© 2014 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. 1–1. The shaft is supported by a smooth thrust bearing at B and a journal bearing at C. Determine the resultant internal loadings acting on the cross section at E. B A 4 ft C E 4 ft 4 ft D 4 ft 400 lb 800 lb Support Reactions: We will only need to compute Cy by writing the moment equation of equilibrium about B with reference to the free-body diagram of the entire shaft, Fig. a. a + ©MB = 0; Cy(8) + 400(4) - 800(12) = 0 Cy = 1000 lb Internal Loadings: Using the result for Cy, section DE of the shaft will be considered. Referring to the free-body diagram, Fig. b, + : ©Fx = 0; NE = 0 + c ©Fy = 0; VE + 1000 - 800 = 0 Ans. Ans. VE = - 200 lb a + ©ME = 0; 1000(4) - 800(8) - ME = 0 ME = - 2400 lb # ft = - 2.40 kip # ft Ans. The negative signs indicates that VE and ME act in the opposite sense to that shown on the free-body diagram. Ans: NE = 0, VE = - 200 lb, ME = - 2.40 kip # ft 1 © 2014 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. a 1–2. Determine the resultant internal normal and shear force in the member at (a) section a–a and (b) section b–b, each of which passes through point A. The 500-lb load is applied along the centroidal axis of the member. b 30⬚ 500 lb 500 lb A b a (a) + : ©Fx = 0; Na - 500 = 0 Na = 500 lb Ans. + T©Fy = 0; Va = 0 Ans. R+ ©Fx = 0; Nb - 500 cos 30° = 0 (b) Ans. Nb = 433 lb +Q©F = 0; y Vb - 500 sin 30° = 0 Vb = 250 lb Ans. Ans: Na = 500 lb, Va = 0, Nb = 433 lb, Vb = 250 lb 2 © 2014 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. 1–3. The beam AB is fixed to the wall and has a uniform weight of 80 lb>ft. If the trolley supports a load of 1500 lb, determine the resultant internal loadings acting on the cross sections through points C and D. 5 ft 20 ft 10 ft 3 ft A B C D 1500 lb Segment BC: + ; ©Fx = 0; NC = 0 + c ©Fy = 0; VC - 2.0 - 1.5 = 0 Ans. VC = 3.50 kip a + ©MC = 0; Ans. -MC - 2(12.5) - 1.5 (15) = 0 MC = - 47.5 kip # ft Ans. Segment BD: + ; ©Fx = 0; ND = 0 Ans. + c ©Fy = 0; VD - 0.24 = 0 VD = 0.240 kip a + ©MD = 0; Ans. -MD - 0.24 (1.5) = 0 MD = - 0.360 kip # ft Ans. Ans: NC = 0, VC = 3.50 kip, MC = - 47.5 kip # ft, ND = 0, VD = 0.240 kip, MD = -0.360 kip # ft 3 © 2014 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. *1–4. The shaft is supported by a smooth thrust bearing at A and a smooth journal bearing at B. Determine the resultant internal loadings acting on the cross section at C. 600 N/m A B D C 1m 1m 1m 1.5 m 1.5 m 900 N Support Reactions: We will only need to compute By by writing the moment equation of equilibrium about A with reference to the free-body diagram of the entire shaft, Fig. a. a + ©MA = 0; By(4.5) - 600(2)(2) - 900(6) = 0 By = 1733.33 N Internal Loadings: Using the result of By, section CD of the shaft will be considered. Referring to the free-body diagram of this part, Fig. b, + Ans. ; ©Fx = 0; NC = 0 VC = - 233 N + c ©Fy = 0; VC - 600(1) + 1733.33 - 900 = 0 a + ©MC = 0; 1733.33(2.5) - 600(1)(0.5) - 900(4) - MC = 0 MC = 433 N # m Ans. Ans. The negative sign indicates that VC act in the opposite sense to that shown on the free-body diagram. 4 © 2014 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. 1–5. Determine the resultant internal loadings in the beam at cross sections through points D and E. Point E is just to the right of the 3-kip load. 3 kip 1.5 kip/ ft A D 6 ft E B 6 ft 4 ft C 4 ft Support Reactions: For member AB a + ©MB = 0; + : ©Fx = 0; + c ©Fy = 0; 9.00(4) - Ay(12) = 0 Ay = 3.00 kip Bx = 0 By + 3.00 - 9.00 = 0 By = 6.00 kip Equations of Equilibrium: For point D + : ©Fx = 0; + c ©Fy = 0; ND = 0 Ans. 3.00 - 2.25 - VD = 0 VD = 0.750 kip a + ©MD = 0; Ans. MD + 2.25(2) - 3.00(6) = 0 MD = 13.5 kip # ft Ans. Equations of Equilibrium: For point E + : ©Fx = 0; + c ©Fy = 0; NE = 0 Ans. - 6.00 - 3 - VE = 0 VE = - 9.00 kip a + ©ME = 0; Ans. ME + 6.00(4) = 0 ME = - 24.0 kip # ft Ans. Negative signs indicate that ME and VE act in the opposite direction to that shown on FBD. Ans: ND = 0, VD = 0.750 kip, MD = 13.5 kip # ft, NE = 0, VE = - 9.00 kip, ME = - 24.0 kip # ft 5 © 2014 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. 1–6. Determine the normal force, shear force, and moment at a section through point C. Take P = 8 kN. B 0.1 m 0.5 m C 0.75 m 0.75 m A 0.75 m P Support Reactions: a + ©MA = 0; 8(2.25) - T(0.6) = 0 T = 30.0 kN + : ©Fx = 0; 30.0 - A x = 0 A x = 30.0 kN + c ©Fy = 0; Ay - 8 = 0 A y = 8.00 kN Equations of Equilibrium: For point C + : ©Fx = 0; - NC - 30.0 = 0 NC = - 30.0 kN + c ©Fy = 0; Ans. VC + 8.00 = 0 VC = - 8.00 kN a + ©MC = 0; Ans. 8.00(0.75) - MC = 0 MC = 6.00 kN # m Ans. Negative signs indicate that NC and VC act in the opposite direction to that shown on FBD. Ans: NC = - 30.0 kN, VC = - 8.00 kN, MC = 6.00 kN # m 6 © 2014 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. 1–7. The cable will fail when subjected to a tension of 2 kN. Determine the largest vertical load P the frame will support and calculate the internal normal force, shear force, and moment at the cross section through point C for this loading. B 0.1 m 0.5 m C 0.75 m 0.75 m A 0.75 m P Support Reactions: a + ©MA = 0; P(2.25) - 2(0.6) = 0 P = 0.5333 kN = 0.533 kN + : ©Fx = 0; 2 - Ax = 0 + c ©Fy = 0; A y - 0.5333 = 0 Ans. A x = 2.00 kN A y = 0.5333 kN Equations of Equilibrium: For point C + : ©Fx = 0; - NC - 2.00 = 0 NC = - 2.00 kN + c ©Fy = 0; Ans. VC + 0.5333 = 0 VC = - 0.533 kN a + ©MC = 0; Ans. 0.5333(0.75) - MC = 0 MC = 0.400 kN # m Ans. Negative signs indicate that NC and VC act in the opposite direction to that shown on FBD. Ans: P = 0.533 kN, NC = - 2.00 kN, VC = - 0.533 kN, MC = 0.400 kN # m 7 © 2014 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. *1–8. Determine the resultant internal loadings on the cross section through point C. Assume the reactions at the supports A and B are vertical. 6 kN 3 kN/m a + ©MB = 0; - Ay(4) + 6(3.5) + 1 (3)(3)(2) = 0 2 Ay = 7.50 kN + c ©Fy = 0; a + ©MC = 0; NC = 0 7.50 - 6 - VC = 0 Ans. VC = 1.50 kN MC + 6(0.5) - 7.5(1) = 0 Ans. MC = 4.50 kN # m 8 C 0.5 m 0.5 m Referring to the FBD of this segment, Fig. b, + : ©Fx = 0; B A Referring to the FBD of the entire beam, Fig. a, Ans. D 1.5 m 1.5 m © 2014 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. 1–9. Determine the resultant internal loadings on the cross section through point D. Assume the reactions at the supports A and B are vertical. 6 kN 3 kN/m B A C 0.5 m 0.5 m D 1.5 m 1.5 m Referring to the FBD of the entire beam, Fig. a, a + ©MA = 0; By(4) - 6(0.5) - 1 (3)(3)(2) = 0 2 By = 3.00 kN Referring to the FBD of this segment, Fig. b, + : ©Fx = 0; + c ©Fy = 0; ND = 0 VD - 1 (1.5)(1.5) + 3.00 = 0 2 a + ©MD = 0; 3.00(1.5) - Ans. VD = - 1.875 kN Ans. 1 (1.5)(1.5)(0.5) - MD = 0 MD = 3.9375 kN # m 2 = 3.94 kN # m Ans. Ans: ND = 0, VD = - 1.875 kN, MD = 3.94 kN # m 9 © 2014 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. 1–10. The boom DF of the jib crane and the column DE have a uniform weight of 50 lb> ft. If the hoist and load weigh 300 lb, determine the resultant internal loadings in the crane on cross sections through points A, B, and C. D 2 ft F A B 8 ft 3 ft 5 ft C 300 lb 7 ft E Equations of Equilibrium: For point A + ; © Fx = 0; + c © Fy = 0; NA = 0 Ans. VA - 150 - 300 = 0 VA = 450 lb a + ©MA = 0; Ans. - MA - 150(1.5) - 300(3) = 0 MA = - 1125 lb # ft = - 1.125 kip # ft Ans. Negative sign indicates that MA acts in the opposite direction to that shown on FBD. Equations of Equilibrium: For point B + ; © Fx = 0; NB = 0 + c © Fy = 0; VB - 550 - 300 = 0 Ans. VB = 850 lb a + © MB = 0; Ans. - MB - 550(5.5) - 300(11) = 0 MB = - 6325 lb # ft = - 6.325 kip # ft Ans. Negative sign indicates that MB acts in the opposite direction to that shown on FBD. Equations of Equilibrium: For point C + ; © Fx = 0; + c © Fy = 0; VC = 0 Ans. - NC - 250 - 650 - 300 = 0 NC = - 1200 lb = - 1.20 kip a + ©MC = 0; Ans. - MC - 650(6.5) - 300(13) = 0 MC = - 8125 lb # ft = - 8.125 kip # ft Ans. Negative signs indicate that NC and MC act in the opposite direction to that shown on FBD. Ans: NA = 0, VA = 450 lb, MA = - 1.125 kip # ft, NB = 0, VB = 850 lb, MB = - 6.325 kip # ft, VC = 0, NC = - 1.20 kip, MC = - 8.125 kip # ft 10 © 2014 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. 1–11. The forearm and biceps support the 2-kg load at A. If C can be assumed as a pin support, determine the resultant internal loadings acting on the cross section of the bone of the forearm at E.The biceps pulls on the bone along BD. D 75⬚ A C E B Support Reactions: In this case, all the support reactions will be completed. Referring to the free-body diagram of the forearm, Fig. a, a + ©MC = 0; FBD sin 75°(0.07) - 2(9.81)(0.3) = 0 + : ©Fx = 0; Cx - 87.05 cos 75° = 0 Cx = 22.53 N + c ©Fy = 0; 87.05 sin 75° - 2(9.81) - Cy = 0 Cy = 64.47 N 230 mm 35 mm 35 mm FBD = 87.05 N Internal Loadings: Using the results of Cx and Cy, section CE of the forearm will be considered. Referring to the free-body diagram of this part shown in Fig. b, + : ©Fx = 0; NE + 22.53 = 0 NE = - 22.5 N Ans. + c ©Fy = 0; - VE - 64.47 = 0 VE = - 64.5 N Ans. a + ©ME = 0; ME + 64.47(0.035) = 0 ME = - 2.26 N # m Ans. The negative signs indicate that NE, VE and ME act in the opposite sense to that shown on the free-body diagram. Ans: NE = - 22.5 N, VE = - 64.5 N, ME = - 2.26 N # m 11 © 2014 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. *1–12. The serving tray T used on an airplane is supported on each side by an arm. The tray is pin connected to the arm at A, and at B there is a smooth pin. (The pin can move within the slot in the arms to permit folding the tray against the front passenger seat when not in use.) Determine the resultant internal loadings acting on the cross section of the arm through point C when the tray arm supports the loads shown. 12 N 9N 15 mm B 60⬚ 500 mm VC C MC NC b+ ©Fx = 0; NC + 9 cos 30° + 12 cos 30° = 0; NC = - 18.2 N Ans. a+ ©Fy = 0; VC - 9 sin 30° - 12 sin 30° = 0; VC = 10.5 N Ans. a + ©MC = 0; - MC - 9(0.5 cos 60° + 0.115) - 12(0.5 cos 60° + 0.265) = 0 MC = - 9.46 N # Ans. m 12 100 mm A 150 mm T © 2014 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. 1–13. The blade of the hacksaw is subjected to a pretension force of F = 100 N. Determine the resultant internal loadings acting on section a–a that passes through point D. a 225 mm 30⬚ b B A D b F 150 mm a F C Internal Loadings: Referring to the free-body diagram of the section of the hacksaw shown in Fig. a, + ; ©Fx = 0; Na - a + 100 = 0 + c ©Fy = 0; Va - a = 0 a + ©MD = 0; - Ma - a - 100(0.15) = 0 Na - a = - 100 N Ans. Ans. Ma - a = - 15 N # m Ans. The negative sign indicates that Na–a and Ma–a act in the opposite sense to that shown on the free-body diagram. Ans: Na - a = - 100 N, Va - a = 0, Ma - a = - 15 N # m 13 © 2014 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. 1–14. The blade of the hacksaw is subjected to a pretension force of F = 100 N. Determine the resultant internal loadings acting on section b–b that passes through point D. a 225 mm 30⬚ b B A D b F 150 mm a F C Internal Loadings: Referring to the free-body diagram of the section of the hacksaw shown in Fig. a, ©Fx¿ = 0; Nb - b + 100 cos 30° = 0 Nb - b = - 86.6 N Ans. ©Fy¿ = 0; Vb - b - 100 sin 30° = 0 Vb - b = 50 N Ans. a + ©MD = 0; - Mb - b - 100(0.15) = 0 Mb - b = - 15 N # m Ans. The negative sign indicates that Nb–b and Mb–b act in the opposite sense to that shown on the free-body diagram. Ans: Nb - b = - 86.6 N, Vb - b = 50 N, Mb - b = - 15 N # m 14 © 2014 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. 1–15. A 150-lb bucket is suspended from a cable on the wooden frame. Determine the resultant internal loadings on the cross section at D. 1 ft 1 ft 2 ft B D H C 2 ft 30⬚ G 1 ft E 3 ft Support Reactions: We will only need to compute Bx, By, and FGH . Referring to the free-body diagram of member BC, Fig. a, a + ©MB = 0: FGH sin 45°(2) - 150(4) = 0 + : ©Fx = 0; 424.26 cos 45° - Bx = 0 Bx = 300 lb + c ©Fy = 0; 424.26 sin 45° - 150 - By = 0 By = 150 lb I A FGH = 424.26 lb Internal Loadings: Using the results of Bx and By , section BD of member BC will be considered. Referring to the free-body diagram of this part shown in Fig. b, + : ©Fx = 0; ND - 300 = 0 ND = 300 lb Ans. + c ©Fy = 0; - VD - 150 = 0 VD = - 150 lb Ans. a + ©MD = 0; 150(1) + MD = 0 MD = -150 lb # ft Ans. The negative signs indicates that VD and MD act in the opposite sense to that shown on the free-body diagram. Ans: ND = 300 lb, VD = - 150 lb, MD = -150 lb # ft 15 © 2014 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. *1–16. A 150-lb bucket is suspended from a cable on the wooden frame. Determine the resultant internal loadings acting on the cross section at E. 1 ft 1 ft 2 ft B D 2 ft 30⬚ G 1 ft E 3 ft Support Reactions: We will only need to compute Ax, Ay, and FBI. Referring to the free-body diagram of the frame, Fig. a, a + ©MA = 0; FBI sin 30°(6) - 150(4) = 0 FBI = 200 lb + : ©Fx = 0; Ax - 200 sin 30° = 0 Ax = 100 lb + c ©Fy = 0; Ay - 200 cos 30° - 150 = 0 Ay = 323.21 lb Internal Loadings: Using the results of Ax and Ay, section AE of member AB will be considered. Referring to the free-body diagram of this part shown in Fig. b, + : ©Fx = 0; NE + 323.21 = 0 NE = - 323 lb Ans. + c ©Fy = 0; 100 - VE = 0 VE = 100 lb Ans. a + ©MD = 0; 100(3) - ME = 0 ME = 300 lb # ft Ans. The negative sign indicates that NE acts in the opposite sense to that shown on the free-body diagram. 16 I A H C © 2014 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. 1–17. Determine resultant internal loadings acting on section a–a and section b–b. Each section passes through the centerline at point C. 5 kN B b a Referring to the FBD of the entire beam, Fig. a, a + ©MA = 0; NB sin 45°(6) - 5(4.5) = 0 1.5 m NB = 5.303 kN C Referring to the FBD of this segment (section a–a), Fig. b, b +b©Fx¿ = 0; Na - a + 5.303 cos 45° = 0 Na - a = - 3.75 kN Va - a = 1.25 kN Ans. +a ©Fy¿ = 0; Va - a + 5.303 sin 45° - 5 = 0 a + ©MC = 0; 5.303 sin 45°(3) - 5(1.5) - Ma - a = 0 Ma - a = 3.75 kN # m Ans. Ans. A 45 1.5 m a 45 3m Referring to the FBD (section b–b) in Fig. c, + ; ©Fx = 0; Nb - b - 5 cos 45° + 5.303 = 0 Nb - b = - 1.768 kN = - 1.77 kN + c ©Fy = 0; a + ©MC = 0; Vb - b - 5 sin 45° = 0 Vb - b = 3.536 kN = 3.54 kN Ans. Ans. 5.303 sin 45° (3) - 5(1.5) - Mb - b = 0 Mb - b = 3.75 kN # m Ans. Ans: Na - a = - 3.75 kN, Va - a = 1.25 kN, Ma - a = 3.75 kN # m, Nb - b = - 1.77 kN, Vb - b = 3.54 kN # m, Mb - b = 3.75 kN # m 17 © 2014 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. 1–18. The bolt shank is subjected to a tension of 80 lb. Determine the resultant internal loadings acting on the cross section at point C. C 6 in. 90 A B Segment AC: + : ©Fx = 0; NC + 80 = 0; + c ©Fy = 0; VC = 0 a + ©MC = 0; NC = - 80 lb MC + 80(6) = 0; Ans. Ans. MC = - 480 lb # in. Ans. Ans: NC = - 80 lb, VC = 0, MC = - 480 lb # in. 18 © 2014 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. 1–19. Determine the resultant internal loadings acting on the cross section through point C. Assume the reactions at the supports A and B are vertical. 6 kip/ft 6 kip/ft A C 3 ft B D 3 ft 6 ft Referring to the FBD of the entire beam, Fig. a, a + ©MB = 0; 1 1 (6)(6)(2) + (6)(6)(10) - Ay(12) = 0 Ay = 18.0 kip 2 2 Referring to the FBD of this segment, Fig. b, + : ©Fx = 0; NC = 0 + c ©Fy = 0; 18.0 - a + ©MC = 0; Ans. 1 (3)(3) - (3)(3) - VC = 0 2 MC + (3)(3)(1.5) + VC = 4.50 kip Ans. 1 (3)(3)(2) - 18.0(3) = 0 2 MC = 31.5 kip # ft Ans. Ans: NC = 0, VC = 4.50 kip, MC = 31.5 kip # ft 19 © 2014 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. *1–20. Determine the resultant internal loadings acting on the cross section through point D. Assume the reactions at the supports A and B are vertical. 6 kip/ft 6 kip/ft A C 3 ft Referring to the FBD of the entire beam, Fig. a, a + ©MB = 0; 1 1 (6)(6)(2) + (6)(6)(10) - Ay(12) = 0 Ay = 18.0 kip 2 2 Referring to the FBD of this segment, Fig. b, + : ©Fx = 0; ND = 0 + c ©Fy = 0; a + ©MA = 0; 18.0 - 1 (6)(6) - VD = 0 2 MD - 18.0 (2) = 0 Ans. VD = 0 Ans. MD = 36.0 kip # ft Ans. 20 B D 3 ft 6 ft © 2014 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. 1–21. The forged steel clamp exerts a force of F = 900 N on the wooden block. Determine the resultant internal loadings acting on section a–a passing through point A. 200 mm F 900 N a 30 F 900 N A a Internal Loadings: Referring to the free-body diagram of the section of the clamp shown in Fig. a, ©Fy¿ = 0; 900 cos 30° - Na - a = 0 Na - a = 779 N Ans. ©Fx¿ = 0; Va - a - 900 sin 30° = 0 Va - a = 450 N Ans. a + ©MA = 0; 900(0.2) - Ma - a = 0 Ma - a = 180 N # m Ans. Ans: Na - a = 779 N, Va - a = 450 N, Ma - a = 180 N # m: 21 © 2014 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. 1–22. The metal stud punch is subjected to a force of 120 N on the handle. Determine the magnitude of the reactive force at the pin A and in the short link BC. Also, determine the internal resultant loadings acting on the cross section passing through the handle arm at D. 120 N 60 50 mm 100 mm 50 mm E B 30 D 100 mm 300 mm A C 200 mm Member: a +©MA = 0; FBC cos 30°(50) - 120(500) = 0 FBC = 1385.6 N = 1.39 kN + c ©Fy = 0; Ans. Ay - 1385.6 - 120 cos 30° = 0 Ay = 1489.56 N + ; ©Fx = 0; Ax - 120 sin 30° = 0; Ax = 60 N FA = 21489.562 + 602 Ans. = 1491 N = 1.49 kN Segment: a+ ©Fx¿ = 0; ND - 120 = 0 ND = 120 N Ans. +Q©F = 0; y¿ VD = 0 Ans. a + ©MD = 0; MD - 120(0.3) = 0 MD = 36.0 N # m Ans. Ans: FBC = 1.39 kN, FA = 1.49 kN, ND = 120 N, VD = 0, MD = 36.0 N # m 22 © 2014 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. 1–23. Solve Prob. 1–22 for the resultant internal loadings acting on the cross section passing through the handle arm at E and at a cross section of the short link BC. 120 N 60 50 mm 100 mm 50 mm E B 30 D 100 mm 300 mm A C 200 mm Member: a + ©MA = 0; FBC cos 30°(50) - 120(500) = 0 FBC = 1385.6 N = 1.3856 kN Segment: +b©F = 0; x¿ NE = 0 a+ ©Fy¿ = 0; VE - 120 = 0; VE = 120 N Ans. a + ©ME = 0; ME - 120(0.4) = 0; ME = 48.0 N # m Ans. Ans. Short link: + ; ©Fx = 0; V = 0 + c ©Fy = 0; 1.3856 - N = 0; a + ©MH = 0; M = 0 Ans. N = 1.39 kN Ans. Ans. Ans: NE = 0, VE = 120 N, ME = 48.0 N # m, Short link: V = 0, N = 1.39 kN, M = 0 23 © 2014 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. *1–24. Determine the resultant internal loadings acting on the cross section of the semicircular arch at C. C w0 r u A p a + ©MA = 0; By (2r) - (w0 r du)(cos u)(r sin u) L0 p - L0 (w0 r du)(sin u)r(1 - cos u) = 0 p By (2r) - w0 r2 sin u du = 0 L0 p By (2r) - w0 r2( -cos u)]0 = 0 By = w0 r + : ©Fx = 0; NC = - w0 r sin u + c ©Fy = 0; - NC - w0 r p 2 L0 p 2 L0 cos u du = 0 Ans. = - w0 r w0 r + VC - w0 r p 2 L0 sin u du = 0 w0 r + VC - w0 r( - cos u) a + ©M0 = 0; p 2 L0 = 0; V2 = 0 Ans. w0 r(r) - MC + ( - w0 r)(r) = 0 MC = 0 Ans. 24 B © 2014 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. z 1–25. Determine the resultant internal loadings acting on the cross section through point B of the signpost. The post is fixed to the ground and a uniform pressure of 7 lb>ft2 acts perpendicular to the face of the sign. 3 ft 2 ft ©Fx = 0; (VB)x - 105 = 0; (VB)x = 105 lb ©Fy = 0; (VB)y = 0 Ans. ©Fz = 0; (NB)z = 0 Ans. ©Mx = 0; (MB)x = 0 ©My = 0; (MB)y - 105(7.5) = 0; ©Mz = 0; (TB)z - 105(0.5) = 0; Ans. 3 ft 2 7 lb/ft Ans. (MB)y = 788 lb # ft 6 ft Ans. (TB)z = 52.5 lb # ft B Ans. A 4 ft x y Ans: (VB)x = 105 lb, (VB)y = 0, (NB)z = 0, (MB)x = 0, (MB)y = 788 lb # ft, (TB)z = 52.5 lb # ft 25 © 2014 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. z 1–26. The shaft is supported at its ends by two bearings A and B and is subjected to the forces applied to the pulleys fixed to the shaft. Determine the resultant internal loadings acting on the cross section located at point C. The 300-N forces act in the - z direction and the 500-N forces act in the + x direction. The journal bearings at A and B exert only x and z components of force on the shaft. A 400 mm 150 mm 200 mm C 250 mm x 300 N 300 N B 500 N 500 N y ©Fx = 0; (VC)x + 1000 - 750 = 0; ©Fy = 0; (NC)y = 0 ©Fz = 0; (VC)z + 240 = 0; (VC)x = - 250 N Ans. Ans. Ans. (VC)z = - 240 N (MC)x = - 108 N # m ©Mx = 0; (MC)x + 240(0.45) = 0; ©My = 0; (TC)y = 0 ©Mz = 0; (MC)z - 1000(0.2) + 750(0.45) = 0; Ans. Ans. (MC)z = - 138 N # m Ans. Ans: (VC)x = - 250 N, (NC)y = 0, (VC)z = - 240 N, (MC)x = - 108 N # m, (TC)y = 0, (MC)z = - 138 N # m 26 © 2014 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. z 1–27. The pipe assembly is subjected to a force of 600 N at B. Determine the resultant internal loadings acting on the cross section at C. 600 N B 60 30 150 mm 150 mm A C 500 mm Internal Loading: Referring to the free-body diagram of the section of the pipe shown in Fig. a, ©Fx = 0; (NC)x - 600 cos 60° sin 30° = 0 (NC)x = 150 N Ans. ©Fy = 0; (VC)y + 600 cos 60° cos 30° = 0 (VC)y = - 260 N Ans. ©Fz = 0; (VC)z + 600 sin 60° = 0 (VC)z = - 520 N Ans. x 400 mm y ©Mx = 0; (TC)x + 600 sin 60°(0.4) - 600 cos 60° cos 30°(0.5) = 0 Ans. (TC)x = -77.9 N # m ©My = 0; (MC)y - 600 sin 60° (0.15) - 600 cos 60° sin 30°(0.5) = 0 (MC)y = 153 N # m Ans. ©Mz = 0; (MC)z + 600 cos 60° cos 30°(0.15) + 600 cos 60° sin 30°(0.4) = 0 (MC)z = -99.0 N # m Ans. The negative signs indicate that (VC)y, (VC)z, (TC)x, and (MC)z act in the opposite sense to that shown on the free-body diagram. Ans: (NC)x = 150 N, (VC)y = - 260 N, (VC)z = - 520 N, (TC)x = -77.9 N # m, (MC)y = 153 N # m, (MC)z = -99.0 N # m 27 © 2014 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. z *1–28. The brace and drill bit is used to drill a hole at O. If the drill bit jams when the brace is subjected to the forces shown, determine the resultant internal loadings acting on the cross section of the drill bit at A. O x Internal Loading: Referring to the free-body diagram of the section of the drill and brace shown in Fig. a, ©Fx = 0; ©Fy = 0; ©Fz = 0; ©Mx = 0; ©My = 0; ©Mz = 0; A VA B x - 30 = 0 A NA B y - 50 = 0 A VA B z - 10 = 0 A MA B x - 10(2.25) = 0 A TA B y - 30(0.75) = 0 A MA B z + 30(1.25) = 0 A VA B x = 30 lb Ans. A NA B y = 50 lb Ans. A VA B z = 10 lb Ans. A MA B x = 22.5 lb # ft A TA B y = 22.5 lb # ft A MA B z = - 37.5 lb # ft Ans. Ans. Ans. The negative sign indicates that (MA)z acts in the opposite sense to that shown on the free-body diagram. 28 3 in. 9 in. Fx 30 lb A Fz 10 lb 9 in. 6 in. 6 in. 6 in. Fy 50 lb y © 2014 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. 1–29. The curved rod AD of radius r has a weight per length of w. If it lies in the vertical plane, determine the resultant internal loadings acting on the cross section through point B. Hint: The distance from the centroid C of segment AB to point O is OC = [2r sin (u>2)]>u. u 2 A C B O u r D R+ ©Fx = 0; NB + wru cos u = 0 NB = - wru cos u +Q©F = 0; y Ans. - VB - wru sin u = 0 VB = -wru sin u Ans. u 2r sin (u/2) a + ©M0 = 0; wru a cos b a b + (NB)r + MB = 0 2 u MB = - NBr - wr2 2 sin (u/2) cos (u/2) MB = wr2(u cos u - sin u) Ans. Ans: NB = - wru cos u, VB = -wru sin u, MB = wr2(u cos u - sin u) 29 © 2014 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. M dM 1–30. A differential element taken from a curved bar is shown in the figure. Show that dN>du = V, dV>du = - N, dM>du = - T, and dT>du = M. V dV N dN M V N T ©Fx = 0; N cos du du du du + V sin - (N + dN) cos + (V + dV) sin = 0 2 2 2 2 (1) ©Fy = 0; N sin du du du du - V cos + (N + dN) sin + (V + dV) cos = 0 2 2 2 2 (2) ©Mx = 0; T cos du du du du + M sin - (T + dT) cos + (M + dM) sin = 0 2 2 2 2 (3) ©My = 0; du du du du - M cos + (T + dT) sin + (M + dM) cos = 0 2 2 2 2 du du du du Since is can add, then sin , cos = = 1 2 2 2 2 T sin Eq. (1) becomes Vdu - dN + dVdu = 0 2 Neglecting the second order term, Vdu - dN = 0 dN = V du Eq. (2) becomes Ndu + dV + QED dNdu = 0 2 Neglecting the second order term, Ndu + dV = 0 dV = -N du Eq. (3) becomes Mdu - dT + QED dMdu = 0 2 Neglecting the second order term, Mdu - dT = 0 dT = M du Eq. (4) becomes Tdu + dM + QED dTdu = 0 2 Neglecting the second order term, Tdu + dM = 0 dM = -T du QED 30 (4) T dT du © 2014 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. 1–31. The supporting wheel on a scaffold is held in place on the leg using a 4-mm-diameter pin as shown. If the wheel is subjected to a normal force of 3 kN, determine the average shear stress developed in the pin. Neglect friction between the inner scaffold puller leg and the tube used on the wheel. 3 kN + c ©Fy = 0; 3 kN # 2V = 0; V = 1.5 kN 3 tavg = 1.5(10 ) V = 119 MPa = p 2 A 4 (0.004) Ans. Ans: tavg = 119 MPa 31 © 2014 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. *1–32. The lever is held to the fixed shaft using a tapered pin AB, which has a mean diameter of 6 mm. If a couple is applied to the lever, determine the average shear stress in the pin between the pin and lever. B 12 mm A 250 mm 20 N a + ©MO = 0; tavg = - F(12) + 20(500) = 0; F = 833.33 N V 833.33 = 29.5 MPa = p 6 2 A 4 (1000 ) Ans. 32 250 mm 20 N © 2014 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. 1–33. The bar has a cross-sectional area A and is subjected to the axial load P. Determine the average normal and average shear stresses acting over the shaded section, which is oriented at u from the horizontal. Plot the variation of these stresses as a function of u 10 … u … 90°2. P P u A Equations of Equilibrium: R+ ©Fx = 0; V - P cos u = 0 V = P cos u Q+ ©Fy = 0; N - P sin u = 0 N = P sin u Average Normal Stress and Shear Stress: Area at u plane, A¿ = savg = P sin u N P = = sin2 u A A¿ A sin u tavg = P cos u V = A A¿ sin u = A . sin u Ans. P P sin u cos u = sin 2u A 2A Ans. Ans: savg = 33 P P sin2 u, tavg = sin 2u A 2A © 2014 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. 1–34. The built-up shaft consists of a pipe AB and solid rod BC. The pipe has an inner diameter of 20 mm and outer diameter of 28 mm. The rod has a diameter of 12 mm. Determine the average normal stress at points D and E and represent the stress on a volume element located at each of these points. 4 kN B A D 6 kN C 8 kN 6 kN E At D: sD = P = A 4(103) p 2 4 (0.028 - 0.022) = 13.3 MPa (C) Ans. At E: sE = P = A 8(103) p 4 (0.0122) = 70.7 MPa (T) Ans. Ans: sD = 13.3 MPa (C), sE = 70.7 MPa (T) 34 © 2014 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. 1–35. If the turnbuckle is subjected to an axial force of P = 900 lb, determine the average normal stress developed in section a–a and in each of the bolt shanks at B and C. Each bolt shank has a diameter of 0.5 in. a 1 in. A 0.25 in. P C B P a Internal Loading: The normal force developed in section a–a of the bracket and the bolt shank can be obtained by writing the force equations of equilibrium along the x axis with reference to the free-body diagrams of the sections shown in Figs. a and b, respectively. + : ©Fx = 0; 900 - 2Na - a = 0 Na - a = 450 lb + : ©Fx = 0; 900 - Nb = 0 Nb = 900 lb Average Normal Stress: The cross-sectional areas of section a–a and the bolt shank p are Aa - a = (1)(0.25) = 0.25 in2 and Ab = (0.52) = 0.1963 in2, respectively. 4 We obtain (sa - a)avg = sb = Na - a 450 = = 1800 psi = 1.80 ksi Aa - a 0.25 Ans. Nb 900 = = 4584 psi = 4.58 ksi Ab 0.1963 Ans. Ans: (sa - a)avg = 1.80 ksi, sb = 4.58 ksi 35 © 2014 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. *1–36. The average normal stresses developed in section a–a of the turnbuckle, and the bolts shanks at B and C, are not allowed to exceed 15 ksi and 45 ksi, respectively. Determine the maximum axial force P that can be applied to the turnbuckle. Each bolt shank has a diameter of 0.5 in. a 0.25 in. P C B a Internal Loading: The normal force developed in section a–a of the bracket and the bolt shank can be obtained by writing the force equations of equilibrium along the x axis with reference to the free-body diagrams of the sections shown in Figs. a and b, respectively. + : ©Fx = 0; P - 2Na - a = 0 Na - a = P>2 + : ©Fx = 0; P - Nb = 0 Nb = P Average Normal Stress: The cross-sectional areas of section a–a and the bolt p shank are Aa - a = 1(0.25) = 0.25 in2 and Ab = (0.52) = 0.1963 in2, respectively. 4 We obtain (sa - a)allow = Na - a ; Aa - a 15(103) = P>2 0.25 P = 7500 lb = 7.50 kip (controls) sb = Nb ; Ab 45(103) = 1 in. A Ans. P 0.1963 P = 8336 lb = 8.84 kip 36 P © 2014 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. 1–37. The plate has a width of 0.5 m. If the stress distribution at the support varies as shown, determine the force P applied to the plate and the distance d to where it is applied. 4m P d x s (15x 1/2) MPa 30 MPa The resultant force dF of the bearing pressure acting on the plate of area dA = b dx = 0.5 dx, Fig. a, 1 1 dF = sb dA = (15x2)(106)(0.5dx) = 7.5(106)x2 dx + c ©Fy = 0; L dF - P = 0 4m 1 L0 7.5(106)x 2 dx - P = 0 P = 40(106) N = 40 MN Ans. Equilibrium requires a + ©MO = 0; L xdF - Pd = 0 4m 1 L0 x[7.5(106)x2 dx] - 40(106) d = 0 d = 2.40 m Ans. Ans: P = 40 MN, d = 2.40 m 37 © 2014 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. 1–38. The two members used in the construction of an aircraft fuselage are joined together using a 30° fish-mouth weld. Determine the average normal and average shear stress on the plane of each weld. Assume each inclined plane supports a horizontal force of 400 lb. N - 400 sin 30° = 0; N = 200 lb 400 cos 30° - V = 0; V = 346.41 lb A¿ = 1.5 in. 800 lb 30 1 in. 1 in. 800 lb 30 1.5(1) = 3 in2 sin 30° s = N 200 = = 66.7 psi A¿ 3 Ans. t = V 346.41 = = 115 psi A¿ 3 Ans. Ans: s = 66.7 psi, t = 115 psi 38 © 2014 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. 1–39. If the block is subjected to the centrally applied force of 600 kN, determine the average normal stress in the material. Show the stress acting on a differential volume element of the material. 150 mm 600 kN 150 mm 150 mm 50 mm 100 mm 100 mm 50 mm 150 mm The cross-sectional area of the block is A = 0.6(0.3) - 0.3(0.2) = 0.12 m2. savg = 600(103) P = = 5(106) Pa = 5 MPa A 0.12 Ans. The average normal stress distribution over the cross section of the block and the state of stress of a point in the block represented by a differential volume element are shown in Fig. a Ans: savg = 5 MPa 39 © 2014 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. *1–40. Determine the average normal stress in each of the 20-mm diameter bars of the truss. Set P = 40 kN. C P 1.5 m B A 2m Internal Loadings: The force developed in each member of the truss can be determined by using the method of joints. First, consider the equilibrium of joint C, Fig. a, + : ©Fx = 0; 4 40 - FBC a b = 0 5 FBC = 50 kN (C) + c ©Fy = 0; 3 50 a b - FAC = 0 5 FAC = 30 kN (T) Subsequently, the equilibrium of joint B, Fig. b, is considered + : ©Fx = 0; 4 50 a b - FAB = 0 5 FAB = 40 kN (T) Average Normal Stress: The cross-sectional area of each of the bars is p A = (0.022) = 0.3142(10 - 3) m2. We obtain, 4 50(103) FBC (savg)BC = = 159 MPa = A 0.3142(10 - 3) Ans. (savg)AC = 30(103) FAC = 95.5 MPa = A 0.3142(10 - 3) Ans. (savg)AB = 40(103) FAB = 127 MPa = A 0.3142(10 - 3) Ans. 40 © 2014 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. 1–41. If the average normal stress in each of the 20-mmdiameter bars is not allowed to exceed 150 MPa, determine the maximum force P that can be applied to joint C. C P 1.5 m B A 2m Internal Loadings: The force developed in each member of the truss can be determined by using the method of joints. First, consider the equilibrium of joint C, Fig. a, + : ©Fx = 0; 4 P - FBC a b = 0 5 FBC = 1.25P(C) + c ©Fy = 0; 3 1.25Pa b - FAC = 0 5 FAC = 0.75P(T) Subsequently, the equilibrium of joint B, Fig. b, is considered + : ©Fx = 0; 4 1.25P a b - FAB = 0 5 FAB = P(T) Average Normal Stress: Since the cross-sectional area and the allowable normal stress of each bar are the same, member BC which is subjected to the maximum normal force is p (0.022) = 4 0.3142(10 - 3) m2. We have, the critical member. The cross-sectional area of each of the bars is A = (savg)allow = FBC ; A 150(106) = 1.25P 0.3142(10 - 3) P = 37 699 N = 37.7 kN Ans. Ans: P = 37.7 kN 41 © 2014 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. 1–42. Determine the average shear stress developed in pin A of the truss. A horizontal force of P = 40 kN is applied to joint C. Each pin has a diameter of 25 mm and is subjected to double shear. C P 1.5 m B A 2m Internal Loadings: The forces acting on pins A and B are equal to the support reactions at A and B. Referring to the free-body diagram of the entire truss, Fig. a, ©MA = 0; By(2) - 40(1.5) = 0 By = 30 kN + : ©Fx = 0; 40 - Ax = 0 Ax = 40 kN + c ©Fy = 0; 30 - Ay = 0 Ay = 30 kN Thus, FA = 2Ax2 + Ay2 = 2402 + 302 = 50 kN Since pin A is in double shear, Fig. b, the shear forces developed on the shear planes of pin A are FA 50 VA = = = 25 kN 2 2 Average Shear Stress: The area of the shear plane for pin A is AA = p (0.0252) = 4 0.4909(10 - 3) m2. We have (tavg)A = 25(103) VA = 50.9 MPa = AA 0.4909(10 - 3) Ans. Ans: (tavg)A = 50.9 MPa 42 © 2014 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. 1–43. The 150-kg bucket is suspended from end E of the frame. Determine the average normal stress in the 6-mm diameter wire CF and the 15-mm diameter short strut BD. 0.6 m 0.6 m D C E 0.6 m 30 B 1.2 m Internal Loadings: The normal force developed in rod BD and cable CF can be determined by writing the moment equations of equilibrium about C and A with reference to the free-body diagram of member CE and the entire frame shown in Figs. a and b, respectively. a + ©MC = 0; FBD sin 45°(0.6) - 150(9.81)(1.2) = 0 FBD = 4162.03 N a + ©MA = 0; FCF sin 30°(1.8) - 150(9.81)(1.2) = 0 FCF = 1962 N F A Average Normal Stress: The cross-sectional areas of rod BD and cable CF are p p ABD = (0.0152) = 0.1767(10 - 3) m2 and ACF = (0.0062) = 28.274(10 - 6) m2. 4 4 We have (savg)BD = FBD 4162.03 = 23.6 MPa = ABD 0.1767(10 - 3) Ans. (savg)CF = FCF 1962 = 69.4 MPa = ACF 28.274(10 - 6) Ans. Ans: (savg)BD = 23.6 MPa, (savg)CF = 69.4 MPa 43 © 2014 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. *1–44. The 150-kg bucket is suspended from end E of the frame. If the diameters of the pins at A and D are 6 mm and 10 mm, respectively, determine the average shear stress developed in these pins. Each pin is subjected to double shear. 0.6 m FBD sin 45°(0.6) - 150(9.81)(1.2) = 0 E 0.6 m 30 Internal Loading: The forces exerted on pins D and A are equal to the support reaction at D and A. First, consider the free-body diagram of member CE shown in Fig. a. a + ©MC = 0; 0.6 m D C B 1.2 m FBD = 4162.03 N Subsequently, the free-body diagram of the entire frame shown in Fig. b will be considered. a + ©MA = 0; FCF sin 30°(1.8) - 150(9.81)(1.2) = 0 FCF = 1962 N + : ©Fx = 0; Ax - 1962 sin 30° = 0 Ax = 981 N + c ©Fy = 0; Ay - 1962 cos 30° - 150(9.81) = 0 Ay = 3170.64 N F A Thus, the forces acting on pins D and A are FD = FBD = 4162.03 N FA = 2Ax2 + Ay2 = 29812 + 3170.642 = 3318.93 N Since both pins are in double shear VD = FD = 2081.02 N 2 VA = FA = 1659.47 N 2 Average Shear Stress: The cross-sectional areas of the shear plane of pins D and A p p (0.012) = 78.540(10 - 6) m2 and AA = (0.0062) = 28.274(10 - 6) m2. 4 4 We obtain VA 1659.47 (tavg)A = = 58.7 MPa = Ans. AA 28.274(10 - 6) are AD = (tavg)D = VD 2081.02 = 26.5 MPa = AD 78.540(10 - 6) Ans. Ans: x = 4 in., y = 4 in., s = 9.26 psi 44 © 2014 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. 1–45. The pedestal has a triangular cross section as shown. If it is subjected to a compressive force of 500 lb, specify the x and y coordinates for the location of point P(x, y), where the load must be applied on the cross section, so that the average normal stress is uniform. Compute the stress and sketch its distribution acting on the cross section at a location removed from the point of load application. 500 lb 12 in. y x A B A B x = 1 12 + 12(6)(12) 12 2 (3)(12) 3 3 1 (9)(12) 2 y = 1 2 1 2 (3)(12)(3) 3 + 2 (6)(12) 1 2 (9)(12) s = 500 P = 9.26 psi = 1 A 2 (9)(12) A B = 4 in. A 3 + 63 B P(x,y) Ans. = 4 in. Ans. Ans. 45 3 in. 6 in. © 2014 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. 1–46. The 20-kg chandelier is suspended from the wall and ceiling using rods AB and BC, which have diameters of 3 mm and 4 mm, respectively. Determine the angle u so that the average normal stress in both rods is the same. C A 30⬚ B u Internal Loadings: The force developed in cables AB and BC can be determined by considering the equilibrium of joint B, Fig. a, + : ©Fx = 0; FBC cos u - FAB cos 30° = 0 (1) Average Normal Stress: The cross-sectional areas of cables AB and BC are p p AAB = (0.0032) = 7.069(10 - 6) m2 and ABC = (0.0042) = 12.566(10 - 6) m2. 4 4 Since the average normal stress in both cables are required to be the same, then (savg)AB = (savg)BC FAB FBC = AAB ABC FAB 7.069(10 - 6) = FBC 12.566(10 - 6) FAB = 0.5625FBC (2) Substituting Eq. (2) into Eq. (1), FBC(cos u - 0.5625 cos 30°) = 0 Since FBC Z 0, then cos u - 0.5625 cos 30° = 0 u = 60.8° Ans. Ans: u = 60.8° 46 © 2014 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. 1–47. The chandelier is suspended from the wall and ceiling using rods AB and BC, which have diameters of 3 mm and 4 mm, respectively. If the average normal stress in both rods is not allowed to exceed 150 MPa, determine the largest mass of the chandelier that can be supported if u = 45. C A 30⬚ B u Internal Loadings: The force developed in cables AB and BC can be determined by considering the equilibrium of joint B, Fig. a, + : ©Fx = 0; FBC cos 45° - FAB cos 30° = 0 (1) + c ©Fy = 0; FBC sin 45° + FAB sin 30° - m(9.81) = 0 (2) Solving Eqs. (1) and (2) yields FAB = 7.181m FBC = 8.795m Average Normal Stress: The cross-sectional areas of cables AB and BC are p p ABC = (0.0042) = 12.566(10 - 6) m2. AAB = (0.0032) = 7.069(10 - 6) m2 and 4 4 We have, (savg)allow = FAB ; AAB 150(106) = 7.181m 7.069(10 - 6) m = 147.64 kg = 148 kg (controls) (savg)allow = FBC ; ABC 150(106) = Ans. 8.795m 12.566(10 - 6) m = 214.31 kg Ans: m = 148 kg 47 © 2014 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. *1–48. The beam is supported by a pin at A and a short link BC. If P = 15 kN, determine the average shear stress developed in the pins at A, B, and C. All pins are in double shear as shown, and each has a diameter of 18 mm. P 0.5 m 4P 1m 4P 1.5 m 1.5 m 2P 0.5 m C 30⬚ B For pins B and C: 82.5 (103) V tB = tC = = p 18 2 = 324 MPa A 4 (1000 ) Ans. For pin A: FA = 2(82.5)2 + (142.9)2 = 165 kN tA = 82.5 (103) V = p 18 2 = 324 MPa A 4 (1000 ) Ans. 48 A © 2014 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. 1–49. The joint is subjected to the axial member force of 6 kip. Determine the average normal stress acting on sections AB and BC. Assume the member is smooth and is 1.5-in. thick. 6 kip 60⬚ A 1.5 in. B + c ©Fy = 0; C 20⬚ 4.5 in. -6 sin 60° + NBC cos 20° = 0 NBC = 5.530 kip + : ©Fx = 0; NAB - 6 cos 60° - 5.530 sin 20° = 0 NAB = 4.891 kip sAB = NAB 4.891 = = 2.17 ksi AAB (1.5)(1.5) Ans. sBC = NBC 5.530 = = 0.819 ksi ABC (1.5)(4.5) Ans. Ans: sAB = 2.17 ksi, sBC = 0.819 ksi 49 © 2014 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. 1–50. The driver of the sports car applies his rear brakes and causes the tires to slip. If the normal force on each rear tire is 400 lb and the coefficient of kinetic friction between the tires and the pavement is mk = 0.5, determine the average shear stress developed by the friction force on the tires. Assume the rubber of the tires is flexible and each tire is filled with an air pressure of 32 psi. 400 lb F = mkN = 0.5(400) = 200 lb p = N ; A tavg = A = 400 = 12.5 in2 32 F 200 = = 16 psi A 12.5 Ans. Ans: tavg = 16 psi 50 © 2014 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. 1–51. During the tension test, the wooden specimen is subjected to an average normal stress of 2 ksi. Determine the axial force P applied to the specimen. Also, find the average shear stress developed along section a–a of the specimen. P a 4 in. a 2 in. 1 in. Internal Loading: The normal force developed on the cross section of the middle portion of the specimen can be obtained by considering the free-body diagram shown in Fig. a. + c ©Fy = 0; P P + - N = 0 2 2 4 in. N = P Referring to the free-body diagram shown in fig. b, the shear force developed in the shear plane a–a is + c ©Fy = 0; P - Va - a = 0 2 Va - a P P = 2 Average Normal Stress and Shear Stress: The cross-sectional area of the specimen is A = 1(2) = 2 in2. We have savg = N ; A 2(103) = P 2 P = 4(103)lb = 4 kip Ans. 4(103) P = = 2(103) lb. The area of the shear plane is 2 2 = 2(4) = 8 in2 . We obtain Using the result of P, Va - a = Aa - a A ta - a B avg = 2(103) Va - a = = 250 psi Aa - a 8 Ans. Ans: P = 4 kip, Ata - aB avg = 250 psi 51 © 2014 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. *1–52. If the joint is subjected to an axial force of P = 9 kN, determine the average shear stress developed in each of the 6-mm diameter bolts between the plates and the members and along each of the four shaded shear planes. P P 100 mm 100 mm Internal Loadings: The shear force developed on each shear plane of the bolt and the member can be determined by writing the force equation of equilibrium along the member’s axis with reference to the free-body diagrams shown in Figs. a. and b, respectively. ©Fy = 0; 4Vb - 9 = 0 Vb = 2.25 kN ©Fy = 0; 4Vp - 9 = 0 Vp = 2.25 kN Average Shear Stress: The areas of each shear plane of the bolt and the member p are Ab = (0.0062) = 28.274(10 - 6) m2 and Ap = 0.1(0.1) = 0.01 m2 , respectively. 4 We obtain A tavg B b = 2.25(103) Vb = 79.6 MPa = Ab 28.274(10 - 6) A tavg B p = Vp Ap = Ans. 2.25(103) = 225 kPa 0.01 Ans. 52 © 2014 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. 1–53. The average shear stress in each of the 6-mm diameter bolts and along each of the four shaded shear planes is not allowed to exceed 80 MPa and 500 kPa, respectively. Determine the maximum axial force P that can be applied to the joint. P P 100 mm 100 mm Internal Loadings: The shear force developed on each shear plane of the bolt and the member can be determined by writing the force equation of equilibrium along the member’s axis with reference to the free-body diagrams shown in Figs. a. and b, respectively. ©Fy = 0; 4Vb - P = 0 Vb = P>4 ©Fy = 0; 4Vp - P = 0 Vp = P>4 Average Shear Stress: The areas of each shear plane of the bolts and the members p are Ab = (0.0062) = 28.274(10 - 6) m2 and Ap = 0.1(0.1) = 0.01 m2, respectively. 4 We obtain A tallow B b = Vb ; Ab 80(106) = P>4 28.274(10 - 6) P = 9047 N = 9.05 kN (controls) A tallow B p = Vp Ap ; 500(103) = Ans. P>4 0.01 P = 20 000 N = 20 kN Ans: P = 9.05 kN 53 © 2014 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. 1–54. When the hand is holding the 5-lb stone, the humerus H, assumed to be smooth, exerts normal forces FC and FA on the radius C and ulna A, respectively, as shown. If the smallest crosssectional area of the ligament at B is 0.30 in2, determine the greatest average tensile stress to which it is subjected. B H FB 75⬚ FC 0.8 in. G C A FA 2 in. a + ©MO = 0; FB sin 75°(2) - 5(14) = 0 14 in. FB = 36.235 lb s = P 36.235 = = 121 psi A 0.30 Ans. Ans: s = 121 psi 54 © 2014 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. 1–55. The 2-Mg concrete pipe has a center of mass at point G. If it is suspended from cables AB and AC, determine the average normal stress developed in the cables.The diameters of AB and AC are 12 mm and 10 mm, respectively. A 30⬚ 45⬚ C B G Internal Loadings: The normal force developed in cables AB and AC can be determined by considering the equilibrium of the hook for which the free-body diagram is shown in Fig. a. FAB = 14 362.83 N (T) ©Fx¿ = 0; 2000(9.81) cos 45° - FAB cos 15° = 0 ©Fy¿ = 0; 2000(9.81) sin 45° - 14 362.83 sin 15° - FAC = 0 FAC = 10 156.06 N (T) Average Normal Stress: The cross-sectional areas of cables AB and AC are p p AAB = (0.0122) = 0.1131(10 - 3) m2 and AAC = (0.012) = 78.540(10 - 6) m2. 4 4 We have, sAB = FAB 14 362.83 = 127 MPa = AAB 0.1131(10 - 3) Ans. sAC = FAC 10 156.06 = 129 MPa = AAC 78.540(10 - 6) Ans. Ans: sAB = 127 MPa, sAC = 129 MPa 55 © 2014 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. *1–56. The 2-Mg concrete pipe has a center of mass at point G. If it is suspended from cables AB and AC, determine the diameter of cable AB so that the average normal stress developed in this cable is the same as in the 10-mm diameter cable AC. A 30⬚ 45⬚ C B Internal Loadings: The normal force in cables AB and AC can be determined by considering the equilibrium of the hook for which the free-body diagram is shown in Fig. a. ©Fx¿ = 0; 2000(9.81) cos 45° - FAB cos 15° = 0 FAB = 14 362.83 N (T) ©Fy¿ = 0; 2000(9.81) sin 45° - 14 362.83 sin 15° - FAC = 0 FAC = 10 156.06 N (T) Average Normal Stress: The cross-sectional areas of cables AB and AC are p p AAB = dAB2 and AAC = (0.012) = 78.540(10 - 6) m2. 4 4 Here, we require sAB = sAC FAB FAC = AAB AAC 10 156.06 14 362.83 = p 2 78.540(10 - 6) 4 dAB dAB = 0.01189 m = 11.9 mm Ans. 56 G © 2014 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. 1–57. If the concrete pedestal has a specific weight of g, determine the average normal stress developed in the pedestal as a function of z. r0 z h 2r0 Internal Loading: From the geometry shown in Fig. a, h¿ + h h¿ = ; r0 2r0 h¿ = h and then r0 r = ; z + h h r = r0 (z + h) h Thus, the volume of the frustrum shown in Fig. b is V = = 2 r0 1 1 e p c (z + h) d f(z + h) - a pr02 bh 3 h 3 pr02 3h2 c (z + h)3 - h3 d The weight of this frustrum is W = gV = pr02g 3h2 c (z + h)3 - h3 d Average Normal Stress: The cross-sectional area the frustrum as a function of z is pr02 (z + h)2. A = pr2 = h2 Also, the normal force acting on this cross section is N = W, Fig. b. We have savg N = = A pr02g 3h2 [(z + h)3 - h3] pr02 h2 (z = + h)2 g (z + h)3 - h3 c d 3 (z + h)2 Ans. Ans: savg = 57 g (z + h)3 - h3 c d 3 (z + h)2 © 2014 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. 1–58. The anchor bolt was pulled out of the concrete wall and the failure surface formed part of a frustum and cylinder. This indicates a shear failure occurred along the cylinder BC and tension failure along the frustum AB. If the shear and normal stresses along these surfaces have the magnitudes shown, determine the force P that must have been applied to the bolt. P A 45⬚ 45⬚ 50 mm 3 MPa 3 MPa B Average Normal Stress: For the frustum, A = 2pxL = 2p(0.025 + 0.025) A 20.05 + 0.05 2 = 0.02221 m P ; s = A 3 A 10 6 B 2 4.5 MPa B 30 mm C 2 25 mm 25 mm F1 = 0.02221 F1 = 66.64 kN Average Shear Stress: For the cylinder, A = p(0.05)(0.03) = 0.004712 m2 F2 V ; 4.5 A 106 B = tavg = A 0.004712 F2 = 21.21 kN Equation of Equilibrium: + c ©Fy = 0; P - 21.21 - 66.64 sin 45° = 0 P = 68.3 kN Ans. Ans: P = 68.3 kN 58 © 2014 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. 1–59. The jib crane is pinned at A and supports a chain hoist that can travel along the bottom flange of the beam, 1 ft … x … 12 ft. If the hoist is rated to support a maximum of 1500 lb, determine the maximum average normal stress in 3 the 4 -in. diameter tie rod BC and the maximum average shear stress in the 58 -in. -diameter pin at B. D B 30⬚ A C x a + ©MA = 0; TBC sin 30°(10) - 1500(x) = 0 1500 lb 10 ft Maximum TBC occurs when x = 12 ft TBC = 3600 lb Ans. s = P = A t = 3600>2 V = p = 5.87 ksi 2 A 4 (5>8) 3600 p 2 4 (0.75) = 8.15 ksi Ans. Ans: s = 8.15 ksi, t = 5.87 ksi 59 © 2014 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. *1–60. If the shaft is subjected to an axial force of 5 kN, determine the bearing stress acting on the collar A. A 5 kN 2.5 mm 60 mm 100 mm 2.5 mm 15 mm Bearing Stress: The bearing area on the collar, shown shaded in Fig. a, is Ab = p a 0.052 - 0.03252 b = 4.536(10 - 3) m2. Referring to the free-body diagram of the collar, Fig. a, and writing the force equation of equilibrium along the axis of the shaft, ©Fy = 0; 5(103) - sb c 4.536(10 - 3) d = 0 sb = 1.10 MPa Ans. 60 © 2014 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. 1–61. If the 60-mm diameter shaft is subjected to an axial force of 5 kN, determine the average shear stress developed in the shear plane where the collar A and shaft are connected. A 5 kN 2.5 mm 60 mm 100 mm 2.5 mm 15 mm Average Shear Stress: The area of the shear plane, shown shaded in Fig. a, is A = 2p(0.03)(0.015) = 2.827(10 - 3) m2. Referring to the free-body diagram of the shaft, Fig. a, and writing the force equation of equilibrium along the axis of the shaft, ©Fy = 0; 5(103) - tavg c 2.827(10 - 3) d = 0 tavg = 1.77 MPa Ans. Ans: tavg = 1.77 MPa 61 © 2014 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. 1–62. The crimping tool is used to crimp the end of the wire E. If a force of 20 lb is applied to the handles, determine the average shear stress in the pin at A. The pin is subjected to double shear and has a diameter of 0.2 in. Only a vertical force is exerted on the wire. 20 lb C E A B D Support Reactions: 5 in. From FBD(a) a + ©MD = 0; + : ©Fx = 0; 1.5 in. 2 in. 1 in. 20(5) - By (1) = 0 20 lb By = 100 lb Bx = 0 From FBD(b) + : ©Fx = 0; a + ©ME = 0; Ax = 0 Ay (1.5) - 100(3.5) = 0 Ay = 233.33 lb Average Shear Stress: Pin A is subjected to double shear. Hence, Ay FA VA = = = 116.67 lb 2 2 (tA)avg = VA 116.67 = p 2 AA 4 (0.2 ) = 3714 psi = 3.71 ksi Ans. Ans: (tA)avg = 3.71 ksi 62 © 2014 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. 1–63. Solve Prob. 1–62 for pin B. The pin is subjected to double shear and has a diameter of 0.2 in. 20 lb Support Reactions: A From FBD(a) a + ©MD = 0; + : ©Fx = 0; C E 20(5) - By (1) = 0 B D By = 100 lb 5 in. Bx = 0 1.5 in. 2 in. 1 in. 20 lb Average Shear Stress: Pin B is subjected to double shear. Hence, By FB VB = = = 50.0 lb 2 2 (tB)avg = VB = AB p 4 50.0 (0.22) = 1592 psi = 1.59 ksi Ans. Ans: (tB)avg = 1.59 ksi 63 © 2014 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. *1–64. A vertical force of P = 1500 N is applied to the bell crank. Determine the average normal stress developed in the 10-mm diameter rod CD, and the average shear stress developed in the 6-mm diameter pin B that is subjected to double shear. D 450 mm 45⬚ 300 mm Internal Loading: Referring to the free-body diagram of the bell crank shown in Fig. a, a + ©MB = 0; FCD (0.3 sin 45°) - 1500(0.45) = 0 FCD = 3181.98 N + : ©Fx = 0; Bx - 3181.98 = 0 Bx = 3181.98 N + c ©Fy = 0; By - 1500 = 0 By = 1500 N Thus, the force acting on pin B is FB = 2Bx2 + By2 = 23181.982 + 15002 = 3517.81 N Pin B is in double shear. Referring to its free-body diagram, Fig. b, VB = FB 3517.81 = = 1758.91 N 2 2 Average Normal and Shear Stress: The cross-sectional area of rod CD is p ACD = (0.012) = 78.540(10 - 6) m2, and the area of the shear plane of pin B is 4 p AB = (0.0062) = 28.274(10 - 6) m2. We obtain 4 (savg)CD = (tavg)B = C FCD 3181.98 = 40.5 MPa = ACD 78.540(10 - 6) Ans. VB 1758.91 = 62.2 MPa = AB 28.274(10 - 6) Ans. 64 P B A © 2014 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. 1–65. Determine the maximum vertical force P that can be applied to the bell crank so that the average normal stress developed in the 10-mm diameter rod CD, and the average shear stress developed in the 6-mm diameter double sheared pin B not exceed 175 MPa and 75 MPa, respectively. D C 450 mm 45⬚ 300 mm P B A Internal Loading: Referring to the free-body diagram of the bell crank shown in Fig. a, a + ©MB = 0; + : ©Fx = 0; FCD (0.3 sin 45°) - P(0.45) = 0 FCD = 2.121P Bx - 2.121P = 0 Bx = 2.121P + c ©Fy = 0; By - P = 0 By = P Thus, the force acting on pin B is FB = 2Bx2 + By2 = 2(2.121P)2 + P2 = 2.345P Pin B is in double shear. Referring to its free-body diagram, Fig. b, VB = FB 2.345P = = 1.173P 2 2 Average Normal and Shear Stress: The cross-sectional area of rod CD is p ACD = (0.012) = 78.540(10 - 6) m2, and the area of the shear plane of pin B 4 p is AB = (0.0062) = 28.274(10 - 6) m2. We obtain 4 (savg)allow = FCD ; ACD (tavg)allow = VB ; AB 175(106) = 2.121P 78.540(10 - 6) P = 6479.20 N = 6.48 kN 75(106) = 1.173P 28.274(10 - 6) P = 1808.43 N = 1.81 kN (controls) Ans. Ans: P = 1.81 kN 65 © 2014 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. 1–66. Determine the largest load P that can be applied to the frame without causing either the average normal stress or the average shear stress at section a–a to exceed s = 150 MPa and t = 60 MPa , respectively. Member CB has a square cross section of 25 mm on each side. B Analyze the equilibrium of joint C using the FBD Shown in Fig. a, + c ©Fy = 0; 4 FBC a b - P = 0 5 FBC = 1.25P 2m a Referring to the FBD of the cut segment of member BC Fig. b. + : ©Fx = 0; + c ©Fy = 0; The 3 Na - a - 1.25Pa b = 0 5 4 1.25Pa b - Va - a = 0 5 cross-sectional area of section Na - a = 0.75P a A Va - a = P a–a is Aa - a = (0.025) a 0.025 b 3>5 C 1.5 m P = 1.0417(10 - 3) m2. For Normal stress, sallow = Na - a ; Aa - a 150(106) = 0.75P 1.0417(10 - 3) P = 208.33(103) N = 208.33 kN For Shear Stress Va - a ; tallow = Aa - a 60(106) = P 1.0417(10 - 3) P = 62.5(103) N = 62.5 kN (Controls!) Ans. Ans: P = 62.5 kN 66 © 2014 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. z 1–67. The pedestal in the shape of a frustum of a cone is made of concrete having a specific weight of 150 lb>ft3. Determine the average normal stress acting in the pedestal at its base. Hint: The volume of a cone of radius r and height h is V = 13pr2h. 1 ft 8 ft z ⫽ 4 ft 1.5 ft h - 8 h = , 1.5 1 V = y h = 24 ft 1 1 p (1.5)2(24) - p (1)2(16); 3 3 x V = 39.794 ft3 W = 150(39.794) = 5.969 kip s = P 5.969 = = 844 psf = 5.86 psi A p(1.5)2 Ans. Ans: s = 5.86 psi 67 © 2014 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. z *1–68. The pedestal in the shape of a frustum of a cone is made of concrete having a specific weight of 150 lb>ft3. Determine the average normal stress acting in the pedestal at its midheight, z = 4 ft. Hint: The volume of a cone of 1 ft radius r and height h is V = 13pr2h. 8 ft z ⫽ 4 ft h h - 8 = , 1.5 1 1.5 ft h = 24 ft y 1 1 W = c p (1.25)2 20 - (p)(12)(16) d (150) = 2395.5 lb 3 3 x + c g Fy = 0; P - 2395.5 = 0 P = 2395.5 lb s = P 2395.5 = = 488 psf = 3.39 psi A p(1.25)2 Ans. 68 © 2014 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. 1–69. Member B is subjected to a compressive force of 800 lb. If A and B are both made of wood and are 38 in. thick, determine to the nearest 14 in. the smallest dimension h of the horizontal segment so that it does not fail in shear. The average shear stress for the segment is tallow = 300 psi. tallow = 300 = B 307.7 13 (38) 12 h 5 h A h = 2.74 in. Use h = 2 800 lb 3 in. 4 Ans. Ans: 3 Use h = 2 in. 4 69 © 2014 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. 1–70. The lever is attached to the shaft A using a key that has a width d and length of 25 mm. If the shaft is fixed and a vertical force of 200 N is applied perpendicular to the handle, determine the dimension d if the allowable shear stress for the key is tallow = 35 MPa. a + ©MA = 0; a A d a 20 mm 500 mm 200 N Fa - a (20) - 200(500) = 0 Fa - a = 5000 N tallow = Fa - a ; Aa - a 35(106) = 5000 d(0.025) d = 0.00571 m = 5.71 mm Ans. Ans: d = 5.71 mm 70 © 2014 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. 1–71. The joint is fastened together using two bolts. Determine the required diameter of the bolts if the failure shear stress for the bolts is tfail = 350 MPa. Use a factor of safety for shear of F.S. = 2.5. 30 mm 80 kN 30 mm 40 kN 40 kN 350(106) = 140(106) 2.5 tallow = 140(106) = 20(103) p 4 d2 d = 0.0135 m = 13.5 mm Ans. Ans: d = 13.5 mm 71 © 2014 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. *1–72. The truss is used to support the loading shown Determine the required cross-sectional area of member BC if the allowable normal stress is sallow = 24 ksi . 800 lb 400 lb 6 ft 6 ft F B 30⬚ A a + © MA = 0; -400(6) - 800(8.485) + 2(8.485)(Dy) = 0 Dy = 541.42 lb a + © MF = 0; 541.42(8.485) - FBC (5.379) = 0 FBC = 854.01 lb s= P ; A 24000 = 854.01 A A = 0.0356 in2 Ans. 72 E 60⬚ 6 ft 45⬚ 6 ft C D © 2014 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. 1–73. The steel swivel bushing in the elevator control of an airplane is held in place using a nut and washer as shown in Fig. (a). Failure of the washer A can cause the push rod to separate as shown in Fig. (b). If the maximum average normal stress for the washer is smax = 60 ksi and the maximum average shear stress is tmax = 21 ksi , determine the force F that must be applied to the bushing that will 1 cause this to happen. The washer is 16 in. thick. tavg = V ; A 21(103) = 0.75 in. F F A (a) (b) F 1 2p(0.375)( 16 ) F = 3092.5 lb = 3.09 kip Ans. Ans: F = 3.09 kip 73 © 2014 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. 1–74. Member B is subjected to a compressive force of 600 lb. If A and B are both made of wood and are 1.5 in. thick, determine to the nearest 18 in. the smallest dimension a of the support so that the average shear stress along the blue line does not exceed tallow = 50 psi. Neglect friction. 600 lb 3 5 4 B a A Consider the equilibrium of the FBD of member B, Fig. a, + : ©Fx = 0; 4 600 a b - Fh = 0 5 Fh = 480 lb Referring to the FBD of the wood segment sectioned through glue line, Fig. b + : ©Fx = 0; 480 - V = 0 V = 480 lb The area of shear plane is A = 1.5(a). Thus, tallow = V ; A 50 = 480 1.5a a = 6.40 in. Use a = 612 in. Ans. Ans: 1 Use a = 6 in. 2 74 © 2014 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. 1–75. The hangers support the joist uniformly, so that it is assumed the four nails on each hanger carry an equal portion of the load. If the joist is subjected to the loading shown, determine the average shear stress in each nail of the hanger at ends A and B. Each nail has a diameter of 0.25 in. The hangers only support vertical loads. 40 lb/ft 30 lb/ft B A 18 ft a + ©MA = 0; + c ©Fy = 0; FB(18) - 540(9) - 90(12) = 0; FB = 330 lb FA + 330 - 540 - 90 = 0; FA = 300 lb For nails at A, FA 300 = p tavg = AA 4( 4 )(0.25)2 Ans. = 1528 psi = 1.53 ksi For nails at B, FB 330 = p tavg = AB 4( 4 )(0.25)2 Ans. = 1681 psi = 1.68 ksi Ans: For nails at A: tavg = 1.53 ksi For nails at B: tavg = 1.68 ksi 75 © 2014 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. *1–76. The hangers support the joists uniformly, so that it is assumed the four nails on each hanger carry an equal portion of the load. Determine the smallest diameter of the nails at A and at B if the allowable stress for the nails is tallow = 4 ksi . The hangers only support vertical loads. 40 lb/ft 30 lb/ft B A 18 ft a + ©MA = 0; FB(18) - 540(9) - 90(12) = 0; FB = 330 lb + c ©Fy = 0; FA + 330 - 540 - 90 = 0; FA = 300 lb For nails at A, tallow = FA ; AA 4(103) = 300 4(p4 )dA2 dA = 0.155 in. Ans. For nails at B, tallow = FB ; AB 4(103) = 330 4(p4 )dB2 dB = 0.162 in. Ans. 76 © 2014 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. 1–77. The tension member is fastened together using two bolts, one on each side of the member as shown. Each bolt has a diameter of 0.3 in. Determine the maximum load P that can be applied to the member if the allowable shear stress for the bolts is tallow = 12 ksi and the allowable average normal stress is sallow = 20 ksi. 60⬚ P P N - P sin 60° = 0 a+ ©Fy = 0; P = 1.1547 N (1) V - P cos 60° = 0 b+ ©Fx = 0; P = 2V (2) Assume failure due to shear: tallow = 12 = V (2) p4 (0.3)2 V = 1.696 kip From Eq. (2), P = 3.39 kip Assume failure due to normal force: sallow = 20 = N (2) p4 (0.3)2 N = 2.827 kip From Eq. (1), P = 3.26 kip (controls) Ans. Ans: P = 3.26 kip 77 © 2014 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. 1–78. The 50-kg flowerpot is suspended from wires AB and BC. If the wires have a normal failure stress of sfail = 350 MPa, determine the minimum diameter of each wire. Use a factor of safety of 2.5. 45⬚ A C 30⬚ B Internal Loading: The normal force developed in cables AB and BC can be determined by considering the equilibrium of joint B, Fig. a. + : ©Fx = 0; FBC cos 45° - FAB cos 30° = 0 (1) + c ©Fy = 0; FAB sin 30° + FBC sin 45° - 50(9.81) = 0 (2) Solving Eqs. (1) and (2), FAB = 359.07 N FBC = 439.77 N Allowable Normal Stress: sfail 350 sallow = = = 140 MPa F.S. 2.5 Using this result, sallow = FAB ; AAB 140(106) = 359.07 p 2 4 dAB dAB = 0.001807 m = 1.81 mm sallow = FBC ; ABC 140(106) = Ans. 439.77 p 2 4 dBC dBC = 0.00200 m = 2.00 mm Ans. Ans: dAB = 1.81 mm, dBC = 2.00 mm 78 © 2014 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. 1–79. The 50-kg flowerpot is suspended from wires AB and BC which have diameters of 1.5 mm and 2 mm, respectively. If the wires have a normal failure stress of sfail = 350 MPa, determine the factor of safety of each wire. 45⬚ A C 30⬚ B Internal Loading: The normal force developed in cables AB and BC can be determined by considering the equilibrium of joint B, Fig. a. + : ©Fx = 0; FBC cos 45° - FAB cos 30° = 0 (1) + c ©Fy = 0; FAB sin 30° + FBC sin 45° - 50(9.81) = 0 (2) Solving Eqs. (1) and (2), FAB = 359.07 N FBC = 439.77 N Average Normal Stress: The cross-sectional area of wires AB and BC are p p AAB = (0.0015)2 = 1.767(10 - 6) m2 and ABC = (0.0022) = 3.142(10 - 6) m2. 4 4 FAB 359.07 (savg)AB = = 203.19 MPa = AAB 1.767(10 - 6) (savg)BC = FBC ABC 439.77 = 3.142(10 - 6) = 139.98 MPa We obtain, (F.S.)AB = sfail 350 = = 1.72 (savg)AB 203.19 Ans. (F.S.)BC = sfail 350 = 2.50 = (savg)BC 139.98 Ans. Ans: (F.S.)AB = 1.72, (F.S.)BC = 2.50 79 © 2014 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. *1–80. The thrust bearing consists of a circular collar A fixed to the shaft B. Determine the maximum axial force P that can be applied to the shaft so that it does not cause the shear stress along a cylindrical surface a or b to exceed an allowable shear stress of tallow = 170 MPa. C b a P B A 30 mm 58 mm a b 35 mm Assume failure along a: tallow = 170(106) = P p(0.03)(0.035) P = 561 kN (controls) Ans. Assume failure along b: tallow = 170(106) = P p(0.058)(0.02) P = 620 kN 80 20 mm © 2014 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. 1–81. The steel pipe is supported on the circular base plate and concrete pedestal. If the normal failure stress for the steel is (sfail)st = 350 MPa, determine the minimum thickness t of the pipe if it supports the force of 500 kN. Use a factor of safety against failure of 1.5. Also, find the minimum radius r of the base plate so that the minimum factor of safety against failure of the concrete due to bearing is 2.5. The failure bearing stress for concrete is (sfail)con = 25 MPa. t 500 kN 100 mm r Allowable Stress: (sallow)st = (sfail)st 350 = = 233.33 MPa F.S. 1.5 (sallow)con = (sfail)con 25 = = 10 MPa F.S. 2.5 The cross-sectional area of the steel pipe and the heaving area of the concrete pedestal are Ast = p(0.12 - ri2 ) and (Acon)b = pr2. Using these results, (sallow)st = P ; Ast 233.33(106) = 500(103) p(0.12 - ri2 ) ri = 0.09653 m = 96.53 mm Thus, the minimum required thickness of the steel pipe is t = rO - ri = 100 - 96.53 = 3.47 mm Ans. The minimum required radius of the bearing area of the concrete pedestal is (sallow)con = P ; (Acon)b 10(106) = 500(103) pr2 r = 0.1262 m = 126 mm Ans. Ans: t = 3.47 mm, r = 126 mm 81 © 2014 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. 1–82. The steel pipe is supported on the circular base plate and concrete pedestal. If the thickness of the pipe is t = 5 mm and the base plate has a radius of 150 mm, determine the factors of safety against failure of the steel and concrete. The applied force is 500 kN, and the normal failure stresses for steel and concrete are (sfail)st = 350 MPa and (sfail)con = 25 MPa, respectively. t 500 kN 100 mm r Average Normal and Bearing Stress: The cross-sectional area of the steel pipe and the bearing area of the concrete pedestal are Ast = p(0.12 - 0.0952) = 0.975(10 - 3)p m2 and (Acon)b = p(0.152) = 0.0225p m2. We have (savg)st = 500(103) P = 163.24 MPa = Ast 0.975(10 - 3)p (savg)con = 500(103) P = = 7.074 MPa (Acon)b 0.0225p Thus, the factor of safety against failure of the steel pipe and concrete pedestal are (F.S.)st = (sfail)st 350 = = 2.14 (savg)st 163.24 (F.S.)con = Ans. (sfail)con 25 = = 3.53 (savg)con 7.074 Ans. Ans: (F.S.)st = 2.14, (F.S.)con = 3.53 82 © 2014 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. 1–83. The 60 mm * 60 mm oak post is supported on the pine block. If the allowable bearing stresses for these materials are soak = 43 MPa and spine = 25 MPa, determine the greatest load P that can be supported. If a rigid bearing plate is used between these materials, determine its required area so that the maximum load P can be supported. What is this load? P For failure of pine block: s = P ; A 25(106) = P (0.06)(0.06) P = 90 kN Ans. For failure of oak post: s = P ; A 43(106) = P (0.06)(0.06) P = 154.8 kN Area of plate based on strength of pine block: 154.8(10)3 P s = ; 25(106) = A A A = 6.19(10 - 3) m2 Ans. Pmax = 155 kN Ans. Ans: P = 90 kN, A = 6.19(10 - 3) m2, Pmax = 155 kN 83 © 2014 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. *1–84. The frame is subjected to the load of 4 kN which acts on member ABD at D. Determine the required diameter of the pins at D and C if the allowable shear stress for the material is tallow = 40 MPa. Pin C is subjected to double shear, whereas pin D is subjected to single shear. 4 kN 1m E 1.5 m C 45 D 1.5 m Referring to the FBD of member DCE, Fig. a, a + ©ME = 0; Dy(2.5) - FBC sin 45° (1) = 0 (1) + : ©Fx = 0 FBC cos 45° - Dx = 0 (2) B 1.5 m Referring to the FBD of member ABD, Fig. b, a + ©MA = 0; 4 cos 45° (3) + FBC sin 45° (1.5) - Dx (3) = 0 (3) Solving Eqs (2) and (3), FBC = 8.00 kN Dx = 5.657 kN Substitute the result of FBC into (1) Dy = 2.263 kN Thus, the force acting on pin D is FD = 2Dx 2 + Dy 2 = 25.6572 + 2.2632 = 6.093 kN Pin C is subjected to double shear white pin D is subjected to single shear. Referring to the FBDs of pins C, and D in Fig c and d, respectively, FBC 8.00 VC = = = 4.00 kN VD = FD = 6.093 kN 2 2 For pin C, tallow = VC ; AC 40(106) = 4.00(103) p 4 dC 2 dC = 0.01128 m = 11.3 mm For pin D, VD ; tallow = AD 40(106) = Ans. 6.093(103) p 4 dD 2 dD = 0.01393 m = 13.9 mm Ans. 84 A © 2014 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. 1–85. The beam is made from southern pine and is supported by base plates resting on brick work. If the allowable bearing stresses for the materials are (spine)allow = 2.81 ksi and (sbrick)allow = 6.70 ksi, determine the required length of the base plates at A and B to the nearest 14 inch in order to support the load shown. The plates are 3 in. wide. 6 kip 200 lb/ft A B 5 ft 5 ft 3 ft The design must be based on strength of the pine. At A: s = P ; A Use lA = 2810 = 1 in. 2 3910 lA(3) lA = 0.464 in. Ans. At B: s = P ; A Use lB = 2810 = 3 in. 4 4690 l(3) lB = 0.556 in. Ans. Ans: Use lA = 85 1 3 in., lB = in. 2 4 © 2014 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. 1–86. The two aluminum rods support the vertical force of P = 20 kN. Determine their required diameters if the allowable tensile stress for the aluminum is sallow = 150 MPa. B C A 45⬚ P + c ©Fy = 0; FAB sin 45° - 20 = 0; FAB = 28.284 kN + : ©Fx = 0; 28.284 cos 45° - FAC = 0; FAC = 20.0 kN For rod AB: sallow = FAB ; AAB 150(106) = 28.284(103) p 2 4 dAB dAB = 0.0155 m = 15.5 mm Ans. For rod AC: sallow = FAC ; AAC 150(106) = 20.0(103) p 2 4 dAC dAC = 0.0130 m = 13.0 mm Ans. Ans: dAB = 15.5 mm, dAC = 13.0 mm 86 © 2014 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. 1–87. The two aluminum rods AB and AC have diameters of 10 mm and 8 mm, respectively. Determine the largest vertical force P that can be supported. The allowable tensile stress for the aluminum is sallow = 150 MPa. B C A 45⬚ P + c ©Fy = 0; FAB sin 45° - P = 0; + : ©Fx = 0; FAB cos 45° - FAC = 0 P = FAB sin 45° (1) (2) Assume failure of rod AB: sallow = FAB ; AAB 150(106) = FAB p 2 4 (0.01) FAB = 11.78 kN From Eq. (1), P = 8.33 kN Assume failure of rod AC: FAC sallow = ; 150(106) = AAC FAC p 2 4 (0.008) FAC = 7.540 kN Solving Eqs. (1) and (2) yields: FAB = 10.66 kN ; P = 7.54 kN Choose the smallest value P = 7.54 kN Ans. Ans: P = 7.54 kN 87 © 2014 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. *1–88. The compound wooden beam is connected together by a bolt at B.Assuming that the connections at A, B, C, and D exert only vertical forces on the beam, determine the required diameter of the bolt at B and the required outer diameter of its washers if the allowable tensile stress for the bolt is 1st2allow = 150 MPa and the allowable bearing stress for the wood is 1sb2allow = 28 MPa. Assume that the hole in the washers has the same diameter as the bolt. 2m FB(4.5) + 1.5(3) + 2(1.5) - FC(6) = 0 (1) FB(5.5) - FC(4) - 3(2) = 0 5.5 FB - 4 FC = 6 (2) Solving Eqs. (1) and (2) yields FC = 4.55 kN For bolt: sallow = 150(106) = 4.40(103) p 2 4 (dB) dB = 0.00611 m = 6.11 mm Ans. For washer: sallow = 28 (104) = 4.40(103) p 2 4 (dw D B From FBD (b): FB = 4.40 kN; 1.5 m C 4.5 FB - 6 FC = - 7.5 a + ©MA = 0; 2m A From FBD (a): a + ©MD = 0; 2 kN 1.5 kN 1.5 m 1.5 m 1.5 m 3 kN - 0.006112) dw = 0.0154 m = 15.4 mm Ans. 88 © 2014 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. 1–89. Determine the required minimum thickness t of member AB and edge distance b of the frame if P = 9 kip and the factor of safety against failure is 2. The wood has a normal failure stress of sfail = 6 ksi, and shear failure stress of tfail = 1.5 ksi. P 3 in. B 3 in. t A 30⬚ b 30⬚ C Internal Loadings: The normal force developed in member AB can be determined by considering the equilibrium of joint A. Fig. a. + : ©Fx = 0; FAB cos 30° - FAC cos 30° = 0 FAC = FAB + c ©Fy = 0; 2FAB sin 30° - 9 = 0 FAB = 9 kip Subsequently, the horizontal component of the force acting on joint B can be determined by analyzing the equilibrium of member AB, Fig. b. + : ©Fx = 0; (FB)x - 9 cos 30° = 0 (FB)x = 7.794 kip Referring to the free-body diagram shown in Fig. c, the shear force developed on the shear plane a–a is + : ©Fx = 0; Va - a = 7.794 kip Va - a - 7.794 = 0 Allowable Normal Stress: sfail 6 = = 3 ksi sallow = F.S. 2 tallow = tfail 1.5 = = 0.75 ksi F.S. 2 Using these results, sallow = FAB ; AAB 3(103) = 9(103) 3t t = 1 in. tallow = Va - a ; Aa - a 0.75(103) = Ans. 7.794(103) 3b b = 3.46 in. Ans. Ans: t = 1 in., b = 3.46 in. 89 © 2014 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. 1–90. Determine the maximum allowable load P that can be safely supported by the frame if t = 1.25 in. and b = 3.5 in. The wood has a normal failure stress of sfail = 6 ksi, and shear failure stress of tfail = 1.5 ksi. Use a factor of safety against failure of 2. P 3 in. B 3 in. t A 30⬚ b 30⬚ C Internal Loadings: The normal force developed in member AB can be determined by considering the equilibrium of joint A. Fig. a. + : ©Fx = 0; FAB cos 30° - FAC cos 30° = 0 FAC = FAB + c ©Fy = 0; 2FAB sin 30° - 9 = 0 FAB = P Subsequently, the horizontal component of the force acting on joint B can be determined by analyzing the equilibrium of member AB, Fig. b. + (FB)x = 0.8660P : ©Fx = 0; (FB)x - P cos 30° = 0 Referring to the free-body diagram shown in Fig. c, the shear force developed on the shear plane a–a is + Va - a = 0.8660P : ©Fx = 0; Va - a - 0.8660P = 0 Allowable Normal and Shear Stress: sallow = sfail 6 = = 3 ksi F.S. 2 tallow = tfail 1.5 = = 0.75 ksi F.S. 2 Using these results, sallow = FAB ; AAB 3(103) = P 3(1.25) P = 11 250 lb = 11.25 kip tallow = Va - a ; Aa - a 0.75(103) = 0.8660P 3(3.5) P = 9093.27 lb = 9.09 kip (controls) Ans. Ans: P = 9.09 kip 90 © 2014 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. 1–91. If the allowable bearing stress for the material under the supports at A and B is 1sb2allow = 1.5 MPa, determine the size of square bearing plates A¿ and B¿ required to support the load. Dimension the plates to the nearest mm. The reactions at the supports are vertical. Take P = 100 kN. 40 kN/m Referring to the FBD of the bean, Fig. a 1.5 m A a + ©MA = 0; NB(3) + 40(1.5)(0.75) - 100(4.5) = 0 NB = 135 kN a + ©MB = 0; 40(1.5)(3.75) - 100(1.5) - NA(3) = 0 NA = 25.0 kN For plate A¿ , NA (sb)allow = ; AA¿ 1.5(106) = P A¿ B¿ 3m B 1.5 m 25.0(103) a2A¿ aA¿ = 0.1291 m = 130 mm Ans. For plate B¿ , sallow = NB ; AB¿ 1.5(106) = 135(103) a2B¿ aB¿ = 0.300 m = 300 mm Ans. Ans: aA¿ = 130 mm, aB¿ = 300 mm 91 © 2014 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. 40 kN/m *1–92. If the allowable bearing stress for the material under the supports at A and B is 1sb2allow = 1.5 MPa, determine the maximum load P that can be applied to the beam. The bearing plates A¿ and B¿ have square cross sections of 150 mm * 150 mm and 250 mm * 250 mm, respectively. A 1.5 m Referring to the FBD of the beam, Fig. a, a + ©MA = 0; NB(3) + 40(1.5)(0.75) - P(4.5) = 0 NB = 1.5P - 15 a + ©MB = 0; 40(1.5)(3.75) - P(1.5) - NA(3) = 0 NA = 75 - 0.5P For plate A¿ , NA (sb)allow = ; AA¿ 1.5(106) = (75 - 0.5P)(103) 0.15(0.15) P = 82.5 kN For plate B¿ , NB ; (sb)allow = AB¿ 1.5(106) = (1.5P - 15)(103) 0.25(0.25) P = 72.5 kN (Controls!) Ans. 92 P A¿ B¿ 3m B 1.5 m © 2014 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. 1–93. The rods AB and CD are made of steel. Determine their smallest diameter so that they can support the dead loads shown. The beam is assumed to be pin connected at A and C. Use the LRFD method, where the resistance factor for steel in tension is f = 0.9, and the dead load factor is gD = 1. 4. The failure stress is sfail = 345 MPa. B D 6 kN 5 kN 4 kN A C 2m 2m 3m 3m Support Reactions: a + ©MA = 0; FCD(10) - 5(7) - 6(4) - 4(2) = 0 FCD = 6.70 kN a + ©MC = 0; 4(8) + 6(6) + 5(3) - FAB(10) = 0 FAB = 8.30 kN Factored Loads: FCD = 1.4(6.70) = 9.38 kN FAB = 1.4(8.30) = 11.62 kN For rod AB 0.9[345(106)] p a dAB 2 b = 11.62(103) 2 dAB = 0.00690 m = 6.90 mm Ans. For rod CD 0.9[345(106)] p a dCD 2 b = 9.38(103) 2 dCD = 0.00620 m = 6.20 mm Ans. Ans: dAB = 6.90 mm, dCD = 6.20 mm 93 © 2014 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. 1–94. The aluminum bracket A is used to support the centrally applied load of 8 kip. If it has a constant thickness of 0.5 in., determine the smallest height h in order to prevent a shear failure. The failure shear stress is tfail = 23 ksi. Use a factor of safety for shear of F.S. = 2.5. A h 8 kip Equation of Equilibrium: + c ©Fy = 0; V - 8 = 0 V = 8.00 kip Allowable Shear Stress: Design of the support size tallow = tfail V = ; F.S A 23(103) 8.00(103) = 2.5 h(0.5) h = 1.74 in. Ans. Ans: h = 1.74 in. 94 © 2014 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. 1–95. The pin support A and roller support B of the bridge truss are supported on concrete abutments. If the bearing failure stress of the concrete is (sfail)b = 4 ksi, determine the required minimum dimension of the square 1 bearing plates at C and D to the nearest 16 in. Apply a factor of safety of 2 against failure. 150 kip 100 kip A B C D 200 kip 300 kip 300 kip 300 kip 300 kip 6 ft 6 ft 6 ft 6 ft 6 ft 6 ft Internal Loadings: The forces acting on the bearing plates C and D can be determined by considering the equilibrium of the free-body diagram of the truss shown in Fig. a, a + ©MA = 0; By(36) - 100(36) - 200(30) - 300(24) - 300(18) - 300(12) - 300(6) = 0 By = 766.67 kip a + ©MB = 0; 150(36) + 300(30) + 300(24) + 300(18) + 300(12) + 200(6) - Ay(36) = 0 Ay = 883.33 kip Thus, the axial forces acting on C and D are FC = Ay = 883.33 kip FD = By = 766.67 kip Allowable Bearing Stress: (sallow)b = (sfail)b 4 = = 2 ksi F.S. 2 Using this result, (sallow)b = FD ; AD 2(103) = 766.67(103) aD2 aD = 19.58 in. = 19 (sallow)b = FC ; AC 2(103) = 5 in. 8 Ans. 1 in. 16 Ans. 883.33(103) aC 2 aC = 21.02 in. = 21 Ans: Use aD = 19 95 5 1 in., aC = 21 in. 8 16 © 2014 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. *1–96. The pin support A and roller support B of the bridge truss are supported on the concrete abutments. If the square bearing plates at C and D are 21 in. * 21 in., and the bearing failure stress for concrete is (sfail)b = 4 ksi, determine the factor of safety against bearing failure for the concrete under each plate. 150 kip A B C D 200 kip 300 kip 300 kip 300 kip 300 kip 6 ft Internal Loadings: The forces acting on the bearing plates C and D can be determined by considering the equilibrium of the free-body diagram of the truss shown in Fig. a, a + ©MA = 0; By(36) - 100(36) - 200(30) - 300(24) - 300(18) - 300(12) - 300(6) = 0 By = 766.67 kips a + ©MB = 0; 150(36) + 300(30) + 300(24) + 300(18) + 300(12) + 200(6) - Ay(36) = 0 Ay = 883.33 kips Thus, the axial forces acting on C and D are FC = Ay = 883.33 kips FD = By = 766.67 kips Allowable Bearing Stress: The bearing area on the concrete abutment is Ab = 21(21) = 441 in2. We obtain (sb)C = FC 883.33 = = 2.003 ksi Ab 441 (sb)D = FD 766.67 = = 1.738 ksi Ab 441 100 kip Using these results, (F.S.)C = (sfail)b 4 = 2.00 = (sb)C 2.003 Ans. (F.S.)C = (sfail)b 4 = = 2.30 (sb)C 1.738 Ans. 96 6 ft 6 ft 6 ft 6 ft 6 ft © 2014 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. 1–97. The beam AB is pin supported at A and supported by a cable BC. A separate cable CG is used to hold up the frame. If AB weighs 120 lb> ft and the column FC has a weight of 180 lb/ft, determine the resultant internal loadings acting on cross sections located at points D and E. Neglect the thickness of both the beam and column in the calculation. C 4 ft 6 ft A B D 12 ft 8 ft E 4 ft G F 12 ft Segment AD: + : ©Fx = 0; ND + 2.16 = 0; ND = - 2.16 kip Ans. + T ©Fy = 0; VD + 0.72 - 0.72 = 0; VD = 0 Ans. a + ©MD = 0; MD - 0.72(3) = 0; MD = 2.16 kip # ft Ans. Segment FE: + ; ©Fx = 0; VE - 0.54 = 0; VE = 0.540 kip Ans. + T ©Fy = 0; NE + 0.72 - 5.04 = 0; NE = 4.32 kip Ans. a + ©ME = 0; - ME + 0.54(4) = 0; ME = 2.16 kip # ft Ans. Ans: ND = - 2.16 kip, VD = 0, MD = 2.16 kip # ft, VE = 0.540 kip, NE = 4.32 kip, ME = 2.16 kip # ft 97 © 2014 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. 1–98. The long bolt passes through the 30-mm-thick plate. If the force in the bolt shank is 8 kN, determine the average normal stress in the shank, the average shear stress along the cylindrical area of the plate defined by the section lines a–a, and the average shear stress in the bolt head along the cylindrical area defined by the section lines b–b. 8 mm a 7 mm b 8 kN 18 mm b a P = ss = A 8 (103) = 208 MPa Ans. (tavg)a = 8 (103) V = = 4.72 MPa A p (0.018)(0.030) Ans. (tavg)b = 8 (103) V = = 45.5 MPa A p (0.007)(0.008) Ans. p 4 (0.007)2 30 mm Ans: ss = 208 MPa, (tavg)a = 4.72 MPa, (tavg)b = 45.5 MPa 98 © 2014 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. 1 1–99. To the nearest 16 in., determine the required thickness of member BC and the diameter of the pins at A and B if the allowable normal stress for member BC is sallow = 29 ksi and the allowable shear stress for the pins is sallow = 10 ksi. C 1.5 in. Referring to the FBD of member AB, Fig. a, a + ©MA = 0; 60⬚ B 2(8)(4) - FBC sin 60° (8) = 0 + : ©Fx = 0; 9.238 cos 60° - Ax = 0 + c ©Fy = 0; 9.238 sin 60° - 2(8) + Ay = 0 8 ft A FBC = 9.238 kip Ax = 4.619 kip 2 kip/ft Ay = 8.00 kip Thus, the force acting on pin A is FA = 2A2x + A2y = 24.6192 + 8.002 = 9.238 kip Pin A is subjected to single shear, Fig. c, while pin B is subjected to double shear, Fig. b. FBC 9.238 VA = FA = 9.238 kip VB = = = 4.619 kip 2 2 For member BC FBC ; sallow = ABC 29 = 9.238 1.5(t) t = 0.2124 in. Use t = For pin A, VA ; tallow = AA 10 = 9.238 p 2 4 dA 1 in. 4 Ans. dA = 1.085 in. 1 Use dA = 1 in. 8 For pin B, VB ; tallow = AB 10 = 4.619 p 2 4 dB Ans. dB = 0.7669 in. Use dB = 13 in. 16 Ans. Ans: Use t = 99 1 1 13 in., dA = 1 in., dB = in. 4 8 16 © 2014 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. *1–100. The circular punch B exerts a force of 2 kN on the lop of the plate A. Determine the average shear stress in the plate due to this loading. 2 kN B 4 mm A Average Shear Stress: The shear area A = p(0.004)(0.002) = 8.00(10 - 6)p m2 tavg = 2(103) V = 79.6 MPa = A 8.00(10 - 6)p Ans. 100 2 mm © 2014 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. 1–101. Determine the average punching shear stress the circular shaft creates in the metal plate through section AC and BD. Also, what is the bearing stress developed on the surface of the plate under the shaft? 40 kN 50 mm A B 10 mm C D 60 mm Average Shear and Bearing Stress: The area of the shear plane and the bearing area on the p punch are AV = p(0.05)(0.01) = 0.5(10 - 3)p m2 and Ab = a0.122 - 0.062 b = 4 2.7(10 - 3)p m2. We obtain tavg = sb = 40(103) P = 25.5 MPa = AV 0.5(10 - 3)p 120 mm Ans. 40(103) P = 4.72 MPa = Ab 2.7(10 - 3)p Ans. Ans: tavg = 25.5 MPa, sb = 4.72 MPa 101 © 2014 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. 1–102. The bearing pad consists of a 150 mm by 150 mm block of aluminum that supports a compressive load of 6 kN. Determine the average normal and shear stress acting on the plane through section a–a. Show the results on a differential volume element located on the plane. 6 kN a 30⬚ a 150 mm Equation of Equilibrium: +Q©Fx = 0; Va - a - 6 cos 60° = 0 Va - a = 3.00 kN a+ ©Fy = 0; Na - a - 6 sin 60° = 0 Na - a = 5.196 kN Average Normal Stress And Shear Stress: The cross sectional Area at section a–a is A = a 0.15 b (0.15) = 0.02598 m2. sin 60° sa - a = 5.196(103) Na - a = = 200 kPa A 0.02598 Ans. ta - a = 3.00(103) Va - a = = 115 kPa A 0.02598 Ans. Ans: sa - a = 200 kPa, ta - a = 115 kPa 102 © 2014 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. 1–103. The yoke-and-rod connection is subjected to a tensile force of 5 kN. Determine the average normal stress in each rod and the average shear stress in the pin A between the members. 5 kN 40 mm 30 mm A 25 mm 5 kN For the 40 – mm – dia rod: s40 = 5 (103) P = p = 3.98 MPa 2 A 4 (0.04) Ans. For the 30 – mm – dia rod: s30 = 5 (103) V = p = 7.07 MPa 2 A 4 (0.03) Ans. Average shear stress for pin A: tavg = 2.5 (103) P = p = 5.09 MPa 2 A 4 (0.025) Ans. Ans: s40 = 3.98 MPa, s30 = 7.07 MPa tavg = 5.09 MPa 103 © 2014 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. *1–104. The cable has a specific weight g (weight>volume) and cross-sectional area A. If the sag s is small, so that its length is approximately L and its weight can be distributed uniformly along the horizontal axis, determine the average normal stress in the cable at its lowest point C. A s C L/2 Equation of Equilibrium: a + ©MA = 0; Ts - gAL L a b = 0 2 4 T = gAL2 8s Average Normal Stress: gAL2 s = B gL2 T 8s = = A A 8s Ans. 104 L/2 © 2014 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. 2–1. An air-filled rubber ball has a diameter of 6 in. If the air pressure within it is increased until the ball’s diameter becomes 7 in., determine the average normal strain in the rubber. d0 = 6 in. d = 7 in. P = pd - pd0 7 - 6 = = 0.167 in.>in. pd0 6 Ans. Ans: P = 0.167 in.>in. 105 © 2014 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. 2–2. A thin strip of rubber has an unstretched length of 15 in. If it is stretched around a pipe having an outer diameter of 5 in., determine the average normal strain in the strip. L0 = 15 in. L = p(5 in.) P = L - L0 5p - 15 = = 0.0472 in.>in. L0 15 Ans. Ans: P = 0.0472 in.>in. 106 © 2014 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. 2–3. The rigid beam is supported by a pin at A and wires BD and CE. If the load P on the beam causes the end C to be displaced 10 mm downward, determine the normal strain developed in wires CE and BD. D E 4m P A B 3m C 2m 2m ¢LBD ¢LCE = 3 7 3 (10) = 4.286 mm 7 ¢LCE 10 = = = 0.00250 mm>mm L 4000 ¢LBD = PCE PBD = Ans. ¢LBD 4.286 = = 0.00107 mm>mm L 4000 Ans. Ans: PCE = 0.00250 mm>mm, PBD = 0.00107 mm>mm 107 © 2014 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. *2–4. The force applied at the handle of the rigid lever causes the lever to rotate clockwise about the pin B through an angle of 2°. Determine the average normal strain developed in each wire. The wires are unstretched when the lever is in the horizontal position. G 200 mm H dA = 200(0.03491) = 6.9813 mm dC = 300(0.03491) = 10.4720 mm dD = 500(0.03491) = 17.4533 mm Average Normal Strain: The unstretched length of wires AH, CG, and DF are LAH = 200 mm, LCG = 300 mm, and LDF = 300 mm. We obtain dA 6.9813 = = 0.0349 mm>mm LAH 200 dC 10.4720 = 0.0349 mm>mm = = LCG 300 (Pavg)AH = Ans. (Pavg)CG Ans. (Pavg)DF = dD 17.4533 = = 0.0582 mm>mm LDF 300 Ans. 108 E C 200 mm 2° bp rad = 0.03491 rad. 180 Since u is small, the displacements of points A, C, and D can be approximated by 200 mm 300 mm 300 mm B A Geometry: The lever arm rotates through an angle of u = a F D © 2014 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. 2–5. The two wires are connected together at A. If the force P causes point A to be displaced horizontally 2 mm, determine the normal strain developed in each wire. C 300 mm 30⬚ 30⬚ 300 A P mm B œ LAC = 23002 + 22 - 2(300)(2) cos 150° = 301.734 mm PAC = PAB = œ - LAC LAC 301.734 - 300 = = 0.00578 mm>mm LAC 300 Ans. Ans: PAC = PAB = 0.00578 mm>mm 109 © 2014 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. 2–6. The rubber band of unstretched length 2r0 is forced down the frustum of the cone. Determine the average normal strain in the band as a function of z. r0 z h 2r0 Geometry: Using similar triangles shown in Fig. a, h¿ h¿ + h = ; r0 2r0 h¿ = h Subsequently, using the result of h¿ r0 r = ; z+h h r = r0 (z + h) h Average Normal Strain: The length of the rubber band as a function of z is 2pr0 (z+ h). With L0 = 2r0, we have L = 2pr = h Pavg L - L0 = = L0 2pr0 (z + h) - 2r0 h p = (z + h) - 1 2r0 h Ans. Ans: Pavg = 110 p (z + h) - 1 h © 2014 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. 2–7. The pin-connected rigid rods AB and BC are inclined at u = 30° when they are unloaded. When the force P is applied u becomes 30.2°. Determine the average normal strain developed in wire AC. P B u u 600 mm A C Geometry: Referring to Fig. a, the unstretched and stretched lengths of wire AD are LAC = 2(600 sin 30°) = 600 mm LAC ¿ = 2(600 sin 30.2°) = 603.6239 mm Average Normal Strain: (Pavg)AC = LAC ¿ - LAC 603.6239 - 600 = = 6.04(10 - 3) mm>mm LAC 600 Ans. Ans: (Pavg)AC = 6.04(10 - 3) mm>mm 111 © 2014 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. u *2–8. Part of a control linkage for an airplane consists of a rigid member CBD and a flexible cable AB. If a force is applied to the end D of the member and causes it to rotate by u = 0.3°, determine the normal strain in the cable. Originally the cable is unstretched. D P 300 mm B 300 mm A C 400 mm AB = 24002 + 3002 = 500 mm AB¿ = 24002 + 3002 - 2(400)(300) cos 90.3° = 501.255 mm PAB = AB¿ - AB 501.255 - 500 = AB 500 = 0.00251 mm>mm Ans. 112 © 2014 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. u 2–9. Part of a control linkage for an airplane consists of a rigid member CBD and a flexible cable AB. If a force is applied to the end D of the member and causes a normal strain in the cable of 0.0035 mm> mm, determine the displacement of point D. Originally the cable is unstretched. D P 300 mm B 300 mm A C 400 mm AB = 23002 + 4002 = 500 mm AB¿ = AB + eABAB = 500 + 0.0035(500) = 501.75 mm 501.752 = 3002 + 4002 - 2(300)(400) cos a a = 90.4185° u = 90.4185° - 90° = 0.4185° = ¢ D = 600(u) = 600( p (0.4185) rad 180° p )(0.4185) = 4.38 mm 180° Ans. Ans: ¢ D = 4.38 mm 113 © 2014 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. 2–10. The corners of the square plate are given the displacements indicated. Determine the shear strain along the edges of the plate at A and B. y 0.2 in. A 10 in. D B x 0.3 in. 0.3 in. 10 in. 10 in. C 10 in. 0.2 in. At A: 9.7 u¿ = tan - 1 a b = 43.561° 2 10.2 u¿ = 1.52056 rad (gA)nt = p - 1.52056 2 = 0.0502 rad Ans. At B: f¿ 10.2 = tan - 1 a b = 46.439° 2 9.7 f¿ = 1.62104 rad (gB)nt = p - 1.62104 2 = -0.0502 rad Ans. Ans: (gA)nt = 0.0502 rad, (gB)nt = - 0.0502 rad 114 © 2014 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. 2–11. The corners of the square plate are given the displacements indicated. Determine the average normal strains along side AB and diagonals AC and DB. y 0.2 in. A 10 in. D B x 0.3 in. 0.3 in. 10 in. 10 in. C 10 in. 0.2 in. For AB: A¿B¿ = 2(10.2)2 + (9.7)2 = 14.0759 in. AB = 2(10)2 + (10)2 = 14.14214 in. PAB = 14.0759 - 14.14214 = - 0.00469 in.>in. 14.14214 Ans. 20.4 - 20 = 0.0200 in.>in. 20 Ans. 19.4 - 20 = - 0.0300 in.>in. 20 Ans. For AC: PAC = For DB: PDB = Ans: PAB = - 0.00469 in.>in., PAC = 0.0200 in.>in., PDB = - 0.0300 in.>in. 115 © 2014 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. *2–12. The piece of rubber is originally rectangular. Determine the average shear strain gxy at A if the corners B and D are subjected to the displacements that cause the rubber to distort as shown by the dashed lines. y 3 mm C D 400 mm A u1 = tan u1 = 2 = 0.006667 rad 300 u2 = tan u2 = 3 = 0.0075 rad 400 gxy = u1 + u2 = 0.006667 + 0.0075 = 0.0142 rad Ans. 116 300 mm B 2 mm x © 2014 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. 2–13. The piece of rubber is originally rectangular and subjected to the deformation shown by the dashed lines. Determine the average normal strain along the diagonal DB and side AD. y 3 mm C D 400 mm A 300 mm B 2 mm x AD¿ = 2(400)2 + (3)2 = 400.01125 mm f = tan - 1 a 3 b = 0.42971° 400 AB¿ = 2(300)2 + (2)2 = 300.00667 w = tan - 1 a 2 b = 0.381966° 300 a = 90° - 0.42971° - 0.381966° = 89.18832° D¿B¿ = 2(400.01125)2 + (300.00667)2 - 2(400.01125)(300.00667) cos (89.18832°) D¿B¿ = 496.6014 mm DB = 2(300)2 + (400)2 = 500 mm 496.6014 - 500 = - 0.00680 mm>mm 500 400.01125 - 400 = = 0.0281(10 - 3) mm>mm 400 PDB = Ans. PAD Ans. Ans: PDB = - 0.00680 mm>mm, PAD = 0.0281(10 - 3) mm>mm 117 © 2014 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. 2–14. The force P applied at joint D of the square frame causes the frame to sway and form the dashed rhombus. Determine the average normal strain developed in wire AC. Assume the three rods are rigid. D P 200 mm 200 mm E C 3 400 mm A B Geometry: Referring to Fig. a, the stretched length of LAC ¿ of wire AC¿ can be determined using the cosine law. LAC ¿ = 24002 + 4002 - 2(400)(400) cos 93° = 580.30 mm The unstretched length of wire AC is LAC = 24002 + 4002 = 565.69 mm Average Normal Strain: (Pavg)AC = LAC ¿ - LAC 580.30 - 565.69 = = 0.0258 mm>mm LAC 565.69 Ans. Ans: (Pavg)AC = 0.0258 mm>mm 118 © 2014 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. 2–15. The force P applied at joint D of the square frame causes the frame to sway and form the dashed rhombus. Determine the average normal strain developed in wire AE. Assume the three rods are rigid. D P 200 mm 200 mm E C 3 400 mm A B Geometry: Referring to Fig. a, the stretched length of LAE¿ of wire AE can be determined using the cosine law. LAE¿ = 24002 + 2002 - 2(400)(200) cos 93° = 456.48 mm The unstretched length of wire AE is LAE = 24002 + 2002 = 447.21 mm Average Normal Strain: (Pavg)AE = LAE ¿ - LAE 456.48 - 447.21 = 0.0207 mm>mm = LAE 447.21 Ans. Ans: (Pavg)AE = 0.0207 mm>mm 119 © 2014 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. *2–16. The triangular plate ABC is deformed into the shape shown by the dashed lines. If at A, eAB = 0.0075, PAC = 0.01 and gxy = 0.005 rad, determine the average normal strain along edge BC. y C 300 mm gxy A Average Normal Strain: The stretched length of sides AB and AC are LAC¿ = (1 + ey)LAC = (1 + 0.01)(300) = 303 mm LAB¿ = (1 + ex)LAB = (1 + 0.0075)(400) = 403 mm Also, u = p 180° - 0.005 = 1.5658 rada b = 89.7135° 2 p rad The unstretched length of edge BC is LBC = 23002 + 4002 = 500 mm and the stretched length of this edge is LB¿C¿ = 23032 + 4032 - 2(303)(403) cos 89.7135° = 502.9880 mm We obtain, PBC = LB¿C¿ - LBC 502.9880 - 500 = = 5.98(10 - 3) mm>mm LBC 500 120 Ans. 400 mm B x © 2014 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. 2–17. The plate is deformed uniformly into the shape shown by the dashed lines. If at A, gxy = 0.0075 rad., while PAB = PAF = 0, determine the average shear strain at point G with respect to the x¿ and y¿ axes. y¿ y F E x¿ 600 mm gxy C D 300 mm Geometry: Here, gxy c = 90° - 0.4297° = 89.5703° G A 180° = 0.0075 rad a b = 0.4297°. Thus, p rad 300 mm 600 mm B b = 90° + 0.4297° = 90.4297° Subsequently, applying the cosine law to triangles AGF¿ and GBC¿, Fig. a, LGF¿ = 26002 + 3002 - 2(600)(300) cos 89.5703° = 668.8049 mm LGC¿ = 26002 + 3002 - 2(600)(300) cos 90.4297° = 672.8298 mm Then, applying the sine law to the same triangles, sin f sin 89.5703° = ; 600 668.8049 f = 63.7791° sin 90.4297° sin a = ; 300 672.8298 a = 26.4787° Thus, u = 180° - f - a = 180° - 63.7791° - 26.4787° = 89.7422° a p rad b = 1.5663 rad 180° Shear Strain: (gG)x¿y¿ = p p - u = - 1.5663 = 4.50(10 - 3) rad 2 2 Ans. Ans: (gG)x¿y¿ = 4.50(10 - 3) rad 121 x © 2014 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. 2–18. The piece of plastic is originally rectangular. Determine the shear strain gxy at corners A and B if the plastic distorts as shown by the dashed lines. y 5 mm 2 mm 2 mm B 4 mm C 300 mm 2 mm D A x 400 mm 3 mm Geometry: For small angles, a = c = 2 = 0.00662252 rad 302 b = u = 2 = 0.00496278 rad 403 Shear Strain: (gB)xy = a + b = 0.0116 rad = 11.6 A 10 - 3 B rad Ans. (gA)xy = u + c = 0.0116 rad = 11.6 A 10 - 3 B rad Ans. Ans: (gB)xy = 11.6(10 - 3) rad, (gA)xy = 11.6(10 - 3) rad 122 © 2014 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. 2–19. The piece of plastic is originally rectangular. Determine the shear strain gxy at corners D and C if the plastic distorts as shown by the dashed lines. y 5 mm 2 mm 2 mm B 4 mm C 300 mm 2 mm D A x 400 mm 3 mm Geometry: For small angles, 2 = 0.00496278 rad 403 2 = 0.00662252 rad b = u = 302 Shear Strain: a = c = (gC)xy = a + b = 0.0116 rad = 11.6 A 10 - 3 B rad Ans. (gD)xy = u + c = 0.0116 rad = 11.6 A 10 - 3 B rad Ans. Ans: (gC)xy = 11.6(10 - 3) rad, (gD)xy = 11.6(10 - 3) rad 123 © 2014 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. *2–20. The piece of plastic is originally rectangular. Determine the average normal strain that occurs along the diagonals AC and DB. y 5 mm 2 mm 2 mm B 4 mm C 300 mm 2 mm D A 400 mm 3 mm Geometry: AC = DB = 24002 + 3002 = 500 mm DB¿ = 24052 + 3042 = 506.4 mm A¿C¿ = 24012 + 3002 = 500.8 mm Average Normal Strain: PAC = A¿C¿ - AC 500.8 - 500 = AC 500 = 0.00160 mm>mm = 1.60 A 10 - 3 B mm>mm PDB = Ans. DB¿ - DB 506.4 - 500 = DB 500 = 0.0128 mm>mm = 12.8 A 10 - 3 B mm>mm Ans. 124 x © 2014 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. 2–21. The rectangular plate is deformed into the shape of a parallelogram shown by the dashed lines. Determine the average shear strain gxy at corners A and B. y 5 mm D C 300 mm 5 mm A B x 400 mm Geometry: Referring to Fig. a and using small angle analysis, u = 5 = 0.01667 rad 300 f = 5 = 0.0125 rad 400 Shear Strain: Referring to Fig. a, (gA)xy = u + f = 0.01667 + 0.0125 = 0.0292 rad Ans. (gB)xy = u + f = 0.01667 + 0.0125 = 0.0292 rad Ans. Ans: (gA)xy = 0.0292 rad, (gB)xy = 0.0292 rad 125 © 2014 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. 2–22. The triangular plate is fixed at its base, and its apex A is given a horizontal displacement of 5 mm. Determine the shear strain, gxy, at A. y 45⬚ 800 mm 45⬚ x¿ A 45⬚ A¿ 5 mm 800 mm x L = 28002 + 52 - 2(800)(5) cos 135° = 803.54 mm sin 135° sin u = ; 803.54 800 gxy = u = 44.75° = 0.7810 rad p p - 2u = - 2(0.7810) 2 2 = 0.00880 rad Ans. Ans: gxy = 0.00880 rad 126 © 2014 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. 2–23. The triangular plate is fixed at its base, and its apex A is given a horizontal displacement of 5 mm. Determine the average normal strain Px along the x axis. y 45⬚ 800 mm 45⬚ x¿ A 45⬚ A¿ 5 mm 800 mm x L = 28002 + 52 - 2(800)(5) cos 135° = 803.54 mm Px = 803.54 - 800 = 0.00443 mm>mm 800 Ans. Ans: Px = 0.00443 mm>mm 127 © 2014 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. *2–24. The triangular plate is fixed at its base, and its apex A is given a horizontal displacement of 5 mm. Determine the average normal strain Px¿ along the x¿ axis. y 45⬚ 800 mm 45⬚ x¿ A 45⬚ 800 mm x L = 800 cos 45° = 565.69 mm Px¿ = 5 = 0.00884 mm>mm 565.69 Ans. 128 A¿ 5 mm © 2014 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. y 2–25. The square rubber block is subjected to a shear strain of gxy = 40(10 - 6)x + 20(10 - 6)y, where x and y are in mm. This deformation is in the shape shown by the dashed lines, where all the lines parallel to the y axis remain vertical after the deformation. Determine the normal strain along edge BC. Shear Strain: Along edge DC, y = 400 mm. Thus, (gxy)DC = 40(10 - 6)x + 0.008. dy Here, = tan (gxy)DC = tan 340(10 - 6)x + 0.0084. Then, dx dc C A B 400 mm 300 mm dy = L0 D dc = - tan [40(10 - 6)x + 0.008]dx L0 1 40(10 - 6) e ln cos c 40(10 - 6)x + 0.008 d f ` 300 mm x 300 mm 0 = 4.2003 mm Along edge AB, y = 0. Thus, (gxy)AB = 40(10 - 6)x. Here, dy = tan (gxy)AB = dx tan [40(10 - 6)x]. Then, 300 mm dB dy = L0 dB = - L0 1 40(10 - 6) tan [40(10 - 6)x]dx e ln cos c 40(10 - 6)x d f ` 300 mm 0 = 1.8000 mm Average Normal Strain: The stretched length of edge BC is LB¿C¿ = 400 + 4.2003 - 1.8000 = 402.4003 mm We obtain, (Pavg)BC = LB¿C¿ - LBC 402.4003 - 400 = = 6.00(10 - 3) mm>mm LBC 400 Ans. Ans: (Pavg)BC = 6.00(10 - 3) mm>mm 129 © 2014 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. 2–26. The square plate is deformed into the shape shown by the dashed lines. If DC has a normal strain Px = 0.004, DA has a normal strain Py = 0.005 and at D, gxy = 0.02 rad, determine the average normal strain along diagonal CA. y y¿ x¿ 600 mm A¿ B¿ B A E 600 mm D C C¿ x Average Normal Strain: The stretched length of sides DA and DC are LDC¿ = (1 + Px)LDC = (1 + 0.004)(600) = 602.4 mm LDA¿ = (1 + Py)LDA = (1 + 0.005)(600) = 603 mm Also, a = p 180° - 0.02 = 1.5508 rad a b = 88.854° 2 p rad Thus, the length of C¿A¿ can be determined using the cosine law with reference to Fig. a. LC¿A¿ = 2602.42 + 6032 - 2(602.4)(603) cos 88.854° = 843.7807 mm The original length of diagonal CA can be determined using Pythagorean’s theorem. LCA = 26002 + 6002 = 848.5281 mm Thus, (Pavg)CA = LC¿A¿ - LCA 843.7807 - 848.5281 = - 5.59(10 - 3) mm>mm Ans. = LCA 848.5281 Ans: (Pavg)CA = - 5.59(10 - 3) mm>mm 130 © 2014 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. 2–27. The square plate ABCD is deformed into the shape shown by the dashed lines. If DC has a normal strain Px = 0.004, DA has a normal strain Py = 0.005 and at D, gxy = 0.02 rad, determine the shear strain at point E with respect to the x¿ and y¿ axes. y y¿ x¿ 600 mm A¿ B¿ B A 600 mm D E C C¿ x Average Normal Strain: The stretched length of sides DC and BC are LDC¿ = (1 + Px)LDC = (1 + 0.004)(600) = 602.4 mm LB¿C¿ = (1 + Py)LBC = (1 + 0.005)(600) = 603 mm Also, a = 180° p - 0.02 = 1.5508 rada b = 88.854° 2 p rad f = 180° p + 0.02 = 1.5908 rad a b = 91.146° 2 p rad Thus, the length of C¿A¿ and DB¿ can be determined using the cosine law with reference to Fig. a. LC¿A¿ = 2602.42 + 6032 - 2(602.4)(603) cos 88.854° = 843.7807 mm LDB¿ = 2602.42 + 6032 - 2(602.4)(603) cos 91.146° = 860.8273 mm Thus, LE¿A¿ = LC¿A¿ = 421.8903 mm 2 LE¿B¿ = LDB¿ = 430.4137 mm 2 Using this result and applying the cosine law to the triangle A¿E¿B¿ , Fig. a, 602.42 = 421.89032 + 430.41372 - 2(421.8903)(430.4137) cos u u = 89.9429° a p rad b = 1.5698 rad 180° Shear Strain: (gE)x¿y¿ = p p - u = - 1.5698 = 0.996(10 - 3) rad 2 2 Ans. Ans: (gE)x¿y¿ = 0.996(10 - 3) rad 131 © 2014 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. *2–28. The wire is subjected to a normal strain that is 2 P ⫽ (x/L)e–(x/L) 2 defined by P = (x>L)e - (x>L) . If the wire has an initial x length L, determine the increase in its length. x L L ¢L = 1 - (x>L)2 xe dx L L0 = -L c = 2 L e - (x>L) L d = 31 - (1>e)4 2 2 0 L 3e - 14 2e Ans. 132 © 2014 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. 2–29. The rectangular plate is deformed into the shape shown by the dashed lines. Determine the average normal strain along diagonal AC, and the average shear strain at corner A. y 6 mm 400 mm 2 mm 2 mm 6 mm C D 300 mm 2 mm A Geometry: The unstretched length of diagonal AC is x B 400 mm 3 mm LAC = 2300 + 400 = 500 mm 2 2 Referring to Fig. a, the stretched length of diagonal AC is LAC¿ = 2(400 + 6)2 + (300 + 6)2 = 508.4014 mm Referring to Fig. a and using small angle analysis, f = 2 = 0.006623 rad 300 + 2 a = 2 = 0.004963 rad 400 + 3 Average Normal Strain: Applying Eq. 2, (Pavg)AC = LAC¿ - LAC 508.4014 - 500 = = 0.0168 mm>mm LAC 500 Ans. Shear Strain: Referring to Fig. a, (gA)xy = f + a = 0.006623 + 0.004963 = 0.0116 rad Ans. Ans: (Pavg)AC = 0.0168 mm>mm, (gA)xy = 0.0116 rad 133 © 2014 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. 2–30. The rectangular plate is deformed into the shape shown by the dashed lines. Determine the average normal strain along diagonal BD, and the average shear strain at corner B. y 6 mm 400 mm 2 mm 2 mm 6 mm C D 300 mm 2 mm A Geometry: The unstretched length of diagonal BD is x B 400 mm 3 mm LBD = 23002 + 4002 = 500 mm Referring to Fig. a, the stretched length of diagonal BD is LB¿D¿ = 2(300 + 2 - 2)2 + (400 + 3 - 2)2 = 500.8004 mm Referring to Fig. a and using small angle analysis, 2 = 0.004963 rad 403 3 = 0.009868 rad a = 300 + 6 - 2 f = Average Normal Strain: Applying Eq. 2, (Pavg)BD = LB¿D¿ - LBD 500.8004 - 500 = = 1.60(10 - 3) mm>mm Ans. LBD 500 Shear Strain: Referring to Fig. a, (gB)xy = f + a = 0.004963 + 0.009868 = 0.0148 rad Ans. Ans: (Pavg)BD = 1.60(10 - 3) mm>mm, (gB)xy = 0.0148 rad 134 © 2014 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. 2–31. The nonuniform loading causes a normal strain in the shaft that can be expressed as Px = kx2, where k is a constant. Determine the displacement of the end B. Also, what is the average normal strain in the rod? L A B x d(¢x) = Px = kx2 dx L (¢x)B = (Px)avg = 3 2 L0 kx = (¢x)B L kL 3 Ans. kL3 kL2 3 = = L 3 Ans. Ans: 3 (¢x)B = 135 kL kL2 , (Px)avg = 3 3 © 2014 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. *2–32 The rubber block is fixed along edge AB, and edge CD is moved so that the vertical displacement of any point in the block is given by v(x) = (v0>b3)x3. Determine the shear strain gxy at points (b>2, a>2) and (b, a). y v (x) A v0 D a B Shear Strain: From Fig. a, b dv = tan gxy dx 3v0 b3 x2 = tan gxy gxy = tan - 1 a 3v0 b3 x2 b Thus, at point (b> 2, a> 2), gxy = tan - 1 c 3v0 b 2 a b d b3 2 3 v0 = tan - 1 c a b d 4 b Ans. and at point (b, a), gxy = tan - 1 c 3v0 b3 = tan - 1 c 3 a C (b2) d v0 bd b Ans. 136 x © 2014 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. 2–33. The fiber AB has a length L and orientation u. If its ends A and B undergo very small displacements uA and vB, respectively, determine the normal strain in the fiber when it is in position A¿B¿ . y B¿ vB B L A uA A¿ u x Geometry: LA¿B¿ = 2(L cos u - uA)2 + (L sin u + vB)2 = 2L2 + u2A + v2B + 2L(vB sin u - uA cos u) Average Normal Strain: LA¿B¿ - L PAB = L = A 1 + 2(vB sin u - uA cos u) u2A + v2B + - 1 L L2 Neglecting higher terms u2A and v2B 1 PAB = B 1 + 2(vB sin u - uA cos u) 2 R - 1 L Using the binomial theorem: PAB = 1 + = 2uA cos u 1 2vB sin u ¢ ≤ + ... - 1 2 L L vB sin u uA cos u L L Ans. Ans. PAB = 137 vB sin u uA cos u L L © 2014 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. 2–34. If the normal strain is defined in reference to the final length, that is, P¿n = lim ¿ a p:p ¢s¿ - ¢s b ¢s¿ instead of in reference to the original length, Eq. 2–2 , show that the difference in these strains is represented as a second-order term, namely, Pn - Pn¿ = Pn Pn¿ . PB = ¢S¿ - ¢S ¢S œ = PB - PA ¢S¿ - ¢S ¢S¿ - ¢S ¢S ¢S¿ ¢S¿ 2 - ¢S¢S¿ - ¢S¿¢S + ¢S2 ¢S¢S¿ ¢S¿ 2 + ¢S2 - 2¢S¿¢S = ¢S¢S¿ = = (¢S¿ - ¢S)2 ¢S¿ - ¢S ¢S¿ - ¢S = ¢ ≤¢ ≤ ¢S¢S¿ ¢S ¢S¿ = PA PBœ (Q.E.D) 138 © 2014 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. 3–1. A tension test was performed on a steel specimen having an original diameter of 0.503 in. and gauge length of 2.00 in. The data is listed in the table. Plot the stress–strain diagram and determine approximately the modulus of elasticity, the yield stress, the ultimate stress, and the rupture stress. Use a scale of 1 in. = 20 ksi and 1 in. = 0.05 in.> in. Redraw the elastic region, using the same stress scale but a strain scale of 1 in. = 0.001 in.> in. A = Load (kip) Elongation (in.) 0 1.50 4.60 8.00 11.00 11.80 11.80 12.00 16.60 20.00 21.50 19.50 18.50 1 p(0.503)2 = 0.1987 in2 4 L = 2.00 in. s(ksi) 0 0.0005 0.0015 0.0025 0.0035 0.0050 0.0080 0.0200 0.0400 0.1000 0.2800 0.4000 0.4600 e(in.>in.) 0 0 7.55 0.00025 23.15 0.00075 40.26 0.00125 55.36 0.00175 59.38 0.0025 59.38 0.0040 60.39 0.010 83.54 0.020 100.65 0.050 108.20 0.140 98.13 0.200 93.10 Eapprox 0.230 48 = = 32.0(103) ksi 0.0015 Ans. Ans: (sult)approx = 110 ksi, (sR)approx = 93.1 ksi, (sY)approx = 55 ksi, Eapprox = 32.0(103) ksi 139 © 2014 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. 3–2. Data taken from a stress–strain test for a ceramic are given in the table. The curve is linear between the origin and the first point. Plot the diagram, and determine the modulus of elasticity and the modulus of resilience. Modulus of Elasticity: From the stress–strain diagram E = 33.2 - 0 = 55.3 A 103 B ksi 0.0006 - 0 S (ksi) P (in./in.) 0 33.2 45.5 49.4 51.5 53.4 0 0.0006 0.0010 0.0014 0.0018 0.0022 Ans. Modulus of Resilience: The modulus of resilience is equal to the area under the linear portion of the stress–strain diagram (shown shaded). ur = 1 lb in. in # lb (33.2) A 103 B ¢ 2 ≤ ¢ 0.0006 ≤ = 9.96 2 in. in in3 Ans. Ans: in # lb E = 55.3 A 103 B ksi, ur = 9.96 in3 140 © 2014 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. 3–3. Data taken from a stress–strain test for a ceramic are given in the table. The curve is linear between the origin and the first point. Plot the diagram, and determine approximately the modulus of toughness. The rupture stress is sr = 53.4 ksi. Modulus of Toughness: The modulus of toughness is equal to the area under the stress-strain diagram (shown shaded). (ut)approx = S (ksi) P (in./in.) 0 33.2 45.5 49.4 51.5 53.4 0 0.0006 0.0010 0.0014 0.0018 0.0022 1 lb in. (33.2) A 103 B ¢ 2 ≤ (0.0004 + 0.0010) ¢ ≤ 2 in. in + 45.5 A 103 B ¢ + 1 lb in. (7.90) A 103 B ¢ 2 ≤ (0.0012) ¢ ≤ 2 in. in + = 85.0 lb in. ≤ (0.0012) ¢ ≤ in. in2 lb in. 1 (12.3) A 103 B ¢ 2 ≤ (0.0004) ¢ ≤ 2 in. in in # lb in3 Ans. Ans: in # lb (ut)approx = 85.0 in3 141 © 2014 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. *3–4. A tension test was performed on a steel specimen having an original diameter of 0.503 in. and a gauge length of 2.00 in. The data is listed in the table. Plot the stress–strain diagram and determine approximately the modulus of elasticity, the ultimate stress, and the rupture stress. Use a scale of 1 in. = 15 ksi and 1 in. = 0.05 in.> in. Redraw the linear-elastic region, using the same stress scale but a strain scale of 1 in. = 0.001 in. A = 1 p(0.503)2 = 0.19871 in2 4 L = 2.00 in. s = P A (ksi) P = ¢L L (in.>in.) 0 0 12.58 0.00045 32.71 0.00125 42.78 0.0020 46.30 0.00325 49.32 0.0049 60.39 0.02 70.45 0.06 72.97 0.125 70.45 0.175 66.43 0.235 Eapprox = 32.71 = 26.2(103) ksi 0.00125 Ans. 142 Load (kip) Elongation (in.) 0 2.50 6.50 8.50 9.20 9.80 12.0 14.0 14.5 14.0 13.2 0 0.0009 0.0025 0.0040 0.0065 0.0098 0.0400 0.1200 0.2500 0.3500 0.4700 © 2014 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. 3–5. A tension test was performed on a steel specimen having an original diameter of 0.503 in. and gauge length of 2.00 in. Using the data listed in the table, plot the stress–strain diagram and determine approximately the modulus of toughness. Modulus of toughness (approx) ut = total area under the curve (1) = 87 (7.5) (0.025) = 16.3 in. # kip Load (kip) Elongation (in.) 0 2.50 6.50 8.50 9.20 9.80 12.0 14.0 14.5 14.0 13.2 0 0.0009 0.0025 0.0040 0.0065 0.0098 0.0400 0.1200 0.2500 0.3500 0.4700 Ans. in3 In Eq.(1), 87 is the number of squares under the curve. s = P A (ksi) P = ¢L L (in.>in.) 0 0 12.58 0.00045 32.71 0.00125 42.78 0.0020 46.30 0.00325 49.32 0.0049 60.39 0.02 70.45 0.06 72.97 0.125 70.45 0.175 66.43 0.235 Ans: in. # kip ut = 16.3 in3 143 © 2014 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. 3–6. A specimen is originally 1 ft long, has a diameter of 0.5 in., and is subjected to a force of 500 lb. When the force is increased from 500 lb to 1800 lb, the specimen elongates 0.009 in. Determine the modulus of elasticity for the material if it remains linear elastic. Normal Stress and Strain: Applying s = s1 = s2 = ¢P = 0.500 p 2 4 (0.5 ) 1.80 p 2 4 (0.5 ) dL P and e = . A L = 2.546 ksi = 9.167 ksi 0.009 = 0.000750 in.>in. 12 Modulus of Elasticity: E = ¢s 9.167 - 2.546 = = 8.83 A 103 B ksi ¢P 0.000750 Ans. Ans: E = 8.83 A 103 B ksi 144 © 2014 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. 3–7. A structural member in a nuclear reactor is made of a zirconium alloy. If an axial load of 4 kip is to be supported by the member, determine its required cross-sectional area. Use a factor of safety of 3 relative to yielding. What is the load on the member if it is 3 ft long and its elongation is 0.02 in.? Ezr = 14(103) ksi, sY = 57.5 ksi. The material has elastic behavior. Allowable Normal Stress: F.S. = 3 = sy sallow 57.5 sallow sallow = 19.17 ksi sallow = P A 19.17 = 4 A A = 0.2087 in2 = 0.209 in2 Ans. Stress–Strain Relationship: Applying Hooke’s law with P = 0.02 d = = 0.000555 in.>in. L 3 (12) s = EP = 14 A 103 B (0.000555) = 7.778 ksi Normal Force: Applying equation s = P . A P = sA = 7.778 (0.2087) = 1.62 kip Ans. Ans: A = 0.209 in2, P = 1.62 kip 145 © 2014 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. *3–8. The strut is supported by a pin at C and an A-36 steel guy wire AB. If the wire has a diameter of 0.2 in., determine how much it stretches when the distributed load acts on the strut. A 60⬚ 200 lb/ft a + ©MC = 0; 1 FAB cos 60°(9) - (200)(9)(3) = 0 2 9 ft FAB = 600 lb The normal stress the wire is sAB = FAB = AAB p 4 600 = 19.10(103) psi = 19.10 ksi (0.22) Since sAB 6 sy = 36 ksi, Hooke’s Law can be applied to determine the strain in wire. sAB = EPAB; 19.10 = 29.0(103)PAB PAB = 0.6586(10 - 3) in>in 9(12) = 124.71 in. Thus, the wire The unstretched length of the wire is LAB = sin 60° stretches dAB = PAB LAB = 0.6586(10 - 3)(124.71) = 0.0821 in. Ans. 146 B C Here, we are only interested in determining the force in wire AB. © 2014 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. s (psi) 3–9. The s -P diagram for elastic fibers that make up human skin and muscle is shown. Determine the modulus of elasticity of the fibers and estimate their modulus of toughness and modulus of resilience. 55 11 1 E = 11 = 5.5 psi 2 Ans. ut = 1 1 (2)(11) + (55 + 11)(2.25 - 2) = 19.25 psi 2 2 Ans. ur = 1 (2)(11) = 11 psi 2 Ans. 2 2.25 P (in./in.) Ans: E = 5.5 psi, ut = 19.25 psi, ur = 11 psi 147 © 2014 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. s (ksi) 3–10. The stress–strain diagram for a metal alloy having an original diameter of 0.5 in. and a gauge length of 2 in. is given in the figure. Determine approximately the modulus of elasticity for the material, the load on the specimen that causes yielding, and the ultimate load the specimen will support. 105 90 75 60 45 30 15 0 0 0 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.001 0.002 0.003 0.004 0.005 0.006 0.007 P (in./in.) From the stress–strain diagram, Fig. a, 60 ksi - 0 E = ; 1 0.002 - 0 sy = 60 ksi E = 30.0(103) ksi Ans. sult = 100 ksi Thus, PY = sYA = 60 C p4 (0.52) D = 11.78 kip = 11.8 kip Ans. Pult = sult A = 100 C p4 (0.52) D = 19.63 kip = 19.6 kip Ans. Ans: E = 30.0(103) ksi, PY = 11.8 kip, Pult = 19.6 kip 148 © 2014 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. s (ksi) 3–11. The stress–strain diagram for a steel alloy having an original diameter of 0.5 in. and a gauge length of 2 in. is given in the figure. If the specimen is loaded until it is stressed to 90 ksi, determine the approximate amount of elastic recovery and the increase in the gauge length after it is unloaded. 105 90 75 60 45 30 15 0 0 0 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.001 0.002 0.003 0.004 0.005 0.006 0.007 P (in./in.) From the stress–strain diagram Fig. a, the modulus of elasticity for the steel alloy is E 60 ksi - 0 = ; 1 0.002 - 0 E = 30.0(103) ksi when the specimen is unloaded, its normal strain recovers along line AB, Fig. a, which has a slope of E. Thus Elastic Recovery = 90 90 ksi = 0.003 in>in. = E 30.0(103) ksi Ans. Thus, the permanent set is PP = 0.05 - 0.003 = 0.047 in>in. Then, the increase in gauge length is ¢L = PPL = 0.047(2) = 0.094 in. Ans. Ans: Elastic Recovery = 0.003 in.>in., ¢L = 0.094 in. 149 © 2014 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. s (ksi) *3–12. The stress–strain diagram for a steel alloy having an original diameter of 0.5 in. and a gauge length of 2 in. is given in the figure. Determine approximately the modulus of resilience and the modulus of toughness for the material. 105 90 75 60 45 30 15 0 The Modulus of resilience is equal to the area under the stress–strain diagram up to the proportional limit. sPL = 60 ksi PPL = 0.002 in.>in. Thus, (ui)r = 1 1 in. # lb sPLPPL = C 60(103) D (0.002) = 60.0 2 2 in3 Ans. The modulus of toughness is equal to the area under the entire stress–strain diagram. This area can be approximated by counting the number of squares. The total number is 38. Thus, C (ui)t D approx = 38 c 15(103) lb in. in. # lb d a 0.05 b = 28.5(103) 2 in. in in3 150 Ans. 0 0 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.001 0.002 0.003 0.004 0.005 0.006 0.007 P (in./in.) © 2014 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. 3–13. A bar having a length of 5 in. and cross-sectional area of 0.7 in.2 is subjected to an axial force of 8000 lb. If the bar stretches 0.002 in., determine the modulus of elasticity of the material. The material has linear-elastic behavior. 8000 lb 8000 lb 5 in. Normal Stress and Strain: s = 8.00 P = = 11.43 ksi A 0.7 P = d 0.002 = = 0.000400 in.>in. L 5 Modulus of Elasticity: E = s 11.43 = = 28.6(103) ksi P 0.000400 Ans. Ans: E = 28.6(103) ksi 151 © 2014 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. 3–14. The rigid pipe is supported by a pin at A and an A-36 steel guy wire BD. If the wire has a diameter of 0.25 in., determine how much it stretches when a load of P = 600 lb acts on the pipe. B 4 ft P A D C 3 ft 3 ft Here, we are only interested in determining the force in wire BD. Referring to the FBD in Fig. a a + ©MA = 0; FBD A 45 B (3) - 600(6) = 0 FBD = 1500 lb The normal stress developed in the wire is sBD = FBD = ABD p 4 1500 = 30.56(103) psi = 30.56 ksi (0.252) Since sBD 6 sy = 36 ksi, Hooke’s Law can be applied to determine the strain in the wire. sBD = EPBD; 30.56 = 29.0(103)PBD PBD = 1.054(10 - 3) in.>in. The unstretched length of the wire is LBD = 232 + 42 = 5 ft = 60 in. Thus, the wire stretches dBD = PBD LBD = 1.054(10 - 3)(60) = 0.0632 in. Ans. Ans: dBD = 0.0632 in. 152 © 2014 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. 3–15. The rigid pipe is supported by a pin at A and an A-36 guy wire BD. If the wire has a diameter of 0.25 in., determine the load P if the end C is displaced 0.15 in. downward. B 4 ft P A D C 3 ft 3 ft Here, we are only interested in determining the force in wire BD. Referring to the FBD in Fig. a FBD A 45 B (3) - P(6) = 0 a + ©MA = 0; FBD = 2.50 P The unstretched length for wire BD is LBD = 232 + 42 = 5 ft = 60 in. From the geometry shown in Fig. b, the stretched length of wire BD is LBD¿ = 2602 + 0.0752 - 2(60)(0.075) cos 143.13° = 60.060017 Thus, the normal strain is PBD = LBD¿ - LBD 60.060017 - 60 = = 1.0003(10 - 3) in.>in. LBD 60 Then, the normal stress can be obtain by applying Hooke’s Law. sBD = EPBD = 29(103) C 1.0003(10 - 3) D = 29.01 ksi Since sBD 6 sy = 36 ksi, the result is valid. sBD = FBD ; ABD 29.01(103) = 2.50 P (0.252) p 4 P = 569.57 lb = 570 lb Ans. Ans: P = 570 lb 153 © 2014 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. *3–16. The wire has a diameter of 5 mm and is made from A-36 steel. If a 80-kg man is sitting on seat C, determine the elongation of wire DE. E W 600 mm D A B 800 mm Equations of Equilibrium: The force developed in wire DE can be determined by writing the moment equation of equilibrium about A with reference to the freebody diagram shown in Fig. a, a + ©MA = 0; 3 FDE a b (0.8) - 80(9.81)(1.4) = 0 5 FDE = 2289 N Normal Stress and Strain: sDE = FDE 2289 = = 116.58 MPa p ADE (0.0052) 4 Since sDE < sY , Hooke’s Law can be applied sDE = EPDE 116.58(106) = 200(109)PDE PDE = 0.5829(10-3) mm>mm The unstretched length of wire DE is LDE = 26002 + 8002 = 1000 mm. Thus, the elongation of this wire is given by dDE = PDELDE = 0.5829(10-3)(1000) = 0.583 mm Ans. 154 C 600 mm © 2014 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. s (ksi) 3–17. A tension test was performed on a magnesium alloy specimen having a diameter 0.5 in. and gauge length 2 in. The resulting stress–strain diagram is shown in the figure. Determine the approximate modulus of elasticity and the yield strength of the alloy using the 0.2% strain offset method. 40 35 30 25 20 15 10 5 0 0.002 0.004 0.006 0.008 0.010 P (in./in.) Modulus of Elasticity: From the stress–strain diagram, when P = 0.002 in.>in., its corresponding stress is s = 13.0 ksi. Thus, Eapprox = 13.0 - 0 = 6.50(103) ksi 0.002 - 0 Ans. Yield Strength: The intersection point between the stress–strain diagram and the straight line drawn parallel to the initial straight portion of the stress–strain diagram from the offset strain of P = 0.002 in.>in. is the yield strength of the alloy. From the stress–strain diagram, sYS = 25.9 ksi Ans. Ans: Eapprox = 6.50(103) ksi, sYS = 25.9 ksi 155 © 2014 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. s (ksi) 3–18. A tension test was performed on a magnesium alloy specimen having a diameter 0.5 in. and gauge length of 2 in. The resulting stress–strain diagram is shown in the figure. If the specimen is stressed to 30 ksi and unloaded, determine the permanent elongation of the specimen. 40 35 30 25 20 15 10 5 0 0.002 0.004 0.006 0.008 0.010 P (in./in.) Permanent Elongation: From the stress–strain diagram, the strain recovered is along the straight line BC which is parallel to the straight line OA. Since 13.0 - 0 = 6.50(103) ksi, then the permanent set for the specimen is Eapprox = 0.002 - 0 30(103) = 0.00318 in.>in. PP = 0.0078 6.5(106) Thus, dP = PPL = 0.00318(2) = 0.00637 in. Ans. Ans: dP = 0.00637 in. 156 © 2014 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. 3–19. The stress–strain diagram for a bone is shown, and can be described by the equation P = 0.45110-62 s ⫹ 0.36110-122 s3, where s is in kPa. Determine the yield strength assuming a 0.3% offset. P s P ⫽ 0.45(10⫺6)s + 0.36(10⫺12)s3 P P P = 0.45(10-6)s + 0.36(10-12)s3, dP = A 0.45(10-6) + 1.08(10-12) s2 B ds E = ds 1 2 = = 2.22(106) kPa = 2.22 GPa dP 0.45(10 - 6) s=0 The equation for the recovery line is s = 2.22(106)(P - 0.003). This line intersects the stress–strain curve at sYS = 2027 kPa = 2.03 MPa Ans. Ans: sYS = 2.03 MPa 157 © 2014 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. *3–20. The stress–strain diagram for a bone is shown and can be described by the equation P = 0.45110-62 s ⫹ 0.36110-122 s3, where s is in kPa. Determine the modulus of toughness and the amount of elongation of a 200-mm-long region just before it fractures if failure occurs at P = 0.12 mm>mm. P s P ⫽ 0.45(10⫺6)s + 0.36(10⫺12)s3 P When P = 0.12 120(10-3) = 0.45 s + 0.36(10-6)s3 Solving for the real root: s = 6873.52 kPa 6873.52 ut = LA dA = L0 (0.12 - P)ds 6873.52 ut = L0 (0.12 - 0.45(10-6)s - 0.36(10-12)s3)ds 6873.52 = 0.12 s - 0.225(10-6)s2 - 0.09(10-12)s4|0 = 613 kJ>m3 Ans. d = PL = 0.12(200) = 24 mm Ans. 158 P © 2014 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. 3–21. The two bars are made of polystyrene, which has the stress–strain diagram shown. If the cross-sectional area of bar AB is 1.5 in2 and BC is 4 in2, determine the largest force P that can be supported before any member ruptures. Assume that buckling does not occur. P 4 ft C B 3 ft A s (ksi) 25 + c g Fy = 0; + ; ©Fx = 0; 3 F - P = 0; 5 AB FBC 4 - (1.6667P) = 0; 5 FAB = 1.6667 P 20 (1) 15 FBC = 1.333 P (2) 0 From the stress–strain diagram (sR)t = 5 ksi FBC ; ABC 5 = tension 5 Assuming failure of bar BC: s = compression 10 FBC ; 4 0 0.20 0.40 0.60 0.80 P (in./in.) FBC = 20.0 kip From Eq. (2), P = 15.0 kip Assuming failure of bar AB: From stress–strain diagram (sR)c = 25.0 ksi s = FAB ; AAB 25.0 = FAB ; 1.5 FAB = 37.5 kip From Eq. (1), P ⫽ 22.5 kip Choose the smallest value P = 15.0 kip Ans. Ans: P = 15.0 kip 159 © 2014 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. 3–22. The two bars are made of polystyrene, which has the stress–strain diagram shown. Determine the cross-sectional area of each bar so that the bars rupture simultaneously when the load P = 3 kip. Assume that buckling does not occur. P 4 ft C B 3 ft A s (ksi) 25 + c ©Fy = 0; + : ©Fx = 0; 3 FBA a b - 3 = 0; 5 -FBC 4 + 5a b = 0; 5 20 FBA = 5 kip 15 FBC = 4 kip compression 10 tension 5 For member BC: 0 (smax)t = 4 kip FBC ; ABC = = 0.8 in2 ABC 5 ksi (smax)c = FBA ; ABA 0 0.20 0.40 0.60 0.80 P (in./in.) Ans. For member BA: ABA = 5 kip = 0.2 in2 25 ksi Ans. Ans: ABC = 0.8 in2, ABA = 0.2 in2 160 © 2014 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. 3–23. The stress–strain diagram for many metal alloys can be described analytically using the Ramberg-Osgood three parameter equation P = s>E + ksn, where E, k, and n are determined from measurements taken from the diagram. Using the stress–strain diagram shown in the figure, take E = 30(103) ksi and determine the other two parameters k and n and thereby obtain an analytical expression for the curve. s (ksi) 80 60 40 20 0.1 0.2 0.3 0.4 0.5 P (10 – 6 ) Choose, s = 40 ksi, e = 0.1 s = 60 ksi, e = 0.3 0.1 = 40 + k(40)n 30(103) 0.3 = 60 + k(60)n 30(103) 0.098667 = k(40)n 0.29800 = k(60)n 0.3310962 = (0.6667)n ln (0.3310962) = n ln (0.6667) n = 2.73 Ans. k = 4.23(10 - 6) Ans. Ans: n = 2.73, k = 4.23(10 - 6) 161 © 2014 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. 3–24. The wires AB and BC have original lengths of 2 ft 3 and 3 ft, and diameters of 18 in. and 16 in., respectively. If these wires are made of a material that has the approximate stress–strain diagram shown, determine the elongations of the wires after the 1500-lb load is placed on the platform. C Equations of Equilibrium: The forces developed in wires AB and BC can be determined by analyzing the equilibrium of joint B, Fig. a, + : ©Fx = 0; + c ©Fy = 0; FBC sin 30° - FAB sin 45° = 0 (1) FBC cos 30° + FAB cos 45° = 1500 (2) A 3 ft 45⬚ 30⬚ 2 ft B Solving Eqs. (1) and (2), FAB = 776.46 lb FBC = 1098.08 lb Normal Stress and Strain: sAB = FAB 776.46 = = 63.27 ksi p AAB (1>8)2 4 s (ksi) sBC = FBC 1098.08 = = 39.77 ksi p ABC 2 (3>16) 4 80 58 The corresponding normal strain can be determined from the stress–strain diagram, Fig. b. 39.77 58 ; = PBC 0.002 PBC = 0.001371 in.>in. 63.27 - 58 80 - 58 = ; PAB - 0.002 0.01 - 0.002 PAB = 0.003917 in.>in. 0.002 Thus, the elongations of wires AB and BC are dAB = PABLAB = 0.003917(24) = 0.0940 Ans. dBC = PBCLBC = 0.001371(36) = 0.0494 Ans. 162 0.01 P (in./in.) © 2014 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. 3–25. The acrylic plastic rod is 200 mm long and 15 mm in diameter. If an axial load of 300 N is applied to it, determine the change in its length and the change in its diameter. Ep = 2.70 GPa, np = 0.4. s = P = A Plong = 300 p 2 4 (0.015) 300 N 300 N 200 mm = 1.678 MPa 1.678(106) s = 0.0006288 = E 2.70(109) d = Plong L = 0.0006288 (200) = 0.126 mm Ans. Plat = - nPlong = - 0.4(0.0006288) = - 0.0002515 ¢d = Platd = - 0.0002515 (15) = - 0.00377 mm Ans. Ans: d = 0.126 mm, ¢d = - 0.00377 mm 163 © 2014 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. 3–26. The thin-walled tube is subjected to an axial force of 40 kN. If the tube elongates 3 mm and its circumference decreases 0.09 mm, determine the modulus of elasticity, Poisson’s ratio, and the shear modulus of the tube’s material. The material behaves elastically. 40 kN 900 mm 10 mm 40 kN 12.5 mm Normal Stress and Strain: s = 40(103) P = 226.35 MPa = A p(0.01252 - 0.012) Pa = 3 d = = 3.3333 (10-3) mm>mm L 900 Applying Hooke’s law, s = EPa; 226.35(106) = E [3.3333(10-3)] E = 67.91(106) Pa = 67.9 GPa Ans. Poisson’s Ratio: The circumference of the loaded tube is 2p(12.5) - 0.09 = 78.4498 mm. Thus, the outer radius of the tube is r = 78.4498 = 12.4857 mm 2p The lateral strain is Plat = r - r0 12.4857 - 12.5 = = - 1.1459(10-3) mm>mm r0 12.5 n = - - 1.1459(10-3) Plat d = 0.3438 = 0.344 = -c Pa 3.3333(10-3) G = Ans. 67.91(109) E = = 25.27(109) Pa = 25.3 GPa 2(1 + n) 2(1 + 0.3438) Ans. Ans: E = 67.9 GPa, v = 0.344, G = 25.3 GPa 164 © 2014 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. 3–27. When the two forces are placed on the beam, the diameter of the A-36 steel rod BC decreases from 40 mm to 39.99 mm. Determine the magnitude of each force P. C P 1m A P 1m 1m 1m B 0.75 m Equations of Equilibrium: The force developed in rod BC can be determined by writing the moment equation of equilibrium about A with reference to the free-body diagram of the beam shown in Fig. a. 4 FBC a b (3) - P(2) - P(1) = 0 5 a + ©MA = 0; FBC = 1.25P Normal Stress and Strain: The lateral strain of rod BC is Plat = d - d0 39.99 - 40 = = - 0.25(10 - 3) mm>mm d0 40 Plat = - nPa; - 0.25(10-3) = - (0.32)Pa Pa = 0.78125(10-3) mm>mm Assuming that Hooke’s Law applies, sBC = EPa; sBC = 200(109)(0.78125)(10-3) = 156.25 MPa Since s 6 sY, the assumption is correct. sBC = FBC ; ABC 156.25(106) = 1.25P p A 0.042 B 4 P = 157.08(103)N = 157 kN Ans. Ans: P = 157 kN 165 © 2014 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. *3–28. If P = 150 kN, determine the elastic elongation of rod BC and the decrease in its diameter. Rod BC is made of A-36 steel and has a diameter of 40 mm. C P 1m A P 1m 1m 1m B 0.75 m Equations of Equilibrium: The force developed in rod BC can be determined by writing the moment equation of equilibrium about A with reference to the freebody diagram of the beam shown in Fig. a. a + ©MA = 0; 4 FBC a b (3) - 150(2) - 150(1) = 0 5 FBC = 187.5 kN Normal Stress and Strain: The lateral strain of rod BC is sBC = 187.5(103) FBC = = 149.21 MPa p ABC A 0.042 B 4 Since s 6 sY, Hooke’s Law can be applied. Thus, sBC = EPBC; 149.21(106) = 200(109)PBC PBC = 0.7460(10-3) mm>mm The unstretched length of rod BC is LBC = 27502 + 10002 = 1250 mm. Thus the elongation of this rod is given by dBC = PBCLBC = 0.7460(10-3)(1250) = 0.933 mm Ans. We obtain, Plat = - nPa ; Plat = - (0.32)(0.7460)(10-3) = - 0.2387(10-3) mm>mm Thus, dd = Plat dBC = - 0.2387(10-3)(40) = - 9.55(10-3) mm 166 Ans. © 2014 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. 3–29. The friction pad A is used to support the member, which is subjected to an axial force of P = 2 kN. The pad is made from a material having a modulus of elasticity of E = 4 MPa and Poisson’s ratio n = 0.4. If slipping does not occur, determine the normal and shear strains in the pad. The width is 50 mm. Assume that the material is linearly elastic. Also, neglect the effect of the moment acting on the pad. P 60⬚ 25 mm A 100 mm Internal Loading: The normal force and shear force acting on the friction pad can be determined by considering the equilibrium of the pin shown in Fig. a. + : ©Fx = 0; V - 2 cos 60° = 0 V = 1 kN + c ©Fy = 0; N - 2 sin 60° = 0 N = 1.732 kN Normal and Shear Stress: t = 1(103) V = = 200 kPa A 0.1(0.05) s = 1.732(103) N = = 346.41 kPa A 0.1(0.05) Normal and Shear Strain: The shear modulus of the friction pad is G = 4 E = = 1.429 MPa 2(1 + n) 2(1 + 0.4) Applying Hooke’s Law, s = EP; 346.41(103) = 4(106)P P = 0.08660 mm>mm Ans. t = Gg; 200(103) = 1.429(106)g g = 0.140 rad Ans. Ans: P = 0.08660 mm>mm, g = 0.140 rad 167 © 2014 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. 3–30. The lap joint is connected together using a 1.25 in. diameter bolt. If the bolt is made from a material having a shear stress–strain diagram that is approximated as shown, determine the shear strain developed in the shear plane of the bolt when P = 75 kip. P 2 P 2 P t (ksi) 75 50 Internal Loadings: The shear force developed in the shear planes of the bolt can be determined by considering the equilibrium of the free-body diagram shown in Fig. a. + : ©Fx = 0; 75 - 2V = 0 0.005 0.05 g (rad) V = 37.5 kip Shear Stress and Strain: t = V 37.5 = = 30.56 ksi p A A 1.252 B 4 Using this result, the corresponding shear strain can be obtained from the shear stress–strain diagram, Fig. b. 30.56 50 = ; g 0.005 g = 3.06(10-3) rad Ans. Ans: g = 3.06(10-3) rad 168 © 2014 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. 3–31. The lap joint is connected together using a 1.25 in. diameter bolt. If the bolt is made from a material having a shear stress–strain diagram that is approximated as shown, determine the permanent shear strain in the shear plane of the bolt when the applied force P = 150 kip is removed. P 2 P 2 P t (ksi) 75 50 Internal Loadings: The shear force developed in the shear planes of the bolt can be determined by considering the equilibrium of the free-body diagram shown in Fig. a. + : ©Fx = 0; 150 - 2V = 0 V = 75 kip 0.005 Shear Stress and Strain: t = 0.05 g (rad) V 75 = = 61.12 ksi p A A 1.252 B 4 Using this result, the corresponding shear strain can be obtained from the shear stress–strain diagram, Fig. b. 61.12 - 50 75 - 50 = ; g - 0.005 0.05 - 0.005 g = 0.02501 rad When force P is removed, the shear strain recovers linearly along line BC, Fig. b, with a slope that is the same as line OA. This slope represents the shear modulus. G = 50 = 10(103) ksi 0.005 Thus, the elastic recovery of shear strain is t = Ggr; 61.12 = (10)(103)gr gr = 6.112(10-3) rad And the permanent shear strain is gP = g - gr = 0.02501 - 6.112(10-3) = 0.0189 rad Ans. Ans: gP = 0.0189 rad 169 © 2014 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. *3–32. A shear spring is made by bonding the rubber annulus to a rigid fixed ring and a plug. When an axial load P is placed on the plug, show that the slope at point y in the rubber is dy>dr = - tan g = - tan1P>12phGr22. For small angles we can write dy>dr = - P>12phGr2. Integrate this expression and evaluate the constant of integration using the condition that y = 0 at r = ro. From the result compute the deflection y = d of the plug. P h ro y d ri r y Shear Stress–Strain Relationship: Applying Hooke’s law with tA g = P = . 2p r h tA P = G 2p h G r dy P = - tan g = - tan a b dr 2p h G r (Q.E.D) If g is small, then tan g = g. Therefore, dy P = dr 2p h G r At r = ro, y = - dr P 2p h G L r y = - P ln r + C 2p h G 0 = - P ln ro + C 2p h G y = 0 C = Then, y = ro P ln r 2p h G At r = ri, y = d d = P ln ro 2p h G ro P ln ri 2p h G Ans. 170 © 2014 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. 3–33. The aluminum block has a rectangular cross section and is subjected to an axial compressive force of 8 kip. If the 1.5-in. side changed its length to 1.500132 in., determine Poisson’s ratio and the new length of the 2-in. side. Eal = 10(103) ksi. s = 2 in. 8 kip 8 kip 3 in. P 8 = = 2.667 ksi A (2)(1.5) Plong = Plat = n = 1.5 in. s - 2.667 = - 0.0002667 = E 10(103) 1.500132 - 1.5 = 0.0000880 1.5 - 0.0000880 = 0.330 - 0.0002667 Ans. h¿ = 2 + 0.0000880(2) = 2.000176 in. Ans. Ans: n = 0.330, h¿ = 2.000176 in. 171 © 2014 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. 3–34. A shear spring is made from two blocks of rubber, each having a height h, width b, and thickness a. The blocks are bonded to three plates as shown. If the plates are rigid and the shear modulus of the rubber is G, determine the displacement of plate A if a vertical load P is applied to this plate. Assume that the displacement is small so that d = a tan g L ag. P d A h Average Shear Stress: The rubber block is subjected to a shear force of V = P . 2 a a P t = V P 2 = = A bh 2bh Shear Strain: Applying Hooke’s law for shear P g = t P 2bh = = G G 2bhG Thus, d = ag = = Pa 2bhG Ans. Ans: d = 172 Pa 2bhG © 2014 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. s (ksi) 3–35. The elastic portion of the tension stress–strain diagram for an aluminum alloy is shown in the figure. The specimen used for the test has a gauge length of 2 in. and a diameter of 0.5 in. When the applied load is 9 kip, the new diameter of the specimen is 0.49935 in. Compute the shear modulus Gal for the aluminum. 70 0.00614 From the stress–strain diagram, P (in./in.) s 70 = = 11400.65 ksi P 0.00614 Eal = When specimen is loaded with a 9 - kip load, s = P = A 0.49935 - 0.5 d¿ - d = = - 0.0013 in.>in. d 0.5 V = - Gal = 9 = 45.84 ksi (0.5)2 45.84 s = = 0.0040205 in.>in. E 11400.65 Plong = Plat = p 4 Plat - 0.0013 = 0.32334 = Plong 0.0040205 11.4(103) Eat = = 4.31(103) ksi 2(1 + v) 2(1 + 0.32334) Ans. Ans: Gal = 4.31(103) ksi 173 © 2014 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. s (ksi) *3–36. The elastic portion of the tension stress–strain diagram for an aluminum alloy is shown in the figure. The specimen used for the test has a gauge length of 2 in. and a diameter of 0.5 in. If the applied load is 10 kip, determine the new diameter of the specimen. The shear modulus is Gal = 3.811032 ksi. P s = = A 70 10 = 50.9296 ksi p 2 (0.5) 4 0.00614 From the stress–strain diagram E = 70 = 11400.65 ksi 0.00614 Plong = G = 50.9296 s = = 0.0044673 in.>in. E 11400.65 E ; 2(1 + v) 3.8(103) = 11400.65 ; 2(1 + v) v = 0.500 Plat = - vPlong = - 0.500(0.0044673) = - 0.002234 in.>in. ¢d = Plat d = - 0.002234(0.5) = - 0.001117 in. d¿ = d + ¢d = 0.5 - 0.001117 = 0.4989 in. Ans. 174 P (in./in.) © 2014 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. 3–37. The rigid beam rests in the horizontal position on two 2014-T6 aluminum cylinders having the unloaded lengths shown. If each cylinder has a diameter of 30 mm. determine the placement x of the applied 80-kN load so that the beam remains horizontal. What is the new diameter of cylinder A after the load is applied? nal = 0.35. 80 kN x A B 220 mm 210 mm 3m a +©MA = 0; FB(3) - 80(x) = 0; a +©MB = 0; - FA(3) + 80(3 - x) = 0; FB = 80x 3 FA = (1) 80(3 - x) 3 (2) Since the beam is held horizontally, dA = dB s = P P ; A s A = E E P = d = PL = a P A E bL = 80(3 - x) 3 dA = dB; PL AE (220) AE = 80x 3 (210) AE 80(3 - x)(220) = 80x(210) x = 1.53 m Ans. From Eq. (2), FA = 39.07 kN sA = 39.07(103) FA = = 55.27 MPa p A (0.032) 4 Plong = sA E 55.27(106) = - 73.1(109) = - 0.000756 Plat = - nPlong = - 0.35(- 0.000756) = 0.0002646 dA¿ = dA + d Plat = 30 + 30(0.0002646) = 30.008 mm Ans. Ans: x = 1.53 m, dA ¿ = 30.008 mm 175 © 2014 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. 3–38. The wires each have a diameter of 12 in., length of 2 ft, and are made from 304 stainless steel. If P = 6 kip, determine the angle of tilt of the rigid beam AB. D C 2 ft P 2 ft A 1 ft B Equations of Equilibrium: Referring to the free-body diagram of beam AB shown in Fig. a, a +©MA = 0; FBC(3) - 6(2) = 0 FBC = 4 kip + c ©MB = 0; 6(1) - FAD(3) = 0 FAD = 2 kip Normal Stress and Strain: sBC = sAD = 4(103) FBC = = 20.37 ksi ABC p 1 2 a b 4 2 2(103) FAD = = 10.19 ksi AAD p 1 2 a b 4 2 Since sBC 6 sY and sA 6 sY, Hooke’s Law can be applied. sBC = EPBC; 20.37 = 28.0(103)PBC PBC = 0.7276(10-3) in.>in. sAD = EPAD; 10.19 = 28.0(103)PAD PAD = 0.3638(10-3) in.>in. Thus, the elongation of cables BC and AD are given by dBC = PBCLBC = 0.7276(10-3)(24) = 0.017462 in. dAD = PADLAD = 0.3638(10-3)(24) = 0.008731 in. Referring to the geometry shown in Fig. b and using small angle analysis, u = dBC - dAD 0.017462 - 0.008731 180° = = 0.2425(10-3) rad a b = 0.0139° 36 36 prad Ans. Ans: u = 0.0139° 176 © 2014 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. 3–39. The wires each have a diameter of 12 in., length of 2 ft, and are made from 304 stainless steel. Determine the magnitude of force P so that the rigid beam tilts 0.015°. D C 2 ft P 2 ft A 1 ft B Equations of Equilibrium: Referring to the free-body diagram of beam AB shown in Fig. a, a +©MA = 0; FBC(3) - P(2) = 0 FBC = 0.6667P + c ©MB = 0; P(1) - FAD(3) = 0 FAD = 0.3333P Normal Stress and Strain: sBC = sAD = FBC 0.6667P = = 3.3953P ABC p 1 2 a b 4 2 FAD 0.3333P = = 1.6977P AAD p 1 2 a b 4 2 Assuming that sBC 6 sY and sAD 6 sY and applying Hooke’s Law, sBC = EPBC; 3.3953P = 28.0(106)PBC PBC = 0.12126(10-6)P sAD = EPAD; 1.6977P = 28.0(106)PAD PAD = 60.6305(10-9)P Thus, the elongation of cables BC and AD are given by dBC = PBCLBC = 0.12126(10-6)P(24) = 2.9103(10-6)P dAD = PADLAD = 60.6305(10-9)P(24) = 1.4551(10-6)P Here, the angle of the tile is u = 0.015° a prad b = 0.2618(10-3) rad. Using small 180° angle analysis, u = dBC - dAD ; 36 0.2618(10-3) = 2.9103(10-6)P - 1.4551(10-6)P 36 P = 6476.93 lb = 6.48 kip Ans. Since sBC = 3.3953(6476.93) = 21.99 ksi 6 sY and sAD = 1.6977(6476.93) = 11.00 ksi 6 sY, the assumption is correct. Ans: P = 6.48 kip 177 © 2014 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. *3–40. The head H is connected to the cylinder of a compressor using six steel bolts. If the clamping force in each bolt is 800 lb, determine the normal strain in the 3 bolts. Each bolt has a diameter of 16 in. If sY = 40 ksi and 3 Est = 29110 2 ksi, what is the strain in each bolt when the nut is unscrewed so that the clamping force is released? C L H Normal Stress: s = P = A 800 A B p 3 2 4 16 = 28.97 ksi 6 sg = 40 ksi Normal Strain: Since s 6 sg, Hooke’s law is still valid. P = s 28.97 = 0.000999 in.>in. = E 29(103) Ans. If the nut is unscrewed, the load is zero. Therefore, the strain P = 0 178 Ans. © 2014 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. s (ksi) 3–41. The stress–strain diagram for polyethylene, which is used to sheath coaxial cables, is determined from testing a specimen that has a gauge length of 10 in. If a load P on the specimen develops a strain of P = 0.024 in.>in., determine the approximate length of the specimen, measured between the gauge points, when the load is removed. Assume the specimen recovers elastically. P 5 4 3 2 1 0 P 0 0.008 0.016 0.024 0.032 0.040 0.048 P (in./in.) Modulus of Elasticity: From the stress–strain diagram, s = 2 ksi when P = 0.004 in.>in. E = 2 - 0 = 0.500(103) ksi 0.004 - 0 Elastic Recovery: From the stress–strain diagram, s = 3.70 ksi when P = 0.024 in.>in. Elastic recovery = s 3.70 = 0.00740 in.>in. = E 0.500(103) Permanent Set: Permanent set = 0.024 - 0.00740 = 0.0166 in.>in. Thus, Permanent elongation = 0.0166(10) = 0.166 in. L = L0 + permanent elongation = 10 + 0.166 = 10.17 in. Ans. Ans: L = 10.17 in. 179 © 2014 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. 3–42. The pipe with two rigid caps attached to its ends is subjected to an axial force P. If the pipe is made from a material having a modulus of elasticity E and Poisson’s ratio n, determine the change in volume of the material. ri ro L P a Section a – a a P Normal Stress: The rod is subjected to uniaxial loading. Thus, slong = P and slat = 0. A dV = AdL + 2prLdr = APlong L + 2prLPlatr Using Poisson’s ratio and noting that AL = pr2L = V, dV = PlongV - 2nPlongV = Plong (1 - 2n)V slong = E (1 - 2n)V Since slong = P>A, dV = = P (1 - 2n)AL AE PL (1 - 2n) E Ans. Ans: dV = 180 PL (1 - 2n) E