14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1159 © 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 sy 14–1 A material is subjected to a general state of plane stress Express the strain energy density in terms of the elastic constants E, G, and n and the stress components sx , sy , and txy txy sx ‘Strain Energy Due to Normal Stresses: We will consider the application of normal stresses on the element in two successive stages For the first stage, we apply only sx on the element Since sx is a constant, from Eq 14-8, we have s2x s2x V dV = 2E Lv 2E (Ui)1 = When sy is applied in the second stage, the normal strain ex will be strained by ex ¿ = -vey = - vsy E Therefore, the strain energy for the second stage is (Ui)2 = = s2y ¢ Lv 2E B s2y Lv 2E + sx ex ¿ ≤ dV + sx a - vsy E b R dV Since sx and sy are constants, (Ui)2 = V (s2 - 2vsx sy) 2E y Strain Energy Due to Shear Stresses: The application of txy does not strain the element in normal direction Thus, from Eq 14–11, we have (Ui)3 = t2xy Lv 2G dV = t2xy V 2G The total strain energy is Ui = (Ui)1 + (Ui)2 + (Ui)3 = t2xy V s2x V V + (s2y - 2vsx sy) + 2E 2E 2G = t2xy V V (s2x + s2y - 2vsx sy) + 2E 2G and the strain energy density is t2xy Ui = (s2x + s2y - 2vsx sy) + V 2E 2G Ans 1159 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1160 © 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 14–2 The strain-energy density must be the same whether the state of stress is represented by sx , sy , and txy , or by the principal stresses s1 and s2 This being the case, equate the strain–energy expressions for each of these two cases and show that G = E>[211 + n2] U = v (s2x + s2y) - sxsy + t R dV E G xy Lv E U = v (s21 + s22) s s R dV B E Lv E B Equating the above two equations yields v v (s2x + s2y) sxsy + txy = (s21 + s22) s s 2E E 2G 2E E However, s1, = sx + sy ; A a sx - sy (1) b + txy Thus, A s21 + s22 B = s2x + s2y + t2xy s1 s2 = sxsy - t2xy Substitute into Eq (1) v v v t = (s2 + s2y + 2t2xy) ss + t A s2 + s2y B - sxsy + 2E x E G xy 2E x E x y E xy t2xy v 2 txy = + t 2G E E xy v = + 2G E E 1 = (1 + v) 2G E G = E 2(1 + v) QED 1160 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1161 © 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 14–3 Determine the strain energy in the stepped rod assembly Portion AB is steel and BC is brass Ebr = 101 GPa, Est = 200 GPa, (sY)br = 410 MPa, (sY)st = 250 MPa 100 mm A B 30 kN 30 kN 1.5 m Referring to the FBDs of cut segments in Fig a and b, + ©F = 0; : x NBC - 20 = + ©F = 0; : x NAB - 30 - 30 - 20 = NBC = 20 kN NAB = 80 kN p The cross-sectional area of segments AB and BC are AAB = (0.12) = 2.5(10 - 3)p m2 and p ABC = (0.0752) = 1.40625(10 - 3)p m2 (Ui)a = © NAB 2LAB NBC 2LBC N2L = + 2AE 2AAB Est 2ABC Ebr = C 80(103) D (1.5) C 2.5(10 - 3)p D C 200(109) D + C 20(103) D 2(0.5) C 1.40625(10 - 3) p D C 101(109) D = 3.28 J Ans This result is valid only if s sy sAB = 80(103) NAB = 10.19(106)Pa = 10.19 MPa (sy)st = 250 MPa = AAB 2.5(10 - 3)p O.K sBC = 20 (103) NBC = 4.527(106)Pa = 4.527 MPa (sy)br = 410 MPa = ABC 1.40625(10 - 3) p O.K 1161 0.5 m 75 mm C 20 kN 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1162 © 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 *14–4 Determine the torsional strain energy in the A-36 steel shaft The shaft has a diameter of 40 mm 900 Nиm 200 Nиm 0.5 m Referring to the FBDs of the cut segments shown in Fig a, b and c, 300 Nиm 0.5 m TAB = 300 N # m ©Mx = 0; TAB - 300 = ©Mx = 0; TBC - 200 - 300 = ©Mx = 0; TCD - 200 - 300 + 900 = TCD = -400 N # m 0.5 m TBC = 500 N # m The shaft has a constant circular cross-section and its polar moment of inertia is p J = (0.024) = 80(10 - 9)p m4 (Ui)t = © TAB LAB TBC 2LBC TCD LCD T2L = + + 2GJ 2GJ 2GJ 2GJ = C 75(10 ) 80 (10 - 9)p D c3002 (0.5) + 5002 (0.5) + (-400)2 (0.5) d = 6.63 J Ans •14–5 Determine the strain energy in the rod assembly Portion AB is steel, BC is brass, and CD is aluminum Est = 200 GPa, Ebr = 101 GPa, and Eal = 73.1 GPa 15 mm A 20 mm kN B 25 mm D kN C kN kN kN 300 mm N2 L Ui = © 2AE [3 (103) ]2 (0.3) = (p4 )(0.0152)(200)(109) [7 (103) ]2 (0.4) + 2(p4 )(0.022)(101)(109) [-3 (103) ]2 (0.2) + (p4 )(0.0252)(73.1)(109) = 0.372 N # m = 0.372 J Ans 1162 400 mm 200 mm 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1163 © 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 14–6 If P = 60 kN, determine the total strain energy stored in the truss Each member has a cross-sectional area of 2.511032 mm2 and is made of A-36 steel 2m B C Normal Forces The normal force developed in each member of the truss can be determined using the method of joints 1.5 m Joint A (Fig a) D + ©F = 0; : x FAD = + c ©Fy = 0; FAB - 60 = FAB = 60 kN (T) P Joint B (Fig b) + c ©Fy = 0; FBD a b - 60 = FBD = 100 kN (C) + ©F = 0; : x 100 a b - FBC = FBC = 80 kN (T) Axial Strain Energy LBD = 222 + 1.52 = 2.5 m (Ui)a = © = A = 2.5 A 103 B mm2 = 2.5 A 10 - B m2 and N2L 2AE C 2.5 A 10 -3 B D C 200 A 109 B D c C 60 A 103 B D (1.5) + C 100 A 103 B D (2.5) + C 80 A 103 B D (2) d = 43.2 J Ans This result is only valid if s sY We only need to check member BD since it is subjected to the greatest normal force sBD = A 100 A 103 B FBD = = 40 MPa sY = 250 MPa A 2.5 A 10 - B O.K 1163 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1164 © 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 14–7 Determine the maximum force P and the corresponding maximum total strain energy stored in the truss without causing any of the members to have permanent deformation Each member has the crosssectional area of 2.511032 mm2 and is made of A-36 steel 2m B C 1.5 m D Normal Forces The normal force developed in each member of the truss can be determined using the method of joints Joint A (Fig a) + ©F = 0; : x FAD = + c ©Fy = 0; FAB - P = FAB = P (T) Joint B (Fig b) + c ©Fy = 0; FBD a b - P = + ©F = 0; : x 1.6667Pa b - FBC = FBD = 1.6667P (C) FBC = 1.3333P(T) Axial Strain Energy A = 2.5 A 103 B mm2 = 2.5 A 10 - B m2 Member BD is critical since it is subjected to the greatest force Thus, sY = FBD A 250 A 106 B = 1.6667P 2.5 A 10 - B P = 375 kN Ans Using the result of P FAB = 375 kN FBD = 625 kN FBC = 500 kN Here, LBD = 21.52 + 22 = 2.5 m (Ui)a = © N2L = 2AE = C 2.5 A 10 - B D C 200 A 109 B D c C 375 A 103 B D (1.5) + C 625 A 103 B D (2.5) + C 500 A 103 B D (2) d = 1687.5 J = 1.6875 kJ Ans 1164 A P 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1165 © 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 *14–8 Determine the torsional strain energy in the A-36 steel shaft The shaft has a radius of 30 mm kNиm kNиm 0.5 m T2L = [02(0.5) + ((3)(103))2(0.5) + ((1)(103))2(0.5)] Ui = © 2JG 2JG = 0.5 m 2.5(106) JG 2.5(106) = 0.5 m 75(109)(p2 )(0.03)4 = 26.2 N # m = 26.2 J Ans •14–9 Determine the torsional strain energy in the A-36 steel shaft The shaft has a radius of 40 mm 12 kNиm kNиm 0.5 m Internal Torsional Moment: As shown on FBD kNиm Torsional Strain Energy: With polar moment p J = A 0.044 B = 1.28 A 10 - B p m4 Applying Eq 14–22 gives of inertia T2L Ui = a 2GJ = C 80002 (0.6) + 20002 (0.4) + 2GJ = 45.0(106) N2 # m3 GJ A -100002 B (0.5) D 45.0(106) = 75(10 )[1.28(10 - 6) p] = 149 J Ans 1165 0.4 m 0.6 m 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1166 © 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 14–10 Determine the torsional strain energy stored in the tapered rod when it is subjected to the torque T The rod is made of material having a modulus of rigidity of G L 2r0 Internal Torque The internal torque in the shaft is constant throughout its length as shown in the free-body diagram of its cut segment, Fig a, Torsional Strain Energy Referring to the geometry shown in Fig b, r = r0 + T r0 r0 (x) = (L + x) L L The polar moment of inertia of the bar in terms of x is J(x) = pr0 p p r0 (L + x)4 r = c (L + x) d = 2 L 2L4 We obtain, L (Ui)t = T2dx dx = L0 2GJ L0 L T2 dx 2G B pr0 2L4 (L + x)4 R L = T2L4 dx pr0 4G L0 (L + x)4 = L T2L4 B R ` pr0 4G 3(L + x)3 = T2L 24pr0 G Ans 1166 r0 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1167 © 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 14–11 The shaft assembly is fixed at C The hollow segment BC has an inner radius of 20 mm and outer radius of 40 mm, while the solid segment AB has a radius of 20 mm Determine the torsional strain energy stored in the shaft The shaft is made of 2014-T6 aluminum alloy The coupling at B is rigid 600 mm 20 mm 600 mm C 40 mm B 60 Nиm Internal Torque Referring to the free-body diagram of segment AB, Fig a, TAB = -30 N # m ©Mx = 0; TAB + 30 = Referring to the free-body diagram of segment BC, Fig b, ©Mx = 0; TBC + 30 + 60 = TAB = -90 N # m p Torsional Strain Energy Here, JAB = A 0.024 B = 80 A 10 - B p m4 p JBC = A 0.044 - 0.024 B = 1200 A 10 - B p m4, (Ui)t = © TAB 2LAB TBC 2LBC T2L = + 2GJ 2GJAB 2GJBC (-30)2(0.6) = and C 27 A 109 B D C 80 A 10 - B p D (-90)2(0.6) + C 27 A 109 B D C 1200 A 10 - B p D = 0.06379 J Ans 1167 A 20 mm 30 Nиm 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1168 © 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 *14–12 Consider the thin-walled tube of Fig 5–28 Use the formula for shear stress, tavg = T>2tAm, Eq 5–18, and the general equation of shear strain energy, Eq 14–11, to show that the twist of the tube is given by Eq 5–20, Hint: Equate the work done by the torque T to the strain energy in the tube, determined from integrating the strain energy for a differential element, Fig 14–4, over the volume of material Ui = t2 dV Lv G but t = T t Am Thus, Ui = T2 dV 2 Lv t AmG L = T2 dV T2 dA TL dA = dx = 2 2 2 A m G Lv t A m G LA t L0 A mG LA t However, dA = t ds Thus, Ui = ds T2L AmG L t Ue = Tf Ue = Ui ds T2L Tf = A2mG L t f = ds TL A2mG L t QED 1168 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1223 © 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 14–78 Determine the vertical displacement of point E Each A-36 steel member has a cross-sectional area of 4.5 in2 F E D ft Virtual-Work Equation: Applying Eq 14–39, we have Member n N L nNL AB 0.6667 3.333 96 213.33 BC 0.6667 3.333 96 213.33 CD 0 72 DE 0 96 EF 0 96 AF 0 72 AE –0.8333 –4.167 120 416.67 CE –0.8333 –4.167 120 416.67 BE 5.00 72 A C B ft ft kip â1260 kip2 # in nNL 1#Â = a AE kip # (¢ E)v = (¢ E)v = 1260 kip2 # in AE 1260 = 0.00966 in T 4.5[29.0(103)] Ans 14–79 Determine the horizontal displacement of joint B of the truss Each A-36 steel member has a cross-sectional area of 400 mm2 kN kN 2m C B Member n N L nNL AB 0 1.5 AC –1.25 –5.00 2.5 15.625 AD 1.00 4.00 2.0 8.000 BC 1.00 4.00 2.0 8.000 CD 0.75 –2.00 1.5 –2.25 1.5 m D A © = 29.375 # ¢ Bh = © nNL AE 29.375(103) ¢ Bh = 400(10 - 6)(200)(109) = 0.3672(10 - 3)m = 0.367 mm Ans 1223 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1224 © 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 *14–80 Determine the vertical displacement of joint C of the truss Each A-36 steel member has a cross-sectional area of 400 mm2 kN 2m kN C B Member n N L nNL AB 0 1.5 AC –5.00 2.5 AD 4.00 2.0 BC 4.00 2.0 CD –1.00 –2.00 1.5 3.00 1.5 m D A © = 3.00 # Â Cv = â nNL AE 3.00 (103) ¢ Cv = 400(10 - 6)(200)(109) = 37.5(10 - 6)m = 0.0375 mm Ans •14–81 Determine the vertical displacement of point A Each A-36 steel member has a cross-sectional area of 400 mm2 E D Virtual-Work Equation: Member n AB –0.750 BC –0.750 AE 1.25 CE –1.25 BE DE 2m N 1.50 –22.5 A 103 B –22.5 A 103 B 37.5 A 10 –62.5 A 10 60.0 A 10 B B 22.0 A 103 B B L 1.5 1.5 2.5 2.5 1.5 nNL 25.3125 A 103 B C A 25.3125 A 103 B 117.1875 A 10 195.3125 A 10 135.00 A 10 B 1.5 m B B 30 kN B © 498.125 A 103 B N2 # m nNL 1#¢ = a AE N # (¢ A)v = (¢ A)v = 498.125(103) N2 # m AE 498.125(103) 0.400(10 - 3)[200(109)] = 6.227 A 10 - B m = 6.23 mm T Ans 1224 1.5 m 20 kN 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1225 © 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 14–82 Determine the vertical displacement of point B Each A-36 steel member has a cross-sectional area of 400 mm2 E D 2m C A B 1.5 m Virtual-Work Equation: Member AB n N BC AE CE –1.25 BE 1.00 DE 0.750 –22.5 A 103 B –22.5 A 10 –62.5 A 10 B 37.5 A 103 B B 22.0 A 103 B 60.0 A 103 B L 30 kN nNL 1.5 1.5 2.5 2.5 1.5 195.3125 A 10 B 40.0 A 103 B 67.5 A 103 B © 302.8125 A 103 B N2 # m nNL 1#¢ = a AE N # (¢ B)v = (¢ B)v = 302.8125(103) N2 # m AE 302.8125(103) 0.400(10 - 3)[200(109)] = 3.785 A 10 - B m = 3.79 mm T Ans 1225 1.5 m 20 kN 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1226 © 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 14–83 Determine the vertical displacement of joint C Each A-36 steel member has a cross-sectional area of 4.5 in2 # Â Cv = â Â Cv = J A nNL AE B 12 ft Ans *14–84 Determine the vertical displacement of joint H Each A-36 steel member has a cross-sectional area of 4.5 in2 ¢ Hv = H G F ft 21 232 = 0.163 in 4.5 (29(103)) # Â Nv = â I C 12 ft 12 ft 12 ft kip kip kip I H G J E D F ft A nNL AE 12 ft 20 368 = 0.156 in 4.5 (29(103)) Ans 1226 C B 12 ft kip 12 ft kip E D 12 ft kip 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1227 © 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 •14–85 Determine the vertical displacement of joint C The truss is made from A-36 steel bars having a cross- sectional area of 150 mm2 G 2m H Member Real Forces N As indicated in Fig a F 2m Member Virtual Forces n As indicated in Fig b 18 A 103 B 0.15 A 10 - B Virtual Work Equation Since smax = Member AB n(N) N(N) A 10 0.375 DE 0.375 BC 0.375 CD 0.375 AH –0.625 EF –0.625 BH DF CH CF GH –0.625 FG –0.625 CG A 10 3 B B A 103 B A 103 B –15 A 103 B –15 A 103 B A 103 B A 10 –3.75 A 10 B –3.75 A 103 B B –11.25 A 103 B –11.25 A 103 B 18 A 103 B nNL(N L(m) # m) 5.0625 A 10 1.5 5.0625 A 10 1.5 3 B B 5.0625 A 103 B 1.5 24.4375 A 103 B 2.5 2.5 24.4375 A 103 B 2 2.5 2.5 17.578125 A 103 B 2.5 17.578125 A 103 B 2.5 72 A 103 B ©174.28125 A 103 B Then 1#Â = â nNL AE 1N # (¢ C)v = 174.28125 A 103 B 0.15 A 10 - B C 200 A 109 B D (¢ C)v = 5.809 A 10 - B m = 5.81 mm T Ans 1227 B 1.5 m 1.5 m kN 5.0625 A 103 B 1.5 E A = 120 MPa sY = 250 MPa, C 1.5 m 12 kN D 1.5 m kN 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1228 © 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 14–86 Determine the vertical displacement of joint G The truss is made from A-36 steel bars having a cross-sectional area of 150 mm2 G 2m H Member Real Forces N As indicated in Fig a F 2m Member Virtual Forces n As indicated in Fig b E A Virtual Work Equation Since 18 A 10 B FCG = = 120 MPa sY = 250 MPa, = A 0.15 A 10 - B smax Member n(N) AB 0.375 DE 0.375 BC 0.375 CD 0.375 AH –0.625 EF –0.625 BH DF CH CF GH FG CG –0.625 –0.625 N(N) A 103 B A 103 B A 103 B A 103 B –15 A 103 B –15 A 103 B A 103 B A 103 B –3.75 A 10 –11.25 A 10 B –3.75 A 103 B –11.25 A 10 18 A 10 3 B B B 5.0625 A 103 B 1.5 5.0625 A 103 B 1.5 5.0625 A 103 B 1.5 5.0625 A 103 B 1.5 24.4375 A 103 B 2.5 2.5 24.4375 A 103 B 2 2.5 2.5 17.578125 A 10 2.5 17.578125 A 10 2.5 3 B B ©102.28125 A 103 B Then 1#Â = â nNL AE 1N # (Â G)v = 1.5 m 1.5 m kN nNL(N # m) L(m) B 102.28125 A 103 B 0.15 A 10 - B C 200 A 109 B D (¢ G)v = 3.409 A 10 - B m = 3.41 mm T Ans 1228 C 1.5 m 12 kN D 1.5 m kN 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1229 © 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 14–87 Determine the displacement at point C EI is constant P P A B a a– C a– a L # ¢C = mM dx L0 EI ¢C = a = a 1 bc a x1 b(Px1)dx1 + EI L0 L0 a>2 (a + x2)(Pa)dx2 d 23Pa3 24EI Ans *14–88 The beam is made of southern pine for which Ep = 13 GPa Determine the displacement at A 15 kN kN/m A L # ¢A = mM L0 EI B 1.5 m 1.5 ¢A = = C 3m c (x1)(15x1)dx1 + (0.5x2)(2x22 + 1.5x2)dx2 d EI L0 L0 43.875(103) 43.875 kN # m3 = 0.0579 m = 57.9 mm = EI 13(10 )(12 )(0.12)(0.18)3 1229 180 mm Ans 120 mm 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1230 © 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 •14–89 Determine the displacement at C of the A-36 steel beam I = 7011062 mm4 kN/m Real Moment Function M(x): As shown on figure(a) C B A Virtual Moment Functions m(x): As shown on figure(b) 10 m 5m Virtual Work Equation: For the displacement at point C L 1#¢ = kN # ¢ C = ¢C = mM dx L0 EI EI L0 10 m 0.500x1 (2.50x1)dx1 + EI L0 5m x2 A x22 B dx2 572.92 kN # m3 EI 572.92(1000) = 200(109)[70(10 - 6)] = 0.04092 m = 40.9 mm T Ans 14–90 Determine the slope at A of the A-36 steel beam I = 7011062 mm4 kN/m C Real Moment Function M(x): As shown on figure(a) B A Virtual Moment Functions mu (x): As shown on figure(b) 10 m Virtual Work Equation: For the slope at point A L 1#u = kN # m # uA = uA = muM dx L0 EI EI L0 10 m (1 - 0.100x1)(2.50x1) dx1 + 5m A 1.00x22 B dx2 41.667 kN # m2 EI 41.667(1000) = EI L0 200(109)[70(10 - 6)] = 0.00298 rad Ans 1230 5m 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1231 © 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 14–91 Determine the slope at B of the A-36 steel beam I = 7011062 mm4 kN/m C Real Moment Function M(x): As shown on figure(a) B A Virtual Moment Functions mU (x): As shown on figure(b) 10 m 5m Virtual Work Equation: For the slope at point B L muM dx L0 EI 1#u = kN # m # uB = uB = EI L0 10 m 0.100x1(2.50x1) dx1 + A 1.00x22 B dx2 5m 83.333 kN # m2 EI 83.333(1000) = EI L0 200(109)[70(10 - 6)] Ans = 0.00595 rad *14–92 Determine the displacement at B of the 1.5-indiameter A-36 steel shaft ft ft A ft D B L # ¢B = mM dx L0 EI 140 lb ¢B 1.5 ft 140 lb C = c (0.5294x1)(327.06x1)dx1 + 0.5294(2 + x2)(654.12 + 47.06x2)dx2 EI L0 L0 1.5 + = L0 (0.4706x3)(592.94x3)dx3 + L0 0.4706(x4 + 1.5)(889.41 - 47.06x4)dx4 d 6437.67(123) 6437.67 lb # ft3 = = 1.54 in EI 29(106) p4 (0.75)4 320 lb 320 lb Ans 1231 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1232 © 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 •14–93 Determine the slope of the 1.5-in-diameter A-36 steel shaft at the bearing support A ft L # uA = muM dx L0 EI uA = (1 - 0.1176x1)(327.06x1)dx1 + B EI L0 L0 L0 ft D 1.5 (0.1176x3)(592.94x3)dx3 140 lb 0.1176(x4 + 1.5)(889.41 - 47.06x4)dx4 R 2387.53(122) 2387.53 lb # ft2 = = 0.0477 rad = 2.73° EI 29(106) A p4 B (0.754) 1.5 ft B + = ft A 140 lb C Ans 320 lb 320 lb 14–94 The beam is made of Douglas fir Determine the slope at C kN Virtual Work Equation: For the slope at point C A L 1#u = muM dx L0 EI kN # m # uC = + EI L0 + uC = B 1.5 m 1.5 m 1.5 m 1.5 m EI L0 (0.3333x2)(4.00x2) dx2 180 mm 1.5 m (1 - 0.3333x3)(4.00x3)dx3 120 mm 4.50 kN # m3 EI 4.50(1000) = - C 13.1(10 ) C 12 (0.12)(0.183) D = 5.89 A 10 - B rad Ans 1232 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1233 © 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 14–95 The beam is made of oak, for which Eo = 11 GPa Determine the slope and displacement at A 200 mm 400 mm kN/m B A 3m Virtual Work Equation: For the displacement at point A, L 1#¢ = kN # ¢ A = + ¢A = L0 mM dx EI EI L0 EI L0 3m 3m x1 a x31 bdx1 (x2 + 3) A 2.00x22 + 6.00x2 + 6.00 B dx2 321.3 kN # m3 EI 321.3(103) = 11(109) C 12 (0.2)(0.43) D Ans = 0.02738 m = 27.4 mm T For the slope at A L muM dx L0 EI 1#u = kN # m # uA = EI L0 3m 3m + uA = = L0 1.00a x31 bdx1 1.00 A 2.00x22 + 6.00x2 + 6.00 B dx2 67.5 kN # m2 EI 67.5(1000) 11(109) C 12 (0.2)(0.43) D = 5.75 A 10 - B rad Ans 1233 3m 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1234 © 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 *14–96 Determine the displacement at point C EI is constant P A C B a a L # ¢C = mM dx L0 EI a ÂC = = 1497 a c (x1)(Px1)dx1 + (x2)(Px2)dx2 d EI L0 L0 2Pa3 3EI Ans Determine the slope at point C EI is constant P A C B a L # uC = L0 a uC = = L0 muMdx EI A a1 B Px1 dx1 x EI a + (1)Px2dx2 EI L0 Pa2 5Pa2 Pa2 + = 3EI 2EI 6EI Ans 1234 a 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1235 © 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 14–98 Determine the slope at point A EI is constant P A L # uA = uA muM dx L0 EI C B a a a x1 Pa2 = a1 (0)(Px2)dx2 R = b (Px1)dx1 + B a EI L0 6EI L0 14–99 Determine the slope at point A of the simply supported Douglas fir beam a Ans kN 0.6 kNиm a A B C Real Moment Function M As indicated in Fig a 1.5 m a 0.5 m Virtual Moment Functions m As indicated in Fig b 75 mm Virtual Work Equation 150 mm L 1#u = mu M dx EI L0 1kN # m # uA = Section a – a B EI L0 2m (1 - 0.3333x1)(0.8x1 + 0.6)dx1 1m + uA = = = B EI L0 2m L0 (0.3333x2)(2.2x2)dx2 R A -0.2667x1 + 0.6x1 + 0.6 B dx1 + 1m L0 0.7333x2 2dx2 R 1.9333 kN # m2 EI 1.9333 A 103 B 13.1 A 109 B c (0.075) A 0.153 B d = 0.00700 rad Ans 1235 1m 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1236 © 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 14–99 Continued 1236 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1237 © 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 *14–100 Determine the displacement at C of the simply supported Douglas fir beam kN 0.6 kNиm a A Real Moment Function M As indicated in Fig a B C Virtual Moment Functions m As indicated in Fig b 1.5 m Virtual Work Equation L 1#¢ = L0 kN # ¢ C = 150 mm B EI L0 1.5 m 1m (0.5x1)(0.8x1 + 0.6)dx1 + L0 Section a – a (0.5x2)(2.2x2)dx2 0.5 m ¢C = B EI L0 1.5 m (0.5x3 + 0.5)(2.2 - 0.8x3)dx3 R L0 A 0.4x1 + 0.3x1 B dx1 + 0.5 m + = = L0 1m L0 1.1x2 dx2 A -0.4x3 + 0.7x3 + 1.1 B dx3 R 1.775kN # m3 EI 1.775 A 103 B 13.1 A 109 B c 0.5 m 75 mm mM dx EI + a (0.075) A 0.153 B d 12 = 6.424 A 10 - B m = 6.42 mm T Ans 1237 1m ... NBC - 20 = + ©F = 0; : x NAB - 30 - 30 - 20 = NBC = 20 kN NAB = 80 kN p The cross-sectional area of segments AB and BC are AAB = (0.12) = 2.5(10 - 3)p m2 and p ABC = (0.0752) = 1.40625(10 - 3)p... 300 N # m ©Mx = 0; TAB - 300 = ©Mx = 0; TBC - 200 - 300 = ©Mx = 0; TCD - 200 - 300 + 900 = TCD = -4 00 N # m 0.5 m TBC = 500 N # m The shaft has a constant circular cross-section and its polar... p>2 = P2r3 2JG L0 p>2 = = P2r3 3p a - 1b JG p>2 [Pr(1 - cos u)]2du (1 - cos u)2 du (1 + cos2 u - cos u)du (1 + cos 2u + - cos u) du Ans 1174 14 Solutions 46060 _Part1 6/11/10 8:18 AM Page 1175 ©