ch04-05 axial load torsion

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Page iii
CONTENTS
To the Instructor
iv
1 Stress
1
2 Strain
73
3 Mechanical Properties of Materials
92
4 Axial Load
122
5 Torsion
214
6 Bending
329
7 Transverse Shear
472
8 Combined Loadings
532
9 Stress Transformation
619
10 Strain Transformation
738
11 Design of Beams and Shafts
830
12 Deflection of Beams and Shafts
883
13 Buckling of Columns
1038
14 Energy Methods
1159
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•4–1. The ship is pushed through the water using an A-36
steel propeller shaft that is 8 m long, measured from the
propeller to the thrust bearing D at the engine. If it has an
outer diameter of 400 mm and a wall thickness of 50 mm,
determine the amount of axial contraction of the shaft
when the propeller exerts a force on the shaft of 5 kN. The
bearings at B and C are journal bearings.
A
Internal Force: As shown on FBD.
B
C
D
5 kN
Displacement:
8m
dA =
PL
=
AE
- 5.00 (103)(8)
p
4
(0.42 - 0.32) 200(109)
= - 3.638(10 - 6) m
= - 3.64 A 10 - 3 B mm
Ans.
Negative sign indicates that end A moves towards end D.
4–2. The copper shaft is subjected to the axial loads
shown. Determine the displacement of end A with respect
to end D. The diameters of each segment are dAB = 3 in.,
dBC = 2 in., and dCD = 1 in. Take Ecu = 1811032 ksi.
50 in.
A
p
The cross-sectional area of segment AB, BC and CD are AAB = (32) = 2.25p in2,
4
p
p
ABC = (22) = p in2 and ACD = (12) = 0.25p in2.
4
4
Thus,
PCD LCD
PiLi
PAB LAB
PBC LBC
=
+
+
AiEi
AAB ECu
ABC ECu
ACD ECu
2.00 (75)
6.00 (50)
=
(2.25p) C 18(10 ) D
3
+
p C 18(10 ) D
3
- 1.00 (60)
+
(0.25p) C 18(103) D
= 0.766(10 - 3) in.
Ans.
The positive sign indicates that end A moves away from D.
122
60 in.
2 kip
6 kip
The normal forces developed in segment AB, BC and CD are shown in the
FBDS of each segment in Fig. a, b and c respectively.
dA>D = ©
75 in.
B 2 kip
1 kip
C
3 kip
D
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4–3. The A-36 steel rod is subjected to the loading shown.
If the cross-sectional area of the rod is 50 mm2, determine
the displacement of its end D. Neglect the size of the
couplings at B, C, and D.
1m
A 9 kN B
The normal forces developed in segments AB, BC and CD are shown in the FBDS
of each segment in Fig. a, b and c, respectively.
The
cross-sectional
areas
of
all
2
1
m
b = 50.0(10 - 6) m2.
A = A 50 mm2 B a
1000 mm
dD = ©
the
segments
are
PiLi
1
=
a PAB LAB + PBC LBC + PCD LCD b
AiEi
A ESC
1
=
50.0(10 ) C 200(109) D
-6
c -3.00(103)(1) + 6.00(103)(1.5) + 2.00(103)(1.25) d
= 0.850(10 - 3) m = 0.850 mm
Ans.
The positive sign indicates that end D moves away from the fixed support.
123
1.5 m
1.25 m
C
4 kN
D
2 kN
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*4–4. The A-36 steel rod is subjected to the loading
shown. If the cross-sectional area of the rod is 50 mm2,
determine the displacement of C. Neglect the size of the
couplings at B, C, and D.
1m
1.5 m
1.25 m
C
A 9 kN B
4 kN
D
2 kN
The normal forces developed in segments AB and BC are shown the FBDS of each
segment in Fig. a and b, respectively. The cross-sectional area of these two segments
2
1m
are A = A 50 mm2 B a
b = 50.0 (10 - 6) m2. Thus,
10.00 mm
dC = ©
PiLi
1
=
A P L + PBC LBC B
AiEi
A ESC AB AB
1
=
50.0(10 - 6) C 200(109) D
c -3.00(103)(1) + 6.00(103)(1.5) d
= 0.600 (10 - 3) m = 0.600 mm
Ans.
The positive sign indicates that coupling C moves away from the fixed support.
4–5. The assembly consists of a steel rod CB and an
aluminum rod BA, each having a diameter of 12 mm. If the rod
is subjected to the axial loadings at A and at the coupling B,
determine the displacement of the coupling B and the end
A. The unstretched length of each segment is shown in the
figure. Neglect the size of the connections at B and C, and
assume that they are rigid. Est = 200 GPa, Eal = 70 GPa.
dB =
PL
=
AE
dA = ©
12(103)(3)
p
4
12(103)(3)
p
4
2
A
B
18 kN
6 kN
3m
= 0.00159 m = 1.59 mm
(0.012)2(200)(109)
PL
=
AE
C
2m
Ans.
18(103)(2)
9
(0.012) (200)(10 )
+
p
2
9
4 (0.012) (70)(10 )
= 0.00614 m = 6.14 mm
Ans.
4–6. The bar has a cross-sectional area of 3 in2, and
E = 3511032 ksi. Determine the displacement of its end A
when it is subjected to the distributed loading.
x
w ⫽ 500x1/3 lb/in.
A
4 ft
x
P(x) =
L0
w dx = 500
x
L0
1
x3 dx =
1500 43
x
4
L
dA =
4(12)
P(x) dx
1
3
1
1500 4
1500
b a b(48)3
=
x3 dx = a
AE
4
(3)(35)(108)(4) 7
(3)(35)(106) L0
L0
dA = 0.0128 in.
Ans.
124
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4–7. The load of 800 lb is supported by the four 304 stainless
steel wires that are connected to the rigid members AB and
DC. Determine the vertical displacement of the load if the
members were horizontal before the load was applied. Each
wire has a cross-sectional area of 0.05 in2.
E
F
4 ft
H
D
C
2 ft
Referring to the FBD of member AB, Fig. a
5 ft
4.5 ft
a + ©MA = 0;
FBC (5) - 800(1) = 0
FBC = 160 lb
a + ©MB = 0;
800(4) - FAH (5) = 0
FAH = 640 lb
800 lb
A
B
1 ft
Using the results of FBC and FAH, and referring to the FBD of member DC, Fig. b
a + ©MD = 0;
FCF (7) - 160(7) - 640(2) = 0
a + ©MC = 0;
640(5) - FDE(7) = 0
FCF = 342.86 lb
FDE = 457.14 lb
Since E and F are fixed,
dD =
457.14(4)(2)
FDE LDE
=
= 0.01567 in T
A Est
0.05 C 28.0 (106) D
dC =
342.86 (4)(12)
FCF LCF
=
= 0.01176 in T
A Est
0.05 C 28.0 (106) D
From the geometry shown in Fig. c,
dH = 0.01176 +
5
(0.01567 - 0.01176) = 0.01455 in T
7
Subsequently,
dA>H =
640(4.5)(12)
FAH LAH
=
= 0.02469 in T
A Est
0.05 C 28.0(106) D
dB>C =
160(4.5)(12)
FBC LBC
=
= 0.006171 in T
A Est
0.05 C 28.0(106) D
Thus,
A + T B dA = dH + dA>H = 0.01455 + 0.02469 = 0.03924 in T
A + T B dB = dC + dB>C = 0.01176 + 0.006171 = 0.01793 in T
From the geometry shown in Fig. d,
dP = 0.01793 +
4
(0.03924 - 0.01793) = 0.0350 in T
5
125
Ans.
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*4–8. The load of 800 lb is supported by the four
304 stainless steel wires that are connected to the rigid
members AB and DC. Determine the angle of tilt of each
member after the load is applied.The members were originally
horizontal, and each wire has a cross-sectional area of 0.05 in2.
E
F
4 ft
H
D
C
2 ft
Referring to the FBD of member AB, Fig. a,
5 ft
4.5 ft
a + ©MA = 0;
FBC (5) - 800(1) = 0
FBC = 160 lb
a + ©MB = 0;
800(4) - FAH (5) = 0
FAH = 640 lb
800 lb
A
B
1 ft
Using the results of FBC and FAH and referring to the FBD of member DC, Fig. b,
a + ©MD = 0;
FCF (7) - 160(7) - 640(2) = 0
FCF = 342.86 lb
a + ©MC = 0;
640(5) - FDE (7) = 0
FDE = 457.14 lb
Since E and F are fixed,
dD =
457.14 (4)(12)
FDE LDE
=
= 0.01567 in T
A Est
0.05 C 28.0(106) D
dC =
342.86 (4)(12)
FCF LCF
=
= 0.01176 in T
A Est
0.05 C 28.0(106) D
From the geometry shown in Fig. c
dH = 0.01176 +
u =
5
(0.01567 - 0.01176) = 0.01455 in T
7
0.01567 - 0.01176
= 46.6(10 - 6) rad
7(12)
Ans.
Subsequently,
dA>H =
640 (4.5)(12)
FAH LAH
=
= 0.02469 in T
A Est
0.05 C 28.0(106) D
dB>C =
160 (4.5)(12)
FBC LBC
=
= 0.006171 in T
A Est
0.05 C 28.0(106) D
Thus,
A + T B dA = dH + dA>H = 0.01455 + 0.02469 = 0.03924 in T
A + T B dB = dC + dB>C = 0.01176 + 0.006171 = 0.01793 in T
From the geometry shown in Fig. d
f =
0.03924 - 0.01793
= 0.355(10 - 3) rad
5(12)
Ans.
126
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4–8. Continued
127
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•4–9.
The assembly consists of three titanium (Ti-6A1-4V)
rods and a rigid bar AC. The cross-sectional area of each rod
is given in the figure. If a force of 6 kip is applied to the ring
F, determine the horizontal displacement of point F.
D
4 ft
C
ACD ⫽ 1 in2
2 ft
E
Internal Force in the Rods:
a + ©MA = 0;
+ ©F = 0;
:
x
FCD(3) - 6(1) = 0
FCD = 2.00 kip
6 - 2.00 - FAB = 0
FAB = 4.00 kip
AAB ⫽ 1.5 in2
6 ft
B
1 ft F
6 kip
2
1 ft AEF ⫽ 2 in
A
Displacement:
dC =
2.00(4)(12)
FCD LCD
= 0.0055172 in.
=
ACD E
(1)(17.4)(103)
dA =
4.00(6)(12)
FAB LAB
= 0.0110344 in.
=
AAB E
(1.5)(17.4)(103)
dF>E =
6.00(1)(12)
FEF LEF
= 0.0020690 in.
=
AEF E
(2)(17.4)(103)
œ
dE
0.0055172
=
;
2
3
œ
dE
= 0.0036782 in.
œ
dE = dC + dE
= 0.0055172 + 0.0036782 = 0.0091954 in.
dF = dE + dF>E
= 0.0091954 + 0.0020690 = 0.0113 in.
Ans.
4–10. The assembly consists of three titanium (Ti-6A1-4V)
rods and a rigid bar AC. The cross-sectional area of each rod
is given in the figure. If a force of 6 kip is applied to the ring
F, determine the angle of tilt of bar AC.
D
4 ft
C
ACD ⫽ 1 in
2
2 ft
E
Internal Force in the Rods:
a + ©MA = 0;
FCD(3) - 6(1) = 0
FCD = 2.00 kip
+ ©F = 0;
:
x
6 - 2.00 - FAB = 0
FAB = 4.00 kip
AAB ⫽ 1.5 in2
B
Displacement:
dC =
2.00(4)(12)
FCD LCD
= 0.0055172 in.
=
ACD E
(1)(17.4)(103)
dA =
4.00(6)(12)
FAB LAB
= 0.0110344 in.
=
AAB E
(1.5)(17.4)(103)
u = tan - 1
dA - dC
0.0110344 - 0.0055172
= tan - 1
3(12)
3(12)
= 0.00878°
Ans.
128
6 ft
A
1 ft F
6 kip
2
1 ft AEF ⫽ 2 in
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4–11. The load is supported by the four 304 stainless steel
wires that are connected to the rigid members AB and DC.
Determine the vertical displacement of the 500-lb load if
the members were originally horizontal when the load was
applied. Each wire has a cross-sectional area of 0.025 in2.
E
F
G
3 ft
5 ft
H
D
C
1 ft
2 ft
1.8 ft
I
Internal Forces in the wires:
A
FBD (b)
FBC(4) - 500(3) = 0
+ c ©Fy = 0;
FAH + 375.0 - 500 = 0
FAH = 125.0 lb
a + ©MD = 0;
FCF(3) - 125.0(1) = 0
FCF = 41.67 lb
+ c ©Fy = 0;
FDE + 41.67 - 125.0 = 0
FBC = 375.0 lb
FBD (a)
FDE = 83.33 lb
Displacement:
dD =
83.33(3)(12)
FDELDE
= 0.0042857 in.
=
ADEE
0.025(28.0)(106)
dC =
41.67(3)(12)
FCFLCF
= 0.0021429 in.
=
ACFE
0.025(28.0)(106)
œ
dH
= 0.0014286 in.
dH = 0.0014286 + 0.0021429 = 0.0035714 in.
dA>H =
125.0(1.8)(12)
FAHLAH
= 0.0038571 in.
=
AAHE
0.025(28.0)(106)
dA = dH + dA>H = 0.0035714 + 0.0038571 = 0.0074286 in.
dB =
375.0(5)(12)
FBGLBG
= 0.0321428 in.
=
ABGE
0.025(28.0)(106)
dlœ
0.0247143
=
;
3
4
1 ft
500 lb
a + ©MA = 0;
œ
dH
0.0021429
=
;
2
3
B
3 ft
dlœ = 0.0185357 in.
dl = 0.0074286 + 0.0185357 = 0.0260 in.
Ans.
129
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*4–12. The load is supported by the four 304 stainless
steel wires that are connected to the rigid members AB and
DC. Determine the angle of tilt of each member after the
500-lb load is applied. The members were originally
horizontal, and each wire has a cross-sectional area of
0.025 in2.
E
F
G
3 ft
5 ft
H
D
C
1 ft
2 ft
1.8 ft
I
Internal Forces in the wires:
A
FBD (b)
FBG(4) - 500(3) = 0
+ c ©Fy = 0;
FAH + 375.0 - 500 = 0
FAH = 125.0 lb
a + ©MD = 0;
FCF(3) - 125.0(1) = 0
FCF = 41.67 lb
+ c ©Fy = 0;
FDE + 41.67 - 125.0 = 0
FBG = 375.0 lb
FBD (a)
FDE = 83.33 lb
Displacement:
dD =
83.33(3)(12)
FDELDE
= 0.0042857 in.
=
ADEE
0.025(28.0)(106)
dC =
41.67(3)(12)
FCFLCF
= 0.0021429 in.
=
ACFE
0.025(28.0)(106)
œ
dH
0.0021429
=
;
2
3
œ
dH
= 0.0014286 in.
œ
+ dC = 0.0014286 + 0.0021429 = 0.0035714 in.
dH = dH
tan a =
0.0021429
;
36
dA>H =
125.0(1.8)(12)
FAHLAH
= 0.0038571 in.
=
AAHE
0.025(28.0)(106)
a = 0.00341°
Ans.
dA = dH + dA>H = 0.0035714 + 0.0038571 = 0.0074286 in.
375.0(5)(12)
FBGLBG
= 0.0321428 in.
=
ABGE
0.025(28.0)(106)
tan b =
1 ft
500 lb
a + ©MA = 0;
dB =
B
3 ft
0.0247143
;
48
b = 0.0295°
Ans.
130
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•4–13.
The bar has a length L and cross-sectional area A.
Determine its elongation due to the force P and its own
weight.The material has a specific weight g (weight>volume)
and a modulus of elasticity E.
d =
=
L
P(x) dx
1
=
(gAx + P) dx
AE L0
L A(x) E
L
gAL2
gL2
1
PL
a
+ PL b =
+
AE
2
2E
AE
Ans.
P
4–14. The post is made of Douglas fir and has a diameter
of 60 mm. If it is subjected to the load of 20 kN and the soil
provides a frictional resistance that is uniformly distributed
along its sides of w = 4 kN>m, determine the force F at its
bottom needed for equilibrium.Also, what is the displacement
of the top of the post A with respect to its bottom B?
Neglect the weight of the post.
20 kN
A
y
w
2m
Equation of Equilibrium: For entire post [FBD (a)]
+ c ©Fy = 0;
F + 8.00 - 20 = 0
B
F = 12.0 kN
Ans.
Internal Force: FBD (b)
+ c ©Fy = 0;
- F(y) + 4y - 20 = 0
F(y) = {4y - 20} kN
Displacement:
L
dA>B =
2m
F(y)dy
1
=
(4y - 20)dy
AE L0
L0 A(y)E
=
1
A 2y2 - 20y B
AE
= -
冷
2m
0
32.0 kN # m
AE
32.0(103)
= -
p
2
4 (0.06 )
13.1 (109)
= - 0.8639 A 10 - 3 B m
= - 0.864 mm
Ans.
Negative sign indicates that end A moves toward end B.
131
F
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4–15. The post is made of Douglas fir and has a diameter
of 60 mm. If it is subjected to the load of 20 kN and the soil
provides a frictional resistance that is distributed along its
length and varies linearly from w = 0 at y = 0 to w =
3 kN>m at y = 2 m, determine the force F at its bottom
needed for equilibrium. Also, what is the displacement of
the top of the post A with respect to its bottom B? Neglect
the weight of the post.
20 kN
A
y
w
2m
B
F
Equation of Equilibrium: For entire post [FBD (a)]
+ c ©Fy = 0;
F + 3.00 - 20 = 0
F = 17.0 kN
Ans.
Internal Force: FBD (b)
+ c ©Fy = 0;
- F(y) +
1 3y
a b y - 20 = 0
2 2
3
F(y) = e y2 - 20 f kN
4
Displacement:
L
dA>B =
2m
F(y) dy
1
3
=
a y2 - 20b dy
AE L0
4
L0 A(y)E
=
2m
y3
1
a
- 20y b 2
AE 4
0
= -
38.0 kN # m
AE
38.0(103)
= -
p
2
4 (0.06 )
13.1 (109)
= - 1.026 A 10 - 3 B m
= - 1.03 mm
Ans.
Negative sign indicates that end A moves toward end B.
132
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*4–16. The linkage is made of two pin-connected A-36
steel members, each having a cross-sectional area of 1.5 in2.
If a vertical force of P = 50 kip is applied to point A,
determine its vertical displacement at A.
P
A
2 ft
B
C
1.5 ft
Analysing the equilibrium of Joint A by referring to its FBD, Fig. a,
+ ©F = 0 ;
:
x
+ c ©Fy = 0
The
3
3
FAC a b - FAB a b = 0
5
5
4
- 2Fa b - 50 = 0
5
initial
FAC = FAB = F
F = - 31.25 kip
length
of
members
AB
and
AC
is
12
in
b = 30 in. The axial deformation of members
L = 21.52 + 22 = (2.50 ft)a
1 ft
AB and AC is
d =
( - 31.25)(30)
FL
=
= - 0.02155 in.
AE
(1.5) C 29.0(103) D
The negative sign indicates that end A moves toward B and C. From the geometry
1.5
shown in Fig. b, u = tan - 1 a
b = 36.87°. Thus,
2
A dA B g =
d
0.02155
=
= 0.0269 in. T
cos u
cos 36.87°
Ans.
133
1.5 ft
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•4–17.
The linkage is made of two pin-connected A-36
steel members, each having a cross-sectional area of 1.5 in2.
Determine the magnitude of the force P needed to displace
point A 0.025 in. downward.
P
A
2 ft
Analysing the equilibrium of joint A by referring to its FBD, Fig. a
+ ©F = 0;
:
x
3
3
FAC a b - FAB a b = 0
5
5
+ c ©Fy = 0;
4
- 2Fa b - P = 0
5
The
initial
B
1.5 ft
F = - 0.625 P
of
members
AB
and
AC
are
12 in
b = 30 in. The axial deformation of members
L = 21.5 + 2 = (2.50 ft)a
1 ft
AB and AC is
2
length
2
d =
C
FAC = FAB = F
- 0.625P(30)
FL
=
= - 0.4310(10 - 3) P
AE
(1.5) C 29.0(103) D
The negative sign indicates that end A moves toward B and C. From the geometry
1.5
shown in Fig. b, we obtain u = tan - 1 a
b = 36.87°. Thus
2
(dA)g =
d
cos u
0.025 =
0.4310(10 - 3) P
cos 36.87°
P = 46.4 kips
Ans.
134
1.5 ft
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4–18. The assembly consists of two A-36 steel rods and a
rigid bar BD. Each rod has a diameter of 0.75 in. If a force
of 10 kip is applied to the bar as shown, determine the
vertical displacement of the load.
C
A
3 ft
2 ft
B
Here, FEF = 10 kip. Referring to the FBD shown in Fig. a,
1.25 ft
a + ©MB = 0;
FCD (2) - 10(1.25) = 0
FCD = 6.25 kip
a + ©MD = 0;
10(0.75) - FAB(2) = 0
FAB = 3.75 kip
The cross-sectional area of the rods is A =
A and C are fixed,
3.75 (2)(12)
FAB LAB
=
= 0.007025 in. T
A Est
0.140625p C 29.0(103) D
dD =
6.25(3)(12)
FCD LCD
=
= 0.01756 in T
A Est
0.140625p C 29.0(103) D
From the geometry shown in Fig. b
dE = 0.007025 +
1.25
(0.01756 - 0.00725) = 0.01361 in. T
2
Here,
dF>E =
10 (1) (12)
FEF LEF
=
= 0.009366 in T
A Est
0.140625p C 29.0(103) D
Thus,
A + T B dF = dE + dF>E = 0.01361 + 0.009366 = 0.0230 in T
Ans.
135
D
0.75 ft
F
p
(0.752) = 0.140625p in2. Since points
4
dB =
E
10 kip
1 ft
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4–19. The assembly consists of two A-36 steel rods and a
rigid bar BD. Each rod has a diameter of 0.75 in. If a force
of 10 kip is applied to the bar, determine the angle of tilt of
the bar.
C
A
Here, FEF = 10 kip. Referring to the FBD shown in Fig. a,
3 ft
a + ©MB = 0;
FCD (2) - 10(1.25) = 0
FCD = 6.25 kip
a + ©MD = 0;
10(0.75) - FAB (2) = 0
FAB = 3.75 kip
2 ft
B
p
The cross-sectional area of the rods is A = (0.752) = 0.140625p in2. Since points
4
A and C are fixed then,
dB =
3.75 (2)(12)
FAB LAB
=
= 0.007025 in T
A Est
0.140625p C 29.0(103) D
dD =
6.25 (3)(12)
FCD LCD
=
= 0.01756 in T
A Est
0.140625p C 29.0(103) D
0.01756 - 0.007025
= 0.439(10 - 3) rad
2(12)
Ans.
136
D
0.75 ft
F
10 kip
From the geometry shown in Fig. b,
u =
1.25 ft
E
1 ft
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*4–20. The rigid bar is supported by the pin-connected
rod CB that has a cross-sectional area of 500 mm2 and is
made of A-36 steel. Determine the vertical displacement of
the bar at B when the load is applied.
C
3m
45 kN/m
Force In The Rod. Referring to the FBD of member AB, Fig. a
a + ©MA = 0;
3
1
1
FBC a b (4) - (45)(4) c (4) d = 0
5
2
3
4m
Displacement. The initial length of rod BC is LBC = 232 + 42 = 5 m. The axial
deformation of this rod is
dBC =
50.0(103)(5)
FBC LBC
=
= 2.50 (10 - 3) m
ABC Est
0.5(10 - 3) C 200(109) D
3
From the geometry shown in Fig. b, u = tan - 1 a b = 36.87°. Thus,
4
(dB)g =
2.50(10 - 3)
dBC
=
= 4.167 (10 - 3) m = 4.17 mm
sin u
sin 36.87°
137
B
A
FBC = 50.0 kN
Ans.
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•4–21.
A spring-supported pipe hanger consists of two
springs which are originally unstretched and have a stiffness
of k = 60 kN>m, three 304 stainless steel rods, AB and CD,
which have a diameter of 5 mm, and EF, which has a
diameter of 12 mm, and a rigid beam GH. If the pipe and
the fluid it carries have a total weight of 4 kN, determine the
displacement of the pipe when it is attached to the support.
F
B
D
k
G
0.75 m
0.75 m
k
H
E
A
C
Internal Force in the Rods:
0.25 m 0.25 m
FBD (a)
a + ©MA = 0;
FCD (0.5) - 4(0.25) = 0
FAB + 2.00 - 4 = 0
+ c ©Fy = 0;
FCD = 2.00 kN
FAB = 2.00 kN
FBD (b)
FEF - 2.00 - 2.00 = 0
+ c ©Fy = 0;
FEF = 4.00 kN
Displacement:
dD = dE =
FEFLEF
=
AEFE
dA>B = dC>D =
4.00(103)(750)
p
4
(0.012)2(193)(109)
PCDLCD
=
ACDE
= 0.1374 mm
2(103)(750)
p
4
(0.005)2(193)(109)
= 0.3958 mm
dC = dD + dC>D = 0.1374 + 0.3958 = 0.5332 mm
Displacement of the spring
dsp =
Fsp
k
=
2.00
= 0.0333333 m = 33.3333 mm
60
dlat = dC + dsp
= 0.5332 + 33.3333 = 33.9 mm
Ans.
138
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4–22. A spring-supported pipe hanger consists of two
springs, which are originally unstretched and have a
stiffness of k = 60 kN>m, three 304 stainless steel rods, AB
and CD, which have a diameter of 5 mm, and EF, which has
a diameter of 12 mm, and a rigid beam GH. If the pipe is
displaced 82 mm when it is filled with fluid, determine the
weight of the fluid.
F
B
D
k
G
0.75 m
0.75 m
k
H
E
Internal Force in the Rods:
A
C
FBD (a)
a + ©MA = 0;
FCD(0.5) - W(0.25) = 0
FCD =
W
- W = 0
2
W
2
FAB +
+ c ©Fy = 0;
FAB =
0.25 m 0.25 m
W
2
FBD (b)
FEF -
+ c ©Fy = 0;
W
W
= 0
2
2
FEF = W
Displacement:
dD = dE =
FEFLEF
=
AEFE
W(750)
p
2
9
4 (0.012) (193)(10 )
= 34.35988(10 - 6) W
dA>B = dC>D =
FCDLCD
=
ACDE
W
2
(750)
p
2
9
4 (0.005) (193)(10 )
= 98.95644(10 - 6) W
dC = dD + dC>D
= 34.35988(10 - 6) W + 98.95644(10 - 6) W
= 0.133316(10 - 3) W
Displacement of the spring
dsp =
W
2
Fsp
k
=
60(103)
(1000) = 0.008333 W
dlat = dC + dsp
82 = 0.133316(10 - 3) W + 0.008333W
W = 9685 N = 9.69 kN
Ans.
139
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4–23. The rod has a slight taper and length L. It is
suspended from the ceiling and supports a load P at its end.
Show that the displacement of its end due to this load is
d = PL>1pEr2r12. Neglect the weight of the material. The
modulus of elasticity is E.
r(x) = r1 +
A(x) =
r2
r1L + (r2 - r1)x
r2 - r1
x =
L
L
L
p
(r1L + (r2 - r1)x)2
L2
r1
L
PL2
Pdx
dx
d =
=
2
A(x)E
pE
[r
L
+
(r
L0
L
1
2 - r1)x]
= -
L
1
PL2
c
dƒ
p E (r2 - r2)(r1L + (r2 - r1)x) 0
= -
=
= -
P
1
1
PL2
c
d
p E(r2 - r1) r1L + (r2 - r1)L
r1L
r1 - r2
PL2
1
1
PL2
c
d = c
d
p E(r2 - r1) r2L
r1L
p E(r2 - r1) r2r1L
r2 - r1
PL2
PL
c
d =
p E(r2 - r1) r2r1L
p E r2r1
QED
*4–24. Determine the relative displacement of one end of
the tapered plate with respect to the other end when it is
subjected to an axial load P.
P
d2
t
w = d1 +
d1 h + (d2 - d1)x
d2 - d1
x =
h
h
h
P(x) dx
P
=
d =
E L0 [d1h
L A(x)E
h
dx
+ ( d 2 - d1 )x ] t
h
h
=
Ph
dx
E t L0 d1 h + (d2 - d1)x
d1
P
h
dx
Ph
=
E t d1 h L0 1 + d2 -
h
d1 h
d2 - d1
Ph
d1
=
a
b c ln a1 +
xb d ƒ
E t d1 h d2 - d1
d1 h
0
d1 h x
=
d2 - d1
d1 + d2 - d1
Ph
Ph
c ln a 1 +
bd =
cln a
bd
E t(d2 - d1)
d1
E t(d2 - d1)
d1
=
d2
Ph
c ln d
E t(d2 - d1)
d1
Ans.
140
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4–25. Determine the elongation of the A-36 steel member
when it is subjected to an axial force of 30 kN. The member
is 10 mm thick. Use the result of Prob. 4–24.
20 mm
30 kN
30 kN
75 mm
0.5 m
Using the result of prob. 4-24 by substituting d1 = 0.02 m, d2 = 0.075 m t = 0.01 m
and L = 0.5 m.
d = 2c
= 2c
d2
PL
ln d
Est t(d2 - d1) d1
30(103) (0.5)
9
200(10 )(0.01)(0.075 - 0.02)
ln a
0.075
bd
0.02
= 0.360(10 - 3) m = 0.360 mm
Ans.
4–26. The casting is made of a material that has a specific
weight g and modulus of elasticity E. If it is formed into a
pyramid having the dimensions shown, determine how far
its end is displaced due to gravity when it is suspended in
the vertical position.
b0
b0
L
Internal Forces:
+ c ©Fz = 0;
P(z) -
1
gAz = 0
3
P(z) =
1
gAz
3
Displacement:
L
d =
P(z) dz
L0 A(z) E
L1
3
=
gAz
L0 A E
dz
=
L
g
z dz
3E L0
=
gL2
6E
Ans.
141
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4–27. The circular bar has a variable radius of r = r0eax
and is made of a material having a modulus of elasticity
of E. Determine the displacement of end A when it is
subjected to the axial force P.
L
Displacements: The cross-sectional area of the bar as a function of x is
A(x) = pr2 = pr0 2e2ax. We have
x
B
L
d =
L
P(x)dx
P
dx
=
2
A(x)E
pr0 E L0 e2ax
L0
r0
r ⫽ r0 eax
L
1
P
2
c
d
=
pr0 2E 2ae2ax 0
= -
A
P
a 1 - e - 2aL b
2apr0 2E
P
Ans.
*4–28. The pedestal is made in a shape that has a radius
defined by the function r = 2>12 + y1>22 ft, where y is
in feet. If the modulus of elasticity for the material is
E = 1411032 psi, determine the displacement of its top
when it supports the 500-lb load.
y
500 lb
0.5 ft
r⫽
2
(2 ⫹ y 1/ 2)
4 ft
d =
=
P(y) dy
L A(y) E
y
4
dy
500
3
14(10 )(144) L0 p(2 + y2-1
2
)
2
1 ft
4
-3
= 0.01974(10 )
L0
r
1
2
(4 + 4y + y) dy
4
2 3
1
= 0.01974(10 - 3) c 4y + 4a y2 b + y2 d
3
2
0
= 0.01974(10 - 3)(45.33)
= 0.8947(10 - 3) ft = 0.0107 in.
Ans.
142
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•4–29.
The support is made by cutting off the two
opposite sides of a sphere that has a radius r0 . If the original
height of the support is r0>2, determine how far it shortens
when it supports a load P. The modulus of elasticity is E.
P
r0
Geometry:
A = p r2 = p(r0 cos u)2 = p r20 cos2 u
y = r0 sin u;
dy = r0 cos u du
Displacement:
L
P(y) dy
L0 A(y) E
d =
= 2B
When y =
u
u
r0 cos u du
P
P
du
=
2
R
B
R
E L0 p r20 cos2 u
p r0 E L0 cos u
=
u
2P
[ln (sec u + tan u)] 2
p r0 E
0
=
2P
[ln (sec u + tan u)]
p r0 E
r0
;
4
u = 14.48°
d =
=
2P
[ln (sec 14.48° + tan 14.48°)]
p r0 E
0.511P
p r0 E
Ans.
Also,
Geometry:
A (y) = p x2 = p (r20 - y2)
Displacement:
L
d =
P(y) dy
L0 A(y) E
0
0
r0 + y p
2P 1
2P p dy
=
ln
=
B
R 2
2
2
pE L0 r0 - y
p E 2r0 r0 - y 0
=
P
[ln 1.667 - ln 1]
p r0 E
=
0.511 P
p r0 E
Ans.
143
r0
2
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4–30. The weight of the kentledge exerts an axial force of
P ⫽ 1500 kN on the 300-mm diameter high strength
concrete bore pile. If the distribution of the resisting skin
friction developed from the interaction between the soil
and the surface of the pile is approximated as shown, and
the resisting bearing force F is required to be zero,
determine the maximum intensity p0 kN>m for equilibrium.
Also, find the corresponding elastic shortening of the pile.
Neglect the weight of the pile.
P
p0
12 m
Internal Loading: By considering the equilibrium of the pile with reference to its
entire free-body diagram shown in Fig. a. We have
1
p (12) - 1500 = 0
2 0
+ c ©Fy = 0;
p0 = 250 kN>m
Ans.
Thus,
p(y) =
250
y = 20.83y kN>m
12
The normal force developed in the pile as a function of y can be determined by
considering the equilibrium of a section of the pile shown in Fig. b.
1
(20.83y)y - P(y) = 0
2
+ c ©Fy = 0;
P(y) = 10.42y2 kN
Displacement: The cross-sectional area of the pile is A =
p
(0.32) = 0.0225p m2.
4
We have
L
d =
12 m
P(y)dy
10.42(103)y2dy
=
0.0225p(29.0)(109)
L0
L0 A(y)E
12 m
=
L0
5.0816(10 - 6)y2dy
= 1.6939(10 - 6)y3 冷 0
12 m
= 2.9270(10 - 3)m = 2.93 mm
Ans.
144
F
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4–31. The column is constructed from high-strength
concrete and six A-36 steel reinforcing rods. If it is subjected
to an axial force of 30 kip, determine the average normal
stress in the concrete and in each rod. Each rod has a
diameter of 0.75 in.
4 in.
30 kip
Equations of Equilibrium:
6Pst + Pcon - 30 = 0
+ c ©Fy = 0;
3 ft
[1]
Compatibility:
dst = dcon
Pcon(3)(12)
Pst(3)(12)
p
4
(0.752)(29.0)(103)
=
[p4 (82) - 6(p4 )(0.75)2](4.20)(103)
Pst = 0.064065 Pcon
[2]
Solving Eqs. [1] and [2] yields:
Pst = 1.388 kip
Pcon = 21.670 kip
Average Normal Stress:
sst =
scon =
Pst
=
Ast
Pcon
=
Acon
1.388
p
2
4 (0.75 )
= 3.14 ksi
21.670
p 2
4 (8 )
- 6 A p4 B (0.752)
Ans.
= 0.455 ksi
Ans.
*4–32. The column is constructed from high-strength
concrete and six A-36 steel reinforcing rods. If it is subjected
to an axial force of 30 kip, determine the required diameter
of each rod so that one-fourth of the load is carried by the
concrete and three-fourths by the steel.
4 in.
30 kip
Equilibrium: The force of 30 kip is required to distribute in such a manner that 3/4
of the force is carried by steel and 1/4 of the force is carried by concrete. Hence
Pst =
3
(30) = 22.5 kip
4
Pcon =
1
(30) = 7.50 kip
4
3 ft
Compatibility:
dst = dcon
PstL
Pcon L
=
AstEst
Acon Econ
Ast =
22.5Acon Econ
7.50 Est
3 C p4 (82) - 6 A p4 B d2 D (4.20)(103)
p
6 a b d2 =
4
29.0(103)
d = 1.80 in.
Ans.
145
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•4–33.
The steel pipe is filled with concrete and subjected
to a compressive force of 80 kN. Determine the average
normal stress in the concrete and the steel due to this
loading. The pipe has an outer diameter of 80 mm and an
inner diameter of 70 mm. Est = 200 GPa, Ec = 24 GPa.
80 kN
Pst + Pcon - 80 = 0
+ c ©Fy = 0;
(1)
500 mm
dst = dcon
Pst L
p
2
4 (0.08
- 0.072) (200) (109)
Pcon L
p
2
(0.07
) (24)
4
=
(109)
Pst = 2.5510 Pcon
(2)
Solving Eqs. (1) and (2) yields
Pst = 57.47 kN
sst =
Pst
=
Ast
scon =
Pcon = 22.53 kN
57.47 (103)
p
4
(0.082 - 0.072)
= 48.8 MPa
Ans.
22.53 (103)
Pcon
= 5.85 MPa
= p
2
Acon
4 (0.07 )
Ans.
4–34. The 304 stainless steel post A has a diameter of
d = 2 in. and is surrounded by a red brass C83400 tube B.
Both rest on the rigid surface. If a force of 5 kip is applied
to the rigid cap, determine the average normal stress
developed in the post and the tube.
5 kip
B
B
A
8 in.
Equations of Equilibrium:
+ c ©Fy = 0;
3 in.
Pst + Pbr - 5 = 0[1]
Compatibility:
d
dst = dbr
Pst(8)
p 2
3
4 (2 )(28.0)(10 )
Pbr(8)
=
p 2
4 (6
- 52)(14.6)(103)
Pst = 0.69738 Pbr
[2]
Solving Eqs. [1] and [2] yields:
Pbr = 2.9457 kip
Pst = 2.0543 kip
Average Normal Stress:
sbr =
sst =
Pbr
=
Abr
2.9457
= 0.341 ksi
- 52)
Ans.
p 2
4 (6
Pst
2.0543
= p 2 = 0.654 ksi
Ast
4 (2 )
Ans.
146
0.5 in.
A
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4–35. The 304 stainless steel post A is surrounded by a red
brass C83400 tube B. Both rest on the rigid surface. If a
force of 5 kip is applied to the rigid cap, determine the
required diameter d of the steel post so that the load is
shared equally between the post and tube.
5 kip
B
B
A
A
8 in.
Equilibrium: The force of 5 kip is shared equally by the brass and steel. Hence
3 in.
Pst = Pbr = P = 2.50 kip
Compatibility:
d
0.5 in.
dst = dbr
PL
PL
=
AstEst
AbrEbr
Ast =
a
p 2
bd =
4
AbrEbr
Est
p
4
(62 - 52)(14.6)(103)
28.0(103)
d = 2.39 in.
Ans.
*4–36. The composite bar consists of a 20-mm-diameter
A-36 steel segment AB and 50-mm-diameter red brass
C83400 end segments DA and CB. Determine the average
normal stress in each segment due to the applied load.
+ ©F = 0;
;
x
250 mm
D
FC - FD + 75 + 75 - 100 - 100 = 0
FC - FD - 50 = 0
+
;
(1)
0 = ¢ D - dD
0 =
50(0.25)
150(0.5)
p
2
9
4 (0.02) (200)(10 )
-
-
FD(0.5)
p
2
9
4 (0.05 )(101)(10 )
p
2
9
4 (0.05 )(101)(10 )
-
500 mm
50 mm
FD(0.5)
p
2
9
4 (0.02 )(200)(10 )
FD = 107.89 kN
From Eq. (1), FC = 157.89 kN
sAD =
107.89(103)
PAD
= 55.0 MPa
= p
2
AAD
4 (0.05 )
Ans.
sAB =
42.11(103)
PAB
= 134 MPa
= p
2
AAB
4 (0.02 )
Ans.
sBC =
157.89(103)
PBC
= 80.4 MPa
= p
2
ABC
4 (0.05 )
Ans.
147
250 mm
20 mm
75 kN 100 kN
A
75 kN
100 kN B
C
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•4–37.
The composite bar consists of a 20-mm-diameter
A-36 steel segment AB and 50-mm-diameter red brass
C83400 end segments DA and CB. Determine
the displacement of A with respect to B due to the
applied load.
250 mm
D
+
;
0 =
-
500 mm
50 mm
250 mm
20 mm
75 kN 100 kN
A
75 kN
100 kN B
C
0 = ¢ D - dD
150(103)(500)
50(103)(250)
p
2
9
4 (0.02 )(200)(10 )
FD(500)
p
2
9
4 (0.05 )(101)(10 )
-
-
p
2
9
4 (0.05 )(101)(10 )
FD(500)
p
2
9
4 (0.02) (200)(10 )
FD = 107.89 kN
Displacement:
dA>B =
42.11(103)(500)
PABLAB
= p
2
9
AABEst
4 (0.02 )200(10 )
= 0.335 mm
Ans.
4–38. The A-36 steel column, having a cross-sectional area
of 18 in2, is encased in high-strength concrete as shown. If
an axial force of 60 kip is applied to the column, determine
the average compressive stress in the concrete and in the
steel. How far does the column shorten? It has an original
length of 8 ft.
16 in.
Pst + Pcon - 60 = 0
+ c ©Fy = 0;
dst = dcon ;
60 kip
Pst(8)(12)
18(29)(103)
(1)
Pcon(8)(12)
=
[(9)(16) - 18](4.20)(103)
Pst = 0.98639 Pcon
(2)
Solving Eqs. (1) and (2) yields
Pst = 29.795 kip;
sst =
Pcon = 30.205 kip
Pst
29.795
=
= 1.66 ksi
Ast
18
scon =
Ans.
Pcon
30.205
=
= 0.240 ksi
Acon
9(16) - 18
Ans.
Either the concrete or steel can be used for the deflection calculation.
d =
29.795(8)(12)
PstL
= 0.0055 in.
=
AstE
18(29)(103)
Ans.
148
9 in.
8 ft
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4–39. The A-36 steel column is encased in high-strength
concrete as shown. If an axial force of 60 kip is applied to
the column, determine the required area of the steel so that
the force is shared equally between the steel and concrete.
How far does the column shorten? It has an original length
of 8 ft.
60 kip
16 in.
9 in.
8 ft
The force of 60 kip is shared equally by the concrete and steel. Hence
Pst = Pcon = P = 30 kip
dcon = dst;
Ast =
PL
PL
=
Acon Econ
Ast Est
[9(16) - Ast] 4.20(103)
AconEcon
=
Est
29(103)
= 18.2 in2
d =
Ans.
30(8)(12)
PstL
= 0.00545 in.
=
AstEst
18.2(29)(103)
Ans.
*4–40. The rigid member is held in the position shown by
three A-36 steel tie rods. Each rod has an unstretched length
of 0.75 m and a cross-sectional area of 125 mm2. Determine
the forces in the rods if a turnbuckle on rod EF undergoes
one full turn. The lead of the screw is 1.5 mm. Neglect the
size of the turnbuckle and assume that it is rigid. Note: The
lead would cause the rod, when unloaded, to shorten 1.5 mm
when the turnbuckle is rotated one revolution.
B
D
0.75 m
E
A
0.5 m
0.5 m
C
0.75 m
a + ©ME = 0;
- TAB(0.5) + TCD(0.5) = 0
F
TAB = TCD = T
+ T ©Fy = 0;
(1)
TEF - 2T = 0
TEF = 2T
(2)
Rod EF shortens 1.5mm causing AB (and DC) to elongate. Thus:
0.0015 = dA>B + dE>F
0.0015 =
T(0.75)
-6
2T(0.75)
9
(125)(10 )(200)(10 )
+
(125)(10 - 6)(200)(109)
2.25T = 37500
T = 16666.67 N
TAB = TCD = 16.7 kN
Ans.
TEF = 33.3 kN
Ans.
149
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•4–41.
The concrete post is reinforced using six steel
reinforcing rods, each having a diameter of 20 mm.
Determine the stress in the concrete and the steel if the post
is subjected to an axial load of 900 kN. Est = 200 GPa,
Ec = 25 GPa.
900 kN
250 mm
375 mm
Referring to the FBD of the upper portion of the cut concrete post shown in Fig. a
Pcon + 6Pst - 900 = 0
+ c ©Fy = 0;
(1)
Since the steel rods and the concrete are firmly bonded, their deformation must be
the same. Thus
0 con = dst
Pcon L
Pst L
=
Acon Econ
Ast Est
C 0.25(0.375) -
Pcon L
6(p4 )(0.022)
D C 25(10 ) D
Pst L
=
9
(p4 )(0.022)
C 200(109) D
Pcon = 36.552 Pst
(2)
Solving Eqs (1) and (2) yields
Pst = 21.15 kN
Pcon = 773.10 kN
Thus,
scon =
sst =
773.10(103)
Pcon
= 8.42 MPa
=
Acon
0.15(0.375) - 6(p4 )(0.022)
21.15(103)
Pst
= 67.3 MPa
= p
2
Ast
4 (0.02 )
Ans.
Ans.
150
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4–42. The post is constructed from concrete and six A-36
steel reinforcing rods. If it is subjected to an axial force of
900 kN, determine the required diameter of each rod so that
one-fifth of the load is carried by the steel and four-fifths by
the concrete. Est = 200 GPa, Ec = 25 GPa.
900 kN
250 mm
375 mm
The normal force in each steel rod is
Pst =
1
5
(900)
6
= 30 kN
The normal force in concrete is
Pcon =
4
(900) = 720 kN
5
Since the steel rods and the concrete are firmly bonded, their deformation must be
the same. Thus
dcon = dst
Pcon L
Pst L
=
Acon Econ
Ast Est
720(103) L
30(103)L
C 0.25(0.375) - 6(p4 d2) D C 25(109) D
=
49.5p d2 = 0.09375
p
4
d2 C 200(109) D
d = 0.02455 m = 24.6 mm
Ans.
4–43. The assembly consists of two red brass C83400
copper alloy rods AB and CD of diameter 30 mm, a stainless
304 steel alloy rod EF of diameter 40 mm, and a rigid cap G.
If the supports at A, C and F are rigid, determine the
average normal stress developed in rods AB, CD and EF.
300 mm
450 mm
40 kN
A
B
E
30 mm
F
40 mm
C
Equation of Equilibrium: Due to symmetry, FAB = FCD = F. Referring to the freebody diagram of the assembly shown in Fig. a,
+ ©F = 0;
:
x
2F + FEF - 2 C 40(103) D = 0
(1)
Compatibility Equation: Using the method of superposition, Fig. b,
+ B 0 = -d + d
A:
P
EF
0 = -
40(103)(300)
p
2
9
4 (0.03 )(101)(10 )
+ cp
FEF (450)
2
9
4 (0.04 )(193)(10 )
+
A
B
FEF>2 (300)
d
p
2
9
4 (0.03 )(101)(10 )
FEF = 42 483.23 N
Substituting this result into Eq. (1),
F = 18 758.38 N
151
30 mm
40 kN
D
G
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4–43. Continued
Normal Stress: We have,
sAB = sCD =
sEF =
F
18 758.38
= 26.5 MPa
= p
2
ACD
4 (0.03 )
Ans.
FEF
42 483.23
= 33.8 MPa
= p
2
AEF
4 (0.04 )
Ans.
152
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*4–44. The two pipes are made of the same material and
are connected as shown. If the cross-sectional area of BC is
A and that of CD is 2A, determine the reactions at B and D
when a force P is applied at the junction C.
B
L
–
2
Equations of Equilibrium:
+ ©F = 0;
;
x
FB + FD - P = 0
[1]
Compatibility:
+ B
A:
0 = dP - dB
0 =
0 =
P A L2 B
2AE
- C
FB
A L2 B
AE
FB
+
A L2 B
2AE
S
3FBL
PL
4AE
4AE
FB =
P
3
Ans.
From Eq. [1]
FD =
C
2
P
3
Ans.
153
D
P
L
–
2
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•4–45.
The bolt has a diameter of 20 mm and passes
through a tube that has an inner diameter of 50 mm and an
outer diameter of 60 mm. If the bolt and tube are made of
A-36 steel, determine the normal stress in the tube and bolt
when a force of 40 kN is applied to the bolt. Assume the end
caps are rigid.
160 mm
40 kN
Referring to the FBD of left portion of the cut assembly, Fig. a
+ ©F = 0;
:
x
40(103) - Fb - Ft = 0
(1)
Here, it is required that the bolt and the tube have the same deformation. Thus
dt = db
Ft(150)
p
2
4 (0.06
- 0.05 ) C 200(10 ) D
2
Fb(160)
=
9
p
2
4 (0.02 )
C 200(109) D
Ft = 2.9333 Fb
(2)
Solving Eqs (1) and (2) yields
Fb = 10.17 (103) N
Ft = 29.83 (103) N
Thus,
sb =
10.17(103)
Fb
= 32.4 MPa
= p
2
Ab
4 (0.02 )
st =
Ft
=
At
29.83 (103)
p
2
4 (0.06
- 0.052)
40 kN
150 mm
Ans.
= 34.5 MPa
Ans.
154
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4–46. If the gap between C and the rigid wall at D is
initially 0.15 mm, determine the support reactions at A and
D when the force P = 200 kN is applied. The assembly
is made of A36 steel.
Equation of Equilibrium: Referring to the free-body diagram of the assembly
shown in Fig. a,
200(103) - FD - FA = 0
(1)
Compatibility Equation: Using the method of superposition, Fig. b,
+ B
A:
d = dP - dFD
0.15 =
200(103)(600)
p
2
9
4 (0.05 )(200)(10 )
- Cp
FD (600)
2
9
4 (0.05 )(200)(10 )
0.15 mm
P
A
+ ©F = 0;
:
x
600 mm
600 mm
+
FD (600)
S
p
2
(0.025
)(200)(109)
4
FD = 20 365.05 N = 20.4 kN
Ans.
Substituting this result into Eq. (1),
FA = 179 634.95 N = 180 kN
Ans.
155
50 mm
D
B
25 mm
C
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4–47. Two A-36 steel wires are used to support the 650-lb
engine. Originally, AB is 32 in. long and A¿B¿ is 32.008 in.
long. Determine the force supported by each wire when the
engine is suspended from them. Each wire has a crosssectional area of 0.01 in2.
B¿ B
A¿ A
TA¿B¿ + TAB - 650 = 0
+ c ©Fy = 0;
(1)
dAB = dA¿B¿ + 0.008
TA¿B¿ (32.008)
TAB (32)
(0.01)(29)(106)
=
(0.01)(29)(106)
+ 0.008
32TAB - 32.008TA¿B¿ = 2320
TAB = 361 lb
Ans.
TA¿B¿ = 289 lb
Ans.
*4–48. Rod AB has a diameter d and fits snugly between
the rigid supports at A and B when it is unloaded. The
modulus of elasticity is E. Determine the support reactions
at A and B if the rod is subjected to the linearly distributed
axial load.
p⫽
A
1
p L - FA - FB = 0
2 0
+ ©F = 0;
:
x
(1)
Compatibility Equation: Using the method of superposition, Fig. b,
+ B
A:
0 = dP - dFA
L
0 =
L0
FA (L)
P(x)dx
AE
AE
L
0 =
L0
B
x
Equation of Equilibrium: Referring to the free-body diagram of rod AB shown in
Fig. a,
P(x)dx - FAL
156
p0
p0
x
L
L
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4–48. Continued
Here, P(x) =
p0 2
1 p0
a xbx =
x . Thus,
2 L
2L
0 =
L
p0
x2 dx - FAL
2L L0
0 =
p0 x3 L
¢ ≤ ` - FAL
2L 3 0
FA =
p0L
6
Ans.
Substituting this result into Eq. (1),
FB =
p0L
3
Ans.
157
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•4–49.
The tapered member is fixed connected at its ends
A and B and is subjected to a load P = 7 kip at x = 30 in.
Determine the reactions at the supports. The material is
2 in. thick and is made from 2014-T6 aluminum.
A
B
3 in.
P
6 in.
x
60 in.
y
1.5
=
120 - x
60
y = 3 - 0.025 x
+ ©F = 0;
:
x
FA + FB - 7 = 0
(1)
dA>B = 0
30
-
L0
60
FA dx
FBdx
+
= 0
2(3 - 0.025 x)(2)(E)
L30 2(3 - 0.025 x)(2)(E)
30
- FA
L0
60
dx
dx
+ FB
= 0
(3 - 0.025 x)
L30 (3 - 0.025x)
60
40 FA ln(3 - 0.025 x)|30
0 - 40 FB ln(3 - 0.025x)|30 = 0
-FA(0.2876) + 0.40547 FB = 0
FA = 1.40942 FB
Thus, from Eq. (1).
FA = 4.09 kip
Ans.
FB = 2.91 kip
Ans.
158
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4–50. The tapered member is fixed connected at its ends
A and B and is subjected to a load P. Determine the location
x of the load and its greatest magnitude so that the average
normal stress in the bar does not exceed sallow = 4 ksi. The
member is 2 in. thick.
A
B
3 in.
P
6 in.
x
60 in.
y
1.5
=
120 - x
60
y = 3 - 0.025 x
+ ©F = 0;
:
x
FA + FB - P = 0
dA>B = 0
x
-
60
FA dx
FBdx
+
= 0
Lx 2(3 - 0.025 x)(2)(E)
L0 2(3 - 0.025 x)(2)(E)
x
- FA
60
dx
dx
+ FB
= 0
L0 (3 - 0.025 x)
Lx (3 - 0.025 x)
FA(40) ln (3 - 0.025 x)|x0 - FB(40) ln (3 - 0.025x)|60
x = 0
FA ln (1 -
0.025 x
0.025x
) = - FB ln (2 )
3
1.5
For greatest magnitude of P require,
4 =
FA
;
2(3 - 0.025 x)(2)
4 =
FB
;
2(3)
FA = 48 - 0.4 x
FB = 24 kip
Thus,
(48 - 0.4 x) ln a 1 -
0.025 x
0.025 x
b = - 24 ln a 2 b
3
1.5
Solving by trial and error,
x = 28.9 in.
Ans.
Therefore,
FA = 36.4 kip
P = 60.4 kip
Ans.
159
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4–51. The rigid bar supports the uniform distributed load
of 6 kip>ft. Determine the force in each cable if each cable
has a cross-sectional area of 0.05 in2, and E = 3111032 ksi.
C
6 ft
6 kip/ft
A
D
B
3 ft
a + ©MA = 0;
u = tan - 1
TCB a
2
25
b (3) - 54(4.5) + TCD a
2
25
b9 = 0
(1)
6
= 45°
6
L2B¿C¿ = (3)2 + (8.4853)2 - 2(3)(8.4853) cos u¿
Also,
L2D¿C¿ = (9)2 + (8.4853)2 - 2(9)(8.4853) cos u¿
(2)
Thus, eliminating cos u¿ .
-L2B¿C¿(0.019642) + 1.5910 = - L2D¿C¿(0.0065473) + 1.001735
L2B¿C¿(0.019642) = 0.0065473 L2D¿C¿ + 0.589256
L2B¿C¿ = 0.333 L2D¿C¿ + 30
But,
LB¿C = 245 + dBC¿ ,
LD¿C = 245 + dDC¿
Neglect squares or d¿ B since small strain occurs.
L2D¿C = ( 245 + dBC)2 = 45 + 2 245 dBC
L2D¿C = ( 245 + dDC)2 = 45 + 2 245 dDC
45 + 2245 dBC = 0.333(45 + 2245 dDC) + 30
2 245 dBC = 0.333(2245 dDC)
dDC = 3dBC
Thus,
TCD 245
TCB 245
= 3
AE
AE
TCD = 3 TCB
From Eq. (1).
TCD = 27.1682 kip = 27.2 kip
Ans.
TCB = 9.06 kip
Ans.
160
3 ft
3 ft
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*4–52. The rigid bar is originally horizontal and is
supported by two cables each having a cross-sectional area
of 0.05 in2, and E = 3111032 ksi. Determine the slight
rotation of the bar when the uniform load is applied.
C
See solution of Prob. 4-51.
6 ft
TCD = 27.1682 kip
dDC =
TCD 245
0.05(31)(103)
27.1682245
= 0.1175806 ft
0.05(31)(103)
=
6 kip/ft
A
D
B
Using Eq. (2) of Prob. 4-51,
3 ft
3 ft
3 ft
(245 + 0.1175806)2 = (9)2 + (8.4852)2 - 2(9)(8.4852) cos u¿
u¿ = 45.838°
Thus,
¢u = 45.838° - 45° = 0.838°
Ans.
•4–53.
The press consists of two rigid heads that are held
together by the two A-36 steel 12-in.-diameter rods. A 6061T6-solid-aluminum cylinder is placed in the press and the
screw is adjusted so that it just presses up against the
cylinder. If it is then tightened one-half turn, determine
the average normal stress in the rods and in the cylinder.
The single-threaded screw on the bolt has a lead of 0.01 in.
Note: The lead represents the distance the screw advances
along its axis for one complete turn of the screw.
12 in.
2 in.
10 in.
+ ©F = 0;
:
x
2Fst - Fal = 0
dst = 0.005 - dal
Fst(12)
p
( 4 )(0.5)2(29)(103)
= 0.005 -
Fal(10)
p(1)2(10)(103)
Solving,
Fst = 1.822 kip
Fal = 3.644 kip
srod =
Fst
1.822
= p
= 9.28 ksi
Ast
( 4 )(0.5)2
Ans.
scyl =
Fal
3.644
=
= 1.16 ksi
Aal
p(1)2
Ans.
161
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–54. The press consists of two rigid heads that are held
together by the two A-36 steel 12-in.-diameter rods. A 6061T6-solid-aluminum cylinder is placed in the press and the
screw is adjusted so that it just presses up against the
cylinder. Determine the angle through which the screw can
be turned before the rods or the specimen begin to yield.
The single-threaded screw on the bolt has a lead of 0.01 in.
Note: The lead represents the distance the screw advances
along its axis for one complete turn of the screw.
12 in.
2 in.
10 in.
+ ©F = 0;
:
x
2Fst - Fal = 0
dst = d - dal
Fst(12)
(p4 )(0.5)2(29)(103)
= d-
Fal(10)
(1)
p(1)2(10)(103)
Assume steel yields first,
sY = 36 =
Fst
(p4 )(0.5)2
;
Fst = 7.068 kip
Then Fal = 14.137 kip;
sal =
14.137
= 4.50 ksi
p(1)2
4.50 ksi 6 37 ksi steel yields first as assumed. From Eq. (1),
d = 0.01940 in.
Thus,
0.01940
u
=
360°
0.01
u = 698°
Ans.
162
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4–55. The three suspender bars are made of A-36 steel
and have equal cross-sectional areas of 450 mm2.
Determine the average normal stress in each bar if the rigid
beam is subjected to the loading shown.
A
2m
1m
+ c ©Fy = 0;
a + ©MD = 0;
FAD + FBE + FCF - 50(103) - 80(103) = 0
FBE(2) + FCF(4) - 50(103)(1) - 80(103)(3) = 0
(1)
(2)
Referring to the geometry shown in Fig. b,
dBE = dAD + a
dBE =
dCF - dAD
b(2)
4
1
A d + dCF B
2 AD
FBE L
FCF L
1 FADL
= a
+
b
AE
2
AE
AE
FAD + FCF = 2 FBE
(3)
Solving Eqs. (1), (2) and (3) yields
FBE = 43.33(103) N
FAD = 35.83(103) N
FCF = 50.83(103) N
Thus,
sBE =
43.33(103)
FBE
= 96.3 MPa
=
A
0.45(10 - 3)
Ans.
sAD =
35.83(103)
FAD
= 79.6 MPa
=
A
0.45(10 - 3)
Ans.
sCF = 113 MPa
Ans.
163
80 kN
50 kN
E
D
Referring to the FBD of the rigid beam, Fig. a,
C
B
1m
1m
F
1m
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*4–56. The rigid bar supports the 800-lb load. Determine
the normal stress in each A-36 steel cable if each cable has a
cross-sectional area of 0.04 in2.
C
12 ft
800 lb
B
A
5 ft
Referring to the FBD of the rigid bar, Fig. a,
FBC a
a + ©MA = 0;
12
3
b (5) + FCD a b (16) - 800(10) = 0
13
5
(1)
The unstretched length of wires BC and CD are LBC = 2122 + 52 = 13 ft and
LCD = 2122 + 162 = 20 ft. The stretches of wires BC and CD are
dBC =
FBC (13)
FBC LBC
=
AE
AE
dCD =
FCD(20)
FCD LCD
=
AE
AE
Referring to the geometry shown in Fig. b, the vertical displacement of the points on
d
12
3
the rigid bar is dg =
. For points B and D, cos uB =
and cos uD = . Thus,
cos u
13
5
the vertical displacement of points B and D are
A dB B g =
FBC (13)>AE
dBC
169 FBC
=
=
cos uB
12>13
12AE
A dD B g =
FCD (20)>AE
dCD
100 FCD
=
=
cos uD
3>5
3 AE
The similar triangles shown in Fig. c give
A dB B g
5
=
A dD B g
16
1 169 FBC
1 100 FCD
b =
b
a
a
5 12 AE
16
3AE
FBC =
125
F
169 CD
(2)
Solving Eqs. (1) and (2), yields
FCD = 614.73 lb
FBC = 454.69 lb
Thus,
sCD =
FCD
614.73
= 15.37(103) psi = 15.4 ksi
=
ACD
0.04
Ans.
sBC =
FBC
454.69
=
= 11.37(103) psi = 11.4 ksi
ABC
0.04
Ans.
164
D
5 ft
6 ft
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4–56. Continued
165
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•4–57.
The rigid bar is originally horizontal and is
supported by two A-36 steel cables each having a crosssectional area of 0.04 in2. Determine the rotation of the bar
when the 800-lb load is applied.
C
12 ft
800 lb
B
A
Referring to the FBD of the rigid bar Fig. a,
a + ©MA = 0;
FBC a
12
3
b (5) + FCD a b (16) - 800(10) = 0
13
5
5 ft
(1)
The unstretched length of wires BC and CD are LBC = 2122 + 52 = 13 ft and
LCD = 2122 + 162 = 20 ft. The stretch of wires BC and CD are
dBC =
FBC (13)
FBC LBC
=
AE
AE
dCD =
FCD(20)
FCD LCD
=
AE
AE
Referring to the geometry shown in Fig. b, the vertical displacement of the points on
d
12
3
the rigid bar is dg =
. For points B and D, cos uB =
and cos uD = . Thus,
cos u
13
5
the vertical displacement of points B and D are
A dB B g =
FBC (13)>AE
dBC
169 FBC
=
=
cos uB
12>13
12AE
A dD B g =
FCD (20)>AE
dCD
100 FCD
=
=
cos uD
3>5
3 AE
The similar triangles shown in Fig. c gives
A dB B g
5
=
A dD B g
16
1 169 FBC
1 100 FCD
a
b =
a
b
5 12 AE
16
3 AE
FBC =
125
F
169 CD
(2)
Solving Eqs (1) and (2), yields
FCD = 614.73 lb
FBC = 454.69 lb
Thus,
A dD B g =
100(614.73)
3(0.04) C 29.0 (106) D
= 0.01766 ft
Then
u = a
0.01766 ft 180°
ba
b = 0.0633°
p
16 ft
Ans.
166
D
5 ft
6 ft
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4–58. The horizontal beam is assumed to be rigid and
supports the distributed load shown. Determine the vertical
reactions at the supports. Each support consists of a wooden
post having a diameter of 120 mm and an unloaded
(original) length of 1.40 m. Take Ew = 12 GPa.
18 kN/m
A
B
C
1.40 m
2m
a + ©MB = 0;
+ c ©Fy = 0;
FC(1) - FA(2) = 0
(1)
FA + FB + FC - 27 = 0
dB - dA
dC - dA
=
;
2
3
1m
(2)
3dB - dA = 2dC
3FBL
FAL
2FCL
=
;
AE
AE
AE
3FB - FA = 2FC
(3)
Solving Eqs. (1)–(3) yields :
FA = 5.79 kN
Ans.
FB = 9.64 kN
Ans.
FC = 11.6 kN
Ans.
4–59. The horizontal beam is assumed to be rigid and
supports the distributed load shown. Determine the angle
of tilt of the beam after the load is applied. Each support
consists of a wooden post having a diameter of 120 mm and
an unloaded (original) length of 1.40 m. Take Ew = 12 GPa.
a + ©MB = 0;
c + ©Fy = 0;
18 kN/m
A
FC(1) - FA(2) = 0
2m
3FB - FA = 2FC
(3)
Solving Eqs. (1)–(3) yields :
FA = 5.7857 kN;
FB = 9.6428 kN;
FC = 11.5714 kN
3
dA =
5.7857(10 )(1.40)
FAL
= 0.0597(10 - 3) m
= p
2
9
AE
(0.12
)12(10
)
4
dC =
11.5714(103)(1.40)
FCL
= 0.1194(10 - 3) m
= p
2
9
AE
(0.12
)12(10
)
4
tan u =
1.40 m
(2)
3dB - dA = 2dC
3FBL
FAL
2FCL
=
;
AE
AE
AE
C
(1)
FA + FB + FC - 27 = 0
dB - dA
dC - dA
=
;
2
3
B
0.1194 - 0.0597
(10 - 3)
3
u = 0.0199(10 - 3) rad = 1.14(10 - 3)°
Ans.
167
1m
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*4–60. The assembly consists of two posts AD and CF
made of A-36 steel and having a cross-sectional area of
1000 mm2, and a 2014-T6 aluminum post BE having a crosssectional area of 1500 mm2. If a central load of 400 kN is
applied to the rigid cap, determine the normal stress in each
post. There is a small gap of 0.1 mm between the post BE
and the rigid member ABC.
400 kN
0.5 m
A
Equation of Equilibrium. Due to symmetry, FAD = FCF = F. Referring to the
FBD of the rigid cap, Fig. a,
FBE + 2F - 400(103) = 0
(1)
Compatibility Equation. Referring to the initial and final position of rods AD (CF)
and BE, Fig. b,
d = 0.1 + dBE
F(400)
1(10 ) C 200(10 ) D
-3
9
= 0.1 +
FBE (399.9)
1.5(10 - 3) C 73.1(109) D
F = 1.8235 FBE + 50(103)
(2)
Solving Eqs (1) and (2) yield
FBE = 64.56(103) N
F = 167.72(103) N
Normal Stress.
sAD = sCF =
sBE =
B
C
0.4 m
D
+ c ©Fy = 0;
0.5 m
167.72(103)
F
= 168 MPa
=
Ast
1(10 - 3)
Ans.
64.56(103)
FBE
= 43.0 MPa
=
Aal
1.5(10 - 3)
Ans.
168
E
F
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•4–61.
The distributed loading is supported by the three
suspender bars. AB and EF are made of aluminum and CD
is made of steel. If each bar has a cross-sectional area of
450 mm2, determine the maximum intensity w of the
distributed loading so that an allowable stress of 1sallow2st =
180 MPa in the steel and 1sallow2al = 94 MPa in the
aluminum is not exceeded. Est = 200 GPa, Eal = 70 GPa.
Assume ACE is rigid.
1.5 m
1.5 m
B
al
D
st
A
F
al
C
2m
E
w
a + ©MC = 0;
FEF(1.5) - FAB(1.5) = 0
FEF = FAB = F
+ c ©Fy = 0;
2F + FCD - 3w = 0
(1)
Compatibility condition :
dA = dC
FCDL
FL
=
;
9
A(70)(10 )
A(200)(109)
F = 0.35 FCD
(2)
Assume failure of AB and EF:
F = (sallow)al A
= 94(106)(450)(10 - 6)
= 42300 N
From Eq. (2) FCD = 120857.14 N
From Eq. (1) w = 68.5 kN>m
Assume failure of CD:
FCD = (sallow)st A
= 180(106)(450)(10 - 6)
= 81000 N
From Eq. (2) F = 28350 N
From Eq. (1) w = 45.9 kN>m
(controls)
Ans.
169
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4–62. The rigid link is supported by a pin at A, a steel wire
BC having an unstretched length of 200 mm and crosssectional area of 22.5 mm2, and a short aluminum block
having an unloaded length of 50 mm and cross-sectional
area of 40 mm2. If the link is subjected to the vertical load
shown, determine the average normal stress in the wire and
the block. Est = 200 GPa, Eal = 70 GPa.
C
200 mm
B
100 mm
D
450(250) - FBC(150) - FD(150) = 0
50 mm
750 - FBC - FD = 0
[1]
Compatibility:
dBC = dD
FD(50)
FBC(200)
22.5(10 - 6)200(109)
=
40(10 - 6)70(109)
FBC = 0.40179 FD
[2]
Solving Eqs. [1] and [2] yields:
FD = 535.03 N
FBC = 214.97 N
Average Normal Stress:
sD =
sBC =
150 mm
450 N
Equations of Equilibrium:
a + ©MA = 0;
A
150 mm
FD
535.03
= 13.4 MPa
=
AD
40(10 - 6)
Ans.
FBC
214.97
= 9.55 MPa
=
ABC
22.5(10 - 6)
Ans.
170
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4–63. The rigid link is supported by a pin at A, a steel wire
BC having an unstretched length of 200 mm and crosssectional area of 22.5 mm2, and a short aluminum block
having an unloaded length of 50 mm and cross-sectional area
of 40 mm2. If the link is subjected to the vertical load shown,
determine the rotation of the link about the pin A. Report
the answer in radians. Est = 200 GPa, Eal = 70 GPa.
C
200 mm
B
100 mm
D
450(250) - FBC(150) - FD(150) = 0
50 mm
750 - FBC - FD = 0
[1]
Compatibility:
dBC = dD
FD(50)
FBC(200)
-6
9
22.5(10 )200(10 )
=
40(10 - 6)70(109)
FBC = 0.40179 FD
[2]
Solving Eqs. [1] and [2] yields :
FD = 535.03 N
FBC = 214.97 N
Displacement:
dD =
535.03(50)
FDLD
= 0.009554 mm
=
ADEal
40(10 - 6)(70)(109)
tan u =
150 mm
450 N
Equations of Equilibrium:
a + ©MA = 0;
A
150 mm
dD
0.009554
=
150
150
u = 63.7(10 - 6) rad = 0.00365°
Ans.
171
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*4–64. The center post B of the assembly has an original
length of 124.7 mm, whereas posts A and C have a length of
125 mm. If the caps on the top and bottom can be
considered rigid, determine the average normal stress in
each post. The posts are made of aluminum and have a
cross-sectional area of 400 mm2. Eal = 70 GPa.
800 kN/m
A
a + ©MB = 0;
100 mm
B
C
- FA(100) + FC(100) = 0
(1)
2F + FB - 160 = 0
(2)
dA = dB + 0.0003
F (0.125)
FB (0.1247)
400 (10 - 6)(70)(106)
=
400 (10 - 6)(70)(106)
+ 0.0003
0.125 F - 0.1247FB = 8.4
(3)
Solving Eqs. (2) and (3)
F = 75.762 kN
FB = 8.547 kN
sA = sC =
sB =
75.726 (103)
400(10 - 6)
8.547 (103)
400 (10 - 6)
= 189 MPa
Ans.
= 21.4 MPa
Ans.
•4–65.
The assembly consists of an A-36 steel bolt and a
C83400 red brass tube. If the nut is drawn up snug against
the tube so that L = 75 mm, then turned an additional
amount so that it advances 0.02 mm on the bolt, determine
the force in the bolt and the tube. The bolt has a diameter of
7 mm and the tube has a cross-sectional area of 100 mm2.
L
Equilibrium: Since no external load is applied, the force acting on the tube and the
bolt is the same.
Compatibility:
0.02 = dt + db
0.02 =
P(75)
P(75)
-6
9
100(10 )(101)(10 )
+
125 mm
800 kN/m
FA = FC = F
+ c ©Fy = 0;
100 mm
p
2
9
4 (0.007 )(200)(10 )
P = 1164.83 N = 1.16 kN
Ans.
172
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4–66. The assembly consists of an A-36 steel bolt and a
C83400 red brass tube. The nut is drawn up snug against
the tube so that L = 75 mm. Determine the maximum
additional amount of advance of the nut on the bolt so that
none of the material will yield. The bolt has a diameter of
7 mm and the tube has a cross-sectional area of 100 mm2.
L
Allowable Normal Stress:
(sg)st = 250 A 106 B =
Pst
p
2
4 (0.007)
Pst = 9.621 kN
(sg)br = 70.0 A 106 B =
Pbr
100(10 - 6)
Pbr = 7.00 kN
Since Pst 7 Pbr, by comparison he brass will yield first.
Compatibility:
a = dt + db
7.00(103)(75)
=
100(10 - 6)(101)(109)
7.00(103)(75)
+
p
2
9
4 (0.007) (200)(10 )
= 0.120 mm
Ans.
173
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4–67. The three suspender bars are made of the same
material and have equal cross-sectional areas A. Determine
the average normal stress in each bar if the rigid beam ACE
is subjected to the force P.
a + ©MA = 0;
D
d
FCD(d) + FEF(2d) - Pa b = 0
2
FCD + 2FEF =
+ c ©Fy = 0;
B
F
L
P
P
2
(1)
FAB + FCD + FEF - P = 0
A
C
d
2
(2)
d
2
E
d
dC - dE
dA - dE
=
d
2d
2dC = dA + dE
2FCDL
FABL
FEFL
=
+
AE
AE
AE
2FCD - FAB - FEF = 0
(3)
Solving Eqs. (1), (2) and (3) yields
P
3
P
12
FAB =
7P
12
sAB =
7P
12A
Ans.
sCD =
P
3A
Ans.
sEF =
P
12A
Ans.
FCD =
FEF =
*4–68. A steel surveyor’s tape is to be used to measure the
length of a line. The tape has a rectangular cross section of
0.05 in. by 0.2 in. and a length of 100 ft when T1 = 60°F and
the tension or pull on the tape is 20 lb. Determine the
true length of the line if the tape shows the reading to
be 463.25 ft when used with a pull of 35 lb at T2 = 90°F. The
ground on which it is placed is flat. ast = 9.60110-62>°F,
Est = 2911032 ksi.
P
P
0.2 in.
0.05 in.
dT = a¢TL = 9.6(10 - 6)(90 - 60)(463.25) = 0.133416 ft
d =
(35 - 20)(463.25)
PL
= 0.023961 ft
=
AE
(0.2)(0.05)(29)(106)
L = 463.25 + 0.133416 + 0.023961 = 463.41 ft
Ans.
174
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•4–69.
Three bars each made of different materials are
connected together and placed between two walls when the
temperature is T1 = 12°C. Determine the force exerted
on the (rigid) supports when the temperature becomes
T2 = 18°C. The material properties and cross-sectional
area of each bar are given in the figure.
Steel
Copper
Brass
Est ⫽ 200 GPa
Ebr ⫽ 100 GPa
Ecu ⫽ 120 GPa
ast ⫽ 12(10⫺6)/⬚C abr ⫽ 21(10⫺6)/°C acu ⫽ 17(10⫺6)/⬚C
Ast ⫽ 200 mm2
300 mm
+ )
(;
Acu ⫽ 515 mm2
Abr ⫽ 450 mm2
200 mm
100 mm
0 = ¢T - d
0 = 12(10 - 6)(6)(0.3) + 21 (10 - 6)(6)(0.2) + 17 (10 - 6)(6)(0.1)
F(0.3)
-
-6
F(0.2)
9
200(10 )(200)(10 )
-
-6
F(0.1)
9
-
450(10 )(100)(10 )
515(10 - 6)(120)(109)
F = 4203 N = 4.20 kN
Ans.
k ⫽ 1000 lb/ in.
4–70. The rod is made of A-36 steel and has a diameter of
0.25 in. If the rod is 4 ft long when the springs are compressed
0.5 in. and the temperature of the rod is T = 40°F, determine
the force in the rod when its temperature is T = 160°F.
k ⫽ 1000 lb/in.
4 ft
Compatibility:
+ B
A:
x = dT - dF
x = 6.60(10 - 6)(160 - 40)(2)(12) -
1.00(0.5)(2)(12)
p
2
3
4 (0.25 )(29.0)(10 )
x = 0.01869 in.
F = 1.00(0.01869 + 0.5) = 0.519 kip
Ans.
4–71. A 6-ft-long steam pipe is made of A-36 steel with
sY = 40 ksi. It is connected directly to two turbines A
and B as shown. The pipe has an outer diameter of 4 in.
and a wall thickness of 0.25 in. The connection was made
at T1 = 70°F. If the turbines’ points of attachment are
assumed rigid, determine the force the pipe exerts on
the turbines when the steam and thus the pipe reach a
temperature of T2 = 275°F.
6 ft
A
Compatibility:
+ B
A:
0 = dT - dF
0 = 6.60(10 - 6)(275 - 70)(6)(12) -
F(6)(12)
p 2
4 (4
- 3.52)(29.0)(103)
F = 116 kip
Ans.
175
B
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*4–72. A 6-ft-long steam pipe is made of A-36 steel with
sY = 40 ksi. It is connected directly to two turbines A and
B as shown. The pipe has an outer diameter of 4 in. and a
wall thickness of 0.25 in. The connection was made at
T1 = 70°F. If the turbines’ points of attachment are
assumed to have a stiffness of k = 8011032 kip>in., determine
the force the pipe exerts on the turbines when the steam
and thus the pipe reach a temperature of T2 = 275°F.
6 ft
A
B
Compatibility:
x = dT - dF
x = 6.60(10 - 6)(275 - 70)(3)(12) -
80(103)(x)(3)(12)
p 2
4 (4
- 3.52)(29.0)(103)
x = 0.001403 in.
F = k x = 80(103)(0.001403) = 112 kip
Ans.
•4–73.
The pipe is made of A-36 steel and is connected to
the collars at A and B. When the temperature is 60° F, there
is no axial load in the pipe. If hot gas traveling through the
pipe causes its temperature to rise by ¢T = 140 + 15x2°F,
where x is in feet, determine the average normal stress in the
pipe. The inner diameter is 2 in., the wall thickness is 0.15 in.
A
Compatibility:
L
0 = dT - dF
0 = 6.60 A 10 - 6 B
Where
dT =
L0
8ft
L0
(40 + 15 x) dx -
0 = 6.60 A 10 - 6 B B 40(8) +
a ¢T dx
F(8)
A(29.0)(103)
15(8)2
F(8)
R 2
A(29.0)(103)
F = 19.14 A
Average Normal Stress:
s =
B
8 ft
19.14 A
= 19.1 ksi
A
Ans.
176
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4–74. The bronze C86100 pipe has an inner radius of
0.5 in. and a wall thickness of 0.2 in. If the gas flowing
through it changes the temperature of the pipe uniformly
from TA = 200°F at A to TB = 60°F at B, determine the
axial force it exerts on the walls. The pipe was fitted between
the walls when T = 60°F.
A
B
8 ft
Temperature Gradient:
T(x) = 60 + a
8 - x
b 140 = 200 - 17.5x
8
Compatibility:
0 = dT - dF
0 = 9.60 A 10 - 6 B
Where
dT = 1 a¢Tdx
2ft
0 = 9.60 A 10 - 6 B
L0
[(200 - 17.5x) - 60] dx 2ft
L0
(140 - 17.5x) dx -
F(8)
p
2
4 (1.4
- 12)15.0(103)
F(8)
p
2
4 (1.4
- 12) 15.0(103)
F = 7.60 kip
Ans.
4–75. The 40-ft-long A-36 steel rails on a train track are laid
with a small gap between them to allow for thermal
expansion. Determine the required gap d so that the rails just
touch one another when the temperature is increased from
T1 = - 20°F to T2 = 90°F. Using this gap, what would be the
axial force in the rails if the temperature were to rise to
T3 = 110°F? The cross-sectional area of each rail is 5.10 in2.
d
40 ft
Thermal Expansion: Note that since adjacent rails expand, each rail will be
d
required to expand on each end, or d for the entine rail.
2
d = a¢TL = 6.60(10 - 6)[90 - ( - 20)](40)(12)
Ans.
= 0.34848 in. = 0.348 in.
Compatibility:
+ B
A:
0.34848 = dT - dF
0.34848 = 6.60(10 - 6)[110 - ( -20)](40)(12) -
d
F(40)(12)
5.10(29.0)(103)
F = 19.5 kip
Ans.
177
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*4–76. The device is used to measure a change in
temperature. Bars AB and CD are made of A-36 steel and
2014-T6 aluminum alloy respectively. When the temperature
is at 75°F, ACE is in the horizontal position. Determine the
vertical displacement of the pointer at E when the
temperature rises to 150°F.
0.25 in.
A
3 in.
C
E
1.5 in.
Thermal Expansion:
A dT B CD = aal ¢TLCD = 12.8(10 - 6)(150 - 75)(1.5) = 1.44(10 - 3) in.
B
D
A dT B AB = ast ¢TLAB = 6.60(10 - 6)(150 - 75)(1.5) = 0.7425(10 - 3) in.
From the geometry of the deflected bar AE shown Fig. b,
dE = A dT B AB + C
= 0.7425(10 - 3) + B
A dT B CD - A dT B AB
0.25
S (3.25)
1.44(10 - 3) - 0.7425(10 - 3)
R (3.25)
0.25
= 0.00981 in.
Ans.
•4–77. The bar has a cross-sectional area A, length L,
modulus of elasticity E, and coefficient of thermal
expansion a. The temperature of the bar changes uniformly
along its length from TA at A to TB at B so that at any point
x along the bar T = TA + x1TB - TA2>L. Determine the
force the bar exerts on the rigid walls. Initially no axial force
is in the bar and the bar has a temperature of TA.
+
:
x
A
TA
0 = ¢ T - dF
(1)
However,
d¢ T = a¢ T dx = a(TA +
TB - TA
x - TA)dx
L
L
¢T = a
= ac
L
TB - TA
TB - TA 2
x dx = a c
x d冷
L
2L
L0
0
TB - TA
aL
Ld =
(TB - TA)
2
2
From Eq.(1).
0 =
FL
aL
(TB - TA) 2
AE
F =
a AE
(TB - TA)
2
B
Ans.
178
TB
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4–78. The A-36 steel rod has a diameter of 50 mm and is
lightly attached to the rigid supports at A and B when
T1 = 80°C. If the temperature becomes T2 = 20°C and an
axial force of P = 200 kN is applied to its center, determine
the reactions at A and B.
0.5 m
FB - FA + 200(103) = 0
(1)
When the rod is unconstrained at B, it has a free contraction of
dT = ast ¢ TL = 12(10 - 6)(80 - 20)(1000) = 0.72 mm. Also, under force P and FB
with unconstrained at B, the deformation of the rod are
dP =
dFB =
PLAC
=
AE
FB LAB
=
AE
200(103)(500)
p
2
4 (0.05 )
C 200(109) D
FB (1000)
p
2
4 (0.05 )
C 200(109) D
= 0.2546 mm
= 2.5465(10 - 6) FB
Using the method of super position, Fig. b,
+ B
A:
B
P
Referring to the FBD of the rod, Fig. a
+ ©F = 0;
:
x
C
A
0 = - dT + dP + dFB
0 = - 0.72 + 0.2546 + 2.5465(10 - 6) FB
FB = 182.74(103) N = 183 kN
Ans.
Substitute the result of FB into Eq (1),
FA = 382.74(103) N = 383 kN
Ans.
179
0.5 m
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4–79. The A-36 steel rod has a diameter of 50 mm and is
lightly attached to the rigid supports at A and B when
T1 = 50°C. Determine the force P that must be applied to
the collar at its midpoint so that, when T2 = 30°C, the
reaction at B is zero.
C
A
B
P
0.5 m
0.5 m
When the rod is unconstrained at B, it has a free contraction of
dT = ast ¢TL = 12(10 - 6)(50 - 30)(1000) = 0.24 mm. Also, under force P with
unconstrained at B, the deformation of the rod is
dP =
PLAC
=
AE
P(500)
p
2
4 (0.05 )
C 200(109) D
= 1.2732(10 - 6) P
Since FB is required to be zero, the method of superposition, Fig. b, gives
+ B
A:
0 = - dT + dP
0 = - 0.24 + 1.2732(10 - 6)P
P = 188.50(103) N = 188 kN
Ans.
*4–80. The rigid block has a weight of 80 kip and is to be
supported by posts A and B, which are made of A-36 steel,
and the post C, which is made of C83400 red brass. If all the
posts have the same original length before they are loaded,
determine the average normal stress developed in each post
when post C is heated so that its temperature is increased
by 20°F. Each post has a cross-sectional area of 8 in2.
A
Equations of Equilibrium:
a + ©MC = 0;
+ c ©Fy = 0;
FB(3) - FA(3) = 0
FA = FB = F
2F + FC - 80 = 0
[1]
Compatibility:
(dC)F - (dC)T = dF
(+ T)
FCL
8(14.6)(103)
- 9.80 A 10 - 5 B (20)L =
FL
8(29.0)(103)
8.5616 FC - 4.3103 F = 196
[2]
Solving Eqs. [1] and [2] yields:
F = 22.81 kip
FC = 34.38 kip
average Normal Sress:
sA = sB =
sC =
F
22.81
=
= 2.85 ksi
A
8
Ans.
FC
34.38
=
= 4.30 ksi
A
8
Ans.
180
C
B
3 ft
3 ft
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•4–81.
The three bars are made of A-36 steel and form
a pin-connected truss. If the truss is constructed when
T1 = 50°F, determine the force in each bar when
T2 = 110°F. Each bar has a cross-sectional area of 2 in2.
t
5f
5f
t
A
4 ft
B
D
3 ft
(dT ¿)AB - (dF ¿)AB = (dT)AD + (dF)AD
(1)
œ
cos u;
However, dAB = dAB
œ
dAB
=
dAB
5
= dAB
cos u
4
Substitute into Eq. (1)
5
5
(dT)AB - (dF)AB = (dT)AD + (dF)AD
4
4
FAB(5)(12)
5
d
c 6.60(10 - 6)(110° - 50°)(5)(12) 4
2(29)(103)
= 6.60(10 - 6)(110° - 50°)(4)(12) +
FAD(4)(12)
2(29)(103)
620.136 = 75FAB + 48FAD
+ ©F = 0;
:
x
3
3
F
- FAB = 0;
5 AC
5
+ c ©Fy = 0;
4
FAD - 2 a FAB b = 0
5
(2)
FAC = FAB
(3)
Solving Eqs. (2) and (3) yields :
FAD = 6.54 kip
Ans.
FAC = FAB = 4.09 kip
Ans.
181
C
3 ft
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4–82. The three bars are made of A-36 steel and form a pinconnected truss. If the truss is constructed when T1 = 50°F,
determine the vertical displacement of joint A when
T2 = 150°F. Each bar has a cross-sectional area of 2 in2.
t
5f
5f
t
A
4 ft
(dT ¿)AB - (dF ¿)AB = (dT)AD + (dF)AD
(1)
œ
cos u;
However, dAB = dAB
œ
dAB
=
B
dAB
5
= dAB
cos u
4
3 ft
Substitute into Eq. (1)
5
5
(d )
- (dT)AB = (dT)AD + (dF)AD
4 T AB
4
FAB(5)(12)
5
d
c 6.60(10 - 6)(150° - 50°)(5)(12) 4
2(29)(103)
= 6.60(10 - 6)(150° - 50°)(4)(12) +
FAD(4)(12)
2(29)(103)
239.25 - 6.25FAB = 153.12 + 4 FAD
4 FAD + 6.25FAB = 86.13
+ © F = 0;
:
x
3
3
F
- FAB = 0;
5 AC
5
+ c © Fy = 0;
4
FAD - 2 a FAB b = 0;
5
(2)
FAC = FAB
FAD = 1.6FAB
(3)
Solving Eqs. (2) and (3) yields:
FAB = 6.8086 kip:
FAD = 10.8939 kip
(dA)r = (dT)AD + (dT)AD
= 6.60(10 - 6)(150° - 50°)(4)(12) +
D
10.8939(4)(12)
2(29)(103)
= 0.0407 in. c
Ans.
182
C
3 ft
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4–83. The wires AB and AC are made of steel, and wire
AD is made of copper. Before the 150-lb force is applied,
AB and AC are each 60 in. long and AD is 40 in. long. If
the temperature is increased by 80°F, determine the
force in each wire needed to support the load. Take
Est = 29(103) ksi, Ecu = 17(103) ksi, ast = 8(10-6)>°F,
acu = 9.60(10-6)>°F. Each wire has a cross-sectional area of
0.0123 in2.
40 in.
60 in.
45⬚
45⬚
A
150 lb
Equations of Equilibrium:
+ ©F = 0;
:
x
FAC cos 45° - FAB cos 45° = 0
FAC = FAB = F
2F sin 45° + FAD - 150 = 0
+ c ©Fy = 0;
[1]
Compatibility:
(dAC)T = 8.0 A 10 - 6 B (80)(60) = 0.03840 in.
(dAC)Tr =
(dAC)T
0.03840
=
= 0.05431 in.
cos 45°
cos 45°
(dAD)T = 9.60 A 10 - 6 B (80)(40) = 0.03072 in.
d0 = (dAC)Tr - (dAD)T = 0.05431 - 0.03072 = 0.02359 in.
(dAD)F = (dAC)Fr + d0
F(60)
FAD (40)
6
0.0123(17.0)(10 )
=
0.0123(29.0)(106) cos 45°
C
D
B
+ 0.02359
0.1913FAD - 0.2379F = 23.5858
[2]
Solving Eq. [1] and [2] yields:
FAC = FAB = F = 10.0 lb
Ans.
FAD = 136 lb
Ans.
183
60 in.
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*4–84. The AM1004-T61 magnesium alloy tube AB is
capped with a rigid plate E. The gap between E and end C of
the 6061-T6 aluminum alloy solid circular rod CD is 0.2 mm
when the temperature is at 30° C. Determine the normal
stress developed in the tube and the rod if the temperature
rises to 80° C. Neglect the thickness of the rigid cap.
25 mm
a
Section a-a
E
B
A
20 mm
C
25 mm
a
0.2 mm
300 mm
Compatibility Equation: If tube AB and rod CD are unconstrained, they will have a
free expansion of A dT B AB = amg ¢TLAB = 26(10 - 6)(80 - 30)(300) = 0.39 mm and
A dT)CD = aal ¢TLCD = 24(10 - 6)(80 - 30)(450) = 0.54 mm. Referring
deformation diagram of the tube and the rod shown in Fig. a,
d =
to
the
C A dT B AB - A dF B AB D + C A dT B CD - A dF B CD D
0.2 = C 0.39 -
F(300)
p A 0.025 - 0.02 B (44.7)(10 )
2
2
9
S + C 0.54 -
F(450)
p
4
A 0.0252 B (68.9)(109)
S
F = 32 017.60 N
Normal Stress:
sAB =
F
32 017.60
=
= 45.3 MPa
AAB
p A 0.0252 - 0.022 B
sCD =
F
32 017.60
=
= 65.2 MPa
p
2
ACD
4 A 0.025 B
Ans.
Ans.
F = 107 442.47 N
184
450 mm
D
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•4–85.
The AM1004-T61 magnesium alloy tube AB is
capped with a rigid plate. The gap between E and end C of
the 6061-T6 aluminum alloy solid circular rod CD is 0.2 mm
when the temperature is at 30° C. Determine the highest
temperature to which it can be raised without causing
yielding either in the tube or the rod. Neglect the thickness
of the rigid cap.
25 mm
a
Section a-a
E
B
A
20 mm
C
25 mm
a
0.2 mm
300 mm
Then
sCD =
F
107 442.47
=
= 218.88MPa 6 (sY)al
p
2
ACD
4 A 0.025 B
(O.K.!)
Compatibility Equation: If tube AB and rod CD are unconstrained, they will have
a free expansion of A dT B AB = amg ¢TLAB = 26(10 - 6)(T - 30)(300) = 7.8(10 - 6)
(T - 30) and
A dT B CD = aal ¢TLCD = 24(10 - 6)(T - 30)(450) = 0.0108(T - 30).
Referring to the deformation diagram of the tube and the rod shown in Fig. a,
d =
C A dT B AB - A dF B AB D + C A dT B CD - A dF B CD D
0.2 = C 7.8(10 - 3)(T - 30) -
+ C 0.0108(T - 30) -
107 442.47(300)
p A 0.0252 - 0.022 B (44.7)(109)
107 442.47(450)
p
4
A 0.0252 B (68.9)(109)
S
S
T = 172° C
Ans.
185
450 mm
D
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4–86. The steel bolt has a diameter of 7 mm and fits
through an aluminum sleeve as shown. The sleeve has an
inner diameter of 8 mm and an outer diameter of 10 mm.
The nut at A is adjusted so that it just presses up against
the sleeve. If the assembly is originally at a temperature
of T1 = 20°C and then is heated to a temperature of
T2 = 100°C, determine the average normal stress in the
bolt and the sleeve. Est = 200 GPa, Eal = 70 GPa, ast =
14(10-6)>°C, aal = 23(10-6)>°C.
A
Compatibility:
(ds)T - (db)T = (ds)F + (db)F
23(10 - 6)(100 - 20)L - 14(10 - 6)(100 - 20)L
=
p
2
4 (0.01
FL
+
- 0.0082)70(109)
FL
p
2
9
4 (0.007 )200(10 )
F = 1133.54 N
Average Normal Stress:
ss =
F
=
As
sb =
F
1133.54
= 29.5 MPa
= p
2
Ab
4 (0.007 )
1133.54
= 40.1 MPa
- 0.0082)
Ans.
p
2
4 (0.01
Ans.
4–87. Determine the maximum normal stress developed
in the bar when it is subjected to a tension of P = 8 kN.
5 mm
40 mm
20 mm
P
P
For the fillet:
r ⫽ 10 mm
20 mm
r
10
=
= 0.5
h
20
w
40
=
= 2
h
20
From Fig. 10-24. K = 1.4
smax = Ksavg
= 1.4 a
8 (103)
b
0.02 (0.005)
= 112 MPa
For the hole:
r
10
=
= 0.25
w
40
From Fig. 4-25. K = 2.375
smax = Ksavg
= 2.375 a
8 (103)
b
(0.04 - 0.02)(0.005)
= 190 MPa
Ans.
186
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*4–88. If the allowable normal stress for the bar is
sallow = 120 MPa, determine the maximum axial force P
that can be applied to the bar.
5 mm
40 mm
20 mm
P
P
Assume failure of the fillet.
r ⫽ 10 mm
20 mm
r
10
=
= 0.5
h
20
w
40
=
= 2;
h
20
From Fig. 4-24. K = 1.4
sallow = smax = Ksavg
120 (106) = 1.4 a
P
b
0.02 (0.005)
P = 8.57 kN
Assume failure of the hole.
r
10
=
= 0.25
w
20
From Fig. 4-25. K = 2.375
sallow = smax = Ksavg
120 (104) = 2.375 a
P
b
(0.04 - 0.02) (0.005)
P = 5.05 kN (controls)
Ans.
•4–89.
The member is to be made from a steel plate that is
0.25 in. thick. If a 1-in. hole is drilled through its center,
determine the approximate width w of the plate so that it
can support an axial force of 3350 lb. The allowable stress is
sallow = 22 ksi.
0.25 in.
w
3350 lb
sallow = smax = Ksavg
1 in.
3.35
d
22 = K c
(w - 1)(0.25)
w =
3.35K + 5.5
5.5
By trial and error, from Fig. 4-25, choose
w =
r
= 0.2;
w
K = 2.45
3.35(2.45) + 5.5
= 2.49 in.
5.5
Since
0.5
r
=
= 0.2
w
2.49
3350 lb
Ans.
OK
187
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4–90. The A-36 steel plate has a thickness of 12 mm. If
there are shoulder fillets at B and C, and sallow = 150 MPa,
determine the maximum axial load P that it can support.
Calculate its elongation, neglecting the effect of the fillets.
r = 30 mm
120 mm
r = 30 mm
60 mm
P
A
w
120
=
= 2
h
60
and
60 mm
P
D
B
Maximum Normal Stress at fillet:
r
30
=
= 0.5
h
60
C
800 mm
200 mm
200 mm
From the text, K = 1.4
smax = sallow = Ksavg
150(106) = 1.4 B
P
R
0.06(0.012)
P = 77142.86 N = 77.1 kN
Ans.
Displacement:
d = ©
PL
AE
77142.86(800)
77142.86(400)
=
9
+
(0.06)(0.012)(200)(10 )
(0.12)(0.012)(200)(109)
= 0.429 mm
Ans.
4–91. Determine the maximum axial force P that can be
applied to the bar. The bar is made from steel and has an
allowable stress of sallow = 21 ksi.
0.125 in.
1.25 in.
1.875 in.
P
Assume failure of the fillet.
r
0.25
=
= 0.2
h
1.25
P
w
1.875
=
= 1.5
h
1.25
0.75 in.
From Fig. 4-24, K = 1.73
sallow = smax = Ksavg
21 = 1.73 a
P
b
1.25 (0.125)
P = 1.897 kip
Assume failure of the hole.
r
0.375
=
= 0.20
w
1.875
From Fig. 4-25, K = 2.45
sallow = smax = Ksavg
21 = 2.45 a
P
b
(1.875 - 0.75)(0.125)
P = 1.21 kip (controls)
Ans.
188
r ⫽ 0.25 in.
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*4–92. Determine the maximum normal stress developed
in the bar when it is subjected to a tension of P = 2 kip.
0.125 in.
1.25 in.
1.875 in.
At fillet:
P
r
0.25
=
= 0.2
h
1.25
P
w
1.875
=
= 1.5
h
1.25
0.75 in.
From Fig. 4-24, K = 1.73
smax = K a
r ⫽ 0.25 in.
P
2
d = 22.1 ksi
b = 1.73 c
A
1.25(0.125)
At hole:
r
0.375
=
= 0.20
w
1.875
From Fig. 4-25, K = 2.45
smax = 2.45 c
2
d = 34.8 ksi
(1.875 - 0.75)(0.125)
(Controls)
Ans.
•4–93.
Determine the maximum normal stress developed
in the bar when it is subjected to a tension of P = 8 kN.
5 mm
60 mm
P
Maximum Normal Stress at fillet:
r
15
=
= 0.5
h
30
P
ht
= 1.4 B
8(103)
R = 74.7 MPa
(0.03)(0.005)
Maximum Normal Stress at the hole:
r
6
=
= 0.1
w
60
From the text, K = 2.65
smax = K savg = K
P
(w - 2r) t
= 2.65 B
8(103)
R
(0.06 - 0.012)(0.005)
= 88.3 MPa
P
r = 15 mm
12 mm
w
60
=
= 2
h
30
and
From the text, K = 1.4
smax = Ksavg = K
30 mm
(Controls)
Ans.
189
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4–94. The resulting stress distribution along section AB
for the bar is shown. From this distribution, determine the
approximate resultant axial force P applied to the bar. Also,
what is the stress-concentration factor for this geometry?
0.5 in.
A
P
4 in.
1 in.
B
12 ksi
P =
L
3 ksi
sdA = Volume under curve
Number of squares = 10
P = 10(3)(1)(0.5) = 15 kip
savg =
K =
Ans.
15 kip
P
=
= 7.5 ksi
A
(4 in.)(0.5 in.)
smax
12 ksi
=
= 1.60
savg
7.5 ksi
Ans.
4–95. The resulting stress distribution along section AB
for the bar is shown. From this distribution, determine the
approximate resultant axial force P applied to the bar. Also,
what is the stress-concentration factor for this geometry?
0.5 in.
A
0.6 in.
0.8 in.
Number of squares = 28
P = 28(6)(0.2)(0.5) = 16.8 kip
savg
P
16.8
=
=
= 28 ksi
A
2(0.6)(0.5)
K =
smax
36
=
= 1.29
savg
28
0.6 in.
Ans.
B
6 ksi
36 ksi
Ans.
*4–96. The resulting stress distribution along section AB
for the bar is shown. From this distribution, determine the
approximate resultant axial force P applied to the bar. Also,
what is the stress-concentration factor for this geometry?
10 mm
A
20 mm
80 mm
B
5 MPa
Number of squares = 19
30 MPa
6
P = 19(5)(10 )(0.02)(0.01) = 19 kN
savg =
K =
P
0.2 in.
Ans.
19(103)
P
=
= 23.75 MPa
A
0.08(0.01)
smax
30 MPa
=
= 1.26
savg
23.75 MPa
Ans.
190
P
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•4–97.
The 300-kip weight is slowly set on the top of a post
made of 2014-T6 aluminum with an A-36 steel core. If both
materials can be considered elastic perfectly plastic,
determine the stress in each material.
Aluminum
1 in.
2 in.
Steel
Equations of Equilibrium:
+ c ©Fy = 0;
Pst + Pal - 300 = 0
[1]
Elastic Analysis: Assume both materials still behave elastically under the load.
dst = dal
Pst L
p
2
(2)
(29)(103)
4
Pal L
=
p 2
4 (4
- 22)(10.6)(103)
Pst = 0.9119 Pal
Solving Eqs. [1] and [2] yields:
Pal = 156.91 kip
Pst = 143.09 kip
Average Normal Stress:
sal =
Pal
=
Aal
156.91
- 22)
p 2
4 (4
(OK!)
= 16.65 ksi 6 (sg)al = 60.0 ksi
sst =
Pst
143.09
= p 2
Ast
4 (2 )
= 45.55 ksi 7 (sg)st = 36.0 ksi
Therefore, the steel core yields and so the elastic analysis is invalid. The stress in the
steel is
sst = (sy)st = 36.0 ksi
Ans.
p
Pst = (sg)stAst = 36.0a b A 22 B = 113.10 kip
4
From Eq. [1] Pal = 186.90 kip
sal =
Pal
=
Aal
186.90
= 19.83 ksi 6 (sg)al = 60.0 ksi
- 22)
p 2
4 (4
Then sal = 19.8 ksi
Ans.
191
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4–98. The bar has a cross-sectional area of 0.5 in2 and is
made of a material that has a stress–strain diagram that
can be approximated by the two line segments shown.
Determine the elongation of the bar due to the applied
loading.
A
B
5 ft
8 kip
C 5 kip
2 ft
s(ksi)
40
20
Average Normal Stress and Strain: For segment BC
sBC =
0.001
PBC
5
=
= 10.0 ksi
ABC
0.5
10.0
20
=
;
eBC
0.001
eBC =
0.001
(10.0) = 0.00050 in.>in.
20
Average Normal Stress and Strain: For segment AB
sAB =
PAB
13
=
= 26.0 ksi
AAB
0.5
40 - 20
26.0 - 20
=
eAB - 0.001
0.021 - 0.001
eAB = 0.0070 in.>in.
Elongation:
dBC = eBCLBC = 0.00050(2)(12) = 0.0120 in.
dAB = eAB LAB = 0.0070(5)(12) = 0.420 in.
dTot = dBC + dAB = 0.0120 + 0.420 = 0.432 in.
Ans.
192
0.021
P (in./in.)
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4–99. The rigid bar is supported by a pin at A and two
steel wires, each having a diameter of 4 mm. If the yield
stress for the wires is sY = 530 MPa, and Est = 200 GPa,
determine the intensity of the distributed load w that can
be placed on the beam and will just cause wire EB to
yield. What is the displacement of point G for this case?
For the calculation, assume that the steel is elastic
perfectly plastic.
E
D
800 mm
A
B
C
G
w
400 mm
Equations of Equilibrium:
a + ©MA = 0;
FBE(0.4) + FCD(0.65) - 0.8w (0.4) = 0
0.4 FBE + 0.65FCD = 0.32w
[1]
Plastic Analysis: Wire CD will yield first followed by wire BE. When both wires yield
FBE = FCD = (sg)A
= 530 A 106 B a
p
b A 0.0042 B = 6.660 kN
4
Substituting the results into Eq. [1] yields:
w = 21.9 kN>m
Ans.
Displacement: When wire BE achieves yield stress, the corresponding yield strain is
eg =
sg
E
530(106)
=
200(109)
= 0.002650 mm>mm
dBE = eg LBE = 0.002650(800) = 2.120 mm
From the geometry
dBE
dG
=
0.8
0.4
dG = 2dBE = 2(2.120) = 4.24 mm
Ans.
193
250 mm
150 mm
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*4–100. The rigid bar is supported by a pin at A and two
steel wires, each having a diameter of 4 mm. If the yield
stress for the wires is sY = 530 MPa, and Est = 200 GPa,
determine (a) the intensity of the distributed load w that
can be placed on the beam that will cause only one of the
wires to start to yield and (b) the smallest intensity of
the distributed load that will cause both wires to yield. For
the calculation, assume that the steel is elastic perfectly
plastic.
E
D
800 mm
A
B
C
G
w
400 mm
Equations of Equilibrium:
a + ©MA = 0;
FBE(0.4) + FCD(0.65) - 0.8w (0.4) = 0
0.4 FBE + 0.65 FCD = 0.32w
[1]
(a) By observation, wire CD will yield first.
p
Then FCD = sg A = 530 A 106 B a b A 0.0042 B = 6.660 kN.
4
From the geometry
dCD
dBE
=
;
0.4
0.65
dCD = 1.625dBE
FBEL
FCDL
= 1.625
AE
AE
FCD = 1.625 FBE
[2]
Using FCD = 6.660 kN and solving Eqs. [1] and [2] yields:
FBE = 4.099 kN
w = 18.7 kN>m
Ans.
(b) When both wires yield
FBE = FCD = (sg)A
p
= 530 A 106 B a b A 0.0042 B = 6.660 kN
4
Substituting the results into Eq. [1] yields:
w = 21.9 kN>m
Ans.
194
250 mm
150 mm
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•4–101.
The rigid lever arm is supported by two A-36 steel
wires having the same diameter of 4 mm. If a force of
P = 3 kN is applied to the handle, determine the force
developed in both wires and their corresponding elongations.
Consider A-36 steel as an elastic-perfectly plastic material.
P
450 mm
150 mm
150 mm
30⬚
A
C
300 mm
B
Equation of Equilibrium. Refering to the free-body diagram of the lever shown
in Fig. a,
FAB (300) + FCD (150) - 3 A 103 B (450) = 0
a + ©ME = 0;
2FAB + FCD = 9 A 103 B
(1)
Elastic Analysis. Assuming that both wires AB and CD behave as linearly elastic,
the compatibility equation can be written by referring to the geometry of Fig. b.
dAB = a
300
bd
150 CD
dAB = 2dCD
(2)
FAB L
FCD L
= 2a
b
AE
AE
FAB = 2FCD
(3)
Solving Eqs. (1) and (3),
FCD = 1800 N
FAB = 3600 N
Normal Stress.
sCD =
FCD
=
ACD
sAB =
FAB
=
AAB
1800
p
4
A 0.0042 B
p
4
A 0.0042 B
3600
= 143.24 MPa 6 (sY)st
(O.K.)
= 286.48 MPa 7 (sY)st
(N.G.)
Since wire AB yields, the elastic analysis is not valid. The solution must be reworked
using
FAB = (sY)st AAB = 250 A 106 B c
p
A 0.0042 B d
4
Ans.
= 3141.59 N = 3.14 kN
Substituting this result into Eq. (1),
FCD = 2716.81 N = 2.72 kN
sCD =
Ans.
FCD
2716.81
=
= 216.20 MPa 6 (sY)st
p
2
ACD
4 A 0.004 B
(O.K.)
195
D
E
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4–101. Continued
Since wire CD is linearly elastic, its elongation can be determined by
dCD =
FCD LCD
=
ACD Est
2716.81(300)
p
4
A 0.0042 B (200) A 109 B
Ans.
= 0.3243 mm = 0.324 mm
From Eq. (2),
dAB = 2dCD = 2(0.3243) = 0.649 mm
Ans.
196
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4–102. The rigid lever arm is supported by two A-36 steel
wires having the same diameter of 4 mm. Determine the
smallest force P that will cause (a) only one of the wires to
yield; (b) both wires to yield. Consider A-36 steel as an
elastic-perfectly plastic material.
P
450 mm
150 mm
150 mm
30⬚
A
C
300 mm
B
Equation of Equilibrium. Refering to the free-body diagram of the lever arm shown
in Fig. a,
a + ©ME = 0;
FAB (300) + FCD (150) - P(450) = 0
2FAB + FCD = 3P
(1)
Elastic Analysis. The compatibility equation can be written by referring to the
geometry of Fig. b.
dAB = a
300
bd
150 CD
dAB = 2dCD
FAB L
FCD L
= 2a
b
AE
AE
FCD =
1
F
2 AB
(2)
Assuming that wire AB is about to yield first,
FAB = (sY)st AAB = 250 A 106 B c
p
A 0.0042 B d = 3141.59 N
4
From Eq. (2),
FCD =
1
(3141.59) = 1570.80 N
2
Substituting the result of FAB and FCD into Eq. (1),
P = 2618.00 N = 2.62 kN
Ans.
Plastic Analysis. Since both wires AB and CD are required to yield,
FAB = FCD = (sY)st A = 250 A 106 B c
p
A 0.0042 B d = 3141.59 N
4
Substituting this result into Eq. (1),
197
D
E
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4–103. The three bars are pinned together and subjected
to the load P. If each bar has a cross-sectional area A, length
L, and is made from an elastic perfectly plastic material, for
which the yield stress is sY, determine the largest load
(ultimate load) that can be supported by the bars, i.e., the
load P that causes all the bars to yield. Also, what is the
horizontal displacement of point A when the load reaches
its ultimate value? The modulus of elasticity is E.
B
L
u
C
L
P
u
L
D
P = 3141.59 N = 3.14 kN
A
Ans.
When all bars yield, the force in each bar is, FY = sYA
+ ©F = 0;
:
x
P - 2sYA cos u - sYA = 0
P = sYA(2 cos u + 1)
Ans.
Bar AC will yield first followed by bars AB and AD.
dAB = dAD =
dA =
FY(L)
sYAL
sYL
=
=
AE
AE
E
dAB
sYL
=
cos u
E cos u
Ans.
*4–104. The rigid beam is supported by three 25-mm
diameter A-36 steel rods. If the beam supports the force of
P = 230 kN, determine the force developed in each rod.
Consider the steel to be an elastic perfectly-plastic material.
D
F
E
600 mm
P
Equation of Equilibrium. Referring to the free-body diagram of the beam shown in
Fig. a,
+ c ©Fy = 0;
FAD + FBE + FCF - 230 A 103 B = 0
(1)
FBE + 3FCF = 460 A 103 B
(2)
FBE(400) + FCF(1200) - 230 A 103 B (800) = 0
a + ©MA = 0;
400 mm
Elastic Analysis. Referring to the deflection diagram of the beam shown in Fig. b,
the compatibility equation can be written as
dBE = dAD + a
dBE =
dCF - dAD
b(400)
1200
2
1
d
+ dCF
3 AD
3
FBEL
2 FCDL
1 FCF L
= a
b + a
b
AE
3 AE
3 AE
FBE =
2
1
F
+ FCF
3 AD
3
(3)
Solving Eqs. (1), (2), and (3)
FCF = 131 428.57 N
FBE = 65 714.29 N FAD = 32 857.14 N
198
A
B
400 mm
C
400 mm
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4–104. Continued
Normal Stress.
sCF =
FCF
131428.57
=
= 267.74 MPa 7 (sY)st
p
2
ACF
4 A 0.025 B
(N.G.)
sBE =
FBE
65714.29
=
= 133.87 MPa 6 (sY)st
p
2
ABE
4 A 0.025 B
(O.K.)
sAD =
FAD
32857.14
=
= 66.94 MPa 6 (sY)st
p
2
AAD
4 A 0.025 B
(O.K.)
Since rod CF yields, the elastic analysis is not valid. The solution must be
reworked using
FCF = (sY)st ACF = 250 A 106 B c
p
A 0.0252 B d = 122 718.46 N = 123 kN
4
Ans.
Substituting this result into Eq. (2),
FBE = 91844.61 N = 91.8 kN
Ans.
Substituting the result for FCF and FBE into Eq. (1),
FAD = 15436.93 N = 15.4 kN
sBE =
FBE
91844.61
=
= 187.10 MPa 6 (sY)st
p
2
ABE
4 A 0.025 B
sAD =
FAD
15436.93
=
= 31.45 MPa 6 (sY)st
p
2
AAD
4 A 0.025 B
Ans.
(O.K.)
(O.K.)
199
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•4–105.
The rigid beam is supported by three 25-mm
diameter A-36 steel rods. If the force of P = 230 kN is
applied on the beam and removed, determine the residual
stresses in each rod. Consider the steel to be an elastic
perfectly-plastic material.
D
600 mm
P
A
400 mm
Equation of Equilibrium. Referring to the free-body diagram of the beam shown in
Fig. a,
+ c ©Fy = 0;
FAD + FBE + FCF - 230 A 103 B = 0
(1)
FBE + 3FCF = 460 A 103 B
(2)
FBE(400) + FCF(1200) - 230 A 103 B (800) = 0
a + ©MA = 0;
Elastic Analysis. Referring to the deflection diagram of the beam shown in Fig. b,
the compatibility equation can be written as
dBE = dAD + a
dBE =
dCF - dAD
b(400)
1200
2
1
d
+ dCF
3 AD
3
(3)
FBE L
2 FCD L
1 FCF L
= a
b + a
b
AE
3 AE
3 AE
FBE =
2
1
F
+ FCF
3 AD
3
(4)
Solving Eqs. (1), (2), and (4)
FCF = 131428.57 N
FBE = 65714.29 N
FAD = 32857.14 N
Normal Stress.
sCF =
FCF
131428.57
=
= 267.74 MPa (T) 7 (sY)st
p
2
ACF
4 A 0.025 B
sBE =
FBE
65714.29
=
= 133.87 MPa (T) 6 (sY)st
p
2
ABE
4 A 0.025 B
sAD =
FAD
32857.14
=
= 66.94 MPa (T) 6 (sY)st
p
2
AAD
4 A 0.025 B
(N.G.)
(O.K.)
(O.K.)
Since rod CF yields, the elastic analysis is not valid. The solution must be
reworked using
sCF = (sY)st = 250 MPa (T)
FCF = sCF ACF = 250 A 106 B c
F
E
p
A 0.0252 B d = 122718.46 N
4
200
B
400 mm
C
400 mm
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4–105. Continued
Substituting this result into Eq. (2),
FBE = 91844.61 N
Substituting the result for FCF and FBE into Eq. (1),
FAD = 15436.93N
sBE =
FBE
91844.61
=
= 187.10 MPa (T) 6 (sY)st
p
2
ABE
4 A 0.025 B
(O.K.)
sAD =
FAD
15436.93
=
= 31.45 MPa (T) 6 (sY)st
p
2
AAD
4 A 0.025 B
(O.K.)
Residual Stresses. The process of removing P can be represented by applying the
force P¿ , which has a magnitude equal to that of P but is opposite in sense, Fig. c.
Since the process occurs in a linear manner, the corresponding normal stress must
have the same magnitude but opposite sense to that obtained from the elastic
analysis. Thus,
œ
sCF
= 267.74 MPa (C)
œ
sBE
= 133.87 MPa (C)
œ
sAD
= 66.94 MPa (C)
Considering the tensile stress as positive and the compressive stress as negative,
œ
= 250 + (- 267.74) = - 17.7 MPa = 17.7 MPa (C)
(sCF)r = sCF + sCF
Ans.
œ
= 187.10 + (- 133.87) = 53.2 MPa (T)
(sBE)r = sBE + sBE
Ans.
œ
(sAD)r = sAD + sAD
= 31.45 + (- 66.94) = - 35.5 MPa = 35.5 MPa (C)
Ans.
201
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4–106. The distributed loading is applied to the rigid
beam, which is supported by the three bars. Each bar has a
cross-sectional area of 1.25 in2 and is made from a
s(ksi)
material having a stress–strain diagram that can be
approximated by the two line segments shown. If a load of
60
w = 25 kip>ft is applied to the beam, determine the stress
in each bar and the vertical displacement of the beam.
4 ft
A
B
5 ft
36
A
0.0012
a + ©MB = 0;
0.2
FC(4) - FA(4) = 0;
FA = FC = F
+ c ©Fy = 0;
2F + FB - 200 = 0
(1)
Since the loading and geometry are symmetrical, the bar will remain horizontal.
Therefore, the displacement of the bars is the same and hence, the force in each bar
is the same. From Eq. (1).
F = FB = 66.67 kip
Thus,
sA = sB = sC =
66.67
= 53.33 ksi
1.25
Ans.
From the stress-strain diagram:
53.33 - 36
60 - 36
=
:
e - 0.0012
0.2 - 0.0012
e = 0.14477 in.>in.
d = eL = 0.14477(5)(12) = 8.69 in.
Ans.
202
4 ft
∋ (in./in.)
B
C
w
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4–107. The distributed loading is applied to the
rigid beam, which is supported by the three bars.
Each bar has a cross-sectional area of 0.75 in2 and is
s(ksi)
made from a material having a stress–strain diagram
that can be approximated by the two line segments
60
shown. Determine the intensity of the distributed
loading w needed to cause the beam to be displaced
36
downward 1.5 in.
0.0012
a + ©MB = 0;
+ c ©Fy = 0;
FC(4) - FA(4) = 0;
4 ft
A
A
0.2
(1)
Since the system and the loading are symmetrical, the bar will remain horizontal.
Hence the displacement of the bars is the same and the force supported by each bar
is the same.
From Eq. (1),
FB = F = 2.6667 w
(2)
From the stress-strain diagram:
e =
1.5
= 0.025 in.>in.
5 (12)
60 - 36
s - 36
=
;
0.025 - 0.0012
0.2 - 0.0012
s = 38.87 ksi
Hence F = sA = 38.87 (0.75) = 29.15 kip
From Eq. (2), w = 10.9 kip>ft
Ans.
203
B
5 ft
FA = FC = F
2F + FB - 8 w = 0
4 ft
∋ (in./in.)
B
C
w
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*4–108. The rigid beam is supported by the three posts A,
B, and C of equal length. Posts A and C have a diameter of
75 mm and are made of aluminum, for which Eal = 70 GPa
and 1sY2al = 20 MPa. Post B has a diameter of 20 mm and
is made of brass, for which Ebr = 100 GPa and
1sY2br = 590 MPa. Determine the smallest magnitude of P
so that (a) only rods A and C yield and (b) all the posts yield.
P
A
B
FA = FC = Fal
Fat + 2Fat - 2P = 0
+ c ©Fy = 0;
(1)
(a) Post A and C will yield,
Fal = (st)alA
= 20(104)(pa )(0.075)2
= 88.36 kN
(Eal)r =
(sr)al
20(104)
= 0.0002857
=
Eal
70(104)
Compatibility condition:
dbr = dal
= 0.0002857(L)
Fbr (L)
p
2
(0.02)
(100)(104)
4
= 0.0002857 L
Fbr = 8.976 kN
sbr =
8.976(103)
p
3
4 (0.02 )
= 28.6 MPa 6 sr
OK.
From Eq. (1),
8.976 + 2(88.36) - 2P = 0
P = 92.8 kN
Ans.
(b) All the posts yield:
Fbr = (sr)brA
= (590)(104)(p4 )(0.022)
= 185.35 kN
Fal = 88.36 kN
From Eq. (1); 185.35 + 2(88.36) - 2P = 0
P = 181 kN
Ans.
204
C
br
al
2m
©MB = 0;
P
2m
2m
al
2m
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•4–109.
The rigid beam is supported by the three posts
A, B, and C. Posts A and C have a diameter of 60 mm
and are made of aluminum, for which Eal = 70 GPa and
1sY2al = 20 MPa. Post B is made of brass, for which
Ebr = 100 GPa and 1sY2br = 590 MPa. If P = 130 kN,
determine the largest diameter of post B so that all the
posts yield at the same time.
P
A
B
2(Fg)al + Fbr - 260 = 0
(1)
(Fal)g = (sg)al A
= 20(106)(p4 )(0.06)2 = 56.55 kN
From Eq. (1),
2(56.55) + Fbr - 260 = 0
Fbr = 146.9 kN
(sg)br = 590(106) =
146.9(103)
p
3
4 (dB)
dB = 0.01779 m = 17.8 mm
Ans.
205
C
br
al
2m
+ c ©Fy = 0;
P
2m
2m
al
2m
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4–110. The wire BC has a diameter of 0.125 in. and the
material has the stress–strain characteristics shown in the
figure. Determine the vertical displacement of the handle at
D if the pull at the grip is slowly increased and reaches a
magnitude of (a) P = 450 lb, (b) P = 600 lb.
C
40 in.
A
D
B
50 in.
30 in.
P
s (ksi)
Equations of Equilibrium:
a + ©MA = 0;
FBC(50) - P(80) = 0
(a) From Eq. [1] when P = 450 lb,
[1]
80
70
FBC = 720 lb
Average Normal Stress and Strain:
sBC =
FBC
=
ABC
720
p
2
4 (0.125 )
P (in./in.)
= 58.67 ksi
0.007
From the Stress–Strain diagram
58.67
70
=
;
eBC
0.007
eBC = 0.005867 in.>in.
Displacement:
dBC = eBCLBC = 0.005867(40) = 0.2347 in.
dBC
dD
=
;
80
50
dD =
8
(0.2347) = 0.375 in.
5
(b) From Eq. [1] when P = 600 lb,
Ans.
FBC = 960 lb
Average Normal Stress and Strain:
sBC =
FBC
=
ABC
960
p
2
4 (0.125)
= 78.23 ksi
From Stress–Strain diagram
78.23 - 70
80 - 70
=
eBC - 0.007
0.12 - 0.007
eBC = 0.09997 in.>in.
Displacement:
dBC = eBCLBC = 0.09997(40) = 3.9990 in.
dBC
dD
=
;
80
50
dD =
8
(3.9990) = 6.40 in.
5
Ans.
206
0.12
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4–111. The bar having a diameter of 2 in. is fixed
connected at its ends and supports the axial load P. If the
material is elastic perfectly plastic as shown by the
stress–strain diagram, determine the smallest load P needed
to cause segment CB to yield. If this load is released,
determine the permanent displacement of point C.
P
A
B
C
2 ft
3 ft
s (ksi)
20
0.001
When P is increased, region AC will become plastic first, then CB will become
plastic. Thus,
FA = FB = sA = 20(p)(1)2 = 62.832 kip
+ ©F = 0;
:
x
FA + FB - P = 0
(1)
P = 2(62.832) = 125.66 kip
P = 126 kip
Ans.
The deflection of point C is,
dC = eL = (0.001)(3)(12) = 0.036 in. ;
Consider the reverse of P on the bar.
FB ¿(3)
FA ¿(2)
=
AE
AE
FA ¿ = 1.5 FB ¿
So that from Eq. (1)
FB ¿ = 0.4P
FA ¿ = 0.6P
dC ¿ =
0.4(P)(3)(12)
0.4(125.66)(3)(12)
FB ¿L
= 0.02880 in. :
=
=
AE
AE
p(1)2(20>0.001)
¢d = 0.036 - 0.0288 = 0.00720 in. ;
Ans.
207
P (in./in.)
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*4–112. Determine the elongation of the bar in
Prob. 4–111 when both the load P and the supports are
removed.
P
A
B
C
2 ft
3 ft
s (ksi)
20
0.001
When P is increased, region AC will become plastic first, then CB will become
plastic. Thus,
FA = FB = sA = 20(p)(1)2 = 62.832 kip
+ ©F = 0;
:
x
FA + FB - P = 0
(1)
P = 2(62.832) = 125.66 kip
P = 126 kip
Ans.
The deflection of point C is,
dC = eL = (0.001)(3)(12) = 0.036 in. ;
Consider the reverse of P on the bar.
FB ¿(3)
FA ¿(2)
=
AE
AE
FA ¿ = 1.5 FB ¿
So that from Eq. (1)
FB ¿ = 0.4P
FA ¿ = 0.6P
The resultant reactions are
FA ¿¿ = FB ¿¿ = -62.832 + 0.6(125.66) = 62.832 - 0.4(125.66) = 12.568 kip
When the supports are removed the elongation will be,
d =
12.568(5)(12)
PL
= 0.0120 in.
=
AE
p(1)2(20>0.001)
Ans.
208
P (in./in.)
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s
•4–113.
A material has a stress–strain diagram that can be
described by the curve s = cP1>2. Determine the deflection
d of the end of a rod made from this material if it has a
length L, cross-sectional area A, and a specific weight g.
L
1
2
A
s2 = c2 e
s = ce ;
s2(x) = c2e(x)
P
(1) d
P(x)
;
A
However s(x) =
e(x) =
dd
dx
From Eq. (1),
P2(x)
= c2
A2
P2(x)
dd
=
dx
A2c2
dd
;
dx
L
d =
1
1
P2(x) dx = 2 2
(gAx)2 dx
2 2
Ac L
A c L0
g2
=
d =
L
c2 L0
x2 dx =
g2 x3 L
冷
c2 3 0
g3L3
Ans.
3c2
4–114. The 2014-T6 aluminum rod has a diameter of 0.5 in.
and is lightly attached to the rigid supports at A and B when
T1 = 70°F. If the temperature becomes T2 = - 10°F, and
an axial force of P = 16 lb is applied to the rigid collar as
shown, determine the reactions at A and B.
A
B
P/2
P/2
5 in.
8 in.
+ 0 = ¢ - ¢ + d
:
B
T
B
0 =
0.016(5)
p
2
3
4 (0.5 )(10.6)(10 )
- 12.8(10 - 6)[70° - ( -10°)](13) +
FB(13)
p
2
(0.5
)(10.6)(103)
4
FB = 2.1251 kip = 2.13 kip
+ ©F = 0;
:
x
Ans.
2(0.008) + 2.1251 - FA = 0
FA = 2.14 kip
Ans.
4–115. The 2014-T6 aluminum rod has a diameter of
0.5 in. and is lightly attached to the rigid supports at A
and B when T1 = 70°F. Determine the force P that must
be applied to the collar so that, when T = 0°F, the
reaction at B is zero.
+
:
A
P/2
5 in.
0 = ¢ B - ¢ T + dB
0 =
P(5)
p
2
3
4 (0.5 )(10.6)(10 )
B
P/2
- 12.8(10 - 6)[(70)(13)] + 0
P = 4.85 kip
Ans.
209
8 in.
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*4–116. The rods each have the same 25-mm diameter
and 600-mm length. If they are made of A-36 steel,
determine the forces developed in each rod when the
temperature increases to 50° C.
C
600 mm
60⬚
B
A
60⬚
600 mm
Equation of Equilibrium: Referring to the free-body diagram of joint A shown in
Fig. a,
FAD sin 60° - FAC sin 60° = 0
+ c ©Fx = 0;
+ ©F = 0;
:
x
FAC = FAD = F
FAB - 2F cos 60° = 0
FAB = F
(1)
Compatibility Equation: If AB and AC are unconstrained, they will have
a free expansion of A dT B AB = A dT B AC = ast ¢TL = 12(10 - 6)(50)(600) = 0.36 mm.
Referring to the initial and final position of joint A,
dFAB - A dT B AB = a dT ¿ b
AC
- dFAC ¿
Due to symmetry, joint A will displace horizontally, and dAC ¿ =
a dT ¿ b
AC
dAC
= 2dAC. Thus,
cos 60°
= 2(dT)AC and dFAC ¿ = 2dFAC. Thus, this equation becomes
dFAB - A dT B AB = 2 A dT B AC - 2dAC
FAB (600)
p
4
A 0.025 B (200)(10 )
2
9
- 0.36 = 2(0.36) - 2 C
F(600)
p
4
A 0.0252 B (200)(109)
FAB + 2F = 176 714.59
S
(2)
Solving Eqs. (1) and (2),
FAB = FAC = FAD = 58 904.86 N = 58.9 kN
Ans.
210
D
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•4–117.
Two A-36 steel pipes, each having a crosssectional area of 0.32 in2, are screwed together using a
union at B as shown. Originally the assembly is adjusted so
that no load is on the pipe. If the union is then tightened
so that its screw, having a lead of 0.15 in., undergoes two full
turns, determine the average normal stress developed in the
pipe. Assume that the union at B and couplings at A and C
are rigid. Neglect the size of the union. Note: The lead would
cause the pipe, when unloaded, to shorten 0.15 in. when the
union is rotated one revolution.
B
A
3 ft
C
2 ft
The loads acting on both segments AB and BC are the same since no external load
acts on the system.
0.3 = dB>A + dB>C
0.3 =
P(2)(12)
P(3)(12)
3
0.32(29)(10 )
+
0.32(29)(103)
P = 46.4 kip
P
46.4
=
= 145 ksi
A
0.32
sAB = sBC =
Ans.
4–118. The brass plug is force-fitted into the rigid casting.
The uniform normal bearing pressure on the plug is
estimated to be 15 MPa. If the coefficient of static friction
between the plug and casting is ms = 0.3, determine the
axial force P needed to pull the plug out. Also, calculate the
displacement of end B relative to end A just before the plug
starts to slip out. Ebr = 98 GPa.
100 mm
B
15 MPa
P - 4.50(106)(2)(p)(0.02)(0.1) = 0
P = 56.549 kN = 56.5 kN
Ans.
Displacement:
PL
dB>A = a
AE
0.1 m
56.549(103)(0.15)
=
2
9
p(0.02 )(98)(10 )
+
L0
P
A
Equations of Equilibrium:
+ ©F = 0;
:
x
150 mm
0.56549(106) x dx
p(0.022)(98)(109)
= 0.00009184 m = 0.0918 mm
Ans.
211
20 mm
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4–119. The assembly consists of two bars AB and CD of
the same material having a modulus of elasticity E1 and
coefficient of thermal expansion a1, and a bar EF having a
modulus of elasticity E2 and coefficient of thermal
expansion a2. All the bars have the same length L and
cross-sectional area A. If the rigid beam is originally
horizontal at temperature T1, determine the angle it makes
with the horizontal when the temperature is increased
to T2.
D
B
L
A
C
d
Equations of Equilibrium:
a + ©MC = 0;
+ c ©Fy = 0;
FAB = FEF = F
FCD - 2F = 0
[1]
Compatibility:
dAB = (dAB)T - (dAB)F
dCD = (dCD)T + (dCD)F
dEF = (dEF)T - (dEF)F
From the geometry
dCD - dAB
dEF - dAB
=
d
2d
2dCD = dEF + dAB
2 C (dCD)T + (dCD)F D = (dEF)T - (dEF)F + (dAB)T - (dAB)F
2 B a1 (T2 - T1) L +
= a2 (T2 - T1) L -
FCD (L)
R
AE1
F(L)
F(L)
+ a1 (T2 - T1) L AE2
AE1
[2]
Substitute Eq. [1] into [2].
2a1 (T2 - T1) L +
4FL
FL
FL
= a2 (T2 - T1)L + a1 (T2 - T1)L AE1
AE2
AE1
F
5F
+
= a2 (T2 - T1) - a1 (T2 - T1)
AE1
AE2
F¢
5E2 + E1
b = (T2 - T1)(a2 - a1) ;
AE1E2
F =
AE1E2 (T2 - T1)(a2 - a1)
5E2 + E1
(dEF)T = a2 (T2 - T1) L
(dEF)F =
AE1E2 (T2 - T1)(a2 - a1)(L)
E1 (T2 - T1)(a2 - a1)(L)
=
AE2 (5E2 + E1)
5E2 + E1
dEF = (dEF)T - (dEF)F =
a2 L(T2 - T1)(5E2 - E1) - E1L(T2 - T1)(a2 - a1)
5E2 + E1
(dAB)T = a1 (T2 - T1) L
(dAB)F =
AE1E2 (T2 - T1)(a2 - a1)(L)
E2 (T2 - T1)(a2 - a1)(L)
=
AE1 (5E2 + E1)
5E2 + E1
dAB = (dAB)T - (dAB)F =
F
a1 L(5E2 + E1)(T2 - T1) - E2 L(T2 - T1)(a2 - a1)
5E2 + E1
212
E
d
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4–119. Continued
dEF - dAB =
L(T2 - T1)
[a2 (5E2 + E1) - E1 (a2 - a1) - a1 (5E2 + E1)
5E2 + E1
+ E2 (a2 - a1)]
=
L(T2 - T1)
C (5E2 + E1)(a2 - a1) + (a2 - a1)(E2 - E1) D
5E2 + E1
=
L(T2 - T1)(a2 - a1)
(5E2 + E1 + E2 - E1)
5E2 + E1
=
L(T2 - T1)(a2 - a1)(6E2)
5E2 + E1
u =
3E2L(T2 - T1)(a2 - a1)
dEF - dAB
=
2d
d(5E2 + E1)
Ans.
*4–120. The rigid link is supported by a pin at A and two
A-36 steel wires, each having an unstretched length of 12 in.
and cross-sectional area of 0.0125 in2. Determine the force
developed in the wires when the link supports the vertical
load of 350 lb.
12 in.
C
5 in.
B
Equations of Equilibrium:
a + ©MA = 0;
4 in.
A
- FC(9) - FB (4) + 350(6) = 0
[1]
Compatibility:
6 in.
dC
dB
=
4
9
350 lb
FC(L)
FB (L)
=
4AE
9AE
9FB - 4FC = 0‚
[2]
Solving Eqs. [1] and [2] yields:
FB = 86.6 lb
Ans.
FC = 195 lb
Ans.
213
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•5–1.
A shaft is made of a steel alloy having an allowable
shear stress of tallow = 12 ksi. If the diameter of the shaft is
1.5 in., determine the maximum torque T that can be
transmitted. What would be the maximum torque T¿ if a
1-in.-diameter hole is bored through the shaft? Sketch the
shear-stress distribution along a radial line in each case.
T
T¿
Allowable Shear Stress: Applying the torsion formula
tmax = tallow =
12 =
Tc
J
T (0.75)
p
2
(0.754)
T = 7.95 kip # in.
Ans.
Allowable Shear Stress: Applying the torsion formula
tmax = tallow =
12 =
T¿c
J
T¿ (0.75)
p
2
(0.754 - 0.54)
T¿ = 6.381 kip # in. = 6.38 kip # in.
tr = 0.5 in =
T¿r
=
J
6.381(0.5)
p
2
(0.754 - 0.54)
Ans.
= 8.00 ksi
214
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5–2. The solid shaft of radius r is subjected to a torque T.
Determine the radius r¿ of the inner core of the shaft that
resists one-half of the applied torque 1T>22. Solve the
problem two ways: (a) by using the torsion formula, (b) by
finding the resultant of the shear-stress distribution.
r¿
r
T
a)
tmax =
t =
Since t =
r¿ =
Tc
Tr
2T
= p 4 =
J
p r3
2 r
(T2 )r¿
p
2
(r¿)4
=
T
p(r¿)3
T
r¿ 2T
=
a
b
r pr3
p(r¿)3
r¿
t
;
r max
r
1
= 0.841 r
Ans.
24
r
2
b)
r¿
dT = 2p
L0
r
2
r¿
dT = 2p
L0
r
2
L0
tr2 dr
L0
r
tmax r2 dr
L0 r
r¿
dT = 2p
r 2T 2
a 3 b r dr
L0 r pr
r¿
4T
T
= 4
r3 dr
2
r L0
r¿ =
r
1
Ans.
= 0.841r
24
215
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5–3. The solid shaft is fixed to the support at C and
subjected to the torsional loadings shown. Determine the
shear stress at points A and B and sketch the shear stress on
volume elements located at these points.
10 kN⭈m
C
A
50 mm
The internal torques developed at Cross-sections pass through point B and A are
shown in Fig. a and b, respectively.
The polar moment of inertia of the shaft is J =
p
(0.0754) = 49.70(10 - 6) m4. For
2
point B, rB = C = 0.075 Thus,
tB =
4(103)(0.075)
TB c
= 6.036(106) Pa = 6.04 MPa
=
J
49.70(10 - 6)
Ans.
From point A, rA = 0.05 m.
tA =
TArA
6(103)(0.05)
= 6.036(106) Pa = 6.04 MPa.
=
J
49.70 (10 - 6)
216
Ans.
B
75 mm
4 kN⭈m
75 mm
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*5–4. The tube is subjected to a torque of 750 N # m.
Determine the amount of this torque that is resisted by the
gray shaded section. Solve the problem two ways: (a) by
using the torsion formula, (b) by finding the resultant of the
shear-stress distribution.
75 mm
100 mm
750 Nm
25 mm
a) Applying Torsion Formula:
tmax =
Tc
=
J
750(0.1)
p
2
(0.14 - 0.0254)
tmax = 0.4793 A 106 B =
= 0.4793 MPa
T¿(0.1)
p
2
(0.14 - 0.0754)
T¿ = 515 N # m
Ans.
b) Integration Method:
r
t = a b tmax
c
dA = 2pr dr
and
dT¿ = rt dA = rt(2pr dr) = 2ptr2 dr
0.1m
T¿ =
L
2ptr2 dr = 2p
r
tmax a br2 dr
c
L0.075m
=
0.1m
2ptmax
r3 dr
c
L0.075m
=
2p(0.4793)(106) r4 0.1 m
c d2
0.1
4 0.075 m
= 515 N # m
Ans.
5–5. The copper pipe has an outer diameter of 40 mm and
an inner diameter of 37 mm. If it is tightly secured to the wall
at A and three torques are applied to it as shown, determine
the absolute maximum shear stress developed in the pipe.
tmax =
Tmax c
=
J
A
30 N⭈m
90(0.02)
p
2
4
4
(0.02 - 0.0185 )
20 N⭈m
= 26.7 MPa
Ans..
80 N⭈m
217
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5–6. The solid shaft has a diameter of 0.75 in. If it is
subjected to the torques shown, determine the maximum
shear stress developed in regions BC and DE of the shaft.
The bearings at A and F allow free rotation of the shaft.
F
E
D
C
B
(tBC)max =
35(12)(0.375)
TBC c
= 5070 psi = 5.07 ksi
= p
4
J
2 (0.375)
Ans.
(tDE)max =
25(12)(0.375)
TDE c
= 3621 psi = 3.62 ksi
= p
4
J
2 (0.375)
Ans.
A
35 lb⭈ft
5–7. The solid shaft has a diameter of 0.75 in. If it is
subjected to the torques shown, determine the maximum
shear stress developed in regions CD and EF of the shaft.
The bearings at A and F allow free rotation of the shaft.
F
E
D
C
B
(tEF)max =
TEF c
= 0
J
(tCD)max =
25 lb⭈ft
40 lb⭈ft
20 lb⭈ft
Ans.
A
25 lb⭈ft
40 lb⭈ft
20 lb⭈ft
35 lb⭈ft
15(12)(0.375)
TCD c
= p
4
J
2 (0.375)
= 2173 psi = 2.17 ksi
Ans.
218
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300 Nm
*5–8. The solid 30-mm-diameter shaft is used to transmit
the torques applied to the gears. Determine the absolute
maximum shear stress on the shaft.
500 Nm
A
200 Nm
Internal Torque: As shown on torque diagram.
C
Maximum Shear Stress: From the torque diagram Tmax = 400 N # m. Then, applying
torsion Formula.
400 Nm
300 mm
abs =
tmax
Tmax c
J
400(0.015)
=
p
2
(0.0154)
= 75.5 MPa
Ans.
The shaft consists of three concentric tubes, each
made from the same material and having the inner and
outer radii shown. If a torque of T = 800 N # m is applied to
the rigid disk fixed to its end, determine the maximum shear
stress in the shaft.
500 mm
T 800 Nm
ri 20 mm
ro 25 mm
2m
p
p
p
((0.038)4 - (0.032)4) + ((0.030)4 - (0.026)4) + ((0.025)4 - (0.020)4)
2
2
2
-6
ri 26 mm
ro 30 mm
4
J = 2.545(10 ) m
tmax =
B
400 mm
•5–9.
J =
D
800(0.038)
Tc
= 11.9 MPa
=
J
2.545(10 - 6)
Ans.
219
ri 32 mm
ro 38 mm
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5–10. The coupling is used to connect the two shafts
together. Assuming that the shear stress in the bolts is
uniform, determine the number of bolts necessary to make
the maximum shear stress in the shaft equal to the shear
stress in the bolts. Each bolt has a diameter d.
T
R
r
n is the number of bolts and F is the shear force in each bolt.
T
T
F =
nR
T - nFR = 0;
T
tavg =
F
4T
nR
= p 2 =
A
( 4 )d
nRpd2
Maximum shear stress for the shaft:
tmax =
Tc
Tr
2T
= p 4 =
J
pr3
2r
4T
2T
=
nRpd2
p r3
tavg = tmax ;
n =
2 r3
Rd2
Ans.
5–11. The assembly consists of two sections of galvanized
steel pipe connected together using a reducing coupling at B.
The smaller pipe has an outer diameter of 0.75 in. and an inner
diameter of 0.68 in., whereas the larger pipe has an outer
diameter of 1 in. and an inner diameter of 0.86 in. If the pipe is
tightly secured to the wall at C, determine the maximum shear
stress developed in each section of the pipe when the couple
shown is applied to the handles of the wrench.
C
B
A
15 lb 6 in.
tAB =
tBC
Tc
=
J
Tc
=
=
J
8 in.
210(0.375)
p
2
(0.3754 - 0.344)
Ans.
15 lb
210(0.5)
p
2
= 7.82 ksi
(0.54 - 0.434)
= 2.36 ksi
Ans.
220
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*5–12. The motor delivers a torque of 50 N # m to the shaft
AB. This torque is transmitted to shaft CD using the gears
at E and F. Determine the equilibrium torque Tⴕ on shaft
CD and the maximum shear stress in each shaft. The
bearings B, C, and D allow free rotation of the shafts.
A
50 mm
30 mm
Equilibrium:
B
a + ©ME = 0;
a + ©MF = 0;
50 - F(0.05) = 0
F = 1000 N
35 mm
T¿
T¿ - 1000(0.125) = 0
T¿ = 125 N # m
C
E
125 mm
D
F
Ans.
Internal Torque: As shown on FBD.
Maximum Shear Stress: Applying torsion Formula.
(tAB)max =
50.0(0.015)
TAB c
= 9.43 MPa
= p
4
J
2 (0.015 )
Ans.
(tCD)max =
125(0.0175)
TCDc
= 14.8 MPa
= p
4
J
2 (0.0175 )
Ans.
•5–13. If the applied torque on shaft CD is T¿ = 75 N # m,
determine the absolute maximum shear stress in each shaft.
The bearings B, C, and D allow free rotation of the shafts,
and the motor holds the shafts fixed from rotating.
A
50 mm
Equilibrium:
30 mm
a + ©MF = 0;
75 - F(0.125) = 0;
a + ©ME = 0;
600(0.05) - TA = 0
B
F = 600 N
35 mm
T¿
TA = 30.0 N # m
Internal Torque: As shown on FBD.
Maximum Shear Stress: Applying the torsion formula
(tEA)max =
30.0(0.015)
TEA c
= 5.66 MPa
= p
4
J
2 (0.015 )
Ans.
(tCD)max =
75.0(0.0175)
TCDc
= 8.91 MPa
= p
4
J
2 (0.0175 )
Ans.
221
C
E
125 mm
D
F
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250 N⭈m
5–14. The solid 50-mm-diameter shaft is used to transmit
the torques applied to the gears. Determine the absolute
maximum shear stress in the shaft.
75 N⭈m
A
325 N⭈m
150 N⭈ m
B
500 mm
The internal torque developed in segments AB , BC and CD of the shaft are shown
in Figs. a, b and c.
C
D
400 mm
500 mm
The maximum torque occurs in segment AB. Thus, the absolute maximum shear
stress occurs in this segment. The polar moment of inertia of the shaft is
p
J = (0.0254) = 0.1953p(10 - 6)m4. Thus,
2
A tmax B abs =
250(0.025)
TAB c
= 10.19(106)Pa = 10.2 MPa
=
J
0.1953p(10 - 6)
222
Ans.
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5–15. The solid shaft is made of material that has an
allowable shear stress of tallow = 10 MPa. Determine the
required diameter of the shaft to the nearest mm.
15 N⭈m
25 N⭈m
A
30 N⭈m
B
60 N⭈m
C
70 N⭈m
D
E
The internal torques developed in each segment of the shaft are shown in the torque
diagram, Fig. a.
Segment DE is critical since it is subjected to the greatest internal torque. The polar
p d 4
p 4
moment of inertia of the shaft is J =
a b =
d . Thus,
2 2
32
tallow
TDE c
=
;
J
d
70a b
2
10(106) =
p 4
d
32
d = 0.03291 m = 32.91 mm = 33 mm
223
Ans.
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*5–16. The solid shaft has a diameter of 40 mm.
Determine the absolute maximum shear stress in the shaft
and sketch the shear-stress distribution along a radial line
of the shaft where the shear stress is maximum.
15 N⭈m
25 N⭈m
A
30 N⭈m
B
The internal torque developed in each segment of the shaft are shown in the torque
diagram, Fig. a.
60 N⭈m
C
70 N⭈m
D
E
Since segment DE subjected to the greatest torque, the absolute maximum shear
p
stress occurs here. The polar moment of inertia of the shaft is J = (0.024)
2
= 80(10 - 9)p m4. Thus,
tmax =
70(0.02)
TDE c
= 5.57(106) Pa = 5.57 MPa
=
J
80(10 - 9)p
Ans.
The shear stress distribution along the radial line is shown in Fig. b.
•5–17.
The rod has a diameter of 1 in. and a weight of
10 lb/ft. Determine the maximum torsional stress in the rod
at a section located at A due to the rod’s weight.
4.5 ft
B
Here, we are only interested in the internal torque. Thus, other components of the
internal loading are not indicated in the FBD of the cut segment of the rod, Fig. a.
©Mx = 0;
TA - 10(4)(2) = 0
TA = 80 lb # ft a
The polar moment of inertia of the cross section at A is J =
12in
b = 960 lb # in.
1ft
p
(0.54) = 0.03125p in4.
2
Thus
tmax =
960 (0.5)
TA c
=
= 4889.24 psi = 4.89 ksi
J
0.03125p
Ans.
224
4 ft
A
1.5 ft
1.5 ft
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5–18. The rod has a diameter of 1 in. and a weight of
15 lb/ft. Determine the maximum torsional stress in the rod
at a section located at B due to the rod’s weight.
4.5 ft
B
4 ft
Here, we are only interested in the internal torque. Thus, other components of the
internal loading are not indicated in the FBD of the cut segment of the rod, Fig. a.
©Mx = 0;
TB - 15(4)(2) = 0
TB = 120 lb # ft a
12 in
b = 1440 lb # in.
1ft
p
The polar moment of inertia of the cross-section at B is J = (0.54)
2
= 0.03125p in4. Thus,
tmax =
1440(0.5)
TB c
=
= 7333.86 psi = 7.33 ksi
J
0.03125p
Ans.
225
A
1.5 ft
1.5 ft
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5–19. Two wrenches are used to tighten the pipe. If P =
300 N is applied to each wrench, determine the maximum
torsional shear stress developed within regions AB and BC.
The pipe has an outer diameter of 25 mm and inner
diameter of 20 mm. Sketch the shear stress distribution for
both cases.
P
B
Internal Loadings: The internal torque developed in segments AB and BC of the
pipe can be determined by writing the moment equation of equilibrium about the x
axis by referring to their respective free - body diagrams shown in Figs. a and b.
©Mx = 0; TAB - 300(0.25) = 0
TAB = 75 N # m
TBC = 150 N # m
Allowable Shear Stress: The polar moment of inertia of the pipe is
p
J =
A 0.01254 - 0.014 B = 22.642(10 - 9)m4.
2
A tmax B AB =
75(0.0125)
TAB c
= 41.4 MPa
=
J
22.642(10 - 9)
A tAB B r = 0.01 m =
A tmax B BC =
Ans.
TAB r
75(0.01)
= 33.1 MPa
=
J
22.642(10 - 9)
150(0.0125)
TBC c
= 82.8 MPa
=
J
22.642(10 - 9)
A tBC B r = 0.01 m =
Ans.
TBC r
150(0.01)
= 66.2 MPa
=
J
22.642(10 - 9)
The shear stress distribution along the radial line of segments AB and BC of the
pipe is shown in Figs. c and d, respectively.
226
A
250 mm
P
And
©Mx = 0; TBC - 300(0.25) - 300(0.25) = 0
250 mm
C
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*5–20. Two wrenches are used to tighten the pipe. If the
pipe is made from a material having an allowable shear stress
of tallow = 85 MPa, determine the allowable maximum force
P that can be applied to each wrench. The pipe has an outer
diameter of 25 mm and inner diameter of 20 mm.
P
250 mm
C
B
A
250 mm
Internal Loading: By observation, segment BC of the pipe is critical since it is
subjected to a greater internal torque than segment AB. Writing the moment
equation of equilibrium about the x axis by referring to the free-body diagram
shown in Fig. a, we have
©Mx = 0; TBC - P(0.25) - P(0.25) = 0
TBC = 0.5P
Allowable Shear Stress: The polar moment of inertia of the pipe is
p
J =
A 0.01254 - 0.014 B = 22.642(10 - 9)m4
2
tallow =
TBC c
;
J
85(106) =
0.5P(0.0125)
22.642(10 - 9)
P = 307.93N = 308 N
Ans.
227
P
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•5–21.
The 60-mm-diameter solid shaft is subjected to the
distributed and concentrated torsional loadings shown.
Determine the absolute maximum and minimum shear
stresses on the outer surface of the shaft and specify their
locations, measured from the fixed end A.
A
2 kN⭈m/m
1.5 m
1200 N⭈m
C
The internal torque for segment BC is Constant TBC = 1200 N # m, Fig. a. However,
the internal for segment AB varies with x, Fig. b.
TAB - 2000x + 1200 = 0
TAB = (2000x - 1200) N # m
The minimum shear stress occurs when the internal torque is zero in segment AB.
By setting TAB = 0,
0 = 2000x - 1200
x = 0.6 m
Ans.
And
d = 1.5 m - 0.6 m = 0.9 m
Ans.
tmin = 0
Ans.
The maximum shear stress occurs when the internal torque is the greatest. This
occurs at fixed support A where
d = 0
Ans.
At this location,
(TAB)max = 2000(1.5) - 1200 = 1800 N # m
The polar moment of inertia of the rod is J =
tmax =
p
(0.034) = 0.405(10 - 6)p. Thus,
2
(TAB)max c
1800(0.03)
= 42.44(106)Pa = 42.4 MPa
=
J
0.405(10 - 6)p
228
Ans.
B
0.8 m
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5–22. The solid shaft is subjected to the distributed and
concentrated torsional loadings shown. Determine the
required diameter d of the shaft to the nearest mm if the
allowable shear stress for the material is tallow = 50 MPa.
A
2 kN⭈m/m
1.5 m
1200 N⭈m
C
The internal torque for segment BC is constant TBC = 1200 N # m, Fig. a. However,
the internal torque for segment AB varies with x, Fig. b.
TAB - 2000x + 1200 = 0 TAB = (2000x - 1200) N # m
For segment AB, the maximum internal torque occurs at fixed support A where
x = 1.5 m. Thus,
A TAB B max = 2000(1.5) - 1200 = 1800 N # m
Since A TAB B max 7 TBC, the critical cross-section is at A. The polar moment of inertia
p d 4
pd4
of the rod is J =
. Thus,
a b =
2 2
32
tallow =
Tc
;
J
50(106) =
1800(d>2)
pd4>32
d = 0.05681 m = 56.81 mm = 57 mm
229
Ans.
B
0.8 m
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*5–24. The copper pipe has an outer diameter of 2.50 in.
and an inner diameter of 2.30 in. If it is tightly secured to the
wall at C and a uniformly distributed torque is applied to it
as shown, determine the shear stress developed at points A
and B. These points lie on the pipe’s outer surface. Sketch
the shear stress on volume elements located at A and B.
B
A
C
125 lbft/ft
4 in.
9 in.
12 in.
Internal Torque: As shown on FBD.
Maximum Shear Stress: Applying the torsion formula
tA =
TA c
J
125.0(12)(1.25)
=
tB =
p
2
(1.254 - 1.154)
Ans.
= 3.02 ksi
Ans.
TB c
J
218.75(12)(1.25)
=
= 1.72 ksi
p
2
(1.254 - 1.154)
•5–25.
The copper pipe has an outer diameter of 2.50 in.
and an inner diameter of 2.30 in. If it is tightly secured to
the wall at C and it is subjected to the uniformly distributed
torque along its entire length, determine the absolute
maximum shear stress in the pipe. Discuss the validity of
this result.
B
A
C
125 lbft/ft
4 in.
9 in.
Internal Torque: The maximum torque occurs at the support C.
Tmax = (125 lb # ft>ft)a
12 in.
25 in.
b = 260.42 lb # ft
12 in.>ft
Maximum Shear Stress: Applying the torsion formula
abs =
tmax
Tmax c
J
260.42(12)(1.25)
=
p
2
(1.254 - 1.154)
= 3.59 ksi
Ans.
According to Saint-Venant’s principle, application of the torsion formula should be
as points sufficiently removed from the supports or points of concentrated loading.
230
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5–26. A cylindrical spring consists of a rubber annulus
bonded to a rigid ring and shaft. If the ring is held fixed and
a torque T is applied to the shaft, determine the maximum
shear stress in the rubber.
ro
ri
T
h
T
r
T
F
t =
=
=
A
2prh
2p r2 h
Shear stress is maximum when r is the smallest, i.e. r = ri. Hence,
tmax =
T
2p ri 2 h
Ans.
300 N⭈m
5–27. The A-36 steel shaft is supported on smooth
bearings that allow it to rotate freely. If the gears are
subjected to the torques shown, determine the maximum
shear stress developed in the segments AB and BC. The
shaft has a diameter of 40 mm.
100 N⭈m
A
The internal torque developed in segments AB and BC are shown in their
respective FBDs, Figs. a and b.
The polar moment of inertia of the shaft is J =
A tAB B max
200 N⭈m
B
p
(0.024) = 80(10-9)p m4. Thus,
2
C
300(0.02)
TAB c
= 23.87(106)Pa = 23.9 MPa
=
=
J
80(10-9)p
A tBC B max =
200(0.02)
TBC c
= 15.92(106) Pa = 15.9 MPa
=
J
80(10-9)p
231
Ans.
Ans.
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300 N⭈m
*5–28. The A-36 steel shaft is supported on smooth
bearings that allow it to rotate freely. If the gears are
subjected to the torques shown, determine the required
diameter of the shaft to the nearest mm if tallover = 60 MPa.
100 N⭈m
The internal torque developed in segments AB and BC are shown in their
respective FBDs, Fig. a and b
A
200 N⭈m
B
Here, segment AB is critical since its internal torque is the greatest. The polar
p d 4
pd4
moment of inertia of the shaft is J =
. Thus,
a b =
2 2
32
C
tallow
TC
=
;
J
60(106) =
300(d>2)
pd4>32
d = 0.02942 m = 30 mm
Ans.
•5–29.
When drilling a well at constant angular velocity,
the bottom end of the drill pipe encounters a torsional
resistance TA . Also, soil along the sides of the pipe creates a
distributed frictional torque along its length, varying
uniformly from zero at the surface B to tA at A. Determine
the minimum torque TB that must be supplied by the drive
unit to overcome the resisting torques, and compute
the maximum shear stress in the pipe. The pipe has an outer
radius ro and an inner radius ri .
TA +
TB
B
L
1
t L - TB = 0
2 A
tA
2TA + tAL
TB =
2
Ans.
Maximum shear stress: The maximum torque is within the region above the
distributed torque.
tmax =
tmax =
Tc
J
(2TA + tAL)
] (r0)
2
p 4
4
(r
r
i)
2 0
[
(2TA + tAL)r0
=
Ans.
p(r40 - r4i )
232
A
TA
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5–30. The shaft is subjected to a distributed torque along
its length of t = 110x22 N # m>m, where x is in meters. If the
maximum stress in the shaft is to remain constant at
80 MPa, determine the required variation of the radius c of
the shaft for 0 … x … 3 m.
x
x
T =
L
t dx =
Tc
t =
;
J
L0
10 x2dx =
6
80(10 ) =
3m
c
10 3
x
3
t ⫽ (10x2) N⭈m/m
3
(10
3 )x c
p
2
c4
c3 = 26.526(10-9) x3
c = (2.98 x) mm
Ans.
5–31. The solid steel shaft AC has a diameter of 25 mm and
is supported by smooth bearings at D and E. It is coupled to
a motor at C, which delivers 3 kW of power to the shaft
while it is turning at 50 rev>s. If gears A and B remove 1 kW
and 2 kW, respectively, determine the maximum shear stress
developed in the shaft within regions AB and BC. The shaft
is free to turn in its support bearings D and E.
TC =
3(103)
P
=
= 9.549 N # m
v
50(2p)
TA =
1
T = 3.183 N # m
3 C
3 kW
2 kW
25 mm
1 kW
A
D
(tAB)max =
3.183 (0.0125)
TC
= 1.04 MPa
= p
4
J
2 (0.0125 )
Ans.
(tBC)max =
9.549 (0.0125)
TC
= 3.11 MPa
= p
4
J
2 (0.0125 )
Ans.
233
B
E
C
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*5–32. The pump operates using the motor that has a
power of 85 W. If the impeller at B is turning at 150 rev>min,
determine the maximum shear stress developed in the
20-mm-diameter transmission shaft at A.
150 rev/min
A
Internal Torque:
v = 150
rev 2p rad 1 min
= 5.00p rad>s
¢
≤
min
rev
60 s
P = 85 W = 85 N # m>s
T =
P
85
=
= 5.411 N # m
v
5.00p
Maximum Shear Stress: Applying torsion formula
tmax =
Tc
J
5.411 (0.01)
=
p
4
2 (0.01 )
Ans.
= 3.44 MPa
•5–33.
The gear motor can develop 2 hp when it turns at
450 rev>min. If the shaft has a diameter of 1 in., determine
the maximum shear stress developed in the shaft.
The angular velocity of the shaft is
v = ¢ 450
rev
2p rad
1 min
≤ ¢
≤ ¢
≤ = 15p rad>s
min
1 rev
60 s
and the power is
P = 2 hp ¢
550 ft # lb>s
≤ = 1100 ft # lb>s
1 hp
Then
T =
P
1100
12 in
=
= 23.34 lb # ft a
b = 280.11 lb # in
v
15p
1ft
The polar moment of inertia of the shaft is J =
tmax =
p
(0.54) = 0.03125p in4. Thus,
2
280.11 (0.5)
Tc
=
= 1426.60 psi = 1.43 ksi
J
0.03125p
Ans.
234
B
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5–34. The gear motor can develop 3 hp when it turns at
150 rev>min. If the allowable shear stress for the shaft is
tallow = 12 ksi, determine the smallest diameter of the shaft
to the nearest 18 in. that can be used.
The angular velocity of the shaft is
v = a 150
rev
2p rad 1 min
ba
ba
b = 5p rad>s
min
1 rev
60 s
and the power is
P = (3 hp) a
550 ft # lb>s
b = 1650 ft # lb>s
1 hp
Then
T =
P
1650
12 in
=
= (105.04 lb # ft)a
b = 1260.51 lb # in
v
5p
1 ft
The polar moment of inertia of the shaft is J =
tallow =
Tc
;
J
12(103) =
p d 4
pd4
a b =
. Thus,
2 2
32
1260.51 (d>2)
pd4>32
d = 0.8118 in. =
7
in.
8
Ans.
5–35. The 25-mm-diameter shaft on the motor is made
of a material having an allowable shear stress of
tallow = 75 MPa. If the motor is operating at its maximum
power of 5 kW, determine the minimum allowable rotation
of the shaft.
Allowable Shear Stress: The polar moment of inertia of the shaft is
p
J =
A 0.01254 B = 38.3495(10-9) m4.
2
tallow =
Tc
;
J
75(106) =
T(0.0125)
38.3495(10-9)
T = 230.10 N # m
Internal Loading:
T =
P
;
v
230.10 =
5(103)
v
v = 21.7 rad>s
Ans.
235
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*5–36. The drive shaft of the motor is made of a material
having an allowable shear stress of tallow = 75 MPa. If the
outer diameter of the tubular shaft is 20 mm and the wall
thickness is 2.5 mm, determine the maximum allowable
power that can be supplied to the motor when the shaft is
operating at an angular velocity of 1500 rev>min.
Internal Loading: The angular velocity of the shaft is
v = a 1500
rev
2p rad 1 min
ba
ba
b = 50p rad>s
min
1 rev
60 s
We have
T =
P
P
=
v
50p
Allowable Shear Stress: The polar moment of inertia of the shaft is
p
J = A 0.014 - 0.00754 B = 10.7379(10-9) m4.
2
tallow =
Tc
;
J
75(106) =
a
P
b (0.01)
50p
10.7379(10-9)
P = 12 650.25 W = 12.7 kW
Ans.
•5–37.
A ship has a propeller drive shaft that is turning at
1500 rev>min while developing 1800 hp. If it is 8 ft long and
has a diameter of 4 in., determine the maximum shear stress
in the shaft caused by torsion.
Internal Torque:
v = 1500
rev 2p rad 1 min
a
b
= 50.0 p rad>s
min 1 rev
60 s
P = 1800 hp a
T =
550 ft # lb>s
b = 990 000 ft # lb>s
1 hp
990 000
P
=
= 6302.54 lb # ft
v
50.0p
Maximum Shear Stress: Applying torsion formula
tmax =
6302.54(12)(2)
Tc
=
p 4
J
2 (2 )
= 6018 psi = 6.02 ksi
Ans.
236
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5–38. The motor A develops a power of 300 W and turns
its connected pulley at 90 rev>min. Determine the required
diameters of the steel shafts on the pulleys at A and B if the
allowable shear stress is tallow = 85 MPa.
60 mm
90 rev/min
A
B
150 mm
Internal Torque: For shafts A and B
vA = 90
rev 2p rad 1 min
a
b
= 3.00p rad>s
min
rev
60 s
P = 300 W = 300 N # m>s
P
300
=
= 31.83 N # m
vA
3.00p
TA =
vB = vA a
rA
0.06
b = 3.00pa
b = 1.20p rad>s
rB
0.15
P = 300 W = 300 N # m>s
TB =
P
300
=
= 79.58 N # m
vB
1.20p
Allowable Shear Stress: For shaft A
tmax = tallow =
85 A 106 B =
TA c
J
31.83 A d2A B
A B
p dA 4
2 2
dA = 0.01240 m = 12.4 mm
Ans.
For shaft B
tmax = tallow =
85 A 106 B =
TB c
J
79.58 A d2B B
A B
p dB 4
2 2
dB = 0.01683 m = 16.8 mm
Ans.
237
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5–39. The solid steel shaft DF has a diameter of 25 mm
and is supported by smooth bearings at D and E. It is
coupled to a motor at F, which delivers 12 kW of power to
the shaft while it is turning at 50 rev>s. If gears A, B, and C
remove 3 kW, 4 kW, and 5 kW respectively, determine the
maximum shear stress developed in the shaft within regions
CF and BC. The shaft is free to turn in its support bearings
D and E.
v = 50
3 kW 4 kW
A
B
D
C
E
F
rev 2p rad
c
d = 100 p rad>s
s
rev
TF =
12(103)
P
=
= 38.20 N # m
v
100 p
TA =
3(103)
P
=
= 9.549 N # m
v
100 p
TB =
4(103)
P
=
= 12.73 N # m
v
100 p
(tmax)CF =
38.20(0.0125)
TCF c
= 12.5 MPa
= p
4
J
2 (0.0125 )
Ans.
(tmax)BC =
22.282(0.0125)
TBC c
= 7.26 MPa
= p
4
J
2 (0.0125 )
Ans.
*5–40. Determine the absolute maximum shear stress
developed in the shaft in Prob. 5–39.
v = 50
12 kW
5 kW
25 mm
3 kW 4 kW
rev 2p rad
c
d = 100 p rad>s
s
rev
TF =
12(103)
P
=
= 38.20 N # m
v
100p
TA =
3(103)
P
=
= 9.549 N # m
v
100p
TB =
4(103)
P
=
= 12.73 N # m
v
100p
A
D
Tmax = 38.2 N # m
38.2(0.0125) = 12.5 MPa
tabs = Tc =
max
p
4
J
2 (0.0125 )
Ans.
238
B
C
From the torque diagram,
12 kW
5 kW
25 mm
E
F
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•5–41.
The A-36 steel tubular shaft is 2 m long and has an
outer diameter of 50 mm. When it is rotating at 40 rad> s, it
transmits 25 kW of power from the motor M to the pump
P. Determine the smallest thickness of the tube if the
allowable shear stress is tallow = 80 MPa.
P
M
P
M
The internal torque in the shaft is
T =
25(103)
P
=
= 625 N # m
v
40
The polar moment of inertia of the shaft is J =
tallow =
Tc
;
J
80(106) =
p
(0.0254 - Ci 4). Thus,
2
625(0.025)
p
4
2 (0.025
- Ci 4)
Ci = 0.02272 m
So that
t = 0.025 - 0.02272
= 0.002284 m = 2.284 mm = 2.5 mm
Ans.
5–42. The A-36 solid tubular steel shaft is 2 m long and has
an outer diameter of 60 mm. It is required to transmit
60 kW of power from the motor M to the pump P.
Determine the smallest angular velocity the shaft can have
if the allowable shear stress is tallow = 80 MPa.
The polar moment of inertia of the shaft is J =
tallow =
Tc
;
J
80(106) =
p
(0.034) = 0.405(10-6)p m4. Thus,
2
T(0.03)
0.405(10-6)p
T = 3392.92 N # m
P = Tv ;
60(103) = 3392.92 v
v = 17.68 rad>s = 17.7 rad>s
Ans.
239
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5–43. A steel tube having an outer diameter of 2.5 in. is
used to transmit 35 hp when turning at 2700 rev>min.
Determine the inner diameter d of the tube to the nearest
1
8 in. if the allowable shear stress is tallow = 10 ksi.
v =
2700(2p)
= 282.74 rad>s
60
d
2.5 in.
P = Tv
35(550) = T(282.74)
T = 68.083 lb # ft
tmax = tallow =
10(103) =
Tc
J
68.083(12)(1.25)
p
4
2 (12.5
- ci 4)
ci = 1.2416 in.
d = 2.48 in.
Use d = 212 in.
Ans.
*5–44. The drive shaft AB of an automobile is made of a
steel having an allowable shear stress of tallow = 8 ksi. If the
outer diameter of the shaft is 2.5 in. and the engine delivers
200 hp to the shaft when it is turning at 1140 rev>min,
determine the minimum required thickness of the shaft’s wall.
v =
B
1140(2p)
= 119.38 rad>s
60
P = Tv
200(550) = T(119.38)
T = 921.42 lb # ft
tallow =
8(103) =
Tc
J
921.42(12)(1.25)
p
4
2 (1.25
- r4i )
,
ri = 1.0762 in.
t = ro - ri = 1.25 - 1.0762
t = 0.174 in.
Ans.
240
A
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•5–45. The drive shaft AB of an automobile is to be
designed as a thin-walled tube. The engine delivers 150 hp
when the shaft is turning at 1500 rev>min. Determine the
minimum thickness of the shaft’s wall if the shaft’s outer
diameter is 2.5 in. The material has an allowable shear stress
of tallow = 7 ksi.
v =
B
A
1500(2p)
= 157.08 rad>s
60
P = Tv
150(550) = T(157.08)
T = 525.21 lb # ft
tallow =
7(103) =
Tc
J
525.21(12)(1.25)
p
4
2 (1.25
- r4i )
ri = 1.1460 in.
,
t = ro - ri = 1.25 - 1.1460
t = 0.104 in.
Ans.
5–46. The motor delivers 15 hp to the pulley at A while
turning at a constant rate of 1800 rpm. Determine to the
nearest 18 in. the smallest diameter of shaft BC if the
allowable shear stress for steel is tallow = 12 ksi. The belt
does not slip on the pulley.
B
C
3 in.
The angular velocity of shaft BC can be determined using the pulley ratio that is
vBC
1.5 in.
rA
1.5
rev
2p rad 1 min
= a b vA = a
b a 1800
ba
ba
b = 30p rad>s
rC
3
min
1 rev
60 s
A
The power is
P = (15 hp) a
550 ft # n>s
b = 8250 ft # lb>s
1 hp
Thus,
T =
P
8250
12 in.
=
= (87.54 lb # ft)a
b = 1050.42 lb # in
v
30p
1 ft
The polar moment of inertia of the shaft is J =
tallow =
Tc
;
J
12(103) =
p d 4
pd4
a b =
. Thus,
2 2
32
1050.42(d>2)
pd4>32
d = 0.7639 in =
7
in.
8
Ans.
241
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5–47. The propellers of a ship are connected to a A-36
steel shaft that is 60 m long and has an outer diameter of
340 mm and inner diameter of 260 mm. If the power output is
4.5 MW when the shaft rotates at 20 rad>s, determine the
maximum torsional stress in the shaft and its angle of twist.
T =
4.5(106)
P
=
= 225(103) N # m
v
20
tmax =
f =
225(103)(0.170)
Tc
= 44.3 MPa
=
p
J
[(0.170)4 - (0.130)4]
2
Ans.
225 A 103 B (60)
TL
= 0.2085 rad = 11.9°
=
p
JG
[(0.170)4 - (0.130)4)75(109)
2
Ans.
*5–48. A shaft is subjected to a torque T. Compare the
effectiveness of using the tube shown in the figure with that
of a solid section of radius c. To do this, compute the percent
increase in torsional stress and angle of twist per unit length
for the tube versus the solid section.
T
c
2
T
c
Shear stress:
For the tube,
(tt)max =
c
Tc
Jt
For the solid shaft,
(ts)max =
Tc
Js
% increase in shear stress =
=
(ts)max - (tt)max
(100) =
(tt)max
Js - Jt
(100) =
Jt
p
2
Tc
Jt
-
Tc
Js
Tc
Js
(100)
c4 - [p2 [c4 - (p2 )4]]
p
2
[c4 - (p2 )4]
= 6.67 %
(100)
Ans.
Angle of twist:
For the tube,
ft =
TL
Jt(G)
For the shaft,
fs =
TL
Js(G)
% increase in f =
ft - fs
(100%) =
fs
=
Js - Jt
(100%) =
Jt
TL
Jt(G)
-
TL
Js(G)
TL
Js(G)
p
2
(100%)
c4 - [p2 [c4 - (p2 )4]]
p
2
[c4 - (p2 )4]
(100%)
= 6.67 %
Ans.
242
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•5–49.
The A-36 steel axle is made from tubes AB and CD
and a solid section BC. It is supported on smooth bearings
that allow it to rotate freely. If the gears, fixed to its ends, are
subjected to 85-N # m torques, determine the angle of twist
of gear A relative to gear D. The tubes have an outer
diameter of 30 mm and an inner diameter of 20 mm. The
solid section has a diameter of 40 mm.
400 mm
250 mm
400 mm
B
fND = ©
TL
JG
A
(85)(0.25)
2(85)(0.4)
=
p
2
4
4
9
(0.015 - 0.01 )(75)(10 )
+
p
2
85 Nm
(0.024)(75)(109)
= 0.01534 rad = 0.879°
Ans.
5–50. The hydrofoil boat has an A-36 steel propeller
shaft that is 100 ft long. It is connected to an in-line diesel
engine that delivers a maximum power of 2500 hp and
causes the shaft to rotate at 1700 rpm. If the outer
diameter of the shaft is 8 in. and the wall thickness is 38 in.,
determine the maximum shear stress developed in the
shaft. Also, what is the “wind up,” or angle of twist in the
shaft at full power?
100 ft
Internal Torque:
v = 1700
rev 2p rad 1 min
a
b
= 56.67p rad>s
min
rev
60 s
P = 2500 hp a
T =
550 ft # lb>s
b = 1 375 000 ft # lb>s
1 hp
P
1 375 000
=
= 7723.7 lb # ft
v
56.67p
Maximum Shear Stress: Applying torsion Formula.
tmax =
Tc
J
7723.7(12)(4)
=
p
2
(44 - 3.6254)
Ans.
= 2.83 ksi
Angle of Twist:
f =
TL
=
JG
7723.7(12)(100)(12)
p
2
(44 - 3.6254)11.0(106)
= 0.07725 rad = 4.43°
Ans.
243
C
D
85 Nm
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5–51. The engine of the helicopter is delivering 600 hp
to the rotor shaft AB when the blade is rotating at
1200 rev>min. Determine to the nearest 18 in. the diameter
of the shaft AB if the allowable shear stress is tallow = 8 ksi
and the vibrations limit the angle of twist of the shaft to
0.05 rad. The shaft is 2 ft long and made from L2 steel.
A
B
1200(2)(p)
= 125.66 rad>s
v =
60
P = Tv
600(550) = T(125.66)
T = 2626.06 lb # ft
Shear - stress failure
tallow =
8(103) =
Tc
J
2626.06(12)c
p
2
c4
c = 1.3586 in.
Angle of twist limitation
f =
0.05 =
TL
JG
2626.06(12)(2)(12)
p
2
c4(11.0)(106)
c = 0.967 in.
Shear - stress failure controls the design.
d = 2c = 2 (1.3586) = 2.72 in.
Use d = 2.75 in.
Ans.
244
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*5–52. The engine of the helicopter is delivering 600 hp
to the rotor shaft AB when the blade is rotating at
1200 rev>min. Determine to the nearest 18 in. the diameter of
the shaft AB if the allowable shear stress is tallow = 10.5 ksi
and the vibrations limit the angle of twist of the shaft to
0.05 rad. The shaft is 2 ft long and made from L2 steel.
A
v =
B
1200(2)(p)
= 125.66 rad>s
60
P = Tv
600(550) = T(125.66)
T = 2626.06 lb # ft
Shear - stress failure
tallow = 10.5(10)3 =
2626.06(12)c
p
2
c4
c = 1.2408 in.
Angle of twist limitation
f =
0.05 =
TL
JG
2626.06(12)(2)(12)
p
2
c4 (11.0)(106)
c = 0.967 in.
Shear stress failure controls the design
d = 2c = 2 (1.2408) = 2.48 in.
Use d = 2.50 in.
Ans.
245
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•5–53. The 20-mm-diameter A-36 steel shaft is subjected to
the torques shown. Determine the angle of twist of the end B.
A
Internal Torque: As shown on FBD.
D
Angle of Twist:
C
B
TL
fB = a
JG
30 Nm
600 mm
200 mm
20 Nm
800 mm
1
[ -80.0(0.8) + ( -60.0)(0.6) + ( - 90.0)(0.2)]
= p
4
9
(0.01
)(75.0)(10
)
2
= - 0.1002 rad = | 5.74° |
80 Nm
Ans.
5–54. The assembly is made of A-36 steel and consists of a
solid rod 20 mm in diameter fixed to the inside of a tube using
a rigid disk at B. Determine the angle of twist at D. The tube
has an outer diameter of 40 mm and wall thickness of 5 mm.
A
B
The internal torques developed in segments AB and BD of the assembly are shown
in Fig. a and b
0.4 m
C
0.1 m
p
The polar moment of inertia of solid rod and tube are JAB = (0.024 - 0.0154)
2
p
= 54.6875(10 - 9)p m4 and JBD = (0.014) = 5(10 - 9)p m4. Thus,
2
fD = ©
Ti Li
TAB LAB
TBD LBD
=
+
Ji Gi
JAB Gst
JBD Gst
-60 (0.4)
90(0.4)
=
54.6875(10 - 9)p [75(109)]
+
5(10 - 9)p [75(109)]
= - 0.01758 rad = 1.01°
Ans.
246
150 N⭈m
D
0.3 m
60 N⭈m
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5–55. The assembly is made of A-36 steel and consists of a
solid rod 20 mm in diameter fixed to the inside of a tube using
a rigid disk at B. Determine the angle of twist at C. The tube
has an outer diameter of 40 mm and wall thickness of 5 mm.
A
B
The polar moment of inertia
= 54.6875 (10 - 9)p m4. Thus,
fC = ©
of
the
tube
is
J =
150 N⭈m
0.4 m
The internal torques developed in segments AB and BC of the assembly are shown
in Figs. a and b.
C
0.1 m
p
(0.024 - 0.0154)
2
D
0.3 m
60 N⭈m
Ti Li
TAB LAB
TBC LBC
=
+
Ji Gi
JGst
J Gst
=
1
C 90(0.4) + 150(0.1) D
54.6875(10 )p [75(109)]
-9
= 0.003958 rad = 0.227°
Ans.
*5–56. The splined ends and gears attached to the A-36
steel shaft are subjected to the torques shown. Determine
the angle of twist of end B with respect to end A. The shaft
has a diameter of 40 mm.
fB>A = ©
300 N⭈m
500 N⭈m
A
200 N⭈m
-300(0.3)
200(0.4)
400(0.5)
TL
=
+
+
JG
JG
JG
JG
C
400 N⭈m
190
=
=
JG
300 mm
190
p
4
(0.02
)(75)(109)
2
D
B
400 mm
= 0.01008 rad = 0.578°
Ans.
500 mm
247
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•5–57.
The motor delivers 40 hp to the 304 stainless steel
shaft while it rotates at 20 Hz. The shaft is supported on
smooth bearings at A and B, which allow free rotation of
the shaft. The gears C and D fixed to the shaft remove 25 hp
and 15 hp, respectively. Determine the diameter of the
shaft to the nearest 18 in. if the allowable shear stress is
tallow = 8 ksi and the allowable angle of twist of C with
respect to D is 0.20°.
External Applied Torque: Applying T =
TM =
40(550)
= 175.07 lb # ft
2p(20)
TD =
15(550)
= 65.65 lb # ft
2p(20)
A
D
10 in.
6 in.
25(550)
= 109.42 lb # ft
2p(20)
Internal Torque: As shown on FBD.
Allowable Shear Stress: Assume failure due to shear stress. By observation, section
AC is the critical region.
Tc
J
tmax = tallow =
175.07(12) A d2 B
8(103) =
p
2
A d2 B
4
d = 1.102 in.
Angle of Twist: Assume failure due to angle of twist limitation.
fC>D =
0.2(p)
=
180
TCDLCD
JG
65.65(12)(8)
p
2
B
8 in.
P
, we have
2pf
TC =
C
A d2 B (11.0)(106)
4
d = 1.137 in. (controls !)
1
Use d = 1 in.
4
Ans.
248
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5–58. The motor delivers 40 hp to the 304 stainless steel
solid shaft while it rotates at 20 Hz. The shaft has a diameter
of 1.5 in. and is supported on smooth bearings at A and B,
which allow free rotation of the shaft. The gears C and D
fixed to the shaft remove 25 hp and 15 hp, respectively.
Determine the absolute maximum stress in the shaft and
the angle of twist of gear C with respect to gear D.
A
C
D
10 in.
B
8 in.
6 in.
External Applied Torque: Applying T =
TM =
40(550)
= 175.07 lb # ft
2p(20)
TD =
15(550)
= 65.65 lb # ft
2p(20)
P
, we have
2pf
TC =
25(550)
= 109.42 lb # ft
2p(20)
Internal Torque: As shown on FBD.
Allowable Shear Stress: The maximum torque occurs within region AC of the shaft
where Tmax = TAC = 175.07 lb # ft.
abs =
tmax
175.07(12)(0.75)
Tmax c
= 3.17 ksi
=
p
4
J
2 (0.75 )
Ans.
Angle of Twist:
fC>D =
TCD LCD
JG
65.65(12)(8)
=
p
2
(0.754)(11.0)(106)
= 0.001153 rad = 0.0661°
Ans.
249
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5–59. The shaft is made of A-36 steel. It has a diameter of
1 in. and is supported by bearings at A and D, which allow
free rotation. Determine the angle of twist of B with
respect to D.
A
B
60 lb⭈ft
C
2 ft
60 lb⭈ft
2.5 ft
The internal torques developed in segments BC and CD are shown in Figs. a and b.
D
3 ft
p
The polar moment of inertia of the shaft is J = (0.54) = 0.03125p in4. Thus,
2
TiLi TBC LBC TCD LCD
=
+
FB/D = a
JiGi
J Gst
J Gst
-60(12)(2.5)(12)
=
(0.03125p)[11.0(106)]
+ 0
= - 0.02000 rad = 1.15°
Ans.
*5–60. The shaft is made of A-36 steel. It has a diameter of
1 in. and is supported by bearings at A and D, which allow
free rotation. Determine the angle of twist of gear C with
respect to B.
A
B
60 lb⭈ft
C
2 ft
60 lb⭈ft
2.5 ft
The internal torque developed in segment BC is shown in Fig. a
D
3 ft
p
The polar moment of inertia of the shaft is J = (0.54) = 0.03125p in4. Thus,
2
fC>B =
- 60(12)(2.5)(12)
TBC LBC
=
J Gst
(0.03125p)[11.0(106)]
= - 0.02000 rad
= 1.15°
Ans.
250
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•5–61.
The two shafts are made of A-36 steel. Each has a
diameter of 1 in., and they are supported by bearings at A,
B, and C, which allow free rotation. If the support at D is
fixed, determine the angle of twist of end B when the
torques are applied to the assembly as shown.
D
10 in.
C
80 lbft
A
30 in.
40 lbft
8 in.
10 in.
12 in.
Internal Torque: As shown on FBD.
Angle of Twist:
TL
fE = a
JG
1
=
p
2
(0.54)(11.0)(105)
[ -60.0(12)(30) + 20.0(12)(10)]
= - 0.01778 rad = 0.01778 rad
fF =
6
6
f = (0.01778) = 0.02667 rad
4 E
4
Since there is no torque applied between F and B then
fB = fF = 0.02667 rad = 1.53°
Ans.
251
4 in.
6 in.
B
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5–62. The two shafts are made of A-36 steel. Each has a
diameter of 1 in., and they are supported by bearings at A,
B, and C, which allow free rotation. If the support at D is
fixed, determine the angle of twist of end A when the
torques are applied to the assembly as shown.
D
10 in.
C
80 lbft
A
30 in.
40 lbft
8 in.
10 in.
12 in.
Internal Torque: As shown on FBD.
Angle of Twist:
TL
fE = a
JG
1
=
p
2
(0.54)(11.0)(106)
[ -60.0(12)(30) + 20.0(12)(10)]
= - 0.01778 rad = 0.01778 rad
fF =
6
6
f = (0.01778) = 0.02667 rad
4 E
4
fA>F =
TGF LGF
JG
- 40(12)(10)
=
p
2
(0.54)(11.0)(106)
= - 0.004445 rad = 0.004445 rad
fA = fF + fA>F
= 0.02667 + 0.004445
= 0.03111 rad = 1.78°
Ans.
252
4 in.
6 in.
B
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5–63. The device serves as a compact torsional spring. It is
made of A-36 steel and consists of a solid inner shaft CB
which is surrounded by and attached to a tube AB using a
rigid ring at B. The ring at A can also be assumed rigid
and is fixed from rotating. If a torque of T = 2 kip # in. is
applied to the shaft, determine the angle of twist at the end C
and the maximum shear stress in the tube and shaft.
12 in.
12 in.
B
T
1 in.
A
0.5 in. C
Internal Torque: As shown on FBD.
Maximum Shear Stress:
(tBC)max =
2.00(0.5)
TBC c
= 10.2 ksi
= p
4
J
2 (0.5 )
TBA c
=
J
(tBA)max =
2.00(1)
p
2
(14 - 0.754)
Ans.
= 1.86 ksi
Ans.
Angle of Twist:
fB =
TBA LBA
JG
(2.00)(12)
=
p
2
4
(1 - 0.754)11.0(103)
fC>B =
TBC LBC
JG
2.00(24)
=
= 0.002032 rad
p
2
(0.54)11.0(103)
= 0.044448 rad
fC = fB + fC>B
= 0.002032 + 0.044448
= 0.04648 rad = 2.66°
Ans.
253
0.75 in.
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*5–64. The device serves as a compact torsion spring. It is
made of A-36 steel and consists of a solid inner shaft CB
which is surrounded by and attached to a tube AB using a
rigid ring at B. The ring at A can also be assumed rigid and
is fixed from rotating. If the allowable shear stress for the
material is tallow = 12 ksi and the angle of twist at C is
limited to fallow = 3°, determine the maximum torque T
that can be applied at the end C.
12 in.
12 in.
B
T
1 in.
A
0.5 in. C
Internal Torque: As shown on FBD.
Allowable Shear Stress: Assume failure due to shear stress.
tmax = tallow =
12.0 =
TBC c
J
T (0.5)
p
2
(0.54)
T = 2.356 kip # in
tmax = tallow =
12.0 =
TBA c
J
T (1)
p
2
(14 - 0.754)
T = 12.89 kip # in
Angle of Twist: Assume failure due to angle of twist limitation.
fB =
TBA LBA
=
JG
T(12)
p
2
(14 - 0.754) 11.0(103)
= 0.001016T
fC>B =
TBC LBC
=
JG
T(24)
p
2
(0.54)11.0(103)
= 0.022224T
(fC)allow = fB + fC>B
3(p)
= 0.001016T + 0.022224T
180
T = 2.25 kip # in (controls !)
Ans.
254
0.75 in.
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•5–65.
The A-36 steel assembly consists of a tube having
an outer radius of 1 in. and a wall thickness of 0.125 in. Using
a rigid plate at B, it is connected to the solid 1-in-diameter
shaft AB. Determine the rotation of the tube’s end C if a
torque of 200 lb # in. is applied to the tube at this end. The
end A of the shaft is fixed supported.
B
C
200 lb⭈in.
4 in.
A
6 in.
fB =
TABL
=
JG
fC>B =
200(10)
p
2
TCBL
=
JG
(0.5)4(11.0)(106)
= 0.001852 rad
-200(4)
p
2
4
(1 - 0.8754)(11.0)(106)
= - 0.0001119 rad
fC = fB + fC>B
= 0.001852 + 0.0001119
= 0.001964 rad = 0.113°
Ans.
255
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5–66. The 60-mm diameter shaft ABC is supported by two
journal bearings, while the 80-mm diameter shaft EH is
fixed at E and supported by a journal bearing at H. If
T1 = 2 kN # m and T2 = 4 kN # m, determine the angle of
twist of gears A and C. The shafts are made of A-36 steel.
E
A
600 mm
D
100 mm
H
T2
600 mm
B
75 mm
900 mm
Equilibrium: Referring to the free - body diagram of shaft ABC shown in Fig. a
©Mx = 0; F(0.075) - 4(103) - 2(103) = 0
F = 80(103) N
Internal Loading: Referring to the free - body diagram of gear D in Fig. b,
©Mx = 0; 80(103)(0.1) - TDH = 0
TDH = 8(103)N # m
Also, from the free - body diagram of gear A, Fig. c,
©Mx = 0; TAB - 4(103) = 0
TAB = 4 A 103 B N # m
And from the free - body diagram of gear C, Fig. d,
©Mx = 0; - TBC - 2 A 103 B = 0
TBC = - 2(103) N # m
Angle of Twist: The polar moment of inertia of segments AB, BC and DH
p
of
the
shaft
are
JAB = JBC =
and
A 0.034 B = 0.405(10 - 6)p m4
2
p
4
-6
4
JDH =
A 0.04 B = 1.28(10 )p m . We have
2
fD =
8(103)(0.6)
TDH LDH
= 0.01592 rad
=
JDHGst
1.28(10 - 6)p(75)(109)
Then, using the gear ratio,
fB = fD a
rD
100
b = 0.02122 rad
b = 0.01592 a
rB
75
Also,
fC>B =
- 2(103)(0.9)
TBC LBC
= - 0.01886 rad = 0.01886 rad
=
JBCGst
0.405(10 - 6)p(75)(109)
fA>B =
4(103)(0.6)
TABLAB
= 0.02515 rad
=
JAB Gst
0.405(10 - 6)p(75)(109)
Thus,
fA = fB + fA>B
fA = 0.02122 + 0.02515
= 0.04637 rad = 2.66°
Ans.
fC = fB + fC>B
fC = 0.02122 + 0.01886
= 0.04008 rad = 2.30°
Ans.
256
T1
C
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5–66. Continued
257
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5–67. The 60-mm diameter shaft ABC is supported by two
journal bearings, while the 80-mm diameter shaft EH is fixed
at E and supported by a journal bearing at H. If the angle
of twist at gears A and C is required to be 0.04 rad,
determine the magnitudes of the torques T1 and T2. The
shafts are made of A-36 steel.
E
A
600 mm
D
100 mm
H
T2
600 mm
B
75 mm
900 mm
Equilibrium: Referring to the free - body diagram of shaft ABC shown in Fig. a
©Mx = 0; F(0.075) - T1 - T2 = 0
T1
F = 13.333 A T1 + T2 B
Internal Loading: Referring to the free - body diagram of gear D in Fig. b,
©Mx = 0; 13.333 A T1 + T2 B (0.1) - TDE = 0
TDE = 1.333 A T1 + T2 B
Also, from the free - body diagram of gear A, Fig. c,
©Mx = 0; TAB - T2 = 0
TAB = T2
and from the free - body diagram of gear C, Fig. d
©Mx = 0; TBC - T1 = 0
TBC = T1
Angle of Twist: The polar moments of inertia of segments AB, BC and DH
p
of
the
shaft
are
and
JAB = JBC =
A 0.034 B = 0.405(10 - 6)pm4
2
p
JDH =
A 0.044 B = 1.28(10 - 6)pm4. We have
2
fD =
1.333 A T1 + T2 B (0.6)
TDE LDH
= 2.6258(10 - 6) A T1 + T2 B
=
JDE Gst
1.28(10 - 6)p (75)(109)
Then, using the gear ratio,
fB = fD ¢
rD
100
b = 3.5368(10 - 6) A T1 + T2 B
≤ = 2.6258(10 - 6) A T1 + T2 B a
rB
75
Also,
fC>B =
T1(0.9)
TBC LBC
= 9.4314(10 - 6)T1
=
JBC Gst
0.405(10 - 6)p(75)(109)
fA>B =
T2(0.6)
TAB LAB
= 6.2876(10 - 6)T2
=
JAB Gst
0.405(10 - 6)p(75)(109)
Here, it is required that fA = fC = 0.04 rad. Thus,
fA = fB + fA>B
0.04 = 3.5368(10 - 6) A T1 + T2 B + 6.2876(10 - 6)T2
T1 + 2.7778T2 = 11309.73
(1)
fC = fB + fC>B
0.04 = 3.5368(10 - 6) A T1 + T2 B + 9.4314(10 - 6)T1
3.6667T1 + T2 = 11309.73
(2)
258
C
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5–67. Continued
Solving Eqs. (1) and (2),
T1 = 2188.98 N # m = 2.19 kN # m
Ans.
T2 = 3283.47 N # m = 3.28 kN # m
Ans.
*5–68. The 30-mm-diameter shafts are made of L2 tool
steel and are supported on journal bearings that allow the
shaft to rotate freely. If the motor at A develops a torque of
T = 45 N # m on the shaft AB, while the turbine at E is fixed
from turning, determine the amount of rotation of gears B
and C.
A
45 Nm
B
1.5 m
50 mm
D
C
Internal Torque: As shown on FBD.
0.5 m
Angle of Twist:
fC =
TCE LCE
JG
67.5(0.75)
=
p
2
(0.0154)75.0(103)
= 0.008488 rad = 0.486°
fB =
75
50
Ans.
fC = 0.729°
Ans.
259
E
75 mm
0.75 m
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•5–69.
The shafts are made of A-36 steel and each has a
diameter of 80 mm. Determine the angle of twist at end E.
0.6 m
A
B
150 mm 10 kN⭈m
C
D
200 mm
0.6 m
E
Equilibrium: Referring to the free - body diagram of shaft CDE shown in Fig. a,
©Mx = 0; 10(103) - 2(103) - F(0.2) = 0
150 mm
F = 40(103) N
0.6 m
2 kN⭈m
Internal Loading: Referring to the free - body diagram of gear B, Fig. b,
©Mx = 0;
- TAB - 40(103)(0.15) = 0
TAB = - 6(103) N # m
Referring to the free - body diagram of gear D, Fig. c,
©Mx = 0; 10(103) - 2(103) - TCD = 0
TCD = 8(103) N # m
Referring to the free - body diagram of shaft DE, Fig. d,
©Mx = 0;
- TDE - 2(103) = 0
Angle of Twist: The polar
p
J =
A 0.044 B = 1.28(10 - 6)p m4.
2
TDE = - 2(103) N # m
moment
of
inertia
of
the
shafts
are
We have
fB =
- 6(103)(0.6)
TAB LAB
= - 0.01194 rad = 0.01194 rad
=
JGst
1.28(10 - 6)p(75)(109)
Using the gear ratio,
fC = fB ¢
rB
150
b = 0.008952 rad
≤ = 0.01194 a
rC
200
fE>C = ©
TiLi
TCD LCD
TDE LDE
=
+
JiGi
JGst
JGst
Also,
0.6
=
-6
1.28(10 )p(75)(109)
b 8(103) + c - 2(103) d r
= 0.01194 rad
Thus,
fE = fC + fE>C
fE = 0.008952 + 0.01194
= 0.02089 rad = 1.20°
Ans.
260
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5–69. Continued
261
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5–70. The shafts are made of A-36 steel and each has a
diameter of 80 mm. Determine the angle of twist of gear D.
0.6 m
Equilibrium: Referring to the free-body diagram of shaft CDE shown in Fig. a,
3
3
A
©Mx = 0; 10(10 ) - 2(10 ) - F(0.2) = 0
F = 40(10 ) N
- TAB - 40(103)(0.15) = 0
150 mm 10 kN⭈m
C
Internal Loading: Referring to the free - body diagram of gear B, Fig. b,
©Mx = 0;
B
3
TAB = - 6(103) N # m
D
200 mm
0.6 m
E
Referring to the free - body diagram of gear D, Fig. c,
©Mx = 0; 10(103) - 2(103) - TCD = 0
150 mm
TCD = 8(103) N # m
0.6 m
2 kN⭈m
Angle of Twist: The polar moment
p
J =
A 0.044 B = 1.28(10 - 6)p m4. We have
2
fB =
of
inertia
of
the
shafts
are
- 6(103)(0.6)
TAB LAB
= - 0.01194 rad = 0.01194 rad
=
JGst
1.28(10 - 6)p(75)(109)
Using the gear ratio,
fC = fB ¢
rB
150
b = 0.008952 rad
≤ = 0.01194 a
rC
200
Also,
fD>C =
8(103)(0.6)
TCD LCD
= 0.01592 rad
=
JGst
1.28(10 - 6)p(75)(109)
Thus,
fD = fC + fD>C
fD = 0.008952 + 0.01592
= 0.02487 rad = 1.42°
Ans.
262
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*5–72. The 80-mm diameter shaft is made of 6061-T6
aluminum alloy and subjected to the torsional loading
shown. Determine the angle of twist at end A.
0.6 m
0.6 m
C
10 kN⭈m/m
B
A
2 kN⭈m
Equilibrium: Referring to the free - body diagram of segment AB shown in Fig. a,
©Mx = 0;
- TAB - 2(103) = 0
TAB = - 2(103)N # m
And the free - body diagram of segment BC, Fig. b,
©Mx = 0;
Angle of Twist: The polar moment
p
J =
A 0.042 B = 1.28(10 - 6)p m4. We have
2
fA = ©
1.28(10 - 6)p(26)(109)
0.6 m -
+
1
= -
of
inertia
of
the
shaft
is
LBC
TiLi
TABLAB
TBC dx
=
+
JiGi
JGal
JGal
L0
- 2(103)(0.6)
=
TBC = - C 10(103)x + 2(103) D N # m
- TBC - 10(103)x - 2(103) = 0
1.28(10 - 6)p(26)(109)
L0
C 10(103)x + 2(103) D dx
1.28(10 - 6)p(26)(109)
b 1200 + C 5(103)x2 + 2(103)x D 2
0.6m
0
r
= - 0.04017 rad = 2.30°
Ans.
263
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•5–73.
The tapered shaft has a length L and a radius r at
end A and 2r at end B. If it is fixed at end B and is subjected
to a torque T, determine the angle of twist of end A. The
shear modulus is G.
B
2r
L
T
Geometry:
r
A
r
rL + rx
r(x) = r +
x =
L
L
p rL + rx 4
p r4
(L + x)4
a
b =
2
L
2L4
J(x) =
Angle of Twist:
L
T dx
L0 J(x)G
f =
L
=
2TL4
dx
p r4G L0 (L + x)4
=
L
1
2TL4
cd2
4
3
pr G
3(L + x) 0
=
7TL
12p r4G
Ans.
5–74. The rod ABC of radius c is embedded into a medium
where the distributed torque reaction varies linearly from
zero at C to t0 at B. If couple forces P are applied to the lever
arm, determine the value of t0 for equilibrium. Also, find the
angle of twist of end A. The rod is made from material
having a shear modulus of G.
L
2
L
2
Equilibrium: Referring to the free-body diagram of the entire rod shown in Fig. a,
1
L
©Mx = 0; Pd - (t0) a b = 0
2
2
to =
Ans.
Internal Loading: The distributed torque expressed as a function of x, measured
4Pd>L
to
8Pd
from the left end, is t = ¢
≤x = ¢
≤ x = ¢ 2 ≤ x. Thus, the resultant
L>2
L>2
L
torque within region x of the shaft is
TR =
1
1
8Pd
4Pd 2
tx = B ¢ 2 ≤ x R x =
x
2
2
L
L2
Referring to the free - body diagram shown in Fig. b,
©Mx = 0; TBC -
4Pd 2
x = 0
L2
d
2
B
P
4Pd
L
TBC =
4Pd 2
x
L2
Referring to the free - body diagram shown in Fig. c,
©Mx = 0; Pd - TAB = 0
TAB = Pd
264
d
2
t0
C
A
P
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5–74. Continued
Angle of Twist:
f = ©
LBC
TAB LAB
TBC dx
TiLi
=
+
JiGi
JG
JG
L0
L>2
=
L0
4Pd 2
x dx
L2
Pd(L>2)
+
p
2
p
2
¢ c4 ≤ G
¢ c4 ≤ G
L>2
8Pd
x3
=
£ ≥3
4 2
pc L G 3
+
PLd
pc4G
0
=
4PLd
3pc4G
Ans.
265
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5–75. When drilling a well, the deep end of the drill pipe
is assumed to encounter a torsional resistance TA .
Furthermore, soil friction along the sides of the pipe creates
a linear distribution of torque per unit length, varying from
zero at the surface B to t0 at A. Determine the necessary
torque TB that must be supplied by the drive unit to turn
the pipe. Also, what is the relative angle of twist of one end
of the pipe with respect to the other end at the instant the
pipe is about to turn? The pipe has an outer radius ro and an
inner radius ri . The shear modulus is G.
TB
B
L
t0
A
1
t L + TA - TB = 0
2 0
TB =
t0L + 2TA
2
T(x) +
t0 2
t0L + 2TA
x = 0
2L
2
T(x) =
t0 2
t0 L + 2TA
x
2
2L
f =
Ans.
T(x) dx
L JG
=
L
t0L + 2TA
t0 2
1
(
x ) dx
J G L0
2
2L
=
t0 3 L
1 t0 L + 2TA
c
x x dƒ
JG
2
6L
0
=
t0 L2 + 3TAL
3JG
However, J =
f =
p
(r 4 - ri 4)
2 o
2L(t0 L + 3TA)
Ans.
3p(ro 4 - ri 4)G
266
TA
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*5–76. A cylindrical spring consists of a rubber annulus
bonded to a rigid ring and shaft. If the ring is held fixed and
a torque T is applied to the rigid shaft, determine the angle
of twist of the shaft. The shear modulus of the rubber is G.
Hint: As shown in the figure, the deformation of the
element at radius r can be determined from rdu = drg. Use
this expression along with t = T>12pr2h2 from Prob. 5–26,
to obtain the result.
ro
r
ri
T
h
gdr rdu
dr
g
du
r
r du = g dr
du =
gdr
r
(1)
From Prob. 5-26,
t =
T
2p r2h
g =
T
2p r2hG
and
g =
t
G
From (1),
du =
T dr
2p hG r3
r
u =
o
dr
T
1 ro
T
=
cd|
3
2p hG Lri r
2p hG
2 r2 ri
=
1
1
T
c- 2 + 2d
2p hG
2ro
2ri
=
1
1
T
c - 2d
4p hG r2i
ro
Ans.
267
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•5–77.
The A-36 steel shaft has a diameter of 50 mm and is
fixed at its ends A and B. If it is subjected to the torque,
determine the maximum shear stress in regions AC and CB
of the shaft.
A
300 Nm
0.4 m
C
0.8 m
Equilibrium:
TA + TB - 300 = 0
[1]
Compatibility:
fC>A = fC>B
TA(0.4)
TB(0.8)
=
JG
JG
TA = 2.00TB
[2]
Solving Eqs. [1] and [2] yields:
TA = 200 N # m
TB = 100 N # m
Maximum Shear stress:
(tAC)max =
200(0.025)
TAc
= 8.15 MPa
= p
4
J
2 (0.025 )
Ans.
(tCB)max =
100(0.025)
TBc
= 4.07 MPa
= p
4
J
2 (0.025 )
Ans.
268
B
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5–78. The A-36 steel shaft has a diameter of 60 mm and is
fixed at its ends A and B.If it is subjected to the torques shown,
determine the absolute maximum shear stress in the shaft.
200 N⭈m
B
500 N⭈m
D
1.5 m
C
A
Referring to the FBD of the shaft shown in Fig. a,
TA + TB - 500 - 200 = 0
©Mx = 0;
(1)
Using the method of superposition, Fig. b
fA = (fA)TA - (fA)T
0 =
500 (1.5)
700 (1)
TA (3.5)
- c
+
d
JG
JG
JG
TA = 414.29 N # m
Substitute this result into Eq (1),
TB = 285.71 N # m
Referring to the torque diagram shown in Fig. c, segment AC is subjected to
maximum internal torque. Thus, the absolute maximum shear stress occurs here.
tAbs =
414.29 (0.03)
TAC c
=
= 9.77 MPa
p
J
(0.03)4
2
Ans.
269
1m
1m
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5–79. The steel shaft is made from two segments: AC has a
diameter of 0.5 in, and CB has a diameter of 1 in. If it is
fixed at its ends A and B and subjected to a torque of
determine the maximum shear stress in the shaft.
Gst = 10.811032 ksi.
A
0.5 in.
C
D 500 lbft
5 in.
1 in.
8 in.
B
12 in.
Equilibrium:
TA + TB - 500 = 0
(1)
Compatibility condition:
fD>A = fD>B
TA(5)
p
2
4
(0.25 )G
TA(8)
+
p
2
4
(0.5 )G
TB(12)
=
p
2
(0.54)G
1408 TA = 192 TB
(2)
Solving Eqs. (1) and (2) yields
TA = 60 lb # ft
TB = 440 lb # ft
tAC =
60(12)(0.25)
TC
= 29.3 ksi
= p
4
J
2 (0.25 )
tDB =
440(12)(0.5)
TC
= 26.9 ksi
=
p
4
J
2 (0.5 )
(max)
Ans.
270
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*5–80. The shaft is made of A-36 steel, has a diameter of
80 mm, and is fixed at B while A is loose and can rotate
0.005 rad before becoming fixed. When the torques are
applied to C and D, determine the maximum shear stress in
regions AC and CD of the shaft.
2 kN⭈m
B
4 kN⭈m
D
600 mm
C
600 mm
A
Referring to the FBD of the shaft shown in Fig. a,
600 mm
TA + TB + 2 - 4 = 0
©Mx = 0;
(1)
Using the method of superposition, Fig. b,
fA = (fA)T - (uA)TA
0.005 = B
p
2
4(103)(0.6)
(0.04 ) C 75(10 ) D
4
9
2(103)(0.6)
+
p
2
(0.04 ) C 75(10 ) D
4
9
R -
TA (1.8)
p
2
(0.044) C 75(109) D
TA = 1162.24 N # m = 1.162 kN # m
Substitute this result into Eq (1),
TB = 0.838 kN # m
Referring to the torque diagram shown in Fig. c, segment CD is subjected to a
maximum internal torque. Thus, the absolute maximum shear stress occurs here.
t$$$ =
2.838 (103)(0.04)
TCD c
=
= 28.23 (106) Pa = 28.2 MPa
p
4
J
2 (0.04)
271
Ans.
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•5–81. The shaft is made of A-36 steel and has a diameter
of 80 mm. It is fixed at B and the support at A has a torsional
stiffness of k = 0.5 MN # m>rad. If it is subjected to the gear
torques shown, determine the absolute maximum shear stress
in the shaft.
2 kN⭈m
B
4 kN⭈m
D
600 mm
C
600 mm
A
600 mm
Referring to the FBD of the shaft shown in Fig. a,
TA + TB + 2 - 4 = 0
©Mx = 0;
(1)
Using the method of superposition, Fig. b,
fA = (fA)T - (fA)TA
TA
6
0.5(10 )
= D
4(103)(0.6)
p
2
(0.04 ) C 75(10 ) D
4
9
2(103)(0.6)
+
p
2
(0.04 ) C 75(10 ) D
4
9
T -
TA(1.8)
p
2
(0.044) C 75(109) D
TA = 1498.01 N # m = 1.498 kN # m
Substituting this result into Eq (1),
TB = 0.502 kN # m
Referring to the torque diagram shown in Fig. c, segment CD subjected to maximum
internal torque. Thus, the maximum shear stress occurs here.
t$$$ =
2.502(103)(0.04)
TCD C
=
= $$$ = 24.9 MPa
p
4
J
2 (0.04)
272
Ans.
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5–82. The shaft is made from a solid steel section AB and
a tubular portion made of steel and having a brass core.
If it is fixed to a rigid support at A, and a torque of
T = 50 lb # ft is applied to it at C, determine the angle of
twist that occurs at C and compute the maximum shear
stress and maximum shear strain in the brass and steel.
Take Gst = 11.511032 ksi, Gbr = 5.611032 ksi.
3 ft
2 ft
A
0.5 in.
B
1 in.
Equilibrium:
Tbr + Tst - 50 = 0
(1)
Both the steel tube and brass core undergo the same angle of twist fC>B
fC>B =
TL
=
JG
Tst (2)(12)
Tst (2)(12)
p
2
(0.54)(5.6)(104)
=
p
2
4
(1 - 0.54)(11.5)(106)
Tbr = 0.032464 Tst
(2)
Solving Eqs. (1) and (2) yields:
Tst = 48.428 lb # ft;
fC = ©
Tbr = 1.572 lb # ft
50(12)(3)(12)
1.572(12)(2)(12)
TL
+ p 4
= p
4
6
6
JG
2 (0.5 )(5.6)(10 )
2 (1 )(11.5)(10 )
= 0.002019 rad = 0.116°
Ans.
(tst)max AB =
50(12)(1)
TABc
= 382 psi
= p 4
J
2 (1 )
(tst)max BC =
48.428(12)(1)
Tst c
= 394.63 psi = 395 psi (Max)
= p 4
4
J
2 (1 - 0.5 )
Ans.
(gst)max =
(tst)max
394.63
= 343.(10 - 6) rad
=
G
11.5(106)
Ans.
(tbr)max =
1.572(12)(0.5)
Tbr c
= 96.07 psi = 96.1 psi (Max)
=
p
4
J
2 (0.5 )
Ans.
(gbr)max =
(tbr)max
96.07
= 17.2(10 - 6) rad
=
G
5.6(106)
Ans.
273
C
T 50 lbft
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5–83. The motor A develops a torque at gear B of 450 lb # ft,
which is applied along the axis of the 2-in.-diameter steel shaft
CD. This torque is to be transmitted to the pinion gears at E
and F. If these gears are temporarily fixed, determine the
maximum shear stress in segments CB and BD of the shaft.
Also, what is the angle of twist of each of these segments? The
bearings at C and D only exert force reactions on the shaft
and do not resist torque. Gst = 1211032 ksi.
B
E
4 ft
(1)
Compatibility condition:
fB>C = fB>D
TC(4)
TD(3)
=
JG
JG
TC = 0.75 TD
(2)
Solving Eqs. (1) and (2), yields
TD = 257.14 lb # ft
TC = 192.86 lb # ft
(tBD)max =
f =
192.86(12)(1)
p
2
(14)
257.14(12)(1)
p
2
(14)
192.86(12)(4)(12)
p
2
(14)(12)(106)
3 ft
D
A
TC + TD - 450 = 0
(tBC)max =
F
C
Equilibrium:
= 1.47 ksi
Ans.
= 1.96 ksi
Ans.
= 0.00589 rad = 0.338°
Ans.
274
450 lb⭈ft
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*5–84. A portion of the A-36 steel shaft is subjected to a
linearly distributed torsional loading. If the shaft has the
dimensions shown, determine the reactions at the fixed
supports A and C. Segment AB has a diameter of 1.5 in. and
segment BC has a diameter of 0.75 in.
300 lb⭈in./in.
A
60 in.
B
C
48 in.
Equilibrium:
TA + TC - 9000 = 0
TR = t x +
1
tx
(300 - t)x = 150x +
2
2
300
t
=
;
60 - x
60
But
(1)
TR = 150 x +
t = 5(60 - x)
1
[5(60 - x)]x
2
= (300x - 2.5x2) lb # in.
Compatibility condition:
fB>A = fB>C
fB>A =
60
T(x) dx
1
=
[TA - (300x - 2.5x2)] dx
JG L0
L JG
60TA - 360 000
p
2
(0.754)G
=
60
1
[TAx - 150x2 + 0.8333x3] |
JG
0
=
60TA - 360 000
JG
TC(48)
=
p
2
(0.3754)G
(2)
60TA - 768TC = 360 000
Solving Eqs. (1) and (2) yields:
TC = 217.4 lb # in. = 18.1 lb # ft
Ans.
TA = 8782.6 lb # in. = 732 lb # ft
Ans.
275
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•5–85.
Determine the rotation of joint B and the absolute
maximum shear stress in the shaft in Prob. 5–84.
300 lb⭈in./in.
A
Equilibrium:
TA + TC - 9000 = 0
TR = tx +
But
(1)
1
tx
(300 - t)x = 150x +
2
2
300
t
=
;
60 - x
60
C
t = 5(60 - x)
= (300x - 2.5x2) lb # in.
Compatibility condition:
fB>A = fB>C
fB>A =
60
T(x) dx
1
=
[TA - (300x - 2.5x2)] dx
JG L0
L JG
60TA - 360 000
p
2
4
(0.75 )G
=
60
1
[TAx - 150x2 + 0.8333x3] |
JG
0
=
60TA - 360 000
JG
TC(48)
=
p
2
(0.3754)G
(2)
60TA - 768TC = 360 000
Solving Eqs. (1) and (2) yields:
TC = 217.4 lb # in. = 18.1 lb # ft
TA = 8782.6 lb # in. = 732 lb # ft
For segment BC:
fB = fB>C =
TCL
=
JG
217.4(48)
p
2
(0.375)4(11.0)(106)
= 0.030540 rad
fB = 1.75°
tmax =
Ans.
217.4(0.375)
TC
= p
= 2.62 ksi
4
J
2 (0.375)
For segment AB,
tmax =
B
48 in.
1
[5(60 - x)]x
2
TR = 150x +
60 in.
8782.6(0.75)
TC
= p
= 13.3 ksi
4
J
2 (0.75)
abs = 13.3 ksi
tmax
Ans.
276
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5–86. The two shafts are made of A-36 steel. Each has a
diameter of 25 mm and they are connected using the gears
fixed to their ends. Their other ends are attached to fixed
supports at A and B. They are also supported by journal
bearings at C and D, which allow free rotation of the shafts
along their axes. If a torque of 500 N # m is applied to the
gear at E as shown, determine the reactions at A and B.
B
F
D
50 mm
0.75 m
100 mm
500 Nm
E
C
1.5 m
A
Equilibrium:
TA + F(0.1) - 500 = 0
[1]
TB - F(0.05) = 0
[2]
TA + 2TB - 500 = 0
[3]
From Eqs. [1] and [2]
Compatibility:
0.1fE = 0.05fF
fE = 0.5fF
TA(1.5)
TB(0.75)
= 0.5 c
d
JG
JG
TA = 0.250TB
[4]
Solving Eqs. [3] and [4] yields:
TB = 222 N # m
Ans.
TA = 55.6 N # m
Ans.
277
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5–87. Determine the rotation of the gear at E in
Prob. 5–86.
B
F
D
50 mm
0.75 m
100 mm
500 Nm
E
C
1.5 m
A
Equilibrium:
TA + F(0.1) - 500 = 0
[1]
TB - F(0.05) = 0
[2]
TA + 2TB - 500 = 0
[3]
From Eqs. [1] and [2]
Compatibility:
0.1fE = 0.05fF
fE = 0.5fF
TA(1.5)
TB(0.75)
= 0.5 c
d
JG
JG
TA = 0.250TB
[4]
Solving Eqs. [3] and [4] yields:
TB = 222.22 N # m
TA = 55.56 N # m
Angle of Twist:
fE =
TAL
=
JG
55.56(1.5)
p
2
(0.01254)(75.0)(109)
= 0.02897 rad = 1.66°
Ans.
278
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*5–88. The shafts are made of A-36 steel and have the
same diameter of 4 in. If a torque of 15 kip # ft is applied to
gear B, determine the absolute maximum shear stress
developed in the shaft.
2.5 ft
2.5 ft
Equilibrium: Referring to the free - body diagrams of shafts ABC and DE shown in
Figs. a and b, respectively, we have
©Mx = 0; TA + F(0.5) - 15 = 0
(1)
and
A
B
6 in.
15 kip⭈ft
C
D
12 in.
©Mx = 0; F(1) - TE = 0
E
(2)
3 ft
Internal Loadings: The internal torques developed in segments AB and BC of shaft
ABC and shaft DE are shown in Figs. c, d, and e, respectively.
Compatibility Equation:
fCrC = fDrD
¢
TABLAB
TDE LDE
TBCLBC
+
≤ rC = ¢
≤ rD
JGst
JGst
JGst
C - TA(2.5) + F(0.5)(2.5) D (0.5) = - TE(3)(1)
TA - 0.5F = 2.4TE
(3)
Solving Eqs. (1), (2), and (3), we have
F = 4.412 kip
TE = 4.412 kip # ft
TA = 12.79 kip # ft
Maximum Shear Stress: By inspection, segment AB of shaft ABC is subjected to the
greater torque.
A tmax B abs =
12.79(12)(2)
TAB c
= 12.2 ksi
=
Jst
p 4
a2 b
2
Ans.
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•5–89.
The shafts are made of A-36 steel and have the
same diameter of 4 in. If a torque of 15 kip # ft is applied to
gear B, determine the angle of twist of gear B.
2.5 ft
Equilibrium: Referring to the free - body diagrams of shafts ABC and DE shown in
Figs. a and b, respectively,
©Mx = 0; TA + F(0.5) - 15 = 0
2.5 ft
A
B
(1)
6 in.
15 kip⭈ft
and
©Mx = 0; F(1) - TE = 0
(2)
Internal Loadings: The internal torques developed in segments AB and BC of shaft
ABC and shaft DE are shown in Figs. c, d, and e, respectively.
Compatibility Equation: It is required that
fCrC = fDrD
¢
TAB LAB
TDE LDE
TBC LBC
+
≤ rC = ¢
≤ rD
JGst
JGst
JGst
C - TA(2.5) + F(0.5)(2.5) D (0.5) = - TE(3)(1)
TA - 0.5F = 2.4TE
(3)
Solving Eqs. (1), (2), and (3),
F = 4.412 kip
TE = 4.412 kip # ft
TA = 12.79 kip # ft
Angle of Twist: Here, TAB = - TA = - 12.79 kip # ft
fB =
-12.79(12)(2.5)(12)
TAB LAB
=
JGst
p 4
a 2 b (11.0)(103)
2
= - 0.01666 rad = 0.955°
Ans.
280
C
D
12 in.
E
3 ft
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5–90. The two 3-ft-long shafts are made of 2014-T6
aluminum. Each has a diameter of 1.5 in. and they are
connected using the gears fixed to their ends. Their other
ends are attached to fixed supports at A and B. They are
also supported by bearings at C and D, which allow free
rotation of the shafts along their axes. If a torque of 600 lb # ft
is applied to the top gear as shown, determine the maximum
shear stress in each shaft.
A
B
E
2 in.
TA + F a
4
b - 600 = 0
12
(1)
TB - Fa
2
b = 0
12
(2)
From Eqs. (1) and (2)
TA + 2TB - 600 = 0
TAL
TBL
= 0.5 a
b;
JG
JG
(3)
fE = 0.5fF
TA = 0.5TB
(4)
Solving Eqs. (3) and (4) yields:
TB = 240 lb # ft;
TA = 120 lb # ft
(tBD)max =
240(12)(0.75)
TB c
= 4.35 ksi
=
p
4
J
2 (0.75 )
Ans.
(tAC)max =
120(12)(0.75)
TA c
= 2.17 ksi
=
p
4
J
2 (0.75 )
Ans.
281
3 ft
D
4 in.
4(fE) = 2(fF);
C
600 lbft
F
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5–91. The A-36 steel shaft is made from two segments: AC
has a diameter of 0.5 in. and CB has a diameter of 1 in. If
the shaft is fixed at its ends A and B and subjected to a uniform
distributed torque of 60 lb # in.>in. along segment CB,
determine the absolute maximum shear stress in the shaft.
A
0.5 in.
C
5 in.
60 lbin./in.
1 in.
20 in.
Equilibrium:
TA + TB - 60(20) = 0
(1)
Compatibility condition:
fC>B = fC>A
fC>B =
20
(TB - 60x) dx
T(x) dx
=
p
JG
L
L0 2 (0.54)(11.0)(106)
= 18.52(10-6)TB - 0.011112
18.52(10-6)TB - 0.011112 =
TA(5)
p
4
6
2 (0.25 )(11.0)(10 )
18.52(10-6)TB - 74.08(10-6)TA = 0.011112
18.52TB - 74.08TA = 11112
(2)
Solving Eqs. (1) and (2) yields:
TA = 120.0 lb # in. ;
TB = 1080 lb # in.
(tmax)BC =
1080(0.5)
TB c
= 5.50 ksi
= p
4
J
2 (0.5 )
(tmax)AC =
120.0(0.25)
TA c
= 4.89 ksi
= p
4
J
2 (0.25 )
abs = 5.50 ksi
tmax
Ans.
282
B
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*5–92. If the shaft is subjected to a uniform distributed
torque of t = 20 kN # m>m, determine the maximum shear
stress developed in the shaft. The shaft is made of 2014-T6
aluminum alloy and is fixed at A and C.
400 mm
20 kN⭈m/m
600 mm
a
A
80 mm
60 mm
B
a
C
Equilibrium: Referring to the free - body diagram of the shaft shown in Fig. a, we
have
©Mx = 0; TA + TC - 20(103)(0.4) = 0
(1)
Compatibility Equation: The resultant torque of the distributed torque within the
region x of the shaft is TR = 20(103)x N # m. Thus, the internal torque developed in
the shaft as a function of x when end C is free is T(x) = 20(103)x N # m, Fig. b. Using
the method of superposition, Fig. c,
fC = A fC B t - A fC B TC
0 =
0 =
0.4 m
T(x)dx
TCL
JG
JG
0.4 m
20(103)xdx
TC(1)
JG
JG
L0
L0
0 = 20(103) ¢
x2 2 0.4 m
- TC
≤
2 0
TC = 1600 N # m
Substituting this result into Eq. (1),
TA = 6400 N # m
Maximum Shear Stress: By inspection, the maximum internal torque occurs at
support A. Thus,
A tmax B abs =
6400(0.04)
TA c
= 93.1 MPa
=
J
p
4
4
a 0.04 - 0.03 b
2
Ans.
283
Section a–a
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•5–93.
The tapered shaft is confined by the fixed supports
at A and B. If a torque T is applied at its mid-point,
determine the reactions at the supports.
2c
T
A
B
Equilibrium:
c
TA + TB - T = 0
[1]
L/2
L/2
Section Properties:
r(x) = c +
J(x) =
c
c
x =
(L + x)
L
L
4
p c
pc4
c (L + x) d =
(L + x)4
2 L
2L4
Angle of Twist:
fT =
Tdx
=
Lp2
L J(x)G
L
Tdx
pc4
2L4
(L + x)4 G
L
=
dx
2TL4
pc4 G Lp2 (L + x)4
= -
=
fB =
L
1
2TL4
c
d 2
4
3
p
3pc G (L + x)
2
37TL
324 pc4 G
Tdx
=
J(x)G
L
L0
L
TBdx
pc4
2L4
(L + x)4G
L
2TBL4
=
dx
pc G L0 (L + x)4
4
2TBL4
= -
L
1
d 2
3
3pc G (L + x)
0
4
c
7TB L
=
12pc4G
Compatibility:
0 = fT - fB
0 =
7TBL
37TL
4
324pc G
12pc4G
TB =
37
T
189
Ans.
Substituting the result into Eq. [1] yields:
TA =
152
T
189
Ans.
284
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5–94. The shaft of radius c is subjected to a distributed
torque t, measured as torque>length of shaft. Determine the
reactions at the fixed supports A and B.
B
t0
(
t t0 1 ( Lx ) 2 )
x
L
2t0
x
x2
x2
T(x) =
t0 a 1 + 2 b dx = t0 a x +
b
L
3L2
L0
(1)
By superposition:
0 = fB - fB
L
0 =
L0
TB =
A
t0 a x +
x
3L2 b
3
2
dx
-
JG
TB(L)
7t0L
=
- TB(L)
JG
12
7t0 L
12
Ans.
From Eq. (1),
TA = t0 a L +
TA +
4t0 L
L3
b =
2
3
3L
7t0 L
4t0 L
= 0
12
3
TA =
3t0 L
4
Ans.
285
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5–95. Compare the values of the maximum elastic shear
stress and the angle of twist developed in 304 stainless steel
shafts having circular and square cross sections. Each shaft
has the same cross-sectional area of 9 in2, length of 36 in.,
and is subjected to a torque of 4000 lb # in.
r
A
Maximum Shear Stress:
For circular shaft
1
A = pc2 = 9;
(tc)max =
9 2
c = a b
p
2(4000)
Tc
Tc
2T
=
= 525 psi
= p 4 =
1
3
J
pc
p A 9x B 2
2 c
Ans.
For rectangular shaft
A = a2 = 9 ;
(tr)max =
a = 3 in.
4.81(4000)
4.81T
=
= 713 psi
3
a
33
Ans.
Angle of Twist:
For circular shaft
fc =
TL
=
JG
4000(36)
p
2
A B 11.0(106)
9 2
p
= 0.001015 rad = 0.0582°
Ans.
For rectangular shaft
fr =
7.10(4000)(36)
7.10 TL
= 4
a4 G
3 (11.0)(106)
= 0.001147 rad = 0.0657°
Ans.
The rectangular shaft has a greater maximum shear stress and angle of twist.
286
a
A
a
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*5–96. If a = 25 mm and b = 15 mm, determine the
maximum shear stress in the circular and elliptical shafts when
the applied torque is T = 80 N # m. By what percentage is the
shaft of circular cross section more efficient at withstanding
the torque than the shaft of elliptical cross section?
b
a
a
For the circular shaft:
(tmax)c =
80(0.025)
Tc
= 3.26 MPa
= p
4
J
2 (0.025 )
Ans.
For the elliptical shaft:
(tmax)c =
2(80)
2T
= 9.05 MPa
=
p a b2
p(0.025)(0.0152)
Ans.
(tmax)c - (tmax)c
(100%)
(tmax)c
% more efficient =
9.05 - 3.26
(100%) = 178 %
3.26
=
Ans.
•5–97.
It is intended to manufacture a circular bar to resist
torque; however, the bar is made elliptical in the process of
manufacturing, with one dimension smaller than the other
by a factor k as shown. Determine the factor by which the
maximum shear stress is increased.
kd
For the circular shaft:
(tmax)c =
d
TA B
Tc
16T
=
=
3
p
J
AdB4 p d
d
2
2 2
For the elliptical shaft:
(tmax)c =
d
2T
2T
16T
=
=
2
p a b2
p k2 d3
p A d2 B A kd
B
2
Factor of increase in shear stress =
=
(tmax)c
=
(tmax)c
16T
p k2 d3
16T
p d3
1
k2
Ans.
287
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5–98. The shaft is made of red brass C83400 and has an
elliptical cross section. If it is subjected to the torsional
loading shown, determine the maximum shear stress within
regions AC and BC, and the angle of twist f of end B
relative to end A.
A
20 Nm
50 Nm
30 Nm
2m
Maximum Shear Stress:
C
(tBC)max =
2(30.0)
2TBC
p a b2
=
B
p(0.05)(0.022)
Ans.
= 0.955 MPa
(tAC)max =
2(50.0)
2TAC
2
50 mm
20 mm
1.5 m
=
pab
p(0.05)(0.022)
Ans.
= 1.59 MPa
Angle of Twist:
fB>A = a
(a2 + b2)T L
p a3b3 G
(0.052 + 0.022)
=
p(0.053)(0.023)(37.0)(109)
[( - 30.0)(1.5) + ( -50.0)(2)]
= - 0.003618 rad = 0.207°
Ans.
5–99. Solve Prob. 5–98 for the maximum shear stress
within regions AC and BC, and the angle of twist f of end B
relative to C.
A
20 Nm
50 Nm
30 Nm
2m
Maximum Shear Stress:
C
(tBC)max =
2(30.0)
2TBC
p a b2
=
B
p(0.05)(0.022)
Ans.
= 0.955 MPa
(tAC)max =
2(50.0)
2TAC
2
=
pab
p(0.05)(0.022)
Ans.
= 1.59 MPa
Angle of Twist:
fB>C =
(a2 + b2) TBC L
p a3 b3 G
(0.052 + 0.022)( -30.0)(1.5)
=
50 mm
20 mm
1.5 m
p(0.053)(0.023)(37.0)(109)
= - 0.001123 rad = | 0.0643°|
Ans.
288
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*5–100. Segments AB and BC of the shaft have circular and
square cross sections, respectively. If end A is subjected to a
torque of T = 2 kN # m, determine the absolute maximum
shear stress developed in the shaft and the angle of twist of
end A. The shaft is made from A-36 steel and is fixed at C.
600 mm
C
600 mm
90 mm
B
30 mm
90 mm
Internal Loadings: The internal torques developed in segments AB and BC are
shown in Figs. a, and b, respectively.
Maximum Shear Stress: For segment AB,
A tmax B AB =
2(103)(0.03)
TAB c
= 47.2 MPa (max)
J
p
a 0.034 b
2
For segment BC,
A tmax B BC =
4.81TBC
3
=
a
4.81 C 2(103) D
(0.09)3
Ans.
= 13.20 MPa
Angle of Twist:
fA =
7.10TBCLBC
TABLAB
+
JG
a4G
2(103)(0.6)
=
p
a 0.034 b (75)(109)
2
7.10(2)(103)(0.6)
+
(0.09)4(75)(109)
= 0.01431 rad = 0.820°
Ans.
289
A
T
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•5–101. Segments AB and BC of the shaft have circular and
square cross sections, respectively. The shaft is made from
A-36 steel with an allowable shear stress of tallow = 75 MPa,
and an angle of twist at end A which is not allowed to exceed
0.02 rad. Determine the maximum allowable torque T that
can be applied at end A.The shaft is fixed at C.
600 mm
C
600 mm
Internal Loadings: The internal torques developed in segments AB and BC are
shown in Figs. a, and b, respectively.
90 mm
B
30 mm
90 mm
Allowable Shear Stress: For segment AB,
A
T
tallow =
TAB c
;
J
75(106) =
T(0.03)
p
a0.034 b
2
T = 3180.86 N # m
For segment BC,
tallow =
4.81TBC
a3
75(106) =
;
4.81T
(0.09)3
T = 11 366.94 N # m
Angle of Twist:
fA =
TABLAB
7.10TBC LBC
+
JG
a4G
0.02 =
T(0.6)
7.10T(0.6)
p
a 0.034 b (75)(109)
2
+
(0.09)4 (75)(109)
T = 2795.90 N # m = 2.80 kN # m (controls)
Ans.
5–102. The aluminum strut is fixed between the two walls
at A and B. If it has a 2 in. by 2 in. square cross section, and
it is subjected to the torque of 80 lb # ft at C, determine the
reactions at the fixed supports. Also, what is the angle of
twist at C? Gal = 3.811032 ksi.
A
C
2 ft
80 lb⭈ft
By superposition:
3 ft
0 = f - fB
0 =
7.10(TB)(5)
7.10(80)(2)
4
-
a G
a4 G
TB = 32 lb # ft
Ans.
TA + 32 - 80 = 0
TA = 48 lb # ft
fC =
7.10(32)(12)(3)(12)
(24)(3.8)(106)
Ans.
Ans.
= 0.00161 rad = 0.0925°
290
B
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5–103. The square shaft is used at the end of a drive cable in
order to register the rotation of the cable on a gauge. If it has
the dimensions shown and is subjected to a torque of 8 N # m,
determine the shear stress in the shaft at point A. Sketch the
shear stress on a volume element located at this point.
5 mm
A
5 mm
Maximum shear stress:
8 Nm
(tmax)A =
4.81(8)
4.81T
=
= 308 MPa
a3
(0.005)3
Ans.
*5–104. The 6061-T6 aluminum bar has a square cross
section of 25 mm by 25 mm. If it is 2 m long, determine the
maximum shear stress in the bar and the rotation of one
end relative to the other end.
C
1.5 m
20 N⭈m
B
Maximum Shear Stress:
tmax =
0.5 m
A
4.81(80.0)
4.81Tmax
a3
=
(0.0253)
= 24.6 MPa
Ans.
60 N·m
25 mm
Angle of Twist:
7.10(- 20.0)(1.5)
7.10( - 80.0)(0.5)
7.10TL
=
+
fA>C = a 4
4
9
aG
(0.025 )(26.0)(10 )
(0.0254)(26.0)(109)
= - 0.04894 rad = | 2.80° |
Ans.
291
80 N⭈m
25 mm
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•5–105.
The steel shaft is 12 in. long and is screwed into
the wall using a wrench. Determine the largest couple
forces F that can be applied to the shaft without causing the
steel to yield. tY = 8 ksi.
1 in.
12 in.
F(16) - T = 0
tmax = tY =
8(103) =
(1)
4.81T
a3
F
8 in.
4.81T
(1)3
1 in.
8 in.
T = 1663.2 lb # in.
F
From Eq. (1),
F = 104 lb
Ans.
5–106. The steel shaft is 12 in. long and is screwed into the
wall using a wrench. Determine the maximum shear stress
in the shaft and the amount of displacement that each
couple force undergoes if the couple forces have a
magnitude of F = 30 lb, Gst = 10.811032 ksi.
1 in.
12 in.
T - 30(16) = 0
F
T = 480 lb # in.
tmax =
4.81(480)
4.18T
=
a3
(1)3
1 in.
8 in.
Ans.
= 2.31 ksi
f =
8 in.
7.10(480)(12)
7.10TL
=
= 0.00379 rad
a4 G
(1)4(10.8)(106)
dF = 8(0.00397) = 0.0303 in.
Ans.
292
F
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5–107. Determine the constant thickness of the rectangular
tube if the average shear stress is not to exceed 12 ksi when a
torque of T = 20 kip # in. is applied to the tube. Neglect stress
concentrations at the corners. The mean dimensions of the
tube are shown.
T
4 in.
Am = 2(4) = 8 in2
tavg =
12 =
T
2 t Am
2 in.
20
2 t (8)
t = 0.104 in.
Ans.
*5–108. Determine the torque T that can be applied to the
rectangular tube if the average shear stress is not to exceed
12 ksi. Neglect stress concentrations at the corners. The
mean dimensions of the tube are shown and the tube has a
thickness of 0.125 in.
T
4 in.
Am = 2(4) = 8 in2
tavg =
T
;
2 t Am
12 =
2 in.
T
2(0.125)(8)
T = 24 kip # in. = 2 kip # ft
Ans.
293
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•5–109.
For a given maximum shear stress, determine the
factor by which the torque carrying capacity is increased if
the half-circular section is reversed from the dashed-line
position to the section shown. The tube is 0.1 in. thick.
1.80 in.
0.6 in.
1.20 in.
0.5 in.
Am
p(0.552)
= 1.4498 in2
= (1.10)(1.75) 2
Am ¿ = (1.10)(1.75) +
tmax =
p(0.552)
= 2.4002 in2
2
T
2t Am
T = 2 t Am tmax
Factor =
=
2t Am ¿ tmax
2t Am tmax
Am ¿
2.4002
=
= 1.66
Am
1.4498
Ans.
5–110. For a given average shear stress, determine the
factor by which the torque-carrying capacity is increased if
the half-circular sections are reversed from the dashed-line
positions to the section shown. The tube is 0.1 in. thick.
1.80 in.
0.6 in.
1.20 in.
0.5 in.
Section Properties:
œ
Am
= (1.1)(1.8) - B
p (0.552)
R (2) = 1.02967 in2
2
Am = (1.1)(1.8) + B
p (0.552)
R (2) = 2.93033 in2
2
Average Shear Stress:
tavg =
Hence,
T
;
2 t Am
T = 2 t Am tavg
œ
tavg
T¿ = 2 t Am
The factor of increase =
Am
2.93033
T
= œ =
T¿
Am
1.02967
= 2.85
Ans.
294
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5–111. A torque T is applied to two tubes having the
cross sections shown. Compare the shear flow developed in
each tube.
t
t
t
Circular tube:
a
T
T
2T
=
=
2Am
2p (a>2)2
p a2
qct =
a
a
Square tube:
qst =
T
T
=
2Am
2a2
qst
T>(2a2)
p
=
=
qct
4
2T>(p a2)
Thus;
qst =
p
q
4 ct
Ans.
*5–112. Due to a fabrication error the inner circle of the
tube is eccentric with respect to the outer circle. By what
percentage is the torsional strength reduced when the
eccentricity e is one-fourth of the difference in the radii?
ab
2
a
Average Shear Stress:
e
2
For the aligned tube
tavg =
T
T
=
2 t Am
2(a - b)(p) A a
T = tavg (2)(a - b)(p) a
B
+ b 2
2
a + b 2
b
2
For the eccentric tube
tavg =
b
T¿
2 t Am
t = a -
= a -
e
e
- a + bb = a - e - b
2
2
1
3
(a - b) - b = (a - b)
4
4
3
a + b 2
b
T¿ = tavg (2) c (a - b) d(p) a
4
2
Factor =
tavg (2) C 34 (ab) D (p) A a
T¿
=
T
tavg (2)(a - b)(p) A a
Percent reduction in strength = a1 -
B
+ b 2
2
B
+ b 2
2
=
3
4
3
b * 100 % = 25 %
4
295
Ans.
e
2
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•5–113.
The mean dimensions of the cross section of an
airplane fuselage are shown. If the fuselage is made of
2014-T6 aluminum alloy having allowable shear stress of
tallow = 18 ksi, and it is subjected to a torque of 6000 kip # ft,
determine the required minimum thickness t of the cross
section to the nearest 1>16 in. Also, find the corresponding
angle of twist per foot length of the fuselage.
t
3 ft
4.5 ft
3 ft
Section Properties: Referring to the geometry shown in Fig. a,
Am = p A 32 B + 4.5(6) = 55.2743 ft2 ¢
F
144 in2
≤ = 7959.50 in2
1 ft2
ds = 2p(3) + 2(4.5) = 27.8496 fta
12 in.
b = 334.19 in.
1 ft
Allowable Average Shear Stress:
A tavg B allow =
T
;
2tAm
18 =
6000(12)
2t(7959.50)
t = 0.2513 in. =
Angle of Twist: Using the result of t =
f = ©
Ans.
5
in,
16
ds
TL
4Am 2G F t
6000(12)(1)(12)
=
5
in.
16
4(7959.502)(3.9)(103)
¢
334.19
≤
5>16
= 0.9349(10 - 3) rad = 0.0536°
Ans.
296
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5–114. The mean dimensions of the cross section of an
airplane fuselage are shown. If the fuselage is made from
2014-T6 aluminum alloy having an allowable shear stress of
tallow = 18 ksi and the angle of twist per foot length of
fuselage is not allowed to exceed 0.001 rad>ft, determine
the maximum allowable torque that can be sustained by the
fuselage. The thickness of the wall is t = 0.25 in.
t
3 ft
4.5 ft
3 ft
Section Properties: Referring to the geometry shown in Fig. a,
Am = p A 32 B + 4.5(6) = 55.2743 ft2 ¢
F
144 in2
≤ = 7959.50 in2
1 ft2
ds = 2p(3) + 2(4.5) = 27.8496 fta
12 in.
b = 334.19 in.
1 ft
Allowable Average Shear Stress:
A tavg B allow =
T
;
2tAm
18 =
T
2(0.25)(7959.50)
T = 71635.54 kip # in a
1ft
b = 5970 kip # ft
12 in.
Angle of Twist:
f =
ds
TL
4Am 2G F t
0.001 =
T(1)(12)
4(7959.502)(3.9)(103)
T = 61610.65 kip # in a
a
334.19
b
0.25
1ft
b = 5134 kip # ft (controls)
12 in.
Ans.
297
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5–115. The tube is subjected to a torque of 750 N # m.
Determine the average shear stress in the tube at points A
and B.
4 mm
6 mm
A
100 mm
Referring to the geometry shown in Fig. a,
6 mm
2
Am = 0.06 (0.1) = 0.006 m
B
750 N⭈m
Thus,
(tavg)A
T
750
=
=
= 15.63(106)Pa = 15.6 MPa
2tA Am
2(0.004)(0.006)
Ans.
T
750
=
= 10.42(106)Pa = 10.4 MPa
2tB Am
2(0.006)(0.006)
Ans.
(tavg)B =
4 mm
60 mm
*5–116. The tube is made of plastic, is 5 mm thick, and has
the mean dimensions shown. Determine the average shear
stress at points A and B if it is subjected to the torque of
T = 5 N # m. Show the shear stress on volume elements
located at these points.
Am = (0.11)(0.08) +
tA = tB = tavg =
A
1
(0.08)(0.03) = 0.01 m2
2
B
50 mm
60 mm
T
5
=
= 50 kPa
2tAm
2(0.005)(0.01)
T
Ans.
30 mm
40 mm
298
40 mm
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•5–117.
The mean dimensions of the cross section of the
leading edge and torsion box of an airplane wing can be
approximated as shown. If the wing is made of 2014-T6
aluminum alloy having an allowable shear stress of
tallow = 125 MPa and the wall thickness is 10 mm,
determine the maximum allowable torque and the
corresponding angle of twist per meter length of the wing.
10 mm
0.5 m
10 mm
Section Properties: Referring to the geometry shown in Fig. a,
Am =
F
p
1
a 0.52 b + A 1 + 0.5 B (2) = 1.8927 m2
2
2
ds = p(0.5) + 2 222 + 0.252 + 0.5 = 6.1019 m
Allowable Average Shear Stress:
A tavg B allow =
T
;
2tAm
125(106) =
T
2(0.01)(1.8927)
T = 4.7317(106)N # m = 4.73 MN # m
Ans.
Angle of Twist:
f =
ds
TL
2
4Am G F t
4.7317(106)(1)
=
4(1.89272)(27)(109)
¢
0.25 m
10 mm
6.1019
≤
0.01
= 7.463(10 - 3) rad = 0.428°>m
Ans.
299
2m
0.25 m
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5–118. The mean dimensions of the cross section of the
leading edge and torsion box of an airplane wing can
be approximated as shown. If the wing is subjected to a
torque of 4.5 MN # m and the wall thickness is 10 mm,
determine the average shear stress developed in the wing
and the angle of twist per meter length of the wing. The
wing is made of 2014-T6 aluminum alloy.
10 mm
0.5 m
10 mm
Section Properties: Referring to the geometry shown in Fig. a,
Am =
F
p
1
a 0.52 b + A 1 + 0.5 B (2) = 1.8927 m2
2
2
ds = p(0.5) + 2 222 + 0.252 + 0.5 = 6.1019 m
Average Shear Stress:
tavg =
4.5(106)
T
=
= 119 MPa
2tAm
2(0.01)(1.8927)
Ans.
Angle of Twist:
f =
ds
TL
4Am 2G F t
4.5(106)(1)
=
4(1.89272)(27)(109)
¢
0.25 m
10 mm
6.1019
≤
0.01
= 7.0973(10 - 3) rad = 0.407°>m
Ans.
300
2m
0.25 m
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5–119. The symmetric tube is made from a high-strength
steel, having the mean dimensions shown and a thickness of
5 mm. If it is subjected to a torque of T = 40 N # m,
determine the average shear stress developed at points A
and B. Indicate the shear stress on volume elements located
at these points.
20 mm
30 mm
60 mm
A
B
40 Nm
Am = 4(0.04)(0.06) + (0.04)2 = 0.0112 m2
tavg =
T
2 t Am
(tavg)A = (tavg)B =
40
= 357 kPa
2(0.005)(0.0112)
Ans.
*5–120. The steel used for the shaft has an allowable shear
stress of tallow = 8 MPa. If the members are connected with
a fillet weld of radius r = 4 mm, determine the maximum
torque T that can be applied.
50 mm
20 mm
T
2
Allowable Shear Stress:
D
50
=
= 2.5
d
20
and
4
r
=
= 0.20
d
20
From the text, K = 1.25
tmax = tallow = K
Tc
J
t
2 (0.01)
4 R
2 (0.01 )
8(10)4 = 1.25 B p
T = 20.1 N # m
Ans.
301
T
20 mm
T
2
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•5–121.
The built-up shaft is to be designed to rotate
at 720 rpm while transmitting 30 kW of power. Is this
possible? The allowable shear stress is tallow = 12 MPa.
v = 720
T =
75 mm
rev 2p rad 1 min
a
b
= 24 p rad>s
min
1 rev
60 s
60 mm
30(103)
P
=
= 397.89 N # m
v
24 p
tmax = K
12(106) = K c
Tc
;
J
397.89(0.03)
p
4
2 (0.03 )
d;
K = 1.28
D
75
=
= 1.25
d
60
From Fig. 5-32,
r
= 0.133
d
r
= 0.133 ;
60
r = 7.98 mm
Check:
D - d
75 - 60
15
=
=
= 7.5 mm 6 7.98 mm
2
2
2
No, it is not possible.
Ans.
5–122. The built-up shaft is designed to rotate at 540 rpm.
If the radius of the fillet weld connecting the shafts
is r = 7.20 mm, and the allowable shear stress for the
material is tallow = 55 MPa, determine the maximum power
the shaft can transmit.
D
75
=
= 1.25;
d
60
75 mm
60 mm
r
7.2
=
= 0.12
d
60
From Fig. 5-32, K = 1.30
tmax = K
v = 540
Tc
;
J
T(0.03)
d;
55(106) = 1.30 c [ p
4
2 (0.03 )
T = 1794.33 N # m
rev 2p rad 1 min
a
b
= 18 p rad>s
min
1 rev
60 s
P = Tv = 1794.33(18p) = 101466 W = 101 kW
Ans.
302
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5–123. The steel shaft is made from two segments: AB and
BC, which are connected using a fillet weld having a radius
of 2.8 mm. Determine the maximum shear stress developed
in the shaft.
C
50 mm
D
100 N⭈m
20 mm B
(tmax)CD =
100(0.025)
TCDc
= p
4
J
2 (0.025 )
40 N⭈m
A
= 4.07 MPa
60 N⭈m
For the fillet:
D
50
=
= 2.5;
d
20
r
2.8
=
= 0.14
d
20
From Fig. 5-32, K = 1.325
(tmax)f = K
60(0.01)
TABc
d
= 1.325 c p
4
J
2 (0.01 )
= 50.6 MPa (max)
Ans.
*5–124. The steel used for the shaft has an allowable shear
stress of tallow = 8 MPa. If the members are connected
together with a fillet weld of radius r = 2.25 mm, determine
the maximum torque T that can be applied.
30 mm
30 mm
15 mm
T
T
2
Allowable Shear Stress:
D
30
=
= 2
d
15
r
2.25
=
= 0.15
d
15
and
From the text, K = 1.30
tmax = tallow = K
Tc
J
8(106) = 1.3 C
A 2r B (0.0075)
p
4
2 (0.0075 )
S
T = 8.16 N # m
Ans.
303
T
2
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The assembly is subjected to a torque of 710 lb # in.
If the allowable shear stress for the material is tallow = 12 ksi,
determine the radius of the smallest size fillet that can be used
to transmit the torque.
•5–125.
tmax = tallow = K
0.75 in.
A
Tc
J
710 lb⭈in.
B
1.5 in.
3
12(10 ) =
K(710)(0.375)
p
4
2 (0.375 )
C
K = 1.40
710 lb⭈ft
D
1.5
=
= 2
d
0.75
From Fig. 5-32,
r
= 0.1;
d
r = 0.1(0.75) = 0.075 in.
Ans.
Check:
D - d
1.5 - 0.75
=
= 0.375 7 0.075 in.
2
2
OK
5–126. A solid shaft is subjected to the torque T, which
causes the material to yield. If the material is elastic plastic,
show that the torque can be expressed in terms of the angle
of twist f of the shaft as T = 43 TY11 - f3Y>4f32, where TY
and fY are the torque and angle of twist when the material
begins to yield.
gY
gL
=
L
r
rY
f =
rY =
gYL
f
(1)
When rY = c, f = fY
From Eq. (1),
c =
gYL
fY
(2)
Dividing Eq. (1) by Eq. (2) yields:
rY
fY
=
c
f
(3)
Use Eq. 5-26 from the text.
T =
r3Y
p tY
2p tYc3
(4 c3 - r3Y) =
a1 )b
6
3
4 c3
Use Eq. 5-24, TY =
T =
p
t c3 from the text and Eq. (3)
2 Y
f3Y
4
TY a 1 b
3
4 f3
QED
304
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5–127. A solid shaft having a diameter of 2 in. is made
of elastic-plastic material having a yield stress of
tY = 16 ksi and shear modulus of G = 1211032 ksi.
Determine the torque required to develop an elastic core
in the shaft having a diameter of 1 in. Also, what is the
plastic torque?
Use Eq. 5-26 from the text:
T =
p (16)
p tY
(4 c3 - rY 3) =
[4(13) - 0.53]
6
6
= 32.46 kip # in. = 2.71 kip # ft
Ans.
Use Eq. 5-27 from the text:
TP =
2p
2p
t c3 =
(16)(13)
3 Y
3
= 33.51 kip # in. = 2.79 kip # ft
Ans.
*5–128. Determine the torque needed to twist a short
3-mm-diameter steel wire through several revolutions if it is
made from steel assumed to be elastic plastic and having a
yield stress of tY = 80 MPa. Assume that the material
becomes fully plastic.
When the material becomes fully plastic then, from Eq. 5-27 in the text,
TP =
2 p (80)(106)
2 p tY 3
c =
(0.00153) = 0.565 N # m
3
3
305
Ans.
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•5–129.
The solid shaft is made of an elastic-perfectly
plastic material as shown. Determine the torque T needed
to form an elastic core in the shaft having a radius of
rY = 20 mm. If the shaft is 3 m long, through what angle
does one end of the shaft twist with respect to the other end?
When the torque is removed, determine the residual stress
distribution in the shaft and the permanent angle of twist.
80 mm
T
T
t (MPa)
160
Elastic-Plastic Torque: Applying Eq. 5-26 from the text
T =
=
0.004
p tY
A 4c3 - r3Y B
6
p(160)(106)
C 4 A 0.043 B - 0.023 D
6
= 20776.40 N # m = 20.8 kN # m
Ans.
Angle of Twist:
gY
0.004
L = a
b (3) = 0.600 rad = 34.4°
rY
0.02
f =
Ans.
When the reverse T = 20776.4 N # m is applied,
G =
160(106)
= 40 GPa
0.004
f¿ =
TL
=
JG
20776.4(3)
p
4
9
2 (0.04 )(40)(10 )
= 0.3875 rad
The permanent angle of twist is,
fr = f - f¿
= 0.600 - 0.3875 = 0.2125 rad = 12.2°
Ans.
Residual Shear Stress:
(t¿)r = c =
20776.4(0.04)
Tc
= 206.67 MPa
=
p
4
J
2 (0.04 )
(t¿)r = 0.02 m =
20776.4(0.02)
Tc
= 103.33 MPa
=
p
4
J
2 (0.04 )
(tr)r = c = - 160 + 206.67 = 46.7 MPa
(tr)r = 0.02m = - 160 + 103.33 = - 56.7 MPa
306
g (rad)
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5–130. The shaft is subjected to a maximum shear strain
of 0.0048 rad. Determine the torque applied to the shaft
if the material has strain hardening as shown by the shear
stress–strain diagram.
2 in.
From the shear - strain diagram,
rY
2
=
;
0.0006
0.0048
T
rY = 0.25 in.
t (ksi)
12
From the shear stress–strain diagram,
t1 =
6
r = 24r
0.25
6
t2 - 6
12 - 6
=
;
r - 0.25
2 - 0.25
t2 = 3.4286 r + 5.1429
0.0006
0.0048
c
T = 2p
L0
t r2 dr
0.25
= 2p
2
24r3 dr + 2p
L0
= 2p[6r4] | + 2p c
0.25
0
L0.25
(3.4286r + 5.1429)r2 dr
3.4286r4
5.1429r3 2
+
d |
4
3
0.25
= 172.30 kip # in. = 14.4 kip # ft
Ans.
5–131. An 80-mm diameter solid circular shaft is made of
an elastic-perfectly plastic material having a yield shear
stress of tY = 125 MPa. Determine (a) the maximum elastic
torque TY; and (b) the plastic torque Tp.
Maximum Elastic Torque.
TY =
=
1 3
pc tY
2
1
pa 0.043 b A 125 B a 106 b
2
= 12 566.37 N # m = 12.6 kN # m
Ans.
Plastic Torque.
TP =
=
2 3
pc tY
3
2
pa 0.043 b A 125 B a 106 b
3
= 16755.16 N # m = 16.8 kN # m
Ans.
307
g (rad)
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*5–132. The hollow shaft has the cross section shown and
is made of an elastic-perfectly plastic material having a
yield shear stress of tY. Determine the ratio of the plastic
torque Tp to the maximum elastic torque TY.
c
c
2
Maximum Elastic Torque. In this case, the torsion formula is still applicable.
tY =
TY c
J
TY =
J
t
c Y
c 4
p 4
B c - a b R tY
2
2
=
=
c
15 3
pc tY
32
Plastic Torque. Using the general equation, with t = tY,
c
TP = 2ptY
Lc>2
r2dr
c
r3
= 2ptY ¢ ≤ `
3 c>2
=
7
pc3tY
12
The ratio is
7
pc3tY
TP
12
=
= 1.24
TY
15 3
pc tY
32
Ans.
5–133. The shaft consists of two sections that are rigidly
connected. If the material is elastic plastic as shown,
determine the largest torque T that can be applied to the
shaft. Also, draw the shear-stress distribution over a radial
line for each section. Neglect the effect of stress
concentration.
T
1 in.
0.75 in.
T
0.75 in. diameter segment will be fully plastic. From Eq. 5-27 of the text:
T = Tp =
2p tY 3
(c )
3
t (ksi)
12
3
=
2p (12)(10 )
(0.3753)
3
= 1325.36 lb # in. = 110 lb # ft
Ans.
For 1 – in. diameter segment:
tmax =
1325.36(0.5)
Tc
=
p
4
J
2 (0.5)
= 6.75 ksi 6 tY
308
0.005
g (rad)
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5–134. The hollow shaft is made of an elastic-perfectly
plastic material having a shear modulus of G and a yield
shear stress of tY. Determine the applied torque Tp when the
material of the inner surface is about to yield (plastic torque).
Also, find the corresponding angle of twist and the maximum
shear strain. The shaft has a length of L.
ci
Plastic Torque. Using the general equation with t = tY,
co
TP = 2ptY
Lci
r2dr
co
r3
= 2ptY ¢ ≤ `
3 ci
=
2
pt A c 3 - ci 3 B
3 Y o
Ans.
Angle of Twist. When the material is about to yield at the inner surface, g = gY at
r = rY = ci. Also, Hooke’s Law is still valid at the inner surface.
gY =
f =
tY
G
gY
tY>G
tYL
L =
L =
rY
ci
ciG
Ans.
Shear Strain. Since the shear strain varies linearly along the radial line, Fig. a,
gmax
gY
=
co
ci
gmax = ¢
co
co tY
cotY
≤ gY = ¢ ≤ a b =
ci
ci
G
ciG
Ans.
309
c0
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5–135. The hollow shaft has inner and outer diameters of
60 mm and 80 mm, respectively. If it is made of an elasticperfectly plastic material, which has the t -g diagram shown,
determine the reactions at the fixed supports A and C.
150 mm
450 mm
B
C
15 kN⭈m
A
t (MPa)
120
Equation of Equilibrium. Refering to the free - body diagram of the shaft shown in
Fig. a,
©Mx = 0; TA + TC - 15 A 103 B = 0
(1)
0.0016
Elastic Analysis. It is required that fB>A = fB>C. Thus, the compatibility equation is
fB>A = fB>C
TCLBC
TALAB
=
JG
JG
TA (0.45) = TC(0.15)
TC = 3TA
(2)
Solving Eqs. (1) and (2),
TA = 3750 N # m
TC = 11 250N # m
The maximum elastic torque and plastic torque in the shaft can be determined from
p
A 0.044 - 0.034 B
J
2
T(120) A 106 B = 8246.68 N # m
TY = tY = D
c
0.04
co
TP = 2ptY
Lci
r2dr
= 2p(120) A 106 B ¢
0.04 m
r3
= 9299.11 N # m
≤`
3 0.03 m
Since TC 7 TY, the results obtained using the elastic analysis are not valid.
Plastic Analysis. Assuming that segment BC is fully plastic,
TC = TP = 9299.11N # m = 9.3kN # m
Ans.
Substituting this result into Eq. (1),
TA = 5700 N # m = 5.70 kN # m
Ans.
310
g (rad)
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5–135. Continued
Since TA 6 TY, segment AB of the shaft is still linearly elastic. Here,
120 A 106 B
= 75GPa.
G =
0.0016
fB>C = fB>A =
fB>C =
gi
L ;
ci BC
5700.89(0.45)
TALAB
= 0.01244 rad
=
p
JG
A 0.044 - 0.034 B (75) A 109 B
2
0.01244 =
gi
(0.15)
0.03
gi = 0.002489 rad
Since gi 7 gY, segment BC of the shaft is indeed fully plastic.
311
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*5–136. The tubular shaft is made of a strain-hardening
material having a t -g diagram as shown. Determine the
torque T that must be applied to the shaft so that the
maximum shear strain is 0.01 rad.
T
0.5 in.
0.75 in.
t (ksi)
15
10
0.005
From the shear–strain diagram,
g
0.01
=
;
0.5
0.75
g = 0.006667 rad
From the shear stress–strain diagram,
15 - 10
t - 10
=
; t = 11.667 ksi
0.006667 - 0.005
0.01 - 0.005
15 - 11.667
t - 11.667
=
;
r - 0.5
0.75 - 0.50
t = 13.333 r + 5
co
T = 2p
tr2 dr
Lci
0.75
= 2p
(13.333r + 5) r2 dr
L0.5
0.75
= 2p
L0.5
= 2p c
(13.333r3 + 5r2) dr
13.333r4
5r3 0.75
+
d |
4
3 0.5
= 8.426 kip # in. = 702 lb # ft
Ans.
312
0.01
g (rad)
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•5–137.
The shear stress–strain diagram for a solid
50-mm-diameter shaft can be approximated as shown in
the figure. Determine the torque T required to cause a
maximum shear stress in the shaft of 125 MPa. If the shaft is
1.5 m long, what is the corresponding angle of twist?
T
1.5 m
T
t (MPa)
125
50
Strain Diagram:
rg
0.0025
=
0.025
;
0.01
0.0025
rg = 0.00625 m
Stress Diagram:
t1 =
50(106)
r = 8(109) r
0.00625
t2 - 50(106)
125(106) - 50(106)
=
r - 0.00625
0.025 - 0.00625
t2 = 4 A 109 B r + 25 A 106 B
The Ultimate Torque:
c
T = 2p
L0
t r2dr
0.00625 m
= 2p
L0
8 A 109 B r3 dr
0.025 m
+ 2p
L0.00625 m
9
6
C 4 A 10 B r + 25 A 10 B D r2dr
m
= 2p C 2 A 109 B r4 D |0.00625
0
+ 2p B 1 A 109 B r4 +
25(106)r3 0.025 m
R 2
3
0.00625 m
= 3269.30 N # m = 3.27 kN # m
Ans.
Angle of Twist:
f =
gmax
0.01
L = a
b (1.5) = 0.60 rad = 34.4°
c
0.025
Ans.
313
0.010
g (rad)
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5–138. A tube is made of elastic-perfectly plastic material,
which has the t -g diagram shown. If the radius of the
elastic core is rY = 2.25 in., determine the applied torque T.
Also, find the residual shear-stress distribution in the shaft
and the permanent angle of twist of one end relative to the
other when the torque is removed.
3 ft
3 in.
T
T
6 in.
t (ksi)
Elastic - Plastic Torque. The shear stress distribution due to T is shown in Fig. a. The
10
linear portion of this distribution can be expressed as t =
r = 4.444r. Thus,
2.25
tr = 1.5 in. = 4.444(1.5) = 6.667 ksi.
T = 2p
L
tr2dr
0.004
2.25 in.
= 2p
L1.5 in.
= 8.889p ¢
4.444r A r2dr B + 2p(10)
3 in.
L2.25 in.
r2dr
r4 2.25 in.
r3 3 in.
+ 20p ¢ ≤ 2
≤2
4 1.5 in.
3 2.25 in.
= 470.50 kip # in = 39.2 kip # ft
Ans.
Angle of Twist.
f =
gY
0.004
L =
(3)(12) = 0.064 rad
rY
2.25
The process of removing torque T is equivalent to the application of T¿ , which is
equal magnitude but opposite in sense to that of T. This process occurs in a linear
10
= 2.5 A 103 B ksi.
manner and G =
0.004
f¿ =
10
T¿L
=
JG
470.50(3)(2)
p
2
A 34 - 1.54 B (2.5) A 103 B
470.50(3)
= 0.0568 rad
trœ = co =
T¿co
=
J
trœ = rY =
T¿rY
470.50(2.25)
=
= 8.875 ksi
p 4
4
J
2 A 3 - 1.5 B
trœ = ci =
470.50(1.5)
T¿ci
=
= 5.917 ksi
p 4
4
J
2 A 3 - 1.5 B
p
2
A 34 - 1.54 B
= 11.83 ksi
Thus, the permanent angle of twist is
fP = f - f¿
= 0.064 - 0.0568
= 0.0072 rad = 0.413°
Ans.
314
g (rad)
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5–138. Continued
And the residual stresses are
(tr)r = co = tr = c + trœ = c = - 10 + 11.83 = 1.83 ksi
(tr)r = rY = tr = rY + trœ = rY = - 10 + 8.875 = - 1.125 ksi
(tr)r = ci = tr = ci + trœ = ci = - 6.667 + 5.917 = - 0.750 ksi
The residual stress distribution is shown in Fig. a.
315
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5–139. The tube is made of elastic-perfectly plastic
material, which has the t -g diagram shown. Determine the
torque T that just causes the inner surface of the shaft to
yield. Also, find the residual shear-stress distribution in the
shaft when the torque is removed.
3 ft
3 in.
T
T
6 in.
t (ksi)
Plastic Torque. When the inner surface of the shaft is about to yield, the shaft is
about to become fully plastic.
T = 2p
L
10
tr2dr
3 in.
= 2ptY
L1.5 in.
= 2p(10)a
r2dr
0.004
r3 3 in.
b2
3 1.5 in.
= 494.80 kip # in. = 41.2 kip # ft
Ans.
Angle of Twist.
f =
gY
0.004
(3)(12) = 0.096 rad
L =
rY
1.5
The process of removing torque T is equivalent to the application of T¿ , which is
equal magnitude but opposite in sense to that of T. This process occurs in a linear
10
= 2.5 A 103 B ksi.
manner and G =
0.004
f¿ =
494.80(3)(12)
T¿L
=
= 0.05973 rad
p 4
JG
A 3 - 1.54 B (2.5) A 103 B
2
trœ = co =
trœ = ci =
494.80(3)
T¿co
=
= 12.44 ksi
p 4
J
4
3
1.5
A
B
2
494.80(1.5)
T¿ci
=
= 6.222 ksi
p 4
J
A 3 - 1.54 B
2
316
g (rad)
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5–139. Continued
And the residual stresses are
(tr)r = co = tr = c + trœ = c = - 10 + 12.44 = 2.44 ksi
Ans.
(tr)r = ci = tr = ci + trœ = ci = - 10 + 6.22 = - 3.78 ksi
Ans.
The shear stress distribution due to T and T¿ and the residual stress distribution are
shown in Fig. a.
317
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*5–140. The 2-m-long tube is made of an elastic-perfectly
plastic material as shown. Determine the applied torque T
that subjects the material at the tube’s outer edge to a shear
strain of gmax = 0.006 rad. What would be the permanent
angle of twist of the tube when this torque is removed?
Sketch the residual stress distribution in the tube.
T
35 mm
30 mm
t (MPa)
210
Plastic Torque: The tube is fully plastic if gi Ú gr = 0.003 rad.
g
0.006
=
;
0.03
0.035
0.003
g = 0.005143 rad
Therefore the tube is fully plastic.
co
TP = 2p
Lci
2p tg
=
=
3
tg r2 dr
A c3o - c3i B
2p(210)(106)
A 0.0353 - 0.033 B
3
= 6982.19 N # m = 6.98 kN # m
Ans.
Angle of Twist:
fP =
gmax
0.006
L = a
b (2) = 0.34286 rad
co
0.035
When a reverse torque of TP = 6982.19 N # m is applied,
G =
fPœ =
210(106)
tY
=
= 70 GPa
gY
0.003
TPL
=
JG
6982.19(2)
p
4
2 (0.035
- 0.034)(70)(109)
= 0.18389 rad
Permanent angle of twist,
fr = fP - fPœ
= 0.34286 - 0.18389 = 0.1590 rad = 9.11°
Ans.
Residual Shear Stress:
6982.19(0.035)
tPœ o =
TP c
=
J
p
4
2 (0.035
tPœ i =
TP r
=
J
p
4
2 (0.035
- 0.034)
6982.19(0.03)
- 0.034)
= 225.27 MPa
= 193.09 MPa
(tP)o = - tg + tPœ o = - 210 + 225.27 = 15.3 MPa
(tP)i = - tg + tPœ i = - 210 + 193.09 = - 16.9 MPa
318
g (rad)
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•5–141.
A steel alloy core is bonded firmly to the copper
alloy tube to form the shaft shown. If the materials have the
t -g diagrams shown, determine the torque resisted by the
core and the tube.
450 mm
A
100 mm
60 mm
B
15 kN⭈m
t (MPa)
Equation of Equilibrium. Refering to the free - body diagram of the cut part of the
assembly shown in Fig. a,
©Mx = 0; Tc + Tt - 15 A 103 B = 0
180
(1)
Elastic Analysis. The shear modulus of steel and copper are Gst =
36 A 106 B
and G q =
= 18 GPa. Compatibility requires that
0.002
180 A 106 B
0.0024
g (rad)
0.0024
= 75 GPa
Steel Alloy
t (MPa)
fC = ft
36
TcL
TtL
=
JcGst
JtG q
0.002
Tc
p
2
A 0.03 B (75) A 10 B
4
Tt
=
9
p
2
Copper Alloy
A 0.05 - 0.034 B (18) A 109 B
4
Tc = 0.6204Tt
(2)
Solving Eqs. (1) and (2),
Tt = 9256.95 N # m
Tc = 5743.05 N # m
The maximum elastic torque and plastic torque of the core and the tube are
(TY)c =
1 3
1
pc (tY)st = p A 0.033 B (180) A 106 B = 7634.07 N # m
2
2
(TP)c =
2 3
2
pc (tY)st = p A 0.033 B (180) A 106 B = 10 178.76 N # m
3
3
and
p
A 0.054 - 0.034 B
J
2
T c(36) A 106 B d = 6152.49 N # m
(TY)t = tY = D
c
0.05
r2dr = 2p(36) A 106 B ¢
co
(TP)t = 2p(tY) q
Lci
r3 0.05 m
= 7389.03 N # m
≤2
3 0.03 m
Since Tt 7 (TY)t, the results obtained using the elastic analysis are not valid.
319
g (rad)
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5–141. Continued
Plastic Analysis. Assuming that the tube is fully plastic,
Tt = (TP)t = 7389.03 N # m = 7.39 kN # m
Ans.
Substituting this result into Eq. (1),
Tc = 7610.97 N # m = 7.61 kN # m
Ans.
Since Tc 6 (TY)c, the core is still linearly elastic. Thus,
ft = ftc =
ft =
gi
L;
ci
TcL
=
JcGst
7610.97(0.45)
p
4
9
2 (0.03 )(75)(10 )
0.3589 =
= 0.03589 rad
gi
(0.45)
0.03
gi = 0.002393 rad
Since gi 7 (gY) q = 0.002 rad, the tube is indeed fully plastic.
320
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5–142. A torque is applied to the shaft of radius r. If the
material has a shear stress–strain relation of t = kg1>6,
where k is a constant, determine the maximum shear stress
in the shaft.
r
r
r
g = gmax = gmax
c
r
T
1
gmax 6 1
b r6
t = kg = ka
r
1
6
r
T = 2p
tr2 dr
L0
r
= 2p
L0
gmax = a
ka
1
1
1
gmax 6 13
gmax 6 6
12p kg6max r3
19
b r 6 dr = 2pk a
b a b r6 =
r
r
19
19
6
19T
b
3
12p kr
19T
12p r3
1
tmax = kg6max =
Ans.
5–143. Consider a thin-walled tube of mean radius r and
thickness t. Show that the maximum shear stress in the tube
due to an applied torque T approaches the average shear
stress computed from Eq. 5–18 as r>t : q .
t
r
t
2r + t
;
ro = r + =
2
2
J =
=
t
2r - t
ri = r - =
2
2
2r - t 4
p 2r + t 4
ca
b - a
b d
2
2
2
p
p
[(2r + t)4 - (2r - t)4] =
[64 r3 t + 16 r t3]
32
32
tmax =
Tc
;
J
c = ro =
2r + t
2
T(2 r 2+ t)
=
p
3
32 [64 r t
2r
T(2r
2 +
=
+ 16 r t3]
2p r t[r2 + 14t2]
t
2 r2 )
2p r t c rr2 +
2
As
T(2 r 2+ t)
=
1 t2
4 r2 d
t
r
: q , then : 0
r
t
tmax =
=
T(1r + 0)
2p r t(1 + 0)
=
T
2p r2 t
T
2 t Am
QED
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*5–144. The 304 stainless steel shaft is 3 m long and has an
outer diameter of 60 mm. When it is rotating at 60 rad>s, it
transmits 30 kW of power from the engine E to the
generator G. Determine the smallest thickness of the shaft
if the allowable shear stress is tallow = 150 MPa and the
shaft is restricted not to twist more than 0.08 rad.
E
Internal Torque:
P = 30(103) W a
T =
1 N # m>s
b = 30(103) N # m>s
W
30(103)
P
=
= 500 N # m
v
60
Allowable Shear Stress: Assume failure due to shear stress.
tmax = tallow =
150(106) =
Tc
J
500(0.03)
p
4
2 (0.03
- r4i )
ri = 0.0293923 m = 29.3923 mm
Angle of Twist: Assume failure due to angle of twist limitation.
f =
0.08 =
TL
JG
500(3)
p
2
A 0.03 - r4i B (75.0)(109)
4
ri = 0.0284033 m = 28.4033 mm
Choose the smallest value of ri = 28.4033 mm
t = ro - ri = 30 - 28.4033 = 1.60 mm
Ans.
322
G
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•5–145.
The A-36 steel circular tube is subjected to a
torque of 10 kN # m. Determine the shear stress at the mean
radius r = 60 mm and compute the angle of twist of the
tube if it is 4 m long and fixed at its far end. Solve the
problem using Eqs. 5–7 and 5–15 and by using Eqs. 5–18
and 5–20.
r 60 mm
4m
t 5 mm
10 kNm
Shear Stress:
Applying Eq. 5-7,
ro = 0.06 +
tr = 0.06 m =
0.005
= 0.0625 m
2
Tr
=
J
ri = 0.06 -
10(103)(0.06)
p
4
2 (0.0625
- 0.05754)
0.005
= 0.0575 m
2
= 88.27 MPa
Ans.
Applying Eq. 5-18,
tavg =
10(103)
T
= 88.42 MPa
=
2 t Am
29(0.005)(p)(0.062)
Ans.
Angle of Twist:
Applying Eq. 5-15,
f =
TL
JG
10(103)(4)
=
p
4
2 (0.0625
- 0.05754)(75.0)(109)
= 0.0785 rad = 4.495°
Ans.
Applying Eq. 5-20,
f =
=
ds
TL
4A2mG L t
TL
ds
4A2mG t L
Where
L
ds = 2pr
2pTLr
=
4A2mG t
2p(10)(103)(4)(0.06)
=
4[(p)(0.062)]2 (75.0)(109)(0.005)
= 0.0786 rad = 4.503°
Ans.
323
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5–146. Rod AB is made of A-36 steel with an allowable
shear stress of 1tallow2st = 75 MPa, and tube BC is made of
AM1004-T61 magnesium alloy with an allowable shear
stress of 1tallow2mg = 45 MPa. The angle of twist of end C is
not allowed to exceed 0.05 rad. Determine the maximum
allowable torque T that can be applied to the assembly.
0.3 m
0.4 m
a
A
C
60 mm
Internal Loading: The internal torque developed in rod AB and tube BC are shown
in Figs. a and b, respectively.
Allowable Shear Stress: The polar moment of inertia of rod AB and tube
p
p
a 0.0154 b = 25.3125(10 - 9)p m4 and JBC =
a 0.034 - 0.0254 b
BC are JAB =
2
2
= 0.2096875(10 - 6)p m4. We have
A tallow B st =
TAB cAB
;
JAB
75(106) =
T(0.015)
25.3125(10 - 9)p
T = 397.61 N # m
and
A tallow B mg =
TBC cBC
;
JBC
45(106) =
T(0.03)
0.2096875(10 - 6)p
T = 988.13 N # m
Angle of Twist:
fB>A =
- T(0.7)
TAB LAB
= - 0.11737(10 - 3)T = 0.11737(10 - 3)T
=
JAB Gst
25.3125(10 - 9)p(75)(109)
and
fC>B =
T(0.4)
TBC LBC
= 0.03373(10 - 3)T
=
JBC Gmg
0.2096875(10 - 6)p(18)(109)
It is required that fC>A = 0.05 rad. Thus,
fC>A = fB>A + fC>B
0.05 = 0.11737(10 - 3)T + 0.03373(10 - 3)T
T = 331 N # m A controls B
Ans.
324
50 mm
30 mm
Section a–a
T
a
B
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5–147. A shaft has the cross section shown and is made of
2014-T6 aluminum alloy having an allowable shear stress of
tallow = 125 MPa. If the angle of twist per meter length is
not allowed to exceed 0.03 rad, determine the required
minimum wall thickness t to the nearest millimeter when
the shaft is subjected to a torque of T = 15 kN # m.
30⬚ 30⬚
75 mm
Section Properties: Referring to the geometry shown in Fig. a,
Am =
C
0.075
1
1
(0.15) ¢
≤ + p A 0.0752 B = 0.01858 m2
2
tan 30°
2
ds = 2(0.15) + p(0.075) = 0.53562 m
Allowable Shear Stress:
A tavg B allow =
T
;
2tAm
125(106) =
15(103)
2t(0.01858)
t = 0.00323 m = 3.23 mm
Angle of Twist:
f =
ds
TL
2
4Am G C t
0.03 =
15(103)(1)
4(0.018582)(27)(109)
a
0.53562
b
t
t = 0.007184 m = 7.18 mm (controls)
Use t = 8 mm
Ans.
325
t
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*5–148. The motor A develops a torque at gear B of
500 lb # ft, which is applied along the axis of the 2-in.diameter A-36 steel shaft CD. This torque is to be
transmitted to the pinion gears at E and F. If these gears are
temporarily fixed, determine the maximum shear stress in
segments CB and BD of the shaft. Also, what is the angle of
twist of each of these segments? The bearings at C and D
only exert force reactions on the shaft.
B
E
F
2 ft
1.5 ft
C
D
A
Equilibrium:
TC + TD - 500 = 0
[1]
Compatibility:
fB>C = fB>D
TC(2)
TD(1.5)
=
JG
JG
TC = 0.75TD
[2]
Solving Eqs. [1] and [2] yields:
TD = 285.71 lb # ft
TC = 214.29 lb # ft
Maximum Shear Stress:
(tCB)max =
214.29(12)(1)
TCc
= 1.64 ksi
=
p
4
J
2 (1 )
Ans.
(tBD)max =
285.71(12)(1)
TDc
= 2.18 ksi
=
p
4
J
2 (1 )
Ans.
Angle of Twist:
fCB = fBD =
TD LBD
JG
285.71(12)(1.5)(12)
=
p
2
(14)(11.0)(106)
= 0.003572 rad = 0.205°
Ans.
326
500 lb·ft
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5–149. The coupling consists of two disks fixed to separate
shafts, each 25 mm in diameter. The shafts are supported on
journal bearings that allow free rotation. In order to limit
the torque T that can be transmitted, a “shear pin” P is used
to connect the disks together. If this pin can sustain an
average shear force of 550 N before it fails, determine the
maximum constant torque T that can be transmitted from
one shaft to the other. Also, what is the maximum shear
stress in each shaft when the “shear pin” is about to fail?
25 mm
P
130 mm
25 mm
T
Equilibrium:
T - 550(0.13) = 0
©Mx = 0;
T = 71.5 N # m
Ans.
Maximum Shear Stress:
tmax =
71.5(0.0125)
Tc
= 23.3 MPa
= p
4
J
2 (0.0125 )
Ans.
5–150. The rotating flywheel and shaft is brought to a
sudden stop at D when the bearing freezes. This causes the
flywheel to oscillate clockwise–counterclockwise, so that a
point A on the outer edge of the flywheel is displaced
through a 10-mm arc in either direction. Determine the
maximum shear stress developed in the tubular 304 stainless
steel shaft due to this oscillation. The shaft has an inner
diameter of 25 mm and an outer diameter of 35 mm. The
journal bearings at B and C allow the shaft to rotate freely.
D
2m
B
A
80 mm
Angle of Twist:
f =
0.125 =
TL
JG
Where
f =
10
= 0.125 rad
80
T(2)
p
4
2 (0.0175
- 0.01254)(75.0)(109)
T = 510.82 N # m
Maximum Shear Stress:
tmax =
Tc
=
J
510.82(0.0175)
p
4
2 (0.0175
- 0.01254)
Ans.
= 82.0 MPa
327
C
T
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5–151. If the solid shaft AB to which the valve handle is
attached is made of C83400 red brass and has a diameter of
10 mm, determine the maximum couple forces F that can be
applied to the handle just before the material starts to fail.
Take tallow = 40 MPa. What is the angle of twist of the
handle? The shaft is fixed at A.
B
A
150 mm
150 mm
F
150 mm
tmax = tallow =
40(106) =
Tc
J
F
0.3F(0.005)
p
4
2 (0.005)
F = 26.18 N = 26.2 N
Ans.
T = 0.3F = 7.85 N # m
f =
TL
=
JG
7.85(0.15)
p
4
9
2 (0.005) (37)(10 )
= 0.03243 rad = 1.86°
Ans.
328