Hibbeler Mechanics of Materials 10th Edition Instructors Solution Manual Chapter 1 Solutions PDF

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Summary

Here is the Instructor's Solution Manual to learn the procedures of analysis to solve a variety of problems explored in Mechanics of Materials/Mechanics of Solids....

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1–1. The shaft is supported by a smooth thrust bearing atB and a journal bearing at C. Determine the resultant internal loadings acting on the cross section at E.

B

A

4 ft

C

E

4 ft

4 ft

D

4 ft

400 lb 800 lb

SOLUTION Support Reactions: We will only need to compute Cy by writing the moment equation of equilibrium about B with reference to the free-body diagram of the entire shaft, Fig. a. a+ Σ MB = 0;

Cy(8) + 400(4) - 800(12) = 0

Cy = 1000 lb

Internal Loadings: Using the result for Cy, section DE of the shaft will be considered. Referring to the free-body diagram, Fig. b, + Σ Fx = 0; S Fy = 0; + cΣ

NE = 0

Ans.

VE + 1000 - 800 = 0

VE = -200 lb

Ans.

a+ Σ ME = 0; 1000(4) - 800(8) - ME = 0

ME = - 2400 lb # ft = - 2.40 kip # ft

Ans.

The negative signs indicates that VE and ME act in the opposite sense to that shown on the free-body diagram.

Ans: NE = 0, VE = -200 lb, ME = - 2.40 kip # ft 1

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1–2. Determine the resultant internal normal and shear force in the member at (a) section a–a and (b) section b–b, each of which passes through the centroid A. The 500-lb load is applied along the centroidal axis of the member.

a

b 30

500 lb

500 lb

A b

a

SOLUTION (a) + Fx = 0; SΣ

Na - 500 = 0 Na = 500 lb

Ans.

Fy = 0; +T Σ

Va = 0

Ans.

Fx = 0; R+ Σ

Nb - 500 cos 30° = 0

(b)

Nb = 433 lb +Q Σ Fy = 0;

Ans.

Vb - 500 sin 30° = 0 Vb = 250 lb

Ans.

Ans: (a) Na = 500 lb, Va = 0, (b) Nb = 433 lb, Vb = 250 lb 2

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1–3. Determine the resultant internal loadings acting on section b–b through the centroid C on the beam. B b

900 lb/ft

b

C

A

30

60 3 ft

6 ft

SOLUTION Support Reaction: a+ Σ MA = 0;

NB(9 sin 30°) -

1 (900)(9)(3) = 0 2

NB = 2700 lb Equations of Equilibrium: For section b–b + Σ Fx = 0; S

F +cΣ

y

= 0;

MC = 0; a+ Σ

1 (300)(3) sin 30° - 2700 = 0 2 Vb - b = 2475 lb = 2.475 kip

Ans.

1 (300)(3) cos 30° = 0 2 Nb - b = 389.7 lb = 0.390 kip

Ans.

Vb - b +

Nb - b -

2700(3 sin 30°) -

1 (300)(3)(1) - Mb - b = 0 2

Mb - b = 3600 lb # ft = 3.60 kip # ft

Ans.

Ans: Vb - b = 2.475 kip, Nb - b = 0.390 kip, Mb - b = 3.60 kip # ft 3

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*1–4. The shaft is supported by a smooth thrust bearing atA and a smooth journal bearing at B. Determine the resultant internal loadings acting on the cross section at C.

600 N/m A

D

B C

1m

1m

1m

1.5 m

1.5 m 900 N

SOLUTION Support Reactions: We will only need to compute By by writing the moment equation of equilibrium about A with reference to the free-body diagram of the entire shaft, Fig. a. a+ Σ MA = 0;

B y(4.5) - 600(2)(2) - 900(6) = 0

B y = 1733.33 N

Internal Loadings: Using the result of By, section CD of the shaft will be considered. Referring to the free-body diagram of this part, Fig. b, + Σ Fx = 0; S F +c Σ

y

= 0;

a+ Σ MC = 0;

NC = 0 VC - 600(1) + 1733.33 - 900 = 0

Ans. VC = -233 N

Ans.

1733.33(2.5) - 600(1)(0.5) - 900(4) - MC = 0

MC = 433 N # m

Ans.

The negative sign indicates that VC acts in the opposite sense to that shown on the free-body diagram.

Ans: NC = 0, VC = -233 N, MC = 433 N # m 4

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1–5. Determine the resultant internal loadings acting on the cross section at point B.

60 lb/ ft

A

C

B 3 ft

12 ft

SOLUTION + Σ Fx = 0; S F +c Σ

y

= 0;

NB = 0 VB -

Ans.

1 (48)(12) = 0 2

VB = 288 lb a+ Σ MB = 0;

-MB -

Ans.

1 (48)(12)(4) = 0 2

MB = -1152 lb # ft = -1.15 kip # ft

Ans.

Ans: NB = 0, VB = 288 lb, MB = -1.15 kip # ft 5

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1–6. Determine the resultant internal loadings on the cross section at point D.

C

1m

F

2m

1.25 kN/m

SOLUTION

A

Support Reactions: Member BC is the two force member. 4 a+ Σ MA = 0; F (1.5) - 1.875(0.75) = 0 5 BC

D

E

0.5 m 0.5 m 0.5 m

B

1.5 m

FBC = 1.1719 kN F +cΣ

y

= 0;

Ay +

4 (1.1719) - 1.875 = 0 5

Ay = 0.9375 kN + Σ Fx = 0; S

3 (1.1719) - Ax = 0 5 Ax = 0.7031 kN

Equations of Equilibrium: For point D + Σ Fx = 0; S Fy = 0; +cΣ a+ Σ MD = 0;

ND - 0.7031 = 0 ND = 0.703 kN

Ans.

0.9375 - 0.625 - VD = 0 VD = 0.3125 kN

Ans.

MD + 0.625(0.25) - 0.9375(0.5) = 0 MD = 0.3125 kN # m

Ans.

Ans: ND = 0.703 kN , VD = 0.3125 kN , MD = 0.3125 kN # m 6

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1–7. Determine the resultant internal loadings at cross sections at points E and F on the assembly.

C

1m

F

2m

1.25 kN/m

SOLUTION

A

Support Reactions: Member BC is the two-force member. 4 a+ Σ F (1.5) - 1.875(0.75) = 0 MA = 0; 5 BC

D

E

0.5 m 0.5 m 0.5 m

B

1.5 m

FBC = 1.1719 kN Fy = 0; +cΣ

Ay +

4 (1.1719) - 1.875 = 0 5

Ay = 0.9375 kN + Σ Fx = 0; S

3 (1.1719) - Ax = 0 5 Ax = 0.7031 kN

Equations of Equilibrium: For point F + bΣ Fx′ = 0;

NF - 1.1719 = 0 NF = 1.17 kN

Ans.

a+Σ Fy′ = 0;

VF = 0

Ans.

MF = 0; a+ Σ

MF = 0

Ans.

Equations of Equilibrium: For point E 3 +Σ Fx = 0; NE - (1.1719) = 0 d 5 NE = 0.703 kN Fy = 0; +cΣ

VE - 0.625 +

Ans.

4 (1.1719) = 0 5

VE = -0.3125 kN a+ Σ ME = 0;

-ME - 0.625(0.25) +

Ans. 4 (1.1719)(0.5) = 0 5

ME = 0.3125 kN # m

Ans.

Negative sign indicates that VE acts in the opposite direction to that shown on FBD.

Ans: NF = 1.17 kN, VF = 0, MF = 0, NE = 0.703 kN, VE = -0.3125 kN, ME = 0.3125 kN # m 7

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*1–8. The beam supports the distributed load shown. Determine the resultant internal loadings acting on the cross section at point C. Assume the reactions at the supports A and B are vertical.

4 kN/m

A

B C 1.5 m

D 3m

1.5 m

SOLUTION Support Reactions: Referring to the FBD of the entire beam, Fig. a, a+ Σ MA = 0;

By(6) -

1 (4)(6)(2) = 0 2

By = 4.00 kN

Internal Loadings: Referring to the FBD of the right segment of the beam sectioned through C, Fig. b, + Σ Fx = 0; S

NC = 0

Fy = 0; +cΣ

VC + 4.00 -

MC = 0; a+ Σ

4.00(4.5) -

Ans. 1 (3)(4.5) = 0 2

VC = 2.75 kN

Ans.

1 (3)(4.5)(1.5) - MC = 0 2

MC = 7.875 kN # m

Ans.

Ans: NC = 0 , VC = 2.75 kN , MC = 7.875 kN # m 8

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1–9. The beam supports the distributed load shown. Determine the resultant internal loadings acting on the cross section at point D. Assume the reactions at the supports A and B are vertical.

4 kN/m

A

B C 1.5 m

D 3m

1.5 m

SOLUTION Support Reactions: Referring to the FBD of the entire beam, Fig. a, a+ Σ MA = 0;

B y(6) -

1 (4)(6)(2) = 0 2

By = 4.00 kN

Internal Loadings: Referring to the FBD of the right segment of the beam sectioned through D, Fig. b, + Σ Fx = 0; S

ND = 0

Ans.

Fy = 0; +c Σ

V D + 4.00 -

a+ Σ MD = 0;

4 .00(1.5) -

1 (1.00)(1.5) = 0 2

VD = -3.25 kN

Ans.

1 (1.00)(1.5)(0.5) - MD = 0 2

MD = 5.625 kN # m

Ans.

The negative sign indicates that VD acts in the sense opposite to that shown on the FBD.

Ans: ND = 0, VD = -3.25 kN, MD = 5.625 kN # m 9

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1–10. The boom DF of the jib crane and the column DE have a uniform weight of 50 lb>ft. If the supported load is 300 lb, determine the resultant internal loadings in the crane on cross sections at points A, B, and C.

D

B 2 ft

A 8 ft

F

3 ft

5 ft C 300 lb 7 ft

SOLUTION Equations of Equilibrium: For point A + Σ Fx = 0; d Fy = 0; +c Σ

E

NA = 0

Ans.

VA - 150 - 300 = 0 VA = 450 lb

a+ Σ MA = 0;

Ans.

-MA - 150(1.5) - 300(3) = 0

MA = -1125 lb # ft = -1.125 kip # ft

Ans.

Negative sign indicates that MA acts in the opposite direction to that shown on FBD. Equations of Equilibrium: For point B + Σ Fx = 0; d F +c Σ

y

= 0;

NB = 0

Ans.

VB - 550 - 300 = 0 VB = 850 lb

a+ Σ MB = 0;

Ans.

-MB - 550(5.5) - 300(11) = 0

MB = -6325 lb # ft = -6.325 kip # ft

Ans.

Negative sign indicates that MB acts in the opposite direction to that shown on FBD. Equations of Equilibrium: For point C + Σ Fx = 0; d Fy = 0; +c Σ

VC = 0

Ans.

-NC - 250 - 650 - 300 = 0 NC = -1200 lb = -1.20 kip

MC = 0; a+ Σ

Ans.

-MC - 650(6.5) - 300(13) = 0 MC = -8125 lb # ft = -8.125 kip # ft

Ans.

Negative signs indicate that NC and MC act in the opposite direction to that shown on FBD.

Ans: NA = 0, VA = 450 lb, MA = -1.125 kip # ft, NB = 0, VB = 850 lb,MB = -6.325 kip # ft, VC = 0, NC = -1.20 kip, MC = -8.125 kip # ft 10

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1–11. Determine the resultant internal loadings acting on the cross sections at points D and E of the frame.

C

1 ft F

4 ft

75 lb/ft A

1 ft

SOLUTION Member AG:

B 1 ft

E

G 2 ft

1 ft

30 150 lb

a+ Σ MA = 0;

4 F (3) - 75(4)(5) - 150 cos 30°(7) = 0; 5 BC

a+ Σ MB = 0;

Ay (3) - 75(4)(2) - 150 cos 30°(4) = 0; Ay = 373.20 lb

+ Σ Fx = 0; S

D 2 ft

Ax -

FBC = 1003.89 lb

3 (1003.89) + 150 sin 30° = 0; Ax = 527.33 l b 5

For point D: + Σ Fx = 0; S

ND + 527.33 = 0 ND = -527 lb

+cΣ Fy = 0;

Ans.

-373.20 - VD = 0 VD = -373 lb

MD = 0; a+ Σ

Ans.

MD + 373.20(1) = 0 MD = -373 lb # ft

Ans.

For point E: + Σ Fx = 0; S

150 sin 30° - NE = 0 NE = 75.0 lb

Fy = 0; +cΣ

Ans.

VE - 75(3) - 150 cos 30° = 0 VE = 355 lb

ME = 0; a+ Σ

Ans.

-ME - 75(3)(1.5) - 150 cos 30°(3) = 0; ME = -727 lb # ft

Ans.

Ans: ND = -527 lb, VD = -373 lb, MD = -373 lb # ft , NE = 75.0 lb, VE = 355 lb , ME = -727 lb # ft 11

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*1–12. Determine the resultant internal loadings acting on the cross sections at points F and G of the frame.

C

1 ft F

4 ft

75 lb/ft A

1 ft

SOLUTION Member AG: a+ Σ MA = 0;

D 2 ft

B 1 ft

E

G 2 ft

1 ft

30 150 lb

4 F (3) - 300(5) - 150 cos 30°(7) = 0 5 BF FBF = 1003.9 lb

For point F: +Q Σ Fx′ = 0; Fy′ = 0; a+Σ

a+ Σ MF = 0;

VF = 0

Ans.

NF - 1003.9 = 0 NF = 1004 lb

Ans.

MF = 0

Ans.

For point G: + Σ Fx = 0; d

NG - 150 sin 30° = 0 NG = 75.0 lb

Fy = 0; + cΣ

Ans.

VG - 75(1) - 150 cos 30° = 0 VG = 205 lb

MG = 0; a+ Σ

Ans.

-MG - 75(1)(0.5) - 150 cos 30°(1) = 0 MG = -167 lb # ft

Ans.

Ans: VF = 0, NF = 1004 lb, MF = 0 , NG = 75.0 lb , VG = 205 lb , MG = -167 lb # ft 12

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1–13. a

The blade of the hacksaw is subjected to a pretension force of F = 100 N. Determine the resultant internal loadings acting on section a–a that passes through point D.

225 mm 30 b B

A D b F E

150 mm a

F C

SOLUTION Internal Loadings: Referring to the free-body diagram of the section of the hacksaw shown in Fig. a, + Σ Fx = 0; d

Na - a + 100 = 0

Fy = 0; +cΣ

Va - a = 0

a+ Σ MD = 0;

- Ma - a - 100(0.15) = 0

Na - a = -100 N

Ans. Ans.

Ma - a = -15 N # m

Ans.

The negative sign indicates that N a - a and Ma - a act in the opposite sense to that shown on the free-body diagram.

Ans: Na - a = -100 N, Va - a = 0, Ma - a = -15 N # m 13

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1–14. a

The blade of the hacksaw is subjected to a pretension force of F = 100 N. Determine the resultant internal loadings acting on section b–b that passes through point D.

225 mm 30 b B

A D b F E

150 mm a

F C

SOLUTION Internal Loadings: Referring to the free-body diagram of the section of the hacksaw shown in Fig. a, Σ F

x′

= 0;

Nb - b + 100 cos 30° = 0

Nb - b = - 86.6 N

Ans.

Σ F

y′

= 0;

Vb - b - 100 sin 30° = 0

Vb - b = 50 N

Ans.

- Mb - b - 100(0.15) = 0

Mb - b = -15 N # m

Ans.

a+ Σ MD = 0;

The negative sign indicates that Nb–b and Mb–b act in the opposite sense to that shown on the free-body diagram.

Ans: Nb - b = - 86.6 N, Vb - b = 50 N , Mb - b = -15 N # m 14

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1–15. The beam supports the triangular distributed load shown. Determine the resultant internal loadings on the cross section at point C. Assume the reactions at the supports A and B are vertical.

800 lb/ft

A D 6 ft

6 ft

C 6 ft

B

E

4.5 ft 4.5 ft

SOLUTION Support Reactions: Referring to the FBD of the entire beam, Fig. a, a+ Σ MB = 0;

1 1 (0.8)(18)(6) - (0.8)(9)(3) - Ay(18) = 0 2 2

Ay = 1.80 kip

Internal Loadings: Referring to the FBD of the left beam segment sectioned through point C, Fig. b, + Σ Fx = 0; S

NC = 0

Fy = 0; +cΣ

1.80 -

MC = 0; a+ Σ

MC +

1 (0.5333)(12) - VC = 0 2

Ans. VC = -1.40 kip

...