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Solutions (8th ed structural analysis) chapter 12

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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

12–1. Determine the moments at B and C. EI is constant.
Assume B and C are rollers and A and D are pinned.

3 k/ft

A

B

C

8 ft

FEMAB = FEMCD = -

wL2
= -16,
12


FEMBC = -

wL2
= -100
12

3EI
,
8

KBC =

KAB =

FEMBA = FEMDC =

FEMCB =
4EI
,
20

20 ft

wL2
= 16
12

wL2
= 100
12

KCD =

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

DF
FEM

1
–16
16

B
BA
0.652
16
54.782
8
4.310
0.750
0.130
0.023

aM

0

84.0

C
BC
0.348
–100

CB
0.348

D
CD
0.652

100

–16

29.218

–29.218

–54.782

–14.609

14.609

–8
–4.310

2.299

–2.299

–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

442

0 k # ft

Ans.

D
8 ft


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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

12–2. Determine the moments at A, B, and C. Assume the
support at B is a roller and A and C are fixed. EI is constant.

3 k/ft
2 k/ft

A

B
36 ft

(DF)AB = 0

(DF)BA =

(DF)BC = 0.6
(FEM)AB =

C
24 ft

I>36
= 0.4
I>36 + I>24

(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

A

Mem.

AB

BA

BC

CB

DF

0

0.4

0.6

0

FEM

–216

216

–144

144

–28.8

–43.2

–14.4
aM

B

R

Joint

–230

C

R
187

–21.6
–122 k # ft

–187

Ans.

12–3. 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 fixed. EI is constant.

900 lb 900 lb

B

A

I>18
=
= 0.5263
I>18 + I>20

(DF)AB = 0

(DF)BA

(DF)CB = 0

(DF)BC = 0.4737

(FEM)AB =

6 ft

-2(0.9)(18)
= -3.60 k # ft
9

(FEM)BA = 3.60 k # ft
(FEM)BC =

–0.4(20)
= –1.00 k # ft
8

(FEM)CB = 1.00 k # ft
Joint

A

Mem.

AB

BA

BC

CB

DF

0

0.5263

0.4737

0

FEM

B

–3.60

C

R

3.60

–1.00

–1.368

–1.232

R

–0.684
aM

–4.28

1.00
–0.616

2.23

–2.23

0.384 k # ft
443

Ans.

400 lb

6 ft

6 ft

C
10 ft

10 ft


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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

*12–4. Determine the reactions at the supports and then
draw the moment diagram. Assume A is fixed. EI is constant.

20 ft

wL2
= –26.67,
12

FEMCB =

wL2
= 26.67
12

MCD = 0.5(15) = 7.5 k # ft
KAB =

4EI
4EI
, KBC =
20
20

DFAB = 0

DFBA = DFBC

4EI
20
=
= 0.5
4EI
4EI
+
20
20

DFCB = 1

Joint

A

B

Member

AB

BA

BC

CB

CD

DF

0

0.5

0.5

1

0

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

444

D

C

B

A

FEMBC = -

500 lb

800 lb/ ft

20 ft

15 ft


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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

12–5. 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 kN
8 kN/m

A

B
6m

Member Stiffness Factor and Distribution Factor.

KBA =

3EI
3EI
EI
=
=
LBA
6
2

(DF)AB = 1
(DF)BC =

(DF)BA =

KBC =

4EI
4EI
EI
=
=
LBC
8
2

EI>2
= 0.5
EI>2 + EI>2

EI>2
= 0.5
EI>2 + EI>2

(DF)CB = 0

Fixed End Moments. Referring to the table on the inside back cover,
(FEM)BA =

8(62)
wL2
=
= 36 kN # m
8
8

(FEM)BC = (FEM)CB =

12(8)
PL
= = -12 kN # m
8
8

12(8)
PL
=
= 12 kN # m
8
8

Moment Distribution. Tabulating the above data,
Joint

A

Member

AB

DF

1

FEM

0

Dist.

B
BA
0.5

C
BC

CB

0.5

36

–12

–12

–12

0
12

R
–6
aM

0

24

–24

6

Using these results, the shear and both ends of members AB and BC are computed
and shown in Fig. a. Subsequently, the shear and moment diagram can be plotted,
Fig. b.

445

C
4m

4m


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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

12–6. 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 kN/m
4m
C
A
4m

3EI
3EI
=
LAB
4

(DF)AB = 1 (DF)BA =

KBC =

6EI
6EI
3EI
=
=
LBC
4
2

3EI>4
1
=
3EI>4 + 3EI>2
3

(DF)BC =

3EI>2
2
=
3EI>4 + 3EI>2
3

Fixed End Moments. Referring to the table on the inside back cover,
(FEM)BA =

12(42)
wL2
=
= 24 kN # m
8
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/m

Member Stiffness Factor and Distribution Factor.

KAB =

B

Using these results, the shear at both ends of members AB, BC, and CD are
computed and shown in Fig. a. Subsequently the shear and moment diagram can be
plotted, Fig. b and c, respectively.

446


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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

12–7. 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 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 = (FEM)BA =

12(52)
wL2
= = -25 kN # m
12
12

12(52)
wL2
=
= 25 kN # m
12
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

–10

–15

15

–15

FEM

B

–25

R

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 = 0

Ay = 33 kN c

MA = 30 kN # m a

By = 27 + 6 = 33 kN c

Ans.

Cy = 6 kN T

Ans.

447

C
2.5 m


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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

*12–8. 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 and D are pins. EI is constant.

12 kN/m

A

B
4m

Member Stiffness Factor and Distribution Factor.

KAB =

3EI
3EI
=
LAB
4

(DF)AB = 1 (DF)BA =

KBC =

2EI
2EI
EI
=
=
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)BA =

(FEM)AB = (FEM)BC = 0

12(42)
wL2
=
= 24 kN # m
8
8

Moment Distribution. Tabulating the above data,
Joint

A

Member

AB

B
BA

DF

1

9
13

FEM

0

24

Dist.
aM

–16.62
0

7.385

BC
4
13
0
–7.385
–7.385

Using these results, the shear at both ends of members AB, BC, and CD are
computed and shown in Fig. a. Subsequently, the shear and moment diagram can be
plotted, Fig. b and c, respectively.

448

12 kN/m

C
6m

D
4m


© 2012 Pearson Education, Inc., Upper Saddle River, NJ. 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.

12–9. 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.

300 lb
200 lb/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

3
0.3EI
=
=
0.3EI + 0.4EI
7

4
0.4EI
=
=
0.3EI + 0.4EI
7

(DF)BC

(DF)CD = 0

(DF)CB = 1

Fixed End Moments. Referring to the table on the inside back cover,
(FEM)CD = -300(8) = 2400 lb # ft
(FEM)BA =

(FEM)BC = (FEM)CB = 0

200(102)
wL2AB
=
= 2500 lb # ft
8
8

Moment Distribution. Tabulating the above data,
Joint

A

Member

AB

DF

1

FEM

0

BA
3/7
2500
–1071.43

CO

C
BC

CB

4/7

1

0

0

0

–2400

–1428.57
1200

–514.29

CO

–685.71
357.15

–153.06

CO

–204.09
171.43

–73.47

CO

–97.96
51.03

–21.87

CO

–29.16
24.99

–10.50

CO

–13.99
7.29

–3.12

–4.17

R

R

Dist.

R

R

Dist.

R

R

Dist.

R

R

Dist.

R

R

Dist.

R

R

Dist.

R

Dist.

B

CD

2400
–714.29
714.29
–342.86
342.86
–102.05
102.05
–48.98
48.98
–14.58
14.58
–7.00
7.00

R

CO

3.50
–1.50

CO

1.04
–0.45

CO

–0.59
0.500

–0.21

CO

–0.29
0.15

–0.06

CO

–0.09
0.07

Dist.
0

–0.03

–0.04

650.01

–650.01

R

R

Dist.

R

R

Dist.

R

R

Dist.

aM

–2.00

R

Dist.

–2.08

R

2.08
–1.00
1.00
–0.30
0.30
–0.15
0.15
–0.04
0.04
2400

–2400

Using these results, the shear at both ends of members AB, BC, and CD are
computed and shown in Fig. a. Subsequently, the shear and moment diagrams can be
plotted, Fig. b and c, respectively.
449

D

C

B

10 ft

8 ft


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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

12–9.

Continued

450


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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

12–10. Determine the moment at B, then draw the
moment diagram for the beam. Assume the supports at A
and C are rollers and B is a pin. EI is constant.

6 kN/m

D D

E
A
2m

Member Stiffness Factor and Distribution Factor.
KAB =

4EI
4EI
=
= EI
LAB
4

KBC =

4EI
4EI
=
= EI
LBC
4

(DF)AB = 1 (DF)AD = 0 (DF)BA = (DF)BC =

EI
= 0.5
EI + EI

(DF)CE = 0

(DF)CB = 1

Fixed End Moments. Referring to the table on the inside back cover,
(FEM)AD = 6(2)(1) = 12 kN # m

(FEM)CE = -6(2)(1) = -12 kN # m

(FEM)AB =

6(42)
-wL2AB
= = -8 kN # m
12
12

(FEM)BA =

6(42)
wL2AB
=
= 8 kN # m
12
12

(FEM)BC =

6(42)
-wL2BC
= = -8 kN # m
12
12

(FEM)CB =

6(42)
wL2BC
=
= 8 kN # m
12
12

Moment Distribution. Tabulating the above data,

Joint

A

B

C

AD

AB

BA

BC

CB

CE

DF

0

1

0.5

0.5

1

0

FEM

12

–8

8

–8

8

–12

–4

0

0

4

–2

2

6

–6

Dist.

R

CO
aM

12

–12

R

Member

12

–12

Using these results, the shear at both ends of members AD, AB, BC, and CE are
computed and shown in Fig. a. Subsequently, the shear and moment diagram can be
plotted, Fig. b and c, respectively.

451

C

B
4m

4m

2m


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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

12–11. Determine the moments at B, C, and D, then draw
the moment diagram for the beam. EI is constant.

1.5 k/ ft
10 kиft

10 kиft

B
10 ft

Member Stiffness Factor and Distribution 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 =

1.5(202)
wL2
= = -50 k # ft
12
12

1.5(202)
wL2
= = 50 k # ft
12
12

Moment Distribution. Tabulating the above data,
Joint

B

C

D

BA

BC

CB

CD

DC

DE

DF

0

1

0.5

0.5

1

0

FEM

10

–50

50

–50

50

–10

40

0

0

–40

20

–20

70

–70

Dist.

R

CO
aM

10

–10

R

Member

10

–10

Using these results, the shear at both ends of members AB, BC, CD, and DE are
computed and shown in Fig. a . Subsequently, the shear and moment diagram can be
plotted, Fig. b and c, respectively.

452

D

C

A
20 ft

20 ft

E
10 ft


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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

*12–12. Determine the moment at B, then draw the
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

B

A
15 ft

FEMAB =

4(152)
wL2
=
= 30 k # ft
30
30

FEMBA =

4(152)
wL2
=
= 45 k # ft
20
20

FEMBC =

(4)(122)
wL2
=
= 48 k # ft
12
12

C
12 ft

FEMCB = 48 k # ft
Joint

A

Member

AB

BA

BC

CB

DF

1

0.375

0.625

0

FEM

–30
30

B

45

C

–48

1.125

48

1.875

15

0.9375

–5.625

–9.375
–4.688

aM

0

55.5

–55.5

44.25

MB = -55.5 k # ft

Ans.

8 kN/m

12–13. Determine the moment at B, then draw the
moment diagram for each member of the frame. Assume
the supports at A and C are pins. EI is constant.

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

453


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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

12–13.

Continued

Moment Distribution. Tabulating the above data,
Joint

A

Member

AB

DF

1

FEM

0

Dist.
aM

0

B
BA

C
BC

CB

6
11

5
11

1

0

–36

0

19.64

16.36

19.64

–19.64

0

Using these results, the shear at both ends of member AB and BC are computed and
shown in Fig. a. Subsequently, the shear and moment diagram can be plotted, Fig. b
and c, respectively.

454


© 2012 Pearson Education, Inc., Upper Saddle River, NJ. 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.

12–14. Determine the moments at the ends of each
member of the frame. Assume the joint at B is fixed, C is
pinned, and A is fixed. The moment of inertia of each
member is listed in the figure. E = 29(103) ksi.

2 k/ft
B
IBC ϭ 800 in4
12 ft

8 ft

4k
IAB ϭ 550 in4

(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 =

A

(DF)CB = 1

-4(16)
= -8 k # ft
8

(FEM)BA = 8 k # ft
(FEM)BC =

-2(122)
= -24 k # ft
12

(FEM)CB = 24 k # ft

Joint

A

B

Mem.

AB

BA

BC

CB

DF

0

0.4047

0.5926

1

FEM

–8.0

8.0
6.518

R

3.259

C

–24.0
9.482

24.0
–24.0

–12.0

R

4.889

7.111

2.444
aM

–2.30

19.4

–19.4

0
Ans.

455

C


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12–15. Determine the reactions at A and D. Assume the
supports at A and D are fixed and B and C are fixed
connected. EI is constant.

8 k/ft
B

C

15 ft

(DF)AB = (DF)DC = 0
(DF)BA = (DF)CD =

I>15
= 0.6154
I>15 + I>24

A

(DF)BC = (DF)CB = 0.3846

24 ft

(FEM)AB = (FEM)BA = 0
(FEM)BC =

-8(24)2
= -384 k # ft
12

(FEM)CB = 384 k # ft
(FEM)CD = (FEM)DC = 0

Joint

A

Mem.

AB

BA

BC

CB

CD

DC

DF

0

0.6154

0.3846

0.3846

0.6154

0

B

FEM

C

-147.69

-236.31

R

R

-73.84

73.84

28.40

-28.40

R

R

45.44

-118.16
-45.44

R

R

22.72

14.20

-14.20
5.46

R

R

8.74

-5.46

-22.72
-8.74

R

R

4.37

-2.73

2.73

1.05

-1.05

R

R

1.68

-4.37
-1.68

R

R

0.84

0.53

-0.53
0.20

R

R

0.32

-0.20

-0.84
-0.33

R

R

0.16

-0.10

0.10

0.04

-0.04

R

R

0.06

-0.17
-0.06

R

R

0.03

146.28

147.69

R

R

236.31

D

384

-384

118.16

aM

D

-0.02

0.02

0.01

0.01

-0.01

-0.01

292.57

-292.57

292.57

-292.57

-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

Ans.

456


© 2012 Pearson Education, Inc., Upper Saddle River, NJ. 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.

*12–16. Determine the moments at D and C, then 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.

5 k/ft

D

C
12 ft

9 ft

Member Stiffness Factor and Distribution Factor.
KAD = KBC

3EI
3EI
EI
=
=
=
L
9
3

(DF)AD = (DF)BC = 1

KCD

A

4EI
4EI
EI
=
=
=
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 =

5(122)
wL2CD
= = - 60 k # ft
12
12

5(122)
wL2CD
=
= 60 k # ft
12
12

Moments Distribution. Tabulating the above data,

Joint

A

Member

AD

DA

DC

CD

CB

BC

DF

1

0.5

0.5

0.5

0.5

1

FEM

0

0

C

B

0

-60

60

0

30

30

-30

-30

R

Dist.

D

R

CO

15

-15
7.50

-7.50

R

7.50

R

Dist.

-1.875

-7.50

R

C0

3.75

-3.75

Dist.

1.875

1.875

-1.875

R

C0

0.9375

-0.9375
0.4688

-0.4688

-0.2344

0.2344

0.1172

R

0.4688

R

Dist.

-0.1172

-0.4688

R

C0
Dist.

0.1172

-0.1172

R

C0

0.0586

-0.0586
0.0293

0.0293

R

Dist.

-0.0293

-0.0293

R

C0
Dist.
aM

0.0073
0

40.00

-0.0146

0.0146

0.0073

-0.0073

-40.00

40.00

-0.0073
-40.00

Using these results, the shear at both ends of members AD, CD, and BC are
computed and shown in Fig. a. Subsequently, the shear and moment diagram can be
plotted.

457

B


© 2012 Pearson Education, Inc., Upper Saddle River, NJ. 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.

12–16.

Continued

458


© 2012 Pearson Education, Inc., Upper Saddle River, NJ. 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.

12–17. Determine the moments at the fixed support A
and joint D and then draw the moment diagram for the
frame. Assume B is pinned.

4 k/ft

A
12 ft

12 ft
12 ft
B

Member Stiffness Factor and Distribution Factor.
KAD =

4EI
4EI
EI
=
=
LAD
12
3

(DF)AD = O

(DF)DA =

(DF)DC = (DF)DB =

KDC = KDB =

3EI
3EI
EI
=
=
L
12
4

EI>3
= 0.4
EI>3 + EI>4 + EI>4

EI>4
= 0.3
EI>3 + EI>4 + EI>4

(DF)CD = (DF)BD = 1
Fixed End Moments. Referring to the table on the inside back cover,
4(122)
wL2AD
= = -48 k # ft
12
12
4(122)
wL2AD
=
=
= 48 k # ft
12
12
4(122)
wL2CD
= = = -72 k # ft
8
8
= (FEM)BD = (FEM)DB = 0

(FEM)AD = (FEM)DA
(FEM)DC
(FEM)CD

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

R

aM

48

-48

Dist.
CO

D

9.60

C

0
7.20

-72

B

7.20

4.80
-43.2

57.6

7.20

-64.8

Using these results, the shears at both ends of members AD, CD, and BD are
computed and shown in Fig. a. Subsequently, the shear and moment diagram can be
plotted, Fig. b and c, respectively.

459

C

D


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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

12–17.

Continued

460


© 2012 Pearson Education, Inc., Upper Saddle River, NJ. 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.

12–18. Determine the moments at each joint of the frame,
then draw the moment diagram for member BCE. Assume
B, C, and E are fixed connected and A and D are pins.
E = 29(103) ksi.

0.5 k/ft
B

2k
8 ft

IBC ϭ 400 in4

8 ft

IAB ϭ 600 in4
A

24 ft

(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>24
= 0.2270
4IBC>24 + 3(1.25IBC)>16 + 4IBC>12

(DF)CD = 0.3191
(DF)CE = 0.4539
-3(16)
= -6 k # ft
8
= 6 k # ft

(FEM)AB =
(FEM)BA

-(0.5)(24)2
= -24 k # ft
12
= 24 k # ft

(FEM)BC =
(FEM)CB

-(0.5)(12)2
= -6 k # ft
12
= 6 k # ft

(FEM)CE =
(FEM)EC

(FEM)CD = (FEM)DC = 0

Joint

A

Mem.

AB

B
BA
0.6279

C
BC

0.2270

0.3191

DF

1
-6.0

6.0

6.0

11.30

6.70

-4.09

3.0

-2.04

3.35

-0.60

-0.36

-0.76

-0.38

-0.18

0.14

0.04

0.02

0.07

-0.01

-0.02

0.24

-0.01

0.3721

CD

FEM

-24.0

E

CB
24.0

CE
0.4539
-6.0

-5.74

D
EC

DC

0

1

6.0

-8.17
-4.09

-1.07

-1.52
-0.76

0.06

0.08
0.04

-0.02

-0.03
-0.02

aM

0

19.9

-19.9

22.4

-6.77

461

-15.6

E
ICE ϭ 400 in4

IDC ϭ 500 in4

3k

(DF)BA =

C

1.18

0

D
12 ft


© 2012 Pearson Education, Inc., Upper Saddle River, NJ. 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.

12–19. The frame is made from pipe that is fixed connected.
If it supports the loading shown, determine the moments
developed at each of the joints. EI is constant.

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

B

Member

AB

BA

BC

CB

CD

DC

DF

0

0.75

0.25

0.25

0.75

0

FEM
36
18
4.5
2.25

-48

48

12

-12

-6

6

1.5
-0.75

0.5625
0.281
0.0704
20.6

C

41.1

-1.5

-36
-18
-4.5

0.75

0.1875

-0.1875

-0.0938

0.0938

0.0234

-0.0234

-41.1

D

41.1

-2.25
-0.5625
-0.281
-0.0704
-41.1

462

-20.6

Ans.

4m

4m


© 2012 Pearson Education, Inc., Upper Saddle River, NJ. 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.

*12–20. Determine the moments at B and C, then draw
the moment diagram for each member of the frame.
Assume the supports at A, E, and D are fixed. EI is constant.

10 k
2 k/ft
8 ft

8 ft

A

C

B
12 ft
16 ft

E

Member Stiffness Factor and Distribution Factor.
KAB =

4EI
4EI
EI
=
=
LAB
12
3

KBC = KBE = KCD =

(DF)AB = (DF)EB = (DF)DC = 0 (DF)BA =

D

4EI
4EI
EI
=
=
L
16
4

EI>3
= 0.4
EI>3 + EI>4 + EI>4

(DF)BC = (DF)BE =

EI>4
= 0.3
EI>3 + EI>4 + EI>4

(DF)CB = (DF)CD =

EI>4
= 0.5
EI>4 + EI>4

Fixed End Moments. Referring to the table on the inside back cover,
2(122)
wL2AB
= = -24 k # ft
12
12
2(122)
wL2AB
=
= 24 k # ft
12
12
10(16)
PLBC
= = -20 k # ft
8
8
10(16)
PLBC
=
= 20 k # ft
8
8
(FEM)EB = (FEM)CD = (FEM)DC = 0

(FEM)AB = (FEM)BA =
(FEM)BC =
(FEM)CB =
(FEM)BE =

Moment Distribution. Tabulating the above data,

Joint

A

Member

AB

DF

0

FEM

0

1.50

–1.20
1.50

–0.045

–0.045
–0.1875

0.05625

0.05625
0.005625

24.41

0.3096

–24.72

DC

EB
0

20

0

0

0

–10

–10
–5

–0.6

R
R

R

–0.0016875 –0.0016875

E

0

R

R

R

0.075

CD

D
0.5

R

R

–0.06

0.5

–20

0.15

–0.00225
–23.79

0.3

R

2.00

CB

R

–1.20

0.0375

Dist.
aM

–1.60

0.3

BC

–5

R

–0.03

Dist.
CO

24

R
1.00

Dist.
CO

0.4

C

BE

–0.80

Dist.
CO

BA

–24

Dist.
CO

B

R

–0.60
0.30

0.30

R

0.75
–0.375

0.15

0.75

–0.1875

–0.0225

–0.375

R
–0.0225
0.01125

0.01125

R

0.028125
–0.01406
10.08

–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, the shear and moment diagram can be
plotted.
463

0.005625

0.028125

–0.01406
–5.031

0.1556


© 2012 Pearson Education, Inc., Upper Saddle River, NJ. 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.

12–20.

Continued

464


© 2012 Pearson Education, Inc., Upper Saddle River, NJ. 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.

12–21. Determine the moments at D and C, then draw the
moment diagram for each member of the frame. Assume
the supports at A and B are pins. EI is constant.

16 kN
1m

3m

D

C

A

B

4m

Moment Distribution. No 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 =

3EI>4
3
=
3EI>4 + EI
7

EI
4
=
3EI>4 + EI
7

(FEM)DC = -

16(32)(1)
Pb2a
=
= -9 kN # m
L2
42

(FEM)CD = -

16(12)(3)
Pa2b
=
= 3 kN # m
L2
42

Joint

A

Member

AD

DF

1

FEM

0

DC
4
7

CD
4
7

CB
3
7

BC

0

–9

3

0

0

3.857

5.143
–0.857

0.367

CO

0.490
–0.735

0.315

CO

0.420
–0.070

0.030

CO

0.040
–0.060

0.026

CO

0.034
–0.006

Dist.
0

R

R

Dist.

R

R

Dist.

R

R

Dist.

R

R

Dist.

B

DA
3
7

CO

aM

C

R

Dist.

D

R

–1.714

1

–1.286

2.572
–1.470

–1.102

0.245
–0.140

–0.105

0.210
–0.120

–0.090

0.020
–0.011

–0.009

0.017

0.003

0.003

–0.010

–0.007

4.598

–4.598

2.599

–2.599

0

Using these results, the shears at A and B are computed and shown in Fig. d. Thus,
for the entire frame
+ ©F
:
x = 0; 1.1495 - 0.6498 - R = 0 R = 0.4997 kN

465


© 2012 Pearson Education, Inc., Upper Saddle River, NJ. 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.

12–21.

Continued

For the frame in Fig. e,
Joint

A

Member

AD

DF

1

FEM

0

D
DA
3
7
–10
4.286

CO

CD
4
7

0

0

5.714
2.857

–1.224

CO

–1.633
–0.817

0.350

CO

0.467
0.234

–0.100

CO

–0.134
–0.067

0.029

CO

0.038
0.019

Dist.
0

–1.633
0.467

–10

BC
1
0

4.286
–1.224
0.350

0.234
–0.134

–0.100

–0.067
0.038

R

Dist.

R

CB
3
7

–0.817

R

R

Dist.

R

B

2.857

R

Dist.

5.714

R

R

Dist.

aM

DC
4
7

R

Dist.

C

R

0.029

0.019

–0.008

–0.011

–0.011

–0.008

–6.667

6.667

6.667

–6.667

0

Using these results, the shears at A and B caused by the application of R¿ are
computed and shown in Fig. f. For the entire frame,
+ ©F = 0; R¿1.667 - 1.667 = 0 R¿ = 3.334 kN
:
x
Thus,
MDA = 4.598 + ( -6.667) a

0.4997
b = 3.60 kN # m
3.334

Ans.

MDC = -4.598 + (6.667) a

0.4997
b = -3.60 kN # m
3.334

Ans.

MCD = 2.599 + (6.667) a

0.4997
b = -3.60 kN # m
3.334

MCB = 2.599 + (-6.667) a

Ans.

0.4997
b = -3.60 kN # m
3.334

Ans.

466


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