# Solutions (8th ed structural analysis) chapter 15

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

15–1. Determine the moments at 1 and 3 . Assume
is a roller and 1 and 3 are fixed. EI is constant.

2

4

5
25 kN/m
2

1
1

1

Member Stiffness Matrices. For member ƒ 1 ƒ ,
12EI

12EI
=
= 0.05556EI
L3
63

6EI
6EI
=
= 0.16667EI
L2
62

4EI
4EI
=
= 0.66667EI
L
6

2EI
2EI
=
= 0.33333EI
L
6

2
0.16667
0.66667
-0.16667
0.33333

4
-0.05556
-0.16667
0.05556
-0.16667

1

0.16667
0.33333
T
-0.16667
0.66667

5
2
4
1

For member ƒ 2 ƒ ,
12EI
12EI
=
= 0.1875EI
L3
43

6EI
6EI
=
= 0.375EI
L2
42

4EI
4EI
=
= EI
L
4

2EI
2EI
=
= 0.5EI
L
4

4
0.1875
k2 = EI 0.375
D
-0.1875
0.375

1
0.375
1.00
-0.375
0.5

6
-0.1875
-0.375
0.1875
-0.375

3
0.375
0.5
T
-0.375
1.00

4
1
6
3

Known Nodal Loads and Deflection. The nodal load acting on the unconstrained
degree of freedom (Code number 1) is shown in Fig. a. Thus;

Qk = [75] 1

and

0
0
Dk = E 0 U
0
0

2
3
4
5
6

Load-Displacement Relation. The structure stiffness matrix is a 6 * 6 matrix
since the highest Code number is 6. Applying Q = KD
75
Q2
Q
F 3 V = EI
Q4
Q5
Q6

1
1.6667
0.33333
0.5
F
0.20833
0.16667
-0.375

2
0.33333
0.66667
0
-0.16667
0.16667
0

3
0.5
0
1.00
0.375
0
-0.375

4
0.20833
-0.16667
0.375
0.24306
-0.05556
-0.1875

5
0.16667
0.16667
0
-0.05556
0.05556
0

From the matrix partition, Qk = K11Du + K12Dk,
75 = 1.66667EID1 + 0

D1 =

45
EI
517

6
-0.375
0
-0.375
V
-0.1875
0
0.1875

3
2

6m

5
0.05556
k1 = EI 0.16667
D
-0.05556
0.16667

6

1 D1
2 0
3 0
F V
4 0
5 0
6 0

2
4m

3

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

15–1.

Continued

Also, Qu = K21Du + K22Dk,
Q2 = 0.33333EIa
Q3 = 0.5EIa

45
b + 0 = 15 kN # m
EI

45
b + 0 = 22.5 kN # m
EI

Superposition of these results and the (FEM) in Fig. b,
M1 = 15 + 75 = 90 kN # m d

Ans.

M3 = 22.5 + 0 = 22.5 kN # m d

Ans.

15–2. Determine the moments at
2 moves upward 5 mm. Assume
3 are fixed. EI = 60(106) N # m2.

1
2

and 3 if the support
is a roller and 1 and

4

5
25 kN/m
2
1

1
1

Member Stiffness Matrices. For member ƒ 1 ƒ ,
12EI
12EI
=
= 0.05556 EI
L3
63

6EI
6EI
=
= 0.16667 EI
L2
62

4EI
4EI
=
= 0.66667EI
L
6

2EI
2EI
=
= 0.33333 EI
L
6

5
0.05556
0.16667
D
-0.05556
0.16667

2
0.16667
0.66667
-0.16667
0.33333

4
-0.05556
-0.16667
0.05556
-0.16667

1
0.16667
0.33333
T
-0.16667
0.66667

5
2
4
1

For member ƒ 2 ƒ ,
12EI
12EI
=
= 0.1875EI
3
L
43

6EI
6EI
=
= 0.375EI
2
L
42

4EI
4EI
=
= EI
L
4

2EI
2EI
=
= 0.5EI
L
4

518

3
2

6m

k1 = EI

6

2
4m

3

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

15–2.

Continued

4
0.1875
0.375
k2 = EI
D
-0.1875
0.375

1
0.375
1.00
-0.375
0.5

6
-0.1875
-0.375
0.1875
-0.375

3
0.375
0.5
T
-0.375
1.00

4
1
6
3

Known Nodal Loads and Deflection. The nodal load acting on the unconstrained
degree of freedoom (code number 1) is shown in Fig. a. Thus,

Qk = [75(103)] 1

and

0
2
0
3
Dk = E 0.005 U 4
0
5
0
6

Load-Displacement Relation. The structure stiffness matrix is a 6 * 6 matrix since
the highest code number is 6. Applying Q = kD
1
75(103)
1.66667
Q2
0.33333
Q3
0.5
F
V = EI F
Q4
0.20833
Q5
0.16667
Q6
-0.375

2
0.33333
0.66667
0
-0.16667
0.16667
0

3
0.5
0
1.00
0.375
0
-0.375

4
0.20833
-0.16667
0.375
0.24306
-0.05556
-0.1875

5
0.16667
0.16667
0
-0.05556
0.05556
0

6
-0.375
0
-0.375
V
-0.1875
0
0.1875

From the matrix partition, Qk = K11Du + K12Dk,
75(103) = [1.6667D1 + 0.20833(0.005)][60(106)]
D1 = 0.125(10 - 3) rad
Using this result and apply, Qu = K21Du + K22Dk,
Q2 = {0.33333[0.125(10-3)] + (-0.16667)(0.005)}[60(106)] = -47.5 kN # m
Q3 = {0.5[0.125(10 - 3)] + 0.375(0.005)}[60(106)] = 116.25 kN # m
Superposition these results to the (FEM) in Fig. b,
M1 = -47.5 + 75 = 27.5 kN # m

Ans.

M3 = 116.25 + 0 = 116.25 kN.m = 116 kN # m

Ans.

519

1 D1
0
2
0
3
V
F
4 0.005
0
5
0
6

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

15–3. Determine the reactions at the supports. Assume
the rollers can either push or pull on the beam. EI is
constant.

6

4
5

1

1

6EI
6EI
=
= 0.041667EI
L2
122

4EI
4EI
=
= 0.333333EI
L
12

2EI
2EI
=
= 0.166667EI
L
12

5
0.041667
0.333333
-0.041667
0.166667

4
-0.006944
-0.041667
0.006944
-0.041667

2
0.041667
0.166667
T
-0.041667
0.333333

6
5
4
2

For member ƒ 2 ƒ ,
12EI
12EI
=
= 0.0234375EI
L3
83

6EI
6EI
=
= 0.09375EI
L2
82

4EI
4EI
=
= 0.5EI
L
8

2EI
2EI
=
= 0.25EI
L
8

4
0.0234375
k2 = EI D 0.09375
-0.0234375
0.09375

2
0.09375
0.5
-0.09375
0.25

3
-0.0234375
-0.09375
0.0234375
-0.09375

1
0.09375
0.25
T
-0.09375
0.5

4
0
2
87
D
T
3
0
1 -3.76

Known Nodal Loads And Deflection. The nodal loads acting on the unconstrained
degree of freedoom (code number 1 and 2) are shown in Fig. a. Thus,
20 1
Qk = c d
72 2

and

0
0
Dk = D T
0
0

3
4
5
6

Load-Displacement Relation. The structure stiffness matrix is a 6 * 6 matrix since
the highest code number is 6. Applying Q = KD
1
20
0.5
72
0.25
Q3
-0.09375
V = EI F
F
Q4
0.09375
Q5
0
Q6
0

2
0.25
0.833333
-0.09375
0.052083
0.166667
0.041667

3
-0.09375
-0.09375
0.0234375
-0.0234375
0
0

4
0.09375
0.052083
-0.0234375
0.0303815
-0.041667
-0.006944

5
0
0.166667
0
-0.041667
0.333333
0.041667

6
0
0.041667
0
V
-0.006944
0.041667
0.006944

From the matrix partition, Qk = K11Du + K12Dk,
20 = EI[0.5D1 + 0.25D2]

(1)

72 = EI[0.25D1 + 0.833333D2]

(2)

520

1 D1
2 D2
3 0
F V
4 0
5 0
6 0

1

2
2

Member Stiffness Matrices. For member ƒ 1 ƒ ,
12EI
12EI
=
= 0.006944EI
L3
123

20 kNиm

2

12 m

6
0.006944
0.041667
k1 = EI D
-0.006944
0.041667

3

6 kN/m

3
8m

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

15–3.

Continued

Solving Eqs. (1) and (2),
D1 = -

3.7647
EI

87.5294
EI

D2 =

Also, Qu = K21Du + K22Dk
Q3
-0.09375
Q4
0.09375
D
T = EI D
Q5
0
Q6
0

0
-0.09375
-3.7647
0
0.052083
1
T
c
d + D T
0.166667 EI 87.5294
0
0.041667
0

Q3 = -0.09375(-3.7647) + (-0.09375)(87.5294) = -7.853 kN
Q4 = 0.09375(-3.7647) + 0.052083(87.5294) = 4.206 kN
Q5 = 0 + 0.166667(87.5294) = 14.59 kN # m
Q6 = 0 + 0.041667(87.5294) = 3.647 kN
Superposition these results with the (FEM) in Fig. b,
R3 = -7.853 + 0 = -7.853 kN = 7.85 kN T

Ans.

R4 = 4.206 + 36 = 40.21 kN = 40.2 kN c

Ans.

M5 = 14.59 + 72 = 86.59 kN # m = 86.6 kN # m c

Ans.

R6 = 3.647 + 36 = 39.64 kN = 39.6 kN c

Ans.

*15–4. Determine the reactions at the supports. Assume
1 is a pin and 2 and 3 are rollers that can either push or
pull on the beam. EI is constant.

7

8
1
1

10 ft

Member Stiffness Matrices. For member ƒ 1 ƒ , ƒ 2 ƒ and ƒ 3 ƒ ,
12EI
12EI
=
= 0.012
L3
103

6EI
6EI
=
= 0.06
L2
102

4EI
4EI
=
= 0.4
L
10

2EI
2EI
=
= 0.2
L
10

k1

8
0.012
0.06
= EI
D
-0.012
0.06

1
0.06
0.4
-0.06
0.2

7
-0.012
-0.06
0.012
-0.06

2
0.06 8
0.2 1
T
-0.06 7
0.4 2

521

3

2
1

3k

6

2

2
10 ft

5
4

3

3
10 ft

4

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

15–4.

Continued

k2

7
0.012
= EI
0.06
D
-0.012
0.06

k3

6
0.012
0.06
= EI
D
- 0.012
0.06

2
0.06
0.4
-0.06
0.2

6
-0.012
-0.06
0.012
-0.06

3
0.06
0.4
- 0.06
0.2

3
0.06
0.2
T
-0.06
0.4

5
- 0.012
- 0.06
0.012
- 0.06

7
2
6
3

4
0.06
0.2
T
- 0.06
0.4

6
3
5
4

Known Nodal Load and Deflection. The nodal loads acting on the unconstrained
degree of freedom (code number 1, 2, 3, 4 and 5 ) is

Qk

0
0
= E 0 U
0
-3

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

Load-Displacement Relation. The structure stiffness matrix is a 8 * 8 matrix since
the highest code number is 8. Applying Q = KD
1
0
0.4
0
0.2
0
0
0
=
EI
0
H
X
H
- 3
0
Q6
0
Q7
-0.06
Q8
0.06

2
0.2
0.8
0.2
0
0
-0.06
0
0.06

3
0
0.2
0.8
0.2
-0.06
0
0.06
0

4
0
0
0.2
0.4
-0.06
0.06
0
0

5
0
0
-0.06
-0.06
0.012
-0.012
0
0

6
0
-0.06
0
0.06
-0.012
0.024
-0.012
0

7
-0.06
0
0.06
0
0
-0.012
0.024
-0.012

8
0.06
0.06
0
0
X
0
0
-0.012
0.012

1
2
3
4
5
6
7
8

D1
D2
D3
D4
H X
D5
0
0
0

From the matrix partition, Qk = k11 Du + k12 Dk,
0 = 04D1 + 0.2D2

(1)

0 = 0.2D1 + 0.8D2 + 0.2D3

(2)

0 = 0.2D2 + 0.8D3 + 0.2D4 - 0.06D5

(3)

0 = 0.2D3 + 0.4D4 – 0.06D5

(4)

-3 = -0.06D3 - 0.06D4 + 0.012D5

(5)

Solving Eq. (1) to (5)
D1 = -12.5 D2 = 25 D3 = -87.5 D4 = -237.5 D5 = -1875
Using these results, Qu = K21 Du + k22 Dk
Q6 = 6.75 kN

Ans.

Q7 = -4.5 kN

Ans.

Q8 = -0.75 kN

Ans.

522

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

15–5. Determine the support reactions. Assume
3 are rollers and 1 is a pin. EI is constant.

2

1

Member Stiffness Matrices. For member ƒ 1 ƒ
6EI
6EI
= 2 = 0.16667EI
2
L
6

4EI
4EI
=
= 0.066667EI
L
6

2EI
2EI
=
= 0.033333EI
L
6

k1

1
0.16667
0.66667
-0.16667
0.33333

5
- 0.05556
-0.16667
0.05556
-0.16667

2
0.16667
0.33333
T
-0.16667
0.66667

1
6m

6
1
5
2

For Member ƒ 2 ƒ ,
12EI
12EI
=
= 0.0234375EI
L3
83

6EI
6EI
= 2 = 0.09375EI
L2
8

4EI
4EI
=
= 0.5EI
L
8

2EI
2EI
=
= 0.025EI
L
8

k2

5
0.0234375
= EI
0.09375
D
-0.0234375
0.09375

2
0.09375
0.5
-0.09375
0.25

4
-0.0234375
-0.09375
0.0234375
-0.09375

3
0.09375
0.25
T
-0.09375
0.5

5
2
4
3

Known Nodal Load and Deflection. The nodal loads acting on the unconstrained
degree of freedom (code number 1, 2, and 3) are shown in Fig. a
0 1
Qk = C 36 S 2
0 3

and

0 4
Dk = C 0 S 5
0 6

Load-Displacement Relation. The structure stiffness matrix is a 6 * 6 matirx
since the highest code number is 6. Applying Q = KD,
1
2
3
4
5
0
0.66667
0.33333
0
0
0.16667
36
0.33333
1.16667
0.25
-0.09375
-0.07292
0
0
0.25
0.5
-0.09375
0.09375
F V = EI F
Q4
0
-0.09375 -0.09375
0.0234375
-0.0234375
Q5
-0.16667 -0.07292
0.09375
-0.0234375
0.0789931
Q6
0.16667
0.16667
0
0
-0.05556
From the matrix partition, Qk = k11 Du + k12 Dk
0 = 0.66667D1 + 0.33333D2
36 = 0.33333D1 + 1.16667D2 + 0.25D3
0 = 0.25D2 + 0.5D3

(1)
(2)
(3)

523

6
0.16667
0.16667
0
V
0
-0.05556
0.05556

1
2
3
4
5
6

4

2

1

12EI
12EI
=
= 0.05556EI
3
L
63

6
0.05556
= EI 0.16667
D
-0.05556
0.16667

15 kN/m 5

6

and

D1
D2
D
F 3V
0
0
0

2

3
2
8m

3

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

15–5.

Continued

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

-20.5714
EI

D2 =

41.1429
EI

D3 =

-20.5714
EI

Using these results and apply Qu = k21 Du +
Q4 = 0 + (-0.09375EI) a
Q5 = -0.16667EI a -

+ k22

Dk

20.5714
41.1429
b + (-0.09375EI)a b = -1.929 kN
EI
EI

20.5714
41.1429
b + ( -0.07292EI) a
b
EI
EI

+ 0.09375EI a -

20.5714
b
EI

= -1.500 kN
Q6 = 0.16667EI a -

20.5714
41.1429
b + 0.16667EI a
b = 3.429 kN
EI
EI

Superposition these results with the FEM show in Fig. b
R4 = -1.929 + 0 = -1.929 kN = 1.93 kN T

Ans.

R5 = -1.500 + 36 = 34.5 kN c
R6 = 3.429 + 9 = 12.43 kN = 12.4 kN c

Ans.

524

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

15–6. Determine the reactions at the supports. Assume
is fixed 2 and 3 are rollers. EI is constant.

5

1

4

10 kN/m

3

1

6

2
1

1

2

6m

8m

Member Stiffness Matrices. For member 1 ,
12EI
12EI
=
= 0.05556EI
L3
63

6EI
6EI
= 2 = 0.16667EI
L2
6

4EI
4EI
=
= 0.066667EI
L
6

2EI
2EI
=
= 0.33333EI
L
6

k1

5
0.05556
= EI
0.16667
D
-0.05556
0.16667

6
0.16667
0.66667
-0.16667
0.33333

4
-0.05556
-0.16667
0.05556
-0.16667

1
0.16667
0.33333
T
-0.16667
0.66667

5
6
4
1

For Member 2 ,
12EI
12EI
=
= 0.0234375EI
L3
83

6EI
8EI
= 2 = 0.09375EI
L2
8

4EI
4EI
=
= 0.5EI
L
8

2EI
2EI
=
= 0.025EI
L
8

k2

4
0.0234375
= EI
0.09375
D
-0.0234375
0.09375

1
0.09375
0.5
-0.09375
0.25

3
-0.0234375
-0.09375
0.0234375
-0.09375

2
0.09375
0.25
T
-0.09375
0.5

4
1
3
2

Known Nodal Load and Deflections. The nodal loads acting on the unconstrained
degree of freedom (code number 1 and 2) are shown in Fig. a

Qk

-50 1
= c
d
0 2

and

0
0
Dk = D T
0
0

3
4
5
6

Load-Displacement Relation. The structure stiffness matrix is a 6 * 6 matirx
since the highest code number is 6. Applying Q = KD,
1
-50
1.16667
0
0.25
Q
-0.09375
F 3 V = EI F
Q4
-0.07292
Q5
0.16667
Q6
0.33333

2
0.25
0.5
-0.09375
0.09375
0
0

3
-0.09375
-0.09375
0.0234375
-0.0234375
0
0

4
-0.07292
0.09375
-0.0234375
0.0789931
-0.05556
-0.16667

525

5
0.16667
0
0
-0.05556
0.05556
0.16667

6
0.33333
0
0
V
-0.16667
0.16667
0.66667

2

1 D1
2 D2
3 0
F V
4 0
5 0
6 0

3

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

15–6.

Continued

From the matrix partition, Qk = K11 Du + K12 Dk
-50 = EI(1.16667D1 + 0.25D2)
0 = EI (0.25D1 + 0.5D2)
Solving Eqs. (1) and (2),
D1 =

48
EI

24
EI

D2 =

Using these results and apply in Qu = K21 Du + K22 Dk
Q3 = -0.09375EI a-

24
48
b + ( -0.09375EI) a
b + 0 = 2.25 kN
EI
EI

Q4 = -0.07292EI a-

48
24
b + 0.09375EI a
b + 0 = 5.75 kN
EI
EI

Q5 = 0.16667EIa-

48
b + 0 + 0 = -8.00 kN
EI

Q6 = (0.33333EI)a-

48
b + 0 + 0 = -16.0 kN
EI

Superposition these results with the FEM show in Fig. b
R3 = 2.25 + 30 = 32.25 kN c

Ans.

R4 = 5.75 + 30 + 50 = 85.75 kN c

Ans.

R5 = - 8.00 + 30 = 22.0 kN c

Ans.

R6 = -16.0 + 30 = 14.0 kN # md

Ans.

526

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

15–7. Determine the reactions at the supports. Assume
and 3 are fixed and 2 is a roller. EI is constant.

1

3

2

5
9 kN/m

6 kN/m
1

6

4
1
1
6m

Member Stiffness Matrices. For member 1 ,
12EI
12EI
=
= 0.05556EI
L3
63

6EI
6EI
=
= 0.16667EI
L2
62

4EI
4EI
=
= 0.66667EI
L
6

2EI
2EI
=
= 0.33333EI
L
6

k1

5
0.05556
= EI
0.16667
D
-0.05556
0.16667

6
0.16667
0.66667
-0.16667
0.33333

2
-0.05556
-0.16667
0.05556
-0.16667

1
0.16667
0.33333
T
-0.16667
0.66667

5
6
2
1

For member 2 ,
12EI
12EI
=
= 0.1875EI
L3
43

6EI
6EI
=
= 0.375EI
L2
42

4EI
4EI
=
= EI
L
4

2EI
2EI
=
= 0.5EI
L
4

k2

2
0.1875
0.375
= EI D
-0.1875
0.375

1
0.375
1.00
-0.375
0.5

3
-0.1875
-0.375
0.1875
-0.375

4
0.375
0.5
T
-0.375
1.00

2
1
3
4

Known Nodal Loads and Deflections. The nodal load acting on the unconstrained
degree of freedom (code number 1) are shown in Fig. a

Qk = [19] 1

and

0
0
Dk = E 0 U
0
0

2
3
4
5
6

527

2

2
4m

3

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

15–7.

Continued

Load-Displacement Relation. The structure stiffness matrix is a 6 * 6 matirx
since the highest code number is 6. Applying Q = KD,
1
19
1.66667
Q2
0.20833
Q3
-0.375
F V = EI F
Q4
0.5
Q5
0.16667
Q6
0.33333

2
0.20833
0.24306
-0.1875
0.375
-0.05556
-0.16667

3
-0.375
-0.1875
0.1875
-0.375
0
0

4
0.5
0.375
-0.375
1.00
0
0

5
0.16667
-0.05556
0
0
0.05556
0.16667

6
0.33333
-0.16667
0
V
0
0.16667
0.66667

1 D1
2 0
3 0
F V
4 0
5 0
6 0

From the matrix partition, Qk = K11Du + K12Dk
19 = 1.66667EID1

D1 =

11.4
EI

Using this result and applying Qu = K21Du + K22Dk
Q2 = 0.20833EIa
Q3 = -0.375EIa
Q4 = 0.5EIa

11.4
b = 2.375 kN
EI

11.4
b = -4.275 kN
EI

11.4
b = 5.70 kN # m
EI

Q5 = 0.16667a

11.4
b = 1.90 kN
EI

Q6 = 0.33333a

11.4
b = 3.80 kN # m
EI

Superposition these results with the FEM shown in Fig. b,
R2 = 2.375 + 27 + 12 = 41.375 kN = 41.4 kN c

Ans.

R3 = -4.275 + 12 = 7.725 kN c

Ans.

R4 = 5.70 - 8 = -2.30 kN # m = 2.30 kN.m b

Ans.

R5 = 1.90 + 27 = 28.9 kN c

Ans.

R6 = 3.80 + 27 = 30.8 kN.m c

Ans.

528

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

*15–8. Determine the reactions at the supports. EI is
constant.

4

6
7

1

1
4m

Member Stiffness Matrices. For member 1
12EI
12EI
=
= 0.1875EI
L3
43

6EI
6EI
=
= 0.375EI
L2
42

4EI
4EI
=
= EI
L
4

2EI
2EI
=
= 0.5EI
L
4

k1

6
0.1875
= EI
0.375
D
-0.1875
0.375

7
0.375
1.00
-0.375
0.5

4
-0.1875
-0.375
0.1875
-0.375

3
0.375
0.5
T
-0.375
1.00

6
7
4
3

For member 2 ,
12EI
12EI
=
= 0.44444EI
3
L
33

6EI
6EI
=
= 0.66667EI
2
L
32

4EI
4EI
=
= 1.33333EI
L
3

2EI
2EI
=
= 0.66667EI
L
3

k2

4
0.44444
= EI
0.66667
D
-0.44444
0.66667

2
0.66667
1.33333
-0.66667
0.66667

5
-0.44444
-0.66667
0.44444
-0.66667

1
0.66667
0.66667
T
-0.66667
1.33333

4
2
5
1

unconstrained degree of freedom (code number 1, 2, 3, and 4) are
shown in Fig. a and b.

Qk

0 1
-9 2
= D
T
0 3
-18 4

and

0 5
Dk = C 0 S 6
0 7

529

5
15 kN/m
2

3

2

1
2
3

3m

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

15–8.

Continued

Load-Displacement Relation. The structure stiffness matrix is a 7 * 7 matirx
since the highest code number is 7. Applying Q = KD,
1
0
1.33333
-9
0.66667
0
0
G -18 W = EI G 0.66667
Q5
-0.66667
Q6
0
Q7
0

2
0.66667
1.33333
0
0.66667
-0.66667
0
0

3
0
0
1.00
-0.375
0
0.375
0.5

4
0.66667
0.66667
-0.375
0.63194
-0.44444
-0.1875
-0.375

5
-0.66667
-0.66667
0
-0.44444
0.44444
0
0

6
0
0
0.375
-0.1875
0
0.1875
0.375

7
0
0
0.5
-0.375 W
0
0.375
1.00

1
2
3
4
5
6
7

D1
D2
D3
G D4 W
0
0
0

From the matrix partition, Qk = K11Du + K12Dk,
0 = EI(1.33333D1 + 0.66667D2 + 0.66667D4)

(1)

-9 = EI(0.66667D1 + 1.33333D2 + 0.66667D4)

(2)

0 = EI(D3 - 0.375D4)

(3)

-18 = EI(0.66667D1 + 0.66667D2 - 0.375D3 + 0.63194D4)

(4)

Solving Eqs. (1) to (4),
D1 =

111.167
EI

D2 =

97.667
EI

D3 = -

120
EI

D4 = -

320
EI

Using these result and applying Qu = K21Du + K22Dk
Q5 = -0.66667EIa
Q6 = 0.375EIa Q7 = 0.5EIa -

111.167
97.667
-320
b + a-0.66667EIb a
b + ( -0.44444EI) a
b + 0 = 3.00 kN
EI
EI
EI

120
320
b + (-0.1875EI) a b + 0 = 15.00 kN
EI
EI

120
320
b + (-0.375EI)a b + 0 = 60.00 kN # m
EI
EI

Superposition of these results with the (FEM),
R5 = 3.00 + 4.50 = 7.50 kN c

Ans.

R6 = 15.00 + 0 = 15.0 kN c

Ans.

R7 = 60.00 + 0 = 60.0 kN # m a

Ans.

530

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

4 kN/ m

15–9. Determine the moments at 2 and 3 . EI is constant.
Assume 1 , 2 , and 3 are rollers and 4 is pinned.

1

1

2

1
12 m

The FEMs are shown on the figure.

D1
D2
Dk = D T
D3
D4

-19.2
-19.2
Qk = D
T
19.2
19.2

k1 = EI c

0.3333
0.16667

0.16667
d
0.3333

k2 = EI c

0.3333
0.16667

0.16667
d
0.3333

k3 = EI c

0.3333
0.16667

0.16667
d
0.3333

K = k1 + k2 + k3
0.3333
0.16667
K = EI D
0
0

0.16667
0.6667
0.16667
0

0
0.16667
0.6667
0.16667

0
0
T
0.16667
0.3333

Q = KD
-19.2
0.3333
-19.2
0.16667
D
T = EI D
19.2
0
19.2
0

0.16667
0.6667
0.16667
0

0
0.16667
0.6667
0.16667

D1
0
0
D
T D 2T
0.16667
D3
0.3333
D4

-19.2 = EI[0.3333D1 + 0.16667D2]
-19.2 = EI[0.16667D1 + 0.6667D2 + 0.16667D3]
19.2 = EI[0.16667D2 + 0.6667D3 + 0.16667D4]
19.2 = EI[0.16667D3 + 0.16667D4]
Solving,
D1 = -46.08>EI
D2 = -23.04>EI
D3 = 23.04>EI
D4 = 46.08>EI
q = k1D

531

3

2

2
12 m

4

3

3
12 m

4

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

15–9.

c

Continued

q1
0.3333
d = EI c
q2
0.16667

0.16667 -46.08>EI
d
d c
0.3333
-23.04>EI

q1 = EI[0.3333(-46.08>EI) + 0.16667(-23.04>EI)]
q1 = -19.2 kN # m
q2 = EI[0.16667(-46.08>EI) + 0.3333(-23.04>EI)]
q2 = -15.36 kN # m
Since the opposite FEM = 19.2 kN # m is at node 1, then
M1 = M4 = 19.2 - 19.2 = 0
Since the FEM = -28.8 kN # m is at node 2, then
M2 = M3 = -28.8 - 15.36 = 44.2 kN # m

15–10. Determine the reactions at the supports. Assume
is pinned and 1 and 3 are rollers. EI is constant.

Ans.

4

2

5

1
1
4 ft

Member 1

k1

0.1875
EI
0.75
=
D
8 -0.1875
0.75

0.75
4
-0.75
2

-0.1875
-0.75
0.1875
- 0.75

0.75
2
T
-0.75
4

532

3 k/ ft 6

2
1

8 ft

2

3
2

8 ft

3
4 ft

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

15–10.

Continued

Member 2

k2

0.1875
EI
0.75
=
D
8 -0.1875
0.75

0.75
4
-0.75
2

0.75
2
T
-0.75
4

-0.1875
-0.75
0.1875
-0.75

Q = KD

8.0
4
0
2
- 8.0
EI
0
F
V =
F
Q4 - 24.0
8
0.75
Q5 - 24.0
-0.75
Q6 - 24.0
0

8.0 =
0 =

-8.0 =

2
8
2
0.75
0
-0.75

0
2
4
0
0.75
-0.75

0.75
0.75
0
0.1875
-0.1875
0

-0.75
0
0.75
-0.1875
0.375
-0.1875

0
D1
D2
-0.75
-0.75
D3
V F V
0
0
-0.1875
0
0
0.1875

EI
[4D1 + 2D2]
8
EI
[2D1 + 8D2 + 2D3]
8
EI
[2D2 + 4D3]
8

Solving:
D1 =

16.0
,
EI

Q4 - 24.0 =

D2 = 0,

D3 = -

16.0
EI

EI
16.0
(0.75)a
b + 0 + 0
8
EI

Q4 = 25.5 k
Q5 - 24.0 =

Ans.
EI
16.0
EI
16.0
( -0.75)a
b + 0 +
(0.75)a b
8
EI
8
EI

Q5 = 21.0 k

Ans.

Q6 - 24.0 = 0 + 0 +

EI
-16.0
(-0.75)a
b
8
EI

Q6 = 25.5 k

Ans.

a + a M2 = 0; 25.5(8) - 25.5(8) = 0
+ c a F = 0;

(Check)

25.5 + 21.0 + 25.5 - 72 = 0

(Check)

533

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

15–11. Determine the reactions at the supports. There is a
smooth slider at 1 . EI is constant.

1

3

30 kN/m

2
4

Member Stiffness Matrix. For member 1 ,

4m

12EI
12EI
=
= 0.1875EI
L3
43

6EI
6EI
= 2 = 0.375EI
L2
4

4EI
4EI
=
= EI
L
4

2EI
2EI
=
= 0.5EI
L
4

k1

3
0.1875
= EI
0.375
D
-0.1875
0.375

4
0.375
1.00
-0.375
0.5

1
-0.1875
-0.375
0.1875
-0.375

2
0.375
0.5
T
-0.375
1.00

3
4
1
2

Known Nodal Loads And Deflections. The nodal load acting on the unconstrained
degree of freedom (code number 1) is shown in Fig. a. Thus,
Qk = [-60] 1

and

0 2
Dk = C 0 S 3
0 4

Load-Displacement Relation. The structure stiffness matrix is a 4 * 4 matirx
since the highest code number is 4. Applying Q = KD,
1
0.1875
-60
-0.375
Q
D 2 T = EI D
-0.1875
Q3
-0.375
Q4

2
-0.375
1.00
-0.375
0.5

3
-0.1875
0.375
0.1875
0.375

4
-0.375
0.5
T
0.375
1.00

1 D1
2 0
D T
3 0
4 0

From the matrix partition, Qk = K11Du + K12Dk,
-60 = 0.1875EID1

D1 = -

320
EI

Using this result, and applying Qu = K21Du + K22Dk,
Q2 = -0.375EIa -

320
b + 0 = 120 kN # m
EI

Q3 = -0.1875EIaQ4 = -0.375EIa-

1

320
b + 0 = 60 kN
EI

320
b + 0 = 120 kN # m
EI

Superposition these results with the FEM shown in Fig. b,
R2 = 120 - 40 = 80 kN # m d

Ans.

R3 = 60 + 60 = 120 kN c

Ans.

R4 = 120 + 40 = 160 kN # m d

Ans.

534

1

2

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