Hibbeler-Mechanics of Materials, Eighth Edition - Instructor's Solutions Manual.pdf-资料库

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01 Solutions 46060 5/6/10 2:43 PM Page 1 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 1–1. Determine the resultant internal normal force acting on the cross section through point A in each column. In (a), segment BC weighs 180 >ft and segment CD weighs 250 m. lb lb > kg >ft. In (b), the column has a mass of 200 + c ©Fy = 0; FA - 1.0 - 3 - 3 - 1.8 - 5 = 0 + c ©Fy = 0; FA - 4.5 - 4.5 - 5.89 - 6 - 6 - 8 = 0 FA = 13.8 kip (a) (b) FA = 34.9 kN 5 kip B 10 ft Ans. 8 in. 8 in. Ans. 3 kip 4 ft 4 ft 3 kip C A D 8 kN 200 mm 200 mm 6 kN 6 kN 3 m 200 mm 4.5 kN 200 mm 4.5 kN A 1 m (a) (b) 1–2. Determine the resultant internal torque acting on the cross sections through points C and D.The support bearings at A and B allow free turning of the shaft. ©Mx = 0; TC - 250 = 0 ©Mx = 0; TD = 0 TC = 250 N # m A 250 N⭈m 300 mm C 150 N⭈m 400 N⭈m Ans. Ans. 200 mm 150 mm 200 mm D B 250 mm 150 mm 1–3. Determine the resultant internal torque acting on the cross sections through points B and C. ©Mx = 0; TB + 350 - 500 = 0 TB = 150 lb # ft ©Mx = 0; TC - 500 = 0 TC = 500 lb # ft A 600 lb⭈ft B 350 lb⭈ft Ans. 3 ft Ans. 1 ft 2 ft C 500 lb⭈ft 2 ft 1

01 Solutions 46060 5/6/10 2:43 PM Page 2 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *1–4. A force of 80 N is supported by the bracket as shown. Determine the resultant internal loadings acting on the section through point A. 0.3 m A 30⬚ 0.1 m 80 N 45⬚ Equations of Equilibrium: +Q©Fx¿ = 0; NA - 80 cos 15° = 0 a+ ©Fy¿ = 0; VA - 80 sin 15° = 0 + ©MA = 0; MA + 80 cos 45°(0.3 cos 30°) a NA = 77.3 N VA = 20.7 N - 80 sin 45°(0.1 + 0.3 sin 30°) = 0 MA = -0.555 N # m or + ©MA = 0; MA + 80 sin 15°(0.3 + 0.1 sin 30°) a -80 cos 15°(0.1 cos 30°) = 0 MA = -0.555 N # m Ans. Ans. Ans. Ans. Negative sign indicates that MA acts in the opposite direction to that shown on FBD. 2

01 Solutions 46060 5/6/10 2:43 PM Page 3 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. •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 B E C 6 ft 6 ft 4 ft 4 ft Support Reactions: For member AB Equations of Equilibrium: For point D :+ + ©MB = 0; 9.00(4) - Ay(12) = 0 Ay = 3.00 kip :+ ©Fx = 0; Bx = 0 + c ©Fy = 0; By + 3.00 - 9.00 = 0 By = 6.00 kip ©Fx = 0; ND = 0 + c ©Fy = 0; 3.00 - 2.25 - VD = 0 + ©MD = 0; MD + 2.25(2) - 3.00(6) = 0 :+ ©Fx = 0; NE = 0 + c ©Fy = 0; -6.00 - 3 - VE = 0 + ©ME = 0; ME + 6.00(4) = 0 ME = -24.0 kip # ft MD = 13.5 kip # ft Equations of Equilibrium: For point E a a a VD = 0.750 kip VE = -9.00 kip Ans. Ans. Ans. Ans. Ans. Ans. Negative signs indicate that ME and VE act in the opposite direction to that shown on FBD. 3

01 Solutions 46060 5/6/10 2:43 PM Page 4 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. B A B A 1–6. Determine the normal force, shear force, and moment at a section through point C. Take P = 8 kN. a Support Reactions: Equations of Equilibrium: For point C + ©MA = 0; 8(2.25) - T(0.6) = 0 T = 30.0 kN :+ ©Fx = 0; 30.0 - Ax = 0 Ax = 30.0 kN + c ©Fy = 0; Ay - 8 = 0 Ay = 8.00 kN :+ ©Fx = 0; -NC - 30.0 = 0 + c ©Fy = 0; VC + 8.00 = 0 + ©MC = 0; 8.00(0.75) - MC = 0 VC = -8.00 kN MC = 6.00 kN # m NC = -30.0 kN a 0.1 m 0.5 m C 0.75 m 0.75 m 0.75 m P Ans. Ans. Ans. Negative signs indicate that NC and VC act in the opposite direction to that shown on FBD. 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. a Support Reactions: P = 0.5333 kN = 0.533 kN Equations of Equilibrium: For point C + ©MA = 0; P(2.25) - 2(0.6) = 0 :+ ©Fx = 0; 2 - Ax = 0 Ax = 2.00 kN + c ©Fy = 0; Ay - 0.5333 = 0 Ay = 0.5333 kN :+ ©Fx = 0; -NC - 2.00 = 0 + c ©Fy = 0; VC + 0.5333 = 0 VC = -0.533 kN + ©MC = 0; 0.5333(0.75) - MC = 0 MC = 0.400 kN # m NC = -2.00 kN a 0.1 m 0.5 m C 0.75 m 0.75 m 0.75 m P Ans. Ans. Ans. Ans. Negative signs indicate that NC and VC act in the opposite direction to that shown on FBD. 4

01 Solutions 46060 5/6/10 2:43 PM Page 5 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *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 2 a Referring to the FBD of this segment, Fig. b, Referring to the FBD of the entire beam, Fig. a, (3)(3)(2) = 0 Ay = 7.50 kN + ©MB = 0; -Ay(4) + 6(3.5) + 1 :+ ©Fx = 0; NC = 0 + c ©Fy = 0; 7.50 - 6 - VC = 0 VC = 1.50 kN + ©MC = 0; MC + 6(0.5) - 7.5(1) = 0 MC = 4.50 kN # m a A C D B 0.5 m 0.5 m 1.5 m 1.5 m Ans. Ans. Ans. •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 Referring to the FBD of the entire beam, Fig. a, a + ©MA = 0; By(4) - 6(0.5) - 1 :+ ©Fx = 0; ND = 0 + c ©Fy = 0; VD - 1 2 2 Referring to the FBD of this segment, Fig. b, (1.5)(1.5) + 3.00 = 0 VD = -1.875 kN a + ©MD = 0; 3.00(1.5) - 1 2 (1.5)(1.5)(0.5) - MD = 0 MD = 3.9375 kN # m = 3.94 kN # m Ans. (3)(3)(2) = 0 By = 3.00 kN A C D B 0.5 m 0.5 m 1.5 m 1.5 m Ans. Ans. 5

01 Solutions 46060 5/6/10 2:43 PM Page 6 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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. Equations of Equilibrium: For point A ;+ © Fx = 0; NA = 0 + c © Fy = 0; VA - 150 - 300 = 0 + ©MA = 0; -MA - 150(1.5) - 300(3) = 0 MA = -1125 lb # ft = -1.125 kip # ft VA = 450 lb a D B A F 2 ft 8 ft 3 ft 300 lb C E 5 ft 7 ft Ans. Ans. 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 + © MB = 0; -MB - 550(5.5) - 300(11) = 0 MB = -6325 lb # ft = -6.325 kip # ft VB = 850 lb a Ans. Ans. Ans. Negative sign indicates that MB acts in the opposite direction to that shown on FBD. Equations of Equilibrium: For point C ;+ © Fx = 0; VC = 0 + c © Fy = 0; -NC - 250 - 650 - 300 = 0 + ©MC = 0; -MC - 650(6.5) - 300(13) = 0 MC = -8125 lb # ft = -8.125 kip # ft NC = -1200 lb = -1.20 kip a Ans. Ans. Ans. Negative signs indicate that NC and MC act in the opposite direction to that shown on FBD. 6

01 Solutions 46060 5/6/10 2:43 PM Page 7 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. F = 80 lb 1–11. The force acts on the gear tooth. Determine the resultant internal loadings on the root of the tooth, i.e., at the centroid point A of section a–a. Equations of Equilibrium: For section a–a +Q©Fx¿ = 0; VA - 80 cos 15° = 0 a+ ©Fy¿ = 0; NA - 80 sin 15° = 0 + ©MA = 0; -MA - 80 sin 15°(0.16) + 80 cos 15°(0.23) = 0 NA = 20.7 lb VA = 77.3 lb a MA = 14.5 lb # in. *1–12. The sky hook is used to support the cable of a scaffold over the side of a building. If it consists of a smooth rod that contacts the parapet of a wall at points A, B, and C, determine the normal force, shear force, and moment on the cross section at points D and E. Support Reactions: F ⫽ 80 lb a 0.16 in. 30⬚ A 0.23 in. Ans. Ans. 45⬚ Ans. a 0.2 m 0.2 m B D A 0.2 m 0.2 m E C 0.2 m 0.3 m 0.3 m 18 kN NA = 90.0 kN :+ Equations of Equilibrium: For point D + c ©Fy = 0; NB - 18 = 0 NB = 18.0 kN d+ ©MC = 0; 18(0.7) - 18.0(0.2) - NA(0.1) = 0 ©Fx = 0; NC - 90.0 = 0 NC = 90.0 kN :+ © Fx = 0; VD - 90.0 = 0 + c © Fy = 0; ND - 18 = 0 d+ © MD = 0; MD + 18(0.3) - 90.0(0.3) = 0 MD = 21.6 kN # m :+ © Fx = 0; 90.0 - VE = 0 + c © Fy = 0; NE = 0 d+ © ME = 0; 90.0(0.2) - ME = 0 Equations of Equilibrium: For point E ND = 18.0 kN VD = 90.0 kN VE = 90.0 kN ME = 18.0 kN # m Ans. Ans. Ans. Ans. Ans. Ans. 7

01 Solutions 46060 5/6/10 2:43 PM Page 8 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. •1–13. The 800-lb load is being hoisted at a constant speed using the motor M, which has a weight of 90 lb. Determine the resultant internal loadings acting on the cross section through point B in the beam. The beam has a weight of 40 lb>ft and is fixed to the wall at A. :+ ©Fx = 0; - NB - 0.4 = 0 + c ©Fy = 0; VB - 0.8 - 0.16 = 0 NB = - 0.4 kip VB = 0.960 kip a + ©MB = 0; - MB - 0.16(2) - 0.8(4.25) + 0.4(1.5) = 0 MB = -3.12 kip # ft 1–14. Determine the resultant internal loadings acting on the cross section through points C and D of the beam in Prob. 1–13. For point C: ;+ ©Fx = 0; NC + 0.4 = 0; NC = - 0.4kip + c ©Fy = 0; VC - 0.8 - 0.04 (7) = 0; VC = 1.08 kip a + ©MC = 0; - MC - 0.8(7.25) - 0.04(7)(3.5) + 0.4(1.5) = 0 MC = -6.18 kip # ft For point D: ;+ ©Fx = 0; ND = 0 + c ©Fy = 0; VD - 0.09 - 0.04(14) - 0.8 = 0; VD = 1.45 kip a + ©MD = 0; - MD - 0.09(4) - 0.04(14)(7) - 0.8(14.25) = 0 MD = -15.7 kip # ft 1.5 ft 4 ft 0.25 ft 1.5 ft 4 ft 0.25 ft A A M D 4 ft C 3 ft B 3 ft 4 ft Ans. Ans. Ans. M D 4 ft C 3 ft B 3 ft 4 ft Ans. Ans. Ans. Ans. Ans. Ans. 8

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