© 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.

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

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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.

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

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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.

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

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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.

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

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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.

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

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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.

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

© 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.

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

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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.

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

© 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.

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

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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.

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

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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.

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

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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.

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

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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.

*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

Known Nodal Loads and Deflections. The nodal loads acting on the

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

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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.

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

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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.

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

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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

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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

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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

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

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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.

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

© 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.

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

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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.

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

© 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.

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

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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.

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

© 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.

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

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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.

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

© 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.

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

© 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.

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

© 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.

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

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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.

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

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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.

*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

Known Nodal Loads and Deflections. The nodal loads acting on the

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

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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.

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

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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.

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

© 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.

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

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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.

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

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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.

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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