05 Solutions 46060 5/25/10 3:53 PM Page 214 © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher •5–1 A shaft is made of a steel alloy having an allowable shear stress of tallow = 12 ksi If the diameter of the shaft is 1.5 in., determine the maximum torque T that can be transmitted What would be the maximum torque T¿ if a 1-in.-diameter hole is bored through the shaft? Sketch the shear-stress distribution along a radial line in each case T T¿ Allowable Shear Stress: Applying the torsion formula tmax = tallow = 12 = Tc J T (0.75) p (0.754) T = 7.95 kip # in Ans Allowable Shear Stress: Applying the torsion formula tmax = tallow = 12 = T¿c J T¿ (0.75) p (0.754 - 0.54) T¿ = 6.381 kip # in = 6.38 kip # in tr = 0.5 in = T¿r = J 6.381(0.5) p (0.754 - 0.54) Ans = 8.00 ksi 214 05 Solutions 46060 5/25/10 3:53 PM Page 215 © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 5–2 The solid shaft of radius r is subjected to a torque T Determine the radius r¿ of the inner core of the shaft that resists one-half of the applied torque 1T>22 Solve the problem two ways: (a) by using the torsion formula, (b) by finding the resultant of the shear-stress distribution r¿ r T a) tmax = t = Since t = r¿ = Tc Tr 2T = p = J p r3 r (T2 )r¿ p (r¿)4 = T p(r¿)3 T r¿ 2T = a b r pr3 p(r¿)3 r¿ t ; r max r = 0.841 r Ans 24 r b) r¿ dT = 2p L0 r r¿ dT = 2p L0 r L0 tr2 dr L0 r tmax r2 dr L0 r r¿ dT = 2p r 2T a br dr L0 r pr r¿ 4T T = r3 dr r L0 r¿ = r Ans = 0.841r 24 215 05 Solutions 46060 5/25/10 3:53 PM Page 216 © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 5–3 The solid shaft is fixed to the support at C and subjected to the torsional loadings shown Determine the shear stress at points A and B and sketch the shear stress on volume elements located at these points 10 kNиm C A 50 mm The internal torques developed at Cross-sections pass through point B and A are shown in Fig a and b, respectively The polar moment of inertia of the shaft is J = p (0.0754) = 49.70(10 - 6) m4 For point B, rB = C = 0.075 Thus, tB = 4(103)(0.075) TB c = 6.036(106) Pa = 6.04 MPa = J 49.70(10 - 6) Ans From point A, rA = 0.05 m tA = TArA 6(103)(0.05) = 6.036(106) Pa = 6.04 MPa = J 49.70 (10 - 6) 216 Ans B 75 mm kNиm 75 mm 05 Solutions 46060 5/25/10 3:53 PM Page 217 © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher *5–4 The tube is subjected to a torque of 750 N # m Determine the amount of this torque that is resisted by the gray shaded section Solve the problem two ways: (a) by using the torsion formula, (b) by finding the resultant of the shear-stress distribution 75 mm 100 mm 750 Nиm 25 mm a) Applying Torsion Formula: tmax = Tc = J 750(0.1) p (0.14 - 0.0254) tmax = 0.4793 A 106 B = = 0.4793 MPa T¿(0.1) p (0.14 - 0.0754) T¿ = 515 N # m Ans b) Integration Method: r t = a b tmax c dA = 2pr dr and dT¿ = rt dA = rt(2pr dr) = 2ptr2 dr 0.1m T¿ = L 2ptr2 dr = 2p r tmax a br2 dr c L0.075m = 0.1m 2ptmax r3 dr c L0.075m = 2p(0.4793)(106) r4 0.1 m c d2 0.1 0.075 m = 515 N # m Ans 5–5 The copper pipe has an outer diameter of 40 mm and an inner diameter of 37 mm If it is tightly secured to the wall at A and three torques are applied to it as shown, determine the absolute maximum shear stress developed in the pipe tmax = Tmax c = J A 30 Nиm 90(0.02) p 4 (0.02 - 0.0185 ) 20 Nиm = 26.7 MPa Ans 80 Nиm 217 05 Solutions 46060 5/25/10 3:53 PM Page 218 © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 5–6 The solid shaft has a diameter of 0.75 in If it is subjected to the torques shown, determine the maximum shear stress developed in regions BC and DE of the shaft The bearings at A and F allow free rotation of the shaft F E D C B (tBC)max = 35(12)(0.375) TBC c = 5070 psi = 5.07 ksi = p J (0.375) Ans (tDE)max = 25(12)(0.375) TDE c = 3621 psi = 3.62 ksi = p J (0.375) Ans A 35 lbиft 5–7 The solid shaft has a diameter of 0.75 in If it is subjected to the torques shown, determine the maximum shear stress developed in regions CD and EF of the shaft The bearings at A and F allow free rotation of the shaft F E D C B (tEF)max = TEF c = J (tCD)max = 25 lbиft 40 lbиft 20 lbиft Ans A 25 lbиft 40 lbиft 20 lbиft 35 lbиft 15(12)(0.375) TCD c = p J (0.375) = 2173 psi = 2.17 ksi Ans 218 05 Solutions 46060 5/25/10 3:53 PM Page 219 © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 300 Nиm *5–8 The solid 30-mm-diameter shaft is used to transmit the torques applied to the gears Determine the absolute maximum shear stress on the shaft 500 Nиm A 200 Nиm Internal Torque: As shown on torque diagram C Maximum Shear Stress: From the torque diagram Tmax = 400 N # m Then, applying torsion Formula 400 Nиm 300 mm abs = tmax Tmax c J 400(0.015) = p (0.0154) Ans = 75.5 MPa The shaft consists of three concentric tubes, each made from the same material and having the inner and outer radii shown If a torque of T = 800 N # m is applied to the rigid disk fixed to its end, determine the maximum shear stress in the shaft 500 mm T ϭ 800 Nиm ri ϭ 20 mm ro ϭ 25 mm 2m p p p ((0.038)4 - (0.032)4) + ((0.030)4 - (0.026)4) + ((0.025)4 - (0.020)4) 2 -6 ri ϭ 26 mm ro ϭ 30 mm J = 2.545(10 ) m tmax = B 400 mm •5–9 J = D 800(0.038) Tc = 11.9 MPa = J 2.545(10 - 6) Ans 219 ri ϭ 32 mm ro ϭ 38 mm 05 Solutions 46060 5/25/10 3:53 PM Page 220 © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 5–10 The coupling is used to connect the two shafts together Assuming that the shear stress in the bolts is uniform, determine the number of bolts necessary to make the maximum shear stress in the shaft equal to the shear stress in the bolts Each bolt has a diameter d T R r n is the number of bolts and F is the shear force in each bolt T T F = nR T - nFR = 0; T tavg = F 4T nR = p = A ( )d nRpd2 Maximum shear stress for the shaft: tmax = Tc Tr 2T = p = J pr3 2r 4T 2T = nRpd2 p r3 tavg = tmax ; n = r3 Rd2 Ans 5–11 The assembly consists of two sections of galvanized steel pipe connected together using a reducing coupling at B The smaller pipe has an outer diameter of 0.75 in and an inner diameter of 0.68 in., whereas the larger pipe has an outer diameter of in and an inner diameter of 0.86 in If the pipe is tightly secured to the wall at C, determine the maximum shear stress developed in each section of the pipe when the couple shown is applied to the handles of the wrench C B A 15 lb in tAB = tBC Tc = J Tc = = J in 210(0.375) p (0.3754 - 0.344) Ans 15 lb 210(0.5) p = 7.82 ksi (0.54 - 0.434) = 2.36 ksi Ans 220 05 Solutions 46060 5/25/10 3:53 PM Page 221 © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher *5–12 The motor delivers a torque of 50 N # m to the shaft AB This torque is transmitted to shaft CD using the gears at E and F Determine the equilibrium torque T on shaft CD and the maximum shear stress in each shaft The bearings B, C, and D allow free rotation of the shafts A 50 mm 30 mm Equilibrium: B a + ©ME = 0; a + ©MF = 0; 50 - F(0.05) = F = 1000 N 35 mm T¿ T¿ - 1000(0.125) = T¿ = 125 N # m C E 125 mm D F Ans Internal Torque: As shown on FBD Maximum Shear Stress: Applying torsion Formula (tAB)max = 50.0(0.015) TAB c = 9.43 MPa = p J (0.015 ) Ans (tCD)max = 125(0.0175) TCDc = 14.8 MPa = p J (0.0175 ) Ans •5–13 If the applied torque on shaft CD is T¿ = 75 N # m, determine the absolute maximum shear stress in each shaft The bearings B, C, and D allow free rotation of the shafts, and the motor holds the shafts fixed from rotating A 50 mm Equilibrium: 30 mm a + ©MF = 0; 75 - F(0.125) = 0; a + ©ME = 0; 600(0.05) - TA = B F = 600 N 35 mm T¿ TA = 30.0 N # m Internal Torque: As shown on FBD Maximum Shear Stress: Applying the torsion formula (tEA)max = 30.0(0.015) TEA c = 5.66 MPa = p J (0.015 ) Ans (tCD)max = 75.0(0.0175) TCDc = 8.91 MPa = p J (0.0175 ) Ans 221 C E 125 mm D F 05 Solutions 46060 5/25/10 3:53 PM Page 222 © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 250 Nиm 5–14 The solid 50-mm-diameter shaft is used to transmit the torques applied to the gears Determine the absolute maximum shear stress in the shaft 75 Nиm A 325 Nиm 150 Nи m B 500 mm The internal torque developed in segments AB , BC and CD of the shaft are shown in Figs a, b and c C D 400 mm 500 mm The maximum torque occurs in segment AB Thus, the absolute maximum shear stress occurs in this segment The polar moment of inertia of the shaft is p J = (0.0254) = 0.1953p(10 - 6)m4 Thus, A tmax B abs = 250(0.025) TAB c = 10.19(106)Pa = 10.2 MPa = J 0.1953p(10 - 6) 222 Ans 05 Solutions 46060 5/25/10 3:53 PM Page 223 © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 5–15 The solid shaft is made of material that has an allowable shear stress of tallow = 10 MPa Determine the required diameter of the shaft to the nearest mm 15 Nиm 25 Nиm A 30 Nиm B 60 Nиm C 70 Nиm D E The internal torques developed in each segment of the shaft are shown in the torque diagram, Fig a Segment DE is critical since it is subjected to the greatest internal torque The polar p d p moment of inertia of the shaft is J = a b = d Thus, 2 32 tallow TDE c = ; J d 70a b 10(106) = p d 32 d = 0.03291 m = 32.91 mm = 33 mm 223 Ans 05 Solutions 46060 5/25/10 3:53 PM Page 314 © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 5–138 A tube is made of elastic-perfectly plastic material, which has the t-g diagram shown If the radius of the elastic core is rY = 2.25 in., determine the applied torque T Also, find the residual shear-stress distribution in the shaft and the permanent angle of twist of one end relative to the other when the torque is removed ft in T T in t (ksi) Elastic - Plastic Torque The shear stress distribution due to T is shown in Fig a The 10 r = 4.444r Thus, linear portion of this distribution can be expressed as t = 2.25 tr = 1.5 in = 4.444(1.5) = 6.667 ksi T = 2p L tr2dr g (rad) 0.004 2.25 in = 2p L1.5 in = 8.889p ¢ 4.444r A r2dr B + 2p(10) in L2.25 in r2dr r4 2.25 in r3 in + 20p ¢ ≤ ≤2 1.5 in 2.25 in = 470.50 kip # in = 39.2 kip # ft Ans Angle of Twist f = gY 0.004 L = (3)(12) = 0.064 rad rY 2.25 The process of removing torque T is equivalent to the application of T¿ , which is equal magnitude but opposite in sense to that of T This process occurs in a linear 10 = 2.5 A 103 B ksi manner and G = 0.004 f¿ = 10 T¿L = JG 470.50(3)(2) p A 34 - 1.54 B (2.5) A 103 B 470.50(3) = 0.0568 rad trœ = co = T¿co = J trœ = rY = T¿rY 470.50(2.25) = = 8.875 ksi p 4 J A - 1.5 B trœ = ci = 470.50(1.5) T¿ci = = 5.917 ksi p 4 J A - 1.5 B p A 34 - 1.54 B = 11.83 ksi Thus, the permanent angle of twist is fP = f - f¿ = 0.064 - 0.0568 = 0.0072 rad = 0.413° Ans 314 05 Solutions 46060 5/25/10 3:53 PM Page 315 © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 5–138 Continued And the residual stresses are (tr)r = co = tr = c + trœ = c = -10 + 11.83 = 1.83 ksi (tr)r = rY = tr = rY + trœ = rY = -10 + 8.875 = -1.125 ksi (tr)r = ci = tr = ci + trœ = ci = -6.667 + 5.917 = -0.750 ksi The residual stress distribution is shown in Fig a 315 05 Solutions 46060 5/25/10 3:53 PM Page 316 © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 5–139 The tube is made of elastic-perfectly plastic material, which has the t-g diagram shown Determine the torque T that just causes the inner surface of the shaft to yield Also, find the residual shear-stress distribution in the shaft when the torque is removed ft in T T in t (ksi) Plastic Torque When the inner surface of the shaft is about to yield, the shaft is about to become fully plastic T = 2p L 10 tr2dr in = 2ptY L1.5 in = 2p(10)a g (rad) r2dr 0.004 r3 in b2 1.5 in = 494.80 kip # in = 41.2 kip # ft Ans Angle of Twist f = gY 0.004 (3)(12) = 0.096 rad L = rY 1.5 The process of removing torque T is equivalent to the application of T¿ , which is equal magnitude but opposite in sense to that of T This process occurs in a linear 10 manner and G = = 2.5 A 103 B ksi 0.004 f¿ = 494.80(3)(12) T¿L = = 0.05973 rad p JG A - 1.54 B (2.5) A 103 B trœ = co = trœ = ci = 494.80(3) T¿co = = 12.44 ksi p J 1.5 A B 494.80(1.5) T¿ci = = 6.222 ksi p J A - 1.54 B 316 05 Solutions 46060 5/25/10 3:53 PM Page 317 © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 5–139 Continued And the residual stresses are (tr)r = co = tr = c + trœ = c = -10 + 12.44 = 2.44 ksi Ans (tr)r = ci = tr = ci + trœ = ci = -10 + 6.22 = -3.78 ksi Ans The shear stress distribution due to T and T¿ and the residual stress distribution are shown in Fig a 317 05 Solutions 46060 5/25/10 3:53 PM Page 318 © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher *5–140 The 2-m-long tube is made of an elastic-perfectly plastic material as shown Determine the applied torque T that subjects the material at the tube’s outer edge to a shear strain of gmax = 0.006 rad What would be the permanent angle of twist of the tube when this torque is removed? Sketch the residual stress distribution in the tube T 35 mm 30 mm t (MPa) 210 Plastic Torque: The tube is fully plastic if gi Ú gr = 0.003 rad g 0.006 = ; 0.03 0.035 0.003 g = 0.005143 rad Therefore the tube is fully plastic co TP = 2p Lci 2p tg = = tg r2 dr A c3o - c3i B 2p(210)(106) A 0.0353 - 0.033 B = 6982.19 N # m = 6.98 kN # m Ans Angle of Twist: fP = gmax 0.006 L = a b(2) = 0.34286 rad co 0.035 When a reverse torque of TP = 6982.19 N # m is applied, G = fPœ = 210(106) tY = = 70 GPa gY 0.003 TPL = JG 6982.19(2) p (0.035 - 0.034)(70)(109) = 0.18389 rad Permanent angle of twist, fr = fP - fPœ = 0.34286 - 0.18389 = 0.1590 rad = 9.11° Ans Residual Shear Stress: 6982.19(0.035) tPœ o = TP c = J p (0.035 tPœ i = TP r = J p (0.035 - 0.034) 6982.19(0.03) - 0.034) = 225.27 MPa = 193.09 MPa (tP)o = -tg + tPœ o = -210 + 225.27 = 15.3 MPa (tP)i = -tg + tPœ i = -210 + 193.09 = -16.9 MPa 318 g (rad) 05 Solutions 46060 5/25/10 3:53 PM Page 319 © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher •5–141 A steel alloy core is bonded firmly to the copper alloy tube to form the shaft shown If the materials have the t-g diagrams shown, determine the torque resisted by the core and the tube 450 mm A 100 mm 60 mm B 15 kNиm t (MPa) Equation of Equilibrium Refering to the free - body diagram of the cut part of the assembly shown in Fig a, ©Mx = 0; Tc + Tt - 15 A 103 B = 180 (1) Elastic Analysis The shear modulus of steel and copper are Gst = 36 A 106 B and G q = = 18 GPa Compatibility requires that 0.002 180 A 106 B 0.0024 g (rad) 0.0024 = 75 GPa Steel Alloy t (MPa) fC = ft 36 TcL TtL = JcGst JtG q g (rad) 0.002 Tc p A 0.03 B (75) A 10 B Tt = p Copper Alloy A 0.05 - 0.034 B (18) A 109 B Tc = 0.6204Tt (2) Solving Eqs (1) and (2), Tt = 9256.95 N # m Tc = 5743.05 N # m The maximum elastic torque and plastic torque of the core and the tube are (TY)c = pc (tY)st = p A 0.033 B (180) A 106 B = 7634.07 N # m 2 (TP)c = pc (tY)st = p A 0.033 B (180) A 106 B = 10 178.76 N # m 3 and p A 0.054 - 0.034 B J T c(36) A 106 B d = 6152.49 N # m (TY)t = tY = D c 0.05 r2dr = 2p(36) A 106 B ¢ co (TP)t = 2p(tY) q Lci r3 0.05 m = 7389.03 N # m ≤2 0.03 m Since Tt (TY)t, the results obtained using the elastic analysis are not valid 319 05 Solutions 46060 5/25/10 3:53 PM Page 320 © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 5–141 Continued Plastic Analysis Assuming that the tube is fully plastic, Tt = (TP)t = 7389.03 N # m = 7.39 kN # m Ans Substituting this result into Eq (1), Tc = 7610.97 N # m = 7.61 kN # m Ans Since Tc (TY)c, the core is still linearly elastic Thus, ft = ftc = ft = gi L; ci TcL = JcGst 7610.97(0.45) p (0.03 )(75)(10 ) 0.3589 = = 0.03589 rad gi (0.45) 0.03 gi = 0.002393 rad Since gi (gY) q = 0.002 rad, the tube is indeed fully plastic 320 05 Solutions 46060 5/25/10 3:53 PM Page 321 © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 5–142 A torque is applied to the shaft of radius r If the material has a shear stress–strain relation of t = kg1>6, where k is a constant, determine the maximum shear stress in the shaft r r r g = gmax = gmax c r T gmax b r6 t = kg = ka r r T = 2p tr2 dr L0 r = 2p L0 gmax = a ka 1 gmax 13 gmax 6 12p kg6max r3 19 b r dr = 2pk a b a b r6 = r r 19 19 19T b 12p kr 19T 12p r3 tmax = kg6max = Ans 5–143 Consider a thin-walled tube of mean radius r and thickness t Show that the maximum shear stress in the tube due to an applied torque T approaches the average shear stress computed from Eq 5–18 as r>t : q t r t 2r + t ; ro = r + = 2 J = = t 2r - t ri = r - = 2 2r - t p 2r + t ca b - a b d 2 p p [(2r + t)4 - (2r - t)4] = [64 r3 t + 16 r t3] 32 32 tmax = Tc ; J c = ro = 2r + t T(2 r 2+ t) = p 32 [64 r t 2r T(2r + = + 16 r t3] 2p r t[r2 + 14t2] t r2 ) 2p r t c rr2 + As T(2 r 2+ t) = t2 r2 d t r : q , then : r t tmax = = T(1r + 0) 2p r t(1 + 0) = T 2p r2 t T t Am QED 321 05 Solutions 46060 5/25/10 3:53 PM Page 322 © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher *5–144 The 304 stainless steel shaft is m long and has an outer diameter of 60 mm When it is rotating at 60 rad>s, it transmits 30 kW of power from the engine E to the generator G Determine the smallest thickness of the shaft if the allowable shear stress is tallow = 150 MPa and the shaft is restricted not to twist more than 0.08 rad E Internal Torque: P = 30(103) W a T = N # m>s b = 30(103) N # m>s W 30(103) P = = 500 N # m v 60 Allowable Shear Stress: Assume failure due to shear stress tmax = tallow = 150(106) = Tc J 500(0.03) p (0.03 - r4i ) ri = 0.0293923 m = 29.3923 mm Angle of Twist: Assume failure due to angle of twist limitation f = 0.08 = TL JG 500(3) p A 0.03 - r4i B (75.0)(109) ri = 0.0284033 m = 28.4033 mm Choose the smallest value of ri = 28.4033 mm t = ro - ri = 30 - 28.4033 = 1.60 mm Ans 322 G 05 Solutions 46060 5/25/10 3:53 PM Page 323 © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher •5–145 The A-36 steel circular tube is subjected to a torque of 10 kN # m Determine the shear stress at the mean radius r = 60 mm and compute the angle of twist of the tube if it is m long and fixed at its far end Solve the problem using Eqs 5–7 and 5–15 and by using Eqs 5–18 and 5–20 r ϭ 60 mm 4m t ϭ mm 10 kNиm Shear Stress: Applying Eq 5-7, ro = 0.06 + tr = 0.06 m = 0.005 = 0.0625 m Tr = J ri = 0.06 - 10(103)(0.06) p (0.0625 - 0.05754) 0.005 = 0.0575 m = 88.27 MPa Ans Applying Eq 5-18, tavg = 10(103) T = 88.42 MPa = t Am 29(0.005)(p)(0.062) Ans Angle of Twist: Applying Eq 5-15, f = TL JG 10(103)(4) = p (0.0625 - 0.05754)(75.0)(109) = 0.0785 rad = 4.495° Ans Applying Eq 5-20, f = = ds TL 4A2mG L t TL ds 4A2mG t L Where L ds = 2pr 2pTLr = 4A2mG t 2p(10)(103)(4)(0.06) = 4[(p)(0.062)]2 (75.0)(109)(0.005) = 0.0786 rad = 4.503° Ans 323 05 Solutions 46060 5/25/10 3:53 PM Page 324 © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 5–146 Rod AB is made of A-36 steel with an allowable shear stress of 1tallow2st = 75 MPa, and tube BC is made of AM1004-T61 magnesium alloy with an allowable shear stress of 1tallow2mg = 45 MPa The angle of twist of end C is not allowed to exceed 0.05 rad Determine the maximum allowable torque T that can be applied to the assembly 0.3 m 0.4 m a A C 60 mm Internal Loading: The internal torque developed in rod AB and tube BC are shown in Figs a and b, respectively Allowable Shear Stress: The polar moment of inertia of rod AB and tube p p a0.0154 b = 25.3125(10 - 9)p m4 and JBC = a0.034 - 0.0254 b BC are JAB = 2 = 0.2096875(10 - 6)p m4 We have A tallow B st = TAB cAB ; JAB 75(106) = T(0.015) 25.3125(10 - 9)p T = 397.61 N # m and A tallow B mg = TBC cBC ; JBC 45(106) = T(0.03) 0.2096875(10 - 6)p T = 988.13 N # m Angle of Twist: fB>A = -T(0.7) TAB LAB = -0.11737(10 - 3)T = 0.11737(10 - 3)T = JAB Gst 25.3125(10 - 9)p(75)(109) and fC>B = T(0.4) TBC LBC = 0.03373(10 - 3)T = JBC Gmg 0.2096875(10 - 6)p(18)(109) It is required that fC>A = 0.05 rad Thus, fC>A = fB>A + fC>B 0.05 = 0.11737(10 - 3)T + 0.03373(10 - 3)T T = 331 N # m A controls B Ans 324 50 mm 30 mm Section a–a T a B 05 Solutions 46060 5/25/10 3:53 PM Page 325 © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 5–147 A shaft has the cross section shown and is made of 2014-T6 aluminum alloy having an allowable shear stress of tallow = 125 MPa If the angle of twist per meter length is not allowed to exceed 0.03 rad, determine the required minimum wall thickness t to the nearest millimeter when the shaft is subjected to a torque of T = 15 kN # m 30Њ 30Њ 75 mm Section Properties: Referring to the geometry shown in Fig a, Am = C 0.075 1 (0.15) ¢ ≤ + p A 0.0752 B = 0.01858 m2 tan 30° ds = 2(0.15) + p(0.075) = 0.53562 m Allowable Shear Stress: A tavg B allow = T ; 2tAm 125(106) = 15(103) 2t(0.01858) t = 0.00323 m = 3.23 mm Angle of Twist: f = ds TL 4Am G C t 0.03 = 15(103)(1) 4(0.018582)(27)(109) a 0.53562 b t t = 0.007184 m = 7.18 mm (controls) Use t = mm Ans 325 t 05 Solutions 46060 5/25/10 3:53 PM Page 326 © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher *5–148 The motor A develops a torque at gear B of 500 lb # ft, which is applied along the axis of the 2-in.diameter A-36 steel shaft CD This torque is to be transmitted to the pinion gears at E and F If these gears are temporarily fixed, determine the maximum shear stress in segments CB and BD of the shaft Also, what is the angle of twist of each of these segments? The bearings at C and D only exert force reactions on the shaft B E F ft 1.5 ft C D A Equilibrium: TC + TD - 500 = [1] Compatibility: fB>C = fB>D TC(2) TD(1.5) = JG JG TC = 0.75TD [2] Solving Eqs [1] and [2] yields: TD = 285.71 lb # ft TC = 214.29 lb # ft Maximum Shear Stress: (tCB)max = 214.29(12)(1) TCc = 1.64 ksi = p J (1 ) Ans (tBD)max = 285.71(12)(1) TDc = 2.18 ksi = p J (1 ) Ans Angle of Twist: fCB = fBD = TD LBD JG 285.71(12)(1.5)(12) = p (14)(11.0)(106) = 0.003572 rad = 0.205° Ans 326 500 lb·ft 05 Solutions 46060 5/25/10 3:53 PM Page 327 © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 5–149 The coupling consists of two disks fixed to separate shafts, each 25 mm in diameter The shafts are supported on journal bearings that allow free rotation In order to limit the torque T that can be transmitted, a “shear pin” P is used to connect the disks together If this pin can sustain an average shear force of 550 N before it fails, determine the maximum constant torque T that can be transmitted from one shaft to the other Also, what is the maximum shear stress in each shaft when the “shear pin” is about to fail? 25 mm P 130 mm 25 mm T Equilibrium: T - 550(0.13) = ©Mx = 0; T = 71.5 N # m Ans Maximum Shear Stress: tmax = 71.5(0.0125) Tc = 23.3 MPa = p J (0.0125 ) Ans 5–150 The rotating flywheel and shaft is brought to a sudden stop at D when the bearing freezes This causes the flywheel to oscillate clockwise–counterclockwise, so that a point A on the outer edge of the flywheel is displaced through a 10-mm arc in either direction Determine the maximum shear stress developed in the tubular 304 stainless steel shaft due to this oscillation The shaft has an inner diameter of 25 mm and an outer diameter of 35 mm The journal bearings at B and C allow the shaft to rotate freely D 2m B A 80 mm Angle of Twist: f = 0.125 = TL JG Where f = 10 = 0.125 rad 80 T(2) p (0.0175 - 0.01254)(75.0)(109) T = 510.82 N # m Maximum Shear Stress: tmax = Tc = J 510.82(0.0175) p (0.0175 - 0.01254) Ans = 82.0 MPa 327 C T 05 Solutions 46060 5/25/10 3:53 PM Page 328 © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 5–151 If the solid shaft AB to which the valve handle is attached is made of C83400 red brass and has a diameter of 10 mm, determine the maximum couple forces F that can be applied to the handle just before the material starts to fail Take tallow = 40 MPa What is the angle of twist of the handle? The shaft is fixed at A B A 150 mm 150 mm F 150 mm tmax = tallow = 40(106) = Tc J F 0.3F(0.005) p (0.005) F = 26.18 N = 26.2 N Ans T = 0.3F = 7.85 N # m f = TL = JG 7.85(0.15) p (0.005) (37)(10 ) = 0.03243 rad = 1.86° Ans 328 ... MPa = J 49.70(10 - 6) Ans From point A, rA = 0 .05 m tA = TArA 6(103)(0 .05) = 6.036(106) Pa = 6.04 MPa = J 49.70 (10 - 6) 216 Ans B 75 mm kNиm 75 mm 05 Solutions 46060 5/25/10 3:53 PM Page 217 ©... inertia of the rod is J = tmax = p (0.034) = 0. 405( 10 - 6)p Thus, (TAB)max c 1800(0.03) = 42.44(106)Pa = 42.4 MPa = J 0. 405( 10 - 6)p 228 Ans B 0.8 m 05 Solutions 46060 5/25/10 3:53 PM Page 229 ©... = Tc ; J 80(106) = p (0.034) = 0. 405( 10-6)p m4 Thus, T(0.03) 0. 405( 10-6)p T = 3392.92 N # m P = Tv ; 60(103) = 3392.92 v v = 17.68 rad>s = 17.7 rad>s Ans 239 05 Solutions 46060 5/25/10 3:53 PM