all chapter Mechanics of materials 6th edition William Riley solutions manual PDF

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Authors: William F. Riley , Leroy D. Sturges , Don H. Morris
Published: Wiley 2006
Edition: 6th
Pages: 1283
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https://gioumeh.com/product/mechanics-of-materials-riley-solutions/ MECHANICS OF MATERIALS, 6th Edition

RILEY, STURGES AND MORRIS

Chapter 1

FOLFNKHUHWRGRZQORDG 1-1* From a free-body diagram of the forearm, the equilibrium equations give

↑ ΣFy = 0 :

T − F − W − 20 = 0

4 ΣM F = 0 :

1.5T − 5.5W − 11.5 ( 20 ) = 0

T = 3.667W + 153.33 lb ............................... Ans. F = 2.667W + 133.33 lb ............................... Ans.

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RILEY, STURGES AND MORRIS

1-2* From a free-body diagram of the ring, the equations of equilibrium

FOLFNKHUHWRGRZQORDG

T2 cos10° − T1 sin10° = 0

→ ΣFx = 0 : ↑ ΣFy = 0 :

T1 cos10° − T2 sin10° − 175 ( 9.81) = 0

are solved to get

T1 = 5.67128T2 T1 = 1799 N .............................................................................. Ans. T2 = 317 N ................................................................................ Ans. T3 = 175 (9.81 ) = 1717 N ....................................................... Ans.

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https://gioumeh.com/product/mechanics-of-materials-riley-solutions/ MECHANICS OF MATERIALS, 6th Edition 1-3 The equations of equilibrium

→ ΣFx = 0 : ↑ ΣFy = 0 :

RILEY, STURGES AND MORRIS

FOLFNKHUHWRGRZQORDG

N A sin 30 ° − N B sin 30° = 0 N A cos 30° + N B cos 30° − 800 = 0

are solved to get

N A = NB

N A = 462 lb

60° ................................................................ Ans.

N B = 462 lb

60° ................................................................ Ans.

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https://gioumeh.com/product/mechanics-of-materials-riley-solutions/ MECHANICS OF MATERIALS, 6th Edition 1-4* The equations of equilibrium

→ ΣFx = 0 :

FOLFNKHUHWRGRZQORDG

Ax − NB sin 45 o = 0

Ay + N B cos 45o − 300 = 0

↑ ΣFy = 0 :

4 ΣM A = 0 :

RILEY, STURGES AND MORRIS

1.5 ( N B cos 45o ) −1.5 (300 ) = 0

are solved to get

N B = 424.264 N ≅ 424 N Ax = 300 N A = 300 N →

45............................. Ans.

Ay = 0 N ........................................................ Ans.

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https://gioumeh.com/product/mechanics-of-materials-riley-solutions/ MECHANICS OF MATERIALS, 6th Edition 1-5 The equations of equilibrium

→ ΣFx = 0 : ↑ ΣFy = 0 :

4 ΣM A = 0 :

RILEY, STURGES AND MORRIS

FOLFNKHUHWRGRZQORDG Ax = 0

Ay − 250 = 0 M A − 3 ( 250 ) = 0

are solved to get

Ax = 0 lb

A = 250 lb ↑

Ay = 250 lb .....................................................Ans.

M A = 750 lb ⋅ ft 4 .................................................Ans.

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RILEY, STURGES AND MORRIS

1-6 From an overall free-body diagram, the equations of equilibrium

FOLFNKHUHWRGRZQORDG Ax = 0

→ ΣFx = 0 : ↑ ΣFy = 0 :

4 ΣM A = 0 :

Ay −10 −15 + N F = 0 9 NF − 3 (10 ) − 6 (15 ) = 0

are solved to get

Ax = 0 kN Ay = 11.6667 kN ↑

N F = 13.3333 kN ↑ Then, from a free-body diagram of the right hand section of the truss, the equations of equilibrium

4 ΣM C = 0 :

3( 13.3333) − 3TDE = 0

4 ΣM D = 0 :

3TBC − 3 (15 ) + 6 (13.3333 ) = 0

↑ ΣFy = 0 :

o 13.3333 −15 − TCD cos 45 = 0

are solved to get

TDE = 13.33 kN (T) ............................................................................................................ Ans. TBC = −11.67 kN = 11.67 kN (C) .................................................................................. Ans. TCD = −2.36 kN = 2.36 kN (C) ...................................................................................... Ans.

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4 ΣM A = 0 :

RILEY, STURGES AND MORRIS

27 N F − 13 ( 33cos15 ° ) + 4 ( 35 sin15°) = 0 FOLFNKHUHWRGRZQORDG

N F = 14.93561 lb ≅ 14.94 lb Z ΣFx = 0 :

P cos 30° − 35 sin15° = 0

P = 10.46005 lb ≅ 10.46 lb ^ ΣFy = 0 :

N R = 24.1 lb

75° ............. Ans. 15° ................ Ans.

N R + N F − P sin 30° − 35 cos15° = 0 75° .......................................... Ans.

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RILEY, STURGES AND MORRIS

1-8

↑ ΣFy = 0 :

P + 4060 − 5210 = 0 FOLFNKHUHWRGRZQORDG

P = 1150 N .......................................................................................Ans. 4 ΣM A = 0 :

C − 16( 4060) − 37 ( 5210 ) = 0

C = 257, 730 N ⋅ mm ≅ 258 N ⋅ m ...............................................Ans.

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RILEY, STURGES AND MORRIS

1-9* From an overall free-body diagram, the equations of equilibrium

FOLFNKHUHWRGRZQORDG

→ ΣFx = 0 :

Ax = 0

↑ ΣFy = 0 :

4 ΣM A = 0 :

Ay − 10 − 20 + N E = 0 45 NE −15 (10 ) − 30 ( 20 ) = 0

are solved to get

Ax = 0 kip Ay = 13.3333 kip ↑

N E = 16.6667 kip ↑ Then, from a free-body diagram of the right hand section of the truss, the equations of equilibrium

4 ΣM F = 0 :

15NE + 15 ( TDE sin 30°) = 0

4 ΣM E = 0 :

15 ( 20 ) − 15 (TCF sin 60°) = 0

4 ΣM C = 0 :

22.5 N E − 7.5 (20 ) − (15 cos 30° )TF G = 0

are solved to get

TCD = −33.3333 kip ≅ 33.3 kip (C) ............................................................................... Ans. TCF = +23.094 kip ≅ 23.1 kip (T) .................................................................................. Ans. TFG = +17.32 kip = 17.32 kip (T) .................................................................................. Ans.

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RILEY, STURGES AND MORRIS

W = 2000 ( 9.81) = 19, 620 N FOLFNKHUHWRGRZQORDG

Z ΣFx = 0 : 4 ΣM A = 0 : ^ ΣFy = 0 :

P − W sin 30° = 0 2 (W sin 30° ) − 2 (W cos 30° ) + 3N F − 1P = 0

N R + N F − W cos 30° = 0

P = 9810 N = 9.81 kN 30° ............................... Ans. N R = 8933.806 N ≅ 8.93 kN 60° .................... Ans.

N F = 8057.612 N ≅ 8.06 kN

60° ................... Ans.

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RILEY, STURGES AND MORRIS

1-11 From a free-body diagram of the brake pedal, the equilibrium FOLFNKHUHWRGRZQORDG equations are solved to get the forces

4 ΣM A = 0 :

5.5Q − (30 cos 30° )(11) − ( 30 sin 30°)( 4) = 0

Q = 62.871 lb → ΣFx = 0 :

Ax − Q + 30 cos 30° = 0

Ax = 36.890 lb ↑ ΣFy = 0 :

Ay − 30 sin 30° = 0

Ay = 15.00 lb A = 39.8 lb 22.13° ........................................................... Ans. Q = 62.9 lb ← ........................................................................ Ans.

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RILEY, STURGES AND MORRIS

1-12* From a free-body diagram of the beam, the equilibrium equations are solved to get the forces and moment

FOLFNKHUHWRGRZQORDG

→ ΣFx = 0 :

Ax = 0

↑ ΣFy = 0 :

Ay − 2 = 0

Ay = 2 kN 4 ΣM A = 0 :

MA − 2 ( 4 ) − 3 = 0

M A = 11 kN ⋅ m

A = 2 kN ↑ ................................................................................................................................ Ans. M A = 11 kN ⋅ m 4 ..................................................................................................................... Ans.

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https://gioumeh.com/product/mechanics-of-materials-riley-solutions/ MECHANICS OF MATERIALS, 6th Edition

RILEY, STURGES AND MORRIS

1-13 From a free-body diagram of the beam, the equilibrium equations are solved to get the forces

FOLFNKHUHWRGRZQORDG

→ ΣFx = 0 :

Ax = 0

4 ΣM A = 0 :

15B − 3 (500 )− 6 (800 ) − 9 ( 700 ) − 12 ( 400 ) = 0

B = 1160 lb ↑ ΣFy = 0 :

Ay + B − 500 − 800 − 700 − 400 = 0

Ay = 1240 lb

A = 1240 lb ↑ ........................................................................................................................... Ans. B = 1160 lb ↑ ............................................................................................................................ Ans.

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RILEY, STURGES AND MORRIS

W = 450 ( 9.81) = 4414.50 N FOLFNKHUHWRGRZQORDG

From a free-body diagram of the lower pulley, vertical equilibrium equation gives the tension

↑ ΣFy = 0 :

2T1 − 4414.50 = 0

T1 = 2207.25 N Then, from a free-body diagram of the upper pulley, moment equilibrium equation gives the force F

4 ΣM axle = 0 :

100T1 − 90T1 − 100 F = 0

F = 221 N ............................................................................................................................... Ans.

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RILEY, STURGES AND MORRIS

1-15* From a free-body diagram of the bracket, the equilibrium equations

FOLFNKHUHWRGRZQORDG

4 ΣM A = 0 :

18B −  (12 )(10 ) 2 ( 12 3) = 0

→ ΣFx = 0 :

Ax + B = 0

↑ ΣFy = 0 :

Ay − ( 12)( 10) 2 = 0

Are solved to get the forces

B = 13.333 lb Ax = −13.333 lb

Ay = 60.0 lb

A = 61.464 lb ≅ 61.5 lb 77.47° ..................................... Ans. B = 13.33 lb → ....................................................................... Ans.

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RILEY, STURGES AND MORRIS

1-16 (a) From a free-body diagram of the plane, the force equilibrium equations are solved to FOLFNKHUHWRGRZQORDG get the forces

→ ΣFx = 0 :

V − 40 cos 70° − 70 cos16° = 0

↑ ΣFy = 0 :

N − 40 sin 70° − 70 sin16° = 0 V = 81.0 N .........................................Ans. N = 56.882 N ≅ 56.9 N .................Ans.

(b)

Then the moment equilibrium equation gives the required location of the normal force

4 ΣM A = 0 :

( 70 cos16°)( 75) − ( 70 sin16° )( 280) + Nd + ( 40 cos 70°) ( 60) − ( 40 sin 70° )( 60) = 0

d = 31.5 mm 31.5 mm (from the left end of plane) ....................................................................... Ans.

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4 ΣM A = 0 :

RILEY, STURGES AND MORRIS

1.25 N D − 9 ( 25) =FOLFNKHUHWRGRZQORDG 0

N D = 180.0 lb ....................................Ans. →ΣFx = 0 : ↑ ΣFy = 0 :

Ax + N D sin 38° = 0 Ay − 25 − N D cos 38 ° = 0

Ax = −110.8 lb A = 200.3 lb

Ay = 166.8 lb 56.4° ................................................................... Ans.

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RILEY, STURGES AND MORRIS

1-18* From a free-body diagram of the upper handle, FOLFNKHUHWRGRZQORDG moment equilibrium gives

FA cos θ + Bx = 0

→ΣFx = 0 :

F A sin θ − 100 + B y = 0

↑ ΣFy = 0 :

4 ΣM B = 0 :

93 (100 ) − 28 ( FA sin θ ) + 5 ( FA cos θ ) = 0

30 = 30.964 ° 50 FA = 919.116 N

θ = tan −1

Bx = −788.136 N

By = − 372.881 N

Then a free body diagram of the upper jaw gives

4 ΣM C = 0 :

Fd ...