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Solutions (8th ed structural analysis) chapter 14

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14–1. Determine the stiffness matrix K for the assembly.
Take A = 0.5 in2 and E = 29(103) ksi for each member.

8
7
2
3 ft

2

3

4
1
3

3 ft


6

1

2

4k

5
1

Member 1:

lx =

4 - 0
= 0.8;
5

0.64
0.48
AE
k1 =
D
60 -0.64
-0.48

Member 2:

lx =

10 - 4
= 1;
6

1
AE 0
k2 =
D
72 –1


0

Member 3:

lx =

0.48
0.36
-0.48
-0.36

0
0
0
0

3 - 0
= 0.6
5

-0.64
-0.48
0.64
0.48

-0.48
-0.36
T
0.48
0.36

4 ft

ly =

3 - 3
= 0
6

ly =

3 - 6
= -0.6
5

-0.64
0.48
0.64
-0.48

0.48
-0.36
T
-0.48
0.36

0
0
T
0
0

-1
0
1
0

4 - 0
= 0.8;
5

0.64
AE -0.48
k3 =
D
60 -0.64
0.48

ly =

-0.48
0.36
0.48
-0.36

Assembly stiffness matrix: K = k1 + k2 + k3
510.72
0
-201.39
0
K = H
-154.67
-116
-154.67
116

0
174
0
0
-116
-87.0
116
-87.0

-201.39
0
201.39
0
0
0
0
0

0
0
0
0
0
0
0
0

-154.67
-116
0
0
154.67
116
0
0

-116
-87.0
0
0
116
87.0
0
0

-154.67
116
0
0
0
0
154.67
-116

494

116
-87.0
0
0
X
0
0
-116
87.0

Ans.

6 ft

3
4


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

14–2. Determine the horizontal and vertical displacements
at joint 3 of the assembly in Prob. 14–1.

8
7
2
3 ft

2

3

4
1
3

3 ft

6

1

2

4k

5
1

0
0
0
Dk = F V .
0
0
0

4 ft

Qk = c

0
d
-4

Use the assembly stiffness matrix of Prob. 14–1 and applying Q = KD
510.72
0
-4
0
Q3
-201.39
Q4
0
H X = H
Q5
-154.67
-116
Q6
Q7
-154.67
116
Q8

0
174
0
0
-116
-87.0
116
-87.0

-201.39
0
201.39
0
0
0
0
0

0
0
0
0
0
0
0
0

-154.67
-116
0
0
154.67
116
0
0

-116
-87.0
0
0
116
87.0
0
-0

-154.67
116
0
0
0
0
154.67
-116

Partition matrix
0 = 510.72(D1) + 0(D2)
-4 = 0(D1) + 174(D2)
Solving
D1 = 0
D2 = -0.022990 in.
Thus,
D1 = 0

Ans.

D2 = -0.0230 in.

Ans.

495

116
D1
-87.0
D2
0
0
0
0
X H X
0
0
0
0
-116
0
87.0
0

6 ft

3
4


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

14–3. Determine the force in each member of the
assembly in Prob. 14–1.

8
7
2
3 ft

2

3

4
1
3

3 ft

6

1

2

4k

5
1
4 ft

From Prob. 14–2.
D1 = D3 = D4 = D5 = D6 = D7 = D8 = 0 D2 = -0.02299
To calculate force in each member, use Eq. 14–23.
AE
qF =
[-lx
L

Member 1:

lx =

-ly

4 - 0
= 0.8;
5

AE
[-0.8
q1 =
L

Member 2:

-0.6

lx

DNx
D
ly] D Ny T
DFx
DFy

ly =

3 - 0
= 0.6
5

0.8

0
0
0.6] D
T
0
-0.02299

q1 =

0.5(29(103))
(0.6)(-0.02299) = -3.33 k = 3.33 k (C)
60

lx =

10 - 4
= 1;
6

AE
[-1
q2 =
L

ly =

0

1

3 - 3
= 0
6

0
-0.02299
0] D
T
0
0

q2 = 0
Member 3:

lx =

Ans.

Ans.

4 - 0
= 0.8;
5

ly =

3 - 6
= -0.6
5
0
0
-0.6] D
T
0
-0.02299

q3 =

AE
[-0.8
L

q3 =

0.5(29(103))
(-0.6)(-0.02299) = 3.33 k (T)
60

0.6

0.8

496

Ans.

6 ft

3
4


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

*14–4. Determine the stiffness matrix K for the truss. Take
A = 0.75 in2, E = 29(103) ksi.

2
500 lb

1 1

1

3

2
4
2

6
3
4 ft

Member 1:

lx =

0 - 4
232

0.08839
0.08839
k1 = AED
-0.08839
-0.08839
Member 2:

lx =

4 - 4
= 0
4

0
0
k2 = AED
0
0
Member 3:

lx =

ly =

= -0.7071

0.08839
0.08839
-0.08839
-0.08839
ly =

0
0.25
0
-0.25

7 - 4
= 0.6
5

= -0.7071

232
-0.08839
-0.08839
0.08839
0.08839

-0.08839
-0.08839
T
0.08839
0.08839

0 - 4
= -1
4

0
0
0
0
ly =

0.072
-0.096
k3 = AED
-0.072
0.096

0 - 4

-0.096
0.128
0.096
-0.128

0
-0.25
T
0
0.25
0 - 4
= -0.8
5
-0.072
0.096
0.072
-0.096

0.096
-0.128
T
-0.096
0.128

Structure stiffness matrix
K = k1 + k2 + k3
0.16039
-0.00761
-0.08839
-0.08839
K = AEH
0
0
-0.072
0.096

-0.00761
0.46639
-0.08839
-0.08839
0
-0.25
0.096
-0.128

-0.08839
-0.08839
0.08839
0.08839
0
0
0
0

-0.08839
-0.08839
0.08839
0.08839
0
0
0
0

0
0
0
0
0
0
0
0

0
-0.25
0
0
0
0.25
0
0

497

-0.072
0.096
0
0
0
0
0.072
-0.096

0.096
-0.128
0
0
X Ans.
0
0
-0.096
0.128

4 ft
8

3

5
3 ft

4

7


© 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.

14–5. Determine the horizontal displacement of joint
and the force in member 2 . Take A = 0.75 in2,
E = 29(103) ksi.

2

1

500 lb

1 1

1

3

2
4
2

6
3
4 ft

0
0
0
Dk = F V
0
0
0

Qk = c

-500
d
0

Use the structure stiffness matrix of Prob. 14–4 and applying Q = KD. We have
-500
0.16039
0
-0.00761
Q3
-0.08839
-0.08839
Q4
H
X = AEH
Q5
0
Q6
0
Q7
-0.072
Q8
0.096

-0.00761
0.46639
-0.08839
-0.08839
0
-0.25
0.096
-0.128

-0.08839
-0.08839
0.08839
0.08839
0
0
0
0

-0.08839
-0.08839
0.08839
0.08839
0
0
0
0

0
0
0
0
0
0
0
0

0
-0.25
0
0
0
0.25
0
0

-0.072
0.096
0
0
0
0
0.072
-0.096

Partition matrix
c

-500
0.16039
d = AEc
0
-0.00761

-0.00761 D1
0
dc d + c d
0.46639 D2
0

-500 = AE(0.16039D1 - 0.00761D2)

(1)

0 = AE( -0.00761D1 + 0.46639D2)

(2)

Solving Eq. (1) and (2) yields:
D1 =

-3119.85(12 in.>ft)
-3119.82
=
= -0.00172 in.
AE
0.75 in2(26)(106) lb>in2

D2 =

-50.917
AE

Ans.

For member 2
lx = 0,

AE
q2 =
[0
4

ly = -1,

1

0

L = 4 ft

1
-1]
D
AE

-3119.85
-50.917
T
0
0

= -12.73 lb = 12.7 lb (C)

Ans.

498

0.096 D1
-0.128 D2
0
0
0
0
XH X
0
0
0
0
-0.096
0
0.128
0

4 ft
8

3

5
3 ft

4

7


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

2

14–6. Determine the force in member 2 if its temperature
is increased by 100°F. Take A = 0.75 in2, E = 29(103) ksi,
a = 6.5(10 - 6)>°F.

500 lb

1 1

1

3

2
4
2

6
3
4 ft

(Q1)0
0
0
(Q2)0
-1
-650
D
T = AE(6.5)(10 - 6)( +100)D
T = AED
T(10 - 4)
0
0
(Q3)0
(Q4)0
1
650
Use the structure stiffness matrix of Prob. 14–4.
-500
0.16039
0
-0.00761
Q3
-0.08839
-0.08839
Q4
H
X = AEH
Q5
0
Q6
0
Q7
-0.072
Q8
0.096

-0.00761
0.46639
-0.08839
-0.08839
0
-0.25
0.096
-0.1280

-0.08839
-0.08839
0.08839
0.08839
0
0
0
0

-0.08839
-0.08839
0.08839
0.08839
0
0
0
0

0
0
0
0
0
0
0
0

0
-0.25
0
0
0
0.25
0
0

-0.072
0.096
0
0
0
0
0.072
-0.096

0
-650
0
650
+ AEH
X (10 - 6)
0
0
0
0
-500
(0.75)(29)(106)

= 0.16039D1 - 0.00761D2 + 0

0 = -0.00761D1 + 0.46639D2 - 650(10 - 6)
Solving yields
D1 = -77.837(10 - 6) ft
D2 = 1392.427(106 - 6) ft
For member 2
lx = 0,

q2

ly = -1,

0.75(29)(106)
=
[0
4

L = 4 ft

1

0

-77.837
1392.427
-1] D
T (10 - 6) - 0.75(29)(106)(6.5)(10 - 6)(100)
0
0

= 7571.32 - 14 137.5 = -6566.18 lb = 6.57 k(C)

Ans.

499

D1
0.096
-0.1280 D2
0
0
0
0
XH X
0
0
0
0
-0.096
0
0.1280
0

4 ft
8

3

5
3 ft

4

7


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

14–7. Determine the stiffness matrix K for the truss.
Take A = 0.0015 m2 and E = 200 GPa for each member.

2m

2m
4

10
5

4

6

3

9

8
2
1

7

The origin of the global coordinate system will be set at joint 1 .
0 - 2
0 - 0
For member 1 , L = 2 m.
lx =
= -1
= 0
ly =
2
2
5
1
0.0015[200(109)]
0
k1 =
D
2
-1
0
5
150(106)
=
0
D
-150(106)
0

6
0
0
0
0

6
0
0
0
0

7
-150(106)
0
150(106)
0

For member 2 , L = 2 m.
1
1
0.0015[200(109)]
0
k2 =
D
2
-1
0
1
150(106)
0
=
D
-150(106)
0

2
0
0
0
0

7
-1
0
1
0
8
0
0
T
0
0

2
0
0
0
0

5
-1
0
1
0

5
-150(106)
0
150(106)
0

1
0.5
0.0015[200(10 )]
k3 =
-0.5
D
2 32
-0.5
0.5
1
53.033(106)
= -53.033(106)
D
-53.033(106)
53.033(106)

5
6
7
8

6
0
0
T
0
0

6
0
0
T
0
0

2
-53.033(106)
53.033(106)
53.033(106)
-53.033(106)

0 - 0
= 0
2

22
2

ly =

1
2
5
6

1
2
5
6

lx =

2
-0.5
0.5
0.5
-0.5

ly =

2 - 4

3
-0.5
0.5
0.5
-0.5

2 22

= -

4
0.5
-0.5
T
-0.5
0.5

3
-53.033(106)
53.033(106)
53.033(106)
-53.033(106)

2 - 0
= 222

1
2
3
4
4
53.033(106)
-53.033(106)
T
-53.033(106)
53.033(106)

500

1
2
3
4

6

2m
2

5

3
2

1
30 kN

5
6
7
8

2 - 4
= -1
2

lx =

For member 3 , L = 222m.

9

8
0
0
T
0
0

1

3

4

5

22
2


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

14–7.

Continued

For member 4 , L = 2 m.
lx =

2 - 2
= 0
2

ly =

5
0
0.0015[200(109)] 0
D
k4 =
2
0
0
5
0
= 0
D
0
0

6
0
150(106)
0
-150(106)

6
0
1
0
-1

3
0
0
0
0

2 - 0
= 1
2

3
0
0
0
0

4
0
-1
T
0
1

4
0
-150(106)
T
0
6
150(10 )

5
6
3
4

5
6
3
4

For member 5 , L = 2 22 m.
lx =

0 - 2
2 22

= -

22
2

ly =

3
0.5
0.0015[200(109)]
0.5
k5 =
D
-0.5
2 22
-0.5
3
53.033(106)
= -53.033(106)
D
-53.033(106)
53.033(106)

4
0.5
0.5
-0.5
-0.5

2 - 0
2 22

= -

7
-0.5
-0.5
0.5
0.5

4
53.033(106)
53.033(106)
-53.033(106)
-53.033(106)

22
2

8
-0.5
-0.5
T
0.5
0.5

7
-53.033(106)
-53.033(106)
53.033(106)
53.033(106)

3
4
7
8
8
-53.033(106)
-53.033(106)
T
53.033(106)
53.033(106)

For member 6 , L = 2 m.
lx =

0 - 2
= -1
2

3
1
0.0015[200(10 )]
k6 =
0
D
2
-1
0
9

3
150(106)
=
0
D
-150(106)
0

4
0
0
0
0

ly =
4
0
0
0
0

9
-150(106)
0
150(106)
0

2 - 2
= 0
2

9
-1
0
1
0

10
0
0
T
0
0

10
0
0
T
0
0

3
4
9
10

3
4
9
10

501

3
4
7
8


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

14–7.

Continued

Structure stiffness matrix is a 10 * 10 matrix since the highest code number is 10. Thus,
1
203.033
-53.033
-53.033
53.033
-150
0
0
0
0
0

2
-53.033
53.033
53.033
-53.033
0
0
0
0
0
0

3
-53.033
53.033
256.066
0
0
0
-53.033
-53.033
-150
0

4
53.033
-53.033
0
256.066
0
-150
-53.033
-53.033
0
0

5
-150
0
0
0
300
0
-150
0
0
0

6
0
0
0
-150
0
150
0
0
0
0

7
0
0
-53.033
-53.033
-150
0
203.033
53.033
0
0

8
0
0
-53.033
-53.033
0
0
53.033
53.033
0
0

9
10
0
0 1
0
0 2
-150 0 3
0
0 4
0
0 5
(106)
0
0 6
0
0 7
0
0 8
150
0 9
0
0 10

Ans.

2m

*14–8. Determine the vertical displacement at joint 2
and the force in member ƒ 5 ƒ . Take A = 0.0015 m2 and
E = 200 GPa.

2m
4

10
5

4

6

3

9

Here,
0
-30(103)
0
Qk = F
V
0
0
0

1
2
3
4
5
6

0
0
Dk = D T
0
0

7
8
9
10

8
2
1

7

3

4

5

6

2m
2

5

1

3
1

2

30 kN

Then, applying Q = KD
0
-30(103)
0
0
0
=
0
Q7
Q8
Q9
Q10

203.033
-53.033
-53.033
53.033
-150
0
0
0
0
0

-53.033
53.033
53.033
-53.033
0
0
0
0
0
0

-53.033
53.033
256.066
0
0
0
-53.033
-53.033
-150
0

53.033
-53.033
0
256.066
0
-150
-53.033
-53.033
0
0

-150
0
0
0
300
0
-150
0
0
0

502

0
0
0
-150
0
150
0
0
0
0

0
0
-53.033
-53.033
-150
0
203.033
53.033
0
0

0
0
-53.033
-53.033
0
0
53.033
53.033
0
0

0
0
-150
0
0
0
0
0
150
0

0
0
0
0
0
(106)
0
0
0
0
0

D1
D2
D3
D4
D5
D6
0
0
0
0


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

14–8.

Continued

From the matrix partition, Qk = K11Du + K12Dk is given by

203.033
0
-53.033
-30(103)
0
-53.033
V = F
F
0
-53.033
0
-150
0
0

-53.033
53.033
53.033
-53.033
0
0

53.033
-53.033
0
256.066
0
-150

-53.033
53.033
256.066
0
0
0

-150
0
0
0
300
0

0
D1
0
0
D2
0
0
D
0
V (106)F 3 V + F V
-150
D4
0
0
D5
0
150
D6
0

Expanding this matrix equality,
0 = [203.033D1 - 53.033D2 - 53.033D3 + 53.033D4 - 150D5](106)

(1)

-30(103) = [-53.033D1 + 53.033D2 + 53.033D3 - 53.033D4](106)

(2)

6

0 = [-53.033D1 + 53.033D2 + 256.066D3](10 )

(3)

0 = [53.033D1 - 53.033D2 + 256.066D4 - 150D6](106)

(4)

6

0 = [-150D4 + 300D5](10 )

(5)

0 = [-150D4 + 150D6](106)

(6)

Solving Eqs (1) to (6),
D1 = -0.0004 m D2 = -0.0023314 m D3 = 0.0004 m D4 = -0.00096569 m
D5 = -0.0002 m D6 = 0.00096569 m
Force in member 5 . Here lx = -

= 0.000966 m

Ans.

22
22
, ly = and L = 2 22 m
2
2

Applying Eqs 14–23,
0.0015[200(109)] 22
(q5)F =
c
2
2 22

22
2

22
2

3
22
4
dD T
2
0 7
0 8

= -42.4 kN

Ans.

503


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

14–9. Determine the stiffness matrix K for the truss.
Take A = 0.0015 m2 and E = 200 GPa for each member.

10

4
9

4

7

1

5
6
4m

1
38.4
28.8
= D
-38.4
-28.8

2
28.8
21.6
-28.8
-21.6

5
-38.4
-28.8
38.4
28.8

5
-0.64
-0.48
0.64
0.48

6
-28.8
-21.6
T
28.8
21.6

1
2
(106)
5
6

6
-0.48
-0.36
T
0.48
0.36

For member 2 , L = 4 m, lx =

4 - 8
= -1 and
4

1
1
0.0015[200(109)]
0
k2 =
D
4
-1
0

3
-1
0
1
0

1
75
0
= D
-75
0

2
0
0
0
0

3
-75
0
75
0

4
0
0
T
0
0

2
0
0
0
0

4
0
0
T
0
0

1
2
5
6

ly =

3 - 3
= 0
0

1
2
3
4

1
2
(106)
3
4

504

1

6
8

2
0.48
0.36
-0.48
-0.36

1
3
20 kN

5

7

1
0.64
0.0015[200(109)]
0.48
k1 =
D
5
-0.64
-0.48

2

4

5

The origin of the global coordinate system will be set at joint 1 .
4 - 8
0 - 3
= -0.8 and ly =
= -0.6
For member 1 , L = 5 m, lx =
5
5

2
3

3

2
4m

3m


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

14–9.

Continued

For member 3 , L = 4 m,

lx =

3
1
0.0015[200(109)]
0
k3 =
D
4
-1
0

4
0
0
0
0

3
75
0
= D
-75
0

4
0
0
0
0

9
-75
0
75
0

10
0
0
T
0
0

9
-1
0
1
0

3
0
0
= D
0
0

4
0
100
0
-100

5
0
0
0
0

4
0
1
0
-1

6
0
-1
T
0
1

3
4
5
6

6
0
-100
T
0
100

5
0.64
0.0015[200(109)] -0.48
D
k5 =
5
-0.64
0.48
6
-28.8
21.6
28.8
-21.6

3
4
9
10

4 - 4
0 - 3
= 0 and ly =
= -1
3
3

5
0
0
0
0

3
4
(106)
5
6

For member 5 , L = 5 m, lx =

5
38.4
-28.8
= D
-38.4
28.8

10
0
0
T
0
0

3
4
(106)
9
10

For member 4 , L = 3 m, lx =
3
0
0.0015[200(109)] 0
k4 =
D
3
0
0

0 - 4
3 - 3
= -1 and ly =
= 0
4
4

9
-38.4
28.8
38.4
-28.8

0 - 4
3 - 0
= -0.8 and ly =
= 0.6
5
5

6
-0.48
0.36
0.48
-0.36

9
-0.64
0.48
0.64
-0.48

10
28.8
-21.6
T
-28.8
21.6

5
6
(106)
9
10

10
0.48
-0.36
T
-0.48
0.36

5
6
9
10

505


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

14–9.

Continued

0 - 4
0 - 0
= -1 and ly =
= 0
4
4

For member 6 , L = 4 m, lx =
5
1
0.0015[200(109)]
0
D
k6 =
4
-1
0
5
75
0
= D
-75
0

6
0
0
0
0

8
-75
0
75
0

7
0
0
T
0
0

6
0
0
0
0

8
-1
0
1
0

8
0
0
= D
0
0

7
0
100
0
-100

9
0
0
0
0

5
6
8
7

5
6
(106)
8
7

For member 7 , L = 3 m, lx =

8
0
0.0015[200(109)] 0
D
k7 =
3
0
0

7
0
0
T
0
0

7
0
1
0
-1

0 - 0
3 - 0
= 0 and ly =
= 1
3
3

9
0
0
0
0

10
0
-1
T
0
1

8
7
9
10

10
0
-100
T
0
100

8
7
(106)
9
10

Structure stiffness matrix is a 10 * 10 matrix since the highest code number is 10.
Thus,
1
113.4
28.8
-75
0
-38.4
-28.8
0
0
0
0

2
28.8
21.6
0
0
-28.8
-21.6
0
0
0
0

3
-75
0
150
0
0
0
0
0
-75
0

4
0
0
0
100
0
-100
0
0
0
0

5
-38.4
-28.8
0
0
151.8
0
0
-75
-38.4
28.8

6
-28.8
-21.6
0
-100
0
143.2
0
0
28.8
-21.6

7
0
0
0
0
0
0
100
0
0
-100

506

8
0
0
0
0
-75
0
0
75
0
0

9
0
0
-75
0
-38.4
28.8
0
0
113.4
-28.8

10
0
1
0
2
0
3
0
4
28.8 5
(106)
-21.6 6
-100
7
0
8
-28.8 9
121.6 10

Ans.


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

10

14–10. Determine the force in member ƒ 5 ƒ . Take
A = 0.0015 m2 and E = 200 GPa for each member.

4
9

2

1
3

4

5

20 kN
5

7

4

7

3m

1

6
8

5
6

1

2

4m

Here,
0
-20(103)
0
Qk = G
0
W
0
0
0

2
3

3

4m

1
2
3
0 8
4 Dk = C 0 S 9
5
0 10
6
7

Then applying Q = KD
0
-20(103)
0
0
0
=
0
0
Q
Q9
Q10

113.4
28.8
-75
0
-38.4
-28.8
0
0
0
0

28.8
21.6
0
0
-28.8
-21.6
0
0
0
0

-75
0
150
0
0
0
0
0
-75
0

0
0
0
100
0
-100
0
0
0
0

-38.4
-28.8
0
0
151.8
0
0
-75
-38.4
28.8

-28.8
-21.6
0
-100
0
143.2
0
0
28.8
-21.6

0
0
0
0
0
0
100
0
0
-100

0
0
0
0
-75
0
0
75
0
0

From the matrix partition, Qk = K11Du + K12Dk is given by
113.4
0
28.8
-20(103)
-75
0
0
W = G 0
G
0
-38.4
0
-28.8
0
0

28.8
21.6
0
0
-28.8
-21.6
0

0
0
0
100
0
-100
0

-75
0
150
0
0
0
0

-38.4
-28.8
0
0
151.8
0
0

28.8
-21.6
0
-100
0
143.2
0

0
D1
0
0
D2
0
0
0
D3
0 W (106)G D4 W + G 0 W
0
D5
0
0
D6
0
100
D7
0

Expanding this matrix equality,
0 = (113.4D1 + 28.8D2 - 75D3 - 384D5 - 28.8D6)(106)

(1)

-20(103) = (28.8D1 + 21.6D2 - 28.8D5 - 21.6D6(106))

(2)

6

(3)

0 = (-75D1 + 150D3)(10 )
0 = (100D4 - 100D6)(106)

(4)
6

0 = (-38.4D1 - 28.8D2 + 151.8D5)(10 )

(5)

0 = (-28.8D1 - 21.6D2 + 100D4 + 143.2D6)(106)

(6)

6

0 = (100D7)(10 )

(7)

507

0
0
-75
0
-38.4
28.8
0
0
113.4
-28.8

0
0
0
0
28.8
(106)
-21.6
-100
0
-28.8
121.6

D1
D2
D3
D4
D5
D6
D7
0
0
0


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

14–10.

Continued

Solving Eqs (1) to (7)
D1 = 0.000711 D2 = -0.00470 D3 = 0.000356 D4 = -0.00187
D5 = -0.000711 D6 = -0.00187 D7 = 0
Force in member ƒ 5 ƒ . Here L = 5 m, lx = -0.8 and
0.0015[200(109)]
c0.8
(q5)F =
5

-0.6

-0.8

ly = 0.6.

-0.000711 5
-0.00187 6
0.6 d D
T
0
9
0
10

= 33.3 kN

Ans.

14–11. Determine the vertical displacement of node
2 if member ƒ 6 ƒ was 10 mm too long before it was fitted
into the truss. For the solution, remove the 20-k load. Take
A = 0.0015 m2 and E = 200 GPa for each member.

10

4
9

1
3
20 kN

5

7

4

7

1

6
8
1

(Q5)0
-1
-0.75 5
0.00015[200(109)](0.001)
(Q6)0
0
0
6
D
T =
D T = D
T (106)
(Q7)0
4
1
0.75 8
0
0
7
(Q8)0

2

4

5

For member ƒ 6 ƒ , L = 4 m, lx = -1, ly = 0 and ¢ L = 0.01 m. Thus,

2
3

3

5
6
4m

2
4m

Also
0 1
0 2
0 3
Qk = G 0 W 4
0 5
0 6
0 7

0 8
and Dk = C 0 S 9
0 10

Applying Q = KD + Q0
0
0
0
0
0
=
0
0
Q8
Q9
Q10

113.4
28.8
-75
0
-38.4
-28.8
0
0
0
0

28.8
21.6
0
0
-28.8
-21.6
0
0
0
0

-75
0
150
0
0
0
0
0
-75
0

0
0
0
100
0
-100
0
0
0
0

-38.4
-28.8
0
0
151.8
0
0
-75
-38.4
28.8

-28.8
-21.6
0
-100
0
143.2
0
0
28.8
-21.6

0
0
0
0
0
0
100
0
0
-100

508

0
0
0
0
-75
0
0
75
0
0

0
0
-75
0
-38.4
28.8
0
0
113.4
-28.8

0
0
0
0
28.8
(106)
-21.6
-100
0
-28.8
121.6

D1
0
D2
0
D3
0
D4
0
D5
-0.75
+
(106)
D6
0
D7
0
0
0.75
0
0
0
0

3m


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

14–11.

Continued

From the matrix partition, Qk = K11Du + K12Dk + (Qk)0
0
113.4
0
28.8
0
-75
G0W = G 0
0
-38.4
0
-28.8
0
0

28.8
21.6
0
0
-28.8
-21.6
0

-75
0
150
0
0
0
0

0
0
0
100
0
-100
0

-38.4
-28.8
0
0
151.8
0
0

-28.8
-21.6
0
-100
0
143.2
0

D1
0
0
0
D2
0
0
0
0
0
D3
0
0 W(106) G D4 W + G 0 W + G 0 W(106)
0
-0.75
D5
0
D6
0
0
0
D7
0
0
100

Expanding this matrix equality,
0 = (113.4D1 + 28.8D2 - 75D3 - 38.4D5 - 28.8D6)(106)

(1)

0 = (28.8D1 + 21.6D2 - 28.8D5 - 21.6D6)(106)

(2)

6

(3)

0 = (-75D1 + 150D3)(10 )
0 = (100D4 - 100D6)(106)

(4)
6

6

0 = (-38.4D1 - 28.8D2 + 151.8D5)(10 ) + [-0.75(10 )]

(5)

0 = (-28.8D1 - 21.6D2 - 100D4 + 143.2D6)(106)

(6)

6

(7)

0 = (100D7)(10 )
Solving Eqs. (1) to (7)
D1 = 0 D2 = 0.02667 D3 = 0 D4 = 0.01333
D5 = 0.01 D6 = 0.01333 D7 = 0
D6 = 0.0133 m

Ans.

*14–12. Determine the stiffness matrix K for the truss.
Take A = 2 in2, E = 29(103) ksi.

The origin of the global coordinate system is set at joint

1

8
4 7

.
4

8 - 8
0 - 6
= 0 and ly =
= -1
6
6

1
0
2[29(103)] 0
k1 =
D
72
0
0

2
0
1
0
-1

5
0
0
0
0

6
0
-1
T
0
1

3

2
1

6

For member ƒ 1 ƒ , L = 6(12) = 72 in.,
lx =

3

1

5

4

6

3k

1
2
5
6

2 5
1

3

2

8 ft

509

6 ft


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

14–12.

1
0
0
=D
0
0

Continued

2
0
805.56
0
-805.56

5
0
0
0
0

6
0
-805.56
T
0
805.56

1
2
5
6

For member ƒ 2 ƒ , L = 8(12) = 96 in., lx =
5
1
2[29(103)]
0
k2 =
D
96
-1
0
5
604.17
0
=D
-604.17
0

6
0
0
0
0

6
0
0
0
0

3
-1
0
1
0

3
-604.17
0
604.17
0

4
0
0
T
0
0
4
0
0
T
0
0

0 - 8
0 - 0
= -1 and ly =
= 0.
8
8

5
6
3
4

5
6
3
4

For member ƒ 3 ƒ , L = 8(12) = 96 in., lx =
1
1
2[29(103)]
0
k3 =
D
96
-1
0
1
604.17
0
=D
-604.17
0

2
0
0
0
0

2
0
0
0
0

7
-1
0
1
0

7
-604.17
0
604.17
0

8
0
0
T
0
0
8
0
0
T
0
0

1
2
7
8

1
2
7
8

For member ƒ 4 ƒ , L = 6(12) = 72 in., lx =
3
0
2[29(103)] 0
k4 =
D
72
0
0

4
0
1
0
-1

7
0
0
0
0

0 - 8
6 - 6
= -1 and ly =
= 0.
8
8

8
0
-1
T
0
1

0 - 0
6 - 0
= 0, and ly =
= 1
6
6

3
4
7
8

510


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

14–12.

Continued

3
0
0
=D
0
0

4
0
805.56
0
-805.56

7
0
0
0
0

8
0
-805.56
T
0
805.56

3
4
7
8

0 - 8
0 - 6
= -0.8 and ly =
= -0.6.
10
10

For member ƒ 5 ƒ , L = 10(12) = 120 in., lx =
1
0.64
2[29(103)]
0.48
k5 =
D
120
-0.64
-0.48
1
309.33
232
=D
-309.33
-232

2
232
174
-232
-174

2
0.48
0.36
-0.48
-0.36

3
-0.64
-0.48
0.64
0.48

3
-309.33
-232
309.33
232

4
-232
-174
T
232
174

4
-0.48
-0.36
T
0.48
0.36
1
2
3
4

For member ƒ 6 ƒ , L = 10(12) = 120 in., lx =
5
0.64
2[29(103)] -0.48
k6 =
D
120
-0.64
0.48
5
309.33
-232
=D
-309.33
232

6
-232
174
232
-174

6
-0.48
0.36
0.48
-0.36

7
-0.64
0.48
0.64
-0.48

7
-309.33
232
309.33
-232

1
2
3
4

0 - 8
6 - 0
= -0.8 and ly =
= 0.6.
10
10

8
0.48
-0.36
T
-0.48
0.36

8
232
-174
T
-232
174

5
6
7
8

5
6
7
8

The structure stiffness matrix is a 8 * 8 matrix since the highest code number is 8. Thus,
1
913.5
232
-309.33
-232
H
0
0
-604.17
0

2
232
979.56
-232
-174
0
-805.66
0
0

3
-309.33
-232
913.5
232
-604.17
0
0
0

4
-232
-174
232
979.56
0
0
0
-805.56

5
0
0
-604.17
0
913.5
-232
-309.33
232

6
0
-805.56
0
0
-232
979.56
232
-174

511

7
-604.17
0
0
0
-309.33
232
913.5
-232

8
0
0
0
-805.56
X
232
-174
-232
979.56

1
2
3
4
5
6
7
8


© 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.

8
4 7

14–13. Determine the horizontal displacement of joint 2
and the force in member ƒ 5 ƒ . Take A = 2 in2, E =
29(103) ksi. Neglect the short link at 2 .

3

3

2
1

6
4

1

5

4

Here,

6

3k

0
0
Qk = E 3 U
0
0

1
2
3
4
5

2 5
3

1

0 6
Dk = C 0 S 7
0 8

2

8 ft

Applying Q = KD,
0
913.5
0
232
3
-309.33
0
-232
H X = H
0
0
Q6
0
Q7
-604.17
0
Q8

232
979.56
-232
-174
0
-805.56
0
0

-309.33
-232
913.5
232
-604.17
0
0
0

-232
-174
232
979.56
0
0
0
-805.56

0
0
-604.17
0
913.5
-232
-309.33
232

0
-805.56
0
0
-232
979.56
232
-174

-604.17
0
0
0
-309.33
232
913.5
-232

From the matrix partition; Qk = K11Du + K12Dk
0
913.5
0
232
E 3 U = E -309.33
0
-232
0
0

232
979.56
-232
-174
0

-309.33
-232
913.5
232
-604.17

-232
-174
232
979.56
0

0
D1
0
0
D2
0
-604.17 U E D3 U + E 0 U
D4
0
0
913.5
D5
0

Expanding this matrix equality,
0 = 913.5D1 + 232D2 - 309.33D3 - 232D4

(1)

0 = 232D1 + 979.59D2 - 232D3 - 174D4

(2)

3 = -309.33D1 - 232D2 + 913.5D3 + 232D4 - 604.17D5

(3)

0 = -232D1 - 174D2 + 232D3 + 979.56D4

(4)

0 = -604.17D3 + 913.5D5

(5)

Solving Eqs. (1) to (5),
D1 = 0.002172

D2 = 0.001222

D3 = 0.008248

D4 = -0.001222

D5 = 0.005455 = 0.00546 m

Ans.

Force in Member ƒ 5 ƒ . Here, L = 10(12) = 120 in., lx = -0.8 and ly = -0.6

2[29(103)]
(q5)F =
120

[0.8

0.6

-0.8

0.002172 D1
0.001222 D2
-0.6] D
T
0.008248 D3
-0.001222 D4

= 1.64 k (C)

Ans.

512

0
D1
0
D2
0
D3
-805.56
D
X H 4X
232
D5
-174
0
-232
0
979.56
0

6 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.

8
4 7

14–14. Determine the force in member ƒ 3 ƒ if this member
was 0.025 in. too short before it was fitted onto the truss.Take
A = 2 in2. E = 29(103) ksi. Neglect the short link at 2 .

3

3

2
1

6
4

1

5

4

For member ƒ 3 ƒ , L = 8(12) = 96 in lx = -1, ly = 0 and
¢L = -0.025. Thus,

6

3k

2 5
2

3

1

(Q1)0
15.10 1
-1
2[29(103)](-0.025) 0
(Q2)0
0
2
T =
T
D
D T = D
(Q7)0
96
1
-15.10 7
(Q8)0
0
0
8

8 ft

Also,
0
0
Qk = E 3 U
0
0

1
2
3
4
5

and

913.5
0
0
232
3
-309.33
0
-232
H X = H
0
0
0
Q6
Q7
-604.17
Q8
0

0 6
Dk = C 0 S 7
0 8

232
979.56
-232
-174
0
-805.56
0
0

-309.33
-232
913.5
232
-604.17
0
0
0

0
0
-604.17
0
913.5
-232
-309.33
232

-232
-174
232
979.56
0
0
0
-805.56

0
-805.56
0
0
-232
979.56
232
-174

-604.17
0
0
0
-309.33
232
913.5
-232

0
D1
15.10
0
0
D2
0
D3
0
-805.56
D4
0
X
X H X + H
0
232
D5
-174
0
0
-15.10
-232
0
-979.56
0
0

Applying Q = KD + Q0
From the matrix partition, Qk = K11Du + K12Dk + (Qk)0,
0
913.5
0
232
E 3 U = E -309.33
0
-232
0
0

232
979.56
-232
-174
0

-309.33
-232
913.5
232
-604.17

-232
-174
232
979.56
0

0
D1
0
15.10
0
D2
0
0
-604.17 U E D3 U + E 0 U + E 0 U
0
0
0
D4
913.5
D5
0
0

Expanding this matrix equality,
0 = 913.5D1 + 232D2 - 309.33D3 - 232D4 + 15.10

(1)

0 = 232D1 + 979.56D2 - 232D3 - 174D4

(2)

3 = -309.33D1 - 232D2 + 913.5D3 + 232D4 - 604.17D5

(3)

0 = -232D1 - 174D2 + 232D3 + 979.56D4

(4)

0 = -604.17D3 + 913.5D5

(5)

Solving Eqs. (1) to (5),
D1 = -0.01912

D2 = 0.003305

D3 = -0.002687 D4 = -0.003305

D5 = -0.001779

513

6 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.

14–14.

Continued

Force in member ƒ 3 ƒ . Here, L = 8(12) = 96 in., lx = -1, ly = 0 and
(qF)0 =

-2[29(103)] (-0.025)
= 15.10 k
96

2[29(103)]
(q3)F =
96

[1

0

-1

0

-0.01912
0.003305
D
T + 15.10
0
0

= 3.55 k (T)

Ans.

14–15. Determine the stiffness matrix K for the truss. AE
is constant.

3m
6
5

3

2
2
1

3
3 kN

The origin of the global coordinate system is set at joint

1

4m

.

1
2

For member ƒ 2 ƒ , L = 5 m. Referring to Fig. a, u–x = 180°- 45°- sin

-1

4
a b = 81.87°
5

u–y = 171.87°. Thus, x–x = cos ux– = cos 81.87° = 0.14142 and

1
4

ly– = cos uy– = cos 171.87° = -0.98995
3

0 - 3
0 - 4
Also, lx =
= -0.6 and ly =
= -0.8
5
5
1
0.072
0.096
k1 = AE D
0.01697
-0.11879

2
0.096
0.128
0.02263
-0.15839

3
0.01697
0.02263
0.004
-0.028

45Њ

4
-0.11879 1
-0.15839 2
T
-0.028
3
0.196
4

For member ƒ 1 ƒ , L = 4 m. Referring to Fig. b, ux– = 45° and uy– = 135°.
22
22
Thus, lx– = cos 45° =
and ly– = cos 135° = .
2
2
Also, lx = 0 and ly = -1.
5
0
k2 = AE 0
D
0
0

6
3
0
0
0.25
0.17678
0.17678 0.125
-0.17678 -0.125

4
0
5
-0.17678 6
T
3
-0.125
0.125
4

For member ƒ 3 ƒ , L = 3 m, lx = 1 and ly = 0.
5
0.33333
k3 = AE
0
D
-0.33333
0

6
0
0
0
0

1
-0.33333
0
0.33333
0

2
0 5
0 6
T
0 1
0 2

514


© 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.

14–15.

Continued

The structure stiffness matrix is a 6 * 6 matrix since the highest code number is 6. Thus,
1
0.40533
0.096
k = AE
0.01697
F
-0.11879
-0.33333
0

2
0.096
0.128
0.02263
-0.15839
0
0

3
0.01697
0.02263
0.129
-0.153
0
0.17678

4
-0.11879
-0.15839
-0.153
0.321
0
-0.17678

*14–16. Determine the vertical displacement of joint
and the support reactions. AE is constant.

5
-0.33333
0
0
0
0.33333
0

6
0
1
0
2
0.17678 3
V
-0.17678 4
0
5
0.25
6

2

3m
6
5

3

2
2
1

3
3 kN
4m

1
2
1

Here,

4

0
1
0 4
3
Qk = C -3(10 ) S 2 and Dk = C 0 S 5
0
3
0 6

3
45Њ

Applying Q = KD
0
0.40533
0.096
-3(103)
0.01697
0
F
V = AE F
Q4
-0.11879
Q5
-0.33333
Q6
0

0.096
0.128
0.02263
-0.15839
0
0

0.01697
0.02263
0.129
-0.153
0
0.17678

-0.11879
-0.15839
-0.153
0.321
0
-0.17678

-0.33333
0
0
0
0.33333
0

From the matrix partition; Qk = K11Du + K12Dk,
0
0.40533
3
C -3(10 ) S = AE C 0.096
0
0.011697

0.096
0.128
0.02263

0.01697
D1
0
0.02263 S C D2 S + C 0 S
0.129
D3
0

Expanding this matrix equality,
0 = AE(0.40533 D1 + 0.096 D2 + 0.01697 D3)

(1)

3

-3(10 ) = AE(0.096 D1 + 0.128 D2 + 0.02263 D3)

(2)

0 = AE(0.01697 D1 + 0.02263 D2 + 0.0129 D3)

(3)
515

0
D1
0
D2
0.17678 D3
VF V
-0.17678
0
0
0
0.25
0


© 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.

14–16.

Continued

Solving Eqs. (1) to (3),
D1 =

6.750(103)
4.2466(103)
D3 =
AE
AE

D2 =

-29.250(103)
29.3(103)
=
AE
AE

Ans.

T

Again, the matrix partition Qu = K21Du + K22Dk gives
Q4
-0.11879
C Q5 S = AE C -0.33333
Q6
0

-0.15839
0
0

Q4 = 3.182(103) N = 3.18 kN

-0.153
6.750(103)
0
1
0
S
C -29.250(103) S + C 0 S
AE
0.17678
4.2466(103)
0
Q5 = -2.250(103) N = -2.25 kN

Q6 = 750 N

Ans.
Ans.

516



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