500
SOLUTION
1
FEM
2
BC =
1
FEM
2
CD =
–21621152
9
=–20 k
#
f
t
1FEM2CB =1FEM2DC =20 k #ft
M
N=3E
aI
Lb1
uN–c
2
+
1
FEM
2N
M
BA =
3EI
12
1
uB
2
M
N=2E
aI
Lb1
2uN+uF–3c
2
+
1
FEM
2
N
M
BC =
2EI
15
1
2uB+uC
2
–2
0
M
CB =
2EI
15
1
2uC+uB
2
+2
0
M
CD =
2EI
15
1
2uC+uD
2
–2
0
M
DC =
2EI
15
1
2uD+uC
2
+2
0
M
N=3E
aI
Lb1
uN–c
2
+
1
FEM
2N
M
DE =
3EI
12
1
uD
2
Equilibrium:
*10–16. Determine the moment at each joint of the gable
frame. The roof load is transmitted to each of the purlins over
simply supported sections of the roof decking. Assume the
supports at A and E are pins and the joints are fixed con-
nected. EI is constant.
12 ft
5 ft
5 ft
5 ft
5 ft
5 ft
5 ft
1200 lb>ft
A
B
C
D
E
12 ft 12 ft
1200 lb>
ft
502
SOLUTION
1
FEM
2
BA =0
1
FEM
2
BC =
–31821202
16
=–30 k
#
ft
1
FEM
2
BD =
1
FEM
2
DB =
0
M
N=3E
aI
Lb1
uN–c
2
+
1
FEM
2N
M
BA =3E
aI
15 b1
uB–0
2
+
0
MBA =0.2EIuB
(1)
M
BC =3E
aI
20 b1
uB–0
2
–3
0
MBC =0.15EIuB–30
(2)
M
N=2E
aI
Lb1
2uN+uF–3c
2
+
1
FEM
2
N
M
BD =2E
aI
12 b1
2uB+0–0
2
+
0
MBD =0.3333EIuB
(3)
M
DB =2E
aI
12 b1
2
1
0
2
+uB–0
2
+
0
MDB =0.1667EIuB
(4)
Equilibrium.
MBA +MBC +MBD =0
(5)
Solving Eqs. 1–5,
u
B=
43.90
EI
MBA =8.78 k #ft
Ans.
MBC =–23.4 k #ft
Ans.
MBD =14.6 k #ft
Ans.
MDB =7.32 k #ft
Ans. Ans.
MBA =8.78 k #ft
MBC =–23.4 k #ft
MBD =14.6 k #ft
MDB =7.32 k #ft
10–17. Determine the moments at B and D, then draw the
moment diagram. Assume A and C are pinned and B and D
are fixed connected. EI is constant. 10 ft 10 ft
12 ft
ABC
D
8 k
15 ft
504
SOLUTION
Fixed End Moments. Referring to the table on the inside back
12 b
12 b
15 b1
5b
10–19. Determine the moment that each member exerts on
the joint at B, then draw the moment diagram for each mem-
ber of the frame. Assume the support at A is fixed and C is a
pin. EI is constant.
A
B
C
6 ft
15 ft
2 k>ft
6 ft
10
k
505
Equilibrium. At joint B,
MBA +MBC =0
aEI
3b
uB+15 +
aEI
5b
uB–56.25 =
0
uB=
77.34375
EI
Substitute this result into Eqs. (1) to (3).
MAB =–2.109 k #ft =–2.11 k #ft
MBA =40.78 k #ft =40.8 k #ft
Ans.
MBC =–40.78 k #ft =–40.8 k #ft
Ans.
The negative signs indicate that MAB and MBC have counter–
clockwise rotational sense. Using these results, the shear at
both ends of member AB and BC are computed and shown in
Figs. a and b, respectively. Subsequently, the shear and moment
diagrams can be plotted, Figs. c and d, respectively.
Ans.
MBA =40.8
k#ft
MBC =–40.8
k#ft
10–19. (Continued)
506
SOLUTION
F
EMCD =–
PL
8
=–375;
FEMDC =
PL
8
=37
5
cAB =cBC =cCD =cDE =cEF =0
Applying Eqs. 11–8 and 11–10,
MBA =
3EI
4
1
uB–0
2
+
0
MBC =
2EI
2.5
1
2uB+uC–0
2
+
0
MCB =
2EI
2.5
1
2uC+uB–0
2
+
0
MCD =
2EI
2
1
2uC+uD–0
2
–37
5
MDC =
2EI
2
1
2uD+uC–0
2
+37
5
MDE =
2EI
2.5
1
2uD+uE–0
2
+
0
MED =
2EI
2.5
1
2uE+uD–0
2
+
0
MEF =
3EI
4
1
uE–0
2
+
0
Moment equilibrium at B, C, and D, E:
MBA +MBC =0
3EI
4
1
uB
2
+
2EI
2.5
1
2uB+uC
2
=
0
uC=–2.9375uB
(1)
MCB +MCD =0
2EI
2.5
1
2uC+uB
2
+
2EI
2
1
2uC+uD
2
–375 =
0
3
.6uC+0.8uB+uD=
375
EI
(2)
*10–20. The frame at the rear of the truck is made by welding
pipe segments together. If the applied load is 1500 lb,
determine the moments at the fixed joints B, C, D, and E.
Assume the supports at A and F are pinned. EI is constant. DC
B
A
E
1500 lb
1.5 ft 1 ft 1 ft 1.5 ft
2 ft
4 ft
F
507
Ans.
MBA =–41.1
lb #ft
MEF =41.1
lb #ft
MBC =41.1
lb #ft
MED =–41.1
lb #ft
MCB =214
lb #ft
MDE =–214 lb #ft
MCD =–214 lb #ft
MDC =214 lb #ft
MDC +MDE =0
2EI
2
1
2uD+uC
2
+
2EI
2.5
1
2uD+uE
2
+375 =
0
3
.6uD+0.8uE+uC=–
375
EI
(3)
MED +MEF =0
2EI
2.5
1
2uE+uD
2
+
3EI
4
1
uE
2
=
0
uD=–2.9375uE
(4)
Solving Eqs. (1) through (4),
u
B=–uE=
–54.845
EI
u
C=–uD=
161.106
EI
Thus,
MBA =–41.1
lb #ft
Ans.
MEF =41.1
lb #ft
Ans.
MBC =41.1
lb #ft
Ans.
MED =–41.1
lb #ft
Ans.
MCB =214 lb #ft
Ans.
MDE =–214 lb #ft
Ans.
MCD =–214 lb #ft
Ans.
MDC =214 lb #ft
Ans.
*10–20. (Continued)
SOLUTION
cBC =cCE =0
cAB =cCD =cEF =c
Applying Eq. 10–10,
M
CB =
3EI
6
1
uC–0
2
+0
M
CE =
3EI
4
1
uC–0
2
+0
M
2
3EI
10–21. Wind loads are transmitted to the frame at joint E.
If A, B, E, D, and F are all pin connected and C is fixed con-
nected, determine the moments at joint C and draw the mo-
ment diagram for the girder BCE. EI is constant.
A
BC
E
F
6 m 4 m
8
m
12 kN
D
509
SOLUTION
3b
3b
10–22. Determine the moments at joints A, B, C, and D, then
draw the moment diagram for each member of the frame.
Assume the supports at A and B are fixed. EI is constant.
A
D
B
C
3 m
3 m
30 kN
>m
512
513
Ans.
MAB =–19.4 k #ft
MBA =–15.0 k #ft
MBC =15.0 k #ft
MCB =20.1 k #ft
MCD =–20.1 k #ft
MDC =–36.9 k #ft
MAB =–19.4 k #ft
MBA =–15.0 k #ft
514
SOLUTION
1
FEM
2
BC =–
wL2
12
=–1600
k
#
in.
1
FEM
2
CB =
wL2
12
=1600 k
#
in.
uA=uD=0
cAB =cCD =
∆
25
=c
cBC =–
1.2∆
20
=–1.5
c
M
N=2E
aI
Lb1
2uN+uF–3c
2
+
1
FEM
2N
MAB =2E
a600
251122b1
0+uB–3c
2
+0=116 000uB–348 000
c
MBA =2E
a600
251122b1
2uB+0–3c
2
+0=232 000uB–348 000
c
MBC =2E
a600
201122b1
2uB+uC–3
1
–1.5c
22
–160
0
=290 00uB+145 000uC+652 500c–1600
MCB =2E
a600
201122b1
2uC+uB–3
1
–1.5c
22
+160
0
=290 000uC+145 000uB+652 500c+1600
MCD =2E
a600
201122b1
2uC+0–3c
2
+
0
=232 000uC–348 000c
MDC =2E
a600
251122b1
0+uC–3c
2
+
0
=116 000uC–348 000c
Moment equilibrium at B and C:
MBA +MBC =0
522 000uB+145 000uC+304 500c=1600
(1)
MCB +MCD =0
145 000uB+522 000uC+304 500c=–1600
(2)
*10–24. Determine the moments acting at the supports A
and D of the battered-column frame. Take
E=29(103)
ksi
,
I=600
in4
.
4 k>ft
20 ft
6 k
B
AD
C
15 ft 15 ft20 ft
10–1P. The roof is supported by joists that rest on two girders.
Each joist can be considered simply supported, and the front
girder can be considered attached to the three columns by a
pin at A and rollers at B and C. Assume the roof will be made
from 3 in.-thick cinder concrete, and each joist has a weight of
550 lb. According to code the roof will be subjected to a snow
loading of 25 psf. The joists have a length of 25 ft. Draw the
shear and moment diagrams for the girder. Assume the
supporting columns are rigid.
SOLUTION
From the text,
W
eight of cinder concrete =
1
108 lb
>
ft3
2a 3
12
ft
b
=27 psf
Live load =25 psf
Total load =52 psf
Load on joist =152
lb>ft2213 ft2=156 lb>ft
550
3 ft
A B
C
3 ft 3 ft 3 ft 3 ft 3 ft 3 ft 3 ft
R
R