《材料力学》R.C. Hibbeler希伯勒 第八版 英文 习题解答14 - Part2

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. Real Moment Function M. As indicated in Fig. a. Virtual Moment Function . As indicated in Fig. b. Virtual Work Equation. Ans.uC = - 13wL3 576EI = 13wL3 576EI uC = 1 EI B w 24L L L 0 A12x1 3 - 11Lx1 2 Bdx1 + w 3L L L>2 0 x2 3 dx2R 1 # uC = 1EI BL L 0 ¢ -x1 L ≤ c w 24 A11Lx1 - 12x1 2 B ddx1 + L L>2 0 (1)¢ w 3L x2 3≤dx2R 1 # u = L L 0 mu M EI dx mu •14–101. Determine the slope of end C of the overhang beam. EI is constant. A C B w D L 2 L 2 L 2 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1238 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. Real Moment Function M. As indicated in Fig. a. Virtual Moment Function m. As indicated in Fig. b. Virtual Work Equation. Ans.¢D = wL4 96EI T ¢D = w 48EI BL L>2 0 A11Lx1 2 - 12x1 3 Bdx1 + L L>2 0 A13Lx2 2 - 12x2 3 - L2x2 Bdx2R + L L>2 0 ¢x2 2 ≤ c w 24 A13Lx2 - 12x2 2 - L2 B ddx2R 1 # ¢D = 1EI BL L>2 0 ¢x1 2 ≤ c w 24 A11Lx1 - 12x1 2 B ddx1 1 # ¢ = L L 0 mM EI dx 14–102. Determine the displacement of point D of the overhang beam. EI is constant. A C B w D L 2 L 2 L 2 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1239 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. Real Moment Functions M. As indicated in Fig. a. Virtual Moment Functions m. As indicated in Fig. b. Virtual Work Equation. Ans. = 0.5569 in. = 0.557 in. T = 33066.67 A123 B 1.90 A106 B c 1 12 (3) A63 B d = 33066.67 lb # ft3 EI ¢C = 1 EI BL 8 ft 0 125x1 2dx1 + L 4 ft 0 A400x2 2 + 400x2 Bdx2R 1 lb # ¢C = 1EI BL 8 ft 0 (0.5x1)(250x1)dx1 + L 4 ft 0 x2(400x2 + 400)dx2R 1 # ¢ = L L 0 mM EI dx 14–103. Determine the displacement of end C of the overhang Douglas fir beam. C A B 8 ft 4 ft 400 lb a a 3 in. 6 in. Section a – a 400 lb�ft 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1240 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. Real Moment Functions M. As indicated in Fig. a. Virtual Moment Functions m. indicated in Fig. b. Virtual Work Equation. Ans. = 0.00508 rad = 0.00508 rad = 2666.67 A122 B 1.940 A106 B c 1 12 (3) A63 B d = 2666.67 lb # ft2 EI uA = 1 EI BL 8 ft 0 A250x1 - 31.25x1 2 Bdx1 + 0R 1 lb # ft # uA = 1EI BL 8 ft 0 (1 - 0.125x1)(250x1)dx1 + L 4 ft 0 0(400x2 + 400)dx2R 1 # u = L L 0 muM EI dx *14–104. Determine the slope at A of the overhang white spruce beam. C A B 8 ft 4 ft 400 lb a a 3 in. 6 in. Section a – a 400 lb�ft 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1241 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. 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. Ans. ¢B = 65wa4 48EI T + 2B 1 2EI L a 0 1 2 (x2 + a)Bwa(a + x2) - w2 x22Rdx2R 1 # ¢B = 2B 1EI L a 0 a x1 2 b(w ax1)dx1R 1 # ¢ = L L 0 mM EI dx •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. aa a a A B G C w D 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1242 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. Ans. = 5 w0L 3 192 E I + L L 2 0 a 1 L x2b a w0L 4 x2 - w0 3 L x2 3bdx2 uA = 1 E I C L L 2 0 a1 - 1 L x1b a w0 L 4 x1 - w0 3 L x1 3bdx1S 1 # uA = L L 0 mu M E I dx 14–107. Determine the slope of the shaft at the bearing support A. EI is constant. L 2 – L 2 – A C w0 B Ans. = w0 L 4 120 E I ¢C = 2a 1 E I bL L 2 0 a 1 2 x1b a w0 L 4 x1 - w0 3 L x31bdx1 1 # ¢C = L L 0 m M E I dx 14–106. Determine the displacement of the shaft at C. EI is constant. L 2 – L 2 – A C w0 B 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1243 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. Real Moment Function M. As indicated in Fig. a. Virtual Moment Functions and M. As indicated in Figs. b and c. Virtual Work Equation. For the slope at C, Ans. For the displacement at C, Ans.¢C = 3PL 16EI T 1 # ¢C = 1EI L L>2 0 x1(Px1)dx1 + 1 2EIL L>2 0 ¢x2 + L2 ≤ BP¢x2 + L2 ≤ Rdx2 1 # ¢ = L L 0 mM EI dx uC = 5PL2 16 EI 1 # uC = 1EI L L>2 0 1(Px1)dx1 + 1 2 EI L L>2 0 1BPax2 + L2 b Rdx2 1 # u = L L 0 mu M EI dx mu *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. A B C P L 2 L 2 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1244 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. Real Moment Function M. As indicated in Fig. a. Virtual Moment Functions m. As indicated in Fig. b. Virtual Work Equation. Ans. = 0.009272 rad = 0.00927 rad = 84.375 A103 B 200 A109 B c45.5 A10-6 B d = 84.375 kN # m2 EI uA = 1 EI BL 3 m 0 Ax1 3 - 11.25x1 2 + 31.5x1 Bdx1 + L 3 m 0 A3.75x2 2 - 0.5x2 3 Bdx2R + L 3 m 0 (0.1667x2) A22.5x2 - 3x2 2 Bdx2R 1kN # m # uA = 1EI BL 3 m 0 (1 - 0.1667x1) A31.5x1 - 6x1 2 Bdx1 1 # u = L L 0 mu M EI dx •14–109. Determine the slope at A of the A-36 steel simply supported beam.W200 * 46 A B C 3 m 12 kN/m 6 kN/m 3 m 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1245 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. Real Moment Functions M. As indicated in Fig. a. Virtual Moment Functions m. As indicated in Figs. b. Virtual Work Equation. Ans. = 0.01669 m = 16.7 mm T = 151.875 A103 B 200 A109 B c45.5 A10-6 B d = 151.875 kN # m3 EI ¢C = 1 EI BL 3 m 0 A15.75x1 2 - 3x1 3 Bdx1 + L 3 m 0 A11.25x2 2 - 1.5x2 3 Bdx2R + L 3m 0 (0.5x2) A22.5x2 - 3x2 2 Bdx2R 1kN # ¢C = 1EI BL 3 m 0 (0.5x1) A31.5x1 - 6x1 2 Bdx1 1 # ¢ = L L 0 mM EI dx 14–110. Determine the displacement at point C of the A-36 steel simply supported beam.W200 * 46 A B C 3 m 12 kN/m 6 kN/m 3 m 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1246 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. For bending and shear, Ans. For bending only, Ans.¢ = 5w 96G a L a b 4 = a w G b a L a b 2B a 5 96 b a L a b 2 + 3 20 R = 20wL4 384Ga4 + 3wL2 20Ga2 ¢ = 5wL4 384(3G) A 112 Ba4 + 3wL2 20(G)a2 = 5wL4 384EI + 3wL2 20 GA = 1 EI a wL 6 x3 - wx4 8 b 2 L>2 0 + A65 B GA a wL 2 x - wx2 2 b 2 L>2 0 ¢ = 2 L L>2 0 A12 x B A wL 2 x - w x2 2 Bdx EI + 2 L L>2 0 A65 B A 1 2 B A wL 2 - wx Bdx GA 1 # ¢ = L L 0 mM EI dx + L L 0 fsvV GA dx 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. L a a w 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1247 1248 © 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. 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 vertical displacement at point C, Ans. (¢C)v = 5wL4 8EI T 1 # (¢C)v = 1EI L L 0 (1.00x1)a w 2 x1 2bdx1 + 1 EIL L 0 (1.00L)a wL2 2 bdx2 1 # ¢ = L L 0 mM EI dx *14–112. The frame is made from two segments, each of length L and flexural stiffness EI. If it is subjected to the uniform distributed load determine the vertical displacement of point C. Consider only the effect of bending. L L A B C w 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1248 1249 © 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. 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 horizontal displacement at point B, Ans. (¢B)h = wL4 4EI : 1 # (¢B)h = 1EI L L 0 (0)a w 2 x21bdx1 + 1 EI L L 0 (1.00L - 1.00x2)a wL2 2 bdx2 1 # ¢ = L L 0 mM EI dx •14–113. The frame is made from two segments, each of length L and flexural stiffness EI. If it is subjected to the uniform distributed load, determine the horizontal displacement of point B. Consider only the effect of bending. L L A B C w Ans. = 4PL3 3EI ¢Av = 1 EI C L L 0 (x1)(Px1)dx1 + L L 0 (1L)(PL)dx2S 1 # ¢Av = L L 0 mM EI dx 14–114. Determine the vertical displacement of point A on the angle bracket due to the concentrated force P. The bracket is fixed connected to its support. EI is constant. Consider only the effect of bending. L L P A 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1249 1250 © 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. 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 displacement at point B, Ans. = 0.04354 m = 43.5 mm T = 66.667(1000) 200(109) C 112 (0.1) A0.13 B D + 55.556(1000) Cp4 A0.012 B D C200 A109 B D ¢B = 66.667 kN # m3 EI + 55.556 kN # m AE + 1.667(16.667)(2) AE + 1 EI L 2 m 0 (1.00x2)(10.0x2)dx2 1 kN # ¢B = 1EI L 3 m 0 (0.6667x1)(6.667x1)dx1 1 # ¢ = L L 0 mM EI dx + nNL AE 14–115. Beam AB has a square cross section of 100 mm by 100 mm. Bar CD has a diameter of 10 mm. If both members are made of A-36 steel, determine the vertical displacement of point B due to the loading of 10 kN. 3 m 2 m 10 kN A D B C 2 m 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1250 1251 © 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. Real Moment Function M(x): As shown on Fig. a. Virtual Moment Functions : As shown on Fig. b. Virtual Work Equation: For the slope at point A, Ans. = 0.00529 rad = 10.0(1000) 200 A109 B C 112 (0.1) A0.13 B D - 11.111(1000) Cp4 A0.012 B D C200 A109 B D uA = 10.0 kN # m2 EI - 11.111 kN AE + 1 EI L 2 m 0 0(10.0x2)dx2 + (-0.3333)(16.667)(2) AE 1 kN # m # uA = 1EI L 3 m 0 (1 - 0.3333x1)(6.667x1)dx1 1 # u = L L 0 muM EI dx + nNL AE mu(x) *14–116. Beam AB has a square cross section of 100 mm by 100 mm. Bar CD has a diameter of 10 mm. If both members are made of A-36 steel, determine the slope at A due to the loading of 10 kN. 3 m 2 m 10 kN A D B C 2 m 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 displacement at point C, Ans. = 0.017947 m = 17.9 mm T = 360(1000) 200 A109 B C 112 (0.1) A0.33 B D + 625(1000) Cp4 A0.022 B D C200 A109 B D ¢C = 360 kN # m3 EI + 625 kN # m AE 1 kN # ¢C = 2 c 1EI L 3 m 0 (1.00x)(20.0x) dx d + 2.50(50.0) (5) AE 1 # ¢ = L L 0 mM EI dx + nNL AE 14–117. Bar ABC has a rectangular cross section of 300 mm by 100 mm. Attached rod DB has a diameter of 20 mm. If both members are made of A-36 steel, determine the vertical displacement of point C due to the loading. Consider only the effect of bending in ABC and axial force in DB. 3 m 20 kN A B C 4 m D 100 mm 300 mm 3 m 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1251 1252 © 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. Real Moment Function M(x): As shown on Fig. a. Virtual Moment Functions : As shown on Fig. b. Virtual Work Equation: For the slope at point A, Ans. = -0.991 A10-3 B rad = 0.991 A10-3 B rad = 30.0(1000) 200 A109 B C 112 (0.1) A0.33 B D - 104.167(1000) Cp4 A0.022 B D C200 A109 B D uA = 30.0 kN # m2 EI - 104.167 kN AE 1 kN # m # uA = 1EI L 3 m 0 (1 - 0.3333x)(20.0x)dx + (-0.41667)(50.0)(5) AE 1 # u = L L 0 muM EI dx + nNL AE mu (x) 14–118. Bar ABC has a rectangular cross section of 300 mm by 100 mm. Attached rod DB has a diameter of 20 mm. If both members are made of A-36 steel, determine the slope at A due to the loading. Consider only the effect of bending in ABC and axial force in DB. 3 m 20 kN A B C 4 m D 100 mm 300 mm 3 m 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1252 1253 © 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. Real Moment Functions M. As indicated in Fig. a. Virtual Moment Functions m. As indicated in Fig. b. Virtual Work Equation. Ans. = 0.01683 m = 16.8 mm T = 239.26 A103 B 200 A109 B c71.1 A10-6 B d = 239.26 kN # m3 EI +L 2.5 m 0 A37.5x3 + 11.25x3 2 - 3.75x3 3 Bdx3R (¢C)v = 1 EI BL 2.5 m 0 A26.25x1 2 - 3.75x1 3 Bdx1 + 0 + L 2.5 m 0 (0.5x3) A75 + 22.5x3 - 7.5x3 2 Bdx2R + L 5 m 0 0(15x2)dx2 1 kN # (¢C)v = 1EI BL 2.5 m 0 (0.5x1) A52.5x1 - 7.5x1 2 Bdx1 1 # ¢ = L L 0 mM EI dx 14–119. Determine the vertical displacement of point C. The frame is made using A-36 steel members. Consider only the effects of bending. W250 * 45 A C D B 5 m 2.5 m 15 kN/m 15 kN 2.5 m 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1253 1254 © 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. Real Moment Functions M. As indicated in Fig. a. Virtual Moment Functions m. As indicated in Fig. b. Virtual Work Equation. Ans. = 0.1154 m = 115 mm : = 1640.625 A103 B 200 A109 B c71.1 A10-6 B d = 1640.625 kN # m3 EI (¢B)h = 1 EI BL 5 m 0 A52.5x1 2 - 7.5x1 3 Bdx1 + L 5 m 0 15x2 2dx2R 1 kN # (¢B)h = 1EI BL 5 m 0 x1 A52.5x1 - 7.5x1 2 Bdx1 + L 5 m 0 x2(15x2)dx2R 1 # ¢ = L L 0 mM EI dx *14–120. Determine the horizontal displacement of end B. The frame is made using A-36 steel members. Consider only the effects of bending. W250 * 45 A C D B 5 m 2.5 m 15 kN/m 15 kN 2.5 m 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1254 1255 © 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. Ans. = 5 M0 a 2 6 E I ¢C = L a 0 (1x) AM0a x B E I dx + L a 0 (1x) M0 E I dx 1 # ¢C = L L 0 m M E I dx •14–121. Determine the displacement at point C. EI is constant. A B a a M 0C Ans. = M0 a 3 E I uB = L a 0 Axa B A M 0 a x B E I dx 1 # uB = L L 0 mu M E I dx 14–122. Determine the slope at B. EI is constant. A B a a M 0C Member N L AB 1.1547 1.1547 800 120 110851.25 BC –0.5774P –0.5774 0 60 0 Ans.¢Bb = ©Na 0N 0P b L AE = 110851.25 AE = 110851.25 (2)(29)(106) = 0.00191 in. © = 110851.25 P + 800 N(0N>0P)LN(P = 0)0N>0P 14–123. Solve Prob. 14–72 using Castigliano’s theorem. 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1255 1256 © 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. Member Force N: Member forces due to external force P and external applied forces are shown on the figure. Castigliano’s Second Theorem: Member N L AB –0.8333P –0.8333 –166.67 10.0 1388.89 BC 0.8333P 0.8333 166.67 10.0 1388.89 AC 0.500P 0.500 100.00 12 600.00 Ans. = 3377.78(12) 2 C29.0 A106 B D = 0.699 A10-3 B in. : (¢B)h = 3377.78 lb # ft AE ¢ = aNa 0N0P b L AE © 3377.78 lb # ft Na 0N 0P bLN(P = 200 lb) 0N 0P *14–124. Solve Prob. 14–73 using Castigliano’s theorem. Member N L AB 1.50P 1.50 45.00 3.0 202.50 AD 0 0 BD –20 0 –20 2.0 0 BC 1.5P 1.5 45.00 3.0 202.50 CD 351.54 DE 468.72 Ans. = 0.02.04 m = 20.4 mm ¢Cv = ©Na 0N 0P b L AE = 1225.26 A103 B 300 A10-6 B(200) A109 B © = 1225.26 213-20213–0.5213– A0.5213P + 5213 B 213-15213–0.5213–0.5213P 21352135213 N(0N>0P)LN(P = 30)0N>0P •14–125. Solve Prob. 14–75 using Castigliano’s theorem. 14 Solutions 46060_Part2 6/11/10 8:19 AM Page 1256 1257 © 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. Member N L AB 45 0 45.00 3 0 AD 58.59 BC 45 0 45 3 0