*
1 in. diameter
11 kips
11 kips
Determine the dilatation e and the change in volume of the 8-in. length
of the rod shown if (a) the rod is made of steel with E 29 × 106 psi
and v 0.30, (b) the rod is made of aluminum with E 10.6 × 106 psi
and v 0.35.
8 in.
SOLUTION
A
d2
4
4
3
P 11 10 lb
Stresses :
(a)
Steel.
(1) 2 0.78540 in 2
P 11 103
14.0056 103 psi
A 0.78540
y z 0
x
E 29 106 psi
x
y
z
v 0.30
1
14.0056 103
( x v y v z ) x
482.95 106
E
E
29 106
v
1
(v x y v z ) x v x (0.30)(482.95 106 )
E
E
144.885 106
v
1
(v x v y z ) x y 144.885 106
E
E
e x y z 193.2 106
v ve Le (0.78540)(8)(193.2 106 ) 1.214 103 in 3
(b)
Aluminum.
E 10.6 106 psi
x
x
E
v 0.35
14.0056 103
1.32128 103
10.6 106
y v x (0.35)(1.32128 103 ) 462.45 106
z y 462.45 106
e x y z 396 106
v ve Le (0.78540)(8)(396 106 ) 2.49 103 in 3
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187
PROBLEM 2.86
Determine the change in volume of the 50-mm gage length segment AB in Prob. 2.64 (a) by computing the
dilatation of the material, (b) by subtracting the original volume of portion AB from its final volume.
PROBLEM 2.64 A 2.75-kN tensile load is applied to a test coupon made from 1.6-mm flat steel plate
(E 200 GPa, v 0.30). Determine the resulting change (a) in the 50-mm gage length, (b) in the width of
portion AB of the test coupon, (c) in the thickness of portion AB, (d) in the cross-sectional area of portion AB.
50 mm
2.75 kN
2.75 kN
A
B
12 mm
SOLUTION
(a)
A0 (12)(1.6) 19.2 mm 2 19.2 106 m 2
Volume
V0 L0 A0 (50)(19.2) 960 mm3
x
P
2.75 103
143.229 106 Pa
A0 19.2 106
x
1
143.229 106
( x y z ) x
716.15 106
E
E
200 109
y z 0
y z x (0.30)(716.15 103 ) 214.84 106
e x y z 286.46 106
v v0 e (960)(286.46 106 ) 0.275 mm3
(b)
From the solution to problem 2.64,
x 0.035808 mm
y 0.0025781
z 0.00034374 mm
The dimensions when under the 2.75-kN load are
Length:
L L0 x 50 0.035808 50.035808 mm
Width:
w w0 y 12 0.0025781 11.997422 mm
Thickness: t t0 z 1.6 0.00034374 1.599656 mm
Volume:
V Lwt (50.03581)(11.997422)(1.599656) 960.275 mm3
V V V0 960.275 960 0.275 mm3
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188
PROBLEM 2.87
P
A vibration isolation support consists of a rod A of radius R1 10 mm and
a tube B of inner radius R 2 25 mm bonded to an 80-mm-long hollow
rubber cylinder with a modulus of rigidity G 12 MPa. Determine the
largest allowable force P that can be applied to rod A if its deflection is not
to exceed 2.50 mm.
R1
A
R2
80 mm
B
SOLUTION
Let r be a radial coordinate. Over the hollow rubber cylinder, R1 r R2 .
Shearing stress acting on a cylindrical surface of radius r is
The shearing strain is
P
P
A 2 rh
G
P
2 Ghr
Shearing deformation over radial length dr:
d
dr
d dr
P dr
2 Gh r
Total deformation.
R2
R1
d
P
2 Gh
R2
P
ln r
R1
2 Gh
R
P
ln 2
R1
2 Gh
Data:
R1 10 mm 0.010 m,
R2
R2 25 mm 0.025 m, h 80 mm 0.080 m
G 12 106 Pa
P
dr
R1 r
P
(ln R2 ln R1 )
2 Gh
2 Gh
or P
ln( R2 / R1 )
2.50 103 m
(2 )(12 106 ) (0.080) (2.50 10 3 )
16.46 103 N
ln (0.025/0.010)
16.46 kN
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189
PROBLEM 2.88
P
A
A vibration isolation support consists of a rod A of radius R1 and a tube B of
inner radius R2 bonded to a 80-mm-long hollow rubber cylinder with a
modulus of rigidity G 10.93 MPa. Determine the required value of the ratio
R2/R1 if a 10-kN force P is to cause a 2-mm deflection of rod A.
R1
R2
80 mm
B
SOLUTION
Let r be a radial coordinate. Over the hollow rubber cylinder, R1 r R2 .
Shearing stress acting on a cylindrical surface of radius r is
P
P
A 2 rh
The shearing strain is
G
P
2 Ghr
Shearing deformation over radial length dr:
d
dr
d dr
dr
P dr
2 Gh r
Total deformation.
R2
R1
d
P
2 Gh
R2
P
ln r
R1
2 Gh
R
P
ln 2
R1
2 Gh
ln
dr
r
P
(ln R2 ln R1 )
2 Gh
R2
R1
R2 2 Gh (2 ) (10.93 106 ) (0.080) (0.002)
1.0988
R1
P
10.103
R2
exp (1.0988) 3.00
R1
R2 /R1 3.00
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190
PROBLEM 2.89
The material constants E, G, k, and v are related by Eqs. (2.33) and (2.43). Show that any one of these constants
may be expressed in terms of any other two constants. For example, show that (a) k GE/(9G 3E) and
(b) v (3k 2G)/(6k 2G).
SOLUTION
k
(a)
1 v
and
G
E
2(1 v)
E
E
1
or v
2G
2G
k
(b)
E
3(1 2v)
2 EG
2 EG
E
E
3[2G 2 E 4G ] 18G 6 E
3 1 2
1
2G
k
EG
9G 6 E
v
3k 2G
6k 2G
k
2(1 v)
G 3(1 2v)
3k 6kv 2G 2Gv
3k 2G 2G 6k
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191
PROBLEM 2.90
Show that for any given material, the ratio G/E of the modulus of rigidity over the modulus of elasticity is
always less than 12 but more than 13 . [Hint: Refer to Eq. (2.43) and to Sec. 2.13.]
SOLUTION
G
E
2(1 v)
Assume v 0 for almost all materials, and v <
E
2(1 v)
G
or
1
2
for a positive bulk modulus.
E
1
< 2 1 3
G
2
Applying the bounds,
2
Taking the reciprocals,
1
G 1
2
E 3
1 G 1
3
E 2
or
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192
y
E x 50 GPa
E y 15.2 GPa
E z 15.2 GPa
PROBLEM 2.91
xz 0.254
xy 0.254
zy 0.428
A composite cube with 40-mm sides and the properties shown is
made with glass polymer fibers aligned in the x direction. The cube
is constrained against deformations in the y and z directions and is
subjected to a tensile load of 65 kN in the x direction. Determine
(a) the change in the length of the cube in the x direction, (b) the
stresses x , y , and z .
Ex 50 GPa
vxz 0.254
E y 15.2 GPa vxy 0.254
Ez 15.2 GPa vzy 0.428
z
x
SOLUTION
Stress-to-strain equations are
x
x
Ex
y
z
v yx y
Ey
vxy x
Ex
y
Ey
vzx z
Ez
vzy z
Ez
vxz x v yz y z
Ex
Ey
Ez
vxy
Ex
v yz
Ey
v yx
Ey
vzy
Ez
vzx vxz
Ez Ex
(1)
(2)
(3)
(4)
(5)
(6)
y 0 and z 0.
The constraint conditions are
Using (2) and (3) with the constraint conditions gives
vzy
vxy
1
y z
x
Ey
Ez
Ex
v yz
Ey
y
V
1
z xz x
Ez
Ex
(7)
(8)
1
0.428
0.254
y
z
x or y 0.428 z 0.077216 x
15.2
15.2
50
0.428
1
0.254
y
z
x or 0.428 y z 0.077216 x
15.2
15.2
50
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193
PROBLEM 2.91 (Continued)
Solving simultaneously,
y z 0.134993 x
Using (4) and (5) in (1),
x
Ex
vxy
v
1
x
y xz z
Ex
Ex
E
1
[1 (0.254)(0.134993) (0.254)(0.134993)] x
Ex
0.93142 x
Ex
A (40)(40) 1600 mm 2 1600 106 m 2
65 103
P
40.625 106 Pa
6
A 1600 10
(0.93142)(40.625 103 )
756.78 106
x
50 109
x
(a)
x Lx x (40 mm)(756.78 106 )
x 0.0303 mm
(b)
x 40.625 106 Pa
x 40.6 MPa
y z (0.134993)(40.625 106 ) 5.48 106 Pa
y z 5.48 MPa
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194
y
E x 50 GPa
E y 15.2 GPa
E z 15.2 GPa
xz 0.254
xy 0.254
zy 0.428
PROBLEM 2.92
The composite cube of Prob. 2.91 is constrained against deformation in
the z direction and elongated in the x direction by 0.035 mm due to a
tensile load in the x direction. Determine (a) the stresses x , y , and z
and (b) the change in the dimension in the y direction.
Ex 50 GPa
vxz 0.254
E y 15.2 GPa vxy 0.254
Ez 15.2 GPa vzy 0.428
z
x
SOLUTION
x
x
Ex
y
z
v yx y
vxy x
Ex
Ey
y
Ey
vzx z
Ez
vzy z
Ez
vxz x v yz y z
Ex
Ey
Ez
vxy
Ex
v yz
Ey
v yx
Ey
vzy
Ez
vzx vxz
Ez Ex
Constraint condition:
Load condition :
(1)
(2)
(3)
(4)
(5)
(6)
z 0
y 0
0
From Equation (3),
z
vxz
1
x z
Ex
Ez
vxz Ez
(0.254)(15.2)
x
0.077216 x
50
Ex
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195
PROBLEM 2.92 (Continued)
From Equation (1) with y 0,
x
1
1
[ x 0.254 z ]
[1 (0.254)(0.077216)] x
Ex
Ex
0.98039
x
Ex
x
But
x
(a)
x
x
Lx
v
v
1
1
x zx z x xz z
Ex
Ez
Ex
Ex
Ex x
0.98039
0.035 mm
875 106
40 mm
(50 109 ) (875 106 )
44.625 103 Pa
0.98039
x 44.6 MPa
y 0
z 3.45 MPa
z (0.077216)(44.625 106 ) 3.446 106 Pa
From (2),
y
vxy
Ex
x
vzy
1
y z
Ey
Ez
(0.254)(44.625 106 )
(0.428)(3.446 106 )
0
50 109
15.2 109
6
323.73 10
(b)
y Ly y (40 mm)( 323.73 106 )
y 0.0129 mm
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196
PROBLEM 2.93
P
Knowing that, for the plate shown, the allowable stress is 125 MPa, determine
the maximum allowable value of P when (a) r 12 mm, (b) r 18 mm.
60 mm
r
120 mm
15 mm
SOLUTION
A (60)(15) 900 mm 2 900 106 m 2
D 120 mm
2.00
d
60 mm
(a)
r 12 mm
From Fig. 2.60b,
P
r 12 mm
0.2
d 60 mm
K 1.92
max K
P
A
A max (900 106 )(125 106 )
58.6 103 N
K
1.92
58.3 kN
(b)
r 18 mm,
P
r 18 mm
0.30
d 60 mm
From Fig 2.60b,
A max (900 106 )(125 106 )
64.3 103 N
K
1.75
K 1.75
64.3 kN
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197
PROBLEM 2.94
P
Knowing that P 38 kN, determine the maximum stress when (a) r 10 mm,
(b) r 16 mm, (c) r 18 mm.
60 mm
r
120 mm
15 mm
SOLUTION
A (60)(15) 900 mm 2 900 106 m 2
D 10 mm
2.00
d 60 mm
(a)
r 10 mm
From Fig. 2.60b,
max
(b)
From Fig. 2.60b,
(c)
max
KP
A
87.0 MPa
r 16 mm
0.2667
d 60 mm
K 1.78
(1.78)(38 103 )
75.2 106 Pa
6
900 10
r 18 mm,
From Fig 2.60b,
max
K 2.06
(2.06)(38 103 )
87.0 106 Pa
900 106
r 16 mm
max
r 10 mm
0.1667
d 60 mm
75.2 MPa
r 18 mm
0.30
d 60 mm
K 1.75
(1.75)(38 103 )
73.9 106 Pa
900 106
73.9 MPa
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198
1
2
d
411
in.
16
A
in.
rf 3
8
PROBLEM 2.95
in.
3 18 in.
P
A hole is to be drilled in the plate at A. The diameters of the
bits available to drill the hole range from 12 to 11/2 in. in
1
-in. increments. If the allowable stress in the plate is 21 ksi,
4
determine (a) the diameter d of the largest bit that can be
used if the allowable load P at the hole is to exceed that at
the fillets, (b) the corresponding allowable load P.
SOLUTION
At the fillets:
D
4.6875
1.5
d
3.125
From Fig. 2.60b,
K 2.10
r
0.375
0.12
d
3.125
Amin (3.125)(0.5) 1.5625 in 2
max K
Pall
Amin all
(1.5625)(21)
15.625 kips
K
2.10
Anet ( D 2r )t , K from Fig. 2.60a
At the hole:
max K
with
Hole diam.
0.5 in.
0.75 in.
1 in.
r
0.25 in.
0.375 in.
0.5 in.
1.25 in.
1.5 in.
Pall
all
Amin
0.625 in.
0.75 in.
P
all
Anet
Pall
D 4.6875 in.
t 0.5 in.
d D 2r
2r/D
4.1875 in.
3.9375 in.
3.6875 in.
3.4375 in.
3.1875 in.
0.107
0.16
0.213
0.267
0.32
Anet all
K
all 21 ksi
K
2.68
2.58
2.49
2.41
2.34
Anet
Pall
2.0938 in2
16.41 kips
1.96875 in
2
16.02 kips
1.84375 in
2
15.55 kips
1.71875 in
2
14.98 kips
1.59375 in
2
14.30 kips
(a)
Largest hole with Pall 15.625 kips is the 34 -in.-diameter hole.
(b)
Allowable load Pall 15.63 kips
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199
d
411
in.
16
1
2
A
in.
3
8
rf PROBLEM 2.96
in.
3 18 in.
P
(a) For P 13 kips and d 12 in., determine the maximum
stress in the plate shown. (b) Solve part a, assuming that the
hole at A is not drilled.
SOLUTION
Maximum stress at hole:
Use Fig. 2.60a for values of K.
2r
0.5
0.017,
D
4.6875
K 2.68
Anet (0.5)(4.6875 0.5) 2.0938 in 2
max K
P
(2.68)(13)
16.64 ksi
Anet
2.0938
Maximum stress at fillets:
Use Fig. 2.60b for values of K.
r
0.375
0.12
d
3.125
D
4.6875
1.5
d
3.125
K 2.10
Amin (0.5)(3.125) 1.5625 in 2
max K
P
(2.10)(13)
17.47 ksi
Amin
1.5625
(a)
With hole and fillets:
17.47 ksi
(b)
Without hole:
17.47 ksi
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200
PROBLEM 2.97
9 mm
Knowing that the hole has a diameter of 9 mm, determine (a) the radius
rf of the fillets for which the same maximum stress occurs at the hole A
and at the fillets, (b) the corresponding maximum allowable load P if
the allowable stress is 100 MPa.
rf
96 mm
A
9 mm
60 mm
P
9 mm
SOLUTION
1
r (9) 4.5 mm
2
For the circular hole,
d 96 9 87 mm
2r
2(4.5)
0.09375
D
96
Anet dt (0.087 m)(0.009 m) 783 106 m 2
K hole 2.72
From Fig. 2.60a,
max
P
(a)
For fillet,
K hole P
Anet
Anet max
(783 106 )(100 106 )
28.787 103 N
K hole
2.72
D 96 mm, d 60 mm
D 96
1.60
d
60
Amin dt (0.060 m)(0.009 m) 540 106 m 2
max
K fillet P
A
(5.40 106 )(100 106 )
K fillet min max
Amin
P
28.787 103
1.876
From Fig. 2.60b,
rf
d
0.19
r f 0.19d 0.19(60)
rf 11.4 mm
(b) P 28.8 kN
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201
PROBLEM 2.98
88 mm
For P 100 kN, determine the minimum plate thickness t required if the
allowable stress is 125 MPa.
rA 5 20 mm
A
rB 5 15 mm
B
t
64 mm
P
SOLUTION
At the hole:
rA 20 mm
d A 88 40 48 mm
2rA
2(20)
0.455
88
DA
From Fig. 2.60a,
K 2.20
max
t
At the fillet:
From Fig. 2.60b,
KP
KP
KP
t
Anet
d At
d A max
(2.20)(100 103 N)
36.7 103 m 36.7 mm
6
(0.048 m)(125 10 Pa)
D
88
1.375
dB
64
D 88 mm,
d B 64 mm
rB 15 mm
rB
15
0.2344
dB
64
K 1.70
max
t
KP
KP
Amin
d Bt
KP
d B max
(1.70)(100 103 N)
21.25 103 m 21.25 mm
(0.064 m)(125 106 Pa)
The larger value is the required minimum plate thickness.
t 36.7 mm
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202
PROBLEM 2.99
2 in.
P
t5
5
8
in.
r5
1
4
in.
P
3 in.
(a) Knowing that the allowable stress is 20 ksi, determine the maximum
allowable magnitude of the centric load P. (b) Determine the percent
change in the maximum allowable magnitude of P if the raised portions
are removed at the ends of the specimen.
SOLUTION
D 3
1.50
d 2
r 0.250
0.125
2
d
K 2.08
From Fig. 2.60b,
Amin td (0.625)(2) 1.25 in 2
KP
Amin
Amin max (1.25)(20)
12.0192 kips
K
2.08
(a)
max
(b)
Without raised section, K 1.00
P
P 12.02 kips
P Amin max (1.25)(20) 25 kips
25 12.02
% change
100%
12.02
108.0%
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203
3
4
PROBLEM 2.100
in.
rf 5
1
2
in.
A centric axial force is applied to the steel bar shown. Knowing that
all 20 ksi, determine the maximum allowable load P.
5 in.
P
6 12 in.
1 in.
SOLUTION
r 0.5 in.
At the hole:
2r 2(0.5)
0.2
d
5
d 5 1 4 in.
From Fig. 2.60a,
K 2.51
Anet td (0.75)(4) 3 in 2
max
P
At the fillet :
KP
Anet
Anet max (3)(20)
23.9 kips
K
2.51
D 6.5
1.3
d
5
D 6.5 in.,
d 5 in.,
r 0.5 in.
r 0.5
0.1
d
5
From Fig. 2.60b,
K 2.04
Amin td (0.75)(5) 3.75 in 2
max
P
KP
Amin
Amin max (3.75)(20)
36.8 kips
K
2.04
P 23.9 kips
Smaller value for P controls.
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204
B
PROBLEM 2.101
The cylindrical rod AB has a length L 5 ft and a 0.75-in. diameter; it is made of a mild steel
that is assumed to be elastoplastic with E 29 × 106 psi and Y 36 ksi . A force P is applied
to the bar and then removed to give it a permanent set P . Determine the maximum value of
the force P and the maximum amount m by which the bar should be stretched if the desired
value of P is (a) 0.1 in., (b) 0.2 in.
L
A
P
SOLUTION
A
4
d2
y L Y
4
(0.75) 2 0.44179 in 2
L 5 ft 60 in.
L Y (60)(36 103 )
0.074483 in.
E
29 103
When m exceeds Y , thus causing permanent stretch p , the maximum force is
Pm A Y (0.44179)(36 103 ) 15.9043 103 lb
P 15.90 kips
p m ' m Y so that
m p Y
(a)
p 0.1 in.
m 0.1 0.074483 0.1745 in.
(b)
p 0.2 in.
m 0.2 0.074483 0.274 in.
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205
PROBLEM 2.102
B
The cylindrical rod AB has a length L 6 ft and a 1.25-in. diameter; it is made of a mild steel
that is assumed to be elastoplastic with E 29 × 106 psi and Y 36 ksi . A force P is
applied to the bar until end A has moved down by an amount m . Determine the maximum
value of the force P and the permanent set of the bar after the force has been removed,
knowing (a) m 0.125 in., (b) m 0.250 in.
L
A
P
SOLUTION
A
4
d2
Y L Y
4
(1.25) 2 1.22718 in 2
L 6 ft 72 in.
L Y (72)(36 103 )
0.089379 in.
E
29 103
If m Y ,
Pm A Y (1.22718)(36 103 )
44.179 103 lb 44.2 kips
(a)
m 0.125 in. >Y
so that Pm 44.2 kips
Pm L Y L
Y 0.089379
AE
E
p m 0.125 0.089379 0.356 in.
(b)
m 0.250 in. >Y
Y
so that Pm 44.2 kips
p m 0.250 0.089379 0.1606 in.
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206
PROBLEM 2.103
A
Rod AB is made of a mild steel that is assumed to be elastoplastic
with E 200 GPa and Y 345 MPa . After the rod has been
attached to the rigid lever CD, it is found that end C is 6 mm too
high. A vertical force Q is then applied at C until this point has
moved to position C . Determine the required magnitude of Q and
the deflection 1 if the lever is to snap back to a horizontal position
after Q is removed.
9-mm diameter
1.25 m
C
B
D
6 mm
d1
C⬘
0.4 m
0.7 m
SOLUTION
AAB
4
(9)2 63.617 mm 2 63.617 106 m 2
Since rod AB is to be stretched permanently,
( FAB )max AAB Y (63.617 106 )(345 106 )
21.948 103 N
M D 0: 1.1Q 0.7 FAB 0
Qmax
AB
0.7
(21.948 103 ) 13.9669 103 N
1.1
13.97 kN
( FAB ) max LAB
(21.948 103 )(1.25)
2.15625 103 m
EAAB
(200 109 )(63.617 106 )
AB
0.7
3.0804 103 rad
1 1.1 3.39 103 m
3.39 mm
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207
PROBLEM 2.104
A
Solve Prob. 2.103, assuming that the yield point of the mild steel is
250 MPa.
9-mm diameter
1.25 m
C
B
PROBLEM 2.103 Rod AB is made of a mild steel that is assumed to
be elastoplastic with E 200 GPa and Y 345 MPa . After the rod
has been attached to the rigid lever CD, it is found that end C is 6 mm
too high. A vertical force Q is then applied at C until this point has
moved to position C . Determine the required magnitude of Q and
the deflection 1 if the lever is to snap back to a horizontal position
after Q is removed.
D
6 mm
d1
C⬘
0.4 m
0.7 m
SOLUTION
AAB
4
(9)2 63.617 mm 2 63.617 10 6 m 2
Since rod AB is to be stretched permanently,
( FAB ) max AAB Y (63.617 10 6 )(250 106 )
15.9043 103 N
M D 0: 1.1Q 0.7 FAB 0
Qmax
AB
0.7
(15.9043 103 ) 10.12 103 N
1.1
10.12 kN
( FAB )max LAB
(15.9043 103 )(1.25)
1.5625 103 m
EAAE
(200 109 )(63.617 106 )
AB
0.7
2.2321 103 rad
1 1.1 2.46 103 m
2.46 mm
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208
PROBLEM 2.105
C
Rod ABC consists of two cylindrical portions AB and BC; it is made of a mild
steel that is assumed to be elastoplastic with E 200 GPa and Y 250 MPa.
A force P is applied to the rod and then removed to give it a permanent set
p 2 mm. Determine the maximum value of the force P and the maximum
amount m by which the rod should be stretched to give it the desired permanent
set.
40-mm
diameter
1.2 m
B
30-mm
diameter
0.8 m
A
P
SOLUTION
AAB
4
(30)2 706.86 mm 2 706.86 106 m 2
(40)2 1.25664 103 mm 2 1.25664 103 m 2
4
Amin Y (706.86 106 )(250 106 ) 176.715 103 N
ABC
Pmax
Pmax 176.7 kN
PLAB PLBC
(176.715 103 )(0.8)
(176.715 103 )(1.2)
9
6
EAAB
EABC (200 10 )(706.86 10 ) (200 109 )(1.25664 103 )
1.84375 103 m 1.84375 mm
p m or m p 2 1.84375
m 3.84 mm
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209
PROBLEM 2.106
C
Rod ABC consists of two cylindrical portions AB and BC; it is made of a mild
steel that is assumed to be elastoplastic with E 200 GPa and Y 250 MPa.
A force P is applied to the rod until its end A has moved down by an amount
m 5 mm. Determine the maximum value of the force P and the permanent
set of the rod after the force has been removed.
40-mm
diameter
1.2 m
B
30-mm
diameter
0.8 m
A
P
SOLUTION
AAB
4
(30)2 706.86 mm 2 706.86 106 m 2
(40)2 1.25664 103 mm 2 1.25644 103 m 2
4
Amin Y (706.86 106 )(250 106 ) 176.715 103 N
ABC
Pmax
Pmax 176.7 kN
PLAB PLBC
(176.715 103 )(0.8)
(176.715 103 )(1.2)
9
6
EAAB
EABC (200 10 )(706.68 10 ) (200 109 )(1.25664 103 )
1.84375 103 m 1.84375 mm
p m 5 1.84375 3.16 mm
p 3.16 mm
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210
PROBLEM 2.107
A
Rod AB consists of two cylindrical portions AC and BC, each with a cross-sectional
area of 1750 mm2. Portion AC is made of a mild steel with E 200 GPa and
Y 250 MPa, and portion CB is made of a high-strength steel with E 200 GPa
and Y 345 MPa. A load P is applied at C as shown. Assuming both steels to be
elastoplastic, determine (a) the maximum deflection of C if P is gradually increased
from zero to 975 kN and then reduced back to zero, (b) the maximum stress in each
portion of the rod, (c) the permanent deflection of C.
190 mm
C
190 mm
P
B
SOLUTION
Displacement at C to cause yielding of AC.
L
(0.190)(250 106 )
0.2375 103 m
C ,Y LAC Y , AC AC Y , AC
9
E
200 10
FAC A Y , AC (1750 106 )(250 106 ) 437.5 103 N
Corresponding force.
FCB
EA C
(200 109 )(1750 106 )(0.2375 103 )
437.5 103 N
0.190
LCB
For equilibrium of element at C,
FAC ( FCB PY ) 0
PY FAC FCB 875 103 N
Since applied load P 975 103 N 875 103 N, portion AC yields.
FCB FAC P 437.5 103 975 103 N 537.5 103 N
(a)
C
FCB LCD
(537.5 103 )(0.190)
0.29179 103 m
EA
(200 109 )(1750 106 )
0.292 mm
(b)
Maximum stresses: AC Y , AC 250 MPa
(c)
FBC
537.5 103
307.14 106 Pa 307 MPa
A
1750 106
Deflection and forces for unloading.
P L
P L
L
PAC
AC PAC
AC AC CB CB
PCB
EA
EA
LAB
250 MPa
BC
307 MPa
PCB
2 PAC
PAC
487.5 103 N
P 975 103 PAC
(487.5 103 )(0.190)
0.26464 103 m
6
9
(200 10 )(1750 10 )
p m 0.29179 103 0.26464 103
0.02715 103 m
0.0272 mm
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211
PROBLEM 2.108
A
190 mm
C
190 mm
P
B
For the composite rod of Prob. 2.107, if P is gradually increased from zero until the
deflection of point C reaches a maximum value of m 0.3 mm and then decreased
back to zero, determine (a) the maximum value of P, (b) the maximum stress in each
portion of the rod, (c) the permanent deflection of C after the load is removed.
PROBLEM 2.107 Rod AB consists of two cylindrical portions AC and BC, each with a
cross-sectional area of 1750 mm2. Portion AC is made of a mild steel with E 200 GPa
and Y 250 MPa, and portion CB is made of a high-strength steel with E 200 GPa
and Y 345 MPa. A load P is applied at C as shown. Assuming both steels to be
elastoplastic, determine (a) the maximum deflection of C if P is gradually increased
from zero to 975 kN and then reduced back to zero, (b) the maximum stress in each
portion of the rod, (c) the permanent deflection of C.
SOLUTION
Displacement at C is m 0.30 mm. The corresponding strains are
AC
m
LAC
CB
m
LCB
0.30 mm
1.5789 103
190 mm
0.30 mm
1.5789 103
190 mm
Strains at initial yielding:
Y, AC
Y, CB
(a)
Y, AC
E
Y, BC
E
250 106
1.25 103
(yielding)
200 109
345 106
1.725 103
(elastic)
200 109
Forces: FAC A Y (1750 106 )(250 106 ) 437.5 103 N
FCB EA CB (200 109 )(1750 106 )(1.5789 103 ) 552.6 103 N
For equilibrium of element at C, FAC FCB P 0
P FAC FCD 437.5 103 552.6 103 990.1 103 N
(b)
Stresses: AC : AC Y, AC
CB : CB
990 kN
250 MPa
FCB
552.6 103
316 106 Pa
6
A
1750 10
316 MPa
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212
PROBLEM 2.108 (Continued)
(c) Deflection and forces for unloading.
LAC
PAC
P L
L
PAC
AC PAC
CB CB PCB
EA
EA
LAB
PCB
2 PAC
990.1 103 N PAC
495.05 103 N
P PAC
(495.05 103 )(0.190)
0.26874 103 m 0.26874 mm
(200 109 )(1750 106 )
p m 0.30 mm 0.26874 mm
0.0313mm
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213
PROBLEM 2.109
E
D
2m
C
B
A
Each cable has a cross-sectional area of 100 mm2 and is made of an
elastoplastic material for which Y 345 MPa and E 200 GPa. A force
Q is applied at C to the rigid bar ABC and is gradually increased from 0 to
50 kN and then reduced to zero. Knowing that the cables were initially taut,
determine (a) the maximum stress that occurs in cable BD, (b) the
maximum deflection of point C, (c) the final displacement of point C.
(Hint: In part c, cable CE is not taut.)
Q
1m
1m
SOLUTION
Elongation constraints for taut cables.
Let rotation angle of rigid bar ABC.
BD
BD
LAB
CE
LAC
LAB
1
CE CE
2
LAC
(1)
Equilibrium of bar ABC.
M A 0 : LAB FBD LAC FCE LAC Q 0
Q FCE
LAB
1
FBD FCE FBD
LAC
2
Assume cable CE is yielded.
FCE A Y (100 106 )(345 106 ) 34.5 103 N
From (2),
FBD 2(Q FCE ) (2)(50 103 34.5 103 ) 31.0 103 N
(2)
Since FBD < A Y 34.5 103 N, cable BD is elastic when Q 50 kN.
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214
PROBLEM 2.109 (Continued)
(a)
CE Y 345 MPa
Maximum stresses.
BD
(b)
FBD 31.0 103
310 106 Pa
6
A
100 10
BD 310 MPa
Maximum of deflection of point C.
BD
From (1),
FBD LBD
(31.0 103 )(2)
3.1 103 m
EA
(200 109 )(100 106 )
C CE 2 BD 6.2 103 m
Permanent elongation of cable CE: ( CE ) p ( CE )
6.20 mm
Y LCE
E
( CE ) P ( CE )max
Y LCE
E
(345 106 )(2)
6.20 103
2.75 103 m
200 109
(c)
Unloading. Cable CE is slack ( FCE 0) at Q 0.
From (2),
FBD 2(Q FCE ) 2(0 0) 0
Since cable BD remained elastic, BD
FBD LBD
0.
EA
0
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215
PROBLEM 2.110
E
D
2m
C
B
A
Q
1m
1m
Solve Prob. 2.109, assuming that the cables are replaced by rods of the
same cross-sectional area and material. Further assume that the rods are
braced so that they can carry compressive forces.
PROBLEM 2.109 Each cable has a cross-sectional area of 100 mm2
and is made of an elastoplastic material for which Y 345 MPa and
E 200 GPa. A force Q is applied at C to the rigid bar ABC and is
gradually increased from 0 to 50 kN and then reduced to zero. Knowing
that the cables were initially taut, determine (a) the maximum stress
that occurs in cable BD, (b) the maximum deflection of point C, (c) the
final displacement of point C. (Hint: In part c, cable CE is not taut.)
SOLUTION
Elongation constraints.
Let rotation angle of rigid bar ABC.
BD
BC
LAB
CE
LAC
LAB
1
CE CE
2
LAC
(1)
Equilibrium of bar ABC.
M A 0: LAB FBD LAC FCE LAC Q 0
Q FCE
LAB
1
FBD FCE FBD
LAC
2
(2)
Assume cable CE is yielded. FCE A Y (100 106 )(345 106 ) 34.5 103 N
From (2),
FBD 2(Q FCE ) (2)(50 103 34.5 103 ) 31.0 103 N
Since FBD A Y 34.5 103 N, cable BD is elastic when Q 50 kN.
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216
PROBLEM 2.110 (Continued)
(a)
CE Y 345 MPa
Maximum stresses.
BD
(b)
FBD 31.0 103
310 106 Pa
6
A
100 10
BD 310 MPa
Maximum of deflection of point C.
BD
FBD LBD
(31.0 103 )(2)
3.1 103 m
EA
(200 109 )(100 106 )
C CE 2 BD 6.2 103 m
From (1),
6.20 mm
C
Unloading. Q 50 103 N, CE
12 C
BD
From (1),
Elastic FBD
FCE
(200 109 )(100 106 )( 12 C )
EA BD
5 106 C
LBD
2
EA CE
(200 109 )(100 106 )( C )
10 106 C
LCE
2
From (2),
12 FBD
12.5 106 C
Q FCE
Equating expressions for Q,
12.5 106 C 50 103
C 4 103 m
(c)
Final displacement.
C ( C ) m C 6.2 103 4 103 2.2 103 m
2.20 mm
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217
PROBLEM 2.111
P'
3
16
1
2
in.
3
16
in.
in.
14 in.
2.0 in.
Two tempered-steel bars, each 163 in. thick, are bonded to a 12 -in. mild-steel
bar. This composite bar is subjected as shown to a centric axial load of
magnitude P. Both steels are elastoplastic with E 29 106 psi and with
yield strengths equal to 100 ksi and 50 ksi, respectively, for the tempered
and mild steel. The load P is gradually increased from zero until the
deformation of the bar reaches a maximum value m 0.04 in. and then
decreased back to zero. Determine (a) the maximum value of P, (b) the
maximum stress in the tempered-steel bars, (c) the permanent set after the
load is removed.
P
SOLUTION
1
For the mild steel, A1 (2) 1.00 in 2
2
Y 1
L Y 1 (14)(50 103 )
0.024138 in.
E
29 106
3
For the tempered steel, A2 2 (2) 0.75 in 2
16
Y 2
L Y 2 (14)(100 103 )
0.048276 in.
E
29 103
Total area: A A1 A2 1.75 in 2
Y 1 m Y 2 . The mild steel yields. Tempered steel is elastic.
(a)
Forces: P1 A1 Y 1 (1.00)(50 103 ) 50 103 lb
P2
EA2 m (29 103 )(0.75)(0.04)
62.14 103 lb
14
L
P P1 P2 112.14 103 lb 112.1 kips
(b)
Stresses: 1
2
Unloading:
(c)
P 112.1 kips
P1
Y 1 50 103 psi 50 ksi
A1
P2 62.14 103
82.86 103 psi 82.86 ksi
A2
0.75
82.86 ksi
PL (112.14 103 )(14)
0.03094 in.
EA
(29 106 )(1.75)
Permanent set: p m 0.04 0.03094 0.00906 in.
0.00906 in.
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218
PROBLEM 2.112
P'
3
16
1
2
in.
3
16
For the composite bar of Prob. 2.111, if P is gradually increased from zero
to 98 kips and then decreased back to zero, determine (a) the maximum
deformation of the bar, (b) the maximum stress in the tempered-steel bars,
(c) the permanent set after the load is removed.
in.
in.
14 in.
PROBLEM 2.111 Two tempered-steel bars, each 163 in. thick, are bonded to
a 12 -in. mild-steel bar. This composite bar is subjected as shown to a centric
axial load of magnitude P. Both steels are elastoplastic with E 29 106 psi
and with yield strengths equal to 100 ksi and 50 ksi, respectively, for the
tempered and mild steel. The load P is gradually increased from zero until
the deformation of the bar reaches a maximum value m 0.04 in. and
then decreased back to zero. Determine (a) the maximum value of P, (b) the
maximum stress in the tempered-steel bars, (c) the permanent set after the
load is removed.
2.0 in.
P
SOLUTION
Areas:
Mild steel:
1
A1 (2) 1.00 in 2
2
Tempered steel:
3
A2 2 (2) 0.75 in 2
16
A A1 A2 1.75 in 2
Total:
Total force to yield the mild steel:
Y1
PY
PY A Y 1 (1.75)(50 103 ) 87.50 103 lb
A
P PY , therefore, mild steel yields.
Let P1 force carried by mild steel.
P2 force carried by tempered steel.
P1 A1 1 (1.00)(50 103 ) 50 103 lb
P1 P2 P, P2 P P1 98 103 50 103 48 103 lb
(a)
m
P2 L
(48 103 )(14)
EA2 (29 106 )(0.75)
(b)
2
P2 48 103
64 103 psi
A2
0.75
Unloading:
(c)
0.0309 in.
64.0 ksi
PL
(98 103 )(14)
0.02703 in.
EA (29 106 )(1.75)
P m 0.03090 0.02703 0.003870 in.
0.00387 in.
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219
PROBLEM 2.113
D
E
1.7 m
1m
C
A
B
The rigid bar ABC is supported by two links, AD and BE, of uniform
37.5 6-mm rectangular cross section and made of a mild steel that is
assumed to be elastoplastic with E 200 GPa and Y 250 MPa.
The magnitude of the force Q applied at B is gradually increased
from zero to 260 kN. Knowing that a 0.640 m, determine (a) the
value of the normal stress in each link, (b) the maximum deflection
of point B.
a
Q
2.64 m
SOLUTION
M C 0 : 0.640(Q PBE ) 2.64 PAD 0
Statics:
A 2.64 , B a 0.640
Deformation:
Elastic analysis:
A (37.5)(6) 225 mm 2 225 106 m 2
PAD
AD
PBE
EA
(200 109 )(225 106 )
A
A 26.47 106 A
LAD
1.7
(26.47 106 )(2.64 ) 69.88 106
P
AD 310.6 109
A
EA
(200 109 )(225 106 )
B
B 45 106 B
LBE
1.0
(45 106 )(0.640 ) 28.80 106
BE
PBE
128 109
A
2.64
PAD PBE 4.125 PAD
0.640
[28.80 106 (4.125)(69.88 106 )] 317.06 106
From statics, Q PBE
Y at yielding of link AD:
AD Y 250 106 310.6 109
Y 804.89 106
QY (317.06 106 )(804.89 106 ) 255.2 103 N
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220
PROBLEM 2.113 (Continued)
(a)
AD 250 MPa
Since Q 260 103 QY , link AD yields.
PAD A Y (225 106 )(250 106 ) 56.25 103 N
From statics, PBE Q 4.125 PAD 260 103 (4.125)(56.25 103 )
PBE 27.97 103 N
BE
(b)
B
PBE 27.97 103
124.3 106 Pa
A
225 106
PBE LBE
(27.97 103 )(1.0)
621.53 106 m
EA
(200 109 )(225 106 )
BE 124.3 MPa
B 0.622 mm
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221
PROBLEM 2.114
D
Solve Prob. 2.113, knowing that a 1.76 m and that the magnitude of
the force Q applied at B is gradually increased from zero to 135 kN.
E
1.7 m
1m
C
A
B
a
Q
2.64 m
PROBLEM 2.113 The rigid bar ABC is supported by two links, AD
and BE, of uniform 37.5 6-mm rectangular cross section and made
of a mild steel that is assumed to be elastoplastic with E 200 GPa
and Y 250 MPa. The magnitude of the force Q applied at B is
gradually increased from zero to 260 kN. Knowing that a 0.640 m,
determine (a) the value of the normal stress in each link, (b) the
maximum deflection of point B.
SOLUTION
M C 0 : 1.76(Q PBE ) 2.64 PAD 0
Statics:
A 2.64 , B 1.76
Deformation:
Elastic Analysis:
A (37.5)(6) 225 mm 2 225 106 m 2
PAD
AD
PBE
EA
(200 109 )(225 106 )
A
A 26.47 106 A
LAD
1.7
(26.47 106 )(2.64 ) 69.88 106
P
AD 310.6 109
A
EA
(200 109 )(225 106 )
B
B 45 106 B
LBE
1.0
(45 106 )(1.76 ) 79.2 106
PBE
352 109
A
2.64
PAD PBE 1.500 PAD
From statics, Q PBE
1.76
BE
[73.8 106 (1.500)(69.88 106 )] 178.62 106
Y at yielding of link BE:
BE Y 250 106 352 109Y
Y 710.23 106
QY (178.62 106 )(710.23 106 ) 126.86 103 N
(a)
BE Y 250 MPa
Since Q 135 103 N QY , link BE yields.
PBE A Y (225 106 )(250 106 ) 56.25 103 N
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222
PROBLEM 2.114 (Continued)
From statics, PAD
1
(Q PBE ) 52.5 103 N
1.500
AD
From elastic analysis of AD,
(b)
PAD 52.5 103
233.3 106
6
A
225 10
PAD
69.88 106
AD 233 MPa
751.29 103 rad
B 1.76 1.322 103 m
B 1.322 mm
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223
PROBLEM 2.115
Solve Prob. 2.113, assuming that the magnitude of the force Q
applied at B is gradually increased from zero to 260 kN and then
decreased back to zero. Knowing that a 0.640 m, determine (a) the
residual stress in each link, (b) the final deflection of point B.
Assume that the links are braced so that they can carry compressive
forces without buckling.
D
E
1.7 m
1m
C
A
B
a
Q
2.64 m
PROBLEM 2.113 The rigid bar ABC is supported by two links, AD
and BE, of uniform 37.5 6-mm rectangular cross section and made
of a mild steel that is assumed to be elastoplastic with E 200 GPa
and Y 250 MPa. The magnitude of the force Q applied at B is
gradually increased from zero to 260 kN. Knowing that
a 0.640 m, determine (a) the value of the normal stress in each
link, (b) the maximum deflection of point B.
SOLUTION
See solution to Problem 2.113 for the normal stresses in each link and the deflection of Point B after loading.
AD 250 106 Pa
BE 124.3 106 Pa
B 621.53 106 m
The elastic analysis given in the solution to Problem 2.113 applies to the unloading.
Q 317.06 106
Q
Q
260 103
820.03 106
317.06 106 317.06 106
AD 310.6 109 (310.6 109 )(820.03 106 ) 254.70 106 Pa
128 109 (128 109 )(820.03 106 ) 104.96 106 Pa
BE
B 0.640 524.82 106 m
(a)
(b)
Residual stresses.
250 106 254.70 106 4.70 106 Pa
AD , res AD AD
4.70 MPa
124.3 106 104.96 106 19.34 106 Pa
BE , res BE BE
19.34 MPa
B , P B B 621.53 106 524.82 106 96.71 106 m
0.0967 mm
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224
PROBLEM 2.116
A
B
L
A uniform steel rod of cross-sectional area A is attached to rigid supports and is
unstressed at a temperature of 45F. The steel is assumed to be elastoplastic
with Y 36 ksi and E 29 106 psi. Knowing that 6.5 106 / F,
determine the stress in the bar (a) when the temperature is raised to 320F,
(b) after the temperature has returned to 45F.
SOLUTION
Let P be the compressive force in the rod.
Determine temperature change to cause yielding.
L
PL
L (T ) Y L (T )Y 0
AE
E
3
36 10
(T )Y Y
190.98F
E (29 106 )(6.5 106 )
But T 320 45 275F (TY )
(a)
Y 36.0 ksi
Yielding occurs.
Cooling:
(T) 275F
P T
PL
L (T ) 0
AE
P
E (T )
A
(29 106 )(6.5 106 )(275) 51.8375 103 psi
(b) Residual stress:
res Y 36 103 51.8375 103 15.84 10 psi
15.84 ksi
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225
PROBLEM 2.117
A 500 mm2
A 300 mm2
A
C
150 mm
The steel rod ABC is attached to rigid supports and is unstressed
at a temperature of 25C. The steel is assumed elastoplastic, with
E 200 GPa and y 250 MPa. The temperature of both
portions of the rod is then raised to 150C . Knowing that
11.7 106 / C, determine (a) the stress in both portions of the
rod, (b) the deflection of point C.
B
250 mm
SOLUTION
AAC 500 106 m 2
LAC 0.150 m
ACB 300 106 m 2
LCB 0.250 m
P T 0
Constraint:
Determine T to cause yielding in portion CB.
PLAC PLCB
LAB (T )
EAAC EACB
T
P
LAB E
LAC LCB
AAC ACB
At yielding, P PY ACB Y (300 106 )(2.50 106 ) 75 103 N
(T )Y
PY
LAB E
LAC LCB
AAC ACB
75 103
0.250
0.150
9
6
6
(0.400)(200 10 )(11.7 10 ) 500 10
300 106
150C 25C 125C (T )Y
Actual T:
Yielding occurs. For T (T )Y ,
(a)
(b)
90.812C
P PY 75 103 N
AC
PY
75 103
150 106 Pa
6
AAC
500 10
CB
PY
Y
ACB
AC 150.0 MPa
CB 250 MPa
For T (T )Y , portion AC remains elastic.
C /A
PY LAC
LAC (T )
EAAC
(75 103 )(0.150)
(0.150)(11.7 106 )(125) 106.9 106 m
9
6
(200 10 )(500 10 )
Since Point A is stationary, C C /A 106.9 106 m
C 0.1069 mm
\
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226
PROBLEM 2.118
A 500 mm2
A
Solve Prob. 2.117, assuming that the temperature of the rod is raised to
150C and then returned to 25C.
A 300 mm2
C
B
150 mm
PROBLEM 2.117 The steel rod ABC is attached to rigid supports and is
unstressed at a temperature of 25C. The steel is assumed elastoplastic,
with E 200 GPa and Y 250 MPa. The temperature of both portions
of the rod is then raised to 150C. Knowing that 11.7 106 / C,
determine (a) the stress in both portions of the rod, (b) the deflection
of point C.
250 mm
SOLUTION
AAC 500 106 m 2
Constraint:
ACB 300 106 m 2
LAC 0.150 m
LCB 0.250 m
P T 0
Determine T to cause yielding in portion CB.
PLAC PLCB
LAB (T )
EAAC EACB
T
P
LAB E
LAC LCB
AAC ACB
At yielding, P PY ACB Y (300 106 )(250 106 ) 75 103 N
LAC LCB
AAC ACB
90.812 C
(T )Y
PY
LAB E
75 103
0.250
0.150
6
6
9
(0.400)(200
10
)(11.7
10
)
500
10
300
106
Actual T : 150C 25C 125C (T )Y
P PY 75 103 N
Yielding occurs. For T (T )Y ,
Cooling:
(T ) 125C P
ELAB (T )
LAC
AAC
LCB
ACB
(200 109 )(0.400)(11.7 106 )(125)
103.235 103 N
0.150
0.250
6
6
500 10
300 10
Residual force: Pres P PY 103.235 103 75 103 28.235 103 N (tension)
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227
PROBLEM 2.118 (Continued)
(a)
(b)
Residual stresses.
AC
Pres 28.235 103
AAC
500 106
AC 56.5 MPa
CB
Pres 28.235 103
ACB
300 106
CB 9.41 MPa
Permanent deflection of point C. C
Pres LAC
EAAC
C 0.0424 mm
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228
PROBLEM 2.119
P'
3
16
1
2
in.
3
16
in.
in.
For the composite bar of Prob. 2.111, determine the residual stresses in
the tempered-steel bars if P is gradually increased from zero to
98 kips and then decreased back to zero.
PROBLEM 2.111 Two tempered-steel bars, each 163 in. thick, are
bonded to a 12 -in. mild-steel bar. This composite bar is subjected as
shown to a centric axial load of magnitude P. Both steels are
elastoplastic with E 29 106 psi and with yield strengths equal to
100 ksi and 50 ksi, respectively, for the tempered and mild steel. The
load P is gradually increased from zero until the deformation of the bar
reaches a maximum value m 0.04 in. and then decreased back to
zero. Determine (a) the maximum value of P, (b) the maximum stress
in the tempered-steel bars, (c) the permanent set after the load is
removed.
14 in.
2.0 in.
P
SOLUTION
Areas.
Mild steel:
1
A1 (2) 1.00 in 2
2
Tempered steel:
3
A2 (2) (2) 0.75 in 2
16
A A1 A2 1.75 in 2
Total:
Total force to yield the mild steel: Y 1
PY
A
PY A Y 1 (1.75)(50 103 ) 87.50 103 lb
P > PY; therefore, mild steel yields.
P1 force carried by mild steel
Let
P2 force carried by tempered steel
P1 A1 Y 1 (1.00)(50 103 ) 50 103 lb
P1 P2 P, P2 P P1 98 103 50 103 48 103 lb
2
Unloading.
P2 48 103
64 103 psi
A2
0.75
P 98 103
56 103 psi
A
1.75
Residual stresses.
Mild steel:
1,res 1 50 103 56 103 6 103 psi 6 ksi
Tempered steel:
2,res 2 1 64 103 56 103 8 103 psi
8.00 ksi
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229
PROBLEM 2.120
P'
3
16
1
2
in.
3
16
in.
in.
14 in.
For the composite bar in Prob. 2.111, determine the residual stresses in
the tempered-steel bars if P is gradually increased from zero until the
deformation of the bar reaches a maximum value m 0.04 in. and is
then decreased back to zero.
PROBLEM 2.111 Two tempered-steel bars, each 163 in. thick, are bonded
to a 12 -in. mild-steel bar. This composite bar is subjected as shown to a
centric axial load of magnitude P. Both steels are elastoplastic
with E 29 106 psi and with yield strengths equal to 100 ksi and 50 ksi,
respectively, for the tempered and mild steel. The load P is gradually
increased from zero until the deformation of the bar reaches a maximum
value m 0.04 in. and then decreased back to zero. Determine (a) the
maximum value of P, (b) the maximum stress in the tempered-steel bars,
(c) the permanent set after the load is removed.
2.0 in.
P
SOLUTION
L
(14)(50 103 )
1
A1 (2) 1.00 in 2 Y 1 Y 1
0.024138 in.
E
29 106
2
For the mild steel,
L
(14)(100 103 )
3
For the tempered steel, A2 2 (2) 0.75 in 2 Y 2 Y 2
0.048276 in.
E
29 106
16
A A1 A2 1.75 in 2
Total area:
Y 1 m Y 2
The mild steel yields. Tempered steel is elastic.
P1 A1Y 1 (1.00)(50 103 ) 50 103 lb
Forces:
P2
Stresses:
1
EA2 m (29 106 )(0.75)(0.04)
62.14 103 lb
L
14
P1
P
62.14 103
Y 1 50 103 psi 2 2
82.86 103 psi
A1
A2
0.75
Unloading:
P 112.14
64.08 103 psi
A
1.75
Residual stresses. 1,res 1 50 103 64.08 103 14.08 103 psi 14.08 ksi
2,res 2 82.86 103 64.08 103 18.78 103 psi 18.78 ksi
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230
PROBLEM 2.121
Narrow bars of aluminum are bonded to the two sides of a thick steel plate as
shown. Initially, at T1 70F, all stresses are zero. Knowing that the
temperature will be slowly raised to T2 and then reduced to T1, determine
(a) the highest temperature T2 that does not result in residual stresses, (b) the
temperature T2 that will result in a residual stress in the aluminum equal to
58 ksi. Assume a 12.8 106 / F for the aluminum and s 6.5 106 / F
for the steel. Further assume that the aluminum is elastoplastic, with
E 10.9 106 psi and Y 58 ksi. (Hint: Neglect the small stresses in the
plate.)
SOLUTION
Determine temperature change to cause yielding.
PL
L a (T )Y L s (T )Y
EA
P
E ( a s )(T )Y Y
A
Y
58 103
(T )Y
844.62F
E ( a s ) (10.9 106 )(12.8 6.5)(106 )
(a)
T2Y T1 (T )Y 70 844.62 915F
915F
After yielding,
Y L
E
L a (T ) L s (T )
Cooling:
PL
L a (T ) L s (T )
AE
The residual stress is
res Y
Set
res Y
Y Y E ( a s )(T )
T
(b)
P
Y E ( a s )(T )
A
2 Y
(2)(58 103 )
1689F
E ( a s ) (10.9 106 )(12.8 6.5)(106 )
T2 T1 T 70 1689 1759F
1759F
If T2 1759F, the aluminum bar will most likely yield in compression.
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231
A
C
B
F
a ⫽ 120 mm
440 mm
PROBLEM 2.122
Bar AB has a cross-sectional area of 1200 mm2 and is made of a steel
that is assumed to be elastoplastic with E 200 GPa and Y 250 MPa.
Knowing that the force F increases from 0 to 520 kN and then
decreases to zero, determine (a) the permanent deflection of point C,
(b) the residual stress in the bar.
SOLUTION
A 1200 mm 2 1200 106 m 2
Force to yield portion AC:
PAC A Y (1200 106 )(250 106 )
300 103 N
For equilibrium, F PCB PAC 0.
PCB PAC F 300 103 520 103
220 103 N
C
PCB LCB (220 103 )(0.440 0.120)
EA
(200 109 )(1200 106 )
0.29333 103 m
PCB
220 103
A 1200 106
183.333 106 Pa
CB
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232
PROBLEM 2.122 (Continued)
Unloading:
C
LAC
) LCB
PAC
P L
( F PAC
CB CB
EA
EA
EA
L FL
L
AC BC CB
PAC
EA
EA
EA
PAC
FLCB
(520 103 )(0.440 0.120)
378.18 103 N
0.440
LAC LCB
PAC
F 378.18 103 520 103 141.820 103 N
PCB
PAC
378.18 103
315.150 106 Pa
6
A
1200 10
P
141.820 103
BC
118.183 106 Pa
BC
6
A
1200 10
(378.18 103 )(0.120)
0.189090 103 m
C
(200 109 )(1200 106 )
AC
(a)
C , p C C 0.29333 10 3 0.189090 10 3 0.104240 10 3 m
0.1042 mm
(b)
AC , res Y AC
250 106 315.150 106 65.150 106 Pa
65.2 MPa
CB, res CB CB
183.333 106 118.183 106 65.150 106 Pa
65.2 MPa
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233
PROBLEM 2.123
A
C
B
F
a ⫽ 120 mm
440 mm
Solve Prob. 2.122, assuming that a 180 mm.
PROBLEM 2.122 Bar AB has a cross-sectional area of 1200 mm2 and
is made of a steel that is assumed to be elastoplastic with E 200 GPa
and Y 250 MPa. Knowing that the force F increases from 0 to
520 kN and then decreases to zero, determine (a) the permanent
deflection of point C, (b) the residual stress in the bar.
SOLUTION
A 1200 mm 2 1200 106 m 2
Force to yield portion AC:
PAC A Y (1200 106 )(250 106 )
300 103 N
For equilibrium, F PCB PAC 0.
PCB PAC F 300 103 520 103
220 103 N
C
PCB LCB (220 103 )(0.440 0.180)
EA
(200 109 )(1200 106 )
0.23833 103 m
PCB
220 103
A
1200 106
183.333 106 Pa
CB
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234
PROBLEM 2.123 (Continued)
Unloading:
LAC
) LCB
PAC
P L
( F PAC
CB CB
EA
EA
EA
L FL
L
AC BC CB
PAC
EA
EA
EA
C
PAC
FLCB
(520 103 )(0.440 0.180)
307.27 103 N
LAC LCB
0.440
PAC
F 307.27 103 520 103 212.73 103 N
PCB
C
(307.27 103 )(0.180)
0.23045 103 m
6
9
(200 10 )(1200 10 )
PAC
307.27 103
256.058 106 Pa
A
1200 106
P
212.73 103
CB
177.275 106 Pa
CB
A
1200 106
AC
(a)
C , p C C 0.23833 10 3 0.23045 10 3 0.00788 10 3 m
0.00788 mm
(b)
AC ,res AC AC
250 106 256.058 106 6.0580 106 Pa
6.06 MPa
CB,res CB CB
183.333 106 177.275 106 6.0580 106 Pa
6.06 MPa
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235
PROBLEM 2.124
l
␦
The uniform wire ABC, of unstretched length 2l, is attached to the
supports shown and a vertical load P is applied at the midpoint B.
Denoting by A the cross-sectional area of the wire and by E the
modulus of elasticity, show that, for l , the deflection at the
midpoint B is
l
A
C
B
P
l3
P
AE
SOLUTION
Use approximation.
sin tan
Statics:
l
FY 0 : 2 PAB sin P 0
PAB
Elongation:
P
Pl
2sin 2
AB
PAB l
Pl 2
AE 2 AE
Deflection:
From the right triangle,
(l AB ) 2 l 2 2
2
2 l 2 2l AB AB
l2
1 AB
2l AB 1
2 l
3
2l AB
Pl 3
AE
Pl 3
P
l3
AE
AE
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236
28 kips
PROBLEM 2.125
28 kips
The aluminum rod ABC ( E 10.1 106 psi), which consists of two
cylindrical portions AB and BC, is to be replaced with a cylindrical steel
rod DE ( E 29 106 psi) of the same overall length. Determine the
minimum required diameter d of the steel rod if its vertical deformation is
not to exceed the deformation of the aluminum rod under the same load
and if the allowable stress in the steel rod is not to exceed 24 ksi.
D
A
1.5 in.
12 in.
B
2.25 in.
d
18 in.
C
E
SOLUTION
Deformation of aluminum rod.
A
PLAB PLBC
AAB E ABC E
P LAB LBC
E AAB ABC
28 103 12
10.1 106 4 (1.5)2
0.031376 in.
Steel rod.
18
(2.25) 2
4
0.031376 in.
PL
PL
(28 103 )(30)
A
0.92317 in 2
EA
E (29 106 )(0.031376)
P
A
A
P
28 103
1.16667 in 2
3
24 10
Required area is the larger value.
A 1.16667 in 2
Diameter:
d
4A
(4)(1.16667)
d 1.219 in.
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237
PROBLEM 2.126
C
Two solid cylindrical rods are joined at B and loaded as shown. Rod AB is
made of steel ( E 29 106 psi), and rod BC of brass ( E 15 106 psi).
Determine (a) the total deformation of the composite rod ABC, (b) the
deflection of point B.
3 in.
30 in.
B
30 kips
30 kips
2 in.
40 in.
A
P ⫽ 40 kips
SOLUTION
Portion AB:
PAB 40 103 lb
LAB 40 in.
d 2 in.
E AB
AB
Portion BC:
d2
(2)2 3.1416 in 2
4
4
29 106 psi
AAB
PAB LAB
(40 103 )(40)
17.5619 103 in.
E AB AAB (29 106 )(3.1416)
PBC 20 103 lb
LBC 30 in.
d 3 in.
EBC
BC
d2
(3) 2 7.0686 in 2
4
4
15 106 psi
ABC
PBC LBC
(20 103 )(30)
5.6588 103 in.
EBC ABC (15 106 )(7.0686)
(a)
AB BC 17.5619 106 5.6588 106
11.90 103 in.
(b)
B BC
B 5.66 103 in.
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238
Brass strip:
E 5 105 GPa
a 5 20 3 1026/8C
PROBLEM 2.127
100 kg
A
40 mm
3 mm
20 mm
The brass strip AB has been attached to a fixed
support at A and rests on a rough support at B.
Knowing that the coefficient of friction is 0.60
between the strip and the support at B, determine
the decrease in temperature for which slipping
will impend.
B
SOLUTION
Brass strip:
E 105 GPa
20 106 / C
Fy 0 : N W 0
N W
Fx 0 : P N 0
Data:
PL
L (T ) 0
EA
P W mg
T
mg
P
EA EA
0.60
A (20)(3) 60 mm 2 60 106 m 2
m 100 kg
g 9.81 m/s 2
E 105 109 Pa
T
(0.60)(100)(9.81)
(105 109 )(60 106 )(20 106 )
T 4.67C
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239
P'
112 -in. diameter
A
1-in. diameter
B
112 -in. diameter
C
2 in.
D
3 in.
P
PROBLEM 2.128
The specimen shown is made from a 1-in.-diameter cylindrical steel rod
with two 1.5-in.-outer-diameter sleeves bonded to the rod as shown.
Knowing that E 29 106 psi, determine (a) the load P so that the
total deformation is 0.002 in., (b) the corresponding deformation of the
central portion BC.
2 in.
SOLUTION
(a)
Pi Li P Li
Ai Ei E Ai
1
L
P E i Ai di2
4
Ai
L, in.
d, in.
A, in2
L/A, in1
AB
2
1.5
1.7671
1.1318
BC
3
1.0
0.7854
3.8197
CD
2
1.5
1.7671
1.1318
6.083
P (29 106 )(0.002)(6.083) 1 9.353 103 lb
(b)
BC
PLBC P LBC 9.535 103
(3.8197)
ABC E E ABC
29 106
sum
P 9.53 kips
1.254 103 in.
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240
PROBLEM 2.129
250 mm
Each of the four vertical links connecting the
two rigid horizontal members is made of
aluminum ( E 70 GPa) and has a uniform
rectangular cross section of 10 40 mm. For
the loading shown, determine the deflection
of (a) point E, (b) point F, (c) point G.
400 mm
A
250 mm
B
40 mm
C
D
E
300 mm
F
G
24 kN
SOLUTION
Statics. Free body EFG:
M F 0 : (400)(2 FBE ) (250)(24) 0
FBE 7.5 kN 7.5 103 N
M E 0 : (400)(2 FCF ) (650)(24) 0
FCF 19.5 kN 19.5 103 N
Area of one link:
A (10)(40) 400 mm 2
400 106 m 2
Length: L 300 mm 0.300 m
Deformations.
BE
FBE L
(7.5 103 )(0.300)
80.357 106 m
6
9
EA
(70 10 )(400 10 )
CF
FCF L
(19.5 103 )(0.300)
208.93 106 m
EA
(70 109 )(400 106 )
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241
PROBLEM 2.129 (Continued)
(a)
Deflection of Point E.
E | BF |
E 80.4 m
(b)
Deflection of Point F.
F CF
F 209 m
Geometry change.
Let be the small change in slope angle.
(c)
E F
LEF
Deflection of Point G.
80.357 106 208.93 106
723.22 106 radians
0.400
G F LFG
G F LFG 208.93 106 (0.250)(723.22 106 )
389.73 106 m
G 390 m
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242
PROBLEM 2.130
P
A 4-ft concrete post is reinforced with four steel bars, each with a 34 -in. diameter.
Knowing that Es 29 106 psi and Ec 3.6 106 psi, determine the normal
stresses in the steel and in the concrete when a 150-kip axial centric force P is
applied to the post.
4 ft
8 in.
8 in.
SOLUTION
3 2
As 4 1.76715 in 2
4 4
Ac 82 As 62.233 in 2
s
Ps L
Ps (48)
0.93663 106 Ps
As Es (1.76715)(29 106 )
c
Pc L
Pc (48)
0.21425 106 Pc
Ac Ec (62.233)(3.6 106 )
But s c : 0.93663 106 Ps 0.21425 106 Pc
Ps 0.22875Pc
Also,
Substituting (1) into (2),
(1)
Ps Pc P 150 kips
(2)
1.22875Pc 150 kips
Pc 122.075 kips
From (1),
Ps 0.22875(122.075) 27.925 kips
s
Ps
27.925
1.76715
As
s 15.80 ksi
c
Pc
122.075
62.233
Ac
c 1.962 ksi
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243
PROBLEM 2.131
A
The steel rods BE and CD each have a 16-mm diameter
( E 200 GPa); the ends of the rods are single-threaded with a
pitch of 2.5 mm. Knowing that after being snugly fitted, the nut at
C is tightened one full turn, determine (a) the tension in rod CD,
(b) the deflection of point C of the rigid member ABC.
150 mm
B
100 mm
D
E
C
2m
3m
SOLUTION
Let be the rotation of bar ABC as shown.
Then
B 0.15
But
C turn
PCD
C 0.25
PCD LCD
ECD ACD
ECD ACD
( turn C )
LCD
(200 109 Pa) 4 (0.016 m) 2
2m
(0.0025 m 0.25 )
50.265 103 5.0265 106
B
PBE
PBE LBE
EBE ABE
or
PBE
EBE ABE
B
LBE
(200 109 Pa) 4 (0.016 m)2
3m
(0.15 )
2.0106 106
From free body of member ABC:
M A 0 : 0.15 PBE 0.25 PCD 0
0.15(2.0106 106 ) 0.25(50.265 103 5.0265 106 ) 0
8.0645 103 rad
(a)
PCD 50.265 103 5.0265 106 (8.0645 103 )
9.7288 103 N
(b)
PCD 9.73 kN
C 0.25 0.25(8.0645 103 )
2.0161 103 m
C 2.02 mm
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244
PROBLEM 2.132
8 in.
Aluminum shell
1.25 in.
0.75 in.
Steel
core
The assembly shown consists of an aluminum shell ( Ea 10.6 106 psi,
a 12.9 106/°F) fully bonded to a steel core ( Es 29 106 psi,
s 6.5 106/°F) and is unstressed. Determine (a) the largest allowable
change in temperature if the stress in the aluminum shell is not to exceed
6 ksi, (b) the corresponding change in length of the assembly.
SOLUTION
Since a s , the shell is in compression for a positive temperature rise.
a 6 ksi 6 103 psi
Let
Aa
As
d
4
4
2
o
di2
d2
4
4
(1.252 0.752 ) 0.78540 in 2
(0.75) 2 0.44179 in 2
P a Aa s As
where P is the tensile force in the steel core.
s
( a s )(T )
(6.4 106 )(T )
(a)
T 145.91F
(b)
a Aa
s
Es
s
Es
As
(6 103 )(0.78540)
10.667 103 psi
0.44179
s (T )
a
Ea
a (T )
a
Ea
10.667 103
6 103
0.93385 103
6
6
29 10
10.6 10
T 145.9F
10.667 103
(6.5 106 )(145.91) 1.3163 103
6
29 10
or
6 103
(12.9 106 )(145.91) 1.3163 103
6
10.6 10
L (8.0)(1.3163 103 ) 0.01053 in.
0.01053 in.
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245
PROBLEM 2.133
3.5 in.
P
5.5 in.
The plastic block shown is bonded to a fixed base and to a horizontal
rigid plate to which a force P is applied. Knowing that for the
plastic used G 55 ksi, determine the deflection of the plate
when P 9 kips.
2.2 in.
SOLUTION
Consider the plastic block. The shearing force carried is P 9 103 lb
The area is A (3.5)(5.5) 19.25 in 2
Shearing stress:
Shearing strain:
But
P 9 103
467.52 psi
A
19.25
G
h
467.52
0.0085006
55 103
h (2.2)(0.0085006)
0.01870 in.
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246
PROBLEM 2.134
P
150
75
15
300
60
r56
150
The aluminum test specimen shown is subjected to two equal and opposite
centric axial forces of magnitude P. (a) Knowing that E 70 GPa
and all 200 MPa, determine the maximum allowable value of P and the
corresponding total elongation of the specimen. (b) Solve part a, assuming that
the specimen has been replaced by an aluminum bar of the same length and a
uniform 60 15-mm rectangular cross section.
75
P⬘
Dimensions in mm
SOLUTION
all 200 106 Pa E 70 109 Pa
Amin (60 mm)(15 mm) 900 mm 2 900 106 m 2
(a)
Test specimen.
D 75 mm, d 60 mm, r 6 mm
D 75
1.25
d 60
From Fig. 2.60b,
K 1.95
P
r
6
0.10
d 60
max K
P
A
A max (900 106 ) (200 106 )
92.308 103 N
K
1.95
P 92.3 kN
Wide area A* (75 mm)(15 mm) 1125 mm 2 1.125 103 m 2
Pi Li
P L 92.308 103
i
Ai Ei E Ai
70 109
0.300
0.150
0.150
1.125 103 900 106 1.125 103
7.91 106 m
(b)
0.791 mm
Uniform bar.
P A all (900 106 )(200 106 ) 180 103 N
(180 103 )(0.600)
PL
1.714 103 m
AE (900 106 )(70 109 )
P 180.0 kN
1.714 mm
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247
PROBLEM 2.135
L
B
C
B'
k
m
C'
P
P
The uniform rod BC has a cross-sectional area A and is made of a
mild steel that can be assumed to be elastoplastic with a modulus of
elasticity E and a yield strength Y . Using the block-and-spring
system shown, it is desired to simulate the deflection of end C of the
rod as the axial force P is gradually applied and removed, that is, the
deflection of points C and C should be the same for all values of P.
Denoting by the coefficient of friction between the block and the
horizontal surface, derive an expression for (a) the required mass m
of the block, (b) the required constant k of the spring.
SOLUTION
Force-deflection diagram for Point C or rod BC.
P PY A Y
For
PL
EA
PY A Y
C
Pmax
EA
C
L
P
Force-deflection diagram for Point C of block-and-spring system.
Fy 0 : N mg 0
N mg
Fx 0 : P F f 0
P Ff
If block does not move, i.e., F f N mg or
c
then
P
K
P mg ,
or
P k c
If P mg, then slip at P Fm mg occurs.
If the force P is the removed, the spring returns to its initial length.
(a)
Equating PY and Fmax,
(b)
Equating slopes,
A Y mg
k
m
EA
L
A Y
g
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248
PROBLEM 2.C1
A rod consisting of n elements, each of which is homogeneous and
of uniform cross section, is subjected to the loading shown. The length
of element i is denoted by Li , its cross-sectional area by Ai , modulus of
elasticity by Ei , and the load applied to its right end by Pi , the magnitude
Pi of this load being assumed to be positive if Pi is directed to the right
and negative otherwise. (a) Write a computer program that can be used
to determine the average normal stress in each element, the deformation
of each element, and the total deformation of the rod. (b) Use this
program to solve Probs. 2.20 and 2.126.
Element 1
Element n
P1
Pn
SOLUTION
For each element, enter
Li ,
Ai , Ei
Compute deformation
P P Pi
Update axial load
Compute for each element
i P/ Ai
i PLi / Ai Ei
Total deformation:
Update through n elements
i
Program Outputs
Problem 2.20
Element
Stress (MPa)
Deformation (mm)
1
19.0986
0.1091
2
12.7324
0.0909
Total Deformation
0.0182 mm
Problem 2.126
Element
Stress (ksi)
Deformation (in.)
1
12.7324
0.0176
2
2.8294
0.0057
Total Deformation
0.01190 in.
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249
PROBLEM 2.C2
A
Element n
Element 1
B
Pn
P2
Rod AB is horizontal with both ends fixed; it consists of n elements, each
of which is homogeneous and of uniform cross section, and is subjected to
the loading shown. The length of element i is denoted by Li , its crosssectional area by Ai , its modulus of elasticity by Ei , and the load applied
to its right end by Pi , the magnitude Pi of this load being assumed to be
positive if Pi is directed to the right and negative otherwise. (Note that
P1 0.) (a) Write a computer program which can be used to determine the
reactions at A and B, the average normal stress in each element, and the
deformation of each element. (b) Use this program to solve Probs. 2.41
and 2.42.
SOLUTION
We Consider the reaction at B redundant and release the rod at B
Compute B with RB 0
For each element, enter
Li , Ai , Ei
Update axial load
P P Pi
Compute for each element
i P/Ai
i PLi /Ai Ei
Update total deformation
B B i
Compute B due to unit load at B
Unit i 1/Ai
Unit i Li /Ai Ei
Update total unit deformation
Unit B Unit B Unit i
Superposition
B0
For total displacement at
B RB
Unit
B 0
Solving:
RB B /Unit B
Then:
RA Pi RB
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250
PROBLEM 2.C2 (Continued)
For each element
i RB Unit i
i RB Unit i
Program Outputs
Problem 2.41
RA 62.809 kN
RB 37.191 kN
Element Stress (MPa) Deformation (mm)
1
52.615
0.05011
2
3.974
0.00378
3
2.235
0.00134
4
49.982
0.04498
Problem 2.42
RA 45.479 kN
RB 54.521 kN
Element Stress (MPa) Deformation (mm)
1
77.131
0.03857
2
20.542
0.01027
3
11.555
0.01321
4
36.191
0.06204
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251
PROBLEM 2.C3
Element n
Element 1
A
␦0
B
Rod AB consists of n elements, each of which is homogeneous and of
uniform cross section. End A is fixed, while initially there is a gap 0
between end B and the fixed vertical surface on the right. The length of
element i is denoted by Li , its cross-sectional area by Ai , its modulus of
elasticity by Ei , and its coefficient of thermal expansion by i . After the
temperature of the rod has been increased by T , the gap at B is closed
and the vertical surfaces exert equal and opposite forces on the rod.
(a) Write a computer program which can be used to determine the
magnitude of the reactions at A and B, the normal stress in each element,
and the deformation of each element. (b) Use this program to solve
Probs. 2.59 and 2.60.
SOLUTION
We compute the displacements at B.
Assuming there is no support at B,
enter
Li ,
Ai , Ei , i
Enter temperature change T. Compute for each element.
i i LiT
Update total deformation.
B B i
Compute B due to unit load at B.
Unit i Li /Ai Ei
Update total unit deformation.
Unit B Unit B Unit i
Compute reactions.
From superposition,
RB ( B 0 )/Unit B
Then
RA RB
For each element,
i RB /Ai
i i LiT RB Li /Ai Ei
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252
PROBLEM 2.C3 (Continued)
Program Outputs
Problem 2.59.
R 52.279 kips
Stress (ksi)
Deformation (10 * 3 in.)
1
21.783
9.909
2
18.671
10.091
Element
Problem 2.60.
R 232.390 kN
Element
Stress (MPa)
Deformation (microm)
1
116.195
363.220
2
290.487
136.780
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253
PROBLEM 2.C4
A 1, E1, (Y)1
L
P
A 2 , E2 , ( Y)2
Plate
Bar AB has a length L and is made of two different materials of given
cross-sectional area, modulus of elasticity, and yield strength. The bar is
subjected as shown to a load P that is gradually increased from zero until
the deformation of the bar has reached a maximum value m and then
decreased back to zero. (a) Write a computer program that, for each of 25
values of m equally spaced over a range extending from 0 to a value
equal to 120% of the deformation causing both materials to yield, can be
used to determine the maximum value Pm of the load, the maximum
normal stress in each material, the permanent deformation p of the bar,
and the residual stress in each material. (b) Use this program to solve
Probs. 2.111 and 2.112.
SOLUTION
( Y )1 < ( Y )2
Note: The following assumes
Displacement increment
m 0.05( Y ) 2 L/E2
Displacements at yielding
A ( Y )1 L/E1 B ( Y ) 2 L/E2
For each displacement
If
m < A:
1 m E1/L
2 m E2 /L
Pm ( m /L) ( A1E1 A2 E2 )
If
A <m < B:
1 ( Y )1
2 m E2 / L
Pm A1 1 ( m /L) A2 E2
If
m > B:
1 ( Y )1
2 ( Y ) 2
Pm A1 1 A2 2
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254
PROBLEM 2.C4 (Continued)
Permanent deformations, residual stresses
Slope of first (elastic) segment
Slope ( A1E1 A2 E2 )/L
P m ( Pm /Slope)
( 1 )res 1 ( E1 Pm /( L Slope))
( 2 )res 2 ( E2 Pm /( L Slope))
Program Outputs
Problems 2.111 and 2.112
DM
10** 3 in.
PM
kips
SIGM (1)
ksi
SIGM (2)
ksi
DP
10** 3 in.
SIGR (1)
ksi
SIG (2)
ksi
0.000
2.414
4.828
7.241
9.655
12.069
14.483
16.897
19.310
21.724
24.138
26.552
0.000
8.750
17.500
26.250
35.000
43.750
52.500
61.250
70.000
78.750
87.500
91.250
0.000
5.000
10.000
15.000
20.000
25.000
30.000
35.000
40.000
45.000
50.000
50.000
0.000
5.000
10.000
15.000
20.000
25.000
30.000
35.000
40.000
45.000
50.000
55.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
1.379
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
2.143
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
2.857
28.966
95.000
50.000
60.000
2.759
4.286
5.714
31.379
98.750
50.000
65.000
4.138
6.429
8.571
33.793
102.500
50.000
70.000
5.517
8.571
11.429
36.207
106.250
50.000
75.000
6.897
10.714
14.286
38.621
110.000
50.000
80.000
8.276
12.857
17.143
41.034
113.750
50.000
85.000
9.655
15.000
20.000
43.448
117.500
50.000
90.000
11.034
17.143
22.857
45.862
121.250
50.000
95.000
12.414
19.286
25.714
48.276
125.000
50.000
100.000
13.793
21.429
28.571
50.690
125.000
50.000
100.000
16.207
21.429
28.571
53.103
125.000
50.000
100.000
18.621
21.429
28.571
55.517
125.000
50.000
100.000
21.034
21.429
57.931
125.000
50.000
100.000
23.448
21.429
28.571
28.571
2.112
2.111
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255
PROBLEM 2.C5
The plate has a hole centered across the width. The stress concentration factor
for a flat bar under axial loading with a centric hole is
P9
2
1
2
d
1
2
d
r
D
2r
2r
2r
K 3.00 3.13 3.66 1.53
D
D
D
P
3
where r is the radius of the hole and D is the width of the bar. Write a computer
program to determine the allowable load P for the given values of r, D, the
thickness t of the bar, and the allowable stress all of the material. Knowing
that t 14 in., D 3.0 in., and all 16 ksi, determine the allowable load P for
values of r from 0.125 in. to 0.75 in., using 0.125 in. increments.
SOLUTION
Enter
r , D, t , all
Compute K
RD 2.0r/D
K 3.00 3.13RD 3.66 RD 2 1.53RD3
Compute average stress
ave all /K
Allowable load
Pall ave ( D 2.0r ) t
Program Output
Radius
(in.)
Allowable Load
(kips)
0.1250
3.9802
0.2500
3.8866
0.3750
3.7154
0.5000
3.4682
0.6250
3.1523
0.7500
2.7794
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256
PROBLEM 2.C6
L
A
B
P
2c
c
A solid truncated cone is subjected to an axial force P as shown. The exact
elongation is ( PL) /(2 c 2 E ). By replacing the cone by n circular cylinders
of equal thickness, write a computer program that can be used to calculate
the elongation of the truncated cone. What is the percentage error in the
answer obtained from the program using (a) n 6, (b) n 12, (c) n 60?
SOLUTION
i 1 to n :
Li (i 0.5)(L/n)
For
ri 2c c(Li /L)
Area:
A ri2
Displacement:
P( L/n)/( AE )
Exact displacement:
exact PL/(2.0 c 2 E )
Percentage error:
Percent = 100( exact )/ exact
Program Output
n
Approximate
Exact
Percent
6
0.15852
0.15915
0.40083
12
0.15899
0.15915
0.10100
60
0.15915
0.15915
0.00405
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257
CHAPTER 4
20
40
PROBLEM 4.1
20
20
A
Knowing that the couple shown acts in a vertical plane, determine
the stress at (a) point A, (b) point B.
M 5 15 kN · m
80
20
B
Dimensions in mm
SOLUTION
For rectangle:
I
1 3
bh
12
Outside rectangle:
I1
1
(80)(120)3
12
I1
11.52 106 mm 4
I2
1
(40)(80)3
12
I2
1.70667 106 mm 4
Cutout:
Section:
(a)
yA
40 mm
I
I1
I2
11.52 10
6
m4
1.70667 10
9.81333 10
0.040 m
6
A
6
m4
m4
My A
I
(15 103 )(0.040)
9.81333 10 6
61.6 106 Pa
61.6 MPa
A
(b)
yB
60 mm
0.060 m
B
MyB
I
(15 103 )( 0.060)
9.81333 10 6
91.7 106 Pa
B
91.7 MPa
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447
2 in. 2 in. 2 in.
PROBLEM 4.2
M ! 25 kip · in.
A
Knowing that the couple shown acts in a vertical plane, determine
the stress at (a) point A, (b) point B.
2 in.
B
1.5 in.
2 in.
SOLUTION
I
For rectangle:
1 3
bh
12
For cross sectional area:
I
I1
I2
I3
1
(2)(1.5)3
12
1
(2)(5.5)3
12
1
(2)(1.5)3
12
(a)
yA
2.75 in.
A
My A
I
(25)(2.75)
28.854
(b)
yB
0.75 in.
B
MyB
I
(25)(0.75)
28.854
28.854 in 4
A
B
2.38 ksi
0.650 ksi
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448
200 mm
PROBLEM 4.3
12 mm
y
C
M
x
Using an allowable stress of 155 MPa, determine the largest bending
moment M that can be applied to the wide-flange beam shown. Neglect the
effect of fillets.
220 mm
8 mm
12 mm
SOLUTION
Moment of inertia about x-axis:
I1
1
(200)(12)3
12
(200)(12)(104)2
25.9872 106 mm 4
I2
1
(8)(196)3
12
I3
I1
25.9872 106 mm 4
I
I1
I2
M
Mx
I3
Mc
with c
I
I
with
c
5.0197 106 mm 4
56.944 106 mm 4
1
(220)
2
110 mm
56.944 10
6
m4
0.110 m
155 106 Pa
(56.944 10 6 )(155 106 )
0.110
80.2 103 N m
Mx
80.2 kN m
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449
200 mm
PROBLEM 4.4
12 mm
y
Solve Prob. 4.3, assuming that the wide-flange beam is bent about the y axis
by a couple of moment My.
C
M
x
220 mm
PROBLEM 4.3. Using an allowable stress of 155 MPa, determine the
largest bending moment M that can be applied to the wide-flange beam
shown. Neglect the effect of fillets.
8 mm
12 mm
SOLUTION
Moment of inertia about y axis:
I1
I2
1
(12)(200)3 8 106 mm 4
12
1
(196)(8)3 8.3627 103 mm 4
12
I3
I1
8 106 mm 4
I
I1
I2
My
My
I3
Mc
with c
I
I
with
c
16.0084 106 mm 4
1
(200)
2
100 mm
16.0084 10
6
m4
0.100 m
155 106 Pa
(16.0084 10 6 )(155 106 )
0.100
24.8 103 N m
My
24.8 kN m
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450
PROBLEM 4.5
0.1 in.
0.5 in.
M1
Using an allowable stress of 16 ksi, determine the largest couple that can be
applied to each pipe.
(a)
0.2 in.
0.5 in.
M2
(b)
SOLUTION
(a)
I
c
ro4 ri4
4
0.6 in.
Mc
:
I
M
4
(0.64
I
c
0.54 )
52.7 10
3
in 4
(16)(52.7 10 3 )
0.6
M
(b)
I
c
(0.74
4
0.7 in.
Mc
:
I
0.54 )
M
139.49 10
I
c
3
1.405 kip in.
in 4
(16)(139.49 10 3 )
0.7
M
3.19 kip in.
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451
PROBLEM 4.6
r 5 20 mm
A
30 mm
B
M = 2.8 kN · m
Knowing that the couple shown acts in a vertical plane,
determine the stress at (a) point A, (b) point B.
30 mm
120 mm
SOLUTION
I
1
(0.120 m)(0.06 m)3
12
2.1391 10
(a)
A
6
2
1
12
4
(0.02 m) 4
mm 4
(2.8 103 N m)(0.03 m)
2.1391 10 6 mm 4
M yA
I
39.3 MPa
A
(b)
B
M yB
I
(2.8 103 N m)(0.02 m)
2.1391 10 6 m 4
B
26.2 MPa
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452
PROBLEM 4.7
y
Two W4 13 rolled sections are welded together as shown. Knowing that for the steel
alloy used Y 36 ksi and U 58 ksi and using a factor of safety of 3.0, determine the
largest couple that can be applied when the assembly is bent about the z axis.
z
C
SOLUTION
Properties of W4
13 rolled section.
(See Appendix C.)
Area
3.83 in 2
Width
4.060 in.
Iy
3.86 in 4
For one rolled section, moment of inertia about axis b-b is
For both sections,
Ad 2
Ib
Iy
Iz
2 Ib
c
all
M all
width
U
F .S .
all I
c
3.86 (3.83)(2.030)2
19.643 in 4
39.286 in 4
4.060 in.
58
19.333 ksi
3.0
(19.333)(39.286)
4.060
Mc
I
M all 187.1 kip in.
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453
PROBLEM 4.8
y
C
Two W4 13 rolled sections are welded together as shown. Knowing that for the steel
alloy used U 58 ksi and using a factor of safety of 3.0, determine the largest couple
that can be applied when the assembly is bent about the z axis.
z
SOLUTION
Properties of W4
13 rolled section.
(See Appendix C.)
Area
3.83 in 2
Depth
4.16 in.
Ix
11.3 in 4
For one rolled section, moment of inertia about axis a-a is
For both sections,
Ad 2
Ia
Ix
Iz
2Ia
c
all
M all
depth
U
F .S .
all I
c
11.3 (3.83)(2.08)2
27.87 in 4
55.74 in 4
4.16 in.
58
19.333 ksi
3.0
(19.333)(55.74)
4.16
Mc
I
M all
259 kip in.
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454
PROBLEM 4.9
3 in. 3 in. 3 in.
6 in.
Two vertical forces are applied to a beam of the cross section shown. Determine
the maximum tensile and compressive stresses in portion BC of the beam.
2 in.
A
15 kips
15 kips
B
C
40 in.
60 in.
D
40 in.
SOLUTION
A
y0
A y0
18
5
90
18
1
18
36
Y0
108
36
108
3 in.
Neutral axis lies 3 in. above the base.
I1
I2
I
ytop
M
M
1
1
b1h13 A1d12
(3)(6)3 (18)(2) 2 126 in 4
12
12
1
1
3
2
b2 h2 A2 d 2
(9)(2)3 (18)(2) 2 78 in 4
12
12
I1 I 2 126 78 204 in 4
5 in. ybot
Pa 0
Pa (15)(40) 600 kip in.
M ytop
top
bot
3 in.
I
(600)(5)
204
M ybot
I
(600)( 3)
204
top
14.71 ksi (compression)
bot
8.82 ksi (tension)
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455
PROBLEM 4.10
8 in.
1 in.
Two vertical forces are applied to a beam of the cross section shown.
Determine the maximum tensile and compressive stresses in portion
BC of the beam.
6 in.
1 in.
1 in.
4 in.
A
25 kips
25 kips
B
C
20 in.
60 in.
D
20 in.
SOLUTION
A
y0
A y0
8
7.5
60
6
4
24
4
0.5
18
86
86
18
4.778 in.
(8)(2.772) 2
59.94 in 4
(6)(0.778) 2
21.63 in 4
(4)(4.278) 2
73.54 in 4
Yo
2
Neutral axis lies 4.778 in. above the base.
I1
I2
I3
I
ytop
1
1
(8)(1)3
b1h13 A1d12
12
12
1
1
(1)(6)3
b2 h23 A2 d 22
12
12
1
1
(4)(1)3
b3 h33 A3 d32
12
12
I1 I 2 I 3 59.94 21.63
4.778 in.
3.222 in. ybot
M
M
Pa 0
Pa (25)(20) 500 kip in.
Mytop
top
bot
73.57 155.16 in 4
I
Mybot
I
(500)(3.222)
155.16
(500)( 4.778)
155.16
top
10.38 ksi (compression)
bot
15.40 ksi (tension)
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456
10 mm
PROBLEM 4.11
10 mm
10 kN
10 kN
B
50 mm
C
A
D
Two vertical forces are applied to a beam of
the cross section shown. Determine the
maximum tensile and compressive stresses
in portion BC of the beam.
10 mm
50 mm
250 mm
150 mm
150 mm
SOLUTION
A, mm 2
y0 , mm
A y0 , mm3
600
30
18 103
600
30
18 103
300
5
1.5 103
37.5 103
1500
Y0
37.5 103
1500
25mm
Neutral axis lies 25 mm above the base.
I1
I3
I
ytop
35 mm 0.035 m
a 150 mm
M
bot
ybot
25 mm
0.025 m
0.150 m P 10 103 N
(10 103 )(0.150) 1.5 103 N m
Pa
Mytop
top
1
(10)(60)3 (600)(5)2 195 103 mm 4 I 2 I1 195 mm 4
12
1
(30)(10)3 (300)(20) 2 122.5 103 mm 4
12
I1 I 2 I 3 512.5 103 mm 4 512.5 10 9 m 4
I
M ybot
I
(1.5 103 )(0.035)
512.5 10 9
102.4 106 Pa
(1.5 103 )( 0.025)
512.5 10 9
73.2 106 Pa
top
102.4 MPa (compression)
bot
73.2 MPa (tension)
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457
PROBLEM 4.12
216 mm
y
z
Knowing that a beam of the cross section shown is bent about a horizontal
axis and that the bending moment is 6 kN m, determine the total force
acting on the shaded portion of the web.
36 mm
54 mm
C
108 mm
72 mm
SOLUTION
The stress distribution over the entire cross section is given by the bending stress formula:
My
I
x
where y is a coordinate with its origin on the neutral axis and I is the moment of inertia of the entire cross
sectional area. The force on the shaded portion is calculated from this stress distribution. Over an area
element dA, the force is
dF
My
dA
I
x dA
The total force on the shaded area is then
F
My
dA
I
dF
M
I
y dA
M * *
y A
I
where y * is the centroidal coordinate of the shaded portion and A* is its area.
d1
54 18 36 mm
d2
54 36 54 36 mm
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458
PROBLEM 4.12 (Continued)
Moment of inertia of entire cross section:
I1
I2
I
1
1
(216)(36)3 (216)(36)(36)2 10.9175 106 mm4
b1h13 A1d12
12
12
1
1
b2 h23 A2 d 22
(72)(108)3 (72)(108)(36)2 17.6360 106 mm 4
12
12
I1 I 2 28.5535 106 mm 4 28.5535 10 6 m 4
For the shaded area,
A*
(72)(90)
y*
45 mm
A* y *
F
6480 mm 2
291.6 103 mm3
MA* y *
I
291.6 10 6 m
(6 103 )(291.6 10 6 )
28.5535 10 6
61.3 103 N
F
61.3 kN
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459
20 mm
12 mm
PROBLEM 4.13
20 mm
y
12 mm
24 mm
z
Knowing that a beam of the cross section shown is bent about a horizontal axis
and that the bending moment is 4 kN m, determine the total force acting on
the shaded portion of the beam.
20 mm
C
20 mm
24 mm
SOLUTION
Dimensions in mm:
Iz
1
1
(12 12)(88)3
(40)(40)3
12
12
1.3629 106 0.213 106
1.5763 106 mm 4
1.5763 10 6 m 4
For use in Prob. 4.14,
Iy
1
1
(88)(64)3
(24 24)(40)3
12
12
1.9224 106 0.256 106
1.6664 106 mm 4
Bending about horizontal axis.
Mz
1.6664 10 6 m 4
4 kN m
A
M zc
Iz
(4 kN m)(0.044 m)
1.5763 10 6 m 4
111.654 MPa
B
M zc
Iz
(4 kN m)(0.020 m)
1.5763 10 6 m 4
50.752 MPa
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460
PROBLEM 4.13 (Continued)
A (44)(12) 528 mm 2
Portion (1):
avg
1
2
Force1
A
avg A
avg
Force2
Total force on shaded area
1
2
B
avg A
6
m2
1
(111.654) 55.83 MPa
2
(55.83 MPa)(528 10
A (20)(20)
Portion (2):
528 10
400 mm 2
1
(50.752)
2
6
400 10
m2 )
6
29.477 kN
m2
25.376 MPa
(25.376 MPa)(400 10
6
m 2 ) 10.150 kN
29.477 10.150 39.6 kN
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461
20 mm
12 mm
PROBLEM 4.14
20 mm
y
12 mm
Solve Prob. 4.13, assuming that the beam is bent about a vertical axis by a
couple of moment 4 kN m.
24 mm
z
PROBLEM 4.13. Knowing that a beam of the cross section shown is bent
about a horizontal axis and that the bending moment is 4 kN m, determine the
total force acting on the shaded portion of the beam.
20 mm
C
20 mm
24 mm
SOLUTION
My
Bending about vertical axis.
Iy
See Prob. 4.13 for sketch and
4 kN m
1.6664 10
Mc
Iy
(4 kN m)(0.032 m)
1.6664 10 6 m 4
76.81 MPa
E
Mc
Iy
(4 kN m)(0.020 m)
1.6664 10 6 m 4
48.01 MPa
avg
Force1
1
(
2
E)
D
avg A
avg
Force2
Total force on shaded area
1
2
E
avg A
528 10
6
1
(76.81 48.01)
2
(62.41 MPa)(528 10
A (20)(20)
Portion (2):
m4
D
A (44)(12) 528 mm 2
Portion (1):
6
400 mm 2
1
(48.01)
2
6
400 10
62.41 MPa
m 2 ) 32.952 kN
6
m2
6
m 2 ) 9.602 kN
24.005 MPa
(24.005 MPa)(400 10
32.952 9.602
m2
42.6 kN
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462
0.5 in.
0.5 in.
0.5 in.
1.5 in.
1.5 in.
PROBLEM 4.15
Knowing that for the extruded beam shown the allowable stress is 12 ksi
in tension and 16 ksi in compression, determine the largest couple M that
can be applied.
1.5 in.
0.5 in.
M
SOLUTION
A
y0
Ay0
2.25
1.25
2.8125
2.25
0.25
0.5625
4.50
Y
3.375
3.375
4.50
0.75 in.
The neutral axis lies 0.75 in. above bottom.
ytop
I1
I2
I
2.0 0.75 1.25 in.,
ybot
0.75 in.
1
1
b1h13 A1d12
(1.5)(1.5)3 (2.25)(0.5)2 0.984375 in 4
12
12
1
1
2
2
b2 h2 A2 d 2
(4.5)(0.5)3 (2.25)(0.5)2 0.609375 in 4
12
12
I1 I 2 1.59375 in 4
My
I
M
I
y
Top: (compression)
M
(16)(1.59375)
1.25
20.4 kip in.
Bottom: (tension)
M
(12)(1.59375)
0.75
25.5 kip in.
M all
Choose the smaller as Mall.
20.4 kip in.
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463
PROBLEM 4.16
40 mm
The beam shown is made of a nylon for which the allowable stress is
24 MPa in tension and 30 MPa in compression. Determine the largest
couple M that can be applied to the beam.
15 mm
d ! 30 mm
20 mm
M
SOLUTION
Σ
Y0
A, mm2
y0 , mm
A y0 , mm3
600
22.5
13.5 103
300
7.5
2.25 103
15.75 103
900
15.5 103
900
17.5 mm
ytop
ybot
I1
I2
I
| |
The neutral axis lies 17.5 mm above the bottom.
30 17.5 12.5 mm
17.5 mm
0.0175 m
1
b1h13 A1d12
12
1
b2 h23 A2 d 22
12
I1 I 2 61.875
My
I
M
0.0125 m
1
(40)(15)3 (600)(5)2 26.25 103 mm 4
12
1
(20)(15)3 (300)(10)2 35.625 103 mm 4
12
103 mm 4 61.875 10 9 m 4
I
y
Top: (tension side)
M
(24 106 )(61.875 10 9 )
0.0125
Bottom: (compression)
M
(30 106 )(61.875 10 9 )
106.1 N m
0.0175
Choose smaller value.
118.8 N m
M
106.1 N m
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464
PROBLEM 4.17
40 mm
Solve Prob. 4.16, assuming that d
15 mm
d ! 30 mm
40 mm.
PROBLEM 4.16 The beam shown is made of a nylon for which the
allowable stress is 24 MPa in tension and 30 MPa in compression.
Determine the largest couple M that can be applied to the beam.
20 mm
M
SOLUTION
A, mm2
y0 , mm
A y0 , mm3
600
32.5
19.5 103
500
12.5
6.25 103
Σ
Y0
25.75 103
1100
25.75 103
1100
ytop
ybot
I1
I2
I
| |
23.41 mm
The neutral axis lies 23.41 mm above the bottom.
40 23.41 16.59 mm
23.41 mm
0.01659 m
0.02341 m
1
1
b1h13 A1d12
(40)(15)3 (600)(9.09) 2 60.827 103 mm 4
12
12
1
1
b2 h22 A2 d 22
(20)(25)3 (500)(10.91) 2 85.556 103 mm 4
12
12
I1 I 2 146.383 103 mm 4 146.383 10 9 m 4
My
I
M
I
y
Top: (tension side)
M
(24 106 )(146.383 10 9 )
0.01659
212 N m
Bottom: (compression)
M
(30 106 )(146.383 10 9 )
0.02341
187.6 N m
Choose smaller value.
M
187.6 N m
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465
PROBLEM 4.18
2.4 in.
0.75 in.
Knowing that for the beam shown the allowable stress is 12 ksi in tension and
16 ksi in compression, determine the largest couple M that can be applied.
1.2 in.
M
SOLUTION
rectangle
semi-circular cutout
A1
2.88 in 2
(2.4)(1.2)
A2
2
(0.75)2
A
2.88
y1
0.6 in.
0.8836 in 2
1.9964 in 2
0.8836
4r
(4)(0.75)
0.3183 in.
3
3
Ay
(2.88)(0.6) (0.8836)(0.3183)
A
1.9964
y2
Y
0.7247 in.
Neutral axis lies 0.7247 in. above the bottom.
Moment of inertia about the base:
1 3
bh
3
Ib
8
r4
1
(2.4)(1.2)3
3
(0.75) 4
1.25815 in 4
1.25815 (1.9964)(0.7247) 2
0.2097 in 4
8
Centroidal moment of inertia:
I
Ib
AY 2
ytop
1.2 0.7247
ybot
0.7247 in.
| |
My
I
M
0.4753 in.,
I
y
Top: (tension side)
M
(12)(0.2097)
0.4753
5.29 kip in.
Bottom: (compression)
M
(16)(0.2097)
0.7247
4.63 kip in.
M
Choose the smaller value.
4.63 kip in.
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466
PROBLEM 4.19
80 mm
Knowing that for the extruded beam shown the allowable stress is
120 MPa in tension and 150 MPa in compression, determine the largest
couple M that can be applied.
54 mm
40 mm
M
SOLUTION
Σ
Y
A, mm2
y0 , mm
2160
27
58,320
3
1080
36
38,880
3
A y0 , mm3
3240
d, mm
97,200
97, 200
3240
30 mm
The neutral axis lies 30 mm above the bottom.
ytop
54 30
24 mm
I1
I2
I
| |
1
b1h13 A1d12
12
1
b2 h22 A2 d 22
36
I1 I 2 758.16
My
I
0.024 m
ybot
30 mm
0.030 m
1
(40)(54)3 (40)(54)(3) 2 544.32 103 mm 4
12
1
1
(40)(54)3
(40)(54)(6)2 213.84 103 mm 4
36
2
103 mm 4 758.16 10 9 m 4
I
y
|M |
Top: (tension side)
M
(120 106 )(758.16 10 9 )
0.024
3.7908 103 N m
Bottom: (compression)
M
(150 106 )(758.16 10 9 )
0.030
3.7908 103 N m
Choose the smaller as Mall.
M all
3.7908 103 N m
M all
3.79 kN m
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467
PROBLEM 4.20
48 mm
Knowing that for the extruded beam shown the allowable stress is
120 MPa in tension and 150 MPa in compression, determine the largest
couple M that can be applied.
48 mm
36 mm
48 mm
36 mm
M
SOLUTION
A, mm2
y0 , mm
Solid rectangle
4608
48
221,184
Square cutout
–1296
30
–38,880
Σ
Y
3312
182,304
3312
ytop
ybot
I1
I2
I
| |
55.04 mm
182,304
Neutral axis lies 55.04 mm above bottom.
96 55.04 40.96 mm
55.04 mm
0.04096 m
0.05504 m
1
b1h13 A1d12
12
1
b2 h32 A2 d 22
12
I1 I 2 2.8147
My
I
A y0 , mm3
M
1
(48)(96)3 (48)(96)(7.04)2 3.7673 106 mm 4
12
1
(36)(36)3 (36)(36)(25.04) 2 0.9526 106 mm 4
12
106 mm 4 2.8147 10 6 m 4
I
y
Top: (tension side)
M
(120 106 )(2.8147 10 6 )
0.04096
Bottom: (compression)
M
(150 106 )(2.8147 106 )
0.05504
Mall is the smaller value.
M
7.67 103 N m
8.25 103 N m
7.67 103 N m
7.67 kN m
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468
PROBLEM 4.21
Straight rods of 6-mm diameter and 30-m length are stored by coiling the
rods inside a drum of 1.25-m inside diameter. Assuming that the yield
strength is not exceeded, determine (a) the maximum stress in a coiled rod,
(b) the corresponding bending moment in the rod. Use E 200 GPa.
SOLUTION
Let
D
inside diameter of the drum,
d
diameter of rod,
1
d,
2
radius of curvature of center line of rods when bent.
1
D
2
I
(a)
(b)
Ec
max
M
EI
4
c4
c
1
d
2
4
(200 109 )(0.003)
0.622
(200 109 )(63.617 10
0.622
1
(1.25)
2
(0.003) 4
1
(6 10 3 )
2
63.617 10
12
0.622 m
m4
965 106 Pa
12
)
965 MPa
20.5 N m
M
20.5 N m
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469
M'
PROBLEM 4.22
M
8 mm
A 900-mm strip of steel is bent into a full circle by two couples applied
as shown. Determine (a) the maximum thickness t of the strip if the
allowable stress of the steel is 420 MPa, (b) the corresponding moment
M of the couples. Use E 200 GPa.
t
r
900 mm
SOLUTION
When the rod is bent into a full circle, the circumference is 900 mm. Since the circumference is equal to
2 times , the radius of curvature, we get
900 mm
2
Stress:
E
Ec
420 MPa and E
For
c
(a)
143.24 mm
0.14324 m
or
c
E
200 GPa,
(0.14324)(420 106 )
200 109
t
Maximum thickness:
0.3008 10
2c
3
m
0.6016 10
3
m
t
0.602 mm
Moment of inertia for a rectangular section.
I
(b)
bt 3
12
Bending moment:
M
(8 10 3 )(0.6016 10 3 )3
12
M
145.16 10
15
m4
EI
(200 109 )(145.16 10
0.14324
15
)
0.203 N m
M
0.203 N m
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470
PROBLEM 4.23
Straight rods of 0.30-in. diameter and 200-ft length are sometimes
used to clear underground conduits of obstructions or to thread wires
through a new conduit. The rods are made of high-strength steel and,
for storage and transportation, are wrapped on spools of 5-ft
diameter. Assuming that the yield strength is not exceeded,
determine (a) the maximum stress in a rod, when the rod, which is
initially straight, is wrapped on a spool, (b) the corresponding
bending moment in the rod. Use E 29 106 psi .
5 ft
SOLUTION
Radius of cross section:
r
Moment of inertia:
I
(a)
(b)
D
5 ft
c
r
60 in.
4
r4
1
(0.30)
2
4
(0.15) 4
1
D
2
30 in.
106 )(0.15)
30
145.0
0.15 in.
397.61 10
6
in 4
0.15 in.
Ec
(29
max
M
1
d
2
EI
103 psi
(29 106 )(397.61 10 6 )
30
max
M
145.0 ksi
384 lb in.
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471
PROBLEM 4.24
12 mm
y
60 N · m
A 60-N m couple is applied to the steel bar shown. (a) Assuming that
the couple is applied about the z axis as shown, determine the maximum
stress and the radius of curvature of the bar. (b) Solve part a, assuming
that the couple is applied about the y axis. Use E 200 GPa.
20 mm
z
SOLUTION
(a)
Bending about z-axis.
I
c
1
(b)
1 3 1
bh
(12)(20)3 8 103 mm 4
12
12
20
10 mm 0.010 m
2
Mc
I
(60)(0.010)
8 10 9
M
EI
60
(200 109 )(8 10 9 )
8 10 9 m 4
75.0 106 Pa
37.5 10 3 m
75.0 MPa
1
26.7 m
Bending about y-axis.
I
c
1
1 3 1
bh
(20)(12)3 2.88 103 mm 4
12
12
12
6 mm 0.006 m
2
Mc (60)(0.006)
125.0 106 Pa
9
I
2.88 10
M
EI
60
(200 10 )(2.88 10 9 )
9
2.88 10 9 m 4
104.17 10 3 m
125.0 MPa
1
9.60 m
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472
10 mm
M
(a) Using an allowable stress of 120 MPa, determine the largest couple
M that can be applied to a beam of the cross section shown. (b) Solve
part a, assuming that the cross section of the beam is an 80-mm square.
80 mm
C
PROBLEM 4.25
10 mm
5 mm
80 mm
5 mm
SOLUTION
(a)
I I1 4 I 2 , where I1 is the moment of inertia of an 80-mm square and I2 is the moment of inertia of
one of the four protruding ears.
I1
I2
I
1 3
bh
12
1
(80)(80)3
12
3.4133 106 mm 4
1 3
1
(5)(10)3 (5)(10)(45)2 101.667 103 mm 4
bh Ad 2
12
12
I1 4 I 2 3.82 106 mm 4 3.82 10 6 mm 4 ,
c 50 mm 0.050 m
Mc
I
I
c
M
(120 106 )(3.82 10 6 )
0.050
9.168 103 N m
9.17 kN m
(b)
Without the ears:
I
M
I1
I
c
3.4133 10
6
m2 ,
c
40 mm
0.040 m
(120 106 )(3.4133 10 6 )
10.24 103 N m
0.040
10.24 kN m
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473
0.1 in.
PROBLEM 4.26
0.2 in.
A thick-walled pipe is bent about a horizontal axis by a couple M. The
pipe may be designed with or without four fins. (a) Using an allowable
stress of 20 ksi, determine the largest couple that may be applied if the
pipe is designed with four fins as shown. (b) Solve part a, assuming that
the pipe is designed with no fins.
1.5 in.
M
0.75 in.
SOLUTION
I x of hollow pipe:
Ix
I x of fins:
Ix
(1.5 in.) 4
4
2
(0.75 in.) 4
1
(0.1)(0.2)3
12
3.7276 in 4
(0.1 0.2)(1.6) 2
2
1
(0.2)(0.1)3
12
0.1026 in 4
(a)
(b)
Pipe as designed, with fins:
Ix
3.8302 in 4 ,
all
20 ksi,
c 1.7 in.
M
all
Ix
c
(20 ksi)
3.8302 in 4
1.7 in.
M
45.1 kip in.
M
49.7 kip in.
Pipe with no fins:
all
M
20 ksi,
all
Ix
c
Ix
3.7276 in 4 ,
(20 ksi)
c 1.5 in.
3.7276 in 4
1.5 in.
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474
PROBLEM 4.27
M
A couple M will be applied to a beam of rectangular cross section that
is to be sawed from a log of circular cross section. Determine the ratio
d/b for which (a) the maximum stress m will be as small as possible,
(b) the radius of curvature of the beam will be maximum.
M'
d
b
SOLUTION
Let D be the diameter of the log.
D2
m
d2
1
bd 3
12
I
(a)
b2
c
is the minimum when
I
c
d
db
D2
1
d
2
I
c
b2
D2
1 2
bd
6
I
is maximum.
c
1
1 2
b( D 2 b 2 )
Db
6
6
I
1 2 3 2
0
D
b
6
6
c
d
(b)
d2
1 2
D
3
1 3
b
6
1
D
3
b
2
D
3
d
b
2
d
b
3
EI
M
is maximum when I is maximum,
(D2
d 2 )d 6 is maximum.
6 D 2d 5
b
1
bd 3 is maximum, or b2d 6 is maximum.
12
D2
8d 7
0
3 2
D
4
d
1
D
2
3
D
2
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475
PROBLEM 4.28
h0
M
h
C
h0
h
A portion of a square bar is removed by milling, so that its cross section
is as shown. The bar is then bent about its horizontal axis by a couple M.
Considering the case where h 0.9h0, express the maximum stress in the
bar in the form m k 0 , where 0 is the maximum stress that would
have occurred if the original square bar had been bent by the same
couple M, and determine the value of k.
SOLUTION
I
4 I1
2I2
1
h h3
12
1 4 4
h
h 0 h3
3
3
h
Mc
Mh
4 h h3
I
3 0
(4)
c
For the original square,
h
h0 , c
0
3M
(4h0 3h0 )h02
1
(2h0 2h)(h3 )
3
4
4
h h3
h 0 h3 h 4
3
3
(2)
h
4
3M
3h) h 2
h0 .
3M
h03
h03
0
(4h 0
(4h0
0.950
h03
3h)h 2
(4h0
(3)(0.9)h0 )(0.9 h02 )
0.950
k
0
0.950
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476
PROBLEM 4.29
h0
M
In Prob. 4.28, determine (a) the value of h for which the maximum stress
m is as small as possible, (b) the corresponding value of k.
h
PROBLEM 4.28 A portion of a square bar is removed by milling, so that
its cross section is as shown. The bar is then bent about its horizontal
axis by a couple M. Considering the case where h 0.9h0, express the
maximum stress in the bar in the form m k 0 , where 0 is the
maximum stress that would have occurred if the original square bar had
been bent by the same couple M, and determine the value of k.
C
h
h0
SOLUTION
I
4 I1
2I 2
1
1
hh3 (2)
(2h0 2h) h3
12
3
1 4 4
4 3 4
h
h 0 h3
h
h 0 h3 h 4
3
3
3
3
I 4
2
3
h
h0 h h
c 3
(4)
c
d 4
I
is maximum at
h0 h2
c
dh 3
h3
0.
8
h 0 h 3h 2
3
I
c
For the original square,
4
8
h0
h0
3
9
h
h0
2
8
h0
9
3
256 3
h0
729
c
h0
I0
c0
0
Mc0
I0
3M
h02
0
0
729 1
256 3
h
Mc
I
8
h0
9
729M
256h03
1 3
h0
3
729
768
0.949
k
0.949
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477
PROBLEM 4.30
For the bar and loading of Concept Application 4.1, determine (a) the radius of curvature , (b) the radius of
of a transverse cross section, (c) the angle between the sides of the bar that were originally
curvature
vertical. Use E 29 106 psi and v 0.29.
SOLUTION
M
From Example 4.01,
(a)
(b)
1
1
1
(c)
M
EI
vc
v
v
(30 103 )
(29 106 )(1.042)
1
v
993 10
6
in.
1.042 in 4
I
1
1007 in.
c
(0.29)(993 10 6 )in.
length of arc
radius
30 kip in.
b
0.8
3470
1
288 10
6
in.
230 10 6 rad
1
3470 in.
0.01320
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478
PROBLEM 4.31
y
A
z
M
A W200 31.3 rolled-steel beam is subjected to a couple M of moment
45 kN m. Knowing that E 200 GPa and v 0.29, determine (a) the
of a transverse cross
radius of curvature , (b) the radius of curvature
section.
C
x
SOLUTION
For W 200 31.3 rolled steel section,
I
31.3 106 mm 4
31.3 10 6 m 4
(a)
(b)
1
1
M
EI
v
1
45 103
(200 109 )(31.3 10 6 )
(0.29)(7.1885 10 3 )
7.1885 10 3 m
2.0847 10 3 m
1
1
139.1 m
480 m
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479
PROBLEM 4.32
y
!
2
y # %c
"y
It was assumed in Sec. 4.1B that the normal stresses y in a
member in pure bending are negligible. For an initially straight
elastic member of rectangular cross section, (a) derive an
approximate expression for y as a function of y, (b) show that
(c /2 )( x ) max and, thus, that y can be neglected in
( y )max
all practical situations. (Hint: Consider the free-body diagram of
the portion of beam located below the surface of ordinate y and
assume that the distribution of the stress x is still linear.)
!
2
"y
"x
"x
!
2
!
2
y # $c
SOLUTION
Denote the width of the beam by b and the length by L.
L
cos
Using the free body diagram above, with
Fy
0:
2
sin
L
2
y
But,
(a)
(
y
x ) max
y
c
c
y dy
The maximum value
y bL
x
(
x ) max
(
x ) max
y2
2
c
y
occurs at y
y
2
c
y
c
2
x
1
x b dy
sin
0
2
y
dy
L
c
x
1
dy
y
c
x
dy
y
c
y
(
y
c
x ) max
2 c
( y2
c2 )
0.
(b)
(
y ) max
(
x ) max c
2 c
2
(
x ) max c
2
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480
PROBLEM 4.33
Brass
6 mm
A bar having the cross section shown has been formed by securely
bonding brass and aluminum stock. Using the data given below,
determine the largest permissible bending moment when the
composite bar is bent about a horizontal axis.
Aluminum
30 mm
6 mm
Aluminum
Brass
70 GPa
105 GPa
100 MPa
160 MPa
Modulus of elasticity
30 mm
Allowable stress
SOLUTION
Use aluminum as the reference material.
n
1.0 in aluminum
n
Eb /Ea
105/70
1.5 in brass
For the transformed section,
I1
n1
b1h13 n1 A1d 12
12
1.5
(30)(6)3 (1.5)(30)(6)(18)3
12
I2
n2
b2h23
12
I3
I1
88.29 103 mm 4
I
I1
I2
1.0
(30)(30)3
12
I3
Aluminum:
n 1.0,
M
Brass:
Choose the smaller value
67.5 103 mm 4
244.08 103 mm 4
244.08 10
nMy
I
M
y 15 mm
9
y
21 mm
m4
I
ny
0.015 m,
(100 106 )(244.08 10 9 )
(1.0)(0.015)
n 1.5,
88.29 103 mm 4
0.021 m,
M
(160 106 )(244.08 10 9 )
(1.5)(0.021)
M
1.240 103 N m
100 106 Pa
1.627 103 N m
160 106 Pa
1.240 103 N m
1.240 kN m
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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481
8 mm
PROBLEM 4.34
8 mm
32 mm
8 mm
32 mm
A bar having the cross section shown has been formed by securely
bonding brass and aluminum stock. Using the data given below,
determine the largest permissible bending moment when the
composite bar is bent about a horizontal axis.
8 mm
Brass
Aluminum
Brass
70 GPa
105 GPa
100 MPa
160 MPa
Modulus of elasticity
Aluminum
Allowable stress
SOLUTION
Use aluminum as the reference material.
For aluminum, n 1.0
For brass, n
Eb /Ea
105/70 1.5
Values of n are shown on the sketch.
For the transformed section,
I1
I2
I3
I
| |
Aluminum:
I1
I2
nMy
I
I 3 141.995 103 mm 4
M
100 106 Pa
0.016 m,
(100 106 )(141.995 10 9 )
(1.0)(0.016)
n 1.5, | y | 16 mm
M
141.995 10 9 m 4
I
ny
n 1.0, | y | 16 mm
M
Brass:
n1
1.5
b1h13
(8)(32)3 32.768 103 mm 4
12
12
n2
1.0
b2 H 23 h23
(32)(323 163 ) 76.459 103 mm 4
12
12
I1 32.768 103 mm 4
887.47 N m
160 106 Pa
0.016 m,
(160 106 )(141.995 10 9 )
(1.5)(0.016)
946.63 N m
Choose the smaller value.
M
887 N m
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on a website, in whole or part.
482
PROBLEM 4.35
Brass
6 mm
For the composite bar indicated, determine the largest permissible
bending moment when the bar is bent about a vertical axis.
30 mm
PROBLEM 4.35. Bar of Prob. 4.33.
Aluminum
Modulus of elasticity
6 mm
Allowable stress
30 mm
Aluminum
Brass
70 GPa
105 GPa
100 MPa
160 MPa
SOLUTION
Use aluminum as reference material.
n
1.0 in aluminum
n
Eb /Ea
105/70
1.5 in brass
For transformed section,
I1
I2
n1
b1h13
12
1.5
(6)(30)3
12
n2
b2h23
12
1.0
(30)(30)3
12
I1
20.25 103 mm 4
I
I1
I2
n 1.0,
M
Brass:
67.5 103 mm 4
I3
nMy
I
Aluminum:
20.25 103 mm 4
108
103 mm 4
108
10
9
m4
I
ny
M
y 15 mm
0.015 m,
(100 106 )(108 10 9 )
(1.0)(0.015)
n 1.5,
M
I3
y 15 mm
720 N m
0.015 m,
(160 106 )(108 10 9 )
(1.5)(0.015)
100 106 Pa
160 106 Pa
768 N m
M
Choose the smaller value.
720 N m
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on a website, in whole or part.
483
8 mm
PROBLEM 4.36
8 mm
32 mm
8 mm
For the composite bar indicated, determine the largest permissible
bending moment when the bar is bent about a vertical axis.
PROBLEM 4.36 Bar of Prob. 4.34.
32 mm
8 mm
Brass
Modulus of elasticity
Allowable stress
Aluminum
Aluminum
70 GPa
100 MPa
Brass
105 GPa
160 MPa
SOLUTION
Use aluminum as the reference material.
For aluminum, n 1.0
For brass, n
Eb /Ea
105/70 1.5
Values of n are shown on the sketch.
For the transformed section,
I1
I2
I3
I
| |
Aluminum:
I1
I2
nMy
I
I3
M
354.99 103 mm 4
354.99 10 9 m 4
I
ny
n 1.0, | y | 16 mm
M
Brass:
n1
1.5
h1 B13 b13
(32)(483 323 ) 311.296 103 mm 4
12
12
n2
1.0
h2b23
(8)(32)3 21.8453 103 mm 4
12
12
I 2 21.8453 103 mm 4
0.016 m,
(100 106 )(354.99 10 9 )
(1.0)(0.016)
n 1.5 | y | 24 mm
0.024 m
M
(160 106 )(354.99 10 9 )
(1.5)(0.024)
M
1.57773 103 N m
100 106 Pa
2.2187 103 N m
160 106 Pa
1.57773 103 N m
Choose the smaller value.
M
1.578 kN m
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
484
PROBLEM 4.37
1
5 3 2 in.
Wooden beams and steel plates are securely bolted together to form the composite
member shown. Using the data given below, determine the largest permissible bending
moment when the member is bent about a horizontal axis.
10 in.
Modulus of elasticity:
1
5 3 2 in.
6 in.
Allowable stress:
Wood
Steel
2 106 psi
29 106 psi
2000 psi
22 ksi
SOLUTION
Use wood as the reference material.
n 1.0 in wood
n
E s /E w
29/2 14.5 in steel
For the transformed section,
I1
n1
b1h13
12
n1 A1d12
3
I2
I3
I
nMy
I
Wood:
n
M
Steel:
n
M
M
I
ny
1.0,
y
14.5
1
1
(5)
(14.5)(5)
(5.25)2
12
2
2
n2
1.0
b2 h22
(6)(10)3 500 in 4
12
12
I1 999.36 in 4
I1
y
2498.7 in 4
2000 psi
999.5 103 lb in.
5.5 in.,
(22 103 )(2499)
(14.5)(5.5)
Choose the smaller value.
I3
5 in.,
(2000)(2499)
(1.0)(5)
14.5,
I2
999.36 in 4
22 ksi
22 103 psi
689.3 103 lb in.
M
689 103 lb in.
M
689 kip in.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
485
PROBLEM 4.38
10 in.
3 in.
1
2
Wooden beams and steel plates are securely bolted together to form the composite
member shown. Using the data given below, determine the largest permissible bending
moment when the member is bent about a horizontal axis.
3 in.
Modulus of elasticity:
in.
Allowable stress:
Wood
Steel
2 106 psi
29 106 psi
2000 psi
22 ksi
SOLUTION
Use wood as the reference material.
n 1.0 in wood
n
E s /E w
29/2 14.5 in steel
For the transformed section,
I1
I2
I3
I
nMy
I
Wood:
n
M
Steel:
n
M
M
I
ny
1.0,
y
n1
1.0
(3)(10)3 250 in 4
b1h13
12
12
n2
14.5 1
b2 h23
(10)3 604.17 in 4
12
12 2
I1
y
I3
1104.2 in 4
2000 psi
441.7 103 lb in.
5 in.,
(22 103 )(1104.2)
(14.5)(5)
Choose the smaller value.
I2
5 in.,
(2000)(1104.2)
(1.0)(5)
14.5,
250 in 4
I1
22 ksi
22 103 psi
335.1 103 lb in.
M
335 103 lb in.
M
335 kip in.
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Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
486
PROBLEM 4.39
A copper strip (Ec = 105 GPa) and an aluminum strip (Ea = 75 GPa)
are bonded together to form the composite beam shown. Knowing
that the beam is bent about a horizontal axis by a couple of moment
M = 35 N m, determine the maximum stress in (a) the aluminum
strip, (b) the copper strip.
SOLUTION
Use aluminum as the reference material.
n 1.0 in aluminum
n
Ec /Ea
105/75 1.4 in copper
Transformed section:
A, mm2
nA, mm 2
y0 , mm
nAy0 , mm3
144
144
9
1296
144
201.6
3
Σ
Y0
345.6
1900.8
345.6
604.8
1900.8
5.50 mm
The neutral axis lies 5.50 mm above the bottom.
I
n1
1.0
b1h13 n1 A1d12
(24)(6)3 (1.0)(24)(6)(3.5) 2 2196 mm 4
12
12
n2
1.4
(24)(6)3 (1.4)(24)(6)(2.5)2 1864.8 mm 4
b2 h23 n2 A2 d 22
12
12
I1 I 2 4060.8 mm 4 4.0608 10 9 m 4
n
1.0,
I1
I2
(a)
Aluminum:
y
nMy
I
12
5.5
6.5 mm
0.0065 m
(1.0)(35)(0.0065)
4.0608 10 9
56.0 106 Pa
56.0 MPa
56.0 MPa
(b)
Copper:
n
1.4,
y
nMy
I
5.5 mm
0.0055 m
(1.4)(35)( 0.0055)
4.0608 10 9
66.4 106 Pa
66.4 MPa
66.4 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
487
PROBLEM 4.40
A copper strip (Ec = 105 GPa) and an aluminum strip (Ea = 75 GPa)
are bonded together to form the composite beam shown. Knowing
that the beam is bent about a horizontal axis by a couple of moment
M = 35 N m, determine the maximum stress in (a) the aluminum
strip, (b) the copper strip.
SOLUTION
Use aluminum as the reference material.
n 1.0 in aluminum
n
Ec /Ea
105/75 1.4 in copper
Transformed section:
A, mm2
nA, mm 2
Ay0 , mm
nAy0 , mm3
216
216
7.5
1620
72
100.8
1.5
Σ
Y0
151.8
316.8
1771.2
316.8
1771.2
5.5909 mm
The neutral axis lies 5.5909 mm above the bottom.
I
n1
1.0
b1h13 n1 A1d12
(24)(9)3 (1.0)(24)(9)(1.9091)2 2245.2 mm 4
12
12
n2
1.4
b2 h23 n2 A2 d 22
(24)(3)3 (1.4)(24)(3)(4.0909) 2 1762.5 mm 4
12
12
I1 I 2 4839 mm 4 4.008 10 9 m 4
n
1.0,
I1
I2
(a)
Aluminum:
y
nMy
I
12
5.5909
6.4091 mm
(1.0)(35)(0.0064091)
4.008 10 9
0.0064091
56.0 10
6
Pa
56.0 MPa
56.0 MPa
(b)
Copper:
n
1.4,
y
nMy
I
5.5909 mm
0.0055909 m
(1.4)(35)( 0.0055909)
4.008 10 9
68.4 106 Pa
68.4 MPa
68.4 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
488
PROBLEM 4.41
6 in.
M
12 in.
53
1
2
The 6 12-in. timber beam has been strengthened by bolting to it the steel
reinforcement shown. The modulus of elasticity for wood is 1.8 106 psi and for
steel, 29 106 psi. Knowing that the beam is bent about a horizontal axis by a
couple of moment M 450 kip in., determine the maximum stress in (a) the
wood, (b) the steel.
in.
SOLUTION
Use wood as the reference material.
For wood,
n 1
For steel,
n
Es /Ew
wood
Transformed section:
Yo
29 / 1.8 16.1111
421.931
112.278
3.758 in.
steel
A, in 2
nA, in 2
72
72
6
40.278
0.25
2.5
y0
112.278
nA y0 , in 3
432
10.069
421.931
The neutral axis lies 3.758 in. above the wood-steel interface.
I1
I2
I
M
(a)
n1
1
b1h13 n1 A1d12
(6)(12)3 (72)(6 3.758) 2 1225.91 in 4
12
12
n2
16.1111
b2 h23 n2 A2 d 22
(5)(0.5)3 (40.278)(3.578 0.25)2
12
12
I1 I 2 1873.77 in 4
nMy
I
450 kip in.
Wood:
n 1,
y 12 3.758 8.242 in.
w
(b)
Steel:
(1)(450)(8.242)
1873.77
n 16.1111,
s
647.87 in 4
y
1.979 ksi
3.758 0.5
1.979 ksi
w
4.258 in.
(16.1111)(450)( 4.258)
16.48 ksi
1873.77
s
16.48 ksi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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489
PROBLEM 4.42
6 in.
M
12 in.
The 6 12-in. timber beam has been strengthened by bolting to it the steel
reinforcement shown. The modulus of elasticity for wood is 1.8 106 psi and for
steel, 29 106 psi. Knowing that the beam is bent about a horizontal axis by a
couple of moment M 450 kip in., determine the maximum stress in (a) the
wood, (b) the steel.
C8 & 11.5
SOLUTION
Use wood as the reference material.
For wood,
n 1
For steel,
n
Es
Ew
29 106
1.8 106
16.1111
For C8 11.5 channel section,
A 3.38 in 2 , tw
0.220 in., x
0.571 in., I y
1.32 in 4
For the composite section, the centroid of the channel (part 1) lies 0.571 in. above the bottom of the section.
The centroid of the wood (part 2) lies 0.220 6.00 6.22 in. above the bottom.
Transformed section:
A, in2
3.38
Part
1
72
2
nA, in2
54.456
y , in.
0.571
nAy , in 3
31.091
d, in.
3.216
72
6.22
447.84
2.433
478.93
126.456
Y0
478.93 in 3
126.456 in 2
d
3.787 in.
y0
Y0
The neutral axis lies 3.787 in. above the bottom of the section.
I1
I2
I
M
(a)
n1 I1
n1 A1d12
(16.1111)(1.32) (54.456)(3.216) 2
n2
1
b2 h23 n2 A2 d 22
(6)(12)3
12
12
I1 I 2 1874.69 in 4
450 kip in
(72)(2.433) 2
1290.20 in 4
n My
I
y 12 0.220 3.787 8.433 in.
Wood:
n 1,
Steel:
(1)(450)(8.433)
2.02 ksi
1874.69
n 16.1111, y
3.787 in.
w
(b)
s
584.49 in 4
(16.1111)(450)( 3.787)
14.65 ksi
1874.67
w
s
2.02 ksi
14.65 ksi
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490
PROBLEM 4.43
6 mm
Aluminum
Copper
6 mm
For the composite beam indicated, determine the radius of curvature
caused by the couple of moment 35 N m.
Beam of Prob. 4.39.
24 mm
SOLUTION
See solution to Prob. 4.39 for the calculation of I.
1
M
Ea I
35
(75 10 )(4.0608 10 9 )
9
0.1149 m
1
8.70 m
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491
PROBLEM 4.44
Aluminum
9 mm
Copper
3 mm
For the composite beam indicated, determine the radius of curvature
caused by the couple of moment 35 N m.
Beam of Prob. 4.40.
24 mm
SOLUTION
See solution to Prob. 4.40 for the calculation of I.
1
M
Ea I
35
(75 10 )(4.008 10 9 )
9
0.1164 m
1
8.59 m
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492
6 in.
PROBLEM 4.45
For the composite beam indicated, determine the radius of curvature caused by the
couple of moment 450 kip in.
M
12 in.
Beam of Prob. 4.41.
53
1
2
in.
SOLUTION
See solution to Prob. 4.41 for calculation of I.
I
1873.77 in 4
M
450 kip in
1
M
EI
Ew
1.8 106 psi
450 103 lb in.
450 103
(1.8 106 )(1873.77)
133.421 10 6 in.
1
7495 in. 625 ft
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493
6 in.
PROBLEM 4.46
M
12 in.
For the composite beam indicated, determine the radius of curvature caused by the
couple of moment 450 kip in.
Beam of Prob. 4.42.
C8 & 11.5
SOLUTION
See solution to Prob. 4.42 for calculation of I.
I
M
1
1874.69 in 4
Ew
1.8 106 psi
450 kip in. 450 103 lb in.
M
EI
450 103
(1.8 106 )(1874.69)
133.355 10 6 in.
1
7499 in. 625 ft
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494
PROBLEM 4.47
5
8
-in. diameter
4 in.
A concrete slab is reinforced by 85 -in.-diameter steel rods
placed on 5.5-in. centers as shown. The modulus of elasticity is
106 psi for the steel.
3 106 psi for the concrete and 29
Using an allowable stress of 1400 psi for the concrete and
20 ksi for the steel, determine the largest bending moment in a
portion of slab 1 ft wide.
5.5 in.
5.5 in.
5.5 in.
6 in.
5.5 in.
SOLUTION
n
29 106
3 106
Es
Ec
9.6667
Consider a section 5.5 in. wide.
As
4
d s2
5
4 8
2
0.3068 in 2
2.9657 in 2
nAs
Locate the natural axis.
5.5 x
x
2
(4
2.75 x 2
x )(2.9657)
0
2.9657 x 11.8628 0
Solve for x.
x 1.6066 in.
I
M
2.3934 in.
1
(5.5) x3 (2.9657)(4 x)2
3
1
(5.5)(1.6066)3 (2.9657)(2.3934) 2
3
nMy
I
Concrete: n
4 x
1,
y
M
I
ny
1.6066 in.,
(24.591)(1400)
(1.0)(1.6066)
24.591 in 4
1400 psi
21.429
103 lb in.
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495
PROBLEM 4.47 (Continued)
Steel:
n
M
9.6667,
y
2.3934 in.,
(24.591)(20 103 )
(9.6667)(2.3934)
21.258
20 ksi=20 103 psi
103 lb in.
Choose the smaller value as the allowable moment for a 5.5 in. width.
M
21.258
103 lb in.
For a 1 ft = 12 in. width,
M
12
(21.258
5.5
M
46.38 kip in.
103 )
46.38
103 lb in.
3.87 kip ft
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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496
PROBLEM 4.48
5
8
-in. diameter
Solve Prob. 4.47, assuming that the spacing of the 85 -in.-diameter
steel rods is increased to 7.5 in.
4 in.
PROBLEM 4.47 A concrete slab is reinforced by 85 -in.-diameter
steel rods placed on 5.5-in. centers as shown. The modulus of
elasticity is 3 × 106 psi for the concrete and 29 106 psi for the
steel. Using an allowable stress of 1400 psi for the concrete and
20 ksi for the steel, determine the largest bending moment in a
portion of slab 1 ft wide.
5.5 in.
5.5 in.
5.5 in.
6 in.
5.5 in.
SOLUTION
29 106 psi
3 106 psi
Es
Ec
n
9.667
Number of rails per foot:
12 in.
1.6
7.5 in.
5
5
Area of -in.-diameter bars per foot: 1.6
8
4 8
2
0.4909 in 2
As
Transformed section, all concrete.
First moment of area:
12 x
x
2
4.745(4 x) 0
x 1.4266 in.
nAs
I NA
For concrete:
all
M
For steel:
all
M
We choose the smaller M.
M
1
(12)(1.4266)3
3
1400 psi c
I
c
steel
n
I
c
4.745(4 1.4266)2
4.745 in 2
43.037 in 4
x 1.4266 in.
(1400 psi)
20 ksi c
9.667(0.4909)
4 x
43.037 in 4
1.4266 in.
M
42.24 kip.in.
M
34.60 kip in.
4 1.4266 2.5734 in.
20 ksi 43.042 in 4
9.667 2.5734 in.
34.60 kip in.
M
Steel controls.
2.88 kip ft
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497
PROBLEM 4.49
540 mm
The reinforced concrete beam shown is subjected to a positive bending
moment of 175 kN m. Knowing that the modulus of elasticity is 25 GPa
for the concrete and 200 GPa for the steel, determine (a) the stress in the
steel, (b) the maximum stress in the concrete.
25-mm
diameter
60 mm
300 mm
SOLUTION
n
As
nAs
Es
Ec
4
200 GPa
25 GPa
4
d2
(4)
8.0
(25) 2
4
1.9635 103 mm 2
15.708 103 mm 2
Locate the neutral axis.
x
(15.708 103 )(480 x) 0
2
15.708 103 x 7.5398 106 0
300 x
150 x 2
Solve for x.
x
15.708 103
x 177.87 mm,
I
(15.708 103 ) 2 (4)(150)(7.5398 106 )
(2)(150)
480 x 302.13 mm
1
(300) x3 (15.708 103 )(480 x)2
3
1
(300)(177.87)3 (15.708 103 )(302.13)2
3
1.9966 109 mm 4 1.9966 10 3 m 4
nMy
I
(a)
Steel:
y
302.45 mm
0.30245 m
(8.0)(175 103 )( 0.30245)
1.9966 10 3
(b)
Concrete:
y 177.87 mm
212 106 Pa
212 MPa
0.17787 m
(1.0)(175 103 )(0.17787)
1.9966 10 3
15.59 106 Pa
15.59 MPa
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498
PROBLEM 4.50
540 mm
Solve Prob. 4.49, assuming that the 300-mm width is increased to 350 mm.
25-mm
diameter
PROBLEM 4.49 The reinforced concrete beam shown is subjected to a
positive bending moment of 175 kN m. Knowing that the modulus of
elasticity is 25 GPa for the concrete and 200 GPa for the steel, determine
(a) the stress in the steel, (b) the maximum stress in the concrete.
60 mm
300 mm
SOLUTION
n
As
Es
Ec
4
4
200 GPa
25 GPa
d2
(4)
4
8.0
(25) 2
1.9635 103 mm 2
nAs
15.708 103 mm 2
Locate the neutral axis.
x
(15.708 103 )(480 x) 0
2
15.708 103 x 7.5398 106 0
350 x
175 x 2
Solve for x.
(15.708 103 ) 2 (4)(175)(7.5398 106 )
(2)(175)
x 167.48 mm, 480 x 312.52 mm
x
I
(a)
Steel:
y
15.708 103
1
(350) x3 (15.708 103 )(480 x)2
3
1
(350)(167.48)3 (15.708 103 )(312.52) 2
3
2.0823 109 mm 4 2.0823 10 3 m 4
nMy
I
312.52 mm
0.31252 m
(8.0)(175 103 )( 0.31252)
2.0823 10 3
(b)
Concrete:
y 167.48 mm
210 106 Pa
210 MPa
0.16748 m
(1.0)(175 103 )(0.16748)
2.0823 10 3
14.08 106 Pa
14.08 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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499
4 in.
24 in.
PROBLEM 4.51
Knowing that the bending moment in the reinforced concrete beam is
100 kip ft and that the modulus of elasticity is 3.625 106 psi for the
concrete and 29 106 psi for the steel, determine (a) the stress in the
steel, (b) the maximum stress in the concrete.
20 in.
1-in.
diameter
2.5 in.
12 in.
SOLUTION
n
Es
Ec
As
(4)
29 106
3.625 106
(1)2
4
8.0
3.1416 in 2
nAs
25.133 in 2
Locate the neutral axis.
96 x
6x2
192
Solve for x.
(24)(4)( x
2)
339.3
25.133x
d3
I1
I2
I3
I
Steel:
17.5
nA3d32
(25.133)(12.350) 2
I1
I3
n
Concrete:
where M
8.0
n
1.0,
c
6x2
4
121.133x
x)
147.3
0
0
1.150 in.
12.350 in.
(24)(4)(3.150) 2
1080.6 in 4
6.1 in 4
3833.3 in 4
4920 in 4
100 kip ft
y
y
1200 kip in.
12.350 in.
(8.0)(1200)( 12.350)
4920
s
(b)
x
1
(12)(1.150)3
3
I2
or
1
(24)(4)3
12
A1d12
(25.133)(17.5
(4)(6)(147.3)
4
1
b1h13
12
1 3
b2 x
3
nMy
I
(a)
0
(121.133)2
(2)(6)
121.133
x
x
2
(12 x)
4
1.150
s
24.1 ksi
5.150 in.
(1.0)(1200)(5.150)
4920
c
1.256 ksi
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500
PROBLEM 4.52
7
8
16 in.
-in. diameter
2 in.
A concrete beam is reinforced by three steel rods placed as shown. The
modulus of elasticity is 3 106 psi for the concrete and 29 106 psi for the
steel. Using an allowable stress of 1350 psi for the concrete and 20 ksi for the
steel, determine the largest allowable positive bending moment in the beam.
8 in.
SOLUTION
As
8x
Locate the neutral axis:
29 10 6
3 106
Es
Ec
n
3
x
2
4
d2
(3)
9.67
4
2
7
8
1.8040 in 2
nAs
17.438 in 2
(17.438)(14 x) 0
4 x 2 17.438 x 244.14 0
Solve for x.
17.438
x
17.4382 (4)(4)(244.14)
(2)(4)
5.6326 in.
14 x 8.3674 in.
I
1 3
8x
3
nMy
I
nAs (14
M
x) 2
1
(8)(5.6326)3
3
(17.438)(8.3674)2
1697.45 in 4
I
ny
n 1.0,
Concrete:
M
Steel:
n
M
Choose the smaller value.
5.6326 in.,
y
(1350)(1697.45)
(1.0)(5.6326)
9.67,
y
(20 103 )(1697.45)
(9.67)(8.3674)
M
1350 psi
406.835 103 lb in.
8.3674 in.,
419.72 lb in.
407 kip in.
407 kip in.
20 103 psi
420 kip in.
M
33.9 kip ft
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501
PROBLEM 4.53
d
The design of a reinforced concrete beam is said to be balanced if the maximum stresses
in the steel and concrete are equal, respectively, to the allowable stresses s and c .
Show that to achieve a balanced design the distance x from the top of the beam to the
neutral axis must be
d
x
1
b
s Ec
c Es
where Ec and Es are the moduli of elasticity of concrete and steel, respectively, and d is
the distance from the top of the beam to the reinforcing steel.
SOLUTION
s
s
c
d
x
x
nM (d x)
Mx
c
I
I
n(d x)
d
n
n
x
x
1
1
n
d
Ec
1
Es
s
1
c
Ec
Es
s
c
s
c
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502
PROBLEM 4.54
d
For the concrete beam shown, the modulus of elasticity is 25 GPa for the concrete and
200 GPa for the steel. Knowing that b = 200 mm and d = 450 mm, and using an
allowable stress of 12.5 MPa for the concrete and 140 MPa for the steel, determine
(a) the required area As of the steel reinforcement if the beam is to be balanced, (b) the
largest allowable bending moment. (See Prob. 4.53 for definition of a balanced beam.)
b
SOLUTION
n
s
s
c
d
x
1
x
0.41667d
bx
Locate neutral axis.
As
(a)
x
2
M
1
c
nAs (d
1 140 106
2.40
8.0 12.5 106
(0.41667)(450) 187.5 mm
x)
(200)(187.5) 2
1674 mm 2
(2)(8.0)(262.5)
M
y 187.5 mm
0.1875 m
(1.3623 10 3 )(12.5 106 )
(1.0)(0.1875)
n 8.0
Steel:
s
bx 2
2n ( d x )
n 1.0
Concrete:
1
n
1 3
1
(200)(187.5)3
bx nAs (d x)2
3
3
9
4
1.3623 10 mm 1.3623 10 3 m 4
nMy
I
M
I
ny
I
(b)
200 109
8.0
25 109
nM (d x)
Mx
c
I
I
n( d x )
d
n
n
x
x
Es
Ec
y
262.5 mm
As
1674 mm 2
(8.0)(1674)(262.5) 2
12.5 106 Pa
90.8 103 N m
0.2625 m
(1.3623 10 3 )(140 106 )
(8.0)(0.2625)
140 106 Pa
90.8 103 N m
Note that both values are the same for balanced design.
M
90.8 kN m
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503
Aluminum
0.5 in.
Brass
0.5 in.
Steel
0.5 in.
Brass
0.5 in.
Aluminum
0.5 in.
1.5 in.
PROBLEM 4.55
Five metal strips, each a 0.5 × 1.5-in. cross section, are bonded together to
form the composite beam shown. The modulus of elasticity is 30 × 106 psi
for the steel, 15 × 106 psi for the brass, and 10 × 106 psi for the aluminum.
Knowing that the beam is bent about a horizontal axis by a couple of
moment 12 kip in., determine (a) the maximum stress in each of the
three metals, (b) the radius of curvature of the composite beam.
SOLUTION
Use aluminum as the reference material.
n
30 106
10 106
Es
Ea
3.0 in steel
15 106
1.5 in brass
10 106
n 1.0 in aluminum
n
Eb
Ea
For the transformed section,
I1
I2
I3
I4
n1
b1h13 n1 A1d12
12
0.7656 in 4
1
(1.5)(0.5)3
12
(0.75)(1.0) 2
n2
1.5
(1.5)(0.5)3 (1.5)(0.75)(0.5)2
b2 h23 n2 A2 d 22
12
12
n3
3.0
(1.5)(0.5)3 0.0469 in 4
b3 h33
12
12
I 2 0.3047 in 4
I 5 I1 0.7656 in 4
5
I
Ii
0.3047 in 4
2.1875 in 4
1
(a)
Aluminum:
nMy
I
(1.0)(12)(1.25)
2.1875
6.86 ksi
Brass:
nMy
I
(1.5)(12)(0.75)
2.1875
6.17 ksi
Steel:
nMy
I
(3.0)(12)(0.25)
2.1875
4.11 ksi
(b)
1
M
Ea I
12 103
(10 106 )(2.1875)
548.57 10
6
in.
1
1823 in. = 151.9 ft
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
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504
Steel
0.5 in.
Aluminum
0.5 in.
Brass
0.5 in.
Aluminum
0.5 in.
Steel
0.5 in.
1.5 in.
PROBLEM 4.56
Five metal strips, each a 0.5 × 1.5-in. cross section, are bonded together to
form the composite beam shown. The modulus of elasticity is 30 × 106 psi
for the steel, 15 × 106 psi for the brass, and 10 × 106 psi for the aluminum.
Knowing that the beam is bent about a horizontal axis by a couple of
moment 12 kip · in., determine (a) the maximum stress in each of the
three metals, (b) the radius of curvature of the composite beam.
SOLUTION
Use aluminum as the reference material.
n
30 106
10 106
Es
Ea
3.0 in steel
15 106
1.5 in brass
10 106
n 1.0 in aluminum
n
Eb
Ea
For the transformed section,
I1
I2
I3
I4
n1
b1h13 n1 A1d12
12
3.0
(1.5)(0.5)3 (3.0)(0.75)(1.0)2
12
2.2969 in 4
n2
1.0
(1.5)(0.5)3 (1.0)(0.75)(0.5)2
b2 h23 n2 A2 d 22
12
12
n3
1.5
(1.5)(0.5)3 0.0234 in 4
b3 h33
12
12
I 2 0.2031 in 4
I 5 I1 2.2969 in 4
5
I
Ii
0.2031 in 4
5.0234 in 4
1
(a)
Steel:
nMy
I
(3.0)(12)(1.25)
5.0234
Aluminum:
nMy
I
(1.0)(12)(0.75)
1.792 ksi
5.0234
Brass:
nMy
I
(1.5)(12)(0.25)
5.0234
(b)
1
M
Ea I
8.96 ksi
12 103
(10 106 )(5.0234)
0.896 ksi
238.89 10
6
in.
1
4186 in. 349 ft
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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505
Brass
PROBLEM 4.57
Aluminum
The composite beam shown is formed by bonding together a brass rod and an
aluminum rod of semicircular cross sections. The modulus of elasticity is
15 106 psi for the brass and 10 106 psi for the aluminum. Knowing that the
composite beam is bent about a horizontal axis by couples of moment
8 kip in., determine the maximum stress (a) in the brass, (b) in the aluminum.
0.8 in.
SOLUTION
r
For each semicircle,
y0
4r
3
I
I base
(4)(0.8)
3
Ay02
0.8 in.
A
r2
2
0.33953 in.
1.00531 in 2 ,
I base
8
r4
0.160850 in 4
(1.00531)(0.33953)2
0.160850
0.044953 in 4
Use aluminum as the reference material.
n 1.0 in aluminum
n
15 106
10 106
Eb
Ea
1.5 in brass
Locate the neutral axis.
A, in2
nA, in2
1.00531
1.50796
0.33953
0.51200
1.00531
1.00531
0.33953
0.34133
y0 , in.
nAy0 , in 3
2.51327
d1
0.06791
I1
n1I
n1 Ad12
I2
n2 I
n2 Ad 22
I
(a)
0.33953
I1
I2
0.27162 in., d 2
(1.5)(0.044957)
0.33953
0.06791
(1.5)(1.00531)(0.27162)
(1.0)(0.044957)
0.17067
2.51327
0.06791 in.
The neutral axis lies 0.06791 in.
above the material interface.
0.17067
2
(1.0)(1.00531)(0.40744)2
0.40744 in.
0.17869 in 4
0.21185 in 4
0.39054 in 4
n
Brass:
1.5,
y
0.8
nMy
I
(b)
Y0
Aluminium:
n
1.0,
y
0.8
nMy
I
0.06791
0.73209 in.
(1.5)(8)(0.73209)
0.39054
0.06791
22.5 ksi
0.86791 in.
(1.0)(8)( 0.86791)
0.39054
17.78 ksi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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506
y
Aluminum
PROBLEM 4.58
3 mm
A steel pipe and an aluminum pipe are securely bonded together to form
the composite beam shown. The modulus of elasticity is 200 GPa for the
steel and 70 GPa for the aluminum. Knowing that the composite beam is
bent by a couple of moment 500 N m, determine the maximum stress
(a) in the aluminum, (b) in the steel.
Steel
6 mm
z
10 mm
38 mm
SOLUTION
Use aluminum as the reference material.
n 1.0 in aluminum
n
Es / Ea
Steel:
Is
ns
Aluminium:
Ia
na
I
Is
200 / 70 2.857 in steel
For the transformed section,
(a)
Aluminum:
4
ro4
ri4
(2.857)
ro4
ri4
(1.0)
4
Ia
c 19 mm
a
na Mc
I
4
4
(164 104 ) 124.62 103 mm 4
(194 164 ) 50.88 103 mm 4
175.50 103 mm 4
175.5 10
9
m4
0.019 m
(1.0)(500)(0.019)
175.5 10 9
54.1 106 Pa
a
(b)
Steel:
c 16 mm
s
ns Mc
I
54.1 MPa
0.016 m
(2.857)(500)(0.016)
175.5 10 9
130.2 106 Pa
s
130.2 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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507
%"
M
PROBLEM 4.59
Et #
100 mm
1
2
Ec
%'
50 mm
Ec
The rectangular beam shown is made of a plastic for which the
value of the modulus of elasticity in tension is one-half of its
value in compression. For a bending moment M 600 N m,
determine the maximum (a) tensile stress, (b) compressive stress.
SOLUTION
n
1
on the tension side of neutral axis
2
n 1 on the compression side
Locate neutral axis.
n1bx
x2
x
h x
(a)
Tensile stress:
x
h x
n2 b(h x)
2
2
1 2 1
bx
b( h x ) 2
2
4
Compressive stress:
0
1
1
(h x) 2 x
(h x)
2
2
1
h 0.41421 h 41.421 mm
2 1
58.579 mm
I1
1
n1 bx3
3
I2
1
n2 b(h x)3
3
I
I1
I2
n
1
,
2
y
nMy
I
(b)
0
n 1,
y
nMy
I
(1)
1
(50)(41.421)3 1.1844 106 mm 4
3
1
2
1
(50)(58.579)3 1.6751 106 mm 4
3
2.8595 106 mm 4
58.579 mm
2.8595 10 6 m 4
0.058579 m
(0.5)(600)( 0.058579)
2.8595 10 6
41.421 mm
6.15 106 Pa
t
6.15 MPa
0.041421 m
(1.0)(600)(0.041421)
2.8595 10 6
8.69 106 Pa
c
8.69 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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508
PROBLEM 4.60*
A rectangular beam is made of material for which the modulus of elasticity is Et in tension and Ec in
compression. Show that the curvature of the beam in pure bending is
1
M
Er I
where
Er
4 Et Ec
Ec ) 2
( Et
SOLUTION
Use Et as the reference modulus.
Then Ec
nEt .
Locate neutral axis.
nbx
nx 2
x
I trans
x
2
b( h x )
(h x) 2
h
nh
h x
n 1
n 3
bx
3
h x
0
2
0
nx (h x)
n 1
1
b ( h x )3
3
M
Et I trans
Er I
Et I trans
Er
Et I trans
I
where I
12
bh3
Et
1
n 1
1
3
bh3
n
n 1
2
bh3
1 3
bh
12
n
3
3
n
n 1
M
Er I
4 Et Ec /Et
Ec /Et
3
1
n
1n 1
bh3
3 n 13
1 n n3/ 2
bh3
3 n 13
1
h
3
n 1
2
bh3
4 Et Ec
2
Ec
Et
2
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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509
PROBLEM 4.61
8 mm
r
M
80 mm
Knowing that M 250 N m, determine the maximum stress in the beam
shown when the radius r of the fillets is (a) 4 mm, (b) 8 mm.
40 mm
SOLUTION
c
1 3
bh
12
20 mm
D
d
80 mm
40 mm
2.00
r
d
4 mm
40 mm
0.10
I
(a)
max
(b)
r
d
K
Mc
I
8 mm
40 mm
max
K
1
(8)(40)3
12
0.020 m
42.667 103 mm 4
0.20
Mc
I
K
From Fig. 4.27,
(1.87)(250)(0.020)
42.667 10 9
K
176.0 106 Pa
9
m4
1.87
219 106 Pa
From Fig. 4.27,
(1.50)(250)(0.020)
42.667 10 9
42.667 10
max
219 MPa
1.50
max
176.0 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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510
PROBLEM 4.62
8 mm
r
M
80 mm
Knowing that the allowable stress for the beam shown is 90 MPa, determine
the allowable bending moment M when the radius r of the fillets is (a) 8 mm,
(b) 12 mm.
40 mm
SOLUTION
c
1 3
bh
12
20 mm
D
d
80 mm
40 mm
2.00
r
d
8 mm
40 mm
0.2
I
(a)
max
(b)
r
d
K
12 mm
40 mm
1
(8)(40)3
12
0.020 m
Mc
I
0.3
42.667 103 mm 4
K
From Fig. 4.27,
M
max I
Kc
9
m4
1.50
(90 106 )(42.667 10 9 )
(1.50)(0.020)
K
From Fig. 4.27,
M
42.667 10
M
128.0 N m
M
142.0 N m
1.35
(90 106 )(42.667 10 9 )
(1.35)(0.020)
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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511
3
4
r
PROBLEM 4.63
in.
Semicircular grooves of radius r must be milled as shown in the sides of
a steel member. Using an allowable stress of 8 ksi, determine the largest
bending moment that can be applied to the member when (a) r = 83 in.,
(b) r = 34 in.
M
4.5 in.
SOLUTION
(a)
d
D
2r
D
d
4.5
3.75
4.5
1.20
From Fig. 4.28,
I
1 3
bh
12
Mc
K
I
(2)
3
8
r
d
3.75 in.
0.375
3.75
K
0.10
2.07
1 3
(3.75)3
12 4
I
M
Kc
3.296 in 4
1
2
c
(8)(3.296)
(2.07)(1.875)
1.875 in.
6.79 kip in.
6.79 kip in.
(b)
d
D
2r
4.5
From Fig. 4.28,
(2)
3
4
3.0
D
d
4.5
3.0
K
1.61
I
c
1
d
2
1.5 in.
r
d
1.5
1 3
bh
12
M
0.75
3.0
1 3
(3.0)3
12 4
I
Kc
0.25
1.6875 in 4
(8)(1.6875)
(1.61)(1.5)
5.59 kip in.
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512
3
4
r
PROBLEM 4.64
in.
Semicircular grooves of radius r must be milled as shown in the sides
of a steel member. Knowing that M = 4 kip · in., determine the
maximum stress in the member when the radius r of the semicircular
grooves is (a) r = 83 in., (b) r = 34 in.
M
4.5 in.
SOLUTION
(a)
From Fig. 4.28,
d
D
D
d
4.5
3.75
K
2.07
2r
4.5
r
d
1.20
1 3
bh
12
Mc
K
I
I
3
8
(2)
3.75 in.
0.375
3.75
0.10
1 3
(3.75)3 3.296 in 4
12 4
(2.07)(4)(1.875)
4.71 ksi
3.296
c
1
2
1.875 in.
4.71 ksi
(b)
d
D
2r
4.5
(2)
3
4
3.00 in.
D
d
4.5
3.00
1.50
r
d
0.75
3.0
c
1
d
2
1.5 in.
0.25
From Fig. 4.28,
K
1.61
I
1 3
bh
12
1 3
(3.00)3
12 4
K
Mc
I
1.6875 in 4
(1.61)(4)(1.5)
1.6875
5.72 ksi
5.72 ksi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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513
M
PROBLEM 4.65
M
100 mm
100 mm
150 mm
150 mm
18 mm
(a)
18 mm
A couple of moment M = 2 kN · m is to be
applied to the end of a steel bar. Determine the
maximum stress in the bar (a) if the bar is
designed with grooves having semicircular
portions of radius r = 10 mm, as shown in Fig. a,
(b) if the bar is redesigned by removing the
material to the left and right of the dashed lines
as shown in Fig. b.
(b)
SOLUTION
For both configurations,
D
150 mm
d
100 mm
r
10 mm
D
d
r
d
150
100
10
100
Fig. 4.28 gives
Ka
2.21.
For configuration (b),
Fig. 4.27 gives K b
1.50
0.10
For configuration (a),
I
c
(a)
KMc
I
1.79.
1 3
1
(18)(100)3 1.5 106 mm 4
bh
12
12
1
d 50 mm 0.05 m
2
(2.21)(2 103 )(0.05)
1.5 10 6
147.0 106 Pa
1.5 10
6
m4
147.0 MPa
147.0 MPa
(b)
KMc
I
(1.79)(2 103 )(0.05)
1.5 10 6
119.0 106 Pa
119.0 MPa
119.0 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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514
M
100 mm
150 mm
PROBLEM 4.66
M
100 mm
150 mm
18 mm
(a)
18 mm
The allowable stress used in the design of a
steel bar is 80 MPa. Determine the largest
couple M that can be applied to the bar (a) if
the bar is designed with grooves having
semicircular portions of radius r 15 mm, as
shown in Fig. a, (b) if the bar is redesigned by
removing the material to the left and right of
the dashed lines as shown in Fig. b.
(b)
SOLUTION
For both configurations,
D
150 mm
r
15 mm
D
d
r
d
150
100
15
100
d
100 mm
1.50
0.15
For configuration (a), Fig. 4.28 gives K a
1.92.
For configuration (b), Fig. 4.27 gives Kb
1.57.
I
c
KMc
I
(a)
M
I
Kc
1 3
1
(18)(100)3 1.5 106 mm 4
bh
12
12
1
d 50 mm 0.050 m
2
(80 106 )(1.5 10 6 )
(1.92) (0.05)
1.5 10
6
m4
1.250 103 N m
1.250 kN m
(a)
M
I
Kc
(80 106 )(1.5 10 6 )
(1.57)(0.050)
1.530 103 N m
1.530 kN m
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
515
M
PROBLEM 4.67
The prismatic bar shown is made of a steel that is assumed to be
elastoplastic with Y 300 MPa and is subjected to a couple M
parallel to the x axis. Determine the moment M of the couple for which
(a) yield first occurs, (b) the elastic core of the bar is 4 mm thick.
x
z
12 mm
8 mm
SOLUTION
(a)
I
1 3
bh
12
1
(12)(8)3
12
c
yY
1
(4)
2
2 mm
512 10 12 m 4
1
h 4 mm
0.004 m
2
(300 106 )(512 10
c
0.004
38.4 N m
YI
MY
(b)
512 mm 4
yY
c
2
4
M
3
1 yY
MY 1
2
3 c
12
)
MY
38.4 N m
M
52.8 N m
0.5
2
3
1
(38.4) 1
(0.5) 2
2
3
52.8 N m
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
516
M
PROBLEM 4.68
Solve Prob. 4.67, assuming that the couple M is parallel to the z axis.
PROBLEM 4.67 The prismatic bar shown is made of a steel that is
assumed to be elastoplastic with Y 300 MPa and is subjected to a
couple M parallel to the x axis. Determine the moment M of the couple
for which (a) yield first occurs, (b) the elastic core of the bar is 4 mm
thick.
x
z
12 mm
8 mm
SOLUTION
(a)
I
1 3
bh
12
1
(8)(12)3
12
c
yY
1
(4)
2
2 mm
1.152 10 9 m 4
1
h 6 mm 0.006 m
2
(300 106 )(1.152 10 9 )
c
0.006
57.6 N m
YI
MY
(b)
1.152 103 mm 4
yY
c
2
6
M
3
1 yY
MY 1
2
3 c
MY
57.6 N m
M
83.2 N m
1
3
3
1 1
(57.6) 1
2
3 3
2
2
83.2 N m
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
517
PROBLEM 4.69
A solid square rod of side 0.6 in. is made of a steel that is assumed to be elastoplastic with E = 29 × 106 psi
and Y 48 ksi. Knowing that a couple M is applied and maintained about an axis parallel to a side of the
cross section, determine the moment M of the couple for which the radius of curvature is 6 ft.
SOLUTION
I
MY
1
(0.6)(0.6)3
12
I
Y
c
c
yY
c
M
Y
M
3
in 4
4.16667 10
1.65517 10
4.16667 10
3
MY 1
2
1 YY
3 c
2
3
1
h
2
c
(10.8 10 3 )(48 103 )
0.3
0.3
72
max
10.8 10
0.3 in.
1728 lb in.
Y
Y
E
48 103
29 106
6 ft
72 in.
1.65517 10
3
3
3
0.39724
3
(1728) 1
2
1
(0.39724)2
3
M
2460 lb in.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
518
PROBLEM 4.70
For the solid square rod of Prob. 4.69, determine the moment M for which the radius of curvature is 3 ft.
PROBLEM 4.69. A solid square rod of side 0.6 in. is made of a steel that is assumed to be elastoplastic with
E = 29 × 106 psi and Y 48 ksi. Knowing that a couple M is applied and maintained about an axis parallel
to a side of the cross section, determine the moment M of the couple for which the radius of curvature is 6 ft.
SOLUTION
I
MY
1
(0.6)(0.6)3
12
I
Y
c
yY
c
M
3
in 4
Y
max
8.3333 10
1.65517 10 3
8.3333 10 3
3
MY 1
2
1 YY
3 c
2
3
1
h
2
c
(10.8 10 3 )(48 103 )
0.3
0.3
36
c
max
10.8 10
0.3 in.
1728 lb in.
Y
48 103
29 106
Y
E
3 ft
36 in.
1.65517 10
3
0.19862
3
(1728) 1
2
1
(0.19862)2
3
M
2560 lb in.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
519
y
PROBLEM 4.71
18 mm
The prismatic rod shown is made of a steel that is assumed to be elastoplastic with
E 200 GPa and Y 280 MPa . Knowing that couples M and M of moment
525 N · m are applied and maintained about axes parallel to the y axis, determine
(a) the thickness of the elastic core, (b) the radius of curvature of the bar.
M
24 mm
M!
x
SOLUTION
I
c
MY
M
1 3
1
bh
(24)(18)3 11.664 103 mm 4
12
12
1
h 9 mm 0.009 m
2
YI
c
(280 106 )(11.664 10 9 )
0.009
3
MY 1
2
yY
c
3
1 yY 2
3 c2
(2)(525)
362.88
or
0.32632
yY
c
yY
11.664 10
3
m4
362.88 N m
3
2
M
MY
0.32632c
2.9368 mm
(a)
(b)
tcore
Y
yY
Y
E
EyY
Y
(200 109 )(2.9368 10 3 )
280 106
2 yY
5.87 mm
2.09 m
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
520
y
PROBLEM 4.72
18 mm
Solve Prob. 4.71, assuming that the couples M and M are applied and
maintained about axes parallel to the x axis.
M
24 mm
PROBLEM 4.71 The prismatic rod shown is made of a steel that is assumed to
be elastoplastic with E 200 GPa and Y 280 MPa . Knowing that couples
M and M of moment 525 N · m are applied and maintained about axes parallel
to the y axis, determine (a) the thickness of the elastic core, (b) the radius of
curvature of the bar.
M!
x
SOLUTION
I
c
MY
M
yY
c
1 3
1
bh
(18)(24)3 20.736 103 mm 4
12
12
1
h 12 mm 0.012 m
2
YI
c
(280 106 )(20.736 10 9 )
0.012
3
MY 1
2
3
1 yY 2
3 c2
(2)(525)
483.84
or
yY
c
0.91097
yY
20.736 10
9
m4
483.84 N m
3
2
M
MY
0.91097
c
10.932 mm
(a)
(b)
tcore
Y
yY
Y
E
EyY
Y
(200 109 )(10.932 10 3 )
280 106
2 yY
21.9 mm
7.81 m
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521
y
z
PROBLEM 4.73
C
90 mm
A beam of the cross section shown is made of a steel that is assumed to be
elastoplastic with E 200 GPa and Y 240 MPa. For bending about the z axis,
determine the bending moment at which (a) yield first occurs, (b) the plastic
zones at the top and bottom of the bar are 30 mm thick.
60 mm
SOLUTION
(a)
I
c
MY
1 3 1
(60)(90)3 3.645 106 mm 4 3.645 10 6 m 4
bh
12
12
1
h 45 mm 0.045 m
2
(240 106 )(3.645 10 6 )
YI
19.44 10 N m
0.045
c
MY
R1
Y
19.44 kN m
(240 106 )(0.060)(0.030)
A1
432 103 N
y1 15 mm 15 mm
R2
1
2
Y
A2
0.030 m
1
(240 106 )(0.060)(0.015)
2
108 103 N
y2
(b)
M
2( R1 y1
R2 y2 )
2
(15 mm) 10 mm
3
0.010 m
2[(432 103 )(0.030) (108 103 )(0.010)]
28.08 103 N m
M
28.1 kN m
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
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522
PROBLEM 4.74
y
A beam of the cross section shown is made of a steel that is assumed to be
elastoplastic with E 200 GPa and Y 240 MPa. For bending about the
z axis, determine the bending moment at which (a) yield first occurs,
(b) the plastic zones at the top and bottom of the bar are 30 mm thick.
30 mm
30 mm
C
z
30 mm
15 mm
30 mm
15 mm
SOLUTION
(a)
1 3 1
bh
(60)(90)3 3.645 106 mm 4
12
12
1 3 1
I cutout
bh
(30)(30)3 67.5 103 mm 4
12
12
I 3.645 106 67.5 103 3.5775 106 mm 4
I rect
3.5775 10
c
MY
45 mm
y1
15 mm
M
0.045 m
(240 106 )(3.5775 10 6 )
0.045
c
3
19.08 10 N m
Y A1
y2
mm 4
YI
R1
R2
(b)
1
h
2
6
(240 106 )(0.060)(0.030)
15 mm
30 mm
19.08 kN m
M
27.0 kN m
432 103 N
0.030 m
1
1
(240 106 )(0.030)(0.015)
Y A2
2
2
2
(15 mm) 10 mm 0.010 m
3
2( R1 y1
MY
54 103 N
R2 y2 )
2[(432 103 )(0.030)
(54 103 )(0.010)]
27.00 103 N m
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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523
PROBLEM 4.75
y
3 in.
3 in.
C
z
A beam of the cross section shown is made of a steel that is assumed to
be elastoplastic with E 29 106 psi and Y 42 ksi. For bending
about the z axis, determine the bending moment at which (a) yield first
occurs, (b) the plastic zones at the top and bottom of the bar are 3 in.
thick.
3 in.
1.5 in.
1.5 in.
3 in.
SOLUTION
(a)
I1
I2
I3
1
1
b1h13 A1d12
(3)(3)3 (3)(3)(3) 2
12
12
1
1
b2 h23
(6)(3)3 13.5 in 4
12
12
I1 87.75 in 4
I 3 188.5 in 4
I
I1
c
4.5 in.
(42)(188.5)
YI
c
4.5
MY
I2
87.75 in 4
MY
R1
Y
A1
1759 kip in.
(42)(3)(3) 378 kip
y1 1.5 1.5 3.0 in.
R2
y2
(b)
M
2( R1 y1
R2 y2 )
1
1
(42)(6)(1.5)
Y A2
2
2
189 kip
2
(1.5) 1.0 in.
3
2[(378)(3.0) (189)(1.0)] 2646 kip in.
M
2650 kip in.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
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524
PROBLEM 4.76
y
3 in.
3 in.
C
z
A beam of the cross section shown is made of a steel that is assumed to
be elastoplastic with E 29 106 psi and Y 42 ksi. For bending
about the z axis, determine the bending moment at which (a) yield first
occurs, (b) the plastic zones at the top and bottom of the bar are 3 in.
thick.
3 in.
1.5 in.
3 in.
1.5 in.
SOLUTION
(a)
I1
I2
I3
1
1
(6)(3)3 (6)(3)(3)2
b1h13 A1d12
12
12
1
1
3
(3)(3)3 6.75 in 4
b2 h2
12
12
I1 175.5 in 4
I
I1
c
4.5 in.
MY
I2
YI
c
I3
175.5 in 4
357.75 in 4
(42)(357.75)
4.5
3339 kip in.
MY
R1
Y
(42)(6)(3)
A1
3340 kip in.
756 kip
y1 1.5 1.5 3 in.
(b)
M
2( R1 y1
R2 y2 )
R2
1
2
y2
2
(1.5) 1.0 in.
3
2[(756)(3) (94.5)(1.0)] 4725 kip in.
Y
A2
1
(42)(3)(1.5)
2
94.5 kip
M 4730 kip in.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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525
PROBLEM 4.77
y
For the beam indicated (of Prob. 4.73), determine (a) the fully plastic moment
M p , (b) the shape factor of the cross section.
z
90 mm
C
60 mm
SOLUTION
From Problem 4.73,
E
200 GPa and
A1
(60)(45)
R
2700 10 6 m 2
Y A1
Y
240 MPa
2700 mm 2
(240 106 )(2700 10 6 )
d
(a)
Mp
(b)
I
c
MY
Rd
(648 103 )(0.045)
1 3 1
bh
(60)(90)3
12
12
45 mm 0.045 m
YI
c
648 103 N
45 mm 0.045 m
29.16 103 N m
3.645 106 mm 4
(240 106 )(3.645 10 6 )
0.045
k
Mp
29.2 kN m
3.645 10 6 m 4
19.44 103 N m
Mp
MY
29.16
19.44
k 1.500
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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526
PROBLEM 4.78
y
For the beam indicated (of Prob. 4.74), determine (a) the fully plastic moment
M p , (b) the shape factor of the cross section.
30 mm
30 mm
C
z
30 mm
15 mm
15 mm
30 mm
SOLUTION
From Problem 4.74,
(a)
R1
y1
Y
200 GPa and
Y
240 MPa
A1
(240 106 )(0.060)(0.030)
432 103 N
15 mm 15 mm 30 mm
0.030 m
R2
y2
E
Y
A2
(240 106 )(0.030)(0.015)
108 103 N
1
(15) 7.5 mm 0.0075 m
2
2( R1 y1
Mp
2[( 432 103 )(0.030) (108 103 )(0.0075)]
R2 y2 )
27.54 103 N m
(b)
I rect
I cutout
I
Mp
1 3 1
(60)(90)3 3.645 106 mm 4
bh
12
12
1 3 1
bh
(30)(30)3 67.5 103 mm 4
12
12
I rect I cutout 3.645 106 67.5 103 3.5775 103 mm 4
3.5775 10
c
MY
k
27.5 kN m
1
h
2
YI
c
Mp
MY
45 mm
9
m4
0.045 m
(240 106 )(3.5775 10 9 )
0.045
27.54
19.08
19.08 103 N m
k 1.443
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
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527
PROBLEM 4.79
y
3 in.
3 in.
C
z
For the beam indicated (of Prob. 4.75), determine (a) the fully plastic
moment M p , (b) the shape factor of the cross section.
3 in.
1.5 in.
1.5 in.
3 in.
SOLUTION
From Problem 4.75,
(a)
R1
Y
A1
E
29 106 psi and
Y
42 ksi.
(42)(3)(3) 378 kip
y1 1.5 1.5 3.0 in.
R2
(b)
Y
A2
(42)(6)(1.5) 378 kip
y2
1
(1.5)
2
0.75 in.
Mp
2( R1 y1
R2 y2 )
I1
I2
I3
1
1
b1h13 A1d12
(3)(3)3 (3)(3)(3) 2
12
12
1
1
b2 h23
(6)(3)3 13.5 in 4
12
12
I1 87.75 in 4
I
I1
c
4.5 in.
MY
k
2[( 378)( 3.0) ( 378)(0. 75)]
YI
c
Mp
MY
I2
Mp
2840 kip in.
87.75 in 4
I 3 188.5 in 4
(42)(188.5)
4.5
2835
1759.3
1759.3 kip in
k 1.611
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528
PROBLEM 4.80
y
For the beam indicated (of Prob. 4.76), determine (a) the fully plastic
moment M p , (b) the shape factor of the cross section.
3 in.
3 in.
C
z
3 in.
1.5 in.
3 in.
1.5 in.
SOLUTION
From Problem 4.76,
(a)
R1
Y
A1
E
(42)(6)(3)
29 106 and
Y
42 ksi
756 kip
y1 1.5 1.5 3.0 in.
R2
(b)
Y
A2
(42)(3)(1.5) 189 kip
y2
1
(1.5)
2
0.75 in.
Mp
2( R1 y1
R2 y2 )
I1
I2
I3
1
1
b1h13 A1d12
(6)(3)3 (6)(3)(3)2
12
12
1
1
3
b2 h2
(3)(3)3 6.75 in 4
12
12
I1 175.5 in 4
I
I1
c
4.5 in.
MY
k
2[( 756)( 3.0) (189)(0. 75)]
I2
YI
c
Mp
MY
I3
Mp
4820 kip in.
175.5 in 4
357.75 in 4
(42)(357.75)
4.5
4819.5
3339
3339 kip in.
k 1.443
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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529
PROBLEM 4.81
r ! 18 mm
Determine the plastic moment M p of a steel beam of the cross section shown,
assuming the steel to be elastoplastic with a yield strength of 240 MPa.
SOLUTION
For a semicircle,
A
2
Resultant force on semicircular section:
r 2;
R
r
4r
3
YA
Resultant moment on entire cross section:
Mp
Data:
Y
Mp
2 Rr
4
3
240 MPa
Yr
3
240 106 Pa,
4
(240 106 )(0.018)3
3
r
18 mm
0.018 m
1866 N m
Mp
1.866 kN m
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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530
PROBLEM 4.82
50 mm
30 mm
Determine the plastic moment M p of a steel beam of the cross section shown,
assuming the steel to be elastoplastic with a yield strength of 240 MPa.
10 mm
10 mm
30 mm
10 mm
SOLUTION
A (50)(90) (30)(30) 3600 mm 2
Total area:
1
A 1800 mm 2
2
1
A 1800
x 2
50
b
A1
(50)(36) 1800 mm 2 ,
A2
(50)(14)
A3
(20)(30)
A4
36 mm
y1 18 mm,
A1 y1
32.4 103 mm3
700 mm 2 ,
y2
7 mm,
A2 y2
4.9 103 mm3
600 mm 2 ,
y3
29 mm,
A3 y3
17.4 103 mm3
(50)(10) 500 mm 2 ,
y4
49 mm,
A4 y4
24.5 103 mm3
A1 y1
Mp
A2 y2
Y
Ai yi
A3 y3
A4 y4
79.2 103 mm3
79.2 10 6 m3
(240 106 )(79.2 10 6 ) 19.008 103 N m
Mp
19.01 kN m
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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531
PROBLEM 4.83
36 mm
Determine the plastic moment M p of a steel beam of the cross section shown,
assuming the steel to be elastoplastic with a yield strength of 240 MPa.
30 mm
SOLUTION
Total area:
A
1
(30)(36) 540 mm 2
2
1
A 270 mm 2
2
Half area:
By similar triangles,
b
y
30
36
b
Since
A1
1
by
2
5 2
y ,
12
A1
5
y
6
y2
12
A1
5
12
(270) 25.4558 mm
5
b 21.2132 mm
1
(21.2132)(25.4558) 270 mm 2 270 10 6 m 2
A1
2
A2 (21.2132)(36 25.4558) 223.676 mm 2 223.676 10 6 m 2
y
A3
Ri
R1
y1
y2
y3
Mp
A
Y
A1
Ai
A2
46.324 mm 2
46.324 10 6 m 2
240 106 Ai
64.8 103 N, R2 53.6822 103 N, R3 11.1178 103 N
1
y 8.4853 mm 8.4853 10 3 m
3
1
(36 25.4558) 5.2721 mm 5.2721 10 3 m
2
2
(36 25.4558) 7.0295 mm 7.0295 10 3 m
3
R1 y1 R2 y2 R3 y3 911 N m
Mp
911 N m
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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532
0.4 in.
PROBLEM 4.84
1.0 in.
Determine the plastic moment Mp of a steel beam of the cross section shown,
assuming the steel to be elastoplastic with a yield strength of 42 ksi.
1.0 in.
0.4 in.
0.4 in.
SOLUTION
A (1.0 in.)(0.4 in.) 2(1.4 in.)(0.4 in.) 1.52 in 2
x
R1
1
2
(1.52)
0.95 in.
2(0.4)
0.4 in 2
(1.0)(0.4)
y1 1.2 0.95 0.25 in.
R2
y2
R3
2(0.4)(1.4 0.95)
0.36 in 2
1
(1.4 0.95) 0.225 in.
2
2(0.4)(0.95) 0.760 in 2
y3
1
(0.95)
2
Mp
( R1 y1
0.475 in.
R2 y 2
R3 y 3 )(
y)
[(0.4)(0.25) (0.36)(0.225) (0.760)(0.475)](42)
Mp
22.8 kip in .
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
533
5 mm
PROBLEM 4.85
5 mm
80 mm
t = 5 mm
120 mm
Determine the plastic moment Mp of the cross section shown when the beam is
bent about a horizontal axis. Assume the material to be elastoplastic with a
yield strength of 175 MPa.
SOLUTION
For M p , the neutral axis divides the area into two equal parts.
Total area
(100 100 120)t
320t
1
(320)t
2
a 80 mm
80
b
(80 mm) 64 mm
100
Shaded area
6
2at
m2
A1
2at
A2
2(100 a )t
2(0.02 m)(0.005 m)
200 10
A3
(120 mm)t
(0.120 m)(0.005 m)
600 10
2(0.08 m)(0.005 m) 800 10
R1
Y
A1
(175 MPa)(800 10
6
m 2 ) 140 kN
R2
Y
A2
(175 MPa)(200 10
6
m 2 ) 35 kN
R3
Y
A3
(175 MPa)(600 10
6
m 2 ) 105 kN
Mz : M p
6
6
m2
m2
R1 (32 mm) R2 (8 mm) R3 (16 mm)
(140 kN)(0.032 m) (35 kN)(0.008 m) (105 kN)(0.016 m)
M p = 6.44 k N m
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Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
534
PROBLEM 4.86
4 in.
1
2
in.
1
2
in.
1
2
in.
3 in.
Determine the plastic moment M p of a steel beam of the cross section shown, assuming
the steel to be elastoplastic with a yield strength of 36 ksi.
2 in.
SOLUTION
Total area:
A (4)
1
2
1
1
(3) (2)
2
2
4.5 in 2
1
A 2.25 in 2
2
A1
2.00 in 2 ,
y1
0.75,
A1 y1 1.50 in 3
A2
0.25 in 2 ,
y2
0.25,
A2 y2
0.0625 in 3
A3 1.25 in 2 ,
y3
1.25,
A3 y3
1.5625 in 3
1.00 in 2 ,
y4
2.75,
A4 y4
2.75 in 3
A4
Mp
Y
( A1 y1
A2 y2
A3 y3
A4 y4 )
(36)(1.50 0.0625 1.5625 2.75)
Mp
212 kip in .
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Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
535
PROBLEM 4.87
y
z
C
90 mm
For the beam indicated (of Prob. 4.73), a couple of moment equal to the full
plastic moment M p is applied and then removed. Using a yield strength of
240 MPa, determine the residual stress at y 45 mm .
60 mm
SOLUTION
29.16 103 N m
Mp
See solutions to Problems 4.73 and 4.77.
I
3.645 10 6 m 4
c
0.045 m
M max y
I
LOADING
Mp c
UNLOADING
(29.16 103 )(0.045)
3.645 10 6
res
Y
360 106
120 106 Pa
I
at y
c
45 mm
RESIDUAL STRESSES
360 106 Pa
240 106
res
120.0 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
536
PROBLEM 4.88
y
30 mm
z
C
30 mm
For the beam indicated (of Prob. 4.74), a couple of moment equal to the
full plastic moment M p is applied and then removed. Using a yield
strength of 240 MPa, determine the residual stress at y 45mm .
30 mm
15 mm
30 mm
15 mm
SOLUTION
Mp
I
27.54 103 N m (See solutions to Problems 4.74 and 4.78.)
3.5775 10 6 m 4 ,
M max y
I
Mp c
I
at
c
0.045 m
y
c
(27.54 103 )(0.045)
3.5775 10 6
LOADING
UNLOADING
res
Y
346.4 106
240 106
346.4 106 Pa
RESIDUAL STRESSES
106.4 106 Pa
res
106.4 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
537
PROBLEM 4.89
y
A bending couple is applied to the bar indicated, causing plastic zones
3 in. thick to develop at the top and bottom of the bar. After the couple
has been removed, determine (a) the residual stress at y 4.5 in.,
(b) the points where the residual stress is zero, (c) the radius of
curvature corresponding to the permanent deformation of the bar.
3 in.
3 in.
C
z
Beam of Prob. 4.75.
3 in.
1.5 in.
1.5 in.
3 in.
SOLUTION
See solution to Problem 4.75 for bending couple and stress distribution during loading.
M
2646 kip in.
42 ksi
Y
Mc
I
MyY
I
(a)
At
y
c,
At
y
yY ,
res
(c)
y0
I Y
M
At
y
188.5 in 4
I
63.167
Y
res
Y
c
E
29 106 psi
29 103 ksi
4.5 in.
42
21.056
21.167 ksi
42
20.944 ksi
UNLOADING
My0
Y
I
(188.5)(42)
2646
0
res
1.5 in.
(2646)(4.5)
63.167 ksi
188.5
(2646)(1.5)
21.056 ksi
188.5
LOADING
(b)
yY
yY ,
Ey
2.99 in.
20.9 ksi
res
RESIDUAL STRESSES
Answer:
y0
2.99 in.,
0,
2.99 in.
20.944 ksi
res
Ey
(29 103 )(1.5)
20.944
2077 in.
173.1 ft
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
538
PROBLEM 4.90
y
A bending couple is applied to the bar indicated, causing plastic zones
3 in. thick to develop at the top and bottom of the bar. After the couple
has been removed, determine (a) the residual stress at y 4.5 in.,
(b) the points where the residual stress is zero, (c) the radius of
curvature corresponding to the permanent deformation of the bar.
3 in.
3 in.
C
z
Beam of Prob. 4.76.
3 in.
1.5 in.
3 in.
1.5 in.
SOLUTION
See solution to Problem 4.76 for bending couple and stress distribution.
M
Y
4725 kip in.
42 ksi
I
yY
1.5 in.
357.75 in 4
At
At
y
yY ,
res
Y
I Y
M
y0
(c)
At
My0
I
0
y
19.8113
59.4 ksi
17.4340 ksi
42
22.189 ksi
UNLOADING
LOADING
res
29 103 ksi
4.5 in.
Mc (4725)(4.5)
59.434 ksi
I
357.75
MyY
(4725)(1.5)
19.8113 ksi
I
357.75
y c,
59.434 42
res
Y
(a)
(b)
c
29 106 psi
E
0
Y
(357.75)(42)
4725
yY ,
Ey
res
RESIDUAL STRESSES
3.18 in.
Answer:
y0
3.18 in., 0, 3.18 in.
22.189 ksi
Ey
(29 103 )(1.5)
22.189
1960.43 in.
163.4 ft.
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539
PROBLEM 4.91
y
z
A bending couple is applied to the beam of Prob. 4.73, causing plastic zones
30 mm thick to develop at the top and bottom of the beam. After the couple has
been removed, determine (a) the residual stress at y 45 mm, (b) the points
where the residual stress is zero, (c) the radius of curvature corresponding to
the permanent deformation of the beam.
90 mm
C
60 mm
SOLUTION
See solution to Problem 4.73 for bending couple and stress distribution during loading.
28.08 103 N m
M
Mc
I
c,
At y
yY ,
res
y0
E
c
(28.08 103 )(0.045)
3.645 10 6
346.67 106 Pa
346.67 MPa
(28.08 103 )(0.015)
3.645 10 6
res
346.67
Y
res
115.556 106 Pa
Y
240
115.556
LOADING
(b)
0.015 m
3.645 10 6 m 4
M yY
I
At y
15 mm
I
240 MPa
Y
(a)
yY
0.045 m
115.556 MPa
106.670 MPa
240
My0
I
I Y
M
(3.645 10 6 )(240 106 )
28.08 103
Y
res
124.444 MPa
UNLOADING
0
200 GPa
res
At
y
yY ,
Ey
res
124.4 MPa
RESIDUAL STRESSES
0
31.15 10 3 m
31.15 mm
Answer: y0
(c)
106.7 MPa
31.2 mm, 0, 31.2 mm
124.444 106 Pa
Ey
(200 109 )(0.015)
124.444 106
24.1 m
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540
PROBLEM 4.92
y
A beam of the cross section shown is made of a steel that is assumed to be
elastoplastic with E 29 × 106 psi and Y = 42 ksi. A bending couple is
applied to the beam about the z axis, causing plastic zones 2 in. thick to
develop at the top and bottom of the beam. After the couple has been removed,
determine (a) the residual stress at y 2 in., (b) the points where the residual
stress is zero, (c) the radius of curvature corresponding to the permanent
deformation of the beam.
1 in.
z
2 in.
C
1 in.
1 in.
1 in.
1 in.
SOLUTION
Flange:
I1
Web:
I2
1
b1h13
12
1
(3)(1)3
12
A1d12
1
1
b2 h23
(1)(2)3
12
12
I1 7.0 in 4
I3
I
I1
I2
c
2 in.
YI
MY
Y
7.0 in 4
0.6667 in 4
14.6667 in 4
(42)(14.6667)
2
c
R1
I3
(3)(1)(1.5)2
A1
MY
308 kip in .
M
406 kip in .
(42)(3)(1) 126 kips
y1 1.0 0.5 1.5 in.
1
Y A2
2
21 kips
R2
M
Y
(a)
y2
2
(1.0) 0.6667 in.
3
M
2( R1 y1
406 kip in. yY
42 ksi
1
(42)(1)(1)
2
I
R2 y 2 )
1.0 in. E
14.6667 in 4 c
Mc (406)(2)
I 14.6667
2[(126)(1.5) (21)(0.6667)]
29 106
29 103 ksi
2 in.
55.364
MyY (406)(1.0)
I
14.662
27.682
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541
PROBLEM 4.92 (Continued)
At y
c,
res
55.364 42 13.3640 ksi
Y
res
At y
yY ,
res
Y
27.682 42
14.3180 ksi
res
(b)
res
y0
0
I Y
M
At y
yY ,
14.32 k si
My0
0
Y
I
(14.667)(42)
1.517 in
406
Answer : y0
(c)
13.36 k si
res
Ey
1.517 in., 0, 1.517 in.
14.3180 ksi
Ey
(29 103 )(1.0)
14.3180
2025.42 in.
168.8 ft
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542
PROBLEM 4.93*
A rectangular bar that is straight and unstressed is bent into an arc of circle of radius by two couples of
moment M. After the couples are removed, it is observed that the radius of curvature of the bar is R .
Denoting by Y the radius of curvature of the bar at the onset of yield, show that the radii of curvature satisfy
the following relation:
1
1
3
2
1
R
2
1
3
1
Y
Y
SOLUTION
1
Y
Let m denote
MY
,
EI
M
3
MY 1
2
1
3
2
2
2
Y
2
Y
2
,
2
Y
M
.
MY
m
M
MY
1
1
M
EI
R
1
3
1
2
1
1
mM Y
EI
1
m
Y
1
3
1
2m
m
Y
3
2
1
Y
1
3
2
2
Y
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543
PROBLEM 4.94
A solid bar of rectangular cross section is made of a material that is assumed to be elastoplastic. Denoting by
M Y and Y , respectively, the bending moment and radius of curvature at the onset of yield, determine (a) the
radius of curvature when a couple of moment M 1.25 M Y is applied to the bar, (b) the radius of curvature
after the couple is removed. Check the results obtained by using the relation derived in Prob. 4.93.
SOLUTION
(a)
1
Y
m
MY
,
EI
M
MY
3
MY 1
2
M
3
1
2
1
3
1
3
2
Let m
2
Y
M
MY
1.25
2
2
Y
3
2m
0.70711
Y
0.707
(b)
1
1
M
EI
R
1
mM Y
EI
1
m
Y
1
0.70711
Y
1.25
Y
Y
0.16421
R
6.09
Y
Y
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544
PROBLEM 4.95
M
The prismatic bar AB is made of a steel that is assumed to be elastoplastic and
for which E 200 GPa . Knowing that the radius of curvature of the bar is
2.4 m when a couple of moment M 350 N m is applied as shown,
determine (a) the yield strength of the steel, (b) the thickness of the elastic
core of the bar.
B
A
20 mm
16 mm
SOLUTION
M
1
3
3 YI
1
2 c
1 2 Y2
3 E 2c 2
3
2
Y b(2c)
bc 2
2
Y
1
2
Y
2 2
3E c
M
Data:
2
Y
3
1 2 Y2
3 E 2c 2
1
12c
1 2 Y2
3 E 2c 2
2
Y bc 1
(a)
2
3
MY 1
2
Cubic equation for
Y
200 109 Pa
E
420 N m
M
2.4 m
b
20 mm
c
1
h
2
8 mm
Y
1
750 10
21
2
Y
350
Y
1
750 10
21
2
Y
273.44 106
(1.28 10 6 )
Solving by trial,
(b)
yY
Y
E
(292 106 )(2.4)
200 109
Y
0.020 m
0.008 m
292 106 Pa
3.504 10 3 m
Y
292 MPa
2 yY
7.01 mm
3.504 mm
thickness of elastic core
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545
PROBLEM 4.96
" (MPa)
300
The prismatic bar AB is made of an aluminum
alloy for which the tensile stress-strain diagram
is as shown. Assuming that the - diagram is
the same in compression as in tension,
determine (a) the radius of curvature of the bar
when the maximum stress is 250 MPa, (b) the
corresponding value of the bending moment.
versus y and use an
(Hint: For part b, plot
approximate method of integration.)
B
40 mm
M'
200
M
60 mm
A
100
0
0.005
0.010
#
SOLUTION
(a)
250 MPa
m
0.0064 from curve
c
b
1
h 30 mm 0.030 m
2
40 mm 0.040 m
1
m
c
(b)
250 106 Pa
m
0.0064
0.030
0.21333 m
1
Strain distribution:
4.69 m
y
c
m
Bending couple:
c
M
where the integral J is given by
c
1
u|
0
where
mu
y b dy
2b
c
0
u
y
1
2bc 2 0 u | | du
y | | dy
2bc 2 J
| du
Evaluate J using a method of numerial integration. If Simpson’s rule is used, the integration formula is
J
u
wu | |
3
where w is a weighting factor.
Using
u
0
0.25
0.5
0.75
1.00
u
0.25, we get the values given in the table below:
| |
0
0.0016
0.0032
0.0048
0.0064
J
M
| |, (MPa)
0
110
180
225
250
(0.25)(1215)
3
u | |, (MPa)
0
27.5
90
168.75
250
101.25 MPa
w
1
4
2
4
1
wu | |, (MPa)
0
110
180
675
250
1215
wu | |
101.25 106 Pa
(2)(0.040)(0.030)2 (101.25 106 )
7.29 103 N m
M
7.29 kN m
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546
PROBLEM 4.97
0.8 in.
B
M
" (ksi)
1.2 in.
A
50
The prismatic bar AB is made of a bronze alloy for which the tensile
stress-strain diagram is as shown. Assuming that the - diagram is
the same in compression as in tension, determine (a) the maximum
stress in the bar when the radius of curvature of the bar is 100 in.,
(b) the corresponding value of the bending moment. (See hint given
in Prob. 4.96.)
40
30
20
10
0
0.004
0.008
#
SOLUTION
(a)
100 in., b
c
0.6
100
m
(b)
0.8 in., c
0.6 in.
0.006
From the curve,
Strain distribution:
Bending couple:
m
c
M
where the integral J is given by
y
c
c
1
u
0
mu
y b dy
where u
2b
c
0
y | | dy
m
43.0 ksi
y
1
2bc 2 0 u | | du
2bc 2 J
| | du
Evaluate J using a method of numerial integration. If Simpson’s rule is used, the integration formula is
J
u
wu| |
3
where w is a weighting factor.
Using
u
0.25, we get the values given the table below:
u
| |
0
0.25
0.5
0.75
1.00
0
0.0015
0.003
0.0045
0.006
| |, ksi
0
25
36
40
43
u | |, ksi
0
6.25
18
30
43
J
M
w
1
4
2
4
1
(0.25)(224)
18.67 ksi
3
(2)(0.8)(0.6) 2 (18.67)
wu | |, ksi
0
25
36
120
43
224
wu | |
M
10.75 kip in.
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547
PROBLEM 4.98
"
A prismatic bar of rectangular cross section is made of an alloy for
which the stress-strain diagram can be represented by the relation
k n for
0 and
k n for
0 . If a couple M is
applied to the bar, show that the maximum stress is
#
M
1
m
2n Mc
I
3n
SOLUTION
Strain distribution:
y
c
m
where
mu
y
c
u
Bending couple:
c
M
c
1
0
2 bc 2
For
K
n
c
0
2b
y b dy
2bc 2
y | | dy
,
K
m
m
u
m
m
M
2bc 2 0 u
1
2bc
Recall that
Then
| |
2
m
mu
u
2
1
n
2 1n
m
2n 1 M
2 bc 2
I
c
1 b(2c)3
12 c
m
y
dy
| |
c
c
u | | du
n
Then
c
0
1
n
du
1
0
2 2
bc
3
2bc 2
mu
1
n
1 1 1n
du
m 0u
2n
bc 2
2n 1
1
bc 2
m
2 c
3 I
2n 1 Mc
3n
I
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548
PROBLEM 4.99
30 mm
Knowing that the magnitude of the horizontal force P is 8 kN,
determine the stress at (a) point A, (b) point B.
B
24 mm
A
D
P
45 mm
15 mm
SOLUTION
A
(30)(24)
720 mm 2
e
45
33 mm
I
c
M
(a)
A
P
A
Mc
I
12
720 10 6 m 2
0.033 m
1 3
1
bh
(30)(24)3 34.56 103 mm 4 34.56 10 9 m 4
12
12
1
(24 mm) 12 mm 0.012 m
P 8 103 N
2
Pe
(8 103 )(0.033)
8 103
720 10
6
264 N m
(264)(0.012)
34.56 10 9
102.8 106 Pa
102.8 MPa
A
(b)
B
P
A
Mc
I
8 103
720 10
6
(264)(0.012)
34.56 10 9
80.6 106 Pa
B
80.6 MPa
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549
PROBLEM 4.100
y
b
3 in.
6 kips
A short wooden post supports a 6-kip axial load as shown. Determine the
stress at point A when (a) b 0, (b) b 1.5 in., (c) b 3 in.
C
A
z
x
SOLUTION
A
I
S
P
(a)
b
0
M
P
A
(b)
b
(3) 2
28.27 in 2
r4
(3) 4 63.62 in 4
4
4
I
63.62
21.206 in 3
c
3
6 kips
M Pb
0
6
28.27
1.5 in. M
P
A
r2
M
S
0.212 ksi
212 psi
(6)(1.5)
9 kip in.
6
28.27
9
21.206
0.637 ksi
637 psi
(c)
b
3 in.
P
A
M
M
S
(6)(3)
18 kip in.
6
28.27
18
21.206
1.061 ksi
1061 psi
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550
P
r
r
P
PROBLEM 4.101
Two forces P can be applied separately or at the same time to a plate that is welded
to a solid circular bar of radius r. Determine the largest compressive stress in the
circular bar, (a) when both forces are applied, (b) when only one of the forces is
applied.
SOLUTION
For a solid section,
A
Both forces applied.
4
r 4, c
r
F Mc
A
I
F
4M
2
r
r3
Compressive stress
(a)
r 2, I
F
2 P, M
0
2P
r2
(b)
One force applied.
F
P, M
F
r2
Pr
4Pr
r2
5P
r2
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551
PROBLEM 4.102
y
30 mm
60 mm
A short 120 × 180-mm column supports the three axial loads
shown. Knowing that section ABD is sufficiently far from the
loads to remain plane, determine the stress at (a) corner A,
(b) corner B.
30 kN
20 kN
100 kN
C
z
x
A
D
90 mm
90 mm
B
120 mm
SOLUTION
A
S
M
(a)
A
(0.120 m)(0.180 m) 21.6 10 3 m 2
1
(0.120 m)(0.180 m)2 6.48 10 4 m 2
6
(30 kN)(0.03 m) (100 kN)(0.06 m)
5.10 kN m
P
A
M
S
150 103 N
21.6 10 3 m 2
5.10 103 N m
6.48 10 4 m3
A
(b)
B
P
A
M
S
3
150 10 N
21.6 10 3 m 2
0.926 MPa
3
5.10 10 N m
6.48 10 4 m3
B
14.81 MPa
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552
PROBLEM 4.103
80 mm
80 mm
P
C
1
As many as three axial loads, each of magnitude P 50 kN, can be applied to
the end of a W200 × 31.1 rolled-steel shape. Determine the stress at point A
(a) for the loading shown, (b) if loads are applied at points 1 and 2 only.
P
P
2
3
A
SOLUTION
For W200
(a)
31.3 rolled-steel shape.
Centric load:
A
3970 mm 2
c
1
d
2
I
31.3 106 mm 4
3P
50
1
(210)
2
50
3P
A
(b)
Ececentric loading:
e
50
50
M
Pe
(50 103 )(0.080)
2P
A
Mc
I
80 mm
50
3
m2
105 mm
0.105 m
31.3 10
150 kN
150 103
3.970 10 3
6
m4
150 103 N
37.783 106 Pa
37.8 MPa
0.080 m
100 103 N
2P
A
100 kN
4.000 10
4.0 103 N m
100 103
3.970 10 3
(4.0 103 )(0.105)
31.3 10 6
38.607 106 Pa
38.6 MPa
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553
PROBLEM 4.104
y
10 mm
10 mm
Two 10-kN forces are applied to a 20 × 60-mm rectangular bar as
shown. Determine the stress at point A when (a) b 0, (b) b 15 mm,
(c) b 25 mm.
A
30 mm
z
30 mm
10 kN
C
b
10 kN
x
25 mm
SOLUTION
A
1.2 10
3
m2
b
1
(0.020 m)(0.060 m)2 12 10 6 m3
6
0, M (10 kN)(0.025 m) 250 N m
A
(20 kN)/(1.2 10
S
(a)
(0.060 m)(0.020 m)
16.667 MPa
3
m2 )
(250 N m)/(12 10
6
m3 )
20.833 MPa
4.17 MPa
A
(b)
b
15 mm, M
A
(20 kN)/(1.2 10
16.667 MPa
(10 kN)(0.025 m
3
m2 )
0.015 m)
100 N m
(100 N m)/(12 10
6
m3 )
8.333 MPa
A
(c)
b
25 mm, M
0:
A
(20 kN)/(1.2 10
3
8.33 MPa
m2 )
A
16.67 MPa
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554
P
P'
P
PROBLEM 4.105
P'
Portions of a 12 12 -in. square bar have been bent
to form the two machine components shown.
Knowing that the allowable stress is 15 ksi,
determine the maximum load that can be applied
to each component.
1 in.
(a)
(b)
SOLUTION
The maximum stress occurs at point B.
B
B
where
ec
I
15 103 psi
15 ksi
P
A
K
1
A
A
(0.5)(0.5)
I
1
(0.5)(0.5)3
12
Mc
I
P
A
Pec
I
KP
1.0 in.
e
0.25 in 2
5.2083 10 3 in 4 for all centroidal axes.
(a)
(a)
(b)
c
(b)
0.25 in.
K
1
0.25
(1.0)(0.25)
5.2083 10 3
P
B
( 15 103 )
52
c
K
0.5
2
52 in
2
P
288 lb
P
209 lb
0.35355 in.
K
1
0.25
(1.0)(0.35355)
5.2083 10 3
P
B
( 15 103 )
71.882
K
71.882 in
2
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555
PROBLEM 4.106
P
Knowing that the allowable stress in section ABD is 80 MPa,
determine the largest force P that can be applied to the bracket shown.
A
D
B
18 mm
40 mm
12 mm
12 mm
SOLUTION
A
(24)(18)
432 mm 2
432 10
1
(24)(18)3 11.664 103 mm 4
12
1
(18) 9 mm 0.009 m
c
2
1
(18) 40 49 mm 0.049 m
e
2
On line BD, both axial and bending stresses are compressive.
I
Therefore
Solving for P gives
max
P
P
P
A
Mc
I
P
A
Pec
I
6
m2
11.664 10
9
m4
max
1
A
ec
I
80 106 Pa
1
432 10 6 m 2
(0.049 m)(0.009 m)
11.664 10 9 m 4
P
1.994 kN
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556
PROBLEM 4.107
P'
a
d
a
P
A milling operation was used to remove a portion of a solid bar
of square cross section. Knowing that a 30 mm, d 20 mm,
and all 60 MPa, determine the magnitude P of the largest
forces that can be safely applied at the centers of the ends of the
bar.
SOLUTION
A
e
Data:
ad ,
I
1
ad 3 ,
12
a d
2 2
P Mc
P
6 Ped
A
I
ad
ad 3
3P (a d )
P
KP
ad
ad 2
a
30 mm
K
1
(0.030)(0.020)
P
c
K
0.030 m
d
1
d
2
K
20 mm
0.020 m
(3)(0.010)
(0.030)(0.020)2
60 106
4.1667 103
1
ad
where
14.40 103 N
3(a d )
ad 2
4.1667 103 m
2
P
14.40 kN
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557
PROBLEM 4.108
P'
A milling operation was used to remove a portion of a solid bar of
square cross section. Forces of magnitude P 18 kN are applied at
the centers of the ends of the bar. Knowing that a 30 mm and
135 MPa, determine the smallest allowable depth d of the
all
milled portion of the bar.
a
d
P
a
SOLUTION
1
ad 3 ,
12
A
ad ,
e
a
2
d
2
P
A
Mc
I
3P
d2
I
2P
ad
Solving for d,
d
1
2
Data:
a
0.030 m,
d
1
(2)(135 106 )
16.04 10
3
P
ad
2P
a
c
1
d
2
Pec
I
P
ad
d2
or
2
18 103 N,
(2)(18 103 )
0.030
3P
P
ad
3P(a d )
ad 2
0
2P
a
12 P
P
2P
d
a
P 1 (a d ) 1 d
2
2
1
3
ad
12
135 106 Pa
2
12(18 103 )(135 106 )
m
(2)(18 103 )
0.030
d
16.04 mm
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558
12 kips
5 in.
PROBLEM 4.109
P
The two forces shown are applied to a rigid plate supported by a steel pipe of 8-in.
outer diameter and 7-in. inner diameter. Determine the value of P for which the
maximum compressive stress in the pipe is 15 ksi.
SOLUTION
15 ksi
all
I NA
A
4
(4 in.)4
4
(4 in.) 2
(3.5 in.)4
(3.5 in.)2
83.2 in 4
11.78 in 2
Max. compressive stress is at point B.
B
Q
A
Mc
I
15 ksi
1.019
13.981
0.325P
12 P
11.78 in 2
0.085P
(5P)(4.0 in.)
83.2 in 4
0.240 P
P
43.0 kips
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559
PROBLEM 4.110
d
P'
P
h
P'
P
An offset h must be introduced into a solid circular rod of
diameter d. Knowing that the maximum stress after the offset
is introduced must not exceed 5 times the stress in the rod
when it is straight, determine the largest offset that can be
used.
d
SOLUTION
For centric loading,
c
P
A
For eccentric loading,
e
P
A
Given
e
P
A
5
Phc
I
c
Phc
I
5
P
A
Phc
I
P
4
A
h
4I
cA
(4)
d
2
64
4
d4
d
2
1
d
2
h
0.500 d
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560
PROBLEM 4.111
d
P'
P
An offset h must be introduced into a metal tube of 0.75-in. outer
diameter and 0.08-in. wall thickness. Knowing that the maximum
stress after the offset is introduced must not exceed 4 times the
stress in the tube when it is straight, determine the largest offset
that can be used.
h
P'
P
d
SOLUTION
c
c1
1
d 0.375 in.
2
c t 0.375 0.08
c2
A
c12
0.295 in.
(0.3752
0.2952 )
0.168389 in 2
I
4
c4
c14
4
(0.3754
0.2954 )
9.5835 10 3 in 4
For centric loading,
cen
P
A
For eccentric loading,
ecc
P
A
ecc
4
hc
I
3
A
Phc
I
or
cen
h
P
A
3I
Ac
Phc
I
4
P
A
(3)(9.5835 10 3 )
(0.168389)(0.375)
h
0.455 in.
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561
PROBLEM 4.112
16 kips
2
4
1
A short column is made by nailing four 1 × 4-in. planks to a 4 × 4-in. timber. Using
an allowable stress of 600 psi, determine the largest compressive load P that can be
applied at the center of the top section of the timber column as shown if (a) the
column is as described, (b) plank 1 is removed, (c) planks 1 and 2 are removed,
(d) planks 1, 2, and 3 are removed, (e) all planks are removed.
3
SOLUTION
(a)
Centric loading:
M
P
A
0
A
(4
4)
32 in 2
4(1)(4)
P
A
(0.600 ksi)(32 in 2 )
P
19.20 kips
(b)
Eccentric loading:
M
P
A
Pe
A
(4)(4)
y
Ay
A
I
(I
(c)
Centric loading:
28 in 2
(3)(1)(3)
(1)(4)(2.5)
28
e
y
0.35714 in.
Ad 2 )
1
(6)(4)3
12
P
Pec
I
1
A
1
(4)(1)3
12
(6)(4)(0.35714)2
ec
I
P
(4)(1)(2.14286)2
(0.600 ksi)
(0.35714)(2.35714)
53.762
1
28
53.762 in 4
11.68 kips
P
A
M
0
A
(6)(4)
P
(0.600 ksi)(24)
24 in 2
14.40 kips
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562
PROBLEM 4.112 (Continued)
(d)
Eccentric loading:
M
Pe
A
(4)(4)
x
2.5
P
Centric loading:
(1)(4)(1)
2
1
(4)(5)3
12
I
(e)
P
A
M
Pec
I
20 in 2
x
0.5 in.
41.667 in 4
(0.600 ksi)
1
(0.5)(2.5)
20
41.667
0
e
7.50 kips
P
A
16 in 2
A
(4)(4)
P
(0.600 ksi)(16.0 in 2 )
9.60 kips
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563
1 in.
a
1.5 in.
PROBLEM 4.113
3 in.
a
0.75 in.
1.5 in.
A
A vertical rod is attached at point A to the cast
iron hanger shown. Knowing that the allowable
stresses in the hanger are all
5 ksi and
12 ksi , determine the largest downward
all
force and the largest upward force that can be
exerted by the rod.
3 in.
0.75 in.
B
Section a–a
SOLUTION
X
Ay
A
X
12.75 in 3
7.5 in 2
A
7.5 in 2
all
5 ksi
(1 3)(0.5)
(1 3)
all
1
(3)(1)3
12
0.25)(2.5)
0.75)
1.700 in.
1 3
bh
12
Ic
2(3
2(3
12 ksi
Ad 2
(3 1)(1.70
0.5)2
1
(1.5)(3)3
12
(1.5
3)(2.5
1.70)2
10.825 in 4
Ic
Downward force.
M
At D:
D
5 ksi
5
At E:
E
12 ksi
12
P(1.5 in.+1.70 in.)
P
A
Mc
I
P
(3.20) P(1.70)
7.5
10.825
P( 0.6359)
P
A
(3.20 in.) P
P
7.86 kips
P
21.95 kips
P
7.86 kips
Mc
I
P
(3.20) P(2.30)
7.5
10.825
P( 0.5466)
We choose the smaller value.
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564
PROBLEM 4.113 (Continued)
Upward force.
M
At D:
D
12 ksi
12
At E:
E
5 ksi
5
P(1.5 in.
P
A
1.70 in.)
Mc
I
P
(3.20) P(1.70)
7.5
10.825
P( 0.6359)
P
A
(3.20 in.) P
P
18.87 kips
Mc
I
P
(3.20) P(2.30)
7.5
10.825
P( 0.5466)
We choose the smaller value.
P
9.15 kips
P
9.15 kips
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565
PROBLEM 4.114
1 in.
a
1.5 in.
3 in.
a
1.5 in.
A
Solve Prob. 4.113, assuming that the vertical
rod is attached at point B instead of point A.
0.75 in.
3 in.
PROBLEM 4.113 A vertical rod is attached at
point A to the cast iron hanger shown. Knowing
that the allowable stresses in the hanger are
5 ksi and all
12 ksi, determine the
all
largest downward force and the largest upward
force that can be exerted by the rod.
0.75 in.
B
Section a–a
SOLUTION
X
Ay
A
X
12.75 in 3
7.5 in 2
A
7.5 in 2
all
5 ksi
(1 3)(0.5)
(1 3)
all
1
(3)(1)3
12
Ic
0.25)(2.5)
0.75)
1.700 in.
1 3
bh
12
Ic
2(3
2(3
12 ksi
Ad 2
(3 1)(1.70
0.5)2
1
(1.5)(3)3
12
(1.5
3)(2.5
1.70)2
10.825 in 4
Downward force.
all
M
At D:
D
12 ksi
12
At E:
E
5 ksi
5
5 ksi
(2.30 in.
P
A
12 ksi
all
1.5 in.)
(3.80 in.) P
Mc
I
P
(3.80) P(1.70)
7.5
10.825
P( 0.4634)
P
A
P
25.9 kips
P
5.32 kips
P
5.32 kips
Mc
I
P
(3.80) P(2.30)
7.5
10.825
P( 0.9407)
We choose the smaller value.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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566
PROBLEM 4.114 (Continued)
Upward force.
all
M
At D:
D
5 ksi
5
At E:
E
12 ksi
12
5 ksi
(2.30 in.
P
A
12 ksi
all
1.5 in.)P
Mc
I
P
(3.80) P(1.70)
7.5
10.825
P( 0.4634)
P
A
(3.80 in.)P
P
10.79 kips
P
12.76 kips
P
10.79 kips
Mc
I
P
(3.80) P(2.30)
7.5
10.825
P( 0.9407)
We choose the smaller value.
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567
PROBLEM 4.115
2 mm radius
A
P
Knowing that the clamp shown has been tightened until
P 400 N, determine (a) the stress at point A,
(b) the stress at point B, (c) the location of the neutral
axis of section a a.
P'
32 mm
20 mm
a
B
a
4 mm
Section a–a
SOLUTION
Cross section:
Rectangle
Circle
(20 mm)(4 mm)
y1
1
(20 mm)
2
10 mm
A2
(2 mm)2
4 mm 2
y2
y
cA
20
d1
11.086
d2
18
y
2
18 mm
(80)(10)
80
(4 )(18)
4
1.086 mm
11.086
6.914 mm
I1
A1d12
1
(4)(20)3
12
I2
I2
A2d 22
(2) 4
I
I1
I2
A
A1
A2
(80)(1.086) 2
(4 )(6.914) 2
3.374 103 mm 4
92.566 mm 2
11.086 mm
8.914 mm
10
I1
4
20
Ay
A
cB
80 mm 2
A1
2.761 103 mm 4
0.613 103 mm 4
3.374 10
92.566 10
6
9
m4
m2
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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568
PROBLEM 4.115 (Continued)
(a)
e
32
8.914
M
Pe
(400 N)(0.040914 m)
16.3656 N m
Mc
I
(16.3656)(8.914 10 3 )
3.374 10 9
P
A
4.321 106
400
92.566 10
6
43.23 106
47.55 106 Pa
A
47.6 MPa
Point B:
B
P
A
Mc
I
4.321 106
(c)
0.040914 m
Point A:
A
(b)
40.914 mm
Neutral axis:
400
92.566 10
6
53.72 106
(16.3656)(11.086)
3.374 10 9
49.45 106 Pa
B
49.4 MPa
By proportions,
a
47.55
a
20
47.55 49.45
9.80 mm
9.80 mm below top of section
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569
PROBLEM 4.116
P
a
a
The shape shown was formed by bending a thin steel plate. Assuming that the
thickness t is small compared to the length a of a side of the shape, determine the
stress (a) at A, (b) at B, (c) at C.
90$
t
B
C
A
P'
SOLUTION
Moment of inertia about centroid:
I
a
1
2 2t
12
3
2
1 3
ta
12
Area:
A
2 2t
a
2
2at , c
Pec
I
P
2at
P
A
P
A
(b)
Pec
I
P
2at
P
B
P
A
(c)
C
(a)
a
2 2
1
ta
12
a
2
a
2 2
a
2 2
3
A
P
2at
B
2P
at
C
P
2at
a
2
1
ta3
12
A
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570
PROBLEM 4.117
P
Three steel plates, each of 25 150-mm cross section, are welded
together to form a short H-shaped column. Later, for architectural
reasons, a 25-mm strip is removed from each side of one of the
flanges. Knowing that the load remains centric with respect to the
original cross section, and that the allowable stress is 100 MPa,
determine the largest force P (a) that could be applied to the original
column, (b) that can be applied to the modified column.
50 mm
50 mm
SOLUTION
(a)
P
A
Centric loading:
A
11.25 103 mm 2
(3)(150)(25)
P
11.25 10 3 m 2
( 100 106 )(11.25 10 3 )
A
1.125 106 N
(b)
P
1125 kN
Eccentric loading (reduced cross section):
A, 103 mm 2
y , mm
A y (103 mm3 )
3.75
87.5
328.125
76.5625
3.75
0
0
10.9375
2.50
87.5
218.75
98.4375
10.00
d , mm
109.375
Y
Ay
A
109.375 103
10.00 103
10.9375 mm
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571
PROBLEM 4.117 (Continued)
The centroid lies 10.9375 mm from the midpoint of the web.
I1
I2
I3
1
b1h13
12
1
b2h23
12
1
b3h33
12
A2d 22
A3d32
I2
I3
I
I1
c
10.9375
M
Pe
P
A
K
P
75
K
54.012 106 mm 4
25
Mc
I
P
A
10.4375 mm
Pec
I
1
10.00 10
( 100 106 )
122.465
3
22.177 106 mm 4
7.480 106 mm 4
24.355 106 mm 4
54.012 10 6 m 4
110.9375 mm
e
where
ec
I
1
A
1
(150)(25)3 (3.75 103 )(76.5625)2
12
1
(25)(150)3 (3.75 103 )(10.9375)2
12
1
(100)(25)3 (2.50 103 )(98.4375) 2
12
A1d12
KP
0.1109375 m
10.4375 10 3 m
A
10.00 10 3 m 2
(101.9375 10 3 )(0.1109375)
54.012 10 6
817 103 N
122.465 m
2
P
817 kN
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572
PROBLEM 4.118
y
P
y
B
3 in.
y
x
3 in.
B
A vertical force P of magnitude 20 kips is applied
at point C located on the axis of symmetry of the
cross section of a short column. Knowing that
y 5 in., determine (a) the stress at point A,
(b) the stress at point B, (c) the location of the
neutral axis.
2 in.
C
A
4 in.
A
2 in.
x
2 in.
1 in.
(a)
(b)
SOLUTION
Locate centroid.
A, in 2
12
8
20
Eccentricity of load: e 5
(a)
I1
1
(6)(2)3
12
I
I1
I2
Stress at A : c A
Ay , in 3
60
16
76
y , in.
5
2
Part
3.8
Stress at B : cB
(12)(1.2) 2
21.28 in 4
3.8 in.
1
(2)(4)3
12
I2
(8)(1.8)2
36.587 in 4
57.867 in 4
3.8 in.
P
A
6
3.8
Pec A
I
20
20
20(1.2)(3.8)
57.867
PecB
I
20
20
20(1.2)(2.2)
57.867
A
0.576 ksi
2.2 in.
P
A
B
(c)
76
20
1.2 in.
A
(b)
Ay
A
y
Location of neutral axis:
0
a
I
Ae
P Pea
0
A
I
57.867
2.411 in.
(20)(1.2)
Neutral axis lies 2.411 in. below centroid or 3.8
2.411
ea
I
B
1.912 ksi
1
A
1.389 in. above point A.
Answer: 1.389 in. from point A
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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573
PROBLEM 4.119
y
P
y
B
3 in.
y
x
3 in.
B
A vertical force P is applied at point C located
on the axis of symmetry of the cross section of
a short column. Determine the range of values
of y for which tensile stresses do not occur in
the column.
2 in.
C
A
4 in.
A
x
2 in.
2 in.
1 in.
(a)
(b)
SOLUTION
Locate centroid.
A, in 2
y , in.
Ay , in 3
5
2
60
16
76
12
8
20
Eccentricity of load:
e
y
I1
1
(6)(2)3
12
I
I1
If stress at A equals zero,
If stress at B equals zero,
y
3.8 in.
3.8 in.
A
P
A
e
I
Ac A
e
y
3.8 in.
3.8 in.
(12)(1.2) 2
cA
B
76
20
21.28 in 4
1
(2)(4)3
12
I2
(8)(1.8) 2
36.587 in 4
57.867 in 4
I2
cB
e
Ai yi
Ai
y
6
Pec A
I
3.8
ec A
I
0
57.867
(20)(3.8)
0.761 in.
y
0.761
3.8
4.561 in.
2.2 in.
P PecB
0
A
I
57.867
I
(20)(2.2)
AcB
1.315
1
A
3.8
ecB
I
1
A
1.315 in.
2.485 in.
Answer: 2.49 in.
y
4.56 in.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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574
PROBLEM 4.120
P
P
The four bars shown have the same cross-sectional area. For the given
loadings, show that (a) the maximum compressive stresses are in the
ratio 4:5:7:9, (b) the maximum tensile stresses are in the ratio 2:3:5:3.
(Note: the cross section of the triangular bar is an equilateral triangle.)
P
P
SOLUTION
Stresses:
At A,
A
P
A
Pec A
I
P
1
A
At B,
B
P
A
PecB
I
P AecB
A
I
A1
1 4
a , c A cB
12
1
1
a
(a 2 ) a
P
2
2
1
1 2
A
a
12
a 2 , I1
A
B
A2
A
B
P
A
c2
(a 2 )
a2
1
1
a
a
2
2
1 2
a
12
c
a
Aec A
I
1
1
a, e
2
1
a
2
A
1
, I2
B
4
c4 , e
P
( c 2 )(c)(c)
1
A2
c4
4
P
A1
2
P
A1
5
P
A2
3
P
A2
c
A
P ( c 2 )(c)(c)
1
A2
4
c
4
4
B
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575
PROBLEM 4.120 (Continued)
A3
A
B
a2
2
a I3
2
c
(a 2 )
P
1
A3
P
A3
1 4
a
12
e
c
2
2
a
a
2
2
1 4
a
12
A
2
2
a
a
2
2
1 4
a
12
1
A4
1
(s)
2
3
s
2
3 2
s
4
I4
1
s
36
3
s
2
cA
2 3
s
3 2
(a 2 )
A
B
3
s
P
A4
cB
5
P
A3
3 2
s
4
s
3
9
P
A4
3
P
A4
s
s
3
A
3 4
s
96
3 2
s
4
P
A3
3 4
s
96
e
3
P
1
A4
B
7
s
3
3 4
s
96
s
2 3
1
B
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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576
PROBLEM 4.121
25 mm
An eccentric force P is applied as shown to a steel bar of
25 90-mm cross section. The strains at A and B have been
measured and found to be
30 mm
A
90 mm
45 mm
B
350
A
P
d
70
B
Knowing that E 200 GPa, determine (a) the distance d, (b) the
magnitude of the force P.
15 mm
SOLUTION
h 15 45 30 90 mm
A bh (25)(90)
I
yA
1
h 45 mm
2
2.25 10 3 m 2
25 mm
b
2.25 103 mm 2
Subtracting,
A
A
E
A
(200 109 )(350 10 6 )
B
E
B
(200 109 )( 70 10 6 )
M
A( y A B yB
y A yB
Pd
d
MyA
I
(1)
B
P
A
MyB
I
(2)
M ( y A yB )
I
M
I( A
B)
y A yB
M
P
14 106 Pa
P
A
B
A)
70 106 Pa
A
(1.51875 10 6 )(84 106 )
0.045
Multiplying (2) by y A and (1) by yB and subtracting,
(a)
0.045 m
1 3 1
(25)(90)3 1.51875 106 mm 4 1.51875 10 6 m 4
bh
12
12
60 45 15 mm 0.015 m
30 mm
0.030 m
yB 15 45
Stresses from strain gages at A and B:
P
c
yA
B
yB
A
( yA
2835 N m
yB )
P
A
(2.25 10 3 )[(0.015)( 14 106 ) ( 0.030)(70 106 )]
0.045
2835
94.5 103
0.030 m
(b)
94.5 103 N
d
P
30.0 mm
94.5 kN
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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577
PROBLEM 4.122
25 mm
Solve Prob. 4.121, assuming that the measured strains are
A
90 mm
B
600
A
30 mm
45 mm
P
PROBLEM 4.121 An eccentric force P is applied as shown to a
steel bar of 25 90-mm cross section. The strains at A and B have
been measured and found to be
d
350
A
15 mm
420
B
B
70
Knowing that E 200 GPa, determine (a) the distance d, (b) the
magnitude of the force P.
SOLUTION
h 15 45 30 90 mm
A bh
b
1
h 45 mm
2
2.25 10 3 m 2
2.25 103 mm 2
(25)(90)
yA
Stresses from strain gages at A and B:
A
E
A
(200 109 )(600 10 6 ) 120 106 Pa
B
E
B
(200 109 )(420 10 6 ) 84 106 Pa
A
P
A
My A
I
(1)
B
P
A
MyB
I
(2)
Subtracting,
A
B
M
M ( y A yB )
I
I( A
B)
y A yB
Multiplying (2) by y A and (1) by yB and subtracting,
(a)
M
A( y A B yB
y A yB
Pd
0.045 m
1 3 1
(25)(90)3 1.51875 106 mm 4 1.51875 10 6 m 4
bh
12
12
60 45 15 mm 0.015 m yB 15 45
30 mm
0.030 m
I
P
c
25 mm
d
A)
(1.51875 10 6 )(36 106 )
0.045
yA
B
yB
A
( yA
yB )
(2.25 10 3 )[(0.015)(84 106 ) ( 0.030)(120 106 )]
0.045
M
1215
5 10 3 m
3
P
243 10
(b)
1215 N m
P
A
243 103 N
d
5.00 mm
P
243 kN
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578
PROBLEM 4.123
P'
The C-shaped steel bar is used as a dynamometer to determine the magnitude P
of the forces shown. Knowing that the cross section of the bar is a square of side
40 mm and that strain on the inner edge was measured and found to be 450 ,
determine the magnitude P of the forces. Use E 200 GPa.
40 mm
80 mm
P
SOLUTION
At the strain gage location,
E
A
(200 109 )(450 10 6 )
1600 mm 2
(40)(40)
90 106 Pa
1600 10
e
1
(40)(40)3 213.33 103 mm 4
12
80 20 100 mm 0.100 m
c
20 mm
I
K
P
6
m2
213.33 10
9
m4
0.020 m
P
A
Mc
I
P
A
Pec
I
1
A
ec
I
K
90 106
10.00 103
KP
1
1600 10
6
(0.100)(0.020)
213.33 10 9
10.00 103 m
9.00 103 N
2
P
9.00 kN
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
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579
P
y
6 in.
PROBLEM 4.124
6 in.
10 in.
Q
B
A
A short length of a rolled-steel column supports a rigid plate
on which two loads P and Q are applied as shown. The strains
at two points A and B on the centerline of the outer faces of the
flanges have been measured and found to be
x
x
z
A
z
A = 10.0 in2
Iz = 273 in4
6
in./in.
29
6
400 10
A
Knowing that E
each load.
B
300 10
6
in./in.
10 psi, determine the magnitude of
SOLUTION
Stresses at A and B from strain gages:
A
E
A
(29 106 )( 400 10 6 )
11.6 103 psi
B
E
B
(29 106 )( 300 10 6 )
8.7 103 psi
Centric force:
F
P Q
Bending couple:
M
6 P 6Q
c
F
A
A
5 in.
Mc
I
11.6 103
F
A
B
Mc
I
8.7 103
P Q
10.0
(6 P 6Q)(5)
273
0.00989 P 0.20989Q
P Q
10.0
(1)
(6 P 6Q )(5)
273
0.20989 P 0.00989Q
(2)
Solving (1) and (2) simultaneously,
P
44.2 103 lb
Q
57.3 103 lb 57.3 kips
44.2 kips
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580
PROBLEM 4.125
y
P
A single vertical force P is applied to a short steel post as shown.
Gages located at A, B, and C indicate the following strains:
z
x
500
A
C
A
1000
200
C
Knowing that E 29 106 psi, determine (a) the magnitude of
P, (b) the line of action of P, (c) the corresponding strain at the
1.5 in.
hidden edge of the post, where x
2.5 in. and z
B
3 in.
5 in.
B
SOLUTION
Ix
Mx
1
(5)(3)3 11.25 in 4
12
Pz
Mz
Px
2.5 in., xB
xA
zA
1.5 in., z B
2.5 in., xC
1.5 in., zC
1.5 in., z D
6
6
E
A
(29 10 )( 500 10 )
B
E
B
(29 106 )( 1000 10 6 )
A
31.25 in 4
2.5 in., xD
A
C
A
(5)(3)
15 in 2
2.5 in.
1.5 in.
14,500 psi
29, 000 psi
14.5 ksi
29 ksi
E C (29 106 )( 200 10 6 )
5800 psi
5.8 ksi
P M x z A M z xA
0.06667 P 0.13333M x 0.08M z
A
Ix
Iz
(1)
B
P
A
M x zB
Ix
M z xB
Iz
0.06667 P
0.13333M x
0.08M z
(2)
C
P
A
M x zC
Ix
M z xC
Iz
0.06667 P
0.13333M x
0.08M z
(3)
Substituting the values for
gives
Mx
87 kip in.
x
Mz
P
Mx
P
z
D
A
,
Mz
B
, and
C
into (1), (2), and (3) and solving the simultaneous equations
(a) P
90.625 kip in.
90.625
152.25
87
152.25
P
A
M x zD
Ix
M z xD
Iz
(0.06667)(152.25)
(c)
1
(3)(5)3
12
Iz
Strain at hidden edge:
0.06667 P
(0.13333)(87)
D
E
0.13333M x
(b) x
0.595 in.
z
0.571 in.
0.08M z
(0.08)( 90.625)
8.70 103
29 106
152.3 kips
8.70 ksi
300
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
581
PROBLEM 4.126
b ! 40 mm
A
a ! 25 mm
d
B
D
P
The eccentric axial force P acts at point D, which must be located
25 mm below the top surface of the steel bar shown. For P 60 kN,
determine (a) the depth d of the bar for which the tensile stress at
point A is maximum, (b) the corresponding stress at point A.
C
20 mm
SOLUTION
1
bd 3
12
1
1
d
e
d a
2
2
P Pec
A
I
1
1
P 1 12 2 d a 2 d
b d
d3
A bd
c
A
A
(a)
Depth d for maximum
d A
dd
(b)
A
I
A:
P
b
60 103
4
3
40 10
75 10
6a
d2
Differentiate with respect to d.
4
d2
3
P 4
b d
12a
d3
d
0
(6)(25 10 3 )
(75 10 3 )2
40 106 Pa
3a
d
A
75 mm
40 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
582
y
M ! 300 N · m
A
z
PROBLEM 4.127
" ! 60#
B
16 mm
C
16 mm
The couple M is applied to a beam of the cross section shown in a plane
forming an angle with the vertical. Determine the stress at (a) point A,
(b) point B, (c) point D.
D
40 mm
40 mm
SOLUTION
1
(80)(32)3 218.45 103 mm 4 218.45 10 9 m 4
12
1
(32)(80)3 1.36533 106 mm 4 1.36533 10 6 m 4
12
yB
yD 16 mm
Iz
Iy
yA
zA
My
(a)
A
300cos 30
M z yA
Iz
zB
zD
40 mm
259.81 N m
M y zA
Iy
Mz
(150)(16 10 3 )
218.45 10 9
300sin 30
150 N m
(259.81)(40 10 3 )
1.36533 10 6
3.37 106 Pa
(b)
B
M z yB
Iz
M y zB
Iy
(150)(16 10 3 )
218.45 10 9
(259.81)( 40 10 3 )
1.36533 10 6
18.60 106 Pa
(c)
D
M z yD
Iz
M y zD
Iy
(150)( 16 10 3 )
218.45 10 9
3.37 MPa
A
18.60 MPa
B
(259.81)( 40 10 3 )
1.36533 10 6
3.37 106 Pa
D
3.37 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
583
PROBLEM 4.128
y
! " 30#
A
The couple M is applied to a beam of the cross section shown in a plane
forming an angle with the vertical. Determine the stress at (a) point A,
(b) point B, (c) point D.
B
M " 400 lb · m
0.6 in.
z
C
0.6 in.
D
0.4 in.
SOLUTION
Iz
Iy
yA
zA
My
(a)
A
400 cos 60
200 lb in.,
M y zA
M z yA
Iz
Iy
1
(0.4)(1.2)3 57.6 10 3 in 4
12
1
(1.2)(0.4)3 6.40 10 3 in 4
12
yB
yD 0.6 in.
zB
Mz
1
(0.4) 0.2 in.
2
zD
400 sin 60
( 346.41)(0.6)
57.6 10 3
346.41 lb in.
(200)(0.2)
6.40 10 3
9.86 103 psi 9.86 ksi
(b)
B
M y zB
M z yB
Iz
Iy
( 346.41)(0.6)
57.6 10 6
2.64 103 psi
(c)
D
M z yD
Iz
M y zD
Iy
( 346.41)( 0.6)
57.6 10 3
9.86 103 psi
(200)( 0.2)
6.4 10 3
2.64 ksi
(200)( 0.2)
6.40 10 3
9.86 ksi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
584
y
A
M ! 25 kN · m
PROBLEM 4.129
" ! 15#
The couple M is applied to a beam of the cross section shown in a
with the vertical. Determine the stress at
plane forming an angle
(a) point A, (b) point B, (c) point D.
B
80 mm
C
z
20 mm
80 mm
D
30 mm
SOLUTION
My
25sin 15
6.4705 kN m
Mz
25cos15
24.148 kN m
1
1
(80)(90)3
(80)(30)3
12
12
5.04 10 6 m 4
Iy
Iy
Iz
1
(90)(60)3
3
M yz
Stress:
(a)
A
1
(60)(20)3
3
Iy
5.04 106 mm 4
1
(30)(100)3 16.64 106 mm 4
3
16.64 10 6 m 4
Mzy
Iz
(6.4705 kN m)(0.045 m)
5.04 10 6 m 4
(24.148 kN m)(0.060 m)
16.64 10 6 m 4
57.772 MPa 87.072 MPa
(b)
B
(6.4705 kN m)( 0.045 m)
5.04 10 6 m 4
A
(24.148 kN m)(0.060 m)
16.64 10 6 m 4
57.772 MPa 87.072 MPa
(c)
D
(6.4705 kN m)( 0.015 m)
5.04 10 6 m 4
29.3 MPa
B
144.8 MPa
D
125.9 MPa
(24.148 kN m)( 0.100 m)
16.64 10 6 m 4
19.257 MPa 145.12 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
585
y
PROBLEM 4.130
" ! 20#
M ! 10 kip · in.
A
z
2 in.
C
B
D
2 in.
3 in.
The couple M is applied to a beam of the cross section shown in a plane
forming an angle with the vertical. Determine the stress at (a) point A,
(b) point B, (c) point D.
3 in.
4 in.
SOLUTION
Locate centroid.
A, in 2
z , in.
Az , in 3
16
1
16
8
2
16
Σ
24
0
The centroid lies at point C.
yA
1
1
(2)(8)3
(4)(2)3 88 in 4
12
12
1
1
(8)(2)3
(2)(4)3 64 in 4
3
3
4 in.
yB 1 in.,
yD
zA
zB
Iz
Iy
(a)
M z yA
Iz
M y zA
A
(b)
M z yB
Iz
M y zB
B
(c)
M z yD
Iz
M y zD
D
Iy
Iy
Iy
zD
4 in.,
0
Mz
10 cos 20
9.3969 kip in.
My
10 sin 20
3.4202 kip in.
(9.3969)(1)
88
(3.4202)( 4)
64
(9.3969)( 1)
88
(3.4202)( 4)
64
(9.3969)( 4)
88
(3.4202)(0)
64
A
0.321 ksi
0.107 ksi
B
D
0.427 ksi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
586
PROBLEM 4.131
y
M 5 60 kip · in.
b 5 508
A
The couple M is applied to a beam of the cross section shown in a
plane forming an angle with the vertical. Determine the stress at
(a) point A, (b) point B, (c) point D.
B
3 in.
z
C
3 in.
D
1 in.
2.5 in. 2.5 in.
5 in.
5 in.
1 in.
SOLUTION
Mz
My
yA
zA
(a)
A
M z yA
Iz
60 sin 40
60 cos 40
yB
yD
zB
zD
38.567 kip in.
45.963 kip in.
3 in.
5 in.
Iz
1
(10)(6)3
12
2
Iy
1
(6)(10)3
12
2
M y zA
Iy
4
4
( 38.567)(3)
178.429
(1) 2
178.429 in 4
(1) 4
(1) 2 (2.5) 2
459.16 in 4
(45.963)(5)
459.16
1.149 ksi
(b)
B
M z yB
Iz
M y zB
Iy
( 38.567)(3)
178.429
A
(45.963)( 5)
459.16
0.1479
(c)
D
M z yD
Iz
M y zD
Iy
( 38.567)( 3)
178.429
1.149 ksi
B
0.1479 ksi
D
1.149 ksi
(45.963)( 5)
459.16
1.149 ksi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
587
PROBLEM 4.132
y
M 5 75 kip · in.
The couple M is applied to a beam of the cross section shown in a
plane forming an angle with the vertical. Determine the stress at
(a) point A, (b) point B, (c) point D.
b 5 758
A
2.4 in.
B
1.6 in. z
C
D
4 in.
4.8 in.
SOLUTION
Iz
Iy
yA
zA
(a)
M z yA
Iz
M y zA
A
(b)
M z yB
Iz
M y zB
B
(c)
M z yD
Iz
M y zD
D
Iy
Iy
Iy
1
1
(4.8)(2.4)3
(4)(1.6)3 4.1643 in 4
12
12
1
1
(2.4)(4.8)3
(1.6)(4)3 13.5851 in 4
12
12
yB
yD 1.2 in.
zB
2.4 in.
zD
Mz
75sin15
19.4114 kip in.
My
75cos15
72.444 kip in.
(19.4114)(1.2)
4.1643
(72.444)(2.4)
13.5851
(19.4114)(1.2)
4.1643
(72.444)( 2.4)
13.5851
(19.4114)( 1.2)
4.1643
(72.444)( 2.4)
13.5851
A
B
D
7.20 ksi
18.39 ksi
7.20 ksi
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588
PROBLEM 4.133
y
! " 30#
M " 100 N · m
The couple M is applied to a beam of the cross section shown in a
plane forming an angle with the vertical. Determine the stress at
(a) point A, (b) point B, (c) point D.
B
z
C
D
A
r " 20 mm
SOLUTION
Iz
8
r4
2
4r
3
r2
(0.109757)(20) 4
Iy
yA
yB
A
8
8
9
17.5611 10
r4
3
mm 4
17.5611 10
(20) 4
62.832 10 3 mm 4 62.832 10
8
8
4r
(4)(20)
yD
8.4883 mm
3
3
20 8.4883 11.5117 mm
r4
zA
(a)
2
20 mm
zD
zB
Mz
100 cos30
86.603 N m
My
100sin 30
50 N m
M z yA
Iz
M y zA
Iy
9
9
m4
m4
0
(86.603)( 8.4883 10 3 )
17.5611 10 9
(50)(20 10 3 )
62.832 10 9
57.8 106 Pa
(b)
B
M z yB
Iz
M y zB
Iy
A
3
(86.603)(11.5117 10 )
17.5611 10 9
(50)(0)
62.832 10
9
56.8 106 Pa
(c)
D
M z yD
Iz
M y zD
Iy
56.8 MPa
B
3
(86.603)( 8.4883 10 )
17.5611 10 9
25.9 106 Pa
57.8 MPa
3
(50)( 20 10 )
62.832 10 9
D
25.9 MPa
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589
W310 $ 38.7 15#
PROBLEM 4.134
B
A
C
310 mm
The couple M is applied to a beam of the cross section shown in a plane
with the vertical. Determine the stress at
forming an angle
(a) point A, (b) point B, (c) point D.
M " 16 kN · m
D
E
165 mm
SOLUTION
For W310 38.7 rolled steel shape,
(a)
tan
Iz
tan
I
84.9 10
7.20 10
Iz
84.9 106 mm 4
84.9 10 6 m 4
Iy
7.20 106 mm 4
7.20 10 6 m 4
yA
yB
1
yD
2
zA
zE
zB
yE
zD
1
(310) 155 mm
2
1
(165) 82.5 mm
2
Mz
(16 103 ) cos 15
15.455 103 N m
My
(16 103 ) sin 15
4.1411 103 N m
6
6
tan 15
3.1596
72.4
72.4 15 57.4
(b)
Maximum tensile stress occurs at point E.
E
M z yE
Iz
M y zE
Iy
(15.455 103 )( 155 10 3 )
84.9 10 6
75.7 106 Pa
(4.1411 103 )(82.5 10 3 )
7.20 10 6
75.7 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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590
PROBLEM 4.135
208
S6 3 12.5
A
M 5 15 kip · in.
The couple M acts in a vertical plane and is applied to a beam oriented
as shown. Determine (a) the angle that the neutral axis forms with the
horizontal, (b) the maximum tensile stress in the beam.
B
C
E
3.33 in.
6 in.
D
SOLUTION
For S6 12.5 rolled steel shape,
Iz
22.0 in 4
Iy
1.80 in 4
zE
yA
(a)
tan
Iz
tan
Iy
zA
zB
yB
yD
zD
yE
1
(3.33) 1.665 in.
2
1
(6) 3 in.
2
Mz
15 sin 20
5.1303 kip in.
My
15 cos 20
14.095 kip in.
22.0
tan (90
1.80
20 ) 33.58
88.29
88.29
(b)
70
18.29
Maximum tensile stress occurs at point D.
D
M z yD
Iz
M y zD
Iy
(5.1303)( 3)
22.0
(14.095)(1.665)
13.74 ksi
1.80
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591
PROBLEM 4.136
5#
C150 ! 12.2
B
A
M " 6 kN · m
C
152 mm
The couple M acts in a vertical plane and is applied to a beam oriented as
shown. Determine (a) the angle that the neutral axis forms with the
horizontal, (b) the maximum tensile stress in the beam.
13 mm
E
D
48.8 mm
SOLUTION
My
6 sin 5
0.52293 kN m
Mz
6 cos 5
5.9772 kN m
C150 12.2
(a)
Iy
0.286 10
Iz
5.45 10
6
6
m4
m4
Neutral axis:
tan
Iz
tan
Iy
tan
1.66718
5.45 10 6 m 4
tan 5
0.286 10 6 m 4
59.044
5
(b)
Maximum tensile stress at E: y
M y zE
E
Iy
Mz yE
Iz
E
54.0
76 mm, z E
13 mm
(0.52293 kN m)(0.013 m)
0.286 10 6 m 4
(5.9772 kN m)( 0.076 m)
5.45 10 6 m 4
E
107.1 MPa
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592
y'
30"
PROBLEM 4.137
B
50 mm
A
5 mm
5 mm
C
M ! 400 N · m
z'
The couple M acts in a vertical plane and is applied to a beam oriented
as shown. Determine (a) the angle that the neutral axis forms with the
horizontal, (b) the maximum tensile stress in the beam.
D
E
18.57 mm
5 mm
50 mm
Iy' ! 281 # 103 mm4
Iz' ! 176.9 # 103 mm4
SOLUTION
Iz
176.9 103 mm 4
Iy
281 103 mm 4
yE
(a)
tan
Iz
tan
Iy
18.57 mm, z E
176.9 10
281 10
m4
m4
25 mm
Mz
400 cos 30
346.41 N m
My
400sin 30
200 N m
176.9 10 9
tan 30
281 10 9
9
9
0.36346
19.97
30
(b)
19.97
10.03
Maximum tensile stress occurs at point E.
E
M z yE
Iz
M y zE
Iy
(346.41)( 18.57 10 3 )
176.9 10 9
36.36 106
17.79 106
(200)(25 10 3 )
281 10 9
54.2 106 Pa
E
54.2 MPa
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593
458
PROBLEM 4.138
y'
B
z'
0.859 in.
M 5 15 kip · in.
C
A
1
2
in.
4 in.
D
4 in.
The couple M acts in a vertical plane and is applied to a beam oriented
as shown. Determine (a) the angle that the neutral axis forms with the
horizontal, (b) the maximum tensile stress in the beam.
4 in.
Iy' 5 6.74 in4
Iz' 5 21.4 in4
SOLUTION
Iz
21.4 in 4
zA
zB
yA
Mz
(a)
0.859 in.
4 in.
My
Iy
yB
zD
4 in.
15 sin 45
15 cos 45
6.74 in 4
4 0.859 in.
yD
3.141 in.
0.25 in.
10.6066 kip in.
10.6066 kip in.
Angle of neutral axis:
tan
Iz
tan
Iy
21.4
tan ( 45 ) 3.1751
6.74
72.5
72.5
(b)
27.5
45
The maximum tensile stress occurs at point D.
D
M z yD
Iz
M y zD
Iy
0.12391 4.9429
(10.6066)( 0.25)
21.4
( 10.6066)( 3.141)
6.74
D
5.07 ksi
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594
y'
20"
PROBLEM 4.139
B
10 mm
A
The couple M acts in a vertical plane and is applied to a beam
oriented as shown. Determine (a) the angle that the neutral
axis forms with the horizontal, (b) the maximum tensile stress
in the beam.
6 mm
M ! 120 N · m
C
z'
10 mm
D
6 mm
Iy' ! 14.77 # 103 mm4
Iz' ! 53.6 # 103 mm4
E
10 mm
10 mm
SOLUTION
(a)
Iz
53.6 103 mm 4
Iy
14.77 103 mm 4
14.77 10 9 m 4
Mz
120 sin 70
112.763 N m
My
120 cos 70
41.042 N m
Angle of neutral axis:
20
tan
(b)
53.6 10 9 m 4
Iz
tan
Iy
53.6 10 9
tan 20
14.77 10 9
52.871
52.871
20
1.32084
32.9
The maximum tensile stress occurs at point E.
yE
zE
E
16 mm
10 mm
M z yE
Iz
0.016 m
0.010 m
M y zE
Iy
(112.763)( 0.016)
53.6 10 9
61.448 106 Pa
(41.042)(0.010)
14.77 10 9
E
61.4 MPa
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595
PROBLEM 4.140
208
A
90 mm
M 5 750 N · m
C
The couple M acts in a vertical plane and is applied to a beam oriented as
shown. Determine (a) the angle that the neutral axis forms with the horizontal,
(b) the maximum tensile stress in the beam.
B
30 mm
25 mm
25 mm
SOLUTION
(a)
My
750 sin 20
My
256.5 N m
Mz
750 cos 20
Mz
704.8 N m
1
(90)(25)3
12
0.2344 106 mm 4
Iy
2
Iy
0.2344 10 6 m 4
Iz
1
(50)(90)3 1.0125 106 mm 4
36
1.0125 10 6 m 4
Neutral axis:
tan
Iz
tan
Iy
tan
1.5724
20
1.0125 10 6 m 4
tan 20
0.2344 106 m 4
57.5
57.5
30
37.5
37.5
(b)
Maximum tensile stress, at D: y D
My zD
D
Iy
Mz y D
Iz
30 mm
z
(256.5 N m)( 0.030 m)
0.2344 10 6 m 4
32.83 MPa 17.40 MPa
50.23 MPa
25 mm
(704.8 N m)(0.025 m)
1.0125 10 6 m 4
D
50.2 MPa
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596
A
PROBLEM 4.141
y
z
M ! 1.2 kN · m
10 mm
40 mm
The couple M acts in a vertical plane and is applied to a beam oriented
as shown. Determine the stress at point A.
10 mm
C
40 mm
70 mm
10 mm
Iy ! 1.894 # 106 mm4
Iz ! 0.614 # 106 mm4
Iyz ! $0.800 # 106 mm4
SOLUTION
Using Mohr’s circle, determine the principal axes and principal moments of inertia.
Y : (1.894, 0.800) 106 mm 4
Z : (0.614, 0.800) 106 mm 4
E : (1.254, 0) 106 mm 4
R
tan 2
EF
2
FZ
2
0.6402
0.8002 10
6
1.0245 106 mm 4
Iv
(1.254 1.0245) 106 mm 4
0.2295 106 mm 4
0.2295 10 6 m 4
Iu
(1.254 1.0245) 106 mm 4
2.2785 106 mm 4
2.2785 10 6 m 4
m
Mv
Mu
FZ 0.800 106
1.25
25.67
m
FE 0.640 106
M cos m (1.2 103 ) cos 25.67 1.0816 103 N m
M sin
(1.2 103 ) sin 25.67
m
0.5198 103 N m
uA
y A cos
m
z A sin
m
45 cos 25.67
45 sin 25.67
21.07 mm
vA
z A cos
m
y A sin
m
45 cos 25.67
45 sin 25.67
60.05 mm
A
M vuA
Iv
M u vA
Iu
(1.0816 103 )(21.07 10 3 )
0.2295 10 6
113.0 106 Pa
( 0.5198 103 )(60.05 10 3 )
2.2785 10 6
A
113.0 MPa
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597
PROBLEM 4.142
y
A
2.4 in.
The couple M acts in a vertical plane and is applied to a beam oriented as
shown. Determine the stress at point A.
2.4 in.
z
M ! 125 kip · in.
C
2.4 in.
2.4 in.
2.4 in. 2.4 in.
SOLUTION
Iy
2
1
(7.2)(2.4)3
3
66.355 in 4
Iz
2
1
(2.4)(7.2)3
12
(2.4)(7.2)(1.2) 2
I yz
2 (2.4)(7.2)(1.2)(1.2)
199.066 in 4
49.766 in 4
Using Mohr’s circle, determine the principal axes and principal moments of inertia.
Y : (66.355 in 4 , 49.766 in 4 )
Z : (199.066 in 4 ,
49.766 in 4 )
E : (132.710 in 4 , 0)
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598
PROBLEM 4.142 (Continued)
tan 2
m
2
m
R
DY 49.766
DE 66.355
36.87
18.435
m
DE
2
DY
2
82.944 in 4
Iu
132.710 82.944
49.766 in 4
Iv
132.710 82.944
215.654 in 4
Mu
125sin18.435
39.529 kip in.
Mv
125cos18.435
118.585 kip in.
uA
4.8 cos 18.435
2.4 sin 18.435
4.8 sin 18.435
A
M uA
I
A
Mu
Iu
2.4 cos 18.435
0.7589 in.
A
(118.585)(5.3126)
215.654
2.32 ksi
5.3126 in.
(39.529)(0.7589)
49.766
A
2.32 ksi
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599
y
PROBLEM 4.143
1.08 in.
0.75 in.
The couple M acts in a vertical plane and is applied to a beam oriented
as shown. Determine the stress at point A.
2.08 in.
z
C
M 5 60 kip · in.
6 in.
0.75 in.
A
4 in.
Iy 5 8.7 in4
Iz 5 24.5 in4
Iyz 5 18.3 in4
SOLUTION
Using Mohr’s circle, determine the principal axes and principal moments of inertia.
Y : (8.7, 8.3) in 4
Z : (24.5,
8.3) in 4
E : (16.6, 0) in 4
EF
7.9 in 4
FZ
8.3 in 4
R
7.92
8.32
11.46 in 4
tan 2
m
FZ 8.3
1.0506
EF 7.9
I v 16.6 11.46 28.06 in 4
23.2
Iu
16.6 11.46 5.14 in 4
Mu
M sin
m
(60) sin 23.2
23.64 kip in.
Mv
M cos
m
(60) cos 23.2
55.15 kip in.
uA
y A cos
m
z A sin
m
3.92 cos 23.2
1.08 sin 23.2
4.03 in.
vA
z A cos
m
y A sin
m
1.08cos 23.2
3.92 sin 23.2
0.552 in.
m
A
M vuA
Iv
M u vA
Iu
(55.15)( 4.03)
28.06
(23.64)(0.552)
5.14
A
10.46 ksi
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600
PROBLEM 4.144
D
B
H 14 kN
G
E
A
F
28 kN
The tube shown has a uniform wall thickness of 12 mm. For the
loading given, determine (a) the stress at points A and B, (b) the point
where the neutral axis intersects line ABD.
125 mm
75 mm
28 kN
SOLUTION
Add y- and z-axes as shown. Cross section is a 75 mm
cutout.
125-mm rectangle with a 51 mm
101-mm rectangular
1
1
(75)(125)3
(51)(101)3 7.8283 106 mm 4 7.8283 10 6 m 4
12
12
1
1
(125)(75)3
(101)(51)3 3.2781 103 mm 4 3.2781 10 6 m 4
Iy
12
12
A (75)(125) (51)(101) 4.224 103 mm 2 4.224 10 3 m 2
Iz
Resultant force and bending couples:
70 103 N
P 14 28 28 70 kN
(a)
A
Mz
(62.5 mm)(14 kN) (62.5 mm)(28kN) (62.5 mm)(28 kN)
2625 N m
My
(37.5 mm)(14 kN) (37.5 mm)(28 kN) (37.5 mm)(28 kN)
525 N m
P
A
M z yA
Iz
M y zA
70 103
4.224 10
Iy
3
(2625)( 0.0625)
7.8283 10 6
( 525)(0.0375)
3.2781 10 6
31.524 106 Pa
B
P
A
M z yB
Iz
A
M y zB
70 103
4.224 10
Iy
3
(2625)(0.0625)
7.8283 10 6
( 525)(0.0375)
3.2781 10 6
10.39 106 Pa
(b)
31.5 MPa
B
10.39 MPa
Let point H be the point where the neutral axis intersects AB.
zH
0
yH
0.0375 m,
P
A
yH
M z yH
Iz
Iz P
Mz A
?,
H
0
M y zH
Mz H
Iy
Iy
7.8283 10
2625
6
70 103
4.224 10
3
( 525)(0.0375)
3.2781 10 6
0.03151 m 31.51 mm
31.51 62.5 94.0 mm
Answer: 94.0 mm above point A.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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601
PROBLEM 4.145
A horizontal load P of magnitude 100 kN is applied to the beam
shown. Determine the largest distance a for which the maximum
tensile stress in the beam does not exceed 75 MPa.
y
20 mm
a
20 mm
O
z
x
P
20 mm
60 mm
20 mm
SOLUTION
Locate the centroid.
A, mm2
y , mm
2000
10
20 103
1200
3200
–10
12 103
Ay , mm3
Y
Ay
A
8 103
3200
2.5 mm
8 103
Move coordinate origin to the centroid.
Coordinates of load point:
XP
a,
Bending couples:
Mx
yP P
Ix
1
(100)(20)3
12
(2000)(7.5)2
yP
2.5 mm
My
1
(60)(20)3
12
aP
(1200)(12.5)2
0.40667 106 mm 4
0.40667 10
Iy
1
1
(20)(100)3
(20)(60)3
12
12
P M x y M yx
A
Ix
Iy
2.0267 106 mm 4
A
6
m4
2.0267 10 6 m 4
75 106 Pa, P
100 103 N
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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602
PROBLEM 4.145 (Continued)
My
Iy P
x A
My
2.0267 10
50 10 3
Mxy
Ix
6
For point A,
100 103
3200 10 6
2.0267 10 6
{31.25
50 10 3
a
My
P
50 mm, y
( 2.5)(100 103 )( 2.5 10 3 )
0.40667 10 6
1.537
(1.7111 103 )
100 103
x
75} 106
2.5 mm
75 106
1.7111 103 N m
17.11 103 m
a
17.11 mm
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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603
1 in.
PROBLEM 4.146
1 in.
4 in.
1 in.
5 in.
P
Knowing that P 90 kips, determine the largest distance a for
which the maximum compressive stress dose not exceed 18 ksi.
a
2.5 in.
SOLUTION
A
Ix
Iz
(5 in.)(6 in.)
2(2 in.)(4 in.)
14 in 2
1
1
(5 in.)(6 in.)3 2 (2 in.)(4 in.)3
12
12
1
1
2 (1 in.)(5 in.)3
(4 in.)(1 in.)3
12
12
21.17 in 4
Force-couple system at C:
P
P
For P
P
90 kips M x
90 kips:
Mx
68.67 in 4
Maximum compressive stress at B:
B
18 ksi
18
1.741
P
A
M x (3 in.)
Ix
90 kips
14 in 2
6.429
B
Mz
Pa
(90 kips)(2.5 in.)
225 kip in.
Mz
(90 kips) a
18 ksi
M z (2.5 in.)
Iz
(225 kip in.)(3 in.)
68.67 in 4
9.830
P(2.5 in.)
(90 kips) a (2.5 in.)
21.17 in 4
10.628a
10.628 a
a
0.1638 in.
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604
1 in.
PROBLEM 4.147
1 in.
4 in.
1 in.
P
5 in.
Knowing that a 1.25 in., determine the largest value of P that
can be applied without exceeding either of the following
allowable stresses:
a
2.5 in.
10 ksi
ten
18 ksi
comp
SOLUTION
A
Ix
Iz
(5 in.)(6 in.)
(2)(2 in.)(4 in.)
14 in 2
1
1
(5 in.)(6 in.)3 2 (2 in.)(4 in.)3 68.67 in 4
12
12
1
1
2 (1 in.)(5 in.)3
(4 in.)(1 in.)3 21.17 in 4
12
12
For a
Force-couple system at C:
P
My
P(2.5 in.)
P Mx
Pa
1.25 in.,
(1.25 in.)
Maximum compressive stress at B:
B
18 ksi
18
18
P
A
M x (3 in.)
Ix
P
14 in 2
0.0714P
0.3282P
P(2.5 in.)(3 in.)
68.67 in 4
0.1092 P
P
10 ksi
10
P
A
M x (3 in.)
Ix
0.0714P
0.1854 P
P(1.25 in.)(2.5 in.)
21.17 in 4
0.1476 P
54.8 kips
D
10 ksi
M z (2.5 in.)
Iz
0.1092 P
P
18 ksi
M z (2.5 in.)
Iz
Maximum tensile stress at D:
D
B
0.1476 P
53.9 kips
The smaller value of P is the largest allowable value.
P
53.9 kips
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605
y
R ! 125 mm
C
z
P ! 4 kN
PROBLEM 4.148
E
%
A rigid circular plate of 125-mm radius is attached to a solid
150 200-mm rectangular post, with the center of the plate directly above
30 ,
the center of the post. If a 4-kN force P is applied at E with
determine (a) the stress at point A, (b) the stress at point B, (c) the point
where the neutral axis intersects line ABD.
x
A
D
B
200 mm 150 mm
SOLUTION
4 103 N (compression)
P
Mx
Mz
Iz
xA
A
(a)
A
P
A
56.25 106 mm 4
100 106 mm 4
zA
zB
4 103
30 10 3
M z xA
Iz
56.25 10
100 10
6
6
B
P
A
M x zB
Ix
30 10
3
m2
( 433)( 100 10 3 )
100 10 6
( 250)(75 10 3 )
56.25 10 6
633 103 Pa
633 kPa
( 433)(100 10 3 )
100 10 6
B
(c)
m4
75 mm
( 250)(75 10 3 )
56.25 10 6
4 103
30 10 3
M z xB
Iz
433 N m
m4
A
(b)
250 N m
3
(4 10 )(125 10 ) cos 30
30 103 mm 2
(200)(150)
M xzA
Ix
3
PR cos 30
1
(200)(150)3
12
1
(150)(200)3
12
xB 100 mm
Ix
(4 103 )(125 10 3 )sin 30
PR sin 30
233 103 Pa
233 kPa
Let G be the point on AB where the neutral axis intersects.
G
0
P
A
G
xG
zG
75 mm
M x zG
Ix
Iz P
Mz A
46.2 10
M z xG
Iz
M x zG
Ix
3
m
xG
?
0
100 10
433
6
4 103
30 10 3
46.2 mm
( 250)(75 10 3 )
56.25 10 6
Point G lies 146.2 mm from point A.
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606
y
R ! 125 mm
C
z
P ! 4 kN
PROBLEM 4.149
E
%
In Prob. 4.148, determine (a) the value of θ for which the stress at D
reaches it largest value, (b) the corresponding values of the stress at A, B,
C, and D.
x
A
D
PROBLEM 4.148 A rigid circular plate of 125-mm radius is attached to a
solid 150 200-mm rectangular post, with the center of the plate directly
above the center of the post. If a 4-kN force P is applied at E with
30 , determine (a) the stress at point A, (b) the stress at point B,
(c) the point where the neutral axis intersects line ABD.
B
200 mm 150 mm
SOLUTION
(a)
4 103 N
P
PR sin
Mx
Iz
xD
For
M zx
Iz
d
d
to be a maximum,
d
d
D
P 0
RxD sin
IZ
tan
I z zD
I x xD
sin
A
P
A
0
Rz D cos
Ix
sin
cos
(b)
1
A
P
M xzA
Ix
M z xA
Iz
PR cos
Mx
30 103 mm 2
(200)(150)
A
M xz
Ix
500sin
500 N m
500 cos
1
(200)(150)3 56.25 106 mm 4 56.25 10 6 m 4
2
1
(150)(200)3 100 106 mm 4 100 10 6 m 4
2
100 mm
75 mm
zD
Ix
P
A
(4 103 )(125 10 3 )
PR
0.8,
Rz sin
Ix
Rx cos
Iz
with z
zD , x
30 10 3 m 2
xD
0
(100 10 6 )( 75 10 3 )
(56.25 10 6 )(100 10 3 )
cos
4 103
30 10 3
4
3
53.1
0.6
(500)(0.8)(75 10 3 )
56.25 10 6
(500)(0.6)( 100 10 3 )
100 10 6
( 0.13333
0.53333
0.300) 106 Pa
0.700 106 Pa
A
700 kPa
B
( 0.13333
0.53333
0.300) 106 Pa
0.100 106 Pa
B
100 kPa
C
( 0.13333
0
D
( 0.13333
0.53333
0) 106 Pa
133.3 kPa
C
0.300) 106 Pa
0.967 106 Pa
D
967 kPa
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607
y
PROBLEM 4.150
0.5 in.
1.43 in.
z
M0
C
5 in.
0.5 in.
1.43 in.
A beam having the cross section shown is subjected to a couple M 0 that
acts in a vertical plane. Determine the largest permissible value of the
moment M 0 of the couple if the maximum stress in the beam is
not to exceed 12 ksi. Given: I y I z 11.3 in 4 , A 4.75 in 2 ,
kmin 0.983 in. (Hint: By reason of symmetry, the principal axes form
2
an angle of 45 with the coordinate axes. Use the relations I min Akmin
and I min I max I y I z .)
5 in.
SOLUTION
Mu
M 0 sin 45
0.70711M 0
Mv
M 0 cos 45
0.70711M 0
I min
2
Akmin
I max
Iy
(4.75)(0.983)2
Iz
11.3
I min
11.3
uB
yB cos 45
z B sin 45
vB
z B cos 45
yB sin 45
B
M vu B
Iv
M u vB
Iu
0.70711M 0
M0
B
0.4124
4.59 in 4
3.57 cos 45
0.93 cos 45
0.70711M 0
( 1.866)
4.59
18.01 in 4
4.59
3.182
18.01
12
0.4124
uB
I min
0.93 sin 45
1.866 in.
( 3.57) sin 45
3.182 in.
vB
I max
0.4124M 0
M0
29.1 kip in.
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608
PROBLEM 4.151
y
0.5 in.
Solve Prob. 4.150, assuming that the couple M 0 acts in a horizontal
plane.
1.43 in.
z
M0
C
5 in.
0.5 in.
1.43 in.
5 in.
PROBLEM 4.150 A beam having the cross section shown is subjected
to a couple M 0 that acts in a vertical plane. Determine the largest
permissible value of the moment M 0 of the couple if the maximum
stress in the beam is not to exceed 12 ksi. Given: I y I z 11.3 in 4,
A 4.75 in 2, kmin 0.983 in. (Hint: By reason of symmetry, the
principal axes form an angle of 45 with the coordinate axes. Use the
2
relations I min Akmin
and I min I max I y I z .)
SOLUTION
Mu
Mv
M 0 cos 45
0.70711M 0
M 0 sin 45
I min
2
Akmin
I max
Iy
0.70711M 0
(4.75)(0.983)2
Iz
11.3
I min
4.59 in 4
11.3
4.59
18.01 in 4
uD
yD cos 45
z D sin 45
0.93 cos 45
( 3.57 sin 45 )
vD
z D cos 45
yD sin 45
( 3.57) cos 45
(0.93) sin 45
D
M vu D
Iv
M u vD
Iu
0.70711M 0
M0
D
0.4124
0.70711M 0
( 1.866)
4.59
3.182
18.01
12
0.4124
uD
I min
1.866 in.
3.182 in.
vD
I max
0.4124M 0
M0
29.1 kip in.
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609
PROBLEM 4.152
y
z
M0
40 mm
10 mm
C
10 mm
40 mm
70 mm
10 mm
The Z section shown is subjected to a couple M 0 acting in a vertical
plane. Determine the largest permissible value of the moment M 0 of the
couple if the maximum stress is not to exceed 80 MPa. Given:
I max 2.28 10 6 mm 4 ,
I min 0.23 10 6 mm 4 , principal axes
25.7
and 64.3
.
SOLUTION
Iv
I max
2.28 106 mm 4
2.28 10 6 m 4
Iu
I min
0.23 106 mm 4
0.23 10 6 m 4
Mv
M 0 cos 64.3
Mu
M 0 sin 64.3
64.3
tan
Iv
tan
Iu
2.28 10
0.23 10
20.597
6
6
tan 64.3
87.22
Points A and B are farthest from the neutral axis.
uB
yB cos 64.3
z B sin 64.3
( 45) cos 64.3
( 35) sin 64.3
yB sin 64.3
( 35) cos 64.3
( 45) sin 64.3
51.05 mm
vB
z B cos 64.3
25.37 mm
B
80 106
M vu B
Iv
M u vB
Iu
(M 0 cos 64.3 )( 51.05 10 3 )
2.28 10 6
(M 0 sin 64.3 )(25.37 10 3 )
0.23 10 6
109.1 103 M 0
M0
80 106
109.1 103
M0
733 N m
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610
PROBLEM 4.153
y
z
M0
40 mm
10 mm
C
10 mm
40 mm
70 mm
10 mm
Solve Prob. 4.152 assuming that the couple M 0 acts in a horizontal
plane.
PROBLEM 4.152 The Z section shown is subjected to a couple M 0
acting in a vertical plane. Determine the largest permissible value of the
moment M 0 of the couple if the maximum stress is not to exceed 80 MPa.
Given: I max 2.28 10 6 mm 4 , I min 0.23 10 6 mm 4 , principal axes
25.7
and 64.3
.
SOLUTION
Iv
I min
0.23 106 mm 4
0.23 106 m 4
Iu
I max
2.28 106 mm 4
2.28 106 m 4
Mv
M 0 cos 64.3
Mu
M 0 sin 64.3
64.3
tan
Iv
tan
Iu
0.23 10
2.28 10
0.20961
6
6
tan 64.3
11.84
Points D and E are farthest from the neutral axis.
uD
yD cos 25.7
z D sin 25.7
( 5) cos 25.7
yD sin 25.7
45cos 25.7
45 sin 25.7
24.02 mm
vD
z D cos 25.7
( 5) sin 25.7
38.38 mm
D
M vu D
Iv
M u vD
Iu
(M D cos 64.3 )( 24.02 10 3 )
0.23 10 6
(M 0 sin 64.3 )(38.38 10 3 )
2.28 10 6
80 106
M0
60.48 103 M 0
1.323 103 N m
M0
1.323 kN m
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611
PROBLEM 4.154
y
0.3 in.
M0
1.5 in.
C
z
0.3 in.
An extruded aluminum member having the cross section shown
is subjected to a couple acting in a vertical plane. Determine the
largest permissible value of the moment M 0 of the couple if the
maximum stress is not to exceed 12 ksi. Given:
I max 0.957 in 4 , I min 0.427 in 4 , principal axes 29.4
and
60.6
0.6 in. 1.5 in. 0.6 in.
SOLUTION
Iu
I max
0.957 in 4
Iv
I min
0.427 in 4
Mu
M 0 sin 29.4 , M v
M 0 cos 29.4
29.4
tan
Iv
tan
Iu
0.427
tan 29.4
0.957
0.2514
14.11
Point A is farthest from the neutral axis.
yA
0.75 in., z A
uA
y A cos 29.4
z A sin 29.4
1.0216 in .
vA
z A cos 29.4
y A sin 29.4
0.2852 in .
A
M vu A
Iv
M uVA
Iu
(M 0 cos 29.4 )( 1.0216)
0.427
0.75 in.
(M 0 sin 29.4 )( 0.2852)
0.957
1.9381 M 0
M0
A
1.9381
12
1.9381
M0
6.19 kip in.
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612
PROBLEM 4.155
y
20 mm
z
A beam having the cross section shown is subjected to a couple M0
acting in a vertical plane. Determine the largest permissible value of
the moment M0 of the couple if the maximum stress is not to exceed
100 MPa. Given: I y I z b 4 /36 and I yz b 4 /72.
M0
C
b = 60 mm
20 mm
b = 60 mm
SOLUTION
Iy
I yz
b 4 604
0.360 106 mm 4
36 36
b 4 604
0.180 106 mm 4
72 72
Iz
Principal axes are symmetry axes.
Using Mohr’s circle, determine the principal moments of inertia.
R
Iv
Iu
Mu
0.180 106 mm 4
I yz
Iy
Iz
R
2
0.540 106 mm 4
I y Iz
R
2
0.180 106 mm 4
M 0 sin 45
45
0.540 10 6 m 4
0.180 10 6 m 4
0.70711M 0 ,
Iv
tan
Iu
tan
Mv
0.540 10
0.180 10
M 0 cos 45
0.70711M 0
6
6
tan 45
3
71.56
Point A:
uA
A
M0
0
vA
M vuA
Iv
20 2 mm
M u vA
Iu
A
111.11 103
0
(0.70711M 0 )( 20 2 103 )
0.180 10
100 106
111.11 103
6
11.11 10 3 M 0
900 N m
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613
PROBLEM 4.155 (Continued)
Point B:
60
uB
2
mm,
M v uB
Iv
B
vB
M u vB
Iu
20
2
mm
(0.70711M 0 )
60
2
6
0.540 10
10
3
(0.70711M 0 )
20
2
0.180 10
10
3
6
111.11 103 M 0
M0
B
111.11 10
3
100 106
111.11 10
3
900 N m
M0
Choose the smaller value.
900 N m
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614
b
PROBLEM 4.156
B
A
Show that, if a solid rectangular beam is bent by a couple applied in a plane containing
one diagonal of a rectangular cross section, the neutral axis will lie along the other
diagonal.
h
M
C
D
E
SOLUTION
tan
Mz
Iz
tan
b
h
M cos , M z
1 3
bh
12
Iz
tan
Iy
M sin
Iy
1 3
hb
12
1 3
bh b
12
1 3 h
hb
12
h
b
The neutral axis passes through corner A of the diagonal AD.
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615
PROBLEM 4.157
y
(a) Show that the stress at corner A of the prismatic
member shown in part a of the figure will be zero if
the vertical force P is applied at a point located on the
line
D
A
A
D
P
B
z
x
z
C
h
6
x
x
b /6
h
B
b
(a)
C
1
(b) Further show that, if no tensile stress is to occur in
the member, the force P must be applied at a point
located within the area bounded by the line found in
part a and three similar lines corresponding to the
condition of zero stress at B, C, and D, respectively.
This area, shown in part b of the figure, is known as
the kern of the cross section.
b
6
(b)
z
h /6
SOLUTION
1 3
hb
Ix
12
h
xA
2
Iz
zA
1 3
bh
12
b
2
A
bh
Let P be the load point.
Mz
A
x
b/6
z
h/6
0,
1
At point E:
z
0
xE
b /6
At point F:
x
0
zF
h/6
(a)
For
(b)
A
0
PxP
Mx
Pz P
P
A
M z xA
Iz
P
bh
( PxP )
1
12
M xzA
Ix
hb
b
2
3
( Pz P
1 bh
12
P
1
bh
xP
b /6
zP
h/6
x
b/6
z
h/6
1
3
h)
2
If the line of action ( xP , z P ) lies within the portion marked TA , a tensile stress will occur at corner A.
By considering
B
0,
C
0, and
D
0, the other portions producing tensile stresses are identified.
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616
PROBLEM 4.158
y
A beam of unsymmetric cross section is subjected to a couple M 0 acting in the
horizontal plane xz. Show that the stress at point A, of coordinates y and z, is
z
A
y
C
z
A
zI z
I yIz
yI yz
2
I yz
My
x
where Iy, Iz, and Iyz denote the moments and product of inertia of the cross section
with respect to the coordinate axes, and My the moment of the couple.
SOLUTION
The stress A varies linearly with the coordinates y and z. Since the axial force is zero, the y- and z-axes are
centroidal axes:
A
C1 y
C2 z
where C1 and C2 are constants.
Mz
y
I zC1
C1
My
I yz
Iz
z
I yzC2
AdA
I yz
C2
C1
A
C2 yz dA
0
C2
I yzC1
IzM y
C1 y 2dA
AdA
C1 yz dA
C2 z 2dA
I y C2
I yz
Iz
C2
I y C2
2
I yz
C2
I yIz
IzM y
I yIz
2
I yz
I yz M y
I yIz
Izz
I yIz
2
I yz
I yz y
2
I yz
My
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617
PROBLEM 4.159
y
A beam of unsymmetric cross section is subjected to a couple M 0 acting in the
vertical plane xy. Show that the stress at point A, of coordinates y and z, is
z
A
y
C
z
A
yI y
zI yz
I yIz
2
I yz
Mz
x
where I y , I z , and I yz denote the moments and product of inertia of the cross
section with respect to the coordinate axes, and M z the moment of the couple.
SOLUTION
The stress A varies linearly with the coordinates y and z. Since the axial force is zero, the y- and z-axes are
centroidal axes:
A
C1 y
C2 z
where C1 and C2 are constants.
My
z
AdA
I yzC1
C2
Mz
I yz
Iy
I y C2
C1
C2
A
C2 z 2dA
0
C1
y
C1 y 2dA
Adz
I z C1
I yM z
C1 yz dA
I yz
I yz
Iy
C2 yz dA
C1
2
I yz
C1
I yIz
I yM z
I yIz
2
I yz
I yz M z
I yIz
2
I yz
Iyy
I yz 2
I yIz
2
I yz
Mz
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618
PROBLEM 4.160
y
(a) Show that, if a vertical force P is applied at point A of the section
shown, the equation of the neutral axis BD is
D
B
P
C
A
xA
z
xA
x
rz2
x
zA
zA
z
rx2
1
where rz and rx denote the radius of gyration of the cross section with
respect to the z axis and the x axis, respectively. (b) Further show that, if
a vertical force Q is applied at any point located on line BD, the stress at
point A will be zero.
SOLUTION
Definitions:
Ix
, rz2
A
rx2
(a)
Mx
Pz A
E
P
A
Mz
Px A
M z xE
Iz
M x zE
Ix
P
1
A
xA
xE
rz2
P
A
zA
zE
rx2
Px A xE
Arz2
Iz
A
Pz A z E
Arx2
0
if E lies on neutral axis.
1
(b)
Mx
Pz E
A
P
A
Mz
M z xA
Iz
xA
x
rz2
zA
z
rx2
0,
xA
x
rz2
P
A
PxE x A
Arz2
Pz E z A
Arx2
zA
z
rx2
1
PxE
M x zA
Iy
0 by equation from part (a).
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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619
24 mm
B
PROBLEM 4.161
B
For the curved bar shown, determine the stress at point A when
(a) h 50 mm, (b) h 60 mm.
h
A
600 N · m
A
50 mm
600 N · m
C
SOLUTION
(a)
h
50 mm,
A
(24)(50)
h
r
ln 2
r1
R
r
e
yA
A
r1
50 mm,
r2
100 mm
1.200 103 mm
50
ln 100
50
72.13475 mm
1
(r1 r2 ) 75 mm
2
r R 2.8652 mm
72.13475
50
22.13475 mm
rA
50 mm
3
My A
AerA
(600)(22.13475 10 )
(1.200 10 3 )(2.8652 10 3 )(50 10 3 )
77.3 106 Pa
77.3 MPa
(b)
h
R
r
yA
A
60 mm,
h
r
ln r2
1
r1
60
ln 110
50
50 mm,
r2
110 mm,
A
(24)(60)
1.440 103 mm 2
76.09796 mm
1
(r1 r2 ) 80 mm
e r R 3.90204 mm
2
76.09796 50 26.09796 mm
rA 50 mm
M yA
AerA
(600)(26.09796 10 3 )
(1.440 10 3 )(3.90204 10 3 )(50 10 3 )
55.7 106 Pa
55.7 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
620
24 mm
B
B
A
A
PROBLEM 4.162
For the curved bar shown, determine the stress at points A and B
when h 55 mm.
h
600 N · m
C
50 mm
600 N · m
SOLUTION
h
A
R
r
e
yA
A
55 mm,
(24)(55)
r1
50 mm,
3
1.320 10 mm
r2
105 mm
2
50
h
74.13025 mm
105
r2
ln
ln
50
r1
1
(r1 r2 ) 77.5 mm
2
r R 3.36975 mm
74.13025
M yA
AerA
50
24.13025 mm
rA
50 mm
(600)(24.13025 10 3 )
(1.320 10 3 )(3.36975 10 3 )(50 10 3 )
65.1 106 Pa
65.1 MPa
yB
B
74.13025
MyB
AerB
105
30.86975 mm
rB
105 mm
3
(600)( 30.8697 10 )
(1.320 10 3 )(3.36975 10 3 )(105 10 3 )
39.7 106 Pa
39.7 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
621
4 kip · in.
PROBLEM 4.163
C
4 kip · in.
3 in.
For the machine component and loading shown, determine the
stress at point A when (a) h 2 in., (b) h 2.6 in.
h
A
0.75 in.
B
SOLUTION
M
4 kip in.
A
Rectangular cross section:
(a)
r
1
(r1
2
h
2 in.
r1
3
R
2
ln 13
2
(b)
M (r R )
Aer
h
2.6 in.
r1
3
R
2.6
3
ln 0.4
2.6
r1
e
2
r
r1
1)
1.8205
2 in.
0.1795 in.
r
e
12.19 ksi
A
12.19 ksi
A
11.15 ksi
1.95 in 2
0.4 in.
M (r R )
Aer
1
(3
2
1 in.
(0.75)(2.6)
1.2904 in.
h
R
( 4)(1 1.8205)
(1.5)(0.1795)(1)
A
At point A:
A
r
1.8205 in.
r2
1.5 in 2
(0.75)(2)
r
3 in. r1
r
1 in.
At point A:
A
h
, e
r
ln 2
r1
r2 ), R
A
bh r2
1.7
1
(3
2
0.4)
1.2904
1.7 in.
0.4906 in.
0.4 in.
( 4)(0.4 1.2904)
(1.95)(0.4096)(0.4)
11.15 ksi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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622
4 kip · in.
4 kip · in.
PROBLEM 4.164
C
3 in.
For the machine component and loading shown, determine the
stress at points A and B when h 2.5 in.
h
A
0.75 in.
B
SOLUTION
M
4 kip in.
Rectangular cross section:
h
2.5 in. b
r2
3 in. r1
r
R
At point A:
At point B:
1
(r1
2
h
ln
r2
r2 )
2.5
ln 3.0
0.5
r2
r1
1.75
e
r
R
r
r1
0.5 in.
A
M (r R)
Aer
r
r2
B
M (r R )
Aer
1.875 in.2
0.75 in. A
h
0.5 in.
1
(0.5
2
3.0)
1.75 in.
1.3953 in.
1.3953
0.3547 in.
( 4 kip in.)(0.5 in. 1.3953 in.)
(0.75 in.)(2.5 in.)(0.3547 in.)(0.5 in.)
A
10.77 ksi
( 4 kip in.)(3 in. 1.3953 in.)
(0.75 in. 2.5 in.)(0.3547 in.)(3 in.)
B
3.22 ksi
3 in.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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623
PROBLEM 4.165
40 mm
r1
The curved bar shown has a cross section of 40 60 mm and an inner
radius r1 15 mm. For the loading shown, determine the largest tensile
and compressive stresses.
60 mm
120 N · m
SOLUTION
h
40 mm, r1
A
(60)(40)
R
At r
h
r
ln 2
r1
r
1
(r1
2
e
r
2400 mm 2
40
55
ln
40
r2 )
R
15 mm,
15 mm, r2
55 mm
2400 10 6 m 2
30.786 mm
35 mm
My
Aer
4.214 mm
y
30.786
15
15.756 mm
(120)(15.786 10 3 )
(2400 10 6 )(4.214 10 3 )(15 10 3 )
12.49 10
6
Pa
12.49 MPa
At r
55 mm,
y
30.786
55
24.214 mm
(120)( 24.214 10 3 )
(2400 10 6 )(4.214 10 3 )(55 10 3 )
5.22 106 Pa
5.22 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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624
PROBLEM 4.166
40 mm
r1
For the curved bar and loading shown, determine the percent
error introduced in the computation of the maximum stress by
assuming that the bar is straight. Consider the case when
(a) r1 20 mm, (b) r1 200 mm, (c) r1 2 m.
60 mm
120 N · m
SOLUTION
h
40 mm, A
I
1 3
bh
12
(60)(40)
1
(60)(40)3
12
2400 mm 2
2400 10
0.32 106 mm 4
6
m2 , M
120 N m
0.32 10 6 mm 4 ,
c
1
h
2
20 mm
Assuming that the bar is straight,
s
(a)
r1
R
(120)(20 10 8 )
(0.32 10 6 )
Mc
I
20 mm r2
h
r
ln 2
r1
1
(r1
2
a
M (r1 R)
Aer
r2 )
% error
7.5 MPa
60 mm
40
60
ln
20
r
7.5 106 Pa
36.4096 mm
e
40 mm
r1
R
r
R
16.4096 mm
3.5904 mm
(120)( 16.4096 10 3 )
(2400 10 6 )(3.5904 10 3 )(20 10 3 )
11.426 ( 7.5)
11.426
100%
11.426 106 Pa
11.426 MPa
34.4%
For parts (b) and (c), we get the values in the table below:
(a)
(b)
(c)
r1, mm
r2 , mm
20
200
2000
60
240
2040
R, mm
36.4096
219.3926
2019.9340
r , mm
40
220
2020
e, mm
3.5904
0.6074
0.0660
, MPa
11.426
7.982
7.546
% error
34.4 %
6.0 %
0.6 %
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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625
0.3 in.
B
0.4 in.
P9
PROBLEM 4.167
B
P
0.4 in.
Steel links having the cross section shown are available with
different central angles . Knowing that the allowable stress is
12 ksi, determine the largest force P that can be applied to a
link for which
90 .
0.8 in.
A
A
0.8 in.
b
1.2 in.
C
C
SOLUTION
Reduce section force to a force-couple system at G, the centroid of the cross section AB.
a
The bending couple is M
r 1
cos
2
Pa.
For the rectangular section, the neutral axis for bending couple only lies at
h
R
Also, e
r
.
r2
r1
ln
R
At point A, the tensile stress is
where
A
K
1
r
1.2 in., r1
A
(0.3)(0.8)
e
1.2
a
1.2(1
K
1
P
(0.24)(12)
4.3940
My A
Aer1
ay A
er1
P
A
Pay A
Aer1
yA
and
R
P
1
A
ay A
er1
K
P
A
r1
A A
K
P
Data:
P
A
0.8 in., r2
1.6 in., h
0.24 in 2
1.154156
cos 45 )
R
0.045844 in.,
yA
0.8 in., b 0.3 in.
0.8
1.154156 in.
ln 1.6
0.8
1.154156
0.8
0.35416 in.
0.35147 in.
(0.35147)(0.35416)
(0.045844)(0.8)
4.3940
P
0.65544 kips
655 lb
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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626
PROBLEM 4.168
0.3 in.
B
B
0.4 in.
P9
P
0.4 in.
A
Solve Prob. 4.167, assuming that
PROBLEM 4.167 Steel links having the cross section shown are
available with different central angles . Knowing that the
allowable stress is 12 ksi, determine the largest force P that can be
applied to a link for which
90 .
A
0.8 in.
b
1.2 in.
C
60 .
0.8 in.
C
SOLUTION
Reduce section force to a force-couple system at G, the centroid of the cross section AB.
a
The bending couple is M
r 1
cos
2
Pa.
For the rectangular section, the neutral axis for bending couple only lies at
h
R
Also, e
r
r2
r1
.
R
At point A, the tensile stress is
r
1.2 in., r1
A
(0.3)(0.8)
e
1.2
a
(1.2)(1
K
1
P
(0.24)(12)
2.5525
P
A
A
where
Data:
ln
My A
Aer1
ay A
er1
K
1
P
A A
K
0.8 in., r2
P
A
and
1.6 in., h
0.24 in 2
1.154156
cos 30 )
0.045844 in.
Pay A
Aer1
P
1
A
yA
R
ay A
er1
K
P
A
r1
0.8 in., b 0.3 in.
0.8
1.154156 in.
R
ln 1.6
0.8
yA
1.154156
0.8
0.35416 in.
0.160770 in.
(0.160770)(0.35416)
(0.045844)(0.8)
2.5525
P
1.128 kips
1128 lb
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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627
5 kN
PROBLEM 4.169
a
The curved bar shown has a cross section of 30 30 mm. Knowing
that the allowable compressive stress is 175 MPa, determine the largest
allowable distance a.
30 mm
B
A
20 mm
20 mm
C
30 mm
5 kN
SOLUTION
Reduce the internal forces transmitted across section AB to a force-couple system at the centroid of the
section. The bending couple is
M
P(a
r)
Also, e
r
h
.
r2
ln
r1
R
For the rectangular section, the neutral axis for bending couple only lies at
R
The maximum compressive stress occurs at point A . It is given by
P
A
A
K
Thus, K
Data:
h
30 mm, r1
e
35
a
r,
50 mm, r
2.2593 mm, b
6
175 10 Pa, P
6
a
r
r
(30.5)(2.2593)(20)
12.7407
a
108.17 mm
P
A
35 mm, R
5 kN
P (a
P
with y A R
A
(a r )( R r1)
er1
30 mm, R
r1
30
ln 50
20
r ) yA
Aer1
r1
(1)
32.7407 mm
12.7407 mm, a
?
3
5 10 N
6
(900 10 )( 175 10 )
5 103
A A
P
K
a
32.7407
175 MPa
A
Solving (1) for
20 mm, r2
1
My A
Aer1
31.5
( K 1)er1
R r1
108.17 mm
35 mm
a
73.2 mm
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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628
PROBLEM 4.170
2500 N
For the split ring shown, determine the stress at (a) point A,
(b) point B.
90 mm
40 mm
B
A
14 mm
SOLUTION
r1
1
40
2
20 mm,
r2
A (14)(25) 350 mm 2
1
(90) 45 mm
2
25
h
R
r2
45
ln r ln 20
h
r2
r1
25 mm
30.8288 mm
1
r
1
(r1
2
r2 ) 32.5 mm
e
r
R 1.6712 mm
Reduce the internal forces transmitted across section AB to a force-couple system at the centroid of the cross
section. The bending couple is
M
(a)
Pa
rA
Point A:
A
P
A
(2500)(32.5 10 3 ) 81.25 N m
Pr
20 mm
My A
AeR
yA
2500
350 10
30.8288 20 10.8288 mm
(81.25)(10.8288 10 3 )
(350 10 6 )(1.6712 10 3 )(20 10 3 )
6
82.4 106 Pa
(b)
Point B:
B
P
A
rB
45 mm
MyB
AerB
82.4 MPa
A
yB
2500
350 10
30.8288 45
6
14.1712 mm
(81.25)( 14.1712 10 3 )
(350 10 6 )(1.6712 10 3 )(45 10 3 )
36.6 106 Pa
B
36.6 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
629
2 in.
PROBLEM 4.171
B
0.5 in.
0.5 in.
2 in.
0.5 in.
A
3 in.
M
M'
3 in.
Three plates are welded together to form the curved
beam shown. For M 8 kip in., determine the stress
at (a) point A, (b) point B, (c) the centroid of the cross
section.
C
SOLUTION
A
dA
R
1
r
Part
b
3.5
5.5
6
(a)
yA
R
yB
R
yC
C
R
3.25
4.875
1.0
0.225993
4.5
4.5
1.0
0.174023
5.75
5.75
3.5
0.862468
0.5
1.5
0.5
2
2
0.5
b ln
3.5
4.05812 in.,
0.862468
r R 0.26331 in.
3
r
M
15.125
15.125
4.32143 in.
3.5
8 kip in.
1.05812 in.
( 8)(1.05812)
(3.5)(0.26331)(3)
4.05812
MyB
Aer2
B
(c)
r2
0.462452
3
4.05812
My A
Aer1
A
(b)
r1
Ar
A
e
ri 1
ri
r
h
R
A
r
bi ln i 1
ri
Ari
A
r
r
3
bi hi
r
bi ln i 1
ri
r
MyC
Aer
6
A
3.06 ksi
1.94188 in.
( 8)( 1.94188)
(3.5)(0.26331)(6)
B
2.81 ksi
C
0.529 ksi
e
Me
Aer
M
Ar
8
(3.5)(4.32143)
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
630
2 in.
PROBLEM 4.172
B
0.5 in.
0.5 in.
2 in.
0.5 in.
A
3 in.
M
M'
3 in.
Three plates are welded together to form the curved beam
shown. For the given loading, determine the distance e
between the neutral axis and the centroid of the cross
section.
C
SOLUTION
A
dA
R
b
Ar
0.462452
3.25
4.875
1.0
0.225993
4.5
4.5
1.0
0.174023
5.75
5.75
3.5
0.862468
A
3
0.5
1.5
5.5
0.5
2
6
2
0.5
3.5
R
e
ri 1
ri
r
h
r
3
A
r
bi ln i 1
ri
Ari
A
r
Part
bi hi
r
bi ln i 1
ri
1
r
b ln
3.5
4.05812 in., r
0.862468
r R 0.26331 in.
15.125
3.5
15.125
4.32143 in.
e
0.263 in.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
631
PROBLEM 4.173
C
M
A
Mʹ
150 mm
A
Knowing that the maximum allowable stress is 45 MPa, determine
the magnitude of the largest moment M that can be applied to the
components shown.
45 mm
135 mm
B
B
36 mm
SOLUTION
A
dA
R
1
r
bi hi
r
bi 2 ln i 1
ri
Ai
r
bi ln i 1
ri
Ai ri
Ai
r
r, mm
150
Part bi , mm h, mm
195
330
R
e
yA
r1
ri 1
, mm
ri
r , mm
Ar , mm3
108
45
4860
28.3353
172.5
838.35 103
36
135
4860
18.9394
262.5
1275.75 103
9720
47.2747
r
2114.1 103
9720
9720
205.606 mm
47.2747
r R 11.894 mm
R
bi ln
A, mm 2
205.606
150
55.606 mm
A Aer1
M
yA
(45 106 )(9720 10 6 )(11.894 10 3 )(150 10 3 )
(55.606 10 3 )
B
M
217.5 mm
My A
Aer1
A
yB
2114.1 103
R
r2
205.606
330
14.03 kN m
124.394 mm
MyB
Aer2
B Aer2
yB
(45 106 )(9720 10 6 )(11.894 10 3 )(330 10 3 )
(124.394 10 3 )
13.80 kN m
M
13.80 kN m
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
632
C
M
A
PROBLEM 4.174
Mʹ
150 mm
A
B
B
Knowing that the maximum allowable stress is 45 MPa, determine
the magnitude of the largest moment M that can be applied to the
components shown.
135 mm
45 mm
36 mm
SOLUTION
A
dA
R
1
r
bi hi
r
bi ln i 1
ri
Ai
r
bi ln i 1
ri
Ai ri
Ai
r
bi , mm h, mm
r, mm
150
285
bi ln
A, mm 2
ri 1
, mm
ri2
135
4860
23.1067
217.5
1.05705 106
108
45
4860
15.8332
307.5
1.49445 106
9720
38.9399
R
9720
38.9399
e
r
R
12.885 mm,
yA
R
r1
249.615
249.615 mm,
M
150
2.5515 106
9720
r
262.5 mm
20 103 N m
99.615 mm
A Aer1
M
yA
(45 106 )(9720 10 6 )(12.885 10 3 )(150 10 3 )
(99.615 10 3 )
M
2.5515 106
My A
Aer1
A
B
Ar , mm3
36
330
yB
r , mm
R
r2
249.615
330
8.49 kN m
80.385 mm
MyB
Aer2
B Aer2
yB
(45 106 )(9720 10 6 )(12.885 10 3 )(330 10 3 )
(80.385 10 3 )
23.1 kN m
M
8.49 kN m
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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633
120 lb
PROBLEM 4.175
120 lb
The split ring shown has an inner radius r1 0.8 in. and a circular cross
section of diameter d 0.6 in. Knowing that each of the 120-lb forces is
applied at the centroid of the cross section, determine the stress (a) at point A,
(b) at point B.
r1
d
A
B
SOLUTION
rA
r1
0.8 in.
rB
rA
d
r
1
(rA
2
A
c2
e
1
r
2
r R
Q
120 lb
R
M0
(a)
0.8
rB )
0.6
1.4 in.
1.1 in.
(0.3)2
c
1
d
2
0.28274 in 2
0.3 in.
for solid circular section
1
1.1
(1.1)2 (0.3)2
2
1.1 1.0791503 0.020850 in.
r2
c2
Fx
0:
2rQ
r
rA
0.8 in.
A
P
A
M (rA R)
AerA
M
0: P
0
120
0.28274
Q
M
1.079150 in.
0
P
120 lb
2rQ
(2)(1.1)(120)
( 264)(0.8 1.079150)
(0.28274)(0.020850)(0.8)
264 lb in.
16.05 103 psi
A
(b)
r
rB
1.4 in.
B
P
A
M (rB R)
AerB
120
0.28274
16.05 ksi
( 264)(1.4 1.079150)
(0.28274)(0.020850)(1.4)
9.84 103 psi
B
9.84 ksi
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634
120 lb
PROBLEM 4.176
120 lb
Solve Prob. 4.175, assuming that the ring has an inner radius r1 = 0.6 in. and
a cross-sectional diameter d 0.8 in.
r1
PROBLEM 4.175 The split ring shown has an inner radius r1 0.8 in. and a
circular cross section of diameter d 0.6 in. Knowing that each of the 120-lb
forces is applied at the centroid of the cross section, determine the stress
(a) at point A, (b) at point B.
d
A
B
SOLUTION
rA
r1
0.6 in.
rB
rA
d
r
1
(rA
2
A
c2
e
1
r
2
r R
Q
120 lb
R
0:
M0
(a)
0.6
rB )
0.8
1.0 in.
(0.4)2
r2
1.0
1.4 in.
c
0.50265 in 2
0.4 in.
for solid circular section.
1
1.0
(1.0) 2 (0.4)2
2
0.958258 0.041742 in.
c2
Fx
2rQ
1
d
2
M
r
rA
0.6 in.
A
P
A
M (rA R)
AerA
0
0
P
M
120
0.50265
Q
2rQ
0
0.958258 in.
P
120 lb
(2)(1.0)(120)
240 lb in.
( 240)(0.6 0.958258)
(0.50265)(0.041742)(0.6)
7.069 103 psi
(b)
r
rB
1.4 in.
B
P
A
M (rB R)
AerB
A
120
0.50265
7.07 ksi
( 240)(1.4 0.958258)
(0.50265)(0.041742)(1.4)
3.37 103 psi
B
3.37 ksi
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635
220 N
B
PROBLEM 4.177
The bar shown has a circular cross section of 14-mm diameter. Knowing
that a 32 mm, determine the stress at (a) point A, (b) point B.
A C
220 N
a
16 mm
12 mm
SOLUTION
c
R
e
(a)
1
d 8 mm
r 12 8 20 mm
2
1
1
r
r 2 c2
20
202 82
2
2
r R 20 19.1652 0.83485 mm
A
r2
(8) 2
P
220 N
M
P (a
201.06 mm 2
r)
220(0.032
yA
R
r1
19.1652
12
yb
R
r2
19.1652
28
A
P
A
My A
Aer1
220
201.06 10
19.1652 mm
6
0.020)
11.44 N m
7.1652 mm
8.8348 mm
( 11.44)(7.1652 10 3 )
(201.06 10 6 )(0.83485 10 3 )(12 10 3 )
A
(b)
B
P
A
41.8 MPa
MyB
Aer2
220
201.06 10
6
( 11.44)(8.8348 10 3 )
(201.06 10 6 )(0.83485 10 3 )(28 10 3 )
B
20.4 MPa
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636
220 N
B
PROBLEM 4.178
The bar shown has a circular cross section of 14-mm diameter. Knowing
that the allowable stress is 38 MPa, determine the largest permissible
distance a from the line of action of the 220-N forces to the plane
containing the center of curvature of the bar.
A C
220 N
a
16 mm
12 mm
SOLUTION
c
R
e
1
d 8 mm
r 12 8 20 mm
2
1
1
r
r 2 c2
20
202 82
2
2
r R 20 19.1652 0.83485 mm
A
r2
(8) 2
P
220 N
M
P (a
R
r1
A
P
A
My A
Aer1
K
a
r
201.06 mm 2
r)
yA
KP
A
19.1652 mm
19.1652
P
A
where
12
P(a
K
7.1652 mm
r ) yA
Aer1
P
1
A
(a
r ) yA
er1
1
(38 106 )(201.06 10 6 )
P
220
( K 1)er1
yA
AA
(34.729
(a
r ) yA
er1
34.729
1)(0.83485 10 3 )(12 10 3 )
(7.1652 10 3 )
0.047158 m
a
0.047158
0.020
0.027158
a
27.2 mm
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637
C
M
PROBLEM 4.179
16 mm
The curved bar shown has a circular cross section of 32-mm
diameter. Determine the largest couple M that can be applied to
the bar about a horizontal axis if the maximum stress is not to
exceed 60 MPa.
12 mm
SOLUTION
c
R
e
max
16 mm
r
1
r
2
1
28
2
r2
r
28
16
28 mm
c2
282
R
12
162
25.4891 mm
25.4891
2.5109 mm
occurs at A, which lies at the inner radius.
It is given by
Also,
A
Data:
yA
M
My A
from which
Aer1
max
c2
R
(16) 2
r1
M
Aer1
yA
max
.
804.25 mm 2
25.4891 12
13.4891 mm
(804.25 10 6 )(2.5109 10 3 )(12 10 3 )(60 106 )
13.4891 10 3
M
107.8 N m
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638
P
90 mm
PROBLEM 4.180
Knowing that P
B
10 kN, determine the stress at (a) point A, (b) point B.
80 mm
A
100 mm
SOLUTION
Locate the centroid D of the cross section.
r
90 mm
3
100 mm
130 mm
Force-couple system at D.
P
M
10 kN
Pr
1300 N m
(10 kN)(130 mm)
Triangular cross section.
A
1
bh
2
1
(90 mm)(80 mm)
2
3600 mm 2
R
(a)
A
Point A:
P
A
M (rA R)
AerA
2.778 MPa
(b)
B
Point B:
P
A
rA
1
R
126.752 mm
e
r
R
1
(90)
2
190 190
ln
90 100
130 mm
1
45 mm
0.355025
126.752 mm
3.248 mm
0.100 m
10 kN
3600 10 6 m 2
(1300 N m)(0.100 m 0.126752 m)
(3600 10 6 m 2 )(3248 10 3 m)(0.100 m)
29.743 MPa
rB
M (rB R)
AerB
2.778 MPa
100 mm
1
h
2
r2 r2
ln
h r1
3600 10 6 m 2
190 mm
A
32.5 MPa
B
34.2 MPa
0.190 m
10 kN
3600 10 6 m 2
(1300 N m)(0.190 m 0.126752 m)
(3600 10 6 m 2 )(3.248 10 3 m)(0.190 m)
37.01 MPa
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639
M
2.5 in.
A
PROBLEM 4.181
B
M
Knowing that M
(b) point B.
C
3 in.
2 in.
2 in.
5 kip in., determine the stress at (a) point A,
3 in.
SOLUTION
r
1
bh
2
2 1
b1
2.5 in., r1
A
Use formula for trapezoid with
b2
1
(2.5)(3) 3.75 in 2
2
3.00000 in.
1 h 2 (b b )
1
2
2
r2
b2r1) ln
h(b1
r1
(b1r2
(a)
yA
R
yB
B
R
5 in.
b2 )
(0.5)(3)2 (2.5
[(2.5)(5) (0)(2)] ln 52
0)
(3)(2.5
r
M
R
0.15452 in.
0)
2.84548 in.
5 kip in.
0.84548 in.
My A
Aer1
A
(b)
r1
0, r2
0.
R
e
2 in., b2
r2
MyB
Aer2
(5)(0.84548)
(3.75)(0.15452)(2)
3.65 ksi
A
2.15452 in.
(5)( 2.15452)
(3.75)(0.15452)(5)
B
3.72 ksi
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640
M
PROBLEM 4.182
B
2.5 in.
A
3 in.
Knowing that M
(b) point B.
M
5 kip in., determine the stress at (a) point A,
C
2 in.
2 in.
3 in.
SOLUTION
r
1 (2.5)(3) 3.75 in 2
2
2 2 4.00000 in.
b1
0, r1
b1
0.
A
Use formula for trapezoid with
2 in.,
b2
1 h 2 (b b )
1
2
2
r2
b2r1) ln
h (b1
r1
R
(b1r2
(0.5)(3) 2 (0
[(0)(5)
e
(a)
yA
R
yB
B
R
R
(2.5)(2)] ln 5
2
0.14534 in.
r2
5 in.
b2 )
2.5)
(3)(0
M
2.5)
3.85466 in.
5 kip in.
1.85466 in.
My A
Aer1
A
(b)
r1
r
2.5 in.,
r2
MyB
Aer2
(5)(1.85466)
(3.75)(0.14534)(2)
A
8.51 ksi
1.14534 in.
(5)( 1.14534)
(3.75)(0.14534)(5)
B
2.10 ksi
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641
PROBLEM 4.183
80 kip · in.
b
B
A
B
A
Knowing that the machine component shown has a trapezoidal cross
section with a 3.5 in. and b 2.5 in., determine the stress at
(a) point A, (b) point B.
C
a
6 in. 4 in.
SOLUTION
Locate centroid.
A, in 2
r , in.
Ar , in 3
10.5
6
63
7.5
8
60
18
r
R
123
123
18
6.8333 in.
1
2
(b1r2
h 2 (b1
b2 r1 ) ln
b2 )
r2
r1
h(b1 b2 )
(0.5)(6)2 (3.5 2.5)
[(3.5)(10) (2.5)(4)]ln 104 (6)(3.5 2.5)
e
(a)
yA
A
(b)
yB
B
r
R
0.4452 in.
M
6.3878 in.
80 kip in.
R r1 6.3878 4 2.3878 in.
MyA
(80)(2.3878)
(18)(0.4452)(4)
Aer1
5.96 ksi
A
R r2 6.3878 10
3.6122 in.
MyB
(80)( 3.6122)
(18)(0.4452)(10)
Aer2
B
3.61 ksi
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642
PROBLEM 4.184
80 kip · in.
b
B
A
B
A
Knowing that the machine component shown has a trapezoidal cross
section with a 2.5 in. and b 3.5 in., determine the stress at
(a) point A, (b) point B.
C
a
6 in. 4 in.
SOLUTION
Locate centroid.
A, in 2
r , in.
Ar , in 3
7.5
6
45
10.5
8
84
18
r
R
e
(a)
yA
A
(b)
yB
B
R r1
129
18
7.1667 in.
1
2
(b1r2
h 2 (b1 b2 )
b2 r1 ) ln
r2
r1
h(b1 b2 )
(0.5)(6) 2 (2.5 3.5)
[(2.5)(10) (3.5)(4)]ln 104 (6)(2.5 3.5)
6.7168 in.
r
80 kip in.
R
M
0.4499 in.
2.7168 in.
My A
Aer1
R r2
129
(80)(2.7168)
(18)(0.4499)(4)
A
6.71 ksi
3.2832 in.
MyB
Aer2
(80)( 3.2832)
(18)(0.4499)(10)
B
3.24 ksi
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643
a
B
PROBLEM 4.185
20 mm
B
A
A
250 N · m
250 N · m
For the curved beam and loading shown, determine the stress at
(a) point A, (b) point B.
30 mm
a
35 mm
40 mm
Section a–a
SOLUTION
Locate centroid.
A, mm 2
r , mm
Ar , mm3
600
45
27 103
300
55
16.5 103
43.5 103
900
r
R
43.5 103
900
1
2
(b1r2
48.333 mm
h 2 (b1
b2r1) ln
r2
r1
b2 )
h(b2
b1)
(0.5)(30) 2 (40 20)
65
[(40)(65) (20)(35)]ln 35
(30)(40
e
(a)
yA
R
yB
B
R
R
1.4725 mm
M
46.8608 mm
250 N m
11.8608 mm
My A
Aer1
A
(b)
r1
r
20)
r2
MyB
Aer2
( 250)(11.8608 10 3 )
(900 10 6 )(1.4725 10 3 )(35 10 3 )
63.9 106 Pa
A
63.9 MPa
18.1392 mm
( 250)( 18.1392 10 3 )
(900 10 6 )(1.4725 10 3 )(65 10 3 )
52.6 106 Pa
B
52.6 MPa
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644
PROBLEM 4.186
35 mm
25 mm
For the crane hook shown, determine the largest tensile stress in
section a-a.
60 mm
40 mm
a
a
60 mm
Section a–a
15 kN
SOLUTION
Locate centroid.
A, mm 2
r , mm
Ar , mm3
1050
60
63 103
750
80
60 103
103 103
1800
r
103 103
1800
63.333 mm
P 15 103 N
Force-couple system at centroid:
M
(15 103 )(68.333 10 3 )
Pr
1
2
R
(b1r2
1.025 103 N m
h 2 (b1 b2 )
b2 r1 ) ln
r2
r1
h(b1 b2 )
(0.5)(60)2 (35 25)
[(35)(100) (25)(40)]ln 100
(60)(35 25)
40
e
r
R
63.878 mm
4.452 mm
Maximum tensile stress occurs at point A.
yA
A
R r1
P
A
23.878 mm
My A
Aer1
15 103
1800 10
6
(1.025 103 )(23.878 10 3 )
(1800 10 6 )(4.452 10 3 )(40 10 3 )
84.7 106 Pa
A
84.7 MPa
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645
PROBLEM 4.187
Using Eq. (4.66), derive the expression for R given in Fig. 4.61 for a circular cross section.
SOLUTION
Use polar coordinate
as shown. Let w be the width as a function of
w 2c sin
r r c cos
dr
c sin
d
dA
w dr
2c 2 sin 2
d
2
dA
r
2c sin
r c cos
0
dA
r
c 2 (1 cos 2 )
d
r c cos
0
r2
2
c 2 cos 2
(r 2
r c cos
0
2
d
0
(r
2r
2(r
2(r 2
c cos ) d
2c sin
0
2
c )
r2
c2
tan
1
0) 2c(0 0) 4 r 2
2r (
2 r
r2
2
d
c2 )
0
r
dr
c cos
0
2
2
c2 )
r2
c2
c 2 tan 1
2
r c
2
0
0
c2
c2
A
c2
A
dA
r
R
1
2
2 r
c2 r
r
2
(r
r2
2
r2
c2
2
2
c )
c2
1
2
c2
1
2r
c2 r
r2
r2
c
2
c2
c2
r
r2
c2
r
r2
c2
1
r
2
r2
c2
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646
PROBLEM 4.188
Using Eq. (4.66), derive the expression for R given in Fig. 4.73 for a trapezoidal cross section.
SOLUTION
The section width w varies linearly with r.
w
w
b1
b2
b2
b1
c1
r2b1
r1b2
c0
dA
r
c0 c1r
b1 at r r1 and w b2 at r
c0 c1r1
c0 c1r2
c1 (r1 r2 )
c1h
b1 b2
h
(r2 r1 )c0 hc0
r2 b1 r1b2
h
r2 w
r2 c
c1r
0
dr
dr
r1 r
r1
r
r2
r2
c0 ln r
c1 r
r1
r2
r1
r
c0 ln 2 c1 (r2 r1 )
r1
r2 b1 r1b2 r2 b1 b2
h
ln
h
r1
h
r2 b1 r1b2 r2
ln
(b1 b2 )
h
r1
1
(b1 b2 ) h
A
2
1 2
h (b1 b2 )
A
2
R
r
dA
( r2b1 r1b2 ) ln r2 h(b1 b2 )
r
1
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647
PROBLEM 4.189
Using Equation (4.66), derive the expression for R given in Fig. 4.73 for a triangular cross section.
SOLUTION
The section width w varies linearly with r.
w
w
b
0
b
c1
dA
r
c0 c1r
b at r r1 and w 0 at r r2
c0 c1r1
c0 c1r2
c1 (r1 r2 )
c1h
br2
b
and c0
c1r2
h
h
r2 w
r2 c
c1r
0
dr
dr
r1 r
r1
r
r2
r2
c0 ln r
c1 r
r1
r1
r
c0 ln 2 c1 (r2 r1 )
r1
br2 r2 b
h
ln
h
r1 h
A
R
br2 r2
ln
h
r1
1
bh
2
A
dA
r
b
b b
1
2
r2
h
r2 r2
ln
h r1
bh
ln
r2
r1
1
r2
h
1
1
2
h
ln
r2
r1
1
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648
PROBLEM 4.190
b2
b3
Show that if the cross section of a curved beam consists of two or more
rectangles, the radius R of the neutral surface can be expressed as
b1
A
R
r1
r2
r1
ln
r2
b1
r3
r2
b2
r4
r3
b3
r3
r4
where A is the total area of the cross section.
SOLUTION
A
1 dA
r
R
A
ri 1
ri
bi ln
A
ln
Note that for each rectangle,
1
dA
r
ri
1
ri
bi
ri
r2
A
ri 1
ri
bi
1
bi
r2
r1
ln
b1
r3
r2
b2
r4
r3
b3
dr
r
dr
r
bi ln
ri
1
ri
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649
PROBLEM 4.191
C
"
2
"
2
r1
!x
!x
For a curved bar of rectangular cross section subjected to a
bending couple M, show that the radial stress at the neutral
surface is
r
M
R
1 1 ln
r
Ae
R
r1
R
b
and compute the value of r for the curved bar of Concept
Applications 4.10 and 4.11. (Hint: consider the free-body
diagram of the portion of the beam located above the neutral
surface.)
!r
!r
SOLUTION
M (r R ) M
Aer
Ae
For portion above the neutral axis, the resultant force is
At radial distance r,
r
H
R
r dA
r b dr
r1
Mb R
MRb R dr
dr
Ae r1
Ae r1 r
Mb
MRb R
( R r1 )
ln
Ae
Ae
r1
Resultant of
n:
Fr
/2
r
/2
r
For equilibrium: Fr
r
cos
/2
2 H sin
2 r bR sin
2
/2
2
r
MbR
1 1
Ae
R
ln
R
r1
dA
cos (bR d )
bR sin
MR
Aer
/2
r bR
r
/2
bR sin
cos d
2
0
2
2
r
MbR
1 1
Ae
R
ln
R
sin
r1
2
r
r
M
1 1
Ae
R
ln
R
r1
0
Using results of Examples 4.10 and 4.11 as data,
M
r
8 kip in.,
A
3.75 in 2 , R
8
1
(3.75)(0.0314)
5.25
5.9686
5.9686 in., e
ln
5.9686
5.25
0.0314 in., r1
5.25 in.
r
0.536 ksi
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650
PROBLEM 4.192
25 mm
25 mm
4 kN
A
Two vertical forces are applied to a beam of the cross section shown. Determine
the maximum tensile and compressive stresses in portion BC of the beam.
4 kN
B
C
300 mm
300 mm
SOLUTION
A1
2
A2
A2 y2
A2
r2
bh
2
(25)2
981.7 mm 2
1250 mm 2
(50)(25)
y
A1 y1
A1
(981.7)(10.610) (1250)( 12.5)
981.7 1250
I1
I x1
d1
y1
r 4 A1 y12
(25)4 (981.7)(10.610) 2
8
8
y 10.610 ( 2.334) 12.944 mm
I1
I1
A1d12
(981.7)(12.944)2
d2
1 3 1
bh
(50)(25)3 65.104 103 mm 4
12
12
y2 y
12.5 ( 2.334) 10.166 mm
I2
I2
A2 d 22
I
I1
I2
I2
ytop
ybot
65.104 103
(1250)(10.166) 2
401.16 103 mm 4
401.16 10
25 2.334
25 2.334
27.334 mm
bot
Mytop
I
Mybot
I
(4)(25)
10.610 mm
3
25
12.5 mm
2
42.886 106 mm 4
207.35 103 mm 4
194.288 103 mm 4
m4
0.027334 m
22.666 mm
0.022666 m
M
top
9
y2
4r
3
h
2
2.334 mm
A1 y12
42.866 103
y1
Pa
0:
M
(1200)(0.027334)
401.16 10 9
81.76 106 Pa
(1200)( 0.022666)
401.16 10 9
67.80 106 Pa
Pa
(4 103 )(300 10 3 )
1200 N m
top
bot
81.8 MPa
67.8 MPa
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651
PROBLEM 4.193
A steel band saw blade that was originally straight passes over 8-in.-diameter pulleys
when mounted on a band saw. Determine the maximum stress in the blade, knowing that
it is 0.018 in. thick and 0.625 in. wide. Use E 29 106 psi.
0.018 in.
SOLUTION
Band blade thickness:
t
0.018 in.
Radius of pulley:
r
1
d
2
4.000 in.
Radius of curvature of centerline of blade:
r
c
1
t
2
1
t
2
4.009 in.
0.009 in.
c
0.009
4.009
Maximum strain:
m
Maximum stress:
m
E
m
65.1 103 psi
m
0.002245
(29 106 )(0.002245)
m
65.1 ksi
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652
PROBLEM 4.194
M
A couple of magnitude M is applied to a square bar of side a. For
each of the orientations shown, determine the maximum stress
and the curvature of the bar.
M
a
(a)
(b)
SOLUTION
1 3
bh
12
a
2
I
c
1
a4
12
Ma
2
a4
12
Mc
I
max
1 3
aa
12
M
EI
max
1
M
4
Ea
12
6M
a3
12M
Ea 4
For one triangle, the moment of inertia about its base is
1 3
bh
12
I1
1
12
2a
a
3
2
a4
24
4
c
I2
I1
a
24
I
I1
I2
a4
12
a
2
max
1
Mc
I
M
EI
Ma/ 2
a 4 /12
6 2M
a3
8.49M
a3
max
1
M
4
Ea
12
12M
Ea 4
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653
40 mm
60 mm
PROBLEM 4.195
Determine the plastic moment M p of a steel beam of the cross section shown, assuming
the steel to be elastoplastic with a yield strength of 240 MPa.
SOLUTION
Let c1 be the outer radius and c2 the inner radius.
A1 y1
Aa ya
2
4c1
3
c12
2 3
c1
3
c2
A1 y1
2 3
c1
3
A2 y2
Y ( A1 y1
Mp
Data:
Ab yb
2
3
c 22
A2 y2 )
4
3
Y
c13
c32
240 106 Pa
Y
240 MPa
c1
60 mm
0.060 m
c2
40 mm
0.040 m
Mp
4c2
3
c22
4
(240 106 )(0.0603
3
48.64 103 N m
0.0403 )
Mp
48.6 kN m
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654
PROBLEM 4.196
M # 300 N · m
In order to increase corrosion resistance, a 2-mm-thick cladding of
aluminum has been added to a steel bar as shown. The modulus
of elasticity is 200 GPa for steel and 70 GPa for aluminum. For a
bending moment of 300 N m, determine (a) the maximum stress
in the steel, (b) the maximum stress in the aluminum, (c) the radius
of curvature of the bar.
26 mm
30 mm
46 mm
50 mm
SOLUTION
Use aluminum as the reference material.
n
1 in aluminum
Es
Ea
n
200
70
2.857 in steel
Cross section geometry:
Steel: As
(46 mm)(26 mm)
Aluminum: Aa
1196 mm 2
1196 mm 2
(50 mm)(30 mm)
1
(50 mm)(30 mm3 )
12
Ia
1
(46 mm)(26 mm)3
12
Is
67,375 mm 4
304 mm 2
67,375 mm 4
45,125 mm 4
Transformed section.
I
na I a
ns I s
(1)(45,125)
M
300 N m
Bending moment.
(a)
ns Mys
I
s
(b)
(c)
(2.857)(300)(0.013)
237.615 10 9
Maximum stress in aluminum:
a
2.857
ns
Maximum stress in steel:
na Mya
I
(1)(300)(0.015)
237.615 10 9
Radius of curvative:
EI
M
237, 615 mm 4
(2.857)(67,375)
na
ys
13 mm
237.615 10 9 m 4
0.013 m
46.9 106 Pa
1,
ya
s
15 mm
46.9 MPa
0.015 m
18.94 106 Pa
(70 109 )(237.615 10 9 )
300
a
18.94 MPa
55.4 m
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655
PROBLEM 4.197
The vertical portion of the press shown consists of a rectangular
tube of wall thickness t 10 mm. Knowing that the press has
been tightened on wooden planks being glued together until P
20 kN, determine the stress at (a) point A, (b) point B.
t
P
a
P'
a
t
A
200 mm
80 mm
60 mm
80 mm
B
Section a–a
SOLUTION
Rectangular cutout is 60 mm
40 mm.
A
(80)(60)
I
1
(60)(80)3
12
2.4 103 mm 2
(60)(40)
1
(40)(60)3
12
2.4 10 3 m 2
1.84 106 mm 4
1.84 10 6 m 4
c
P
M
(a)
(b)
40 mm
0.040 m e
200
40
240 mm
0.240 m
3
20 10 N
Pe
(20 103 )(0.240)
4.8 103 N m
A
P
A
Mc
I
20 103
2.4 10 3
(4.8 103 )(0.040)
1.84 10 6
112.7 106 Pa
A
112.7 MPa
B
P
A
Mc
I
20 103
2.4 10 3
(4.8 103 )(0.040)
1.84 10 6
96.0 106 Pa
B
96.0 MPa
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656
PROBLEM 4.198
P
P y
P
P
B
C
D
a
A
x
z
The four forces shown are applied to a rigid plate supported by a solid steel
post of radius a. Knowing that P 24 kips and a 1.6 in., determine the
maximum stress in the post when (a) the force at D is removed, (b) the forces
at C and D are removed.
SOLUTION
For a solid circular section of radius a,
Centric force:
(a)
(b)
A
a2
I
F
4P,
Mx
F
3P,
Mx
F
A
M xz
I
2P,
Mx
M
M x2
4
a4
F
A
4P
a2
( Pa)( a)
a2
4
7P
a2
0
Mz
Force at D is removed.
Pa,
Mz
3P
a2
0
Forces at C and D are removed.
F
Resultant bending couple:
F
A
2P
a2
P
Numerical data:
Answers:
Mc
I
Pa,
Mz
M z2
2 Paa
a2
4
2 Pa
2
24.0 kips,
Pa
4 2 P
a2
a
2.437 P/a 2
1.6 in.
(a)
(7)(24.0)
(1.6)2
20.9 ksi
(b)
(2.437)(24.0)
(1.6)2
22.8 ksi
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657
a
r # 20 mm
PROBLEM 4.199
P # 3 kN
The curved portion of the bar shown has an inner radius of 20 mm.
Knowing that the allowable stress in the bar is 150 MPa, determine the
largest permissible distance a from the line of action of the 3-kN force to
the vertical plane containing the center of curvature of the bar.
25 mm
25 mm
SOLUTION
Reduce the internal forces transmitted across section AB to a force-couple system at the centroid of the
section. The bending couple is
M
P(a
r)
For the rectangular section, the neutral axis for bending couple only lies at
R
Also, e
r
h
r2
r1
ln
R
The maximum compressive stress occurs at point A. It is given by
P
A
A
My A
Aer1
P
A
with
yA
R
Thus,
K
1
Data:
25 mm, r1
R
25
45
ln 20
b
25 mm,
K
a
r
a
r ) yA
Aer1
r )( R
er1
P
A
r1)
20 mm, r2
30.8288 mm, e
A
3
K
r1
(a
h
P
P (a
3 10 N m,
bh
45 mm, r
32.5
(25)(25)
A
30.8288
625 mm 2
32.5 mm
1.6712 mm
625 10 6 m 2
R
r1
10.8288 mm
6
150 10 Pa
( 150 106 )(625 10 6 )
31.25
P
3 103
( K 1)er1 (30.25)(1.6712)(20)
93.37 mm
R r1
10.8288
AA
93.37
a
32.5
60.9 mm
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658
Dimensions in inches
400 lb
400 lb
PROBLEM 4.200
400 lb
2.5
Determine the maximum stress in each of the two machine
elements shown.
400 lb
2.5
3
0.5
3
r 5 0.3
r 5 0.3
1.5
0.5
0.5
1.5
0.5
SOLUTION
For each case, M
(400)(2.5)
1000 lb in.
At the minimum section,
1
(0.5)(1.5)3
12
0.75 in.
I
c
(a)
D/d
r/d
3/1.5
0.140625 in 4
2
0.3/1.5
0.2
K
From Fig 4.32,
max
(b)
D/d
KMc
I
3/1.5
(1.75)(1000)(0.75)
0.140625
2
From Fig. 4.31,
max
1.75
KMc
I
r /d
0.3/1.5
K
9.33 103 psi
max
9.33 ksi
max
8.00 ksi
0.2
1.50
(1.50)(1000)(0.75)
0.140625
8.00 103 psi
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659
PROBLEM 4.201
120 mm
10 mm
M
120 mm
10 mm
Three 120 10-mm steel plates have been welded together to form the
beam shown. Assuming that the steel is elastoplastic with
E 200 GPa and Y 300 MPa, determine (a) the bending moment
for which the plastic zones at the top and bottom of the beam are
40 mm thick, (b) the corresponding radius of curvature of the beam.
10 mm
SOLUTION
A1
R1
A2
R2
A3
R3
y1
(a)
M
2( R1 y1
(120)(10) 1200 mm 2
Y
(300 106 )(1200 10 6 ) 360 103 N
A1
(30)(10) 300 mm 2
Y
(300 106 )(300 10 6 ) 90 103 N
A2
(30)(10)
1
Y A2
2
65 mm
R2 y2
300 mm 2
1
(300 106 )(300 10 6 ) 45 103 N
2
65 10 3 m y2 45 mm 45 10 3 m
R3 y3 )
yY
Y
E
EyY
Y
20 mm
20 10 3 m
2{(360)(65) (90)(45) (45)(20)}
56.7 103 N m
(b)
y3
M
(200 109 )(30 10 3 )
300 106
56.7 kN m
20.0 mm
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660
P
4.5 in.
PROBLEM 4.202
Q
4.5 in.
A short length of a W8 31 rolled-steel shape supports a rigid plate on which two
loads P and Q are applied as shown. The strains at two points A and B on the
centerline of the outer faces of the flanges have been measured and found to be
A
B
550 10
A
6
in./in.
B
680 10
6
in./in.
29 106 psi, determine the magnitude of each load.
Knowing that E
SOLUTION
Strains:
A
C
550 10
1
(
2
6
B
1
( 550
2
B)
A
680)10
6
615 10
6
E
A
(29 106 psi)( 550 10
6
in./in.)
15.95 ksi
C
E
C
(29 106 psi)( 615 10
6
in./in.)
17.835 ksi
W8
M
31:
(4.5 in.)( P
P
C
Q
A
;
A
9.12 in 2
S
27.5 in 3
680 10
P
A
15.95 ksi
Solve simultaneously.
Q
A
in./in.
Q)
17.835 ksi
P Q
9.12 in 2
P
At point A:
6
in./in.
A
Stresses:
At point C:
in./in.
Q
162.655 kips (1)
M
S
17.835 ksi
P
(4.5 in.)( P Q)
;
27.5 in 3
75.6 kips Q
P
Q
11.5194 kips (2)
87.1 kips
P
75.6 kips
Q
87.1 kips
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661
PROBLEM 4.203
!1
M1
A
M'1
M1
!1
M'1
B
C
!1
!1
D
Two thin strips of the same material and same cross
section are bent by couples of the same magnitude
and glued together. After the two surfaces of
contact have been securely bonded, the couples are
removed. Denoting by 1 the maximum stress and
by 1 the radius of curvature of each strip while
the couples were applied, determine (a) the final
stresses at points A, B, C, and D, (b) the final radius
of curvature.
SOLUTION
Let b
width and t
thickness of one strip.
Loading one strip, M
I1
1
1
1
M1
1 3
bt , c
12
M1c
I
M1
EI1
1
t
2
M1
bt 2
12 M 1
Et 3
After M1 is applied to each of the strips, the stresses are those given in the sketch above. They are
A
1,
1,
B
1,
C
D
1
The total bending couple is 2M1.
After the strips are glued together, this couple is removed.
M
1
b(2t )3
12
2 M 1, I
2 3
bt
3
c
t
The stresses removed are
My
I
A
2M 1 y
2 bt 3
3
3M1
bt 2
1
2
1,
3M1 y
bt 2
B
C
0,
D
3M1
bt 2
1
2
1
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662
PROBLEM 4.203 (Continued)
(a)
Final stresses:
A
1
(
1
2
1)
1
2
A
B
1
C
D
1
(b)
M
EI
2M 1
E 23 bt 3
Final radius:
3M 1
Et 3
1
1
1
2
D
1
1
1
1
2
1
4
3
1
1 1
4
1
1
1
1
1
1 1
4 1
3 1
4 1
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663
PROBLEM 4.C1
Two aluminum strips and a steel strip are to be bonded together to form
a composite member of width b 60 mm and depth h 40 mm. The
modulus of elasticity is 200 GPa for the steel and 75 GPa for the
aluminum. Knowing that M 1500 N m, write a computer program
to calculate the maximum stress in the aluminum and in the steel for
values of a from 0 to 20 mm using 2-mm increments. Using appropriate
smaller increments, determine (a) the largest stress that can occur in the
steel, (b) the corresponding value of a.
Aluminum
a
Steel
h " 40 mm
a
b " 60 mm
SOLUTION
Transformed section:
(all steel)
I
At Point 1:
At Point 2:
Esteel
Ealum
n
1 3
bh
12
h
2
M
alum
steel
1
1
2
(nb b) (h 2a )3
12
2
I
n
M
h
2
a
I
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664
PROBLEM 4.C1 (Continued)
For a
0 to 20 mm using 2-mm intervals: compute: n, I ,
b
Moduli of elasticity:
Steel
60 mm h
alum ,
40 mm M
200 GPa Aluminum
steel .
1500 N m
75 GPa
Program Output
a
mm
I
m 4 /106
Sigma
Aluminum
MPa
Sigma
Steel
MPa
0.000
0.8533
35.156
93.750
2.000
0.7088
42.325
101.580
4.000
0.5931
50.585
107.914
6.000
0.5029
59.650
111.347
8.000
0.4352
68.934
110.294
10.000
0.3867
77.586
103.448
12.000
0.3541
84.714
90.361
14.000
0.3344
89.713
71.770
16.000
0.3243
92.516
49.342
18.000
0.3205
93.594
24.958
20.000
0.3200
93.750
0.000
Find ‘a’ for max. steel stress and the corresponding aluminum stress.
6.600
0.4804
62.447
111.572083
6.610
0.4800
62.494
111.572159
6.620
0.4797
62.540
111.572113
Max. steel stress 111.6 MPa occurs when a
Corresponding aluminum stress
6.61 mm.
62.5 MPa
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665
tf
PROBLEM 4.C2
y
x
tw
d
bf
A beam of the cross section shown, made of a steel that is assumed to be
elastoplastic with a yield strength Y and a modulus of elasticity E, is bent
about the x axis. (a) Denoting by yY the half thickness of the elastic core, write
a computer program to calculate the bending moment M and the radius of
curvature
for values of yY from 12 d to 16 d using decrements equal
1
to 2 t f . Neglect the effect of fillets. (b) Use this program to solve Prob. 4.201.
SOLUTION
Compute moment of inertia I x .
1
bf d 3
12
Ix
Maximum elastic moment:
MY
Y
1
(b f
12
tw )(d
2t f )3
Ix
(d/2)
For yielding in the flanges,
(Consider upper half of cross section.)
d
2
c
Stress at junction of web and flange:
(d/2) t f
A
Y
yY
Resultant forces:
Detail of stress diagram:
a1
1
(c
2
a2
yY
a3
yY
a4
2
(c t f )
3
yY )
1
yY
3
2
[ yY
3
(c t f )
(c t f ) ]
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666
PROBLEM 4.C2 (Continued)
Bending moment.
4
2
M
Rn an
n 1
Radius of curvature.
yY
Y
Y
E
yY E
;
Y
(Consider upper half of cross section.)
For yielding in the web,
a5
a6
a7
1
tf
2
c
1
[ yY
2
2
yY
3
(c t f )]
7
Bending moment.
M
2
Rn an
n 5
Radius of curvature.
yY
Y
yY E
Y
E
Y
Program: Key in expressions for an and Rn for n 1 to 7.
For yY
c to (c tf ) at
tf /2 decrements, compute M
2 Rn an for n 1 to 4 and
yY E
, then print.
Y
For yY
(c tw ) to c /3 at
tf /2 decrements, compute M
2 Rn an for n 5 to 7 and
yY E
, then print.
Y
Input numerical values and run program.
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667
PROBLEM 4.C2 (Continued)
Program Output
For a beam of Problem 4.201,
Depth d
140.00 mm
Thickness of flange t f
I
10.00 mm
Width of flange b f
120.00 mm
Thickness of web tw
10.00 mm
0.000011600 m to the 4th
Yield strength of steel sigmaY
Yield moment M Y
300 MPa
49.71 kip in.
yY (mm)
M (kN m)
(m)
For yielding still in the flange,
70.000
49.71
46.67
65.000
52.59
43.33
60.000
54.00
40.00
For yielding in the web,
60.000
54.00
40.00
55.000
54.58
36.67
50.000
55.10
33.33
45.000
55.58
30.00
40.000
56.00
26.67
35.000
56.38
23.33
30.000
56.70
20.00
25.000
56.97
16.67
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668
PROBLEM 4.C3
y
#
0.4
0.4
A
B
z
C
1.2
0.4
#
M
1.2
D
E
0.8 0.4
1.6
An 8 kip in. couple M is applied to a beam of the cross section shown
in a plane forming an angle
with the vertical. Noting that the
centroid of the cross section is located at C and that the y and z axes
are principal axes, write a computer program to calculate the stress at
A, B, C, and D for values of
from 0 to 180° using 10° increments.
(Given: I y 6.23 in 4 and I z 1.481 in 4 . )
0.4 0.8
Dimensions in inches
SOLUTION
Input coordinates of A, B, C, D.
zA
z (1)
2
yA
y (1) 1.4
zB
z (2)
2
yB
y (2) 1.4
zC
z (3)
1
yC
y (3)
1.4
zD
z (4) 1
yD
y (4)
1.4
My
M sin
Components of M.
Mz
Equation 4.55, Page 305:
Program: For
( n)
M cos
M z y ( n)
Iz
M y z ( n)
Iy
0 to 180 using 10° increments.
For n 1 to 4 using unit increments.
Evaluate Equation 4.55 and print stresses.
Return
Return
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669
PROBLEM 4.C3 (Continued)
Program Output
Moment of couple:
M
8.00 kip in.
Moments of inertia:
Iy
6.23 in 4
Iz
1.481 in 4
Coordinates of Points A, B, D, and E:
Point A: z (1)
Point B: z (2)
2:
y (1) 1.4
2: y (2) 1.4
Point D: z (3)
1:
Po int E: z (4) 1:
Beta
y (3)
y (4)
---Stress at Points--A
B
ksi
ksi
1.4
1.4
D
ksi
E
ksi
0
–7.565
–7.565
7.565
7.565
10
–7.896
–7.004
7.673
7.227
20
–7.987
–6.230
7.548
6.669
30
–7.836
–5.267
7.193
5.909
40
–7.446
–4.144
6.621
4.970
50
–6.830
–2.895
5.846
3.879
60
–6.007
–1.558
4.895
2.670
70
–5.001
–0.174
3.794
1.381
80
–3.843
1.216
2.578
0.049
90
–2.569
2.569
1.284
–1.284
100
–1.216
3.843
–0.049
–2.578
110
0.174
5.001
–1.381
–3.794
120
1.558
6.007
–2.670
–4.895
130
2.895
6.830
–3.879
–5.846
140
4.144
7.446
–4.970
–6.621
150
5.267
7.836
–5.909
–7.193
160
6.230
7.987
–6.669
–7.548
170
7.004
7.896
–7.227
–7.673
180
7.565
7.565
–7.565
–7.565
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670
PROBLEM 4.C4
b
B
B
A
A
h
M'
M
r1
C
Couples of moment M 2 kN m are applied as shown to a curved bar
having a rectangular cross section with h 100 mm and b 25 mm. Write
a computer program and use it to calculate the stresses at points A and B for
values of the ratio r1/h from 10 to 1 using decrements of 1, and from 1
to 0.1 using decrements of 0.1. Using appropriate smaller increments,
determine the ratio r1/h for which the maximum stress in the curved bar is
50% larger than the maximum stress in a straight bar of the same cross
section.
SOLUTION
Input:
h 100 mm,
b 25 mm,
M 2 kN m
For straight bar,
straight
M
S
6M
h 2b
48 MPa
Following notation of Section 4.15, key in the following:
r2
Stresses:
A
h r1 ; R
1
M (r1
h /ln (r2
r1 ); r
R)( Aer1 )
r1
2
B
r2 : e
M (r2
r
R; A bh
R)/(Aer2 )
2500
(I)
(II)
Since h 100 mm, for r1/h 10, r1 1000 mm. Also, r1/h 10, r1 100
Program:
For r1 1000 to 100 at
100 decrements,
using equations of Lines I and II, evaluate r2 , R, r , e,
Also evaluate ratio
1/
Return and repeat for r1
1,
and
2
straight
100 to 10 at
10 decrements.
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671
PROBLEM 4.C4 (Continued)
Program Output
M
Bending moment
Stress in straight beam
2 kN m h 100.000 in. A 2500.00 mm 2
48.00 MPa
r1
mm
rbar
mm
R
mm
e
mm
MPa
MPa
r1 /h
–
ratio
–
1000
1050
1049
0.794
–49.57
46.51
10.000
–1.033
900
950
949
0.878
–49.74
46.36
9.000
–1.036
800
850
849
0.981
–49.95
46.18
8.000
–1.041
700
750
749
1.112
–50.22
45.95
7.000
–1.046
600
650
649
1.284
–50.59
45.64
6.000
–1.054
500
550
548
1.518
–51.08
45.24
5.000
–1.064
400
450
448
1.858
–51.82
44.66
4.000
–1.080
300
350
348
2.394
–53.03
43.77
3.000
–1.105
200
250
247
3.370
–55.35
42.24
2.000
–1.153
100
150
144
5.730
–61.80
38.90
1.000
–1.288
1
2
=====================================================
100
150
144
5.730
–61.80
38.90
1.000
–1.288
90
140
134
6.170
–63.15
38.33
0.900
–1.316
80
130
123
6.685
–64.80
37.69
0.800
–1.350
70
120
113
7.299
–66.86
36.94
0.700
–1.393
60
110
102
8.045
–69.53
36.07
0.600
–1.449
50
100
91
8.976
–73.13
35.04
0.500
–1.523
40
90
80
10.176
–78.27
33.79
0.400
–1.631
30
80
68
11.803
–86.30
32.22
0.300
–1.798
20
70
56
14.189
–100.95
30.16
0.200
–2.103
10
60
42
18.297
–138.62
27.15
0.100
–2.888
===================================================================
Find r1/h for (
max ) /( straight )
1.5
52.70
52.80
103
103
94
94
8.703
8.693
–72.036
–71.998
35.34
35.35
0.527
0.528
–1.501
–1.500
52.90
103
94
8.683
–71.959
35.36
0.529
–1.499
Ratio of stresses is 1.5 for r1
52.8 mm or r1/h
0.529.
[Note: The desired ratio r1/h is valid for any beam having a rectangular cross section.]
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672
bn
hn
h2
PROBLEM 4.C5
M
The couple M is applied to a beam of the cross
section shown. (a) Write a computer program that,
for loads expressed in either SI or U.S. customary
units, can be used to calculate the maximum tensile
and compressive stresses in the beam. (b) Use this
program to solve Probs. 4.9, 4.10, and 4.11.
b2
h1
b1
SOLUTION
Input:
Bending moment M.
For n 1 to n,
Enter bn and hn
bn hn
Area
an
an
(Print)
(hn 1 )/2 hn /2
1
[Moment of rectangle about base]
m
( Area)an
m
m
[For whole cross section]
m; Area
Area
Area
Location of centroid above base.
y
m/Area
(Print)
Moment of inertia about horizontal centroidal axis.
For n 1 to n,
an
an
(hn 1 )/2 hn /2
1
I
bn hn3 /12 (bn hn )( y
I
I
I
an )2
(Print)
Computation of stresses.
For n 1 to n,
Total height:
H
H
hn
Stress at top:
M top
M
H
y
I
(Print)
Stress at bottom:
M bottom
M
y
I
(Print)
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673
PROBLEM 4.C5 (Continued)
Problem 4.9
Summary of cross section dimensions:
Width (in.)
Height (in.)
9.00
2.00
3.00
6.00
Bending moment 600.000 kip in.
Centroid is 3.000 mm above lower edge.
Centroidal moment of inertia is 204.000 in4.
Stress at top of beam
14.706 ksi
Stress at bottom of beam 8.824 ksi
Problem 4.10
Summary of cross section dimensions:
Width (in.)
Height (in.)
4.00
1.00
1.00
6.00
8.00
1.00
Bending moment 500.000 kip in.
Centroid is 4.778 in. above lower edge.
Centroidal moment of inertia is 155.111 in4.
Stress at top of beam
10.387 ksi
Stress at bottom of beam 15.401 ksi
Problem 4.11
Summary of cross section dimensions:
Width (mm)
Height (mm)
50
10
20
50
Bending moment 1500.0000 N m
Centroid is 25.000 mm above lower edge.
Centroidal moment of inertia is 512,500 mm4.
Stress at top of beam
102.439 MPa
Stress at bottom of beam 72.171 MPa
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674
PROBLEM 4.C6
y
Dy
y
c
M
z
A solid rod of radius c 1.2 in. is made of a steel that is assumed to be
elastoplastic with E 29,000 ksi and Y 42 ksi. The rod is subjected to a
couple of moment M that increases from zero to the maximum elastic
moment M Y and then to the plastic moment M p . Denoting by yY the half
thickness of the elastic core, write a computer program and use it to calculate
the bending moment M and the radius of curvature
for values of yY from
1.2 in. to 0 using 0.2-in. decrements. (Hint: Divide the cross section into 80
horizontal elements of 0.03-in. height.)
SOLUTION
MY
Y
Mp
Y
c3 (42 ksi) (1.2 in.)3 57 kip in.
4
4
4 3
4
c (42 ksi) (1.2 in.)3 96.8 kip in.
3
3
Consider top half of rod.
Let
i
Number of elements in top half.
c
Height of each element: h
L
h
For n
0 to i 1, Step 1:
y
n( h)
c 2 {(n 0.5) h}2
z
If y
z at midheight of element
yY go to 100
(n 0.5) h
E
Y
Stress in elastic core
y
go to 200
100
E
200
Area
Force
Moment
M
P
Y
Stress in plastic zone
2 z ( h)
E ( Area)
Repeat for yY 1.2 in.
to yY 0
At 0.2-in. decrements
Force (n 0.5) h
Moment
M
yY E / Y
Print yY , M , and .
Next
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675
PROBLEM 4.C6 (Continued)
Program Output
Radius of rod 1.2 in.
Yield point of steel
Yield moment
Plastic moment
Number of elements in half of the rod
42 ksi
57.0 kip in.
96.8 kip in.
40
For yY
1.20 in.,
M
57.1 kip in.
Radius of curvature
828.57 in.
For yY
1.00 in.,
M
67.2 kip in.
Radius of curvature
690.48 in.
For yY
0.80 in.,
M
76.9 kip in.
Radius of curvature
552.38 in.
For yY
0.60 in.,
M
85.2 kip in.
Radius of curvature
414.29 in.
For yY
0.40 in.,
M
91.6 kip in.
Radius of curvature
276.19 in.
For yY
0.20 in.,
M
95.5 kip in.
Radius of curvature
138.10 in.
For yY
0.00 in.,
M
infinite
Radius of curvature
zero
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676
PROBLEM 4.C7
2 in.
C
3 in.
The machine element of Prob 4.178 is to be redesigned by removing part
of the triangular cross section. It is believed that the removal of a small
triangular area of width a will lower the maximum stress in the element.
In order to verify this design concept, write a computer program to
calculate the maximum stress in the element for values of a from 0 to
1 in. using 0.1-in. increments. Using appropriate smaller increments,
determine the distance a for which the maximum stress is as small as
possible and the corresponding value of the maximum stress.
B
A
2.5 in.
a
SOLUTION
See Figure 4.79, Page 289.
M
For a
5 kip in. r2
5 in. b2
2.5 in.
0 to 1.0 at 0.1 intervals,
h 3 a
r1 2 a
b1
b2 (a /(h a ))
Area
(b1 b2 )(h/2)
x
a+
r
r2
R
e
1
b1h(h/3)
2
Area
(h x )
1
2
(b1r2
r
1
b2 h 2h/3
2
h 2 (b1 b2 )
b2 r1 ) ln
r2
r1
h(b1 b2 )
R
D
M (r1
R)/[Area (e)(r1 )]
B
M (r2
R) /[Area (e)(r2 )]
Print and Return
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677
PROBLEM 4.C7 (Continued)
Program Output
a
in.
R
in.
D
B
ksi
ksi
b1
r
e
0.00
3.855
8.5071
2.1014
0.00
4.00
0.145
0.10
3.858
7.7736
2.1197
0.08
4.00
0.144
0.20
3.869
7.2700
2.1689
0.17
4.01
0.140
0.30
3.884
6.9260
2.2438
0.25
4.02
0.134
0.40
3.904
6.7004
2.3423
0.33
4.03
0.127
0.50
3.928
6.5683
2.4641
0.42
4.05
0.119
0.60
3.956
6.5143
2.6102
0.50
4.07
0.111
0.70
3.985
6.5296
2.7828
0.58
4.09
0.103
0.80
4.018
6.6098
2.9852
0.67
4.11
0.094
0.90
4.052
6.7541
3.2220
0.75
4.14
0.086
1.00
4.089
6.9647
3.4992
0.83
4.17
0.078
Determination of the maximum compressive stress that is as small as possible.
a
in.
R
in.
D
B
ksi
ksi
b1
r
e
0.620
3.961
6.51198
2.6425
0.52
4.07
0.109
0.625
3.963
6.51185
2.6507
0.52
4.07
0.109
0.630
3.964
6.51188
2.6591
0.52
4.07
0.109
Answer: When a
625 in., the compressive stress is 6.51 ksi.
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678
CHAPTER 5
PROBLEM 5.1
w
B
A
L
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the equations of the shear and bending-moment
curves.
SOLUTION
Reactions:
M B 0: AL wL
L
0
2
A
wL
2
M A 0:
L
0
2
B
wL
2
BL wL
Free body diagram for determining reactions:
Over whole beam,
0 x L
Place section at x.
Replace distributed load by equivalent concentrated load.
Fy 0:
wL
wx V 0
2
L
V w x
2
M J 0:
M
wL
x
x wx M 0
2
2
w
( Lx x 2 )
2
M
Maximum bending moment occurs at x
w
x ( L x)
2
L
.
2
M max
wL2
8
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681
PROBLEM 5.2
P
A
B
C
a
b
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the equations of the shear and bending-moment
curves.
L
SOLUTION
Reactions:
M C 0: LA bP 0
A
Pb
L
M A 0: LC aP 0
C
Pa
L
0 xa
From A to B,
Fy 0:
Pb
V 0
L
Pb
L
V
M J 0: M
Pb
x0
L
M
Pbx
L
a x L
From B to C,
Fy 0: V
Pa
0
L
V
M K 0: M
Pa
( L x) 0
L
M
Pa( L x)
L
M
At section B,
Pa
L
Pab
L2
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682
PROBLEM 5.3
w0
A
B
L
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the equations of the shear and bendingmoment curves.
SOLUTION
Free body diagram for determining reactions.
Reactions:
Fy 0: RA
w0 L
0
2
RA
w0 L
2
w L 2L
M A 0: M A 0
0
2 3
MA
w0 L2
w L2
0
3
3
Use portion to left of the section as the free body.
Replace distributed load with equivalent concentrated load.
Fy 0:
w0 L 1 w0 x
x V 0
2
2 L
V
w0 L w0 x 2
2
2L
M J 0:
w0 L2 w0 L
1 w0 x x
x M 0
( x)
3
2
2 L
3
M
w0 L2 w0 Lx w0 x3
3
2
6L
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683
PROBLEM 5.4
w
B
A
L
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the equations of the shear and bendingmoment curves.
SOLUTION
Free body diagram for determining reactions.
Reactions:
Fy 0: RA wL 0
RA wL
L
M A 0: M A (wL) 0
2
MA
w0 L2
2
Use portion to the right of the section as the free body.
Replace distributed load by equivalent concentrated load.
Fy 0: V w( L x) 0
V w( L x)
L x
M J 0: M w( L x)
0
2
M
w
( L x) 2
2
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684
P
PROBLEM 5.5
P
B
C
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the equations of the shear and bending-moment
curves.
A
a
a
SOLUTION
0 xa
From A to B:
Fy 0 :
P V 0
V P
M J 0 :
Px M 0
M Px
a x 2a
From B to C:
Fy 0 :
P P V 0
V 2 P
M J 0 :
Px P( x a) M 0
M 2 Px Pa
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685
w
B
A
PROBLEM 5.6
w
C
a
D
a
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the equations of the shear and bending-moment
curves.
L
SOLUTION
Reactions:
A D wa
From A to B,
0 xa
Fy 0:
wa wx V 0
V w(a x)
M J 0: wax (wx)
x
M 0
2
x2
M w ax
2
a x La
From B to C,
Fy 0:
wa wa V 0
V 0
a
wax wa x M 0
2
M J 0:
From C to D,
Fy 0:
M
1 2
wa
2
La x L
V w( L x) wa 0
V w( L x a)
L x
M w( L x)
wa( L x) 0
2
M J 0:
1
M w a ( L x) ( L x ) 2
2
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686
3 kN
A
C
0.3 m
5 kN
2 kN
PROBLEM 5.7
E
B
Draw the shear and bending-moment diagrams for the beam and
loading shown, and determine the maximum absolute value (a) of
the shear, (b) of the bending moment.
2 kN
D
0.3 m
0.3 m
0.4 m
SOLUTION
Origin at A:
Reaction at A:
Fy 0: RA 3 2 5 2 0
RA 2 kN
M A 0: M A (3 kN)(0.3 m) (2 kN)(0.6 m) (5 kN)(0.9 m) (2 kN)(1.3 m) 0
M A 0.2 kN m
From A to C:
Fy 0:
V 2 kN
M1 0:
0.2 kN m (2 kN)x M 0
M 0.2 2 x
From C to D:
Fy 0: 2 3 V 0
V 1 kN
M 2 0:
0.2 kN m (2 kN)x (3 kN)( x 0.3) M 0
M 0.7 x
From D to E:
Fy 0: V 5 2 0
M 3 0:
V 3 kN
M 5(0.9 x) (2)(1.3 x) 0
M 1.9 3x
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687
PROBLEM 5.7 (Continued)
From E to B:
Fy 0: V 2 kN
M 4 0:
M 2(1.3 x) 0
M 2.6 2 x
(a)
(b) M
V
max
max
3.00 kN
0.800 kN m
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688
100 lb
250 lb
C
100 lb
D
E
B
A
15 in.
20 in.
25 in.
PROBLEM 5.8
Draw the shear and bending-moment diagrams for the beam
and loading shown, and determine the maximum absolute value
(a) of the shear, (b) of the bending moment.
10 in.
SOLUTION
Reactions:
M C 0:
RE (45 in.) 100 lb(15 in.) 250 lb(20 in.) 100 lb(55 in.) 0
RE 200 lb
Fy 0:
RC 200 lb 100 lb 250 lb 100 lb 0
RC 250 lb
At any point, V is the sum of the loads and reactions to the left (assuming + ) and M the sum of their moments
about that point (assuming ).
(a) Vmax 150.0 lb
(b) M max 1500 lb in.
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689
PROBLEM 5.8 (Continued)
Detailed computations of moments:
MA 0
M C (100 lb)(15 in.) 1500 lb in.
M D (100 lb)(35 in.) (250 lb)(20 in.) 1500 lb in.
M E (100 lb)(60 in.) (250 lb)(45 in.) (250 lb)(25 in.) 1000 lb in.
M B (100 lb)(70 in.) (250 lb)(55 in.) (250 lb)(35 in.) (200 lb)(10 in.) 0
(Checks)
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690
PROBLEM 5.9
25 kN/m
C
D
B
A
40 kN
0.6 m
40 kN
1.8 m
Draw the shear and bending-moment diagrams for the beam and loading
shown, and determine the maximum absolute value (a) of the shear,
(b) of the bending moment.
0.6 m
SOLUTION
The distributed load is replaced with an equivalent concentrated load of 45 kN to compute the reactions.
(25 kN/m)(1.8 m) 45 kN
M A 0: (40 kN)(0.6 m) 45 kN(1.5 m) 40 kN(2.4 m) RB (3.0 m) 0
RB 62.5 kN
Fy 0:
RA 62.5 kN 40 kN 45 kN 40 kN 0
RA 62.5 kN
At C:
Fy 0: V 62.5 kN
M1 0:
M (62.5 kN)(0.6 m) 37.5kN m
At centerline of the beam:
Fy 0:
62.5 kN 40 kN (25 kN/m)(0.9 m) V 0
V 0
M 2 0:
M (62.5 kN)(1.5 m) (40 kN)(0.9 m) (25 kN/m)(0.9 m)(0.45 m) 0
M 47.625 kN m
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691
PROBLEM 5.9 (Continued)
Shear and bending-moment diagrams:
(a)
(b)
M
V
max
max
62.5 kN
47.6 kN m
From A to C and D to B, V is uniform; therefore M is linear.
From C to D, V is linear; therefore M is parabolic.
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692
2.5 kips/ft
PROBLEM 5.10
15 kips
C
D
B
A
6 ft
3 ft
Draw the shear and bending-moment diagrams for the beam and loading
shown, and determine the maximum absolute value (a) of the shear,
(b) of the bending moment.
6 ft
SOLUTION
M B 0: 15RA (12)(6)(2.5) (6)(15) 0
RA 18 kips
M A 0: 15RB (3)(6)(2.5) (9)(15) 0
RB 12 kips
Shear:
VA 18 kips
VC 18 (6)(2.5) 3 kips
C to D :
V 3 kips
D to B :
V 3 15 12 kips
Areas under shear diagram:
A to C :
V dx 2 (6)(18 3) 63 kip ft
C to D :
V dx (3)(3) 9 kip ft
D to B :
V dx (6)(12) 72 kip ft
1
MA 0
Bending moments:
M C 0 63 63 kip ft
M D 63 9 72 kip ft
M B 72 72 0
V
M
max
max
18.00 kips
72.0 kip ft
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693
3 kN
3 kN
PROBLEM 5.11
E
Draw the shear and bending-moment diagrams for the beam and loading
shown, and determine the maximum absolute value (a) of the shear,
(b) of the bending moment.
450 N ? m
A
C
D
300 mm
B
300 mm
200 mm
SOLUTION
M B 0: (700)(3) 450 (300)(3) 1000 A 0
A 2.55 kN
M A 0: (300)(3) 450 (700)(3) 1000B 0
B 3.45 kN
At A:
V 2.55 kN
A to C:
V 2.55 kN
M 0
M C 0:
At C:
(300)(2.55) M 0
M 765 N m
C to E:
V 0.45 N m
M D 0:
At D:
(500)(2.55) (200)(3) M 0
M 675 N m
M D 0:
At D:
(500)(2.55) (200)(3) 450 M 0
M 1125 N m
E to B:
V 3.45 kN
M E 0:
At E:
M (300)(3.45) 0
M 1035 N m
At B:
V 3.45 kN,
M 0
(a)
(b)
M
V
max
max
3.45 kN
1125 N m
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694
400 lb
1600 lb
PROBLEM 5.12
400 lb
G
D
E
8 in.
F
A
B
8 in.
Draw the shear and bending-moment diagrams for the beam and
loading shown, and determine the maximum absolute value (a) of
the shear, (b) of the bending moment.
C
12 in.
12 in.
12 in.
12 in.
SOLUTION
M G 0: 16C (36)(400) (12)(1600)
(12)(400) 0
C 1800 lb
Fx 0: C Gx 0
Gx 1800 lb
Fy 0: 400 1600 G y 400 0
G y 2400 lb
A to E:
V 400 lb
E to F:
V 2000 lb
F to B:
V 400 lb
At A and B,
M 0
At D ,
M D 0: (12)(400) M 0
At D +,
M D 0: (12)(400) (8)(1800) M 0
M 9600 lb in.
At E,
M E 0: (24)(400) (8)(1800) M 0
M 4800 lb in.
M 4800 lb in.
At F,
M F 0: M (8)(1800) (12)(400) 0 M 19, 200 lb in.
At F ,+
M F 0: M (12)(400) 0
(a)
(b)
M 4800 lb in.
Maximum |V | 2000 lb
Maximum |M | 19, 200 lb in.
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695
1.5 kN
1.5 kN
C
D
A
PROBLEM 5.13
B
0.3 m
0.9 m
0.3 m
Assuming that the reaction of the ground is uniformly distributed, draw
the shear and bending-moment diagrams for the beam AB and determine
the maximum absolute value (a) of the shear, (b) of the bending
moment.
SOLUTION
Over the whole beam,
Fy 0: 1.5w 1.5 1.5 0
A to C:
w 2 kN/m
0 x 0.3 m
Fy 0: 2 x V 0
V (2 x) kN
x
M J 0: (2 x) M 0
2
At C ,
M ( x 2 ) kN m
x 0.3 m
V 0.6 kN, M 0.090 kN m
90 N m
C to D:
0.3 m x 1.2 m
Fy 0: 2 x 1.5 V 0
V (2 x 1.5) kN
x
M J 0: (2 x) (1.5)( x 0.3) M 0
2
M ( x 2 1.5x 0.45) kN m
At the center of the beam, x 0.75 m
V 0
M 0.1125 kN m
112.5 N m
At C +,
x 0.3 m,
V 0.9 kN
(a) Maximum |V | 0.9 kN 900 N
(b)
Maximum |M | 112.5 N m
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696
24 kips
2 kips/ft
C
A
3 ft
D
3 ft
PROBLEM 5.14
2 kips/ft
E
3 ft
B
Assuming that the reaction of the ground is uniformly distributed, draw
the shear and bending-moment diagrams for the beam AB and determine
the maximum absolute value (a) of the shear, (b) of the bending moment.
3 ft
SOLUTION
Over the whole beam,
Fy 0: 12w (3)(2) 24 (3)(2) 0
A to C:
w 3 kips/ft
(0 x 3 ft)
Fy 0: 3x 2 x V 0
M J 0: (3x)
At C,
V ( x) kips
x
x
(2 x) M 0
2
2
M (0.5x 2 ) kip ft
x 3 ft
V 3 kips, M 4.5 kip ft
C to D:
(3 ft x 6 ft)
Fy 0: 3x (2)(3) V 0
V (3x 6) kips
3
x
MK 0: (3x) (2)(3) x M 0
2
2
M (1.5 x 2 6 x 9) kip ft
At D ,
x 6 ft
V 12 kips,
D to B:
M 27 kip ft
Use symmetry to evaluate.
(a)
|V |max 12.00 kips
(b)
|M |max 27.0 kip ft
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697
10 kN
PROBLEM 5.15
100 mm
3 kN/m
For the beam and loading shown, determine the maximum
normal stress due to bending on a transverse section at C.
C
A
B
1.5 m
1.5 m
200 mm
2.2 m
SOLUTION
Using CB as a free body,
M C 0: M (2.2)(3 103 )(1.1) 0
M 7.26 103 N m
Section modulus for rectangle:
S
1 2
bh
6
1
(100)(200)2 666.7 103 mm3
6
666.7 106 m3
Normal stress:
M
7.26 103
10.8895 106 Pa
S
666.7 106
10.89 MPa
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698
PROBLEM 5.16
750 lb
750 lb
150 lb/ft
A
C
4 ft
B
D
4 ft
3 in.
For the beam and loading shown, determine the maximum
normal stress due to bending on a transverse section at C.
12 in.
4 ft
SOLUTION
C A by symmetry.
Reactions:
Fy 0:
A C (2)(750) (12)(150) 0
A C 1650 lb
Use left half of beam as free body.
M E 0:
(1650)(6) (750)(2) (150)(6)(3) M 0
M 5700 lb ft 68.4 103 lb in.
Section modulus:
S
1 2 1
bh (3)(12)2 72 in 3
6
6
Normal stress:
M
68.4 103
950 psi
S
72
950 psi
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699
PROBLEM 5.17
150 kN 150 kN
90 kN/m
C
D
For the beam and loading shown, determine the maximum
normal stress due to bending on a transverse section at C.
E
A
B
W460 ⫻ 113
2.4 m
0.8 m
0.8 m
0.8 m
SOLUTION
Use entire beam as free body.
M B 0:
4.8 A (3.6)(216) (1.6)(150) (0.8)(150) 0
A 237 kN
Use portion AC as free body.
M C 0:
M (2.4)(237) (1.2)(216) 0
M 309.6 kN m
For W460 113, S 2390 106 mm3
Normal stress:
M
309.6 103 N m
S
2390 106 m3
129.5 106 Pa
129.5 MPa
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700
PROBLEM 5.18
30 kN 50 kN 50 kN 30 kN
W310 3 52
a
B
A
a
For the beam and loading shown, determine the maximum normal
stress due to bending on section a-a.
2m
5 @ 0.8 m 5 4 m
SOLUTION
Reactions:
A B
By symmetry,
Fy 0 :
A B 80 kN
Using left half of beam as free body,
M J 0:
(80)(2) (30)(1.2) (50)(0.4) M 0
M 104 kN m 104 103 N m
For
W310 52, S 747 103 mm3
747 106 m3
Normal stress:
M
104 103
139.2 106 Pa
S
747 106
139.2 MPa
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701
PROBLEM 5.19
8 kN
3 kN/m
For the beam and loading shown, determine the maximum
normal stress due to bending on a transverse section at C.
C
A
B
W310 ⫻ 60
1.5 m
2.1 m
SOLUTION
Use portion CB as free body.
M C 0:
M (3)(2.1)(1.05) (8)(2.1) 0
M 23.415 kN m 23.415 103 N m
For W310 60, S 844 103 mm3
844 106 m3
Normal stress:
M
23.415 103
27.7 106 Pa
S
844 106
27.7 MPa
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702
PROBLEM 5.20
5 5 2
2 2
kips kips kips kips kips
C
D
E
F
For the beam and loading shown, determine the maximum normal
stress due to bending on a transverse section at C.
G
B
A
S8 3 18.4
6 @ 15 in. 5 90 in.
SOLUTION
Use entire beam as free body.
M B 0:
90 A (75)(5) (60)(5) (45)(2) (30)(2) (15)(2) 0
A 9.5 kips
Use portion AC as free body.
M C 0: M (15)(9.5) 0
M 142.5 kip in.
For S 8 18.4, S 14.4 in 3
Normal stress:
M
142.5
S
14.4
9.90 ksi
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703
25 kips
25 kips
25 kips
C
D
E
A
PROBLEM 5.21
B
S12 ⫻ 35
1 ft 2 ft
6 ft
Draw the shear and bending-moment diagrams for the beam
and loading shown and determine the maximum normal stress
due to bending.
2 ft
SOLUTION
M B 0:
(11)(25) 10C (8)(25) (2)(25) 0
C 52.5 kips
M C 0:
(1)(25) (2)(25) (8)(25) 10B 0 B 22.5 kips
Shear:
A to C :
V 25 kips
C to D:
V 27.5 kips
D to E:
V 2.5 kips
E to B:
V 22.5 kips
Bending moments:
At C,
M C 0: (1)(25) M 0
M 25 kip ft
At D,
M D 0: (3)(25) (2)(52.5) M 0
M 30 kip ft
At E,
M E 0:
M (2)(22.5) 0 M 45 kip ft
max M 45 kip ft 540 kip in.
For S12 35 rolled steel section,
Normal stress:
S 38.1 in 3
M
540
14.17 ksi
S
38.1
14.17 ksi
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704
PROBLEM 5.22
160 kN
80 kN/m
B
C
D
A
E
W310 ⫻ 60
Hinge
2.4 m
1.5 m
Draw the shear and bending-moment diagrams for the beam
and loading shown and determine the maximum normal
stress due to bending.
1.5 m
0.6 m
SOLUTION
Statics: Consider portion AB and BE separately.
Portion BE:
M E 0:
(96)(3.6) (48)(3.3) C (3) (160)(1.5) 0
C 248kN
E 56 kN
MA MB ME 0
At midpoint of AB:
Fy 0:
V 0
M 0:
M (96)(1.2) (96)(0.6) 57.6 kN m
Just to the left of C:
Fy 0:
V 96 48 144 kN
M C 0: M (96)(0.6) (48)(0.3) 72 kN
Just to the left of D:
Fy 0:
M D 0:
V 160 56 104 kN
M (56)(1.5) 84 kN m
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705
PROBLEM 5.22 (Continued)
From the diagram,
M
max
84 kN m 84 103 N m
For W310 60 rolled-steel shape,
S x 844 103 mm3
844 106 m3
Stress: m
m
M
max
S
84 103
99.5 106 Pa
844 106
m 99.5 MPa
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706
300 N
B
300 N
C
D
40 N
E
300 N
F
G
30 mm
H
A
PROBLEM 5.23
20 mm
Hinge
Draw the shear and bending-moment diagrams for the beam
and loading shown, and determine the maximum normal
stress due to bending.
7 @ 200 mm ⫽ 1400 mm
SOLUTION
Free body EFGH. Note that M E 0 due to hinge.
M E 0: 0.6 H (0.2)(40) (0.40)(300) 0
H 213.33 N
Fy 0: VE 40 300 213.33 0
VE 126.67 N
Shear:
E to F :
V 126.67 N m
F to G :
V 86.67 N m
G to H :
V 213.33 N m
Bending moment at F:
M F 0: M F (0.2)(126.67) 0
M F 25.33 N m
Bending moment at G:
M G 0: M G (0.2)(213.33) 0
M G 42.67 N m
Free body ABCDE.
M B 0: 0.6 A (0.4)(300) (0.2)(300)
(0.2)(126.63) 0
A 257.78 N
M A 0: (0.2)(300) (0.4)(300) (0.8)(126.67) 0.6D 0
D 468.89 N
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707
PROBLEM 5.23 (Continued)
Bending moment at B.
max M 51.56 N m
M B 0: (0.2)(257.78) M B 0
M B 51.56 N m
S
1 2 1
bh (20)(30) 2
6
6
3 103 mm3 3 106 m3
Bending moment at C.
Normal stress:
M C 0: (0.4)(257.78) (0.2)(300)
MC 0
51.56
17.19 106 Pa
3 106
17.19 MPa
M C 43.11 N m
V
Bending moment at D.
M
M D 0: M D (0.2)(213.33) 0
max
max
342 N
516 N m
M D 25.33 N m
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708
64 kN ? m
C
PROBLEM 5.24
24 kN/m
D
A
B
S250 ⫻ 52
2m
2m
SOLUTION
Draw the shear and bending-moment diagrams for the beam
and loading shown and determine the maximum normal stress
due to bending.
2m
Reactions:
M D 0: 4 A 64 (24)(2)(1) 0
A 28 kN
Fy 0: 28 D (24)(2) 0 D 76 kN
A to C:
0 x 2m
Fy 0: V 28 0
V 28 kN
M J 0: M 28 x 0
M (28 x) kN m
C to D:
2m x 4m
Fy 0: V 28 0
V 28 kN
M J 0:
M 28 x 64 0
M (28 x 64) kN m
D to B:
4m x 6m
Fy 0:
V 24(6 x) 0
V (24 x 144) kN
M J 0:
6 x
M 24(6 x)
0
2
M 12(6 x)2 kN m
max M 56 kN m 56 103 N m
S 482 103 mm3
For S250 52 section,
Normal stress:
M
S
56 103 N m
482 10 6 m 3
116.2 106 Pa
116.2 MPa
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709
5 kips
PROBLEM 5.25
10 kips
C
D
A
B
W14 ⫻ 22
5 ft
8 ft
Draw the shear and bending-moment diagrams for the beam
and loading shown, and determine the maximum normal stress
due to bending.
5 ft
SOLUTION
Reaction at C:
M B 0: (18)(5) 13C +(5)(10) 0
C 10.769 kips
Reaction at B:
M C 0: (5)(5) (8)(10) 13B 0
B 4.231 kips
Shear diagram:
A to C :
V 5 kips
C to D :
V 5 10.769 5.769 kips
D to B :
V 5.769 10 4.231 kips
At A and B,
M 0
At C,
M C 0: (5)(5) M C 0
M C 25 kip ft
At D,
M D 0: M D (5)(4.231)
M D 21.155 kip ft
V
max
5.77 kips
|M |max 25 kip ft 300 kip in.
|M |max occurs at C.
For W14 22 rolled-steel section,
S 29.0 in 3
Normal stress:
M
300
S
29.0
10.34 ksi
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710
PROBLEM 5.26
Knowing that W 12 kN , draw the shear and bending-moment
diagrams for beam AB and determine the maximum normal
stress due to bending.
W
8 kN
C
8 kN
D
W310 ⫻ 23.8
E
B
A
1m
1m
1m
1m
SOLUTION
By symmetry, A B
Fy 0: A 8 12 8 B 0
A B 2 kN
Shear:
A to C :
V 2 kN
C to D :
V 6 kN
D to E :
V 6 kN
E to B :
V 2 kN
V
Bending moment:
max
6.00 kN
M C 0: M C (1)(2) 0
At C,
M C 2 kN m
At D,
M D 0: M D (2)(2) (8)(1) 0
By symmetry,
M 2 kN m at D.
M D 4 kN m
M E 2 kN m
max|M | 4.00 kN m occurs at E.
For W310 23.8,
Normal stress:
S x 280 103 mm3 280 106 m3
max
|M |max
4 103
Sx
280 106
14.29 106 Pa
max 14.29 MPa
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711
PROBLEM 5.27
W
8 kN
C
8 kN
D
W310 ⫻ 23.8
E
Determine (a) the magnitude of the counterweight W for which
the maximum absolute value of the bending moment in the beam
is as small as possible, (b) the corresponding maximum normal
stress due to bending. (Hint: Draw the bending-moment diagram
and equate the absolute values of the largest positive and
negative bending moments obtained.)
B
A
1m
1m
1m
1m
SOLUTION
By symmetry,
AB
Fy 0: A 8 W 8 B 0
A B 8 0.5W
Bending moment at C:
M C 0: (8 0.5W )(1) M C 0
M C (8 0.5W ) kN m
Bending moment at D:
M D 0: (8 0.5W )(2) (8)(1) M D 0
M D (8 W ) kN m
M D M C
Equate:
W 8 8 0.5W
W 10.67 kN
(a)
W 10.6667 kN
M C 2.6667 kN m
M D 2.6667 kN m 2.6667.103 N m
|M |max 2.6667 kN m
For W310 23.8 rolled-steel shape,
S x 280 103 mm3 280 106 m3
(b)
max
|M |max
2.6667 103
9.52 106 Pa
Sx
280 106
max 9.52 MPa
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712
5 kips
PROBLEM 5.28
10 kips
C
Determine (a) the distance a for which the maximum absolute
value of the bending moment in the beam is as small as
possible, (b) the corresponding maximum normal stress due to
bending. (See hint of Prob. 5.27.)
D
A
B
W14 ⫻ 22
a
8 ft
5 ft
SOLUTION
Reaction at B:
M C 0: 5a (8)(10) 13RB 0
RB
1
(80 5a)
18
Bending moment at D:
M D 0: M D 5RB 0
M D 5RB
5
(80 5a)
13
Bending moment at C:
M C 0 5a M C 0
M C 5a
Equate:
M C M D
5a
5
(80 5a)
13
(a) a 4.44 ft
a 4.4444 ft
Then
M C M D (5)(4.4444) 22.222 kip ft
|M |max 22.222 kip ft 266.67 kip in.
For W14 22 rolled-steel section, S 29.0 in 3
Normal stress:
M
266.67
9.20 ksi
S
29.0
(b) 9.20 ksi
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713
P
500 mm
Q
C
D
A
PROBLEM 5.29
12 mm
500 mm
18 mm
B
a
Knowing that P Q 480 N, determine (a) the distance a for
which the absolute value of the bending moment in the beam is
as small as possible, (b) the corresponding maximum normal
stress due to bending. (See hint of Prob. 5.27.)
SOLUTION
P 480 N
Q 480 N
M D 0: Aa 480(a 0.5)
Reaction at A:
480(1 a) 0
720
A 960
N
a
Bending moment at C:
M C 0: 0.5A M C 0
360
M C 0.5A 480
Nm
a
Bending moment at D:
M D 0: M D 480(1 a) 0
M D 480(1 a) N m
(a)
M D M C
Equate:
480(1 a) 480
360
a
a 0.86603 m
A 128.62 N
(b)
For rectangular section, S
S
max
M C 64.31 N m
a 866 mm
M D 64.31 N m
1 2
bh
6
1
(12)(13) 2 648 mm3 648 109 m3
6
|M |max
64.31
99.2 106 Pa
S
6.48 109
max 99.2 MPa
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714
PROBLEM 5.30
P
500 mm
Q
12 mm
500 mm
C
D
A
Solve Prob. 5.29, assuming that P 480 N and Q 320 N.
18 mm
B
a
PROBLEM 5.29 Knowing that P Q 480 N, determine
(a) the distance a for which the absolute value of the bending
moment in the beam is as small as possible, (b) the
corresponding maximum normal stress due to bending. (See
hint of Prob. 5.27.)
SOLUTION
P 480 N
Reaction at A:
Q 320 N
M D 0: Aa 480(a 0.5) 320(1 a) 0
560
A 800
N
a
Bending moment at C:
M C 0: 0.5A M C 0
280
M C 0.5A 400
Nm
a
Bending moment at D:
M D 0: M D 320 (1 a) 0
M D (320 320a) N m
(a)
M D M C
Equate:
280
a
a 0.81873 m, 1.06873 m
320 320a 400
320a 2 80a 280 0
a 819 mm
Reject negative root.
A 116.014 N
(b)
M C 58.007 N m
M D 58.006 N m
1 2
bh
6
1
S (12)(18) 2 648 mm3 648 109 m3
6
For rectangular section, S
max
|M |max
58.0065
89.5 106 Pa
9
S
648 10
max 89.5 MPa
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715
PROBLEM 5.31
4 kips/ft
B
C
A
a
W14 ⫻ 68
Hinge
18 ft
Determine (a) the distance a for which the absolute value
of the bending moment in the beam is as small as possible,
(b) the corresponding maximum normal stress due to bending.
(See hint of Prob. 5.27.)
SOLUTION
S x 103 in 3
For W14 68,
Let b (18 a) ft
Segment BC:
By symmetry,
VB C
Fy 0: VB C 4b 0
VB 2b
x
M J 0: VB x (4 x) M 0
2
M VB x 2 x 2 2bx 2 x 2 lb ft
dM
1
2b xm 0
xm b
dx
2
1
1
M max b 2 b 2 b 2
2
2
Segment AB:
(a x )
2
VB (a x) M 0
M K 0: 4(a x)
M 2(a x)2 2b (a x)
|M max | occurs at x 0.
|M max | 2a 2 2ab 2a 2 2a(18 a) 36a
(a)
Equate the two values of |M max |:
36a
1 2 1
1
b (18 a) 2 162 18a a 2
2
2
2
1 2
a 54a 162 0
a 54 (54) 2 (4)
2
a 54 50.9118 3.0883 ft
(b)
12 (162)
a 3.09 ft
|M |max 36a 111.179 kip ft 1334.15 kip in.
|M |max 1334.15
12.95 kips/in 2
Sx
103
m 12.95 ksi
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716
PROBLEM 5.32
d
A
B
A solid steel rod of diameter d is supported as shown. Knowing that for
steel 490 lb/ft 3 , determine the smallest diameter d that can be
used if the normal stress due to bending is not to exceed 4 ksi.
L ⫽ 10 ft
SOLUTION
Let W total weight.
W AL
4
d 2 L
Reaction at A:
A
1
W
2
Bending moment at center of beam:
W L W L
M C 0: M 0
2 2 2 4
2 2
WL
M
d L
8
32
For circular cross section, c 1 d
2
I
4
c4 ,
S
I
3
c3
d
c
4
32
Normal stress:
Solving for d,
Data:
d
M
S
32
d 2 L2
32
d
3
L2
d
L2
L 10 ft (12)(10) 120 in.
490 lb/ft 3
490
0.28356 lb/in 3
123
4 ksi 4000 lb/in 2
d
(120)2 (0.28356)
4000
d 1.021 in.
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717
PROBLEM 5.33
b
A
C
D
B
A solid steel bar has a square cross section of side b and is
supported as shown. Knowing that for steel 7860 kg / m3 ,
determine the dimension b for which the maximum normal stress
due to bending is (a) 10 MPa, (b) 50 MPa.
b
1.2 m
1.2 m
1.2 m
SOLUTION
Weight density: g
Let L total length of beam.
W AL g b 2 L g
Reactions at C and D:
C D
W
2
Bending moment at C:
L W
M C 0: M 0
6 3
WL
M
18
Bending moment at center of beam:
L W L W
M E 0: M 0
4 2 6 2
max|M |
S
For a square section,
Normal stress:
Solve for b:
Data:
M
WL
24
WL b 2 L2 g
18
18
1 3
b
6
|M | b 2 L2 g /18 L2 g
S
3b
b3 /6
b
L 3.6 m 7860 kg/m3
L2 g
3
g 9.81 m/s 2
(a) 10 106 Pa
(b) 50 106 Pa
(a)
b
(3.6) 2 (7860)(9.81)
33.3 103 m
(3)(10 106 )
b 33.3 mm
(b)
b
(3.6) 2 (7860)(9.81)
6.66 103 m
6
(3)(50 10 )
b 6.66 mm
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718
PROBLEM 5.34
w
B
A
L
Using the method of Sec. 5.2, solve Prob. 5.1a.
PROBLEM 5.1 For the beam and loading shown, (a) draw the shear and
bending-moment diagrams, (b) determine the equations of the shear and
bending-moment curves.
SOLUTION
M B 0: AL wL
M A 0: BL wL
L
0
2
L
0
2
A
wL
2
B
wL
2
dV
w
dx
x
V VA 0 w dx wx
V VA wx A wx
V
wL
wx
2
dM
V
dx
x
x wL
M M A 0 V dx 0
wx dx
2
wLx wx 2
2
2
M MA
Maximum M occurs at x
V
wLx wx 2
2
2
M
w
( Lx x 2 )
2
1
, where
2
dM
0
dx
|M |max
wL2
8
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719
PROBLEM 5.35
P
A
B
C
a
Using the method of Sec. 5.2, solve Prob. 5.2a.
PROBLEM 5.2 For the beam and loading shown, (a) draw the shear and
bending-moment diagrams, (b) determine the equations of the shear and
bending-moment curves.
b
L
SOLUTION
At A,
M C 0: LA bP 0
A
Pb
L
M A 0: LC aP 0
C
Pa
L
V A
A to B:
Pb
L
M 0
0 xa
x
0 w dx 0
w0
V VA 0
a Pb
a
M B M A 0 V dx 0
At B +,
V AP
B to C:
Pb
L
V
L
dx
Pba
L
MB
Pba
L
Pb
Pa
P
L
L
a x L
w0
x
a w dx 0
VC VB 0
MC M B
V
Pa
Pa
L
Pab
L
a V dx L ( L a) L
MC M B
Pab
Pba Pab
0
L
L
L
|M | max
Pab
L
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720
PROBLEM 5.36
w0
Using the method of Sec. 5.2, solve Prob. 5.3a.
A
B
L
PROBLEM 5.3 For the beam and loading shown, (a) draw the shear
and bending-moment diagrams, (b) determine the equations of the shear
and bending-moment curves.
SOLUTION
Free body diagram for determining reactions.
Reactions:
Fy 0 : VA
w0 L
0
2
VA
w0 L
2
w L 2 L
M A 0 : M A 0
0
2 3
MA
w w0
w0 L2
3
x
wL
w L2
, VA 0 , M A 0
2
3
L
dV
wx
w 0
dx
L
x
V VA 0
w0 x
w x2
dx 0
2L
L
V
w0 L w0 x 2
2
2L
dM
w L w x2
V o o
2
2L
dx
w x2
x
x w L
M M A 0 V dx 0 0 0 dx
2L
2
w0 L
w x3
x 0
2
6L
M
w0 L2 w0 L
w x3
x 0
3
2
6L
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721
PROBLEM 5.37
w
Using the method of Sec. 5.2, solve Prob. 5.4a.
B
A
L
PROBLEM 5.4 For the beam and loading shown, (a) draw the shear
and bending-moment diagrams, (b) determine the equations of the shear
and bending-moment curves.
SOLUTION
Fy 0: VA wL 0
VA wL
L
M A 0: M (wL) 0
2
MA
wL2
2
dV
w
dx
x
V VA 0 w dx wx
V wL wx
dM
V wL wx
dx
x
M M A 0 (wL wx)dx wLx
wx 2
2
M
V
M
max
wL
max
wL2
wx 2
wLx
2
2
wL2
2
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722
P
PROBLEM 5.38
P
B
C
A
a
a
Using the method of Sec. 5.2, solve Prob. 5.5a.
PROBLEM 5.5 For the beam and loading shown, (a) draw the shear and
bending-moment diagrams, (b) determine the equations of the shear and
bending-moment curves.
SOLUTION
At A+:
VA P
Over AB:
dV
w 0
dx
dM
V VA P
dx
M Px C
M 0 at x 0
C1 0
M Px
At point B:
xa
M Pa
At point B+:
V P P 2P
Over BC:
dV
w 0
dx
dM
V 2 P
dx
M 2 Px C2
xa
At B:
M Pa
Pa 2 Pa C2
C2 Pa
M 2 Px Pa
x 2a
At C:
M 3Pa
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723
w
B
A
PROBLEM 5.39
w
C
a
D
a
L
Using the method of Sec. 5.2, solve Prob. 5.6a.
PROBLEM 5.6 For the beam and loading shown, (a) draw the shear and
bending-moment diagrams, (b) determine the equations of the shear and
bending-moment curves.
SOLUTION
Reactions:
A D wa
A to B:
0 xa
ww
VA A wa,
MA 0
x
V VA 0 w dx wx
V w(a x)
dM
V wa wx
dx
M MA
x
x
0 V dx 0 (wa wx)dx
M wax
VB 0
B to C:
MB
1 2
wx
2
1 2
wa
2
a x La
V 0
dM
V 0
dx
x
M M B a V dx 0
M MB
M
1 2
wa
2
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724
PROBLEM 5.39 (Continued)
La x L
C to D:
x
V VC L a w dx w[ x ( L a)]
V w[ L x a)]
x
x
M M C L a V dx L a w [x ( L a)]dx
x2
w ( L a) x
2
x
La
x2
( L a) 2
w ( L a) x
( L a) 2
2
2
x2
( L a) 2
w ( L a) x
2
2
M
x2
1 2
( L a) 2
wa w ( L a) x
2
2
2
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725
3 kN
A
2 kN
C
0.3 m
D
0.3 m
5 kN
2 kN
E
B
0.3 m
0.4 m
PROBLEM 5.40
Using the method of Sec. 5.2, solve Prob. 5.7.
PROBLEM 5.7 Draw the shear and bending-moment diagrams
for the beam and loading shown, and determine the maximum
absolute value (a) of the shear, (b) of the bending moment.
SOLUTION
Free body diagram for determining reactions.
Reactions:
Fy 0: VA 3 kN 2 kN 5 kN 2 kN 0
VA 2 kN
M A 0: M A (3 kN)(0.3 m) (2 kN)(0.6 m) (5 kN)(0.9 m) (2 kN)(1.3 m) 0
M A 0.2 kN m
Between concentrated loads and the vertical reaction,
the scope of the shear diagram is , i.e., the shear is
constant. Thus, the area under the shear diagram is equal
to the change in bending moment.
A to C:
V 2 kN M C M A 0.6 M C 0.4 kN
C to D:
V 1 kN M D M C 0.3 M D 0.1 kN m
D to E:
V 3 kN M E M D 0.9 M E 0.8 kN m
E to B:
V 2 kN M B M E 0.8 M B 0 (Checks)
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726
100 lb
250 lb
C
100 lb
D
PROBLEM 5.41
Using the method of Sec. 5.2, solve Prob. 5.8.
E
B
A
15 in.
20 in.
25 in.
10 in.
PROBLEM 5.8 Draw the shear and bending-moment diagrams
for the beam and loading shown, and determine the maximum
absolute value (a) of the shear, (b) of the bending moment.
SOLUTION
Free body diagram for determining reactions.
Reactions:
FY 0 : VC VE 100 lb 250 lb 100 lb 0
VC VE 450 lb
M C 0 : VE (45 in.) (100 lb)(15 in.) (250 lb)(20 in.) (100 lb)(55 in.) 0
VE 200 lb
VC 250 lb
Between concentrated loads and the vertical reaction, the
scope of the shear diagram is , i.e., the shear is constant.
Thus, the area under the shear diagram is equal to he change
in bending moment.
A to C:
V 100 lb, M C M A 1500, M C 1500 lb in.
C to D:
V 150 lb M D M C 3000, M D 1500 lb in.
D to E:
V 100 lb, M E M D 2500, M E 1000 lb in.
E to B:
V 100 lb, M B M E 1000, M B 0 (Checks)
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727
PROBLEM 5.42
25 kN/m
C
D
B
A
40 kN
0.6 m
PROBLEM 5.9 Draw the shear and bending-moment diagrams for the
beam and loading shown, and determine the maximum absolute value
(a) of the shear, (b) of the bending moment.
40 kN
1.8 m
Using the method of Sec. 5.2, solve Prob. 5.9.
0.6 m
SOLUTION
Free body diagram to determine reactions:
M A 0:
VB (3.0 m) 45 kN(1.5 m) (40 kN)(0.6 m) (40 kN)(2.4 m) 0
VB 62.5 kN
Fy 0: VA 40 kN 45 kN 40 kN 62.5 kN 0
VA 62.5 kN
Change in bending moment is equal to area under shear diagram.
A to C:
(62.5 kN)(0.6 m) 37.5 kN m
C to E:
1
(0.9 m)(22.5 kN) 10.125 kN m
2
E to D:
1
(0.9 m)( 22.5 kN) 10.125 kN m
2
D to B:
(62.5 kN)(0.6 m) 37.5 kN m
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728
2.5 kips/ft
PROBLEM 5.43
15 kips
C
D
B
A
6 ft
3 ft
6 ft
Using the method of Sec. 5.2, solve Prob. 5.10.
PROBLEM 5.10 Draw the shear and bending-moment diagrams for the
beam and loading shown, and determine the maximum absolute value
(a) of the shear, (b) of the bending moment.
SOLUTION
Reactions at supports A and B:
M B 0: 15( RA ) (12)(6)(2.5) (6)(15) 0
RA 18 kips
M A 0: 15RB (3)(6)(2.5) (9)(15) 0
RB 12 kips
Areas under shear diagram:
1
(6)(15) 63 kip ft
2
A to C:
(6)(3)
C to D:
(3)(3) 9 kip ft
D to B:
(6)(12) 72 kip ft
Bending moments:
MA 0
M C 0 63 63 kip ft
M D 63 9 72 kip ft
M B 72 72 0
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729
PROBLEM 5.44
4 kN
F
C
A
D
B
Draw the shear and bending-moment diagrams for the beam and loading
shown, and determine the maximum absolute value (a) of the shear, (b) of
the bending moment.
E
4 kN
1m
1m
0.5 m 0.5 m
SOLUTION
M B 0:
3 A (1)(4) (0.5)(4) 0
A 2 kN
M A 0: 3B (2)(4) (2.5)(4) 0
B 6 kN
Shear diagram:
A to C:
V 2 kN
C to D:
V 2 4 2 kN
D to B:
V 2 4 6 kN
Areas of shear diagram:
A to C:
C to D:
D to E:
V dx (1)(2) 2 kN m
V dx (1)(2) 2 kN m
V dx (1)(6) 6 kN m
Bending moments:
MA 0
M C 0 2 2 kN m
M C 2 4 6 kN m
M D 6 2 4 kN m
M D 4 2 6 kN m
MB 6 6 0
(a)
(b)
V
M
max
max
6.00 kN
6.00 kN m
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730
E
PROBLEM 5.45
F
75 mm
B
A
C
300 N
200 mm
D
300 N
200 mm
Draw the shear and bending-moment diagrams for the beam and
loading shown, and determine the maximum absolute value (a) of
the shear, (b) of the bending moment.
200 mm
SOLUTION
M A 0:
0.075 FEF (0.2)(300) (0.6)(300) 0
FEF 3.2 103 N
Fx 0:
Ax FEF 0
Ax 3.2 103 N
Fy 0: Ay 300 300 0
Ay 600 N
Couple at D: M D (0.075)(3.2 103 )
240 N m
Shear:
A to C:
V 600 N
C to B:
V 600 300 300 N
Areas under shear diagram:
A to C:
C to D:
D to B:
V dx (0.2)(600) 120 N m
V dx (0.2)(300) 60 N m
V dx (0.2)(300) 60 N m
Bending moments:
MA 0
M C 0 120 120 N m
M D 120 60 180 N m
M D 180 240 60 N m
M B 60 60 0
Maximum V 600 N
Maximum M 180.0 N m
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731
10 kN
100 mm
3 kN/m
Using the method of Sec. 5.2, solve Prob. 5.15.
C
A
B
1.5 m
1.5 m
PROBLEM 5.46
2.2 m
200 mm
PROBLEM 5.15 For the beam and loading shown, determine
the maximum normal stress due to bending on a transverse
section at C.
SOLUTION
M C 0:
3 A (1.5)(10) (1.1)(2.2)(3) 0
A 2.58 kN
M A 0:
(1.5)(10) 3C (4.1)(2.2)(3) 0
C 14.02 kN
Shear:
A to D:
D to C :
V 2.58 kN
V 2.58 10 7.42 kN
C:
V 7.42 14.02 6.60 kN
B:
V 6.60 (2.2)(3) 0
Areas under shear diagram:
A to D:
V dx (1.5)(2.58) 3.87 kN m
D to C:
V dx (1.5)(7.42) 11.13 kN m
C to B:
V dx 2 (2.2)(6.60) 7.26 kN m
1
Bending moments:
MA 0
M D 0 3.87 3.87 kN m
M C 3.87 11.13 7.26 kN m
M B 7.26 7.26 0
M C 7.26 kN m 7.26 103 N m
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732
PROBLEM 5.46 (Continued)
For rectangular cross section,
S
1 2 1
bh (100)(200)2
6
6
666.67 103 mm3 666.67 106 m 2
Normal stress:
MC
S
7.26 103
10.89 106 Pa
666.67 106
10.89 MPa
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733
PROBLEM 5.47
750 lb
750 lb
150 lb/ft
A
C
4 ft
B
D
4 ft
3 in.
Using the method of Sec. 5.2, solve Prob. 5.16.
PROBLEM 5.16 For the beam and loading shown, determine the
maximum normal stress due to bending on a transverse section at C.
12 in.
4 ft
SOLUTION
C A by symmetry
Reactions:
Fy 0:
A C (2)(750) (12)(150) 0
A C 1650 lb
Shear:
VA 1650 lb
VC 1650 (4)(150) 1050 lb
VC 1050 750 300 lb
VE 300 (2)(150) 0
Areas under shear diagram:
V dx 2 (1650 1050)(4)
1
A to C:
5400 lb ft
V dx 2 (300)(2) 300 lb ft
1
C to E:
Bending moments:
MA 0
M C 0 5400 5400 lb ft
M E 5400 300 5700 lb ft
M E 5700 lb ft 68.4 103 lb in.
For rectangular cross section,
S
Normal stress:
1 2 1
bh (3)(12)2 72 in 3
6
6
M
S
68.4 103
950 psi
72
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734
PROBLEM 5.48
30 kN 50 kN 50 kN 30 kN
a
W310 3 52
Using the method of Sec. 5.2, solve Prob. 5.18.
B
A
a
PROBLEM 5.18 For the beam and loading shown, determine
the maximum normal stress due to bending on section a-a.
2m
5 @ 0.8 m 5 4 m
SOLUTION
Reactions:
By symmetry,
A B.
Fy 0: A B 80 kN
Shear diagram:
A to C:
V 80 kN
C to D:
V 80 30 50 kN
D to E:
V 50 50 0
Areas of shear diagram:
A to C:
V dx (80)(0.8) 64 kN m
C to D:
V dx (50)(0.8) 40 kN m
D to E:
V dx 0
Bending moments:
MA 0
M C 0 64 64 kN m
M D 64 40 104 kN m
M E 104 0 104 kN m
M
max
104 kN m 104 103 N m
For W310 52, S 747 103 mm3 747 106 m3
Normal stress:
M
S
104 103
139.2 106 Pa
747 106
139.2 MPa
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735
PROBLEM 5.49
5 5 2
2 2
kips kips kips kips kips
C
D
E
F
Using the method of Sec. 5.2, solve Prob. 5.20.
G
B
A
PROBLEM 5.20 For the beam and loading shown, determine the
maximum normal stress due to bending on a transverse section at C.
S8 3 18.4
6 @ 15 in. 5 90 in.
SOLUTION
Use entire beam as free body.
M B 0:
90 A (75)(5) (60)(5) (45)(2) (30)(2) (15)(2) 0
A 9.5 kips
Shear A to C:
V 9.5 kips
V dx (15)(9.5)
Area under shear curve A to C:
142.5 kip in.
MA 0
M C 0 142.5 142.5 kip in.
For S8 18.4, S 14.4 in 3
Normal stress:
M
142.5
S
14.4
9.90 ksi
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736
PROBLEM 5.50
w
w 5 w0 [x/L] 1/2
B
x
Determine (a) the equations of the shear and bending-moment curves for
the beam and loading shown, (b) the maximum absolute value of the
bending moment in the beam.
A
L
SOLUTION
1
dV
w x1/2
x 2
w w0 01/2
dx
L
L
V
2 wo x3/2
C1
3 L1/2
V 0 at x L
2
0 w0 L C1
3
C1
2
w0 L
3
V
dM
V
dx
M 0 at
M
M C2
x L
2
2 w x3/2
w0 L 01/2
3
3 L
2
2 2 w0 x5/2
w0 Lx
3
3 5 L1/2
0C
2
4
w0 L2
w0 L2
3
15
2
C2 w0 L2
5
2
4 w0 x5/2 2
w0 Lx
w0 L2
3
15 L1/2
5
M
max
occurs at x 0
M
max
2
w0 L2
5
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737
w
PROBLEM 5.51
w ⫽ w0 cos x
2L
A
x
B
Determine (a) the equations of the shear and bending-moment curves for
the beam and loading shown, (b) the maximum absolute value of the
bending moment in the beam.
L
SOLUTION
x
dV
w w0 cos
2L
dx
x
2 Lw0
dM
C1
sin
V
2L
dx
M
4L2 w0
x
V 0 at
C1x C2
2L
x 0. Hence, C1 0.
M 0 at
x 0. Hence, C2
2
cos
4 L2 w0
2
.
(a) V (2 Lw0 / sin( x /2 L)
M (4 L2 w0 / )[1 cos( x /2L)]
(b)
M
max
4w0 L2 / 2
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738
w
w ⫽ w0 sin x
L
PROBLEM 5.52
B
A
x
Determine (a) the equations of the shear and bending-moment curves for
the beam and loading shown, (b) the maximum absolute value of the
bending moment in the beam.
L
SOLUTION
x
dV
w w0 sin
dx
L
w0 L
x
dM
V
cos
C1
L
dx
2
wL
x
M 0 2 sin
C1 x C2
L
M 0 at x 0
C2 0
M 0 at x L
0 0 C1 L 0
C1 0
V
(a)
M
dM
V 0 at
dx
(b)
M max
w0 L2
2
sin
x
w0 L
w0 L2
2
cos
sin
x
L
x
L
L
2
M max
2
w0 L2
2
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739
w
PROBLEM 5.53
w ⫽ w0 x
L
B
A
Determine (a) the equations of the shear and bending-moment curves for
the beam and loading shown, (b) the maximum absolute value of the
bending moment in the beam.
x
L
SOLUTION
dV
x
w w0
dx
L
1
x2
dM
C1
V w0
2
L
dx
3
1
x
M w0
C1 x C2
L
6
(a)
(b)
M 0 at
x0
C2 0
M 0 at
xL
1
0 w0 L2 C1 L
6
C1
1
w0 L
6
1
x2 1
V w0
w0 L2
2
L 6
V
1
x3 1
M w0
w0 Lx
6
L 6
M
M max occurs when
1
w0 ( L2 3x 2 )/L
6
1
w0 ( Lx x3 /L)
6
dM
V 0. L2 3xm2 0
dx
xm
L
3
M max
1 L2
L2
w0
6 3 3 3
M max 0.0642 w0 L2
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740
PROBLEM 5.54
3 kips/ft
A
B
C
2 ft
D
10 ft
S10 3 25.4
Draw the shear and bending-moment diagrams for the beam and
loading shown and determine the maximum normal stress due to
bending.
3 ft
SOLUTION
M C 0: RD (10) (5.5)(15)(3) 0
RD 24.75 kips
M D 0: (4.5)(15)(3) RC (10) 0
RC 20.25 kips
Shear:
VA 0
VC 0 (2)(3) 6 kips
VC 6 20.50 14.25 kips
VD 14.25 (10)(3) 15.75 kips
VD 15.75 24.75 9 kips
VB 9 (3)(3) 0
Locate point E where V 0:
10 e
e
14.25
15.75
e 4.75 ft
10 e 5.25 ft
Areas of shear diagram:
A to C:
V dx 2 (2)(6) 6 kip ft
C to E:
V dx 2 (4.75)(14.25) 33.84375 kip ft
E to D:
V dx 2 (5.25)(15.75) 41.34375 kip ft
D to B:
V dx 2 (3)(9) 13.5 kip ft
1
1
1
1
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741
PROBLEM 5.54 (Continued)
Bending moments:
MA 0
M C 0 6 6 kip ft
M E 6 33.84375 27.84375 kip ft
M D 27.84375 41.34375 13.5 kip ft
M B 13.5 13.5 0
Maximum M 27.84375 kip ft 334.125 kip in.
For S10 25.4, S 24.6 in 3
Normal stress:
M
S
334.125
13.58 ksi
24.6
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742
PROBLEM 5.55
16 kN/m
C
A
B
S150 ⫻ 18.6
1.5 m
1m
Draw the shear and bending-moment diagrams for the beam
and loading shown, and determine the maximum normal
stress due to bending.
SOLUTION
M B 0: 2.5 A (1.75)(1.5)(16) 0
A 16.8 kN
M A 0: (0.75) (1.5)(16) 2.5B 0
B 7.2 kN
Shear diagram:
VA 16.8 kN
VC 16.8 (1.5)(16) 7.2 kN
VB 7.2 kN
Locate point D where V 0.
d
1.5 d
24d 25.2
16.8
7.2
d 1.05 m 1.5 d 0.45 m
Areas of the shear diagram:
1
A to D:
V dx 2 (1.05)(16.8) 8.82 kN m
D to C:
V dx 2 (0.45)(7.2) 1.62 kN m
V dx (1)(7.2) 7.2 kN m
C to B:
1
Bending moments:
MA 0
M D 0 8.82 8.82 kN m
M C 8.82 1.62 7.2 kN m
M B 7.2 7.2 0
Maximum |M | 8.82 kN m 8.82 103 N m
For S150 18.6 rolled-steel section, S 120 103 mm3 120 106 m3
Normal stress:
|M | 8.82 103
73.5 106 Pa
S
120 106
73.5 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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743
9 kN
PROBLEM 5.56
12 kN/m
A
B
C
0.9 m
W200 3 19.3
Draw the shear and bending-moment diagrams for the beam
and loading shown and determine the maximum normal
stress due to bending.
3m
SOLUTION
M C 0: (0.9)(9) (1.5)(3)(12) 3B 0
B 15.3 kN
M B 0: (3.9)(9) 3C (1.5)(3)(12) 0
C 29.7 kN
Shear:
A to C:
V 9 kN
C :
V 9 29.7 20.7 kN
B:
V 20.7 (3)(12) 15.3 kN
V
max
20.7 kN
Locate point E where V 0.
3e
e
20.7
15.3
e 1.725 ft
36e (20.7)(3)
3 e 1.275 ft
Areas under shear diagram:
A to C:
V dx (0.9)(9) 8.1 kN m
C to E:
V dx 2 (1.725)(20.7) 17.8538 kN m
E to B:
V dx 2 (1.275)(15.3) 9.7538 kN m
1
1
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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744
PROBLEM 5.56 (Continued)
Bending moments:
MA 0
M C 0 8.1 8.1 kN m
M E 8.1 17.8538 9.7538 kN m
M B 9.7538 9.7538 0
M
max
9.7538 103 N m at point E
For W200 19.3 rolled-steel section, S 162 103 mm3 162 106 m3
Normal stress:
M
S
9.7538 103
60.2 106 Pa 60.2 MPa
162 106
60.2 MPa
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745
PROBLEM 5.57
1600 lb
80 lb/ft
A
1.5 in.
B
11.5 in.
Draw the shear and bending-moment diagrams for the beam
and loading shown, and determine the maximum normal
stress due to bending.
9 ft
1.5 ft
SOLUTION
M A 0: (1600 lb)(1.5 ft) [(80 lb/ft)(9 ft)](7.5 ft) 12B 0
B 650 lb
B 650 lb
Fy 0: A 1600 lb [(80 lb/ft)(9 ft)] 650 lb 0
A 1670 lb
9x
x
70 lb 650 lb
x 0.875 ft
A 1670 lb
2641 lb ft M max
c
1
(11.5 in.) 5.75 in.
2
I
1
(1.5 in.)(11.5 in.)3 190.1 in 4
12
M max 2641 lb ft 31, 690 lb in.
m
M max c (31, 690 lb in.)(5.75 in.)
I
190.1 in 4
m 959 psi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
746
PROBLEM 5.58
500 lb
25 lb/in.
A
C
B
16 in.
S4 ⫻ 7.7
Draw the shear and bending-moment diagrams for the beam and
loading shown and determine the maximum normal stress due to
bending.
24 in.
SOLUTION
M B 0:
RA (16) (4)(40)(25) (24)(500) 0 RA 1000 lb
M A 0:
RB (16) (20)(40)(25) (40)(500) 0 RB 2500 lb
Shear:
VA 1000 lb
VB 1000 (16)(25) 1400 lb
VB 1400 2500 1100 lb
VC 1100 (24)(25) 500 lb
Areas of shear diagram:
1
A to B:
V dx 2 (1000 1400)(16) 19, 200 lb in.
B to C:
V dx 2 (1100 500)(24) 19, 200 lb in.
1
Bending moments:
MA 0
M B 0 19, 200 19, 200 lb in.
M C 19, 200 19, 200 0
Maximum M 19.2 kip in.
For S4 7.7 rolled-steel section, S 3.03 in 3
Normal stress:
M
S
19.2
6.34 ksi
3.03
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747
PROBLEM 5.59
80 kN/m
60 kN · m
C
D
12 kN · m
A
B
W250 ⫻ 80
1.2 m
1.6 m
Draw the shear and bending-moment diagrams for the
beam and loading shown and determine the maximum
normal stress due to bending.
1.2 m
SOLUTION
Reaction:
M B 0: 4 A 60 (80)(1.6)(2) 12 0
A 76 kN
Shear:
VA 76 kN
V 76.0 kN
A to C :
VD 76 (80)(1.6) 52 kN
V 52 kN
D to C :
Locate point where V 0.
V ( x) 80 x 76 0
x 0.95 m
Areas of shear diagram:
A to C: V dx (1.2)(76) 91.2 kN m
1
(0.95)(76) 36.1 kN m
2
1
E to D: V dx (0.65)(52) 16.9 kN m
2
C to E: V dx
D to B: V dx (1.2)(52) 62.4 kN m
Bending moments:
M A 60 kN m
M C 60 91.2 31.2 kN m
M E 31.2 36.1 67.3 kN m
M D 67.3 16.9 50.4 kN m
M B 50.4 62.4 12 kN m
For W250 80, S 983 103 mm3
Normal stress:
max
M
S
67.3 103 N m
68.5 106 Pa
983 106 m3
m 68.5 MPa
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748
PROBLEM 5.60
400 kN/m
A
C
D
B
w0 W200 3 22.5
0.3 m
0.4 m
0.3 m
Knowing that beam AB is in equilibrium under the loading
shown, draw the shear and bending-moment diagrams and
determine the maximum normal stress due to bending.
SOLUTION
Fy 0: (1)(w0 ) (0.4)(400) 0
w0 160 kN/m
VA 0
Shear diagram:
VC 0 (0.3)(160) 48 kN
VD 48 (0.3)(400) (0.3)(160) 48 kN
VB 48 (0.3)(160) 0
Locate point E where V 0.
By symmetry, E is the midpoint of CD.
Areas of shear diagram:
A to C:
1
(0.3)(48) 7.2 kN m
2
C to E:
1
(0.2)(48) 4.8 kN m
2
E to D:
1
(0.2)(48) 4.8 kN m
2
D to B:
1
(0.3)(48) 7.2 kN m
2
Bending moments:
MA 0
M C 0 7.2 7.2 kN
M E 7.2 4.8 12.00 kN
M D 12.0 4.8 7.2 kN
M B 7.2 7.2 0
M
max
12.00 kN m 12.00 103 N m
For W200 22.5 rolled-steel shape, S x 193 103 mm3 193 106 m3
Normal stress:
M
S
12.00 103
62.2 106 Pa
193 106
62.2 MPa
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749
w0 50 lb/ft
3
4
T
A
PROBLEM 5.61
in.
Knowing that beam AB is in equilibrium under the loading shown,
draw the shear and bending-moment diagrams and determine the
maximum normal stress due to bending.
B
C
w0
1.2 ft
1.2 ft
SOLUTION
0 x 1.2 ft
A to C:
x
w 50 1
50 41.667 x
1.2
dV
w 41.667 x 50
dx
V VA
x
0
(41.667 x 50)dx
0 20.833 x 2 50 x
M MA
0
x
0
dM
dx
x
V dx
0
(20.833x 2 50 x)dx
6.944 x3 25 x 2
x 1.2 ft,
At C to B, use symmetry conditions.
V 30.0 lb
M 24.0 lb in.
Maximum |M | 24.0 lb ft 288 lb in.
Cross section:
c
I
Normal stress:
d 1
(0.75) 0.375 in.
2 2
4
c 4 (0.375) 15.532 103 in 4
4
|M | c (2.88)(0.375)
6.95 103 psi
I
15.532 103
6.95 ksi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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750
0.2 m
0.5 m
P
C
A
PROBLEM 5.62*
0.5 m
24 mm
Q
D
E
0.4 m
F
B
60 mm
The beam AB supports two concentrated loads P and Q. The
normal stress due to bending on the bottom edge of the beam
is 55 MPa at D and 37.5 MPa at F. (a) Draw the shear and
bending-moment diagrams for the beam. (b) Determine the
maximum normal stress due to bending that occurs in the beam.
0.3 m
SOLUTION
At D,
1
(24)(60)3 432 103 mm 4 c 30 mm
12
I
S 14.4 103 mm3 14.4 106 m3 M S
c
M D (14.4 106 )(55 106 ) 792 N m
At F,
M F (14.4 106 )(37.5 106 ) 540 N m
(a)
I
M F 0: 540 0.3B 0
Using free body FB,
B
540
1800 N
0.3
M D 0: 792 3Q (0.8)(1800) 0
Using free body DEFB,
Q 2160 N
M A 0: 0.2 P (0.7)(2160) (1.2)(1800) 0
Using entire beam,
P 3240 N
Fy 0: A 3240 2160 1800 0
A 3600 N
Shear diagram and its areas:
A to C :
V 3600 N
AAC (0.2)(3600) 720 N m
C to E :
V 3600 3240 360 N
ACE (0.5)(360) 180 N m
V 360 2160 1800 N
AEB (0.5)(1800) 900 N m
E to B:
Bending moments:
MA 0
M C 0 720 720 N m
M E 720 180 900 N m
|M |max 900 N m
M B 900 900 0
(b)
Normal stress:
max
|M |max
900
62.5 106 Pa
S
14.4 106
max 62.5 MPa
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751
P
PROBLEM 5.63*
Q
480 lb/ft
A
The beam AB supports a uniformly distributed load of 480 lb/ft and
two concentrated loads P and Q. The normal stress due to bending
on the bottom edge of the lower flange is 14.85 ksi at D and
10.65 ksi at E. (a) Draw the shear and bending-moment diagrams
for the beam. (b) Determine the maximum normal stress due to
bending that occurs in the beam.
B
C
D
E
1 ft
F
W8 31
1 ft
1.5 ft
1.5 ft
8 ft
SOLUTION
(a) For W8 31 rolled-steel section, S 27.5 in 3
M S
At D,
M D (27.5)(14.85) 408.375 kip in.
At E,
M E (27.5)(10.65) 292.875 kip in.
M D 34.03 kip ft
M E 24.41 kip ft
Use free body DE.
M E 0: 34.03 24.41 (1.5)(3)(0.48) 3VD 0
VD 2.487 kips
M D 0: 34.03 24.41 (1.5)(3)(0.48) 3VE 0
VE 3.927 kips
Use free body ACD.
M A 0: 1.5P (1.25)(2.5)(0.48) (2.5)(2.487) 34.03 0
P 25.83 kips
Fy 0: A (2.5)(0.48) 2.487 25.83 0
A 24.54 kips
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752
PROBLEM 5.63* (Continued)
Use free body EFB.
M B 0: 1.5Q (1.25)(2.5)(0.48) (2.5)(3.927) 24.41 0
Q 8.728 kips
Fy 0: B 3.927 (2.5)(0.48) 8.7 0
B 13.855 kips
Areas of load diagram:
A to C:
(1.5)(0.48) 0.72 kip ft
C to F:
(5)(0.48) 2.4 kip ft
F to B:
(1.5)(0.48) 0.72 kip ft
VA 24.54 kips
Shear diagram:
VC 24.54 0.72 23.82 kips
VC 23.82 25.83 2.01 kips
VF 2.01 2.4 4.41 kips
VF 4.41 8.728 13.14 kips
VB 13.14 0.72 13.86 kips
Areas of shear diagram:
A to C:
1
(1.5)(24.52 23.82) 36.23 kip ft
2
1
(5)(2.01 4.41) 16.05 kip ft
2
C to F:
F to B:
1
(1.5)(13.14 13.86) 20.25 kip ft
2
Bending moments:
MA 0
M C 0 36.26 36.26 kip ft
M F 36.26 16.05 20.21 kip ft
M B 20.21 20.25 0
Maximum M occurs at C: M
(b) Maximum stress:
max
36.26 kip ft 435.1 kip in.
M
max
S
435.1
27.5
15.82 ksi
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753
18 mm
2 kN/m
A
C
0.1 m
PROBLEM 5.64*
Q
P
B
D
0.1 m
36 mm
0.125 m
Beam AB supports a uniformly distributed load of 2 kN/m and two
concentrated loads P and Q. It has been experimentally determined
that the normal stress due to bending in the bottom edge of the beam
is 56.9 MPa at A and 29.9 MPa at C. Draw the shear and bendingmoment diagrams for the beam and determine the magnitudes of the
loads P and Q.
SOLUTION
1
(18)(36)3 69.984 103 mm 4
12
1
c d 18 mm
2
I
S 3.888 103 mm3 3.888 106 m3
c
I
At A,
M A S A (3.888 106 )(56.9) 221.25 N m
At C,
M C S C (3.888 106 )(29.9) 116.25 N m
M A 0: 221.23 (0.1)(400) 0.2 P 0.325Q 0
0.2 P 0.325Q 181.25
(1)
M C 0: 116.25 (0.05)(200) 0.1P 0.225Q 0
0.1P 0.225Q 106.25
Solving (1) and (2) simultaneously,
(2)
P 500 N
Q 250 N
RA 400 500 250 0 RA 1150 N m
Reaction force at A:
VA 1150 N
VD 250
M A 221.25 N m
M C 116.25 N m
M D 31.25 N m
|V |max 1150 N
|M |max 221 N m
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754
1.8 kN
PROBLEM 5.65
3.6 kN
40 mm
B
A
0.8 m
C
h
D
0.8 m
For the beam and loading shown, design the cross section of the
beam, knowing that the grade of timber used has an allowable
normal stress of 12 MPa.
0.8 m
SOLUTION
Reactions:
M D 0: 2.4 A (1.6)(1.8) (0.8)(3.6) 0
A 2.4 kN
M A 0: (0.8)(1.8) (1.6)(3.6) 2.4 D 0
D 3 kN
Construct shear and bending moment diagrams:
|M |max 2.4 kN m 2.4 103 N m
all 12 MPa
12 106 Pa
S min
|M |max
all
2.4 103
12 106
200 106 m3
200 103 mm3
1
1
S bh 2 (40)h 2
6
6
200 103
h2
(6)(200 103 )
40
30 103 mm 2
h 173.2 mm
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755
PROBLEM 5.66
120 mm
10 kN/m
A
h
B
5m
For the beam and loading shown, design the cross section of the
beam, knowing that the grade of timber used has an allowable normal
stress of 12 MPa.
SOLUTION
Reactions:
A = B by symmetry
Fy 0:
A B (5)(10) 0
A B 25 kN
From bending moment diagram,
|M |max 31.25 kN m 31.25 103 N m
all 12 MPa 12 106 Pa
S min
M
max
all
31.25 103
2.604 103 m3
6
12 10
2.604 106 mm3
1
1
S bh 2 (120)h 2 2.604 106
6
6
h2
(6)(2.064 106 )
130.21 103 mm 2
120
h 361 mm
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756
PROBLEM 5.67
B
a
a
6 ft
For the beam and loading shown, design the cross section of the beam,
knowing that the grade of timber used has an allowable normal stress of
1750 psi.
A
1.2 kips/ft
SOLUTION
Equivalent concentrated load:
1
P (6)(1.2) 3.6 kips
2
Bending moment at A:
M A (2)(3.6) 7.2 kip ft = 86.4 kip in.
S min
M
max
all
For a square section,
a
3
6S
amin
3
(6)(49.37)
86.4
49.37 in 3
1.75
S
1 3
a
6
amin 6.67 in.
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757
4.8 kips
2 kips
PROBLEM 5.68
4.8 kips
2 kips
B C
b
D E
A
For the beam and loading shown, design the cross section of the
beam, knowing that the grade of timber used has an allowable
normal stress of 1750 psi.
F
9.5 in.
2 ft 2 ft
3 ft
2 ft 2 ft
SOLUTION
B E 2.8 kips
For equilibrium,
Shear diagram:
A to B :
V 4.8 kips
V 4.8 2.8 2 kips
V 2 2 0
D to E :
V 0 2 2 kips
E to F:
V 2 2.8 4.8 kips
B to C :
C to D :
Areas of shear diagram:
A to B:
(2)(4.8) 9.6 kip ft
(2)(2) 4 kip ft
B to C:
C to D:
(3)(0) 0
D to E:
(2)(2) 4 kip ft
E to F:
(2)(4.8) 9.6 kip ft
MA 0
Bending moments:
M B 0 9.6 9.6 kip ft
M C 9.6 4 13.6 kip ft
M D 13.6 0 13.6 kip ft
M E 13.6 4 9.6 kip ft
M F 9.6 9.6 0
|M |max 13.6 kip ft 162.3 kip in. 162.3 103 lb in.
Required value for S:
S
For a rectangular section, I
|M |max
all
162.3 103
93.257 in 3
1750
1 3
1
bh , c h
12
2
Equating the two expressions for S,
S
I bh 2 (b)(9.5)2
15.0417b
6
6
c
15.0417b 93.257
b 6.20 in.
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758
2.5 kN
A
PROBLEM 5.69
2.5 kN
100 mm
6 kN/m
B
C
For the beam and loading shown, design the cross section of the
beam, knowing that the grade of timber used has an allowable
normal stress of 12 MPa.
h
D
3m
0.6 m
0.6 m
SOLUTION
By symmetry,
BC
Fy 0: B C 2.5 2.5 (3)(6) 0
B C 6.5 kN
Shear:
V 2.5 kN
VB 2.5 6.5 9 kN
VC 9 (3)(6) 9 kN
A to B:
V 9 6.5 2.5 kN
C to D:
Areas of the shear diagram:
V dx (0.6)(2.5) 1.5 kN m
1
V dx 2 (1.5)(9) 6.75 kN m
V dx 6.75 kN m
V dx 1.5 kN m
A to B:
B to E:
E to C:
C to D:
MA
MB
ME
MC
MD
Bending moments:
0
0 1.5 1.5 kN m
1.5 6.75 8.25 kN m
8.25 6.75 1.5 kN m
1.5 1.5 0
Maximum |M | 8.25 kN m 8.25 103 N m
all 12 MPa 12 106 Pa
S min
For a rectangular section,
|M |max
all
8.25 103
687.5 106 m3 687.5 103 mm3
12 106
1
S bh 2
6
1
687.5 103 (100) h 2
6
(6)(687.5 103 )
41.25 103 mm 2
h2
100
h 203 mm
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759
3 kN/m
PROBLEM 5.70
b
A
150 mm
C
B
2.4 m
1.2 m
For the beam and loading shown, design the cross section of
the beam, knowing that the grade of timber used has an
allowable normal stress of 12 MPa.
SOLUTION
M B 0: 2.4 A (0.6)(3.6)(3) 0
A 2.7 kN
M A 0: (1.8)(3.6)(3) 2.4 B 0
B 8.1 kN
VA 2.7 kN
Shear:
VB 2.7 (2.4)(3) 4.5 kN
VB 4.5 8.1 3.6 kN
VC 3.6 (1.2)(3) 0
Locate point D where V 0.
2.4 d
d
2.7
4.5
d 0.9 m
7.2 d 6.48
2.4 d 1.5 m
Areas of the shear diagram:
1
A to D:
V dx 2 (0.9)(2.7) 1.215 kN m
D to B:
V dx 2 (1.5)(4.5) 3.375 kN m
B to C:
V dx 2 (1.2)(3.6) 2.16 kN m
Bending moments:
1
1
MA 0
M D 0 1.215 1.215 kN m
M B 1.215 3.375 2.16 kN m
M C 2.16 2.16 0
all 12 MPa 12 106 Pa
Maximum |M | 2.16 kN m 2.16 103 N m
S min
For rectangular section,
|M |
all
2.16 103
180 106 m3 180 103 mm3
12 106
1
1
S bh 2 b(150)2 180 103
6
6
b
(6)(180 103 )
1502
b 48.0 mm
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760
20 kips
11 kips/ft
20 kips
B
E
A
F
C
2 ft 2 ft
D
6 ft
PROBLEM 5.71
Knowing that the allowable normal stress for the steel used is 24 ksi,
select the most economical wide-flange beam to support the loading
shown.
2 ft 2 ft
SOLUTION
By symmetry, RA RF
Fy 0:
RA 20 (6)(11) 20 RF 0
RA RF 50 kips
Maximum bending moment occurs at center of beam.
M J 0:
(7)(53) (5)(20) (1.5)(3)(11) M J 0
M J 221.5 kip.ft 2658 kip in.
S min
Shape
S (in2)
W24 68
154
W21 62
127
W18 76
146
W16 77
134
W14 82
123
W12 96
131
M max
all
2658
110.75 in 3
24
Use W21 62.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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761
PROBLEM 5.72
24 kips
Knowing that the allowable normal stress for the steel used is 24 ksi, select
the most economical wide-flange beam to support the loading shown.
2.75 kips/ft
C
A
B
9 ft
15 ft
SOLUTION
M C 0: 24 A (12)(24)(2.75) (15)(24) 0
A 48 kips
M A 0: 24 C (12)(24)(2.75) (9)(24) 0
C 42 kips
VA 48
Shear:
VB 48 (9)(2.75) 23.25 kips
VB 23.25 24 0.75 kips
VC 0.75 (15)(2.75) 42 kips
Areas of the shear diagram:
1
A to B:
V dx 2 (9)(48 23.25) 320.6 kip ft
B to C:
V dx 2 (15)(0.75 42) 320.6 kip ft
1
Bending moments:
MA 0
M B 0 320.6 320.6 kip ft
M C 320.6 320.6 0
Maximum |M | 320.6 kip ft 3848 kip in.
all 24 ksi
Smin
Shape
W30 99
W27 84
W24 104
W21 101
W18 106
|M |
all
3848
160.3 in 3
24
S , (in 3)
269
213
258
227
204
Lightest wide flange beam: W27 84 @ 84 lb/ft
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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762
PROBLEM 5.73
5 kN/m
D
A
B
C
70 kN
70 kN
5m
3m
Knowing that the allowable normal stress for the steel used is 160 MPa,
select the most economical wide-flange beam to support the loading shown.
3m
SOLUTION
Shape
Section modulus
all 160 Mpa
Smin
M max
all
286 kN m
1787 106 m3
160 MPa
1787 103 mm3
S, (103 mm3)
W610 101
2520
W530 92
2080
W460 113
2390
W410 114
2200
W360 122
2020
W310 143
2150
Use W530 92.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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763
PROBLEM 5.74
50 kN/m
C
A
D
B
0.8 m
2.4 m
Knowing that the allowable normal stress for the steel used is 160 MPa,
select the most economical wide-flange beam to support the loading
shown.
0.8 m
SOLUTION
M D 0: 3.2 B (24)(3.2)(50) 0
B 120 kN
M B 0: 3.2 D (0.8)(3.2)(50) 0
D 40 kN
VA 0
Shear:
VB 0 (0.8)(50) 40 kN
VB 40 120 80 kN
VC 80 (2.4)(50) 40 kN
VD 40 0 40 kN
Locate point E where V 0.
e 2.4 e
80
40
e 1.6 m
Areas:
120e 192
2.4 e 0.8 m
1
A to B :
V dx 2 (0.8)(40) 16 kN m
B to E :
V dx 2 (1.6)(80) 64 kN m
E to C :
V dx 2 (0.8)(40) 16 kN m
C to D :
V dx (0.8)(40) 32 kN m
Bending moments:
1
1
MA 0
M B 0 16 16 kN m
M E 16 64 48 kN m
M C 48 16 32 kN m
M D 32 32 0
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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764
PROBLEM 5.74 (Continued)
Maximum |M | 48 kN m 48 103 N m
all 160 MPa 160 106 Pa
S min
Shape
W310 32.7
W250 28.4
W200 35.9
S (103 mm3 )
415
308
342
|M |
all
48 103
300 106 m3 300 103 mm3
160 106
Lightest wide flange beam:
W250 28.4 @ 28.4 kg/m
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765
18 kips
PROBLEM 5.75
3 kips/ft
B
C
D
A
6 ft
6 ft
Knowing that the allowable normal stress for the steel used is 24 ksi,
select the most economical S-shape beam to support the loading shown.
3 ft
SOLUTION
M C 0: 12 A (9)(6)(3) (3)(18) 0
A 9 kips
M A 0: 12C (3)(6)(3) (15)(18) 0
C 27 kips
VA 9 kips
Shear:
B to C:
V 9 (6)(3) 9 kips
C to D:
V 9 27 18 kips
Areas:
A to E:
(0.5)(3)(9) 13.5 kip ft
E to B:
(0.5)(3)(9) 13.5 kip ft
B to C:
(6)(9) 54 kip ft
C to D:
(3)(18) 54 kip ft
Bending moments: M A 0
M E 0 13.5 13.5 kip ft
M B 13.5 13.5 0
M C 0 54 54 kip ft
M D 54 54 0
Maximum M 54 kip ft 648 kip in. all 24 ksi
S min
Shape
S (in 3 )
S12 31.8
36.2
S10 35
29.4
648
27 in 3
24
Lightest S-shaped beam: S12 31.8
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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766
48 kips
48 kips
B
PROBLEM 5.76
48 kips
C
D
A
E
2 ft
2 ft
6 ft
Knowing that the allowable normal stress for the steel used is 24 ksi,
select the most economical S-shape beam to support the loading shown.
2 ft
SOLUTION
M E 0: (12)(48) 10B (8)(48) (2)(48) 0
B 105.6 kips
M B 0: (2)(48) (2)(48) (8)(48) 10E 0
E 38.4 kips
Shear:
Areas:
A to B:
V 48 kips
B to C:
V 48 105.6 57.6 kips
C to D:
V 57.6 48 9.6 kips
D to E:
V 9.6 48 38.4 kips
A to B:
(2)(48) 96 kip ft
B to C:
(2)(57.6) 115.2 kip ft
C to D:
(6)(9.6) 57.6 kip ft
D to E:
(2)(38.4) 76.8 kip ft
Bending moments: M A 0
M B 0 96 96 kip ft
M C 96 115.2 19.2 kip ft
M D 19.2 57.2 76.8 kip ft
M E 76.8 76.8 0
Maximum M 96 kip ft 1152 kip in.
all 24 ksi
S min
Shape
S15 42.9
S12 50
M
all
1152
48 in 3
24
S (in 3 )
59.4
Lightest S-shaped beam: S15 42.9
50.6
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767
80 kN
100 kN/m
C
A
PROBLEM 5.77
Knowing that the allowable normal stress for the steel used is 160 MPa,
select the most economical S-shape beam to support the loading shown.
B
0.8 m
1.6 m
SOLUTION
M B 0:
0.8 A (0.4)(2.4)(100) (1.6)(80) 0
A 280 kN
M A 0:
0.8 B (1.2)(2.4)(100) (2.4)(80) 0
B 600 kN
VA 280 kN
Shear:
VB 280 (0.8)(100) 360 kN
VB 360 600 240 kN
VC 240 (1.6)(100) 80 kN
Areas under shear diagram:
A to B:
1
(0.8)(280 360) 256 kN m
2
B to C:
1
(1.6)(240 80) 256 kN m
2
Bending moments:
MA 0
M B 0 256 256 kN m
M C 256 256 0
Maximum M 256 kN m 256 103 N m
all 160 MPa 160 106 Pa
S min
Shape
S (103 mm3)
S510 98.2
1950
S460 104
1685
M
all
256 103
1.6 103 m3 1600 103 mm3
160 106
Lightest S-section:
S510 98.2
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768
60 kN
PROBLEM 5.78
40 kN
C
B
A
D
2.5 m
2.5 m
Knowing that the allowable normal stress for the steel used is 160 MPa,
select the most economical S-shape beam to support the loading shown.
5m
SOLUTION
Reaction:
M D 0: 10 A (7.5)(60) (5)(40) 0
A 65kN
Shear diagram:
A to B:
V 65 kN
B to C:
V 65 60 5 kN
C to D:
V 5 40 35 kN
Areas of shear diagram:
A to B: (2.5)(65) 162.5 kN m
B to C: (2.5)(5) 12.5 kN m
C to D: (5)(35) 175 kN m
Bending moments: M A 0
M B 0 162.5 162.5 kN m
M C 162.5 12.5 175 kN m
M D 175 175 0
M
max
175 kN m 175 103 N m
all 160 MPa 160 106 Pa
Shape
Sx, (103 mm3)
S610 119
2870
S510 98.2
1950
S460 81.4
1460
S min
M
all
175 103
1093.75 106 m3
160 106
1093.75 103 mm3
Lightest S-section:
S460 81.4
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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769
PROBLEM 5.79
1.5 kN 1.5 kN 1.5 kN
t
B
A
1m
C
D
100 mm
0.5 m 0.5 m
A steel pipe of 100-mm diameter is to support the loading shown.
Knowing that the stock of pipes available has thicknesses varying
from 6 mm to 24 mm in 3-mm increments, and that the allowable
normal stress for the steel used is 150 MPa, determine the minimum
wall thickness t that can be used.
SOLUTION
M A 0: M A (1)(1.5) (1.5)(1.5) (2)(1.5) 0
M
max
M A 6.75 kN m
M A 6.75 kN m
Smin
Smin
I min
M
max
all
6.75 103 N m
45 106 m3 45 103 mm3
150 106 Pa
I min
c2
4
I min c2 Smin (50)(45 103 ) 2.25 106 mm 4
c
4
2
4
c24
c1max
4
c1max
4
I min (50)4
4
(2.25 106 ) 3.3852 106 mm 4
c1max 42.894 mm
tmin c2 c1max 50 42.894 7.106 mm
t 9 mm
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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770
PROBLEM 5.80
20 kN 20 kN 20 kN
B
C
D
A
E
4 @ 0.675 m = 2.7 m
SOLUTION
Two metric rolled-steel channels are to be welded along their edges
and used to support the loading shown. Knowing that the allowable
normal stress for the steel used is 150 MPa, determine the most
economical channels that can be used.
By symmetry, A = E
Fy 0: A E 20 20 20 0
A E 30 kN
Shear:
A to B:
V 30 kN
B to C:
V 30 20 10 kN
C to D:
V 10 20 10 kN
D to E:
V 10 20 30 kN
Areas:
A to B:
(0.675)(30) 20.25 kN m
B to C:
(0.675)(10) 6.75 kN m
C to D:
(0.675)(10) 6.75 kN m
D to E:
(0.675)(30) 20.25 kN m
Bending moments: M A 0
M B 0 20.25 20.25 kN m
M C 20.25 6.75 27 kN m
M D 27 6.75 20.25 kN m
M E 20.25 20.25 0
Maximum M 27 kN m 27 103 N m
all 150 MPa 150 106 Pa
|M |
S min
For each channels,
1
S min (180 103 ) 90 103 mm3
2
Shape
S (103 mm3 )
C180 14.6
100
C150 19.3
all
94.7
27 103
180 106 m3 180 103 mm3
6
150 10
For a section consisting of two channels,
Lightest channel section:
C180 14.6
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771
PROBLEM 5.81
20 kips
2.25 kips/ft
B
Two rolled-steel channels are to be welded back to back and used to
support the loading shown. Knowing that the allowable normal stress
for the steel used is 30 ksi, determine the most economical channels that
can be used.
C
D
A
6 ft
3 ft
12 ft
SOLUTION
M D 0: 12 A 9(20) (6)(2.25)(3) 0
Reaction:
A 18.375 kips
Shear diagram:
A to B:
V 18.375 kips
B to C:
V 18.375 20 1.625 kips
VD 1.625 (6)(2.25) 15.125 kips
Areas of shear diagram:
A to B:
(3)(18.375) 55.125 kip ft
B to C:
(3)(1.625) 4.875 kip ft
C to D:
0.5(6)(1.625 15.125) 50.25 kip ft
Bending moments: M A 0
M B 0 55.125 55.125 kip ft
M C 55.125 4.875 50.25 kip ft
Shape
S (in)3
C10 15.3
13.5
C9 15
11.3
C8 18.7
11.0
M D 50.25 50.25 0
|M |max 55.125 kip ft 661.5 kip in.
all 30 ksi
For double channel, S min
For single channel,
|M |
all
661.5
22.05 in 3
30
S min 0.5(22.05) 11.025 in 3
Lightest channel section: C9 15
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772
PROBLEM 5.82
2000 lb
300 lb/ft
6 in.
C
A
B
3 ft
4 in.
Two L4 × 3 rolled-steel angles are bolted together and used to support
the loading shown. Knowing that the allowable normal stress for the
steel used is 24 ksi, determine the minimum angle thickness that can
be used.
3 ft
SOLUTION
By symmetry, A = C
Fy 0: A C 2000 (6)(300) 0
A C 1900 lb
Shear:
VA 1900 lb 1.9 kips
VB 1900 (3)(300) 1000 lb 1 kip
VB 1000 2000 1000 lb 1 kip
VC 1000 (3)(300) 1900 lb 1.9 kip
Areas:
A to B:
1
(3)(1.9 1) 4.35 kip ft
2
B to C:
1
(3)(1 1.9) 4.35 kip ft
2
Bending moments: M A 0
M B 0 4.35 4.35 kip ft
M C 4.35 4.35 0
Maximum M 4.35 kip ft 52.2 kip in.
all 24 ksi
For section consisting of two angles,
S min
|M |
all
52.2
2.175 in 3
24
For each angle,
1
S min (2.175) 1.0875 in 3
2
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773
PROBLEM 5.82 (Continued)
Angle section
1
2
3
L4 3
8
1
L4 3
4
L4 3
S (in 3 )
1.87
1.44
Smallest allowable thickness:
0.988
t
3
in.
8
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774
PROBLEM 5.83
Total load ⫽ 2 MN
B
C
D D
A
1m
0.75 m
Assuming the upward reaction of the ground to be uniformly distributed and
knowing that the allowable normal stress for the steel used is 170 MPa,
select the most economical wide-flange beam to support the loading shown.
0.75 m
SOLUTION
Downward distributed load:
w
2
2 MN/m
1.0
Upward distributed reaction:
q
2
0.8 MN/m
2.5
Net distributed load over BC:
1.2 MN/m
VA 0
Shear:
VB 0 (0.75)(0.8) 0.6 MN
VC 0.6 (1.0)(1.2) 0.6 MN
VD 0.6 (0.75)(0.8) 0
Areas:
1
(0.75)(0.6)
2
1
(0.5)(0.6)
2
1
(0.5)(0.6)
2
1
(0.75)(0.6)
2
A to B:
B to E:
E to C:
C to D:
0.225 MN m
0.150 MN m
0.150 MN m
0.225 MN m
MA 0
Bending moments:
M B 0 0.225 0.225 MN m
M E 0.225 0.150 0.375 MN m
M C 0.375 0.150 0.225 MN m
M D 0.225 0.225 0
Maximum |M | 0.375 MN m 375 103 N m
all 170 MPa 170 106 Pa
S min
|M |
all
375 103
2.206 103 m3 2206 103 mm3
170 106
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
775
PROBLEM 5.83 (Continued)
Shape
S (103 mm3 )
W690 125
3490
W610 101
2520
W530 150
3720
W460 113
2390
Lightest wide flange section: W610 101
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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776
200 kips
PROBLEM 5.84
200 kips
B
C
A
D D
4 ft
4 ft
Assuming the upward reaction of the ground to be uniformly distributed and
knowing that the allowable normal stress for the steel used is 24 ksi, select
the most economical wide-flange beam to support the loading shown.
4 ft
SOLUTION
q
Distributed reaction:
Shear:
400
33.333 kip/ft
12
VA 0
VB 0 (4)(33.333) 133.33 kips
VB 133.33 200 66.67 kips
VC 66.67 4(33.333) 66.67 kips
VC 66.67 200 133.33 kips
VD 133.33 (4)(33.333) 0 kips
Areas:
A to B:
1
(4)(133.33) 266.67 kip ft
2
B to E:
1
(2)(66.67) 66.67 kip ft
2
E to C:
1
(2)(66.67) 66.67 kip ft
2
C to D:
1
(4)(133.33) 266.67 kip ft
2
Bending moments:
MA 0
M B 0 266.67 266.67 kip ft
M E 266.67 66.67 200 kip ft
M C 200 66.67 266.67 kip ft
M D 266.67 266.67 0
Maximum |M | 266.67 kip ft 3200 kip in.
all 24 ksi
S min
|M |
all
3200
133.3 in 3
24
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
777
PROBLEM 5.84 (Continued)
Shape
S (in 3 )
W27 84
213
W24 68
154
W21 101
227
W18 76
146
W16 77
134
W14 145
232
Lightest W-shaped section: W24 68
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
778
PROBLEM 5.85
60 mm
w
20 mm
D
A
B
0.2 m
C
0.5 m
60 mm
20 mm
Determine the largest permissible distributed load w for the
beam shown, knowing that the allowable normal stress is
80 MPa in tension and 130 MPa in compression.
0.2 m
SOLUTION
By symmetry, B C
Reactions:
Fy 0: B C 0.9w 0
B C 0.45w
VA 0
Shear:
VB 0 0.2w 0.2w
VB 0.2w 0.45w 0.25w
VC 0.25w 0.5w 0.25w
VC 0.25w 0.45w 0.2w
VD 0.2w 0.2w 0
Areas:
1
(0.2)(0.2 w) 0.02 w
2
A to B:
1
(0.25)(0.25w) 0.03125w
2
B to E:
Bending moments:
MA 0
M B 0 0.02 w 0.02 w
M E 0.02 w 0.03125w 0.01125w
Centroid and moment of inertia:
Part
A, mm 2
y , mm
Ay (103 mm3 )
d , mm
Ad 2 (103 mm 4 )
I (103 mm 4 )
1200
70
84
20
480
40
1200
30
36
20
480
360
2400
960
400
Y
120
120 103
50 mm
2400
I Ad 2 I 1360 103 mm 4
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779
PROBLEM 5.85 (Continued)
Top:
I /y (1360 103 )/30 45.333 103 mm3 45.333 106 m3
Bottom:
I /y (1360 103 )/(50) 27.2 103 mm3 27.2 106 m3
Bending moment limits ( M I / y ) and load limits w.
Tension at B and C:
0.02 w (80 106 )(45.333 106 )
w 181.3 103 N/m
Compression at B and C:
0.02 w (130 106 )(27.2 106 )
w 176.8 103 N/m
Tension at E:
0.01125w (80 106 )(27.2 106 )
w 193.4 103 N/m
Compression at E:
0.01125w (130 10)(45.333 106 )
w 523.8 103 N/m
w 176.8 103 N/m
The smallest allowable load controls.
w 176.8 kN/m
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
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780
PROBLEM 5.86
60 mm
w
20 mm
60 mm
D
A
B
0.2 m
C
0.5 m
20 mm
Solve Prob. 5.85, assuming that the cross section of the beam is
inverted, with the flange of the beam resting on the supports at
B and C.
PROBLEM 5.85 Determine the largest permissible distributed
load w for the beam shown, knowing that the allowable normal
stress is 80 MPa in tension and 130 MPa in compression.
0.2 m
SOLUTION
By symmetry, B C
Reactions:
Fy 0: B C 0.9w 0
B C 0.45w
VA 0
Shear:
VB 0 0.2w 0.2 w
VB 0.2w 0.45w 0.25w
VC 0.25w 0.5w 0.25w
VC 0.25w 0.45w 0.2w
VD 0.2w 0.2w 0
Areas:
1
(0.2)(0.2 w) 0.02w
2
1
(0.25)(0.25w) 0.03125w
2
A to B:
B to E:
MA 0
Bending moments:
M B 0 0.02 w 0.02 w
M E 0.02 w 0.03125w 0.01125w
Centroid and moment of inertia:
Part
A, mm 2
y , mm
Ay , (103 mm3 )
d , mm
Ad 2 (103 mm 4 )
I , (103 mm 4 )
1200
50
60
20
480
360
1200
10
12
20
480
40
2400
960
400
72
Y
72 103
30 mm
2400
I Ad 2 I 1360 103 mm3
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
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781
PROBLEM 5.86 (Continued)
Top:
I /y (1360 103 )/(50) 27.2 103 mm3 27.2 106 m3
Bottom:
I /y (1360 103 )/(30) 45.333 108 mm3 45.333 106 m3
Bending moment limits ( M I / y ) and load limits w.
Tension at B and C:
0.02 w (80 106 )(27.2 106 )
w 108.8 103 N/m
Compression at B and C:
0.02 w (130 106 )(45.333 106 )
w 294.7 103 N/m
Tension at E:
0.01125w (80 106 )(45.333 106 )
w 322.4 103 N/m
Compression at E:
0.01125w (130 106 )(27.2 106 )
w 314.3 103 N/m
w 108.8 103 N/m
The smallest allowable load controls.
w 108.8 kN/m
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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782
P
P
10 in.
P
PROBLEM 5.87
1 in.
10 in.
A
E
B
C
60 in.
5 in.
D
7 in.
60 in.
Determine the largest permissible value of P for the beam and
loading shown, knowing that the allowable normal stress is
8 ksi in tension and 18 ksi in compression.
1 in.
SOLUTION
B D 1.5P
Reactions:
Shear diagram:
A to B :
V P
B to C :
V P 1.5P 0.5 P
C to D :
V 0.5 P P 0.5P
V 0.5 P 1.5P P
D to E :
Areas:
(10)( P ) 10 P
A to B :
B to C :
(60)(0.5 P) 30 P
C to D :
(60)(0.5 P) 30 P
(10)( P) 10 P
D to E :
Bending moments:
MA
MB
MC
MD
ME
0
0 10 P 10 P
10 P 30 P 20 P
20 P 30 P 10 P
10 P 10 P 0
Largest positive bending moment: 20P
Largest negative bending moment: 10P
Centroid and moment of inertia:
Part
A, in 2
y0 , in.
Ay0 , in 3
d , in.
Ad 2 , in 4
I , in 4
5
3.5
17.5
1.75
15.3125
10.417
7
0.5
3.5
1.25
10.9375
0.583
12
Y
21
21
1.75 in.
12
Top: y 4.25 in.
26.25
11.000
I Ad 2 I 37.25 in 4
Bottom: y 1.75 in.
My
I
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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783
PROBLEM 5.87 (Continued)
Top, tension:
Top, comp.:
Bottom. tension:
Bottom. comp.:
(10 P)(4.25)
37.25
(20 P)(4.25)
18
37.25
(20 P)(1.75)
8
37.25
(10 P)(1.75)
18
37.25
8
P 7.01 kips
P 7.89 kips
P 8.51 kips
P 38.3 kips
P 7.01 kips
Smallest value of P is the allowable value.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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784
P
P
10 in.
P
PROBLEM 5.88
1 in.
10 in.
A
Solve Prob. 5.87, assuming that the T-shaped beam is inverted.
E
B
C
60 in.
5 in.
D
7 in.
60 in.
PROBLEM 5.87 Determine the largest permissible value of P
for the beam and loading shown, knowing that the allowable
normal stress is 8 ksi in tension and 18 ksi in compression.
1 in.
SOLUTION
B D 1.5 P
Reactions:
Shear diagram:
A to B :
V P
B to C :
V P 1.5P 0.5 P
C to D :
V 0.5 P P 0.5P
D to E :
V 0.5 P 1.5P P
A to B :
B to C :
C to D :
D to E :
(10)( P ) 10 P
(60)(0.5 P) 30 P
(60)(0.5 P) 30 P
(10)( P) 10 P
Areas:
Bending moments:
MA
MB
MC
MD
ME
0
0 10 P 10 P
10 P 30 P 20 P
20 P 30 P 10 P
10 P 10 P 0
Largest positive bending moment 20P
Largest negative bending moment 10P
Centroid and moment of inertia:
Part
A, in 2
y0 , in.
Ay0 , in 3
d , in.
Ad 2 , in 4
I , in 4
5
2.5
12.5
1.75
15.3125
10.417
7
5.5
38.5
1.25
10.9375
0.583
12
51
26.25
51
4.25 in.
12
Top:
y 1.75 in.
Y
Bottom:
y 4.25 in.
I Ad 2 I 37.25 in 4
11.000
My
I
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785
PROBLEM 5.88 (Continued)
Top, tension:
Top, compression:
Bottom, tension:
Bottom, compression:
(10 P)(1.75)
37.25
(20 P)(1.75)
18
37.25
(20 P)(4.25)
8
37.25
(10 P)(4.25)
18
37.25
8
P 17.03 kips
P 19.16 kips
P 3.51 kips
P 15.78 kips
P 3.51 kips
Smallest value of P is the allowable value.
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786
PROBLEM 5.89
12.5 mm
200 mm
w
150 mm
A
B
C
a
D
a
7.2 m
12.5 mm
Beams AB, BC, and CD have the cross section shown
and are pin-connected at B and C. Knowing that the
allowable normal stress is 110 MPa in tension and
150 MPa in compression, determine (a) the largest
permissible value of w if beam BC is not to be
overstressed, (b) the corresponding maximum distance
a for which the cantilever beams AB and CD are not
overstressed.
SOLUTION
M B MC 0
1
VB VC (7.2) w 3.6 w
2
Area B to E of shear diagram:
1
2 (3.6)(3.6w) 6.48w
M E 0 6.48w 6.48w
Centroid and moment of inertia:
Part
A (mm 2 )
y (mm)
Ay (mm3 )
d (mm)
Ad 2 (mm 4 )
I (mm 4 )
2500
156.25
390,625
34.82
3.031 106
0.0326 106
1875
140,625
46.43
4.042 106
3.516 106
4375
7.073 106
3.548 106
75
531,250
531, 250
121.43 mm
4375
I Ad 2 I 10.621 106 mm 4
Y
Location
Top
Bottom
y (mm)
41.07
121.43
I / y (103 mm3 )
also (106 m3 )
258.6
87.47
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787
PROBLEM 5.89 (Continued)
M I / y
Bending moment limits:
Tension at E:
Compression at E:
Tension at A and D:
(110 106 )(87.47 106 ) 9.622 103 N m
(150 106 )(258.6 106 ) 38.8 103 N m
(110 106 )(258.6 106 ) 28.45 103 N m
Compression at A and D: (150 106 )(87.47 106 ) 13.121 103 N m
(a)
Allowable load w:
Shear at A:
6.48w 9.622 103
w 1.485 103 N/m
w 1.485 kN/m
VA (a 3.6) w
Area A to B of shear diagram:
1
1
a (VA VB ) a (a 7.2) w
2
2
1
Bending moment at A (also D): M A a (a 7.2) w
2
1
a(a 7.2)(4.485 103 ) 13.121 103
2
(b)
Distance a:
1 2
a 3.6a 8.837 0
2
a 1.935 m
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788
PROBLEM 5.90
12.5 mm
P
A
P
Beams AB, BC, and CD have the cross section shown
and are pin-connected at B and C. Knowing that the
allowable normal stress is 110 MPa in tension and
150 MPa in compression, determine (a) the largest
permissible value of P if beam BC is not to be
overstressed, (b) the corresponding maximum
distance a for which the cantilever beams AB and CD
are not overstressed.
200 mm
B
C
D
150 mm
a
a
2.4 m 2.4 m 2.4 m
12.5 mm
SOLUTION
M B MC 0
VB VC P
Area B to E of shear diagram: 2.4P
M E 0 2.4 P 2.4 P M F
Centroid and moment of inertia:
Part
A (mm 2 )
y (mm)
Ay (mm3 )
d (mm)
Ad 2 (mm 4 )
I (mm 4 )
2500
156.25
390,625
34.82
3.031 106
0.0326 106
1875
140,625
46.43
4.042 106
3.516 106
4375
7.073 106
3.548 106
75
531,250
531, 250
121.43 mm
4375
I Ad 2 I 10.621 106 mm 4
Y
Location
Top
Bottom
y (mm)
I / y (103 mm3 )
41.07
258.6
121.43
also (106 m3 )
87.47
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789
PROBLEM 5.90 (Continued)
M I / y
Bending moment limits:
Tension at E and F:
(110 106 )(87.47 106 ) 9.622 103 N m
Compression at E and F:
(150 106 )(258.6 106 ) 38.8 103 N m
Tension at A and D:
(110 106 ) (258.6 106 ) 28.45 103 N m
Compression at A and D: (150 106 )(87.47 106 ) 13.121 103 N m
(a)
Allowable load P:
Shear at A:
(b)
2.4 P 9.622 103
P 4.01 103 N
VA P
Area A to B of shear diagram:
aVA aP
Bending moment at A:
M A aP 4.01 103a
Distance a:
P 4.01 kN
4.01 103 a 13.121 103
a 3.27 m
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790
PROBLEM 5.91
C
12 ft
D
3
2
4 ft
A
8 ft
1
B
8 ft
Each of the three rolled-steel beams
shown (numbered 1, 2, and 3) is to
carry a 64-kip load uniformly
distributed over the beam. Each of
these beams has a 12-ft span and is
to be supported by the two 24-ft
rolled-steel girders AC and BD.
Knowing that the allowable normal
stress for the steel used is 24 ksi,
select (a) the most economical S
shape for the three beams, (b) the
most economical W shape for the
two girders.
4 ft
SOLUTION
For beams 1, 2, and 3,
1
Maximum M (6)(32) 96 kip ft 1152 kip in.
2
S min
M
all
1152
48 in 3
24
Shape
S15 42.9
S (in 3 )
59.4
S12 50
50.6
(a) Use S15 42.9.
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791
PROBLEM 5.91 (Continued)
For beams AC and BC,
areas under shear digram:
(4)(48) 192 kip ft
(8)(16) 128 kip ft
Maximum M 192 128 320 kip ft 3840 kip in.
S min
M
all
Shape
W30 99
W27 84
W24 104
W21 101
W18 106
3840
160 in 3
24
S (in 3 )
269
213
258
227
204
(b) Use W27 84.
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792
PROBLEM 5.92
54 kips
l/2
W12 3 50
l/2
C
D
B
A
L 516 ft
A 54-kip is load is to be supported at the center of the 16-ft span shown.
Knowing that the allowable normal stress for the steel used is 24 ksi,
determine (a) the smallest allowable length l of beam CD if the
W12 50 beam AB is not to be overstressed, (b) the most economical
W shape that can be used for beam CD. Neglect the weight of both
beams.
SOLUTION
(a)
d 8ft
l
2
l 16ft 2d
(1)
Beam AB (Portion AC):
For W12 50,
S x 64.2 in 3 all 24 ksi
M all all S x (24)(64.2) 1540.8 kip in. 128.4 kip ft
M C 27d 128.4 kip ft
Using (1),
(b)
d 4.7556 ft
l 6.49 ft
l 16 2d 16 2(4.7556) 6.4888 ft
Beam CD:
l 6.4888 ft all 24 ksi
M max
(87.599 12) kip in.
Smin
all
24 ksi
43.800 in 3
S (in 3 )
Shape
W18 35
W16 31
W14 38
W12 35
W10 45
57.6
47.2
54.6
45.6
49.1
←
W16 31
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793
66 kN/m
PROBLEM 5.93
66 kN/m
W460 3 74
A
B
C
D
l
L56m
A uniformly distributed load of 66 kN/m is to be supported over
the 6-m span shown. Knowing that the allowable normal stress for
the steel used is 140 MPa, determine (a) the smallest allowable
length l of beam CD if the W460 74 beam AB is not to be
overstressed, (b) the most economical W shape that can be used
for beam CD. Neglect the weight of both beams.
SOLUTION
For W460 74,
S 1460 103 mm3 1460 106 m3
all 140 MPa 140 106 Pa
M all S all (1460 106 )(140 106 )
204.4 103 N m 204.4 kN m
A B,
Reactions: By symmetry,
CD
Fy 0: A B (6)(66) 0
A B 198 kN 198 103 N
Fy 0: C D 66l 0
C D (33l ) kN
(1)
Shear and bending moment in beam AB:
0 x a,
V 198 66 x kN
M 198 x 33x 2 kN m
At C, x a.
M M max
M 198a 33a 2 kN m
Set M M all .
198a 33a 2 204.4
33a 2 198a 204.4 0
a 4.6751 m ,
(a)
By geometry,
From (1),
1.32487 m
l 6 2a 3.35 m
l 3.35 m
C D 110.56 kN
Draw shear and bending moment diagrams for beam CD. V 0 at point E, the midpoint of CD.
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794
PROBLEM 5.93 (Continued)
1
1
V dx 2 (110.560) 2 l 92.602 kN m
Area from A to E:
M E 92.602 kN m 92.602 103 N m
S min
ME
all
92.602 103
661.44 106 m3
140 106
661.44 103 mm3
Shape
S (103 mm3 )
W410 46.1
773
W360 44
688
W310 52
747
W250 58
690
W200 71
708
(b) Use W360 44.
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795
PROBLEM 5.94
wD wL
b
A
B
h
C
1
2
1
2
L
L
P
A roof structure consists of plywood and roofing material
supported by several timber beams of length L 16 m. The
dead load carried by each beam, including the estimated weight
of the beam, can be represented by a uniformly distributed load
wD 350 N/m. The live load consists of a snow load,
represented by a uniformly distributed load wL 600 N/m, and
a 6-kN concentrated load P applied at the midpoint C of each
beam. Knowing that the ultimate strength for the timber used is
U 50 MPa and that the width of the beam is b 75 mm,
determine the minimum allowable depth h of the beams, using
LRFD with the load factors D 1.2, L 1.6 and the
resistance factor 0.9.
SOLUTION
L 16 m, wD 350 N/m 0.35 kN/m
wL 600 N/m 0.6 kN/m, P 6 kN
Dead load:
1
RA (16)(0.35) 2.8 kN
2
Area A to C of shear diagram:
1
2 (8)(2.8) 11.2 kN m
Bending moment at C:
11.2 kN m 11.2 103 N m
Live load:
1
RA [(16)(0.6) 6] 7.8 kN
2
Shear at C :
V 7.8 (8)(0.6) 3 kN
Area A to C of shear diagram:
1
2 (8)(7.8 3) 43.2 kN m
Bending moment at C:
43.2 kN m 43.2 103 N m
Design:
D M D L M L M U U S
S
D M D L M L (1.2)(11.2 103 ) (1.6)(43.2 103 )
U
(0.9)(50 106 )
1.8347 103 m3 1.8347 106 mm3
For a rectangular section,
1
S bh 2
6
h
6S
(6)(1.8347 106 )
b
75
h 383 mm
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796
PROBLEM 5.95
Solve Prob. 5.94, assuming that the 6-kN concentrated load P
applied to each beam is replaced by 3-kN concentrated loads P1
and P2 applied at a distance of 4 m from each end of the beams.
wD wL
b
A
B
C
1
2
1
2
L
P
L
h
PROBLEM 5.94 A roof structure consists of plywood and
roofing material supported by several timber beams of length
L 16 m. The dead load carried by each beam, including the
estimated weight of the beam, can be represented by a uniformly
distributed load wD 350 N/m. The live load consists of a snow
load, represented by a uniformly distributed load wL 600 N/m,
and a 6-kN concentrated load P applied at the midpoint C of each
beam. Knowing that the ultimate strength for the timber used is
U 50 MPa and that the width of the beam is b 75 mm,
determine the minimum allowable depth h of the beams, using
LRFD with the load factors D 1.2, L 1.6 and the resistance
factor 0.9.
SOLUTION
L 16 m,
a 4 m,
wD 350 N/m 0.35 kN/m
wL 600 N/m 0.6 kN/m, P 3 kN
Dead load:
1
RA (16)(0.35) 2.8 kN
2
Area A to C of shear diagram:
1
2 (8)(2.8) 11.2 kN m
Bending moment at C:
11.2 kN m 11.2 103 N m
Live load:
1
RA [(16)(0.6) 3 3] 7.8 kN
2
Shear at D :
7.8 (4)(0.6) 5.4 kN
Shear at D :
5.4 3 2.4 kN
Area A to D:
1
2 (4)(7.8 5.4) 26.4 kN m
Area D to C:
1
2 (4)(2.4) 4.8 kN m
Bending moment at C:
26.4 4.8 31.2 kN m
31.2 103 N m
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797
PROBLEM 5.95* (Continued)
D M D L M L M U U S
Design:
S
D M D L M L (1.2)(11.2 103 ) (1.6)(31.2 103 )
U
(0.9)(50 106 )
1.408 103 m3 1.408 106 mm3
For a rectangular section,
1
S bh 2
6
h
6S
(6)(1.408 106 )
b
75
h 336 mm
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798
PROBLEM 5.96
x
P1
P2
a
A
B
L
A bridge of length L 48 ft is to be built on a secondary road whose
access to trucks is limited to two-axle vehicles of medium weight. It
will consist of a concrete slab and of simply supported steel beams
with an ultimate strength of U 60 ksi. The combined weight of
the slab and beams can be approximated by a uniformly distributed
load w 0.75 kips/ft on each beam. For the purpose of the design, it
is assumed that a truck with axles located at a distance a 14 ft from
each other will be driven across the bridge and that the resulting
concentrated loads P1 and P2 exerted on each beam could be as
large as 24 kips and 6 kips, respectively. Determine the most
economical wide-flange shape for the beams, using LRFD with the
load factors D 1.25, L 1.75 and the resistance factor 0.9.
[Hint: It can be shown that the maximum value of |M L | occurs under
the larger load when that load is located to the left of the center of the
beam at a distance equal to aP2 /2( P1 P2 ) .]
SOLUTION
L 48 ft a 14 ft
P1 24 kips
P2 6 kips W 0.75 kip/ft
Dead load:
1
RA RB (48)(0.75) 18 kips
2
Area A to E of shear diagram:
1
(8)(18) 216 kip ft
2
M max 216 kip ft 2592 kip in. at point E.
u
Live load:
aP2
(14)(6)
1.4 ft
2( P1 P2 ) (2)(30)
L
u 24 1.4 22.6 ft
2
x a 22.6 14 36.6 ft
x
L x a 48 36.6 11.4 ft
M B 0: 48RA (25.4)(24) (11.4)(6) 0
RA 14.125 kips
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799
PROBLEM 5.96* (Continued)
Shear:
A to C:
V 14.125 kips
C to D:
V 14.125 24 9.875 kips
D to B:
V 15.875 kips
Area:
(22.6)(14.125) 319.225 kip ft
A to C:
M C 319.225 kip ft 3831 kip in.
Bending moment:
Design:
D M D L M L M U U Smin
Smin
DM D LM L
U
(1.25)(2592) (1.75)(3831)
(0.9)(60)
184.2 in 3
Shape
S (in 3 )
W30 99
269
W27 84
213
W24 104
258
W21 101
227
W18 106
204
Use W27 84.
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800
PROBLEM 5.97*
Assuming that the front and rear axle loads remain in the same ratio
as for the truck of Prob. 5.96, determine how much heavier a truck
could safely cross the bridge designed in that problem.
x
P1
a
PROBLEM 5.96 A bridge of length L 48 ft is to be built on a
secondary road whose access to trucks is limited to two-axle vehicles
of medium weight. It will consist of a concrete slab and of simply
supported steel beams with an ultimate strength of U 60 ksi. The
combined weight of the slab and beams can be approximated by a
uniformly distributed load w 0.75 kips/ft on each beam. For the
purpose of the design, it is assumed that a truck with axles located at
a distance a 14 ft from each other will be driven across the bridge
and that the resulting concentrated loads P1 and P2 exerted on each
beam could be as large as 24 kips and 6 kips, respectively. Determine
the most economical wide-flange shape for the beams, using LRFD
with the load factors D 1.25, L 1.75 and the resistance factor
0.9. [Hint: It can be shown that the maximum value of |M L |
occurs under the larger load when that load is located to the left of
the center of the beam at a distance equal to aP2 /2( P1 P2 ) .]
P2
A
B
L
SOLUTION
L 48 ft
a 14 ft
P1 24 kips
P2 6 kips W 0.75 kip/ft
See solution to Prob. 5.96 for calculation of the following:
M D 2592 kip in.
M L 3831 kip in.
For rolled-steel section W27 84, S 213 in 3
Allowable live load moment M L* :
D M D L M L* M U U S
S D M D
M L* U
L
(0.9)(60)(213) (1.25)(2592)
1.75
4721 kip in.
Ratio:
M L* 4721
1.232 1 0.232
M L 3831
Increase: 23.2%
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801
PROBLEM 5.98
w0
B
A
a
(a) Using singularity functions, write the equations defining the shear
and bending moment for the beam and loading shown. (b) Use the
equation obtained for M to determine the bending moment at point C and
check your answer by drawing the free-body diagram of the entire beam.
C
a
SOLUTION
w w0 w0 x a 0
(a)
V w0 x w0 x a1
dV
dx
dM
dx
1
1
M w0 x 2 w0 x a 2
2
2
At point C,
(b)
x 2a
1
1
M C w0 (2a ) 2 w0 a 2
2
2
Check:
3
M C w0 a 2
2
3a
M C 0: ( w0 a ) M C 0
2
3
M C w0 a 2
2
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802
PROBLEM 5.99
w0
B
A
a
C
a
(a) Using singularity functions, write the equations defining the shear
and bending moment for the beam and loading shown. (b) Use the
equation obtained for M to determine the bending moment at point C
and check your answer by drawing the free-body diagram of the entire
beam.
SOLUTION
w0 x
w
w0 x a 0 0 x a1
a
a
dV
dx
w
(a)
V
w0 x 2
w
dM
w0 x a1 0 x a 2
dx
2a
2a
M
At point C,
(b)
MC
x 2a
w0 (2a )3 w0 a 2 w0 a3
6a
2
6a
Check:
w0 x3 w0
w
x a 2 0 x a 3
6a
2
6a
2
M C w0 a 2
3
4a 1
M C 0: w0 a M C 0
3 2
2
M C w0 a 2
3
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803
PROBLEM 5.100
w0
B
A
C
a
a
(a) Using singularity functions, write the equations defining the shear
and bending moment for the beam and loading shown. (b) Use the
equation obtained for M to determine the bending moment at point C
and check your answer by drawing the free-body diagram of the entire
beam.
SOLUTION
w w0
(a)
V w0 x
w0 x w0
x a1
a
a
dV
dx
w0 x 2 w0
dM
x a 2
dx
2a
2a
M
At point C,
(b)
MC
x 2a
w0 (2a )2 w0 (2a )3 w0 a 3
2
6a
6a
Check:
w0 x 2 w0 x3 w0
x a 3
2
6a
6a
5
M C w0 a 2
6
5 1
M C 0: a w0 a M C 0
3 2
5
M C w0 a 2
6
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804
w
PROBLEM 5.101
w
B
C
E
A
D
a
a
a
a
(a) Using singularity functions, write the equations defining the shear
and bending moment for the beam and loading shown. (b) Use the
equation obtained for M to determine the bending moment at point E,
and check your answer by drawing the free-body diagram of the portion
of the beam to the right of E.
SOLUTION
w w w x a 0 w x 3a 0
(a)
V wa w dx
wa wx w x a1 w x 3a1
M V dx
wax wx 2 /2 ( w/2) x a 2 ( w/2) x 3a 2
(b)
At point E, x 3a
M E wa(3a ) w(3a) 2 /2 ( w/2)(2a )2
a
M E 0: wa (a ) ( wa ) M E 0
2
M E wa 2 /2
wa 2 /2
(Checks)
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805
PROBLEM 5.102
w0
B
A
D
C
a
a
E
a
a
(a) Using singularity functions, write the equations defining the shear
and bending moment for the beam and loading shown. (b) Use the
equation obtained for M to determine the bending moment at point E,
and check your answer by drawing the free-body diagram of the portion
of the beam to the right of E.
SOLUTION
a
M C 0: 2aA (3aw0 ) 0
2
3
A w0 a
4
5a
M A 0: 2aC (3aw0 ) 0
2
C
w w0 x a 0
15
w0 a
4
dV
dx
(a)
V w0 x a1
3
15
dM
w0 a w0 a x 2a 0
4
4
dx
1
3
15
M w0 x a 2 w0 ax w0 a x 2a1 0
2
4
4
At point E ,
(b)
x 3a
1
3
15
M E w0 (2a )2 w0 a(3a ) w0 a(a )
2
4
4
1
M E w0 a 2
2
Check:
M E 0: M E
a
( w0 a ) 0
2
1
M E w0 a 2
2
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806
P
B
A
a
PROBLEM 5.103
P
C
a
E
a
D
a
(a) Using singularity functions, write the equations defining the shear
and bending moment for the beam and loading shown. (b) Use the
equation obtained for M to determine the bending moment at point E,
and check your answer by drawing the free-body diagram of the portion
of the beam to the right of E.
SOLUTION
M D 0: 4aA 3aP 2aP 0
(a)
A 1.25 P
V 1.25 P P x a 0 P x 2a 0
M 1.25Px P x a1 P x 2a1
(b)
x 3a
At point E ,
M E 1.25 P(3a) P(2a) P( a) 0.750 Pa
Fy 0: A P P D 0
Reaction:
D 0.750 P
M E 0: M E 0.750 Pa 0
M E 0.750 Pa
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807
P
C
A
B
a
PROBLEM 5.104
(a) Using singularity functions, write the equations for the shear and
bending moment for beam ABC under the loading shown. (b) Use the
equation obtained for M to determine the bending moment just to the right
of point B.
a
P
SOLUTION
(a)
M C 0:
RA
(2a ) P aP 2( Pa ) 2aRA 0
1
P
2
V ( RA P ) P x a 0
1
P P x a 0
2
1
dM
P P x a 0
dx
2
1
M Px P x a1 Pa Pa x a 0
2
(b) Just to the right of point B, x a
1
3
M Pa 0 Pa Pa Pa
2
2
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808
PROBLEM 5.105
P
(a) Using singularity functions, write the equations for the shear and
bending moment for beam ABC under the loading shown. (b) Use the
equation obtained for M to determine the bending moment just to the
right of point B.
A
B
a
C
a
SOLUTION
(a)
V P x a 0
dM
P x a 0
dx
M P x a1 Pa x a 0
Just to the right of B, x a1.
(b)
M 0 Pa
M Pa
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809
48 kN
B
60 kN
PROBLEM 5.106
60 kN
C
D
A
E
1.5 m
(a) Using singularity functions, write the equations for the shear and
bending moment for the beam and loading shown. (b) Determine the
maximum value of the bending moment in the beam.
1.5 m
0.6 m
0.9 m
SOLUTION
M E 0: 4.5 RA (3.0)(48) (1.5)(60) (0.9)(60) 0
RA 40 kN
(a)
V 40 48 x 1.5 0 60 x 3.0 0 60 x 3.6 0 kN
M 40 x 48 x 1.51 60 x 3.01 60 x 3.61 kN m
Pt.
x(m)
M (kN m)
A
0
0
B
1.5
(40)(1.5) 60 kN m
C
3.0
(40)(3.0) (48)(1.5) 48 kN m
D
3.6
(40)(3.6) (48)(2.1) (60)(0.6) 7.2 kN m
E
4.5
(40)(4.5) (48)(3.0) (60)(1.5) (60)(0.9) 0
M max 60 kN m
(b)
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810
3 kips
6 kips
C
PROBLEM 5.107
6 kips
D
E
A
B
3 ft
4 ft
4 ft
(a) Using singularity functions, write the equations for the shear and
bending moment for the beam and loading shown. (b) Determine the
maximum value of the bending moment in the beam.
4 ft
SOLUTION
M B 0: (15)(3) 12C (8)(6)C (4)(6) 0
C 9.75 kips
(a)
V 3 9.75 x 3 0 6 x 7 0 6 x 11 0 kips
M 3x 9.75 x 31 6 x 71 6 x 111 kip ft
M (kip ft)
Pt.
x(ft)
A
0
C
3
(3)(3) 9
D
7
(3)(7) (9.75)(4) 18
E
11
(3)(11) (9.75)(8) (6)(4) 21
B
15
(3)(15) (9.75)(12) (6)(8) (6)(4) 0
0
maximum
|M |max 21.0 kip ft
(b)
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811
PROBLEM 5.108
25 kN/m
B
C
A
D
40 kN
0.6 m
40 kN
1.8 m
(a) Using singularity functions, write the equations for the shear and
bending moment for the beam and loading shown. (b) Determine the
maximum value of the bending moment in the beam.
0.6 m
SOLUTION
(a)
By symmetry, RA RD
Fy 0:
RA RD 40 (1.8)(25) 40 0
RA RD 62.5 kN
w 25 x 0.6 0 25 x 2.4 0
(b)
dV
dx
V 62.5 25 x 0.61 25 x 2.41 40 x 0.6 0 40 x 2.4 0 kN
M 62.5 x 12.5 x 0.6 2 12.5 x 2.4 2 40 x 0.61 40 x 2.41 kN m
Locate point where V 0 .
Assume 0.6 x* 1.8
0 62.5 25( x* 0.6) 0 40 0
x* 1.5 m
M (62.5)(1.5) (12.5)(0.9) 2 0 (40)(0.9) 0 47.6 kN m
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812
8 kips
3 kips/ft
C
D
E
A
3 ft
4 ft
PROBLEM 5.109
3 kips/ft
4 ft
B
(a) Using singularity functions, write the equations for the shear and
bending moment for the beam and loading shown. (b) Determine the
maximum value of the bending moment in the beam.
3 ft
SOLUTION
M B 0: 14 A (12.5)(3)(3) (7)(8) (1.5)(3)(3) 0
A 13 kips
w 3 3 x 3 0 3 x 11 0
(a)
dV
dx
V 13 3x 3 x 31 8 x 7 0 3 x 111 kips
M 13x 1.5 x 2 1.5 x 3 2 8 x 71 1.5 x 11 2 kip ft
VC 13 (3)(3) 4 kips
VD 13 (3)(7) (3)(4) 4 kips
VD 13 (3)(7) (3)(4) 8 4 kips
VE 13 (3)(11) (3)(8) 8 4 kips
VB 13 (3)(14) (3)(11) 8 (3)(3) 13 kips
(b)
Note that V changes sign at D.
|M |max M D (13)(7) (1.5)(7) 2 (1.5)(4)2 0 0
|M |max 41.5 kip ft
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813
24 kN
24 kN
B
C
E
D
A
W250 28.4
F
4 @ 0.75 m 3 m
PROBLEM 5.110
24 kN
24 kN
(a) Using singularity functions, write the equations for
the shear and bending moment for the beam and
loading shown. (b) Determine the maximum normal
stress due to bending.
0.75 m
SOLUTION
M E 0: 3RA (2.25)(24) (1.5)(24) (0.75)(24) (0.75)(24) 0
RA 30 kips
M A 0: (0.75)(24) (1.5)(24) (2.25)(24) 3RE (3.75)(24) 0
RE 66 kips
(a)
V 30 24 x 0.75 0 24 x 1.5 0 24 x 2.25 0 66 x 3 0 kN
M 30 x 24 x 0.751 24 x 1.51 24 x 2.251 66 x 31 kN m
Point
x(m)
M (kN m)
B
0.75
(30)(0.75) 22.5 kN m
C
1.5
(30)(1.5) (24)(0.75) 27 kN m
D
2.25
(30)(2.25) (24)(1.5) (24)(0.75) 13.5 kN m
E
3.0
(30)(3.0) (24)(2.25) (24)(1.5) (24)(0.75) 18 kN m
F
3.75
(30)(3.75) (24)(3.0) (24)(2.25) (24)(1.5) (66)(0.75) 0
Maximum |M | 27 kN m 27 303 N m
For rolled-steel section W250 28.4, S 308 103 mm3 308 106 m3
(b)
Normal stress:
|M |
27 103
87.7 106 Pa
S
308 106
87.7 MPa
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814
50 kN
125 kN
B
PROBLEM 5.111
50 kN
C
D
A
S150 18.6
E
0.3 m
0.4 m
0.5 m
(a) Using singularity functions, write the equations for
the shear and bending moment for the beam and loading
shown. (b) Determine the maximum stress due to bending.
0.2 m
SOLUTION
M D 0: (1.2)(50) 0.9 B (0.5)(125) (0.2)(50) 0
B 125 kN
M B 0: (0.3)(50) (0.4)(125) 0.9 D (1.1)(50) 0
D 100 kN
(a)
V 50 125 x 0.3 0 125 x 0.7 0 100 x 1.2 0 kN
M 50 x 125 x 0.31 125 x 0.71 100 x 1.21 kN m
Point
x(m)
M (kN m)
B
0.3
(50)(0.3) 0 0 0 15 kN m
C
0.7
(50)(0.7) (125)(0.4) 0 0 15 kN m
D
1.2
(50)(1.2) (125)(0.9) (125)(0.5) 0 10 kN m
E
1.4
(50)(1.4) (125)(1.1) (125)(0.7) (100)(0.2) 0 (checks)
Maximum M 15 kN m 15 103 N m
For S150 18.6 rolled-steel section, S 120 103 mm3 120 106 m3
(b)
Normal stress:
M
S
15 103
125 106 Pa
6
120 10
125.0 MPa
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815
PROBLEM 5.112
40 kN/m
18 kN ? m
27 kN ? m
B
S310 52
C
A
1.2 m
2.4 m
(a) Using singularity functions, find the
magnitude and location of the maximum
bending moment for the beam and loading
shown. (b) Determine the maximum normal
stress due to bending.
SOLUTION
M c 0: 18 3.6 A (1.2)(2.4)(40) 27 0
A 29.5 kN
1
V 29.5 40 x 1.2 kN
Point D.
V 0
29.5 40( xD 1.2) 0
xD 1.9375 m
M 18 29.5 x 20 x 1.2
2
kN m
M A 18 kN m
M D 18 (29.5)(1.9375) (20)(0.7375)2 28.278 kN m
M E 18 (29.5)(3.6) (20)(2.4)2 27 kN m
(a)
Maximum M 28.3 kN m at x 1.938 m
S 624 103 mm3
For S310 52 rolled-steel section,
624 106 m3
(b)
Normal stress:
M
S
28.278 103
45.3 106 Pa
624 106
45.3 MPa
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816
60 kN
PROBLEM 5.113
60 kN
40 kN/m
B
A
C
1.8 m
D
1.8 m
W530 66
(a) Using singularity functions, find the magnitude and
location of the maximum bending moment for the beam and
loading shown. (b) Determine the maximum normal stress
due to bending.
0.9 m
SOLUTION
M B 0: 4.5A (2.25)(4.5)(40) (2.7)(60) (0.9)(60) 0
A 138 kN
M A 0: (2.25)(4.5)(40) (1.8)(60) (3.6)(60) 4.5 B 0
B 162 kN
w 40 kN/m
dV
dx
V 40 x 138 60 x 1.8 0 60 x 3.6 0
dM
dx
M 20 x 2 138 x 60 x 1.81 60 x 3.61
VC (40)(1.8) 138 60 6 kN
VD (40)(3.6) 138 60 66 kN
Locate point E where V 0. It lies between C and D.
VE 40 xE 138 60 0 0
xE 1.95 m
M E (20)(1.95) 2 (138)(1.95) (60)(1.95 1.8) 184 kN m
|M |max 184 kN m 184 103 N m at
(a)
x 1.950 m
For W530 66 rolled-steel section, S 1340 103 mm3 1340 106 m3
(b)
Normal stress:
|M |max
184 103
137.3 106 Pa
S
1340 106
137.3 MPa
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817
12 kips
PROBLEM 5.114
12 kips
2.4 kips/ft
A
B
D
C
6 ft
6 ft
A beam is being designed to be supported and loaded as shown. (a) Using
singularity functions, find the magnitude and location of the maximum
bending moment in the beam. (b) Knowing that the allowable normal
stress for the steel to be used is 24 ksi, find the most economical wideflange shape that can be used.
3 ft
SOLUTION
M C 0: 15RA (7.5)(15)(2.4) (9)(12) (3)(12) 0
RA 27.6 kips
w 2.4 kips/ft
dV
dx
V 27.6 2.4 x 12 x 6 0 12 x 12 0 kips
VB 27.6 (2.4)(6) 13.2 kips
VB 27.6 (2.4)(6) 12 1.2 kips
VC
Point where V 0
27.6 (2.4)(12) 12 13.2 kips lies between B and C.
Locate point E where V 0.
0 27.6 2.4 xE 12 0
xE 6.50 ft
M 27.6 x 1.2 x 2 12 x 61 12 x 121 kip ft
M (27.6)(6.5) (1.2)(6.5) 2 (12)(0.5) 0
( x 6.5 ft)
At point E,
122.70 kip ft 1472.40 kip in.
Maximum |M | 122.7 kip ft at
S min
M
all
Shape
x 6.50 ft
1472.40
61.35 in 3
24
S (in 3 )
W21 44
W18 50
81.6
88.9
W16 40
64.7
W14 43
W12 50
W10 68
62.6
64.2
75.7
Answer: W16 40
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818
PROBLEM 5.115
22.5 kips
3 kips/ft
C
A
B
3 ft
12 ft
A beam is being designed to be supported and loaded as shown. (a) Using
singularity functions, find the magnitude and location of the maximum
bending moment in the beam. (b) Knowing that the allowable normal
stress for the steel to be used is 24 ksi, find the most economical wideflange shape that can be used.
SOLUTION
M C 0: 15 RA (7.5)(15)(3) (12)(22.5) 0
R A 40.5 kips
w 3 kips/ft
dV
dx
V 40.5 3x 22.5 x 3 0 kips
M 40.5 x 1.5 x 2 22.5 x 31 kip ft
(a)
Location of point D where V 0.
Assume 3 ft xD 12 ft.
0 40.5 3xD 22.5
At point D, ( x 6 ft).
xD 6 ft
M (40.5)(6) (1.5)(6) 2 (22.5)(3)
121.5 kip ft 1458 kip in.
|M |max 121.5 kip ft at
Maximum |M |:
S min
(b)
Shape
W21 44
W18 50
W16 40
W14 43
W12 50
W10 68
M
all
x 6.00 ft
1458
60.75 in 3
24
S (in 3 )
81.6
88.9
64.7
62.6
64.2
75.7
Wide-flange shape: W16 40
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819
PROBLEM 5.116
480 N/m
30 mm
A
C
A timber beam is being designed with supports and loads as
shown. (a) Using singularity functions, find the magnitude and
location of the maximum bending moment in the beam.
(b) Knowing that the available stock consists of beams with an
allowable normal stress of 12 MPa and a rectangular cross section
of 30-mm width and depth h varying from 80 mm to 160 mm in
10-mm increments, determine the most economical cross section
that can be used.
h
C
B
1.5 m
2.5 m
SOLUTION
480 N/m 0.48 kN/m
M C 0:
1
4 RA (3) (1.5)(0.48) (1.25)(2.5)(0.48) 0
2
R A 0.645 kN
0.48
0.48
dV
x
x 1.51 0.32 x 0.32 x 1.51 kN/m
1.5
1.5
dx
2
2
V 0.645 0.16 x 0.16 x 1.5 kN
w
M 0.645 x 0.053333x3 0.053333 x 1.5 3 kN m
(a)
Locate point D where V 0.
Assume 1.5 m xD 4 m.
0 0.645 0.16 xD2 0.16( xD 1.5) 2
0.645 0.16 xD2 0.16 xD2 0.48 xD 0.36
xD 2.0938 m
At point D,
xD 2.09 m
3
M D (0.645)(2.0938) (0.053333)(2.0938) (0.053333)(0.59375)3
M D 0.87211
M D 0.872 kN m
S min
MD
all
1
S bh 2
6
For a rectangular cross section,
hmin
(b)
0.87211 103
72.676 106 m3 72.676 103 mm3
6
12 10
h
6S
b
(6)(72.676 103 )
120.562 mm
30
h 130 mm
At next larger 10-mm increment,
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820
PROBLEM 5.117
500 N/m
30 mm
A
C
C
h
B
1.6 m
2.4 m
A timber beam is being designed with supports and loads as
shown. (a) Using singularity functions, find the magnitude and
location of the maximum bending moment in the beam.
(b) Knowing that the available stock consists of beams with an
allowable stress of 12 MPa and a rectangular cross section of
30-mm width and depth h varying from 80 mm to 160 mm in
10-mm increments, determine the most economical cross section
that can be used.
SOLUTION
500 N/m 0.5 kN/m
1
M C 0: 4 RA (3.2)(1.6)(0.5) (1.6) (2.4)(0.5) 0
2
R A 0.880 kN
0.5
dV
x 1.61 0.5 0.20833 x 1.61 kN/m
2.4
dx
2
V 0.880 0.5 x 0.104167 x 1.6 kN
w 0.5
VA 0.880 kN
VB 0.880 (0.5)(1.6) 0.080 kN
Sign change
VC 0.880 (0.5)(4) (0.104167)(2.4) 0.520 kN
2
Locate point D (between B and C) where V 0.
0 0.880 0.5 xD 0.104167 ( xD 1.6)2
0.104167 xD2 0.83333xD 1.14667 0
xD
0.83333 (0.83333)2 (4)(0.104167)(1.14667)
6.2342 , 1.7658 m
(2)(0.104167)
M 0.880 x 0.25 x 2 0.347222 x 1.6 3 kN m
M D (0.880)(1.7658) (0.25)(1.7658)2 (0.34722)(0.1658)3 0.77597 kN m
M max 0.776 kN m at
(a)
S min
M max
all
0.77597 103
64.664 106 m3 64.664 103 mm3
6
12 10
1
For a rectangular cross section, S bh 2
6
(b)
x 1.766 m
h
6S
b
hmin
(6)(64.664 103 )
113.7 mm
30
h 120 mm
At next higher 10-mm increment,
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821
12 kN
L 0.4 m
PROBLEM 5.118
16 kN/m
Using a computer and step functions, calculate the shear and bending
moment for the beam and loading shown. Use the specified increment
ΔL, starting at point A and ending at the right-hand support.
B
C
A
1.2 m
4m
SOLUTION
M C 0: (5.2)(12) 4 B (2)(4)(16) 0
B 47.6 kN
M B 0: (1.2)(12) (2)(4)(16) 4C 0
dV
w 16 x 1.2
dx
C 28.4 kN
0
V 16 x 1.21 12 47.6 x 1.2 0
M 8 x 1.2 2 12 x 47.6 x 1.21
x
V
M
m
kN
kN m
0.0
12.0
0.00
0.4
12.0
4.80
0.8
12.0
9.60
1.2
35.6
14.40
1.6
29.2
1.44
2.0
22.8
8.96
2.4
16.4
16.80
2.8
10.0
22.08
3.2
3.6
24.80
3.6
2.8
24.96
4.0
9.2
22.56
4.4
15.6
17.60
4.8
22.0
10.08
5.2
28.4
0.00
V
M
max
max
35.6 kN
25.0 kN m
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
822
L 0.25 m
120 kN
B
A
2m
PROBLEM 5.119
36 kN/m
C
1m
Using a computer and step functions, calculate the shear and bending
moment for the beam and loading shown. Use the specified increment ΔL,
starting at point A and ending at the right-hand support.
D
3m
SOLUTION
1
M D 0: 6 RA (4)(120) (1) (3)(36) 0
2
RA 89 kN
36
w x 31 12 x 31
3
x
m
0.0
0.3
0.5
0.8
1.0
1.3
1.5
1.8
2.0
2.3
2.5
2.8
3.0
3.3
3.5
3.8
4.0
4.3
4.5
4.8
V
kN
89.0
89.0
89.0
89.0
89.0
89.0
89.0
89.0
31.0
31.0
31.0
31.0
31.0
31.4
32.5
34.4
37.0
40.4
44.5
49.4
V 89 120 x 2 0 6 x 3 2 kN
M 89 x 120 x 21 2 x 3 3 kN m
M
kN m
0.0
22.3
44.5
66.8
89.0
111.3
133.5
155.8
178.0
170.3
162.5
154.8
147.0
139.2
131.3
122.9
114.0
104.3
93.8
82.0
x
V
M
m
kN
kN m
5.0
55.0
69.0
5.3
61.4
54.5
5.5
68.5
38.3
5.8
76.4
20.2
6.0
85.0
0.0
V
M
max
max
89.0 kN
178.0 kN m
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
823
3.6 kips/ft
L 0.5 ft
1.8 kips/ft
A
C
B
6 ft
PROBLEM 5.120
Using a computer and step functions, calculate the shear and bending
moment for the beam and loading shown. Use the specified increment
ΔL, starting at point A and ending at the right-hand support.
6 ft
SOLUTION
1
M C 0: 12 RA (6)(12)(1.8) (10) (6)(1.8) 0
2
RA 15.3 kips
1.8
1.8
w 3.6
x
x 61
6
6
3.6 0.3x 0.3 x 61
V 15.3 3.6 x 0.15 x 2 0.15 x 6 2 kips
x
ft
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
8.5
9.0
V
kips
15.30
13.54
11.85
10.24
8.70
7.24
5.85
4.54
3.30
2.14
1.05
0.04
0.90
1.80
2.70
3.60
4.50
5.40
6.30
M 15.3x 1.8 x 2 0.05 x3 0.05 x 6 3 kip ft
M
kip ft
0.0
7.2
13.6
19.1
23.8
27.8
31.1
33.6
35.6
37.0
37.8
38.0
37.8
37.1
36.0
34.4
32.4
29.9
27.0
x
V
M
ft
kips
kip ft
9.5
7.20
23.6
10.0
8.10
19.8
10.5
9.00
15.5
11.0
9.90
10.8
11.5
10.80
5.6
12.0
11.70
0.0
V
M
max
max
15.30 kips
38.0 kip ft
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
824
DL 5 0.5 ft
PROBLEM 5.121
4 kips
3 kips/ft
B
A
4.5 ft
1.5 ft
C
Using a computer and step functions, calculate the shear and bending
moment for the beam and loading shown. Use the specified increment ΔL,
starting at point A and ending at the right-hand support.
D
3 ft
SOLUTION
3
3
x 3 x 4.5 0
x 4.51
4.5
4.5
2
2
dV
x 3 x 4.5 0 x 4.51
3
3
dx
1
1
V x 2 3 x 4.51 x 4.5 2 4 x 6 0
3
3
1
3
1
M x3 x 4.5 2 x 4.5 3 4 x 61
9
2
9
w
x
V
ft
kips
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
8.5
9.0
0.00
0.08
0.33
0.75
1.33
2.08
3.00
4.08
5.33
6.75
6.75
6.75
10.75
10.75
10.75
10.75
10.75
10.75
10.75
M
kip ft
0.00
0.01
0.11
0.38
0.89
1.74
3.00
4.76
7.11
10.13
13.50
16.88
20.25
25.63
31.00
36.38
41.75
47.13
52.50
V
M
max
max
10.75 kips
52.5 kip ft
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
825
PROBLEM 5.122
5 kN/m
3 kN/m
A
D
B
2m
C
1.5 m
1.5 m
W200 22.5
L5m
L 0.25 m
3 kN
For the beam and loading shown, and using a computer and
step functions, (a) tabulate the shear, bending moment, and
maximum normal stress in sections of the beam from x 0
to x L, using the increments L indicated, (b) using
smaller increments if necessary, determine with a 2
accuracy the maximum normal stress in the beam. Place the
origin of the x axis at end A of the beam.
SOLUTION
M D 0:
5 RA (4.0)(2.0)(3) (1.5)(3)(5) (1.5)(3) 0
RA 10.2 kN
w 3 2 x 2 0 kN/m
dV
dx
V 10.2 3x 2 x 21 3 x 3.5 0 kN
M 10.2 x 1.5 x 2 x 2 2 3 x 3.51 kN m
(b) For rolled-steel section W200 22.5,
S 193 103 mm3 193 106 m3
max
M max 16.164 103
83.8 106 Pa
6
S
193 10
83.8 MPa
x
V
M
m
kN
kN m
MPa
0.00
10.20
0.00
0.0
0.25
9.45
2.46
12.7
0.50
8.70
4.73
24.5
0.75
7.95
6.81
35.3
1.00
7.20
8.70
45.1
1.25
6.45
10.41
53.9
1.50
5.70
11.93
61.8
1.75
4.95
13.26
68.7
2.00
4.20
14.40
74.6
2.25
2.95
15.29
79.2
2.50
1.70
15.88
82.3
2.75
0.45
16.14
83.6
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
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826
PROBLEM 5.122 (Continued)
x
V
M
m
3.00
3.25
3.50
3.75
4.00
4.25
4.50
4.75
5.00
kN
0.80
2.05
6.30
7.55
8.80
10.05
11.30
12.55
13.80
kN m
16.10
15.74
15.08
13.34
11.30
8.94
6.28
3.29
0.00
MPa
83.4
81.6
78.1
69.1
58.5
46.3
32.5
17.1
0.0
2.83
2.84
2.85
0.05
0.00
0.05
16.164
16.164
16.164
83.8
83.8
83.8
(a)
V
M
max
max
13.80 kN
16.16 kN m
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
827
PROBLEM 5.123
5 kN
20 kN/m
B
50 mm
C
A
D
2m
3m
300 mm
L6m
L 0.5 m
1m
For the beam and loading shown, and using a computer and
step functions, (a) tabulate the shear, bending moment, and
maximum normal stress in sections of the beam from x 0
to x L, using the increments ΔL indicated, (b) using
smaller increments if necessary, determine with a 2%
accuracy the maximum normal stress in the beam. Place the
origin of the x axis at end A of the beam.
SOLUTION
M D 0: 4 RB (6)(5) (2.5)(3)(20) 0
dV
w 20 x 2 0 20 x 5 0 kN/m
dx
RB 45 kN
V 5 45 x 2 0 20 x 21 20 x 51 kN
1
2
2
M 5 x 45 x 2 10 x 2 10 x 5 kN m
(a)
(b) Maximum |M | 30.0 kN m at
x
V
M
stress
m
kN
kN m
MPa
0.00
5
0.00
0.0
0.50
5
2.50
3.3
1.00
5
5.00
6.7
1.50
5
7.50
10.0
2.00
40
10.00
13.3
2.50
30
7.50
10.0
3.00
20
20.00
26.7
3.50
10
27.50
36.7
4.00
0
30.00
40.0
4.50
10
27.50
36.7
5.00
20
20.00
26.7
5.50
20
10.00
13.3
6.00
20
0.00
0.0
x 4.0 m
←
Maximum V 40.0 kN
1
1
For rectangular cross section, S bh 2 (50)(300)2 750 103 mm3 750 106 m3
6
6
max
M max
30 103
40 106 Pa
S
750 106
max 40.0 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
828
PROBLEM 5.124
2 kips/ft
2 in.
1.2 kips/ft
A
D
B
1.5 ft
12 in.
C
2 ft
L 5 ft
L 0.25 ft
1.5 ft
300 lb
For the beam and loading shown, and using a computer and
step functions, (a) tabulate the shear, bending moment, and
maximum normal stress in sections of the beam from x 0 to
x L, using the increments ΔL indicated, (b) using smaller
increments if necessary, determine with a 2% accuracy the
maximum normal stress in the beam. Place the origin of the
x axis at end A of the beam.
SOLUTION
300 lb 0.3 kips
M D 0: 5 RA (4.25)(1.5)(2) (2.5)(2)(1.2) (1.5)(0.3) 0
RA 3.84 kips
w 2 0.8 x 1.5 0 1.2 x 3.5 0 kip/ft
V 3.84 2 x 0.8 x 1.51 1.2 x 3.51 0.3 x 3.5 0 kips
M 3.84 x x 2 0.4 x 1.5 2 0.6 x 3.5 2 0.3 x 3.51 kip ft
x
V
M
stress
ft
kips
kip ft
ksi
0.00
3.84
0.00
0.000
0.25
3.34
0.90
0.224
0.50
2.84
1.67
0.417
0.75
2.34
2.32
0.579
1.00
1.84
2.84
0.710
1.25
1.34
3.24
0.809
1.50
0.84
3.51
0.877
1.75
0.54
3.68
0.921
2.00
0.24
3.78
0.945
2.25
0.06
3.80
0.951
2.50
0.36
3.75
0.937
2.75
0.66
3.62
0.906
3.00
0.96
3.42
0.855
3.25
1.26
3.14
0.786
3.50
1.86
2.79
0.697
3.75
1.86
2.32
0.581
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
829
PROBLEM 5.124 (Continued)
x
V
M
stress
ft
kips
kip ft
ksi
4.00
1.86
1.86
0.465
4.25
1.86
1.39
0.349
4.50
1.86
0.93
0.232
4.75
1.86
0.46
0.116
5.00
1.86
0.00
0.000
2.10
0.12
3.80
0.949
2.20
0.00
3.80
0.951 ←
2.30
0.12
3.80
0.949
Maximum |M | 3.804 kip ft 45.648 kip in. at
x 2.20 ft
Maximum V 3.84 kip
Rectangular section:
2 in. 12 in.
1
1
S bh 2 (2)(12) 2
6
6
48 in 3
M 45.648
S
48
0.951 ksi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
830
PROBLEM 5.125
4.8 kips/ft
3.2 kips/ft
D W12 30
L 15 ft
L 1.25 ft
A
B
C
10 ft
2.5 ft 2.5 ft
For the beam and loading shown, and using a computer and
step functions, (a) tabulate the shear, bending moment, and
maximum normal stress in sections of the beam from x 0
to x L, using the increments ΔL indicated, (b) using
smaller increments if necessary, determine with a 2%
accuracy the maximum normal stress in the beam. Place the
origin of the x axis at end A of the beam.
SOLUTION
M D 0: 12.5 RB (12.5)(5.0)(4.8) (5)(10)(3.2) 0
RB 36.8 kips
w 4.8 1.6 x 5 0 kips/ft
V 4.8 x 36.8 x 2.5 0 1.6 x 51 kips
M 2.4 x 2 36.8 x 2.51 0.8 x 5 2 kip ft
x
V
M
stress
ft
kips
kip ft
ksi
0.00
0.00
0.00
0.00
1.25
6.0
3.75
1.17
2.50
24.8
15.00
4.66
3.75
18.8
12.25
3.81
5.00
12.8
32.00
9.95
6.25
8.8
45.50
14.15
7.50
4.8
54.00
16.79
8.75
0.8
57.50
17.88
10.00
3.2
56.00
17.41
11.25
7.2
49.50
15.39
12.50
11.2
38.00
11.81
13.75
15.2
21.50
6.68
15.00
19.2
0.00
0.00
8.90
0.32
57.58
17.90
9.00
0.00
57.60
17.91 ←
9.10
0.32
57.58
17.90
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
831
PROBLEM 5.125 (Continued)
Maximum |V | 24.8 kips
Maximum |M | 57.6 kip ft 691.2 kip in. at
x 9.0 ft
For rolled-steel section W12 30,
S 38.6 in 3
Maximum normal stress:
M 691.2
S
38.6
17.91 ksi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
832
PROBLEM 5.126
w
A
h
B
h0
x
L/2
L/2
The beam AB, consisting of a cast-iron plate of uniform thickness b and
length L, is to support the load shown. (a) Knowing that the beam is to
be of constant strength, express h in terms of x, L, and h0. (b) Determine
the maximum allowable load if L 36 in., h0 12 in., b 1.25 in., and
all 24 ksi.
SOLUTION
Fy 0: RA RB wL 0
M J 0:
RA RB
wL
x
x wx M 0
2
2
S
For a rectangular cross section,
Equating,
wL
2
|M |
all
wx( L x)
2 all
M
w
x( L x)
2
1
S bh 2
6
1/2
3wx( L x)
h
all b
1 2 wx( L x)
bh
6
2 all
1/2
(a)
L
At x ,
2
2
3wL
h h0
4 all b
(b)
Solving for w,
w
4 all bh02
3L2
(4)(24)(1.25)(12)2
(3)(36)2
1/2
x
x
h h0 1
L
L
w 4.44 kip/in.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
833
PROBLEM 5.127
P
A
h
h0
B
x
L
The beam AB, consisting of a cast-iron plate of uniform thickness b and
length L, is to support the load shown. (a) Knowing that the beam is to
be of constant strength, express h in terms of x, L, and h0. (b) Determine
the maximum allowable load if L 36 in., h0 12 in., b 1.25 in., and
all 24 ksi.
SOLUTION
V P
M Px
S
|M |
all
|M | Px
P
all
x
1
S bh 2
6
For a rectangular cross section,
1 2 Px
bh
all
6
Equating,
1/2
6 Px
h
all b
(1)
1/2
6PL
h h0
all b
At x L,
(a)
Divide Eq. (1) by Eq. (2) and solve for h.
(b)
Solving for P,
P
allbh02
6L
(2)
h h0 ( x/L)1/2
(24)(1.25)(12)2
(6)(36)
P 20.0 kips
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
834
PROBLEM 5.128
w 5 w0 Lx
A
h
h0
B
x
L
The beam AB, consisting of a cast-iron plate of uniform thickness b and
length L, is to support the distributed load w(x) shown. (a) Knowing
that the beam is to be of constant strength, express h in terms of x, L,
and h0 . (b) Determine the smallest value of h0 if L 750 mm,
b 30 mm, w0 300 kN/m, and all 200 MPa.
SOLUTION
wx
dV
w 0
dx
L
2
wx
dM
V 0
2L
dx
3
wx
M 0
6L
w x3
|M |
0
S
all 6 L all
For a rectangular cross section,
At x L,
w0 x3
6L
1
S bh 2
6
w x3
1 2
bh 0
6
6 L all
Equating,
|M |
h h0
h
w0 x3
all bL
w0 L2
all b
x
h h0
L
(a)
Data:
3/2
L 750 mm 0.75 m, b 30 mm 0.030 m
w0 300 kN/m 300 103 N/m, all 200 MPa 200 106 Pa
(b)
h0
(300 103 )(0.75) 2
167.7 103 m
6
(200 10 )(0.030)
h0 167.7 mm
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
835
PROBLEM 5.129
w 5 w0 sin p2 Lx
A
h
h0
B
x
L
The beam AB, consisting of a cast-iron plate of uniform thickness b and
length L, is to support the distributed load w(x) shown. (a) Knowing
that the beam is to be of constant strength, express h in terms of x, L, and
h0 . (b) Determine the smallest value of h0 if L 750 mm, b 30 mm,
w0 300 kN/m, and all 200 MPa.
SOLUTION
dV
x
w w0 sin
dx
2L
2w0 L
x
V
cos
C1
2L
V 0 at
x 0 C1
2w0 L
2w L
x
dM
V 0 1 cos
2 L
dx
2w0 L
2w L
x
x
2L
2L
sin
|M | 0 x
sin
x
2L
L
x
|M | 2w0 L
2L
sin
S
x
all all
2L
M
1
S bh 2
6
For a rectangular cross section,
1 2 2w0 L
2L
x
bh
x
sin
6
2 L
all
Equating,
1/2
12w0 L
x
2L
sin
h
x
2 L
all b
1/ 2
At x L,
12w0 L2
h h0
all b
2
1
(a)
x 2
x 2
h h0 sin
1
2 L
L
Data:
L 750 mm 0.75 m, b 30 mm 0.030 m
1.178
w0 L2
allb
1/2
1/2
x
x 2
h 1.659 h0 sin
2 L
L
w0 300 kN/m 300 103 N/m, all 200 MPa 200 106 Pa
(b)
h0 1.178
(300 103 )(0.75) 2
197.6 103 m
(200 106 )(0.030)
h0 197.6 mm
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836
PROBLEM 5.130
P
C
A
h
B
h0
x
L/2
L/2
The beam AB, consisting of an aluminum plate of uniform thickness b and
length L, is to support the load shown. (a) Knowing that the beam is to be
of constant strength, express h in terms of x, L, and h0 for portion AC of the
beam. (b) Determine the maximum allowable load if L 800 mm,
h0 200 mm, b 25 mm, and all 72 MPa.
SOLUTION
RA RB
M J 0:
M
S
For a rectangular cross section,
Equating,
(a)
At x
L
,
2
For x
(b)
P
2
P
xM 0
2
L
0 x 2
Px
2
M
all
Px
2 all
1
S bh 2
6
1 2
Px
bh
6
2 all
h
h h0
3PL
2 all b
3Px
all b
h h0
2x
L
, 0 x
L
2
L
, replace x by L x.
2
Solving for P,
P
2 allbh02 (2)(72 106 )(0.025)(0.200)2
60.0 103 N
3L
(3)(0.8)
P 60.0 kN
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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837
w 5 w0 sin pLx
PROBLEM 5.131
C
A
h
The beam AB, consisting of an aluminum plate of uniform thickness b and
length L, is to support the load shown. (a) Knowing that the beam is to be of
constant strength, express h in terms of x, L, and h0 for portion AC of the
beam. (b) Determine the maximum allowable load if L = 800 mm,
h0 = 200 mm, b = 25 mm, and all 72 MPa.
B
h0
x
L/2
L/2
SOLUTION
x
dV
w w0 sin
dx
L
w0 L
x
dM
V
cos
C1
dx
L
w L2
x
M 0 2 sin
C1 x C2
L
At A,
x0
M 0
0 0 0 C2
At B,
xL
M 0
0
w0 L2
M
w0 L2
2
For constant strength,
S
For a rectangular section,
I
sin C1 L
2
M
all
L
,
2
h h0
sin
L
all
2
c
sin
x
L
h
,
2
S
I 1 2
bh
C 6
1 2 w0 L2
x
sin
bh 2
6
L
all
(1)
1 2 w0 L2
bh0 2
6
all
(2)
(a) Dividing Eq. (1) by Eq. 2,
h2
x
sin
2
L
h0
(b) Solving Eq. (1) for w0,
w0
Data:
C1 0
x
w0 L2
1 3
bh ,
12
Equating the two expressions for S,
At x
C2 0
1
x 2
h h0 sin
L
2 allbh02
6 L2
all 72 106 Pa, L 800 mm 0.800 m, h0 200 mm 0.200 m,
b 25 mm 0.025 m
w0
2 (72 106 )(0.025)(0.200)2
(6)(0.800) 2
185.1 103 N/m
185.1 kN/m
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838
PROBLEM 5.132
P
B
A
6.25 ft
(a)
A
C
D
B
l2
l1
A preliminary design on the use of a cantilever prismatic timber
beam indicated that a beam with a rectangular cross section 2 in.
wide and 10 in. deep would be required to safely support the load
shown in part a of the figure. It was then decided to replace that
beam with a built-up beam obtained by gluing together, as shown in
part b of the figure, five pieces of the same timber
as the original beam and of 2 2-in. cross section. Determine the
respective lengths l1 and l2 of the two inner and outer pieces of
timber that will yield the same factor of safety as the original
design.
(b)
SOLUTION
M J 0: Px M 0
M Px
|M | Px
At B,
|M |B M max
At C,
|M |C M max xC /6.25
At D,
|M |D M max xD /6.25
SB
1 2 1
25 3
bh b(5b)2
b
6
6
6
A to C:
SC
1
1
b(b)2 b3
6
6
C to D:
SD
1
9
b(3b) 2 b3
6
6
|M |C
x
S
1
C C
|M |B 6.25 S B 25
xC
(1)(6.25)
0.25 ft
25
l1 6.25 0.25
| M |D
x
S
9
D D
|M |B 6.25 S B 25
xD
l1 6.00 ft
(9)(6.25)
2.25 ft
25
l2 6.25 2.25
l2 4.00 ft
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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839
PROBLEM 5.133
w
B
A
6.25 ft
(a)
A
C
D
B
l2
A preliminary design on the use of a cantilever prismatic timber beam
indicated that a beam with a rectangular cross section 2 in. wide and
10 in. deep would be required to safely support the load shown in part
a of the figure. It was then decided to replace that beam with a built-up
beam obtained by gluing together, as shown in part b of the figure, five
pieces of the same timber as the original beam and of 2 2-in. cross
section. Determine the respective lengths l1 and l2 of the two inner
and outer pieces of timber that will yield the same factor of safety as
the original design.
l1
(b)
SOLUTION
M J 0: wx
M
wx 2
2
x
M 0
2
wx 2
|M |
2
At B,
|M |B |M |max
At C,
|M |C |M |max ( xC /6.25)2
At D,
|M |D |M |max ( xD /6.25)2
At B,
SB
1 2 1
25 3
bh b(5b)2
b
6
6
6
A to C:
SC
1 2 1
1
bh b(b) 2 b3
6
6
6
C to D:
SD
1 2 1
9
bh b (3b) 2 b3
6
6
6
2
|M |C xC
S
1
C
|M |B 6.25
S B 25
xC
6.25
25
1.25 ft
l1 6.25 1.25 ft
2
| M | D xD
S
9
D
|M |B 6.25
S B 25
xD
6.25 9
25
l1 5.00 ft
3.75 ft
l2 6.25 3.75 ft
l2 2.50 ft
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840
PROBLEM 5.134
P
1.2 m
1.2 m
A preliminary design on the use of a simply supported prismatic timber
beam indicated that a beam with a rectangular cross section 50 mm wide
and 200 mm deep would be required to safely support the load shown in
part a of the figure. It was then decided to replace that beam with a
built-up beam obtained by gluing together, as shown in part b of the
figure, four pieces of the same timber as the original beam and of
50 50-mm cross section. Determine the length l of the two outer
pieces of timber that will yield the same factor of safety as the original
design.
C
A
B
(a)
A
B
l
(b)
SOLUTION
P
2
RA RB
0 x
1
2
P
xM 0
2
M x
or M max
1.2
M J 0:
M
Px
2
Bending moment diagram is two straight lines.
At C,
SC
1 2
bhC
6
M C M max
Let D be the point where the thickness changes.
At D,
SD
1 2
bhD
6
MD
M max xD
1.2
2
S D hD2 100 mm
1 M D xD
2
SC hC 200 mm
4 M C 1.2
l
1.2 xD 0.9
2
xD 0.3 m
l 1.800 m
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841
PROBLEM 5.135
w
C
D
A
A preliminary design on the use of a simply supported prismatic timber
beam indicated that a beam with a rectangular cross section 50 mm
wide and 200 mm deep would be required to safely support the load
shown in part a of the figure. It was then decided to replace that beam
with a built-up beam obtained by gluing together, as shown in part b of
the figure, four pieces of the same timber as the original beam and of
50 50-mm cross section. Determine the length l of the two outer
pieces of timber that will yield the same factor of safety as the original
design.
B
0.8 m
0.8 m
0.8 m
(a)
A
B
l
(b)
SOLUTION
RA RB
0.8 N
0.4 w
2
Shear:
A to C:
V 0.4w
D to B:
V 0.4w
Areas:
A to C:
(0.8)(0.4) w 0.32 w
C to E:
1
(0.4)(0.4) w 0.08w
2
Bending moments:
M C 0.40 w
At C,
M 0.40wx
A to C:
At C,
SC
1 2
bhC
6
M C M max 0.40 w
Let F be the point where the thickness changes.
At F,
SF
1 2
bhF
6
M F 0.40 wxF
2
S F hF2 100 mm
1 M F 0.40 wxF
SC hC2 200 mm
4 MC
0.40 w
xF 0.25 m
l
1.2 xF 0.95 m
2
l 1.900 m
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842
PROBLEM 5.136
P
A
d
A machine element of cast aluminum and in the shape of a solid of
revolution of variable diameter d is being designed to support the load
shown. Knowing that the machine element is to be of constant strength,
express d in terms of x, L, and d 0 .
B
d0
C
x
L/2
L/2
SOLUTION
Draw shear and bending moment diagrams.
0 x
L
,
2
Px
2
P( L x)
M
2
M
L
x L,
2
For a solid circular section, c
I
1
d
2
4
c4
64
d4
S
For constant strength design, constant.
For
For
S
L
,
2
32
L
x L,
2
32
0 x
Dividing Eq. (1b) by Eq. (2),
L
,
2
L
x L,
2
Px
2
(1a)
d3
P( L x)
2
(1b)
d 03
PL
4
32
0 x
M
d3
At point C,
Dividing Eq. (1a) by Eq. (2),
I 3
d
c 32
d 3 2x
L
d 03
d 3 2( L x)
L
d03
(2)
d d 0 (2 x/L)1/3
d d0 [2( L x)/L]1/3
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843
PROBLEM 5.137
w
A
d
A machine element of cast aluminum and in the shape of a solid of
revolution of variable diameter d is being designed to support the load
shown. Knowing that the machine element is to be of constant strength,
express d in terms of x, L, and d 0 .
B
d0
C
x
L/2
L/2
SOLUTION
RA RB
M J 0:
M
S
For a solid circular cross section,
Equating,
L
At x ,
2
c
wL
2
wL
x
x wx M 0
2
2
w
x ( L x)
2
|M |
all
d
2
wx( L x)
2 all
I
4
c3
S
I d3
32
c
1/ 3
16 wx ( L x)
d
all
d3
wx( L x)
32
2 all
1/ 3
2
4wL
d d0
all
1/3
x
x
d d0 4 1
L
L
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844
PROBLEM 5.138
H
d0
B
A transverse force P is applied as shown at end A of the conical taper AB.
Denoting by d 0 the diameter of the taper at A, show that the maximum
normal stress occurs at point H, which is contained in a transverse section
of diameter d 1.5d 0 .
A
P
SOLUTION
V P
dM
dx
Let
d d0 k x
For a solid circular section,
I
4
c4
64
M Px
d3
d
I 3
S
d ( d 0 k x )3
2
32
c 32
3 2
dS 3
(d 0 k x) 2 k
d k
32
dx 32
c
Stress:
At H,
|M | Px
S
S
d
1
dS
2 PS PxH
0
dx S
dx
dS 3
3 2
S xH
d xH
d k
dx 32
32
1
1
1
k xH d ( d 0 k H xH ) k xH d 0
3
3
2
d d0
1
3
d0 d0
2
2
d 1.5d 0
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845
PROBLEM 5.139
b0
w
B
b
A
x
h
L
A cantilever beam AB consisting of a steel plate of uniform depth h and variable
width b is to support the distributed load w along its centerline AB. (a) Knowing
that the beam is to be of constant strength, express b in terms of x, L, and b0 .
(b) Determine the maximum allowable value of w if L 15 in., b0 8 in.,
h 0.75 in., and all 24 ksi.
SOLUTION
Lx
0
2
w ( L x) 2
|M |
2
M J 0: M w ( L x)
w ( L x) 2
2
|M | w ( L x ) 2
S
2 all
all
M
For a rectangular cross section,
1
S bh 2
6
1 2 w( L x)2
bh
6
2 all
(a)
At
(b)
Solving for w,
b
3w( L x) 2
all h 2
3wL2
x 0, b b0
all h2
w
all b0 h 2
3L2
(24)(8)(0.75) 2
0.160 kip/in.
(3)(15) 2
2
x
b b0 1
L
w 160.0 lb/in.
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846
PROBLEM 5.140
Assuming that the length and width of the cover plates used with the beam of Sample Prob. 5.12 are,
respectively, l 4 m and b 285 mm, and recalling that the thickness of each plate is 16 mm, determine the
maximum normal stress on a transverse section (a) through the center of the beam, (b) just to the left of D.
SOLUTION
A B 250 kN
M J 0: 250 x M 0
M 250 x kN m
At center of beam,
x 4m
M C (250)(4) 1000 kN m
At D,
x
At center of beam,
I I beam 2I plate
1
1
(8 l ) (8 4) 2 m
2
2
M 0 500 kN m
2
1
678 16
1190 106 2 (285)(16)
(285)(16)3
2
12
2
2288 106 mm 4
c
678
16 355 mm
2
S
I
6445 103 mm3
c
6445 10 6 m 3
(a)
(b)
M
1000 103
155.2 106 Pa
6
S
6445 10
Normal stress:
At D,
S 3490 103 mm 3 3510 10 6 m 3
Normal stress:
M
500 103
143.3 106 Pa
6
S
3490 10
155.2 MPa
143.3 MPa
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847
PROBLEM 5.141
160 kips
D
C
b
E
A
1
2
in.
B
1
2
l
1
2
9 ft
W27 × 84
l
Two cover plates, each 12 in. thick, are welded to a
W27 84 beam as shown. Knowing that l 10 ft and
b 10.5 in., determine the maximum normal stress on a
transverse section (a) through the center of the beam, (b) just
to the left of D.
9 ft
SOLUTION
R A R B 80 kips
M J 0: 80 x M 0
M 80 x kip ft
At C,
x 9 ft
M C 720 kip ft 8640 kip in.
At D,
x 9 5 4 ft
M D (80)(4) 320 kip ft 3840 kip in.
At center of beam,
I I beam 2 I plate
2
1
26.71 0.500
I 2850 2 (10.5)(0.500)
(10.5)(0.500)3
2 12
2
4794 in 3
26.71
c
0.500 13.855 in.
2
(a)
(b)
Mc (8640)(13.855)
I
4794
Normal stress:
At point D,
S 213 in 3
Normal stress:
M 3840
S
213
25.0 ksi
18.03 ksi
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Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
848
PROBLEM 5.142
160 kips
D
C
b
E
A
1
2
in.
Two cover plates, each 12 in. thick, are welded to a W27 84
beam as shown. Knowing that all 24 ksi for both the beam
and the plates, determine the required value of (a) the length
of the plates, (b) the width of the plates.
B
1
2
l
9 ft
1
2
W27 × 84
l
9 ft
SOLUTION
RA RB 80 kips
M J 0: 80 x M 0
M 80 x kip ft
S 213 in 3
At D,
Allowable bending moment:
M all all S (24)(213) 5112 kip in.
426 kip ft
Set M D M all .
80 xD 426
xD 5.325 ft
l 7.35 ft
l 18 2 xD
(a)
At center of beam,
M (80)(9) 720 kip ft 8640 kip in.
S
c
M
all
8640
360 in 3
24
26.7
0.500 13.85 in.
2
Required moment of inertia:
I Sc 4986 in 4
But
I I beam 2 I plate
2
1
26.7 0.500
(b)(0.500)3
4986 2850 2 (b)(0.500)
2 12
2
2850 184.981b
b 11.55 in.
(b)
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849
P
18 ⫻ 220 mm
C
A
B
D
PROBLEM 5.143
Knowing that all 150 MPa, determine the largest concentrated
load P that can be applied at end E of the beam shown.
E
W410 ⫻ 85
2.25 m 1.25 m
4.8 m
2.2 m
SOLUTION
M C 0: 4.8 A 2.2 P 0
A 0.45833P
A 0.45833P
M A 0: 4.8D 7.0 P 0
D 1.45833P
Shear:
A to C:
V 0.45833P
C to E:
V P
Bending moments:
MA 0
M C 0 (4.8)(0.45833P) 2.2 P
M E 2.2 P 2.2 P 0
4.8 2.25
MB
(2.2 P ) 1.16875 P
48
2.2 1.25
MD
(2.2P) 0.95P
2.2
M D | |M B |
<figure>
For W410 85,
S 1510 103 mm 3 1510 106 m 3
Allowable value of P based on strength at B.
150 106
1.16875P
1510 106
|M B |
S
P 193.8 103 N
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850
PROBLEM 5.143 (Continued)
Section properties over portion BCD:
W410 85: d 417 mm,
Plate:
1
d 208.5 mm, I x 316 106 mm 4
2
A (18)(220) 3960 mm 2
I
1
d 208.5 (18) 217.5 mm
2
1
(220)(18)3 106.92 103 mm 4
12
Ad 2 187.333 106 mm 4
I x I Ad 2 187.440 106 mm 4
For section,
I 316 106 (2)(187.440 106 ) 690.88 106 mm 4
c 208.5 18 226.5 mm
S
690.88 106
I
3050.2 103 mm3 3050.2 106 m3
226.5
c
Allowable load based on strength at C:
150 106
The smaller allowable load controls.
|M C |
S
2.2P
3050.2 106
P 208.0 103 N
P 193.8 103 N
P 193.8 kN
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851
PROBLEM 5.144
40 kN/m
b
7.5 mm
B
A
D
E
l
W460 × 74
Two cover plates, each 7.5 mm thick, are welded to
a W460 74 beam as shown. Knowing that l 5 m and
b 200 mm, determine the maximum normal stress on
a transverse section (a) through the center of the beam,
(b) just to the left of D.
8m
SOLUTION
RA RB 160 kN
x
M 0
2
M 160 x 20 x 2 kN m
M J 0: 160 x (40 x)
At center of beam,
x4m
M C 320 kN m
At D,
x
At center of beam,
I I beam 2 I plate
1
(8 l ) 1.5 m
2
M D 195 kN m
2
1
457 7.5
3
333 106 2 (200)(7.5)
(200)(7.5)
2 12
2
494.8 106 mm 4
457
7.5 236 mm
2
I
S 2097 103 mm3 2097 106 m3
c
c
(a)
Normal stress:
M
320 103
152.6 106 Pa
S
2097 106
152.6 MPa
(b)
At D,
S 1460 103 mm 3 1460 10 6 m 3
Normal stress:
M
195 103
133.6 106 Pa
S 1460 106
133.6 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
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852
PROBLEM 5.145
40 kN/m
b
7.5 mm
B
A
D
E
W460 × 74
l
Two cover plates, each 7.5 mm thick, are welded to a
W460 74 beam as shown. Knowing that all 150 MPa
for both the beam and the plates, determine the required
value of (a) the length of the plates, (b) the width of the
plates.
8m
SOLUTION
RA RB 160 kN
x
M J 0: 160 x (40 x) M 0
2
M 160 x 20 x 2 kN m
For W460 74 rolled-steel beam,
S 1460 103 mm 3 1460 10 6 m 3
Allowable bending moment:
M all all S (150 106 )(1460 106 )
219 103 N m 219 kN m
M M all
To locate points D and E, set
160 x 20 x 2 219
x
(a)
20 x 2 160 x 219 0
160 1602 (4)(20)(219)
(2)(20)
xD 1.753 ft
At center of beam,
x 1.753 m and x 6.247 m
xE 6.247 ft
l xE xD 4.49 m
M 320 kN m 320 103 N m
S
M
all
c
457
7.5 236 mm 4
2
320 103
2133 106 m3 2133 103 mm3
6
150 10
Required moment of inertia:
I Sc 503.4 106 mm 4
But
I I beam 2 I plate
2
1
457 7.5
(b)(7.5)3
503.4 106 333 106 2 (b)(7.5)
2 12
2
(b)
333 106 809.2 103b
b 211 mm
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853
PROBLEM 5.146
30 kips/ft
5
8
A
in.
b
B
E
D
l
W30 × 99
Two cover plates, each 85 in. thick, are welded to a W30 99 beam
as shown. Knowing that l 9 ft and b 12 in., determine the
maximum normal stress on a transverse section (a) through the
center of the beam, (b) just to the left of D.
16 ft
SOLUTION
A B 240 kips
M J 0: 240 x 30 x
x
M 0
2
M 240 x 15 x 2 kip ft
x 8 ft
At center of beam,
M C 960 kip ft 11,520 kip in.
x
At point D,
1
(16 9) 3.5 ft
2
M D 656.25 kip ft 7875 kip in.
At center of beam,
I I beam 2 I plate
2
1
29.7 0.625
3
4
I 3990 2 (12)(0.625)
(12)(0.625)
7439 in
2
2
12
c
(a)
(b)
29.7
0.625 15.475 in.
2
Mc
(11,520)(15.475)
I
7439
Normal stress:
At point D,
S 269 in 3
Normal stress:
M
7875
S
269
24.0 ksi
29.3 ksi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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854
PROBLEM 5.147
30 kips/ft
5
8
A
in.
b
Two cover plates, each 85 in. thick, are welded to a W30 99
beam as shown. Knowing that all 22 ksi for both the beam
and the plates, determine the required value of (a) the length
of the plates, (b) the width of the plates.
B
E
D
W30 × 99
l
16 ft
SOLUTION
RA RB 240 kips
M J 0: 240 x 30 x
x
M 0
2
M 240 x 15x 2 kip ft
For W30 99 rolled-steel section, S 269 in 3
Allowable bending moment:
M all all S (22)(269) 5918 kip in. 493.167 kip ft
To locate points D and E, set M M all.
240 x 15 x 2 493.167
x
15 x 2 240 x 493.167 0
240 (240)2 (4)(15)(493.167)
2.42 ft, 13.58 ft
(2)(15)
l 11.16 ft
l xE xD 13.58 2.42
(a)
M 960 kip ft 11,520 kip in.
Center of beam:
S
M
all
11,520
523.64 in 3
22
Required moment of inertia:
But
c
29.7
0.625 15.475 in.
2
I Sc 8103.3 in 4
I I beam 2 I plate
2
1
29.7 0.625
(b)(0.625)3
8103.3 3990 2 (b)(0.625)
2
12
2
3990 287.42b
b 14.31 in.
(b)
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855
A
120 mm
PROBLEM 5.148
20 mm
w
B
C
h 300 mm
h
x
0.6 m
0.6 m
For the tapered beam shown, determine (a) the transverse
section in which the maximum normal stress occurs, (b) the
largest distributed load w that can be applied, knowing
that all 140 MPa.
SOLUTION
1
wL
L 1.2 m
2
1
x
M J 0: wL wx M 0
2
2
w
M ( Lx x 2 )
2
w
x( L x)
2
RA RB
h a kx
For the tapered beam,
a 120 mm
300 120
k
300 mm/m
0.6
For rectangular cross section,
S
1 2 1
bh b (a kx)2
6
6
Bending stress:
M 3w Lx x 2
S
b (a kx) 2
To find location of maximum bending stress, set
d
0.
dx
d 3w d Lx x 2 3w (a kx)2 ( L 2 x) ( Lx x 2 )2(a kx)k
dx
b dx (a kx) 2 b
(a kx)3
3w (a kx)( L 2 x) 2k ( Lx x 2 )
b
(a kx)3
3w aL kLx 2ax 2kx2 2kLx 2kx2
b
(a kx)3
3w aL (2a kL) x
0
b (a kx)3
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856
PROBLEM 5.148 (Continued)
(a)
xm
aL
(120)(1.2)
0.240 m
2a kL (2)(120) (300)(1.2)
xm 240 mm
hm a kxm 120 (300)(0.24) 192 mm
1
1
Sm bhm2 (20)(192)2 122.88 103 mm3 122.88 106 m3
6
6
Allowable value of M m: M m Sm all (122.88 106 )(140 106 )
17.2032 103 N m
(b)
Allowable value of w:
w
2M m
(2)(17.2032 103 )
xm ( L xm )
(0.24)(0.96)
149.3 103 N/m
w 149.3 kN/m
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857
A
120 mm
PROBLEM 5.149
20 mm
w
B
C
h 300 mm
h
x
0.6 m
For the tapered beam shown, knowing that w 160 kN/m,
determine (a) the transverse section in which the maximum
normal stress occurs, (b) the corresponding value of the
normal stress.
0.6 m
SOLUTION
1
wL
2
1
x
M J 0: wLx wx M 0
2
2
w
M ( Lx x 2 )
2
w
x( L x)
2
RA RB
where w 160 kN/m and L 1.2 m.
h a kx
For the tapered beam,
a 120 mm
k
300 120
300 mm/m
0.6
1
1
For a rectangular cross section, S bh 2 b(a kx) 2
6
6
Bending stress:
M 3w Lx x 2
S
b (a kx) 2
To find location of maximum bending stress, set
d
0.
dx
d 3w d Lx x 2 3w (a kx) 2 ( L 2 x) ( Lx x 2 )2(a kx)k
dx
b dx (a kx)2 b
(a kx) 4
3w (a kx)( L 2 x) 2k ( Lx x 2 )
b
(a kx)3
3w aL kLx 2ax 2kx2 2kLx 2kx2
b
(a kx)3
3w aL 2ax kLx
0
b (a k x)3
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858
PROBLEM 5.149 (Continued)
(a)
xm
aL
(120)(1.2)
0.240 m
2a kL (2)(120) (300)(1.2)
xm 240 mm
hm a k xm 120 (300)(0.24) 192 mm
1
1
Sm bhm2 (20)(192) 2 122.88 103 mm3 122.88 106 m3
6
6
w
160 103
M m xm ( L xm )
(0.24)(0.96) 18.432 103 N m
2
2
(b)
Maximum bending stress:
m
M m 18.432 103
150 106 Pa
6
Sm 122.88 10
m 150.0 MPa
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859
3
4
w
A
B
C
4 in.
h
PROBLEM 5.150
in.
h
8 in.
x
30 in.
For the tapered beam shown, determine (a) the transverse section
in which the maximum normal stress occurs, (b) the largest
distributed load w that can be applied, knowing that
all 24 ksi.
30 in.
SOLUTION
RA RB
1
wL L 60 in.
2
1
x
M J 0: wLx wx M 0
2
2
w
2
M (Lx x )
2
w
x( L x)
2
For the tapered beam,
h a kx
a 4 in. k
84 2
in./in.
30
15
For a rectangular cross section,
S
1 2 1
bh b(a kx) 2
6
6
Bending stress:
M 3w Lx x 2
S
b (a kx)2
To find location of maximum bending stress, set
d
0.
dx
d 3w d Lx x 2
dx
b dx (a kx) 2
3w (a kx)2 ( L 2 x) ( Lx x 2 )2(a kx)k
b
(a kx) 4
3w (a kx)( L 2 x) 2k ( Lx x 2 )
b
(a kx)3
3w aL kLx 2ax 2kx 2 2kLx 2kx 2
b
(a kx)3
3w aL (2a kL) x
0
b (a kx)3
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860
PROBLEM 5.150 (Continued)
xm
(a)
aL
(4)(60)
2a kL (2)(4) 152 (6.0)
xm 15.00 in.
2
hm a kxm 4 (15) 6.00 in.
15
1
1 3
Sm bhm3 (6.00)2 4.50 in 3
6
6 4
Allowable value of M m : M m S m all (4.50)(24) 180.0 kip in.
(b)
Allowable value of w:
w
2M m
(2)(108.0)
0.320 kip/in.
xm ( L xm )
(15)(45)
w 320 lb/in.
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861
P
A
3
4
C
4 in.
h
PROBLEM 5.151
in.
B
h
8 in.
x
30 in.
For the tapered beam shown, determine (a) the transverse
section in which the maximum normal stress occurs, (b) the
largest concentrated load P that can be applied, knowing
that all 24 ksi.
30 in.
SOLUTION
P
2
Px
M J 0:
M 0
2
Px
L
M
0 x 2
2
RA RB
For a tapered beam,
h a kx
For a rectangular cross section,
S
1 2 1
bh b (a kx)2
6
6
Bending stress:
M
3Px
S b(a kx)2
To find location of maximum bending stress, set
d
0.
dx
3P (a kx)2 x 2(a kx)k
d 3P d
x
dx
b dx (a kx) 2 b
(a kx)4
3P a kx
a
0
xm
b (a kx)3
k
a 4 in.,
Data:
(a)
xm
k
84
0.13333 in./in.
30
4
30 in.
0.13333
xm 30.0 in.
hm a kxm 8 in.
1
1 3
Sm bhm2 (8)2 8 in 3
6
6 4
M m all Sm (24)(8) 192 kip in.
(b)
P
2 M m (2)(192)
12.8 kips
30
xm
P 12.80 kips
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862
250 mm
250 mm
PROBLEM 5.152
250 mm
A
B
C
D
50 mm
50 mm
75 N
Draw the shear and bending-moment diagrams for the beam and loading
shown, and determine the maximum absolute value (a) of the shear, (b) of
the bending moment.
75 N
SOLUTION
M B 0: 0.750 RA (0.550)(75) (0.300)(75) 0
Reaction at A:
RA 85 N
RB 65 N
Also,
A to C :
V 85 N
C to D :
V 10 N
D to B :
V 65 N
At A and B,
M 0
Just to the left of C,
M C 0: (0.25)(85) M 0
M 21.25 N m
Just to the right of C,
M C 0:
(0.25)(85) (0.050)(75) M 0
M 17.50 N m
Just to the left of D,
M D 0:
(0.50)(85) (0.300)(75) M 0
M 20 N m
Just to the right of D,
M D 0:
M (0.25)(65) 0
M 16.25 kN
(a)
(b)
|V |max 85.0 N
|M |max 21.3 N m
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863
40 kN ? m
PROBLEM 5.153
25 kN/m
C
A
B
W200 31.3
1.6 m
Draw the shear and bending-moment diagrams for the
beam and loading shown and determine the maximum
normal stress due to bending.
3.2 m
SOLUTION
Reaction at A:
M B 0 : 4.8 A 40 (25)(3.2)(1.6) 0
A 35 kN
0 x 1.6 m
A to C:
Fy 0: 35 V 0 V 35 kN
M J 0: M 40 35x 0
M (30 x 40) kN m
1.6 m x 4.8 m
C to B:
Fy 0: 35 25( x 1.6) V 0
V (25x 75) kN
M K 0: M 40 35 x
x 1.6
(25)( x 1.6)
0
2
M (12.5x 2 75x 72) kN m
Normal stress:
For W200 31.3, S 298 103 mm3
M
40.5 103 N m
135.9 106 Pa
S
298 106 m3
135.9 MPa
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864
PROBLEM 5.154
1.2 kips
0.8 kips
C
1.2 kips
D
E
B
A
S3 5.7
a
1.5 ft
Determine (a) the distance a for which the absolute value of the
bending moment in the beam is as small as possible, (b) the
corresponding maximum normal stress due to bending. (See hint of
Prob. 5.27.)
1.2 ft 0.9 ft
SOLUTION
M C 0: 0.8a (1.5)(1.2) (2.7)(1.2) (3.6) B 0
B 1.4 0.22222a
M B 0: (0.8)(3.6 a) 3.6C (2.1)(1.2) (0.9)(1.2) 0
C 1.8 0.22222a
Bending moment at C:
M C 0: M C (0.8)(a) 0
M C 0.8a
Bending moment at D:
M D 0:
M D (0.8)(a 1.5) 1.5C 0
M D 1.5 0.46667a
Bending moment at E:
M E 0: M E 0.9 B 0
M E 1.26 0.2a
Assume
MC M E :
M C 1.008 kip ft
Note that M D 1.008 kip ft
For rolled-steel section S3 5.7:
Normal stress:
0.8a 1.26 0.2a
a 1.260 ft
M E 1.008 kip ft M D 0.912 kip ft
max M 1.008 kip ft 12.096 kip in.
S 1.67 in 3
M
12.096
S
1.67
7.24 ksi
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865
PROBLEM 5.155
w
w0
For the beam and loading shown, determine the equations of the shear
and bending-moment curves, and the maximum absolute value of the
bending moment in the beam, knowing that (a) k 1, (b) k 0.5.
x
– kw0
L
SOLUTION
w0 x kw0 ( L x)
wx
(1 k ) 0 kw
L
L
L
wx
w kw0 (1 k ) 0
L
2
wx
kw0 x (1 k ) 0 C1
2L
0 at x 0 C1 0
w
dV
dx
V
V
w x2
dM
V kw0 x (1 k ) 0
2L
dx
kw0 x 2
w0 x3
M
(1 k )
C2
2
6L
M 0 at x 0 C2 0
M
(a)
w0 x 2
L
w x 2 w x3
M 0 0
2
3L
k 1.
V w0 x
x L.
Maximum M occurs at
(b)
kw0 x 2 (1 k ) w0 x3
2
6L
k
1
.
2
V 0 at
At
At
M
2
x L,
3
x L,
M
x
w0 32 L
4
2
w0 L2
6
V
w0 x 3w0 x 2
2
4L
M
w0 x 2 w0 x3
4
4L
2
L
3
w0 23 L
3
max
4L
w0 L2
0.03704 w0 L2
27
M 0
|M |max
w0 L2
27
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866
250 kN
C
A
PROBLEM 5.156
150 kN
D
B
W410 114
2m
2m
Draw the shear and bending-moment diagrams for the beam
and loading shown and determine the maximum normal stress
due to bending.
2m
SOLUTION
w0
M D 0:
4 RA (2)(250) (2)(150) 0
R A 50 kN
M A 0:
4RD (2)(250) (6)(150) 0
R D 350 kN
Shear:
VA 50 kN
A to C:
V 50 kN
C to D:
V 50 250 200 kN
D to B:
V 200 350 150 kN
Areas of shear diagram:
A to C:
V dx (50)(2) 100 kN m
C to D:
V dx (200)(2) 400 kN m
D to B:
V dx (150)(2) 300 kN m
Bending moments: M A 0
M C M A V dx 0 100 100 kN m
M D M C V dx 100 400 300 kN m
M B M D V dx 300 300 0
Maximum V 200 kN
Maximum M 300 kN m 300 103 N m
For W410 114 rolled-steel section,
m
M
max
Sx
S x 2200 103 mm3 2200 106 m3
300 103
136.4 106 Pa
2200 106
m 136.4 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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867
PROBLEM 5.157
w0
Beam AB, of length L and square cross section of side a, is
supported by a pivot at C and loaded as shown. (a) Check that the
beam is in equilibrium. (b) Show that the maximum normal stress
due to bending occurs at C and is equal to w0 L2 /(1.5a)3.
a
A
a
B
C
2L
3
L
3
SOLUTION
(a)
Replace distributed load by equivalent concentrated load at the centroid of the area of the load diagram.
2L
.
3
For the triangular distribution, the centroid lies at x
(a)
Fy 0: RD W 0
RD
1
w0 L
2
W
1
w0 L
2
M C 0: 0 0 equilibrium
V 0, M 0, at x 0
0 x
2L
,
3
dV
wx
w 0
dx
L
dM
w x2
w x2
V 0 C1 0
dx
2L
2L
M
w0 x3
w x3
C2 0
6L
6L
Just to the left of C,
V
w0 (2L / 3)2
2
w0 L
2L
9
Just to the right of C,
2
5
V w0 L RD
w0 L
9
18
Note sign change. Maximum M occurs at C.
Maximum M
m
M
max
I
w0 (2 L / 3)3
4
w0 L2
6L
81
4
w0 L2
81
For square cross section,
(b)
MC
c
I
1 4
a
12
c
1
a
2
3
4 w0 L2 6
8 w0 L2 2 w0 L2
3
3
81 a
27 a3
3 a
m
w0 L2
(1.5a)3
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868
PROBLEM 5.158
1.5 kips/ft
5 in.
A
D
B
3 ft
C
6 ft
h
For the beam and loading shown, design the cross section of the
beam, knowing that the grade of timber used has an allowable
normal stress of 1750 psi.
3 ft
SOLUTION
By symmetry, A D.
Reactions:
1
1
(3)(1.5) (6)(1.5) (3)(1.5) D 0
2
2
A D 6.75 kips
Fy 0: A
Shear diagram:
VB 6.75
VA 6.75 kips
1
(3)(1.5) 4.5 kips
2
VC 4.5 (6)(1.5) 4.5 kips
VD 4.5
1
(3)(1.5) 6.75 kips
2
Locate point E where V 0.
By symmetry, E is the midpoint of BC.
Areas of the shear diagram:
A to B : (3)(4.5)
2
(3)(2.25) 18 kip ft
3
1
(3)(4.5) 6.75 kip ft
2
1
(3)(4.5) 6.75 kip ft
E to C :
2
B to E :
C to D : By antisymmetry, 18 kip ft
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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869
PROBLEM 5.158 (Continued)
Bending moments: M A 0
M B 0 18 18 kip ft
M E 18 6.75 24.75 kip ft
M C 24.75 6.75 18 kip ft
M D 18 18 0
max
M
max
S
S
M
max
max
For a rectangular section,
(24.75 kip ft)(12 in./ft)
169.714 in 3
1.750 ksi
S
1 2
bh
6
h
6S
b
6(169.714)
14.27 in.
5
h 14.27 in.
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870
PROBLEM 5.159
62 kips
B
C
A
D
62 kips
12 ft
5 ft
Knowing that the allowable normal stress for the steel used is 24 ksi,
select the most economical wide-flange beam to support the loading
shown.
5 ft
SOLUTION
M C 0: (17)(62) 12 B (5)(62) 0
B 113.667 kips
M B 0: (5)(62) 12C (17)(62) 0
C 113.667 kips or C 113.667 kips
Shear diagram:
A to B :
V 62 kips
B to C :
V 62 113.667 51.667 kips
V 51.667 113.667 62 kips
C to D:
Areas of shear diagram:
A to B:
(5)(62) 310 kip ft
B to C:
(12)(51.667) 620 kip ft
(5)(62) 310 kip ft
C to D:
MA 0
Bending moments:
M B 0 310 310 kip ft
M C 310 620 310 kip ft
M D 310 310 0
|M |max 310 kip ft 3.72 103 kip in.
S min
Required Smin :
Shape
S (in 3 )
W27 84
213
W21 101
227
W18 106
204
W14 145
232
|M |max
all
3.72 103
155 in 3
24
Use W27 84.
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871
8 kips
32 kips
32 kips
B
C
D
b
1 in.
E
A
4.5 ft
14 ft
14 ft
3
4
in.
19 in.
1 in.
9.5 ft
PROBLEM 5.160
Three steel plates are welded together to form the
beam shown. Knowing that the allowable normal stress
for the steel used is 22 ksi, determine the minimum
flange width b that can be used.
SOLUTION
M E 0: 42 A (37.5)(8) (23.5)(32) (9.5)(32) 0
Reactions:
A 32.2857 kips
M A 0: 42 E (4.5)(8) (18.5)(32) (32.5)(32) 0
E 39.7143 kips
Shear:
A to B :
32.2857 kips
B to C :
32.2857 8 24.2857 kips
C to D :
24.2857 32 7.7143 kips
D to E :
7.7143 32 39.7143 kips
Areas:
(4.5)(32.2857) 145.286 kip ft
A to B :
B to C :
(14)(24.2857) 340 kip ft
C to D :
(14)(7.7143) 108 kip ft
(9.5)(39.7143) 377.286 kip ft
D to E :
MA 0
Bending moments:
M B 0 145.286 145.286 kip ft
M C 145.286 340 485.29 kip ft
M D 485.29 108 377.29 kip ft
M E 377.29 377.286 0
all 22 ksi
Maximum |M | 485.29 kip ft 5.2834 103 kip in.
Smin
|M |
all
5.2834 103
264.70 in 3
22
1 3
1
(19)3 2 (b)(1)3 (b)(1)(10)2 428.69 200.17b
12 4
12
c 9.5 1 10.5 in.
I
S min
I
40.828 19.063b 264.70
c
b 11.74 in.
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872
PROBLEM 5.161
10 kN
80 kN/m
B
A
D
C
W530 150
(a) Using singularity functions, find the magnitude and
location of the maximum bending moment for the beam
and loading shown. (b) Determine the maximum normal
stress due to bending.
4m
1m 1m
SOLUTION
M D 0: (6)(10) 5RB (2)(4)(80) 0
RB 140 kN
w 80 x 2
0
kN/m dV /dx
V 10 140 x 1
0
A to B:
V 10 kN
B to C:
V 10 140 130 kN
( x 6)
D:
1
80 x 2 kN
V 10 140 80(4) 190 kN
V changes sign at B and at point E ( x xE ) between C and D.
V 0 10 140 xE 1
0
10 140 80( xE 2)
80 xE 2
1
xE 3.625 m
1
2
M 10 x 140 x 1 40 x 2 kN m
At pt. B,
x 1
At pt. E,
x 3.625
M B (10)(1) 10 kN m
M E (10)(3.625) (140)(2.625) (40)(1.625)2 225.6 kN m
(a)
M
For W530 150,
S 3720 103 mm3 3720 106 m3
(b)
Normal stress:
max
225.6 kN m at x 3.63 m
M
225.6 103
60.6 106 Pa
S
3720 106
60.6 MPa
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on a website, in whole or part.
873
PROBLEM 5.162
M0
A
C
h
The beam AB, consisting of an aluminum plate of uniform thickness b and
length L, is to support the load shown. (a) Knowing that the beam is to be
of constant strength, express h in terms of x, L, and h0 for portion AC of the
beam. (b) Determine the maximum allowable load if L 800 mm,
h0 200 mm, b 25 mm, and all 72 MPa.
B
h0
x
L/2
L/2
SOLUTION
A M 0 /L
B M 0 /L
M0
L
M 0x
M
L
M J 0:
S
M
all
M 0x
all L
S
1 2
bh
6
1 2 M0 x
bh
6
all L
Equating,
L
,
2
(a)
At x
(b)
Solving for M 0 , M 0
h h0
allbh02
3
L
0 x
2
L
0 x
2
for
For a rectangular cross section,
xM 0
h
6M 0 x
allbL
3M 0
allb
h h0 2 x/L
(72 106 )(0.025)(0.200)2
24 103 N m
3
M 0 24.0 kN m
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874
PROBLEM 5.163
b0
P
B
b
A
x
h
L
A cantilever beam AB consisting of a steel plate of uniform depth h and variable
width b is to support the concentrated load P at point A. (a) Knowing that the beam
is to be of constant strength, express b in terms of x, L, and b0 . (b) Determine the
smallest allowable value of h if L 300 mm, b0 375 mm, P 14.4 kN, and
all 160 MPa.
SOLUTION
M J 0: M P ( L x) 0
M P( L x)
|M | P ( L x)
|M | P ( L x)
S
all
For a rectangular cross section,
At x 0,
b b0
h
Solving for h,
Data:
1
S bh 2
6
1 2 P ( L x)
bh
all
6
Equating,
(a)
all
b
6 P ( L x)
all h 2
6PL
all h 2
x
b b0 1
L
6PL
all b0
L 300 mm 0.300 m, b0 375 mm 0.375 m
P 14.4 kN 14.4 103 N m, all 160 MPa 160 106 Pa
(b)
h
(6)(14.4 103 )(0.300)
20.8 103 m
6
(160 10 )(0.375)
h 20.8 mm
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875
PROBLEM 5.C1
xn
x2
x1
xi
P1
P2
Pi
A
a
Pn
B
L
Several concentrated loads Pi (i 1, 2, , n) can be applied to a beam as
shown. Write a computer program that can be used to calculate the shear,
bending moment, and normal stress at any point of the beam for a given
loading of the beam and a given value of its section modulus. Use this
program to solve Probs. 5.18, 5.21, and 5.25. (Hint: Maximum values will
occur at a support or under a load.)
b
SOLUTION
Reactions at A and B
M A 0: RB L
P ( x a)
i
i
i
1
RB
L
RA
P ( x a)
i
i
i
P R
i
B
i
We use step functions (See bottom of Page 348 of text.)
We define:
If x a
Then STP A 1
Else STP A 0
If x a L
Then STP B 1
Else STP B 0
If x xi
Then STP (I ) 1
Else STP (I) 0
V RA STP A RB STP B
P STP (I)
i
i
M RA (x a) STP A RB ( x a L) STP B
P ( x x ) STP (I)
i
i
i
M/S , where S is obtained from Appendix C.
Program Outputs
Problem 5.18
RA 80.0 kN RB 80.0 kN
X
m
V
kN
M
kN · m
MPa
2.00
0.00
104.00
139.0
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876
PROBLEM 5.C1 (Continued)
Program Outputs (Continued)
Problem 5.21
R1 52.5 kips R2 22.5 kips
X
ft
V
kips
ksi
M
kip ft
0.00
25.00
0.00
0.00
1.00
27.50
25.00
7.85
3.00
2.50
30.00
9.42
9.00
22.50
45.00
14.14
11.00
0.00
0.00
0.00
Problem 5.25
R1 10.77 kips R2 4.23 kips
X
ft
V
kips
M
kip in.
0
5.00
5
5.69
13
4.23
253.8
18
4.23
0
ksi
0
0
300.00
10.34
8.75
0
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877
PROBLEM 5.C2
x4
x2
x3
x1
w
P1
P2
t
h
A
B
L
a
b
A timber beam is to be designed to support a distributed load and
up to two concentrated loads as shown. One of the dimensions of
its uniform rectangular cross section has been specified and the
other is to be determined so that the maximum normal stress in
the beam will not exceed a given allowable value all . Write a
computer program that can be used to calculate at given intervals
L the shear, the bending moment, and the smallest acceptable
value of the unknown dimension. Apply this program to solve
the following problems, using the intervals L indicated:
(a) Prob. 5.65 ( L 0.1 m), (b) Prob. 5.69 ( L 0.3 m),
(c) Prob. 5.70 ( L 0.2 m).
SOLUTION
Reactions at A and B:
x x3
M A 0: RB L P1 ( x1 a) P2 ( x2 a) w( x4 x3 ) 4
a 0
2
1
1
P1 ( x1 a ) P2 ( x2 a) w( x4 x3 )( x4 x3 2a )
L
2
RA P1 P2 w( x4 x3 ) RB
RB
We use step functions. (See bottom of Page 348 of text.)
Set
n (a b L)/ L
For
i 0 to n: x ( L)i
We define:
If x a
Then STP A 1
Else STP A 0
If x a L
Then STP B 1
Else STP B 0
If x x1
Then STP 1 1
Else STP 1 0
If x x2
Then STP 2 1
Else STP 2 0
If x x3
Then STP 3 1
Else STP 3 0
If x x4
Then STP 4 1
Else STP 4 0
V RA Step A RB STP B P1 STP1 P2 STP 2 w( x x3 )STP 3 w( x x4 )STP 4
M RA (x a )STP A RB (x a L)STP B P1 ( x x1 )STP1
P2 ( x x2 )STP 2
Smin
1
1
w( x x3 )2 STP 3 w( x x4 )4 STP 4
2
2
|M |
all
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878
PROBLEM 5.C2 (Continued)
If unknown dimension is h:
1
S th 2 , we have h 6S/t
6
From
If unknown dimension is t:
1
S th 2 , we have t 6S/h 2
6
From
Program Outputs
RA 2.40 kN RB 3.00 kN
Problem 5.65
X
m
V
kN
M
kN m
H
mm
0.00
2.40
0.000
0.00
0.10
2.40
0.240
54.77
0.20
2.40
0.480
77.46
0.30
2.40
0.720
94.87
0.40
2.40
0.960
109.54
0.50
2.40
1.200
122.47
0.60
2.40
1.440
134.16
0.70
2.40
1.680
144.91
0.80
0.60
1.920
154.92
0.90
0.60
1.980
157.32
1.00
0.60
2.040
159.69
1.10
0.60
2.100
162.02
1.20
0.60
2.160
164.32
1.30
0.60
2.220
166.58
1.40
0.60
2.280
168.82
1.50
0.60
2.340
171.03
1.60
3.00
2.400
173.21
1.70
3.00
2.100
162.02
1.80
3.00
1.800
150.00
1.90
3.00
1.500
136.93
2.00
3.00
1.200
122.47
2.10
3.00
0.900
106.07
2.20
3.00
0.600
86.60
2.30
3.00
0.300
61.24
2.40
0.00
0.000
0.05
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879
PROBLEM 5.C2 (Continued)
Program Outputs (Continued)
Problem 5.69
RA 6.50 kN RB 6.50 kN
X
m
V
kN
M
kN m
H
mm
0.00
2.50
0.000
0.00
0.30
2.50
0.750
61.24
0.60
9.00
1.500
86.60
0.90
7.20
3.930
140.18
1.20
5.40
5.820
170.59
1.50
3.60
7.170
189.34
1.80
1.80
7.980
199.75
2.10
0.00
8.250
203.10
2.40
1.80
7.980
199.75
2.70
3.60
7.170
189.34
3.00
5.40
5.820
170.59
3.30
7.20
3.930
140.18
3.60
2.50
1.500
86.60
3.90
2.50
0.750
61.24
4.20
0.00
0.000
0.06
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on a website, in whole or part.
880
PROBLEM 5.C2 (Continued)
Program Outputs (Continued)
Problem 5.70
RA 2.70 kN RB 8.10 kN
X
m
V
kN
M
kN m
T
mm
0.00
2.70
0.000
0.00
0.20
2.10
0.480
10.67
0.40
1.50
0.840
18.67
0.60
0.90
1.080
24.00
0.80
0.30
1.200
26.67
1.00
0.30
1.200
26.67
1.20
0.90
1.080
24.00
1.40
1.50
0.840
18.67
1.60
2.10
0.480
10.67
1.80
2.70
0.000
0.00
2.00
3.30
0.600
13.33
2.20
3.90
1.320
29.33
2.40
3.60
2.160
48.00
2.60
3.00
1.500
33.33
2.80
2.40
0.960
21.33
3.00
1.80
0.540
12.00
3.20
1.20
0.240
5.33
3.40
0.60
0.060
1.33
3.60
0.00
0.000
0.00
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881
PROBLEM 5.C3
Two cover plates, each of thickness t, are to be welded to a wideflange beam of length L that is to support a uniformly distributed
load w. Denoting by all the allowable normal stress in the beam
and in the plates, by d the depth of the beam, and by I b
and Sb , respectively, the moment of inertia and the section
modulus of the cross section of the unreinforced beam about a
horizontal centroidal axis, write a computer program that can
be used to calculate the required value of (a) the length a of the
plates, (b) the width b of the plates. Use this program to solve
Prob. 5.145.
w
t
A
b
B
E
D
a
L
SOLUTION
(a)
Required length of plates:
FB AD :
x
M D 0: M D wx RA x 0
2
But
RA
Divide by
1
2
1
wL and M D S all .
2
w:
x 2 Lx (2S all /w) 0
Set k
2S all
:
w
x 2 Lx k 0
(b)
L L2 4k
2
Solving the quadratic,
x
Compute x and
a L 2x
Required width of plates:
At midpoint C of beam:
FB AC:
M C 0: M C
Compute
wL L wL L
0
2 4
2 2
1
M C wL2
8
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882
PROBLEM 5.C3 (Continued)
Compute
From
C t
all
1
d
2
MCC
I
compute I
MCC
all
But
2
1
d t
I I beam I plates I b 2 bt 3 bt
2
12
Solving for b,
b
6( I I b )
2
t[t 3(d t )2 ]
Program Outputs
Problems 5.155:
W460 74, all 150 MPa
w 40 kN/m, L 8 m, t 7.5 mm
d 457 mm, I b 333 106 mm 4 , S 416 103 mm3
Problem 5.145
a 4.49 m
b 211 mm
Problem 5.157:
W30 99, all 22 ksi
w 30 kips/ft, L 16 ft, t 5/8 in.
d 29.65 in., I b 3990 in.4 , S 269 in.3
Problem 5.143
a 11.16 ft
b 14.31 in.
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883
25 kips
PROBLEM 5.C4
25 kips
6 ft
C
A
B
9 ft
x
18 ft
Two 25-kip loads are maintained 6 ft apart as they are moved
slowly across the 18-ft beam AB. Write a computer program and
use it to calculate the bending moment under each load and at the
midpoint C of the beam for values of x from 0 to 24 ft at intervals
x 1.5 ft.
SOLUTION
Length of beam L 18 ft
Notation:
P1 P2 P 25 kips
Loads:
Distance between loads d 6 ft
We note that d L /2.
(1)
From
x 0 to x d :
M B 0: P( L x) RA L 0
RA P( L x)/L
Under P1:
At C:
(2)
From
M 1 RA x
L
L
M C RA P x
2
2
xd
to x L:
M B 0: P(L x) P( L x d )
RA L 0
RA P(2 L 2 x d )/L
Under P1:
M1 RA x Pd
Under P2 :
M 2 RA ( x d )
(2A) From
xd
to x L/2:
L
L
M C RA P x
2
2
L
P x d
2
RA ( L/2) P( L 2 x d )
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Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
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884
PROBLEM 5.C4 (Continued)
x L/2 to x L/2 d :
(2B) From
L
M C RA ( L/2) P x d
2
x L/2 d
(2C) From
to
x L:
M C RA L/2
(3)
x L to x L d :
From
M B 0: P( L x d ) RA L 0
RA P( L x d )/L
Under P2 :
M z RA ( x d )
At C:
M C RA ( L/2)
Program Output
P 25 kips, L 18 ft, D 6 ft
X
ft
MC
kip ft
M1
kip ft
M2
kip ft
0.0
0.00
0.00
0.00
1.5
18.75
34.38
0.00
3.0
37.50
62.50
0.00
4.5
56.25
84.38
0.00
6.0
75.00
100.00
0.00
7.5
112.50
131.25
56.25
9.0
150.00
150.00
100.00
10.5
150.00
156.25
131.25
12.0
150.0
150.00
150.00
13.5
150.00
131.25
156.25
15.0
150.00
100.00
150.00
16.5
112.50
56.25
131.25
18.0
75.00
0.00
100.00
19.5
56.25
0.00
84.38
21.0
37.50
0.00
62.50
22.5
18.75
0.00
34.38
24.0
0.00
0.00
0.00
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885
PROBLEM 5.C5
a
w
B
A
Write a computer program that can be used to plot the shear and
bending-moment diagrams for the beam and loading shown. Apply this
program with a plotting interval L 0.2 ft to the beam and loading
of (a) Prob. 5.72, (b) Prob. 5.115.
P
b
L
SOLUTION
Reactions at A and B:
Using FB diagram of beam,
M A 0: RB L Pb wa(a/2) 0
1
RB (1/L) Pb wa 2
2
RA P wa RB
We use step functions (see bottom of Page 348 of text).
set n L/L. For
i 0 to n: x ( L)i
We define:
If x a
Then STP A 1
Else STP A 0
If x b
Then STP B 1
Else STP B 0
V RA wx w( x a )STP A P STP B
M RA x
Locate and Print
1 2 1
wx w( x a) 2 STP A P( x b)STP B
2
2
( x, v) and (x, M )
See next pages for program outputs
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886
PROBLEM 5.C5 (Continued)
Program Outputs
Problem 5.72
RA 48.00 kips RB 42.00 kips
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
887
PROBLEM 5.C5 (Continued)
Program Outputs (Continued)
Problem 5.115
RA 40.50 kips RB 27.00 kips
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
888
PROBLEM 5.C6
b
a
w
MA
Write a computer program that can be used to plot the shear and
bending-moment diagrams for the beam and loading shown. Apply
this program with a plotting interval L 0.025 m to the beam and
loading of Prob. 5.112.
MB
B
A
L
SOLUTION
Reactions at A and B:
1
M A 0: RB L M B M A w(b a) (a b) 0
2
1
RB (1/L) M A M B w(b2 a 2 )
2
RA w(b a) RB
We use step functions (see bottom of Page 348 of text).
L
L
Set
n
For
i 0 to n : x (L)i
We define:
If x a Then STPA 1 Else STPA 0
If x b Then STPB 1 Else STPB 0
V RA w( x a )STPA w ( x b) STPB
M M A RA x
1
1
w( x a ) 2 STPA w( x b) 2 STPB
2
2
Locate and print ( x, V ) and ( x, M ).
Program output on next page
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
889
PROBLEM 5.C6 (Continued)
Program Output
Problem 5.112
RA 29.50 kips RB 66.50 kips
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
890