# Solutions (8th ed structural analysis) chapter 11

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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

© 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.

11–11.

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

© 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.

11–14. 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

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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

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