Ch04 05 axial load torsion (2)

FM_TOC 46060 6/22/10 11:26 AM Page iii CONTENTS To the Instructor iv Stress Strain 73 Mechanical Properties of Materials 92 Axial Load 122 Torsion 214 Bending 329 Transverse Shear 472 Combined Loadings 532 Stress Transformation 619 10 Strain Transformation 738 11 Design of Beams and Shafts 830 12 Deflection of Beams and Shafts 883 13 Buckling of Columns 1038 14 Energy Methods 1159 04 Solutions 46060 5/25/10 3:19 PM Page 122 © 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 •4–1 The ship is pushed through the water using an A-36 steel propeller shaft that is m long, measured from the propeller to the thrust bearing D at the engine If it has an outer diameter of 400 mm and a wall thickness of 50 mm, determine the amount of axial contraction of the shaft when the propeller exerts a force on the shaft of kN The bearings at B and C are journal bearings A Internal Force: As shown on FBD B C D kN Displacement: 8m dA = PL = AE - 5.00 (103)(8) p (0.42 - 0.32) 200(109) = - 3.638(10 - 6) m = - 3.64 A 10 - B mm Ans Negative sign indicates that end A moves towards end D 4–2 The copper shaft is subjected to the axial loads shown Determine the displacement of end A with respect to end D The diameters of each segment are dAB = in., dBC = in., and dCD = in Take Ecu = 1811032 ksi 50 in A p The cross-sectional area of segment AB, BC and CD are AAB = (32) = 2.25p in2, p p ABC = (22) = p in2 and ACD = (12) = 0.25p in2 4 Thus, PCD LCD PiLi PAB LAB PBC LBC = + + AiEi AAB ECu ABC ECu ACD ECu 2.00 (75) 6.00 (50) = (2.25p) C 18(10 ) D + p C 18(10 ) D - 1.00 (60) + (0.25p) C 18(103) D = 0.766(10 - 3) in Ans The positive sign indicates that end A moves away from D 122 60 in kip kip The normal forces developed in segment AB, BC and CD are shown in the FBDS of each segment in Fig a, b and c respectively dA>D = © 75 in B kip kip C kip D 04 Solutions 46060 5/25/10 3:19 PM Page 123 © 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 4–3 The A-36 steel rod is subjected to the loading shown If the cross-sectional area of the rod is 50 mm2, determine the displacement of its end D Neglect the size of the couplings at B, C, and D 1m A kN B The normal forces developed in segments AB, BC and CD are shown in the FBDS of each segment in Fig a, b and c, respectively The cross-sectional areas of all m b = 50.0(10 - 6) m2 A = A 50 mm2 B a 1000 mm dD = © the segments are PiLi = a PAB LAB + PBC LBC + PCD LCD b AiEi A ESC = 50.0(10 ) C 200(109) D -6 c -3.00(103)(1) + 6.00(103)(1.5) + 2.00(103)(1.25) d = 0.850(10 - 3) m = 0.850 mm Ans The positive sign indicates that end D moves away from the fixed support 123 1.5 m 1.25 m C kN D kN 04 Solutions 46060 5/25/10 3:19 PM Page 124 © 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 *4–4 The A-36 steel rod is subjected to the loading shown If the cross-sectional area of the rod is 50 mm2, determine the displacement of C Neglect the size of the couplings at B, C, and D 1m 1.5 m 1.25 m C A kN B kN D kN The normal forces developed in segments AB and BC are shown the FBDS of each segment in Fig a and b, respectively The cross-sectional area of these two segments 1m are A = A 50 mm2 B a b = 50.0 (10 - 6) m2 Thus, 10.00 mm dC = © PiLi = A P L + PBC LBC B AiEi A ESC AB AB = 50.0(10 - 6) C 200(109) D c -3.00(103)(1) + 6.00(103)(1.5) d = 0.600 (10 - 3) m = 0.600 mm Ans The positive sign indicates that coupling C moves away from the fixed support 4–5 The assembly consists of a steel rod CB and an aluminum rod BA, each having a diameter of 12 mm If the rod is subjected to the axial loadings at A and at the coupling B, determine the displacement of the coupling B and the end A The unstretched length of each segment is shown in the figure Neglect the size of the connections at B and C, and assume that they are rigid Est = 200 GPa, Eal = 70 GPa dB = PL = AE dA = © 12(103)(3) p 12(103)(3) p A B 18 kN kN 3m = 0.00159 m = 1.59 mm (0.012)2(200)(109) PL = AE C 2m Ans 18(103)(2) (0.012) (200)(10 ) + p (0.012) (70)(10 ) = 0.00614 m = 6.14 mm Ans 4–6 The bar has a cross-sectional area of in2, and E = 3511032 ksi Determine the displacement of its end A when it is subjected to the distributed loading x w ⫽ 500x1/3 lb/in A ft x P(x) = L0 w dx = 500 x L0 x3 dx = 1500 43 x L dA = 4(12) P(x) dx 1500 1500 b a b(48)3 = x3 dx = a AE (3)(35)(108)(4) (3)(35)(106) L0 L0 dA = 0.0128 in Ans 124 04 Solutions 46060 5/25/10 3:19 PM Page 125 © 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 4–7 The load of 800 lb is supported by the four 304 stainless steel wires that are connected to the rigid members AB and DC Determine the vertical displacement of the load if the members were horizontal before the load was applied Each wire has a cross-sectional area of 0.05 in2 E F ft H D C ft Referring to the FBD of member AB, Fig a ft 4.5 ft a + ©MA = 0; FBC (5) - 800(1) = FBC = 160 lb a + ©MB = 0; 800(4) - FAH (5) = FAH = 640 lb 800 lb A B ft Using the results of FBC and FAH, and referring to the FBD of member DC, Fig b a + ©MD = 0; FCF (7) - 160(7) - 640(2) = a + ©MC = 0; 640(5) - FDE(7) = FCF = 342.86 lb FDE = 457.14 lb Since E and F are fixed, dD = 457.14(4)(2) FDE LDE = = 0.01567 in T A Est 0.05 C 28.0 (106) D dC = 342.86 (4)(12) FCF LCF = = 0.01176 in T A Est 0.05 C 28.0 (106) D From the geometry shown in Fig c, dH = 0.01176 + (0.01567 - 0.01176) = 0.01455 in T Subsequently, dA>H = 640(4.5)(12) FAH LAH = = 0.02469 in T A Est 0.05 C 28.0(106) D dB>C = 160(4.5)(12) FBC LBC = = 0.006171 in T A Est 0.05 C 28.0(106) D Thus, A + T B dA = dH + dA>H = 0.01455 + 0.02469 = 0.03924 in T A + T B dB = dC + dB>C = 0.01176 + 0.006171 = 0.01793 in T From the geometry shown in Fig d, dP = 0.01793 + (0.03924 - 0.01793) = 0.0350 in T 125 Ans 04 Solutions 46060 5/25/10 3:19 PM Page 126 © 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 *4–8 The load of 800 lb is supported by the four 304 stainless steel wires that are connected to the rigid members AB and DC Determine the angle of tilt of each member after the load is applied.The members were originally horizontal, and each wire has a cross-sectional area of 0.05 in2 E F ft H D C ft Referring to the FBD of member AB, Fig a, ft 4.5 ft a + ©MA = 0; FBC (5) - 800(1) = FBC = 160 lb a + ©MB = 0; 800(4) - FAH (5) = FAH = 640 lb 800 lb A B ft Using the results of FBC and FAH and referring to the FBD of member DC, Fig b, a + ©MD = 0; FCF (7) - 160(7) - 640(2) = FCF = 342.86 lb a + ©MC = 0; 640(5) - FDE (7) = FDE = 457.14 lb Since E and F are fixed, dD = 457.14 (4)(12) FDE LDE = = 0.01567 in T A Est 0.05 C 28.0(106) D dC = 342.86 (4)(12) FCF LCF = = 0.01176 in T A Est 0.05 C 28.0(106) D From the geometry shown in Fig c dH = 0.01176 + u = (0.01567 - 0.01176) = 0.01455 in T 0.01567 - 0.01176 = 46.6(10 - 6) rad 7(12) Ans Subsequently, dA>H = 640 (4.5)(12) FAH LAH = = 0.02469 in T A Est 0.05 C 28.0(106) D dB>C = 160 (4.5)(12) FBC LBC = = 0.006171 in T A Est 0.05 C 28.0(106) D Thus, A + T B dA = dH + dA>H = 0.01455 + 0.02469 = 0.03924 in T A + T B dB = dC + dB>C = 0.01176 + 0.006171 = 0.01793 in T From the geometry shown in Fig d f = 0.03924 - 0.01793 = 0.355(10 - 3) rad 5(12) Ans 126 04 Solutions 46060 5/25/10 3:19 PM Page 127 © 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 4–8 Continued 127 04 Solutions 46060 5/25/10 3:19 PM Page 128 © 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 •4–9 The assembly consists of three titanium (Ti-6A1-4V) rods and a rigid bar AC The cross-sectional area of each rod is given in the figure If a force of kip is applied to the ring F, determine the horizontal displacement of point F D ft C ACD ⫽ in2 ft E Internal Force in the Rods: a + ©MA = 0; + ©F = 0; : x FCD(3) - 6(1) = FCD = 2.00 kip - 2.00 - FAB = FAB = 4.00 kip AAB ⫽ 1.5 in2 ft B ft F kip ft AEF ⫽ in A Displacement: dC = 2.00(4)(12) FCD LCD = 0.0055172 in = ACD E (1)(17.4)(103) dA = 4.00(6)(12) FAB LAB = 0.0110344 in = AAB E (1.5)(17.4)(103) dF>E = 6.00(1)(12) FEF LEF = 0.0020690 in = AEF E (2)(17.4)(103) œ dE 0.0055172 = ; œ dE = 0.0036782 in œ dE = dC + dE = 0.0055172 + 0.0036782 = 0.0091954 in dF = dE + dF>E = 0.0091954 + 0.0020690 = 0.0113 in Ans 4–10 The assembly consists of three titanium (Ti-6A1-4V) rods and a rigid bar AC The cross-sectional area of each rod is given in the figure If a force of kip is applied to the ring F, determine the angle of tilt of bar AC D ft C ACD ⫽ in 2 ft E Internal Force in the Rods: a + ©MA = 0; FCD(3) - 6(1) = FCD = 2.00 kip + ©F = 0; : x - 2.00 - FAB = FAB = 4.00 kip AAB ⫽ 1.5 in2 B Displacement: dC = 2.00(4)(12) FCD LCD = 0.0055172 in = ACD E (1)(17.4)(103) dA = 4.00(6)(12) FAB LAB = 0.0110344 in = AAB E (1.5)(17.4)(103) u = tan - dA - dC 0.0110344 - 0.0055172 = tan - 3(12) 3(12) = 0.00878° Ans 128 ft A ft F kip ft AEF ⫽ in 04 Solutions 46060 5/25/10 3:19 PM Page 129 Edited by Foxit Reader Copyright(C) by Foxit Software Company,2005-2007 For Evaluation Only © 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 4–11 The load is supported by the four 304 stainless steel wires that are connected to the rigid members AB and DC Determine the vertical displacement of the 500-lb load if the members were originally horizontal when the load was applied Each wire has a cross-sectional area of 0.025 in2 E F G ft ft H D C ft ft 1.8 ft I Internal Forces in the wires: A FBD (b) FBC(4) - 500(3) = + c ©Fy = 0; FAH + 375.0 - 500 = FAH = 125.0 lb a + ©MD = 0; FCF(3) - 125.0(1) = FCF = 41.67 lb + c ©Fy = 0; FDE + 41.67 - 125.0 = FBC = 375.0 lb FBD (a) FDE = 83.33 lb Displacement: dD = 83.33(3)(12) FDELDE = 0.0042857 in = ADEE 0.025(28.0)(106) dC = 41.67(3)(12) FCFLCF = 0.0021429 in = ACFE 0.025(28.0)(106) œ dH = 0.0014286 in dH = 0.0014286 + 0.0021429 = 0.0035714 in dA>H = 125.0(1.8)(12) FAHLAH = 0.0038571 in = AAHE 0.025(28.0)(106) dA = dH + dA>H = 0.0035714 + 0.0038571 = 0.0074286 in dB = 375.0(5)(12) FBGLBG = 0.0321428 in = ABGE 0.025(28.0)(106) dlœ 0.0247143 = ; ft 500 lb a + ©MA = 0; œ dH 0.0021429 = ; B ft dlœ = 0.0185357 in dl = 0.0074286 + 0.0185357 = 0.0260 in Ans 129 04 Solutions 46060 5/25/10 3:19 PM Page 130 © 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 *4–12 The load is supported by the four 304 stainless steel wires that are connected to the rigid members AB and DC Determine the angle of tilt of each member after the 500-lb load is applied The members were originally horizontal, and each wire has a cross-sectional area of 0.025 in2 E F G ft ft H D C ft ft 1.8 ft I Internal Forces in the wires: A FBD (b) FBG(4) - 500(3) = + c ©Fy = 0; FAH + 375.0 - 500 = FAH = 125.0 lb a + ©MD = 0; FCF(3) - 125.0(1) = FCF = 41.67 lb + c ©Fy = 0; FDE + 41.67 - 125.0 = FBG = 375.0 lb FBD (a) FDE = 83.33 lb Displacement: dD = 83.33(3)(12) FDELDE = 0.0042857 in = ADEE 0.025(28.0)(106) dC = 41.67(3)(12) FCFLCF = 0.0021429 in = ACFE 0.025(28.0)(106) œ dH 0.0021429 = ; œ dH = 0.0014286 in œ + dC = 0.0014286 + 0.0021429 = 0.0035714 in dH = dH tan a = 0.0021429 ; 36 dA>H = 125.0(1.8)(12) FAHLAH = 0.0038571 in = AAHE 0.025(28.0)(106) a = 0.00341° Ans dA = dH + dA>H = 0.0035714 + 0.0038571 = 0.0074286 in 375.0(5)(12) FBGLBG = 0.0321428 in = ABGE 0.025(28.0)(106) tan b = ft 500 lb a + ©MA = 0; dB = B ft 0.0247143 ; 48 b = 0.0295° Ans 130 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 linear portion of this distribution can be expressed as t = r = 4.444r Thus, 2.25 tr = 1.5 in = 4.444(1.5) = 6.667 ksi T = 2p L tr2dr 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 g (rad) 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 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 = 2.5 A 103 B ksi manner and G = 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 g (rad) 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 kN⭈m 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 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 g (rad) 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  60 mm 4m t  mm 10 kNm 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 a 0.0154 b = 25.3125(10 - 9)p m4 and JBC = a 0.034 - 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 30⬚ 30⬚ 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