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文库首页行业制造Hibbeler-Mechanics of Materials, Eighth Edition - Instructor's Solutions Manual

Hibbeler-Mechanics of Materials, Eighth Edition - Instructor's S...

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Hibbeler-材料力学解决方案，第8版钢筋混凝土Hibbeler毕业于伊利诺伊大学厄巴纳分校，获得土木工程学士学位（结构专业）和核工程硕士学位。他获得了西北大学理论与应用力学博士学位。 Hibbeler教授的专业经验包括在阿贡国家实验室进行反应堆安全和分析的博士后工作，以及芝加哥桥和铁的结构和应力分析工作，以及芝加哥的Sargent和Lundy。他曾在俄亥俄州，纽约州和路易斯安那州从事工程学。 Hibbeler教授目前在路易斯安那大学拉斐特分校教授土木和机械工程课程。在过去，他曾在伊利诺伊大学厄巴纳分校，扬斯敦州立大学，伊利诺伊理工学院和联合学院任教。

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(a)

Ans.

(b)

Ans. F

= 34.9 kN

©F

= 0;

- 4.5 - 4.5 - 5.89 - 6 - 6 - 8 = 0

= 13.8 kip

©F

= 0;

- 1.0 - 3 - 3 - 1.8 - 5 = 0

1–1. Determine the resultant internal normal force acting

on the cross section through point A in each column. In

(a), segment BC weighs 180

ft and segment CD weighs

250

ft. In (b), the column has a mass of 200 >m.kglb

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

8 kN

3 m

1 m

6 kN6 kN

4.5 kN4.5 kN

200 mm200 mm

(b)

200 mm200 mm

3 kip3 kip

5 kip

10 ft

4 ft

8 in.8 in.

(a)

1–2. Determine the resultant internal torque acting on the

cross sections through points C and D.The support bearings

at A and B allow free turning of the shaft.

Ans.

Ans.©M

= 0;

= 0

= 250 N

©M

= 0;

- 250 = 0

300 mm

200 mm

150 mm

200 mm

250 mm

150 mm

400 Nm

150 Nm

250 Nm

Ans.

Ans. T

= 500 lb

©M

= 0;

- 500 = 0

= 150 lb

©M

= 0;

+ 350 - 500 = 0

1–3. Determine the resultant internal torque acting on the

cross sections through points B and C.

3 ft

2 ft

1 ft

500 lbft

350 lbft

600 lbft

01 Solutions 46060 5/6/10 2:43 PM Page 1

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

*1–4. A force of 80 N is supported by the bracket as

shown. Determine the resultant internal loadings acting on

the section through point A.

Equations of Equilibrium:

Ans.

Negative sign indicates that M

acts in the opposite direction to that shown on FBD.

=-0.555 N

-80 cos 15°(0.1 cos 30°) = 0

©M

= 0;

+ 80 sin 15°(0.3 + 0.1 sin 30°)

=-0.555 N

80 sin 45°(0.1 + 0.3 sin 30°) = 0

©M

= 0;

+ 80 cos 45°(0.3 cos 30°)

= 20.7 N

©F

y¿

= 0;

- 80 sin 15° = 0

= 77.3 N

Q©F

x¿

= 0;

- 80 cos 15° = 0

0.1 m

0.3 m

30

80 N

45

01 Solutions 46060 5/6/10 2:43 PM Page 2

Support Reactions: For member AB

Equations of Equilibrium: For point D

Ans.

Equations of Equilibrium: For point E

Ans.

Negative signs indicate that M

and V

act in the opposite direction to that shown

on FBD.

=-24.0 kip

+©M

= 0;

+ 6.00(4) = 0

=-9.00 kip

©F

= 0;

-6.00 - 3 - V

= 0

©F

= 0;

= 0

= 13.5 kip

+©M

= 0;

+ 2.25(2) - 3.00(6) = 0

= 0.750 kip

©F

= 0;

3.00 - 2.25 - V

= 0

©F

= 0;

= 0

©F

= 0;

+ 3.00 - 9.00 = 0

= 6.00 kip

©F

= 0;

= 0

+ ©M

= 0;

9.00(4) - A

(12) = 0

= 3.00 kip

•1–5. Determine the resultant internal loadings in the

beam at cross sections through points D and E. Point E is

just to the right of the 3-kip load.

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

6 ft

4 ft

6 ft

3 kip

1.5 kip/

01 Solutions 46060 5/6/10 2:43 PM Page 3

Support Reactions:

Equations of Equilibrium: For point C

Ans.

Negative signs indicate that N

and V

act in the opposite direction to that shown

on FBD.

= 6.00 kN

+©M

= 0;

8.00(0.75) - M

= 0

=-8.00 kN

©F

= 0;

+ 8.00 = 0

=-30.0 kN

©F

= 0;

-N

- 30.0 = 0

©F

= 0;

- 8 = 0

= 8.00 kN

©F

= 0;

30.0 - A

= 0

= 30.0 kN

+©M

= 0;

8(2.25) - T(0.6) = 0

T = 30.0 kN

1–6. Determine the normal force, shear force, and moment

at a section through point C.Take P = 8

kN.

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

0.75 m

0.5 m

0.1 m

0.75 m 0.75 m

Support Reactions:

Ans.

Equations of Equilibrium: For point C

Ans.

Negative signs indicate that N

and V

act in the opposite direction to that shown

on FBD.

= 0.400 kN

+©M

= 0;

0.5333(0.75) - M

= 0

=-0.533 kN

©F

= 0;

+ 0.5333 = 0

=-2.00 kN

©F

= 0;

-N

- 2.00 = 0

©F

= 0;

- 0.5333 = 0

= 0.5333 kN

©F

= 0;

2 - A

= 0

= 2.00 kN

P = 0.5333 kN = 0.533 kN

+©M

= 0;

P(2.25) - 2(0.6) = 0

1–7. The cable will fail when subjected to a tension of 2 kN.

Determine the largest vertical load P the frame will support

and calculate the internal normal force, shear force, and

moment at the cross section through point C for this loading.

0.75 m

0.5 m

0.1 m

0.75 m 0.75 m

01 Solutions 46060 5/6/10 2:43 PM Page 4

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

Referring to the FBD of the entire beam, Fig. a,

Referring to the FBD of this segment, Fig. b,

Ans.

a Ans.+©M

= 0;

+ 6(0.5) - 7.5(1) = 0

= 4.50 kN

©F

= 0;

7.50 - 6 - V

= 0

= 1.50 kN

©F

= 0;

= 0

+©M

= 0;

-A

(4) + 6(3.5) +

(3)(3)(2) = 0

= 7.50 kN

*1–8. Determine the resultant internal loadings on the

cross section through point C. Assume the reactions at

the supports A and B are vertical.

0.5 m

1.5 m1.5 m

3 kN/m

6 kN

Referring to the FBD of the entire beam, Fig. a,

Referring to the FBD of this segment, Fig. b,

Ans.

Ans. = 3.94 kN

+©M

= 0;

3.00(1.5) -

(1.5)(1.5)(0.5) - M

= 0

= 3.9375 kN

©F

= 0;

(1.5)(1.5) + 3.00 = 0

=-1.875 kN

©F

= 0;

= 0

+©M

= 0;

(4) - 6(0.5) -

(3)(3)(2) = 0

= 3.00 kN

•1–9. Determine the resultant internal loadings on the

cross section through point D. Assume the reactions at

the supports A and B are vertical.

0.5 m

1.5 m1.5 m

3 kN/m

6 kN

01 Solutions 46060 5/6/10 2:43 PM Page 5

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