632
Solution
a
A=
6
20
=0.3
aB=
4
20
=0.2
r
A=rB=
4–2
2
=
1
From Table 13–1,
for span AB,
CAB =0.622
CBA =0.748
KAB =10.06
KBA =8.37
K
BA =
K
BA
EI
C
L
=
8.37EI
C
20
=0.4185EI
C
1FEM2AB =–0.108918212022=–348.48
k#ft
1FEM2BA =0.094218212022=301.44
k#ft
For span BC,
CBC =0.748
CCB =0.622
KBC =0.4185EIC
1FEM2BC =–301.44
k#ft
1FEM2CB =348.48
k#ft
Joint A B C
Mem. AB BA BC CB
K0.4185EIC0.4185EIC
DF 00.5 0.5 0
COF 0.622 0.748 0.748 0.622
FEM
–348.48
301.44
–301.44
348.48
0 0
–348.48
–301.44
Ans.
MAB =–348.48
k#ft;
MBA =301.44
k#ft;
MBC =–301.44
k#ft;
MCB =348.48
k#ft
13–1. Determine the moments at A, B, and C by the moment-
distribution method. Assume the supports at A and C are fixed
and a roller support at B is on a rigid base. The girder has a
thickness of 4 ft. Use Table 13.1. E is constant. The haunches
are tapered. 6 ft
4 ft 4 ft
2 ft
AC
4 ft 4 ft
4 ft
6 ft
B
20 ft 20 ft
8 k>ft
Solution
The necessary data for member BC can be found from Table 13–2.
Here,
a
C=
8
40
=0.2
aB=
12
40
=0.3
rC=
4–2
2
=1.0
rB=
5–2
2
=1.5
12
13–3. Apply the moment-distribution method to determine
the moment at each joint of the parabolic haunched frame.
Supports A and B are fixed. Use Table 13.2. The members are
each 1 ft thick. E is constant.
5 ft
40 ft
40 ft
2 ft
2 ft
8 ft 12 ft
C
B
A
1.5 k>ft
4 ft
C
635
5 ft
40 ft
40 ft
2 ft
2 ft
8 ft 12 ft
C
B
A
1.5 k>ft
4 ft
Solution
40
*13–4. Solve Prob. 13–3 using the slope-deflection equations.
636
Ans.
MCB =–75.1
k#ft
MBC =369
k#ft
MAC =37.5
k#ft
MCA =75.1
k#ft
MCA +MCB =0
MCB =–75.09
MAC =37.546
MCA =75.09
637
Solution
For span AB,
a
A=
6
30
=0.2
aB=
9
30
=0.3
rA=rB=
4–2
2
=
1
From Table 13.2,
CAB
=0.683
CBA =0.598
kAB
=6.73
kBA =7.68
K
AB
=
6.73EI
30
=0.2243EI
K
BA
=
7.68EI
30
=0.256E
I
K′BA =0256EI[1 –10.683210.5982]
=0.15144
EI
1FEM2AB =–0.091114213022=–327.96
k#ft
1FEM2BA =0.104214213022=375.12
k#ft
For span CD,
CDC =0.683
CCD =0.598
KDC =6.73
KCD =7.68
KDC =0.2243EI
KCD =0.256EI
K′CD =0.15144EI
1FEM2CD =–375.12
k#ft
1FEM2DC =327.96
k#ft
For span BC,
a
B=aC=
8
40
=0.2
rB=rC=
4–2
2
=1
From Table 13.2,
CBC =CCB =0.619
kBC =kCB =6.41
K
BC =KCB =
6.41EI
40
=0.16025E
I
1FEM2BC =–0.095614214022=–611.84
k#ft
1FEM2CB =611.84
k#ft
13–5. Use the moment-distribution method to determine the
moment at each joint of the symmetric bridge frame. Supports
at F and E are fixed and B and C are fixed connected. Use
Table 13–2. The modulus of elasticity is constant and the mem-
bers are each 1 ft thick. The haunches are parabolic.
9 ft
6 ft 8 ft
2 ft 4 ft
4 ft
25 ft
2 ft
ABC
D
FE
30 ft 40 ft 30 ft
4 k>ft
638
For span BF,
CBF =0.5
K
BF =
4EI
25
=0.16E
I
1FEM2BF =1FEM2FB =0
For span CE,
CCE =0.5
KCE =0.16EI
1FEM2CE =1FEM2EC =0
Joint A F B C E D
Member AB FB BF BA BC CB CD CE EC DC
DF 1 0 0.3392 0.3211 0.3397 0.3397 0.3211 0.3392 0 1
COF 0.683 0.5 0.598 0.619 0.619 0.598 0.5 0.683
FEM
–327.96
375.12
–611.84
611.84
–375.12
327.96
Dist. 327.96 80.30 76.01 80.41
–80.41
–76.01
80.30 327.96
CO 40.15 224.00
–49.77
49.77
–224.00
–40.15
Dist.
–59.09
–55.95
–59.19
59.19 55.95 59.09
CO
–29.55
36.64
–36.64
29.55
Dist.
–12.42
–11.77
–12.45
12.45 11.77 12.42
CO
–6.21
7.71
–7.71
6.21
Dist.
–2.61
–2.48
–2.62
2.62 2.48 2.61
CO
–1.31
1.62
–1.62
1.31
Dist.
–0.55
–0.52
–0.55
0.55 0.52 0.55
CO
–0.27
0.34
–0.34
0.27
Dist.
–0.11
–0.11
–0.12
0.12 0.11 0.11
CO
–0.05
0.07
–0.07
0.05
Dist.
–0.03
–0.02
–0.02
0.02 0.02 0.03
Σ
02.76 5.49 604
–609
609
–604
–5.49
–2.76
0
Ans:
Σ
02.76 5.49 604
–609
609
–604
–5.49
–2.76
0
13–5. (Continued)
k#ft
k#ft
Ans.
639
Solution
13–6. Solve Prob. 13–5 using the slope-deflection equations.
9 ft
6 ft 8 ft
2 ft 4 ft
4 ft
25 ft
2 ft
ABC
D
FE
30 ft 40 ft 30 ft
4 k>ft
642
Solution
a
B=aC=
3.2
16
=0.
2
r
B=rC=
5–2.5
2.5
=
1
CBC =CCB =0.619
kBC =kCB =6.41
1FEM2BC =0.14591221162=–4.6688
k#ft
1FEM2CB =–4.6688
k#ft
K
BC =KCB =
kBCEIc
L
=
6.41
1
E
2a1
12
b1
1
21
2.5
2
2
16
=0.5216
E
MN=KN[uN+CNuF–c11+CN2]+1FEM2N
M
AB =
2EI
15
1
0+uB–0
2
+
0
M
BA =
2EI
15
1
2uB+0–0
2
+
0
M
CD =
2EI
15
1
2uC+0–0
2
+
0
M
DC =
2EI
15
1
0+uC–0
2
+
0
MBC =0.5216E1uB+0.6191uC2–02–4.6688
MCB =0.5216E1uC+0.6191uB2–02+4.6688
Equilibrium.
MBA +MBC =0
MCB +MCD =0
Or
2E
a1
12
b1
1
21
3
2
3
15
1
2uB
2
+0.5216E[uB+0.619uC]–4.6688 =
0
1
.1216uB+0.32287uC=
4.6688
E
(1)
2E
a1
12
b1
1
21
3
2
3
15
1
2uC
2
+0.5216E[uC+0.619uB]+4.6688 =
0
1
.1216uC+0.32287uB=–
4.6688
E
(2)
*13–8. Solve Prob. 13–7 using the slope-deflection equations.
9 ft
2 ft
2 k>ft
3 ft
5 ft
2 ft
3 ft
5 ft
2 ft
A
BC
643
Ans.
MAB =1.75
k#ft
MBA =3.51
k#ft
MBC =–3.51
k#ft
MCB =3.51
k#ft
MCD =–3.51
k#ft
MDC =–1.75
k#ft
Solving Eqs. 1 and 2:
u
B=–uC=
5.84528
E
MAB =1.75
k#ft
Ans.
MBA =3.51
k#ft
Ans.
MBC =–3.51
k#ft
Ans.
MCB =3.51
k#ft
Ans.
MCD =–3.51
k#ft
Ans.
MDC =–1.75
k#ft
Ans.
*13–8. (Continued)
Solution
13–11. Use the moment-distribution method to determine
the moment at each joint of the frame. The supports at A and
C are pinned and the joints at B and D are fixed connected.
Assume that E is constant and the members have a thickness
of 1 ft. The haunches are tapered, so use Table 13.1.
20 ft
6 ft
1 ft
6 ft
18 ft
2.5 ft
1 ft 1 ft
A
B
C
D
500 lb>ft
2.5 ft
647
Ans.
ΣM
028.3
–28.3
28.3
–28.3
0 k·ft
Solution
See Prob. 13-11 for the tabular data.
For span AB,
I
*13–12. Solve Prob. 13–11 using the slope-deflection
equations.
20 ft
6 ft
1 ft
6 ft
18 ft
2.5 ft
1 ft 1 ft
A
B
C
D
500 lb>ft
2.5 ft
I