14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1238 © 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 •14–101 Determine the slope of end C of the overhang beam EI is constant w C A Real Moment Function M As indicated in Fig a B D L Virtual Moment Function mu As indicated in Fig b Virtual Work Equation L 1#u = # uC = uC = mu M dx L0 EI L L>2 x1 w w (1) ¢ x ≤ dx2 R B ¢ - ≤ c A 11Lx1 - 12x1 B ddx1 + EI L0 L 24 3L L0 w B EI 24L L0 uC = - L A 12x1 - 11Lx1 B dx1 + w 3L L0 L>2 x2 dx2 R 13wL3 13wL3 = 576EI 576EI Ans 1238 L L 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1239 © 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 14–102 Determine the displacement of point D of the overhang beam EI is constant w L Virtual Work Equation L # ¢D = L0 mM dx EI B EI L0 L>2 ¢ x1 w ≤ c A 11Lx1 - 12x1 B ddx1 24 L>2 + L0 ¢D = w B 48EI L0 ¢D = wL4 T 96EI L>2 ¢ x2 w ≤ c A 13Lx2 - 12x2 - L2 B ddx2 R 24 A 11Lx1 - 12x1 B dx1 + L>2 L0 B D Virtual Moment Function m As indicated in Fig b 1#¢ = C A Real Moment Function M As indicated in Fig a A 13Lx2 - 12x2 - L2x2 B dx2 R Ans 1239 L L 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1240 © 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 14–103 Determine the displacement of end C of the overhang Douglas fir beam 400 lb a Real Moment Functions M As indicated in Fig a 400 lbиft A B a Virtual Moment Functions m As indicated in Fig b ft Virtual Work Equation ft in L mM 1#¢ = dx L0 EI lb # ¢ C = ¢C = = = B EI L0 B EI L0 in ft ft ft 125x1 2dx1 + Section a – a ft (0.5x1)(250x1)dx1 + L0 L0 x2(400x2 + 400)dx2 R A 400x2 + 400x2 B dx2 R 33066.67 lb # ft3 EI 33066.67 A 12 B 1.90 A 106 B c (3) A 63 B d 12 = 0.5569 in = 0.557 in T Ans 1240 C 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1241 © 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 *14–104 Determine the slope at A of the overhang white spruce beam 400 lb a 400 lbиft A ft Virtual Moment Functions m indicated in Fig b in L muM 1#u = dx L0 EI uA = = = B EI L0 B EI L0 ft Section a – a ft ft (1 - 0.125x1)(250x1)dx1 + L0 0(400x2 + 400)dx2 R A 250x1 - 31.25x1 B dx1 + R 2666.67 lb # ft2 EI 2666.67 A 12 B 1.940 A 106 B c (3) A 63 B d 12 = 0.00508 rad = 0.00508 rad Ans 1241 C ft in Virtual Work Equation lb # ft # uA = B a Real Moment Functions M As indicated in Fig a 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1242 © 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 •14–105 Determine the displacement at point B The moment of inertia of the center portion DG of the shaft is 2I, whereas the end segments AD and GC have a moment of inertia I The modulus of elasticity for the material is E w A C D a Real Moment Function M(x): As shown on Fig a Virtual Moment Functions m(x): As shown on Fig b Virtual Work Equation: For the slope at point B, apply Eq 14–42 L 1#¢ = mM dx L0 EI # ¢B = B a x1 a b(w ax1)dx1 R EI L0 a + 2B ¢B = 1 w (x + a) B wa(a + x2) x R dx2 R 2EI L0 2 2 65wa4 48EI Ans T 1242 B a G a a 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1243 © 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 14–106 Determine the displacement of the shaft at C EI is constant w0 A B C L – L – L # ¢C = mM dx L0 E I L ¢C w0 L w0 1 = 2a b a x1 b a x1 x bdx1 E I L0 3L = w0 L4 120 E I Ans 14–107 Determine the slope of the shaft at the bearing support A EI is constant w0 A B C L – L # uA = mu M dx L0 E I L uA w0 L w0 1 = C a1 x ba x1 x bdx1 S E I L0 L 3L L + = L0 a w0L w0 x ba x x bdx2 L 3L w0L3 192 E I Ans 1243 L – 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1244 © 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 *14–108 Determine the slope and displacement of end C of the cantilevered beam The beam is made of a material having a modulus of elasticity of E The moments of inertia for segments AB and BC of the beam are 2I and I, respectively P L Real Moment Function M As indicated in Fig a Virtual Moment Functions mu and M As indicated in Figs b and c Virtual Work Equation For the slope at C, L 1#u = L0 # uC = uC = mu M dx EI EI L0 L>2 1(Px1)dx1 + EI L0 L>2 B Pa x2 + L b R dx2 5PL2 16 EI Ans For the displacement at C, L 1#¢ = # ¢C = ¢C = mM dx L0 EI EI L0 L>2 x1(Px1)dx1 + 2EI L0 L>2 ¢ x2 + L L ≤ B P ¢ x2 + ≤ R dx2 2 3PL T 16EI Ans 1244 C B A L 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1245 © 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 •14–109 Determine the slope at A of the A-36 steel W200 * 46 simply supported beam 12 kN/m kN/m Real Moment Function M As indicated in Fig a A 3m Virtual Work Equation L 1#u = mu M dx EI L0 1kN # m # uA = B EI L0 3m (1 - 0.1667x1) A 31.5x1 - 6x1 B dx1 3m + uA = = = B EI L0 3m L0 (0.1667x2) A 22.5x2 - 3x2 B dx2 R A x1 - 11.25x1 + 31.5x1 B dx1 + 3m L0 B C Virtual Moment Functions m As indicated in Fig b A 3.75x2 - 0.5x2 B dx2 R 84.375 kN # m2 EI 84.375 A 103 B 200 A 109 B c45.5 A 10 - B d = 0.009272 rad = 0.00927 rad Ans 1245 3m 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1246 © 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 14–110 Determine the displacement at point C of the A-36 steel W200 * 46 simply supported beam 12 kN/m kN/m Real Moment Functions M As indicated in Fig a A 3m Virtual Work Equation L 1#¢ = mM dx L0 EI 1kN # ¢ C = B EI L0 (0.5x1) A 31.5x1 - 6x1 B dx1 3m 3m + ¢C = = = B EI L0 3m L0 (0.5x2) A 22.5x2 - 3x2 B dx2 R A 15.75x1 - 3x1 B dx1 + 3m L0 B C Virtual Moment Functions m As indicated in Figs b A 11.25x2 - 1.5x2 B dx2 R 151.875 kN # m3 EI 151.875 A 103 B 200 A 109 B c45.5 A 10 - B d = 0.01669 m = 16.7 mm T Ans 1246 3m 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1247 © 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 14–111 The simply supported beam having a square cross section is subjected to a uniform load w Determine the maximum deflection of the beam caused only by bending, and caused by bending and shear Take E = 3G w a L For bending and shear, L 1#¢ = L fsvV mM dx + dx EI L0 GA L0 L>2 ¢ = L0 x A 12 x B A wL x - w B dx EI L>2 + L0 A 65 B A 12 B A wL - wx B dx GA A B wL wx2 L>2 wx4 L>2 wL a x b + a x b EI GA 2 0 = = 5wL4 3wL2 + 384EI 20 GA 5wL4 ¢ = = 384(3G) A 12 B a4 + 3wL2 20(G)a2 20wL4 3wL2 + 384Ga 20Ga2 = a L L w ba b Ba ba b + R a a G 96 20 Ans For bending only, ¢ = 5w L a b 96G a Ans 1247 a 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1261 © 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 14–134 Solve Prob 14–84 using Castiglianos theorem  Hv = âN a 20368 20368 0N L b = = 0P AE AE 4.5 (29) A 103 B = 0.156 in 14–135 Ans Solve Prob 14–87 using Castigliano’s theorem 0M1 x1 = 0P¿ 0M2 x2 a = + 0P¿ 2 Set P¿ = M1 = Px1 M2 = Pa a ¢C = L0 Ma 0M dx b 0P EI a = (2) = 1 (Px1) a x1 b dx + B EI L0 L0 a>2 (Pa) a a + x2 bdx2 R 2 23Pa3 24 EI Ans 1261 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1262 © 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 *14–136 Solve Prob 14–88 using Castigliano’s theorem 0M1 = x1 0P 0M2 = -0.5 x2 0P Set P = 15 kN M2 = -1.5x2 - 2x22 M1 = 15x1 L ¢A = Ma L0 0M dx b 0P EI 1.5 A -1.5x2 - 2x22 B (-0.5x2)dx2 R = B EI L0 = 43.875(103) 43.875 kN # m3 = = 0.0579 m EI 13(10 ) 12 (0.12)(0.18)3 (15x1)(x1)dx + L0 = 57.9 mm •14–137 Ans Solve Prob 14–90 using Castigliano’s theorem Internal Moment Function M(x): The internal moment function in terms of the couple moment M¿ and the applied load are shown on the figure Castigliano’s Second Theorem: The slope at A can be determined with 0M(x1) 0M(x2) = - 0.100x1, = and setting M¿ = 0M¿ 0M¿ L u = uA = = = L0 Ma EI L0 0M dx b 0M¿ EI 10 m (2.50x1)(1 - 0.100x1)dx1 + EI L0 5m A 1.00x22 B (0)dx2 41.667 kN # m2 EI 41.667 A 103 B 200 A 109 B C 70 A 10 - B D Ans = 0.00298 rad 1262 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1263 © 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 14–138 Solve Prob 14–92 using Castigliano’s theorem 0M1 = 0.5294x1 0P 0M2 = 0.5294x2 + 1.0588 0P 0M3 = 0.4706x3 0P 0M4 = 0.4706x4 + 0.7059 0P Set P = M1 = 327.06x1 M2 = 47.06x2 + 654.12 M3 = 592.94x3 M4 = 889.41 - 47.06x4 L ¢B = L0 Ma 0M dx b 0P EI (327.06x1)(0.5294x1)dx1 c EI L0 = + (47.06x2 + 654.12)(0.5294x2 + 1.0588)dx2 L0 1.5 + (592.94x3)(0.4706x3)dx3 L0 + = 14–139 L0 (889.41 - 47.06x4)(0.4706x4 + 0.7059)dx4 d 6437.69 A 12 B 6437.69 lb # ft3 = = 1.54 in EI 29 A 106 B A p4 B A 0.754 B Ans Solve Prob 14–93 using Castigliano’s theorem 0M1 = - 0.1176 x1 0M¿ 0M2 = 0.1176 x2 0M¿ 0M3 = 0.1176x3 + 0.1764 0M¿ Set M¿ = M1 = 327.06x1 uA = L Ma M2 = 592.94x2 M3 = 889.41 - 47.06x3 0M dx b = c (327.06x1)(1 - 0.1176x1)dx1 0M¿ EI EI L0 1.5 + (592.94x2)(0.1176x2)dx2 + L0 + = L0 (889.41 - 47.06x3)(0.1176x3 + 0.1764)dx3 d 2387.54(12 2) 2387.54 lb # ft2 = 0.0477 rad = 2.73° = EI 29(106)(p4 )(0.754) 1263 Ans 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1264 © 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 *14–140 Solve Prob 14–96 using Castigliano’s theorem 0M1 = x1 0P¿ 0M2 = x2 0P¿ Set P = P¿ M1 = Px1 M2 = Px2 L ¢C = = a Ma L0 a 0M bdx = (Px1)(x1)dx1 + (Px2)(x2)dx2 R B 0P¿ EI L0 L0 2Pa3 3EI 14–141 Ans Solve Prob 14–89 using Castigliano’s theorem Set M¿ = L uC = Ma L0 0M dx b 0M¿ EI (Px1)a x1 b dx1 a (Px2)(1)dx2 a + = EI EI L0 L0 a = Pa2 5Pa2 Pa2 + = 3EI 2EI 6EI 14–142 Ans Solve Prob 14–98 using Castigliano’s theorem 0M1 x1 = a 0M¿ 0M2 = 0M¿ Set M¿ = M1 = -Px1 L uA = L0 Ma M2 = Px2 0M dx b 0M¿ EI = a a x1 -Pa2 bdx1 + ( -Px1)a (Px2)(0)dx2 R = B a EI L0 6EI L0 = Pa2 6EI Ans 1264 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1265 © 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 14–143 Solve Prob 14–112 using Castigliano’s theorem Internal Moment Function M(x): The internal moment function in terms of the load P and external applied load are shown on the figure Castigliano’s Second Theorem: The vertical displacement at C can be determined 0M(x1) 0M(x2) with = 1.00x1, = 1.00L and setting P = 0P 0P L ¢ = Ma L0 0M dx b 0P EI L (¢ C)v = = *14–144 L w wL a x1 b(1.00x1) dx1 + a b(1.00L) dx2 EI L0 EI L0 5wL4 8EI Ans T Solve Prob 14–114 using Castigliano’s theorem Castigliano’s Second Theorem: The horizontal displacement at A can be determined 0M(x1) 0M(x2) using = 1.00x1, = 1.00L and setting P¿ = P 0P¿ 0P¿ L ¢ = L0 Ma L (¢ A)h = = 0M dx b 0P EI L 1 (Px1)(1.00x1) dx1 + (PL)(1.00L) dx2 EI L0 EI L0 4PL3 3EI Ans 1265 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1266 © 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 •14–145 Solve Prob 14–121 using Castigliano’s theorem L 0M dx b = = Ma 0P E I L0 L0 ¢C = A a M0 a a M0 (1x) x B (1x) dx + dx EI EI L0 M0 a2 6EI Ans 14–146 Determine the bending strain energy in the beam due to the loading shown EI is constant P a L Ui = = a a M dx = c2 (Px1)2 dx1 + (Pa)2 dx2 d 2EI 2EI L0 L0 L0 5P2a3 6EI Ans 1266 P a a 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1267 © 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 14–147 The 200-kg block D is dropped from a height h = m onto end C of the A-36 steel W200 * 36 overhang beam If the spring at B has a stiffness k = 200 kN>m, determine the maximum bending stress developed in the beam D h A B 4m Equilibrium The support reactions and the moment functions for regions AB and BC of the beam under static conditions are indicated on the free-body diagram of the beam, Fig a, Ue = Ui L M 2dx P st = â L0 2EI 1 P¢ st = B 2EI L0 ¢ st = 4m a P x2 b dx + L0 2m (Px1)2 dx R 8P EI Here, I = 34.4 A 106 B mm4 = 34.4 A 10 - B m4 (see the appendix) and E = Est = 200 GPa Then, the equivalent spring constant can be determined from P = kb ¢ st P = kb a 8P b EI EI = kb = 200 A 109 B c34.4 A 10 - B d = 860 A 103 B N>m From the free-body diagram, Fsp = P ksp ¢ sp = ¢ sp = (k ¢ ) b b 3 kb 860 A 10 B ≥ ¢ b = 6.45 ¢ b ¢ ≤ ¢b = £ ksp 200 A 103 B (1) Conservation of Energy mga h + ¢ b + 1 ¢ b = ksp ¢ sp + kb ¢ b 2 sp 2 1267 C k 2m 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1268 © 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 14–147 Continued Substiuting Eq (1) into this equation 200(9.81)c1 + ¢ b + 1 (6.45¢ b) d = c200 A 103 B d(6.45¢ b)2 + c860 A 103 B d ¢ b 2 2 4590.25 A 103 B ¢ b - 20944.35¢ b - 1962 = Solving for the positive root ¢ b = 0.02308 m Maximum Stress The maximum force on the beam is Pmax = kb ¢ b = 860 A 103 B (0.02308) = 19.85 A 103 B N The maximum moment occurs at the Mmax = Pmax L = 19.85 A 103 B (2) = 39.70 A 103 B N # m 0.201 d Applying the flexure formula with c = = = 0.1005 m 2 supporting smax = spring, where 39.70 A 103 B (0.1005) Mmax c = = 115.98 MPa = 116 MPa I 34.4 A 10 - B Since smax sY = 250 MPa, this result is valid 1268 Ans 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1269 © 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 *14–148 Determine the maximum height h from which the 200-kg block D can be dropped without causing the A-36 steel W200 * 36 overhang beam to yield The spring at B has a stiffness k = 200 kN>m D h A B 4m Equilibrium The support reactions and the moment functions for regions AB and BC of the beam under static conditions are indicated on the free-body diagram of the beam, Fig a, Ue = Ui L M 2dx P st = â L0 2EI 1 P¢ st = B 2EI L0 ¢ st = 4m a P x2 b dx + L0 2m (Px1)2 dx R 8P EI Here, I = 34.4 A 106 B mm4 = 34.4 A 10 - B m4 (see the appendix) and E = Est = 200 GPa Then, the equivalent spring constant can be determined from P = kb ¢ st P = kb a 8P b EI EI kb = = 200 A 109 B c34.4 A 10 - B d = 860 A 103 B N>m From the free-body diagram, Fsp = P ksp ¢ sp = ¢ sp = (k ¢ ) b b 3 kb 860 A 10 B ≥ ¢ b = 6.45¢ b ¢ ≤ ¢b = £ ksp 200 A 103 B (1) 1269 C k 2m 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1270 © 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 *14–148 Continued Maximum Stress The maximum force on the beam is Pmax = kb ¢ b = 860 A 103 B ¢ b The maximum moment occurs at the supporting spring, where Mmax = Pmax L = 860 A 103 B ¢ b(2) = 1720 A 103 B ¢ b 0.201 d = = 0.1005 m, c = 2 smax = Applying the flexure formula with Mmaxc I 250 A 106 B = 1720 A 103 B ¢ b(0.1005) 34.4 A 10 - B ¢ b = 0.04975 m Substituting this result into Eq (1), ¢ sp = 0.3209 m Conservation of Energy mga h + ¢ b + 1 ¢ b = ksp ¢ sp + kb ¢ b 2 sp 2 200(9.81)ch + 0.04975 + (0.3209) d = c200 A 103 B d(0.3209)2 2 + c860 A 103 B d(0.04975)2 h = 5.26 m Ans 1270 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1271 © 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 •14–149 The L2 steel bolt has a diameter of 0.25 in., and the link AB has a rectangular cross section that is 0.5 in wide by 0.2 in thick Determine the strain energy in the link AB due to bending, and in the bolt due to axial force The bolt is tightened so that it has a tension of 350 lb Neglect the hole in the link in in 0.2 in A B in Bending strain energy: L (Ub)i = = M dx = c (140x1)2dx1 + (210x2)2dx2 d 2EI 2EI L0 L0 L0 1.176(106) 1.176(106) = 122 in # lb = 10.1 ft # lb = EI 29(106)(12 )(0.5)(0.2 3) Ans Axial force strain energy: L (Ua)i = (350)2(8) N dx N 2L = 0.344 in # lb = = 2AE 2(29)(106)(p4 )(0.252) L0 2EA Ans 14–150 Determine the vertical displacement of joint A Each bar is made of A-36 steel and has a cross-sectional area of 600 mm2 Use the conservation of energy B C Joint A: 2m + c ©Fy = 0; F - = AB FAB = 6.25 kN D A + ©F = 0; ; x FAD - (6.25) = 1.5 m FAD = 3.75 kN kN Joint B: + c ©Fy = 0; 4 F - (6.25) = BD + ©F = 0; : x FBC - 2a b (6.25) = FBD = 6.25 kN FBC = 7.5 kN Conservation of energy: Ue = Ui N 2L P = â 2AE 1 (5) A 103 B ¢ A c A 6.25 A 103 B B 2(2.5) + A 3.75 A 103 B B 2(3) 2AE + A 6.25 A 103 B B 2(2.5) + A 7.5 A 103 B B 2(1.5) d ¢A = 64 375 64 375 = = 0.5364 A 10 - B m = 0.536 mm -6 AE 600 A 10 B (200) A 109 B 1271 Ans 1.5 m 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1272 © 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 14–151 Determine the total strain energy in the A-36 steel assembly Consider the axial strain energy in the two 0.5-in.-diameter rods and the bending strain energy in the beam for which I = 43.4 in4 ft 500 lb Support Reactions: As shown FBD(a) Internal Moment Function: As shown on FBD(b) Total Strain Energy: L (Ui)T = ft NL M dx + 2EI 2AE L0 = 2B 2EI L0 ft (250x)2 dx R + B = 1.3333 A 106 B lb2 # ft3 = 1.3333 A 106 B A 12 B + EI 29.0 A 106 B (43.4) + p 2502(3) R 2AE 0.1875 A 106 B lb2 # ft AE 0.1875 A 106 B (12) A 0.52 B C 29.0 A 106 B D = 2.23 in # lb Ans *14–152 Determine the vertical displacement of joint E For each member A = 400 mm2, E = 200 GPa Use the method of virtual work Member n N L F 1.5 AE – 0.8333 – 37.5 2.5 78.125 AB 0.6667 30.0 2.0 40.00 EF 0 2.0 EB – 0.50 22.5 1.5 –16.875 ED – 0.6667 – 30.0 2.0 40.00 BC 0 2.0 BD 0.8333 37.5 2.5 78.125 CD – 0.5 – 22.5 1.5 16.875 A 2m nNL AE 400 A 10 - B (200) A 109 B C B 45 kN © = 236.25 236.25 A 103 B D 1.5 m ¢ Bv = E nNL AF # ¢ Bv = © ft = 2.95 A 10 - B = 2.95 mm Ans 1272 2m 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1273 © 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 •14–153 Solve Prob 14–152 using Castigliano’s theorem F E D 1.5 m A C B 45 kN 2m Member N 0N>0P N(P = 45) L 2m N(0N>0P)L AF 0 1.5 AE –(0.8333P + 37.5) – 0.8333 –37.5 2.5 78.125 AB 0.6667P + 30 0.6667 30.0 2.0 40.00 BE 22.5–0.5P – 0.5 22.5 1.5 –16.875 BD 0.8333P + 37.5 0.8333 37.5 2.5 78.125 BC 0 2.0 CD –(0.5P + 22.5) – 0.5 –22.5 1.5 16.875 DE –(0.6667P + 30) – 0.6667 –30.0 2.0 40.00 EF 0 2.0 © = 236.25  Bv = âN = 236.25 0N L = 0P AE AE 236.25 A 103 B 400 A 10 - B (200) A 109 B = 2.95 A 10 - B m = 2.95 mm Ans 14–154 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 Use the method of virtual work M0 A B L L uB = = L (1) M0 muM dx = dx EI L0 EI L0 M0L EI Ans 1273 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1274 © 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 14–155 Solve Prob 14–154 using Castigliano’s theorem M0 A L uB = = L0 ma B L M0(1) dm dy b = dx dm¿ EI EI L0 L M0L EI Ans *14–156 Determine the displacement of point B on the aluminum beam Eal = 10.6110 ksi Use the conser in kip C B 12 ft L 12(12) M dx = (2) 2EI 2EI L0 L0 Ue = 1 P¢ = (3)¢ B = 1.5¢ B 2 (1.5x)2 dx = 2239488 EI Conservation of energy: Ue = Ui 1.5¢ B = 239 488 EI ¢B = 492 992 EI y = 0.5(7)(1) + (4)(6)(1) = 2.1154 in 7(1) + 6(1) I = 1 (7) A 13 B + (7)(1)(2.1154 - 0.5)2 + (1) A 63 B + (1)(6)(4 - 2.1154)2 12 12 = 58.16 in4 ¢B = in in in A Ui = in 492 992 = 2.42 in (10.6)(103)(58.16) Ans 1274 12 ft 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1275 © 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 14–157 A 20-lb weight is dropped from a height of ft onto the end of a cantilevered A-36 steel beam If the beam is a W12 * 50, determine the maximum stress developed in the beam ft 12 ft From Appendix C: ¢ st = 20(12(12))3 PL3 = = 1.742216 A 10 - B in 3EI 3(29) A 106 B (394) n = + A + 2a h 4(12) b = 235.74 b = + + 2a ¢ st A 1.742216(10 - 3) smax = nsst = 235.74 £ 20(12)(12) A 12.19 B 394 ≥ = 10503 psi = 10.5 ksi sg O.K Ans 1275 ... publisher *14 136 Solve Prob 14 88 using Castigliano’s theorem 0M1 = x1 0P 0M2 = -0 .5 x2 0P Set P = 15 kN M2 = -1 .5x2 - 2x22 M1 = 15x1 L ¢A = Ma L0 0M dx b 0P EI 1.5 A -1 .5x2 - 2x22 B (-0 .5x2)dx2... 3m (1 - 0.3333x)(20.0x)dx + 30.0 kN # m2 104.167 kN EI AE 104.167(1000) 30.0(1000) = ( -0 .41667)(50.0)(5) AE 200 A 10 B C (0.1) A 0.3 B D 12 - C A 0.02 B D C 200 A 109 B D p = -0 .991 A 10 - B rad... –0.25213 -2 0 213 213 234.36 – A 0.25 213P + 20 213 B DE ¢ Dv = ©Na © = 292.95 292.95 A 103 B 292.95 0N L b = = 0P AE AE 300 A 10 - B (200) A 109 B = 4.88 A 10 - B m = 4.88 mm 14 127 Ans Solve Prob 14 77