Barry S. Onouye, Kevin Kane - Instructors Solutions for Statics and Strength of Materials for Architecture and Building Construction-Prentice Hall (2011)

Instructor's M anual to Accompany
Statics and Strength of Materials
For Architecture and Building Construction
Fourth Edition
Barry Onouye
Pearson/Prentice Hall
Upper Saddle River, New Jersey Columbus, Ohio
Chapter 2 Problem Solutions
2.3
y
2.5
2.6
F=1000 lb.
By similar triangles:
4
3
5
Fx = £ f = £(1000#) = 800#
Fy = i F = i(l0 0 0 # ) = 600#
3
sin0 = —
5
and
4
co s0 = —
5
Fx = F co s6 = (l0 0 0 # )(-|) = 800#
Fy = F sin 0 = (1000#)(|) = 600#
Tx = T sin 10°
Ty = T cosl0°
T=
Ty _ _250N _ 254 N
cos 10° 0.985
2.8
6 = tan"1^ ) = 18.43°
Px = P(i2fe) = (300#Xa3 1 6 ) = 94-9#
Py = P ( ^ ) = (300#X0.949) = 285#
Purlin Detail
2.9
Fly = +F, cos30° = 10k(0.866) = 8.66k
Flx = +F, sin 30° = 10k(0.50) = 5k
F2 = -F 2x= -1 2 k
1
18k
V2
V2
18k
F3 y = - ^ ( F 3) = - ^
18 k
R =2F„ = +5k -1 2 k + - ^ = +5.73k
V2
18k
R = 2 F = +8.66k = -4.07k
y
y
V2
tan6 = ^ = — = 0.710
Rx 5.73
6 = tarf‘(0.710) = 35.4°
from horizontal
-4.07
k
x
sin0 =
R„
R„
sin0
sin 35.4°
R = i ^ L = 7.03k
(0.579)
Graphical solution using
the tip-to-tail method
2.10
-T A C ,
-T ac
-T,A Cy
-T Ac sin 60° = -0.866T ac
;
cos
60° = -0 .5T,AC
Rx = 59.6N
+TAg = +TAg cos40° = +0.766TAB
“ T ab, = -T ab sin40° = -0.642T ab
0 = 86.8
Rx = 2FX = -(0.5X800N) + (0.766)(600N) = 59.6N
<j>= 3.2”
Ry = £Fy =-(0.866)(800N )-(0.642)(600N ) = -1078N
0 = ta n '1( ^ ) =
Rv
<)>= tan
tan-1(l8 .l) = 86.8°
Rx |
,/5 9 .6 \
—— =: tantan“‘l ------ I= ttan_1(0.055) = 3.2°
vRy
=V59.62 + 1078 = 1079N
R = 1079N
Ry = 1078N
I
ivl
ii
i
2.10 cont’d
G raphical Solution:
2.11
-W x = -W cos 30° = -0.866W
-W y = -W sin 30° = -0.50W
-F x = - F cos 40° = -0.766F
+F„ = +F sin 40° = +0.642F
Ry = SFy = 0;
• 0.50(200#)+ 0.642F = 0
,
F = i ° 2 i = 156#
0.642
R = Rx = 2FX= -0.866(200#) - 0.766(156#)
R = -173#-120#= -293#
by
Scale: 1” = 100#
2.12
-F2, = -F2cos 25°
-F2y = -F2sin 25°
Since the resultant must be vertical,
Then: Rx =ZF x =0
-F2, + F i = 0
F2cos2 5 ‘ = F,
From this equation, it is seen that F,
Is only a fraction of F2 , therefore,
F2 = 7kN.
Then, F, = F2cos25° = (7kN)(0.706)
F, = 6.34kN and F2 = 7kN
R = F2y = (7kN)(sin25°) = 2.95kN.
Graphical solution
2.13
Tlx = - T 1cos30° = -0.866T1
Tly = -T[ sin30° = -0.50TJ
T2x = -T 2 cos 60° = -0 .50T2
T2y = -T 2 sin 60° = -0.866T2
+FX = +F cos 45° = +0.707(8k) = +5.65k
-F y = -F sin 45° = - 707(8k) = -5.65k
But Tj = T2
For the resultant to be vertical,
Rx =ZFx = 0
-0 .8 6 6 T -0 .5 0 T + 5.65k = 0
T = 4 .14k
R = Ry = 2Fy = -0 .5 0 (4 .14k)-0 .8 6 6 (4 .1 4 k -5 .6 5 k ) = -11.3k
2.14
Ma = -20#(5’) + 25#(4’) = -100 #-ft + 100 #-ft = 0
The box is just on the verge of tipping over.
2.15
W=700N
SOON
2m
(2M a = 0)
+800N (lm )-700N (x) = 0
_ (800N)(lm)
- = 1.14m
(700N)
Since x > 1m, the man is OK.
1m
2.16
2 M a = -3 6 # (l3 ') + 15#(8") =
-(5 4 0 # -in ) + (l2 0 # -in ) = -4 2 0 # -in.
2.17
W = 100#
P
2 M a = -W (l8.8") = -100#(l8.8") = -1 8 8 0 # -in. (clockwise)
[SM A = 0] -1 0 0 # (l 8.8") + P(45.1") = 0
p _ 1880#-in. _ 41 7jy
45.1 in.
2.18
[ZM a = 0] -2 0 0 # (l2 ") = F(26") = 0;
F = 92.3#
[2M b =0] + F (4")-P (36") = 0;
P = 10.3#
2.19
M a = -(5kN )(24m ) - (10kN)(20m) - (9kN )(l 6m) - (8kN)(l 2m)
-(8 k N )(8 m )-(8 k N )(4 m )
M a = -( l 20kN - m) - (200kN - m) - (144kN - m)
-(96kN - m) - (64kN - m) - (32kN - m)
M a = -6 5 6 k N - m
2.2 0
Fk = {§(1300#) = 1200#
Fy = j|(l300#) = 500#
M b = -F x(5') + Fy(0) = -1200# (5') = -6 0 0 0 # - f t
Mc = —Fx(5') + Fy(12') = -1200# (5') + 500# (12')
Mc = -(6 00 0# -f t) + (6000#-ft) = 0
2.21
M a = Fy(6") = (27.2#)(4“) = 108.8'“in
CCW
M e = -F y(6") - Fx(2") = -(12.7#)(6") - (27.2#)(2") = -130.6'""
2.22
F = 1.5kN
30°
"
----
Fx
dy
Fx = F cos 30° = (1.5kN)(0.866) = 1.3kN
Fy = Fsin30° = (1.5kN)(0.50) = 0.75kN
dx = dcos60° = (200mm)(0.50) = 100mm
dy = dsin60° = (200mm)(0.866) = 173mm
MA= -F x(dy) + Fy(120mm + dx)
MA =-1.3kN(173mm) + 0.75kN(220mm)
M a == -60kN - mm = -0.06kN - m
2.23
Tx = T cos 30° = 2000# (.866) = 1732#
Ty = T sin 30° = 2000#(0.50) = 1000#
d d = 7' cos 60° = 7 (0 .5 0 ) = 3.5
d x2 = 1 0 (0.50) = 5
dy = 7' sin 60° = 7' (.866) = 6.06'
[2 M C = 0]
Tx(dy ) - T y (dxl) - W ( d ![2) = 0
(1732#)(6.06') - (1000#)(3.5I) - W (5) = 0
5W = 10,500#_ft - 3500#_ft
2.24
M a = -(10k)(l 1.3,)-(4k)(14') = -113k ft-5 6 kft =-169k
M b = -(10k)(l 1.3') - (4k)(14') = -169k' fL
2.25
R Ax and R Bx fo rm a couple
M couple
1
2 5 # (l2 ') = + 300, "ft
RAy and 150# m an fo rm a cou p le
M couple2
Since the moment due to a couple is constant,
MA= MB= Mc = +300"" - 300#ft = 0
-(1 5 0 # )(2 ') = - 3 0 0 #_ft
2.26
90k N
125mm -
-180mm
= 90kN(0.305m) = 27.45kN-m
90kN
2.27
Fx = (85 lb.) cos 55° = 48.8 lb.
Fy = (85 lb.) sin 55° = 69.6 lb.
M A = - F y(l5 ') + Fx(4") = -(6 9 .6 lb.)(l5') + (48.8 lb.)(4") = -8 4 9 lb .- in .
M B = - F y(l2 ,,) + Fx(4") = -(6 9 .6 lb.)(l2")+ (48.8 lb.)(4") = - 6 4 0 l b . - i n
2.28
A x = A cos6 0 ° = 0.50A
A y = A sin 6 0 ° = 0.866A
C x = C eos 45° = 0.707C
C y = C s in 4 5 ° = 0.707C
[2FX=0] - Cx + Ax = 0
-0.707C + 0.50A = 0
[2Fy = 0 ] + C y + A y -1 0 0 0 # = 0
0.707(0.707A) + 0.866A = 1000#
.-. 1.37A = 1000#
A = 732#
C = 0.707(732#) = 518#
2.29
Free-body diagram of joint C
Force
Magnitude
F ____
E*-----------
F
500N
-500cos20° = -470N
-500sin20°: -171 N
AC
?
+ACsin10p = +0.174AC
+ACcos10° : +0.984AC
+BCsin30P = +0.50BC
-BCcos30°: -0.066BC
BC
?
[EFX= 0] - 470N + 0 .174AC + 0.50BC = 0 ... Eq (1)
[2Fy = 0 ] -171N + 0.984A C -0.866B C = 0 ... Eq (2)
Solving equations (1) and (2) simultaneously,
0.866 x [0.50BC + 0.174AC = 470N] ... Eq (1)
0.50 x [-0.866BC -0.984A C = 171N] ... Eq (2)
+0.433BC + 0 .15AC = 407N ... Eq (1)
-0.433BC + 0.492AC = 86N ... Eq (2)
Adding the two equations;
AC = +768N (compression)
Substitute and solve for BC;
BC = 672N (tension)
2.30
F2 = 1 5 0 #
F1 = 5 0 #
2 5 ) \ s<^
x
■ W = 60#
Force
Maanitude
Fx
Fv
F1
50*
-50*COS25° = -45.3*
+50#sin25° = +21.1*
F2
150*
+150*(3/5) = +90*
+150*(4/5) = +120*
W
60*
0
-60*
P
?
+PCOSa
+Psina
[2F X = 0 ] - 4 5 .3 # + 9 0 # + P c o s a = 0
(l)
[2Fy = 0 ] + 21. l# + 1 2 0 # -6 0 # + P s in a = 0
....(2 )
—44.7#
P=-
E quating:
....
(1)
—44.7#
-8 1 .1 #
co sa
sin a
s in a
-8 1 .1 #
------- = ta n a = -----------= +1.81
-4 4 .7 #
a = tan_1(l.8 l) = 61.1°
_
-r-r. ITT
cos 61.1°
_
t t .
rr _
0.483
Note that the negative sign for P
indicates that it was initially assumed
in the wrong direction.
Final Free Body Diagram
Graphical check
2.31
[2FX= 0 ] A c o s7 5 ° -B sin 6 0 ° = 0
A .B W
0.259
[2Fy = 0 ] Asdn75° + B co s6 0 °-2 .5 k N = 0
3.346B(0.966) + 0.50B = 2.5kN
3.732B = 2.5kN
B = 0.67kN
A =2.24kN
Forces exerted bt the sphere
onto the smooth surface.
2.32
Tx = Tsin5° ==0.087T
Ty = T cos5°: = 0.996T
Px = P cos 20° = 0.940P
Py = Psin20° = 0.342P
[2FX= 0] -0 .0 8 7 T + 0.94P = 0
T = 0u ^ f r94P
=10gp
0.087
[2Fy = 0] + 0.996T - 0.342P - 2000# = 0
substituting;
0.996(10.8P) - 0.342P = 2000#
P = 192#
AB = T = 10.8(192#) = 2074#
2.33
Solving FBD(a) first:
F o r c e _________Fx
-FJL
DC
- —DC = -0.80DC
+—DC = +0.60DC
5
DE
+DE cos 15° = +0.966DE
+DEsinl5° = +0.259DE
W
0
-200 lb.
^ F x = - 0.80DC + 0.966DE = 0 ;
D C-1.21D E
Y F = + 0.60(1.21DE)+ 0.259DE- 2001b. = 0
y
(DC,)
0.985DE= 2001b. ;
DE-+2031b.;
DC = 1.2l(2031b.)- 2461b.
Writing equations of equilibrium for FBD(b);
Force
CD
+i(2461b.) = +1971b.
—|(2461b.) - -1481b.
CA
-CA cos 60° = -0.50CA
-CAsin 60° = -0.866CA
BC
0
+BC
J F , = - 0.50CA +1971b. = 0 ;
CA-3941b. (T)
2 Fy = -0.866(3941b.) + BC-1481b. = 0 ;
BC = +4891b. (C)
2.34
[2M a = 0] + 100'(4') - 300#(3') + Bx(10') = 0
.-. Bx = +50'
[SFx =0] + A X- 1 0 0 '- B x = 0
+AX- 1 0 0 ' - 5 0 ' =0
Ax = +150'
[EFy =0] + Ay - 30 0 ' = 0
Ay = +300'
2.35
40kN
50kN
[2 M a = 0 ] -4 0 k N (2 .5 m )-5 0 k N (5 .0 m ) + By(7.5m ) = 0
By = 46.7kN
[2Fy = 0 ] + A - 4 0 k N - 5 0 k N + 46.7kN = 0
= 43.3KN
No horizontal reaction is necessary for this load case.
2.36
[2M a = 0] - 2k(20') - 3k(40') - 4k(60') + By(80') = 0
40k_fl + 120k_fl + 240k_fl
Cl
Bv = ------------------------------- = +5k
y
80'
[£M b = 0] +4k(20') + 3k(40') + 2 k (6 0 ')-A y(80') = 0
Ay = +4k
2.37
1500#
^90°
/
\
90y V ,
i/ M "
30>\ _
Dx
• Dy
■Ay
30’
[EM a = 0] - 1500#(17.33') - 3000'(8.67') + Dy(30') = 0
.-. D„ = +1733'
[sF y = 0] -1 5 0 0 # cos30° - 3000' cos30° -1 5 0 0 ' cos30° + Ay +1733' = 0
.-. Ay = +3463'
[2FX=0] + 1500sin30° + 3000sin30° + 1500sin30°-D x =0
.-. Dx = +3000'
2.38
1kN
(i)(lk N ) = 0.385kN
(§)(lkN ) = 0.923kN
Ax
Ay
[2 M A = 0 ] + B (l2m )-(0 .9 2 3 k N )(2 .5 m ) + (0.385kN )(2.5m )(0.385kN + 0.923kN)(6m) = 0
Solving for B;
B = 0.767kN
Bx = (1y)(0.767kN ) = 0.295kN
By = (f)(0.767kN ) =0.708kN
Reverting back to the unresolved forces;
[2FX= 0 ] + A x + lk N -(0 .2 9 5 k N ) = 0
A x = +0.705kN
[2Fy = 0 ] + Ay - lk N + (0.708kN) = 0
B,
.-. A y = +0.292kN
2.39
8k
12’ ’ ■ 12’
)
Upper beam:
C y + D y = 8 k;
C y= D y= 4k
Left beam:
[2 M a = 0] - 20k(2 4 ') - 4 k (6 0 ') + By(48') = 0
By = + 15k
[2 F y = 0] + A y - 20k + 15k - 4 k = 0
A y = +9k
Right beam:
[2 M e = 0] + 4 k (l2 ')-1 6 k (2 4 ') + Fy ( 4 8 ) - 4 k ( 6 0 ') = 0
Fy = +12k
[2Fy = 0 ] - 4 k + E y - 1 6 k + 1 2 k - 4 k = 0
Ey = + 12k
2.40
Upper beam:
[ 2 M d = 0] + 4 0 0 * (l2 ') - C y(l0 ') = 0
C y = +480*
[2 F y = 0 ] - 400* + C y + D y = 0
D y = -4 8 0 * + 400* = -8 0 * (l)
[2 F X= 0 ] + 300* + D x = 0
D x = -3 0 0 * (« -)
Lower beam:
[ 2 M b = 0] - C y( 5 ') - A y(l0 ') = 0
A y = -2 4 0 * (i)
[2 F y = 0] - 240* + By - 480* = 0
B y = +720* (T)
[2 „ = 0 ]
Bx = 0
2.41
FD
Ax
[ 2 M a = 0] + F D X( 4 ') + F D y ( 2 0 ') - 2 k ( 1 6 ') - 8 k ( 3 2 ') = 0
( 4/ 5) F D ( 4 ‘) + ( % ) F D ( 2 0 ') - 3 2 k_ft - 2 5 6 k- ft = 0
E D = 1 8 .9 k
F D x = ( / 5) ( l 8 .9 k ) = 1 5 .2 k
E D y = ( % ) ( l 8 . 9 k ) = 1 1 .3 k
[2 F X = 0]
+ A x -F D x = 0
A x = + 1 5 .2 k H
[2 F y = 0]
+ A y - 2 k - 8 k + 1 1 .3 k = 0
A y = - 1 .3 k ( i)
y
+ 1 5 .2 k = 0
* (dfk)
(D C „)
(B D .)
(D C y)
(BDy )
Solving the two equations simultaneously;
BD = 17.9k
DC = 19.7k
x
= (% )F D
F D y = (% )F D
2.42
Parallelogram Method
Tip-to-Tail Method
2.43
Tip-to-Tail Method
2.44
2.45
Rx
=EFX
= -(6k) cos 60° + (3k) cos 30°
Rx = -3 k + 2.6k = -0.4 k
Ry =EFy = -(6 k )sin 6 0 °-(3 k )sin 3 0 °
R., = -5 .2 k -1 .5 k = -6.7k
R x = - 0 .4 k
|
) = 86.6^*^
IR \
/ 'J \
■tan- 1 —— = tan”M ^— = tan_1(l6.75) = 86.6°
Rv
10 .4 1
v
’
R = ^ R 2a + Ry = ^(0 .4 0 )2 +(6.7)2 = 6.71k
R y = - 6 .7 k
R = 6 .7 1 k
2.46
y
Fly
F1 = 800#
Force
-500#
P=500#
F,
+F1cos30°=(800#)(0.866)=692.8#
+ F1sin30°=(800#)(0.50)=400#
+ F2cos30°=(1200#)(0.866)=1039.2# - F2sin30°=-(1200#)(0.50)=-600#
R x = 2 F x = + 6 9 2 .8 # + 1 0 3 9 .2 # = + 17 3 2 # (-» )
R y = EFy = - 5 0 0 # + 4 0 0 # - 6 0 0 # = - 7 0 0 # (1)
R = -^(1732#)2 + (-7 0 0 # )2 = 1868#
Alternate way to find the resultant R:
Rx= 1732#
R„
sin 0
0 = tan
= tan _1(0 .4 0 4 ) = 22
700
700
sin 2 2 °
0.375
=1867#
2.47
y
x
d
Component
F
F
y
-(-J=)(9°kN) - -63.6kN
AD = 90kN
- ( ^ H = - T = - 63-6kN
BD = 45kN
-% (45kN) - -27kN
-% (45kN) - -36kN
+{l 10kN) cos 30° = +95.3kN
-(H 0kN )sin30° = -55kN
Rx = 2 F x =+4.7kN
Ry = D Fy =_1546kN
CD = 11 OkN
R=
+ Ry = ^(4.7)2 +(l54.6)2 = 154.7kN
0 = tan -i| R y
■tan
0 = tan-1(32.9) = 88.3°
Resultant
154.6
4.7
2.48
100#
d2x = d j cos 45° = 14'(0.707) = 9.9'
d2y = d 2 sin 45° = 14' (0.707) = 9.9
dlx = d! cos 20° = Iff (0.94) = 9.4'
[2 M C = 0] + 250# ( dlx) - 1 00# (d2y) - F (d2x) = 0
250#(9 .4 ')- 1 0 0 #(9 .9 )
^
^
9.9
2.49
[2F X = 0 ] + A X -3 .8 3 k N = 0
A x =3.83kN
(F»)
[2Fy = 0 ] + A y -3 .2 1 k N = 0
A y =3.21kN
(Fy)
M a = -3 .2 1 k N (3 .1 m )-3 .8 3 k N (l.3 3 m )
M a = - 9 . 9 5 k N - m - 5 . 0 9 k N - m = - 1 5 .0 4 k N - m
2.50
T ,y
T 1 = 500#
Force
T, = 500#
(500#)cos15° = 483#
(500#)sin15o =129.5#
J2 = 700@#
(700#)cos10° = 689.5#
(700#)sin10o = 121.8#
Ma = +T£x(30’) + T2y(6*) - T 1y(35') - T1y(4') =0
M. = 689.5#(30') + 121 8#(6’) - 483#(35’) - 129.5#(4’) = +3990#-ft
2.51
R = 2 F y = 10 # + 7# + 6# - 18# = + 5 #
M0 = +(7#)(4”) + (6#)(9”) - (18#)(17” = - 224#-in.
R(x) = 224 #-in;
2 2 4 #_in
: 44.8"
R = 5#
44.8”
origin
2.52
150N
Weight of wood member:
100N
co = 30N/m
Assume the member weight is
located at the
center of the length.
30N/m
o
m
m
M
M
M
I
origin
1m
2m
150N
a
100N
1m
o
origin
1.5m
i i
W
= 90N (beam
w t.)
R = 2 F y = -100N - 90N + 150N = -40N
M0 = - (100N)(1m) - (90N)(1.5m) + (150N)(3m) = +215 N-m
R(x) = Mo
215N - m
, _
x = -------------- = 5 .4 m
40N
For a 40N force to produce a moment directed counter-clockwise,
the R = 40N force will be at an imaginary location where
x = 5.4m to the left of the origin.
2.53
y
400#
200#
11
100#
20#/ft
TT~T I
I I
l
I I H r ~ r r
origin
4'
Total beam weight equals (20%)(16') = 320#
at the center of the beam length.
For the beam to remain stationary and horizontal,
the moments taken about points A and B should
be balanced by the opposing moments due to B
and A respectively, resulting in no resultant moment.
[Z M a = 0] + 200*(4') + 100*(8') - 320*(8') - 40 0 *(l2 ') - B (l6 ') = 0
B = 360*
[ 2 M b = 0] + 400*(4 1) + 320*(8') - 100*(8') - 2 00*(l2') - A (l6 ') = 0
A = 60*
R y = 2 F y = + A +100* + 100* - 320* - 400* + B = 0
= +60* + 200* + 100* - 320* - 400* + 360* = 0
2.54
-------------
Force
—*--------------------
F
»n
AB
-T T AB
13
AC
4
--A C
5
W
0
F
—j f -
+ — AB
13
3
--A C
5
-5k
19
4
[SF =01 - — A B - - A C = 0;
L x
1 13
5
13
.-. AB = - — AC
15
19
fs F = 0 l + — A C - - A C - 5 k = 0;
L y
J
13
5
= 5k;
4
— AB = - - A C = 0
13
5
+ — [ - — A C |- - A C = +5k
13 V 15
/ 5
AC = -5.36k (compression)
AB = —j^ (-5 .3 6 k ) = +4.64k (tension)
2.55
—x
Force
AC
BC
F-2kN
[2FX = 0]
-AC
--A C
5
-BC cos60° = -0.50BC
+ - ( 2 k N ) - + 1 .2 k N
+BCsin 60° = +0.866BC
- - i ( 2 k N ) = - 1 .6 k N
AC -0 .5 0 B C + 1.2kN = 0;
[2Fy = 0 ] - —AC + 0.866B C - 1.6kN = 0;
Solving simultaneously;
B C = 2 . lkN (compression)
AC = 0.27kN (tension)
- A C + 0.50BC = 1.2kN
- - A C + 0.866BC = 1.6kN
2.56
y
!
Force
x
________________
BA
F_
- — BA
13
DB
- — BA
13
+ D B s in 3 0 ° = + 0 .5 0 D B
W
+ D B cos30°:
0
-8 0 0 #
n
94
fR
L x = 2 F x =01j
1- 3—
[Rv = 2FV = ol
- — B A + 0 .8 6 6 D B -8 0 0 # = 0
L y
13
y
BA + 0.50DB = 0;D B = — 1BA
3
J 13
BA + 0.866| — BA | = 800#
(13
]
BA = 658.2#
DB = |^ -j(6 5 8 .2 #) = 1215.2#
+ 0 .8 6 6 D B
2.57
i
y
For ce
------
F
_x-------------------------
F
—H------------------------
CA
-CA cos 45° = -0.707CA
CA sin45° = +0.707CA
CB
+CB cos 30° = +0.866CB
+CBsin30° = +0.50CB
W
O
r2FX
L x = olj - 0.707CA + 0.866CB = 0;
-W
n
CA = 0--------1.22CB
70?
This relationship indicates the CA > CB, therefore, CA = 1 8kN
Then, CB = (1 8kN)/1.22 = 1 48kN
[2Fy = 0] 0.707CA + 0.50CB - W = 0
W = 0.707(1.8kN) + 0.50(l.48kN) = 2.0kN
2.58
Joint B:
y
Force
AB=1560#
+— (l560#) = 1440#
13'
>
BE
0
BC
fl560#') = - 600*
13'
1
+BE
4
BC
5
[2FX=0] + 1440# - ^ B C = 0;
3
BC
5
BC = 1800#
[2Fy = 0] - 600# + B E - |( l 8 0 0 #) = 0;
BE = 1680#
2.58 cont’d
Joint C:
Force
CD
-0.707CD
+0.707CD
CB = 1800"
+~ (l8 0 0 #) = 1440#
+ |( l 8 0 0 #) = +1080#
W
0
-W
[2FX = 0] - 0.707CD +1440* = 0;
CD = 2037#
[2Fy = 0] + 0.707(2037#) + 1 080# - W = 0;
W = 2520#
2.59
500#
[2 M A = 0] - 5 0 0 #( l 0 ' ) 3 ( l 0 ' ) + ^ ( 2 4 ' ) = 0
(B*)
50B 288B
# »
— + ------- = 5000
;
13
13
B = 192.3
(By)
#
B„x = 14* \Bv = 111.5*
y
[2F
= 0lJ + A x - 7(BJ
4 # =0;
L
f 2 F V = ol
L y
J
Ax =74#
+ A v - 5 0 0 # + 1 7 7 .5 # = 0;
y
(r> \
A
y
= 3 2 2 .5 #
2.60
Cx
1.8kN
2m
,, 2.5m
Bx
1
i
By
2.7kN
By
M
3m
Bx
3m
— MB
rc
-,A
Cx
B eam AB:
[ZFx =0]
Bx = 0
[2M b = 0] - A y(4.5m) + 1.8kN(2.5m) = 0;
[2Fy =0] + lkN - 1.8kN + By = 0;
Ay =l kN
By = 0.8kN
B eam BC:
[ZFx =0] Cx = 0
[2Fy =o] - 0.8kN - 2.7kN +C y = Q
Cy =3.5kN
[2 M C = 0] - M rc + 2.7kN(3m) + 0.8kN(6m) = 0
M RC = 8. lkN - m + 4.8kN - m = 12.9kN - m
2.61
2k
[2F X = 0] - A x + 4 k = 0;
A x = +4 k ( - )
[2M A = 0] - 2k(20') - 3k(40l) - 2k(60l) - 3k(40') + 4 k (2 0 ) + 65,( 80-) = 0
By = 4 4 k (t)
[ 2 FV = ol + A v - 2k - 3k - 2k - 3k + 4 k = 0
L
J
A y = +6 k (t)
(By)
2.62
Dy
Cy
U pper Beam:
[ZM a = o] - 300#(s) + By(8') = 0;
[ZFx =0] + A X-1 8 0 # =0;
By = +187.5# (f)
A x =+180# (->)
[SFy = 0] +187.5# - 240# + A y = 0;
A y = +52.5# (f)
L ow er Beam:
[ZM d = 0] +200#(4')+ C y(6')-187.5#(9l) - 8 0 #(l3l) = 0
Cy = +322#(f)
[ZFx = 0] + Dx - 60# = 0;
Dx = +60#( ^ )
[2Fy = 0] - 200# + Dy + 322# - 187.5# - 80# = 0
Dy = +145.5#(f)
Chapter 3 Problem Solutions
:
■
E
X _V 29’
[Z M
a
= 0]
y “ V 29
+ ^ - ^ - J ( 4 0 ,) - ( 3 0 0 # ) ( 3 0 ' ) - ( 3 0 0 ^) ( 2 0 ,) - ( 3 0 0 # )( 1 0 ,) = 0
E = +22SV 29 = +1212*
E x = 1125#;
Ey =450#
[Z F x = 0 ]
C B X = E X = 1125*
[Z F y = 0 ]
+ C B y -3 0 0 * -3 0 0 * + 450* = 0
C B y = +150*
.. (15QJX10') n3.
CBy
CBX'
h c = 4 '+ y ; h = 5.33'
1125*
3.2
Assume E„,m = 1200#
Cable forcelbE = E
[ZM a = 0] - (3 0 0 ' )(1 O') + (300* )(20') + (300* )(3 a ) + E y(40') = 0
E y =+450*
e x=
1/ e 2 -
X =i ^ ;
10'
1112
e^
E x = 1112
Ey=450#
y = 4.05'
E=1200#
Ex=1112#
[SFX = 0] CBX = 1112*
[ZFy = 0] + C B y - 3 0 0 * - 3 0 0 * + 450* = 0
CBy = + 150*
CB = 1112* <1200*
Y' CB
— = ---- y'= 1.35'
10' CBX
hc = y '+4.05'= 5.4'
.-.OK
(31.1 ftKN)
Original FBD
[X M a = 0] - 3 k (1 2 ') - 7 k ( 3 2 ') - 2 k ( 4 2 ') + E y ( 6 0 ') = 0
E y = 5 .7 3 k ( 2 5 .5 k N )
[Z F y = 0] + A y - 3 k - 7 k - 2 k + 5 .7 3 k = 0
A y = 6 .2 7 k (2 7 .9 k N
FBD (a)
[X M C = 0] - 2 k ( 1 0 ,) + 5 .7 3 k ( 2 8 ,) - E K(1 2 ,) = 0
E x = 1 1 .7 k ( 5 2 . lk N );
E = E D = 1 3 .0 3 k ( 5 8 .0 k N )
From the original FBD:
[X Fx = 0 ] A x = 1 1 .7 k (5 2 .1 k N )
A = A B = 1 3 .2 7 k ( 5 9 .1 k N )
FBD(a)
[XFx = 0] CB„ = E „ = 11.7k (52.1kN)
[SFy =0] + CBy - 7k- 2k + 5.73k = 0
CBy = 3.27k (14.4kN)
CB = 12.15k (54.1kN)
12
(3.67m)'
V .—
20 ’
(6.1m)
....
Ax
A=13.28k (59.1kN)
Ax=11.7k (52.1kN)
Ay=6.27k (27.9kN)
CDy
CDx
3k’
(13.35kN)
7k ’
(31.14kN)
[SFX = 0] CDX= 11.7k (52. lkN)
[ZFy = 0] CDy = 3.73k (l6.6kN)
.-. CD = 12.28k (54.5kN)
3.4
From the solution for Prob. 3.3, it appears that
the force in cable AB is maximum. Therefore,
assume the following:
reaction @ A = 20k and AB = 20k.
Using the original FBD of Prob. 3.3, A„ = 6.27k
A=20k
Ay=6.27k
A x =-1jA 2 - A 2y - 19k
A=20k
Ay=6.27k
[ZFx = 0 ]-1 9 k + C D x =0;
CDX= 19k
(A»)
[ZFy = 0] + 6 .2 7 k -3 k -7 k + CDy = 0
CDy = 3.73k
il£. = — ;
1 3.03
h B= 3.96'
B
[E F x = 0 ]
C B X = 19k
[S F y = 0 ] + C B y + 3 . 7 3 k - 7 k = 0
C B y = + 3 .2 7 k
hc - h B
2 ff
CBy
3 ,2 7 k
“ C B X ~~ 1 9 k
(h‘= -hB) = 2{ l ^ i ) =3-44,
h c = h B + 3 . 4 4 '= 3 .9 6 '+ 3 .4 4 '= 7 .4 0 '
3.5
p
1
A
1200#
[ZM a = 0] - 3 0 0 % ( 1 8')(9')- 1200#(18') + B(12') = 0
B = 5850# (T)
[ZFy = 0] + A -3 0 0 % (1 8 ') -1 2 0 0 # + 5850# = 0
A = 750#
(T)
3.6
BOON
(0 = 145 N/m
1.52m
2.13m
[EM a = 0] -80 0 N (1 .5 2 m )-(1 4 5 % )(1 .5 2 m + 2.13m)(1.825m)+B(3.65m) = 0
B = 598 N
[lF y = 0 ] + A - 800N - (145 % )(3.65m ) + 598N = 0
A = 731N
3.7
2k
2k
CO = 0.4 k/ft
5’ ,
’ 4'
3 ’,
|
“
A T
UUUI
p
B f
[ I M a = 0 ] - 2 k (3 ')-2 k (8 l) - 0 .4 ^ f t (4')(141) - B(12') = 0
B = 3.7k
[ZFy = 0 ] + A - 2 k - 2 k - ( 0 . 4 % ) ( 4 ') + 3.7k = 0
A = 1.9k
3.8
1000#
900#
Ax
t
-
A
►
.B y
Ay'
6'
6’
I
6’
[I F x = 0] A x = 0
[ I M a = 0 ] - 1000#(6 ') - 9 0 0 #(14')+ By (12') = 0
By =+1550
[ZFy = 0] + Ay - 1000# - 900# + 1550# = 0
Ay = +350
3.9
R=60kN
1
C0 = 15kN/m
Ax
Ayi
:
.By
1m
1m
2m
T------ ------- 71------ 7
2m
[IF k = 0 ] A x = 0
[EM a = 0 ] By(3m) - 60kN (4m ) = 0;
[SFy = 0] - A y + 80kN - 60kN = 0;
By = 80 k N
A y = 20kN
3.10
[ZFk = 0] a x = o
[ZM A = 0] - 900* (2'J - 600*(31) + By(lO') = 0
By = 360*
[SM B = 0] + 600* (7') + 900* (81) - A y(1 O') = 0
A y = 1140*
3.11
Ri
R2
Ra
8 . 2’
13.375’
16.375'
Ri = GDO'H15^ ^ 6') + l 100/ ^ ^ 6') = 1200#
r 2 = (iD(i')(150%>)(&®)+ t100^
1' ) ^ 6') = 1400#
R3 = ( ^ ) ( l ,) ( 150% 3 )(6 ')+ (1 0 0 % 2)(1')(6') = 1200#
[SM a = 0] - 1200*(3') -1400*(8.2') -1200*(13.375) + E(16.37S) = 0
E =1900*
[2Fy = 0] A -1200* -1400* -1200* +1900* = 0
A = 1900*
3.12
1000#
Step 1: FBD of the entire truss.
Step 2: Solve for the external
support reactions.
[£M D= 0] + A , (20') - 866* (10') -1000' (20 I - 0
A.,, = 1-1433*
[ZFX... 0]
Dz 500' - 1433’ - 0
Dx = +933*
|IT; = 0] + D y - 8 6 6 # - 1 0 0 0 ' = 0
Step 3: Isolate a joint and solve two
unknown member forces.
ACX - 0.707AC:
ACy = 0.707AC
Joint A:
AB
AC
[EF;, - 0] +1433"' + 0.707AC - 0
A C - 2030' (comp.)
[ZFy = 0] + AB + A C y =0
A B - -(-0.707 x 2030') = +1433* (tension)
Joint B:
[IF x = 0] D C - 0
BD,
[XF, —0 1 -D D
J 30
BA *
BD = • 1433' (T)
1-133'
0
Joint D:
[XF,, - 0] BC - 0
D"
[ZFV- O] i-BD
D
BD - -i-1433# (T)
1433# ™0
'D C
1<133 #
BD =
Joint E:
DE ::: 5001?
1000#
30/
% /
E J &X£
&------- ►eef
[SPy - 0 ]
866* liC
EC = - 866' (C)
[2F;; =0] -5 0 0 * -5 0 0 *
EF = +1000* (T)
' EC
Joint F:
I t -1 [IMP
<.
F
[SFV= 0] -1000* - 0.707CF = 0
CF
M l5* (C)
CF
1000#
1000#
3.13
3.14
3.15
3.16
3.17
LLJQ-
3.18
| 2lX-1 - 0] + ^ £ ( 3 ' | + ^ B (6 ' i 4k(6')
V5 '
(A C „)
V5,'
4k(12'|
(A C „)
AC • :9V5 ' -20. Ik (T)
ODp
[Y U „ - 0] +
BC •
V5
■p/“«
+ - ^ r (6'i + 4k(6') 2k(6') - 0
V -5
V5
(BC.,)
(B C .)
' 2.24k (C)
[EM.. = 0] - BD(3') - 4k(6') - 2k(l 2') = 0
BD .. 16k (C)
2k(18') - 0
3.19
[IM
a
= 0] II. (2 5'j ■■■■(3kN)(5m) + (1 .2 k N )(2 m )- (1 .6 k N )(7.5m ) = 0
H. = 9 .8 4 k N
[2 F y = 0 ] - A y - 3 - 1 . 6 t 9 .8 4 = 0
A.. - 5 2 lkN
[2Fx = 0] - A,, +1.2=0
A „ = + 1.2kN
f2 F „
1 J
= 0 l -3 k N -1 .6 k N -------- C H
J
6.4
I
h
=0
.;
CH - -7.35kN (C)
[ZM h = 0] + BC(2m) -■3(2.5in) -1.6(5m) = 0
BC - +7.75kN (T)
[SM; - 0] -FH(2m) : 1.2(2m)~ l.<5(2.5m) = 0
F1I- O.SkN
(C )
3.20
J y - 4600'
[ i f ; = 0] ••• II.,...2000' ...500'... 800# + 4600# = 0
Hy -1 3 0 0 '
f£p; = 0] - 11., + 600# = 0
II, •• 600#
Section a-a:
B
BC
A
600#
[2Ma = 0] + 500'(5’) + BB(5’) = 0
BE = -500' (C)
BE' '
EA
’ 800#
Section b-b:
500#
'[EM* = 0 ] + 500* (5') - CEy(5') - C E X(2.5') = 0
But; CE = 4 ^ C E :
V5
C E ,.= -j~ C E
'
V5
—J—-CEj'5') -T-- ^ C E i^ .S ') ..500' 15')
V5
V5
CE = 250V5#
S e c tio n c-c:
[EM G—0] —600'* (5') + FJ(5') + 4600* (5‘) = 0
FJ - 600' 4600* - 4000' (C)
3.21
A B ,, - % A B
AB..
[SMH= 0] AB=
(AB)(12') - 6 k(1 2') = 0
- 6( 12).- 51
, .
- A - = -10k (<’|
3(12)
' '
[£M a = 0] BH(12') = ft
BH = 0
= 0] ■i-HG(9,) - 6 k ( l 2 ' ) = 0
H G - 18k (T )
AB
Section a-a:
I Vi '..
, 4 . 7 3 k - 2 k - H H ,. -
0
EE-I = %(: !.73k) —3.41k
EI-Iy ~ + 2 ,7 3 k ;
(T)
Section b-b:
[SFv = 0] + 4 ,7 3 k -2 k - 3
HCy --10.27k;
+H CV- 0
HC-- • % ( 0 . 2 7 k )
= + 0 .3 4 k f t )
Section c-c:
|EF = 0 + 2.27k-BI„ =■0
(A ,)
Blv -
2.27k:
BI = %, (2.27k) - ' 2.84 k (T)
3.23
Section a-a:
~
\
D
-----------------------7
\
/'
\
/
C E \
/ DB
/
/
/•
/
/
/
S
/
/
\
2k N
[XF.. = 0] ■■■2 k N + C E X - 0
C l i... = + 2 k N ;
1
\
\
\
\
\
X
C E = + 2 i / 2 k N (T )
E
CT
Section b-b:
[ZF. =0] + 6k N - FBV= 0
FB„ = + 6kN
FB ■■■■■+6->/2kN (T)
F» = 3.5k
Ay = 3.5k
Section a-a:
I- o
A
[2!M,,
2
/
7
OD = 1
/
0’
D
1k
Only DB can resist the rotational tendency
(counter-clockwise) of the 1 -k applied load.
DBy " —7— ;
1
|Z M
vC \
B ir
iS
0|
V t3
q
~ u ] -f 1^(16*^
D B - = I lk ;
V l3
D B X(I 6 'J ~ U
S e c tio n b-b:
*
o
[ZMq =0] + lk(28'| + 3k(36') EA J36') = 0
EA* —-!-3.8k
5
..........
HA -3EA^
EA - -4 "k (T)
Zero Force Members:
3.25
3.26
3.27
3.28
' 200#
- ^ 4
[EM B= 0] + Ax(9‘) - 1 5 0 '( 6 ') - 200* (16') = 0
A x “ +455* (->)
for member A C :
[EFX= 0] C x = 455*
For member BC:
[EFx = 0] Bx = 455* (<-)
200#
Member AC:
[£Ma = 0] - 200*(16') +Cy(12') = 0
By - 150* -Cy = 0
Cy = 267*
Ay =Cy-200#;
For equilibrium in member BC:
Ay =-67* (4-)
By =412* (T)
3.29
(0 = 400#/ft
BCX=%BC;
[ZMa = 0] -
BCy = %BC
400*1ft(1 ffV 5) + %
B C ( 6') = 0
(B Cy)
BC = 4167*
BCX=%{4167) = 2500*;
BCy = %(4167) = 3334*
[EFX= 0] - Ax + 2500* = 0
Ax = 2500* («-)
[ZFy =0] + A y + 3334*-400#,ft(10') = 0
Ay = 666* (T)
6 .6 7 m
3.30
[DM B= 0] + I80kN(8m j + 135kN(22.67m) - Ay (29.34m) = 0
Ay = 153.4kN
[ZFV= 0] + 153.4kN I35kN
ISO lcN
B „ = 1 6 1 .6 k N
135kN
[XF,, - 0]
+ 1 7 1 .6 k N - C x
=0
C ., = 1 7 I . 6 k N
[lF y = 0] i 153.4kN - 135kN-Cv = ()
C y = 1 8 .4 k N
[XFx = 0] + 1 7 1 .6 k N - B.. = 0
Bx = 1 7 1 .6 k N
+ Bv = 0
3.31
[EM,, - 0 ] + Bx(14') - 500J(10') - 200* (16') = 0
Bx - 586* (< )
[EE, = 0] + Ax - 586# = 0
A, = 586* (•••>)
[EF, - 0] + A., - 500’ - 200'* - 0
A..
Bx = 5 8 6 #
500#
*1
Ax = 586#
t
■Ay = 700#
C„ =820*;
C y = 820*
D x = 234 #;
D v = 820#
E „ = 820 f ;
E y = 820#
Ey
Dx
(c)
200#
'
~00? (T)
3.32
10kN
2m
FBD(a):
1.33m
[£ M a = 0] - 10kN(2m) + By(2.5m) = 0
By = 8kN
[XFy = 0] - lOkN + A y + 8kN = 0
2.5m
A y = 2kN
A.
Bx
(a)
t
Ay
I
I
FBD(c):
[Z M C = 0] - Bx(2m) + 8kN(2.5m) = 0
Bx = lOkN
[XFy = 0 ] - C y + 8kN = 0
Cy = 8kN
10KN
[SFx = 0 ] + Cx - lOkN = 0
Cx =10kN
2m
or, since member CB is a two-force
member, use the
slope relationship for C and C .
Cy
C
Cx
2.5m
2-lorce
member
Cx
C yl
(b)
2m
(c )
Bx
Ax
Ay
:
3.33
FBD of the entire frame:
[ZMc = 0] -3 k (4 ')-lk (l2 ')+ A x(6') = 0
Ax = +4k (<-)
[ZMa = 0] -3 k (4 ')-lk (1 2 ')+ C x(6') = 0
Cx = ^ k (-»)
FBD(a):
[ZM b = 0] + 3 k (2 ')-A y(6') = 0
Ay = lk (t)
[ZFy = 0] + lk + Cy - 3 k - l k = 0
Cy = 3 k ( t)
(b)
2-force
members
DB
DB,
Cx
'
.1
"t
C y|
D
(d)
FBD(d):
[SMe = 0] +BD(6')-3k(12') = 0
BD = +6k
[SFy = 0] + 3 k -6 k + Ey - l k = 0
E y = +4k
[SFx =0] + 4 k - E x = 0
E=+4k
E
, 1k
FBD(c):
[ZFy = 0] +By - 4 k = 0
By =+4k
P F X= 0] - B x +4k = 0
Bv = +4k
3.34
1m Ri = 4.5kN
R2 = 9k N
1 kN/m-(r
IkN/m
Bx
Ay
By.
3.5m
R i= 4 .5 k N
R2 = 9kN
,1 m
Bx
By
Ay
3.5m
Equivalent FBD
PF* = 0] Bx = 0
[ Z M B = 0] + 9 k N ( 3 .5 m ) + 4 . 5 k N ( 5 m ) - A y ( 6 m ) = 0
A y = 9kN
[ZFy = 0] + 9 k N
- 4 .5 k N - 9 k N
+ By = 0
By = 4 .5 k N
3.35
Poor reaction
p = Yso„ x h = ( 3 5 % 3) x 7 . 5 ' = 2 6 2 . 5 % 2
o o= p x l ' = ( 2 6 2 . 5 ^ 2) x l ' = 2 6 2 . 5 %
F = ^ _ ( ^ % X Z i L 984*
2
2
[ I M a = 0 ] + B x (8 ') - 9 8 4 * ( 2 .3 ) = 0
B x = 308#
[ZFX = 0] - A x + 9 8 4 *
Ax=
676*
- 308*
=0
3.36
I
R = 20k
Bx
Ay
4k
Ay
Cx
_ 3 k ._ j
•Cy
Upper Beam:
P F * = 0] Bx = o
[ I M B = 0 ] - A y (2 0 ') + R (1 5 ') = 0
[IM
a
= 0 ] - 2 0 k ( 5 ) + B y ( 2 0 l) = 0
By = 5k
Lower Beam:
[IFx =0] +CX+3k = o
Cx = -3 k (<-)
[ZMC = 0] - 4 k ( S ) - 15k(10') + Dy(20') = 0
Dy =8.5k
[XFy = 0] + Cy - 4 k - 15k + 8.5k= 0
Cy = 10.5k
VT Bv
3.37
Overhang beam simple beam
} DL= 15psf
SL = 25psf
DL+SL = 40psf
CO= 320 #/ft
l l l l l l l l l l l l l H i l l III IT U T
Cl) = 3 2 0 m
TTTTU
20’
Dy = 3 2 0 0 #
13 2 0 0 #
co = 3 2 0 #/ft
[C y = 3 2 0 0 #
3200#
111 H i 111 i
co = 3 2 0 # /f t
.h
i ’ F
h
h
u
u
i ^ i
[ZMA =0] + (3200')(30')+(320%)(30')(15')-By(25') = 0
By =+9600'
[ZFy = 0] - 3200' - (320%)(30') + 9600' + Ay = 0
Ay = 3200’
4.8 k
9 .6 k
24’
1 4 .4k
9 .6 k
9 .6 k
|J
2 8 .8 k
9 .6 k .
24’
|J
24’
2 8 .8 k
Girder with Columns @ 24’ o.c.
3.38
[ZMA=0] +3k(12') + 2k(l7')-}fB(10,) = 0
B = 7.6k;
Bx = 2.91k (-»);
By =7k(J.)
[IF k =0] - 2 k - A x + 2.91k = 0
Ax = +0.91k (<-)
[!Fy =0] - 3 k + A y - 7 k = 0
Ay = + 10k (T)
Joint E:
EDx =j§ED;
EDy = ^ E D
[ZFy = 0] + -^-ED-3k = 0
ED = +7.8k (T)
[IFX= 0] +|f(7.8k) + EF = 0
EF = -7 .2k (C)
Joint D:
DCX= 0.707DC;
[IFx = 0] ED=7.8k
DCy = 0.707DC
(7.8k) - 2k + 0.707DC = 0
DC = +13k (T)
[!Fy = 0] —||(7.8k)-D F -0.707(13k) = 0
DF =-12.2k
3.38 - cont’d
Joint F:
DF=12k
[E F„--0] + 7.2k + CF - 0
CF - -7.2k (C)
[2FV = 0] - 12.2k -F A = 0
FA —-12.2k (C)
Joint C:
[SFk = 0] +7.2k
0.707(13k)i-C A r - f CB = 0
[lF y = 0 ] + 0.707(13k) - f C A - jf CB = 0
CB = +7.6k (T)
C A - + 2 .4 k ( T )
Force Summation Diagram
3.39
Force Summation Diagram
Solving for the support reactions;
[ I M E = 0]
10k(14') + + ^ ( 1 0 ' ) + - ^ ( 4 ' ) ^ 0
l> « )
A = loV 2k;
A., = 10k;
[ i f ., - o ] - r : .
in k
o
E x = -H-10k (->)
[IF ,,
- C'|
Ey = 0
E,.
I I Ilk
10k
(A y)
Ay - 10k
=0
3.40
Solving for the support reactions;
[E M a = 0] + D v (I2 m )
1 8 k N (8 m )9 k N (4 m )-0
D „ = +15kN (T)
[2 F y = 0] t A y - 1 8kN l 15kN - 0
A ,. ' l.'k N
(T)
[2 T « = 0)
A x + 9kN - 0
A, " 19kN (<-)
Solving for the support reactions;
|I.V1, =
0]
-2 iV
- I 5 u '( 8 ' j : r .,(6 ') =
0
( :•j
|SFX= 0j +200# - A ;: = 0
Ax = +200* (<—)
[Z F v = 0 ]
150# + A „ = 0
A v = + 1 5 0 * (T)
3.42
8k
3.43
Support reactions for the entire FBD:
[ir , - o] c ,
i)
|X M , = 0 ) - 1 0 k ( 2 4 ') + C 1,(<1«,) = 0
Cy
[IF,,
5k
=
0]
(T)
5k
10k + 5k
+ A,, = 0
Av = +10k (T)
Using the left half of the section cut;
DGy = ys DG;
DG_ = y* DGF
[IF.
n] • / D G 5 k + 10k - 0
(Dcy
DG-
8 .3 3 k (C );
|IM .
DGX= 6.66k;
0] -5k(2-r)
AB= 14.8k (T)
[IFX=
0] +
FG + (- 6.66k) + (4,8 k ) = 0
' (DC-,) '
FG = +1.87k (T)
DGV= 5k
lllki 21') AB| 2f?)
' (A B )'
0
3.44
Support reactions for the truss:
|i!H.
0| A x = 0
[EM. - 0] - 2k(25‘) - 4k(35') + Fy(40) - 0
Fy .. 14.75k 0
[ill;, = 0|
A., - 2 k - 4k t 4.75k = 0
A., ••• ■1.25k |1|
Using the left side of the section cut;
| I'M D= 0] ~ 1.25k(2.5‘) ••• Gil |5V-Sk‘i = 0
■
’
GH -+3.61k (T)
[EMH- 0]
C'D|5V3) 1.2.>k(20l) - 0
CD = -2 .89k (C)
|EM,.
0] - 1.25k.(15')+ 3.61k|5V3) t X(HD)|5V3) t ^ ( H D ) ( 5 | = 0
'•GH-i
I ID
-1.44k (Cl
'
'
(K D J
'
'
(H D .)
3.45
Solve for BG:
R ( i ;: — '/■ i H Ci:B G y = % 4 B G
[1M m-■ 0] -3000*(20')-
4BG)(12') ■H000#(2tf) ‘ 1000#(10') = 0
B G - 2700' (C)
Solve for HE:
[EM b = 0] -3 0 0 0 ’ (30') ! HE{16') I 1000'(10') I 1000*(20'i I 1000*(30') -■ 0
HE = +1875# (T)
Solve for HB:
I IB.. = %.4:,(HB)
[ZM n --0] + 3000*(10‘) + -rfr(HB)(30')
HB = +1179* (T)
1000*(10') 1000*(20') 1000*(3ff) = 0
3.46
|IM g = 0] = 2 kNj-l m) + 4kN(2 m) - DE(3m) - 0
DK - • 5..13UN (C)
[XMZ = 0] + 2kN(2m) - AB(3m) - 0
AB = -rl.33kN (T)
[XF„ = ()] -2 k N -4 k N -- )<BC = 0
BC - -10kN (T)
3.47
In this example, a moment equation will be used to
determine the effective tension counter.
From (ZM,.=0), A,causes a counter-clockwise rotation
about point O. The only tension counter capable of
resisting in the clockwise direction about O is member DG.
Therefore:
r a v u = 01 -
O'i - S ^ 3 6 0 ^ + A„ (312') = 0
V II "
'
(D O ,)
DG
v 11
'
(DC-y)
-25k (T)
To solve for DF:
[2M o - o]
^ ( 3 0 ')
iDKj
J ^ C 24') 24k(72') = 0
(DK;)
DF = 54.2k (C)
To solve for EG:
[ZMd = 0] - 24k(48') + EG(3O') = 0
ECj —-i-38.4k
3.48
Ax - •20k:
Av - 20k;
A - 20V2k
[XFy = 0] : 20k - 5k 5k 1Ok - B.; = C
B, - 0
HD
[2 3 ;= 0] ' Bx 20k -
c < -
\
B. = 20k
[XI-;. ... 0| + E D y
ED., = 5k;
5k - 0
ir> -5 V 2 k (T)
/
o-
A
EF
5k
[XFV- ()] + FCy -1 0 k - 5 k = 0
FCV= 15k;
5k
10k
FC -- 15>/2k
3.49
800#
,400#
[EFy = 0] + Ay + 400# - 800# = 0
Ay =+400# (T)
[ 2 ^ = 0 ] + AX- 693# = 0
Ax
-h
Ax = 693* (-»)
-
M ra
[ZMA= 0] - M r ^ + 693#(81) +400#(6') -8 0 0 #(6') = 0
clockwise
Ay
MRa = 3144#-ft
Ay = 400#
3.50
FBD of the entire frame:
[SM 0 = 0] +A lc(9,)-3 6 0 #(8') = 0
A x =+320# (<-)
[2 F* = 0] G* - 3(20# = °
G x =320# (-»)
FBD of the horizontal beam DBE:
[XFx =0] Bx = o
[XMB= 0] + D (2')-360#(4') = 0
D = F = 720*
[XF =0l + BV-7 2 0 #-360* =0
1
y
1
y
(D)
By = 1080*
Using the result for By = 1080"
proceed to FBD of the inclined
member ABC:
[ZMC= 0] + 1080'(2') - Ay( 6') + 320'(6') = 0
Ay = +680' (T)
[SFy = 0] (+ 680')-1080'+ Cy =0
(A y)
< B ,)
Cy = +400'
Return to the FBD of the entire
frame and solve for Gy;
[SFy = 0] + (6 8 0 ')- Gy - 360' = 0
(A,)
Gy =+320'(l)
Use the FBD of the inclined
member CFG;
[EFx =0] + (3 2 0 ')-C x = 0
Cx = 320'
(Q .)
fZF = 0l -3 2 0 ' + 720' - Cv = 0
v
1 (o,)
y
C = 400'
y
3.51
FBD (a)
FBD(b):
[XFk = 0] b x = 0
[XMa = 0] -100% (4')(2') + By(10') = 0
By = +80#
[lF y = 0] + Ay -100% (4') + 80* = 0
Ay = +320*
FBD(c):
[ZFx = 0] Cx = 0
[ZFy = 0] - 80# - 200* + Cy = 0
Cy = 280*
[ZMc = 0] +80#(6') + 200#(4 ')- M Rc = 0
M Rc =+1280*“ft
3.52
FBD (a)
Support reactions:
[XMa = 0] - 8kN(6m) + B(4m) = 0
B = 12kN (->)
[SF
=0
L
x =01J - Ax +12kN
|,Bj
A k = 12kN (<-)
[XFy = 0] + Ay - 8kN = 0
Ay = 8kN (t)
Joint F:
[SFx = 0] - 0.707DF + 0.707EF = 0
DF = EF
[ZFX= 0] + 0.707DF + 0.707EF - 8kN = 0
2(0.707DF) = 8kN
DF = 5.66kN (T);
EF = 5.66kN (C)
3.52 - c o n t’d
[EMC = 0] +12kN(2m) - 8kN(2m) + DE(2kK) - 5.66kN(2V2m) = 0
DE = 8kN (T)
[XFx =0] Cx-12kN-(0.707)(5.66kN) = 0
Cx =16kN
[XFy = 0] + 8kN - Cy + 8kN - (0.707)(5.66kN) = 0
Cy=12kN
3.53
Support reactions:
[XFx =0] Ax = o
[EMa= 0] - Dy(12') + 500'(4') = 0
Dy= 166.7' (T)
[SFy = 0] +166.7' - 5 0 0 ' + Ay = 0
Ay = 333.3'(1)
Figure (b):
[ s m a = 0] - c y(9‘) + 500'(4') = 0
Cy = 222.2*
[£Fy = 0] + 222.2' - 500' + 333.3' - ABy =0
ABy = +55.5'
Figure (c):
[£MB = 0] + 55.5'(4')-ABX(5') = 0
ABX= +44.4'
[2FX=0] + BAx- A B x =0
BAX= +44.4#
[2Fy =0] - B A y + (55.5') = 0
BAy = +55.5'
Figure (b):
p Fx = o] - c x + (44.4#) = o
Cx = +44.4'
3.54
R, = 400 */h (181) = 7200#
R 2 = 600 Jft (6')3600#
[XMA =0] Ey(20')-7.2k(4,)-3.6k(22') = 0
Ey = 7.56k
[XFy = 0] + Ay -7 .2 k -3 .6 k + 7.56k = 0
Ay = 3.24k
[XFx = 0] A x = 0
FBD (b):
[IM D= 0] - By (18') + 7.2k(4') - 3.6k(3') = 0
By =3k
[lF y =0] + 3k + Dy -3 .6 k -7 .2 k = 0
D y = 7.8k
FBD (a):
[XFy = 0] + 3 .2 4 k - 3 k - C y =0
Cy = 0.24k
[IM b = 0] + Cx(16.2k) - 3.24k(l') + (-0.24k)(9')
Cx = 0.333k
[IF x = 0] + Bx - 0.333k = 0
Bx = 0.333k
FBD (b):
[IF X= 0] - 0.333k+ DX= 0
Dx = 0.333k
3.55
Ri = 4243#
3000#
'600#
R2 = 1560#
L
1440#
R, = (100 %)(30V2') = 4242.6’
R, =
R,
(100%)(30V2')
V2
V2
= 3000
R, = 4Ri^ = 3000*
VT
R2 = (6 0 % )(% x 2 4 ') = 1560#
R2,= % (R 2 )= 1 4 4 0 #
R2 =X3(R2) = 600#
From FBD (b):
[XMA = 0] -3 0 0 0 #(15')-3000#(15)+B![(30,)+ B y(30,) = 0
From FBD (c):
[£MC = 0] -6 0 0 #(5,)-1440*(12,) - B x(24,)+ B y(10,) = 0
Solving the two equations for (FBDb):
simultaneously,
B = 286# ( ^ )
By = 2714# (T)
The same forces are equal and opposite for FBD(c).
From FBD (b):
[E F X = 0 ] + 3 0 0 0 * - 2 8 6 * - A x = 0
A x = 2 7 1 4 * (« -)
[Z F y = 0 ] - 3 0 0 0 * + 2 7 1 4 * - A y = 0
A y = 2 8 6 * (T)
From FBD (c):
[E F X = 0 ] + 1 4 4 0 * + 2 8 6 * - C x = 0
C x = 17 2 6 * ( « - )
[Z F y = 0 ] - 2 7 1 4 * + 6 0 0 * + C y = 0
C y = 2 1 1 4 * (T)
3.56
2kN
FBD (a):
[ Z M A = 0 ] + 2 k N ( 4 m ) + lk N ( 2 m ) - lk N (1 .3 3 m ) - D y (2 .6 6 m ) = 0
D y = 3 .2 6 k N (4)
[Z F y = 0 ] + A y - ik N - 3 . 2 6 k N = 0
A y = 4 .2 6 k N (T)
3.56 - cont’d
Using FBD (c):
[ lF y = 0] + B y - l k N - 3 .2 6 k N = 0
B y = 4 .2 6 k N
[Z M
d
= 0] - 4 .2 6 k N ( 2 .6 6 m ) + lk N ( 1 .3 3 m ) + B x(3 m ) = 0
B x = 3 .3 3 k N
[S F x = 0] - 3 . 3 3 k N + D x = 0
D x = 3 .3 3 k N
Go back to FBD (b) and solve for Ax.
[ I F x = 0] - 2 k N - l k N + 3 .3 3 k N - A x = 0
A x = 0 .3 3 k N
3.57
[ZFy = 0] - A y +152* = 0
G x = 3 7 6 # (« -)
P F X= 0] DX=376#H
[ZFy =
0]
+752#- D
y =0
( Gy)
Dy =152* (I)
From FBD (b):
[z fx = o]
+1900# - a x -3 7 6 #
A x = 1 5 2 4 ' (< -)
=o
3.58
1’
W = (r)(3')(l')(l50% 3) = 450#
Pmax=CD'xh=(35%,)(3') = 105%2
p = O0P™ x h x 1'= (/2)(105%,)(3')(1') = 157.5#
M om ents about the toe @ A:
m o tm „
= P x 1 - (157.5#) X
l '=
157.5#—ft.
M RMi = W x 0 .5 '= (450#)(0.5') = 2 2 5 # - f t.
SP = j ^
=
u ,,,.
2 2 5 * - f l - = 1 .4 3 < 1 .5
157.5#-ft.
R etaining wall does not m eet the
overturning requirem ent.
R
=
tJ
( W ) 2 + ( p f
=
t/(450)2 + (157.5)2 = 477#
P=157.5#
<toe) 4 r
pmax-
x = ° , 5'
Bearing pressure check:
M rm. ~ M 0TMa
(2 2 5 * -ft.) - (1 5 7 .5 # -f t.)
W
(450#)
X
n , p|
x = 0 . 1 5 ' < ^ = % = 0.33'
M axim um bearing pressure at the toe is:
= 2000 P sf < 3 0 0 0 p sf (allow able)
Pmax = ^ - =
OK
3.59
W ,= X (1.5')(1.5')(1')(150% 3) = 168.75#
W2 = ( 4 ') ( l.5 ') ( r ) ( l5 0 % ,) = 9 0 0 #
W3 =(1')(5.5')(1')(150% 3) = 8 2 5 #
WT = 1 6 8 .7 5 # + 9 0 0 # + 8 2 5 # = 1894#
Pmax= a > x h = (3 0 % 3 )(5 .5 ')= 1 6 5 % 2
P = >^pmax x h x 1'= > ^ (l6 5 % 2)(5 .5 ')(r) = 4 5 4 #
m o tm = p x %
= ( 4 5 4 # ) ( 1 . 8 3 ' ) = 8 3 1 # —ft.
M fl = (W 1)(l') + (W 2)(0.75l) + (W 3)(2')
M a = (I 68 .8 # )( r) + (900#)(0.75') + (8 2 5 # )(2 ‘)
M a = 2 4 9 4 # -f t.
3.59 c o n t’d
But, M A = WTx b
,
(2 4 9 4 # -ft.)
b = j —--------- -^ - = 1.32
(1894#)
M rm = M a = 2 4 9 4 # -f t.
£r
M rm
(2 4 9 4 # -ft.;
M otm
(8 3 1 # -f t.)
O K, w all is stab le
x=O.80’
R = V(W T )2 + (P )2 = ^ (1 8 9 4 )2 + (4 5 4 )2 = 1948#
..
M rm - M otm
(2 4 9 4 # —ft.) - (83 l# - f t.)
WT
1894#
=
=2
0 . 88 '
= 0 .8 3 '< x = 0 .8 8 '< 2 ^ / = 1.76'
w ith in the m iddle third
P max = - ^ r (4a - 6 x) = ^ J ( 4
a
(2 .5 )
p max = 1430psf < 3 0 0 0 p s f
X 2.5 '—6 x 0 .8 8 1)
OK
= 3.0 > 1.5
3.60
Pma* = “ 'Xh
P = ^Pn»*X hxr
= {*>%,)(*) =
P = K (3 2 0 % 2)(8Xl') = 1280#
W1=(l')(8')(r)(l50% ,) = 1200#
W2 =(3')(1')(1')(150%3) = 450#
W3 =(3')(7')(l')(l50% 3) = 2520#
WT =
R = -^ (4 1 7 0 f
1200#+450#+2520# = 4170#
+ (1 2 8 0 )2 = 4 3 6 2 #
Wr=4170#
b=1.92-
M rm = MA = W,(0.5') + W2(2.5') + W3(2.5)
MA =(l200#)(0.5')+(450#)(2.5') +(2520#) =8025#-ft.
■ U
1280#
b = 8025 # - f t .= 1 9 2 ,
4170#
2.67’
M 0Tm = P X %
:r
M .m
M
otm
= ( 1 2 8 0 # ) ( 2 .6 7 ‘) = 3 4 1 8 # - f t .
8025#-fl.
M
rm
-M
OK
2^ = 2 67'
^ = % = 1-33'
:L
= 2 .3 4 > 1 .5
3 4 1 8 # -ft.
otm
(8 0 2 5
WT
3418)
^ ^
4170
X<M
2W
2 (4 1 7 0 )
Pmax=1 7 = 3(U) =2527^ 2<30°° PSf
•■•°K
3.61
W i = ( 1') ( 3 ,) ( 1') ( 1 5 0 % 3 ) = 4 5 0 #
W 2 = ( l ' ) ( 8 ') ( l 5 0 % 3) = 1 2 0 0 #
W T = 4 5 0 # + 1 2 0 0 # = 1650#
P maX = ( 4 0 % , ) ( 8 ') = 3 2 0 % t2
P = X ( 3 2 0 % 2)(8 ')(1 ') = 1 2 8 0 #
R = ^ (1 6 5 0 )2 + (1 2 8 0 )2 = 2 0 8 8 #
M rm = M a = W ,( 1 .5 ') + W 2( 3 .5 ')
M
a
= (4 5 0 # ) ( 1 .5 ') + ( 1 2 0 0 # ) ( 3 .5 ) = 4 8 7 5 # - f t .
M A = W T x b = 4 8 7 5 # -ft.
b = 4 8 7 5 # -f t. = 1 9 5 ,
1650#
M o tm = P
x
%
= ( 1 2 8 0 # ) ( 2 .6 T ) = 3 4 1 8 # - f t .
S.F. = - M b m _ = 4 8 7 5 # - f t . = L 4 3 < L 5
M o tm
3 4 1 8 # - ft.
N o t s ta b le f o r o v e r tu r n in g
t
m rm
~
m otm
(4 8 7 5 )
WT
(3 4 1 8 )
^
1650
= 1.33'
x < /4=
2W
2 (1 6 5 0 )
p- = i ^ = i ( ^ ) =
OK
fo r
b e a rin g
^ = % = 133'
1 2 5 0 p sf< 3 0 0 0 p sf
p re s s u r e
2^ = 2.67-
3.62
pnax = co'xh= (5 .5 y m!)(5.5m) = 30.25*%i
P = >£pmax(h)(lO = X (30.25% )(5.5m )(l') = 83.2kN
W, = (0.5m)(5m)(lm)(23.5“/ n,) = 58.8kN
2m
W2 = (0.5m)(3m)(lm)(23.5kl// ,) = 35.3kN
0.75m-
W3 = (2m )(5m )(lm )(l8% !) = 180kN
Wi
1.5m--
m
Wa
Pm ax
R = -yjw* + P2 = t/(274)2 + (83.3)2 = 286kN
= Ma = W, (0.75m) + W2(1.5m) + W 3(2m)
M a = (58.8kN)(0.75m) + (35.3kN)(1.5m) + (180kN)(2m)
b = 1,67m -
W t
= 274kN
P = 83.3kN
f R = 286kN
u 457kN - m
b = -------------- = 1.67m
274kN
Mn
:Px
= (83.2kN)(l.83m) = 152kN - m
S.F. = -^a*L= 457kN~ m = 3 ,0 > 1.5
Mn
152kN -m
x = 1.11m
^ = 3 n ^ = im
/. OK
2^ / = 2 m
(4 5 7 -1 5 2 )k N -m
= 1. 11m
x = 1.1 lm < 2^
Pmax =
(4a - 6x) =
(4 x 3m - 6 x 1.1 lm)
P .„ = !©■%■ (3400psf)> 1 4 3 .6 % , (3000psf)
Overstressed in bearing
3.63
W, =(1.33,)(14.67')(1,)(150%,) = 2927#
W2 = (1.33')(8')(r)(l50%,) = 1596#
W3 = (5.33')(14.67')(1')(150X1») = 8992#
WT = 2927#+ 1596#+8992#= 13,515#
P - = ( 4 0 /n,) (l 60 = 640% ,
P = ^ (6 4 0 % !)(16')(1')=5120#
R = ^/w 2 + P2 = ^/(13,520)2 + (5120)2 = 14,450#
= MA = W,(2') + W2(4') + W3(5.33')
MA = (2927#)(2') + (1596#)(4‘) + (8992#)(5.33')
Ma =60,200#-ft.
Ma = WTx b = 60,200#-ft.
, 60,200#-ft. ^ _
b = --------------- = 4.45'
13,520#
m ctim
=
P X /'j
Mb
S.F.=
M01m
=
(5120#)(5.33') =27,300#-ft.
60,200#—ft.
= 2.2 >1.5
27,300#-ft.
OK for overturning
^ = % = 2.67'
2)4 = 5 .3 3 '
RM
CTIM 60,200-27,300 = 2 .43'< a^
WT
13,520
Kmax
™ = 3 1M 202 =
3x
3(2.43')
P
K
.'. exceed the allowable bearing pressure
Chapter 4 Problem Solutions
4.1
CO= (50 psf)x(5') = 250 lb /fj
R
(1.250 lb .)
R
(1,250 lb.)
FBD of beams B-1, B-2, B-3
Beam B-1
(1,250 lb.)
Beam B-1
(1,250 lb.)
10'
10'
(COl. Ar 1)
(1.250 lb.)
FBD of girder G-1
Girder G-1 (partial framing)
CD
(Col. D-1)
(1.250 lb.)
1250lb.
= 250 lb /fl
10'
t
FBD of girder G-2
-------------------------------------------------------------- 1
T C o l D-3
Col.A-3
37501b.
FBD o f girder G-3
(B)
! 1250
!
(C)
G 1
(5)
!
1250 !
N\ p e o / /
.1250
<D—
Col. D -1 1
2500 lb.
Col. D-3
3750 lb.
(A)
G >-
CD=250lb/ft
1- u u i u i
1250
o
3750
B1 X
B1
a
/
b
1 250
3
\
1250
-
B2
1250
3750..
(tySummary o f Column Loads
4.2
®snow= (25 lb ./It 2)x(2‘) = 50 lb /It
Slope conversion:
11111 n n 11 n 1 1 1 1 1 1 1 n i n n n
u DL = ( l 0 1 b . / f t . 2) x
12.37
12
( 2 ') = 2 0 lb . / f t .
fihb \
^width)
coSnDW = ( 2 5 1 b . / f t . 2 ) x
( 2 ') = 5 0 lb ./ f t.
(Inb.
I width I
“ •n*i - “ s^ + ^ dl = (501b./ft.)+(20.61b./ft.) = 70.6lb./ft.
Snow loads are assum ed to be on the horizontal projection of the roof while the dead loads
are applied along the length of the rafter.
Both load conditions are combined to simplify the computations. Generally, the dead load is
converted into an equivalent horizontal load and added to the snow load. Determination of
the dead load a s an equivalent horizontal load requires a slope conversion.
Ceiling Joist: DL + LL= 15 lb.At.2
CO = (15 lb /ft 2)x(2") = 30 lb /It
• W W U U H
10'
Wall A
1501b.
Ceiling
Joists
U i U U H U U i
[interior Wall
150+180=
330 lb.
\
Wall B|
1801b.
co = 151 b./ft. x 2ft. = 3 0 lb./ft.
3rd floor walls:
Wall A:
Roof =
1463 lb /It
.478 lb /ft
I 165 lb /ft
7771b./2' = 388 lb./ft.
Ceiling = 150 lb./2' = 75 lb./ft.
463 lb./ft.
W alls
Wall - 80 lb /It
Wall B:
W all— 80 lb /ft
Roof = 777 lb./2' = 388 lb./ft.
Ceiling =1801b./2' = 90 lb./ft.
Wall A
543 Ib./ft.
l Int.
Wall
245 Ib./ft.
LWall B
558 lb /ft
Interior Walls:
Ceiling = 330 lb./2' = 1651b./ft.
3rd Floor:
Wall B = co x 6'= (60 lb./ft.) x (&) = 360 lb./ft.
Interior Wall = (co)x(S + 61) = (6 0 lb ./ft.)x (l 1') = 6 6 0 lb./ft.
478 lb./ft.
Wall A:
543lb./ft.+300lb./ft. = 84-3lb /ft.
(wall above)
Wall B:
(3rd floor)
5581b./ft.+3601b./ft. = 918lb./ft.
(u/all above)
(3rd floor)
Interior W all: 2451b./ft.+ 660 lb./ft. = 905lb./ft.
(u/all above)
(3rd floor)
CD= (60 lb./ft.2)x(1') = 60 lb./ft
■u w
m
1
10'
m
.
-----------------Tz---------------- 1
Hr
f
Wall A
Inter. Wall
Wall B 1
300 lb.
660 lb.
360 lb.
2nd floor:
Wall A = cox 5 = (6 0 lb./ft.2) x ( 3 ) = 3 0 0 lb./ft.
Wall B = co x 6'= (6 0 lb./ft.) x ( 6') = 360 lb./ft.
Interior Wall = (co)x(5'+6') = (60lb./ft.) x (11') = 6 6 0 lb./ft.
4.2 cont’d.
RA=12,660 lb.
R,nt =23,880 lb.
Beam - 20’ span:
(0„U = w tolal =(923 lb./ft.)+(3001b./ft.)+(431b./ft.) = 1266 lb/ft.
( wall above)
(2nd floor)
x L = (1266lb./ft.)x(20) =12 60Q[b
End reaction R =
A
(beam weighl)
2
2
Beam - 28’ span:
“ umi = “ total =(9851b./ft.)+(660 lb./ft.)+(61 lb./ft.) = 1706 lb./ft.
(wall abov*)
(2nd floor)
(beam weight)
a)
xL
(1706 lb./ft.) x (28')
End reaction R. . = —— ------= -------------— -—-— - = 23,8801b.
2
2
4.3
Roof Loads:
co = 6 6 lb /ft
cn
764 lb per 2‘
Front wall
Roof
Rafters
12'
14'
424 lb 396 lb
Roof beam
Snow = 20 psf
Roofing = 5 psf
Sheathing = 3 psf
Rafters = 3 psf
Ceiling = 2 psf
Total DL= 13 psf
396 lb. per 2'
Back wall
Design Load = SL + DL = 20 psf + 13 psf
= 33 psf
coSL+DL =(331b./ft.2)x
(2ft.) = 66 lb./ft.
(rafter spcg)
P Front wall
“ ►wall = 6 4
Floor Loads:
Occupancy (LL) = 40 psf
Floor = 5 psf
Joists = 3 psf
Total DL= 8 psf
Design Floor Load = LL + DL = 40 psf + 8 psf = 48 psf
w d u l l =(48lb./ft.2)x (2ft.) = 96lb./ft.
( joist spcg)
co = 96 ib./ft.
r
672 lb. per 2'
(front footing)
f1248 lb. per 2' n
(floor bea m )
:
576 lb. per 2‘
(back footing)
4.3 co n t’d
Front footing:
co = (446 lb./ft.)+(336 lb./ft.) = 782 lb./ft.
(Wall load)
(floor joisls)
Back footing:
co = (262 lb./ft.)+(288 lb./ft.) = 550 lb./ft.
(Wall load)
4920 lb.
(Col.)
2460 lb.
(C0L)
(D = 624 lb./ft.
' H
i i i n
L
6'
i i w
1
6'
u
,
,
|
4332 lb.
3744 lb.
6664 lb.
37441b.
Ext. post
Int. post
Int. post
Int post
8664 lb.
(Int. post)
Critical footing
(floor joisls)
4.4
Roof Loads:
Snow = 25 psf
Dead Load = 15 psf
—(S L + D L ) x
(8 1)
+ w beam
(Tnb. widlh)
c o ^ = (40 lb./ft.2)x ( 81) + 151b./ft. = 335lb./ft.
Beam Reaction: (24’ span)
(Wall or
beam)
co^xL
(335lb./ft.) x 24'
R,■beam
: 40201b.
or u/all
2x4020 lb. = 80401b.(every 8')
Girder G-1:
Spacing of beam
should be treated
as concentrated
loads every 8'
Girder G-1 supports concentrated beam
reactions from both sides plus its own self
(0
Wall
(17k)
^ g ird e r
= 50 lb./ft.
weight. Since the beam reactions occur at
8’ o.c., they must be represented as
concentrated loads and not reduced to an
Girders
Column
(42.2k)
equivalent uniform load.
Column:
The column load includes the girder
Wall
(17k)
reactions from both sides plus the tributary
weight of the girder.
P = 42.2k
fPl
h = 30'
Column
0 = 255 lb /ft. (trib. width = 6‘)
L= 16'
Beam B-2
G -2
(2040 lb.)
Beam B-2: (Tributary width = 6’)
+
>
Wall
(2040 lb.)
“ snow = (251b./ft.2) x (61) = 150 lb./ft.
coDL = (l5 1 b ./ft.2)x (6 ')+ (l5 1 b ./ft.) = 1051b./ft.
(b e am w t.)
codl +sl
= 150 lb./ft. + 1051b./ft. = 2551b./ft.
4.4 co n t’d
Girder G-2:
Beam reactions are treated as
concentrated loads spaced at 6’-0” o.c.
Truss joists are spaced close together
(2' or less),therefore, the reactions may
be treated as an equivalent distributed
load on girder G-2.
20401b.
20401b.
20401b.
20401b.
^otal - *
I 1 T T '- t r h
■ m i ' 1I H
1- I W i
L = 30'
12,5701b.
Girder G-2
12,5701b.
The tributary width of load from the
truss joists onto the girder equals half
of the span or 12.
“ snow = (251b./ ft.2) x 12' = 300 lb./ft.
coDL = (l81b./ft.2) x 12'+(501b./ft.) = 266lb./ft.
(girder weighl)
“ toul = “ sr™ + w DL = 566 lb/ft.
4.5
=20 psf x ( 16/ l 2 ft ) = 26.7 plf
Critical roof joist: (16” o.c. spacing)
Loads:
D L = ( l 2 1 b . / f t . 2 ) x ( 1 6 / 1 2 ) = 1 6 lb . / f t .
Joist Wt. = 4 Ib./ft.
coSL = ( 2 0
FBD of the critical
inclined roof joist
c o 'd l
lb ./ft.)
= (5 /4 )
x ( 1 6 /1 2 ) ' = 2 6 .7
X
(2 0
lb ./ft.)
= 25
lb ./fL
lb ./ft.
(slope adj)
“ total = cos l + cod l = 2 6 .7 lb ./f L + 2 5 1 b ./f t . = 5 1 . 7 lb ./ f t.
c o 'D l = ( 5 / 4 ) x ( 2 0
plf) = 25 plf
“ total = “ SL + ® DL
“ total = 26.7 p lf + 25 p lf = 5 1 .7 plf
17'
439 lb.
439 lb.
FBD of the equivalent,
horizontally projected roof joist
co = 2x(439 ib/16")x(i2/16ft.)
Ridge Beam:
The equivalent concentrated load
from the triangular
load distribution is equal to:
R = Y 2 x ( 3 4 ') x ( 6 5 9 l b ./ f t.) = 1 1 ,2 0 0
2 M a = - ( 1 1 , 2001b.)(17') - ( 4 0 lb ./ft.)(3 4 ,)(17l) + B y (2 2 .6 7 ') - 0
.-. B y - 9 ,4 2 0 lb.
^ F y -
- ( 1 1 , 2 0 0 1 b .)- ( 4 0 lb ./ft.)(3 4 ') + (9 ,4 2 0 1 b .) + A y - 0
A y - 3 ,1 4 0
lb.
lb.
4.6
1). Rafters
®SL = 6 0 'b4
0>SL = 60lb/ft
Ridge
beam
Slope adjusted deald load:
Slope adjusted deald load:
ffl,DL = ( 3 6 lb ./ „ . ) x ( 1 2 .6 5 / 12) = 3 0 lb ./ft
Q)’DL = ( 3 6 'b /,) x ( 1 3 /12)= 3 9 'b /f
a>SL= 60|b/ft
aiSL = 60 lb /ft
“ Total = 3 8 'b 4
+ 60 'h /ft = 9 8 'b / ft
“hotal:= 39 lb / ft
L = 14'
686 lb./24„
Bearing wall
+ 60 'b / ft =99 lb / ft
L = 16'
686 Ib/24„
Ridge beam
792 lb724"
Bearing wall
FBD of an equivalent
horizontal rafter - left
792 lb./24„
Ridge beam
FBD of an equivalent
horizontal rafter - right
2) Short wall/roof beam
col.
4.6 co nt’d
(3) Walls:
(4 ) In te rio r C o lu m ns: S e e (2) above.
(5) Floor Joists
0) =(50 psf )x(
1
ffl = 66.7
= 66.7 *>/ft
t T*
± jL= 14'
L= 16'
534 /•
467
(Wall)
(Beam)
FBD o f th e flo o r jo is t - le ft
(6) a n d (7) F loo r Beam and P o st
7545 lb.
(Beam)
534
/-;
(Wall)
FBD of the floor jo is t-rig h t
4.6 c o n t’d
(8) Foundation Walls:
(8) Continuous Footings
m.;j, = («/13)'x(3')x(150 "•/1 .•) = 300 it
Footing A
(i
25:;
Footing B
(W- .)'x(150
.______...
= 100 • /■■■
q = 2000psf;
qnet = q - base wt.
qnet = 2000psf - 10Opsf = 1900psf
Footing A
P/a = (323
/j + 300 "
)/(1.25' x 1') = 898 psf < 1900 psf (OK)
Footing B
P/a = (926|b /|t + 300 Ib./j )/(1.25'x 1■) = 981 psf < 1900 psf (OK)
(9) Critical Pier Footing
P = 15,090 lb.
Base = (!V|'))‘x(150 I',' ■)
= 100 ■'! :
qM.| = 2000psf - 10Opsf = 1900psf
P/A = (15,090 lb.)/(x2) = 1900 » /,2
x- = 7.94 ft 2;
x = 2.82' = 2-10" square
4.7
Jack Rafter (Typical span):
= 25psfx (24/12 ft.) = 50 plf
Roof DL:
coDL = (l21b./ft 2 Jx (2'j = 24 lb./ft.
tH t
“
dl
(honz pp.'])
= 0 V 12) ( 2 4
= 3 0 lb / f t -
Snow:
coSL = 12 5 1b./ft 2) X ( 2‘) = 5 0 1b./f't.
FBD of a typical jack rafter
= 50 plf + 30 plf = 80 plf
i i u u n m
CD,,,*,
m
= wSL + w'DL = 50 lb./ft. +30 lb./ft. = 80Ib./ft.
(horiz pro))
101
400 lb.
400 lb.
FBD of the equivalent, horizontally
projected jack rafter
This rafter represents the maximum load
condition onto the hip rafter since other rafters
Hip Rafter
diminish in length.
The spacing of the jack rafter along the length of
the hip rafter is:
/. to =
2,xV 2 = 2 8'>1
x (4001b.) = 284 lb./ft.
4.7 cont’d
Ceiling Joists: (2’ o.c. spacing)
o'< d i = ( 7 ( b ./f t..2 ) X ( 2 ') = 14-1 b . / f 1:.
Ci) = 54 p(f
t
1
111
t
TT T
t
2161b.
Wall
1
./ft2 ) x
(2'j = 40 l b . / f t .
cotota| = 14 l b . / f t . + 40 l b . / f t . = 54 l b . / f t
270 /-.
1u
i
(20 l b
Beam B-1:
B eam B-1:
ic e o ib .
Col.
=
1
324 lb.
Wall
5401b.
Beam B-1
co =
i
12'
81
n
T
540 lb.
1 i »
= 270 lb./ft.
2 ft.
■■ jreo ib.
itre
wall
B e a m B -2:
(!) = 1 0 8 1 1 '/, + 2 7
/
= 135 I '/
Beam B-2:
Joist load: (span = 8 ')
L = 12'
CD
810 1b.
wall
( 4 1)
= 127 lb ./ft. ) X
= 108 lb ./ft.
(tjrib w id th ;
810 1b.
Col.
B e a m B -3 :
Beam B-3:
co = 135 ib /ft
The load condition on beam B-3 is
t r r t i
identical to beam B-2.
I
540 Ib.
540 Ib.
w D.lL = 1351b/ft.
4.7 co n t’d
Interior Column:
Loads to the column Ceiling joist:
1081b.
Beam B-1:
10801b.
Beam B-2:
8101b.
Beam B-3:
540 lb.
P = 2538 lb.
In addition to the vertically applied loads on
both the jack and hip rafters, truss action
develops due to the ceiling tie condition. An
examination of the truss action for each rafter
case will be performed. The three dimensional
truss solution for the hip rafter was not
covered in the text but can be done relatively
easily using readily available structural
software.
4.8
600#
H
G
F
E
[EFX = 0] + 600#-G A X- ECX= 0
Assuming GAX= ECX
Then GAV= ECV= 300#
600#
H
600
G
300
F
300
E
4.9
r
„
I200#(16‘)
[l'M A =0] By = ----- —----- = 1 3/0# I ; i
[2Fy - 0 ] Ay =1370# (!)
Assume Ax = B,
Then; A.x = Bx = 600#
Gy
1200#
Dyj
r
Gk
K
FH
n
H
D
j2 M „ = 0] FH.,(4'i -6 0 0 # (’Iff) = 0
Di
!;11. = 2400#;
CE
FHV= 2400#
* \ c
[SE. =0] - Gv + 2400#-1370#= 0
G y = 1030#
[vR =0] +1200#-G x +2400#-600#= 0
Gy = 3000#
A
Ax - 600#
— v•A » =
. . .
. ,
[SMr; =0] C E ,(4 ')-6 0 0 # (1 6 I) = 0
1370#
C E ,= 2 4 0 0 # ;
CEV= 2400#
[2 Fz = 0] - Dx + 2400# -600# = 0
D, =1800[2Fy = 0] + Dy - 2+00#+1370# = 0
D. =1030.7
1030
11030
3000
F
G
IT J2400
2-100
2136
2400
D 1800
p,
,
Bx = 600#
By = 1 3 7 0 #
|v F = 0 | + 4k-G K . =0
GK, = 4k;
GK„ = 4 k
[SFX= 0] + 4k + 6k + CE,
Assume AQX ■ CEX
AGv = CEX= 5k
4.10 co n t’d
4.11
Roof
(20psfx10'=200«ft)
F„ = 200^rt(40')=8000#/rt
2nd
(20psfx10'=200#ffi)
F, = 200#>Tt(40’)=8000#/(t
A= 0] - 2000#(1 O') +1000#(5') + T(IO') = 0
„ 20,000-5,000
T = —:-------- :---- = 1500#
10'
4.11 c o n t’d
8000# A nnt.
V- 2= —
= 400%
2000#
W =1000#
, M= 20,000#-ft
M = 2,000#xl0'=20,000#-ft.
4000#
[2M B =0] -2 0 0 0 # (2 0 ')-2000#(1 O') + 2000#(51) + T(10') = 0
= 40,000 + 20,000-10,000 = 500Q#
101
[2M b =0] -20,000, “n-4000#(10l) + 1000#(5')+ 1000#(5') + T(10I) = 0
(M)
T =5,000#
aien bs „ £ - ,Z = ,lZ Z = y
W SLSl
"b
#L0Z6
d
zm6'p = ■
■= x = v
7'% SL 8i = ( [’% o g i ) [ y ^ J - j^ o o o z : = ” nb
y,Li.8 =
#X?L9\
((pe jdojs)
% V V 9 = % v v \ + Y .o s =
+ '^ os =
Z
#W9l
J9 d
w t = .2xjsd z= n a
W#09 = ,2XJSd92 = ns
f t m t t t t m t t t t t m t t m m l
js d 9 J = PEOI MOUS
ZVfr
Chapter 5 Problem Solutions
5.1
[EMC - 0] i 500# (3')
c«
. ,p
BA._ g .
Cy’
I'A
>
P :::: 500#
B Av = 50(^ ( 3 ') = 7 5 0 #
, .
liAV
BA - -50V 2r
p - 750 V2 #
A - K " x 2"=1 in.3
BAX(2') - 0
5 .2
Total are of marquee = 20'xl0'= 200ft/'
Total load = 200ft2 x 100%, -■=20,000#
Since the framing is symmetrical,
each rod carries an equal amount
of the load.
T.j, = T ^ sin 3 0° = 0.5TAR
[EM,. = 0] i 10k(5')
0.5X^(10*) - 0
T„,v - 10k
P
f.r = —;
‘ A
10k 0.-ki
r
in.^ ■
A_= _ P= ----‘-1'
T 22%,,
To the nearest Xs“'~ % " <l)
For % " $ rod, A -0 .5 1 8 5 in.2
10k
1 = ---, =19.2ksi< 22ksi
(stasl) 0.5185 in.
(dtorable)
stress is OK, within the stress range
5 .3
. p p
.
120k
on . 2
a) f = —; A
= ---------- = 8.9 in.
A
rcqd 13.5%,*
From the steel tables in the appendix;
Use: W 8x31 (A = 9.12 in.2
P
.
120k
b) fp = 4-:
\Ab=-----=„
=266.7 in.2
—;
A
b
0.45%
For a square base plate:
U se: 16.3"xl6.3"
or
16J£"xl6)£" plate.
c)*
r
p
f= P—;
A
A
Ah
b
120k
.-J . 2
= -= 40ft
3% ,
Use : 6.32' square or 6'-4-" square footing
5.4
a) Taking a 1' (12") strip;
f = | - ; P = A x f = (48 in.2)x l5 0 % nJ=7200#
But: P = Ybrx A x h
h
P
7200#
ISO'
- ^ - ( n o x ^ f t 2) -
b) Taking a 1' (12") strip;
f = —; P = A x f = 7 2 in.2x l 5 0 y , =10,800#
A
But:
h=
/,n
P = yxA xh
P
yxh
10,800#
(no*/n, ) x ( & e )
,
5.5
a) P = 20,000#;
P
20,000#
A
64 in
A = 64 in.2
= 312.5j/ ;
JlD
b) P = 40,000#; A ^ ----4
since rod is threaded
A = 0.3 0 2 in.2
.
P =
A
4000#
0.302 in.2
c) P = 40,000#;
A = 4 "x4"—
- = 15.4 i n 2
_ 4000#
br*
15.4in.2
A = 8"xl2'= 96 in 2
d) P = 15,000#;
f = 1510 0 0 # = l
96 in.2
A = 8"xL;
e) P = 16,000#;
Fv =120psi
.
P
16,000#
2
A„0., = — = ------- — = 133.3 in
Fv 120%,«
133.3 in.2 = 8"xL;
L = 16.7"
5.6
,
„
P
A
10,000#
•>
71X2
D B = 1 0 ,0 0 0 #
= 4,273psi
7 tX l2
I 4 JI 4 J
b) [ZFy = 0] @ jo in t D ;
-1 0 ,0 0 0 # + A D y + CD y = 0
But, A D y = CDy = 5000#
.-.CD = 5,000V 5#= 11,180#
P _ 11,180#
A ” 1 in 2
, ,
P
11,180#
.
c) fv = — = —----------- r—r- = 12,660psi
2A 2 [rc x 0 .7 5 ^ /
AD
CDy
ADy
A D ,.
**•* 1
CD
5 .7
S
L
0.0024"
2.0"
.
e = — = -----------= 0.0012'“/,,
5.8
" r w
a“ “ *
5.9
*—
^----PT \
5 = 0.0033in./in.
II
Q
8 = ^-;
5 = eL = (0.0033'^n)x (8") = 0.0264 in
5.10
e= £ ;
5 = eL = 0 .0 0 5 '/n(50 0 'x l 2'»/() = 30 in
5.11
Consider a 1’ length of wall;
Roof = 1’x10’x100psf= 1,000#
Snow = 1’x10'x30psf = 300#
Dead Load + Snow = 1300#/ft.
Brick = 1 ’x(4/12)’x120#flt2 = 480#
Tool lead at the base of the wall = 1780#
Bearing area = 4”x12” = 48 in.2
,
P
1780#
.^
•
f = — = -------- t-= 37.1 psi < 125 p a
p A 48 in.2
.'. OK, w ithin stress allowable
5.12
a) wire weight(total) = 0.042% X 300'= 12.6#
A , « t f , ,( O .I 2 5 ) - , 0 0 1 2 3 .n ,
4
,
P
4
12.6#
A
0.0123 in.2
„
f. = — = ------------rr = 1024.4pSl
_ F,n,
*■allow
.
65ksi
q
(S.F.)
Paiiow = (2 1 .6 7 % t)(0.0123m 2) = 266.5#
Wire wt. = 12.6#
M aximum W = P - 12.6#= 254#
5 .1 3
,)
= ft|7„
8=I k ,
AE
71X1.5
UN S - PL. A
AE'
rerd
2 9 x l 0 6%,i )
- PL
5E
(2 9 ,0 0 0 # ) ( 2 5 'x l2 ^ ) _ o . 2
re,,<1
(0 .1 0 )(2 9 x l0 6%ii)
A = fD l; D = M
4
Vn
Use : 2" (b rod
=m
. = 1.9 S ,
(3 .1 4
5.14
A = 0.006 in .2
„
5 = pl
AE
P = 16#
E = 30x10s psi
, 0.106T.
( o .o o e in /^ ^ o x io 6; ^ )
LN
P
16#
.
b) f = — = -----------5- = 2667psi
A 0.006in
5 .1 5
L = 90’-10” = 1090”;
P = 60k
F = 20 ksi; Upset rods,
E = 29x10® ksi
,
a)
.
reqi
P
Ft
60k
20%,
= — = --------- = 3 in.
2
Ttd
A = — = 3 in 2
4
,----- turnbuckle
U se: d = , — =1.95"
Era
b) 5 = P L = a L = (20y . . ) C ^ ) =075„
AE
E
29x10
Each turn = X " movement per rod (one thread)
one turn on the tum buckle =
movement
0.75"
Number of turns = ------- = 1.5 turns
0.5"
5.16
5 = aLAT = (6 x 1O'6 /°F)(2 x 20'xl 2 %)(60°F)
5 = 0.173"
5.17
A ssum ing unrestrained movement;
5a, = cca,LAT = (12.8X 10"6/°F)(L)(55°F) = 704x 10^(L)
Sco„c = « concLAT = (6.0 x Iff* /°F)(L)(l 5°F) = 90 x 1(T“(L)
R estrained deformation in the alum inum panel:
5 „ = 5 A1- 5 _ = 6 1 4 x lO ^ ( L )
Stress required to restrain the alum inum by 6 1 4 x 1 0 ^ (L ):
g_PL_fL
” AE_ E ’
r
f _5E
” L
614xlO-(L)(lOxlO6X„0
_
.
f = ---------------- ---------------L= 6140 p a
L
5.18
a) 5 = aLAT
Set 8 = 0.25"
• AT = — = -----------------------------------= 53 4°F
aL
(6.5x10^ /°F)(60,xl2% )
AT = Tflnll- T m8 = 53.4° F
Tfmll = 53.4°F + 70°F = 123.4°F (no stress condition)
b)
@T = 150°F
5 = aLAT = (6.5x 1(T6/°F)(720")(150° -123.4°) = 0.124"
(restrained deformation)
fL
8= — ;
E
5E (0.1 2 4 ")(2 9 x l0 * /1\
f = — = ----------^ ----------- / n i = 4994 psi
L
720"
5 .1 9
From the equilibrium condition;
P = 100k
[SFy = 0 ] fsAs + fcA c = 100k
8s = 5c =0.01"
=
5 =—
E
f . 5E1. (0 01")(20X1 0 ^ .,). 2 42fai
3 Lc
f
120"
SE„
(0.01")(3xlO >X , )
c
Lc
120"
A
rc(15j— A
q
= 0.25 ksi
n 6 6 itl2 _ A
4
Substituting in to the equilibrium equation;
2.42( As) + 0.25(176.6 - As) = 100k
2.42(AS) + 4 4 .15k -0 .2 5 A S= 100k
A s = 25.8 i n 2
A =150.8 in.2
0 . 01 "
5 .2 0
P = 180k
A„„ =
7iD2 _ 71(12.75"-2x 0.375")
4
7CD2
Ail = ^ - A
a 13 in.2
con= ^ ^ ^ - - 1 1 3 i n 2 = 14.7 i n 2
L= 30”
From equilibrium;
[ZFy = 0 ] fcA c + fsAs = 180k
fc(l 13 i n 2) + fs(l4.7 i n 2) = 180k ........ (Eq. 1)
From the deformation and strain relationship;
5= 5 =5s
c
5= —
E
£k=Lk.
ec
e,
’
E
2 9 x l0 5
f = —- X f = -------- r- X f = 9.67f
s E„
3x10
.......... (Eq. 2)
c
Substituting Eq. 2 in to Eq. 1;
113fc +14.7(9.67fc) = 180k
113fc + 142fc = 180k
fc = 0.71 ksi
f =9.67(0.71) = 6.87 ksi
s=s.=s.=Y
^ , = 6 .8 7 ./ (3 0 ') = o o o 7 1 ,
s
Es
29x10
cv'
5.21
a) From equilibrium;
P = 50k
[ZFy = 0] Ps + P0 = 50k
but; Ps = f sAs and P0 = f 0A 0
fsAs + f0A 0 = 50k
As = 4 in 2
.... (Eq. 1)
A 0 = 32 in .2
From elastic deformation;
5s = 5o (since the load is symmetrically applied)
s PL
5
o = ----- and E = — and L„ = L„
AE
L
3
0
f
E
8 = —
SO
fs
—
Es
fo
E0
= —
f = E , x f = 3 0 x l 0 ^ x f =155
3 E0 0 2x10
Substituting in to Eq. 2:
15f0(4 in.2) + f0(32 in.2) = 50k
f0 = 0.543 ksi;
fs = 8 .1 5 k s i
b) 5 = 5S = 5 0
fL
8 .15V ',(8")
8 = — = ------ 7 V
= 0.002"
E
3 0 x 1 0 /yi ni
Chapter 6 Problem Solutions
6.1
,Y
CG
/
X
y = 5.67'
Rot.
I'....*
C om ponent
5.33'
AA
X
XAA
20 inr
5”
100 in."
9”
100 in.a
16 in
1”
16 in.:<
4"
64 in.3
24 in.2
8.5"
204 in.3
4"
96 in.3
y
yAA
10"
•
•
h"
■ \ 0“
YAA = 60 in.2
ZxAA
A
320in.
= 5.33"
60in."
- = M A = 3 4 0 i^ =567„
A
60in.
Ex A A = 320 in.3
ly A A
340 in.
6.2
y
.Y
........................-k
4’
C om ponent
AA
X
xAA
y
yAA
A"
«■
0 2"
s.. \ \
\
24 In:'
2”
+48 in.-'
3"
+72 In.3
-2 in.2
2.5'
-5 in’
3“
-6in.“
9 In.2
5"
+45 In.J
2’
18 In.3
\
2AA = 31 in.2
-
ZyAA
84 in.3
A
31in.2
LxA A = 88 in3
XyAA » 84in*
6.3
C om ponent
AA
-
y=
1000ft.J
A
132 ft.
XyAA _ 702ft.A
132ft/
= 7.6'
= 5.3'
xAA
y
160ft.2
8’
1280ft.3
5’
800ft3
-28ft.2
10’
-280ft.3
3.5'
-98 ft.3
XAA = 1 32ft.2
ZxAA
x
Ix A A = 1000 ft3
ZyAA = 702 ft.3
6.4
C15x40
(cenlerted)
S'
Vi
d+x
C om ponent
X
11 .Sin?
o
o
14.02+ .78
=14.8”
174.6 in.s
26.5 in.2
0
o
14.02
2
= 7.01'
185.8 in.3
.
■'■T ...........
!J11
1----
xAA
y
---- 1
___ ___
XAA = 3 8 3 in ^
x=0
-
yAA
AA
360.4m .3
y = ------- - r = 9.4"
38.3in.
XxAA = 0
lyAA . =360.4 in:
6.5
3”
3”
Ref.
Y
C om ponent
U
m
AA
x
6.09in.2
0
^ - + 1 0 ”+.28”-.7’
= 9.7’
59 in.3
2x5 =
10in.2
0
■22 . 10 - 5 1 1 ”
2 + 2 - 311
51.1 in.3
4.71 in.2
0
XAA =20.8in.2
y=
i 20.8m
^ 2
= 5.3„
xAA
X xA A = 0
yAA
XyAA =110.1in.3
6.6
Component
AA
X
xAA
(in 2)
(in.)
(in3)
2.75
0.25
0.69
2.0
2.0
4.0
y
yAA
(in )
(in 3 )
3.25
8.94
ft■
•
5ife”
-
i» i
•
4”
1
ZAA = 4.75 in.2
- = ScAA = 469_m^ = Q9
ZAA
4.75 in.2
ExAA = 4.69 in.3
- = M 4^ = 2ffl
4.75in.2
0.25
0.50
EyAA = 9.44 in.3
6.6 co n t’d
Comp.
•
4A
I,, (in.*)
dv(in.)
Adv'
(in.')
IVC(in.*)
'
7
dy (in.)
A d/
2.75
6.93
y-y =
4.3
0.06
x -x =
1.5
0.74
1.25
5li"
2
0.04
4"
6.1
2.7
1 A .I - =
10 4 in 4
I I yc =
') S in 4
y - y 2=
1.75
I 1. =
7.0 in.4
I x = Z I „ + ZAdJ = 7.0 + 10.4 = 17.4in.4
I y = £1^ + XAdx = 2.7 in.4 + 3.5 in.4 =6.2in.4
x2 - x =
1.01
2.0
3 5 in 4
6.7
yc y
^-d>e = I”
dyi = 5"
^
,
dysi = 3"
12“
Ref.
Y
C om ponent
2|
■
1
AA
lin'1;
12
lyi n «;
9
yAA
(in3;
108
Ixo
4
dy
5
Ady2
300
bh3
12
12
5
60
36
1
12
24
1
24
8
3
216
2
12
IAA = 48 in.2
XyAA = 192 in"
IAdy-= 528ln?
IL, = 48in:;
Component
AA
lyc
12
35
dx
Adx‘
- EyAA 192 in.3
y = —---- + ---------t- = 4"
A
48 in.2
12
24
288
il « "328inJ
IAdx2 = 0
I, = ZI„ + EAd2' = 48 in.4 = 528 in.4 = 576 in4
I y = £1yc + EAd2* = 328 m 4 + 0 = 328 i n 4
6.8
yt
y=
-------- --------- ■—j— -------,
= 6T
|
;
j
-B
__
II
4*
1
X
\ dx = 4”
.
CL
\
Y
Component
1
\
1...............j .............. ]1
1
2!
•
o
21
•
'SO
1
1
AA
Ixc
dy
Ady2
20
6.67
+6
720
40
333.33
0
0
20
6.67
-6
720
2 lxc = 346.7 in:1
EAdy2= 1440 in?
Ix = ZIXC+ SAd^ = 346.7 +1440 = 1787 in.4
By fo rm u la :
bh; - b , h ? _ (1 0 )(1 4 f-(6 )(1 0 )3
12
12
6.8 c o n t’d
Component
AA
lye
dx
Adx
2 [I
20
166.7
0
0
40
13.3
4
640
20
166.7
0
0
10
Slyc - 346.7 in:5
Iy = E l,, + £ A 2 = 3 4 7 + 640 = 987 i n 4
ZAdx’ = 640 in/1
6.9
-X
dyi =yi-y
dy2 = y-ya
dja = y-ya
Component
AA
yAA
Ixc
dy
Ady
5.25
10.05
55.13
0.98
4.76
119.0
2x16.88
= 33.76
5.63
190.1
356.0
0.11
0.4
5.25
1.75
9.19
0.98
3.99
83.6
y.AA = 44.3 in2
SyAA = 254.4 in.'
Slxc= 357.9 in.'1
y -
^yA _ 2 5 4 M n ^ _ 5 7 4 „
A
44.3 in.2
Ix = £ I „ + EAdJ = 358+ 203 = 561 in.4
ZAdy2 = 203 in.'1
6.10
^ rr .
groove
tongue
r a
Vj
Component
r
K. y ......... ’v 7............
AA
Ixc
432 in.2.. ^ ^ p . = 2 915 in4
O ....(■>...<J-
3x12.56 = 37.7 in2
-3 x(tt)(4)... = -38 in4
64
Zlxc = 2878 in 4
Ix = ZI X0+ ZAdJ
ZAd 2 =0
Ix = ZIXC= 2878 in.4
No transfer is necessary since the solid slab and the
three holes all have their component centroids on the
major entroidal x-axis.
6,11
Component
AA
y
yAA
Ixc
16
14.31+1
=15.31
245
5.33
yi - y
=4.9”
384
24.1
7.16
172.6
882
y-
247
YAA = 40.1 in.')
dy
Ady®
XyAA - 417.6 inf
Xlxc - 887.3 in4
- XyA
417.6 n,
y = — — = ------------£- = 10.4
3
A
40.1 in.2
Ix =X Ixc + XAdy =887 + 631 = 1518 in.4
M y 2 = 631 in '
6.12
Y
Component
AA
x
(in?5
( in .)
3.38
4.49
0
-0.634
LAA=7.87in.J
(in3)
0
-2.85
ZxAA =-?.8;>in."'
- = - 2 8 5 ^ =Q 3611
7.87 m 2
- = 5S0Sia»= 7 O ,
7.87 in.2
d
xAA
= y 1- y = 9.65,,-7 "= 2.65"
dy2 = y - y 2 = 7"-5 "= 2 "
dxl= x = -0.36"
dx2 = x2 - x = 0.634"-0.36"= 0.27"
y
yAA
1tW.22-.57
= 9.63
32.6
iin.)
Jin. )
22.45
ZyAA =55.05in.3
6 .1 2 c o n t’d
AA
Ixc
dy
3.38
1.32
2.65
4.49
6/1
Component
23.7
18
il.: = 68.7 In:
Component
Ady
SAdyJ = 41.7in4
AA
ly
dx
Adx
3.38
32.6
-0.36
0.44
4.49
2.28
-0.27
0.33
_l_
Xlyc =34.9111*
lA d ;- = 0.77 in?
- = -2 8 5 _ m _ l = _ 0 3 6 „
7.87 in.2
I
55.05 in.3
7.87 in.2
= 7.0"
Ix = Z IX0+ S A d2 = 68.7 + 41.7 = 110.4 in.4
Iy = 21^ + ZA d2 = 34.9 + 0.8 = 35.7 in.4
Component
A
It
...
'1
Adx"1
Ixc
dy
Ady
lyc
4.44
6 - 0.93
- 5.07
141.4
4.44
0.93+0.25
-1.18
0
0
0.13
0
0
5.07
141.4
4.44
1.18
7.66
6.0
72
s.s
4. .14
E lxc = 8 0 .9
E A d y 2 = 2 8 2 .8
Ix = ZIXC+ SAdJ = 80.9 + 282.8 = 363.7 in.4
Iy = E I ^ + S A < = 9 .0 + 1 5 .3 = 24.3 i n 4
dx
X ly r - 9 0
7.66
X A d x 2 = 1 5 .3
6.14
Component
AA
Ixo
dy
A dy2
12
0 .5 6
7 .5 + .3 7 5
= 7 .8 8
745
12
0 .5 6
1 4 .7
404
1 4 .7
404
y=
CXa
ilx c
Component
7 .8 8
E A d y '= 1 4 9 0 In:1
8 0 9 In.4
AA
ly
12
256
12
256
745
dx
A dx2
1 4 .7
11
dx
14.7dx‘
-
1 4 .7
11
dx
14.7d*2
E lyc = 5 3 4 ln 4
X A dx2 = 2 9 .4 d x '
6.14 co n t’d
Ix = 8 0 9 + 1490 = 2299m 4
Iy = 5 3 4 + 29.4d^
B u t: Ix = I y
2299 in.4 = 534 i n 4 + 29.4d^
,2 1 / 03
„ .
d = -------= 60.2 in.
29.4
= 7.8"+
d v = 7.8"
0.8"
'from A ISC
T ables
w = 15.8"
- 0 .7 " = 7 .9 '
Chapter 7 Problem Solutions
7.1
Sect. E: x = 0 to x = 5'
X Fy=0:
i),
V = +10k (constant + shear)
XMe = 0: -10k(x) = M; M = 10x
@ x = 0, M = 0; @ x = 5’, M = +50k-ft
Section F: x = 5 'to x = 10’
X Fy= 0: +10k - 10k - V = 0; V = 0 (no shear)
EMf = 0: -10k(x) + 10k(x - 5') + M = 0
M = +50k-ft. (constant for x = 5’ to x = 10’)
Section G: x = 10’ to x = 15’
XFy = 0: +10k - 10k- 10k + V = 0;
V = 1 0 k (constant -shear)
ZM g = 0: -10k(x) + 10k(x - 5 ’) + 10k(x - 1 0’) + M
M = 150k-ft. - 10x
(varies with x; between x = 10’ to x = 15’)
7.1 co n t’d
7.2
Ma = 200k-ft.
(0= 1 k/ft
L^ )
Lo ad
V B = 20k
20’
■A
Section cut C:
(for x = 0 to x = 20)
EFy = 0: -1k/p,(x) + V = 0
V = 1k/p,(x) (1st degree function)
Shear varies as a function of x (linear)
@ x = 0, V = 0;
@ x = 20’, V = 20k (negative shear)
ZM d = 0: 1%(x)(x/2)-M = 0
M = x2/2 (2nd degree function)
@ x = 0; M = 0
7.3
FBD
Solve lor Ihe external reactions al B and D.
XlVb =+(10k)(15')+(20k)(5')-RB = 0
Rb = +25k
EFy = -1 Ok - 20k + 25k + Rd = 0
Rd = +5k
Cut sections E, F and G and write equations
of equilibrium XM = Oand
ZFy = 0 to determine the internal shear and
moment developed at each of the respective
sections.
,.vG
h
Section cut E:
XFy = 0: -1 Ok + Ve = 0;
VE=10k
The shear between A and B remains
constant.
XM e = 0: +10k(x) - Me = 0; Me = 10x
The moment M increases as a function of x,
between A and B; x = 0 to x = 5’.
Section F:
XFy = -1 Ok + 25k - Vf = 0; VF = 15k
The shear remains constant (positive)
between B and C.
XlVt = (10k)(x) - (25k)(x - 5 ’) + IVt = 0
Mf = 15x -1 2 5
The moment varies linearly from x = 5’ to
x = 10’.
Section G:
XFy = -10k + 25k - 20k + Vg = 0;
Vg = 5k
Shear is constant between C aqnd D.
XM g = (10k)(x) - (25k)(x - 5’) - (20k)(x -1 0 ’+
Mg = 0;
Mg = -5x + 75
Moment varies linearly for x = 10’ to x = 15’.
7.3 co nt’d
Load
V
(Shear)
M
(Moment)
7.4
b|
al
00 = 2 k/fl i
20k.
t
m
i
I
a;
10’
cd =
O
n
CD
n
j
ci
I
i
b;
<—
20k J
M
I)
CD
= 2 k/ll
A i i i i n r i B
20k
=
Secction a-a: @ x= 0 to x = 10’
EFy = 0: +20k - oox - V = 0
V = 20k - 2x
(as x increases, V decreases)
IMa-a = 0: -20(x) +(<ox)(x/2) + M = 0
M = 20x - x2
(M increases with x; 2nd degree curve)
2 k/ll I
20k
Id
i
120k
™ —
Af
co = 2 k/ll
a h 1111 riB
co = 2 k/ll
III 1 I I I
i
Section b-b: @ x = 10’ to x = 20’
EFy = 0: +20k - 2(k*t)(1 O’) - V = 0
V=0
(No shear between x = 10' to x = 20’)
EM>b = 0: -20x + 2(k«)(10’)(x - 5’) - M = 0
M = -10Ok-ft.
(assumed direction of M on the FBD is
incorrect; it should be counter-clock)
Section c-c: @ x = 20’ to x = 30’
EFy = 0: +20k - 2(k«)(1 O’) - 2(km)(x - 20’) + V = 0
V = 2x - 40k (V increases with x)
EM = 0: -20(x) + 2(10)(x-5’) + 2(x-20')(x - 20')/2
+ M = 0
M = -x2 + 40x -300
(M is a function of x ....2nd degree)
7.4 co n t’d
“V” Diagram:
@ x= 10’; V = 0 (V = 20k - 2x )
Between x = 10’ to x = 20’; V = 0
Between x = 20’ to x = 30’;
V goes from 0 to -20k
‘M’ Diagram:
@ x = 0 M= 0
(no moment at the hinge)
Betweenx = 0 tox = 10', Mis
increasing.
@ x = 10’, M = 100 k-ft.
(positive bending)
Between x = 20’ to x = 30’ M is
deer easing.
@ x = 30’, M = 0.
7.5
7.6
15k
a)=3k/ft
30k [•-
10’
20’
45k
7.7
4k
co=2k/ft
4 D
B
9.5kJ
Tl 0.5k
t—
*—
*
8’
L
6’
7.8
7.9
(D=600lb/ft
MA=10k-H.
<N
1
....... ...........
(j i A
<N
\ 3k
4
lC
B
5k*
5'
<
5'
3k
»<
5'
f
0 VL
U l
7.12
(D = 6 k/ft
7.12 co n t’d
(
A=1/2(XX2/3X) = *73
6k/ft
X
9'
From sim ilar triangles: °VX = 6k/ft/g, ; (C = 2Xf.
AV = 9k = x2/3; * = 5 -2
zero slope
—A=2/3(5.2 i)(9i)=31 .2
x = 5.2'
E
x = 5.2'
AEBD _ Aacd - Aacbe 81 k-ft - 49.8k-ft = 31,2k-ft
AACD = ( 1/3)(9')(27k) = 81k-ft
Aacbo = (9k)(9‘) = 81 k-ft
AAE0 = (2/3)(5.2')(9k)=31.2k-ft
= AaCBO - a a e o =
81 k-ft - 31,2kft. = 49.8k-ft.
a acbe
-
7.13
co=4k/ft
16k
T
bI
26k
(a)
70k
c
<D=4k/ft
i
FBD's of beam components
mu III
c
JL
16k
16k
(b)
<D=4k/ft
I ......................i D
C ((hinge)
1
L
fJ 26k
I -------------------- .
16k
70k
4'
---------- * ------- J
'M'
7.14
7.15
(o2=3k/ft
f
B
30k
®1 = 2
^ k/»
' ft
MA=400k-ft
B
iA
" ■Ra
30k
SC t
30k
30k
m i ~ 2 k / <t
Mn=400k-ft
Kt
50k
= 50k
FBDs of beam components
Chapter 8 Problem Solutions
W 8 x 1 8 (S x = 1 5 . 2 in3)
M
(1 8 k — ft.) x (12 in ./f t.)
, ,
< 2 2 k si
fb = — = i ------------- ^ — -— - = 14 .2 k / i n
b
Sx
1 5 .2 in .
'
O.K., the beam
is not overstressed.
'O+
8.2
11V
3V
.o +
4x12 S 4S
Sx = 73.8 in .3
ffc
Me _ M _ 4 .2 8 k -f t. x 12 in./ft. _
= 0.696ksi
I
S„
73.8in.;
fb = 696psi < l,300psi
The beam is safe.
/. OK
8.3
P = 6k
w = 3^/ft
1
j- I -
It
L
■ H H
B
,1
36k
C
18k
'
4'
|
8'
W8x35; (Ix = 127 in.4, Sx = 31.2 in.3, d = 8.12", c = 4.06")
^
Ix
OK
= M = (5 4 k -ft.)x (1 2 ,r ^ :) = 2
Sx
31.2in.
0 8 k . < 3 0 k ,.
o +
8.4
M m ax = 1 7 .2 k - f t .
fb= — ;
b
Sx
m,n
Fb
22k/in.
S can th e W 8x s e ctio n s until you find one th a t has an Sxeq u al to
o r g re a te r th a n th e Sfoin. va lu e above.
Use: W 8 x1 3
= 9 .9 1 in3)
8.5
W 8 x1 8 (d = 8 . 1 4 ’ , c = d /2 = 4 .0 7 ” , (, = 6 1 .9 in", q, = 15.2 in?)
^
= M = ( 2 4 2 k - » ) « ( l 2 l„/H .) = i 9 | t / i | | 2
I,
S,
15.2in.
fb = 19.1ksi <Fb = 30ksi
OK
8.6
co = 2 k/ft
5k
M _ (16.67k- ft.)(l2in./ft.) _
Sx
1 l.Sin.3
17ksi < Ft, = 22ksi
b) Timber beam required:
o
req'd
_ ^max
c
(■6.«7t - f , J ( n ln / f , . ) s i K .| | ,
1.6k/in.
Use 8 x 12 S4S (Sx = 165.3 in.3)
•. OK
8.7
M (25.6k —ft.)(l2 in./ft.)
— =-i------------- \ — -— - = 1.9ksi < 2.4ksi
S„
162in.
OK
8.9
oo = 1 k/ft
=
M
Sx
(50k -ft.)(l2 in /ft.)
i
1
= A------------------ A
30k/in
Use: W 14xl8 (Sx =21.1in3)
. ,
_ 20m 3
8.10
W 18x40 (Sx = 68.4 in.3)
M
fb = —
w
r-
„
max = Fb ^ x
M = (22k/in.2)(68.4in3) = 1504.8k -in . = 125.4k - ft
M max= 8 P = 1 2 5 . 4 k - f t .
p _ 1 2 5 . 4 k - f t . _ 5-68k
8.11
- V yA 43in 3
y = - = - — = —- ^ - = 3.07"
A
14in.
|y
c = 5.93"
= 2 !Xc + 2 Ady = 43-2in 4 + VOin.4 = 113.2in.4
N .A .
y = 3.07"
t
Ref.
8.11b
= Mc = (43.75k - f , ) ( l 2 , n / f t.) ( 5 .^ ) = 275
b
L
113.2in.
fh = 27.5kksi < Fh = 30ksi
'
/. OK.
-A = 8 in.2
= 5.93 in.2
J y = 2.97"
N .A .
s
_
n a
J
shear plane
____ £ y = 1.93'
□
t : shear plane
VQ _ (8.75k)( 5.93in.z x 2.97" = 1.36 k/in.2 < Fv = 20ksi
f,
=
(8.75k )(8in. xl.93")
------- 77------- = 1. 19ksi
(ll3.2in.4)(l")
8.12
(d = 0.4 k / ft )
« i§
f
T 6.4k
L = 32'
Log beam
6.4k
Fb = 1200psi; Fv = lOOpsi
V
* max =
coL _ (4001b/ft.)(32')
toL2 _ (4001b/ft.)(32')
8
c = d/2 = R
f
Me
(51,2001b.- ft.)(l2 ia /ft.)R
fb = — = ------------- ^ 7 -------ji R
:64001b.
= 51,2 0 0 1 b .-ft.
8
I = rtD4 _ j i R4
64
4
3
= 12001b./in
Due to the bending requirement; R = 8.67” (say 9”)
_ (5 1 ,2 0 0x 12)(4) _
R -
3 14(1200)
-653
8.12b
„4R_
N.A.
3it
Shear:
VQ
(64001b)(”RX)(4K )
lb
Ri
= 1001b./in.2
( " R% |(2R)
(4K6400)
T2
(3)(3.14)(100)
/. R = 5.2"
Bending governs the design. Use an 18” diameter log.
8.13
co
•
L = 20'
V
v mat
ff>L
oaL
mL
2
2
©L
M max =
C om ponent
A
y
yA
I*
i = g— n ^
10
8.5
85
0.833
2.15
46
9.13
4
36.4
110
2.35
50.3
121.4
ill
— --
2
- ,I > ^ = 1 2 1 .4 .n ;
A
19.13
ft35,
19.13in
= 2 I ^ + E A y = l l l i n -4 + 9 6 3 m 4 = 2 0 7 '3 in 4
A dy2
ye>.3
8.13b
Bending:
jy
Me
fb =
M
I
allow
[22 k / in.2 ^207.3in.4 j
c
633'
toL
8
7 2 0 k - in .
720k - in.
12in./ft.
(0
Shear stress at the flange:
f„ =
VQ
Ib
< Q L = ( l- 2 k /f t.) ( 2 0 ') = i 2 k
2
2
(
A
\
y
(12k) lOin.2 2.15in.
\
/\
/ = 0.156k/in.2 < Fv =14.5ksi
f., =(207.3in.4)(8in.)
Shear stress is not critical.
8.14
C om ponent
A
y
yA
10.5
10.5
110.5
12
6
72
U
3 -5(3)3 = 7.89
12
2M (123)_144
dy
3.77
149.5
0.73
6.4
5.98
188
12
5.25
IS ]
27.75
z
V =
=
A
186 5m 3 = 6 73"
27.75in.2
0.75
3.9
3.5(l.53)
'
' = .99
12
186.5
152.9
344
Ix = 21*. + SAdJ = 152.9 + 344 = 469.9in 4
b
Me
(7,2001b- ft)(l2in./ft.)(673")
Ix
469.9in.4
n
8.14b
Shear stress:
5.98*
Component
ESI
A
y
Q=yA
6.75in.2
3.37"
22.7in.3
5.25in.2
5.98"
£
Q = Ay
VQ
f„ =
lb
31.4in.4
Q=541in.
(1 8 0 0 lb .)(.54. lin .= 196
( 4 9 6 .9 in .4 ) (l" )
lb./ in .2
8.15
co = (lb/ft)
>t__6^_
+3C0
Beam cross section
V m, , = 3 c o
M max = 6 a) + ){ (3 )(3 < o ) = 1 0 .5 U
A = 7 2 i n .2
t
(6)(12?) = 8 6 4 in .
12
c = 6"
I
8 6 4 in .
S. = — = ---------- = 144in.
c
Fb = 1 ,6 0 0 p s i
_ Me _ M
3
6 in.
Fv = 8 5 p s i
M aiiow = F b * s x = ( 1 6 k/ in 2) * ( l 4 4 i n . 3 ) = 2 3 0 k - in. = 1 9 .2 k - ft.
b = ix = sx
1 9 .2 k - f t .
0»,bending “
- 1.83 k / f t .
10.5
VQ
fv = SN A
-y = 3"
_ F vI b _ (-085Xn2)(8 6 4 i,4)(6'')_ io g ^
Q
4 .0 8 k
( 3 6 in .2 x 3 ”l
,
,
0) = ——— = 1.36k ft.
(shear)
3 f t.
Shear governs.
8.16
29
| P
6'
i
6
,
f
6'
a) Vmx = 1.67P
1.67P
4
x 12 S4S (Sx = 73.8 in .3, A = 39.4 in.2)
fb = Sx
M _ = F b x S , = ( l . 6 k / i n . 2) x ( 7 3 . 8 i n . 3)
o +
= 118k —in. = 9.85k —ft.
E q u a tin g t h e t w o m o m e n t e q u a tio n s ,
10P = 9.85k- ft.
P = 0.985k = 9851b.
1.67P
_
c
h
~
1 5V
Yma x
A
(@ N .A.)
FV(A)
A
(851b./in.2)(39.4in.2)
= -i----------- &----------= 2,2301b.
1.5
1.5
E q u a t i n g th e s h e a r e q u a tio n s ;
22301b. = 1.67P ;
P = 13401b.
B e n d i n g g o v e r n s : P = 9 8 5 lb.
b)
V = 1 3 3 P = 1.33(9851b.) = 1 ,3 1 51b.
M = 1 .3 3 P (4 ') = 5 .3 3 P = 5.33( 9851b.) = 5 ,2501b.
© 4'
1 5V
1.5(13 IS lb.)
fv = ------- = — i-------- r ^ = 5 0 . 2 p s i
A
3 9 .4 m .2
fb = M = (52501b . - ft.)(12„./ft.)=854ps.
73 .8 in .
8.17
a>L
M max
=
Mmax
Fb
* s x = (2 2 k/i:n-2) X(79i:n.3) = 173 5k - in.
1 7 3 5 k -in .
: = 1 4 5 k -f t.
12in./ft.
E q u a tin g th e m o m e n t e q u a tio n s ,
V
fiiT2
8
- = 1 4 5 k -ft.
145(8)
« *--
— -2 .9 k /ft.
K)
1/2“ x 10“ steel
cover plale (lop & bottom)
- (2) C10x20
N.A.
Adv'
Component
0.1
157.8
5.25
138
11.8
][
0.1
5.25
138
276
158
I* = 2 'xc+2 Ady = 158 + 276 = 434in 4
c = 5.5"
^ = ^ = 4 3 4 1 ,^ ^
= 7 9 in .‘
, / _ <D L _ ( 2 . 9 k / f t . ) ( 2 0 ') _ 2 ? i:
2
2
N.A.
b = 2 x 2 . 7 4 ” = 5 .4 8 "
Q = A y = ( 5 i n 2) ( 5 .2 5 “) = 2 6 .3 in .3
vo
lb
2
( 4 3 4 in .4) ( 5 .4 8 in .)
8.18
M m,v = 32k - ft. = 384k -in .
M
req'd
(384k-in.)
Fb "
22 k/in.2
1 7 .5 in .
From the Appendix Tables;
Use: W12x19 (Sx= 21.3 in.3)
does not account for the beam’s weight.
orW10x10 (Sx = 18.8 in.3) Same weight
as the W12x19 but shallower in depth.
f
(ave)
_
V
v max
t,„d
14.7k
.
(0.235")(12.16")
fv = 5 . 1 4 k s i <
(aw)
-32k-ft.
1 4 .5 k s i
OK
8.19
Fb = 1200 psi
Fv = 10Opsi
12”
Plank Cross Sectio9n
bh! __(12)h1 = h3
x
12
12
A =12h
Bending :
_ Mc
b _ TAX'
h9
h2
h = 2.2"
Shear:
ijv
A
, „ =a
h. L 2 ! M
.w .
v
1200
Bending controls:
5
12h
h = 2 .2 “
8.20
5k
2Tx4'
rough cut
P (pilch)
t ± ____
"j" ~ ~ ~ ~ - - - T - ~ - T
20 -
1
L= 16'
+2.5K
‘V ’
_(5")(20")3
x
-2.5K
(4")(16")3
12
1965in.4
12
Q = Ay = (8in.2)(9") = 72in.3
20k-ll.
V = 2500#
r
'M '
1
F=
2 shear
planes
y = 9”
f vb p = ( b p ) ^
F = capacity of 2 nails at the flange
representing 2shear surfaces
( 2 x 8 0 # = 160#)
A lternate m ethod:
VO (2 5 0 0 # )(7 2 in 3)
f =— =
/ = 22.8psi
Ib
(l9 6 5 in )(4")
A
ff v
A = 2"x2p = 4p
F
160#
4 p = — = ----------- ;
fv 2 2 . 8 ^ ,
(160#)(l965in.4)
= 1.75"
P ~ V Q ~ (2500#)(72in.3)
U se: 1y4" spacing
p = pitch or spacing
f . = ^ - A =
FI
p = 1.75"
8.21
F b = 22 ksi
E = 29 x 10 ksi
Vmax=10k;
S
Fv = 14.5 ksi
Mmax=48k-ft.
= M = (4 8 k -ftJ (l2 ir^ =
reqd
Fb
22k/in.
Try: W8x31 (Sx = 27.5 in .3,
d = 8", t„ = 0.285", Ix = 110 in.4)
<i)hmL2 (31 lb/ft.)(8')
2
2
adJ
(9921b. - ft.)(l 2 in/ft.)
22k/in.2
San = 0.54in.3
= Sreq.d + S ,^ = 26.2 + 0.54
Stolal = 26.74in.3 < 27.5in.3
/. OK
10k
f = X = __________
<™> t„d (0.285")(8")
fv = 4.38k/in.2 < Fv = 14.5k/ir
(ave)
.-.OK
Deflection:
!!!!!!!!!!!!!!!!!!!!!
* total
0>L4
8 EI
PL3
3EI
(1.031 k/ft.)(8ft.)4 (1728 in.3/f t.3)
(2kX8ft.)3(l728in.3/ft.
(8)(29xl03 k /in 2)( ll0 in 4)
3 (2 9 x l0 3 k/in.2)(l 10in.4)
Absl = 0.286"+0.185"= 0.47"
3
8.22
mpL = 300 lb/ft
(Qll = 400 lb/ft
_ wL2 _ (700lb./ft.)(16')2 _
= 22,4001b.ft.
B
5600 lb.
5600 lb.
L= 16'
^req'd
M _ (22,4001b. - ft)(12in./ft.) _
= 207in.
—
1300 lb./in.2
b.)(16ft.)
~2
"
L5V _ 1.5(5,6001b.)
"*req'd
851b./ in.2
T r y : 8 x 14 S 4 S
( A = 1 0 1 .2 5 m .2 Sx = 2 2 7 .3 i n . 3, Ix = 1 5 3 8 i n . 4,
* N o te :
a) = 0 .2 5 2 x A ( f o r D o u g l a s F i r a n d S o u t h e r n P in e )
_a>bmL_(25.51b./ft.)(16')
_ 1.5Vad, _ 1.5(2041b.) _
Aadd “
c d = 2 5 . 5 lb ./f t.* )
Fv
2041b.
2
~~ 851b./in.2 -
Atotai = Areq d
= 98.8in.2 +3.6in.2 = 102.4in.2 > 101.3in.3
T h e b e a m is o v e r s t r e s s e d i n s h e a r .
T ry : 8 x 1 6 S 4 S
(A = 1 1 6 .2 5 i n . 2, S x = 3 0 0 .3 i n . 3, I., = 2 3 2 7 i n . 4, a = 2 9 .3 l b . / f t . )
-L /
_ ( I 6 f t . ) ( l 2 in ./ft.)
’ / 3 60 _
5toL4
*
actual
(LL)
384El
360
: 0.53"
5(400lb/ft.)(16ft.)4(l728in.Vft.3) _ o
3 8 4 ( l . 6 x l 0 6 lb ./in .2) ( 2 3 2 7 in .4 )
O K f o r d e fle c tio n .
U se: 8 x l 6 S 4 S
= 98.8in/
8.23
B
V m, v = 20001b.
2k
Veq'd '
1 .5 V
1 .5 (2 0 0 0 lb .)
F„
110 lb ./in .2
= 27.3in.
M1max
max = 1 2 ,0001b. —ft.
_ M _ (1 2 ,0001b. — ft.)(l 2 in./ft.) _
Sreq'd _
_
9 2 .9in.
1 5 5 0 lb ./ in.
T r y : 4 x 14 S 4 S
(A = 4 6 .4 i n . 2, S = 1 0 2 .4 i n . 3,
L = 6 7 8 .5 i n . 4, u = 1 2 l b . / f t . )
0>hmL 2
c
(1 2 lb ./f t.) ( 1 6 f t.) 2
= 3 8 4 1 b .-ft.
. 3
(3 8 4 1 b . - f t. ) ( l 2 i n . / f t . )
_—
^add
—
----
1 5 5 0 lb ./ i n .2
Stotal —Sreq,d +
~~
—92.9in. + 3in.
,3 v i A i a:„ 3
allow
L
(1 6 ')(1 2 in ./ft.)
PL
240
240
2 0 .1 H
rr
( l k ) ( l 6 ') 3 (1 7 2 8 )
p
2 0 . l ( l . 6 x l 0 3) ( 6 7 8 .5 )
PL
5a>L
48EI
384El
- + ------ + -
.
^eam
M H I U U U I U H
rt
( l k ) ( l 6 ')3 (1 7 2 8 )
5(. 0 1 2 )(1 6 ')4 (1 7 2 8 )
4 8 ( l . 6 x 103) ( 6 7 8 .5 )
3 8 4 ( l . 6 x 1 0 3 )(6 7 8 .5 )
A a c t= 0 .3 2 4 " + 0 .1 3 6 " + 0 .0 1 6 " = 0 .4 8 " < 0 .8 0 "
OK
C h e c k th e b e a r in g s tre s s .
1 2 0 001b. + 961b. |
j. _
P
_ \ ___________ (bm.wt)/
P “ A b rg “
- = 1 0 9 p s i < Fo l = 4 1 0 p s i
(5 .5 " x 3 .5 " )
OK
U se : 4 x 1 4 S4S
t
8.24
4 “ concrete slab
jL ._______
CO = 9 9 lb /f t2 x f = 7 9 2 lb /ft
n m n n in in n n u j
0 7 1 2 lb
B
I 8 7 1 2 lb
I
^ M e la l
L o ad s:
Section A-A
ft .j * (1 5 0 l b / f t . 3) = 50 lb ./ f t.2
C one. = ( %
t2
M e ta l d e c k = 4 lb ./f t.2
. 2
P l a s t e r c e ili n g = 5 lb ./ft.
59 p s f
D ead L oad =
o>DL = (5 9 lb ./f t 2)(8') = 4 7 2 lb./ft.
L iv e L o a d = 4 0 p s f
o>LL = (40 lb ./ f t.2)(8') = 320 l b / f t.
99 p s f
D L + LL =
to = 99 lb ./ f t.2 x 8'= 7 9 2 lb ./ft.
M
. , = ^ = ( , , ; ' b / f ‘ -X22')
8
rc,d
8
Fb
= 48,0001b— ft.
22k/in.
Try: W14 x 22 (Sx = 29 in.3, A = 6.49 in.2 Ix = 199 in.4)
Deflection check:
_
^
B eam B-1
decking
L
_ (2 2 ') ( 1 2 in ./f t.) _
360
360
r4
384EI
384^29 x 103)( 199)
8.24b
Beam
SB1:
Mmax= 95.6k-ft.
(95.6x12)
^req'd “
22
= 52. lin."
Try: W 16 x 36
<SX = 56.5in.3, A =10.6in.2, Ix = 448in.4)
(D+L)
L
(24')(12 in/ft.)
240
240
5o>L
384 El
_ 5(.336)(24')4(1728) + (8.95)(24')3(1728)
ao> 384(29 x 103)(448) 28.2(29xl03j(448)
A„, = 0 . 193"+0.584" = 0.78" < 1.20"
(12.55k)
(aF;e) twd (0.293')(15.86")
OK in shear
U se: W16x36 for SB1
.-. OK
= 2.7ksi < F„ = 14.5ksi
PL3
28.2EI
1.2 "
Chapter_9_Problem Solutions
9.1
W8x31
Iv
= 37. lin .4
/min.
L = 20'
A = 9.13 in 2
j i 2E I
3 - 1 4 i( 2 9 x 103 k /in .2)( 3 7 .lin .4)
L2
(2 0 '* 1 2 in ./ft.)2
P°r
^
cr
184.2k
= i 8 4 2 k = 2 0 2 k s.
A
9 . 13in.
9.2
2-31/2” <t>standard pipe
Iy = 2 x l = 2(4.79in.4) = 9.58in4
(min)
L = 24'= 288"
°r
n 2EI
(3.14)2(2 9 x l0 3)(9.58in4)
L2
(288)2
33k
9.3
Pcr = 250k
W10x54;
Imln = 103 in .4
P., =
l 2 « 2EI
3 . 1 4 2 ( 2 9 x 103 k / i n . 2 ) ( l 0 3 i n 4 )
L = ?
250k
L = 343" =28.6'
9.4
8" diameter pole
A = ------ = ——— = 50.2"
4
4
; _ nD4 _ n(8")4 _
64
64
2 0 lin.
P„ = 25 k
P,,=
« 2EI
( k l )2 = —
(KL)2
, 3.14 (lxlO k/in. )(201in. )
^ ,
(KL)2 = ------- i--------- ------^ ---------L = 7.93 x 104 in.2
v ’
25k
KL = 282" ;
'IR'l"
L = ------ = 403”= 33.6'
0.7
8“x6"x%" rectangular tube.
(A = 9 .5 8 in .2;
ry = 2 .3 6 ” ;
KL
( l ) ( 3 8 'x l 2 i n ./ f t .)
ry
2 .3 6 '
I y = 5 3 .5 in .4 )
^
;
j t 2E I y
3 .1 4 2( 2 9 x l 0 3 k / i n . 2) (5 3 .5 in .4)
(K L ) 2
(4 5 6 " )2
P = ^ 6 4 | = 77ks.
A
9 .58in.
73.64k
9.6
W8x28 (W200x42)
(A = 8.25in.2; Iy =21.7in.4; rx =3.45"; ry = 1.62")
W eak Axis:
Le = 16'
Le _ 16'xl2in./ft. _ 119S
ry
1.62"
Strong Axis:
L = Le = 26'
Le _ 26'xl 2 in./ft.
3.45'
Weak axis buckling
= 9 0 .4
The weak axis governs.
n 2EIy
3 .1 4 2 ( 2 9 x 10 3
k/in.2)(21.7in.4)
= 168k
?Cr
L2
(16'xl 2 in/ft.)2
for =^ = J ^
= 20.4ksi
or A
8 .2 5 in .
CD
(N
II
Ifc
Strong axis buckling
W12x65 (A = 19.lin.2; ry = 3.02")
Case a):
KL = (0.65)(18'xl2in/ft.)
ry
3.02in.
Fa = 18.66 ksi
Pa =F a x A = (l8 .6 6 k /in 2)x(l9.1in.2) = 356k
Case b):
KL _ (0.8)(18'xl2in/ft.)
ry
3.02"
g?
Fa =17.69 ksi
Pa = (l7.69k/in.2)x (l9 . lin 2) = 338k
Case c):
KL
ry
(l)(18fxl2in/ft.) _ ?15
3.02”
Fa = 16.28 ksi
Pa = (16.28 k/in.2) x(l9. lin.2) = 31 lk
9.8
Pa
WM M.
Vci
\
\
1
K = 1.0
J
I ye
2 - C 1 2 x 2 0 .7 ( A 3 6 s t e e l )
o
<N
II
_l
d x = b f - x = 2 .9 4 " - 0 .7 0 " = 2 .2 4 "
X
1
i
[M r-
A
Component
A
li—'
4
A d/
6.09
129
3.88
2.24
30.6
6.09
129
3.88
2.24
30.6
1Z18 in .12
258 in.4
7.76 in.4
Iy = S I^ +S Adx = 7.8 + 61.2=69i„4
ry =
K L _ (l)(20'xl2in./ft.)_ ioog
ry
2.38”
Fa =12.88 ksi
=t ]
1
61.2 in.4
i x = 2 I- = 258in-4
P a = F a x A = ( l 2 . 8 8 k / i n . 2) x ( l 2 . 2 i n . 2) = 157 k
=
=2.38''
9.9
L5x3>£x>£.
(A = 4 in.2, rz = 0.755")
Since trasses are assum ed to be pin connected,
it is reasonable to assume K = 1.0.
KL
rz
(l)(7'x!2 in/ft.)
0.755"
Fa = 11.54-ksi
Pa =Fa xA = (ll.5 4 k /in .2)x (4in.2) = 46k
9.10
P = 60k
5" $ Std. wt.pipe
A = 4.3in.2; r = 1.88"; P = 60k
O
F, = — = 60k =13.95ksi
A 4.3 in.
II
KT
— =92.2
r
(from Table 10.1)
L = — x 92.2 =
x 92.2 = 216.7"= 18'
K
0.80
9.11
Weak Axis:
Ptotal = 500k
12' !
\
kl
= 14.
JL
W 12x 106; A = 31.2in.2; rx = 5.47"; r = 3.11*
Bracing
(2nd fir beams)
t
i
I
KL
14'x 12 in/ft.
ry ”
3.11”
From the Table 10.1; Fa = 17.99ksi
yM
/ M .
Pa = 17.99 k/in.2 x31.2in.2 = 561k > 500k
.-.OK
Weak axis
Ptotal - 500k
Strong Axis:
Assumes that the 2nd floor beam loads are applied
at the top of the column. This is a bit conservative.
KL=26'
K L _(l)(2 6 'x l2 in ./ft.)_
= 57
"77“
5AT
From the Table 10.1;
Fa = 17.71ksi
Pa = 17.71 k/in.2 x31.2in.2 =553k > 500k
Strong axis
OK
9.12
L = 20'
K L= 20'
P = 30k
Try a 5”(|) pipe: A = 4.3 in .2 ; r = 1.88”
KL
(1)(20'x 1 2 ^ ) _ 1277
r
1.88"
Enter Table C-36;
^allow = 9.17 ksi
Pallow = ^ allow X A = 9.17 x 4.3 in .2 = 39.4k
Pa How > ^actual
9.13
P = 30k
L = 20’; KL = 20'; P = 30k
Try: W8 x 18 (A = 5.26 in.2, ry = 1.23”)
- = 195
\
1.23"
\
I KL = 20'
! Fy = 36 ksi
Enter Table C-36;
Faiow = 3.93 ksi
Paiow = Fao„ x A = 3.93 ksi x 5.26 in.2 = 20.7k
Paiow< P = 30k;
Inadequate design
Try W8 x 24 (A= 7.06in.2; ry = 1.61”)
KL _ (20x12)"
r
~
1.61"
F*,™ = 6.73 ksi
Use: W8x24
-149
9.14
p
P = 350k
T r y a W 1 4 x 7 4 (A = 21.8 in .2; r x = 6.0 4 "; ry = 2 .4 8 ")
m m
W e a k A xis:
L = 12'; K L = 12'
KL
(l)(1 2 'x l2 in ./ft.)
12' !
2.4 8 “
KL= 12'
-Bracing
( 2 n d fir
b eam s)
= 58
S tro n g A xis:
L = 2 4 '; K L = 24'
KL
( l) ( 2 4 'x l2 in ./f t.)
rx
6.04"
T h e w e a k a x is g o v e r n s th e d e s ig n .
KL
E n te r th e s le n d e r n e s s r a t io ta b le w i t h — = 58.
Weak axis
ry
Fa = 17.62ksi
Pa = F a x A = 17.62 k / i n .2 x 21 .8 in .2 = 3 8 4 k
P = 350k
Pa = 3 8 4 k > P * , = 3 5 0 k
.-. O K
P
E fficien c y : — i£!- x l 0 0 % = ----------x 100% = 91%
384k
T r y f o r a m o r e e ff ic ie n t c o l u m n s e c t io n
W 14x68: (A = 20 .0 i n .2; rx = 6 .0 1 "; r y = 2 .4 6 ")
W e a k A xis:
K L _ 1Z . KL
'
(1X12' x 1 2 in /f t.)
rv
58.5
2.4 6 “
S tr o n g A xis:
Strong axis
K L - 2 4 ''
KL
rx
P X ^ ' x 1 2 in / f t.) 47.9
6 .0 1 1
T h e w e a k a x is g o v e rn s .
TCT
— = 58.5; Fa = 1 7 .57 k s i
P a = Fa x A = 17.57 k / i n .2 x 2 0.0in .2 = 3 5 1.4k
Pa = 3 5 1.4k > P ^ = 3 5 0 k
P
E fficien c y :
.-. O K
3501c
x 100% = ^ ^
x 100% = 99.6%
Pa
3 5 1 .4 k
Use: W14x68
9.15
P = 397.5k
R oof L oad: D L = 8 0 p sf
JL
LL = 4 0 p s f
D L + LL = 120psf
x
5 0 0 f t .2 = 6 0 k
x
5 0 0 f t .2 = H Z 5 k
F lo o r L o a d : D L = 100 p s f
L L = 125 p s f
D L + LL = 2 2 5 p sf
T h e t h i r d f l o o r c o l u m n s u p p o r t s t h e 4 lh, 5 lh, a n d 6 lh f l o o r s p l u s t h e
r o o f . T h e c o l u m n l o a d is t h e r e f o r e : P = 3
x
1 1 2 .5 k + 6 0 k = 3 9 7 .5 k
T r y : W 1 2 x 7 9 (K L = 1 6 '; A = 2 3 .2 i n . 2; r y = 3 .0 5 " )
w/w/A
3rd Floor Column
KL
( 1 6 'x l 2 in ./ft.)
------ —-------------------------— 6 3 ;
r„
F , = 1 7 .1 4 k s i
3.OS'
P a = F a x A = ( l 7 . 1 4 k / i n 2 ) x ( 2 3 . 2 i n . 2) = 3 9 7 .6 k
P = 622.5k
JL
P a = 3 9 7 .6 k
>
P a o t= 3 9 7 .5 k
OK
T h e g r o u n d f lo o r s u p p o r t s a n a d d i t i o n a l t w o f lo o r s o f lo a d .
wmm,
P = 5 x 1 1 2 .5 k + 6 0 k = 6 2 2 .5 k
K L = 20'
U se a W 12x s e c tio n f o r a b e t t e r tr a n s i tio n to th e W 12x79 a b o v e .
T r y : W 1 2 x l 3 6 ( A = 3 9 .9 i n . 2; r y = 3 .1 6 " )
K L _ ( 2 0 'x l 2 i n . / f t . ) _ 7 6
F , = 1 5 .7 9 k s i
fy “
Pa =
P,
1st Floor Column
3 .1 6 ”
“
;
F a x A = 1 5 .7 9 k / i n .2 x 3 9 .9 in .2 = 6 3 0 k
= 630k
>
P„„f
= 6 2 2 .5 k
OK
6 x 6 S4S S o u th e r n P in e
A = 3 0 .2 5 i n . ^ E = 1 ,6 0 0 k s i; F c = 9 7 5 p s i
L
(I4 'x l2 in/ft.)
^ = i --------------- -— - = 3 0 .5
d
5 .5 '
6x 6 S4S
So. Pine
(Dense No. 1)
/
03E
—
0 .3 (l.6 xl0 6 lb ./in2)
_ — -------------- 2---------
516psi
(30.5)‘
(y
F0* = F 0C d = ( 9 7 5 l b . / i n . 2) ( l.2 5 ) = 1 2 1 9 p s i
I g = 5 1 6 Psi = 0 .4 2 3
Fc
1219psi
E n t e r A p p e n d i x T a b le 14.
C p = 0 .3 7 7
F ^ = F ’ x C p = 1 2 1 9 p s ix 0 .3 7 7 = 4 6 0 p s i
P a = Fc x A = ( 4 6 0 lb ./ in .2) x ( 3 0 . 2 5 i n 2 ) = 1 3 ,9001b.
9.17
8 x 8 S 4 S D o u g la s F ir
( A = 5 6 .2 5 i n . 2; E = 1 .6 x 1 0 6 p s i ; F,. = 1 0 0 0 p s i )
KL = 13.5'
le _ 1 3 .5 x 1 2 i n / f t .
F ’ = F cC
d
= ( I 0 0 0 p s i) ( 1 .0 ) = lOOOpsi
FqE _ 1 0 2 9 p s i _ l Q3
F*
lOOOpsi
F r o m A p p e n d i x T a b l e 14;
8x8 S4S
Douglas Fir
(No. 1)
C p = 0 .7 0 1
F^ = F * C p = (1 0 0 0 p s i)( 0 .7 0 1 ) = 7 0 1 p s i
P , = F j x A = ( 7 0 lp s i) x (5 6 .2 5 in .2) = 3 9 , 4001b.
P a = 3 9 .4 k
>
32k;
OK
U s i n g A p p e n d i x T a b l e 12, i n t e r p o l a t e b e t w e e n 1 3 ' a n d 1 4 ';
P a = 3 9 .4 k
9.18
G lu -L a m C o lu m n :
6 % " x l0 ^ "
( A = 7 0 .8 8 i n . 2; E = 1 .8 x 1 0 6 p s i ; F,. = 1 6 5 0 p s i )
i . •
W e a k A x is :
KL= 11'
t. i
1<l l 'x l 2 i n . / f t .
-S- = --------------- — = 1 9 .5 5
d
6 .7 S '
A .
L
2 2 'x 1 2 in ./f t.
,,
S t r o n g A x i s : — = -----------------— = 2 5 .1 5
&
d
1 0 .5 '
Bracing
The
@ mid-height
s tr o n g a x is g o v e r n s t h e d e s ig n .
,
0 .4 1 8 E
, ,,,
0 .4 1 8 (1 .8 x 106 lb ./in .2 )
= 119 0psi
&
F* = F cC
d
(25.15)"
= ( 1 6 5 0 p s i)(1 .0 0 ) = 1 6 5 0 p s i
F sE_= 1 1 9 0 p s i = 0 7 2
F*
16 5 0 p si
F r o m A p p e n d i x T a b le 14;
C p = 0 .6 1 9
F^ = F * C p = (1 6 5 0 p s i)(0 .6 1 9 ) = 1 0 2 1 p si
P , = F t x A = (1 0 2 lp s i) x (7 0 .8 8 in .2) = 7 2 , 370 1 b . = 7 2 . 4 k
9.19
H e m -fir: E = 1400ksi;
Fo=1050psi; A = 5.25in.2
Le _ (7.5')(12^,)
Fj = FcC d = (l050psi)(l .0) = 1050psi
C =0.504
F'c = F*Cp = (I050psi)(0.504) = 529.2psi
Pai,ow =Fc x A = (529.2)(5.25) = 2778# (per stud, every 16")
co = 2778# x % = 2084%
Bearing stress:
Foi = 405psi
Panow = Fcl x A = (405psi)(5.25in2) = 2126#
p aiiow <
(b e an n g )
Pahow
(com pression)
Bearing stress governs
9.20
4x8 S4S D ouglas-Fir
(A = 25.38m 2, E = 1600ksi, Fc = lOOOpsi)
Weak Axis:
Le =K L =0.8(8')(l2'"/t)= 76.8”
Slenderness ratio =
Strong Axis:
L /
76 8"
yA = — 1— = 21.9
/d
3.5"
Le = K L = 1.0(10')(12"/t) = 120"
Slenderness ratio =
L /
120"
yA = -------= 16.6
/d
725"
Weak axis governs.
0 3E
0 .3 (l.6 x l0 6)
FcE = 7 T ^ r =
= 1OOlpsi
(21.9)
L.
vdy
Fc' = FcC d = (lOOOpsi) (1.0) = lOOOpsi
F ^ _ 1 0 ° 1p si _ i qq
Fc
lOOOpsi
Enter Table 9 - E d
Cp =0.691
F^ = Fc* x Cp = (I000psi)(0.69l) = 691psi
Paiiow = Fo x A = 69 lpsi x 25.38m 2 = 17,540#
A
^
pa«°w = 17,540# _ 351 ft 2
DL + LL
50j5(t2
9.21
Southern Pine:
E = 1600ksi, F0=975psi
Try: 6x6 S4S (A = 30.25in.2)
L. / _ (I6')(l2'n/ft)
/d
5.5"
0.3E
F cE =
~ T ~
0.3(l.6x106psi)
■= 394psi
(34.9 f
Fc* = FcC d = (975psi)(l .0) = 975psi
cE
394psi
975psi
f:
= 0.404
Cp = 0.363
F; = Fc* x Cp = (975psi)(0.363) = 354psi
Pallow = F > A = (354psi)(30.25in2) = 10.7k < 25k
Insufficient capacity in the 6x6
Try: 8x8 S4S (A = 56.25in.2)
_ (16Q(12%)
U
d
FcE =
7.5"
0.3E
1 cE _
732psi
Fc
975psi
= 25.6
0.3(1.6xl06)
— i-------t—- = 732psi
(25.6)
= 0.75
Cp =0.585
F,; = Fj x Cp = (975psi)(0.585) = 570psi
Paiiow = F' x A = (0.570ksi)(56.25in2) = 32k > 25k
.-.OK
9.22
G lu -L a m :
L /
6% "x______ ; E = 1.8xl06psi;
(I8')(l2>»/t )
-= 32
0.418E
I ^
0.418(1.8xl06psi)
r —
^
.
—
Fc* = FcC d = (1650psi)(1.0) = 1650psi
FqE _ 7 3 5 P S1
Fc*
-=0.45
1650psi
Cp = 0.420
F; = Fc* x Cp = (1650psi)(0.420) = 693psi
P 15,000#
2
= — = — -------- = 21.6m 2
^
Fc
693psi
Use: 6X "x7X " G lu-Lam ;
A = 50.63m2
F:c =1650psi
Chapter 10 Problem Solutions
10.1
Shear: (double) Pv =10.4%oltx 2 bolts = 20.8k
(Table 10 1)
Bearing: Thickness = %"
Ap = ^ ,,x K ”= 0.469in2
Fp = 1.2FU= 1.2(58ksi) = 69.6ksi
Pp = 2 bolts x (0.469m 2j x (69.6ksi) = 65.2k
Net Tension: d = % "+ & "= % "= 0.688"
Anet = X" (4"-0.688") = 2.48m 2
Ft = 0.5FU= 0.5(58ksi) = 29ksi
Pt = 29% i2x2.48in.2 =71.9k
Plate tension:
Agross =4"x% "=3.Qin.2; Ft = 0.6Fy = 22ksi
Pplate= 2 2 / n2x 3 .0 m 2 =66k
Shear governs the design; Pall01u = 20.8k
10.2
P = 28k;
A 3 2 5 -X bolts in double shear
a) Based on shear : 14%olt requirement
A 3 2 5 -X bolts (Pv = 18.4%olt) (Table 10.1)
Capacity of 2-%"<|> bolts in bearing:
Pp = 2 x {%" x%")(69.6 X .0 = 32.6k;
b) Based on net tension:
.-. OK
d = % “+X6U= 0.688"
Anet = t( W - d ) = % " (W - 0.688")
Ptrei = Ft x Anet = 29X„, x X ”(W -0.688") = 28k
W = 3.26"
Agr0ss = t x W = X" xW ;
Ft
=0.6Fy = 22ksi
Pgross = Ft x Agross = 2 2 ) / 2 x (%" xW) = 28k
W = 3.4" *- this condition governs
10.3
Group A bolts:
3-%"<|> A 3 2 5 -X in double shear
Shear : P = 3 x 2 6 .5 %o,t = 79.5k
(Table 10 1)
v
Bearing: P = 3 x ( 2 6 . 1 ^ ) = 78.3k
(l=X')
(Table 102)
Net tension: Ft = 0.5FU= 0.5(58ksi) = 29ksi
A net = (2 plates x y 6" ) U - 2 x % , = 3.98m.2
V
!d+Xs'i J
Pt = Ft x Anet = 2 9 ) / 2 x 3.98in2 = 115.5k
Group B bolts:
2-%"<|> A 3 2 5 -X in double shear
Shear : PL=2x36.1% olt = 72.2k
(Table 10 1)
v
Bearing: P = 2 x 30.5%olt = 61k
(Table 10.2)
( t= # )
Net Tension: Ft =29ksi; d + X6- = % "
Pt = 2 9 ) / 2 x (y2“) x (3■-% ») = 29.9k « - governs
Tension capacity of the y x3" bar:
Pt = Ft x A = 2 2 ) / 2 x (%" x3") = 33k
(0.6Fy)
Net tension in the yi "x3' bar is critical;
10.4
Each member will be checked for shear and bearing only.
Member a : P = 105k - 63k = 42k
% "<|>A325-X(NSL)-A36 steel
49 k
Shear: n = ---------- = 1 .6 - 2 bolts
(double)
2 6 . 5 %
bolt
42k
Bearing: n = ---------- = 2 .1 4 - 3 bolts * - governs
19-6% olt
Member b : P = 26k
26k
Shear: n = ------------ = 1 .9 5 - 2 bolts
(angle)
13.3%olt
-
26k
B earing: n = = 2.65 - 3 bolts *—governs
(t=K«-)
9 8 Yboh
Member c : P = 26k
Shear: n =
96k
= 1 .9 5 - 2 bolts
1 3 .3 % olt
26k
Bearing: n = ----------= 1 .9 8 - 2 bolts * - governs
(t=X")
13. l^ o it
Member d : P = 42k
42k
Shear: n = ---------- = 3 . 1 6 - 4 bolts
(angle)
13.3%olt
42k
Bearing: n = ----------= 3 .2 1 - 4 bolts
W )
13.1%olt
governs
10.5
4 - %" <|>A325 - SC(STD) w / A 36 steel
Shear: Pv = 15%.^ x 4bolts= 60k «—governs
(double)
( T ab,e 10 j)
Bearing: P =26.1%oltx 4 bolts = 104.4k
( t - X ')
(T able 10 2)
Net tension: Anet = (8 "-2 x % ")(X ") =3.19in.2
Ft =0.5FU= 29ksi
Pt = Ft x A net = ( 2 9 X ») x (3.19in.2) = 89.9k
10.6
1-1%"<|» A490X
S hear: Pv = 119k
(double)
(Table 10.1)
Bearing: P = F xA
(t=i-)
Fp = 1.2F„ =1.2(58ksi) =69.6ksi
Ap = (2 x
= 1.375in.2
Pp = 6 9 . 6/ i n/ . 2 x 1,375m2 = 95.7k
Net tension:
Hole diameter = 1%"+X6"= ljfe
Anet = (2 x X ")(5 X "-lX 6") = 4.06in.2
Pt = Ft x Anet = (2 9 )/ 0 x (4 06in.2) = 117.8k
Bearing governs the design.
= 95.7k
10.7
Beam reaction= 210k
5 -A 4 9 0 X bolts
Using Table 10.3;
5 -%"<(> A490X bolts carry 242k in shear.
Clip angle thickness is %"
Angle length = 1 4 / / (it fits within the beam flanges)
10.8
%"<|> A 3 2 5 -S C @3" o.c.
a) Maximum clearance:
= 21.62" -2(0.93") - 2(1") = 17.76"
From Table 10.3;
using L=17J4"
n = 6 bolts
Pv =90. Ik
A 3 2 5 -S C
10.9
Plate capacity: PPL = F t x A = (22% 2)x(6"x%„) = 49.5k
Minimum weld size = /3 6"
Maximum weld size = Xe"
T total = 1 2 "
495k
= 4 13 /
12"
Use: Xe" weld (s = 4 .6 4 % ) Table 10.4
10.10
PpL= 2 2 X ,2x (5 x X 6") = 34.88k
Minimum weld size = Xe"
(s = 2.78% )
Maximum weld size = %"
(s = 3.71% )
L mn = 5" (distance between longitudinal welds)
Total minimum weld length = 15"
'I A OOU
Required strength:
s = — —— = 2.29%
U se: YVj. weld with L = 5"
10.11
Using the maximum weld size: t - Y \/= Y a '-Y ^ '= Y n"
Pallow
= L x 3.71 & =3.71L
(based on w eld)
(T able 10.4)
Plate capacity: PPL = FT x A = 22% 2 x(% "x3") = 20.6k
Equating: 3.71L = 20.6k
L = 5.6"
10.12
A = 3.59 in.2
! 4” !
Fillet weld:
X" weld = 3.71%
Total weld length = 4x4"= 16"
Weld capacity = 3.71%, x 16"= 59.4k
Fillet Weld
Full penetration groove weld:
Capacity is equal to the tensile capacity
of the square tube.
.-. Pt = Ft x A = 2 2 % 2 x3.59in.2 = 79k
l ~
i
//
Groove Weld
//
10.13
Y
____
Ri
,
Re
a = 3.34"
0.84”
P
O
*
r2
[ZFx =0] R1+ R2 + R e = P
P = F, x A = 22% 2 x 4.0in.2 = 88k
[2M 0 = 0] ^(3.34") + Re(0.84") - R 2(l .66") = 0
Minimum weld: Xs"
Maximum weld: 1/ 6"
Try
weld: s = 2.78%
Re = s x L t = 2.78% x 5" = 13.9k
Returning to the moment equilibrium equation;
^(3 .3 4 ) + (13.9k)(0.84") = R2(l .66")
R, =
2
R, (3.34)+11.68
----------= 2.01R, + 7.04
1.66
1
Substituting into the 2F, = 0 equation;
Rj + (2.01^ + I M ) +13.9 = 88k
Rj = 22.28k
R2 = 51.82k
R = 13.9k
Rj = s x Lj
L
2228k =8.01"
2.78 %
T
R, = s x L ,
T
51.82k1 0 ,.„
L, = --------- = 18.64"
2.78%
X
b - y = 1.66”
—6L—} —1