Mechanics of Materials 6th edition beer solution chapter 3 PDF

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CHAPTER 3

PROBLEM 3.1 (a) Determine the maximum shearing stress caused by a 4.6-kN ⋅ m torque T in the 76-mm-diameter shaft shown. (b) Solve part a, assuming that the solid shaft has been replaced by a hollow shaft of the same outer diameter and of 24-mm inner diameter.

SOLUTION (a)

Solid shaft:

c = J =

τ=

d = 38 mm = 0.038 m 2

π 2

c4 =

π 2

(0.038)4 = 3.2753 × 10 −6 m4

Tc (4.6 × 103 )(0.038) = = 53.4 × 106 Pa J 3.2753 × 10−6

τ = 53.4 MPa  (b)

Hollow shaft:

do = 0.038 m 2 1 c1 = di = 12 mm = 0.012 m 2 c2 =

J =

τ=

π 2

(c

4 2

)

− c14 =

π 2

(0.0384 − 0.0124 ) = 3.2428 × 10−6 m4

Tc (4.6 × 103 )(0.038) = = 53.9 × 106 Pa J 3.2428 × 10−6

τ = 53.9 MPa 

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PROBLEM 3.2 (a) Determine the torque T that causes a maximum shearing stress of 45 MPa in the hollow cylindrical steel shaft shown. (b) Determine the maximum shearing stress caused by the same torque T in a solid cylindrical shaft of the same cross-sectional area.

SOLUTION (a)

Given shaft:

J =

π 2

(c

4 2

− c14

)

π

(454 − 304 ) = 5.1689 × 106 mm4 = 5.1689 × 10−6 m4 2 Tc Jτ τ = T = J c −6 (5.1689 × 10 )(45 × 106 ) = 5.1689 × 103 N ⋅ m T = − 45 × 10 3 J =

T = 5.17 kN ⋅ m 

(b)

Solid shaft of same area:

(

)

A = π c 22 − c12 = π (45 2 − 30 2 ) = 3.5343 × 103 mm 2

π c2 = A or c = J =

τ =

π 2

c4 , τ =

A

π

= 33.541 mm

2T Tc = 3 J πc

(2)(5.1689 × 103 ) = 87.2 × 106 Pa 3 (0.033541) π

τ = 87.2 MPa 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 3.3 Knowing that d = 1.2 in., determine the torque T that causes a maximum shearing stress of 7.5 ksi in the hollow shaft shown.

SOLUTION c2 =

1 1 d = (1.6) = 0.8 in. 2 2  2 

c1 =

1  1 d1 =   (1.2) = 0.6 in. 2  2

J =

π 2

(c

4 2

)

− c14 =

π 2

c = 0.8 in.

(0.84 − 0.64 ) = 0.4398 in4

Tc J (0.4398)(7.5) J τmax T = = c 0.8

τ max =

T = 4.12 kip ⋅ in 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 3.4 Knowing that the internal diameter of the hollow shaft shown is d = 0.9 in., determine the maximum shearing stress caused by a torque of magnitude T = 9 kip ⋅ in.

SOLUTION c2 =

1  1 d2 =   (1.6) = 0.8 in. 2  2

c1 =

1  1 d1 =   (0.9) = 0.45 in. 2  2

(

)

π

c24 − c14 = (0.8 4 − 0.45 4) = 0.5790 in 4 2 2 Tc (9)(0.8) = = J 0.5790

J =

τ max

π

c = 0.8 in.

τmax = 12.44 ksi 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 3.5 A torque T = 3 kN ⋅ m is applied to the solid bronze cylinder shown. Determine (a) the maximum shearing stress, (b) the shearing stress at point D which lies on a 15-mm-radius circle drawn on the end of the cylinder, (c) the percent of the torque carried by the portion of the cylinder within the 15 mm radius.

SOLUTION (a)

c= J =

1 d = 30 mm = 30 × 10− 3 m 2

π

c4 =

π

−

−

(30 × 10 3 )4 = 1.27235 × 10 6 m 4

2 2 T = 3 kN = 3 × 103 N

τm =

Tc (3 × 103 )(30 × 10 −3 ) = = 70.736 × 106 Pa J 1.27235 × 10−6

τ m = 70.7 MPa  (b)

ρD = 15 mm = 15 × 10 −3 m τD =

(c)

τD = TD =

ρD c

τ =

TD ρ D JD

π

(15 × 10−3 )(70.736 × 10−6 ) (30 × 10 −3) TD =

J Dτ D

ρD

=

π 2

τ D = 35.4 MPa 

ρ D3 τ D

(15 × 10−3 )3 (35.368 × 106 ) = 187.5 N ⋅ m

2 TD 187.5 × 100% = (100%) = 6.25% T 3 × 103

6.25%

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 3.6 (a) Determine the torque that can be applied to a solid shaft of 20-mm diameter without exceeding an allowable shearing stress of 80 MPa. (b) Solve Part a, assuming that the solid shaft has been replaced by a hollow shaft of the same cross-sectional area and with an inner diameter equal to half of its own outer diameter.

SOLUTION (a)

Solid shaft:

c= J = T=

(b)

Hollow shaft:

1 1 d = (0.020) = 0.010 m 2 2

π 2

c4 −

π 2

(0.10) 4 = 15.7080 × 10 −9 m 4

J τ max (15.7080 × 10−9 )(80 × 106 ) = = 125.664 0.010 c

T = 125.7 N ⋅ m 

Same area as solid shaft. 2  1   3 A = π c22 − c12 = π c22 −  c2   = π c22 = π c2   2   4 2 2 c2 = c= (0.010) = 0.0115470 m 3 3

(

c1 = J = T =

)

1 c = 0.0057735 m 2 2

π

(c 2

4 2

τ max J c2

)

− c14 = =

π 2

(0.01154704 − 0.00577354 ) = 26.180× 10−9 m4

(80 × 10 6)(26.180 × 10− 9) = 181.38 0.0115470

T = 181.4 N ⋅ m 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 3.7 The solid spindle AB has a diameter ds= 1.5 in. and is made of a steel with an allowable shearing stress of 12 ksi, while sleeve CD is made of a brass with an allowable shearing stress of 7 ksi. Determine the largest torque T that can be applied at A.

SOLUTION Analysis of solid spindle AB:

c=

1 d s = 0.75 in. 2

τ = Tc

T=

J

T =

Analysis of sleeve CD:

c2 =

π 2

π Jτ = τ c3 c 2

(12 ×10 3)(0.75) 3 = 7.95 × 103 lb ⋅ in

1 1 d = (3) =1.5 in. 2 o 2

c 1 = c 2 − t = 1.5 − 0.25 = 1.25 in. J = T=

The smaller torque governs: 

π 2

(c

4 2

)

− c14 =

π 2

(1.54 − 1.254 ) = 4.1172 in4

Jτ (4.1172)(7 × 103 ) = = 19.21× 103 lb ⋅ in 1.5 c2

T = 7.95 × 10 3 lb ⋅ in T = 7.95 kip ⋅ in 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 3.8 The solid spindle AB is made of a steel with an allowable shearing stress of 12 ksi, and sleeve CD is made of a brass with an allowable shearing stress of 7 ksi. Determine (a) the largest torque T that can be applied at A if the allowable shearing stress is not to be exceeded in sleeve CD, (b) the corresponding required value of the diameter d s of spindle AB.

SOLUTION (a)

Analysis of sleeve CD:

1 1 do = (3) = 1.5 in. 2 2 c1 = c2 − t =1.5 − 0.25 =1.25 in.

c2 =

J = T=

π

(c 2

4 2

)

− c14 =

π 2

(1.54 − 1.254 ) = 4.1172 in4

J τ (4.1172)(7 × 10 3) = = 19.21 × 103 lb ⋅ in 1.5 c2 T = 19.21 kip ⋅ in 

(b)

Analysis of solid spindle AB:

τ=

Tc J

π J T 19.21 × 103 = c3= = = 1.601 in 3 c 2 12 × 103 τ c =

3

(2)(1.601)

π

= 1.006 in. d s = 2c d = 2.01in. 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 3.9 The torques shown are exerted on pulleys A and B. Knowing that both shafts are solid, determine the maximum shearing stress (a) in shaft AB, (b) in shaft BC.

SOLUTION (a)

Shaft AB:

T AB = 300 N ⋅ m, d = 0.030 m, c = 0.015 m

τ max =

Tc 2T (2)(300) = = J π c 3 π (0.015) 3

= 56.588 × 10 6Pa

(b)

Shaft BC:

τ max = 56.6 MPa 

T BC = 300 + 400 = 700 N ⋅ m d = 0.046 m, c = 0.023 m

τ max =

Tc 2T (2)(700) = = 3 J πc π (0.023)3

= 36.626 × 106 Pa

τ max = 36.6 MPa 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 3.10 In order to reduce the total mass of the assembly of Prob. 3.9, a new design is being considered in which the diameter of shaft BC will be smaller. Determine the smallest diameter of shaft BC for which the maximum value of the shearing stress in the assembly will not increase.

SOLUTION TAB = 300 N ⋅ m, d = 0.030 m, c = 0.015 m

Shaft AB:

τ

max

=

Tc 2T (2)(300) = = J π c 3 π (0.015) 3

= 56.588 × 106 Pa = 56.6 MPa

Shaft BC:

TBC = 300 + 400 = 700 N ⋅ m d = 0.046 m, c = 0.023 m

τ max =

Tc 2T (2)(700) = = 3 J πc π (0.023) 3

= 36.626 × 106 Pa = 36.6 MPa

The largest stress (56.588 × 106 Pa) occurs in portion AB. Reduce the diameter of BC to provide the same stress. Tc 2T = J π c3 (2)(700) = = 7.875 × 10− 6 m3 π (56.588 × 10 6 )

TBC = 700N ⋅ m c3 =

2T

πτ max

τ max =

c = 19.895 × 10− 3 m

d = 2c = 39.79 × 10− 3 m d = 39.8 mm 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 3.11 Knowing that each portion of the shafts AB, BC, and CD consist of a solid circular rod, determine (a) the shaft in which the maximum shearing stress occurs, (b) the magnitude of that stress.

SOLUTION Shaft AB:

T = 48 N ⋅ m 1 d = 7.5 mm = 0.0075 m 2 Tc 2T = = 3 J πc (2) (48) = = 72.433MPa π (0.0075)3

c=

τ max τ max Shaft BC:

T = − 48 + 144 = 96 N ⋅ m c=

Shaft CD:

τ max =

Tc 2 T (2) (96) = = = 83.835 MPa J π c3 π (0.009) 3

T = − 48 + 144 + 60 = 156 N⋅ m c=

Answers:

1 d = 9 mm 2

1 d = 10.5 mm 2

τ max =

Tc 2 T (2 × 156) = = = 85.79 MPa 3 J πc π (0.0105) 3

(a) shaft CD

(b) 85.8 MPa 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 3.12 Knowing that an 8-mm-diameter hole has been drilled through each of the shafts AB, BC, and CD, determine (a) the shaft in which the maximum shearing stress occurs, (b) the magnitude of that stress.

SOLUTION 1 d1 = 4 mm 2

Hole:

c1 =

Shaft AB:

T = 48 N ⋅ m c2 = J =

τ max = Shaft BC:

τ max

(c 2

4 2

)

− c14 =

π 2

(0.00754 − 0.0044 ) = 4.5679× 10−9 m4

Tc2 (48)(0.0075) = = 78.810 MPa J 4.5679 × 10− 9

π

(c 2

4 2

)

− c14 =

π

c2 =

−9

T = − 48 + 144 + 60 = 156 N ⋅ m

τ max =

π 2

(c

4 2

)

− c 14 =

π 2

1 d 2 = 9 mm 2

(0.0094 − 0.0044 ) = 9.904× 10

2 Tc 2 (96)(0.009) = = = 87.239 MPa J 9.904 × 10− 9

J =

Answers:

π

T = − 48 + 144 = 96 N ⋅ m J =

Shaft CD:

1 d2 = 7.5 mm 2

c2 =

m4

1 d2 = 10.5 mm 2

(0.01054 − 0.0044 ) = 18.691× 10−9 m4

Tc2 (156)(0.0105) = = 87.636 MPa J 18.691 × 10− 9

(a) shaft CD

(b) 87.6 MPa 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 3.13 Under normal operating conditions, the electric motor exerts a 12-kip ⋅ in. torque at E. Knowing that each shaft is solid, determine the maximum shearing in (a) shaft BC, (b) shaft CD, (c) shaft DE.

SOLUTION (a)

Shaft BC:

From free body shown: TBC = 3 kip ⋅ in

τ=

τ=

(b)

Shaft CD:

Tc Tc 2 T = = π 4 π c3 J c 2

3 kip ⋅ in

2

π 1

  × 1.75 in.  2 

3

(1)

τ = 2.85 ksi 

From free body shown: TCD = 3 + 4= 7 kip ⋅ in

From Eq. (1):

τ= (c)

Shaft DE:

2 T

π c3

=

2 7 kip ⋅ in

π (1 in .) 3

τ = 4.46 ksi 

From free body shown: TDE = 12 kip ⋅ in

From Eq. (1):

τ =

2 T 2 12 kip ⋅ in = 3 π c 3 π 1   × 2.25 in.  2 

τ = 5.37 ksi 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 3.14 Solve Prob.3.13, assuming that a 1in.-diameter hole has been drilled into each shaft. PROBLEM 3.13 Under normal operating conditions, the electric motor exerts a 12-kip ⋅ in. torque at E. Knowing that each shaft is solid, determine the maximum shearing in (a) shaft BC, (b) shaft CD, (c) shaft DE.

SOLUTION (a)

Shaft BC:

From free body shown: c2 =

J =

τ = (b)

Shaft CD:

J =

τ= (c)

Shaft DE:

1 (1.75) = 0.875 in . 2

π

(c 2

4 2

)

− c14 =

π 2

Tc (3 kip ⋅ in)(0.875 in .) = J 0.82260 in 4

J =

τ =

τ = 3.19 ksi 

TCD = 3 + 4 = 7 kip ⋅ in

1 (2.0) = 1.0 in . 2

π

(c 2

4 2

)

− c14 =

π 2

(1.04 − 0.54 ) = 1.47262 in4

Tc (7 kip ⋅ in)(1.0 in.) = J 1.47262 in 4

From free body shown: c2 =

1 c1 = (1) = 0.5 in . 2

(0.8754 − 0.54 ) = 0.82260 in4

From free body shown: c2 =

TBC = 3 kip ⋅ in

τ = 4.75 ksi 

TDE = 12 kip ⋅ in

2.25 = 1.125 in. 2

π 2

(c

4 2

)

− c14 =

π 2

(1.1254 − 0.54 ) = 2.4179 in4

Tc (12 kip ⋅ in)(1.125 in.) = J 2.4179 in 4

τ = 5.58 ksi 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without per...