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

11–1. Determine the moments at A, B, and C and then

draw the moment diagram. EI is constant. Assume the

support at B is a roller and A and C are fixed.

3k

3k

B

A

3 ft

3 ft

Fixed End Moments. Referring to the table on the inside back cover

2(3)(9)

2PL

= = - 6 k # ft

9

9

2(3)(9)

2PL

=

= 6 k # ft

=

9

9

(FEM)AB = (FEM)BA

4(20)

PL

= = -10 k # ft

8

8

4(20)

PL

=

=

= 10 k # ft

8

8

(FEM)BC = (FEM)BC

Slope-Deflection Equations. Applying Eq. 11–8,

MN = 2Ek(2uN + uF - 3c) + (FEM)N

For span AB,

I

2EI

MAB = 2Ea b[2(0) + uB - 3(0)] + (-6) = a

b uB - 6

9

9

(1)

I

4EI

MBA = 2Ea b [2uB + 0 - 3(0)] + 6 = a

b uB + 6

9

9

(2)

For span BC,

MBC = 2Ea

I

EI

b [2uB + 0 - 3(0)] + (-10) = a

b uB - 10

20

5

(3)

MCB = 2Ea

I

EI

b [2(0) + uB - 3(0)] + (10) = a

b uB + 10

20

10

(4)

Equilibrium. At Support B,

MBA + MBC = 0

(5)

Substitute Eq. 2 and 3 into (5),

a

EI

4EI

b uB + 6 + a

b uB - 10 = 0

9

5

uB =

180

29EI

Substitute this result into Eqs. 1 to 4,

MAB = -4.621 k # ft = -4.62 k # ft

Ans.

MBA = 8.759 k # ft = 8.76 k # ft

Ans.

MBC =

MCB =

-8.759 k # ft = -8.76 k # ft

10.62 k # ft = 10.6 k # ft

Ans.

Ans.

The Negative Signs indicate that MAB and MBC have the counterclockwise

rotational sense. Using these results, the shear at both ends of span AB and BC are

computed and shown in Fig. a and b, respectively. Subsequently, the shear and

moment diagram can be plotted, Fig. c and d respectively.

406

4k

3 ft

C

10 ft

10 ft

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

11–1.

Continued

11–2. Determine the moments at A, B, and C, then draw

the moment diagram for the beam. The moment of inertia

of each span is indicated in the figure. Assume the support

at B is a roller and A and C are fixed. E = 29(103) ksi.

C

A

IAB ϭ 900 in.4

24 ft

Fixed End Moments. Referring to the table on the inside back cover,

2(242)

wL2

= = -96 k # ft

12

12

2(242)

wL2

=

=

= 96 k # ft

12

12

(FEM)AB = (FEM)BA

30(16)

PL

= = -60 k # ft

8

8

30(16)

PL

=

=

= 60 k # ft

8

8

(FEM)BC = (FEM)CB

30 k

2 k/ ft

407

B I ϭ 1200 in.4

BC

8 ft

8 ft

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

11–2.

Continued

Slope-Deflection Equations. Applying Eq. 11–8,

MN = 2Ek (2uN + uF - 3c) + (FEM)N

For span AB,

MAB = 2Ec

900 in4

d[2(0) + uB - 3(0)] + [-96(12) k # in]

24(12) in

MAB = 6.25EuB – 1152

MBA = 2Ec

(1)

900 in4

d[2uB + 0 - 3(0)] + 96(12) k # in

24(12) in

MBA = 12.5EuB + 1152

(2)

For span BC,

MBC = 2Ec

1200 in4

d[2uB + 0 - 3(0)] + [-60(12) k # in]

16(12) in

MBC = 25EuB - 720

MCB = 2Ec

(3)

1200 in4

d[2(0) + uB - 3(0)] + 60(12) k # in

16(12) in

MCB = 12.5EuB + 720

(4)

Equilibrium. At Support B,

MBA + MBC = 0

(5)

Substitute Eqs. 3(2) and (3) into (5),

12.5EuB + 1152 + 25EuB - 720 = 0

uB = -

11.52

E

Substitute this result into Eqs. (1) to (4),

MAB = -1224 k # in = -102 k # ft

Ans.

MBA = 1008 k # in = 84 k # ft

Ans.

MBC =

MCB =

-1008 k # in = -84 k # ft

576 k # in = 48 k # ft

Ans.

Ans.

The negative signs indicate that MAB and MBC have counterclockwise rotational

senses. Using these results, the shear at both ends of spans AB and BC are computed

and shown in Fig. a and b, respectively. Subsequently, the shear and moment

diagram can be plotted, Fig. c and d respectively.

408

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

11–2.

Continued

11–3. Determine the moments at the supports A and C,

then draw the moment diagram. Assume joint B is a roller.

EI is constant.

25 kN

15 kN/m

A

B

3m

MN = 2Ea

I

b(2uN + uF - 3c) + (FEM)N

L

MAB =

(25)(6)

2EI

(0 + uB) 6

8

MBA =

(25)(6)

2EI

(2uB) +

6

8

MBC =

(15)(4)2

2EI

(2uB) 4

12

MCB =

(15)(4)2

2EI

(uB) +

4

12

Equilibrium.

MBA + MBC = 0

25(6)

15(4)2

2EI

2EI

(2uB) +

+

(2uB) = 0

6

8

4

12

uB =

0.75

EI

MAB = -18.5 kN # m

Ans.

MCB = 20.375 kN # m = 20.4 kN # m

Ans.

MBA

MBC

= 19.25 kN # m

= -19.25 kN # m

Ans.

Ans.

409

3m

C

4m

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

*11–4. Determine the moments at the supports, then draw

the moment diagram. Assume B is a roller and A and C are

fixed. EI is constant.

15 kN 15 kN 15 kN

25 kN/m

A

B

3m

6m

(FEM)AB = -

11(25)(6)2

= -51.5625 kN # m

192

(FEM)BA =

5(25)(6)2

= 23.4375 kN # m

192

(FEM)BC =

-5(15)(8)

= -37.5 kN # m

16

(FEM)CB = 37.5 kN # m

MN = 2Ea

I

b(2uN + uF - 3c) + (FEM)N

L

I

MAB = 2Ea b(2(0) + uB - 0) - 51.5625

6

MAB =

EIuB

- 51.5625

3

(1)

I

MBA = 2Ea b(2uB + 0 - 0) + 23.4375

6

MBA =

2EIuB

+ 23.4375

3

(2)

I

MBC = 2Ea b(2uB + 0 - 0) - 37.5

8

MBC =

EIuB

- 37.5

2

(3)

I

MCB = 2Ea b(2(0) + uB - 0) + 37.5

8

MCB =

EIuB

+ 37.5

4

(4)

Equilibrium.

MBA + MBC = 0

(5)

Solving:

uB =

12.054

EI

MAB = -47.5 kN # m

Ans.

MBA = 31.5 kN # m

MBC = -31.5 kN # m

MCB = 40.5 kN # m

Ans.

Ans.

Ans.

410

2m

C

2m

2m

2m

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

11–5. Determine the moment at A, B, C and D, then draw

the moment diagram for the beam. Assume the supports at

A and D are fixed and B and C are rollers. EI is constant.

20 kN/m

A

5m

Fixed End Moments. Referring to the table on the inside back cover,

(FEM)AB = 0

(FEM)BC = (FEM)CB =

(FEM)BA = 0

(FEM)CD = 0

(FEM)DC = 0

20(32)

wL2

= = -15 kN # m

12

12

20(32)

wL2

=

= 15 kN # m

12

12

Slope-Deflection Equation. Applying Eq. 11–8,

MN = 2Ek(2uN + uF - 3c) + (FEM)N

For span AB,

I

2EI

MAB = 2Ea b[2(0) + uB - 3(0)] + 0 = a

b uB

5

5

(1)

I

4EI

MBA = 2Ea b[2uB + 0 - 3(0)] + 0 = a

buB

5

5

(2)

For span BC,

I

4EI

2EI

MBC = 2Ea b [2uB + uC - 3(0)] + (-15) = a

buB + a

b uC - 15

3

3

3

(3)

I

4EI

2EI

MCB = 2Ea b[2uC + uB - 3(0)] + 15 = a

buC + a

buB + 15

3

3

3

(4)

For span CD,

I

4EI

MCD = 2Ea b[2uC + 0-3(0)] + 0 = a

buC

5

5

(5)

I

2EI

MDC = 2Ea b[2(0) + uC -3(0)] + 0 = a

buC

5

5

(6)

Equilibrium. At Support B,

MBA + MBC = 0

a

4EI

4EI

2EI

b uB + a

buB + a

b uC - 15 = 0

5

3

3

a

2EI

32EI

buB + a

b uC = 15

15

3

(7)

At Support C,

MCB + MCD = 0

a

4EI

2EI

4EI

b uC + a

buB + 15 + a

buC = 0

3

3

5

411

C

B

3m

D

5m

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

11–5.

Continued

a

2EI

32EI

b uB + a

buC = -15

3

15

(8)

Solving Eqs. (7) and (8)

uB =

225

22EI

uC = -

225

22EI

Substitute these results into Eqs. (1) to (6),

MAB = 4.091 kN # m = 4.09 kN # m

Ans.

MBA = 8.182 kN # m = 8.18 kN # m

MBC = -8.182 kN # m = -8.18 kN # m

MCB = 8.182 kN # m = 8.18 kN # m

MCD = -8.182 kN # m = -8.18 kN # m

MDC = -4.091 kN # m = -4.09 kN # m

Ans.

Ans.

Ans.

Ans.

Ans.

The negative sign indicates that MBC, MCD and MDC have counterclockwise

rotational sense. Using these results, the shear at both ends of spans AB, BC, and

CD are computed and shown in Fig. a, b, and c respectively. Subsequently, the shear

and moment diagram can be plotted, Fig. d, and e respectively.

412

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

11–6. Determine the moments at A, B, C and D, then

draw the moment diagram for the beam. Assume the

supports at A and D are fixed and B and C are rollers. EI is

constant.

9k

A

Fixed End Moments. Referring to the table on the inside back cover,

2(15)2

wL2

= = -37.5 k # ft

12

12

2(152)

wL2

=

= 37.5 k # ft

=

12

12

(FEM)AB = -

(FEM)BC = (FEM)CB = 0

2(9)(15)

-2PL

= = -30 k # ft

9

9

2(9)(15)

2PL

=

=

= 30 k # ft

9

9

(FEM)CD =

(FEM)DC

Slope-Deflection Equation. Applying Eq. 11–8,

MN = 2Ek(2uN + uF - 3c) + (FEM)N

For span AB,

MAB = 2Ea

I

2EI

b[2(0) + uB - 3(0)] + (-37.5) = a

buB - 37.5

15

15

(1)

MBA = 2Ea

I

4EI

b[2uB + 0 - 3(0)] + 37.5 = a

buB + 37.5

15

15

(2)

For span BC,

MBC = 2Ea

I

4EI

2EI

b[2uB + uC - 3(0)] + 0 = a

buB + a

buC

15

15

15

(3)

MCB = 2Ea

I

4EI

2EI

b[2uC + uB - 3(0)] + 0 = a

buC + a

buB

15

15

15

(4)

413

C

B

15 ft

(FEM)BA

9k

2 k/ft

15 ft

5 ft

D

5 ft

5 ft

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

11–6.

Continued

For span CD,

MCD = 2Ea

I

4EI

b[2uC + 0 - 3(0)] + (-30) = a

buC - 30

15

15

(5)

MDC = 2Ea

I

2EI

b[2(0) + uC - 3(0)] + 30 = a

buC + 30

15

15

(6)

Equilibrium. At Support B,

MBA + MBC = 0

a

4EI

4EI

2EI

b uB + 37.5 + a

b uB + a

buC = 0

15

15

15

a

8EI

2EI

buB + a

b uC = -37.5

15

15

(7)

At Support C,

MCB + MCD = 0

a

2EI

4EI

4EI

buC + a

b uB + a

b uC - 30 = 0

15

15

15

a

8EI

2EI

buC + a

b uB = 30

15

15

(8)

Solving Eqs. (7) and (8),

uC =

78.75

EI

uB = -

90

EI

Substitute these results into Eqs. (1) to (6),

MAB = -49.5 k # ft

Ans.

MBA = 13.5 k # ft

MBC = -13.5 k # ft

MCB = 9 k # ft

MCD = -9 k # ft

MDC = 40.5 k # ft

Ans.

Ans.

Ans.

Ans.

Ans.

The negative signs indicate that MAB, MBC and MCD have counterclockwise

rotational sense. Using these results, the shear at both ends of spans AB, BC, and

CD are computed and shown in Fig. a, b, and c respectively. Subsequently, the shear

and moment diagram can be plotted, Fig. d, and e respectively.

414

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

11–7. Determine the moment at B, then draw the moment

diagram for the beam. Assume the supports at A and C are

pins and B is a roller. EI is constant.

40 kN

20 kN

A

6m

Fixed End Moments. Referring to the table on the inside back cover,

(FEM)BA = a

22(6)

P

a2b

40

2

2

b

a

b

a

+

b

c6

(2)

+

b

=

a

d = 52.5 kN # m

2

2

L2

82

(FEM)BC = -

3(20)(8)

3PL

= = -30 kN # m

16

16

Slope-Deflection Equations. Applying Eq. 11–10 Since one of the end’s

support for spans AB and BC is a pin.

MN = 3Ek(uN - c) + (FEM)N

For span AB,

I

3EI

MBA = 3Ea b(uB - 0) + 52.5 = a

b uB + 52.5

8

8

(1)

For span BC,

I

3EI

MBC = 3Ea b(uB - 0) + (-30) = a

buB - 30

8

8

(2)

Equilibrium. At support B,

MBA + MBC = 0

a

3EI

3EI

b uB + 52.5 + a

b uB - 30 = 0

8

8

a

3EI

buB = -22.5

4

uB = -

30

EI

415

C

B

2m

4m

4m

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

11–7.

Continued

Substitute this result into Eqs. (1) and (2)

MBA = 41.25 kN # m

Ans.

MBC = -41.25 kN # m

Ans.

The negative sign indicates that MBC has counterclockwise rotational sense. Using

this result, the shear at both ends of spans AB and BC are computed and shown in

Fig. a and b respectively. Subsequently, the shear and Moment diagram can be

plotted, Fig. c and d respectively.

*11–8. Determine the moments at A, B, and C, then draw

the moment diagram. EI is constant. Assume the support at

B is a roller and A and C are fixed.

6k

0.5 k/ft

B

A

8 ft

(FEM)AB = (FEM)BA =

PL

= -12,

8

PL

= 12,

8

(FEM)BC = (FEM)CB =

wL2

= -13.5

12

wL2

= 13.5

12

uA = uC = cAB = cBC = 0

MN = 2Ea

I

b(2uN + uF - 3c) + (FEM)N

L

MAB =

2EI

(u ) - 12

16 B

MBA =

2EI

(2uB) + 12

16

MBC =

2EI

(2uB) - 13.5

18

MCB =

2EI

(uB) + 13.5

18

Moment equilibrium at B:

MBA + MBC = 0

2EI

2EI

(2uB) + 12 +

(2uB) - 13.5 = 0

16

18

uB =

3.1765

EI

416

8 ft

C

18 ft

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

11–8.

Continued

Thus

MAB = -11.60 = -11.6 k # ft

Ans.

MBA = 12.79 = 12.8 k # ft

Ans.

MBC = -12.79 = -12.8 k # ft

MCB = 13.853 = 13.9 k # ft

Ans.

Ans.

Left Segment

c + a MA = 0;

-11.60 + 6(8) + 12.79 - VBL(16) = 0

VBL = 3.0744 k

+ c a Fy = 0;

Ay = 2.9256 k

Right Segment

c + a MB = 0;

-12.79 + 9(9) - Cy(18) + 13.85 = 0

Cy = 4.5588 k

+ c a Fy = 0;

VBK = 4.412 k

At B

By = 3.0744 + 4.4412 = 7.52 k

11–9. Determine the moments at each support, then draw

the moment diagram. Assume A is fixed. EI is constant.

B

A

20 ft

MN = 2Ea

I

b(2uN + uF - 3c) + (FEM)N

L

MAB =

4(20)2

2EI

(2(0) + uB - 0) 20

12

MBA =

4(20)2

2EI

(2uB + 0 - 0) +

20

12

MBC =

2EI

(2uB + uC - 0) + 0

15

MCB =

2EI

(2uC + uB - 0) + 0

15

MN = 3Ea

MCD =

12 k

4 k/ft

I

b(uN - c) + (FEM)N

L

3(12)16

3EI

(uC - 0) 16

16

417

D

C

15 ft

8 ft

8 ft

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

11–9.

Continued

Equilibrium.

MBA + MBC = 0

MCB + MCD = 0

Solving

uC =

178.08

EI

uB = -

336.60

EI

MAB = -167 k # ft

Ans.

MBA = 66.0 k # ft

Ans.

MBC = -66.0 k # ft

Ans.

MCB = 2.61 k # ft

Ans.

MCD = -2.61 k # ft

Ans.

11–10. Determine the moments at A and B, then draw the

moment diagram for the beam. EI is constant.

2400 lb

200 lb/ ft

A

B

30 ft

(FEM)AB = -

C

10 ft

1

1

(w)(L2) = - (200)(302) = -15 k # ft

12

12

MAB =

2EI

(0 + uB - 0) - 15

30

MBA =

2EI

(2uB + 0 - 0) + 15

30

a MB = 0;

MBA = 2.4(10)

Solving,

uB =

67.5

EI

MAB = -10.5 k # ft

Ans.

MBA = 24 k # ft

Ans.

418

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

11–11. Determine the moments at A, B, and C, then draw

the moment diagram for the beam. Assume the support at

A is fixed, B and C are rollers, and D is a pin. EI is constant.

6k

6k

B

A

4 ft

4 ft

Fixed End Moments. Referring to the table on the inside back cover,

(FEM)AB = (FEM)BA =

2(6)(12)

2PL

= = -16 k # ft

9

9

2(6)(12)

2PL

=

= 16 k # ft

9

9

(FEM)BC = (FEM)CB = 0 (FEM)CD = -

3(122)

wL2

= = -54 k # ft

8

8

Slope-Deflection Equations. Applying Eq. 11–8, for spans AB and BC.

MN = 2Ek(2uN + uF - 3c) + (FEM)N

For span AB,

MAB = 2Ea

I

EI

b[2(0) + uB - 3(0)] + (-16) = a

buB - 16

12

6

(1)

MBA = 2Ea

I

EI

b[2uB + 0 - 3(0)] + 16 = a

buB + 16

12

3

(2)

For span BC,

MBC = 2Ea

I

EI

EI

b[2uB + uC - 3(0)] + 0 = a

buB + a

buC

12

3

6

(3)

MCB = 2Ea

I

EI

EI

b[2uC + uB - 3(0)] + 0 = a

buC + a

buB

12

3

6

(4)

Applying Eq. 11–10 for span CD,

MN = 3Ek(uN - c) + (FEM)N

MCD = 3Ea

I

EI

b (uC - 0) + (-54) = a

b uC - 54

12

4

(5)

Equilibrium. At support B,

MBA + MBC = 0

a

EI

EI

EI

b uB + 16 + a

b uB + a

buC = 0

3

3

6

a

2EI

EI

b uB + a

buC = -16

3

6

(6)

At support C,

MCB + MCD = 0

a

EI

EI

EI

b uC + a

buB + a

b uC - 54 = 0

3

6

4

a

7EI

EI

b uC + a

buB = 54

12

6

(7)

419

3 k/ft

4 ft

C

12 ft

D

12 ft

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

11–11.

Continued

Solving Eqs. (6) and (7)

uC =

1392

13EI

uB = -

660

13EI

Substitute these results into Eq. (1) to (5)

MAB = -24.46 k # ft = -24.5 k # ft

Ans.

MBA = -0.9231 k # ft = -0.923 k # ft

MBC = 0.9231 k # ft = 0.923 k # ft

MCB = 27.23 k # ft = 27.2 k # ft

MCD = -27.23 k # ft = -27.2 k # ft

Ans.

Ans.

Ans.

Ans.

The negative signs indicates that MAB, MBA, and MCD have counterclockwise

rotational sense. Using these results, the shear at both ends of spans AB, BC, and

CD are computed and shown in Fig. a, b, and c respectively. Subsequently, the shear

and moment diagram can be plotted, Fig. d and e respectively.

420

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

*11–12. Determine the moments acting at A and B.

Assume A is fixed supported, B is a roller, and C is a pin.

EI is constant.

20 kN/ m

3m

9m

wL2

= -54,

30

(FEM)BA =

wL2

= 81

20

(FEM)BC =

C

B

A

(FEM)AB =

80 kN

3m

3PL

= -90

16

Applying Eqs. 11–8 and 11–10,

MAB =

2EI

(uB) - 54

9

MBA =

2EI

(2uB) + 81

9

MBC =

3EI

(uB) - 90

6

Moment equilibrium at B:

MBA + MBC = 0

4EI

EI

(uB) + 81 +

u - 90 = 0

9

2 B

uB =

9.529

EI

Thus,

MAB = -51.9 kN # m

Ans.

MBA = 85.2 kN # m

Ans.

MBC = -85.2 kN # m

Ans.

11–13. Determine the moments at A, B, and C, then draw

the moment diagram for each member. Assume all joints

are fixed connected. EI is constant.

4 k/ft

B

A

18 ft

9 ft

C

(FEM)AB

-4(18)2

= -108 k # ft

=

12

(FEM)BA = 108 k # ft

(FEM)BC = (FEM)CB = 0

421

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

11–13.

Continued

MN = 2Ea

I

b(2uN + uF - 3c) + (FEM)N

L

MAB = 2Ea

I

b(2(0) + uB - 0) - 108

18

MAB = 0.1111EIuB - 108

MBA = 2Ea

(1)

I

b(2uB + 0 - 0) + 108

18

MBA = 0.2222EIuB + 108

(2)

I

MBC = 2Ea b(2uB + 0 - 0) + 0

9

MBC = 0.4444EIuB

(3)

I

MCB = 2Ea b(2(0) + uB - 0) + 0

9

MCB = 0.2222EIuB

(4)

Equilibrium

MBA + MBC = 0

(5)

Solving Eqs. 1–5:

uB =

-162.0

EI

MAB = -126 k # ft

Ans.

MBA = 72 k # ft

Ans.

MBC = -72 k # ft

Ans.

MCB = -36 k # ft

Ans.

422

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

11–14. Determine the moments at the supports, then draw

the moment diagram. The members are fixed connected at

the supports and at joint B. The moment of inertia of each

member is given in the figure. Take E = 29(103) ksi.

20 k

8 ft

8 ft

B

A

IAB ϭ 800 in4

6 ft

IBC ϭ 1200 in4

15 k

6 ft

(FEM)AB

-20(16)

=

= -40 k # ft

8

C

(FEM)BA = 40 k # ft

(FEM)BC =

-15(12)

= -22.5 k # ft

8

(FEM)CB = 22.5 k # ft

MN = 2Ea

MAB =

I

b(2uN + uF - 3c) + (FEM)N

L

2(29)(103)(800)

(2(0) + uB - 0) - 40

16(144)

MAB = 20,138.89uB - 40

(1)

3

MBA =

2(29)(10 )(800)

(2uB + 0 - 0) + 40

16(144)

MBA = 40,277.78uB + 40

MBC =

(2)

2(29)(103)(1200)

(2uB + 0 - 0) - 22.5

12(144)

MBC = 80,555.55uB - 22.5

MCB =

(3)

2(29)(103)(1200)

(2(0) + uB - 0) + 22.5

12(144)

MCB = 40,277.77uB + 22.5

(4)

Equilibrium.

MBA + MBC = 0

(5)

Solving Eqs. 1–5:

uB = -0.00014483

MAB = -42.9 k # ft

Ans.

MBA = 34.2 k # ft

MBC = -34.2 k # ft

MCB = 16.7 k # ft

Ans.

423

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

11–15. Determine the moment at B, then draw the moment

diagram for each member of the frame. Assume the support

at A is fixed and C is pinned. EI is constant.

2 kN/m

A

B

Fixed End Moments. Referring to the table on the inside back cover,

3m

2(32)

wL2

= = -1.50 kN # m

12

12

2(32)

wL2

=

=

= 1.50 kN # m

12

12

(FEM)AB = (FEM)BA

4m

(FEM)BC = 0

Slope-Deflection Equations. Applying Eq. 11–8 for member AB,

C

MN = 2Ek(2uN + uF - 3c) + (FEM)N

I

2EI

MAB = 2Ea b[2(0) + uB - 3(0)] + (-1.50) = a

b uB - 1.50

3

3

(1)

I

4EI

MBA = 2Ea b[2uB + 0 - 3(0)] + 1.50 = a

buB + 1.50

3

3

(2)

Applying Eq. 11–10 for member BC,

MN = 3Ek(uN - c) + (FEM)N

I

3EI

MBC = 3Ea b(uB - 0) + 0 = a

buB

4

4

(3)

Equilibrium. At Joint B,

MBA + MBC = 0

a

4EI

3EI

b uB + 1.50 + a

b uB = 0

3

4

uB = -

0.72

EI

Substitute this result into Eqs. (1) to (3)

MAB = -1.98 kN # m

Ans.

MBA = 0.540 kN # m

Ans.

MBC = -0.540 kN # m

Ans.

The negative signs indicate that MAB and MBC have counterclockwise rotational

sense. Using these results, the shear at both ends of member AB and BC are

computed and shown in Fig. a and b respectively. Subsequently, the shear and

moment diagram can be plotted, Fig. c and d respectively.

424

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

*11–16. Determine the moments at B and D, then draw

the moment diagram. Assume A and C are pinned and B

and D are fixed connected. EI is constant.

8k

15 ft

A

10 ft

B

C

12 ft

D

(FEM)BA = 0

-3(8)(20)

= -30 k # ft

16

(FEM)BC =

(FEM)BD = (FEM)DB = 0

MN = 3Ea

I

b(uN - c) + (FEM)N

L

MBA = 3Ea

I

b(uB - 0) + 0

15

MBA = 0.2EIuB

MBC = 3Ea

(1)

I

b(uB - 0) - 30

20

MBC = 0.15EIuB - 30

MN = 2Ea

(2)

I

b(2uN + uF - 3c) + (FEM)N

L

MBD = 2Ea

I

b(2uB + 0 - 0) + 0

12

MBD = 0.3333EIuB

MDB = 2Ea

(3)

I

b(2(0) + uB - 0) + 0

12

MDB = 0.1667EIuB

(4)

Equilibrium.

MBA + MBC + MBD = 0

(5)

Solving Eqs. 1–5:

uB =

43.90

EI

MBA = 8.78 k # ft

Ans.

MBC = -23.41 k # ft

Ans.

MBD = 14.63 k # ft

Ans.

MDB = 7.32 k # ft

Ans.

425

10 ft

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

11–17. Determine the moment that each member exerts

on the joint at B, then draw the moment diagram for each

member of the frame. Assume the support at A is fixed and

C is a pin. EI is constant.

2 k/ ft

B

C

15 ft

6 ft

10 k

6 ft

A

Fixed End Moments. Referring to the table on the inside back cover,

(FEM)AB = -

10(12)

PL

= = -15 k # ft

8

8

(FEM)BC = -

2(152)

wL2

= = -56.25 k # ft

8

8

(FEM)BA =

10(12)

PL

=

= 15 k # ft

8

8

Slope Reflection Equations. Applying Eq. 11–8 for member AB,

MN = 2Ek(2uN + uF - 3c) + (FEM)N

MAB = 2Ea

I

EI

b [2(0) + uB - 3(0)] + (–15) = a

b uB - 15

12

6

(1)

I

EI

b[2uB + 0 - 3(0)] + 15 = a

buB + 15

12

3

(2)

MBA = 2Ea

For member BC, applying Eq. 11–10

MN = 3Ek(uN - c) + (FEM)N

MBC = 3Ea

I

EI

b (uB - 0) + (-56.25) = a

buB - 56.25

15

5

(3)

Equilibrium. At joint B,

MBA + MBC = 0

a

EI

EI

b uB + 15 + a

b uB - 56.25 = 0

3

5

uB =

77.34375

EI

Substitute this result into Eqs. (1) to (3)

MAB = -2.109 k # ft = -2.11 k # ft

Ans.

MBA = 40.78 k # ft = 40.8 k # ft

Ans.

MBC = -40.78 k # ft = -40.8 k # ft

Ans.

The negative signs indicate that MAB and MBC have counterclockwise rotational

sense. Using these results, the shear at both ends of member AB and BC are

computed and shown in Fig. a and b respectively. Subsequently, the shear and

Moment diagram can be plotted, Fig. c and d respectively.

426

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

11–17.

Continued

427

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

11–18. Determine the moment that each member exerts

on the joint at B, then draw the moment diagram for each

member of the frame. Assume the supports at A, C, and D

are pins. EI is constant.

B

D

C

6m

6m

8m

12 kN/ m

Fixed End Moments. Referring to the table on the inside back cover,

12(82)

wL

=

=

= 96 kN # m

8

8

A

2

(FEM)BA

(FEM)BC = (FEM)BD = 0

Slope-Reflection Equation. Since the far end of each members are pinned, Eq. 11–10

can be applied

MN = 3Ek(uN - c) + (FEM)N

For member AB,

I

3EI

MBA = 3Ea b(uB - 0) + 96 = a

buB + 96

8

8

(1)

For member BC,

I

EI

MBC = 3Ea b(uB - 0) + 0 = a

buB

6

2

(2)

For member BD,

I

EI

MBD = 3Ea b (uB - 0) + 0 =

u

6

2 B

(3)

Equilibrium. At joint B,

MBA + MBC + MBD = 0

a

3EI

EI

EI

b uB + 96 + a

b uB +

u = 0

8

2

2 B

uB = -

768

11EI

Substitute this result into Eqs. (1) to (3)

MBA = 69.82 kN # m = 69.8 kN # m

Ans.

MBC = -34.91 kN # m = -34.9 kN # m

Ans.

MBD = -34.91 kN # m = -34.9 kN # m

Ans.

The negative signs indicate that MBC and MBD have counterclockwise rotational

sense. Using these results, the shear at both ends of members AB, BC, and BD are

computed and shown in Fig. a, b and c respectively. Subsequently, the shear and

moment diagrams can be plotted, Fig. d and e respectively.

428

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

11–18.

Continued

429

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

11–19. Determine the moment at joints 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.

3 k/ft

C

D

12 ft

B

A

5 ft

Fixed End Moments. Referring to the table on the inside back cover,

3(102)

3(102)

wL2

wL2

(FEM)DC = = = -25 k # ft (FEM)CD =

=

= 25 k # ft

12

12

12

12

(FEM)DA = (FEM)CB = 0

Slope-Deflection Equations. For member CD, applying Eq. 11–8

MN = 2Ek(2uN + uF - 3c) + (FEM)N

MDC = 2Ea

I

2EI

EI

b [2uD + uC - 3(0)] + (–25) = a

buD + a

buC - 25

10

5

5

(1)

I

2EI

EI

b [2uC + uD - 3(0)] + 25 = a

b uC + a

b uD + 25

10

5

5

(2)

MCD = 2Ea

For members AD and BC, applying Eq. 11–10

MN = 3Ek(uN - c) + (FEM)N

MDA = 3Ea

I

3EI

b(uD - 0) + 0 = a

b uD

13

13

(3)

MCB = 3Ea

I

3EI

b(uC - 0) + 0 = a

buC

13

13

(4)

Equilibrium. At joint D,

MDC + MDA = 0

a

2EI

EI

3EI

buD + a

buC - 25 + a

buD = 0

5

5

13

a

41EI

EI

buD + a

buC = 25

65

5

(5)

At joint C,

MCD + MCB = 0

a

2EI

EI

3EI

b uC + a

buD + 25 + a

buC = 0

5

5

13

a

41EI

EI

buC + a

buD = –25

65

5

(6)

Solving Eqs. (5) and (6)

uD =

1625

28EI

uC = -

1625

28EI

Substitute these results into Eq. (1) to (4)

MDC = -13.39 k # ft = -13.4 k # ft

Ans.

MCD = 13.39 k # ft = 13.4 k # ft

Ans.

MDA = 13.39 k # ft = 13.4 k # ft

MCB = -13.39 k # ft = -13.4 k # ft

Ans.

Ans.

430

10 ft

5 ft

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

11–1. Determine the moments at A, B, and C and then

draw the moment diagram. EI is constant. Assume the

support at B is a roller and A and C are fixed.

3k

3k

B

A

3 ft

3 ft

Fixed End Moments. Referring to the table on the inside back cover

2(3)(9)

2PL

= = - 6 k # ft

9

9

2(3)(9)

2PL

=

= 6 k # ft

=

9

9

(FEM)AB = (FEM)BA

4(20)

PL

= = -10 k # ft

8

8

4(20)

PL

=

=

= 10 k # ft

8

8

(FEM)BC = (FEM)BC

Slope-Deflection Equations. Applying Eq. 11–8,

MN = 2Ek(2uN + uF - 3c) + (FEM)N

For span AB,

I

2EI

MAB = 2Ea b[2(0) + uB - 3(0)] + (-6) = a

b uB - 6

9

9

(1)

I

4EI

MBA = 2Ea b [2uB + 0 - 3(0)] + 6 = a

b uB + 6

9

9

(2)

For span BC,

MBC = 2Ea

I

EI

b [2uB + 0 - 3(0)] + (-10) = a

b uB - 10

20

5

(3)

MCB = 2Ea

I

EI

b [2(0) + uB - 3(0)] + (10) = a

b uB + 10

20

10

(4)

Equilibrium. At Support B,

MBA + MBC = 0

(5)

Substitute Eq. 2 and 3 into (5),

a

EI

4EI

b uB + 6 + a

b uB - 10 = 0

9

5

uB =

180

29EI

Substitute this result into Eqs. 1 to 4,

MAB = -4.621 k # ft = -4.62 k # ft

Ans.

MBA = 8.759 k # ft = 8.76 k # ft

Ans.

MBC =

MCB =

-8.759 k # ft = -8.76 k # ft

10.62 k # ft = 10.6 k # ft

Ans.

Ans.

The Negative Signs indicate that MAB and MBC have the counterclockwise

rotational sense. Using these results, the shear at both ends of span AB and BC are

computed and shown in Fig. a and b, respectively. Subsequently, the shear and

moment diagram can be plotted, Fig. c and d respectively.

406

4k

3 ft

C

10 ft

10 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.

11–1.

Continued

11–2. Determine the moments at A, B, and C, then draw

the moment diagram for the beam. The moment of inertia

of each span is indicated in the figure. Assume the support

at B is a roller and A and C are fixed. E = 29(103) ksi.

C

A

IAB ϭ 900 in.4

24 ft

Fixed End Moments. Referring to the table on the inside back cover,

2(242)

wL2

= = -96 k # ft

12

12

2(242)

wL2

=

=

= 96 k # ft

12

12

(FEM)AB = (FEM)BA

30(16)

PL

= = -60 k # ft

8

8

30(16)

PL

=

=

= 60 k # ft

8

8

(FEM)BC = (FEM)CB

30 k

2 k/ ft

407

B I ϭ 1200 in.4

BC

8 ft

8 ft

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

11–2.

Continued

Slope-Deflection Equations. Applying Eq. 11–8,

MN = 2Ek (2uN + uF - 3c) + (FEM)N

For span AB,

MAB = 2Ec

900 in4

d[2(0) + uB - 3(0)] + [-96(12) k # in]

24(12) in

MAB = 6.25EuB – 1152

MBA = 2Ec

(1)

900 in4

d[2uB + 0 - 3(0)] + 96(12) k # in

24(12) in

MBA = 12.5EuB + 1152

(2)

For span BC,

MBC = 2Ec

1200 in4

d[2uB + 0 - 3(0)] + [-60(12) k # in]

16(12) in

MBC = 25EuB - 720

MCB = 2Ec

(3)

1200 in4

d[2(0) + uB - 3(0)] + 60(12) k # in

16(12) in

MCB = 12.5EuB + 720

(4)

Equilibrium. At Support B,

MBA + MBC = 0

(5)

Substitute Eqs. 3(2) and (3) into (5),

12.5EuB + 1152 + 25EuB - 720 = 0

uB = -

11.52

E

Substitute this result into Eqs. (1) to (4),

MAB = -1224 k # in = -102 k # ft

Ans.

MBA = 1008 k # in = 84 k # ft

Ans.

MBC =

MCB =

-1008 k # in = -84 k # ft

576 k # in = 48 k # ft

Ans.

Ans.

The negative signs indicate that MAB and MBC have counterclockwise rotational

senses. Using these results, the shear at both ends of spans AB and BC are computed

and shown in Fig. a and b, respectively. Subsequently, the shear and moment

diagram can be plotted, Fig. c and d respectively.

408

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

11–2.

Continued

11–3. Determine the moments at the supports A and C,

then draw the moment diagram. Assume joint B is a roller.

EI is constant.

25 kN

15 kN/m

A

B

3m

MN = 2Ea

I

b(2uN + uF - 3c) + (FEM)N

L

MAB =

(25)(6)

2EI

(0 + uB) 6

8

MBA =

(25)(6)

2EI

(2uB) +

6

8

MBC =

(15)(4)2

2EI

(2uB) 4

12

MCB =

(15)(4)2

2EI

(uB) +

4

12

Equilibrium.

MBA + MBC = 0

25(6)

15(4)2

2EI

2EI

(2uB) +

+

(2uB) = 0

6

8

4

12

uB =

0.75

EI

MAB = -18.5 kN # m

Ans.

MCB = 20.375 kN # m = 20.4 kN # m

Ans.

MBA

MBC

= 19.25 kN # m

= -19.25 kN # m

Ans.

Ans.

409

3m

C

4m

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

*11–4. Determine the moments at the supports, then draw

the moment diagram. Assume B is a roller and A and C are

fixed. EI is constant.

15 kN 15 kN 15 kN

25 kN/m

A

B

3m

6m

(FEM)AB = -

11(25)(6)2

= -51.5625 kN # m

192

(FEM)BA =

5(25)(6)2

= 23.4375 kN # m

192

(FEM)BC =

-5(15)(8)

= -37.5 kN # m

16

(FEM)CB = 37.5 kN # m

MN = 2Ea

I

b(2uN + uF - 3c) + (FEM)N

L

I

MAB = 2Ea b(2(0) + uB - 0) - 51.5625

6

MAB =

EIuB

- 51.5625

3

(1)

I

MBA = 2Ea b(2uB + 0 - 0) + 23.4375

6

MBA =

2EIuB

+ 23.4375

3

(2)

I

MBC = 2Ea b(2uB + 0 - 0) - 37.5

8

MBC =

EIuB

- 37.5

2

(3)

I

MCB = 2Ea b(2(0) + uB - 0) + 37.5

8

MCB =

EIuB

+ 37.5

4

(4)

Equilibrium.

MBA + MBC = 0

(5)

Solving:

uB =

12.054

EI

MAB = -47.5 kN # m

Ans.

MBA = 31.5 kN # m

MBC = -31.5 kN # m

MCB = 40.5 kN # m

Ans.

Ans.

Ans.

410

2m

C

2m

2m

2m

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

11–5. Determine the moment at A, B, C and D, then draw

the moment diagram for the beam. Assume the supports at

A and D are fixed and B and C are rollers. EI is constant.

20 kN/m

A

5m

Fixed End Moments. Referring to the table on the inside back cover,

(FEM)AB = 0

(FEM)BC = (FEM)CB =

(FEM)BA = 0

(FEM)CD = 0

(FEM)DC = 0

20(32)

wL2

= = -15 kN # m

12

12

20(32)

wL2

=

= 15 kN # m

12

12

Slope-Deflection Equation. Applying Eq. 11–8,

MN = 2Ek(2uN + uF - 3c) + (FEM)N

For span AB,

I

2EI

MAB = 2Ea b[2(0) + uB - 3(0)] + 0 = a

b uB

5

5

(1)

I

4EI

MBA = 2Ea b[2uB + 0 - 3(0)] + 0 = a

buB

5

5

(2)

For span BC,

I

4EI

2EI

MBC = 2Ea b [2uB + uC - 3(0)] + (-15) = a

buB + a

b uC - 15

3

3

3

(3)

I

4EI

2EI

MCB = 2Ea b[2uC + uB - 3(0)] + 15 = a

buC + a

buB + 15

3

3

3

(4)

For span CD,

I

4EI

MCD = 2Ea b[2uC + 0-3(0)] + 0 = a

buC

5

5

(5)

I

2EI

MDC = 2Ea b[2(0) + uC -3(0)] + 0 = a

buC

5

5

(6)

Equilibrium. At Support B,

MBA + MBC = 0

a

4EI

4EI

2EI

b uB + a

buB + a

b uC - 15 = 0

5

3

3

a

2EI

32EI

buB + a

b uC = 15

15

3

(7)

At Support C,

MCB + MCD = 0

a

4EI

2EI

4EI

b uC + a

buB + 15 + a

buC = 0

3

3

5

411

C

B

3m

D

5m

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

11–5.

Continued

a

2EI

32EI

b uB + a

buC = -15

3

15

(8)

Solving Eqs. (7) and (8)

uB =

225

22EI

uC = -

225

22EI

Substitute these results into Eqs. (1) to (6),

MAB = 4.091 kN # m = 4.09 kN # m

Ans.

MBA = 8.182 kN # m = 8.18 kN # m

MBC = -8.182 kN # m = -8.18 kN # m

MCB = 8.182 kN # m = 8.18 kN # m

MCD = -8.182 kN # m = -8.18 kN # m

MDC = -4.091 kN # m = -4.09 kN # m

Ans.

Ans.

Ans.

Ans.

Ans.

The negative sign indicates that MBC, MCD and MDC have counterclockwise

rotational sense. Using these results, the shear at both ends of spans AB, BC, and

CD are computed and shown in Fig. a, b, and c respectively. Subsequently, the shear

and moment diagram can be plotted, Fig. d, and e respectively.

412

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

11–6. Determine the moments at A, B, C and D, then

draw the moment diagram for the beam. Assume the

supports at A and D are fixed and B and C are rollers. EI is

constant.

9k

A

Fixed End Moments. Referring to the table on the inside back cover,

2(15)2

wL2

= = -37.5 k # ft

12

12

2(152)

wL2

=

= 37.5 k # ft

=

12

12

(FEM)AB = -

(FEM)BC = (FEM)CB = 0

2(9)(15)

-2PL

= = -30 k # ft

9

9

2(9)(15)

2PL

=

=

= 30 k # ft

9

9

(FEM)CD =

(FEM)DC

Slope-Deflection Equation. Applying Eq. 11–8,

MN = 2Ek(2uN + uF - 3c) + (FEM)N

For span AB,

MAB = 2Ea

I

2EI

b[2(0) + uB - 3(0)] + (-37.5) = a

buB - 37.5

15

15

(1)

MBA = 2Ea

I

4EI

b[2uB + 0 - 3(0)] + 37.5 = a

buB + 37.5

15

15

(2)

For span BC,

MBC = 2Ea

I

4EI

2EI

b[2uB + uC - 3(0)] + 0 = a

buB + a

buC

15

15

15

(3)

MCB = 2Ea

I

4EI

2EI

b[2uC + uB - 3(0)] + 0 = a

buC + a

buB

15

15

15

(4)

413

C

B

15 ft

(FEM)BA

9k

2 k/ft

15 ft

5 ft

D

5 ft

5 ft

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

11–6.

Continued

For span CD,

MCD = 2Ea

I

4EI

b[2uC + 0 - 3(0)] + (-30) = a

buC - 30

15

15

(5)

MDC = 2Ea

I

2EI

b[2(0) + uC - 3(0)] + 30 = a

buC + 30

15

15

(6)

Equilibrium. At Support B,

MBA + MBC = 0

a

4EI

4EI

2EI

b uB + 37.5 + a

b uB + a

buC = 0

15

15

15

a

8EI

2EI

buB + a

b uC = -37.5

15

15

(7)

At Support C,

MCB + MCD = 0

a

2EI

4EI

4EI

buC + a

b uB + a

b uC - 30 = 0

15

15

15

a

8EI

2EI

buC + a

b uB = 30

15

15

(8)

Solving Eqs. (7) and (8),

uC =

78.75

EI

uB = -

90

EI

Substitute these results into Eqs. (1) to (6),

MAB = -49.5 k # ft

Ans.

MBA = 13.5 k # ft

MBC = -13.5 k # ft

MCB = 9 k # ft

MCD = -9 k # ft

MDC = 40.5 k # ft

Ans.

Ans.

Ans.

Ans.

Ans.

The negative signs indicate that MAB, MBC and MCD have counterclockwise

rotational sense. Using these results, the shear at both ends of spans AB, BC, and

CD are computed and shown in Fig. a, b, and c respectively. Subsequently, the shear

and moment diagram can be plotted, Fig. d, and e respectively.

414

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

11–7. Determine the moment at B, then draw the moment

diagram for the beam. Assume the supports at A and C are

pins and B is a roller. EI is constant.

40 kN

20 kN

A

6m

Fixed End Moments. Referring to the table on the inside back cover,

(FEM)BA = a

22(6)

P

a2b

40

2

2

b

a

b

a

+

b

c6

(2)

+

b

=

a

d = 52.5 kN # m

2

2

L2

82

(FEM)BC = -

3(20)(8)

3PL

= = -30 kN # m

16

16

Slope-Deflection Equations. Applying Eq. 11–10 Since one of the end’s

support for spans AB and BC is a pin.

MN = 3Ek(uN - c) + (FEM)N

For span AB,

I

3EI

MBA = 3Ea b(uB - 0) + 52.5 = a

b uB + 52.5

8

8

(1)

For span BC,

I

3EI

MBC = 3Ea b(uB - 0) + (-30) = a

buB - 30

8

8

(2)

Equilibrium. At support B,

MBA + MBC = 0

a

3EI

3EI

b uB + 52.5 + a

b uB - 30 = 0

8

8

a

3EI

buB = -22.5

4

uB = -

30

EI

415

C

B

2m

4m

4m

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

11–7.

Continued

Substitute this result into Eqs. (1) and (2)

MBA = 41.25 kN # m

Ans.

MBC = -41.25 kN # m

Ans.

The negative sign indicates that MBC has counterclockwise rotational sense. Using

this result, the shear at both ends of spans AB and BC are computed and shown in

Fig. a and b respectively. Subsequently, the shear and Moment diagram can be

plotted, Fig. c and d respectively.

*11–8. Determine the moments at A, B, and C, then draw

the moment diagram. EI is constant. Assume the support at

B is a roller and A and C are fixed.

6k

0.5 k/ft

B

A

8 ft

(FEM)AB = (FEM)BA =

PL

= -12,

8

PL

= 12,

8

(FEM)BC = (FEM)CB =

wL2

= -13.5

12

wL2

= 13.5

12

uA = uC = cAB = cBC = 0

MN = 2Ea

I

b(2uN + uF - 3c) + (FEM)N

L

MAB =

2EI

(u ) - 12

16 B

MBA =

2EI

(2uB) + 12

16

MBC =

2EI

(2uB) - 13.5

18

MCB =

2EI

(uB) + 13.5

18

Moment equilibrium at B:

MBA + MBC = 0

2EI

2EI

(2uB) + 12 +

(2uB) - 13.5 = 0

16

18

uB =

3.1765

EI

416

8 ft

C

18 ft

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

11–8.

Continued

Thus

MAB = -11.60 = -11.6 k # ft

Ans.

MBA = 12.79 = 12.8 k # ft

Ans.

MBC = -12.79 = -12.8 k # ft

MCB = 13.853 = 13.9 k # ft

Ans.

Ans.

Left Segment

c + a MA = 0;

-11.60 + 6(8) + 12.79 - VBL(16) = 0

VBL = 3.0744 k

+ c a Fy = 0;

Ay = 2.9256 k

Right Segment

c + a MB = 0;

-12.79 + 9(9) - Cy(18) + 13.85 = 0

Cy = 4.5588 k

+ c a Fy = 0;

VBK = 4.412 k

At B

By = 3.0744 + 4.4412 = 7.52 k

11–9. Determine the moments at each support, then draw

the moment diagram. Assume A is fixed. EI is constant.

B

A

20 ft

MN = 2Ea

I

b(2uN + uF - 3c) + (FEM)N

L

MAB =

4(20)2

2EI

(2(0) + uB - 0) 20

12

MBA =

4(20)2

2EI

(2uB + 0 - 0) +

20

12

MBC =

2EI

(2uB + uC - 0) + 0

15

MCB =

2EI

(2uC + uB - 0) + 0

15

MN = 3Ea

MCD =

12 k

4 k/ft

I

b(uN - c) + (FEM)N

L

3(12)16

3EI

(uC - 0) 16

16

417

D

C

15 ft

8 ft

8 ft

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

11–9.

Continued

Equilibrium.

MBA + MBC = 0

MCB + MCD = 0

Solving

uC =

178.08

EI

uB = -

336.60

EI

MAB = -167 k # ft

Ans.

MBA = 66.0 k # ft

Ans.

MBC = -66.0 k # ft

Ans.

MCB = 2.61 k # ft

Ans.

MCD = -2.61 k # ft

Ans.

11–10. Determine the moments at A and B, then draw the

moment diagram for the beam. EI is constant.

2400 lb

200 lb/ ft

A

B

30 ft

(FEM)AB = -

C

10 ft

1

1

(w)(L2) = - (200)(302) = -15 k # ft

12

12

MAB =

2EI

(0 + uB - 0) - 15

30

MBA =

2EI

(2uB + 0 - 0) + 15

30

a MB = 0;

MBA = 2.4(10)

Solving,

uB =

67.5

EI

MAB = -10.5 k # ft

Ans.

MBA = 24 k # ft

Ans.

418

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

11–11. Determine the moments at A, B, and C, then draw

the moment diagram for the beam. Assume the support at

A is fixed, B and C are rollers, and D is a pin. EI is constant.

6k

6k

B

A

4 ft

4 ft

Fixed End Moments. Referring to the table on the inside back cover,

(FEM)AB = (FEM)BA =

2(6)(12)

2PL

= = -16 k # ft

9

9

2(6)(12)

2PL

=

= 16 k # ft

9

9

(FEM)BC = (FEM)CB = 0 (FEM)CD = -

3(122)

wL2

= = -54 k # ft

8

8

Slope-Deflection Equations. Applying Eq. 11–8, for spans AB and BC.

MN = 2Ek(2uN + uF - 3c) + (FEM)N

For span AB,

MAB = 2Ea

I

EI

b[2(0) + uB - 3(0)] + (-16) = a

buB - 16

12

6

(1)

MBA = 2Ea

I

EI

b[2uB + 0 - 3(0)] + 16 = a

buB + 16

12

3

(2)

For span BC,

MBC = 2Ea

I

EI

EI

b[2uB + uC - 3(0)] + 0 = a

buB + a

buC

12

3

6

(3)

MCB = 2Ea

I

EI

EI

b[2uC + uB - 3(0)] + 0 = a

buC + a

buB

12

3

6

(4)

Applying Eq. 11–10 for span CD,

MN = 3Ek(uN - c) + (FEM)N

MCD = 3Ea

I

EI

b (uC - 0) + (-54) = a

b uC - 54

12

4

(5)

Equilibrium. At support B,

MBA + MBC = 0

a

EI

EI

EI

b uB + 16 + a

b uB + a

buC = 0

3

3

6

a

2EI

EI

b uB + a

buC = -16

3

6

(6)

At support C,

MCB + MCD = 0

a

EI

EI

EI

b uC + a

buB + a

b uC - 54 = 0

3

6

4

a

7EI

EI

b uC + a

buB = 54

12

6

(7)

419

3 k/ft

4 ft

C

12 ft

D

12 ft

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

11–11.

Continued

Solving Eqs. (6) and (7)

uC =

1392

13EI

uB = -

660

13EI

Substitute these results into Eq. (1) to (5)

MAB = -24.46 k # ft = -24.5 k # ft

Ans.

MBA = -0.9231 k # ft = -0.923 k # ft

MBC = 0.9231 k # ft = 0.923 k # ft

MCB = 27.23 k # ft = 27.2 k # ft

MCD = -27.23 k # ft = -27.2 k # ft

Ans.

Ans.

Ans.

Ans.

The negative signs indicates that MAB, MBA, and MCD have counterclockwise

rotational sense. Using these results, the shear at both ends of spans AB, BC, and

CD are computed and shown in Fig. a, b, and c respectively. Subsequently, the shear

and moment diagram can be plotted, Fig. d and e respectively.

420

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

*11–12. Determine the moments acting at A and B.

Assume A is fixed supported, B is a roller, and C is a pin.

EI is constant.

20 kN/ m

3m

9m

wL2

= -54,

30

(FEM)BA =

wL2

= 81

20

(FEM)BC =

C

B

A

(FEM)AB =

80 kN

3m

3PL

= -90

16

Applying Eqs. 11–8 and 11–10,

MAB =

2EI

(uB) - 54

9

MBA =

2EI

(2uB) + 81

9

MBC =

3EI

(uB) - 90

6

Moment equilibrium at B:

MBA + MBC = 0

4EI

EI

(uB) + 81 +

u - 90 = 0

9

2 B

uB =

9.529

EI

Thus,

MAB = -51.9 kN # m

Ans.

MBA = 85.2 kN # m

Ans.

MBC = -85.2 kN # m

Ans.

11–13. Determine the moments at A, B, and C, then draw

the moment diagram for each member. Assume all joints

are fixed connected. EI is constant.

4 k/ft

B

A

18 ft

9 ft

C

(FEM)AB

-4(18)2

= -108 k # ft

=

12

(FEM)BA = 108 k # ft

(FEM)BC = (FEM)CB = 0

421

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

11–13.

Continued

MN = 2Ea

I

b(2uN + uF - 3c) + (FEM)N

L

MAB = 2Ea

I

b(2(0) + uB - 0) - 108

18

MAB = 0.1111EIuB - 108

MBA = 2Ea

(1)

I

b(2uB + 0 - 0) + 108

18

MBA = 0.2222EIuB + 108

(2)

I

MBC = 2Ea b(2uB + 0 - 0) + 0

9

MBC = 0.4444EIuB

(3)

I

MCB = 2Ea b(2(0) + uB - 0) + 0

9

MCB = 0.2222EIuB

(4)

Equilibrium

MBA + MBC = 0

(5)

Solving Eqs. 1–5:

uB =

-162.0

EI

MAB = -126 k # ft

Ans.

MBA = 72 k # ft

Ans.

MBC = -72 k # ft

Ans.

MCB = -36 k # ft

Ans.

422

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

11–14. Determine the moments at the supports, then draw

the moment diagram. The members are fixed connected at

the supports and at joint B. The moment of inertia of each

member is given in the figure. Take E = 29(103) ksi.

20 k

8 ft

8 ft

B

A

IAB ϭ 800 in4

6 ft

IBC ϭ 1200 in4

15 k

6 ft

(FEM)AB

-20(16)

=

= -40 k # ft

8

C

(FEM)BA = 40 k # ft

(FEM)BC =

-15(12)

= -22.5 k # ft

8

(FEM)CB = 22.5 k # ft

MN = 2Ea

MAB =

I

b(2uN + uF - 3c) + (FEM)N

L

2(29)(103)(800)

(2(0) + uB - 0) - 40

16(144)

MAB = 20,138.89uB - 40

(1)

3

MBA =

2(29)(10 )(800)

(2uB + 0 - 0) + 40

16(144)

MBA = 40,277.78uB + 40

MBC =

(2)

2(29)(103)(1200)

(2uB + 0 - 0) - 22.5

12(144)

MBC = 80,555.55uB - 22.5

MCB =

(3)

2(29)(103)(1200)

(2(0) + uB - 0) + 22.5

12(144)

MCB = 40,277.77uB + 22.5

(4)

Equilibrium.

MBA + MBC = 0

(5)

Solving Eqs. 1–5:

uB = -0.00014483

MAB = -42.9 k # ft

Ans.

MBA = 34.2 k # ft

MBC = -34.2 k # ft

MCB = 16.7 k # ft

Ans.

423

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

11–15. Determine the moment at B, then draw the moment

diagram for each member of the frame. Assume the support

at A is fixed and C is pinned. EI is constant.

2 kN/m

A

B

Fixed End Moments. Referring to the table on the inside back cover,

3m

2(32)

wL2

= = -1.50 kN # m

12

12

2(32)

wL2

=

=

= 1.50 kN # m

12

12

(FEM)AB = (FEM)BA

4m

(FEM)BC = 0

Slope-Deflection Equations. Applying Eq. 11–8 for member AB,

C

MN = 2Ek(2uN + uF - 3c) + (FEM)N

I

2EI

MAB = 2Ea b[2(0) + uB - 3(0)] + (-1.50) = a

b uB - 1.50

3

3

(1)

I

4EI

MBA = 2Ea b[2uB + 0 - 3(0)] + 1.50 = a

buB + 1.50

3

3

(2)

Applying Eq. 11–10 for member BC,

MN = 3Ek(uN - c) + (FEM)N

I

3EI

MBC = 3Ea b(uB - 0) + 0 = a

buB

4

4

(3)

Equilibrium. At Joint B,

MBA + MBC = 0

a

4EI

3EI

b uB + 1.50 + a

b uB = 0

3

4

uB = -

0.72

EI

Substitute this result into Eqs. (1) to (3)

MAB = -1.98 kN # m

Ans.

MBA = 0.540 kN # m

Ans.

MBC = -0.540 kN # m

Ans.

The negative signs indicate that MAB and MBC have counterclockwise rotational

sense. Using these results, the shear at both ends of member AB and BC are

computed and shown in Fig. a and b respectively. Subsequently, the shear and

moment diagram can be plotted, Fig. c and d respectively.

424

© 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.

*11–16. Determine the moments at B and D, then draw

the moment diagram. Assume A and C are pinned and B

and D are fixed connected. EI is constant.

8k

15 ft

A

10 ft

B

C

12 ft

D

(FEM)BA = 0

-3(8)(20)

= -30 k # ft

16

(FEM)BC =

(FEM)BD = (FEM)DB = 0

MN = 3Ea

I

b(uN - c) + (FEM)N

L

MBA = 3Ea

I

b(uB - 0) + 0

15

MBA = 0.2EIuB

MBC = 3Ea

(1)

I

b(uB - 0) - 30

20

MBC = 0.15EIuB - 30

MN = 2Ea

(2)

I

b(2uN + uF - 3c) + (FEM)N

L

MBD = 2Ea

I

b(2uB + 0 - 0) + 0

12

MBD = 0.3333EIuB

MDB = 2Ea

(3)

I

b(2(0) + uB - 0) + 0

12

MDB = 0.1667EIuB

(4)

Equilibrium.

MBA + MBC + MBD = 0

(5)

Solving Eqs. 1–5:

uB =

43.90

EI

MBA = 8.78 k # ft

Ans.

MBC = -23.41 k # ft

Ans.

MBD = 14.63 k # ft

Ans.

MDB = 7.32 k # ft

Ans.

425

10 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.

11–17. Determine the moment that each member exerts

on the joint at B, then draw the moment diagram for each

member of the frame. Assume the support at A is fixed and

C is a pin. EI is constant.

2 k/ ft

B

C

15 ft

6 ft

10 k

6 ft

A

Fixed End Moments. Referring to the table on the inside back cover,

(FEM)AB = -

10(12)

PL

= = -15 k # ft

8

8

(FEM)BC = -

2(152)

wL2

= = -56.25 k # ft

8

8

(FEM)BA =

10(12)

PL

=

= 15 k # ft

8

8

Slope Reflection Equations. Applying Eq. 11–8 for member AB,

MN = 2Ek(2uN + uF - 3c) + (FEM)N

MAB = 2Ea

I

EI

b [2(0) + uB - 3(0)] + (–15) = a

b uB - 15

12

6

(1)

I

EI

b[2uB + 0 - 3(0)] + 15 = a

buB + 15

12

3

(2)

MBA = 2Ea

For member BC, applying Eq. 11–10

MN = 3Ek(uN - c) + (FEM)N

MBC = 3Ea

I

EI

b (uB - 0) + (-56.25) = a

buB - 56.25

15

5

(3)

Equilibrium. At joint B,

MBA + MBC = 0

a

EI

EI

b uB + 15 + a

b uB - 56.25 = 0

3

5

uB =

77.34375

EI

Substitute this result into Eqs. (1) to (3)

MAB = -2.109 k # ft = -2.11 k # ft

Ans.

MBA = 40.78 k # ft = 40.8 k # ft

Ans.

MBC = -40.78 k # ft = -40.8 k # ft

Ans.

The negative signs indicate that MAB and MBC have counterclockwise rotational

sense. Using these results, the shear at both ends of member AB and BC are

computed and shown in Fig. a and b respectively. Subsequently, the shear and

Moment diagram can be plotted, Fig. c and d respectively.

426

© 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.

11–17.

Continued

427

© 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.

11–18. Determine the moment that each member exerts

on the joint at B, then draw the moment diagram for each

member of the frame. Assume the supports at A, C, and D

are pins. EI is constant.

B

D

C

6m

6m

8m

12 kN/ m

Fixed End Moments. Referring to the table on the inside back cover,

12(82)

wL

=

=

= 96 kN # m

8

8

A

2

(FEM)BA

(FEM)BC = (FEM)BD = 0

Slope-Reflection Equation. Since the far end of each members are pinned, Eq. 11–10

can be applied

MN = 3Ek(uN - c) + (FEM)N

For member AB,

I

3EI

MBA = 3Ea b(uB - 0) + 96 = a

buB + 96

8

8

(1)

For member BC,

I

EI

MBC = 3Ea b(uB - 0) + 0 = a

buB

6

2

(2)

For member BD,

I

EI

MBD = 3Ea b (uB - 0) + 0 =

u

6

2 B

(3)

Equilibrium. At joint B,

MBA + MBC + MBD = 0

a

3EI

EI

EI

b uB + 96 + a

b uB +

u = 0

8

2

2 B

uB = -

768

11EI

Substitute this result into Eqs. (1) to (3)

MBA = 69.82 kN # m = 69.8 kN # m

Ans.

MBC = -34.91 kN # m = -34.9 kN # m

Ans.

MBD = -34.91 kN # m = -34.9 kN # m

Ans.

The negative signs indicate that MBC and MBD have counterclockwise rotational

sense. Using these results, the shear at both ends of members AB, BC, and BD are

computed and shown in Fig. a, b and c respectively. Subsequently, the shear and

moment diagrams can be plotted, Fig. d and e respectively.

428

© 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.

11–18.

Continued

429

© 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.

11–19. Determine the moment at joints 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.

3 k/ft

C

D

12 ft

B

A

5 ft

Fixed End Moments. Referring to the table on the inside back cover,

3(102)

3(102)

wL2

wL2

(FEM)DC = = = -25 k # ft (FEM)CD =

=

= 25 k # ft

12

12

12

12

(FEM)DA = (FEM)CB = 0

Slope-Deflection Equations. For member CD, applying Eq. 11–8

MN = 2Ek(2uN + uF - 3c) + (FEM)N

MDC = 2Ea

I

2EI

EI

b [2uD + uC - 3(0)] + (–25) = a

buD + a

buC - 25

10

5

5

(1)

I

2EI

EI

b [2uC + uD - 3(0)] + 25 = a

b uC + a

b uD + 25

10

5

5

(2)

MCD = 2Ea

For members AD and BC, applying Eq. 11–10

MN = 3Ek(uN - c) + (FEM)N

MDA = 3Ea

I

3EI

b(uD - 0) + 0 = a

b uD

13

13

(3)

MCB = 3Ea

I

3EI

b(uC - 0) + 0 = a

buC

13

13

(4)

Equilibrium. At joint D,

MDC + MDA = 0

a

2EI

EI

3EI

buD + a

buC - 25 + a

buD = 0

5

5

13

a

41EI

EI

buD + a

buC = 25

65

5

(5)

At joint C,

MCD + MCB = 0

a

2EI

EI

3EI

b uC + a

buD + 25 + a

buC = 0

5

5

13

a

41EI

EI

buC + a

buD = –25

65

5

(6)

Solving Eqs. (5) and (6)

uD =

1625

28EI

uC = -

1625

28EI

Substitute these results into Eq. (1) to (4)

MDC = -13.39 k # ft = -13.4 k # ft

Ans.

MCD = 13.39 k # ft = 13.4 k # ft

Ans.

MDA = 13.39 k # ft = 13.4 k # ft

MCB = -13.39 k # ft = -13.4 k # ft

Ans.

Ans.

430

10 ft

5 ft

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