Solution Manual for Mechanics of Materials 4th Edition by Philpot PDF

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P1.1 A steel bar of rectangular cross section, 15 mm by 60 mm, is loaded by a compressive force of 110 kN that acts in the longitudinal direction of the bar. Compute the average normal stress in the bar. Solution A (15 = mm )( 60 mm ) 900 mm 2 The cross-sectional area of the steel bar is F (110 kN )...

Description

P1.1 A steel bar of rectangular cross section, 15 mm by 60 mm, is loaded by a compressive force of 110 kN that acts in the longitudinal direction of the bar. Compute the average normal stress in the bar.

Solution

The cross-sectional area of the steel bar is A (15 = mm )( 60 mm ) 900 mm 2 The normal stress in the bar is F (110 kN )(1,000 N/kN ) σ= = = 122.222 MPa = 122.2 MPa A 900 mm 2

Ans.

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P1.2 A circular pipe with outside diameter of 4.5 in. and wall thickness of 0.375 in. is subjected to an axial tensile force of 42,000 lb. Compute the average normal stress in the pipe.

Solution The outside diameter D, the inside diameter d, and the wall thickness t are related by D= d + 2t Therefore, the inside diameter of the pipe is d = D − 2t = 4.5 in. − 2 ( 0.375 in.) = 3.75 in. The cross-sectional area of the pipe is

A=

(D 4

π

− d 2 )=

π 2 2 ( 4.5 in.) − ( 3.75 in.) = 4.8597 in.2

 4 The average normal stress in the pipe is F 42,000 lb σ= = = 8,642.6 psi = 8,640 psi A 4.8597 in.2 2

Ans.

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P1.3 A circular pipe with an outside diameter of 80 mm is subjected to an axial compressive force of 420 kN. The average normal stress may not exceed 130 MPa. Compute the minimum wall thickness required for the pipe.

Solution From the definition of normal stress, solve for the minimum area required to support a 420 kN load without exceeding a normal stress of 130 MPa F F ( 420 kN )(1,000 N/kN ) = σ ∴ Amin= ≥ = 3, 230.77 mm 2 2 A 130 N/mm σ The cross-sectional area of the pipe is given by π 2 = A (D − d 2 ) 4 Set this expression equal to the minimum area and solve for the maximum inside diameter d

π 2 (80 mm ) − d 2  ≥ 3, 230.77 mm 2 4

(80 mm )



2

− d2 ≥

( 3, 230.77 mm ) π 4

2

∴ d max ≤ 47.8169 mm

The outside diameter D, the inside diameter d, and the wall thickness t are related by D= d + 2t Therefore, the minimum wall thickness required for the aluminum tube is D − d 80 mm − 47.8169 mm tmin ≥ = = 16.092 mm = 16.09 mm 2 2

Ans.

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P1.4 Three solid bars, each with square cross sections, make up the axial assembly shown in Figure P1.4/5. Two loads of P = 30 kN are applied to the assembly at flange B, two loads of Q = 18 kN are applied at C, and one load of R = 42 kN is applied at end D. The bar dimensions are b 1 = 60 mm, b 2 = 20 mm, and b 3 = 40 mm. Determine the normal stress in each bar.

FIGURE P1.4/5

Solution Cut an FBD through bar (1). The FBD should include the free end of the assembly at D. We will assume that the internal force in bar (1) is tension. From equilibrium, the force in bar (1) is ΣFx = − F1 − 2 P + 2Q − R = 0

∴ F1 = −2 P + 2Q − R = −2 ( 30 kN ) + 2 (18 kN ) − 42 kN = −66 kN = 66 kN (C)

From the given width of bar (1), the cross-sectional area of bar (1) is 2 2 A= b= ( 60 mm )= 3,600 mm 2 1 1 and thus, the normal stress in bar (1) is F1 ( −66 kN )(1,000 N/kN ) σ1 = 18.33 MPa (C) = = −18.333 MPa = 3,600 mm 2 A1

Ans.

Cut an FBD through bar (2). The FBD should include the free end of the assembly at D. We will assume that the internal force in bar (2) is tension. From equilibrium, the force in bar (2) is ΣFx = − F2 + 2Q − R = 0

∴ F2 = 2Q − R = 2 (18 kN ) − 42 kN = −6 kN = 6 kN (C)

From the given width of bar (2), the cross-sectional area of bar (2) is 2 2 A= b= ( 20 mm )= 400 mm 2 2 2 The normal stress in bar (2) is

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( −6 kN )(1,000 N/kN ) = −15.000 MPa = 15.00 MPa (C)

2 = σ2 =

F A2

400 mm 2

Ans.

Cut an FBD through bar (3). The FBD should include the free end of the assembly at D. We will assume that the internal force in bar (3) is tension. From equilibrium, the force in bar (3) is ΣFx = − F3 − R = 0

∴ F3 =− R =−42 kN =42 kN (C)

The cross-sectional area of bar (3) is 2 2 A= b= ( 40 mm )= 1, 600 mm 2 3 3 The normal stress in bar (3) is F2 ( −42 kN )(1, 000 N/kN ) σ2 = 26.3 MPa (C) = = −26.250 MPa = 1, 600 mm 2 A2

Ans.

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P1.5 Three solid bars, each with square cross sections, make up the axial assembly shown in Figure P1.4/5. Two loads of P = 25 kN are applied to the assembly at flange B, two loads of Q = 15 kN are applied at C, and one load of R = 35 kN is applied at end D. Bar (1) has a width of b 1 = 90 mm. Calculate the width b 2 required for bar (2) if the normal stress magnitude in bar (2) must equal the normal stress magnitude in bar (1).

FIGURE P1.4/5

Solution Cut an FBD through bar (1). The FBD should include the free end of the assembly at D. We will assume that the internal force in bar (1) is tension. From equilibrium, the force in bar (1) is ΣFx = − F1 − 2 P + 2Q − R = 0

∴ F1 = −2 P + 2Q − R = −2 ( 25 kN ) + 2 (15 kN ) − 35 kN = −55 kN = 55 kN (C)

From the given width of bar (1), the cross-sectional area of bar (1) is 2 2 A= b= ( 90 mm )= 8,100 mm 2 1 1 and thus, the normal stress in bar (1) is F ( −55 kN )(1, 000 N/kN ) = −6.7901 MPa σ1 = 1 = A1 8,100 mm 2

Cut an FBD through bar (2). The FBD should include the free end of the assembly at D. We will assume that the internal force in bar (2) is tension. From equilibrium, the force in bar (2) is ΣFx = − F2 + 2Q − R = 0

∴ F2 = 2Q − R = 2 (15 kN ) − 35 kN = −5 kN

The normal stress in bar (2) must equal the normal stress in bar (1). Thus, σ 2 = σ 1 = −6.7901 MPa Solve for the required area of bar (2): Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

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σ2 =

F2 A2

F2 ( −5 kN )(1, 000 N/kN ) = 736.364 mm 2 2 −6.7901 N/mm σ2 The width of bar (2) is therefore:

= ∴ A2

= b2

736.364 = mm 2 27.136 = mm 27.1 mm

Ans.

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P1.6 Axial loads are applied with rigid bearing plates to the solid cylindrical rods shown in Figure P1.6/7. One load of P = 1,500 lb is applied to the assembly at A, two loads of Q = 900 lb are applied at B, and two loads of R = 1,300 lb are applied at C. The diameters of rods (1), (2), and (3) are d 1 = 0.625 in., d 2 = 0.500 in., and d 3 = 0.875 in. Determine the axial normal stress in each of the three rods.

FIGURE P1.6/7

Solution Cut an FBD through rod (1). The FBD should include the free end of the assembly at A. We will assume that the internal force in rod (1) is tension. From equilibrium, the force in rod (1) is ΣFx =− P + F1 =0

∴ F1 = P = 1,500 lb = 1,500 lb (T)

π

π

= = d12 ( 0.625 in. ) 0.3068 in.2 4 4 The normal stress in rod (1) is F1 1,500 lb σ= = = 4,889.24 psi = 4,890 psi (T) 1 A1 0.3068 in.2

Use the given diameter to calculate the cross-sectional area of rod (1):

= A1

2

Ans.

Cut an FBD through rod (2). The FBD should include the free end of the assembly at A. We will assume that the internal force in rod (2) is tension. From equilibrium, the force in rod (2) is ΣFx =− P + 2Q + F2 =0

1,500 lb − 2 ( 900 lb ) = 300 lb (C) ∴ F2 =− P 2Q = −300 lb =

π

π

= d 22 = ( 0.500 in. ) 0.1963 in.2 4 4 The normal stress in rod (2) is F2 −300 lb σ2 = 1,528 psi (C) == −1,527.89 psi = A2 0.1963 in.2

Use the given diameter to calculate the cross-sectional area of rod (2):

= A2

2

Ans.

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Cut an FBD through rod (3). The FBD should include the free end of the assembly at A. We will assume that the internal force in rod (3) is tension. From equilibrium, the force in rod (3) is ΣFx =− P + 2Q − 2 R + F3 =0

∴ F3 = P − 2Q + 2 R =1,500 lb − 2 ( 900 lb ) + 2 (1,300 lb ) = 2,300 lb = 2,300 lb (T)

π

π

= d32 = ( 0.8750 in. ) 0.6013 in.2 4 4 The normal stress in rod (3) is F3 2,300 lb σ= = = 3,824.92 psi = 3,820 psi (T) 3 A3 0.6013 in.2

Use the given diameter to calculate the cross-sectional area of rod (3):

= A3

2

Ans.

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P1.7 Axial loads are applied with rigid bearing plates to the solid cylindrical rods shown in Figure P1.6/7. One load of P = 30 kips is applied to the assembly at A, two loads of Q = 25 kips are applied at B, and two loads of R = 35 kips are applied at C. The normal stress magnitude in aluminum rod (1) must be limited to 20 ksi. The normal stress magnitude in steel rod (2) must be limited to 35 ksi. The normal stress magnitude in brass rod (3) must be limited to 25 ksi. Determine the minimum diameter required for each of the three rods.

FIGURE P1.6/7

Solution Cut an FBD through aluminum rod (1). The FBD should include the free end of the assembly at A. We will assume that the internal force in rod (1) is tension. From equilibrium, the force in rod (1) is ΣFx =− P + F1 =0

∴ F1 = P = 30 kips = 30 kips (T)

The normal stress magnitude in aluminum rod (1) must be limited to 20 ksi. Therefore, the minimum cross-sectional area of rod (1) must be F1 30 kips A1 ≥ = = 1.500 in.2 σ1 20 ksi The diameter must be

A1 ≤

π

4

d12

∴ d1 ≥

π

4

1.382 in. 1.500 in.2 =

Ans.

Cut an FBD through steel rod (2). The FBD should include the free end of the assembly at A. We will assume that the internal force in rod (2) is tension. From equilibrium, the force in rod (2) is ΣFx =− P + 2Q + F2 =0

∴ F2 =− P 2Q = 30 kips − 2 ( 25 kips ) = −20 kips = 20 kips (C)

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The normal stress magnitude in steel rod (2) must be limited to 35 ksi. Therefore, the minimum crosssectional area of rod (2) must be F2 −20 kips = 0.5714 in.2 A2 ≥= σ2 35 ksi The diameter of rod (2) must be

A2 ≤

π

4

d 22

∴ d2 ≥

π

4

0.5714 in.2 = 0.853 in.

Ans.

Cut an FBD through brass rod (3). The FBD should include the free end of the assembly at A. We will assume that the internal force in rod (3) is tension. From equilibrium, the force in rod (3) is ΣFx =− P + 2Q − 2 R + F3 =0

∴ F3 = P − 2Q + 2 R =30 kips − 2 ( 25 kips ) + 2 ( 35 kips ) =50 kips =50 kips (T)

The normal stress magnitude in brass rod (3) must be limited to 25 ksi. Therefore, the minimum cross-sectional area of rod (3) must be F3 50 kips = 2.0000 in.2 A3 ≥ = 25 ksi σ3 The diameter of rod (3) must be

A3 ≤

π

4

d32

∴ d3 ≥

π

4

2.0000 in.2 = 1.596 in.

Ans.

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P1.8 Determine the normal stress in rod (1) for the mechanism shown in Figure P1.8. The diameter of rod (1) is 8 mm, and load P = 2,300 N. Use the following dimensions: a = 120 mm, b = 200 mm, c = 170 mm, and d = 90 mm.

FIGURE P1.8

Solution First, consider an FBD of the pulley to determine the reaction forces exerted on the pulley by the mechanism. ΣFx = Ax − P − P cos ( 60° )= 0

= ( 2,300 N ) + ( 2,300 N ) cos ( 60° ) 3, 450.000 N ΣFy = Ay − P sin ( 60° )= 0 N ) sin ( 60° ) 1,991.858 N = ∴ Ay ( 2,300 = = ∴ Ax

FBD of pulley

FBD of mechanism

Next, consider an FBD of the mechanism to determine the force in rod (1). Rod (1) is oriented at an angle of: c + d 170 mm + 90 mm tan = = 1.30 b = b 200 mm ∴= b 52.431°

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Rod (1) is a two-force member, and its axial force can be calculated from: ΣM C = Ax c + Ay a − ( F1 cos b )( c + d ) = 0

= ∴ F1

π

Ax c + Ay a = ( c + d ) cos b

π

mm ) ( 3, 450.000 N )(170 mm ) + (1,991.858 N )(120 = (170 mm + 90 mm ) cos ( 52.431° )

5, 207.523 N

= d12 = (8 mm ) 50.265 mm 2 4 4

The area of rod (1) is

= A1

2

The normal stress in the rod is thus F1 5, 207.532 N = = 103.601 MPa 103.6 MPa = σ1 = A1 50.265 mm 2

Ans.

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P1.9 Determine the normal stress in bar (1) for the mechanism shown in Figure P1.9. The area of bar (1) is 2,600 mm2. The distributed load intensities are w C = 12 kN/m and w D = 30 kN/m. Use the following dimensions: a = 7.5 m and b = 3.0 m.

FIGURE P1.9

Solution

Consider an FBD of the mechanism. Determine the angle b between rod (1) and the horizontal axis: a 7.5 m tan b= = = 2.5 b 3.0 m ∴= b 68.199° Write an equilibrium equation for the sum of moments about C to compu...