Sample Mechanics of materials an integrated learning system Philpot 4th edition solutions PDF

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Authors: Timothy A. Philpot
Published: Wiley 2019
Edition: 4th
Pages: 1490
Type: word/pdf
Size: 114MB
Content: 4th edition solutions...

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Mechanics of Materials: An Integrated Learning System, 4th Ed. Timothy A. Philpot ps://gioumeh.com/product/mechanics-of-materials-an-integrated-learning-system-solut 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.

FOLFNKHUHWRGRZQORDG

Solution The cross-sectional area of the steel bar is A = (15 mm )( 60 mm ) = 900 mm2 The normal stress in the bar is F (110 kN )(1, 000 N/kN )  = = = 122.222 MPa = 122.2 MPa 2 A 900 mm

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Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed. Timothy A. Philpot ps://gioumeh.com/product/mechanics-of-materials-an-integrated-learning-system-solut 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.

FOLFNKHUHWRGRZQORDG

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 − 2 t = 4.5 in. − 2( 0.375 in.) = 3.75 in. The cross-sectional area of the pipe is

A=



(D 4

2

− d2 ) =

 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  = = = 8,642.6 psi = 8,640 psi A 4.8597 in.2

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Mechanics of Materials: An Integrated Learning System, 4th Ed. Timothy A. Philpot ps://gioumeh.com/product/mechanics-of-materials-an-integrated-learning-system-solut 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 FOLFNKHUHWRGRZQORDG 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)  Amin  = = 3, 230.77 mm2  = A 130 N/mm2  The cross-sectional area of the pipe is given by

A=



2 2 (D − d )

4 Set this expression equal to the minimum area and solve for the maximum inside diameter d



4

(80 mm )2 − d 2   3,230.77 mm 2 ( 80 mm ) 2 − d 2 

4



(3,230.77 mm ) 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 t min  = = 16.092 mm = 16.09 mm 2 2

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Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed. Timothy A. Philpot ps://gioumeh.com/product/mechanics-of-materials-an-integrated-learning-system-solut 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 FOLFNKHUHWRGRZQORDG applied at C, and one load of R = 42 kN is applied at end D. The bar dimensions are b1 = 60 mm, b2 = 20 mm, and b3 = 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 Fx = − 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 A1 = b12 = ( 60 mm ) = 3,600 mm2 and thus, the normal stress in bar (1) is F ( −66 kN )(1, 000 N/kN ) = −18.333 MPa = 18.33 MPa (C) 1 = 1 = 3, 600 mm 2 A1

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 Fx = − F2 + 2 Q − 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 A2 = b22 = ( 20 mm ) = 400 mm 2 The normal stress in bar (2) is 2

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Mechanics of Materials: An Integrated Learning System, 4th Ed. Timothy A. Philpot ps://gioumeh.com/product/mechanics-of-materials-an-integrated-learning-system-solut

2 =

F2 (−6 kN )(1,000 N/kN ) = = −15.000 MPa = 15.00 MPa (C) A2 400 mm2 FOLFNKHUHWRGRZQORDG

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 Fx = − F3 − R = 0  F3 = − R = − 42 kN = 42 kN (C)

The cross-sectional area of bar (3) is 2 A3 = b32 = ( 40 mm ) = 1,600 mm2 The normal stress in bar (3) is F ( −42 kN)(1, 000 N/kN) = −26.250 MPa = 26.3 MPa (C) 2 = 2 = A2 1, 600 mm2

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Mechanics of Materials: An Integrated Learning System, 4th Ed. Timothy A. Philpot ps://gioumeh.com/product/mechanics-of-materials-an-integrated-learning-system-solut 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 FOLFNKHUHWRGRZQORDG applied at C, and one load of R = 35 kN is applied at end D. Bar (1) has a width of b1 = 90 mm. Calculate the width b2 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 Fx = − 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 A1 = b12 = ( 90 mm ) = 8,100 mm2 and thus, the normal stress in bar (1) is F ( −55 kN )(1, 000 N/kN ) = −6.7901 MPa 1 = 1 = 8,100 mm2 A1

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 Fx = − F2 + 2 Q − 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,  2 = 1 = −6.7901 MPa Solve for the required area of bar (2): @solutionmanual1

Mechanics of Materials: An Integrated Learning System, 4th Ed. Timothy A. Philpot ps://gioumeh.com/product/mechanics-of-materials-an-integrated-learning-system-solut

2 =

F2 A2

FOLFNKHUHWRGRZQORDG

F ( −5 kN )(1, 000 N/kN )  A2 = 2 = 736.364 mm2  2 −6.7901 N/mm2

The width of bar (2) is therefore: b2 = 736.364 mm2 = 27.136 mm = 27.1 mm

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Mechanics of Materials: An Integrated Learning System, 4th Ed. Timothy A. Philpot ps://gioumeh.com/product/mechanics-of-materials-an-integrated-learning-system-solut 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 areFOLFNKHUHWRGRZQORDG applied at C. The diameters of rods (1), (2), and (3) are d1 = 0.625 in., d2 = 0.500 in., and d3 = 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 Fx = − P + F1 = 0

 F1 = P = 1,500 lb = 1,500 lb (T)

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

A1 =



2 d1 =



( 0.625 in.)

2

= 0.3068 in. 2

4 4 The normal stress in rod (1) is F 1, 500 lb = 4,889.24 psi = 4,890 psi (T ) 1 = 1 = A1 0.3068 in.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 Fx = − P + 2Q + F2 = 0

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

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





d22 = ( 0.500 in.) = 0.1963 in. 2 4 4 The normal stress in rod (2) is −300 lb F = −1, 527.89 psi = 1, 528 psi ( C) 2 = 2 = A2 0.1963 in.2 A2 =

2

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Mechanics of Materials: An Integrated Learning System, 4th Ed. Timothy A. Philpot ps://gioumeh.com/product/mechanics-of-materials-an-integrated-learning-system-solut 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 FOLFNKHUHWRGRZQORDG Fx = − 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)

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

A3 =



d32 =



( 0.8750 in.)

2

= 0.6013 in.2 4 4 The normal stress in rod (3) is F 2,300 lb 3 = 3 = = 3,824.92 psi = 3, 820 psi (T ) 0.6013 in.2 A3

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Mechanics of Materials: An Integrated Learning System, 4th Ed. Timothy A. Philpot ps://gioumeh.com/product/mechanics-of-materials-an-integrated-learning-system-solut 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 FOLFNKHUHWRGRZQORDG 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 Fx = − 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 F 30 kips A1  1 = = 1.500 in.2 20 ksi 1 The diameter must be

A1 

 4

2

d1

 d1 

(1.500 in. ) = 1.382 in.  4

2

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 Fx = − P + 2Q + F2 = 0

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

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Mechanics of Materials: An Integrated Learning System, 4th Ed. Timothy A. Philpot ps://gioumeh.com/product/mechanics-of-materials-an-integrated-learning-system-solut The normal stress magnitude in steel rod (2) must be limited to 35 ksi. Therefore, the minimum crosssectional area of rod (2) must be FOLFNKHUHWRGRZQORDG F2 −20 kips 2 A2  = = 0.5714 in. 2 35 ksi The diameter of rod (2) must be

A2 

 4

d22

d 2 

( 0.5714 in. ) = 0.853 in.  4

2

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 Fx = − 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 A3  = = 2.0000 in. 3 25 ksi The diameter of rod (3) must be

A3 

 4

d32

d 3 

4



( 2.0000 in. ) = 1.596 in. 2

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Mechanics of Materials: An Integrated Learning System, 4th Ed. Timothy A. Philpot ps://gioumeh.com/product/mechanics-of-materials-an-integrated-learning-system-solut P1.8 Determine the normal stress in rod (1) for the mechanism shown in Figure P1.8. TheFOLFNKHUHWRGRZQORDG 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.  Fx = Ax − P − P cos ( 60 ) = 0  Ax = ( 2,300 N) +( 2,300 N) cos( 60) = 3, 450.000 N  Fy = Ay − P sin ( 60 ) = 0  Ay = (2,300 N ) sin ( 60 ) = 1, 991.858 N

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 = = 1.30 tan  = b 200 mm   = 52.431

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