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. 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 . sy 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 V s2x 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 s2y (Ui)2 = Lv 2E = Lv 2E ¢ Since sx and sy are constants, B (Ui)2 = s2y + sx ex ¿ ≤ dV + sx a - vsy E b R dV 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 1 = (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 = U = v 1 2 1 (s2x + s2y) - sxsy + t R dV E 2 G xy Lv 2 E B 1 v (s21 + s22) s s R dV B E 1 2 Lv 2 E Equating the above two equations yields. 1 v 1 2 1 v (s2x + s2y) sxsy + txy = (s21 + s22) s s 2E E 2G 2E E 1 2 However, s1, 2 = sx + sy 2 ; A a sx - sy 2 (1) 2 2 b + txy Thus, A s21 + s22 B = s2x + s2y + 2 t2xy s1 s2 = sxsy - t2xy Substitute into Eq. (1) v 1 2 1 v v 2 1 t = (s2 + s2y + 2t2xy) ss + t A s2 + s2y B - sxsy + 2E x E 2 G xy 2E x E x y E xy t2xy v 2 1 2 txy = + t 2G E E xy 1 v 1 = + 2G E E 1 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 = 0 + ©F = 0; : x NAB - 30 - 30 - 20 = 0 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 4 p ABC = (0.0752) = 1.40625(10 - 3)p m2. 4 (Ui)a = © NAB 2LAB NBC 2LBC N2L = + 2AE 2AAB Est 2ABC Ebr = C 80(103) D 2 (1.5) 2 C 2.5(10 - 3)p D C 200(109) D = 3.28 J + C 20(103) D 2(0.5) 2 C 1.40625(10 - 3) p D C 101(109) D Ans. This result is valid only if s 6 sy. sAB = 80(103) NAB = 10.19(106)Pa = 10.19 MPa 6 (sy)st = 250 MPa = AAB 2.5(10 - 3)p O.K. sBC = 20 (103) NBC = 4.527(106)Pa = 4.527 MPa 6 (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 N⭈m 200 N⭈m 0.5 m Referring to the FBDs of the cut segments shown in Fig. a, b and c, TAB = 300 N # m 300 N⭈m 0.5 m ©Mx = 0; TAB - 300 = 0 ©Mx = 0; TBC - 200 - 300 = 0 ©Mx = 0; TCD - 200 - 300 + 900 = 0 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. 2 (Ui)t = © TAB 2 LAB TBC 2LBC TCD LCD T2L = + + 2GJ 2GJ 2GJ 2GJ = 1 2 C 75(10 ) 80 (10 - 9)p D 9 c3002 (0.5) + 5002 (0.5) + (-400)2 (0.5) d = 6.63 J Ans. 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. •14–5. 15 mm A 20 mm 2 kN B 25 mm D 5 kN C 2 kN 5 kN 3 kN 300 mm N2 L Ui = © 2AE = [3 (103) ]2 (0.3) 2 (p4 )(0.0152)(200)(109) + [7 (103) ]2 (0.4) 2(p4 )(0.022)(101)(109) + [-3 (103) ]2 (0.2) 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) + ©F = 0; : x FAD = 0 + c ©Fy = 0; FAB - 60 = 0 D FAB = 60 kN (T) P Joint B (Fig. b) + c ©Fy = 0; 3 FBD a b - 60 = 0 5 FBD = 100 kN (C) + ©F = 0; : x 4 100 a b - FBC = 0 5 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 - 3 B m2 and N2L 2AE 2 C 2.5 A 10 1 -3 B D C 200 A 109 B D c C 60 A 103 B D 2 (1.5) + C 100 A 103 B D 2 (2.5) + C 80 A 103 B D 2 (2) d = 43.2 J Ans. This result is only valid if s 6 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 6 sY = 250 MPa A 2.5 A 10 - 3 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 = 0 + c ©Fy = 0; FAB - P = 0 FAB = P (T) Joint B (Fig. b) + c ©Fy = 0; 3 FBD a b - P = 0 5 + ©F = 0; : x 4 1.6667Pa b - FBC = 0 5 FBD = 1.6667P (C) FBC = 1.3333P(T) Axial Strain Energy. A = 2.5 A 103 B mm2 = 2.5 A 10 - 3 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 - 3 B P = 375 kN Ans. Using the result of P FBD = 625 kN FAB = 375 kN FBC = 500 kN Here, LBD = 21.52 + 22 = 2.5 m. (Ui)a = © = N2L = 2AE 1 2 C 2.5 A 10 - 3 B D C 200 A 109 B D c C 375 A 103 B D 2 (1.5) + C 625 A 103 B D 2 (2.5) + C 500 A 103 B D 2 (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. 4 kN⭈m 3 kN⭈m 0.5 m T2L 1 Ui = © = [02(0.5) + ((3)(103))2(0.5) + ((1)(103))2(0.5)] 2JG 2JG = = 0.5 m 0.5 m 2.5(106) JG 2.5(106) 75(109)(p2 )(0.03)4 = 26.2 N # m = 26.2 J Ans. Determine the torsional strain energy in the A-36 steel shaft. The shaft has a radius of 40 mm. •14–9. 12 kN⭈m 6 kN⭈m Internal Torsional Moment: As shown on FBD. Torsional Strain Energy: With polar moment p J = A 0.044 B = 1.28 A 10 - 6 B p m4. Applying Eq. 14–22 gives 2 of inertia T2L Ui = a 2GJ = 1 C 80002 (0.6) + 20002 (0.4) + 2GJ = 45.0(106) N2 # m3 GJ = 0.5 m 8 kN⭈m A -100002 B (0.5) D 45.0(106) 75(10 )[1.28(10 - 6) p] 9 = 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) = 4 pr0 4 p 4 p r0 (L + x)4 r = c (L + x) d = 2 2 L 2L4 We obtain, L (Ui)t = T2dx dx = L0 2GJ L0 L = = = L dx T2L4 pr0 4G L0 (L + x)4 T2 dx 2G B pr0 4 2L4 (L + x)4 R L T2L4 1 B R ` pr0 4G 3(L + x)3 0 7 T2L 24pr0 4 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 N⭈m Internal Torque. Referring to the free-body diagram of segment AB, Fig. a, ©Mx = 0; TAB = -30 N # m TAB + 30 = 0 Referring to the free-body diagram of segment BC, Fig. b, ©Mx = 0; TBC + 30 + 60 = 0 TAB = -90 N # m p Torsional Strain Energy. Here, JAB = A 0.024 B = 80 A 10 - 9 B p m4 2 p JBC = A 0.044 - 0.024 B = 1200 A 10 - 9 B p m4, 2 (Ui)t = © = and TAB 2LAB TBC 2LBC T2L = + 2GJ 2GJAB 2GJBC (-30)2(0.6) 2 C 27 A 109 B D C 80 A 10 - 9 B p D = 0.06379 J + (-90)2(0.6) 2 C 27 A 109 B D C 1200 A 10 - 9 B p D Ans. 1167 A 20 mm 30 N⭈m 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 2 G but t = T 2 t Am Thus, Ui = T2 dV 2 2 Lv 8 t AmG L = 2 dV dA dA T2 T2 TL = dx = 2 2 2 2 2 2 8 A m G Lv t 8 A m G LA t L0 8 A mG LA t However, dA = t ds. Thus, Ui = ds T2L 2 8 AmG L t Ue = 1 Tf 2 Ue = Ui ds T2L 1 Tf = 2 8 A2mG L t f = ds TL 4 A2mG L t QED 1168 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1169 © 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. Determine the ratio of shearing strain energy to bending strain energy for the rectangular cantilever beam when it is subjected to the loading shown. The beam is made of material having a modulus of elasticity of E and Poisson’s ratio of n. •14–13. P L a a b h Section a – a Internal Moment. Referring to the free-body diagram of the left beam’s cut segment, Fig. a, + c ©Fy = 0; -V - P = 0 V = -P + ©MO = 0; M + Px = 0 M = -Px Shearing Strain Energy. For the rectangular cross section, the form factor is fs = 6 5 6 2 2 L (-P) dx L fsV dx 3P2L 5 3P2 (Ui)v = dx = = = 2GA 2G(bh) 5bhG L0 5bhG L0 L0 L However, G = E , then 2(1 + v) (Ui)v = 6(1 + v)P2L 5bhE Bending Strain Energy. L (Ui)b = M2dx = L0 2EI L0 L (-Px)2dx2 2Ea 1 bh3 b 12 6P2 6P2 x3 L 2P2L3 2 x dx = = ¢ ≤ ` bh3E L0 bh3E 3 0 bh3E L = Then, the ratio is 6(1 + v)P2L 3(1 + v) h 2 (Ui)v 5bhE = a b = 2 3 (Ui)b 5 L 2P L bh3E Ans. From this result, we can conclude that the proportion of the shearing strain energy stored in the beam increases if the depth h of the beam’s cross section increases but (Ui)v 1 = 0.009. the decreases if L increases. Suppose that v = and L = 10h, then 2 (Ui)b shearing strain energy is only 0.09% of the bending strain energy. Therefore, the effect of the shearing strain energy is usually neglected if L 7 10h. 1169 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1170 © 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. 14–14. Determine the bending strain-energy in the beam due to the loading shown. EI is constant. M0 A L0 Ui = L C B 2 M dx 2EI = 1 c 2EI L0 = M20L 24EI L>2 L — 2 a L — 2 L>2 2 2 -M0 M0 a x1 b dx1 + x2 b dx2] L L L0 Ans. Note: Strain energy is always positive regardless of the sign of the moment function. 14–15. Determine the bending strain energy in the beam. EI is constant. P P Referring to the FBD of the entire beam, Fig. a, a + ©MB = 0; Pa L 3L b + Pa b - Ay (L) = 0 4 4 L 4 Ay = P Using the coordinates, x1 and x2, the FBDs of the beam’s cut segments in Figs. b and c are drawn. For coordinate x1, a + ©Mc = 0; M(x1) - Px1 = 0 M(x1) = Px1 For coordinate x2 coordinate, a + ©Mc = 0; M(x2) - Pa L b = 0 4 L (Ui)b = © 1 M2dx = c2 2EI 2EI L0 L0 M(x2) = L>4 (Px1)2dx1 + L 4 PL 4 L0 L>2 a PL 2 b dx2 d 4 L = 1 P2 3 P2L2 2 c2 a x1 b ` + x ` d 2EI 3 16 2 0 0 = P2L3 48EI Ans. 1170 L 2 L 4 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1171 © 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. *14–16. Determine the bending strain energy in the A-36 structural steel W10 * 12 beam. Obtain the answer using the coordinates 1a2 x1 and x4, and 1b2 x2 and x3. 6 kip x1 x4 x2 12 ft Support Reactions: As shown on FBD(a). Internal Moment Function: As shown on FBD(b), (c), (d) and (e). Bending Strain Energy: Using coordinates x1 and x4 and applying Eq. 14–17 gives L Ui = M2dx L0 2EI 1 c 2EI L0 12ft = 1 c 2EI L0 12ft = = L0 (-3.00x1)2 dx1 + 9.00x21dx1 + 3888 kip2 # ft3 EI L0 6ft (-6.00x4)2 dx4 d 6ft 36.0x24 dx4 d For W10 * 12 wide flange section, I = 53.8 in4. Ui = 3888(123) 29.0(103)(53.8) = 4.306 in # kip = 359 ft # lb Ans. b) Using coordinates x2 and x3 and applying Eq. 14–17 gives L Ui = M2dx L0 2EI 1 c 2EI L0 12ft = 1 c 2EI L0 12ft = = (3.00x2 - 36.0)2dx2 + L0 6ft (6.00x3 - 36.0)2 dx3 d A 9.00x22 - 216x + 1296 B dx2 + 3888 kip2 # ft3 EI L0 6ft A 36.0x23 - 432x + 1296 B dx3 d For W 10 * 12 wide flange section, I = 53.8 in4. Ui = 3888(123) 29.0(103)(53.8) = 4.306 in # kip = 359 ft # lb Ans. 1171 x3 6 ft 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1172 © 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. Determine the bending strain energy in the A-36 steel beam. I = 99.2 (106) mm4. •14–17. 9 kN/m 6m Referring to the FBD of the entire beam, Fig. a, 1 (9)(6)(2) - Ay (6) = 0 2 a + ©MB = 0; Ay = 9 kN Referring to the FBD of the beam’s left cut segment, Fig. b, a + ©M0 = 0; M(x) + 1 3 a xb (x) (x>3) - 9x = 0 2 2 M(x) = a 9x 1 M2 dx = 2EI L0 L0 2EI 6m a 9x - 1 2EI L0 6m a 81x2 + L (Ui)b = = = = For A 36 1 3 x b 4 steel, 1 3 2 x b dx 4 1 6 9 x - x4 bdx 16 2 1 7 9 5 2 6m 1 d c a27x3 + x x b 2EI 112 10 0 666.51 kN2 # m3 EI E = 200 GPa. Here, = 99.2(10 - 6) m4. Then (Ui)b = kN # m 666.51 (10002) 200(109) C 99.2(10 - 6) D I = C 99.2 (106) mm4 D a = 33.6 J 4 1m b 1000 mm Ans. 1172 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1173 © 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. 14–18. Determine the bending strain energy in the A-36 steel beam due to the distributed load. I = 122 (106) mm4. 15 kN/m A B 3m Referring to the FBD of the entire beam, Fig. a, + c ©Fy = 0; 1 (15)(3) = 0 2 Ay - a + ©MA = 0; MA - Ay = 22.5 kN 1 (15)(3)(2) = 0 2 MA = 45 kN # m Referring to the FBD of the beam’s left cut segment, Fig. b, a + ©M0 = 0; M(x) + 1 (5x)(x)(x>3) - 22.5x + 45 = 0 2 M(x) = (22.5x - 0.8333x3 - 45) kN # m L (Ui)b = = M2dx 1 = c 2EI 2EI L0 L0 1 c 2EI L0 3m (22.5x - 0.8333x3 - 45)2 dx 3m 0.6944x6 - 37.5x4 + 75x3 + 506.25x2 - 2025x + 2025)dx d = 1 a0.09921x7 - 7.5x5 + 18.75x4 + 168.75x3 2EI - 1012.5x2 + 2025xb 2 715.98 kN2 # m2 = EI For A 36 steel, E = 200 GPa. Here, 3m 0 I = c122(106) mm4 d a = 122(10 - 6) m4. Thus, (Ui)b = 715.98 (10002) 200(109) C 122 (10 - 6) D = 29.3 J 4 1m b 1000 mm Ans. 1173 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1174 © 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. 14–19. Determine the strain energy in the horizontal curved bar due to torsion. There is a vertical force P acting at its end. JG is constant. r 90⬚ P T = Pr(1 - cos u) Strain energy: L Ui = T2 ds L0 2JG However, s = ru; ds = rdu u Ui = T2rdu r = 2JG L0 L0 2JG P2r3 2JG L0 p>2 = P2r3 2JG L0 p>2 = P2r3 2JG L0 p>2 = = P2r3 3p a - 1b JG 8 p>2 [Pr(1 - cos u)]2du (1 - cos u)2 du (1 + cos2 u - 2 cos u)du (1 + cos 2u + 1 - 2 cos u) du 2 Ans. 1174 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1175 © 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. *14–20. Determine the bending strain energy in the beam and the axial strain energy in each of the two rods. The beam is made of 2014-T6 aluminum and has a square cross section 50 mm by 50 mm. The rods are made of A-36 steel and have a circular cross section with a 20-mm diameter. 2m 8 kN Support Reactions: As shown on FBD(a). 8 kN Internal Moment Function: As shown on FBD(b) and (c). Axial Strain Energy: Applying Eq. 14–16 gives 1m N2L (Ui)a = 2AE C 8.00(103) D 2 (2) = 2AE 64.0(106) N2 # m AE = 64.0(106) = p 4 (0.022) [200(109)] = 1.02 J Ans. Bending Strain Energy: Applying Eq. 14–17 gives L (Ui) b = = = = M2dx L0 2EI 1 B2 2EI L0 1m (8.00x1)2 dx 1 + 85.333 kN2 # m3 EI L0 2m 8.002 dx2 R 85.333(106) 1 (0.05) (0.053) D 73.1(109) C 12 = 2241.3 N # m = 2.24 kJ Ans. 1175 2m 1m 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1176 © 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. The pipe lies in the horizontal plane. If it is subjected to a vertical force P at its end, determine the strain energy due to bending and torsion. Express the results in terms of the cross-sectional properties I and J, and the material properties E and G. •14–21. z L C x B L — 2 A P Ui = M2 dx T2 dx + L 2E I L 2JG L 2 = L L PL 2 ( 2 ) dx (P x)2 dx (P x)2 dx + + 2 E I 2 EI L0 L0 2 J G L0 = L 31 P2 L3 P2L2 P2 a b + + (L) 2EI 2 3 2EI 3 8JG = P2 L3 3 P2 L3 + 16 E I 8JG = P2 L3 c 1 3 + d 16 E I 8JG Ans. 1176 y 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1177 © 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. 14–22. The beam shown is tapered along its width. If a force P is applied to its end, determine the strain energy in the beam and compare this result with that of a beam that has a constant rectangular cross section of width b and height h. b h P L Moment of Inertia: For the beam with the uniform section, I = bh3 = I0 12 For the beam with the tapered section, I = I0 bh3 1 b a xb A h3 B = x = x 12 L 12L L Internal Moment Function: As shown on FBD. Bending Strain Energy: For the beam with the tapered section, applying Eq. 14–17 gives L UI = M2 dx L0 2EI = L (-Px)2 1 dx I0 2E L0 L x = P2L xdx 2EI0 L0 = 3P2 L3 P2L3 = 4EI0 bh3 E L Ans. For the beam with the uniform section, L Ui = M2dx L0 2EI L = 1 (-Px)2 dx 2EI0 L0 = P3 L3 6EI0 The strain energy in the capered beam is 1.5 times as great as that in the beam having a uniform cross section. Ans. 1177 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1178 © 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. 14–23. Determine the bending strain energy in the cantilevered beam due to a uniform load w. Solve the problem two ways. (a) Apply Eq. 14–17. (b) The load w dx acting on a segment dx of the beam is displaced a distance y, where y = w1-x4 + 4L3x - 3L42>124EI2, the equation of the elastic curve. Hence the internal strain energy in the differential segment dx of the beam is equal to the external work, i.e., dUi = 121w dx21-y2. Integrate this equation to obtain the total strain energy in the beam. EI is constant. w dx w dx Internal Moment Function: As shown on FBD. Bending Strain Energy: a) Applying Eq. 14–17 gives L Ui = M2dx L0 2EI L = w 2 1 c - x2 d dx R B 2EI L0 2 L = = b) Integrating dUi = dUi = dUi = Ui = = w2 x4 dx R B 8EI L0 w2 L5 40EI Ans. 1 (wdx)( -y) 2 w 1 (wdx) B A -x4 + 4L3x - 3L4 B R 2 24EI w2 A x4 -4L3x + 3L4 B dx 48EI w2 48EI L0 L x L A x4 - 4L3x + 3L4 B dx w2L5 40EI Ans. 1178 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1179 © 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. *14–24. Determine the bending strain energy in the simply supported beam due to a uniform load w. Solve the problem two ways. (a) Apply Eq. 14–17. (b) The load w dx acting on the segment dx of the beam is displaced a distance y, where y = w1-x4 + 2Lx3 - L3x2>124EI2, the equation of the elastic curve. Hence the internal strain energy in the differential segment dx of the beam is equal to the external work, i.e., dUi = 211w dx21-y2. Integrate this equation to obtain the total strain energy in the beam. EI is constant. w dx w x Support Reactions: As shown on FBD(a). Internal Moment Function: As shown on FBD(b). Bending Strain Energy: a) Applying Eq. 14–17 gives L Ui = M2dx L0 2EI L = 2 w 1 2 c (Lx - x ) d dx R B 2EI L0 2 L = = b) Integrating dUi = dUi = dUi = w2 (L2x2 + x4 - 2Lx3) dx R B 8EI L0 w2L5 240EI Ans. 1 (wdx) (-y) 2 1 w (wdx) B (-x4 + 2Lx3 - L3x) R 2 24EI w2 (x4 - 2Lx3 + L3x) dx 48EI L Ui = = w2 (x4 - 2Lx3 + L3x) dx 48EI L0 w2L5 240EI Ans. 1179 dx L 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1180 © 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. Determine the horizontal displacement of joint A. Each bar is made of A-36 steel and has a cross-sectional area of 1.5 in2. •14–25. 2 kip A 3 ft D 3 ft C B 4 ft Member Forces: Applying the method of joints to joint at A, we have + ©F = 0; : x 4 F - 2 = 0 5 AD + c ©Fy = 0; FAB - FAD = 2.50 kip (T) 3 (2.50) = 0 5 FAB = 1.50 kip (C) At joint D + ©F = 0; : x 4 4 F - (2.50) = 0 5 DB 5 + c ©Fy = 0; 3 3 ( 2.50) + (2.50) - FDC = 0 5 5 FDB = 2.50 kip (C) FDC = 3.00 kip (T) Axial Strain Energy: Applying Eq. 14–16, we have N2L Ui = a 2AE = 1 [2.502 (5) + (-1.50)2 (6) + (-2.50)2 (5) + 3.002(3)] 2AE = 51.5 kip2 # ft AE = 51.5(12) 1.5[29.0(103)] = 0.014207 in # kip External Work: The external work done by 2 kip force is Ue = 1 (2) (¢ A)h = (¢ A)h 2 Conservation of Energy: Ue = Ui (¢ A)h = 0.014207 = 0.0142 in. Ans. 1180 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1181 © 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. 14–26. Determine the horizontal displacement of joint C. AE is constant. C P L L A B L Member Forces: Applying the method of joints to C, we have + c ©Fy = 0; FBC cos 30° - FAC cos 30° = 0 + ©F = 0; : x P - 2F sin 30° = 0 Hence, FBC = P (C) FBC = FAC = F F = P FAC = P (T) Axial Strain Energy: Applying Eq. 14–16, we have N2L Ui = a 2AE = 1 C P2L + (-P)2 L D 2AE = P2L AE External Work: The external work done by force P is Ui = 1 P(¢ c)k 2 Conservation of Energy: Ue = Ui P2L 1 P(¢ C)k = 2 AE (¢ C)k = 2PL AE Ans. 1181 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1182 © 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. 14–27. Determine the vertical displacement of joint C. AE is constant. A L L C L B Joint C: + ©F = 0 : x FCB cos 30° - FCA cos 30° = 0 FCB = FCA + c © Fy = 0 FCA sin 30° + FCB sin 30° - P = 0 FCB = FCA = P Conservation of energy: Ue = Ui N2L 1 P¢ C = © 2 2EA 1 L 2 P¢ C = [F 2 + FCA ] 2 2EA CB P¢ C = ¢C = L (P2 + P2) EA 2PL AE Ans. 1182 P 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1183 © 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. *14–28. Determine the horizontal displacement of joint D. AE is constant. D C P Joint B: L + c ©Fy = 0; FBC = 0.75P + ©F = 0; ; x FBA = P 0.6 L B A Joint D: 0.8 L + T ©Fy = 0; FDA = 0 + ©Fx = 0; : FDC = P Joint A: + T ©Fy = 0; 3 F - 0.75P = 0 5 AC FAC = 1.25P Conservation of energy: Ue = Ui N2L 1 P¢ D = © 2 2AE 1 1 P¢ D = [(0.75P)2(0.6L) + (P)2(0.8L) + (02)(0.6L) 2 2AE + (P2)(0.8L) + (1.25P)2(L)] ¢D = 3.50PL AE Ans. The cantilevered beam is subjected to a couple moment M0 applied at its end. Determine the slope of the beam at B. EI is constant. •14–29. M0 A B Ui = L 2 M20L 1 M dx M20 dx = = 2EI L0 2EI L0 2EI Ue = 1 (M0 uB) 2 L L Conservation of energy: Ue = Ui M0 2L 1 M0 uB = 2 2EI uB = M0L EI Ans. 1183 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1184 © 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. 14–30. Determine the vertical displacement of point C of the simply supported 6061-T6 aluminum beam. Consider both shearing and bending strain energy. 100 kip a B A C 1.5 ft Support Reactions. Referring to the free-body diagram of the entire beam, Fig. a, a + ©MB = 0; 100(1.5) - Ay(3) = 0 Ay = 50 kip + c ©Fy = 0; 50 - V = 0 a + ©MO = 0; V = 50 kip M - 50x = 0 M = 50x Shearing Strain Energy. For the rectangular beam, the form factor is fs = 6 5 6 A 502 B dx 18 in. fsV2dx 5 (Ul)v = = 2 = 0.3041 in # kip L0 2GA L0 2 C 3.7 A 103 B D [4(12)] L Bending Strain Energy. I = 1 (4) A 123 B = 576 in4. We obtain 12 L (50x)2 dx M2dx = 2 L0 2 C 10.0 A 103 B D (576) L0 2EI L (Ui)b = = 0.4340 A 10-3 B L0 = 0.4340 A 10-3 B ¢ 18 in. x2 dx x3 18in. ≤` 3 0 = 0.84375 in # kip Thus, the total strain energy stored in the beam is Ui = (Ui)v + (Ui)b = 0.3041 + 0.84375 = 1.1478 in # kip Ans. External Work. The external work done by the external force (100 kip) is Ue = 1 1 P¢ = (100)¢ C = 50¢ C 2 2 Conservation of Energy. Ue = Ui 50¢ C = 1.1478 ¢ C = 0.02296 in. = 0.0230 in. Ans. 1184 1.5 ft 4 in. 12 in. Internal Loading. Referring to the free-body diagram of the beam’s left cut segment, Fig. b, a Section a – a 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1185 © 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. 14–31. Determine the slope at the end B of the A-36 steel beam. I = 8011062 mm4. 6 kN⭈m A B M = -750x 8m L 2 1 M dx M uB = 2 L0 2EI B ( -750x)2dx 1 (6(103)) uB = 2 2EI L0 uB = 16000 200 (109)(80)(10-6) = 1 (10-3) rad Ans. *14–32. Determine the deflection of the beam at its center caused by shear. The shear modulus is G. P b Support Reactions: As shown on FBD(a). h Shear Functions: As shown on FBD(b). Shear Strain Energy: Applying 14–19 with fe = L 2 6 for a rectangular section, we have 5 L Ui = feV2dx L0 2GA L 2 6 P 2 1 a b a b dx R = B2 2bhG 2 L0 5 = 3P2L 20bhG External Work: The external work done by force P is Ue = 1 (P) ¢ 2 Conservation of Energy: Ue = Ui 3P2L 1 (P)¢ = 2 20bhG ¢ = 3PL 10bhG Ans. 1185 L 2 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1186 © 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. •14–33. The A-36 steel bars are pin connected at B and C. If they each have a diameter of 30 mm, determine the slope at E. 3 2 C D A E 2m 3m L 300 N⭈m B 2m 3m 2 Ui = M dx 1 1 65625 = (2) (75x1)2dx1 + (2) (-75x2)2dx2 = 2EI 2EI 2EI EI L0 L0 L0 Ue = 1 1 (M¿)u = (300) uE = 150 uE 2 2 Conservation of energy: Ue = Ui 150 uE = uE = 65625 EI 473.5 473.5 = 0.0550 rad = 3.15° = EI (200)(109)(p4 )(0.0154) 14–34. The A-36 steel bars are pin connected at B. If each has a square cross section, determine the vertical displacement at B. Ans. 800 lb 2 in. A B C D 2 in. Support Reactions: As shown on FBD(a). 8 ft Moment Functions: As shown on FBD(b) and (c). Bending Strain Energy: Applying 14–17, we have L Ui = = = = M2dx L0 2EI 1 B 2EI L0 4ft ( -800x1)2 dx1 + 23.8933(106) lb2 # ft3 EI 23.8933(106)(123) 1 (2)(23) D 29.0(106) C 12 L0 10ft (-320x2)2 dx2 R = 1067.78 in # lb External Work: The external work done by 800 lb force is Ue = 1 (800)(¢ B) = 400¢ B 2 Conservation of Energy: Ue = Ui 400¢ B = 1067.78 ¢ B = 2.67 in. Ans. 1186 4 ft 10 ft 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1187 © 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. 14–35. Determine the displacement of point B on the A-36 steel beam. I = 8011062 mm4. 20 kN A C B 3m 3 L 5m 5 1 M2dx = [(12.5)(103)(x1)]2dx1 + [(7.5)(103)(x2)]2 dx2 R Ui = B 2EI L0 L0 L0 2EI = Ue = 1.875(109) EI 1 1 P¢ = (20)(103)¢ B = 10(103)¢ B 2 2 Conservation of energy: Ue = Ui 10(103)¢ B = ¢B = 1.875(109) EI 187500 187500 = 0.0117 m = 11.7 mm = EI 200(109)(80)(10 - 6) Ans. *14–36. The rod has a circular cross section with a moment of inertia I. If a vertical force P is applied at A, determine the vertical displacement at this point. Only consider the strain energy due to bending. The modulus of elasticity is E. r A Moment function: P a + ©MB = 0; P[r(1 - cos u)] - M = 0; M = P r (1 - cos u) Bending strain energy: s Ui = M2 ds L0 2 E I ds = r du p u = r M2 r du = [P r (1 - cos u) ]2 du 2 E I L0 L0 2 E I = P2 r3 (1 + cos2 u - 2 cos u)du 2 E I L0 = 1 cos 2u P2 r3 a1 + + - 2 cos u b du 2 E I L0 2 2 = P2 r3 3 cos 2u P2 r3 3 3 p P2 r3 a + - 2 cos ub du = a pb = 2 E I L0 2 2 2 EI 2 4 EI p p p Conservation of energy: Ue = Ui ; ¢A = 1 3 p P2 r3 P ¢A = 2 4 EI 3 p P r3 2 EI Ans. 1187 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1188 © 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. The load P causes the open coils of the spring to make an angle u with the horizontal when the spring is stretched. Show that for this position this causes a torque T = PR cos u and a bending moment M = PR sin u at the cross section. Use these results to determine the maximum normal stress in the material. •14–37. P R d u T = P R cos u; M = P R sin u Bending: smax = Mc P R sin u d = d4 I ) 2 (p4 )(16 tmax = P R cos u d2 Tc = p d4 J ( ) P 2 16 smax = sx + sy 2 ; C a sx - sy 2 2 b + t2xy = 16 P R cos u 2 16 P R sin u 2 16 P R sin u ; b + a b a C p d3 pd3 p d3 = 16 P R [sin u + 1] p d3 Ans. 14–38. The coiled spring has n coils and is made from a material having a shear modulus G. Determine the stretch of the spring when it is subjected to the load P. Assume that the coils are close to each other so that u L 0° and the deflection is caused entirely by the torsional stress in the coil. P R u Bending Strain Energy: Applying 14–22, we have Ui = P2R2L 16P2R2L T2L = = p 2GJ pd4G (d4) D 2G C 32 However, L = n(2pR) = 2npR. Then Ui = P 32nP2R3 d4G External Work: The external work done by force P is Ue = 1 P¢ 2 Conservation of Energy: Ue = Ui 1 32nP2R3 P¢ = 2 d4G ¢ = 64nPR3 d4G Ans. 1188 d 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1189 © 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. 14–39. The pipe assembly is fixed at A. Determine the vertical displacement of end C of the assembly. The pipe has an inner diameter of 40 mm and outer diameter of 60 mm and is made of A-36 steel. Neglect the shearing strain energy. A 800 mm 600 N B 400 mm Internal Loading: Referring to the free-body diagram of the cut segment BC, Fig. a, ©My = 0; My + 600x = 0 My = -600x Referring to the free-body diagram of the cut segment AB, Fig. b, ©Mx = 0; Mx - 600y = 0 Mx = 600y ©My = 0; 600(0.4) - Ty = 0 Ty = 240 N # m p A 0.034 - 0.024 B = 0.325 A 10 - 6 B pm4. We obtain 2 Torsional Strain Energy. J = L (Ui)t = T2dx = L0 2GJ L0 Bending Strain Energy. I = 0.8 m 2 C 75 A 109 B D C 0.325 A 10 - 6 B p D = = = = 0.3009 J p A 0.034 - 0.024 B = 0.1625 A 10 - 6 B pm4. We obtain 4 L (Ui)b = 2402 dx 1 M2dx = B 2EI 2EI L0 L0 0.4 m (-600x)2 dx + L0 0.4 m 0.8 m 1 + 120 A 103 B y3 2 B 120 A 103 B x3 2 R 2EI 0 0 0.8 m A 600y)2 dy R 34 560 N2 # m3 EI 34 560 200 A 10 B c0.1625 A 10 9 -6 Bp d = 0.3385 J Thus, the strain energy stored in the pipe is Ui = (Ui)t + (Ui)b = 0.3009 + 0.3385 = 0.6394 J External Work. The work done by the external force P = 600 N is Ue = 1 1 P¢ = (600)¢ C = 300¢ C 2 2 Conservation of Energy. Ue = Ut 300¢ C = 0.6394 ¢ C = 2.1312 A 10 - 3 B = 2.13 mm Ans. 1189 C 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1190 © 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. *14–40. The rod has a circular cross section with a polar moment of inertia J and moment of inertia I. If a vertical force P is applied at A, determine the vertical displacement at this point. Consider the strain energy due to bending and torsion. The material constants are E and G. z y r P T = Pr(1 - cos u); M = Pr sin u Torsion strain energy: x s Ui = u T2 ds T2 rdu = L0 2GJ L0 2GJ p = r [Pr(1 - cos u)]2 du 2GJ L0 = P2 r3 (1 + cos2 u - 2 cos u)du 2GJ L0 = cos 2u + 1 P2 r3 a1 + - 2 cos ubdu 2GJ L0 2 = 3P2r3 p 4GJ p p Bending strain energy: s Ui = M2ds L0 2EI u p = M2r du r = [Pr sin u]2 du 2EI 2EI L0 L0 = P2 r3 1 - cos 2u P2 r3 p a bdu = 2EI L0 2 4EI p Conservation of energy: Ue = Ui 1 3P2 r3 p P2 r3 p P¢ = + 2 4GJ 4EI ¢ = 1 Pr3 p 3 a + b 2 GJ EI Ans. 1190 A 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1191 © 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. •14–41. Determine the vertical displacement of end B of the frame. Consider only bending strain energy. The frame is made using two A-36 steel W460 * 68 wide-flange sections. 3m B Internal Loading. Using the coordinates x1 and x2, the free-body diagrams of the frame’s segments in Figs. a and b are drawn. For coordinate x1, -M1 - 20 A 103 B x1 = 0 + ©MO = 0; For coordinate x2, + ©MO = 0; M1 = -20 A 103 B x1 M2 - 20 A 103 B (3) = 0 M2 = 60 A 103 B N # m 20 kN A Bending Strain Energy. L (Ub)i = M2dx 1 = B 2EI L0 L0 2EI 400 A 10 1 D£ = 2EI 3 6 = B 3m c -20 A 103 B x1 d dx1 + 3m x1 ≥ 3 3 0 2 + 3.6 A 109 B x 2 4m 0 L0 4m c60 A 103 B d dx2 R 2 T 9 A 109 B N2 # m2 EI For a W460 * 68, I = 297 A 106 B mm4 = 297 A 10 - 6 B m4. Then (Ub)i = 9 A 109 B 200 A 109 B (297) A 10 - 6 B = 151.52 J External Work. The work done by the external force P = 20 kN is Ue = 4m 1 1 P¢ = c20 A 103 B d ¢ B = 10 A 103 B ¢ B 2 2 Conservation of Energy. Ue = Ui 10 A 103 B ¢ B = 151.52 ¢ B = 0.01515 m = 15.2 mm Ans. 1191 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1192 © 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. 14–42. A bar is 4 m long and has a diameter of 30 mm. If it is to be used to absorb energy in tension from an impact loading, determine the total amount of elastic energy that it can absorb if (a) it is made of steel for which Est = 200 GPa, sY = 800 MPa, and (b) it is made from an aluminum alloy for which Eal = 70 GPa, sY = 405 MPa. a) eg = 800(106) sY = 4(10 - 3) m>m = E 200(109) ur = 1 1 (s )(e ) = (800)(106)(N>m2)(4)(10 - 3)m>m = 1.6 MJ>m3 2 Y g 2 V = p (0.03)2(4) = 0.9(10 - 3)p m2 4 ui = 1.6(106)(0.9)(10 - 3)p = 4.52 kJ Ans. b) eg = 405(106) sY = 5.786(10 - 3) m>m = E 70(109) ur = 1 1 (s )(e ) = (405)(106)(N>m2)(5.786)(10 - 3)m>m = 1.172 MJ>m3 2 Y g 2 V = p (0.03)2 (4) = 0.9(10 - 3)p m3 4 ui = 1.172(106)(0.9)(10 - 3)p = 3.31 kJ Ans. 14–43. Determine the diameter of a red brass C83400 bar that is 8 ft long if it is to be used to absorb 800 ft # lb of energy in tension from an impact loading. No yielding occurs. Elastic Strain Energy: The yielding axial force is PY = sgA. Applying Eq. 14–16, we have Ui = (sgA)2L s2gAL N2L = = 2AE 2AE 2E Substituting, we have Ui = 0.8(12) = s2gAL 2E 11.42 C p4 (d2) D (8)(12) 2[14.6(103)] d = 5.35 in. Ans. 1192 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1193 © 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. *14–44. A steel cable having a diameter of 0.4 in. wraps over a drum and is used to lower an elevator having a weight of 800 lb. The elevator is 150 ft below the drum and is descending at the constant rate of 2 ft兾s when the drum suddenly stops. Determine the maximum stress developed in the cable when this occurs. Est = 2911032 ksi, sY = 50 ksi. k = AE = L p 4 (0.42)(29)(103) 150 (12) 150 ft = 2.0246 kip>in. Ue = Ui 1 1 mv2 + W ¢ max = k¢ 2max 2 2 1 800 1 c d[(12) (2)]2 + 800 ¢ max = (2.0246)(103)¢ 2max 2 32.2 (12) 2 596.27 + 800 ¢ max = 1012.29 ¢ 2max ¢ max = 1.2584 in. Pmax = k¢ max = 2.0246 (1.2584) = 2.5477 kip smax = Pmax 2.5477 = 20.3 ksi 6 sg = p 2 A 4 (0.4) O.K. Ans. The composite aluminum bar is made from two segments having diameters of 5 mm and 10 mm. Determine the maximum axial stress developed in the bar if the 5-kg collar is dropped from a height of h = 100 mm. Eal = 70 GPa, sY = 410 MPa. •14–45. 5 mm 200 mm 300 mm ¢ st = © WL = AE Pmax = W B 1 + 5(9.81)(0.2) p 4 (0.0052)(70)(109) C 1 + 2a = 5(9.81) B 1 + smax = Pmax = A p 4 C h bR ¢ st 1 + 2a + 5(9.81)(0.3) p 4 (0.012)(70)(109) 0.1 9.8139(10 - 6) 7051 = 359 MPa 6 sy (0.0052) h = 9.8139(10 - 4) m 10 mm b R = 7051 N O.K. 1193 Ans. 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1194 © 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. 14–46. The composite aluminum bar is made from two segments having diameters of 5 mm and 10 mm. Determine the maximum height h from which the 5-kg collar should be dropped so that it produces a maximum axial stress in the bar of s max = 300 MPa, Eal = 70 GPa, sY = 410 MPa. 5 mm 200 mm 300 mm h 10 mm ¢ st = © WL = AE Pmax = W B 1 + 5(9.81)(0.2) p 2 9 4 (0.005 )(70)(10 ) C 1 + 2a + 5(9.81)(0.3) p 4 (0.012)(70)(109) = 9.8139(10 - 6) m h bR ¢ st p h bR 300(106)a b (0.0052) = 5(9.81) B 1 + 1 + 2a 4 C 9.8139(10 - 6) 120.1 = 1 + 21 + 203791.6 h h = 0.0696 m = 69.6 mm Ans. 1194 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1195 © 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. 14–47. The 5-kg block is traveling with the speed of v = 4 m>s just before it strikes the 6061-T6 aluminum stepped cylinder. Determine the maximum normal stress developed in the cylinder. C B 40 mm Equilibrium. The equivalent spring constant for segments AB and BC are kAB kBC p A 0.022 B c68.9 A 109 B d AAB E 4 = = = 72.152 A 106 B N>m LAB 0.3 p A 0.042 B c68.9 A 109 B d ABC E 4 = = = 288.608 A 106 B N>m LBC 0.3 Equilibrium requires FAB = FBC kAB ¢ AB = kBC ¢ BC 72.152 A 106 B ¢ AB = 288.608 A 106 B ¢ BC ¢ BC = 1 ¢ 4 AB (1) Conservation of Energy. Ue = Ui 1 1 1 mv2 = kAB ¢ AB 2 + kBC ¢ BC 2 2 2 2 (2) Substituting Eq. (1) into Eq. (2), 2 1 1 1 1 mv2 = kAB ¢ AB 2 + kBC a ¢ AB b 2 2 2 4 1 1 1 mv2 = kAB ¢ AB 2 + k ¢ 2 2 2 32 BC AB 1 1 1 (5) A 42 B = c 72.152 A 106 B d ¢ AB 2 + c288.608 A 106 B d ¢ AB 2 2 2 32 ¢ AB = 0.9418 A 10 - 3 B m Maximum Stress. The force developed in segment AB is FAB = kAB ¢ AB = 72.152 A 106 B c0.9418 A 10 - 3 B d = 67.954 A 103 B N. Thus, smax = sAB = 300 mm 300 mm 67.954 A 103 B FAB = = 216.30 MPa = 216 MPa p AAB 2 0.02 A B 4 Since smax 6 sY = 255 MPa, this result is valid. 1195 Ans. v A 20 mm 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1196 © 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. *14–48. Determine the maximum speed v of the 5-kg block without causing the 6061-T6 aluminum stepped cylinder to yield after it is struck by the block. C B 40 mm Equilibrium. The equivalent spring constant for segments AB and BC are kAB kBC p A 0.022 B C 68.9 A 109 B D AAB E 4 = = = 72.152 A 106 B N>m LAB 0.3 p A 0.042 B C 68.9 A 109 B D ABC E 4 = = = 288.608 A 106 B N>m LBC 0.3 Equilibrium requires FAB = FBC kAB ¢ AB = kBC ¢ BC 72.152 A 106 B ¢ AB = 288.608 A 106 B ¢ BC ¢ BC = 1 ¢ 4 AB (1) Conservation of Energy. Ue = Ui 1 1 1 mv2 = kAB ¢ AB 2 + kBC ¢ BC 2 2 2 2 (2) Substituting Eq. (1) into Eq. (2), 2 1 1 1 1 mv2 = kAB ¢ AB 2 + kBC a ¢ AB b 2 2 2 4 1 1 1 mv2 = kAB ¢ AB 2 + k ¢ 2 2 2 32 BC AB 1 1 1 (5)v2 = c72.152 A 106 B d ¢ AB 2 + c288.608 A 106 B d ¢ AB 2 2 2 32 ¢ AB = 0.23545 A 10 - 3 B v Maximum Stress. The force developed in segment AB is FAB = kAB ¢ AB = 72.152 A 106 B C 0.23545 A 10 - 3 B v D = 16988.46v. Thus, smax = sAB = 255 A 106 B = 300 mm 300 mm FAB AAB 16988.46v p A 0.022 B 4 v = 4.716 m>s = 4.72 m>s Ans. 1196 v A 20 mm 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1197 © 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. •14–49. The steel beam AB acts to stop the oncoming railroad car, which has a mass of 10 Mg and is coasting towards it at v = 0.5 m>s . Determine the maximum stress developed in the beam if it is struck at its center by the car. The beam is simply supported and only horizontal forces occur at A and B. Assume that the railroad car and the supporting framework for the beam remains rigid. Also, compute the maximum deflection of the beam. Est = 200 GPa,sY = 250 MPa. 200 mm v ⫽ 0.5 m/s k = 1m B 10(103)(9.81)(23) PL3 = 0.613125(10 - 3) m = 1 48EI )(0.2)(0.23) 48(200)(104)(12 10(103)(9.81) W = 160(106) N>m = ¢ st 0.613125(10 - 3) ¢ max = 0.613125(10 - 3)(0.52) ¢ st v2 = = 3.953(10 - 3) m = 3.95 mm C g C 9.81 Ans. W¿ = k¢ max = 160(106)(3.953)(10 - 3) = 632455.53 N M¿ = 632455.53(2) w¿L = = 316228 N # m 4 4 smax = 316228(0.1) M¿c = 237 MPa 6 sg = 1 3 I 12 (0.2)(0.2 ) A 1m From Appendix C: ¢ st = 200 mm O.K. Ans. 1197 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1198 © 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. 14–50. The aluminum bar assembly is made from two segments having diameters of 40 mm and 20 mm. Determine the maximum axial stress developed in the bar if the 10-kg collar is dropped from a height of h = 150 mm. Take Eal = 70 GPa, sY = 410 MPa. C 1.2 m 40 mm B 0.6 m kAB = kBC = AAB E = LAB p (0.012) C 70(109) D 0.6 = 11.667(106) p N>m A p (0.022) C 70(109) D ABC E = = 23.333 (106) p N>m LBC 1.2 Equilibrium requires FAB = FBC kAB ¢ AB = kBC ¢ BC 11.667(106) p ¢ AB = 23.333(106) p ¢ BC ¢ BC = 0.5 ¢ AB (1) Ue = Ui mg (h + ¢ AB + ¢ BC) = 1 1 k ¢ 2 + kBC ¢ 2BC 2 AB AB 2 (2) Substitute Eq. (1) into (2), mg (h + ¢ AB + 0.5 ¢ AB) = mg (h + 1.5¢ AB) = 1 1 k ¢ 2 + kBC (0.5¢ AB)2 2 AB AB 2 1 k ¢ 2 + 0.125 kBC ¢ 2AB 2 AB AB 10(9.81)(0.15 + 1.5¢ AB) = 1 C 11.667(106)p D ¢ 2AB + 0.125 C 23.333(106)p D ¢ 2AB 2 27.4889 (106)¢ 2AB - 147.15 ¢ AB - 14.715 = 0 ¢ AB = 0.7343(10 - 3) m The force developed in segment AB C 11.667(106)p D C 0.7343(10 - 3) D = 26.915(103) N. Thus smax = sAB = h 20 mm is FAB = kAB ¢ AB = 26.915(103) FAB = 85.67(106)Pa = 85.7 MPa = AAB p (0.012) Since smax 6 sy = 410 MPa, this result is valid 1198 Ans. 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1199 © 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. 14–51. The aluminum bar assembly is made from two segments having diameters of 40 mm and 20 mm. Determine the maximum height h from which the 60-kg collar can be dropped so that it will not cause the bar to yield. Take Eal = 70 GPa, sY = 410 MPa. C 1.2 m 40 mm B 0.6 m kAB = kBC = p(0.012) C 70(109) D AAB E = = 11.667(106) p N>m LAB 0.6 A p(0.022) C 70(109) D ABC E = = 23.333(106) p N>m LBC 1.2 Here, FAB = kAB ¢ AB = C 11.667(106)p D ¢ AB. It is required that smax = sAB = sy. sy = FAB ; AAB 410(106) = C 11.667(106)p D ¢ AB p(0.012) ¢ AB = 0.003514 m Equilibrium requires that FAB = FBC kAB ¢ AB = kBC ¢ BC 11.6667(106)p ¢ AB = 23.333(106)p ¢ BC ¢ BC = 0.5 ¢ B = 0.5(0.003514) = 0.001757 m Ue = Ui mg(h + ¢ AB + ¢ BC) = 1 1 k ¢ 2 + kBC ¢ 2BC 2 AB AB 2 60(9.81)(h + 0.003514 + 0.001757) = 1 C 11.667(106)p D (0.0035142) 2 + h 20 mm The equivalent spring constants for segment AB and BC are 1 C 23.333(106)p D (0.0017572) 2 h = 0.571 m Ans. 1199 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1200 © 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. *14–52. The 50-lb weight is falling at 3 ft>s at the instant it is 2 ft above the spring and post assembly. Determine the maximum stress in the post if the spring has a stiffness of k = 200 kip>in. The post has a diameter of 3 in. and a modulus of elasticity of E = 6.8011032 ksi. Assume the material will not yield. 3 ft/s 2 ft k Equilibrium: This requires Fsp = FP. Hence ksp ¢ sp = kP ¢ P ¢ sp = = and P ksp 2 ft ¢P [1] Conservation of Energy: The equivalent spring constant for the post is kp = AE = L p 4 (32) C 6.80(103) D 2(12) = 2.003 A 106 B lb>in. Ue = Ui 1 1 1 my2 + W(h + ¢ max) = kP ¢ 2P + ksp ¢ 2sp 2 2 2 [2] However, ¢ max = ¢ P + ¢ sp. Then, Eq. [2] becomes 1 1 1 my2 + W A h + ¢ P + ¢ sp B = kP ¢ 2P + ksp ¢ 2sp 2 2 2 [3] Substituting Eq. [1] into [3] yields kp 1 1 1 k2P 2 my2 + W ¢ h + ¢ P + ¢ P ≤ = kP¢ 2P + ¢ ¢ ≤ 2 ksp 2 2 ksp P 2.003(106) 1 50 ¢P R ¢ ≤ A 32 B (12) + 50 B 24 + ¢ p + 2 32.2 200(103) = 1 [2.003(106)]2 1 ≤ ¢ 2P C 2.003 A 106 B D ¢ 2P + ¢ 2 2 200(103) 11.029 A 106 B ¢ 2P - 550.69¢ P - 1283.85 = 0 Solving for positive root, we have ¢ P = 0.010814 in. Maximum Stress: The maximum axial force for the post is Pmax = kp ¢ p = 2.003 A 106 B (0.010814) = 21.658 kip. smax = Pmax 21.658 = p 2 = 3.06 ksi A 4 (3 ) Ans. 1200 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1201 © 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. The 50-kg block is dropped from h = 600 mm onto the bronze C86100 tube. Determine the minimum length L the tube can have without causing the tube to yield. •14–53. A 30 mm 20 mm Maximum Stress. A = p A 0.03 - 0.02 2 B = 0.5 A 10 B p 50(9.81)L WL = 3.0317 A 10 - 6 B L = AE C 0.5 A 10 - 3 B p D C 103 A 109 B D ¢ st = sst = h ⫽ 600 mm -3 2 Section a – a a a L B 50(9.81) W = = 0.3123 MPa A 0.5 A 10 - 3 B p Using these results, n = 1 + C 1 + 2a Then, h 0.6 395 821.46 b = 1 + 1 + 2B R = 1 + 1 + -6 ¢ st C C L 3.0317 A 10 B L smax = sY = nsst 345 = ¢ 1 + A 1 + 395 821.46 ≤ (0.3123) L L = 0.3248 m = 325 mm Ans. 14–54. The 50-kg block is dropped from h = 600 mm onto the bronze C86100 tube. If L = 900 mm, determine the maximum normal stress developed in the tube. A 30 mm 20 mm Maximum Stress. A = p A 0.03 - 0.02 2 ¢ st = sst = 2 h ⫽ 600 mm B = 0.5 A 10 B p -3 50(9.81)(0.9) WL = = 2.7285 A 10 - 6 B AE C 0.5 A 10 - 3 B p D C 103 A 109 B D Section a – a Thus, C B 50(9.81) W = = 0.3123 MPa A 0.5 A 10 - 3 B p 1 + 2a 0.6 h b = 1 + 1 + 2B R = 664.18 ¢ st C 2.7285 A 10 - 6 B smax = nsst = 664.18(0.3123) = 207.40 MPa = 207 MPa Since smax 6 sY = 345 MPa, this result is valid. 1201 a L Using these results, n = 1 + a Ans. 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1202 © 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. 14–55. The steel chisel has a diameter of 0.5 in. and a length of 10 in. It is struck by a hammer that weighs 3 lb, and at the instant of impact it is moving at 12 ft兾s. Determine the maximum compressive stress in the chisel, assuming that 80% of the impacting energy goes into the chisel. Est = 2911032 ksi, sY = 100 ksi. k = AE = L p 4 (0.52)(29)(103) 10 = 569.41 kip>in. 10 in. 0.8 Ue = Ui 0.8c 3 1 1 a b((12)(12))2 + 3¢ max d = (569.41)(103)¢ 2max 2 (32.2)(12) 2 ¢ max = 0.015044 in. P = k¢ max = 569.41(0.015044) = 8.566 kip smax = Pmax 8.566 = p = 43.6 ksi 6 sg 2 A 4 (0.5) O.K. Ans. *14–56. The sack of cement has a weight of 90 lb. If it is dropped from rest at a height of h = 4 ft onto the center of the W10 * 39 structural steel A-36 beam, determine the maximum bending stress developed in the beam due to the impact. Also, what is the impact factor? h 12 ft Impact Factor: From the table listed in Appendix C, ¢ st = 90[24(12)]3 PL3 = 7.3898 A 10 - 3 B in. = 48EI 48[29.0(106)](209) n = 1 + = 1 + C 1 + 2a C 1 + 2a h b ¢ st 4(12) 7.3898(10 - 3) b = 114.98 = 115 Ans. Maximum Bending Stress: The maximum moment occurs at mid-span where 90(24)(12) PL = = 6480 lb # in. Mmax = 4 4 sst = 6480(9.92>2) Mmax c = = 153.78 psi I 209 Thus, smax = nsst = 114.98(153.78) = 17.7 ksi Ans. Since smax 6 sg = 36 ksi, the above analysis is valid. 1202 12 ft 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1203 © 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. The sack of cement has a weight of 90 lb. Determine the maximum height h from which it can be dropped from rest onto the center of the W10 * 39 structural steel A-36 beam so that the maximum bending stress due to impact does not exceed 30 ksi. •14–57. h 12 ft Maximum Bending Stress: The maximum moment occurs at mid-span where 90(24)(12) PL Mmax = = = 6480 lb # in. 4 4 sst = 6480(9.92>2) Mmax c = = 153.78 psi I 209 However, smax = nsst 30 A 103 B = n(153.78) n = 195.08 Impact Factor: From the table listed in Appendix C, ¢ st = 90[24(12)]3 PL3 = 7.3898 A 10 - 3 B in. = 48EI 48[29.0(106)](209) n = 1 + 195.08 = 1 + C C 1 + 2a h b ¢ st 1 + 2a h b 7.3898(10 - 3) h = 139.17 in. = 11.6 ft Ans. 1203 12 ft 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1204 © 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. 14–58. The tugboat has a weight of 120 000 lb and is traveling forward at 2 ft兾s when it strikes the 12-in.-diameter fender post AB used to protect a bridge pier. If the post is made from treated white spruce and is assumed fixed at the river bed, determine the maximum horizontal distance the top of the post will move due to the impact. Assume the tugboat is rigid and neglect the effect of the water. 3 ft A C 12 ft B From Appendix C: Pmax = 3EI(¢ C)max (LBC)3 Conservation of energy: 1 1 mv2 = Pmax (¢ C)max 2 2 1 1 3EI(¢ C)2max b mv2 = a 2 2 (LBC)3 (¢ C)max = (¢ C)max = Pmax = uC = mv2L3BC C 3EI (120 000>32.2)(2)2(12)3 C (3)(1.40)(106)(144)(p4 )(0.5)4 3[1.40(106)](p4 )(6)4(11.177) (144)3 = 0.9315 ft = 11.177 in. = 16.00 kip 16.00(103)(144)2 PmaxL2BC = = 0.11644 rad 2EI 2(1.40)(106)(p4 )(6)4 (¢ A)max = (¢ C)max + uC(LCA) (¢ A)max = 11.177 + 0.11644(36) = 15.4 in. Ans. 1204 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1205 © 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. 14–59. The wide-flange beam has a length of 2L, a depth 2c, and a constant EI. Determine the maximum height h at which a weight W can be dropped on its end without exceeding a maximum elastic stress s max in the beam. W h A 2c B L L 1 1 (-Px)2 dx P¢ C = 2 a b 2 2EI L0 ¢C = 2PL3 3EI ¢ st = 2WL3 3EI n = 1 + C 1 + 2a h b ¢ st smax = n(sst)max smax = B 1 + a C 1 + 2a (sst)max = WLc I h WLc b R ¢ st I 2 smax I 2h - 1b = 1 + WLc ¢ st h = = 2 ¢ st smax I - 1b - 1 R Ba 2 WLc smax I 2 2smaxI smax L2 smax I WL3 b - 2R Ba R = B 3EI WLc WLc 3Ec WLc Ans. 1205 L 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1206 © 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. *14–60. The 50-kg block C is dropped from h = 1.5 m onto the simply supported beam. If the beam is an A-36 steel W250 * 45 wide-flange section, determine the maximum bending stress developed in the beam. C h Equilibrium. Referring to the free-body diagram of the beam under static A condition, Fig. a a + ©MA = 0; By(6) - P(4) = 0 By = 2 P 3 4m Then, the maximum moment in the beam occurs at the position where P is applied. Thus, Mmax = By(2) = 2 4 P(2) = P 3 3 Impact Factor. From the table listed in the appendix, the deflection of the beam at Pba the point of application of P is ¢ = A L2 - b2 - a2 B , where P = 50(9.81) 6EIL = 490.5 N, L = 6 m, a = 4 m, and b = 2 m. From the table listed in the appendix, the necessary section properties for a W250 * 45 are d = 266 mm = 0.266 m and Ix = 71.1 A 106 B mm4 = 71.1 A 10 - 6 B m4. Then ¢ st = 490.5(2)(4) 6c 200 A 10 B d c 71.1 A 10 9 -6 B d(6) A 62 - 22 - 42 B = 0.1226 A 10 - 3 B m We have, n = 1 + Maximum C 1 + 2¢ Stress. h 1.5 1 + 2C S = 157.40 ≤ = 1 + ¢ st Q 0.1226 A 10 - 3 B The maximum = 157.40(490.5) = 77.21 A 10 B N. 3 = 102.94 A 10 smax = 3 force Then, on B N # m. Applying the flexure formula, beam is Pmax = nP 4 4 = Pmax = C 77.21 A 103 B D 3 3 the Mmax 102.94 A 103 B (0.266>2) Mmaxc = = 192.56 MPa = 193 MPa I 71.1 A 10 - 6 B Since smax 6 sY = 250 MPa, this result is valid. 1206 B Ans. 2m 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1207 © 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. Determine the maximum height h from which the 50-kg block C can be dropped without causing yielding in the A-36 steel W310 * 39 wide flange section when the block strikes the beam. •14–61. C h A B 4m Equilibrium. Referring to the free-body diagram of the beam under static condition, Fig. a a + ¢MA = 0; By(6) - P(4) = 0 By = 2 P 3 Then, the maximum moment in the beam occurs at the position where P is applied. Thus, Mmax = By(2) = 4 2 P(2) = P 3 3 Maximum Stress. Since P = 50(9.81) = 490.5 N. Then the maximum force on the 4 4 beam is Pmax = nP = 490.5n and Mmax = P = (490.5n) = 654n. From the 3 3 table listed in the appendix, the necessary section properties for a W310 * 39 are d = 310 mm = 0.31 m and Ix = 84.8 A 106 B mm4 = 84.8 A 10 - 6 B m4. Applying the flexure formula, smax = Mmax c I 250 A 106 B = 654n(0.31>2) 84.8 A 10 - 6 B n = 209.13 Impact Factor. From the table listed in the appendix, the deflection of the beam at Pba the point of where P is applied is ¢ = A L2 - b2 - a2 B , where L = 6 m, 6EIL a = 4 m, and b = 2 m. Then ¢ st = 490.5(2)(4) 6 C 200 A 109 B D C 84.8 A 10 - 6 B D (6) A 62 - 22 - 42 B = 0.1028 A 10 - 3 B m We have, n = 1 + C 1 + 2¢ 209.13 = 1 + S h ≤ ¢ st 1 + 2C h = 2.227 m = 2.23 m h 0.1028 A 10 - 3 B S Ans. Since smax 6 sY = 250 MPa, this result is valid. 1207 2m 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1208 © 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. 14–62. The diver weighs 150 lb and, while holding himself rigid, strikes the end of a wooden diving board 1h = 02 with a downward velocity of 4 ft>s. Determine the maximum bending stress developed in the board. The board has a thickness of 1.5 in. and width of 1.5 ft. Ew = 1.811032 ksi, sY = 8 ksi. v h 4 ft Static Displacement: The static displacement at the end of the diving board can be determined using the conservation of energy. L 1 M2 dx P¢ = 2 L0 2EI 1 1 (150)¢ st = c 2 2EI L0 ¢ st = = 4 ft (-375x1)2 dx1 + 70.0(103) lb # ft3 EI L0 10 ft (-150x2) dx2 d 70.0(103)(123) 1 (18)(1.53) D 1.8(106) C 12 = 13.274 in. Conservation of Energy: The equivalent spring constant for the board is W 150 k = = = 11.30 lb>in., ¢ st 13.274 Ue = Ui 1 1 my2 + W¢ max = k¢ 2max 2 2 c 1 150 1 a b A 42 B d (12) + 150¢ max = (11.30)¢ 2max 2 32.2 2 Solving for the positive root, we have ¢ max = 29.2538 in. Maximum Stress: The maximum force on to the beam is Pmax = k¢ max = 11.30(29.2538) = 330.57 lb. The maximum moment occurs at the middle support Mmax = 330.57(10)(12) = 39668.90 lb # in. smax = 39668.90(0.75) Mmax c = 5877 psi = 5.88 ksi = 1 3 I 12 (18)(1.5 ) Ans. Note: The result will be somewhat inaccurate since the static displacement is so large. 1208 10 ft 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1209 © 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. 14–63. The diver weighs 150 lb and, while holding himself rigid, strikes the end of the wooden diving board. Determine the maximum height h from which he can jump onto the board so that the maximum bending stress in the wood does not exceed 6 ksi. The board has a thickness of 1.5 in. and width of 1.5 ft. Ew = 1.811032 ksi. v Static Displacement: The static displacement at the end of the diving board can be determined using the conservation of energy. h L M2 dx 1 P¢ = 2 L0 2EI 1 1 (150)¢ st = c 2 2EI L0 ¢ st = = 4 ft 4 ft (-375x1)2 dx1 + 70.0(103) lb # ft3 EI L0 10 ft (-150x2) dx2 d 70.0(103)(123) 1 (18)(1.53) D 1.8(106) C 12 = 13.274 in. Maximum Stress: The maximum force on the beam is Pmax. The maximum moment occurs at the middle support Mmax = Pmax (10)(12) = 120Pmax. smax = 6 A 103 B = Mmax c I 120Pmax (0.75) 1 12 (18)(1.53) Pmax = 337.5 lb Conservation of Energy: The equivalent spring constant for the board is 150 W = = 11.30 lb>in.. The maximum displacement at the end of the k = ¢ st 13.274 Pmax 337.5 board is ¢ max = = = 29.687 in. k 11.30 Ue = Ui W(h + ¢ max) = 150(h + 29.867) = 1 k¢ 2max 2 1 (11.30) A 29.8672 B 2 h = 3.73 in. Ans. Note: The result will be somewhat at inaccurate since the static displacement is so large. 1209 10 ft 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1210 © 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. *14–64. The weight of 175 lb is dropped from a height of 4 ft from the top of the A-36 steel beam. Determine the maximum deflection and maximum stress in the beam if the supporting springs at A and B each have a stiffness of k = 500 lb>in. The beam is 3 in. thick and 4 in. wide. 4 ft A k 8 ft 8 ft From Appendix C: ¢ beam = PL3 48EI kbeam = 1 )(4)(33) 48(29)(103)(12 48EI = = 1.7700 kip>in. 3 L (16(12))3 From equilibrium (equivalent system): 2Fsp = Fbeam 2ksp ¢ sp = kbeam ¢ beam ¢ sp = 1.7700(103) ¢ beam 2(500) ¢ sp = 1.7700¢ beam (1) Conservation of energy: Ue = Ui W(h + ¢ sp + ¢ beam) = 1 1 k ¢2 + 2 a bksp ¢ 2sp 2 beam beam 2 From Eq. (1): 175[(4)(12) + 1.770¢ beam + ¢ beam] = 1 (1.7700)(103)¢ 2beam + 500(1.7700¢ beam)2 2 2451.5¢ 2beam - 484.75¢ beam - 8400 = 0 ¢ beam = 1.9526 in. From Eq. (1): ¢ sp = 3.4561 in. ¢ max = ¢ sp + ¢ beam = 3.4561 + 1.9526 = 5.41 in. Ans. Fbeam = kbeam ¢ beam = 1.7700(1.9526) = 3.4561 kip Mmax = smax = 3.4561(16)(12) Fbeam L = = 165.893 kip # in. 4 4 165.893(1.5) Mmax c = 27.6 ksi 6 sg = 1 3 I 12 (4)(3 ) O.K. Ans. 1210 3 in. B k 4 in. 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1211 © 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. The weight of 175 lb, is dropped from a height of 4 ft from the top of the A-36 steel beam. Determine the load factor n if the supporting springs at A and B each have a stiffness of k = 300 lb/in. The beam is 3 in. thick and 4 in. wide. •14–65. 4 ft A k 8 ft 8 ft From Appendix C: ¢ beam = PL3 48EI kbeam = 1 )(4)(33) 48(29)(103)(12 48EI = = 1.7700 kip>in. L3 (16(12))3 From equilibrium (equivalent system): 2Fsp = Fbeam 2ksp ¢ sp = kbeam ¢ beam ¢ sp = 1.7700(103) ¢ beam 2(300) ¢ sp = 2.95¢ beam (1) Conservation of energy: Ue = Ui W(h + ¢ beam + ¢ sp) = 1 1 k ¢2 + 2a b ksp ¢ 2sp 2 beam beam 2 From Eq. (1): 175[(4)(12) + ¢ beam + 2.95¢ beam] = 1 (1.7700)(103)¢ 2beam + 300(2.95¢ beam)2 2 3495.75¢ 2beam - 691.25¢ beam - 8400 = 0 ¢ beam = 1.6521 in. Fbeam = kbeam ¢ beam = 1.7700(1.6521) = 2.924 kip n = 2.924(103) = 16.7 175 smax = n(sst)max = na M = Ans. Mc b I 175(16)(12) = 8.40 kip # in. 4 smax = 16.7 a 8.40(1.5) 1 12 (4)(33) b = 23.4 ksi 6 sg O.K. 1211 3 in. B k 4 in. 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1212 © 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. 14–66. Block C of mass 50 kg is dropped from height h = 0.9 m onto the spring of stiffness k = 150 kN>m mounted on the end B of the 6061-T6 aluminum cantilever beam. Determine the maximum bending stress developed in the beam. C h a k 100 mm B Conservation of Energy. From the table listed in the appendix, the PL . Thus, the = 3EI equivalent spring constant for the beam is kb = 1 (0.1) A 0.23 B = 66.6667 A 10 - 6 B m4, 12 a 3 displacement of end B under static conditions is ¢ st I = A L = 3 m, 3m 3EI , where L3 and E = Eal = 68.9 GPa. Thus, 3EI kb = = L3 3 c68.9 A 109 B d c66.6667 A 10 - 6 B d 33 = 510.37 A 103 B N>m Equilibrium requires, Fsp = P ksp ¢ sp = kb ¢ b 150 A 103 B ¢ sp = 510.37 A 103 B ¢ b ¢ sp = 3.4025¢ b (1) We have, Ue = Ui mg A h + ¢ sp + ¢ b B = 1 1 k ¢ 2 + ksp ¢ sp 2 2 b b 2 Substituting Eq. (1) into this equation, 50(9.81)(0.9 + 3.4025¢ b + ¢ b) = 1 1 c510.37 A 103 B d ¢ b 2 + c150 A 103 B d(3.4025¢ b)2 2 2 1123444.90¢ b 2 - 2159.41¢ b - 441.45 = 0 Solving for the positive root, ¢ b = 0.020807 m Maximum Stress. The maximum force Pmax = kb ¢ b = 510.37 A 10 B (0.020807) = 10.619 A 10 3 occurs at = 31.858 A 103 B smax = 3 on the beam is B N. The maximum moment Mmax = Pmax L = 10.619 A 103 B (3) 0.2 N # m. Applying the flexure formula with c = = 0.1 m, 2 fixed support A, where 31.858 A 103 B (0.1) Mmax c = = 47.79 MPa = 47.8 MPa I 66.6667 A 10 - 6 B Since smax 6 sY = 255 MPa, this result is valid. 1212 Ans. 200 mm Section a – a 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1213 © 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. 14–67. Determine the maximum height h from which 200-kg block C can be dropped without causing the 6061-T6 aluminum cantilever beam to yield. The spring mounted on the end B of the beam has a stiffness of k = 150 kN>m. C h a k 100 mm B A a 3m Maximum Stress. From the table listed in the appendix, the displacement of end B PL3 under static conditions is ¢ st = . Thus, the equivalent spring constant for the 3EI 3EI 1 beam is kb = , where I = (0.1) A 0.23 B = 66.6667 A 10 - 6 B m4, L = 3 m, and 12 L3 E = Eal = 68.9 GPa. Thus, 3EI = kb = L3 3 c68.9 A 109 B d c66.6667 A 10 - 6 B d 33 = 510.37 A 103 B N>m The maximum force on the beam is Pmax = kb ¢ b = 510.37 A 103 B ¢ b. The maximum moment occurs at the fixed support A, where Mmax = Pmax L = 510.37 A 103 B ¢ b(3) = 1.5311 A 106 B ¢ b. Applying the flexure formula with smax = sY = 255 MPa and 0.2 c = = 0.1 m, 2 smax = sY = 255 A 106 B = Mmax c I 1.5311 A 106 B ¢ b(0.1) 66.6667 A 10 - 6 B ¢ b = 0.11103 m Equilibrium requires, Fsp = P ksp ¢ sp = kb ¢ b 150 A 103 B ¢ sp = 510.37 A 103 B (0.11103) ¢ sp = 0.37778 m Conservation of Energy. Ue = Ui mg A h + ¢ sp + ¢ b B = 1 1 kb ¢ b 2 + ksp ¢ sp 2 2 2 200(9.81)(h + 0.37778 + 0.11103) = 1 1 c510.37 A 103 B d(0.11103)2 + c150 A 103 B d(0.37778)2 2 2 h = 6.57 m Ans. 1213 200 mm Section a – a 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1214 © 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. *14–68. The 2014-T6 aluminum bar AB can slide freely along the guides mounted on the rigid crash barrier. If the railcar of mass 10 Mg is traveling with a speed of v = 1.5 m>s, determine the maximum bending stress developed in the bar. The springs at A and B have a stiffness of k = 15 MN>m. 300 mm k A v 2m 2m a a k B Equilibrium. Referring to the free-body diagram of the bar for static conditions, Fig. a, + ©F = 0; : x 2Fsp - P = 0 Fsp = P 2 (1) 300 mm 400 mm Referring to the table listed in the appendix, the displacement of the bar at the PL3 position where P is applied under static conditions is ¢ st = . Thus, the 48EI 1 48EI equivalent spring constant for the bar is kb = , where I = (0.4) A 0.33 B 12 L3 and Thus, = 0.9 A 10 - 3 B m4, L = 4 m, E = Eal = 73.1 GPa. kb = 48 c73.1 A 109 B d c0.9 A 10 - 3 B d 43 = 49.3425 A 106 B N>m Using Eq. (1) Fsp = P 2 ksp ¢ sp = 1 k ¢ 2 b b 1 49.3425 A 10 1 kb ¢ ≤ ¢b = C 2 ksp 2 15 A 106 B 6 ¢ sp = B S ¢ b = 1.64475¢ b Conservation of Energy. 1 1 1 mv2 = kb ¢ b 2 + 2 c ksp ¢ sp 2 d 2 2 2 Substituting Eq. (2) into this equation, 1 1 mv2 = kb ¢ b 2 + ksp (1.64475¢ b)2 2 2 1 1 mv2 = kb ¢ b 2 + 2.7052ksp ¢ b 2 2 2 1 1 c10 A 103 B d A 1.52 B = c49.3425 A 106 B d ¢ b 2 + 2.7052c15 A 106 B d ¢ b 2 2 2 ¢ b = 0.01313 m 1214 (2) Section a – a 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1215 © 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. 14–68. Continued Maximum Stress. The maximum force on the bar is (Pb)max = kb ¢ b = 49.3425 A 106 B (0.01313) = 647.90 A 103 B N. The maximum moment occurs at the midspan of the bar, where Mmax = = 647.90 A 103 B N # m. Applying the flexure formula, smax = 647.90 A 103 B (4) (Pb)max L = 4 4 647.90 A 103 B (0.15) Mmax c = = 107.98 MPa = 108 MPa I 0.9 A 10 - 3 B Since smax 6 sY = 414 MPa, this result is valid. 1215 Ans. 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1216 © 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. •14–69. The 2014-T6 aluminum bar AB can slide freely along the guides mounted on the rigid crash barrier. Determine the maximum speed v the 10-Mg railcar without causing the bar to yield when it is struck by the railcar. The springs at A and B have a stiffness of k = 15 MN>m. 300 mm k A v 2m 2m a a k B Equilibrium. Referring to the free-body diagram of the bar for static conditions, Fig. a, + ©F = 0; : x 2Fsp - P = 0 Fsp P = 2 (1) 300 mm 400 mm Referring to the table listed in the appendix, the displacement of the bar at the PL3 position where P is applied under static conditions is ¢ st = . Thus, the 48EI 1 48EI equivalent spring constant for the bar is kb = , where I = (0.4) A 0.33 B 12 L3 and Thus, = 0.9 A 10 - 3 B m4, L = 4 m, E = Eal = 73.1 GPa. kb = 48 c73.1 A 109 B d c0.9 A 10 - 3 B d 43 = 49.3425 A 106 B N>m Using Eq. (1) Fsp = P 2 ksp ¢ sp = 1 k ¢ 2 b b 1 49.3425 A 10 1 kb ¢ ≤ ¢b = C 2 ksp 2 15 A 106 B 6 B S ¢ b = 1.64475¢ b (2) Maximum Stress. The maximum force on the bar is (Pb)max = kb ¢ b ¢ sp = = 49.3425 A 106 B ¢ b. The maximum moment occurs at the midspan of the bar, where 49.3425 A 106 B ¢ b(4) (Pb)max L = = 49.3425 A 106 B ¢ b. Applying the flexure Mmax = 4 4 formula with smax = sY = 414 MPa, smax = Mmax c I 414 A 106 B = 49.3425 A 106 B ¢ b (0.15) 0.9 A 10 - 3 B ¢ b = 0.050342 m 1216 Section a – a 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1217 © 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. 14–69. Continued Substituting this result into Eq. (2), ¢ sp = 0.0828 m Conservation of Energy. 1 1 1 mv2 = kb ¢ b 2 + 2 B ksp ¢ sp 2 R 2 2 2 1 1 1 c 10 A 103 B dv2 = c49.3425 A 106 B d A 0.0503422 B + 2 B c15 A 106 B d A 0.08282 B R 2 2 2 v = 5.75 m>s Ans. 1217 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1218 © 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. 14–70. The simply supported W10 * 15 structural A-36 steel beam lies in the horizontal plane and acts as a shock absorber for the 500-lb block which is traveling toward it at 5 ft兾s. Determine the maximum deflection of the beam and the maximum stress in the beam during the impact. The spring has a stiffness of k = 1000 lb>in. For W 10 * 15: I = 68.9 in4 12 ft v ⫽ 5 ft/s d = 9.99 in. k From Appendix C: ¢ beam = PL3 48EI kbeam = 48(29)(103)(68.9) 48EI = = 4.015 kip>in. L3 (24(12))3 12 ft Equilibrium (equivalent system): Fsp = Fbeam ksp ¢ sp = kbeam ¢ beam ¢ sp = 4.015(103) ¢ beam 1000 ¢ sp = 4.015¢ beam (1) Conservation of energy: Ue = Ui 1 1 1 mv2 = kbeam ¢ 2beam + ksp ¢ 2sp 2 2 2 From Eq. (1): 500 1 1 1 a b (5(12))2 = (4.015)(103)¢ 2beam + (1000)(4.015¢ beam)2 2 32.2(12) 2 2 10067.6¢ 2beam = 2329.2 ¢ beam = 0.481 in. Ans. Fbeam = kbeam ¢ beam = 4.015(0.481) = 1.931 kip Mmax = a smax = 1.931 b (12) (12) = 139.05 kip # in. 2 139.05(9.99 Mmax c 2 ) = = 10.1 ksi 6 sg I 68.9 O.K. Ans. 1218 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1219 © 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. 14–71. The car bumper is made of polycarbonatepolybutylene terephthalate. If E = 2.0 GPa, determine the maximum deflection and maximum stress in the bumper if it strikes the rigid post when the car is coasting at v = 0.75 m/s. The car has a mass of 1.80 Mg, and the bumper can be considered simply supported on two spring supports connected to the rigid frame of the car. For the bumper take I = 30011062 mm4, c = 75 mm, sY = 30 MPa and k = 1.5 MN>m. 0.9 m k 0.9 m k v ⫽ 0.75 m/s Equilibrium: This requires Fsp = ksp ¢ sp = k¢ beam 2 Pbeam . Then 2 or ¢ sp = k ¢ 2ksp beam [1] Conservation of Energy: The equivalent spring constant for the beam can be determined using the deflection table listed in the Appendix C. k = 48 C 2(109) D C 300(10 - 6) D 48EI = = 4 938 271.6 N>m L3 1.83 Thus, Ue = Ui 1 1 1 my2 = k¢ 2beam + 2 a ksp ¢ 2sp b 2 2 2 [2] Substitute Eq. [1] into [2] yields 1 1 k2 mv2 = k¢ 2beam + ¢2 2 2 4ksp beam (4 93 8271.6)2 2 1 1 ¢ beam (1800) A 0.752 B = (493 8271.6) ¢ 2beam + 2 2 4[1.5(106)] ¢ beam = 8.8025 A 10 - 3 B m Maximum Displacement: From Eq. [1] ¢ sp = 4 938 271.6 2[1.5(106)] C 8.8025 A 10 - 3 B D = 0.014490 m. ¢ max = ¢ sp + ¢ beam = 0.014490 + 8.8025 A 10 - 3 B = 0.02329 m = 23.3 mm Ans. Maximum Stress: The maximum force on the beam is Pbeam = k¢ beam = 4 938 271.6 C 8.8025 A 10 - 3 B D = 43 469.3 N. The maximum moment occurs at mid-span. Mmax = smax = 43 469.3(1.8) Pbeam L = = 19 561.2 N # m. 4 4 19 561.2(0.075) Mmax c = 4.89 MPa = I 300(10 - 6) Ans. Since smax 6 sg = 30 MPa, the above analysis is valid. 1219 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1220 © 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. *14–72. Determine the horizontal displacement of joint B on the two-member frame. Each A-36 steel member has a cross-sectional area of 2 in2. 800 lb B 30⬚ 60⬚ C A Member n N L nNL AB 1.1547 800 120 11085.25 BC –0.5774 0 60 0 5 ft © = 110851.25 1 # ¢ Bh = © ¢ Bh = nNL AE 110851.25 110851.25 = 0.00191 in. = AE 29(106)(2) Ans. Determine the horizontal displacement of point B. Each A-36 steel member has a cross-sectional area of 2 in2. •14–73. B 200 lb Member Real Forces N: As shown on figure(a). 8 ft Member Virtual Forces n: As shown on figure(b). nNL 1#¢ = a AE 1 lb # (¢ B)h = 6 ft 1 [0.8333(166.67)(10)(12) AE +( -0.8333)(-166.67)(10)(12) +0.500(100)(12)(12)] 1 lb # (¢ B)h = (¢ B)h = 40533.33 lb2 # in AE 40533.33 2[29.0(106)] A C Virtual-Work Equation: = 0.699 A 10 - 3 B in. : Ans. 1220 6 ft 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1221 © 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. 14–74. Determine the vertical displacement of point B. Each A-36 steel member has a cross-sectional area of 2 in2. B 200 lb Member Real Forces N: As shown on figure(a). 8 ft Member Virtual Forces n: As shown on figure(b). A C Virtual-Work Equation: nNL 1#¢ = a AE 1 lb # (¢ B)v = 6 ft 6 ft 1 [(-0.625)(166.67)(10)(12) AE +(-0.625)(-166.67)(10)(12) +0.375(100)(12)(12)] 1 lb # (¢ B)v = (¢ B)v = 5400 lb2 # in AE 5400 2[29.0(106)] = 0.0931 A 10 - 3 B in. T Ans. 14–75. Determine the vertical displacement of joint C on the truss. Each A-36 steel member has a cross-sectional area of A = 300 mm2. Member n N L 3m C A 1.50 45.0 3 202.5 AD 0 18.03 213 0 BC 1.50 45.0 3 202.5 BD 0 –20.0 2 0 CD –1.803 –54.08 213 351.56 DE –1.803 –72.11 213 468.77 4m D E © = 1225.33 ¢ Cv = nNL AE 1225.33(103) 300(10 - 6)(200)(109) 3m B nNL AB 1 # ¢ Cv = © 30 kN 20 kN = 0.0204 m = 20.4 mm Ans. 1221 14 Solutions 46060_Part1 6/11/10 8:18 AM Page 1222 © 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. *14–76. Determine the vertical displacement of joint D on the truss. Each A-36 steel member has a cross-sectional area of A = 300 mm2. 30 kN 20 kN 3m 3m B C A Member n N L nNL 4m AB 0 45.0 3 0 AD 0.9014 18.03 213 58.60 BC 0 45.0 3 0 BD 0 –20.0 2 0 CD 0 –54.08 213 0 DE –0.9014 –72.11 213 234.36 D E © = 292.96 1 # ¢ Dv = © ¢ Dv = nNL AE 292.96(103) 300(10 - 6)(200)(109) = 4.88(10 - 3) m = 4.88 mm Ans. Determine the vertical displacement of point B. Each A-36 steel member has a cross-sectional area of 4.5 in2. •14–77. F E D Virtual-Work Equation: Applying Eq. 14–39, we have Member n AB BC 6 ft N L nNL 0.6667 3.333 96 213.33 0.6667 3.333 96 213.33 A C B CD 0 0 72 0 DE 0 0 96 0 EF 0 0 96 0 AF 0 0 72 0 AE –0.8333 –4.167 120 416.67 CE –0.8333 –4.167 120 416.67 BE 1.00 5.00 72 360.00 8 ft ©1620 kip2 # in. nNL 1#¢ = a AE 1 kip # (¢ B)v = (¢ B)v = 8 ft 5 kip 1620 kip2 # in. AE 1620 = 0.0124 in. T 4.5[29.0(103)] Ans. 1222 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 6 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 0 72 0 DE 0 0 96 0 EF 0 0 96 0 AF 0 0 72 0 AE –0.8333 –4.167 120 416.67 CE –0.8333 –4.167 120 416.67 BE 0 5.00 72 0 A C B 8 ft 8 ft 5 kip ©1260 kip2 # in. nNL 1#¢ = a AE 1 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. 5 kN 4 kN 2m C B Member n N L nNL AB 0 0 1.5 0 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 1 # ¢ Bh = © ¢ Bh = nNL AE 29.375(103) 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. 5 kN 2m 4 kN C B Member n N L nNL AB 0 0 1.5 0 AC 0 –5.00 2.5 0 AD 0 4.00 2.0 0 BC 0 4.00 2.0 0 CD –1.00 –2.00 1.5 3.00 1.5 m D A © = 3.00 1 # ¢ Cv = © ¢ Cv = nNL AE 3.00 (103) 400(10 - 6)(200)(109) = 37.5(10 - 6)m = 0.0375 mm Ans. Determine the vertical displacement of point A. Each A-36 steel member has a cross-sectional area of 400 mm2. •14–81. E D Virtual-Work Equation: Member n AB –0.750 BC –0.750 AE 1.25 CE –1.25 BE 0 DE 2m N 1.50 –22.5 A 103 B –22.5 A 103 B 37.5 A 10 3 –62.5 A 10 3 60.0 A 10 3 B B 22.0 A 103 B B nNL 1#¢ = a AE 1 N # (¢ A)v = (¢ A)v = L 1.5 1.5 2.5 2.5 2 1.5 nNL 25.3125 A 103 B C A 25.3125 A 103 B 117.1875 A 10 3 195.3125 A 10 3 135.00 A 10 3 B 1.5 m B B 30 kN 0 B © 498.125 A 103 B N2 # m 498.125(103) N2 # m AE 498.125(103) 0.400(10 - 3)[200(109)] = 6.227 A 10 - 3 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 0 BC 0 AE 0 CE –1.25 BE 1.00 DE 0.750 –22.5 A 103 B –22.5 A 10 3 –62.5 A 10 3 B 37.5 A 103 B B 22.0 A 103 B 60.0 A 103 B nNL 1#¢ = a AE 1 N # (¢ B)v = (¢ B)v = L 30 kN nNL 1.5 0 1.5 0 2.5 2.5 2 1.5 195.3125 A 10 0 3 B 40.0 A 103 B 67.5 A 103 B © 302.8125 A 103 B N2 # m 302.8125(103) N2 # m AE 302.8125(103) 0.400(10 - 3)[200(109)] = 3.785 A 10 - 3 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. 1 # ¢ Cv = © ¢ 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 9 ft 21 232 = 0.163 in. 4.5 (29(103)) 1 # ¢ Nv = © I C 12 ft 12 ft 12 ft 6 kip 8 kip 6 kip I H G J E D F 9 ft A nNL AE 12 ft 20 368 = 0.156 in. 4.5 (29(103)) Ans. 1226 C B 12 ft 6 kip 12 ft 8 kip E D 12 ft 6 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. Determine the vertical displacement of joint C. The truss is made from A-36 steel bars having a cross- sectional area of 150 mm2. •14–85. 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 Virtual Work Equation. Since smax = Member AB n(N) N(N) 9 A 10 0.375 DE 0.375 BC 0.375 CD 0.375 AH –0.625 EF –0.625 BH 0 DF 0 CH 0 CF 0 GH –0.625 FG –0.625 CG 1 0.15 A 10 - 3 B 9 A 10 3 3 B B 9 A 103 B 9 A 103 B –15 A 103 B –15 A 103 B 6 A 103 B 6 A 10 3 –3.75 A 10 3 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) 2 # m) 5.0625 A 10 1.5 5.0625 A 10 1.5 3 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 0 2 0 2.5 0 2.5 0 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#¢ = © 1N # (¢ C)v nNL AE = 174.28125 A 103 B 0.15 A 10 - 3 B C 200 A 109 B D (¢ C)v = 5.809 A 10 - 3 B m = 5.81 mm T Ans. 1227 B 1.5 m 1.5 m 6 kN 5.0625 A 103 B 1.5 4 E A = 120 MPa 6 sY = 250 MPa, C 1.5 m 12 kN D 1.5 m 6 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 Since Equation. 18 A 10 B FCG = 120 MPa 6 sY = 250 MPa, = = A 0.15 A 10 - 3 B 3 smax Member n(N) AB 0.375 DE 0.375 BC 0.375 CD 0.375 AH –0.625 EF –0.625 BH 0 DF 0 CH 0 CF 0 GH FG CG –0.625 –0.625 N(N) 9 A 103 B 9 A 103 B 9 A 103 B 9 A 103 B –15 A 103 B –15 A 103 B 6 A 103 B 6 A 103 B –3.75 A 10 3 –11.25 A 10 3 B –3.75 A 103 B –11.25 A 10 18 A 10 0 3 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 0 2 0 2.5 0 2.5 17.578125 A 10 2.5 17.578125 A 10 2.5 4 0 3 3 B B 0 ©102.28125 A 103 B Then 1#¢ = © nNL AE 1N # (¢ G)v = 1.5 m 1.5 m 6 kN nNL(N 2 # m) L(m) B 102.28125 A 103 B 0.15 A 10 - 3 B C 200 A 109 B D (¢ G)v = 3.409 A 10 - 3 B m = 3.41 mm T Ans. 1228 C 1.5 m 12 kN D 1.5 m 6 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 1 # ¢C = mM dx L0 EI a 1 1 a x1 b(Px1)dx1 + bc EI L0 2 L0 a>2 1 (a + x2)(Pa)dx2 d 2 23Pa3 24EI Ans. *14–88. The beam is made of southern pine for which Ep = 13 GPa. Determine the displacement at A. 1 # ¢A = 15 kN 4 kN/m A L mM L0 EI B 1.5 m 1.5 ¢A = = a L ¢C = 2 a = a– 2 C a– 2 C 3m 3 1 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 = 1 9 EI )(0.12)(0.18)3 13(10 )(12 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. 2 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. 1#¢ = 1 kN # ¢ C = ¢C = = L mM dx L0 EI 1 EI L0 10 m 0.500x1 (2.50x1)dx1 + 1 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. 2 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. 1#u = 1 kN # m # uA = uA = = L muM dx L0 EI 1 EI L0 10 m (1 - 0.100x1)(2.50x1) dx1 + 1 EI L0 5m 0 A 1.00x22 B dx2 41.667 kN # m2 EI 41.667(1000) 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. 2 kN/m Real Moment Function M(x): As shown on figure(a). C 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 1#u = 1 kN # m # uB = uB = = muM dx L0 EI 1 EI L0 10 m 0.100x1(2.50x1) dx1 + 1 EI L0 0 A 1.00x22 B dx2 5m 83.333 kN # m2 EI 83.333(1000) 200(109)[70(10 - 6)] = 0.00595 rad Ans. *14–92. Determine the displacement at B of the 1.5-indiameter A-36 steel shaft. 2 ft 2 ft A 3 ft D 1 # ¢B = B L mM dx L0 EI 140 lb 2 ¢B 1.5 ft 140 lb C 2 1 = c (0.5294x1)(327.06x1)dx1 + 0.5294(2 + x2)(654.12 + 47.06x2)dx2 EI L0 L0 + = L0 1.5 (0.4706x3)(592.94x3)dx3 + L0 3 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. Determine the slope of the 1.5-in-diameter A-36 steel shaft at the bearing support A. •14–93. 1 # uA = L muM dx L0 EI 1 (1 - 0.1176x1)(327.06x1)dx1 + B EI L0 L0 + = 2 ft A 2 uA = 2 ft L0 3 ft D 1.5 (0.1176x3)(592.94x3)dx3 5 140 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 140 lb C Ans. 320 lb 320 lb 14–94. The beam is made of Douglas fir. Determine the slope at C. 8 kN Virtual Work Equation: For the slope at point C. 1#u = muM dx L0 EI 1 kN # m # uC = 0 + 1 EI L0 + uC = A L B 1.5 m 1.5 m 1.5 m 1.5 m 1 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 = - C 4.50(1000) 1 (0.12)(0.183) D 13.1(10 ) C 12 9 = 5.89 A 10 - 3 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 4 kN/m A B 3m Virtual Work Equation: For the displacement at point A, 1#¢ = 1 kN # ¢ A = + ¢A = = L0 L mM dx EI 1 EI L0 1 EI L0 3m 3m 2 x1 a x31 bdx1 9 (x2 + 3) A 2.00x22 + 6.00x2 + 6.00 B dx2 321.3 kN # m3 EI 321.3(103) 1 (0.2)(0.43) D 11(109) C 12 = 0.02738 m = 27.4 mm T Ans. For the slope at A. L 1#u = 1 kN # m # uA = muM dx L0 EI 1 EI L0 + uA = = L0 3m 3m 2 1.00a x31 b dx1 9 1.00 A 2.00x22 + 6.00x2 + 6.00 B dx2 67.5 kN # m2 EI 67.5(1000) 1 (0.2)(0.43) D 11(109) C 12 = 5.75 A 10 - 3 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 1 # ¢C = L mM dx L0 EI a a ¢C = = •14–97. a 1 (x2)(Px2)dx2 d (x1)(Px1)dx1 + c EI L0 L0 2Pa3 3EI Ans. Determine the slope at point C. EI is constant. P A C B a 1 # uC = uC = = L0 L L0 a muMdx EI A a1 B Px1 dx1 x EI a + (1)Px2dx2 EI L0 Pa2 Pa2 5Pa2 + = 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. 1 # uA = uA Determine the slope at point A. EI is constant. P A L muM dx L0 EI C B a a a x1 1 Pa2 a1 = b (Px1)dx1 + (0)(Px2)dx2 R = B a EI L0 6EI L0 14–99. Determine the slope at point A of the simply supported Douglas fir beam. a Ans. 3 kN 0.6 kN⭈m 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. 1#u = 150 mm L mu M dx EI L0 1kN # m # uA = Section a – a 1 B EI L0 2m + uA = = = 1 B EI L0 2m (1 - 0.3333x1)(0.8x1 + 0.6)dx1 L0 1m (0.3333x2)(2.2x2)dx2 R A -0.2667x1 2 + 0.6x1 + 0.6 B dx1 + 1.9333 kN # m2 EI L0 1m 0.7333x2 2dx2 R 1.9333 A 103 B 1 13.1 A 109 B c (0.075) A 0.153 B d 2 = 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. 3 kN 0.6 kN⭈m Real Moment Function M. As indicated in Fig. a. a A B C Virtual Moment Functions m. As indicated in Fig. b. 1.5 m Virtual Work Equation. 1#¢ = L0 L 1 kN # ¢ C = ¢C = 1.5 m = 1.5 m (0.5x1)(0.8x1 + 0.6)dx1 + L0 1.775kN # m3 EI L0 1m Section a – a (0.5x2)(2.2x2)dx2 0.5 m (0.5x3 + 0.5)(2.2 - 0.8x3)dx3 R A 0.4x1 2 + 0.3x1 B dx1 + + = 150 mm 1 B EI L0 1 B EI L0 L0 0.5 m L0 1m 1.1x2 2 dx2 A -0.4x3 2 + 0.7x3 + 1.1 B dx3 R 1.775 A 103 B 13.1 A 109 B c 0.5 m 75 mm mM dx EI + a 1 (0.075) A 0.153 B d 12 = 6.424 A 10 - 3 B m = 6.42 mm T Ans. 1237 1m