# Chapter 12

#### Description

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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

4 42

#

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

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

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

–4.28

2.23

0

–1.00 –1.232

–0.684

aM

CB

–2.23

1.00

R

–0.616

#

0.384 k ft 443

Ans.

400 lb

6 ft

6 ft

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

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

4 44

D

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

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.

445

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..

4 46

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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

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.

447

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

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..

4 48

12 kN/ m

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

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.. 449

D

B

10 ft

8 ft

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12–9. 12– 9.

Conti Co ntinue nued d

4 50

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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

6(42) = = - 8 kN m 12

#

12 wL2AB

12

=

- wL2BC

12

6(42) = 8 kN m 12

#

= -

12 wL2BC

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

=

6(42) = - 8 kN m 12

#

6(42) = 8 kN m 12

#

Moment Distribution. Tabulating the above data,

Joint

A

B

Member

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

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..

451

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

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 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..

4 52

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

4(152) = 30 k ft 30

#

FEMAB =

wL2 30

=

FEMBA =

wL2 20

4(152) = = 45 k ft 20

FEMBC =

wL2 12

=

12 ft

#

(4)(122) = 48 k ft 12

#

#

FEMCB = 48 k ft Joint

A

Member

AB

BA

BC

CB

DF

1

0.375

0.625

0

FEM

B

–30

45

30

–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.

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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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 c and , respe respectively ctively..

4 54

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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

–24.0 9.482

24.0 –24.0

–12.0 4.889 19.4

7.111 –19.4

0 Ans.

455

800 in4

550 in 4

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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

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

8.74 1.68

0.32

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

-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

D

= 146 k # ft

Ans.

4 56

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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

12 ft

9 ft

Member Stiffness Factor and Distribution Factor. KAD

= KBC =

EI 3EI 3EI = = L 9 3

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

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.

457

B

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12–16. 12–1 6.

Continue Cont inued d

4 58

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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

A

D

12 ft

12 ft 12 ft

B

Member Stiffness Factor and Distribution Factor. KAD

=

EI 4EI 4EI = = LAD 12 3

(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

12

= -

4(122)

12 2 4(12 )

= - 48 k # ft

#

= = 48 k ft 12 12 4(122) wL2 = - CD = = - 72 k ft 8 8 = (FEM)BD = (FEM)DB = 0

#

Moments Distribution. Tabulating the above data,

Joint

A

Member

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..

459

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12–17. 12–1 7.

Continue Cont inued d

4 60

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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

24 ft

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

461

-15.6

1.18

500 in4

(DF)AB = (DF)DC = 1 (DF)DC = 0

>

400 in 4

A

(DF)BA =

E  I CE

3k 8 ft

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

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

4 62

-20.6

Ans.

4m

4m

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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

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

2(122)

#

= - 24 k ft 12 2(122) wL2AB = = 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 12

= -

#

#

#

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

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

Dist.

aM

BA

–24

Dist. CO

B

10.08

R

0.005625

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. 463

0.15

R –0.1875

0.01125

0.028125

–0.01406

R

–0.6

–5.031

0.1556

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12–20. 12– 20.

Continue Con tinued d

4 64

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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

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

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

:

465

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12–21. 12– 21.

Continue Con tinued d

For the frame in Fig. e, Joint

A

Member

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.

4 66

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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

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

467

R  R  R  R  R R

-78.55 -19.01 -4.32 -1.04 -0.24 -0.06 -103.22

= 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

=

a

6EIAB 152

ba6

100(102)

E(0.75IAB)

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

0.65

-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

b = 59.26 k # ft

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.

4 68

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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

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

469

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 +

Ans.

Ans.

Ans.

Ans.

= MDC = 0

Ans.

4 70

© 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. 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.

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

(DF)CB =

12

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

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

–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

471

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

DF

0

FEM CO CO

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 2 L 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

4 72

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 +

Ans.

Ans.

Ans.

Ans.

Ans.

Ans.

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.

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

473

10 ft

5 ft

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12–25. 12– 25.

Continue Con tinued d

4 74

© 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

=

Ans.

Ans.

Ans.

Ans.

475

© 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 = 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

=

5 13

1200 EI

¢¿ =

B

#

3EI(1200 EI) 3EI ¢ ¿ = = 25 k ft 2 LCB 122

4 76

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12–26. 12– 26.

Continue Cont inued d

Joint

A

Member

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

=

Ans.

Ans.

Ans.

Ans.

477