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Determine the moments at B and C . EI is constant. Assume B and C are rollers and A and D are pinned. 12–1.
3 k/ k/ ft ft
A
B
C
8 ft
FEMAB = FEMCD = wL2 12
FEMBC = KAB
=
3EI , 8
wL2 12
= - 100 KBC
= - 16,
FEMBA = FEMDC =
FEMCB =
=
4EI , 20
wL2 12
KCD
=
wL2 12
20 ft
= 16
= 100
3EI 8
DFAB = 1 = DFDC
DFBA = DFCD
3EI 8 = = 0.652 3EI 4EI + 8 20
DFBA = DFCB = 1 - 0.652 = 0.348
Joint
A
Member
AB
BA
BC
CB
DF
1
0.652
0.348
0.348
FEM
–16 16
B
16 54.782 8 4.310 0.750 0.130 0.023
aM
0
84.0
C
–100
D CD
0.652
100
–16
29.218
–29.218
–54.782
–14.609
14.609
–8
2.299
–2.299
–4.310
–1.149
1.149
0.400
–0.400
–0.200
0.200
0.070
–0.070
–0.035
0.035
0.012
–0.012
–84.0
84.0
DC
1 16 –16
–0.750 –0.130 –0.023 –84.0
#
0 k ft
Ans.
D
8 ft
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Determine the moments at A, B, an and d C . Assum Assume e the support at B is a roller and A and C are fixed. EI is constant. 12–2.
3 k/ ft ft 2 k/ ft ft
A
B
36 ft
(DF)AB = 0
(DF)BA =
(DF)BC = 0.6 (FEM)AB =
I>36 I>36
+ I>24
C
24 ft
= 0.4
(DF)CB = 0
- 2(36)2
= - 216 k # ft
12
#
(FEM)BA = 216 k ft (FEM)BC =
- 3(24)2
= - 144 k # ft
12
#
(FEM)CB = 144 k ft
Joint
A
Mem.
AB
BA
BC
CB
DF
0
0.4
0.6
0
FEM
–216
216
–144
144
R –28.8
–43.2
B
–14.4
a M
–230
187
C
R
–187
–21.6
#
–122 k ft
Ans.
Determine the moments at A, B, an and d C , the then n draw draw the moment diagram. Assume the support at B is a roller and A and C are fixed. EI is constant. 12–3.
900 90 0 lb lb 90 900 0 lb lb
B
A
(DF)AB = 0 (DF)CB = 0 (FEM)AB =
(DF)BA =
I>18 I>18
= 0.5263
+ I>20
6 ft
(DF)BC = 0.4737
- 2(0.9)(18) 9
= - 3.60 k # ft
#
(FEM)BA = 3.60 k ft (FEM)BC =
= –1.00 k # ft
–0.4(20) 8
#
(FEM)CB = 1.00 k ft Joint
A
Mem.
AB
BA
BC
DF
0
0.5263
0.4737
FEM
–3.60
B
3.60 R –1.368
C
–4.28
2.23
0
–1.00 –1.232
–0.684
aM
CB
–2.23
1.00
R
–0.616
#
0.384 k ft
Ans.
400 lb
6 ft
6 ft
C
10 ft
10 ft
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*12–4. Determine the reactions at the supports and then draw the moment diagram.Assume A is fixed. EI is constant.
20 ft
wL2 12
= –26.67, FEMCB =
MCD
= 0.5(15) = 7.5 k # ft
KAB
=
4EI , 20
KBC
=
wL2 12
= 26.67
4EI 20
DFAB = 0
DFBA = DFBC
4EI 20 = = 0.5 4EI 4EI + 20 20
DFCB = 1
Joint
A
Member
AB
BA
BC
CB
CD
DF
0
0.5
0.5
1
0
B
FEM 13.33 6.667
–26.67
26.67
13.33
–19.167
–9.583 4.7917
2.396 1.667 0.8333 0.5990 0.2994 0.2083 0.1042 0.07485 10.4
C
20.7
4.7917
6.667 –6.667
–3.333
2.396
1.667
–2.396
–1.1979
0.8333
0.5990
–0.8333
–0.4167
0.2994
0.2083
–0.2994
–0.1497
0.1042
0.07485 –20.7
–7.5
–0.1042 7.5
#
–7.5 k ft
D
C
B
A
FEMBC = -
500 lb
800 lb/ lb/ ft ft
20 ft
15 ft
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Determine the moments at B and C , then draw the moment diagram for the beam.Assume C is a fixed support. EI is constant. 12–5.
12 kN 8 kN/ m
A
B 6m
Member Stiffness Factor and Distribution Factor.
KBA
=
EI 3EI 3EI = = LBA 6 2
(DF)AB = 1 (DF)BC =
(DF)BA =
>
>
EI 2
EI 2
>
+ EI 2
=
KBC
>
>
EI 2
>
+ EI 2
EI 2
= 0.5
EI 4EI 4EI = = LBC 8 2
= 0.5
(DF)CB = 0
Fixed End Moments. Referring to the table on the inside back cover,
(FEM)BA
=
wL2 8
(FEM)BC = (FEM)CB =
=
PL 8
PL 8
=
8(62) = 36 kN m 8
#
= -
#
12(8) = - 12 kN m 8
#
12(8) = 12 kN m 8
Moment Distribution. Tabulating the above data,
Joint
A
Member
AB
BA
BC
CB
DF
1
0.5
0.5
0
FEM
0
Dist.
aM
0
B
C
36
–12
–12
–12
24
–24
12
R
–6 6
Using these results, the shear and both ends of members AB and BC are computed and shown in Fig. a. Subse Subsequently quently,, the shear and moment moment diagram can be plotted, Fig. b.
C 4m
4m
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Determine the moments at B and C , then draw the moment diagram for the beam. All connections are pins. Assume the horizontal reactions are zero. EI is constant. 12–6.
12 kN/ kN/ m 4m C
A
4m
=
3EI 3EI = 4 LAB
(DF)AB = 1
(DF)BA =
KBC
6EI = LBC
=
>
6EI = 4
3EI 4 1 = 3EI 4 + 3EI 2 3
>
>
3EI 2
(DF)BC =
>
3EI 2 2 = 3EI 4 + 3EI 2 3
>
>
Fixed End Moments. Referring to the table on the inside back cover,
(FEM)BA =
wL2 8
12(42) = = 24 kN m 8
#
(FEM)BC = 0
Moment Distribution. Tabulating the above data,
Joint
A
Member
AB
BA
BC
DF
1
1/3
2/3
FEM
0
Dist.
aM
0
B
24
0
–8
–16
16
–16
D
4m 12 kN/ kN/ m
Member Stiffness Factor and Distribution Factor. Factor.
KAB
B
Using these these results, results, the shear at both ends ends of members AB, BC , and CD are computed and shown in Fig. a. Subsequently the shear and moment diagram can be plotted,, Fig plotted Fig.. b and c, respec respectively tively..
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Determine the reactions at the supports. Assume A is fixed and B and C are rollers that can either push or pull on the beam. EI is constant. 12–7.
12 kN/ kN/ m
A
B
5m
Member Stiffness Factor and Distribution Factor.
KAB
=
4EI 4EI = = 0.8EI LAB 5
(DF)AB = 0
(DF)BA =
KBC
=
3EI 3EI = = 1.2EI LBC 2.5
0.8EI = 0.4 0.8EI + 1.2EI
1.2.EI = 0.6 0.8EI + 1.2EI =1
(DF)BC = (DF)CB
Fixed End Moments. Referring to the table on the inside back cover,
(FEM)AB =
wL2 12
(FEM)BA =
wL2 12
=
12(52) = = - 25 kN m 12
#
12(52) = 25 kN m 12
#
(FEM)BC = (FEM)CB = 0 Moment Distribution. Tabulating the above data,
Joint
A
Member
AB
BA
BC
CB
DF
0
0.4
0.6
1
25
0
0
R –10
–15
15
–15
FEM
B
–25
Dist. CO
–5
aM
–30
C
Ans.
Using these results, the shear at both ends of members AB and BC are computed and shown in Fig. a. From this figure, Ax MA
=0
Ay
= 33 kN c
= 30 kN # m a
By Cy
= 27 + 6 = 33 kN c = 6 kN T
Ans. Ans.
C
2.5 m
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*12–8. Determine the moments at B and C , then draw draw the the moment diagram diagram for the beam. Assum Assume e the supports supports at B and C are rollers and A and D are pins. EI is constant.
12 kN/ m
A
B 4m
Member Stiffness Factor and Distribution Factor. Factor.
KAB
=
3EI 3EI = LAB 4
(DF)AB = 1
(DF)BA =
KBC
=
EI 2EI 2EI = = LBC 6 3
>
3EI 4 9 = 3EI 4 + 3EI 3 13
>
(DF)BC =
>
>
EI 3
>
>
3EI 4 + EI 3
=
4 13
Fixed End Moments. Referring to the table on the inside back cover,
(FEM)AB = (FEM)BC = 0
(FEM)BA =
wL2 8
=
12(42) = 24 kN m 8
#
Moment Distribution. Tabulating the above data,
Joint
A
Member
AB
DF
1
9 13
4 13
FEM
0
24
0
Dist.
aM
B BA
–16.62 0
7.385
12 kN/ m
BC
–7.385 –7.385
Using these results results,, the shear at both ends of members members AB, BC , and CD are computed and shown in Fig. a. Subsequently Subsequently,, the shear and moment diagram can be plotted,, Fig plotted Fig.. b and c, respec respectively tively..
C
6m
D
4m
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Determine the moments at B and C , then draw the moment diagram for the beam. Assume the supports at B and C are rollers and A is a pin. EI is constant. 12–9.
300 lb 200 lb/ ft ft
Member Stiffness Factor and Distribution Factor. KAB
3EI 3EI = = = 0.3EI LAB 10
KBC
4EI 4EI = = = 0.4EI. LBC 10
A
10 ft
(DF)BA =
0.3EI 3 = 0.3EI + 0.4EI 7
(DF)CB = 1
(DF)BC =
0.4EI 4 = 0.3EI + 0.4EI 7
(DF)CD = 0
Fixed End Moments. Referring to the table on the inside back cover,
#
(FEM)CD = - 300(8) = 2400 lb ft (FEM)BA =
wL2AB
8
=
200(102) 8
(FEM)BC = (FEM)CB = 0
= 2500 lb # ft
Moment Distribution. Tabulating the above data,
Joint
A
Member
AB
BA
BC
CB
CD
DF
1
3/7
4/7
1
0
FEM
0
0
0
–2400
Dist.
B
2500 –1071.43
CO
–1428.57 R 2400 1200
Dist.
–514.29
CO –153.06
CO –73.47
CO –21.87
CO Dist.
–10.50
CO Dist.
–3.12
CO –1.50
CO Dist.
–0.45
CO Dist.
–0.21
CO Dist.
–0.06
CO Dist. 0
R
–29.16 R
R 24.99
–13.99 R
R 7.29
–4.17 R 3.50
Dist.
R –102.05
–97.96 R 102.05 51.03
Dist.
R –342.86
–204.09 R 342.86 171.43
Dist.
R –714.29
–685.71 R 714.29 357.15
Dist.
aM
C
R
–2.00 R
R 1.04
–0.59 R
R 0.500
–0.29 R
R 0.15
–0.09 R
R 0.07
–0.03
–0.04
650.01
–650.01
–48.98
48.98 –14.58 14.58 –7.00 7.00 –2.08 2.08 –1.00 1.00 –0.30 0.30 –0.15 0.15 –0.04 0.04 2400
–2400
Using these these results, results, the shear at both both ends of members AB, BC , and CD are computed and shown in Fig. a. Subsequently Subsequently,, the shear and moment diagrams can be plotted, plotte d, Fig Fig.. b and c, respec respectively tively..
D
C
B
10 ft
8 ft
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12–9. 12– 9.
Conti Co ntinue nued d
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Determine the moment at B, th then en dr draw aw th the e moment diagram for the beam. Assume the supports at A and C are rollers and B is a pin. EI is constant. 12–10.
6 kN / m
D D
E A 2m
Member Stiffness Factor and Distribution Factor. KAB
=
4EI 4EI = = EI LAB 4
KBC
(DF)AB = 1 (DF)AD = 0 (DF)CB = 1
4EI 4EI = = EI LBC 4
=
(DF)BA = (DF)BC =
EI EI
+ EI
= 0.5
(DF)CE = 0
Fixed End Moments. Referring to the table on the inside back cover,
#
#
(FEM)AD = 6(2)(1) = 12 kN m (FEM)AB = (FEM)BA = (FEM)BC = (FEM)CB =
- wL2AB
= -
12 wL2AB
12
=
- wL2BC
12
6(42) = - 8 kN m 12
#
6(42) = 8 kN m 12
#
6(42) = = - 8 kN m 12
#
12 wL2BC
(FEM)CE = - 6(2)(1) = - 12 kN m
=
6(42) = 8 kN m 12
#
Moment Distribution. Tabulating the above data,
Joint
A
B
C
Member
AD
AB
BA
BC
CB
CE
DF
0
1
0.5
0.5
1
0
FEM
12
–8
8
–12
Dist.
–4
CO
aM
12
–12
R
8
–8
0
0
–2
2
6
–6
R
4
12
–12
Using these results, results, the shear at both both ends of members AD, AB, BC , an and d CE are computed and shown in Fig. a. Subsequently Subsequently,, the shear and moment diagram can be plotted, plotte d, Fig Fig.. b and c, respec respectively tively..
C
B 4m
4m
2m
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Determine the moments at B, C , an and d D, the then n draw draw the moment diagram for the beam. EI is constant. 12–11.
1.5 k/ ft ft 10 k ft
10 k ft
B
D
C
E
A
10 ft
Member Stiffness Factor and Distribution Factor. Factor. KBC
=
4EI 4EI = = 0.2 EI LBC 20
(DF)BA = (DF)DE = 0 (DF)CB = (DF)CD =
KCD
=
4EI 4EI = = 0.2 EI LCD 20
(DF)BC = (DF)DC = 1
0.2EI = 0.5 0.2EI + 0.2EI
Fixed End Moments. Referring to the table on the inside back cover,
#
#
(FEM)BA = 10 k ft
(FEM)DE = - 10 k ft
(FEM)BC = (FEM)CD = (FEM)CB = (FEM)DC =
wL2 12
wL2 12
= -
= -
1.5(202) = - 50 k ft 12
#
1.5(202) = 50 k ft 12
#
Moment Distribution. Tabulating the above data,
Joint
B
C
D
Member
BA
BC
CB
CD
DC
DE
DF
0
1
0.5
0.5
1
0
FEM
10
–50
–10
Dist.
40
CO
aM
10
–10
R
50
–50
50
0
0
20
–20
R –40
70
–70
10
–10
Using these results, results, the shear at both ends of members AB, BC , CD, an and d DE are a computed and shown in Fig. . Subsequently Subsequently,, the shear and moment diagram can be plotted,, Fig plotted Fig.. b and c, respec respectively tively..
20 ft
20 ft
10 ft
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*12–12. Determine the moment at B, th then en dr draw aw th the e moment diagram for the beam. Assume the support at A is pinned, B is a roller and C is fixed. EI is constant.
4 k/ ft ft
B
A
15 ft
=
wL2 30
=
4(152) = 30 k ft 30
FEMBA =
wL2 20
=
4(152) = 45 k ft 20
FEMBC =
wL2 12
=
(4)(122) = 48 k ft 12
FEMAB
#
C
12 ft
#
#
#
FEMCB = 48 k ft Joint
A
Member
AB
BA
BC
CB
DF
1
0.375
0.625
0
FEM
B
–30
45
30
C
–48
1.125
48
1.875
15
0.9375
–5.625
–9.375 –4.688
aM MB
0
55.5
–55.5
44.25
= - 55.5 k # ft
Ans.
8 kN/ m
Determine the moment at B, th then en dr draw aw th the e moment diagram for each member of the frame. frame. Assume the supports at A and C are pins. EI is constant. 12–13.
C
B
6m
Member Stiffness Factor and Distribution Factor. KBC
=
3EI 3EI = = 0.5 EI LBC 6
KBA
=
3EI 3EI = = 0.6 EI LAB 5
5m
A
(DF)AB = (DF)CB = 1 (DF)BA =
(DF)BC
0.5EI 5 = = 0.5EI + 0.6EI 11
0.6EI 6 = 0.5EI + 0.6EI 11
Fixed End Moments. Referring to the table on the inside back cover,
(FEM)CB = (FEM)AB = (FEM)BA = 0 (FEM)BC = -
wL2BC
8
8(62) = = - 36 kN m 8
#
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12–13. 12–1 3.
Continue Cont inued d
Moment Distribution. Tabulating the above data,
Joint
A
Member
AB
DF
1
6 11
5 11
1
FEM
0
0
–36
0
Dist.
aM
0
B BA
C BC
19.64
16.36
19.64
–19.64
CB
0
Using these results, the shear at both ends of member AB and BC are computed and shown in Fig. a. Subse Subsequently quently,, the shear and moment diagram diagram can be plotted, Fig Fig.. b and c, respe respectively ctively..
© 2012 Pearson Education, Inc., Upper Saddle River, NJ NJ.. All rights reserved.This material is protected under under all copyright laws as they currently currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. publisher.
Determine the moments at the ends of each member of the frame. Assume the joint at B is fixed, C is pinn pi nned ed,, an and d A is fixed. The moment moment of inertia inertia of each member is listed in the figure. E = 29(103) ksi. 12–14.
2 k/ k/ ft ft B I BC
12 ft
8 ft
4k I AB
(DF)AB = 0 (DF)BA =
8 ft
4(0.6875IBC)>16 = 0.4074 4(0.6875IBC)>16 + 3IBC>12
(DF)BC = 0.5926 (FEM)AB =
- 4(16) 8
A
(DF)CB = 1
= - 8 k # ft
#
(FEM)BA = 8 k ft (FEM)BC =
- 2(122) 12
= - 24 k # ft
#
(FEM)CB = 24 k ft
Joint
A
Mem.
AB
BA
BC
CB
DF
0
0.4047
0.5926
1
FEM
–8.0
B
8.0 6.518
3.259 2.444
aM
–2.30
R R
C
–24.0 9.482
24.0 –24.0
–12.0 4.889 19.4
7.111 –19.4
0 Ans.
800 in4
550 in 4
C
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Determine the reactions at A and D. Ass Assume ume the the supports at A and D are fixed and B and C are fixed connected. EI is constant. 12–15.
8 k/ k/ ft ft B
C
15 ft
(DF)AB = (DF)DC = 0 (DF)BA = (DF)CD =
>
I 15
>
I 15
>
+ I 24
(DF)BC = (DF)CB = 0.3846
= 0.6154
A
24 ft
(FEM)AB = (FEM)BA = 0 (FEM)BC =
- 8(24)2 12
= - 384 k # ft
#
(FEM)CB = 384 k ft (FEM)CD = (FEM)DC = 0 Joint
A
Mem.
AB
BA
BC
CB
DF
0
0.6154
0.3846
0.3846
B
R 236.31
118.16 45.44
R
8.74 1.68
R
0.32
R
0.06
146.28
DC
0.6154
R
R
R
-45.44 -8.74
2.73
-1.05
-1.68
0.53
-0.33
0.10
0.04 R -0.04
R
-236.31
14.20
0.20 R -0.20
-0.02
0.03
R
1.05 R
-0.10
0.16
CD
73.84
5.46 R -5.46
-0.53
0.84
R
28.40 R -28.40
-2.73
4.37 R
-73.84
D
0
384
147.69 R -147.69
-14.20
22.72
aM
C
-384
FEM
R
-0.06
0.02
0.01
0.01
-0.01
-0.01
292.57
-292.57
292.57
-292.57
R R R R R R
-118.16 -22.72 -4.37 -0.84 -0.17 -0.03 -146.28
Thus from the free-body diagrams: Ax
= 29.3 k
Ans.
Ay
= 96.0 k
Ans.
MA
= 146 k # ft
Ans.
Dx
= 29.3 k
Ans.
Dy
= 96.0 k
Ans.
MD
= 146 k # ft
D
Ans.
© 2012 Pearson Education, Inc., Upper Saddle River, NJ NJ.. All rights reserved.This material is protected under under all copyright laws as they currently currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. publisher.
Determine the moments at D and C , th then en draw draw the moment diagram for each member of the frame. Assume the supports at A and B are pins and D and C are fixed joints. EI is constant. *12–16.
5 k/ ft ft
D
C
12 ft
9 ft
Member Stiffness Factor and Distribution Factor. KAD
= KBC =
EI 3EI 3EI = = L 9 3
(DF)AD = (DF)BC = 1
KCD
=
A
EI 4EI 4EI = = L 12 3
(DF)DA = (DF)DC = (DF)CD EI 3 1 = DFCB = = EI 3 + EI 3 2
>
>
>
Fixed End Moments. Referring to the table on the inside back cover,
(FEM)AD = (FEM)DA = (FEM)BC = (FEM)CB = 0 (FEM)DC = (FEM)CD =
wL2CD
12
wL2CD
12
=
5(122) = = - 60 k ft 12
#
5(122) = 60 k ft 12
#
Moments Distribution. Tabulating the above data,
Joint
A
Member
AD
DA
DC
CD
CB
DF
1
0.5
0.5
0.5
0.5
1
FEM
0
0
-60
60
0
0
30
30
R -30
-30
Dist.
D
-15
CO Dist.
7.50
7.50
-3.75
C0 Dist.
1.875
R
Dist.
0.4688
Dist.
0.1172
Dist.
0.0293
C0 Dist.
0.0073 0
40.00
0.9375
0.1172 R -0.1172
-0.1172
R
-0.0586
C0
-1.875 -0.4688
-0.2344
C0
-7.50
0.4688R -0.4688
R
0.2344 0.0586
0.0293 R -0.0293
R
-0.0146
0.0146
0.0073
-0.0073
-40.00
BC
3.75
1.875 R -1.875
R
B
15
R -7.50
R
-0.9375
C0
aM
C
40.00
-0.0293 -0.0073 -40.00
Using these these results, results, the shear at both ends of members members AD, CD, and BC are computed and shown in Fig. a. Subsequently Subsequently,, the shear and moment diagram can be plotted.
B
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12–16. 12–1 6.
Continue Cont inued d
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Determine the moments at the fixed support A and joint D and then draw the moment diagram for the frame.. Assum frame Assume e B is pinned. 12–17.
4 k/ ft ft
C
A
D
12 ft
12 ft 12 ft
B
Member Stiffness Factor and Distribution Factor. KAD
=
EI 4EI 4EI = = LAD 12 3
(DF)AD = O
(DF)DA =
(DF)DC = (DF)DB =
>
> + >4 >4 +
EI 3 EI
EI 3
= KDB =
KDC
+ EI
EI 3EI 3EI = = L 12 4
> >4 + >4 = 0.4 >4 = 0.3
EI 3 EI
EI
EI
(DF)CD = (DF)BD = 1 Fixed End Moments. Referring to the table on the inside back cover,
(FEM)AD = (FEM)DA = (FEM)DC (FEM)CD
wL2AD
12
wL2AD
= -
4(122)
12 2 4(12 )
= - 48 k # ft
#
= = 48 k ft 12 12 4(122) wL2CD = = = - 72 k ft 8 8 = (FEM)BD = (FEM)DB = 0
#
Moments Distribution. Tabulating the above data,
Joint
A
Member
AD
DA
DB
DC
CD
BD
DF
0
0.4
0.3
0.3
1
1
0
0
0
0
FEM
-48
Dist. CO
aM
D
48 R
9.60
C
0 7.20
-72
B
7.20
4.80
-43.2
57.6
7.20
-64.8
Using these results, results, the shears at both ends of of members AD, CD , and BD are computed and shown in Fig. a. Subsequently Subsequently,, the shear and moment diagram can be plotted, plotte d, Fig Fig.. b and c, respec respectively tively..
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12–17. 12–1 7.
Continue Cont inued d
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Determine the moments at each joint of the frame, then draw the moment diagram for member BCE. Ass Assume ume B, C , and E are fixed connected and A and D are pins. E = 29(103) ksi. 12–18.
0.5 k/ k/ ft ft B
2k
I BC
8 ft
400 in 4
I DC I AB
600 in 4 D
(DF)AB = (DF)DC = 1 (DF)DC = 0 3(A1.5IBC) 16 = 0.6279 3(1.5IBC) 16 + 4IBC 24
>
(DF)BC = 0.3721 (DF)CB =
>
4IBC
>
>
4IBC 24 = 0.2270 24 + 3(1.25IBC) 16 + 4IBC 12
>
>
(DF)CD = 0.3191 (DF)CE = 0.4539 (FEM)AB =
- 3(16) 8
= - 6 k # ft
#
(FEM)BA = 6 k ft (FEM)BC = (FEM)CB
12 = 24 k ft
= - 24 k # ft
#
(FEM)CE = (FEM)EC
- (0.5)(24)2
- (0.5)(12)2
12 = 6 k ft
= - 6 k # ft
#
(FEM)CD = (FEM)DC = 0
Joint
A
Mem.
AB
BA
BC
CB
DF
1
0.6279
0.3721
0.2270
FEM
-6.0 6.0
B
6.0 11.30 3.0
-0.60 0.24
-0.01
C
-24.0
E CD
0.3191
24.0
6.70
-4.09
-5.74
-2.04 -0.36 -0.38
3.35
-0.76 -0.18
-1.07
0.14
0.04
0.06
0.02
0.07
-0.01
-0.02
CE
0.4539
-6.0 -8.17
D EC
0
DC
1
6.0
-4.09 -1.52 -0.76 0.08 0.04
-0.02
-0.03 -0.02
aM
0
19.9
-19.9
22.4
-6.77
-15.6
1.18
500 in4
24 ft
>
400 in 4
A
(DF)BA =
E I CE
3k 8 ft
C
0
12 ft
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The frame is made from pipe that is fixed fi xed connected. If it supports the loading loading shown, determi determine ne the moments developed at each of the joints. EI is constant. 12–19.
18 kN
18 kN
C
B
4m D
A
4m
FEMBC = KAB
= KCD
2PL 2PL = - 48, FEMCB = = 48 9 9 4EI 4EI = , KBC = 4 12
DFAB = DFDC = 0 DFBA
= DFCD =
4EI 4
4EI 5
+
4EI 12
= 0.75
DFBC = DFCB = 1 - 0.75 = 0.25
Joint
A
Member
AB
BA
BC
CB
DF
0
0.75
0.25
0.25
B
FEM 36 18 4.5
-48
48
12
-12
-6
6
1.5
-0.75
2.25 0.5625 0.281 0.0704 20.6
C
41.1
-1.5
CD
0.75
-0.1875
-0.0938
0.0938
0.0234
-0.0234 41.1
DC
0
-36 -18 -4.5 -2.25
0.75
0.1875
-41.1
D
-0.5625 -0.281 -0.0704 -41.1
-20.6
Ans.
4m
4m
© 2012 Pearson Education, Inc., Upper Saddle River, NJ NJ.. All rights reserved.This material is protected under under all copyright laws as they currently currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. publisher.
Determine the moments at B and C , th then en dra draw w the moment diagram for each member of the frame. Assume the supports at A, E, and D are fixed. EI is constant. *12–20.
10 k 2 k/ ft ft 8 ft
A
8 ft
C
B
12 ft 16 ft
Member Stiffness Factor and Distribution Factor. KAB
=
4EI 4EI EI = = LAB 12 3
= KBE = KCD =
KBC
(DF)AB = (DF)EB = (DF)DC = 0
E
(DF)BA =
4EI 4EI EI = = L 16 4
> >4 + >4 = 0.4
EI 3
>
EI 3
> >3 + >4 + >4 = 0.3 >4 = 0.5 >4 + >4
D
+ EI
EI
EI 4
(DF)BC = (DF)BE =
EI
(DF)CB = (DF)CD =
EI
EI
EI
EI
EI
Fixed End Moments. Referring to the table on the inside back cover,
(FEM)AB = (FEM)BA = (FEM)BC (FEM)CB (FEM)BE
wL2AB
12
= -
2(122)
12 2 2(12 )
wL2AB
= - 24 k # ft
#
= = 24 k ft 12 12 10(16) PL = - BC = = - 20 k ft 8 8 10(16) PLBC = = = 20 k ft 8 8 = (FEM)EB = (FEM)CD = (FEM)DC = 0
#
#
Moment Distribution. Tabulating the above data,
Joint
A
Member
AB
DF
0
FEM
–0.80
Dist. CO
1.00
Dist. CO
–0.03
Dist. CO
0.4 24
0.3 0 –1.20
2.00
1.50
R
BC
0.075
–0.045
–1.20 1.50
–0.00225 24.41
–0.045 –0.1875
0.05625
0.5
DC
EB
0
20
0
0
0
R –10
–10
R –5
R
–0.60
R 0.30
0.30
R –0.375
–0.375
R
R
0.75
–0.0225
0.05625 R 0.01125
R 0.005625 –24.72
E
0
–0.0016875 –0.0016875 0.3096
CD
D
0.5
–20
0.15
R –0.06
CB
0.3
–5
0.0375
–23.79
C
BE
R –1.60
R
Dist.
aM
BA
–24
Dist. CO
B
10.08
0.15
R –0.1875
0.01125
0.028125
–0.01406
R
R
0.005625
–0.6 0.75 –0.0225 0.028125
–0.01406 –10.08
Using these results, the shear at both ends of members AB, BC, BE, and CD are computed and shown in Fig. a. Subsequently Subsequently,, the shear and moment diagram can be plotted.
–5.031
0.1556
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12–20. 12– 20.
Continue Con tinued d
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Determine the moments at D and C , then draw draw the moment diagram for each member of the frame. frame. Assume the supports at A and B are pins. EI is constant. 12–21.
16 kN
1m
3m
D
C
A
B
4m
Moment Distribution. No sidesway, sidesway, Fig. b. KDA
= KCB =
3EI 3EI = L 4
KCD
=
4EI 4EI = = EI L 4
(DF)AD = (DF)BC = 1 (DF)DA = (DF)CB = (DF)DC = (DF)CD = (FEM)DC = (FEM)CD = -
Pb2a L2 Pa2b L2
Joint
A
Member
AD
DF
1
FEM
0
Dist.
EI 3EI 4 + EI
>
= = -
16(32)(1) 42 16(12)(3) 42
DA 3 7
0
–9
CD 4 7
CB 3 7
BC
3
0
0
–0.105
0.040 R –0.120
–0.090
0.034 R –0.011
–0.009
0.003
0.003
–0.010
–0.007
4.598
–4.598
2.599
–2.599
0.030
CO 0.026
CO
0
DC 4 7
B
0.420 R –0.140
CO
aM
C
–1.102
0.315
Dist.
= 3 kN # m
0.490 R –1.470
CO
Dist.
= - 9 kN # m
–1.286
0.367
Dist.
4 7
5.143 R –1.714
CO
Dist.
>
D
3.857
Dist.
=
>
3EI 4 3 = 3EI 4 + EI 7
R –0.857
R –0.735 R –0.070 R –0.060
R –0.006
1
2.572 0.245 0.210 0.020 0.017
0
Using these results, results, the shears at A and B are computed and shown in Fig. d. Thus Thus,, for the entire frame
+ © F = 0; 1.1495 - 0.6498 - R = 0 R = 0.4997 kN x
:
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12–21. 12– 21.
Continue Con tinued d
For the frame in Fig. e, Joint
A
Member
AD
DF
1
FEM
0
D DA 3 7
–10
Dist.
4.286
CO Dist.
–1.224
CO Dist.
0.350
CO Dist.
–0.100
CO Dist.
0.029
CO Dist.
aM
0
C DC 4 7
CD 4 7
0
0
5.714 R
5.714
R 2.857
0.467 R
R 0.234
–0.134 R
R –0.067
0.038 R
R 0.019
CB 3 7
–10
BC
1 0
4.286
2.857
–1.633 R –1.633
R –0.817
B
–1.224
–0.817
0.467
0.350
0.234 –0.134
–0.100
–0.067 0.038
0.029
0.019
–0.008
–0.011
–0.011
–0.008
–6.667
6.667
6.667
–6.667
0
Using these these results, results, the shears shears at A and B caused by the application of R ¿ are computed and shown in Fig. f . Fo Forr the entire frame frame,,
+ © F = 0; R¿ 1.667 - 1.667 = 0 R ¿ = 3.334 kN x
:
Thus, MDA MDC MCD MCB
a 0.4997 b = 3.60 kN # m 3.334 0.4997 = - 4.598 + (6.667) a b = - 3.60 kN # m 3.334 0.4997 = 2.599 + (6.667) a b = - 3.60 kN # m 3.334 0.4997 = 2.599 + ( - 6.667) a b = - 3.60 kN # m 3.334 = 4.598 + ( - 6.667)
Ans.
Ans.
Ans.
Ans.
© 2012 Pearson Education, Inc., Upper Saddle River, NJ NJ.. All rights reserved.This material is protected under under all copyright laws as they currently currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. publisher.
Determine the moments acting at the ends of each member.. Assum member Assume e the supports at A and D are fixed. The moment of inertia of each member is indicated in the figure. E = 29(103) ksi. 12–22.
6 k/ k/ ft ft
B
I BC = 1200 in I CD
15 ft
I AB
24 ft
(DF)AB = (DF)DC = 0 (DF)BA =
>
1 IBC) 15 (12 1 IBC) (12
>15 + >24 = 0.5161 IBC
(DF)BC = 0.4839 (DF)CB =
>
IBC 24
>
>
0.5IBC 10 + IBC 24
= 0.4545
(DF)CD = 0.5455 (FEM)AB = (FEM)BA = 0 (FEM)BC =
- 6(24)2 12
= - 288 k # ft
#
(FEM)CB = 288 k ft (FEM)CD = (FEM)DC = 0
Joint
A
Mem.
AB
BA
BC
CB
DF
0
0.5161
0.4839
0.4545
B
-288
FEM 74.32 16.89 4.09 0.93 0.22 0.05
aM
C
96.50
R 148.64 R R
R R
R
33.78
38.01
7.66 R -7.20
-8.64
1.74 R -1.74
-2.09
0.42 R -0.40
-0.47
0.10 R -0.10
-0.11
-0.87 0.45
0.20 0.10
-0.05
0
288
31.67 R -31.67
-65.45
DC
0.5455
-157.10
-3.60 1.86
CD
139.36 R -130.90
-15.84 8.18
D
R
R
R
R
R R
69.68 15.84 3.83 0.87
0.21
0.05
0.02
0.02
-0.02
-0.03
193.02
-193.02
206.46
-206.46
R R R R R R
-78.55 -19.01 -4.32 -1.04 -0.24 -0.06 -103.22
C
= 600 in 4 D
800 in4
A
Consider no sideway
4
10 ft
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12–22. 12– 22.
+
:
Continue Con tinued d
a Fx = 0 (for the frame without sideway)
+ 19.301 - 30.968 = 0 R = 11.666 k R
(FEM)CD = (FEM)DC = 100 =
6E(0.75IAB) ¢ ¿ 102
100(102) ¢¿ = 6E(0.75IAB) (FEM)AB = (FEM)BA =
6EIAB ¢ ¿ 152
=
2
) a 6 15 b a 6 100(10 b = 59.26 k # ft (0.75 ) EIAB 2
E
Joint
A
Mem.
AB
BA
BC
CB
DF
0
0.5161
0.4839
0.4545
FEM
59.26
-15.29 5.87
-0.84 0.32
-0.05 0.02
aM R¿ MAB MBA MBC MCB MCD MDC
B
49.28
C
59.26 R -30.58 R
11.73
R
0.65
R
-28.68 R –45.45 R -14.34 -22.73 11.00 R
0.04
-125
CD
DC
0.5455
R
7.82
-3.00
-0.79 0.36
R
0.43
0.30
-0.09 R -0.14 R -0.04 -0.07 0.03 R
-39.31
-54.55
5.50
0.60 R
0.01 39.31
6.52
R
–1.58 R -2.50
0.18
R -0.09
D
100
3.26
R -1.68
IAB
0.02
R
-0.16 0.02
0.02
-50.55
50.55
0 100
R R R R R R
-27.28 3.91
-1.50 0.22
-0.08 0.01 75.28
= 5.906 + 12.585 = 18.489 k
a 11.666 b (49.28) = 128 k # ft 18.489 11.666 = 193.02 - a b (39.31) = 218 k # ft 18.489 11.666 = - 193.02 + a b ( - 39.31) = 218 k # ft 18.489 11.666 = 206.46 - a b ( - 50.55) = 175 k # ft 18.489 11.666 = - 206.46 + a b (50.55) = 175 k # ft 18.489 11.666 = - 103.21 + a b (75.28) = - 55.7 k # ft 18.489 = 96.50 +
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
© 2012 Pearson Education, Inc., Upper Saddle River, NJ NJ.. All rights reserved.This material is protected under under all copyright laws as they currently currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. publisher.
Determine the moments acting at the ends of each member of the frame. EI is the constant. 12–23.
1.5 k/ k/ ft ft 15 k B
C
A
D
20 ft
Consider no sideway 24 ft
(DF)AB = (DF)DC = 1 (DF)BA = (DF)CD =
>
3I 20 = 0.4737 3I 20 + 4I 24
>
>
(DF)BC = (DF)CB = 0.5263 (FEM)AB = (FEM)BA = 0 (FEM)BC =
- 1.5(24)2 12
= - 72 k # ft
#
(FEM)CB = 72 k ft (FEM)CD = (FEM)DC = 0
Joint
A
Member
AB
BA
BC
CB
DF
1
0.4737
0.5263
0.5263
B
C
-72.0
FEM 34.41
-8.98
4.98
R
-2.36
1.31
-0.62
0.18 R -0.18
-0.16
0.05 R -0.05
-0.04 -0.01 -46.28
-0.02
aM
R
18.95
0.69 R -0.69
-0.09 0.04
R
2.62 R -2.62
-0.35 0.16
72.0
9.97 R -9.97
-1.31 0.62
0.4737
-34.11
-4.98 2.36
CD
37.89 R -37.89
-18.95 8.98
D
R
R
R
0.35 0.09 0.02
0.01
0.01
-0.01
46.28
-46.28
46.28
DC
1
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12–23. 12– 23.
Continue Con tinued d
+ © F = 0 (for the frame without sidesway) x
;
R
+ 2.314 – 2.314 – 15 = 0
R
= 15.0 k
Joint
A
Mem.
AB
BA
BC
CB
DF
1
0.4737
0.5263
0.5263
FEM
B
C
–100 47.37
–0.86
47.37
–13.85 R –13.85
–12.47
R
1
–6.93
1.82
1.82
R
–0.96 R –0.96
R
0.25 R
R
R
3.28 –0.86
–0.48 0.25
0.23
0.13
–0.07 R –0.07
–0.03
R¿
DC
26.32
3.64
0.13 –0.06
R
3.64 R
–0.48 0.23
0.4737
52.63 R 52.63
–6.93 3.28
CD
–100 26.32
–12.47
D
–0.06
–0.03
0.02
0.02
0.02
0.02
–62.50
62.50
62.50
–62.50
= 3.125 + 3.125 = 6.25 k
MBA MBC MCB MCD MAB
15 a 6.25 b ( - 62.5) = - 104 k # ft 15 = - 46.28 + a b (62.5) = 104 k # ft 6.25 15 = 46.28 + a b (62.5) = 196 k # ft 6.25 15 = - 46.28 + a b ( - 62.5) = - 196 k # ft 6.25
= 46.28 +
= MDC = 0
Ans.
Ans.
Ans.
Ans.
Ans.
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*12–24. Determine the moments acting at the ends of each member. Assume the joints are fixed connected and A and B are fixed supports. EI is constant.
C
D
12 ft 0.2 k/ ft ft
18 ft B
Moment Distribution. No sidesway, sidesway, Fig Fig.. b, KAD
=
4EI 4EI 2EI = = LAD 18 9
KBC
=
EI 4EI 4EI = = 12 3 LBC
(DF)CD =
A
EI 4EI 4EI = = LCD 20 5
=
20 ft
(DF)AD = (DF)BC = 0 (DF)DC =
KCD
(DF)DA =
> =9 >9 + >5 19 >5 = 3 >5 + >3 8
>
2EI 59 10 = 2EI 9 + EI 5 9
>
>
EI 5
2EI
EI
EI
EI
EI
(FEM)AD = (FEM)DA =
(DF)CB =
wL2AD
12
wL2AD
12
=
= -
0.2(182) 12
0.2(182) 12
>
>
EI 3
EI 5
>
+ EI 3
=
5 8
= - 5.40 k # ft
= 5.40 k # ft
(FEM)DC = (FEM)CD = (FEM)CB = (FEM)BC = 0
Joint
A
Member
AD
DF
0
FEM
–5.40
Dist. CO
D DC
CD
CB
BC
10 19
9 19
3 8
5 8
0
0
0
0
0
–2.842
–2.558
–1.421
Dist. CO
0.240
Dist. CO
–0.063
R
–0.126
–0.114
Dist. CO
0.010
Dist.
aM
–6.884
B
DA
5.40 R
C
–0.005
–0.005
2.427
–2.427
R R
R R
–1.279 0.480
0.799
R
0.400
–0.057 0.021
–0.835
0.036
0.835
R
0.018 0.418
© 2012 Pearson Education, Inc., Upper Saddle River, River, NJ.All rights reserved. This material material is protected under all copyright copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. publisher.
12–24. 12– 24.
Continue Con tinued d
Using these results, results, the shears at A and B are computed and shown in Fig. d. Thu Thus, s, for the entire frame,
+ © F = 0; 0.2(18) + 0.104 – 2.048 - R = 0 R = 1.656 k x
:
For the frame in Fig. e,
#
(FEM)BC = (FEM)CB = - 10 k ft; (FEM)AD = (FEM)DA = -
Joint
A
Member
AD
DF
0
FEM CO CO
R
CO CO
–0.022
Dist. CO
0.003
Dist.
aM
–3.735
C
R R
B
DC
CD
CB
BC
10 19
9 19
3 8
5 8
0
–10
–10
2.339 –0.987 0.104
0.052
Dist.
#
DA
–0.494 R
240 EI
6EI(240/EI) 6EI ¢ ¿ = = - 4.444 k ft L2 182
1.170
Dist.
¢¿ =
–4.444 R
Dist.
6EI ¢ ¿ = –10 L2
D
–4.444
Dist.
-
0.044 0.005
2.105 R
R 1.875
3.75 1.053
–0.888 R –0.395
R –0.198
0.094 R
R 0.084
0.004 R
R 0.004
–0.658
–0.444 0.767
0.277
0.047
–0.040 R –0.018
R –0.009
6.25
–0.029
–0.020 0.008
0.012
0.002
–0.002
–0.002
–0.001
–0.001
–3.029
3.029
4.149
–4.149
R R R R R
3.125 –0.329 0.139 –0.015 0.006
–7.074
© 2012 Pearson Education, Inc., Upper Saddle River, NJ NJ.. All rights reserved.This material is protected under under all copyright laws as they currently currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. publisher.
12–24. 12– 24.
Continued Cont inued
Using these results, results, the shears at both ends of members AD and BC are computed and shown in Fig. f . For the entire frame,
+
:
a Fx = 0;
R¿
- 0.376 – 0.935 = 0 R ¿ = 1.311 k
Thus, MAD MDA MDC MCD MCB MCD
a 1.656 b ( - 3.735) = 11.6 k # ft 1.311 1.656 = 2.427 + a b ( - 3.029) = - 1.40 k # ft 1.311 1.656 = - 2.427 + a b (3.029) = 1.40 k # ft 1.311 1.656 = - 0.835 + a b (4.149) = 4.41 k # ft 1.311 1.656 = 0.835 + a b ( - 4.149) = - 4.41 k # ft 1.311 1.656 = 0.418 + a b ( - 7.074) = - 8.52 k # ft 1.311 = - 6.884 +
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8k
Determine the moments at joints B and C , then draw the moment diagram for each member of the frame. The supports at A and D are pinned. EI is constant. 12–25.
C
B
12 ft
A
D
Moment Distribution. For the frame with P acting at C , Fi Fig. g. a, KAB
= KCD
3EI 3EI = = L 13
KBC
(DF)AB = (DF)DC = 1 (DF)BA = (DF)CD = (DF)BC = (DF)CB =
5 ft
4EI 2EI = = 10 5
>
>
3EI 13
>
>
3EI 13 + 2EI 5
=
15 41
2EI 5 26 = 3EI 13 + 2EI 5 41
>
#
(FEM)BA = (FEM)CD = 100 k ft;
>
3EI ¢ ¿ = 100 L2
¢¿ =
16900 3EI
From the geometry shown in Fig. b,
¢ ¿ BC =
5 5 10 ¢¿ + ¢¿ = ¢¿ 13 13 13
Thus
(FEM)BC = (FEM)CB = -
6EI ¢ ¿ BC L2BC
6EI
= -
a 1013 b a 16900 b 3 EI
10
2
= - 260 k # ft
10 ft
5 ft
© 2012 Pearson Education, Inc., Upper Saddle River, River, NJ.All rights reserved. This material material is protected under all copyright copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. publisher.
12–25. 12– 25.
Continue Con tinued d
© 2012 Pearson Education, Inc., Upper Saddle River, NJ NJ.. All rights reserved.This material is protected under under all copyright laws as they currently currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. publisher.
12–25. 12– 25.
Continued Cont inued
Joint
A
Member
AB
BA
BC
CB
CD
DF
1
15/41
26/41
26/41
15/41
FEM
0
Dist.
B
100 58.54
-260
18.56
-32.17 R -32.17 R -16.09 -16.09
-18.56
5.89
10.20 R 10.20
5.89
CO Dist. CO
5.10
-1.87
Dist. CO Dist.
0.59
CO
-0.19
CO Dist.
0.06
CO Dist. 0
R
R
1.03 R
R
1 0
5.10
1.03
-1.87 0.59
0.51
-0.32 R -0.32 R -0.16 -0.16 0.10 R
0.10
0.05
0.05
R
DC
50.73
-3.23 R -3.23 R -1.62 -1.62 0.51
Dist.
100 58.54
50.73
Dist.
-260
D
101.46 R 101.46
CO
aM
C
-0.19 0.06
-0.02
-0.03
-0.03
-0.02
144.44
-144.44
-144.44
-144.44
0
Using these results, results, the shears at A and D are computed and shown i n Fig. c. Thus for the entire frame,
+ a Fx = 0; 24.07 + 24.07 - P = 0 P = 48.14 k
:
Thus, Thu s, for P = 8 k, MBA MBC MCB MCD
8 a 48.14 b (144.44) = 24.0 k # ft 8 = a b ( - 144.44) = - 24.0 k # ft 48.14 8 = a b ( - 144.44) = - 24.0 k # ft 48.14 8 = a b (144.44) = 24.0 k # ft 48.14
=
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© 2012 Pearson Education, Inc., Upper Saddle River, River, NJ.All rights reserved. This material material is protected under all copyright copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. publisher.
Determine the moments at C and D, then draw draw the moment diagram for each member of the frame. frame. Assume the supports at A and B are pins. EI is constant. 12–26.
3k C
12 ft D
6 ft
A
Moment Distribution. For the frame with P acting at C , Fig ig.. a, KAD
=
EI 3EI 3EI = = LAD 6 2
KCD
=
4EI 4EI 2EI = = LCD 10 5
(DF)AD = (DF)BC = 1 (DF)DC = (DF)CD =
KBC
=
(DF)DA =
EI 3EI 3EI = = LBC 12 4
>
>
EI 2
EI 2
>
+ 2EI 5
=
8 ft
5 9
2EI/5 4 = EI 2 + 2EI 5 9
>
> 2 >5 8 = >5 + >4 13 EI
2EI
EI
#
(FEM)DA = 100 k ft; (FEM)CB =
(DF)CB =
3EI ¢ ¿ = 100 L2DA
>
>
EI 4
>
>
2EI 5 + EI 4 1200 EI
¢¿ =
#
3EI(1200 EI) 3EI ¢ ¿ = = 25 k ft 2 LCB 122
B
=
5 13
© 2012 Pearson Education, Inc., Upper Saddle River, NJ NJ.. All rights reserved.This material is protected under under all copyright laws as they currently currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. publisher.
12–26. 12– 26.
Continue Cont inued d
Joint
A
Member
AD
DF FEM
D
B
DA
DC
CD
1
5 9
4 9
8 13
5 13
1
0
100
0
0
25
0
Dist. Dist.
-44.44 R -15.38 R -22.22 -7.69
-9.62
4.27
3.42 R 13.67
8.55
CO
6.84
-3.80
Dist. CO Dist.
0.29
CO
-0.26
CO
-0.02
Dist. 0
44.96
R
0.24 R
R
BC
1.71
-3.04 R -1.05 R -1.52 -0.53 0.47
Dist.
CB
-55.56
CO
aM
C
0.94
-0.66 0.58
0.12
-0.21 R -0.07 R -0.04 -0.11 -0.02 0.07 -44.96 -23.84
-0.05 0.04 23.84
0
Using the results, results, the shears at A and B are computed and shown in Fig. c. Thus Thus,, for the entire frame,
+ a FX = 0; 7.493 + 1.987 – P = 0 P = 9.480 k
:
Thus, Thu s, for P = 3 k, MDA MDC MCD MCB
3 a 9.480 b (44.96) = 14.2 k # ft 3 = a b ( - 44.96) = - 14.2 k # ft 9.480 3 = a b ( - 23.84) = - 7.54 k # ft 9.480 3 = a b (23.84) = 7.54 k # ft 9.480
=
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