Solution: 11-1. Determine The Moments at A, B, and C, Then Draw The
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Determine the moments at A, B, and C , then draw the moment diagram or the beam. The moment o inertia o each span is indicated in the igure. Assume the support at B is a roller and A and C are are ixed. E 29(103) ksi. 11–1.
30 k
800 lb>ft
C
=
A
I AB
4
900 in
5
B I BC
24 ft
4
1200 in
5
8 ft
8 ft
Solution
(DF) AB
=
0 (DF)BA
(DF)BC
=
0.6667
(FEM) AB
=
(FEM)BA
=
(FEM)BC
=
(FEM)CB
=
>
- 0.8(24)2 12
=
>
0.75 I BC BC 24 + I BC BC 16 (DF)CB
>
0.75 I BC BC 24 =
=
- 38.4 k # t
38.4 k # t
-
30(16) 8
=
- 60.0 k # t
60.0 k # t
A
Mem.
AB AB
BA
BC
CB
0
0.3333
0.6667
0
FEM
- 38.4
C
B
38.4 - 0.0 7.20
a M - 34.8
=
M BA
=
M BC BC
=
M CB
=
- 34.8 k # t 45.6 k # t
- 45.6 k # t 67.2 k # t
60.0
14.40
3.60
M AB
0.3333
0
Joint DF
=
7.20 45.6 - 45.6
67.2 k # t
Ans. Ans. Ans. Ans.
Ans. M AB M BA M BC M CB 5 18
=
=
=
=
- 34.8 k # t 45.6 k # t - 45. k # t 67.2 k # t
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Determine the moments at B and C . EI is constant. Assume B and C are are rollers and A and D are pinned. 11–2.
3 k>ft
A
B
8 ft
C
20 ft
D
8 ft
Solution FEM AB
=
FEMCD
=
FEMBA
=
FEMDC
=
FEMBC
=
wL2
-
=
wL2 12 wL2 12
=
DF AB
3EI , 8
=
1
=
=
- 16
16
K BC
=
4EI , 20
wL2 =
=
100
12 K CD
=
3EI 8
DF DC
DFBA
=
DFCD
DFBC
=
DFCB
Joint
=
- 100 FEMCB
12
K AB
=
=
3EI 8 3EI 4EI + 8 20 1 - 0. 52
=
A
0.652
0.348 B
Member AB
BA
DF 1 FEM - 16
0.652 16
Dist.
54.782
16
=
C BC
0.348 - 100
CB
CD
0.348 10 0
0.652 - 16
8
- 14.609
14 14.609
Dist.
4.310
2.299
- 2.299
- 1.149
1.149
0.40 0
- 0.400
- 0.200
0.20 0
0.070
- 0.070
CO 0.750
CO Dist.
0.130
- 0.035
CO Dist.
a M
0.023 0
84.0
DC
1 16
29.218 - 29.218 - 54.782 - 1
CO
Dist.
D
0.0 12
- 84.0
-8 - 4.310 - 0.750 - 0.130
0.035
- 0.012 84.0
- 0.023 - 84.0
0 k # t
Ans. Ans. M BA BA M BC BC M CB M CD
519
=
=
=
=
84.0 k # t - 84.0 k # t 84.0 k # t - 84.0 k # t
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Determine the reactions at the supports. Assume A is ixed and B and C are are rollers that can either push or pull on the beam. EI is is constant. 11–3.
12 kN>m
A
B
C
5m
2.5 m
Solution Member Stiffness Factor and Distribution Factor. Factor. K AB
4EI L AB
=
(DF) AB
=
(DF)BC
=
(DF)CB
=
=
4EI 5
0
=
0.8EI
(DF)BA
=
=
3EI LBC
0.8EI 0.8EI + 1.2EI
=
1.2 EI 0.8EI + 1.2EI
K BC BC
=
=
3EI 2.5
=
1.2EI
0.4
0.6
1
Fixed End Moments. Reerring to the table on the inside back cover,
(FEM) AB
=
(FEM)BA
=
(FEM)BC
=
wL2 12 wL2 12
12(152)
=
=
12
=
12(52)
(FEM)CB
12 =
=
- 25 kN # m
25 kN # m
0
Moment Distribution. T Tabulating abulating the above data, A Joint Member AB
DF
B
C
BA
0
BC
0.4
0.6
1
0
0
FEM - 25
25
- 10
- 15
15
- 15 kN # m
Dist. CO
CB
-5
a M - 30
Using these results, the shear at both ends o members AB and BC are are computed and shown in Fig. a. From this fgure, A x
=
M A
=
0 A y
=
33 kN
30 kN # ma
C y
=
B y
=
27 + 6
6 kN T
=
33 kN
Ans. 0 A x =
Ans.
A y B y
Ans.
=
=
M A C y 5 20
=
=
33 kN c 33 kN
30 kN # m B
6 kN T
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all 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.
500 lb
Determine the reactions at the supports and then draw the moment diagram. Assume Assume A is ixed. EI is is constant. *11–4.
800 lb>ft
B
A
20 ft
D
C
20 ft
15 ft
500 lb 800 lb>ft
10.4 k · ft B
A
20 ft
Solution FEMBC
K AB
=
=
wL2 12
=
- 26.67, FEMCB
4EI , K BC BC 20
=
4EI DF AB 20
=
=
wL2 12
=
0 DFCB
Joint
A
Member
AB
BA
BC
0
0.5
0.5
DF Dist. CO
6.667
Dist. CO Dist. CO
0.8333
Dist. CO Dist. CO
- 9.583
0.1042
1.667
1.667
- 1.1979
=
=
1
0
- 0.41 7
4EI 20 4EI 4EI + 20 20
=
0.5
6.667 7
2.396
- 2.39 0.8333 0.2994
M A
=
10.4 k # t
Ans.
A x
=
0
Ans.
0.1042
A y
=
1.56 k
Ans.
0.07485 - 0.1042
B y
=
10.2 k
Ans.
C y
=
7.84 k
Ans.
- 0.1497
- 7.5 k # t
- 19.1 7
0.2083 - 0.2994
7.84 k
- 7.5
0.5990 - 0.8333
20.7
DFBC
CD
26.67
- 3.333
0.07485 10.4
=
CB
4.7917 - .
0.2083
Dist.
a M
13.33
- 0.5(15)
10.215 k
15 ft
C
13.33
0.5990 0.2994
1 DFBA
- 26.67
4.7917 2.396
=
=
B
FEM
26.67 M CD CD
D
C
20 ft
1.555 k
- 20.7
7.5 - 7.5 k # t
Ans. M A 10.4 k t A
#
=
A x A y
521
=
=
0 1.56 k T
B y
=
10.2 k
C y
=
7.84 k
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Determine the moments at A, B, and C . Assume the support at B is a roller and A and C are are ixed. EI is is constant. 11–5.
3 k>ft 2 k>ft
A
B
3 6 ft
C
24 ft
Solution (DF) AB
=
(DF)BC
=
> 36 > 36 + > 24 0.6 (DF) 0 0 (DF)BA
I
=
CB
(FEM) AB
=
(FEM)BA
=
(FEM)BC
=
(FEM)CB
=
- 2(36)2 12
=
I
I
=
0.4
=
- 216 k # t
216 k # t
- 3(24)2 12
=
- 144 k # t
144 k # t
Joint
A
Mem.
AB
BA
BC
CB
0
0.4
0.6
0
- 21
216
- 144
144
DF
FEM
Dist. CO
a M
B
- 28.8
- 14.4
- 230
C
- 43.2
187
- 21.6 - 187
- 122 k # t
Ans.
Ans. M AB M BA BA M BC BC M CB CB 5 22
=
=
=
=
- 230 k # t 187 k # t - 187 k # t - 122 k # t
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all 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.
Determine the moments at B and C , then draw the moment diagram or the beam. The supports at A, B, C , and D are pins. Assume the horizontal reactions are zero. EI is constant. 11–6.
12 kN>m 4m C A
4m
D
B 4m 12 kN>m
Solution Member Stiffness Factor and Distribution Factor. Factor. K AB
=
(DF) AB
3EI L AB =
=
3EI K BC BC 4
1 (DF)BA
6EI LBC
=
>
=
3EI 4
=
>
6EI 4
>
3EI 4 + 3EI 2
=
=
3EI 2 1 (DF)BC 3
>
3EI 2
=
>
>
3EI 4 + 3EI 2
=
2 3
Fixed End Moments. Reerring to the table on the inside back cover,
(FEM)BA
=
wL2 8
12(42) =
8
=
24 kN # m (FEM)BC
=
0
Moment Distribution. T Tabulating abulating the above data,
Joint
A
Member
AB
BA
BC
DF
1
1/3
2/3
FEM
0
24
0
Dist.
-8
- 16
a M
0
16
- 16
B
Ans.
Using these results, the shear at both ends o members m embers AB, BC , and CD are computed and shown in i n Fig. a. Subsequently the shear and moment diagrams can be plotted, Figs. b and c, respectively.
Ans. M B - 16.0 kN m M C 16.0 kN m =
=
523
#
#
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all 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.
Determine the moments at A, B, and C , then draw the moment diagram. Assume the support at B is a roller and A and C are are ixed. EI is is constant. 11–7.
900 90 0 lb lb 90 900 0 lb lb
B
A
6 ft
400 lb
6 ft
6 ft
C
10 ft
10 ft
Solution (DF) AB
=
(DF )CB
=
(FEM) AB (FEM)BA
0 (
0 (DF)BA
(FEM)BC
=
(FEM)CB
=
=
- 2(0.9)(18) 9
=
=
DF )BC
=
>
>
I 18
>
I 18 + I 20
=
0.5263
0.4737 =
- 3.60 k # t
#
3.60 k t - 0.4(20) 8
=
- 1.00 k # t
1.00 k # t
Joint
A
Mem.
AB
BA
BC
CB
0
0.5263
0.4737
0
DF FEM
Dist. CO
a M
B
- 3. 0
3.60
- 1.368
- 0.684 - 4.28
C
- 1.00
- 1.232
2.23
1.00
- 0.616
- 2.23
0.384 k # t
Ans.
Ans.
M AB M BA M BA M CB
5 24
# =
=
=
=
- 4.28 k t 2.23 k # t - 2.23 k # t 0.384 k # t
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all 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.
Determine the moments at B and C , then draw the moment diagram or the beam. Assume the supports at B and C are are rollers and A and D are pins. EI is is constant. *11–8.
12 kN>m
A
12 kN>m
B
4m
C
D
6m
4m
Solution Member Stiffness Factor and Distribution Factor. Factor. K AB
=
(DF)BA
3EI L AB
3EI 4
=
K BC BC
>
3EI 4
>
=
>
3EI 4 + 3EI 3
=
=
2EI LBC
=
2EI 6
9 (DF)BC 13
=
EI 3
(DF) AB
>
EI 3
=
>
>
=
3EI 4 + EI 3
=
1 4 13
Fixed End Moments. Reerring to the table on the inside back cover, 12(42) wL2 (FEM) AB (FEM)BC 0 (FEM)BA 24 kN # m 8 8 =
=
=
=
=
Moment Distribution. T Tabulating abulating the above data,
Joint
A
B
Member AB DF
1
FEM
0
Dist.
a M
BA
9 13
4 13
24 24
0
- 6.62 0
7.38
Due to symmetry, M CB 7.38 kN # m MCD CB =
BC
- 7.38 - 7.38 kN # m
=
Ans.
- 7.38 kN # m
Ans.
Using these results, the shear at both ends o members AB AB,, BC , and CD are computed and an d shown in Fig. a. Subsequently, the shear and moment diagram can be plotted, Figs. b and c, respectively.
Ans. M CB M CD 525
=
=
7.38 kN # m - 7.38 kN # m
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all 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.
The bar is pin supported at points A, B, C , and D. I the normal orce in the bar can be neglected, determine the vertical reaction at each pin. EI is is constant. 11–9.
16 kN
A
B 4m
4m
C 8m
D 4m
4m
16 kN
Solution Use antisymmetric load and symmetric beam. K BA BA
3EI 8
=
(DF)BA
(DF)BC
K BC BC
=
6EI 8
=
3EI 8 3EI 6EI + 8 8
=
0.3333
=
6EI 8 3EI 6EI + 8 8
=
0.6667
(3)(16)(8)
FEMBA
=
Joint
16
24 kN # m
A
B
Member AB DF
=
BA
1
BC
0.3333 0.6667
FEM
24 0
M
-8
- 16
16
- 16
#
kN m
Segment AB: a + M B
+ F y
=
=
0;
A y(8) + 16(4) - 16
0;
V BL BL + 6 - 16
=
0
Segment BC : a + M C C
+ F y
=
=
0; 0;
V BR BR(8) + 16 + 16 V CL CL + 4
=
=
+ F y B y
=
C y
=
=
0;
-
0
D y(8) + 16(4) - 16
CR - 6 + 16 0 - V CR V BL + V BR 10 + 4 14 kN =
0;
V CL + V CR
=
=
=
=
4 + 10
=
=
0
Segment CD: a + M C C
0 A y
=
=
6 kN
V BL BL
=
10 kN
V BR BR
=
4 kN
V CL CL
=
4 kN
0 D y
Ans.
Ans. =
CR V CR
6 kN
=
Ans.
10 kN
14 kN 5 26
A y
=
kN
D y
=
kN
Ans.
B y
=
14 kN
Ans.
C y
=
14 kN
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all 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.
300 lb
Determine the moments at B and C , then draw the moment diagram or the beam. Assume the supports at B and C are are rollers and A is a pin. EI is is constant. 11–10.
200 lb>ft
Solution
A
B
D
C
Member Stiffness Factor and Distribution Factor. Factor. K AB
=
3EI L AB
(DF)BA
=
(DF)CB
=
3EI 10
=
=
0.3EI
0.3EI 0.3EI + 0.4EI
=
1 (DF)CD
=
K BC BC
=
4EI LBC
3 (DF)BC 7
=
4EI 10
=
10 ft =
0.4EI
0.4EI 0.3EI + 0.4EI
4 7
=
0
Fixed End Mome Moments. nts. Reerring to the table on the inside back cover, (FEM)CD - 300(8) - 2400 lb # t (FEM)BC (FEM)CB =
(FEM)BA
=
=
2 wL AB
8
200(102) =
8
=
=
0
2500 lb # t
=
Moment Distribution. T Tabulating abulating the above data,
Joint
A
B
Member AB
C
BA
DF
1
3 7
FEM
0
250 0
Dist.
BC
- 514.29
1
0
0
0
357.15
- 153.0
- 73.47
51.03
Dist.
- 21.87
- 29.1
CO Dist.
- 10.50
24.49 - 13.99
CO
7.29
- 3.12
CO
Dist.
- 1.50
CO
- 0.45
CO
- 0.21
CO
- 0.0
342.86
- 102.05
102.05
- 48.98 48.98
- 14.58 14.58
7.00
a M
0
- 2.08 2.08
- 1.00 1.00
- 0.30 0.30
- 0.15
- 0.09
0.15
CO Dist.
- 7.00
- 0.29 0.15
Dist.
- 342.86
- 0.59 0.50 0
Dist.
- 2.00 1.04
Dist.
714.29
- 4.17 3.50
- 0.03
0.07 - 0.04
650
- 650
- 2400
- 714.29
- 97.9
CO
Dist.
- 204.09 171.43
2400
- 85.71
CO Dist.
4 7
120 0
CO Dist.
CD
- 1071.43 - 1428.57
CO Dist.
CB
0.04 0.04 2400 - 2400 lb # t
527
Ans.
10 ft
8 ft
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all 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.
11–10.
(Continued)
Using these results, the shear at both ends o members AB, BC , and CD are computed and shown in Fig. a. Subsequently, the shear and moment diagrams can be plotted, Figs. b and c, respectively.
Ans. M BA M BC M CB M CD 5 28
=
=
=
=
650 lb # t - 650 lb # t 2400 lb # t - 2400 lb # t
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all 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.
Determine the moments at A and B, then draw the moment diagram. Assume the support at B is a roller, C is a pin, and A is ixed. 11–11.
12 kN>m C B
A
5m
2.5 m
Solution (DF) AB
=
(DF)BC
=
(FEM) AB (FEM)BA (FEM)BC
(DF) 0.6 (DF) 0
=
=
=
BA
CB
- 12(5)2 12
25 kN m # (FEM)CB
Joint A Mem. AB DF
0
FEM - 25
=
=
=
>
0.4
>
4 I 5 + 3 I 2.5
=
0.4
1
- 25 kN # m
=
0
B BA
>
4 I 5
C
BC
0.6
CB
1
25
- 10 - 15 -5 M
- 30
M AB
=
M BA BA
=
M BC BC
=
15 - 15
- 30 kN # m 15 kN # m
- 15 kN # m
0 Ans. Ans. Ans.
Ans. M AB
529
=
M BA
=
M BC BC
=
- 30 kN # m 15 kN # m
- 15 kN # m
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all 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.
12 kN
Determine the moments at B and C , then draw the moment diagram or the beam. Assume C is is a ixed support. EI is constant. *11–12.
8 kN>m
A
B
C
6m
Member Stiffness Factor and Distribution Factor. Factor. =
(DF) AB
(DF)BC
3EI LBA =
=
=
3EI 6
EI 2
=
1 (DF)BA
K BC BC
>
=
>
EI 2
=
>
EI 2 + EI 2
> > >
EI 2 EI 2 + EI 2
=
4EI LBC =
0.5 (DF)CB
Fixed End Moments.
=
Joint
4EI 8
=
=
(FEM)BC
=
(FEM)CB
=
8(62)
wL2 8
-
=
PL 8
PL 8
8
=
-
=
0.5
8
12(8) =
8
=
=
A
DF
1
FEM
0
Dist.
B BA
=
0
a M
C BC
0.5
- 12
- 12
- 12
0
24
CB
0.5
36
CO
0 12
-6
- 24
6 kN # m
Ans.
Using these results, the shear and both ends o members AB and BC are are computed and shown in Fig. a. Subsequently, the shear and moment m oment diagrams can be plotted, Fig. b.
36 kN # m
12(8)
Member AB
EI 2
Reerring to the table on the inside back cover, (FEM)BA
4m
Moment Distribution. T Tabulating abulating the above data,
Solution
K BA BA
4m
- 12 kN # m
12 kN # m
Ans. M AB M BA M BA M CB 5 30
=
=
=
=
0 kN # m 24 kN # m - 24 kN # m - 6 kN # m
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all 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.
Determine the moments at the ends o each member o the rame. Assume the joint at B is ixed, C is is pinned, and A is ixed. The moment o inertia o each member is listed in the igure. E 29(103) ksi. 11–13.
2 k>ft B I BC
=
4
800 in
5
C
12 ft
8 ft
4k
I AB
550 in4
5
8 ft
A
Solution (DF) AB
=
(DF)BA
=
(DF)BC
=
0
>
4(0.6875 I BC BC ) 16
>
0.5926 (DF)CB
- 4(16) 8
=
1
=
(FEM)BC
=
Joint
A
Mem.
AB
BA
0
0.4074
FEM
- 2(122) 12
- 8.0
Dist. CO
a M
0.4074
=
=
- 24 k # t (FEM)CB
8 k # t 24 k # t
=
B
1
- 24.0 9.482
3.259
CB
0.5926
8.0
C BC
6.518
Dist. CO
=
=
- 8 k # t (FEM)BA
(FEM) AB
DF
>
4(0.6875 I BC ) 16 + 3 I BC 12
24.0
- 24.0
- 12.0
4.889
7.111
2.444
- 2.30
19.4
- 19.4
0 k # t
Ans.
Ans. M AB M BA M BC M CB 531
=
=
=
=
- 2.30 k # t 19.4 k # t - 19.4 k # t 0 k # t
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4 k>ft
Determine the internal moments acting at each joint. Assume A, D, and E are pinned and B and C are are ixed joints. The moment o inertia o each member is listed in the igure. Take E 29(103) ksi. 11–14.
A
=
B
I ABC
800 in
5
6 ft
C
4
6 ft
18 ft
20 k I BD
600 in
5
4
I CE
1000 in
5
D
Solution (DF) AB
=
(DF)BA
=
(DF)BC
=
(DF)BD
1
>
3( I ABC ) 6
>
>
>
3( I ABC ) 6 + 4( I ABC ) 18 + 3(0.75 I ABC ) 12
=
0.5496
0.2443
=
0.2061 4( I ABC ) 18
(DF)CB (DF)CE (DF)DB
=
0.5844
=
=
(DF)EC
(FEM) AB
=
(FEM)BA
=
(FEM)BC
=
(FEM)CB
=
(FEM)CE
=
-
>
=
12
=
0.4156
1
4(6)2 =
- 12.0 k # t
-
4(18)2
12
=
- 108 k # t
108 k # t
-
=
20(12) 8
=
Mem. AB 1
FEM - 12.0
- 30.0 k # t
#
30.0 k t (FEM)DB
Joint A DF
>
12.0 k # t
=
(FEM)EC (FEM)BD
>
4( I ABC ) 18 + 3(1.25 I ABC ) 12
=
0 B
C
E
D
BA
BD
BC
CB
CE
EC
DB
0.5496
0.2061
0.2443
0.4156
0.5844
1
1
12.0
- 108
108
- 30
30
12.0 52.76 6.0 5.61
19.79
2.10
23.45 - 32.42 - 45.58 - 30
- 16.21
11.73 - 15.0
2.49
1.36
0.68
1.25
1.91
- 0.37 - 0.14 - 0.17 - 0.52 - 0.73 - 0.26 - 0.08 M
0
0.14
0.05
0.06
76.2
21.8 - 98.0
0.03
0.05
89.4 - 89.4
0
5 32
0
6 ft
4
E
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11–14.
(Continued)
M AB
=
0
M BA BA
=
M BD BD
=
M BC
=
M CB
=
M CE CE
=
M EC M DB
=
=
Ans.
7 .2 k # t
Ans.
21.8 k # t
Ans.
- 98.0 k # t 89.4 k # t
- 89.4 k # t 0 0
Ans. Ans. Ans. Ans. Ans.
Ans. M AB
M BA BA
=
M BD BD
=
M BC M CB CB M CE CE M EC EC M DB DB 533
=
=
=
=
=
=
0 7 .2 k # t 21.8 k # t - 98.0 k # t 89.4 k t # - 89.4 k # t 0 0
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Determine the reactions at A and D. Assume the supports at A and D are ixed and B and C are are ixed connected. EI is is constant. 11–15.
8 k>ft B
C
15 ft
A
D
24 ft
Solution (DF) AB
=
(DF)DC
=
(DF)BA
=
(DF)CD
=
(DF)BC
=
(DF)CB
=
(FEM)BA
=
- 8(24)2 12 384 k # t
(FEM) AB (FEM)BC (FEM)CB (FEM)CD
=
=
=
0
>
I 15
>
# =
(FEM)DC
- 384 k t
=
0
A
Mem.
AB
BA
0
0.6154
B
118.16
C BC
22.72
147.69
- 73.84 28.40
4.37
5.46
1.05
0.16
0.20
0.03
0.04
146.28
- 28.40
292.57
0.0 1
- 5.46
- 1.05
- 0.20
- 0.04
- 292.57
- 0.01
- 0.17
- 0.0
0.02
- 0.84
- 0.33
0.10
- 4.37
- 1.68
0.53
- 22.72
- 8.74
2.73
- 118.1
- 45.44
14.20
- 0.02
0.0 1
a M
- 0.10
0.06
73.84
- 0.53
0.32
0
- 147.69 - 236.31
- 2.73
1.68 0.84
- 14.20
8.74
0.6154
DC
384
236.31
CD
0.3846
- 384
45.44
D
CB
0.3846
FEM
0.6154
=
0
=
Joint DF
>
I 15 + I 24 0.3846
- 0.03
- 0.01
292.57 - 292.57 - 146.28
Ans. A x 29.3 k =
A y
Thus, rom the ree-body diagrams,
=
M A
=
96.0 k
146 k # t
A x
=
29.3 k
D x
=
29.3 k
Ans.
D x
=
29.3 k
A y
=
96.0 k
D y
=
96.0 k
Ans.
D y
=
96.0 k
Ans.
M D
M A
=
146 k # t
M D
=
146 k # t
5 34
=
146 k # t
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18 kN
The rame is made rom pipe that tha t is ixed connected. I it supports the loading shown, determine the moments developed at each o the joints. EI is is constant. *11–16.
18 kN
C
B
4m
D
A
4m
4m
4m
Solution FEMBC
K AB
=
DF AB
=
-
K CD CD
=
2PL 9
DFDC
=
DFCD
DFBC
=
DFCB
- 48, FEMCB
4EI , K BC BC 4
=
DFBA
=
=
=
0
=
4EI 4 4EI 4EI 4 + 12
=
1 - 0.75
2PL 9
=
48
4EI 12
0.75
=
=
=
0.25
Joint
A
Member
AB
BA
BC
CB
CD
0
0.75
0.25
0.25
0.75
DF
FEM Dist. CO
18
Dist. CO
0.281
Dist. M
12
20.6
1.5
- 0.75
- 12
- 1.5
0.0938 - 0.0938 0.0234 - 0.0234 41.1
- 18
- 4.5
0.75
0.0704
- 41.1
0
6
0.1875 - 0.1875
DC
- 36
0.5625
41.1
D
48
-6 4.5
2.25
C
- 48 36
Dist. CO
B
- 0.5625
- 0.0704
- 41.1
- 2.25 - 0.281
- 20.6 kN # m
Ans.
Ans. M AB
M BA BA M BC M CB CB
535
=
=
=
=
M CD
=
M DC DC
=
20.6 kN # m 41.1 kN # m
#
m - 41.1 kN 41.1 kN # m
- 41.1 kN # m - 20. kN # m
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11–17. Determine the moments at the ixed support A and joint D and then draw the moment diagram or the rame. Assume B is pinned.
4 k>ft
C
A
D 12 ft
12 ft 12 ft
B
Solution Member Stiffness Factor and Distribution Factor. Factor. K AD
=
4EI L
=
4EI 12
(DF) AD
=
0
(DF)DC
=
(DF)DB
EI 3
=
(DF)DA
=
K DC DC
=
3EI L
=
> >4 + >4
=
EI 3
>
=
K DB DB
EI 3 + EI
EI
>
EI 4 EI 3 + EI 4 + EI 4
>
>
=
=
3EI 12
=
EI 4
0.4
0.3 (DF)CD
=
(DF)BD
=
1
Fixed End Momen Moments. ts. Reerring to the table on the inside back cover,
(FEM) AD
(FEM)DA
wL2 12
=
wL2 12
=
(FEM)DC
=
(FEM)CD
=
=
wL2 8
=
-
-
=
12
4(122) 12
-
(FEM)BD
4(122)
=
=
48 k # t
4(122) 8
=
- 48 k # t
=
- 72 k # t
(FEM)DB
=
0
Moments Distribution. T Tabulating abulating the above data,
Joint
A
Member DF FEM
a M
C
B
AD
DA
DB
DC
CD
BD
0
0.4
0.3
0.3
1
1
0
0
0
0
- 48
Dist. CO
D
48
0
9.60
7.20
57.6
7.20
- 72 7.20
4.80
- 43.2
- 4.8 k # t
5 36
Ans.
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11–17.
(Continued)
Using these results, the shears at both ends o members AD, CD, and BD are computed and shown in Fig. a. Subsequently, the shear and moment diagrams can be plotted, Figs. b and c, respectively.
Ans. M AD
=
M DA DA
=
M
=
M DC DC
=
M CD
=
M BD BD
=
DB
537
- 43.2 k # t 57.6 k # t
#
7.20 k t - 4.8 k # t 0 k # t 0 k # t
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6 kN>m
Determine the moments at A, C , and D, then draw the moment diagram or each member o the rame. Support A and joints C and D are ixed connected. EI is constant. 11–18.
C
D
E
6m
A
B
8m
7m
Solution
FEM AC
=
0 FEMBD
FEMCD
=
-
wL2 12
FEMDE
=
-
wL2 8
Joint Member
DF
=
-
=
-
0
6(82)
12
=
6(72) 8
=
- 32 kN # m
=
- FEMDC
- 36.75 kN # m
A
B
AC
BD BD
CA CA
CD
DC
DB
DE
ED
0
1
0.5714
0.4286
0.35
0.35
0.3
1
C
- 32
32
- 36.75
13.714
1.6625
1.6625
1.4250
0.83125
6.8571
FEM Dist. CO
=
18.286 9.1429
D
Dist.
0.4750 0 - 0.35625 - 2.4000 - 2.4000
CO - 0.23750
- 1.2000 - 0.17813
Dist.
0.68571
0.51429
a M M AC
=
=
BD M BD
M CA CA
=
M CD
=
M DC
=
M DB
=
M DE
=
M ED
=
8.905
18.50 - 18.50
0.062344 0.062344
Ans.
0 18.5 kN # m
Ans.
38 kN # m
- 0.675 kN # m - 37.3 kN # m
- 2.0571 0.053438
38.0 0 - 0.6752 - 37.33 kN # m
8.91 kN # m
- 18.5 kN # m
E
Ans. Ans. Ans. Ans.
0
Ans. M AC
M CA
=
=
M CD M DC
5 38
=
=
M DB
=
M DE DE
=
8.91 kN # m 18.5 kN # m
-
#
18.5 kN m 38 kN # m
- 0. 75 k kN N#m - 37.3 kN # m
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Determine the moment at B, then draw the moment diagram or each member o the rame. Support Support A is pinned. EI is constant. 11–19.
1.5 k>ft
A
B
12 ft
C
12 ft
5 ft
Solution FEMBC
=
FEMBA
=
Joint
wL2 8
Member DF
0 1.5(122) =
8
=
A
27 k # t B
C
AB
BA
BC
CB
1
0.44828
0.55172
0
FEM
27
- 12.103 - 14.897 - 7.4485 M
M B
=
0
- 14.9 k # t
14.897 - 14.897 - 7.4485 Ans.
Ans. M B - 14.9 k t =
539
#
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10 k
Determine the moments at B and C , then draw the moment diagram or each member o the rame. Assume the supports at A, E, and D are ixed. EI is is constant. *11–20.
2 k>ft 8 ft
A
8 ft
C
B 12 ft 16 ft
E
Solution
D
Member Stiffness Factor and Distribution Factor. Factor. K AB
=
4EI L AB
4EI 12
=
EI 3
=
(DF) AB
=
(DF)EB
=
(DF)BC
=
(DF)BE
=
(DF)CB
=
(DF)CD
K BC BC
(DF)DC
=
=
K BE BE
K CD CD
0 (DF)BA
> >4 + >4
=
EI 4
>
EI 3 + EI
>
EI 4
>
=
=
EI
>
EI 4 + EI 4
=
4EI L
=
4EI 16
=
EI 4
=
> >4 + >4
EI 3
>
EI 3 + EI
EI
=
0.4
0.3
0.5
=
Fixed End Moments. Reerring to the table on the inside back cover,
(FEM) AB
=
(FEM)BA
=
(FEM)BC
=
(FEM)CB
=
(FEM)BE
=
-
2 wL AB
12
-
=
PLBC
8
PLBC
8
-
2(122)
12
2(122)
2 wL AB
12
=
12
=
-
(FEM)EB
10(16)
8
8 =
- 24 k # t
24 k # t
=
=
- 20 k # t
20 k # t
10(16) =
=
=
(FEM)CD
=
(FEM)DC
=
0
Moment Distribution. T Tabulating abulating the above data,
Joint A
B
Member AB DF
0
CO
24
- 1.60
BC
0.3 0
0.3
- 1.20
- 1.20
1.50
- 0.06
- 0.045
- 0.03
Dist.
0.075 0.0375
a M - 23.79
0.05625
20
24.41
0.3096
- 10
1.50
0.30
0.15
0.75
- 0.045
- 0.375
- 0.1875
- 0.0225
0.0 05625 - 0.0016875
5 40
DC
EB
0
0
0
0
0
-5
- 0.6
0.30 0.15
0.75
- 0.375
0.028125 - 0.01406 10.08
E
0.5
0.0 1125
- 24.72
D
- 10
- 0.60
0.05625
- 0.00225 - 0.0016875
CD
0.5
- 20
1.0 0
Dist.
CB
-5 2.0 0
CO
CO Dist.
BE
- 0.80
Dist.
CO
BA
0.4
FEM - 24 Dist.
C
- 0.1875 - 0.0225
0.0 1125
- 0.01406
0.0 05625
- 10.08 k # t - 5.032
0.028125 0.1556
Ans.
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*11–20.
(Continued)
Using these results, the shear at both ends o members AB, BC , BE, and CD are computed and shown in Fig. a. Subsequently, the shear and moment diagrams can be plotted.
541
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*11–20.
(Continued)
Ans. M
- 23.79
24.41 0.3096 - 24.72 10.08 - 10.08 k # t - 5.032 0.1556 5 42
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Determine the moments at D and C , then draw the moment diagram or each member o the rame. Assume the supports at A and B are pins. EI is is constant. 11–21.
16 kN
1m
3m
D
C
A
B
4m
Solution K DA DA
=
K CB CB
=
3EI L
=
(DF) AD
=
(DF)BC
=
(DF)
=
(DF)
=
DC
CD
3EI 4
1 (DF)DA
K CD CD
=
=
>
4EI L
(DF)CB
EI
3EI 4 + EI
=
4
=
4EI 4
=
>
EI
3EI 4 =
>
=
3EI 4 + EI
3 7
7
543
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11–21.
(Continued)
Moment Distribution. No sidesway, Fig. b.
Joint
A
D
Member AD
DC
CD
CB
4 7
4 7
3 7
1
3
0
0
1
3 7
FEM
0
0
-9 5.143
3.857
CO
- 0.735
Dist.
- 0.070
Dist.
- 0.060
Dist.
- 0.006
Dist.
0.017
0.0 03 - 0.010 - 0.007
0.0 03
a M
0.020
0.034 - 0.011 - 0.009
0.026
CO
0.210
0.040 - 0.120 - 0.090
0.030
CO
0.245
0.420 - 0.140 - 0.105
0.315
CO
2.572
0.490 - 1.470 - 1.102
0.367
CO
4.598 - 4.598
0
BC
- 1.714 - 1.286
- 0.857
Dist.
B
DA
DF
Dist.
C
2.599
- 2.599 kN # m
0
Using these results, the shears at A and B are computed and shown in Fig. d. Thus, or the entire rame,
+
F x
=
(FEM)DC
(FEM)CD
0;
=
1.1495 - 0.6498 - R
-
Pb2a L2
L2
-
16(32)(1) 42
16(12)(3)
Pa2b =
=
=
42
=
=
=
0 R
=
0.4997 kN
- 9 kN # m
3 kN # m
5 44
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11–21.
(Continued)
For the rame in Fig. e, Joint
A
D
Member AD DF
1
FEM
0
C
DA
DC
CD
CB
3
4
4
3
7 - 10
Dist.
4.286
CO
Dist.
7 0
7 0
5.714
5.714
2.857
2.857
1
7 - 10
0
4.286
- 0.817 - 0.817
Dist.
0.350
CO
0.467
0.467
0.234
0.234
0.350
- 0.100 - 0.134 - 0.134 - 0.100
CO
- 0.0 7 - 0.0 7
Dist.
0.029
CO
-
0
0.008 - . 7
Dist.
a M
BC
- 1.224 - 1.633 - 1.633 - 1.224
CO
Dist.
B
0.038
0.038
0.0 19
0.0 19
-
0.011 6.667
0.029
0.011 0.008 - 6.667 6.667 kN # m
0
Using these results, the shears at A and B caused by the application o R are computed and shown in Fig. f . For the entire rame,
+
F x
=
0;
R - 1.667 - 1.667
0
=
R
=
3.334 kN
Thus, M DA DA
=
M DC DC
=
M CD
M CB
=
=
4.598 + ( - 6.667)
a 0.4997 b 3.334
- 4.598 + (6.667)
=
=
0.4997
3.334 a0.4997 b 2.599 + (6.667) a b 3.334
=
- 2.599 + ( - 6.667)
a 0.4997 b 3.334
3.60 kN # m
Ans.
- 3.60 kN # m
Ans.
3.60 kN # m
=
- 3.60 kN # m
Ans.
Ans.
Ans. M DA M DC M CD M CB
=
=
=
=
545
3. 0 kN # m - 3.60 kN # m 3.60 kN # m - 3.60 kN # m
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1.5 k>ft
Determine the moments acting at the ends o each member o the rame. EI is is the constant. 11–22.
15 k B
C
20 ft
D
A
24 ft
(DF) AB
=
(DF)DC
=
(DF)BA
=
(DF)CD
=
(DF)BC
=
(DF)CB
1
>
>
3 I 20
>
3 I 20 + 4 I 24
=
0.4737
0.5263
=
Consider no sidesway Joint
A
B
Member AB DF 1
BA 0.4737
BC 0.5263
Dist.
34.11
37.89
CO Dist. Dist.
- 9.97
- 4.98
4.98
2.62
- 2.62
- 1.31
1.31
0.69
- 0.69
- 0.35
0.35
0.16
0.18
0.04
CO 0.62
CO Dist. CO Dist. CO Dist.
0.0 1
a M
46.28
=
(FEM)BC
=
(FEM)CB
=
(FEM)BA
- 1.5(24) 12 72 k # t
=
2 =
- 8.98
- 2.36
- 0.62
- 0.18
- 0.16
- 0.09 0.05
0.09 - 0.05
- 0.04
- 0.02
0.02
0.0 1
- 0.01
- 0.01
- 4 .28
46.28
0
- 72 k # t
(FEM)CD (FEM)DC 0 + F d 0 (or the rame without sidesway): x =
=
=
R + 2.314 - 2.314 - 15 R 15.0 k
=
0
CD 0.4737
- 37.89 - 34.11
9.97
2.36
Dist.
18.95
8.98
D
72.0
- 18.95
CO
=
CB 0.5263
- 72.0
FEM
(FEM) AB
C
- 4 .28 k # t
DC 1
5 46
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11–22.
(Continued)
Joint
A
Mem.
AB
DF
B
1
FEM
BA
BC
CB
0.4737
0.5263
0.5263
- 100
Dist. CO
- 12.47
Dist. CO
- 0.8
0.23
CO Dist.
a M
M BC
=
M CB
=
M CD
=
M AB
=
- 0.06
Dist.
=
26.32
3.125 + 3.125
=
0
- 12.47 3.28
- 0.8
- 0.48 0.25
0.13
0.13
- 0.07
- 0.07
- 0.03
- 0.03
0.02
0.02
- 62.50
62.50
62.50
0.23 - 0.06 0.02
- 62.50 k # t
6.25 k =
Ans.
=
Ans.
=
=
- 0.9
0.02
=
M DC DC
47.37
- .93 3.64
0.25
15 a 6.25 b ( - 62.5) - 104 k # t 15 b (62.5) 104 k # t - 46.28 + a 6.25 15 46.28 + a b (62.5) 196 k # t 6.25 15 b ( - 62.5) - 196 k # t - 46.28 + a 6.25 46.28 +
1
1.82
- 0.48
CO
M BA
26.32
- 0.9
Dist.
=
52.63
1.82
CO
R
52.63
- .93 3.64
3.28
Dist.
0.4737
DC
- 100
- 13.85 - 13.85
CO
D CD
47.37
Dist.
C
Ans.
Ans. Ans. Ans.
M BA M BC M CB M CD CD M AB
# = =
=
=
=
k t 104 - 104 k # t
19 k # t - 19 k # t 0 M DC DC =
547
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Determine the moments at the ends o each member o the rame. The The members are ixed connected at the supports and joints. EI is is the same or each member. 11–23.
4 k>ft
B
C
20 k
6 ft
15 ft
Solution 15 k
PL wL2 FEMs or AB use , and or BA use . Each member 8 12 4EI has a K L .
6 ft
=
A
Joint
A
Member
AB
BA
BC
CB
CD
0
0.5556
0.4444
0.4444
0.5556
- 22.5
22.5
- 75
75
DF
FEM
B
14.583 d 29.167
C
23.333
- 33.333
- 16.667 4.630 d
9.259
7.407
1.440
1.152
- 5.182
- 1.646
d 0.457
0.366 - 0.128
- 0.256 0.183
0.036 d
0.071
0.057
- 0.081
- 0.041 - 2.303 Member AB: b + M B
=
0;
62.917
b + M C
=
0;
=
+
F x
=
0;
- 2.917
50.643
=
0
2.449 k
- 50.643 - 25.314 + D x(12) D x .330 k
R
=
- 6.482
d
- 2.058
d
-
d
0.320
- 0.013
=
0
=
Frame:
- 20.833
d
- 3.241
- 1.029
- 0.102
- 0.016
- 0.160
- 0.051
d
0.028
62.917 - 2.303 - 15(6) + A x(12) A x
Member CD:
- 41.667
0.576
0.229
0.0 18
0
3.704
- 0.823
0.023
DC
11.667
- 2.593 0.720 d
D
20 + 15 - 2.449 - 6.330
=
26.221
- 50. 43
- 25.314
D
5 48
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11–23.
(Continued)
Joint
A
Member
AB
BA
BC
CB
CD
0
0.5556
0.4444
0.4444
0.5556
DF FEM
B
- 100
C
- 100
27.778 d - 55.556
- 6.173 d - 12.346 1.372 d
2.744
- 0.305 d - 0.610 0.068
0.135
- 0.030 - 77.260 - 54.551
44.444
22.222
22.222
- 9.877
- 9.877
- 4.938
- 4.938
2.195
2.195
1.097
1.097
0.108
0.108
0.054
0.054
- 0.024
- 0.488 - 0.244
0
54.551
- 100
0.5556
d
- 12.346
d
2.744
d
- 0.610
27.778
- 6.173 1.372
- 0.305
d
0.135
- 0.024
54.551
DC
- 100
44.444
- 0.488 - 0.244
D
0.068
- 0.030 - 54.551
- 77.260
Member AB and CD: b + M B
+
F x
=
=
0;
0;
A x(12) - 54.55 - 77.260 A x
=
=
D x
=
0
=
10.984 k
Frame: R
10.984 + 10.984
=
21. 89 k
Thus M AB
=
M BA
=
M BC BC
=
M CB CB
=
M CD CD
M DC
=
=
- 2.303 + ( - 77.260) 62.917 + ( - 54.551)
a 26.221 b 21.968
a 26.221 b 21.968
- 62.917 + ( + 54.551) 50.643 + (54.551)
=
=
a 26.221 b 21.968
a 26.221 b 21.968
- 50.643 + ( - 54.551)
=
- 94.5 k # t
- 2.19 k # t
Ans.
2.19 k # t
Ans.
=
116 k # t
26.221
a 21.968 b 26.221 - 25.314 + ( - 77.260) a b 21.968
Ans.
=
=
- 116 k # t - 118 k # t
Ans.
Ans. M AB
M BA BA
=
=
- 94.5 k # t - 2.19 k # t
Ans.
M BC M CB Ans.
=
=
M CD CD
=
M DC
=
2.19 k t 116 k # #t
- 11 k # t - 118 k # t
549
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6 k>ft
Determine the moments acting at the ends o each member. Assume the supports at A and D are ixed. The moment o inertia o each member is indicated in the igure. 29(103) ksi. E *11–24.
B
=
I BC
1200 in
5
I CD
15 ft I AB
800 in
5
(DF) AB
24 ft
(DF)DC
=
0
=
a 23 b > 15 a 23 b > 15 + > 24 I BC I BC
(DF)BA
=
I BC I BC
(DF)BC
=
(DF)CB
=
(DF)CD
0.5161
I BC BC
0.4839
>
I BC BC 24
>
>
0.5 I BC 10 + I BC 24
=
=
=
0.4545
0.5455
Consider no sidesway Joint Mem.
DF
A AB
B
0
CO
74.32
Dist.
0.5161
0.4839
0.4545
0.5455
148.64
139.36
- 5.45 31.67
16.89
Dist.
7.66
4.09
Dist.
- 130.90
1.74
- 31.67
- 38.01
15.84
- 7.20
- 19.01
- 8.64
3.83
- 78.55
- 1.74
0.93
- 0.87
Dist. CO
0.45
0.42 - 0.20
- 0.40 0.21
0.22
- 0.47
0.10
0.10
- 0.10
- 0.11
CO
0.05
Dist.
0.02
a M
96.50
(FEM) AB
=
(FEM)BC
=
(FEM)CB
=
(FEM)
=
(FEM)BA
- 6(24)2 12
=
=
(FEM)
- 193.02
0
- 288 k # t
=
0
DC
0 (or the rame without sidesway):
R + 19.301 - 30.968
11.
=
288 k # t
CD
F x
193.02
0.02
k
=
0
0.87
- 0.05
- 0.02 206.46
- 1.04 - 0.24
0.05
- 4.32
- 2.09
CO
Dist.
0
- 157.10
69.68
- 3. 0
1.86
D DC
288
- 15.84
8.18
CO
=
CD
33.78
CO
R
CB
- 288
Dist.
+
BC
FEM
C
BA
- 0.0
- 0.03 - 206.46
- 103.22 k # t
C
600 in
5
4
D
4
A
Solution
4
10 ft
5 50
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*11–24.
(Continued)
Joint
A
Mem.
AB
BA
BC
CB
CD
DC
0
0.5161
0.4839
0.4545
0.5455
0
DF FEM
59.26
Dist. CO
B
59.26
- 30.58
- 15.29
Dist.
- 1. 8
- 0.84
Dist. 0.32
Dist.
- 0.09
- 0.05
Dist.
a M
0.65
CO
0.04 49.28
39.31
10 0
- 28.68 - 45.45 - 54.55
11.73
Dist.
CO
- 22.73 - 14.34
5.87
D
10 0
CO CO
C
11.0 0
6.52
3.26
5.50
- 1.58
- 2.50
- 1.25
- 0.79
0.60
0.36
0.18
0.30
- 0.09
- 0.14
- 0.07
- 0.04
0.03
- 27.28
7. 82 3.91
- 3.00
- 1.50
0.43 0.22
- 0.16
0.02
0.02
- 39.31 - 50.55
50.55
- 0.08 75.28
k # t
551
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*11–24.
(Continued)
(FEM)CD
(FEM) AB R
=
M AB
=
M BA
=
M BC
=
M CB
=
M CD CD
=
M DC DC
=
=
=
(FEM)DC
(FEM)BA
=
5.90 + 12.585 96.50 +
=
=
100
6E(0.75 I AB) =
6EI AB 152
=
a 11.666 b (39.31) 18.489
- 193.02 +
=
100(102) =
6E(0.75 I AB)
100(102)
a 15 b a 6 (0.75 ) b 6EI AB 2
E
128 k # t
=
218 k # t
a 11.666 b ( - 39.31) 18.489
=
a 11.666 b ( - 50.55) 18.489
=
- 206.46 +
a 11.666 b (50.55) 18.489
=
- 103.22 +
a 11.666 b (75.28) 18.489
206.46 +
I AB
=
59.26 k # t
18.489 k
a 11.666 b (49.28) 18.489
193.02 +
102
=
- 218 k # t
Ans.
Ans.
Ans.
175 k # t
Ans.
- 175 k # t
Ans.
- 55.7 k # t
Ans.
Ans. M AB
=
=
BA M M BC M CB M CD M DC
=
=
=
=
128 k # t
#
218 - 218k kt# t 175 k # t - 175 k # t - 55.7 k # t
5 52
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Determine the moments at joints B and C , then draw the moment diagram or each member o the rame. The supports at A and D are pinned. EI is is constant. 11–25.
8k
B
C
12 ft
A
5 ft
Solution K AB
=
K CD CD
=
3EI L
(DF) AB
=
(DF)DC
(DF)BC
=
(DF)CB
(FEM)BA 3EI L2
=
=
3EI K BC BC 13
=
1 (DF)BA
=
=
=
=
(DF)CD
>
2EI 5 =
3EI 13 + 2EI 5
(FEM)CD
100
4EI 10
=
=
2EI 5
26 41
>100 k # t; >
=
1 900 3EI
From the geometry shown in Fig. b, BC
=
5 5 + 13 13
=
10 13
Thus, (FEM)BC
=
(FEM)CB
-
=
=
-
6EI BC L2BC
a 1013 b a 16900 b 3
6EI
EI
2
10
=
>
3EI 13
=
- 260 k # t
>
>
3EI 13 + 2EI 5
=
15 41
10 ft
D
5 ft
553
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all 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.
11–25.
(Continued)
Moment Distribution. For For the rame with P acting at C, Fig. a,
Joint
A
Member
AB
BA
DF
1
15 41
26 41
FEM
0
-2 0
Dist.
B
>
10 0
BC
58.54
CO Dist.
- 18.5
Dist.
5.89
CO
CO Dist.
-2 0
10 1.46
10 1.46
50.73
50.73
100
DC
1 0
58.54
- 32.17 - 32.17 - 18.5 10.20
10.20
5.10
5.10
- 3.23 - 1.87
- 1. 2 1.03
0.59
- 1. 2 1.03
- 0.19 0.06
CO
- 0.02 0
5.89
0.51
Dist.
a M
>
15 41
- 3.23
CO
Dist.
>
26 41
CD
- 1.87
CO Dist.
>
CB
D
- 16.09 - 16.09
CO
Dist.
C
0.59
0.51
- 0.32
- 0.32 - 0.19
- 0.16
- 0.16
0.10
0.10
0.05
0.05
- 0.03
0.06
- 0.03 - 0.02
144.44 - 144.44 - 144.44
144.44
0
5 54
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11–25.
(Continued)
Using these results, the shears at A and D are computed and shown in Fig. c. Thus, or the entire rame,
+
F x
=
0; 24.07 + 24.07 - P
Thus, or P
=
=
8 a 48.14 b (144.44)
24.0 k # t
=
M BC
=
8 a 48.14 b ( - 144.44)
=
M CB CB
=
8 a 48.14 b ( - 144.44)
=
=
=
48.14 k
8 k,
M BA
M CD CD
0 P
8 a 48.14 b (144.44)
=
=
Ans.
- 24.0 k # t
Ans.
- 24.0 k # t
Ans.
24.0 k # t
Ans.
Ans. M BA M BC M CB CB M CD CD
=
=
=
=
24.0 k # t - 24.0 k # t - 24.0 k # t 24.0 k # t
555
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all 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.
4 k>ft
11–26. Determine the moments acting at the ixed supports A and D o the battered-column rame. EI is is constant. 6k B
C
20 ft
A
15 ft
Solution (DF) AB
=
(DF)DC
=
(DF)BA
=
(DF)CD
=
(DF)BC
=
(DF)CB
0
>
I 25
>
>
I 25 + I 20
=
0.4444
=
0.5556
Consider no sidesway sidesway.. Joint
A
Mem.
AB
DF
B BA
0
- 133.33
29.63
Dist.
2.29
Dist.
- 0.35
=
(FEM)
=
82.03
=
6k
- 0.12
0.03
(FEM)
- 0.03
- 82.03
=
(FEM) CD
=
0
=
(FEM) DC
=
0
- 0.64 - 0.18
- 0.10
- 0.05
- 0.03
82.03 - 82.03
- 133.33 k # t
BA
0.06
133.33 k # t
R + 36.15 - 36.15 - 6 =
- 2.29
0.22
- 0.06
0.03
(FEM)CB AB
0.12
0.05
2.86
0.79 - 0.44
0.10
- 8.23
- 4.57
- 0.22
=
- 1.27
(FEM)BC
- 5.72
0.18
- 4(20)2 12
- 1.59
CO
a M 41.0 0
- 0.79 0.44
Dist.
- 29.63
1.59
0.35
0
1.27 0.64
CO
0.4444
10.29
- 2.86
CO Dist. Dist.
R
5.72
0.5556
DC
- 20.58 - 16.46
- 10.29
4.57
CD
37.0 4
20.58
8.23
CB
- 74.08 - 59.25
- 37.04
16.46
D
133.33
74.08
59.25
Dist.
CO
0.5556
Dist.
CO
BC
0.4444
FEM CO
C
- 41.00
20 ft
D
15 ft
5 56
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all 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.
11–26.
(Continued)
(FEM) AB (FEM)BC (FEM)CD
(FEM)BA
=
(FEM)CB
=
=
A
Mem.
AB
DF
FEM - 100
CO
=
=
CB
0.4444
0.5556
0.5556
- 100
Dist.
- 3.00
- 1.50
Dist. 0.42
Dist.
CO
6.75
6.75
- 3.75
- 3.75
- 1.88
- 1.88
- 0.12
0.06
1.04
1.04
0.52
0.52
- 0.29
- 0.29
- 0.14 0.08
- 0.14 0.08
0.04
- 0.02
- 115.21 - 130.44
- 0.02 130.44
22.065 k + 22.065 k 44.130 k 6 41.00 + ( - 115.21) 44.130
a
(100)(25)2 6EI
b
- 100
- 0.02
=
187.5 k # t
DC
0
- 100
- 38.88
- 19.44
10.80 5.40
- 3.00
- 1.50
0.84 0.42
- 0.23 0.06
0.04
D
0.4444
- 0.12 0.03
- 0.02
130.44 - 130.44 - 115.21
=
=
M DC
- 0.23
a M
M AB
13.51
0.03
Dist.
187. 5
13.51
0.84
CO
=
6EI
CD
- 24.31 - 24.31
5.40
R
187. 5
(100)(25)2 =
C BC
CO
(20)2
BA
10.80
CO Dist.
=
- 100 k # t
- 19.44
(20)2
(6EI )(2)(0.6) )(2)(0.6)
- 38.88 - 48.62 - 48.62
Dist.
CO
- 100 k # t
=
B
0
Dist.
(6EI )2 )2 cos u
(FEM)DC
Joint
- 6EI (25)2
=
=
a
- 41.00 +
b
=
6 a 44.130 b ( - 115.21)
25.3 k # t
=
- 56.7 k # t
Ans.
Ans.
Ans. M AB M DC DC
=
=
25.3 k # t - 56.7 k # t
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