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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 (b) Solve part a, assuming that the solid shaft has been replaced by a hollow shaft of the same outer diameter and of 24-mm inner diameter.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 m 4 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 (c 2 4 2 ) − c14 = π 2 (0.0384 − 0.0124 ) = 3.2428 × 10−6 m 4 Tc (4.6 × 103 )(0.038) = = 53.9 × 106 Pa −6 J 3.2428 × 10 τ = 53.9 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 3.2 (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 = J = π 2 π (c 4 2 − c14 ) (454 − 304 ) = 5.1689 × 106 mm 4 = 5.1689 × 10−6 m 4 2 Tc τ = J T = T = Jτ c (5.1689 × 10−6 )(45 × 106 ) = 5.1689 × 103 N ⋅ m −3 45 × 10 T = 5.17 kN ⋅ m (b) Solid shaft of same area: ( ) A = π c22 − c12 = π (452 − 302 ) = 3.5343 × 103 mm 2 π c 2 = A or c = J = τ = π 2 c4, τ = A π = 33.541 mm 2T Tc = J π c3 (2)(5.1689 × 103 ) = 87.2 × 106 Pa π (0.033541)3 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 2 = (1.6) = 0.8 in. 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 in 4 Tc J Jτ max (0.4398)(7.5) = 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 = 0.9in., determine the maximum shearing stress caused by a torque of magnitude T = 9 kip ⋅ in. SOLUTION c2 = 1 1 d 2 = (1.6) = 0.8 in. 2 2 c1 = 1 1 d1 = (0.9) = 0.45 in. 2 2 ( ) π c24 − c14 = (0.84 − 0.454 ) = 0.5790 in 4 2 2 (9)(0.8) Tc = = 0.5790 J J = τ max π c = 0.8 in. τ max = 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 = 2 π 2 (30 × 10−3 ) 4 = 1.27235 × 10−6 m 4 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) ρ D = 15 mm = 15 × 10−3 m τD = (c) τD = TD = ρD c τ = TD ρ D JD π 2 (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 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 c4 − 2 π 2 (0.10) 4 = 15.7080 × 10−9 m 4 Jτ 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 c 2 2 4 2 2 (0.010) = 0.0115470 m c2 = c= 3 3 1 c1 = c2 = 0.0057735 m 2 ( J = T = π 2 ) (c 4 2 τ max J c2 ) − c14 = = π 2 (0.01154704 − 0.00577354 ) = 26.180 × 10−9 m 4 (80 × 106 )(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 J T = Analysis of sleeve CD: c2 = π 2 T= π Jτ = τ c3 c 2 (12 × 103 )(0.75)3 = 7.95 × 103 lb ⋅ in 1 1 d o = (3) = 1.5 in. 2 2 c1 = c2 − t = 1.5 − 0.25 = 1.25 in. J= T= The smaller torque governs: π (c 2 4 2 ) − c14 = π 2 (1.54 − 1.254 ) = 4.1172 in 4 Jτ (4.1172)(7 × 103 ) = = 19.21 × 103 lb ⋅ in c2 1.5 T = 7.95 × 103 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 in 4 (4.1172)(7 × 103 ) Jτ = = 19.21 × 103 lb ⋅ in 1.5 c2 T = 19.21 kip ⋅ in (b) Analysis of solid spindle AB: τ= Tc J π 19.21 × 103 J T = c3 = = = 1.601 in 3 τ 2 c 12 × 103 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: TAB = 300 N ⋅ m, d = 0.030 m, c = 0.015 m τ max = Tc 2T (2)(300) = = 3 J πc π (0.015)3 = 56.588 × 106 Pa (b) Shaft BC: τ max = 56.6 MPa 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 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 Shaft AB: TAB = 300 N ⋅ m, d = 0.030 m, c = 0.015 m τ max = 2T (2)(300) Tc = = 3 J πc 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 = 2T (2)(700) Tc = = 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 × 106 ) 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 2T (2) (96) = 3 = = 83.835 MPa J πc 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 2T (2 × 156) = 3 = = 85.79 MPa 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 2 (c 4 2 ) − c14 = π 2 (0.00754 − 0.0044 ) = 4.5679 × 10−9 m 4 Tc2 (48)(0.0075) = = 78.810 MPa J 4.5679 × 10−9 π (c 2 ) c2 = 1 d 2 = 9 mm 2 π (0.0094 − 0.0044 ) = 9.904 × 10−9 m 4 2 Tc (96)(0.009) = 2 = = 87.239 MPa J 9.904 × 10−9 4 2 − c14 = T = − 48 + 144 + 60 = 156 N ⋅ m J = τ max = Answers: π T = − 48 + 144 = 96 N ⋅ m J = Shaft CD: 1 d 2 = 7.5 mm 2 π (c 2 4 2 ) − c14 = π 2 c2 = 1 d 2 = 10.5 mm 2 (0.01054 − 0.0044 ) = 18.691 × 10−9 m 4 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 = = 3 π J c4 π 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 2 7 kip ⋅ in = 3 π c 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 3 π c 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 = τ = (b) Shaft CD: (c 2 ) − c14 = π 2 2 (c 4 2 ) − c14 = π 2 (1.04 − 0.54 ) = 1.47262 in 4 From free body shown: τ = TCD = 3 + 4 = 7 kip ⋅ in Tc (7 kip ⋅ in)(1.0 in.) = J 1.47262 in 4 c2 = τ = 3.19 ksi 1 (2.0) = 1.0 in. 2 π τ = J = 1 (1) = 0.5 in. 2 (0.8754 − 0.54 ) = 0.82260 in 4 From free body shown: J = Shaft DE: 4 2 c1 = Tc (3 kip ⋅ in)(0.875 in.) = J 0.82260 in 4 c2 = (c) 1 (1.75) = 0.875 in. 2 π J = TBC = 3 kip ⋅ in τ = 4.75 ksi TDE = 12 kip ⋅ in 2.25 = 1.125 in. 2 π (c 2 4 2 ) − c14 = π 2 (1.1254 − 0.54 ) = 2.4179 in 4 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 permission. PROBLEM 3.15 The allowable shearing stress is 15 ksi in the 1.5-in.-diameter steel rod AB and 8 ksi in the 1.8-in.-diameter brass rod BC. Neglecting the effect of stress concentrations, determine the largest torque that can be applied at A. SOLUTION τ max = Rod AB: τ max = 15 ksi T = Rod BC: π π Tc , J = c 4 , T = c3τ max J 2 2 π 2 π 2 The allowable torque is the smaller value. 1 d = 0.75 in. 2 (0.75)3 (15) = 9.94 kip ⋅ in τ max = 8 ksi T = c= c= 1 d = 0.90 in. 2 (0.90)3 (8) = 9.16 kip ⋅ in T = 9.16 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 3.16 The allowable shearing stress is 15 ksi in the steel rod AB and 8 ksi in the brass rod BC. Knowing that a torque of magnitude T = 10 kip ⋅ in. is applied at A, determine the required diameter of (a) rod AB, (b) rod BC. SOLUTION τ max = (a) Rod AB: Tc π 2T , J = , c3 = J 2 πτ max T = 10 kip ⋅ in c3 = τ max = 15 ksi (2)(10) = 0.4244 in 3 π (15) c = 0.7515 in. (b) Rod BC: T = 10 kip ⋅ in c3 = d = 2c = 1.503 in. τ max = 8 ksi (2)(10) = 0.79577 in 2 π (8) c = 0.9267 in. d = 2c = 1.853 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 3.17 The allowable stress is 50 MPa in the brass rod AB and 25 MPa in the aluminum rod BC. Knowing that a torque of magnitude T = 1250 N ⋅ m is applied at A, determine the required diameter of (a) rod AB, (b) rod BC. SOLUTION τ max = (a) Rod AB: c3 = Tc J J = π 2 c4 c3 = 2T πτ max (2)(1250) = 15.915 × 10−6 m3 π (50 × 106 ) c = 25.15 × 10−3 m = 25.15 mm d AB = 2c = 50.3 mm (b) Rod BC: c3 = (2)(1250) = 31.831 × 10−6 m3 6 π (25 × 10 ) c = 31.69 × 10−3 m = 31.69 mm d BC = 2c = 63.4 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 3.18 The solid rod BC has a diameter of 30 mm and is made of an aluminum for which the allowable shearing stress is 25 MPa. Rod AB is hollow and has an outer diameter of 25 mm; it is made of a brass for which the allowable shearing stress is 50 MPa. Determine (a) the largest inner diameter of rod AB for which the factor of safety is the same for each rod, (b) the largest torque that can be applied at A. SOLUTION Solid rod BC: τ = Tc J J = π 2 c4 τ all = 25 × 106 Pa c= Tall = Hollow rod AB: 1 d = 0.015 m 2 π 2 c3τ all = π 2 (0.015)3 (25 × 106 ) = 132.536 N ⋅ m τ all = 50 × 106 Pa Tall = 132.536 N ⋅ m 1 1 d 2 = (0.025) = 0.0125 m 2 2 c1 = ? c2 = Tall = Jτ all π 4 τ c2 − c14 all = c2 2 c2 c14 = c24 − ( 2Tallc2 πτ all = 0.01254 − (2)(132.536)(0.0125) = 3.3203 × 10−9 m 4 6 π (50 × 10 ) c1 = 7.59 × 10−3 m = 7.59 mm (a) (b) ) Allowable torque. d1 = 2c1 = 15.18 mm Tall = 132.5 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 3.19 The solid rod AB has a diameter dAB = 60 mm. The pipe CD has an outer diameter of 90 mm and a wall thickness of 6 mm. Knowing that both the rod and the pipe are made of a steel for which the allowable shearing stress is 75 Mpa, determine the largest torque T that can be applied at A. SOLUTION τ all = 75 × 106 Pa Tall = Rod AB: c= Tall = 1 d = 0.030 m 2 π 2 c3τ all = π 2 Jτ all c J= π 2 c4 (0.030)3 (75 × 106 ) = 3.181 × 103 N ⋅ m Pipe CD: c2 = J = Tall = 1 d 2 = 0.045 m 2 π (c 2 4 2 ) − c14 = c1 = c2 − t = 0.045 − 0.006 = 0.039 m π 2 (0.0454 − 0.0394 ) = 2.8073 × 10−6 m 4 (2.8073 × 10−6 )(75 × 106 ) = 4.679 × 103 N ⋅ m 0.045 Allowable torque is the smaller value. Tall = 3.18 × 103 N ⋅ m 3.18 kN ⋅ 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 3.20 The solid rod AB has a diameter dAB = 60 mm and is made of a steel for which the allowable shearing stress is 85 Mpa. The pipe CD, which has an outer diameter of 90 mm and a wall thickness of 6 mm, is made of an aluminum for which the allowable shearing stress is 54 MPa. Determine the largest torque T that can be applied at A. SOLUTION Rod AB: τ all = 85 × 106 Pa Tall = = Pipe CD: c= 1 d = 0.030 m 2 π Jτ all = c3 τ all 2 c π 2 (0.030)3 (85 × 106 ) = 3.605 × 103 N ⋅ m τ all = 54 × 106 Pa c2 = 1 d 2 = 0.045 m 2 c1 = c2 − t = 0.045 − 0.006 = 0.039 m J = Tall = π (c 2 4 2 ) − c14 = π 2 (0.0454 − 0.0394 ) = 2.8073 × 10−6 m 4 Jτ all (2.8073 × 10−6 )(54 × 106 ) = = 3.369 × 103 N ⋅ m 0.045 c2 Allowable torque is the smaller value. Tall = 3.369 × 103 N ⋅ m 3.37 kN ⋅ 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 3.21 A torque of magnitude T = 1000 N ⋅ m is applied at D as shown. Knowing that the diameter of shaft AB is 56 mm and that the diameter of shaft CD is 42 mm, determine the maximum shearing stress in (a) shaft AB, (b) shaft CD. SOLUTION TCD = 1000 N ⋅ m TAB = (a) (b) Shaft AB: Shaft CD: rB 100 TCD = (1000) = 2500 N ⋅ m rC 40 c= 1 d = 0.028 m 2 τ= Tc 2T (2) (2500) = 3= = 72.50 × 106 J π c π (0.028)3 c= 1 d = 0.020 m 2 τ= Tc 2T (2) (1000) = 3= = 68.7 × 106 J π c π (0.020)3 72.5 MPa 68.7 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 3.22 A torque of magnitude T = 1000 N ⋅ m is applied at D as shown. Knowing that the allowable shearing stress is 60 MPa in each shaft, determine the required diameter of (a) shaft AB, (b) shaft CD. SOLUTION TCD = 1000 N ⋅ m TAB = (a) Shaft AB: rB 100 TCD = (1000) = 2500 N ⋅ m rC 40 τ all = 60 × 106 Pa τ = Tc 2T = 3 J πc c3 = 2T πτ = (2) (2500) = 26.526 × 10−6 m3 π (60 × 106 ) c = 29.82 × 10−3 = 29.82 mm (b) Shaft CD: d = 2c = 59.6 mm τ all = 60 × 106 Pa τ = Tc 2T = 3 J πc c3 = 2T πτ = (2) (1000) = 10.610 × 10−6 m3 π (60 × 106 ) c = 21.97 × 10−3 m = 21.97 mm d = 2c = 43.9 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 3.23 Under normal operating conditions, a motor exerts a torque of magnitude TF = 1200 lb ⋅ in. at F. Knowing that rD = 8 in., rG = 3 in., and the allowable shearing stress is 10.5 ksi ⋅ in each shaft, determine the required diameter of (a) shaft CDE, (b) shaft FGH. SOLUTION TF = 1200 lb ⋅ in TE = rD 8 TF = (1200) = 3200 lb ⋅ in rG 3 τ all = 10.5 ksi = 10500 psi τ = (a) 2T Tc = 3 J πc c3 = 2T πτ Shaft CDE: c3 = (2) (3200) = 0.194012 in 3 π (10500) c = 0.5789 in. (b) or d DE = 2c d DE = 1.158 in. Shaft FGH: c3 = (2) (1200) π (10500) c = 0.4174 in. = 0.012757 in 3 d FG = 2c d FG = 0.835 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 3.24 Under normal operating conditions, a motor exerts a torque of magnitude TF at F. The shafts are made of a steel for which the allowable shearing stress is 12 ksi and have diameters dCDE = 0.900 in. and dFGH = 0.800 in. Knowing that rD = 6.5 in. and rG = 4.5 in., determine the largest allowable value of TF. SOLUTION τ all = 12 ksi Shaft FG: 1 d = 0.400 in. 2 π Jτ = all = c3τ all c 2 c= TF , all = Shaft DE: c= TE , all = π 2 (0.400)3 (12) = 1.206 kip ⋅ in 1 d = 0.450 in. 2 π 2 c3τ all π (0.450)3 (12) = 1.7177 kip ⋅ in 2 r 4.5 TF = G TE TF , all = (1.7177) = 1.189 kip ⋅ in rD 6.5 = Allowable value of TF is the smaller. TF = 1.189 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 3.25 The two solid shafts are connected by gears as shown and are made of a steel for which the allowable shearing stress is 8500 psi. Knowing that a torque of magnitude TC = 5 kip ⋅ in. is applied at C and that the assembly is in equilibrium, determine the required diameter of (a) shaft BC, (b) shaft EF. SOLUTION τ max = 8500 psi = 8.5 ksi (a) Shaft BC: TC = 5 kip ⋅ in τ max = c= Tc 2T = J π c3 3 c= 3 2T πτ max (2)(5) = 0.7208 in. π (8.5) d BC = 2c = 1.442 in. (b) Shaft EF: TF = c= rD 2.5 TC = (5) = 3.125 kip ⋅ in rA 4 3 2T πτ max = 3 (2)(3.125) = 0.6163 in. π (8.5) d EF = 2c = 1.233 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 3.26 The two solid shafts are connected by gears as shown and are made of a steel for which the allowable shearing stress is 7000 psi. Knowing the diameters of the two shafts are, respectively, d BC = 1.6 in. and d EF = 1.25 in., determine the largest torque TC that can be applied at C. SOLUTION τ max = 7000 psi = 7.0 ksi Shaft BC: d BC = 1.6 in. 1 d = 0.8 in. 2 π Jτ max TC = = τ max c3 2 c c= = Shaft EF: π 2 (7.0)(0.8)3 = 5.63 kip ⋅ in d EF = 1.25 in. 1 d = 0.625 in. 2 π Jτ max TF = = τ max c3 c 2 c= = By statics, TC = π 2 (7.0)(0.625)3 = 2.684 kip ⋅ in rA 4 TF = (2.684) = 4.30 kip ⋅ in rD 2.5 Allowable value of TC is the smaller. TC = 4.30 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 3.27 A torque of magnitude T = 100 N ⋅ m is applied to shaft AB of the gear train shown. Knowing that the diameters of the three solid shafts are, respectively, d AB = 21 mm, dCD = 30 mm, and d EF = 40 mm, determine the maximum shearing stress in (a) shaft AB, (b) shaft CD, (c) shaft EF. SOLUTION Statics: Shaft AB: Gears B and C: Force on gear circles. TAB = TA = TB = T rB = 25 mm, rC = 60 mm FBC = TC = Shaft CD: Gears D and E: Force on gear circles. rC 60 TB = T = 2.4 T rB 25 TCD = TC = TD = 2.4 T rD = 30 mm, rE = 75 mm FDE = TE = Shaft EF: TB T = C rB rC TD T = E rD rE rE 75 TD = (2.4 T ) = 6 T rD 30 TEF = TE = TF = 6T Maximum Shearing Stresses. τ max = Tc 2T = J π c3 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 3.27 (Continued) (a) (b) Shaft AB: Shaft CD: T = 100 N ⋅ m c= 1 d = 10.5 mm = 10.5 × 10−3 m 2 τ max = (2)(100) = 55.0 × 106 Pa π (10.5 × 10−3 )3 T = (2.4)(100) = 240 N ⋅ m c= τ max = (c) Shaft EF: τ max = 55.0 MPa 1 d = 15 mm = 15 × 10−3 m 2 (2)(240) = 45.3 × 106 Pa −3 3 π (15 × 10 ) τ max = 45.3 MPa T = (6)(100) = 600 N ⋅ m c= τ max = 1 d = 20 mm = 20 × 10−3 m 2 (2)(600) π (20 × 10−3 )3 = 47.7 × 106 Pa τ max = 47.7 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 3.28 A torque of magnitude T = 120 N ⋅ m is applied to shaft AB of the gear train shown. Knowing that the allowable shearing stress is 75 MPa in each of the three solid shafts, determine the required diameter of (a) shaft AB, (b) shaft CD, (c) shaft EF. SOLUTION Statics: Shaft AB: Gears B and C: Force on gear circles. TAB = TA = TB = T rB = 25 mm, rC = 60 mm FBC = TC = Shaft CD: Gears D and E: Force on gear circles. rC 60 TB = T = 2.4 T rB 25 TCD = TC = TD = 2.4 T rD = 30 mm, rE = 75 mm FDE = TE = Shaft EF: TB T = C rB rC TD T = E rD rE rE 75 TD = (2.4 T ) = 6 T rD 30 TEF = TE = TF = 6 T Required Diameters. τ max = c= 2T Tc = J π c3 3 2T πτ d = 2c = 2 3 2T πτ max τ max = 75 × 106 Pa 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 3.28 (Continued) (a) Shaft AB: TAB = T = 120 N ⋅ m d AB = 2 3 (b) Shaft CD: Shaft EF: d AB = 20.1 mm TCD = (2.4)(120) = 288 N ⋅ m dCD = 2 3 (c) 2(120) = 20.1 × 10−3 m π (75 × 106 ) (2)(288) π (75 × 106 ) = 26.9 × 10−3 m dCD = 26.9 mm TEF = (6)(120) = 720 N ⋅ m d EF = 2 3 (2)(720) = 36.6 × 10−3 m 3 π (75 × 10 ) d EF = 36.6 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 3.29 (a) For a given allowable shearing stress, determine the ratio T/w of the maximum allowable torque T and the weight per unit length w for the hollow shaft shown. (b) Denoting by (T /w)0 the value of this ratio for a solid shaft of the same radius c2 , express the ratio T/w for the hollow shaft in terms of (T /w)0 and c1 /c2. SOLUTION w = weight per unit length, ρ g = specific weight, W = total weight, L = length w= Tall = ( ) (a) T = c12 + c22 τ all W c1 = 0 for solid shaft: (b) (T /w) h c2 = 1 + 12 (T /w)0 c2 W ρ gLA = = ρ gA = ρ gπ c22 − c12 L L ( ( )( ) ) 2 2 2 2 π c24 − c14 π c2 + c1 c2 − c1 Jτ all τ all = τ all = 2 c2 2 c2 c2 ( ) c12 + c22 τ all T (hollow shaft) = 2ρ gc2 w c2τ all T (solid shaft) = 2 pg w 0 c12 T T = 1 + c22 w w 0 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 3.30 While the exact distribution of the shearing stresses in a hollow-cylindrical shaft is as shown in Fig. a, an approximate value can be obtained for τmax by assuming that the stresses are uniformly distributed over the area A of the cross section, as shown in Fig. b, and then further assuming that all of the elementary shearing forces act at a distance from O equal to the mean radius ½(c1 + c2 ) of the cross section. This approximate value τ 0 = T /Arm , where T is the applied torque. Determine the ratio τ max /τ 0 of the true value of the maximum shearing stress and its approximate value τ 0 for values of c1 /c2 , respectively equal to 1.00, 0.95, 0.75, 0.50, and 0. SOLUTION For a hollow shaft: τ max = τ0 = By definition, Tc2 2Tc2 2Tc2 2Tc2 = = = 4 4 2 2 2 2 J π c2 − c1 π c2 − c1 c2 + c1 A c22 + c12 ( ) ( )( ) ( T 2T = Arm A(c2 + c1) τ max c (c + c ) 1 + (c1 /c2 ) = 2 2 2 21 = τ0 c2 + c1 1 + (c1 /c2 ) 2 Dividing, ) c1 /c2 1.0 0.95 0.75 0.5 0.0 τ max /τ 0 1.0 1.025 1.120 1.200 1.0 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 3.31 (a) For the solid steel shaft shown (G = 77 GPa), determine the angle of twist at A. (b) Solve part a, assuming that the steel shaft is hollow with a 30-mm-outer diameter and a 20-mm-inner diameter. SOLUTION (a) c= 1 d = 0.015 m, 2 −9 J = π 2 c4 = π 2 (0.015) 4 J = 79.522 × 10 m , L = 1.8 m, G = 77 × 109 Pa T = 250 N ⋅ m ϕ = ϕ = (b) 4 ϕ = TL GJ (250)(1.8) = 73.49 × 10−3 rad 9 −9 (77 × 10 )(79.522 × 10 ) (73.49 × 10−3 )180 ϕ = 4.21° π 1 π 4 c2 = 0.015 m, c1 d1 = 0.010 m, J = c2 − c14 2 2 π TL J = (0.0154 − 0.0104 ) = 63.814 × 10−9 m 4 ϕ 2 GJ ( ϕ = ) (250)(1.8) 180 = 91.58 × 10−3 rad = (91.58 × 10−3 ) 9 −9 π (77 × 10 )(63.814 × 10 ) ϕ = 5.25° 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 3.32 For the aluminum shaft shown (G = 27 GPa), determine (a) the torque T that causes an angle of twist of 4°, (b) the angle of twist caused by the same torque T in a solid cylindrical shaft of the same length and cross-sectional area. SOLUTION TL GJ ϕ = (a) T = GJ ϕ L ϕ = 4° = 69.813 × 10−3 rad, L = 1.25 m G = 27 GPa = 27 × 109 Pa J = π (c 2 4 2 ) − c14 = π 2 (0.0184 − 0.0124 ) = 132.324 × 10−9 m 4 (27 × 109 )(132.324 × 109 )(69.813 × 10−3 ) 1.25 = 199.539 N⋅ m T = (b) Matching areas: ( A = π c 2 = π c22 − c12 T = 199.5 N ⋅ m ) c = c22 − c12 = 0.0182 − 0.0122 = 0.013416 m J = ϕ = π 2 c4 = π 2 (0.013416) 4 = 50.894 × 10−9 m 4 TL (195.539)(1.25) = = 181.514 × 10−3 rad GJ (27 × 109 )(50.894 × 109 ) ϕ = 10.40° 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 3.33 Determine the largest allowable diameter of a 10-ft-long steel rod (G = 11.2 × 106 psi) if the rod is to be twisted through 30° without exceeding a shearing stress of 12 ksi. SOLUTION L = 10 ft = 120 in. ϕ = 30° = 30π = 0.52360 rad 180 τ = 12 ksi = 12 × 103 psi TL GJ ϕ Tc GJ ϕ c Gϕ c τL , T = , τ = , c= = = ϕ = GJ L J JL L Gϕ c= (12 × 103 )(120) = 0.24555 in. (11.2 × 106 )(0.52360) d = 2c = 0.491 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 3.34 While an oil well is being drilled at a depth of 6000 ft, it is observed that the top of the 8-in.-diameter steel drill pipe rotates though two complete revolutions before the drilling bit starts to rotate. Using G = 11.2 × 106 psi, determine the maximum shearing stress in the pipe caused by torsion. SOLUTION For outside diameter of 8 in., c = 4 in. For two revolutions, ϕ = 2(2π ) = 4π radians. G = 11.2 × 106 psi L = 6000 ft = 72000 in. From text book, TL GJ Tc τm = J ϕ = Divide (2) by (1). (1) (2) τ m Gc = L ϕ τm = Gcϕ (11 × 106 )(4)(4π ) = = 7679 psi 72000 L τ m = 7.68 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 3.35 The electric motor exerts a 500 N ⋅ m torque on the aluminum shaft ABCD when it is rotating at a constant speed. Knowing that G = 27 GPa and that the torques exerted on pulleys B and C are as shown, determine the angle of twist between (a) B and C, (b) B and D. SOLUTION (a) Angle of twist between B and C. TBC = 200 N ⋅ m, LBC = 1.2 m c= J BC = ϕ B/C = (b) 1 d = 0.022 m, G = 27 × 109 Pa 2 π 2 c 4 = 367.97 × 10−9 m TL (200)(1.2) = = 24.157 × 10−3 rad GJ (27 × 109 )(367.97 × 109 ) ϕ B /C = 1.384° Angle of twist between B and D. TCD = 500 N ⋅ m, LCD = 0.9 m, c = J CD = ϕC /D π 2 c4 = π 1 d = 0.024 m, G = 27 × 109 Pa 2 (0.024) 4 = 521.153 × 10−9 m 4 2 (500)(0.9) = = 31.980 × 10−3 rad (27 × 109 )(521.153 × 109 ) ϕ B /D = ϕ B /C + ϕC /D = 24.157 × 10−3 + 31.980 × 10−3 = 56.137 × 10−3 rad ϕ B /D = 3.22° 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 3.36 The torques shown are exerted on pulleys B, C, and D. Knowing that the entire shaft is made of steel (G = 27 GPa), determine the angle of twist between (a) C and B, (b) D and B. SOLUTION (a) Shaft BC: c= J BC = 1 d = 0.015 m 2 π 4 c 4 = 79.522 × 10−9 m 4 LBC = 0.8 m, G = 27 × 109 Pa ϕ BC = (b) Shaft CD: TL (400)(0.8) = = 0.149904 rad GJ (27 × 109 )(79.522 × 10−9 ) ϕ BC = 8.54° 1 π d = 0.018 m J CD = c 4 = 164.896 × 10−9 m 4 2 4 = 1.0 m TCD = 400 − 900 = −500 N ⋅ m c= LCD ϕCD = (−500)(1.0) TL = = −0.11230 rad GJ (27 × 109 )(164.896 × 10−9 ) ϕ BD = ϕ BC + ϕCD = 0.14904 − 0.11230 = 0.03674 rad ϕ BD = 2.11° 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 3.37 The aluminum rod BC (G = 26 GPa) is bonded to the brass rod AB (G = 39 GPa). Knowing that each rod is solid and has a diameter of 12 mm, determine the angle of twist (a) at B, (b) at C. SOLUTION Both portions: c= 1 d = 6 mm = 6 × 10−3 m 2 π π c 4 = (6 × 10−3 )4 = 2.03575 × 10−9 m 4 2 2 T = 100 N ⋅ m J = Rod AB: (a) ϕ B = ϕ AB = Rod BC: GAB = 39 × 109 Pa, LAB = 0.200 m TLAB (100)(0.200) = = 0.25191 rad G AB J (39 × 109 )(2.03575 × 10−9 ) GBC = 26 × 109 Pa, LBC = 0.300 m ϕ BC = (b) ϕ B = 14.43° TLBC (100)(0.300) = = 0.56679 rad GBC J (26 × 109 )(2.03575 × 10−9 ) ϕC = ϕ B + ϕ BC = 0.25191 + 0.56679 = 0.81870 rad ϕC = 46.9° 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 3.38 The aluminum rod AB (G = 27 GPa) is bonded to the brass rod BD (G = 39GPa). Knowing that portion CD of the brass rod is hollow and has an inner diameter of 40 mm, determine the angle of twist at A. SOLUTION G = 27 × 109 Pa, L = 0.400 m Rod AB: T = 800 N ⋅ m c = Part BC: π (0.018) 4 = 164.896 × 10−9 m 2 2 TL (800)(0.400) = = = 71.875 × 10−3 rad 9 −9 GJ (27 × 10 )(164.896 × 10 ) J = ϕ A/B π 1 d = 0.018 m 2 G = 39 × 109 Pa c4 = L = 0.375 m, c = T = 800 + 1600 = 2400 N ⋅ m, J = ϕ B /C 1 d = 0.030 m 2 π c4 = π (0.030) 4 = 1.27234 × 10−6 m 4 2 2 (2400)(0.375) TL = = = 18.137 × 10−3 rad GJ (39 × 109 )(1.27234 × 10−6 ) 1 d1 = 0.020 m 2 1 c2 = d 2 = 0.030 m, L = 0.250 m 2 c1 = Part CD: ϕC / D Angle of twist at A. π ( ) π c24 − c14 = (0.0304 − 0.0204 ) = 1.02102 × 10−6 m 4 2 2 (2400)(0.250) TL = = = 15.068 × 10−3 rad 9 −6 GJ (39 × 10 )(1.02102 × 10 ) J = ϕ A = ϕ A/B + ϕ B/C + ϕC/D = 105.080 × 10−3 rad ϕ A = 6.02° 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 3.39 The solid spindle AB has a diameter d s =1.5 in. and is made of a steel with G = 11.2 × 106 psi and τ all = 12 ksi, while sleeve CD is made of a brass with G = 5.6 × 106 psi and τ all = 7 ksi. Determine the angle through end A can be rotated. SOLUTION Stress analysis of solid spindle AB: c= 1 ds = 0.75 in. 2 τ = Tc J (12 × 103 )(0.75)3 = 7.95 × 103 lb ⋅ in 2 1 1 c2 = do = (3) = 1.5in. 2 2 c1 = c2 − t = 1.5 − 0.25 = 1.25in. J = π T = The smaller torque governs. Deformation of spindle AB: (c 2 ) − c14 = π 2 (1.54 − 1.254 ) = 4.1172in 4 T = 7.95 × 103 lb ⋅ in c = 0.75 in. ϕ AB = π 2 c 4 = 0.49701 in 4 , L = 12 in., G = 11.2 × 106 psi TL (7.95 × 103 ) (12) = = 0.017138 radians GJ (11.2 × 106 )(0.49701) J = 4.1172 in 4 , L = 8 in., G = 5.6 × 106 psi ϕCD = Total angle of twist: 4 2 Jτ (4.1172)(7 × 10−3 ) = = 19.21 × 103 lb ⋅ in c2 1.5 J = Deformation of sleeve CD: Jτ π = τ c3 c 2 π T = Stress analysis of sleeve CD: T = TL (7.95 × 103 ) (8) = = 0.002758 radians GJ (5.6 × 106 ) (4.1172) ϕ AD = ϕ AB + ϕCD = 0.019896 radians ϕ AD = 1.140° 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 3.40 The solid spindle AB has a diameter d s = 1.75in. and is made of a steel with G = 11.2 × 106 psi and τ all = 12 ksi, while sleeve CD is made of a brass with G = 5.6 × 106 psi and τ all = 7 ksi. Determine (a) the largest torque T that can be applied at A if the given allowable stresses are not to be exceeded and if the angle of twist of sleeve CD is not to exceed 0.375°, (b) the corresponding angle through which end A rotates. SOLUTION Spindle AB: c= J = Sleeve CD: π 2 c4 = π 2 L = 12 in., τ all = 12 ksi, G = 11.2 × 106 psi (0.875)4 = 0.92077in 4 c1 = 1.25 in., c2 = 1.5 in., L = 8 in., τ all = 7 ksi J = (a) 1 (1.75in.) = 0.875 in. 2 π 2 (c 4 2 ) − c14 = 4.1172 in 4 , G = 5.6 × 106 psi Largest allowable torque T. Ciriterion: Stress in spindle AB. Critrion: Stress in sleeve CD. Jτ c τ = Tc J T= (0.92077)(12) = 12.63 kip ⋅ in 0.875 T= Criterion: Angle of twist of sleeve CD T = Jτ 4.1172 in 4 = (7 ksi) c2 1.5 in. T = 19.21 kip ⋅ in φ = 0.375° = 6.545 × 10−3 rad φ = TL JG T = JG (4.1172)(5.6 × 106 ) φ = (6.545 × 10−3 ) L 8 T = 18.86 kip ⋅ in T = 12.63 kip ⋅ in The largest allowable torque is (b) Angle of rotation of end A. φ A = φ A / D = φ A / B + φC / D = Ti Li =T J i Gi Li J i Gi 12 8 = (12.63 × 103 ) + 6 6 (0.92077)(11.2 × 10 ) (4.1172)(5.6 × 10 ) = 0.01908 radians ϕ A = 1.093° 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 3.41 Two shafts, each of 78 -in. diameter, are connected by the gears shown. Knowing that G = 11.2 × 106 psi and that the shaft at F is fixed, determine the angle through which end A rotates when a 1.2 kip ⋅ in. torque is applied at A. SOLUTION Calculation of torques. Circumferential contact force between gears B and E. F = TAB T = EF rB rE TEF = rE TAB rB TAB = 1.2 kip ⋅ in = 1200 lb ⋅ in TEF = 6 (1200) = 1600 lb ⋅ in 4.5 Twist in shaft FE. L = 12 in., c = J = ϕ E /F = π 2 4 c = ϕB = −3 4 = 57.548 × 10 in 2 16 δ = rEϕ E = rBϕ B rE 6 (29.789 × 10−3 ) = 39.718 × 10−3 rad ϕE = 4.5 rB L = 8 + 6 = 14 in. ϕ A/B = Rotation at A. 4 ϕ E = ϕ E/F = 29.789 × 10−3 rad Tangential displacement at gear circle. Twist in shaft BA. π7 TL (1600)(12) = = 29.789 × 10−3 rad −3 6 GJ (11.2 × 10 )(57.548 × 10 ) Rotation at E. Rotation at B. 1 7 d = in., G = 11.2 × 106 psi 2 16 J = 57.548 × 10−3 in 4 (1200)(14) TL = = 26.065 × 10−3 rad GJ (11.2 × 106 )(57.548 × 10−3 ) ϕ A = ϕ B + ϕ A/B = 65.783 × 10−3 rad ϕ A = 3.77° 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 3.42 Two solid shafts are connected by gears as shown. Knowing that G = 77.2 GPa for each shaft, determine the angle through which end A rotates when TA = 1200 N ⋅ m. SOLUTION Calculation of torques: Circumferential contact force between gears B and C. TAB = 1200 N ⋅ m Twist in shaft CD: c= J = ϕC/D Rotation angle at C. TCD = 1 d = 0.030 m, 2 π c4 = π F = TAB T = CD rB rC Twist in shaft AB: L = 1.2 m, G = 77.2 × 109 Pa (0.030) 4 = 1.27234 × 10−6 m 4 2 2 (3600)(1.2) TL = = = 43.981 × 10−3 rad GJ (77.2 × 109 )(1.27234 × 10−9 ) ϕC = ϕC /D = 43.981 × 10−3 rad ϕB = c= J = ϕ A/B Rotation angle at A. rC TAB rB 240 (1200) = 3600 N ⋅ m 80 Circumferential displacement at contact points of gears B and C. Rotation angle at B. TCD = δ = rCϕC = rBϕ B rC 240 (43.981 × 10−3 ) = 131.942 × 10−3 rad ϕC = 80 rB 1 d = 0.021m, L = 1.6 m, G = 77.2 × 109 Pa 2 π c4 = π (0.021)4 = 305.49 × 109 m 4 2 2 (1200)(1.6) TL = = = 81.412 × 10−3 rad GJ (77.2 × 109 )(305.49 × 10−9 ) ϕ A = ϕ B + ϕ A/B = 213.354 × 10−3 rad ϕ A = 12.22° 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 3.43 A coder F, used to record in digital form the rotation of shaft A, is connected to the shaft by means of the gear train shown, which consists of four gears and three solid steel shafts each of diameter d. Two of the gears have a radius r and the other two a radius nr. If the rotation of the coder F is prevented, determine in terms of T, l, G, J, and n the angle through which end A rotates. SOLUTION TAB = TA TCD = rC T T TAB = AB = A rB n n TEF = rE T T TCD = CD = A2 rD n n TEF lEF T l = 2A GJ n GJ ϕ r Tl = E ϕE = E = 3 A rD n n GJ ϕ E = ϕ EF = ϕD TCDlCD T l = A GJ nGJ 1 Tl T l T l 1 ϕC = ϕ D + ϕCD = 3 A + A = A 3 + GJ n n n GJ nGJ ϕCD = 1 ϕ rC Tl 1 ϕC = C = 4 + 2 rB n GJ n n T l T l = AB AB = A GJ GJ ϕB = ϕ AB ϕ A = ϕ B + ϕ AB = TAl 1 1 + 2 + 1 4 GJ n n 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 3.44 For the gear train described in Prob. 3.43, determine the angle through which end A rotates when T = 5lb ⋅ in., l = 2.4in., d = 1/16in., G = 11.2 × 106 psi, and n = 2. PROBLEM 3.43 A coder F, used to record in digital form the rotation of shaft A, is connected to the shaft by means of the gear train shown, which consists of four gears and three solid steel shafts each of diameter d. Two of the gears have a radius r and the other two a radius nr. If the rotation of the coder F is prevented, determine in terms of T, l, G, J, and n the angle through which end A rotates. SOLUTION See solution to Prob. 3.43 for development of equation for ϕ A. ϕA = Data: Tl 1 1 1+ 2 + 4 GJ n n T = 5 lb ⋅ in, l = 2.4 in., c = n = 2, J = ϕA = π 2 c4 = π 1 1 1 d = in., G = 11.2 × 106 psi 2 32 4 −6 4 = 1.49803 × 10 in 2 32 (5)(2.4) 1 1 1 + + = 938.73 × 10−3 rad −6 4 16 (11.2 × 10 )(1.49803 × 10 ) 6 ϕ A = 53.8° 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 3.45 The design of the gear-and-shaft system shown requires that steel shafts of the same diameter be used for both AB and CD. It is further required that τ max ≤ 60 MPa, and that the angle φD through which end D of shaft CD rotates not exceed 1.5°. Knowing that G = 77 GPa, determine the required diameter of the shafts. SOLUTION TCD = TD = 1000 N ⋅ m For design based on stress, use larger torque. TAB = rB 100 TCD = (1000) = 2500 N ⋅ m rC 40 TAB = 2500 N ⋅ m Tc 2T = J π c3 2T (2)(2500) c3 = = = 26.526 × 10−6 m3 πτ π (60 × 106 ) τ = c = 29.82 × 10−3 m = 29.82 mm , d = 2c = 59.6 mm Design based on rotation angle. Shaft AB: ϕ D = 1.5° = 26.18 × 10−3 rad TAB = 2500 N ⋅ m, L = 0.4 m TL (2500)(0.4) 1000 = = GJ GJ GJ 1000 ϕ B = ϕ AB = GJ Gears ϕC = rB ϕ B = 100 1000 = 2500 rC GJ 40 GJ ϕ AB = Shaft CD: TCD = 1000 N ⋅ m, L = 0.6 m TL (1000) (0.6) 600 = = GJ GJ GJ 2500 600 3100 3100 = ϕC + ϕCD = + = = π 4 GJ GJ GJ G2c ϕCD = ϕD c4 = (2) (3100) (2)(3100) = = 979.06 × 10−9 m 4 π Gϕ D π (77 × 109 )(26.18 × 10−3 ) c = 31.46 × 10−3 m = 31.46 mm, Design must use larger value for d. d = 2c = 62.9 mm d = 62.9 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 3.46 The electric motor exerts a torque of 800 N ⋅ m on the steel shaft ABCD when it is rotating at constant speed. Design specifications require that the diameter of the shaft be uniform from A to D and that the angle of twist between A to D not exceed 1.5°. Knowing that τ max ≤ 60 MPa and G = 77 GPa, determine the minimum diameter shaft that can be used. SOLUTION Torques: TAB = 300 + 500 = 800 N ⋅ m TBC = 500 N ⋅ m, TCD = 0 τ = 60 × 106 Pa Design based on stress. τ = Tc 2T = J π c3 c3 = 2T πτ = (2)(800) = 8.488 × 10−6 m3 6 π (60 × 10 ) −3 c = 20.40 × 10 m = 20.40 mm, Design based on deformation. d = 2c = 40.8 mm ϕ D/ A = 1.5° = 26.18 × 10−3 rad ϕD / C = 0 (500)(0.6) 300 TBC LBC = = GJ GJ GJ (800) (0.4) 320 T L = AB AB = = GJ GJ GJ 620 620 (2)(620) = ϕ D / C + ϕC / B + ϕ B / A = = π 4 = GJ π Gc 4 G2c ϕ C/ B = ϕ B/ A ϕ D/ A c4 = (2)(620) (2)(620) = = 195.80 × 10−9 m 4 π Gϕ D/ A π (77 × 109 )(26.18 × 10−3 ) c = 21.04 × 10−3 m = 21.04 mm, Design must use larger value of d. d = 2c = 42.1 mm d = 42.1 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 3.47 The design specifications of a 2-m-long solid circular transmission shaft require that the angle of twist of the shaft not exceed 3° when a torque of 9 kN ⋅ m is applied. Determine the required diameter of the shaft, knowing that the shaft is made of (a) a steel with an allowable shearing stress of 90 MPa and a modulus of rigidity of 77 GPa, (b) a bronze with an allowable shearing of 35 MPa and a modulus of rigidity of 42 GPa. SOLUTION ϕ = 3° = 52.360 × 10−3 rad, T = 9 × 103 N ⋅ m L = 2.0 m 2TL 2TL TL = ∴ c4 = based on twist angle. 4 π Gϕ GJ πc G 2T 2T Tc τ = = 2 ∴ c3 = based on shearing stress. πτ J πc ϕ = (a) Steel shaft: Based on twist angle, τ = 90 × 106 Pa, G = 77 × 109 Pa c4 = (2)(9 × 103 )(2.0) = 2.842 × 10−6 m 4 π (77 × 109 )(52.360 × 10−3 ) c = 41.06 × 10−3 m = 41.06 mm Based on shearing stress, c3 = d = 2c = 82.1 mm (2)(9 × 103 ) = 63.662 × 10−6 m3 π (90 × 106 ) c = 39.93 × 10−3 m = 39.93 mm d = 2c = 79.9 mm d = 82.1 mm Required value of d is the larger. (b) Bronze shaft: Based on twist angle, τ = 35 × 106 Pa, G = 42 × 109 Pa c4 = (2)(9 × 103 )(2.0) = 5.2103 × 10−6 m 4 9 −3 π (42 × 10 )(52.360 × 10 ) c = 47.78 × 10−3 m = 47.78 mm Based on shearing stress, c3 = (2)(9 × 103 ) = 163.702 × 10−6 m3 π (35 × 106 ) c = 54.70 × 10−3 m = 54.70 mm Required value of d is the larger. d = 2c = 95.6 mm d = 2c = 109.4 mm d = 109.4 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 3.48 A hole is punched at A in a plastic sheet by applying a 600-N force P to end D of lever CD, which is rigidly attached to the solid cylindrical shaft BC. Design specifications require that the displacement of D should not exceed 15 mm from the time the punch first touches the plastic sheet to the time it actually penetrates it. Determine the required diameter of shaft BC if the shaft is made of a steel with G = 77 GPa and τ all = 80 MPa. SOLUTION T = rP = (0.300 m)(600 N) = 180 N ⋅ m Torque Shaft diameter based on displacement limit. ϕ = δ r = 15 mm = 0.005 rad 300 mm ϕ = TL 2TL = GJ π Gc 4 c4 = 2TL (2)(180)(0.500) = = 14.882 × 10−9 m 4 π Gϕ π (77 × 109 )(0.05) c = 11.045 × 10−3 m = 11.045 m d = 2c = 22.1 mm Shaft diameter based on stress. τ = 80 × 106 Pa c3 = 2T πτ = τ = Tc 2T = J π c3 (2)(180) = 1.43239 × 10−6 m3 π (80 × 106 ) c = 11.273 × 10−3 m = 11.273 mm Use the larger value to meet both limits. d = 2c = 22.5 mm d = 22.5 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 3.49 The design specifications for the gear-and-shaft system shown require that the same diameter be used for both shafts, and that the angle through which pulley A will rotate when subjected to a 2-kip ⋅ in. torque TA while pulley D is held fixed will not exceed 7.5°. Determine the required diameter of the shafts if both shafts are made of a steel with G = 11.2 × 106 psi and τ all = 12 ksi. SOLUTION Statics: Gear B. ΣM B = 0: rB F − TA = 0 F = TB /rB ΣM C = 0: Gear C. rC F − TD = 0 TD = rC F = n= rC 5 = = 2.5 rB 2 Torques in shafts. TAB = TA = TB Deformations: ϕC /D = ϕ A/B Kinematics: ϕD = 0 rC TA = nTB rB TCD L GJ T L = AB GJ TCD = TC = nTB = nTA nTAL GJ T L = A GJ = ϕC = ϕ D + ϕC/D = 0 + rBϕ B = −rCϕ B ϕB = − ϕ A = ϕC + ϕ B /C = nTAL GJ rC ϕC = −nϕC rB ϕB n 2TA L GJ n 2TA L TA L (n2 + 1)TA L + = GJ GJ GJ 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 3.49 (Continued) Diameter based on stress. Tm = TCD = nTA Largest torque: τm = c= Tmc 2nTA = J π c3 3 2nTA πτ m = 3 τ m = τ all = 12 × 103 psi, TA = 2 × 103 lb ⋅ in (2)(2.5)(2 × 103 ) = 0.6425 in., d = 2c = 1.285 in. π (12 × 103 ) Diameter based on rotation limit. ϕ = 7.5° = 0.1309 rad ϕ = c= (n 2 + 1)TA L (2)(7.25)TAL = GJ π c 4G 4 (2)(7.25)TA L = π Gϕ Choose the larger diameter. 4 L = 8 + 16 = 24 in. (2)(7.25)(2 × 103 )(24) = 0.62348 in., d = 2c = 1.247 in. π (11.2 × 106 )(0.1309) d = 1.285 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 3.50 Solve Prob. 3.49, assuming that both shafts are made of a brass with G = 5.6 × 106 psi and τ all = 8 ksi. PROBLEM 3.49 The design specifications for the gearand-shaft system shown require that the same diameter be used for both shafts, and that the angle through which pulley A will rotate when subjected to a 2-kip ⋅ in. torque TA while pulley D is held fixed will not exceed 7.5°. Determine the required diameter of the shafts if both shafts are made of a steel with G = 11.2 × 106 psi and τ all = 12 ksi. SOLUTION Statics: Gear B. ΣM B = 0: rB F − TA = 0 F = TB /rB ΣM C = 0: Gear C. rC F − TD = 0 TD = rC F = n= rC 5 = = 2.5 rB 2 Torques in shafts. TAB = TA = TB Deformations: ϕC /D = ϕ A/B Kinematics: ϕD = 0 rC TA = nTB rB TCD L GJ T L = AB GJ TCD = TC = nTB = nTA nTAL GJ T L = A GJ = ϕC = ϕ D + ϕC/D = 0 + rBϕ B = −rCϕ B ϕB = − ϕ A = ϕ C + ϕ B /C = nTAL GJ rC ϕC = −nϕC rB ϕB n 2TA L GJ n 2TA L TA L (n2 + 1)TA L + = GJ GJ GJ 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 3.50 (Continued) Diameter based on stress. Tm = TCD = nTA Largest torque: τm = c= Tmc 2nTA = J π c3 3 2nTA πτ m = 3 τ m = τ all = 8 × 103 psi, TA = 2 × 103 lb ⋅ in (2)(2.5)(2 × 103 ) = 0.7355 in., d = 2c = 1.471 in. π (8 × 103 ) Diameter based on rotation limit. ϕ = 7.5° = 0.1309 rad ϕ = c= (n 2 + 1)TA L (2)(7.25)TAL = GJ π c 4G 4 (2)(7.25)TA L = π Gϕ Choose the larger diameter. 4 L = 8 + 16 = 24 in. (2)(7.25)(2 × 103 )(24) = 0.7415 in., d = 2c = 1.483 in. π (5.6 × 106 )(0.1309) d = 1.483 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 3.51 A torque of magnitude T = 4 kN⋅ m is applied at end A of the composite shaft shown. Knowing that the modulus of rigidity is 77 GPa for the steel and 27 GPa for the aluminum, determine (a) the maximum shearing stress in the steel core, (b) the maximum shearing stress in the aluminum jacket, (c) the angle of twist at A. SOLUTION c1 = Steel core: 1 d1 = 0.027 m 2 J1 = π 2 c14 = π 2 (0.027) 4 = 834.79 × 10−9 −9 9 G1J1 = (77 × 10 )(834.79 × 10 ) = 64.28 × 103 N ⋅ m 2 Torque carried by steel core. T1 = G1J1ϕ /L Aluminum jacket: c1 = J2 = 1 d1 = 0.027 m, 2 π (c 2 4 2 ) − c14 = π 2 c2 = 1 d 2 = 0.036 m 2 (0.0364 − 0.027 4 ) = 1.80355 × 10−6 m 4 G2 J 2 = (27 × 10 )(1.80355 × 10−6 ) = 48.70 × 103 N ⋅ m 2 9 Torque carried by aluminum jacket. T2 = G2 J 2ϕ /L T = T1 + T2 = (G1J1 + G2 J 2 ) ϕ /L Total torque: ϕ L (a) = T 4 × 103 = = 35.406 × 10−3 rad/m G1 J1 + G2 J 2 64.28 × 103 + 48.70 × 103 Maximum shearing stress in steel core. τ = G1γ = G1c1 (b) ϕ = 73.6 × 106 Pa 73.6 MPa = 34.4 × 106 Pa 34.4 MPa Maximum shearing stress in aluminum jacket. τ = G2γ = G2c2 (c) = (77 × 109 )(0.027)(35.406 × 10−3 ) L Angle of twist. ϕ L = (27 × 109 )(0.036)(35.406 × 10−3 ) ϕ =L ϕ L = (2.5)(35.406 × 10−3 ) = 88.5 × 10−3 rad ϕ = 5.07° 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 3.52 The composite shaft shown is to be twisted by applying a torque T at end A. Knowing that the modulus of rigidity is 77 GPa for the steel and 27 GPa for the aluminum, determine the largest angle through which end A can be rotated if the following allowable stresses are not to be exceeded: τ steel = 60 MPa and τ aluminum = 45 MPa . SOLUTION τ max = Gγ max = Gcmax ϕall L Steel core: τ all Gcmax L for each material. τ all = 60 × 106 Pa , cmax = ϕall L Aluminum Jacket: = ϕ = 60 × 106 = 28.860 × 10−3 rad/m (77 × 109 )(0.027) τ all = 45 × 106 Pa , cmax = ϕall L = 1 d = 0.036 m, G = 27 × 109 Pa 2 45 × 106 = 46.296 × 10−3 rad/m (27 × 109 )(0.036) Smaller value governs: ϕall Allowable angle of twist: ϕall = L L 1 d = 0.027 m, G = 77 × 109 Pa 2 = 28.860 × 10−3 rad/m ϕall L = (2.5) (28.860 × 10−3 ) = 72.15 × 10−3 rad ϕall = 4.13° 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 3.53 The solid cylinders AB and BC are bonded together at B and are attached to fixed supports at A and C. Knowing that the modulus of rigidity is 3.7 × 106 psi for aluminum and 5.6 × 106 psi for brass, determine the maximum shearing stress (a) in cylinder AB, (b) in cylinder BC. SOLUTION The torques in cylinders AB and BC are statically indeterminate. Match the rotation ϕ B for each cylinder. Cylinder AB: c= 1 d = 0.75 in. 2 L = 12 in. π c 4 = 0.49701 in 4 2 T L TAB (12) = 6.5255 × 10−6TAB ϕ B = AB = 6 GJ (3.7 × 10 )(0.49701) J = Cylinder BC: c= J = 1 d = 1.0 in. 2 π c4 = π L = 18 in. (1.0) 4 = 1.5708 in 4 2 2 TBC L TBC (18) ϕB = = = 2.0463 × 10−6TBC GJ (5.6 × 106 )(1.5708) Matching expressions for ϕΒ 6.5255 × 10−6TAB = 2.0463 × 10−6TBC TBC = 3.1889 TAB Equilibrium of connection at B : TAB + TBC − T = 0 (1) T = 12.5 × 103 lb ⋅ in TAB + TBC = 12.5 × 103 (2) 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 3.53 (Continued) 4.1889 TAB = 12.5 × 103 Substituting (1) into (2), TAB = 2.9841 × 103 lb ⋅ in (a) Maximum stress in cylinder AB. τ AB = (b) TBC = 9.5159 × 103 lb ⋅ in TABc (2.9841 × 103 ) (0.75) = = 4.50 × 103 psi J 0.49701 τ AB = 4.50 ksi Maximum stress in cylinder BC. τ BC = TBC c (9.5159 × 103 ) (1.0) = = 6.06 × 103 psi J 1.5708 τ BC = 6.06 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 3.54 Solve Prob. 3.53, assuming that cylinder AB is made of steel, for which G = 11.2 × 106 psi. PROBLEM 3.53 The solid cylinders AB and BC are bonded together at B and are attached to fixed supports at A and C. Knowing that the modulus of rigidity is 3.7 × 106 psi for aluminum and 5.6 × 106 psi for brass, determine the maximum shearing stress (a) in cylinder AB, (b) in cylinder BC. SOLUTION The torques in cylinders AB and BC are statically indeterminate. Match the rotation ϕ B for each cylinder. Cylinder AB: Cylinder BC. 1 π d = 0.75 in. L = 12 in. J = c 4 = 0.49701 in 4 2 2 TAB L TAB (12) = = 2.1557 × 10−6 TAB ϕB = GJ (11.2 × 106 )(0.49701) c= 1 π π d = 1.0 in. L = 18 in. J = c 4 = (1.0)4 = 1.5708 in 4 2 2 2 T L TBC (18) = 2.0463 × 10−6TBC ϕ B = BC = 6 GJ (5.6 × 10 )(1.5708) c= Matching expressions for ϕB 2.1557 × 10−6TAB = 2.0463 × 10−6TBC Equilibrium of connection at B : TAB + TBC − T = 0 TAB = 6.0872 × 103 lb ⋅ in (2) TBC = 6.4128 × 103 lb ⋅ in Maximum stress in cylinder AB. τ AB = (b) TAB + TBC = 12.5 × 103 (1) 2.0535 TAB = 12.5 × 103 Substituting (1) into (2), (a) TBC = 1.0535 TAB TABc (6.0872 × 103 ) (0.75) = = 9.19 × 103 psi J 0.49701 τ AB = 9.19 ksi Maximum stress in cylinder BC. τ BC = TBC c (6.4128 × 103 )(1.0) = = 4.08 × 103 psi J 1.5708 τ BC = 4.08 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 3.55 Two solid steel shafts are fitted with flanges that are then connected by bolts as shown. The bolts are slightly undersized and permit a 1.5° rotation of one flange with respect to the other before the flanges begin to rotate as a single unit. Knowing that G = 11.2 × 106 psi, determine the maximum shearing stress in each shaft when a torque of T of magnitude 420 kip ⋅ ft is applied to the flange indicated. PROBLEM 3.55 The torque T is applied to flange B. SOLUTION Shaft AB: T = TAB , LAB = 2 ft = 24 in., c = π 1 d = 0.625 in. 2 π c 4 = (0.625) 4 = 0.23968 in 4 2 2 T L ϕ B = AB AB GJ AB J AB = TAB = GJ ABϕ B (11.2 × 106 ) (0.23968) ϕB − LAB 24 = 111.853 × 103ϕ B Shaft CD: Applied torque: T = 420 kip ⋅ ft = 5040 lb ⋅ in T = TCD , LCD = 3ft = 36 in., c = J CD = π c4 = π 2 2 T L ϕC = CD CD GJ CD TCD = 1 d = 0.75 in. 2 (0.75) 4 = 0.49701 in 4 (11.2 × 106 )(0.49701) GJ CD ϕC ϕC = 154.625 × 103ϕC = 36 LCD 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 3.55 (Continued) Clearance rotation for flange B: ϕ ′B = 1.5° = 26.18 × 10−3 rad Torque to remove clearance: ′ = (111.853 × 103 )(26.18 × 10−3 ) = 2928.3 lb ⋅ in TΑΒ Torque T ′′ to cause additional rotation ϕ ′′: T ′′ = 5040 − 2928.3 = 2111.7 lb ⋅ in ′′ + T CD ′′ T ′′ = TΑΒ 2111.7 = (111.853 × 103 )ϕ ′′ + (154.625 × 103 )ϕ ′′ ∴ ϕ ′′ = 7.923 × 10−3 rad ′′ = (111.853 × 103 ) (7.923 × 10−3 ) = 886.21 lb ⋅ in TΑΒ ′′ = (154.625 × 103 ) (7.923 × 10−3 ) = 1225.09 lb ⋅ in TCD Maximum shearing stress in AB. τ AB = TAB c (2928.3 + 886.21)(0.625) = = 9950 psi J AB 0.23968 τ AB = 9.95 ksi TCDc (1225.09)(0.75) = = 1849 psi J CD 0.49701 τ CD = 1.849 ksi Maximum shearing stress in CD. τ CD = 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 3.56 Two solid steel shafts are fitted with flanges that are then connected by bolts as shown. The bolts are slightly undersized and permit a 1.5° rotation of one flange with respect to the other before the flanges begin to rotate as a single unit. Knowing that G = 11.2 × 106 psi, determine the maximum shearing stress in each shaft when a torque of T of magnitude 420 kip ⋅ ft is applied to the flange indicated. PROBLEM 3.56 The torque T is applied to flange C. SOLUTION Shaft AB: T = TAB , LAB = 2ft = 24 in., c = π 1 d = 0.625 in. 2 π c 4 = (0.625)4 = 0.23968 in 4 2 2 T L ϕ B = AB AB GJ AB J AB = TAB = GJ ABϕ B (11.2 × 106 ) (0.23968) ϕB − LAB 24 = 111.853 × 103ϕ B Shaft CD: Applied torque: T = 420 kip ⋅ ft = 5040 lb ⋅ in T = TCD , J CD = π c4 = LCD = 3 ft = 36 in., c = π 2 2 T L ϕC = CD CD GJ CD TCD = Clearance rotation for flange C: 1 d = 0.75 in. 2 (0.75) 4 = 0.49701 in 4 GJ CD ϕC (11.2 × 106 )(0.49701) ϕC = 154.625 × 103 ϕC = LCD 36 ϕC′ = 1.5° = 26.18 × 10−3 rad 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 3.56 (Continued) Torque to remove clearance: ′ = (154.625 × 103 )(26.18 × 10−3 ) = 4048.1 lb ⋅ in TCD Torque T′′ to cause additional rotation ϕ ′′: T ′′ = 5040 − 4048.1 = 991.9 lb ⋅ in ′′ + T CD ′′ T ′′ = TΑΒ 991.9 = (111.853 × 103 )ϕ ′′ + (154.625 × 103 )ϕ ′′ ∴ ϕ ′′ = 3.7223 × 10 −3 rad ′′ = (111.853 × 10−3 ) (3.7223 × 10−3 ) = 416.35 lb ⋅ in TAB ′′ = (154.625 × 10−3 ) (3.7223 × 10−3 ) = 575.56 lb ⋅ in TCD Maximum shearing stress in AB. τ AB = TAB c (416.35) (0.625) = = 1086 psi J AB 0.23968 τ AB = 1.086 ksi Maximum shearing stress in CD. τ CD = TCDc (4048.1 + 575.56)(0.75) = = 6980 psi J CD 0.49701 τ CD = 6.98 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 3.57 Ends A and D of two solid steel shafts AB and CD are fixed, while ends B and C are connected to gears as shown. Knowing that a 4-kN ⋅ m torque T is applied to gear B, determine the maximum shearing stress (a) in shaft AB, (b) in shaft CD. SOLUTION Gears B and C: φB = rC 40 φC = φC rB 100 φB = 0.4 φC (1) M C = 0 : TCD = rC F (2) M B = 0 : T − TAB = rB F (3) Solve (2) for F and substitute into (3): T − TAB = rB TCD rC T = TAB + 100 TCD 40 (4) T = TAB + 2.5 TCD L = 0.3 m, c = 0.030 m Shaft AB: φB = φB A = TAB L TAB (0.3) T = = 235.79 × 103 AB π JG G 4 (0.030) 6 2 (5) L = 0.5 m, c = 0.0225 m Shaft CD: φC = φC D = TCD L JG π 2 TCD (0.5) (0.0225) 4 G = 1242 × 103 TCD G (6) Substitute from (5) and (6) into (1): φB = 0.4φC : 235.79 × 103 TCD = 0.47462 = TAB TAB T = 0.4 × 1242 × 103 CD G G (7) 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 3.57 (Continued) Substitute for TCD from (7) into (4): T = TAB + 2.5 (0.47462 TAB ) T = 2.1865 TAB (8) 4000 N ⋅ m = 2.1865TAB TAB = 1829.4 N ⋅ m For T = 4 kN ⋅ m, Eq. (8) yields TCD = 0.47462(1829.4) = 868.3 N ⋅ m Substitute into (7): (a) Stress in AB: τ AB = (b) TAB c 2 TAB 2 1829.4 = = = 43.1 × 106 3 J π c π (0.030)3 τ AB = 43.1 MPa Stress in CD: τ CD = TCDc 2 TCD 2 868.3 = = = 48.5 × 106 3 3 J π c π (0.0225) τ CD = 48.5 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 3.58 Ends A and D of the two solid steel shafts AB and CD are fixed, while ends B and C are connected to gears as shown. Knowing that the allowable shearing stress is 50 MPa in each shaft, determine the largest torque T that may be applied to gear B. SOLUTION Gears B and C: φB = rC 40 φC = φC rB 100 φB = 0.4 φC (1) ΣM C = 0: TCD = rC F (2) ΣM B = 0: T − TAB = rB F (3) Solve (2) for F and substitute into (3): T − TAB = rB TCD rC T = TAB + 100 TCD 40 (4) T = TAB + 2.5 TCD L = 0.3 m, c = 0.030 m Shaft AB: φB = φB / A = TAB L TAB (0.3) T = = 235.77 × 103 AB π JG G (0.030) 4 G 2 (5) L = 0.5 m, c = 0.0225 m Shaft CD: φC = φC / D = TCD L TCD (0.5) T = = 1242 × 103 CD π JG G (0.0225) 4 G 2 (6) 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 3.58 (Continued) Substitute from (5) and (6) into (1) : φB = 0.4 φC : 235.79 × 103 TAB T = 0.4 × 1242 × 103 CD G G TCD = 0.47462 TAB (7) Substitute for TCD from (7) into (4): T = TAB + 2.5 (0.47462 TAB ) T = 2.1865 TAB (8) T = 4.6068 TCD (9) Solving (7) for TAB and substituting into (8), TCD T = 2.1865 0.47462 Stress criterion for shaft AB: τ AB = τ all = 50 MPa: τ AB = = TABc J π TAB = τ AB = c3τ AB 2 J c π 2 (0.030 m)3 (50 × 106 Pa) = 2120.6 N ⋅ m T = 2.1865(2120.6 N ⋅ m) = 4.64 kN ⋅ m From (8): Stress criterion for shaft CD: τ CD = τ all = 50 MPa: τ CD = From (7): TCD c π π TCD = c3τ CD = (0.0225 m)3 (50 × 106 Pa) 2 2 J = 894.62 N ⋅ m T = 4.6068(894.62 N ⋅ m) = 4.12 kN ⋅ m The smaller value for T governs. T = 4.12 kN ⋅ 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 3.59 The steel jacket CD has been attached to the 40-mm-diameter steel shaft AE by means of rigid flanges welded to the jacket and to the rod. The outer diameter of the jacket is 80 mm and its wall thickness is 4 mm. If 500 N ⋅ m torques are applied as shown, determine the maximum shearing stress in the jacket. SOLUTION c= Solid shaft: JS = 1 d = 0.020 m 2 π 2 c4 = π 2 (0.020) 4 = 251.33 × 10−9 m 4 1 d = 0.040 m 2 c1 = c2 − t = 0.040 − 0.004 = 0.036 m c2 = Jacket: JJ = π (c 2 4 2 ) − c14 = π 2 (0.0404 − 0.0364 ) −6 = 1.3829 × 10 m 4 Torque carried by shaft. TS = GJ S ϕ /L Torque carried by jacket. TJ = GJ J ϕ /L Total torque. T = TS + TJ = ( J S + J J ) G ϕ /L TJ = ∴ Gϕ T = L JS + JJ (1.3829 × 10−6 ) (500) JJ = 423.1 N ⋅ m T = JS + JJ 1.3829 × 10−6 + 251.33 × 10−6 Maximum shearing stress in jacket. τ = TJ c2 (423.1) (0.040) = = 12.24 × 106 Pa −6 JJ 1.3829 × 10 12.24 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 3.60 A solid shaft and a hollow shaft are made of the same material and are of the same weight and length. Denoting by n the ratio c1 / c2 , show that the ratio Ts /Th of the torque Ts in the solid shaft to the torque Th in the hollow shaft is (a) (1 − n 2 ) /(1 + n 2 ) if the maximum shearing stress is the same in each shaft, (b) (1 − n)/ (1 + n 2 ) if the angle of twist is the same for each shaft. SOLUTION For equal weight and length, the areas are equal. ( ) π c02 = π c22 − c12 = π c22 (1 − n 2 ) Js = (a) π 2 c04 = 2 c24 (1 − n 2 ) 2 Jh = π (c 2 4 2 ) − c14 = π 2 c24 (1 − n 4 ) For equal stresses. τ = Ts c0 Tc = h 2 Js Jh Ts Jc = s 2 = Th J hc0 (b) π c0 = c2 (1 − n 2 )1/2 ∴ c 4 (1 − n 2 ) 2 c2 2 2 π c 4 (1 − n 4 )c (1 − n 2 )1/2 2 2 2 π = 1 − n2 (1 − n 2 )1/2 = (1 + n 2 ) (1 − n 2 )1/2 1 + n2 For equal angles of twist. ϕ = Ts L TL = h GJ s GJ h Ts J = s = Th Jh c 4 (1 − n 2 )2 2 2 π c 4 (1 − n 4 ) 2 2 π = (1 − n 2 ) 2 1 − n2 = 1 − n4 1 + n2 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 3.61 A torque T is applied as shown to a solid tapered shaft AB. Show by integration that the angle of twist at A is φ = 7TL 12π Gc 4 SOLUTION Introduce coordinate y as shown. r = cy L Twist in length dy: dϕ = Tdy Tdy 2TL4dy = = π GJ π Gc 4 y 4 G r4 2 2L ϕ = L 2TL4 dy 2TL 2L dy = 4 4 π Gc y π Gc 4 L y 4 2L 2TL4 1 2TL4 = − = π Gc 4 3 y 3 L π Gc 4 = 1 1 + 3 − 3 3L 24L 2TL4 7 7TL = 4 3 π Gc 24L 12π Gc 4 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 3.62 The mass moment of inertia of a gear is to be determined experimentally by using a torsional pendulum consisting of a 6-ft steel wire. Knowing that G = 11.2 × 106 psi, determine the diameter of the wire for which the torsional spring constant will be 4.27 lb ⋅ ft/rad. SOLUTION Torsion spring constant K = 4.27 lb ⋅ ft/rad = 51.24 lb ⋅ in/rad K = c4 = T ϕ = π Gc 4 T GJ = = 2L TL /GJ L 2 LK (2) (72) (51.24) = = 209.7 × 10−6 in 4 6 πG π (11.2 × 10 ) c = 0.1203 in. d = 2c = 0.241 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 3.63 An annular plate of thickness t and modulus G is used to connect shaft AB of radius r1 to tube CD of radius r2. Knowing that a torque T is applied to end A of shaft AB and that end D of tube CD is fixed, (a) determine the magnitude and location of the maximum shearing stress in the annular plate, (b) show that the angle through which end B of the shaft rotates with respect to end C of the tube is ϕ BC = T 1 1 2 − 2 4π Gt r1 r2 SOLUTION Use a free body consisting of shaft AB and an inner portion of the plate BC, the outer radius of this portion being ρ . The force per unit length of circumference is τ t. ΣM = 0 τ t (2πρ ) ρ − T = 0 T 2π t ρ 2 τ = (a) Maximum shearing stress occurs at ρ = r1 γ = Shearing strain: τ G τ max = = T 2π tr12 T 2π GT ρ 2 The relative circumferential displacement in radial length d ρ is d δ = γ d ρ = ρ dϕ dϕ = γ dϕ = (b) ϕ B/C = = r2 r1 T dρ T = 3 2π Gt 2π Gt ρ r2 r1 dρ ρ3 dρ ρ T dρ T dρ = 2 2π Gt ρ ρ 2π GT ρ 3 = T 1 r2 − 2π Gt 2 ρ 2 r1 1 1 T 1 T 1 − 2 + 2 = 2 − 2 2π Gt 2r2 2r1 4π Gt r1 r2 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 3.64 Determine the maximum shearing stress in a solid shaft of 12-mm diameter as it transmits 2.5 kW at a frequency of (a) 25 Hz, (b) 50 Hz. SOLUTION c= (a) (b) f = 25 Hz f = 50 Hz T = 1 d = 6 mm = 0.006 m 2 P 2π f = P = 2.5 kW = 2500 W 2500 = 15.9155 N ⋅ m 2π (25) τ = Tc 2T 2(15.9155) = = = 46.9 × 106 Pa 3 J πc π (0.006)3 T = 2500 = 7.9577 N ⋅ m 2π (50) τ = 2(7.9577) = 23.5 × 106 Pa π (0.006)3 τ = 46.9 MPa τ = 23.5 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 3.65 Determine the maximum shearing stress in a solid shaft of 1.5-in. diameter as it transmits 75 hp at a speed of (a) 750 rpm, (b) 1500 rpm. SOLUTION c= (a) f = T = (b) 1 d = 0.75 in. 2 P = 75 hp = (75)(6600) = 495 × 103 lb ⋅ in/s 750 = 12.5 Hz 60 P 2π f = 495 × 103 = 6.3025 × 103 lb ⋅ in 2π (12.5) τ = Tc 2T (2) (6.3025 × 103 ) = = = 9.51 × 103 psi J π c3 π (0.75)3 f = 1500 = 25 Hz 60 T = 495 × 103 = 3.1513 × 103 lb ⋅ in 2π (25) τ = (2) (3.1513 × 103 ) = 4.76 × 103 psi 3 π (0.75) τ = 9.51 ksi τ = 4.76 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 3.66 Design a solid steel shaft to transmit 0.375 kW at a frequency of 29 Hz, if the shearing stress in the shaft is not to exceed 35 MPa. SOLUTION τ all = 35 × 106 Pa P = 0.375 × 103 W T= τ = P 2π f = f = 29 Hz 0.375 × 103 = 2.0580 N ⋅ m 2π (29) Tc 2T = J π c3 ∴ c3 = 2T πτ = (2) (2.0580) π (35 × 106 ) = 37.43 × 10−9 m3 c = 3.345 × 10−3 m = 3.345 mm d = 2c d = 6.69 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 3.67 Design a solid steel shaft to transmit 100 hp at a speed of 1200 rpm, if the maximum shearing stress is not to exceed 7500 psi. SOLUTION τ all = 7500 psi P = 100 hp = 660 × 103 lb ⋅ in/s f = τ = 1200 = 20 Hz 60 Tc 2T = J π c3 c = 0.7639in. T = ∴ c3 = 2T πτ d = 2c P 2π f = = 660 × 103 = 5.2521 × 103 lb ⋅ in 2π (20) (2) (5.2521 × 103 ) = 0.4458 in 3 π (7500) d = 1.528 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 3.68 Determine the required thickness of the 50-mm tubular shaft of Example 3.07 if is to transmit the same power while rotating at a frequency of 30 Hz. SOLUTION P = 100 kW = 100 × 103 W From Example 3.07, τ all = 60 MPa = 60 × 106 Pa c2 = 1 d = 25 mm = 0.025 m 2 f = 30 Hz T = P = 530.52 N ⋅ m 2π f π (c 2 − c14 ) c14 = c24 − 2Tc2 = 0.0254 − J = 4 2 πτ τ = Tc2 2Tc2 = J π c24 − c14 ( ) (2)(530.52) (0.025) = 249.90 × 10−9 m 4 π (60 × 106 ) c1 = 22.358 × 10−3 m = 22.358 mm t = c2 − c1 = 25 mm − 22.358 mm = 2642 mm t = 2.64 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 3.69 While a steel shaft of the cross section shown rotates at 120 rpm, a stroboscopic measurement indicates that the angle of twist is 2° in a 12-ft length. Using G = 11.2 × 106 psi, determine the power being transmitted. SOLUTION ϕ = 2° = 34.907 × 10−3 rad c2 = J = 1 d o = 1.5in. 2 π (c 2 4 2 ) − c14 = c1 = π 2 L = 12ft =144 in. 1 di = 0.6in. 2 (1.54 − 0.64 ) = 7.7486 in 4 120 f = = 2 Hz 60 T = GJ ϕ (11.2 × 106 ) (7.7486) (34.907 × 10−3 ) = = 21.037 × 103 lb ⋅ in L 144 P = 2π fT = 2π (2) (21.037 × 103 ) = 264.36 × 103 lb ⋅ in/s Since 1 hp = 6600 lb ⋅ in/s, P = 40.1 hp 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 3.70 The hollow steel shaft shown (G = 77.2 GPa, τ all = 50 MPa) rotates at 240 rpm. Determine (a) the maximum power that can be transmitted, (b) the corresponding angle of twist of the shaft. SOLUTION 1 d 2 = 30 mm 2 1 c1 = d1 = 12.5 mm 2 c2 = J = π (c 2 4 2 ) − c14 = π 2 [(30) 4 − (12.5) 4 ] = 1.234 × 106 mm 4 = 1.234 × 10−6 m 4 τ m = 50 × 106 Pa τm = (a) (50 × 106 )(1.234 × 10−6 ) τ J Tc T = m = = 2056.7 N ⋅ m J c 30 × 10−3 Angular speed. f = 240 rpm = 4 rev/sec = 4 Hz Power being transmitted. P = 2π f T = 2π (4)(2056.7) = 51.7 × 103 W P = 51.7 kW (b) Angle of twist. ϕ = TL (2056.7)(5) = = 0.1078 rad GJ (77.2 × 109 )(1.234 × 10−6 ) ϕ = 6.17° 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 3.71 As the hollow steel shaft shown rotates at 180 rpm, a stroboscopic measurement indicates that the angle of twist of the shaft is 3°. Knowing that G = 77.2 GPa, determine (a) the power being transmitted, (b) the maximum shearing stress in the shaft. SOLUTION 1 d 2 = 30 mm 2 1 c1 = d1 = 12.5 mm 2 c2 = J = π (c 2 4 2 ) − c14 = π 2 [(30)4 − (12.5)4 )] = 1.234 × 10 mm 4 = 1.234 × 10−6 m 4 6 ϕ = 3° = 0.05236 rad (a) ϕ = TL GJ T = GJ ϕ (77.2 × 109 )(1.234 × 10−6 )(0.0536) = = 997.61 N ⋅ m L 5 Angular speed: f = 180 rpm = 3 rev/sec = 3 Hz Power being transmitted. P = 2π f T = 2π (3)(997.61) = 18.80 × 103 W P = 18.80 kW (b) Maximum shearing stress. τm = Tc2 (997.61)(30 × 10−3 ) = J 1.234 × 10−6 = 24.3 × 106 Pa τ m = 24.3 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 3.72 The design of a machine element calls for a 40-mm-outer-diameter shaft to transmit 45 kW. (a) If the speed of rotation is 720 rpm, determine the maximum shearing stress in shaft a. (b) If the speed of rotation can be increased 50% to 1080 rpm, determine the largest inner diameter of shaft b for which the maximum shearing stress will be the same in each shaft. SOLUTION (a) f = 720 = 12 Hz 60 P = 45 kW = 45 × 103 W T = (b) P 2π f = 45 × 103 = 596.83 N ⋅ m 2π (12) c= 1 d = 20 mm = 0.020 m 2 τ = Tc 2T (2)(596.83) = = = 47.494 × 106 Pa 3 J πc π (0.020)3 f = 1080 = 18 Hz 60 T = 45 × 103 = 397.89 N ⋅ m 2π (18) τ = Tc2 2Tc2 = J π c24 − c14 c14 = c24 − ( τ max = 47.5 MPa ) 2Tc2 πτ c14 = 0.0204 − (2)(397.89)(0.020) = 53.333 × 10−9 π (47.494 × 106 ) c1 = 15.20 × 10−3 m = 15.20 mm d 2 = 2c1 = 30.4 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 3.73 A steel pipe of 3.5-in. outer diameter is to be used to transmit a torque of 3000 lb ⋅ ft without exceeding an allowable shearing stress of 8 ksi. A series of 3.5-in.-outer-diameter pipes is available for use. Knowing that the wall thickness of the available pipes varies from 0.25 in. to 0.50 in. in 0.0625-in. increments, choose the lightest pipe that can be used. SOLUTION T = 3000 lb ⋅ ft = 36 × 103 lb ⋅ in 1 do = 1.75 in. 2 Tc 2Tc2 τ= 2= J π c24 − c14 c2 = ( c14 = c24 − ) 2Tc2 (2)(36 × 103 )(1.75) = 1.754 − = 4.3655 in 4 πτ π(8 × 103 ) c1 = 1.4455 in. Required minimum thickness: t = c2 − c1 t = 1.75 − 1.4455 = 0.3045 in. Available thicknesses: 0.25 in., 0.3125 in., 0.375 in., etc. Use t = 0.3125 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 3.74 The two solid shafts and gears shown are used to transmit 16 hp from the motor at A, operating at a speed of 1260 rpm, to a machine tool at D. Knowing that the maximum allowable shearing stress is 8 ksi, determine the required diameter (a) of shaft AB, (b) of shaft CD. SOLUTION (a) Shaft AB: P = 16 hp = (16)(6600) = 105.6 × 103 lb ⋅ in/sec f = 1260 = 21 Hz 60 τ = 8 ksi = 8 × 103 psi TAB = P 2π f = 105.6 × 103 = 800.32 lb ⋅ in 2π (21) τ = 2T Tc = J π c3 c= 3 c= 3 2T πτ (2)(800.32) = 0.399 in. π (8 × 103 ) d AB = 2c = 0.799 in. (b) Shaft CD: TCD = c= d AB = 0.799 in. rC 5 TAB = (800.32) = 1.33387 × 103 lb ⋅ in rB 3 3 2T πτ = 3 (2)(1.33387 × 103 ) = 0.473 in. π (8 × 103 ) dCD = 2c = 0.947 in. dCD = 0.947 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 3.75 The two solid shafts and gears shown are used to transmit 16 hp from the motor at A operating at a speed of 1260 rpm to a machine tool at D. Knowing that each shaft has a diameter of 1 in., determine the maximum shearing stress (a) in shaft AB, (b) in shaft CD. SOLUTION (a) Shaft AB: P = 16 hp = (16)(6600) = 105.6 × 103 lb ⋅ in/sec f = TAB = 1260 = 21 Hz 60 P 2π f = 105.6 × 103 = 800.32 lb ⋅ in 2π (21) 1 d = 0.5 in. 2 2T Tc τ = = J π c3 (2)(800.32) = = 4.08 × 103 psi 3 π (0.5) c= (b) Shaft CD: TCD = τ = τ AB = 4.08 ksi rC 5 TAB = (800.32) = 1.33387 × 103 lb ⋅ in rB 3 2T (2)(1.33387 × 103 ) = = 6.79 × 103 psi π c3 π (0.5)3 τ CD = 6.79 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 3.76 Three shafts and four gears are used to form a gear train that will transmit 7.5 kW from the motor at A to a machine tool at F. (Bearings for the shafts are omitted in the sketch.) Knowing that the frequency of the motor is 30 Hz and that the allowable stress for each shaft is 60 MPa, determine the required diameter of each shaft. SOLUTION P = 7.5 kW = 7.5 × 103 W Shaft AB: f AB = 30 Hz τ= c3AB = TAB = τ all = 60 MPa = 60 × 106 Pa P 7.5 × 103 = = 39.789 N ⋅ m 2π f AB 2π(30) Tc AB 2T = 3 ∴ J AB πc AB c3AB = 2T πτ (2)(39.789) = 422.17 × 10−9 m3 π(60 × 106 ) c AB = 7.50 × 10−3 m = 7.50 mm Shaft CD: d AB = 2c AB = 15.00 mm P 7.5 × 103 = = 99.472 N ⋅ m 2πfCD 2π(12) fCD = rB 60 f AB = (30) = 12 Hz rC 150 τ CD = TcCD 2T 2T 2(99.472) 3 = 3 ∴ cCD = CD = = 1.05543 × 10−6 m3 πτ J CD πcCD π(60 × 106 ) TCD = cCD = 10.18 × 10−3 m = 10.18 mm Shaft EF: f EF = τ= dCD = 2cCD = 20.4 mm rD 60 fCD = (12) = 4.8 Hz rE 150 TcEF 2T = 3 J EF πcEF 3 ∴ cEF = TEF = P 7.5 × 103 = = 248.68 N ⋅ m 2πf EF 2 π (4.8) (2)(248.68) = 2.6886 × 10−6 m3 6 π(60 × 10 ) cEF = 13.82 × 10−3 m = 13.82 mm d EF = 2cEF = 27.6 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 3.77 Three shafts and four gears are used to form a gear train that will transmit power from the motor at A to a machine tool at F. (Bearings for the shafts are omitted in the sketch.) The diameter of each shaft is as follows: d AB =16 mm, dCD = 20 mm, d EF = 28 mm. Knowing that the frequency of the motor is 24 Hz and that the allowable shearing stress for each shaft is 75 MPa, determine the maximum power that can be transmitted. SOLUTION τ all = 75 MPa = 75 × 106 Pa Shaft AB: c AB = Tall = 1 d AB = 0.008 m 2 π 2 c3AB τ all = π 2 τ = Tc AB 2T = J AB π c3AB (0.008)3 (75 × 106 ) = 60.319 N ⋅ m f AB = 24 Hz Pall = 2π f ABTall = 2π (24) (60.319) = 9.10 × 103 W Shaft CD: 1 dCD = 0.010 m 2 π 3 π 2T Tc τ = CD = 3 ∴ Tall = cCD τ all = (0.010)3 (75 × 106 ) = 117.81 N ⋅ m 2 2 J CD π cCD cCD = fCD = Shaft EF: 60 rB (24) = 9.6 Hz f AB = 150 rC cEF = Pall = 2π fCDTall = 2π (9.6) (117.81) = 7.11 × 103 W 1 d EF = 0.014 m 2 π 3 cEF τ all = π (0.014)3 (75 × 106 ) = 323.27 N ⋅ m 2 2 r 60 f EF = D fCD = (9.6) = 3.84 Hz rE 150 Tall = Pall = 2π f EFTall = 2π (3.84) (323.27) = 7.80 ×103 W Maximum allowable power is the smallest value. Pall = 7.11× 103 W = 7.11 kW 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 3.78 A 1.5-m-long solid steel of 48 mm diameter is to transmit 36 kW between a motor and a machine tool. Determine the lowest speed at which the shaft can rotate, knowing that G = 77.2 GPa, that the maximum shearing stress must not exceed 60 MPa, and the angle of twist must not exceed 2.5°. SOLUTION P = 36 × 103 W, c= 1 d = 0.024 m, L = 1.5 m, G = 77.2 × 109 Pa 2 Torque based on maximum stress: τ = Tc J T = Jτ π π = c3 τ = (0.024)3 (60 × 106 ) = 1.30288 × 103 N ⋅ m c 2 2 Torque based on twist angle: ϕ= TL GJ T = τ = 60 MPa = 60 × 106 Pa ϕ = 2.5° = 43.633 × 10−3 rad GJ ϕ π c 4Gϕ π (0.024)4 (77 × 109 ) (43.633 × 10−3 ) = = L 2L (2) (1.5) = 1.17033 × 103 N ⋅ m Smaller torque governs, so T =1.17033× 103 N ⋅ m P = 2π fT f = P 36 × 103 = 2π T 2π (1.17033 × 103 ) f = 4.90 Hz 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 3.79 A 2.5-m-long steel shaft of 30-mm diameter rotates at a frequency of 30 Hz. Determine the maximum power that the shaft can transmit, knowing that G = 77.2 GPa, that the allowable shearing stress is 50 MPa, and that the angle of twist must not exceed 7.5°. SOLUTION c= 1 d = 15 mm = 0.015 m L = 2.5 m 2 τ = 50 × 106 Pa τ = Stress requirement. T = τJ c = π 2 τ c3 = π 2 Tc J (50 × 106 )(0.015)3 = 265.07 N ⋅ m ϕ = 7.5° = 130.90 × 10−3 rad G = 77.2 × 109 Pa Twist angle requirement. ϕ = T = TL 2TL = GJ π Gc 4 π 2 Gc 4ϕ = π 2 (77.2 × 109 )(0.015)4 (130.90 × 10−3 ) = 803.60 N ⋅ m Smaller value of T is the maximum allowable torque. T = 265.07 N ⋅ m Power transmitted at f = 30 Hz. P = 2π f T = 2π (30)(265.07) = 49.96 × 103 W P = 50.0 kW 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 3.80 A steel shaft must transmit 210 hp at a speed of 360 rpm. Knowing that G = 11.2 × 106 psi, design a solid shaft so that the maximum shearing stress will not exceed 12 ksi, and the angle of twist in a 8.2-ft length must not exceed 3°. SOLUTION Power: P = (210 hp)(6600 in ⋅ lb/s/hp) = 1.336 × 106 in ⋅ lb/s Angular speed: f = (360 rpm) Torque: T = Stress requirement: τ = 12 ksi, τ = c= Angle of twist requirement: P 2π f 3 2T πτ = = 1 min. = 6 Hz 60 sec 1.386 × 106 = 36.765 × 103 lb ⋅ in (2π )(6) 3 Tc 2T = 3 J πc (2)(36.765 × 103 ) = 1.2494 in. π (12 × 103 ) ϕ = 3° = 52.36 × 10−3 rad L = 8.2 ft = 98.4 in. ϕ = TL 2TL = GJ π Gc 4 c= 4 2TL = π Gϕ 4 (2)(36.765 × 103 )(98.4) = 1.4077 in. π (11.2 × 106 )(52.36 × 10−3 ) The larger value is the required radius. c = 1.408 in. d = 2c d = 2.82 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 3.81 The shaft-disk-belt arrangement shown is used to transmit 3 hp from point A to point D. (a) Using an allowable shearing stress of 9500 psi, determine the required speed of shaft AB. (b) Solve part a, assuming that the diameters of shafts AB and CD are, respectively, 0.75 in. and 0.625 in. SOLUTION τ = 9500 psi τ = Tc 2T = J π c3 P = 3 hp = (3)(6600) = 19800 lb ⋅ in/s T = π 2 c3τ Allowable torques. 3 in. diameter shaft: c= 3 − 4 in. diameter shaft: c = 83 in., Tall = Statics: Allowable torques. 3 (9500) = 786.9 lb ⋅ in 2 8 TC = rC ( F1 − F2 ) rB 1.125 TC = TC = 0.25TC rC 4.5 TB,all = 455.4 lb ⋅ in TC ,all = 786.9 lb ⋅ in Assume TC = 786.9 lb ⋅ in Then TB = (0.25)(786.9) = 196.73 lb ⋅ in < 455.4 lb ⋅ in (okay) P = 2π fT (b) π 3 TB = rB ( F1 − F2 ) TB = (a) 5 π5 in., Tall = (9500) = 455.4 lb ⋅ in 16 2 16 5 − 8 Allowable torques. P 19800 = 2π TB 2π (196.73) f AB = TB,all = 786.9 lb ⋅ in TC ,all = 455.4 lb ⋅ in Assume TC = 455.4 lb ⋅ in Then TB = (0.25)(455.4) = 113.85 lb ⋅ in < 786.9 lb ⋅ in P = 2π fT f AB = 16.02 Hz f AB = P 19800 = 2π TB 2π (113.85) f AB = 27.2 Hz 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 3.82 A 1.6-m-long tubular steel shaft of 42-mm outer diameter d1 is to be made of a steel for which τ all = 75 MPa and G = 77.2 GPa. Knowing that the angle of twist must not exceed 4° when the shaft is subjected to a torque of 900 N ⋅ m, determine the largest inner diameter d2 that can be specified in the design. SOLUTION c1 = 1 d1 = 0.021 m L = 1.6 m 2 Based on stress limit: τ = 75 MPa = 75 × 106 Pa τ = Tc1 ∴ J J = Tc1 τ = (900)(0.021) = 252 × 10−9 m 4 75 × 106 Based on angle of twist limit: ϕ = 4° = 69.813 × 10−3 rad ϕ = TL GJ ∴ J = TL (900)(1.6) = = 267.88 × 10−9 m 4 Gϕ (77 × 109 )(69.813 × 10−3 ) Larger value for J governs. J = π 2 (c 4 1 c24 = c14 − − c24 2J π J = 267.88 × 10−9 m 4 ) = 0.0214 − (2)(267.88 × 10−9 ) c2 = 12.44 × 10−3 m = 12.44 mm π = 23.943 × 10−9 m 4 d 2 = 2c2 = 24.9 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 3.83 A 1.6-m-long tubular steel shaft ( G = 77.2 GPa ) of 42-mm outer diameter d1 and 30-mm inner diameter d2 is to transmit 120 kW between a turbine and a generator. Knowing that the allowable shearing stress is 65 MPa and that the angle of twist must not exceed 3°, determine the minimum frequency at which the shaft can rotate. SOLUTION c1 = J = 1 1 d1 = 0.021 m, c2 = d 2 = 0.015 m 2 2 π 2 (c 4 1 ) − c24 = π 2 (0.0214 − 0.0154 ) = 225.97 × 10−9 m 4 Based on stress limit: τ = 65 MPa = 65 × 106 Pa τ = Tc1 Jτ (225.97 × 10−9 )(65 × 106 ) or T = = = 699.43 N ⋅ m J G 0.021 Based on angle of twist limit: ϕ = 3° = 52.36×10−3 rad ϕ = TL GJ ϕ (77 × 109 )(225.97 × 10−9 )(52.36 × 10−3 ) or T = = GJ L 1.6 = 569.40 N ⋅ m T = 569.40 N ⋅ m Smaller torque governs. P = 120 kW = 120 × 103 W P = 2π fT so f = P 120 × 103 = 2π T 2π (569.40) f = 33.5 Hz or 2010 rpm 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 3.84 Knowing that the stepped shaft shown transmits a torque of magnitude T = 2.50 kip ⋅ in., determine the maximum shearing stress in the shaft 3 when the radius of the fillet is (a) r = 18 in., (b) r = 16 in. SOLUTION D 2 = = 1.33 d 1.5 D = 2 in. d = 1.5 in. 1 d = 0.75 in. T = 2.5 kip ⋅ in 2 2T (2)(2.5) Tc = = = 3.773 ksi 3 J πc π (0.75)3 c= (a) r = 1 in. r = 0.125 in. 8 r 0.125 = = 0.0833 d 1.5 K = 1.42 From Fig. 3.32, τ max = K (b) r = Tc = (1.42)(3.773) J 3 in. 16 τ max = 5.36 ksi r = 0.1875 in. r 0.1875 = = 0.125 d 1.5 From Fig. 3.32, τ max = K K = 1.33 Tc = (1.33)(3.773) J τ max = 5.02 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 3.85 Knowing that the allowable shearing stress is 8 ksi for the stepped shaft shown, determine the magnitude T of the largest torque that can be 3 transmitted by the shaft when the radius of the fillet is (a) r = 16 in., (b) r = 14 in. SOLUTION D = 2 in. c= 1 d = 0.75 in. 2 τ max = K (a) r = D = 1.33 d d = 1.5 in. Tc J 3 in. 16 τ max = 8 ksi or T = Jτ max πτ c3 = max Kc 2K r = 0.1875 in. r 0.1875 = = 0.125 d 1.5 From Fig. 3.32, T = (b) r = K = 1.33 π (8)(0.75)3 T = 3.99 kip ⋅ in (2)(1.33) 1 in. 4 r = 0.25 in. r 0.25 = = 0.1667 d 1.5 From Fig. 3.32, T = π (8)(0.75)3 (2)(1.27) K = 1.27 T = 4.17 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 3.86 The stepped shaft shown must transmit 40 kW at a speed of 720 rpm. Determine the minimum radius r of the fillet if an allowable stress of 36 MPa is not to be exceeded. SOLUTION Angular speed: 1 Hz f = (720 rpm) = 12 Hz 60 rpm Power: P = 40 × 103 W Torque: T = P 2π f = 40 × 103 = 530.52 N ⋅ m 2π (12) In the smaller shaft, d = 45 mm, c = 22.5 mm = 0.0225 m τ = Tc 2T (2)(530.52) = = = 29.65 × 106 Pa 3 3 J πc π (0.0225) Using τ max = 36 MPa = 36 × 106 Pa results in K = τ max 36 × 106 = = 1.214 τ 29.65 × 106 From Fig 3.32 with D 90 mm r = = 2, = 0.24 d 45 mm d r = 0.24 d = (0.24)(45 mm) r = 10.8 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 3.87 The stepped shaft shown must transmit 45 kW. Knowing that the allowable shearing stress in the shaft is 40 MPa and that the radius of the fillet is r = 6 mm, determine the smallest permissible speed of the shaft. SOLUTION r 6 = = 0.2 d 30 D 60 = =2 d 30 From Fig. 3.32, For smaller side, K = 1.26 c= 1 d = 15 mm = 0.015 m 2 τ = KTc 2 KT = J π c3 T = π c3τ 2K = π (0.015)3 (40 × 106 ) (2)(1.26) P = 45 kW = 45 × 103 f = = 168.30 N ⋅ m P = 2π fT P 45 × 103 = = 42.6 Hz 2π T 2π (168.30 × 103 ) f = 42.6 Hz 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 3.88 The stepped shaft shown must rotate at a frequency of 50 Hz. Knowing that the radius of the fillet is r = 8 mm and the allowable shearing stress is 45 MPa, determine the maximum power that can be transmitted. SOLUTION τ = KTc 2 KT = J π c3 T = π c3τ 2K 1 d = 15 mm = 15 × 10−3 m 2 D = 60 mm, r = 8 mm d = 30 mm c = 60 D = = 2, 30 d 8 r = = 0.26667 30 d From Fig. 3.32, K = 1.18 Allowable torque. T = Maximum power. P = 2π f T = (2π )(50)(202.17) = 63.5 × 103 W π (15 × 10−3 )3 (45 × 106 ) (2)(1.18) = 202.17 N ⋅ m P = 63.5 kW 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 3.89 In the stepped shaft shown, which has a full quarter-circular fillet, D = 1.25 in. d = 1 in. Knowing that the speed of the shaft is and 2400 rpm and that the allowable shearing stress is 7500 psi, determine the maximum power that can be transmitted by the shaft. SOLUTION D 1.25 = = 1.25 d 1.0 1 r = ( D − d ) = 0.15 in. 2 r 0.15 = = 0.15 1.0 d From Fig. 3.32, For smaller side, K = 1.31 c= 1 d = 0.5 in. 2 τ = π c3τ KTc Jτ = T = 2K J Kc T = π (0.5)3 (7500) (2)(1.31) = 1.1241 × 103 lb ⋅ in f = 2400 rpm = 40 Hz P = 2π f T = 2π (40)(1.1241 × 103 ) = 282.5 × 103 lb ⋅ in/s P = 42.8 hp 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 3.90 A torque of magnitude T = 200 lb ⋅ in. is applied to the stepped shaft shown, which has a full quarter-circular fillet. Knowing that D = 1 in., determine the maximum shearing stress in the shaft when (a) d = 0.8 in., (b) d = 0.9 in. SOLUTION (a) D 1.0 = = 1.25 0.8 d 1 r = ( D − d ) = 0.1 in. 2 0.1 r = = 0.125 d 0.8 From Fig. 3.32, K = 1.31 For smaller side, c= 1 d = 0.4 in. 2 KTc 2KT = J π c3 (2)(1.31)(200) = = 2.61 × 103 psi π (0.4)3 τ = (b) τ = 2.61 ksi D 1.0 = = 1.111 0.9 d 1 r = ( D − d ) = 0.05 2 r 0.05 = = 0.05 d 1.0 From Fig. 3.32, K = 1.44 For smaller side, c= 1 d = 0.45 in. 2 τ = 2 KT (2)(1.44)(200) = = 2.01 × 103 psi π c3 π (0.45)3 τ = 2.01 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 3.91 In the stepped shaft shown, which has a full quarter-circular fillet, the allowable shearing stress is 80 MPa. Knowing that D = 30 mm, determine the largest allowable torque that can be applied to the shaft if (a) d = 26 mm, (b) d = 24 mm. SOLUTION τ = 80 × 106 Pa (a) D 30 = = 1.154 d 26 r = 1 ( D − d ) = 2 mm 2 From Fig. 3.32, K = 1.36 Smaller side, c= 1 d = 13 mm = 0.013 m 2 τ = 2KT KTc = J π c3 T = (b) r 2 = = 0.0768 d 26 D 30 = = 1.25 d 24 From Fig. 3.32, r = π c3τ 2K = 1 ( D − d ) = 3 mm 2 K = 1.31 c= T = π (0.013)3 (80 × 106 ) (2)(1.36) = 203 N ⋅ m T = 203 N ⋅ m r 3 = = 0.125 d 24 1 d = 12 mm = 0.012 m 2 π c3τ 2K = π (0.012)3 (80 × 106 ) (2)(1.31) = 165.8 N ⋅ m T = 165.8 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 3.92 A 30-mm diameter solid rod is made of an elastoplastic material with τ Y = 3.5 MPa. Knowing that the elastic core of the rod is 25 mm in diameter, determine the magnitude of the applied torque T. SOLUTION τ Y = 3.5 × 106 Pa, c = 1 (30 mm) = 15 mm = 0.015 m 2 1 (25 mm) = 12.5 mm = 0.0125 m 2 J π π TY = τ Y = c3τ Y = (0.015)3 (3.5 × 106 ) = 18.555 N ⋅ m c 2 2 ρY = T = ρ3 4 4 (0.0125)3 TY 1 − Y3 = (18.555) 1 − 3 c 3 (0.015)3 = 21.2 N ⋅ m T = 21.2 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 3.93 The solid circular shaft shown is made of a steel that is assumed to be elastoplastic with τ Y = 21 ksi. Determine the magnitude T of the applied torques when the plastic zone is (a) 0.8 in. deep, (b) 1.2 in. deep. SOLUTION τ Y = 21 ksi TY = τY J c = τY π 2 c3 π (1.5 in.)3 2 TY = 111.3 kip ⋅ in = (21 ksi) (a) For t = 0.8 in. ρY = 1.5 − 0.8 = 0.7 in. Eq. (3.32) T = 4 1 ρY3 4 1 (0.7 in.)3 TY 1 − = ⋅ − (111.3 kip in) 1 3 4 c3 3 4 (1.5 in.)3 T = 144.7 kip ⋅ in (b) For t = 1.2 in. T = ρY = 1.5 − 1.2 = 0.3 in. 4 1 ρY3 4 1 (0.3 in.)3 TY 1 − = ⋅ − (111.3 kip in) 1 3 4 c3 3 4 (1.5 in.)3 T = 148.1 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 3.94 The solid circular shaft shown is made of a steel that is assumed to be elastoplastic with τ Y = 145 MPa. Determine the magnitude T of the applied torques when the plastic zone is (a) 16 mm deep, (b) 24 mm deep. SOLUTION c = 32 mm = 0.032 m τ Y = 145 × 106 Pa TY = π π Jτ Y = c3τ Y = (0.032)3 (145 × 106 ) c 2 2 = 7.4634 × 103 N ⋅ m (a) t P = 16 mm = 0.016 m ρY = c − tP = 0.032 − 0.016 = 0.016 m 4 1 ρY3 4 1 0.0163 3 = × − T = TY 1 − (7.4634 10 ) 1 4 0.0323 3 4 c3 3 = 9.6402 × 103 N ⋅ m (b) T = 9.64 kN ⋅ m t P = 24 mm = 0.024 m ρY = c − tP = 0.032 − 0.024 = 0.008 m T = 4 1 ρY 3 4 1 0.0083 3 = × − TY 1 − (7.4634 10 ) 1 3 4 c3 3 4 0.0323 = 9.9123 × 103 N ⋅ m T = 9.91 kN ⋅ 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 3.95 The solid shaft shown is made of a mild steel that is assumed to be elastoplastic with G = 11.2 × 106 psi and τ Y = 21 ksi. Determine the maximum shearing stress and the radius of the elastic core caused by the T = 100 kip ⋅ in., application of torque of magnitude (a) (b) T = 140 kip ⋅ in. SOLUTION c = 1.5 in., J = (a) π 2 c 4 = 7.9522 in 4 , τ Y = 21 ksi T = 100 kip ⋅ in τm = Tc (100 kip ⋅ in)(1.5 in.) = J 7.9522 in 4 τ m = 18.86 ksi Since τ m < τ Y , shaft remains elastic. c = 1.500 in. Radius of elastic core: (b) T = 140 kip ⋅ in τm = (140)(1.5) = 26.4 ksi. 7.9522 Impossible: τ m = τ Y = 21.0 ksi Plastic zone has developed. Torque at onset of yield is TY = Eq. (3.32): T = J 7.9522 (21 ksi) = 111.33 kip ⋅ in τY = c 1.5 4 1 ρY3 TY 1 − 3 4 c3 3 T 140 ρY c = 4 − 3 T = 4 − 3 111.33 = 0.22743 Y ρY = 0.6104c = 0.6104(1.5 in.) ρY c = 0.6104 ρY = 0.916 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 3.96 It is observed that a straightened paper clip can be twisted through several revolutions by the application of a torque of approximately 60 mN ⋅ m. Knowing that the diameter of the wire in the paper clip is 0.9 mm, determine the approximate value of the yield stress of the steel. SOLUTION c= 1 d = 0.45 mm = 0.45 × 10−3 m 2 TP = 60 mN ⋅ m = 60 × 10−3 N ⋅ m TP = 4 4 Jτ Y 4 π 2π 3 = ⋅ c3 TY = TY = c τY 3 3 c 3 2 3 τY = 3TP (3)(60 × 10−3 ) = = 314 × 106 Pa 2π c3 2π (0.45 × 10−3 )3 τ Y = 314 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 3.97 The solid shaft shown is made of a mild steel that is assumed to be elastoplastic with τ Y = 145MPa. Determine the radius of the elastic core caused by the application of a torque equal to 1.1 TY, where TY is the magnitude of the torque at the onset of yield. SOLUTION c= ρY c = 1 d = 15 mm 2 3 4−3 T = TY T = 3 4 ρY TY 1 − 3 c 4 − (3)(1.1) = 0.88790 ρY = 0.88790 c = (0.88790)(15 mm) ρY = 13.32 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 3.98 For the solid circular shaft of Prob. 3.95, determine the angle of twist caused by the application of a torque of magnitude (a) T = 80 kip ⋅ in., (b) T = 130 kip ⋅ in. SOLUTION 1 1 d = (3) = 1.5 in. τ Y = 21 × 103 psi 2 2 c= L = 4 ft = 48 in. J = Torque at onset of yielding: Tc J τ = τY J TY = (a) T = c = 2 c4 = π 2 (1.5)4 = 7.9522 in 4 τJ c (21 × 103 )(7.9522) = 111.330 × 103 lb ⋅ in 1.5 T = 80 kip ⋅ in = 80 × 103 lb ⋅ in Since T < TY , the shaft is fully elastic. ϕ = TL GJ (80 × 103 )(48) = 43.115 × 10−3 rad (11.2 × 106 )(7.9522) ϕ = (b) π 3 T = 130 kip ⋅ in = 130 × 10 lb ⋅ in ϕY = T > TY 3 4 ϕY T = TY 1 − 3 ϕ TY L (111.330 × 103 )(48) = = 60.000 × 10−3 rad 6 GJ (11.2 × 10 )(7.9522) T ϕY = 34−3 = TY ϕ ϕ = ϕ = 2.47° ϕY 0.79205 = 3 4− (3)(130 × 103 ) = 0.79205 111.330 × 103 60.000 × 10−3 = 75.75 × 10−3 rad 0.79205 ϕ = 4.34° 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 3.99 The solid shaft shown is made of a mild steel that is assumed to be elastoplastic with G = 77.2 GPa and τ Y = 145 MPa. Determine the angle of twist caused by the application of a torque of magnitude (a) T = 600 N ⋅ m, (b) T = 1000 N ⋅ m. SOLUTION 1 d = 15 mm = 15 × 10−3 m 2 c= Torque at onset of yielding: τ = TY = (a) π c3τ Y 2 = π (15 × 10−3 )3 (145 × 106 ) 2 = 768.71 N ⋅ m T = 600 N ⋅ m. Since T ≤ TY , the shaft is elastic. ϕ= (b) Tc 2T = J π c3 T = 1000 N ⋅ m. TL 2TL (2)(600)(1.2) = = = 0.11728 rad GJ π c 4G π (15 × 10−3 ) 4 (77.2 × 109 ) ϕ = 6.72° T > TY A plastic zone has developed. 3 T 4 ϕY ϕY = 3 4 − 3 TY 1 − 3 ϕ ϕ TY T L 2T L (2)(768.71)(2.1) ϕY = Y = Y4 = = 0.15026 rad GJ π c G π (15 × 10−3 ) 4 (77.2 × 109 ) T = ϕY (3)(1000) = 34− = 0.46003 ϕ 768.71 ϕY 0.15026 = = 0.32663 rad ϕ = 0.46003 0.46003 ϕ = 18.71° 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 3.100 A 3-ft-long solid shaft has a diameter of 2.5 in. and is made of a mild steel that is assumed to be elastoplastic with τ Y = 21 ksi and G = 11.2 × 106 psi. Determine the torque required to twist the shaft through an angle of (a) 2.5°, (b) 5°. SOLUTION L = 3 ft = 36 in., c = J = (a) π 2 π c4 = 2 1 d = 1.25 in., τ Y = 21 × 103 psi 2 (1.25) 4 = 3.835 in4 (3.835)(21 × 103 ) Jτ Y = = 64.427 × 103 lb ⋅ in 1.25 c τY = TY c J ϕY = (64.427 × 103 )(36) TY L = = 53.999 × 10−3 rad = 3.0939° GJ (11.2 × 106 )(3.835) TY = ϕ = 2.5° = 43.633 × 10−3 rad ϕ < ϕY ϕ = T = The shaft remains elastic. TL GJ GJ ϕ (11.2 × 106 )(3.835)(43.633 × 10−3 ) = L 36 = 52.059 × 103 lb ⋅ in (b) ϕ = 5° = 87.266 × 10−3 rad ϕ > ϕY T = T = 52.1 kip ⋅ in A plastic zone occurs. 3 4 1ϕ TY 1 − Y 3 4 ϕ 3 4 1 53.999 × 10−3 3 = (64.427 × 10 ) 1 − 3 4 87.266 × 10−3 = 80.814 × 103 lb ⋅ in T = 80.8 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 3.101 For the solid shaft of Prob. 3.99, determine (a) the magnitude of the torque T required to twist the shaft through an angle of 15°, (b) the radius of the corresponding elastic core. PROBLEM 3.99 The solid shaft shown is made of a mild steel that is assumed to be elastoplastic with G = 77.2 GPa and τ Y = 145 MPa. Determine the angle of twist caused by the application of a torque of magnitude (a) T = 600 N ⋅ m, (b) T = 1000 N ⋅ m. SOLUTION 1 d = 15 mm = 15 × 10−3 m 2 ϕ = 15° = 0.2618 rad c= ϕY = (a) Lγ Y Lτ (1.2)(145 × 106 ) = Y = = 0.15026 rad c cG (15 × 10−3 )(77.2 × 109 ) Since ϕ > ϕY , there is a plastic zone. τY = TY = TY c 2T = J π c3 π c3τ Y 2 = π (15 × 10−3 )3 (145 × 106 ) 2 = 768.71 N ⋅ m 3 3 4 1ϕ 4 1 0.15026 T = TY 1 − Y = (768.71) 1 − 3 4 ϕ 3 4 0.2618 = 976.5 N ⋅ m (b) T = 977 N ⋅ m Lγ Y = ρY ϕ = cϕY ρY = cϕY ϕ = (15 × 10−3 )(0.15026) = 8.61 × 10−3 m 0.2618 ρY = 8.61 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 3.102 The shaft AB is made of a material that is elastoplastic with τ Y = 12 ksi and G = 4.5 × 106 psi. For the loading shown, determine (a) the radius of the elastic core of the shaft, (b) the angle of twist at end B. SOLUTION (a) Radius of elastic core. τ Y = 12 × 103 psi c = 0.5in. TY = Jτ Y π π = c3τ Y = (0.5)3 (12 × 103 ) 2 2 c = 2356.2 lb ⋅ in T = 2560 lb ⋅ in > TY (plastic region with elastic core) T = ρY 3 c c (b) =4− 3 ρY 3 4 1 ρY TY 1 − 3 4 c3 3T (3)(2560) = 4− = 0.74051 2356.2 TY ρY = (0.9047)(0.5) = 0.9047 ρY = 0.452 in. Angle of twist. L = 6.4 ft = 76.8 in. ϕY = G = 4.5 × 106 psi TY L 2T L (2)(2356.2)(76.8) = Y4 = = 0.4096 radians JG π c G π (0.5) 4 (4.5 × 106 ) ϕY ρ = Y ϕ c ϕ = ϕY c 0.4096 = = 0.4527 radians ρY 0.9047 ϕ = 25.9° 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 3.103 A 1.25-in.-diameter solid circular shaft is made of a material that is assumed to be elastoplastic with τ Y = 18 ksi and G = 11.2 × 106 psi. For an 8-ft length of the shaft, determine the maximum shearing stress and the angle of twist caused by a 7.5 kip ⋅ in. torque. SOLUTION c= 1 d = 0.625 in., G = 11.2 × 106 psi, τ Y = 18 ksi = 18000 psi 2 L = 8 ft = 96 in. T = 7.5 kip ⋅ in = 7.5 × 103 lb ⋅ in TY = Jτ Y π π = c3τ Y = (0.625)3 (18000) = 6.9029 × 103 lb ⋅ in c 2 2 T > TY : plastic region with elastic core γY = cϕY L ∴ τ max = τ Y = 18 ksi ∴ ϕY = τ max = 18 ksi Lγ Y Lτ (96)(18000) = Y = = 246.86 × 10−3 rad c cG (0.625)(11.2 × 106 ) 4 1 ϕ Y3 TY 1 − 3 4 ϕ 3 1 1 ϕ = = = 1.10533 3 ϕY 3T (3)(7.5 10 ) × 34 − 34 − TY 6.9029 × 103 T = ϕ = 1.10533ϕY = (1.10533)(246.86 × 10−3 ) = 272.86 × 10−3 rad ϕ = 15.63° 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 3.104 A 18-mm-diameter solid circular shaft is made of a material that is assumed to be elastoplastic with τ Y = 145 MPa and G = 77 GPa. For a 1.2-m length of the shaft, determine the maximum shearing stress and the angle of twist caused by a 200 N ⋅ m torque. SOLUTION τ Y = 145 × 106 Pa, c = TY = 1 d = 0.009 m, L = 1.2 m, T = 200 N ⋅ m 2 Jτ Y π π = c3τ Y = (0.009)3 (145 × 106 ) = 166.04 N ⋅ m c 2 2 τ max = τ Y = 145MPa T > TY (plastic region with elastic core) ϕY = T = TY L 2T L (2)(166.04)(1.2) = Y4 = = 251.08 × 10−3 radians GJ π c G π (0.009)4 (77 × 109 ) 4 1 ϕ3 TY 1 − 3 4 ϕY3 3 ϕY 3T (3)(200) =4− = 0.38641 = 4− 166.04 ϕ T Y ϕ = ϕ 0.72837 = ϕY = 0.72837 ϕ 251.08 × 10−3 = 344.7 × 10−3 radians 0.72837 ϕ = 19.75° 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 3.105 A solid circular rod is made of a material that is assumed to be elastoplastic. Denoting by TY and φY, respectively, the torque and the angle of twist at the onset of yield, determine the angle of twist if the torque is increased to (a) T = 1.1 TY , (b) T = 1.25 TY , (c) T = 1.3 TY . SOLUTION T = 4 1 ϕY 3 TY 1 − 3 4 ϕ 3 ϕY 3T = 34− TY ϕ (a) T = 1.10 TY ϕ = ϕY (b) T = 1.25 TY ϕ = ϕY (c) T = 1.3 TY ϕ = ϕY 3 3 3 or ϕ = ϕY 1 3 4− 3T TY 1 = 1.126 4 − (3)(1.10) ϕ = 1.126 ϕY 1 = 1.587 4 − (3)(1.25) ϕ = 1.587 ϕY 1 = 2.15 4 − (3)(1.3) ϕ = 2.15 ϕY 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 3.106 The hollow shaft shown is made of a steel that is assumed to be elastoplastic with τ Y = 145 MPa and G = 77.2 GPa. Determine the magnitude T of the torque and the corresponding angle of twist (a) at the onset of yield, (b) when the plastic zone is 10 mm deep. SOLUTION (a) At the onset of yield, the stress distribution is the elastic distribution with τ max = τ Y . c2 = J = 1 1 d 2 = 0.030 m, c1 = d1 = 0.0125 m 2 2 π (c 2 4 2 τ max = τ Y = ) − c14 = π 2 (0.0304 − 0.01254 ) = 1.2340 × 10−6 m 4 TY c2 Jτ (1.2340 × 10−6 )(145 × 106 ) ∴ TY = Y = = 5.9648 × 103 N ⋅ m J c2 0.030 TY = 5.96 kN ⋅ m TY L (5.9643 × 103 )(5) = = 313.04 × 10−3 rad GJ (77.2 × 109 )(2.234010−6 ) ϕY = (b) ρϕ L ϕ = ρY = c2 − t = 0.030 − 0.010 = 0.020 m t = 0.010 m γ = = ϕY = 17.94° ρY ϕ L = γY = τY G τY L (145 × 106 )(5) = = 469.56 × 10−3 rad G ρY (77.2 × 109 )(0.020) Torque T1 carried by elastic portion: τ = τ Y at ρ = ρY . τY = T1ρY J1 ϕ = 26.9° c1 ≤ ρ ≤ ρY where J1 = π 2 (ρ 4 Y − c14 ) π (0.0204 − 0.01254 ) = 212.978 × 10−9 m 4 2 Jτ (212.978 × 10−9 )(145 × 106 ) T1 = 1 Y = = 1.5441 × 103 N ⋅ m ρY 0.020 J1 = 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 3.106 (Continued) Torque T2 carried by plastic portion: c2 2 T2 = 2π ρ τ Y ρ d ρ = 2π τ Y Y = ρ3 3 c2 = ρY 2π τ Y c23 − ρY3 3 ( ) 2π (145 × 106 )(0.0303 − 0.0203 ) = 5.7701 × 103 N ⋅ m 3 Total torque: T = T1 + T2 = 1.5541 × 103 + 5.7701 × 103 = 7.3142 × 103 N ⋅ m T = 7.31 kN ⋅ 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 3.107 For the shaft of Prob. 3.106, determine (a) angle of twist at which the section first becomes fully plastic, (b) the corresponding magnitude T of the applied torque. Sketch the T − φ curve for the shaft. PROBLEM 3.106 The hollow shaft shown is made of a steel that is assumed to be elastoplastic with τ Y = 145 MPa and G = 77.2 GPa. Determine the magnitude T of the torque and the corresponding angle of twist (a) at the onset of yield, (b) when the plastic zone is 10 mm deep. SOLUTION 1 d1 = 0.0125 m 2 c1 = (a) 1 d 2 = 0.030 m 2 For onset of fully plastic yielding, ρY = c1 τ = τY ∴ γ = TP = 2π = τY G = ρY ϕ L = c1ϕ L Lτ Y (25)(145 × 106 ) = = 751.295 × 10−3 rad c1G (0.0125)(77.2 × 109) ϕf = (b) c2 = c2 c1 2 τ Y ρ d ρ = 2π τ Y ρ3 3 c2 = c1 2π τ Y c23 − c13 3 ( ) 2π (145 × 106 )(0.0303 − 0.01253 ) = 7.606 × 103 N ⋅ m 3 From Prob. 3.101, ϕY = 17.94° Also from Prob. 3.101, ϕ f = 43.0° TP = 7.61 kN ⋅ m TY = 5.96 kN ⋅ m ϕ = 26.9° T = 7.31 kN ⋅ m Plot T vs ϕ using the following data. ϕ , deg T kN ⋅ m 0 17.94 26.9 43.0 >43.0 0 5.96 7.31 7.61 7.61 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 3.108 A steel rod is machined to the shape shown to form a tapered solid shaft to which torques of magnitude T = 75 kip ⋅ in. are applied. Assuming the steel to be elastoplastic with τ Y = 21 ksi and G = 11.2 × 106 psi, determine (a) the radius of the elastic core in portion AB of the shaft, (b) the length of portion CD that remains fully elastic. SOLUTION (a) c= In portion AB. TY = T = ρY c = 1 d = 1.25 in. 2 JABτ Y π π = c3τ Y = (1.25)3 (21 × 103 ) = 64.427 × 103 lb ⋅ in 2 2 c ρ3 4 TY 1 − Y3 3 c 3 4− 3T = TY 3 4− (3)(75 × 103 ) = 0.79775 64.427 × 103 ρY = 0.79775c = (0.79775)(1.25) = 0.99718 in. (b) For yielding at point C. ρY = 0.997 in. τ = τ Y , c = cx , T = 75 × 103 lb ⋅ in T = J Cτ Y π = cx3τ Y cx 2 cx = 3 2T πτ Y = 3 (2)(75 × 103 ) = 1.31494 in. π (21 × 103 ) Using proportions from the sketch, 1.50 − 1.31494 x = 1.50 − 1.25 5 x = 3.70 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 3.109 If the torques applied to the tapered shaft of Prob. 3.108 are slowly increased, determine (a) the magnitude T of the largest torques that can be applied to the shaft, (b) the length of the portion CD that remains fully elastic. PROBLEM 3.108 A steel rod is machined to the shape shown to form a tapered solid shaft to which torques of magnitude T = 75 kip ⋅ in. are applied. Assuming the steel to be elastoplastic with τ Y = 21 ksi and G = 11.2 × 106 psi, determine (a) the radius of the elastic core in portion AB of the shaft, (b) the length of portion CD that remains fully elastic. SOLUTION (a) The largest torque that may be applied is that which makes portion AB fully plastic. c= In portion AB, TY = 1 d = 1.25 in. 2 Jτ Y π π = c3τ Y = (1.25)3 (21 × 103 ) = 64.427 × 103 lb ⋅ in 2 2 c For fully plastic shaft, ρY = 0 T = T = (b) 4 1 ρY3 4 TY 1 − = T 3 4 c3 3 4 (64.427 × 103 ) = 85.903 × 103 lb ⋅ in 3 For yielding at point C, τ = τ Y , c = cx , τY = Tcx 2T = Jx π cx3 cx = 3 2T πτ Y = 3 T = 85.9 kip ⋅ in T = 85.903 × 103 lb ⋅ in (2)(85.903 × 103 ) = 1.37580 in. π (21 × 103 ) Using proportions from the sketch, 1.50 − 1.37580 x = 1.50 − 1.25 5 x = 2.48 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 3.110 A hollow shaft of outer and inner diameters respectively equal to 0.6 in. and 0.2 in. is fabricated from an aluminum alloy for which the stressstrain diagram is given in the diagram shown. Determine the torque required to twist a 9-in. length of the shaft through 10°. SOLUTION ϕ = 10° = 174.53 × 10−3 rad c1 = 1 1 d1 = 0.100 in, c2 = d 2 = 0.300 in. 2 2 γ max = c2ϕ (0.300)(174.53 × 10−3 ) = = 0.00582 L 9 γ min = c1ϕ (0.100)(174.53 × 10−3 ) = = 0.00194 L 9 z= Let γ = γ max T = 2π c2 c1 ρ z1 = c2 ρ 2τ d ρ = 2π c23 I= where the integral I is given by c1 1 = c2 3 1 1/3 1 z1 z 2τ dz = 2π c23 I z 2τ dz Evaluate I using a method of numerical integration. If Simpson’s rule is used, the integration formula is I = Δz Σ w z2 τ 3 where w is a weighting factor. Using Δ z = 16 , we get the values given in the table below. z 1/3 1/2 2/3 5/6 1 γ 0.00194 0.00291 0.00383 0.00485 0.00582 I = τ, ksi 8.0 10.0 11.5 13.0 14.0 z 2τ , ksi 0.89 2.50 5.11 9.03 14.0 w 1 4 2 4 1 wz 2τ , ksi 0.89 10.00 10.22 36.11 14.00 71.22 ←⎯ ⎯ Σwz 2τ (1/6)(71.22) = 3.96 ksi 3 T = 2π c23I = 2π (0.300)3 (3.96) = 0.671 kip ⋅ in T = 671 lb ⋅ in Note: Answer may differ slightly due to differences of opinion in reading the stress-strain curve. 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 3.111 Using the stress-strain diagram shown, determine (a) the torque that causes a maximum shearing stress of 15 ksi in a 0.8-in.-diameter solid rod, (b) the corresponding angle of twist in a 20-in. length of the rod. SOLUTION (a) τ max = 15 ksi c= 1 d = 0.400 in. 2 γ max = 0.008 From the stress-strain diagram, z = Let γ γ max T = 2π I= where the integral I is given by 1 c 0 = ρ c ρ 2τ d ρ = 2π c3 1 z τ dz = 2π c I 2 3 0 z τ dz 2 0 Evaluate I using a method of numerical integration. If Simpson’s rule is used, the integration formula is I = Δz Σ w z2 τ 3 where w is a weighting factor. Using Δ z = 0.25, we get the values given in the table below. wz 2τ , ksi γ τ, ksi z 2τ , ksi w 0 0.000 0 0.000 1 0.00 0.25 0.002 8 0.500 4 2.00 0.5 0.004 12 3.000 2 6.00 0.75 0.006 14 7.875 4 31.50 1.0 0.008 15 15.000 1 15.00 z 54.50 I = (0.25)(54.50) = 4.54 ksi 3 T = 2π c3I = 2π (0.400)3 (4.54) (b) γ max = ϕ = ←⎯ ⎯ Σwz 2τ T = 1.826 kip ⋅ in cϕ L Lγ m (20)(0.008) = = 400 × 10−3 rad c 0.400 ϕ = 22.9° Note: Answers may differ slightly due to differences of opinion in reading the stress-strain curve. 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 3.112 A 50-mm-diameter cylinder is made of a brass for which the stressstrain diagram is as shown. Knowing that the angle of twist is 5° in a 725-mm length, determine by approximate means the magnitude T of torque applied to the shaft SOLUTION ϕ = 5° = 87.266 × 10−3 rad γ max 1 d = 0.025m, L = 0.725 m 2 cϕ (0.025)(87.266 × 10−3 ) = = = 0.00301 L 0.725 z = Let c= γ γ max T = 2π c 0 = ρ c ρ 2τ d ρ = 2π c3 1 z τ dz = 2π c I 2 3 0 where the integral I is given by I = 1 z τ dz 2 0 Evaluate I using a method of numerical integration. If Simpson’s rule is used, the integration formula is I = Δz wz 2τ 3 where w is a weighting factor. Using Δz = 0.25, we get the values given in the table below. γ z τ , MPa z 2τ , MPa w wz 2τ , MPa 0 0 0 0 1 0 0.25 0.00075 30 1.875 4 7.5 0.5 0.0015 55 13.75 2 27.5 0.75 0.00226 75 42.19 4 168.75 1.0 0.00301 80 80 1 80 ← wz 2τ = 283.75 × 106 Pa 283.75 I = (0.25)(283.75 × 106 ) = 23.65 × 106 Pa 3 T = 2π c3I = 2π (0.025)3 (23.65 × 106 ) = 2.32 × 103 N ⋅ m T = 2.32 kN ⋅ 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 3.113 Three points on the nonlinear stress-strain diagram used in Prob. 3.112 are (0,0), (0.0015,55 MPa), and (0.003,80 MPa). By fitting the polynomial τ = A + Bγ + Cγ 2 through these points, the following approximate relation has been obtained. T = 46.7 × 109 γ − 6.67 × 1012 γ 2 Solve Prob. 3.113 using this relation, Eq. (3.2), and Eq. (3.26). PROBLEM 3.112 A 50-mm diameter cylinder is made of a brass for which the stress-strain diagram is as shown. Knowing that the angle of twist is 5° in a 725-mm length, determine by approximate means the magnitude T of torque applied to the shaft. SOLUTION ϕ = 5° = 87.266 × 10−3 rad, c = γ max = Let z= 1 d = 0.025 m, L = 0.725 m 2 cϕ (0.025)(87.266 × 10−3 ) = = 3.009 × 10−3 0.725 L γ ρ = γ max c T = 2π c 0 ρ 2τ d ρ = 2π c3 1 z τ dz 2 0 The given stress-strain curve is 2 τ = A + Bγ + Cγ 2 = A + Bγ max z + Cγ max z2 T = 2π c3 z ( A + Bγ 1 2 0 max z ) 2 + Cγ max z 2 dz 1 1 2 = 2π c3 A z 2 dz + Bγ max z 3 dz + Cγ max 0 0 1 1 2 1 = 2π c 2 A + Bγ max + Cγ max 4 5 3 Data: A = 0, 1 A = 0, 3 1 z dz 4 0 B = 46.7 × 109 , C = −6.67 × 1012 1 1 Bγ max = (46.7 × 109 )(3.009 × 10−3 ) = 35.13 × 103 4 4 1 2 1 Cγ max = − (6.67 × 1012 )(3.009 × 10−3 )2 = −12.08 × 103 5 5 T = 2π (0.025)3(0 + (35.13 × 103 − 12.08 × 103 ) = 2.26 × 103 N ⋅ m T = 2.26 kN ⋅ 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 3.114 The solid circular drill rod AB is made of a steel that is assumed to be elastoplastic with τ Y = 22 ksi and G = 11.2 × 106 psi. Knowing that a torque T = 75 kip ⋅ in. is applied to the rod and then removed, determine the maximum residual shearing stress in the rod. SOLUTION c = 1.2 in. π L = 35 ft = 420 in. π c 4 = (1.2) 4 = 3.2572 in 4 2 2 (3.2572)(22) Jτ = 59.715 kip ⋅ in TY = Y = 1.2 c J = Loading: T = 75 kip ⋅ in T = ρY3 c 3 ρY c 4 1 ρY3 TY 1 − 3 4 c3 =4− 3T (3)(75) = 4− = 0.23213 59.715 TY = 0.61458, ρY = 0.61458c = 0.73749 in. Unloading: τ′ = Tρ J At ρ = c τ′ = (75)(1.2) = 27.63 ksi 3.2572 At ρ = ρY τ′ = (75)(0.73749) = 16.98 ksi 3.2572 where T = 75 kip ⋅ in Residual: τ res = τ load − τ ′ At ρ = c τ res = 22 − 27.63 = −5.63 ksi At ρ = ρY τ res = 22 − 16.98 = 5.02 ksi maximum τ res = 5.63 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 3.115 In Prob. 3.114, determine the permanent angle of twist of the rod. PROBLEM 3.114 The solid circular drill rod AB is made of steel that is assumed to be elastoplastic with τ Y = 22 ksi and G = 11.2 × 106 psi. Knowing that a torque T = 75kip ⋅ in. is applied to the rod and then removed, determine the maximum residual shearing stress in the rod. SOLUTION From the solution to Prob. 3.114, c = 1.2 in. J = 3.2572 in 4 ρY c = 0.61458 ρY = 0.73749 in. After loading, γ = ρϕ L ∴ ϕ = ϕload = During unloading, Lγ ρ = Lγ Y ρY = Lτ Y ρY G L = 35 ft = 420 in. (420)(22 × 103 ) = 1.11865 rad = 64.09° (0.73749)(11.2 × 106 ) ϕ′ = TL GJ ϕ′ = (75 × 103 )(420) = 0.86347 rad = 49.47° (11.2 × 106 )(3.2572) (elastic) T = 5 × 103 N ⋅ m Permanent angle of twist. ϕperm = ϕload − ϕ ′ = 1.11865 − 0.86347 = 0.25518 ϕ = 14.62° 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 3.116 The solid shaft shown is made of a steel that is assumed to be elastoplastic with τ Y = 145 MPa and G = 77.2 GPa. The torque is increased in magnitude until the shaft has been twisted through 6°; the torque is then removed. Determine (a) the magnitude and location of the maximum residual shearing stress, (b) the permanent angle of twist. SOLUTION ϕ = 6° = 104.72 × 10−3 rad c = 0.016 m γ max = γY = ρY c = (0.016)(104.72 × 10−3 ) cϕ = = 0.0027925 0.6 L τY G = γY γ max 145 × 106 = 0.0018782 77.2 × 109 0.0018 = = 0.67260 0.0027925 π π c 4 = (0.016)4 = 102.944 × 10−9 m 4 2 2 Jτ π π TY = Y = c3τ Y = (0.016)3 (145 × 106 ) = 932.93 N ⋅ m 2 2 c J = At end of loading. Unloading: At ρ = c At ρ = ρY Residual: (a) Tload = 4 1 ρ Y3 4 1 TY 1 − = (932.93) 1 − (0.67433)3 = 1.14855 × 103 N ⋅ m 3 3 4 c 3 4 T ′ = 1.14855 × 103 N ⋅ m elastic τ′ = T ′c (1.14855 × 103 )(0.016) = = 178.52 × 106 Pa J 102.944 × 10−9 τ′ = T ′c ρY = (178.52 × 106 )(0.67433) = 120.38 × 106 Pa J c ϕ′ = (1.14855 × 103 )(0.6) T ′L = = 86.71 × 10−3 rad = 4.97° GJ (77.2 × 109 )(102.944 × 10−9 ) τ res = τ load − τ ′ ϕ perm = ϕload − ϕ ′ At ρ = c τ res = 145 × 106 − 178.52 × 106 = −33.52 × 106 Pa τ res = −33.5 MPa At ρ = ρY τ res = 145 × 106 − 120.38 × 106 = 24.62 × 106 Pa τ res = 24.6 MPa Maximum residual stress: (b) ϕperm = 104.72 × 10−3 − 86.71 × 10−3 = 17.78 × 10−3 rad 33.5 MPa at ρ = 16 mm ϕperm = 1.032° 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 3.117 After the solid shaft of Prob. 3.116 has been loaded and unloaded as described in that problem, a torque T1 of sense opposite to the original torque T is applied to the shaft. Assuming no change in the value of ϕY , determine the angle of twist ϕ1 for which yield is initiated in this second loading and compare it with the angle ϕY for which the shaft started to yield in the original loading. PROBLEM 3.116 The solid shaft shown is made of a steel that is assumed to be elastoplastic with τ Y = 145 MPa and G = 77.2 GPa. The torque is increased in magnitude until the shaft has been twisted through 6°; the torque is then removed. Determine (a) the magnitude and location of the maximum residual shearing stress, (b) the permanent angle of twist. SOLUTION From the solution to Prob. 3.116, c = 0.016 m, L = 0.6 m τ Y = 145 × 106 Pa, J = 102.944 × 10−9 m 4 The residual stress at ρ = c is τ res = 33.5 MPa For loading in the opposite sense, the change in stress to produce reversed yielding is τ1 = τ Y − τ res = 145 × 106 − 33.5 × 106 = 111.5 × 106 Pa (102.944 × 10−9 )(111.5 × 106 ) T1c Jτ ∴ T1 = 1 = 0.016 J c = 717 N ⋅ m τ1 = Angle of twist at yielding under reversed torque. ϕ1 = T1L (717 × 103 )(0.6) = = 54.16 × 10−3 rad GJ (77.2 × 109 )(102.944 × 10−9 ) ϕ1 = 3.10° Angle of twist for yielding in original loading. γ = ϕY = τY G = cϕY L Lτ Y (0.6)(145 × 106 ) = = 70.434 × 10−3 rad cG (0.016)(77.2 × 109 ) ϕY = 4.04° 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 3.118 The hollow shaft shown is made of a steel that is assumed to be elastoplastic with τ Y = 145 MPa and G = 77.2 GPa. The magnitude T of the torques is slowly increased until the plastic zone first reaches the inner surface of the shaft; the torques are then removed. Determine the magnitude and location of the maximum residual shearing stress in the rod. SOLUTION 1 d1 = 12.5 mm 2 1 c2 = d 2 = 30 mm 2 c1 = When the plastic zone reaches the inner surface, the stress is equal to τ Y . The corresponding torque is calculated by integration. dT = ρτ dA = ρτ Y (2πρ d ρ ) = 2π τ Y ρ 2 d ρ T = 2π τ Y C2 ρ2 dρ = C1 = Unloading. 2π τ Y c23 − c13 3 ( ) 2π (145 × 106 )[(30 × 10−3 )3 − (12.5 × 10−3 )3 ] = 7.6064 × 103 N ⋅ m 3 T ′ = 7.6064 × 103 N ⋅ m J = π (c 2 4 2 ) − c14 = π 2 [(30)4 − (12.5)4 ] = 1.234 × 106 mm 4 = 1.234 × 10−6 m 4 τ1′ = (7.6064 × 103 )(12.5 × 10−3 ) T ′c1 = = 77.050 × 106 Pa = 77.05 MPa J 1.234 × 10−6 τ 2′ = (7.6064 × 103 )(30 × 10−3 ) T ′c2 = = 192.63 × 106 Pa = 192.63 MPa J 1.234 × 10−6 Residual stress. Inner surface: τ res = τ Y − τ1′ = 145 − 77.05 = 67.95 MPa Outer surface: τ res = τ Y − τ 2′ = 145 − 192.63 = −47.63 MPa Maximum residual stress: 68.0 MPa at inner surface. 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 3.119 In Prob. 3.118, determine the permanent angle of twist of the rod. PROBLEM 3.118 The hollow shaft shown is made of a steel that is and assumed to be elastoplastic with τ Y = 145 MPa G = 77.2 GPa. The magnitude T of the torques is slowly increased until the plastic zone first reaches the inner surface of the shaft; the torques are then removed. Determine the magnitude and location of the maximum residual shearing stress in the rod. SOLUTION 1 d1 = 12.5 mm 2 1 c2 = d 2 = 30 mm 2 c1 = When the plastic zone reaches the inner surface, the stress is equal to τ Y . The corresponding torque is calculated by integration. dT = ρτ dA = ρτ Y (2πρ d ρ ) = 2πτ Y ρ 2 d ρ T = 2πτ Y = c2 c1 ρ2 dρ = 2π τ Y c23 − c13 3 ( ) 2π (145 × 106 )[(30 × 10−3 )3 − (12.5 × 10−3 )3 ] = 7.6064 × 103 N ⋅ m 3 Rotation angle at maximum torque. c1ϕmax τ = γY = Y L G ϕmax = Unloading. τY L Gc1 = (145 × 106 )(5) = 0.75130 rad (77.2 × 109 )(12.5 × 10−3 ) T ′ = 7.6064 × 103 N ⋅ m J = ϕ′ = π (c 2 4 2 ) − c14 = π 2 [(30)4 − (12.5)4 ] = 1.234 × 106 mm 4 = 1.234 × 10−6 m 4 (7.6064 × 103 )(5) T ′L = = 0.39922 rad GJ (77.2 × 109 )(1.234 × 10−6 ) Permanent angle of twist. ϕperm = ϕmax − ϕ ′ = 0.75130 − 0.39922 = 0.35208 rad ϕperm = 20.2° 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 3.120 A torque T applied to a solid rod made of an elastoplastic material is increased until the rod is fully plastic and them removed. (a) Show that the distribution of residual shearing stresses is as represented in the figure. (b) Determine the magnitude of the torque due to the stresses acting on the portion of the rod located within a circle of radius c0. SOLUTION (a) ρY = 0, Tload = After loading: Tc 2T 2(Tload ) 4 = = = τY 3 3 J πc π c3 4 ρ τ ′ = τY 3 c τ′ = Unloading: 4 3 4ρ = τ Y 1 − 3c c τ res = τ Y − τ Y To find c0 set, τ res = 0 and ρ = c0 T0 = 2π c0 0 ρ 2τ d ρ = 2π ρ3 4 ρ4 = 2πτ Y − 3 3 4c (3/4) c 0 (3/4) c 0 at ρ = c ρ Residual: 0 =1− (b) 4 4π 3 2π 3 TY = c τY = c τY 3 32 3 4c0 3 ∴ c0 = c 3c 4 c0 = 0.150c 4ρ dρ 3 c 1 3 3 4 1 3 4 = 2πτ Y c3 − 3 4 3 4 4 ρ 2τ Y 1 − 27 9π 9 τ Y c3 = 0.2209 τ Y c3 = 2πτ Y c3 − = 64 256 128 T0 = 0.221τ Y c3 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 3.121 Determine the largest torque T that can be applied to each of the two brass bars shown and the corresponding angle of twist at B, knowing that τ all = 12 ksi and G = 5.6 × 106 psi. SOLUTION L = 25 in., G = 5.6 × 106 psi, τ all = 12 × 103 psi τ max = ϕ = (a) (b) T c1ab 2 or T = c1ab 2τ max TL c2ab3G or ϕ = a = 4 in., b = 1 in., a = 4.0 b c1Lτ max c2bG From Table 3.1: c1 = 0.282, From (1): T = (0.282)(4)(1)2 (12 × 103 ) = 13.54 × 103 From (2): ϕ = a = 2.4in., (1) (2) c2 = 0.281 T = 13.54 kip ⋅ in (0.282)(25)(12 × 103 ) = 0.05376 radians (0.281)(1)(5.6 × 106 ) b = 1.6in., a = 1.5 b From Table 3.1: c1 = 0.231, ϕ = 3.08° c2 = 0.1958 From (1): T = (0.231)(2.4)(1.6)2 (12 × 103 ) = 17.03 × 103 T = 17.03 kip ⋅ in From (2): ϕ = (0.231)(25)(12 × 103 ) = 0.0395 radians (0.1958)(1.6)(5.6 × 106 ) ϕ = 2.26° 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 3.122 Each of the two brass bars shown is subjected to a torque of magnitude T = 12.5 kip · in. Knowing that G = 5.6 × 106 psi, determine for each bar the maximum shearing stress and the angle of twist at B. SOLUTION (a) L = 25 in., G = 5.6 × 106 psi, a = 4 in, b = 1 in, From Table 3.1: τ max = ϕ = T = 12.5 × 103 lb ⋅ in a = 4.0 b c1 = 0.282, c2 = 0.281 T 12.5 × 103 = = 11.08 × 103 (0.282)(4)(1) 2 c1ab 2 τ max = 11.08 ksi TL (12.5 × 103 )(25) = = 0.04965 radians (0.282)(4)(1)3 (5.6 × 106 ) c2ab3G ϕ = 2.84° (b) a = 2.4 in., From Table 3.1: τ max = ϕ = b = 1.6 in., a = 1.5 b c1 = 0.231, c2 = 0.1958 T 12.5 × 103 = = 8.81 × 103 2 2 (0.231)(2.4)(1.6) c1ab τ max = 8.81 ksi TL (12.5 × 106 )(25) = = 0.02899 radians 3 (0.1958)(2.4)(1.6)3 (5.6 × 106 ) c2ab G ϕ = 1.661° 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 3.123 Each of the two aluminium bars shown is subjected to a torque of magnitude T = 1800 N ⋅ m. Knowing that G = 26 GPa, determine for each bar the maximum shearing stress and the angle of twist at B. SOLUTION T = 1800 N ⋅ m (a) τ max = ϕ = c1 = 0.208, c2 = 0.1406 T 1800 = = 40.1 × 106 Pa 2 2 c1ab (0.208)(0.060)(0.060) TL (1800)(0.300) = = 0.011398 radians 3 c2ab G (0.1406)(0.060)(0.060)3 (26 × 109 ) a = 95 mm = 0.095 m, From Table 3.1: τ max = ϕ = G = 26 × 109 Pa a = 1.0 b a = b = 60 mm = 0.060 m From Table 3.1: (b) L = 0.300 m b = 38 mm = 0.038 m, c1 = 0.258, τ max = 40.1 MPa ϕ = 0.653° a = 2.5 b c2 = 0.249 T 1800 = = 50.9 × 106 Pa 2 2 c1ab (0.258)(0.095)(0.038) TL (1800)(0.300) = = 0.01600 radians 3 c2ab G (0.249)(0.095)(0.038)3 (26 × 109 ) τ max = 50.9 MPa ϕ = 0.917° 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 3.124 Determine the largest torque T that can be applied to each of the two aluminium bars shown and the corresponding angle of twist at B, knowing τ all = 50 MPa and G = 26 GPa. SOLUTION L = 0.300 m, G = 26 × 109 Pa, τ all = 50 × 106 Pa τ max = ϕ = (a) T c1ab 2 or TL c2ab 4G or a = b = 60 mm = 0.060 m, From Table 3.1: (b) c1 = 0.208, T = c1ab 2τ max ϕ = (1) c1Lτ max c2bG (2) a = 1.0 b c2 = 0.1406 From (1): T = (0.208)(0.060)(0.060)2 (50 × 106 ) = 2246 N ⋅ m From (2): ϕ = (0.208)(0.300)(50 × 106 ) = 0.01422 radians (0.1406)(0.060)(26 × 109 ) a = 95 mm = 0.095 m, b = 38 mm = 0.038 m, From Table 3.1: T = 2.25 kN ⋅ m ϕ = 0.815° a = 2.5 b c1 = 0.258, c2 = 0.249 From (1): T = (0.258)(0.095)(0.038) 2 (50 × 106 ) = 1770 N ⋅ m From (2): ϕ = (0.258)(0.300)(50 × 106 ) = 0.01573 radians (0.249)(0.038)(26 × 109 ) T = 1.770 kN ⋅ m ϕ = 0.901° 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 3.125 Determine the largest allowable square cross section of a steel shaft of length 20 ft if the maximum shearing stress is not to exceed 10 ksi when the shaft is twisted through one complete revolution. Use G = 11.2 × 106 psi. SOLUTION L = 20 ft = 240 in. τ max = 10 ksi = 10 × 103 psi ϕ = 1 rev = 2π radians τ max = T c1ab 2 (1) TL c2ab3G (2) ϕ = Divide (2) by (1) to eliminate T. ϕ τ max = c1ab 2 L cL = 1 3 c2bG c2ab G c1Lτ max c2Gϕ Solve for b. b= For a square section, a = 1.0 b From Table 3.1, c1 = 0.208, c2 = 0.1406 b= (0.208)(240)(10 × 103 ) (0.1406)(11.2 × 106 )(2π ) b = 0.0505 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 3.126 Determine the largest allowable length of a stainless steel shaft of 3 3 × 8 4 -in. cross section if the shearing stress is not to exceed 15 ksi when the shaft is twisted through 15°. Use G = 11.2 × 106 psi. SOLUTION 3 in. = 0.75 in. 4 3 b = in. = 0.375 in. 8 a= τ max = 15 ksi = 15 × 103 psi ϕ = 15° = τ max = ϕ = Divide (2) by (1) to eliminate T. Solve for L. ϕ τ max = L= 15π rad = 0.26180 rad 180 T c1ab 2 (1) TL c2ab3G (2) c1ab 2 L cL = 1 3 c2bG c2ab G c2bGϕ c1τ max a 0.75 = =2 0.375 b Table 3.1 gives c1 = 0.246, c2 = 0.229 L= (0.229)(0.375)(11.2 × 106 )(0.26180) = 68.2 in. (0.246)(15 × 103 ) L = 68.2 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 3.127 The torque T causes a rotation of 2° at end B of the stainless steel bar shown. Knowing that b = 20 mm and G = 75GPa, determine the maximum shearing stress in the bar. SOLUTION a = 30 mm = 0.030 m b = 20 mm = 0.020 m ϕ = 2° = 34.907 × 10−3 rad ϕ = TL c2ab3Gϕ T ∴ = L c2ab3G τ max = T c2ab3Gϕ c bGϕ = = 2 c1L c1ab 2 c1ab 2 L 30 a = = 1.5. 20 b From Table 3.1, c1 = 0.231 c2 = 0.1958 τ max = (0.1958)(20 × 10−3 )(75 × 109 )(34.907 × 10−3 ) = 59.2 × 106 Pa (0.231)(750 × 10−3 ) τ max = 59.2 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 3.128 The torque T causes a rotation of 0.6° at end B of the aluminum bar shown. Knowing that b = 15 mm and G = 26 GPa, determine the maximum shearing stress in the bar. SOLUTION a = 30 mm = 0.030 m b = 15 mm = 0.015 m ϕ = 0.6° = 10.472 × 10−3 rad ϕ= c2 ab3Gϕ TL ∴ = T c1 L c2 ab3G τ max = c2 ab3Gϕ c2 bGϕ T = = c1 L c1ab 2 c1ab 2 L a 30 = = 2.0 b 15 From Table 3.1, c1 = 0.246 c2 = 0.229 τ max = (0.229)(15 × 10−3 )(26 × 109 )(10.472 × 10−3 ) (0.246)(750 × 10−3 ) = 5.07 × 106 Pa τ max = 5.07 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 3.129 Two shafts are made of the same material. The cross section of shaft A is a square of side b and that of shaft B is a circle of diameter b. Knowing that the shafts are subjected to the same torque, determine the ratio τ A /τ B of maximum shearing stresses occurring in the shafts. SOLUTION a = 1, c1 = 0.208 (Table 3.1) b T T = τA = 2 0.208b3 c1ab A. Square: B. Circle: c= Ratio: τA 1 π = ⋅ = 0.3005π τ B 0.208 16 1 Tc 2T 16T b τB = = = 2 J π c3 π b3 τA = 0.944 τB 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 3.130 Shafts A and B are made of the same material and have the same cross-sectional area, but A has a circular cross section and B has a square cross section. Determine the ratio of the maximum shearing stresses occurring in A and B, respectively, when the two shafts are subjected to the same torque (TA = TB ). Assume both deformations to be elastic. SOLUTION Let c be the radius of circular section A and b be the side square section B. For equal areas, π c2 = b2 TAC 2T = A3 J πc Circle: τA = Square: a =1 b τB = b=c π c1 = 0.208 from Table 3.1 TB T = B3 c1ab 2 c1b Ratio: τ A 2TA c1b3 2c1b3 TA T = = = 2c1 π A 3 3 TB τ B π c TB π c TB For TA = TB , τA = (2)(0.208) π τB τA = 0.737 τB 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 3.131 Shafts A and B are made of the same material and have the same crosssectional area, but A has a circular cross section and B has a square cross section. Determine the ratio of the maximum torques TA and TB that can be safely applied to A and B, respectively. SOLUTION Let c = radius of circular section A and b = side of square section B. For equal areas π c 2 = b 2 , c= Circle: τA = b π TAc 2TA = J π c3 ∴ TA = π 2 c3τ A Square: From Table 3.1, c1 = 0.208 τB = TA T = B3 c1ab 2 c1b π c3τ B Ratio: TA = 2 3 TB c1b τ B For the same stresses, τB = τ A ∴ ∴ TB = c1b3τ B π ⋅ b3 τ 1 τA 2 π 3/2 B = = 3 c1b τ B 2c1 π τ B TA 1 = TB (2) (0.208) π TA = 1.356 TB 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 3.132 Shafts A and B are made of the same material and have the same length and cross-sectional area, but A has a circular cross section and B has a square cross section. Determine the ratio of the maximum values of the angles ϕ A and ϕ B through which shafts A and B, respectively, can be twisted. SOLUTION Let c = radius of circular section A and b = side of square section B. For equal areas, π c2 = b2 ∴ b = π c Circle: γ max = Square: From Table 3.1, Ratio: For equal stresses, τ A = τ B τA G = cϕ A L c1 = 0.208, ∴ ϕA = Lτ A cG c2 = 0.1406 τB = TB TB = 2 c1ab 0.208 b3 ϕB = 0.208 b3τ B L 1.4794 Lτ B TB L = = bG 0.1406 b 4G c2ab3G ∴ TB = 0.208 b3τ B ϕ A Lτ A bG bτ τ = ⋅ = 0.676 A = 0.676 π A cG 1.4794Lτ B cτ B ϕB τB ϕB = 0.676 π ϕA ϕB = 1.198 ϕA 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 3.133 Each of the three aluminum bars shown is to be twisted through an angle of 2°. Knowing that b = 30 mm, τ all = 50 MPa, and G = 27 GPa, determine the shortest allowable length of each bar. SOLUTION ϕ = 2° = 34.907 × 10−3 rad, τ = 50 × 106 Pa τ = For square and rectangle, T c1ab2 Divide to eliminate T; then solve for L. (a) Square: L= (b) a = 1.0 b From Table 3.1, c= cGϕ τ Rectangle: L= TL c2ab3G ϕ c ab2 L = 1 3 τ c2ab G L= = c2bGϕ c1τ c1 = 0.208, c2 = 0.1406 L = 382 mm 1 Tc TL b = 0.015 m τ = ϕ = 2 J GJ Divide to eliminate T; then solve for L. (c) ϕ = (0.1406)(0.030)(27 × 109 )(34.907 × 10−3 ) = 382 × 10−3 m 6 (0.208)(50 × 10 ) Circle: L= G = 27 × 109 Pa, b = 30 mm = 0.030 m ϕ JL L = = cGJ cG τ (0.015)(27 × 109 )(34.907 × 10−3 ) = 283 × 10−3 m 50 × 106 a = 1.2b a = 1.2 b From Table 3.1, (0.1661)(0.030)(27 × 109 )(34.907 × 10−3 ) = 429 × 10−3 m (0.219)(50 × 106 ) L = 283 mm c1 = 0.219, c2 = 0.1661 L = 429 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 3.134 Each of the three steel bars is subjected to a torque as shown. Knowing that the allowable shearing stress is 8 ksi and that b = 1.4 in., determine the maximum torque T that can be applied to each bar. SOLUTION τ max = 8 ksi, b = 1.4 in. (a) a = b = 1.4 in. Square: a = 1.0 b c1 = 0.208 From Table 3.1, τ max = T c1ab 2 T = c1ab 2τ max T = (0.208)(1.4)(1.4)(1.4) 2 (8) (b) Circle: 1 b = 0.7 in. 2 2T π Tc = = T = c3τ max 3 J 2 πc c= τ max T = (c) T = 4.57 kip ⋅ in π 2 (0.7)3 (8) Rectangle: a = (1.2)(1.4) = 1.68 in. From Table 3.1, c1 = 0.219 T = 4.31 kip ⋅ in a = 1.2 b T = c1ab 2τ max = (0.219)(1.68)(1.4)2 (8) T = 5.77 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 3.135 A 36-kip ⋅ in. torque is applied to a 10-ft-long steel angle with an L8 × 8 × 1 cross section. From Appendix C, we find that the thickness of the section is 1 in. and that its area is, 15.00 in2. Knowing that G = 11.2 × 106 psi, determine (a) the maximum shearing stress along line a-a, (b) the angle of twist. SOLUTION a = (a) A 15 in 2 = = 15 in., b = 1 in., t 1 in. Since a 1 b > 5, c1 = c2 = 1 − 0.630 b a 3 or c1 = c2 = 1 0.630 = 0.3193 1− 3 15 T = 36 × 103 lb ⋅ in; L = 120 in.; G = 11.2 × 106 psi T Maximum shearing stress: τ max = c1ab 2 τ max = (b) a = 15 b Angle of twist: 36 × 103 = 7.52 × 103 psi 2 (0.3193)(15)(1) ϕ = TL c2ab 2G ϕ = (36 × 103 )(120) = 0.08052 radians (0.3193)(15)(1)3 (11.2 × 106 ) τ max = 7.52 ksi ϕ = 4.61° 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 3.136 A 3-m-long steel angle has an L203 × 152 × 12.7 cross section. From Appendix C, we find that the thickness of the section is 12.7 mm and that its area is 4350 mm2. Knowing that τ all = 50 MPa and that G = 77.2 GPa, and ignoring the effect of stress concentration, determine (a) the largest torque T that can be applied, (b) the corresponding angle of twist. SOLUTION A = 4350 mm 2 b = 12.7 mm a = ? a = Equivalent rectangle. A 4350 = = 342.52 mm b 12.7 a = 26.97 b b 1 c1 = c2 = 1 − 0.630 = 0.32555 a 3 (a) τ max = T c1ab 2 τ max = 50 × 106 Pa T = c1ab2τ max = (0.32555) (26.97 × 10−3 ) (12.7 × 10−3 ) 2 (50 × 106 ) = 70.807 N ⋅ m (b) ϕ = T = 70.8 N ⋅ m TL (70.807) (3) = 3 c2ab G (0.32555)(26.97 × 10−3 )(12.7 × 10−3 ) (77.2 × 109 ) = 0.15299 rad ϕ = 8.77° 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 3.137 An 8-ft-long steel member with a W8 × 31 cross section is subjected to a 5-kip ⋅ in. torque. The properties of the rolled-steel section are given in Appendix C. Knowing that G = 11.2 × 106 psi, determine (a) the maximum shearing stress along line a-a, (b) the maximum shearing stress along line b-b, (c) the angle of twist. (Hint: consider the web and flanges separately and obtain a relation between the torques exerted on the web and a flange, respectively, by expressing that the resulting angles of twist are equal.) SOLUTION a 7.995 = = 18.38 b 0.435 Tf L 1 c1 = c2 = 1 − 0.630 b = 0.3219 ϕf = a 3 c2ab3G Gϕ f Gϕ T f = c2ab3 = Kf where K f = c2ab3 L L K f = (0.3219) (7.995) (0.435)3 = 0.2138 in 3 a = 7.995 in., b = 0.435 in., Flange: ) ( a = 8.0 − (2)(0.435) = 7.13 in., b = 0.285 in., Web: a 7.13 = = 25.02 b 0.285 b Tw L c1 = c2 = 1 1 − 0.630 = 0.3249 ϕw = 3 a c2ab3G Gϕ w Gϕ Tw = c2ab3 K w = c2ab3 = Kw where L L K w = (0.3249) (7.13) (0.285)3 = 0.0563 in 4 ϕ f = ϕw = ϕ For matching twist angles: T = 2T f + Tw = (2 K f + K w ) Total torque. Gϕ T = , 2K p + K w L Tf = Tf = KfT 2K f + K w , K wT 2K f + K w (0.2138)(5000) (0.0563) (5000) = 2221 lb ⋅ in; Tw = = 557 lb ⋅ in (2) (0.2138) + 0.0563 (2) (0.2138) + 0.0563 Tf 2221 = 4570 psi (0.3219)(7.995)(0.435) 2 τ f = 4.57 ksi Tw 557 = = 2960 psi 2 c1ab (0.3249)(7.13)(0.285)2 τ w = 2.96 ksi (a) τf = (b) τw = (c) Gϕ T TL = ∴ ϕ = L 2K f + K w G(2 K f + K w ) ϕ = Tw = Gϕ L c1ab 2 = where L = 8 ft = 96 in. (5000)(96) = 88.6 × 10−3 rad (11.2 × 10 )[(2)(0.2138) + 0.563] 6 ϕ = 5.08° 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 3.138 A 4-m-long steel member has a W310 × 60 cross section. Knowing that G = 77.2 GPa and that the allowable shearing stress is 40 MPa, determine (a) the largest torque T that can be applied, (b) the corresponding angle of twist. Refer to Appendix C for the dimensions of the cross section and neglect the effect of stress concentrations. (See hint of Prob. 3.137.) SOLUTION W310 × 60, L = 4 m, G = 77.2 G Pa, τ all = 40 MPa For one flange: From App. C, Eq. (3.45): c1 = c2 = Eq. (3.44): φf = Tf L 3 c2ab G = a = 203 mm, b = 13.1 mm, a/b = 15.50 1 0.630 = 0.320 1− 3 15.50 T f (4) 0.320(0.203)(0.0131)3 (77.2 × 109 ) (1) φ f = 355.04 × 10−6T f For web: From App. C, Eq. (3.45): c1 = c2 = Eq. (3.44): φw = a = 303 − 2(13.1) = 276.8 mm, b = 7.5 mm, a/b = 36.9 1 0.630 = 0.328 1− 3 36.9 Tw (4) 0.328(0.2768)(0.0075)3 (77.2 × 109 ) φw = 1.354.2 × 10−6Tw (2) Since angle of twist is the same for flanges and web: φ f = φ w: 355.04 × 10−6T f = 1354.2 × 10−6Tw T f = 3.814Tw (3) But the sum of the torques exerted on the two flanges and on the web is equal to the torque T applied to the member: 2T f + Tw = T (4) 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 3.138 (Continued) Substituting for T f from (3) into (4): 2(3.814Tw ) + Tw = T From (3): T f = 3.814(0.11589T ) Tw = 0.11589T T f = 0.44205T (5) (6) For one flange: From Eq. (3.43): T f = c1ab 2τ max = 0.320(0.203)(0.0131) 2 (40 × 106 ) = 445.91 N ⋅ m Eq. (6): 445.91 = 0.44205T For web: Tw = c1ab 2τ max = 0.328(0.2768)(0.0075)2 (40 × 106 ) T = 1009 N ⋅ m = 204.28 N ⋅ m Eq. (5): 204.28 = 0.11589T (a) Largest allowable torque: (b) Angle of twist: Use T f , which is critical. Eq. (1): Use the smaller value. φ = φ f = (355.04 × 10−6 )(445.91) = 0.15831 rad T = 1763 N ⋅ m T = 1009 N ⋅ m φ = 9.07° 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 3.139 A torque T = 750 kN ⋅ m is applied to a hollow shaft shown that has a uniform 8-mm wall thickness. Neglecting the effect of stress concentrations, determine the shearing stress at points a and b. SOLUTION Detail of corner. 1 t = e tan 30° 2 t e= 2 tan 30° = 8 = 6.928 mm 2 tan 30° b = 90 − 2e = 76.144 mm Area bounded by centerline. a = 1 3 3 2 3 (76.144)2 b b= b = 2 2 4 4 = 2510.6 mm 2 = 2510.6 × 10−6 m 2 t = 0.008 m τ = T 750 = = 18.67 × 106 Pa −6 2ta (2)(0.008) (2510 × 10 ) τ = 18.67 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 3.140 A torque T = 5 kN ⋅ m is applied to a hollow shaft having the cross section shown. Neglecting the effect of stress concentrations, determine the shearing stress at points a and b. SOLUTION T = 5 × 103 N ⋅ m Area bounded by centerline. a = bh = (69) (115) = 7.935 × 103 mm 2 = 7.935 × 10−3 m 2 At point a: t = 6 mm = 0.006 m τ = T 5 × 103 = 2ta (2) (0.006) (7.935 × 10−3 ) = 52.5 × 106 Pa At point b: τ = 52.5 MPa t = 10 mm = 0.010 m τ = T 5 × 103 = = 31.5 × 106 Pa −3 2ta (2)(0.010)(7.935 × 10 ) τ = 31.5 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 3.141 A 90-N ⋅ m torque is applied to a hollow shaft having the cross section shown. Neglecting the effect of stress concentrations, determine the shearing stress at points a and b. SOLUTION Area bounded by centerline. a = 52 × 52 − 39 × 39 + π 4 (39)2 = 2378 mm 2 = 2.378 × 10−3 m 2 T = 90 N ⋅ m τa = T 90 N ⋅ m = −3 2ta 2(4 × 10 m)(2.378 × 10−3 m 2 ) τ a = 4.73 MPa τb = T 90 N ⋅ m = −3 2ta 2(2 × 10 m)(2.378 × 10−3m 2 ) τ b = 9.46 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 3.142 A 5.6 kN ⋅m torque is applied to a hollow shaft having the cross section shown. Neglecting the effect of stress concentrations, determine the shearing stress at points a and b. SOLUTION Area bounded by centerline. a = (96 mm)(95mm) + π 2 (47.5mm)2 = 12.664 × 103 mm2 = 12.664 × 10−3 m 2 t = 5 mm = 0.005m At point a, τ = 5.6 × 103 T = = 44.2 × 106 Pa −3 2at (2)(12.664 × 10 )(0.005) τ = 44.2 MPa t = 8 mm = 0.008m At point b, τ = T 5.6 × 103 = = 27.6 × 106 Pa 2at (2)(12.664 × 10−3 )(0.008) τ = 27.6 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 3.143 A hollow member having the cross section shown is formed from sheet metal of 2-mm thickness. Knowing that the shearing stress must not exceed 3 MPa, determine the largest torque that can be applied to the member. SOLUTION Area bounded by centerline. a = (48)(18) + (30)(18) = 1404 mm 2 = 1404 × 10−6 m 2 t = 0.002 m τ = T 2ta or T = 2taτ = (2)(0.002)(1404 × 10−6 )(3 × 106 ) T = 16.85 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 3.144 A hollow brass shaft has the cross section shown. Knowing that the shearing stress must not exceed 12 ksi and neglecting the effect of stress concentrations, determine the largest torque that can be applied to the shaft. SOLUTION Calculate the area bounded by the center line of the wall cross section. The area is a rectangle with two semicircular cutouts. b = 5 − 0.2 = 4.8 in. h = 6 − 0.5 = 5.5 in. r = 1.5 + 0.1 = 1.6 in. π a = bh − 2 r 2 = (4.8)(5.5) − π (1.6) 2 = 18.3575 in 2 2 T τ max = τ max = 12 × 103 psi tmin = 0.2 in. 2atmin T = 2atmin τ max = (2)(18.3575) (0.2)(12 × 103 ) = 88.116 × 103 lb ⋅ in T = 88.1kip ⋅ in = 7.34 kip ⋅ ft 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 3.145 A hollow member having the cross section shown is to be formed from sheet metal of 0.06 in. thickness. Knowing that a 1250 lb ⋅ in.-torque will be applied to the member, determine the smallest dimension d that can be used if the shearing stress is not to exceed 750 psi. SOLUTION Area bounded by centerline. a = (5.94)(2.94 − d ) + 1.94 d = 17.4636 − 4.00 d t = 0.06 in., τ = 750 psi, T = 1250 lb ⋅ in T τ= 2ta a= 17.4636 − 4.00d = d = T 2tτ 1250 = 13.8889 (2) (0.06) (750) 3.5747 = 0.894 in. 4.00 d = 0.894 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 3.146 A hollow member having the cross section shown is to be formed from sheet metal of 0.06 in. thickness. Knowing that a 1250 lb ⋅ in.-torque will be applied to the member, determine the smallest dimension d that can be used if the shearing stress is not to exceed 750 psi. SOLUTION Area bounded by centerline. a = (5.94)(2.94) − 2.06 d = 17.4636 − 2.06 d t = 0.06 in., τ = 750 psi, T = 1250 lb ⋅ in τ = T 2ta a= 17.4636 − 2.06d = d = T 2tτ 1250 = 13.8889 (2) (0.06) (750) 3.5747 = 1.735 in. 2.06 d = 1.735 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 3.147 A hollow cylindrical shaft was designed to have a uniform wall thickness of 0.1 in. Defective fabrication, however, resulted in the shaft having the cross section shown. Knowing that a 15 kip ⋅ in.-torque is applied to the shaft, determine the shearing stresses at points a and b. SOLUTION Radius of outer circle = 1.2 in. Radius of inner circle = 1.1 in. Mean radius = 1.15 in. Area bounded by centerline. a = π rm2 = π (1.15) 2 = 4.155 in 2 t = 0.08 in. At point a, τ = T 15 = 2ta (2)(0.08)(4.155) τ = 22.6 ksi t = 0.12 in. At point b, τ = T 15 = 2ta (2)(0.12)(4.155) τ = 15.04 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 3.148 A cooling tube having the cross section shown is formed from a sheet of stainless steel of 3-mm thickness. The radii c1 = 150 mm and c2 = 100 mm are measured to the center line of the sheet metal. Knowing that a torque of magnitude T = 3 kN ⋅ m is applied to the tube, determine (a) the maximum shearing stress in the tube, (b) the magnitude of the torque carried by the outer circular shell. Neglect the dimension of the small opening where the outer and inner shells are connected. SOLUTION Area bounded by centerline. ( ) a = π c12 − c22 = π (1502 − 1002 ) = 39.27 × 103 mm 2 = 39.27 × 10−3 m 2 t = 0.003 m T 3 × 103 = = 12.73 × 106 Pa 2ta (2)(0.003)(39.27 × 10−3 ) (a) τ = (b) T1 = (2π c1t τ c1) = 2π c12 t τ = 2π (0.150) 2 (0.003) (12.73 × 106 ) = 5.40 × 103 N ⋅ m τ = 12.76 MPa T1 = 5.40 kN ⋅ 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 3.149 A hollow cylindrical shaft of length L, mean radius cm , and uniform thickness t is subjected to a torque of magnitude T. Consider, on the one hand, the values of the average shearing stress τ ave and the angle of twist ϕ obtained from the elastic torsion formulas developed in Sections 3.4 and 3.5 and, on the other hand, the corresponding values obtained from the formulas developed in Sec. 3.13 for thin-walled shafts. (a) Show that the relative error introduced by using the thinwalled-shaft formulas rather than the elastic torsion formulas is the same for τ ave and ϕ and that the relative error is positive and proportional to the ratio t/cm . (b) Compare the percent error corresponding to values of the ratio t /cm of 0.1, 0.2, and 0.4. SOLUTION Let c2 = outer radius = cm + 1 2 t and c1 = inner radius = cm − 12 t π (c 2 4 2 ) − c14 = π ( ) c22 + c12 (c2 + c1)(c2 − c1) 2 1 1 π = cm2 + cmt + t 2 + cm2 − cmt + t 2 (2cm ) t 2 4 4 J = 1 = 2π cm2 + t 2 cmt 4 Tc T τm = m = 2 1 2 J 2π cm + t t 4 TL TL = ϕ1 = 2 1 2 JG 2π cm + t cmt G 4 Area bounded by centerline. a = π cm2 τ ave = ϕ2 = (a) Ratios: T T = 2ta 2π cm2 t TL ds TL(2π cm /t ) TL = = 2 2 2 4a G t 4(π cm ) G 2π cm3 t G ) ( 2π cm2 + 1 t 2 t 1 t2 T τ ave 4 1 = × = + T τm 4 cm2 2π cm2 t ( ) 2π cm2 + 1 t 2 cmtG 1 t2 TL ϕ2 4 1 = × = + TL ϕ1 2π cm3 tG 4 cm 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 3.149 (Continued) 2 (b) τ ave ϕ 1 t −1 = 2 −1 = τm ϕ1 4 cm2 t cm 0.1 0.2 0.4 1 t2 4 cm2 0.0025 0.01 0.04 % 0.25% 1% 4% 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 3.150 Equal torques are applied to thin-walled tubes of the same length L, same thickness t, and same radius c. One of the tubes has been slit lengthwise as shown. Determine (a) the ratio τ b / τ a of the maximum shearing stresses in the tubes, (b) the ratio ϕb /ϕ a of the angles of twist of the shafts. SOLUTION Without slit: Area bounded by centerline. a = π c 2 τa = T T = 2ta 2π c 2t J ≈ 2π c3t With slit: ϕa = a = 2π c, b = t , c1 = c2 = TL TL = GJ 2π c3t G a 2π c = >> 1 b t 1 3 τb = 3T T = 2 2π ct 2 c1ab ϕb = 3TL T = 3 2π ct 3G c2ab G (a) Stress ratio: τb 3T 2π c 2t 3c = ⋅ = 2 τa T t 2π ct (b) Twist ratio: ϕb 3TL 2π c3t G 3c 2 = ⋅ = 2 ϕa TL 2π ct 3G t τ b 3c = t τa ϕb 3c 2 = 2 ϕa t 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 3.151 The ship at A has just started to drill for oil on the ocean floor at a depth of 5000 ft. Knowing that the top of the 8-in.-diameter steel drill pipe (G = 11.2 × 106 psi) rotates through two complete revolutions before the drill bit at B starts to operate, determine the maximum shearing stress caused in the pipe by torsion. SOLUTION TL GJ ϕ T = GJ L Tc GJ ϕ c Gϕ c τ = = = J JL L ϕ = ϕ = 2 rev = (2)(2π ) = 12.566 rad, c= 1 d = 4.0 in. 2 L = 5000 ft = 60000 in. τ = (11.2 × 106 )(12.566)(4.0) = 9.3826 × 103 psi 60000 τ = 9.38 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 3.152 The shafts of the pulley assembly shown are to be redesigned. Knowing that the allowable shearing stress in each shaft is 8.5 ksi, determine the smallest allowable diameter of (a) shaft AB, (b) shaft BC. SOLUTION (a) Shaft AB: TAB = 3.6 × 103 lb ⋅ in τ max = 8.5 ksi = 8.5 × 103 psi J = c= (b) Shaft BC: π 2 3 c4 2TAB πτ max τ max = = 3 2T Tc = J π c3 (2)(3.6 × 103 ) = 0.646 in. π (8.5 × 103 ) d AB = 2c = 1.292 in. TBC = 6.8 × 103 lb ⋅ in τ max = 8.5 × 103 psi c= 3 2TBC πτ max = 3 (2)(6.8 × 103 ) = 0.7985 in. π (8.5 × 103 ) d BC = 2c = 1.597 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 3.153 A steel pipe of 12-in. outer diameter is fabricated from 1 4 -in. -thick plate by welding along a helix which forms an angle of 45° with a plane perpendicular to the axis of the pipe. Knowing that the maximum allowable tensile stress in the weld is 12 ksi, determine the largest torque that can be applied to the pipe. SOLUTION From Eq. (3.14) of the textbook, σ 45 = τ max hence, τ max = 12 ksi = 12 × 103 psi 1 1 d0 = (12) = 6.00 in. 2 2 c1 = c2 − t = 6.00 − 0.25 = 5.75 in. c2 = J = τ max = T = π (c 2 Tc J 4 2 ) − c14 = T = π 2 [(6.00)4 − (5.75)4 ] = 318.67 in. τ max J c 3 (12 × 10 )(318.67) = 637 × 103 lb ⋅ in 6.00 T = 637 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 3.154 For the gear train shown, the diameters of the three solid shafts are: d AB = 20 mm dCD = 25mm d EF = 40 mm Knowing that for each shaft the allowable shearing stress is 60 MPa, determine the largest torque T that can be applied. SOLUTION Statics: TAB = T TCD T = AB rC rB TCD = rC 75 TAB = T = 2.5T rB 30 TEF T = CD rF rD TEF = rF 90 TCD = (2.5T ) = 7.5T rD 30 Determine the magnitude of T so that the stress is 60 MPa = 60 × 106 Pa. Shaft AB: Tc J c= 1 d AB = 10 mm = 0.010 m 2 TAB = T = Shaft CD: c= Tshaft = π 2 c= (60 × 106 )(0.010)3 T = 94.2 N ⋅ m 1 dCD = 12.5 mm = 0.0125 m 2 TCD = 2.5T = Shaft EF: π Jτ = τ c3 2 c τ = π 2 (60 × 106 )(0.0125)3 T = 73.6 N ⋅ m 1 d EF = 20 mm = 0.020 m 2 TEF = 7.5T = π 2 (60 × 106 )(0.020)3 T = 100.5 N ⋅ m The smallest value of T is the largest torque that can be applied. T = 73.6 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 3.155 Two solid steel shafts (G = 77.2GPa) are connected to a coupling disk B and to fixed supports at A and C. For the loading shown, determine (a) the reaction at each support, (b) the maximum shearing stress in shaft AB, (c) the maximum shearing stress in shaft BC. SOLUTION Shaft AB: 1 d = 25 mm = 0.025 m 2 π π T L = c 4 = (0.025) 4 = 613.59 × 10−9 m 4 ϕ B = AB AB 2 2 GJ AB T = TAB , LAB = 0.200 m, c = J AB TAB = Shaft BC: (77.2 × 109 )(613.59 × 10−9 ) GJ AB ϕB = ϕ B = 236.847 × 103 ϕ B 0.200 LAB 1 d = 19 mm = 0.019 m 2 π π T L ϕ B = BC BC = c 4 = (0.019) 4 = 204.71 × 10−9 m 4 2 2 GJ BC T = TBC , LBC = 0.250 m, c = J BC TBC = GJ BC (77.2 × 109 )(204.71 × 10−9 ) = 63.214 × 103ϕ B ϕB = 0.250 LBC Equilibrium of coupling disk. T = TAB + TBC 1.4 × 103 = 236.847 × 103ϕ B + 63.214 × 103ϕ B ϕ B = 4.6657 × 10−3 rad. TAB = (236.847 × 103 )(4.6657 × 10−3 ) = 1.10506 × 103 N ⋅ m TBC = (63.214 × 103 )(4.6657 × 10−3 ) = 294.94 N ⋅ m (a) TA = TAB = 1105 N ⋅ m Reactions at supports. TC = TBC = 295 N ⋅ m (b) Maximum shearing stress in AB. τ AB = (c) TABc (1.10506 × 103 )(0.025) = = 45.0 × 106 Pa J AB 613.59 × 10−9 τ AB = 45.0 MPa Maximum shearing stress in BC. τ BC = TBC c (294.94)(0.019) = = 27.4 × 106 Pa J BC 204.71 × 10−9 τ BC = 27.4 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 3.156 In the bevel-gear system shown, α = 18.43°. Knowing that the allowable shearing stress is 8 ksi in each shaft and that the system is in equilibrium, determine the largest torque TA that can be applied at A. SOLUTION Using stress limit for shaft A: τ = 8 ksi, TA = c= 1 d = 0.25 in. 2 π π Jτ = τ c3 = (8)(0.25)3 = 0.1963 kip ⋅ in 2 2 c Using stress limit for shaft B: τ = 8 ksi, c = From statics, 1 d = 0.3125in. 2 TB = π π Jτ = τ c3 = (8)(0.3125)3 = 0.3835 kip ⋅ in 2 2 c TA = rA TB = (tan α )TB rB TA = (tan18.43°)(0.3835) = 0.1278 kip ⋅ in The allowable value of TA is the smaller. TA = 0.1278 kip ⋅ in TA = 127.8 lb ⋅ 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 3.157 Three solid shafts, each of 3 4 -in. diameter, are connected by the gears shown. Knowing that G = 11.2 × 106 psi, determine (a) the angle through which end A of shaft AB rotates, (b) the angle through which end E of shaft EF rotates. SOLUTION Geometry: rB = 1.5 in., rC = 6 in., rF = 2in. LAB = 48 in., LCD = 36 in., LEF = 48 in. Statics: TA = 100 lb ⋅ in TE = 200 lb ⋅ in Gear B. ΣM B = 0: −rB F1 + TA = 0 −1.5F1 + 100 = 0 F1 = 67.667 lb Gear F. ΣM F = 0: −rF F2 + TE = 0 −2F2 + 200 = 0 F2 = 100 lb Gear C. ΣM C = 0: −rC F1 − rC F2 + TC = 0 −(6)(66.667) − (6)(100) + TC = 0 TC = 1000 lb ⋅ 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 3.157 (Continued) Deformations: c= For all shafts, J = Kinematics: 1 d = 0.375 in. 2 π 2 c 4 = 0.031063 in 4 ϕ A/B = TAB LAB (100)(48) = = 0.013797 rad GJ (11.2 × 106 )(0.031063) ϕ E /F = TEF LEF (200)(48) = = 0.027594 rad GJ (11.2 × 106 )(0.031063) ϕ C /D = TCD LCD (1000)(36) = = 0.103476 rad GJ (11.2 × 106 )(0.031063) ϕC = ϕC/D = 0.103476 rad rBϕ B = rCϕC (a) rC 6 (0.103476) = 0.41390 rad ϕC rB 1.5 ϕ A = ϕ B + ϕ A/B = 0.41390 + 0.01397 = 0.42788 rad rFϕ F = rCϕC (b) ϕB = ϕF = ϕ A = 24.5° rC 6 ϕC = (0.103476) = 0.31043 rad rF 2 ϕ E = ϕ F + ϕ E/F = 0.31043 + 0.027594 = 0.33802 rad ϕ E = 19.37° 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 3.158 The design specifications of a 1.2-m-long solid transmission shaft require that the angle of twist of the shaft not exceed 4° when a torque of 750 N ⋅ m is applied. Determine the required diameter of the shaft, knowing that the shaft is made of a steel with an allowable shearing stress of 90 MPa and a modulus of rigidity of 77.2 GPa. SOLUTION T = 750 N ⋅ m, ϕ = 4° = 69.813 × 10−3 rad, L = 1.2 m, J = π 2 c4 τ = 90 MPa = 90 × 106 Pa G = 77.2 GPa = 77.2 × 109 Pa Based on angle of twist. ϕ = c= Based on shearing stress. τ = c= Use larger value. TL 2TL = GJ π Gc 4 4 2TL = π Gϕ 4 (2)(750)(1.2) = 18.06 × 10−3 m π (77.2 × 109 )(69.813 × 10−3 ) Tc 2T = J π c3 3 2T πτ = 3 (2)(750) = 17.44 × 10−3 m π (90 × 106 ) c = 18.06 × 10−3 m = 18.06 mm d = 2c = 36.1 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 3.159 The stepped shaft shown rotates at 450 rpm. Knowing that r = 0.5in., determine the maximum power that can be transmitted without exceeding an allowable shearing stress of 7500 psi. SOLUTION d = 5 in. D = 6 in. r = 0.5 in. D 6 = = 1.20 d 5 r 0.5 = = 0.10 d 5 K = 1.33 From Fig. 3.32, 1 d = 2.5 in. 2 KTc 2KT τ = = J π c3 c= For smaller side, T = π c3τ 2K = π (2.5)3 (7500) (2)(1.33) = 138.404 × 103 lb ⋅ in f = 450 rpm = 7.5 Hz Power. P = 2π f T = 2π (7.5)(138.404 × 103 ) = 6.52 × 106 in ⋅ lb/s Recalling that 1 hp = 6600 in ⋅ lb/s, P = 988 hp 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 3.160 A 750-N ⋅ m torque is applied to a hollow shaft having the cross section shown and a uniform 6-mm wall thickness. Neglecting the effect of stress concentrations, determine the shearing stress at points a and b. SOLUTION Area bounded by centerline. a =2 π 2 (33) 2 + (60)(66) = 7381 mm 2 = 7381 × 10−6 m 2 t = 0.006 m at both a and b, Then at points a and b, τ = T 750 = = 8.47 × 106 Pa 2ta (2) (0.006) (7381 × 10−6 ) τ = 8.47 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 3.161 The composite shaft shown is twisted by applying a torque T at end A. Knowing that the maximum shearing stress in the steel shell is 150 MPa, determine the corresponding maximum shearing stress in the aluminum core. Use G = 77.2 GPa for steel and G = 27 GPa for aluminum. SOLUTION Let G1, J1, and τ1 refer to the aluminum core and G2 , J 2 , and τ 2 refer to the steel shell. At the outer surface on the steel shell, γ2 = c2ϕ ϕ γ2 τ ∴ = = 2 L L c2 c2G2 At the outer surface of the aluminum core, γ1 = Matching ϕ L c1ϕ ϕ γ1 τ ∴ = = 1 L L c1 c1G for both components, τ2 c2G2 Solving for τ 2 , τ2 = = τ1 c1G1 0.030 27 × 109 c2 G2 ⋅ ⋅ ⋅ 150 × 106 = 39.3 × 106 Pa τ1 = 9 0.040 77.2 × 10 c1 G1 τ 2 = 39.3 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 3.162 Two solid brass rods AB and CD are brazed to a brass sleeve EF. Determine the ratio d 2 /d1 for which the same maximum shearing stress occurs in the rods and in the sleeve. SOLUTION Let c1 = 1 d1 2 Shaft AB: τ1 = Tc1 2T = J1 π c13 Sleeve EF. τ2 = Tc2 2Tc2 = J2 π c24 − c14 and ( 2T 2Tc2 = 3 π c1 π c24 − c14 For equal stresses, ( c2 = 1 d2 2 ) ) c24 − c14 = c13c2 Let x = c2 c1 x 4 − 1 = x or x = 41 + x Solve by successive approximations starting with x0 = 1.0. x1 = 4 2 = 1.189, x2 = x4 = 4 2.220 = 1.221, x5 = x = 1.221 4 2.189 = 1.216, x3 = c2 = 1.221 c1 4 4 2.216 = 1.220 2.221 = 1.221 (converged). d2 = 1.221 d1 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.