Chapter 01

01 Solutions 46060 5/6/10 2:43 PM Page © 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 lb>ft and segment CD weighs 250 lb>ft In (b), the column has a mass of 200 kg>m (a) + c ©Fy = 0; kip kN B 200 mm 200 mm kN kN FA - 1.0 - - - 1.8 - = 10 ft FA = 13.8 kip in Ans 3m in 200 mm (b) + c ©Fy = 0; kip FA - 4.5 - 4.5 - 5.89 - - - = FA = 34.9 kN kip 200 mm 4.5 kN 4.5 kN C Ans ft A A 1m ft D (a) 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; A 250 Nиm 300 mm TC - 250 = TC = 250 N # m ©Mx = 0; (b) 400 Nиm 200 mm Ans TD = 150 Nиm C 150 mm Ans 200 mm B D 250 mm 150 mm 1–3 Determine the resultant internal torque acting on the cross sections through points B and C A ©Mx = 0; TB = 150 lb # ft ©Mx = 0; 600 lbиft B 350 lbиft TB + 350 - 500 = Ans TC - 500 = TC = ft C 500 lbиft ft 500 lb # ft Ans ft ft 01 Solutions 46060 5/6/10 2:43 PM Page © 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 30Њ 0.1 m 80 N Equations of Equilibrium: + Q©Fx¿ = 0; NA - 80 cos 15° = NA = 77.3 N a+ ©Fy¿ = 0; Ans VA - 80 sin 15° = VA = 20.7 N a+ ©MA = 0; Ans MA + 80 cos 45°(0.3 cos 30°) - 80 sin 45°(0.1 + 0.3 sin 30°) = MA = - 0.555 N # m Ans or a+ ©MA = 0; MA + 80 sin 15°(0.3 + 0.1 sin 30°) -80 cos 15°(0.1 cos 30°) = MA = - 0.555 N # m Ans Negative sign indicates that MA acts in the opposite direction to that shown on FBD 45Њ 01 Solutions 46060 5/6/10 2:43 PM Page © 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 kip 1.5 kip/ ft A D ft Support Reactions: For member AB a + ©MB = 0; + : ©Fx = 0; + c ©Fy = 0; 9.00(4) - A y(12) = A y = 3.00 kip Bx = By + 3.00 - 9.00 = By = 6.00 kip Equations of Equilibrium: For point D + : ©Fx = 0; + c ©Fy = 0; ND = Ans 3.00 - 2.25 - VD = VD = 0.750 kip a + ©MD = 0; Ans MD + 2.25(2) - 3.00(6) = MD = 13.5 kip # ft Ans Equations of Equilibrium: For point E + : ©Fx = 0; + c ©Fy = 0; NE = Ans - 6.00 - - VE = VE = - 9.00 kip a + ©ME = 0; Ans ME + 6.00(4) = ME = - 24.0 kip # ft Ans Negative signs indicate that ME and VE act in the opposite direction to that shown on FBD E B ft ft ft C 01 Solutions 46060 5/6/10 2:43 PM Page © 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 1–6 Determine the normal force, shear force, and moment at a section through point C Take P = kN B 0.1 m Support Reactions: a + ©MA = 0; 8(2.25) - T(0.6) = 0.5 m C T = 30.0 kN + : ©Fx = 0; 30.0 - A x = A x = 30.0 kN + c ©Fy = 0; Ay - = A y = 8.00 kN 0.75 m 0.75 m A 0.75 m P Equations of Equilibrium: For point C + : ©Fx = 0; - NC - 30.0 = NC = - 30.0 kN + c ©Fy = 0; Ans VC + 8.00 = VC = - 8.00 kN a + ©MC = 0; Ans 8.00(0.75) - MC = MC = 6.00 kN # m 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 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 Support Reactions: a + ©MA = 0; 0.75 m P(2.25) - 2(0.6) = P P = 0.5333 kN = 0.533 kN + : ©Fx = 0; - Ax = + c ©Fy = 0; A y - 0.5333 = Ans A x = 2.00 kN A y = 0.5333 kN Equations of Equilibrium: For point C + : ©Fx = 0; - NC - 2.00 = NC = - 2.00 kN + c ©Fy = 0; Ans VC + 0.5333 = VC = - 0.533 kN a + ©MC = 0; Ans 0.5333(0.75) - MC = MC = 0.400 kN # m Ans Negative signs indicate that NC and VC act in the opposite direction to that shown on FBD 0.75 m A 0.75 m 01 Solutions 46060 5/6/10 2:43 PM Page © 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 kN kN/m Referring to the FBD of the entire beam, Fig a, a + ©MB = 0; - A y(4) + 6(3.5) + (3)(3)(2) = + c ©Fy = 0; a + ©MC = 0; C A y = 7.50 kN NC = D 1.5 m 0.5 m 0.5 m Referring to the FBD of this segment, Fig b, + : ©Fx = 0; B A 1.5 m Ans 7.50 - - VC = VC = 1.50 kN MC + 6(0.5) - 7.5(1) = Ans MC = 4.50 kN # m 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 kN kN/m Referring to the FBD of the entire beam, Fig a, B A a + ©MA = 0; By(4) - 6(0.5) - (3)(3)(2) = By = 3.00 kN 0.5 m 0.5 m Referring to the FBD of this segment, Fig b, + : ©Fx = 0; + c ©Fy = 0; ND = VD - (1.5)(1.5) + 3.00 = a + ©MD = 0; 3.00(1.5) - C Ans VD = - 1.875 kN Ans (1.5)(1.5)(0.5) - MD = MD = 3.9375 kN # m = 3.94 kN # m Ans D 1.5 m 1.5 m 01 Solutions 46060 5/6/10 2:43 PM Page © 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 D ft F A B ft ft ft C 300 lb ft E Equations of Equilibrium: For point A + ; © Fx = 0; + c © Fy = 0; NA = Ans VA - 150 - 300 = VA = 450 lb a + ©MA = 0; Ans - MA - 150(1.5) - 300(3) = 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 = + c © Fy = 0; VB - 550 - 300 = Ans VB = 850 lb a + © MB = 0; Ans - MB - 550(5.5) - 300(11) = 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 = Ans - NC - 250 - 650 - 300 = NC = - 1200 lb = - 1.20 kip a + ©MC = 0; Ans - MC - 650(6.5) - 300(13) = 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 01 Solutions 46060 5/6/10 2:43 PM Page © 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 1–11 The force F = 80 lb 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 a F ϭ 80 lb 30Њ Equations of Equilibrium: For section a–a + Q©Fx¿ = 0; VA - 80 cos 15° = 0.23 in VA = 77.3 lb a+ ©Fy¿ = 0; Ans A NA - 80 sin 15° = 0.16 in NA = 20.7 lb a + ©MA = 0; Ans - MA - 80 sin 15°(0.16) + 80 cos 15°(0.23) = MA = 14.5 lb # in 45Њ Ans *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 a 0.2 m 0.2 m B 0.2 m 0.2 m D E Support Reactions: + c ©Fy = 0; NB - 18 = d+ ©MC = 0; 18(0.7) - 18.0(0.2) - NA(0.1) = 0.2 m 0.3 m NB = 18.0 kN A NA = 90.0 kN + : ©Fx = 0; NC - 90.0 = 0.3 m NC = 90.0 kN Equations of Equilibrium: For point D + : © Fx = 0; 18 kN VD - 90.0 = VD = 90.0 kN + c © Fy = 0; Ans ND - 18 = ND = 18.0 kN d+ © MD = 0; Ans MD + 18(0.3) - 90.0(0.3) = MD = 21.6 kN # m Ans Equations of Equilibrium: For point E + : © Fx = 0; 90.0 - VE = VE = 90.0 kN + c © Fy = 0; d + © ME = 0; Ans NE = Ans 90.0(0.2) - ME = ME = 18.0 kN # m Ans C 01 Solutions 46060 5/6/10 2:43 PM Page © 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 M 1.5 ft A D ft + : ©Fx = 0; ft ft ft 0.25 ft Ans VB - 0.8 - 0.16 = VB = 0.960 kip a + ©MB = 0; B - NB - 0.4 = NB = - 0.4 kip + c ©Fy = 0; ft C Ans - MB - 0.16(2) - 0.8(4.25) + 0.4(1.5) = MB = - 3.12 kip # ft Ans 1–14 Determine the resultant internal loadings acting on the cross section through points C and D of the beam in Prob 1–13 M 1.5 ft A D ft ft C B ft ft ft For point C: + ; ©Fx = 0; + c ©Fy = 0; a + ©MC = 0; NC + 0.4 = 0; NC = - 0.4kip VC - 0.8 - 0.04 (7) = 0; Ans VC = 1.08 kip Ans - MC - 0.8(7.25) - 0.04(7)(3.5) + 0.4(1.5) = MC = - 6.18 kip # ft Ans ND = Ans For point D: + ; ©Fx = 0; + c ©Fy = 0; a + ©MD = 0; VD - 0.09 - 0.04(14) - 0.8 = 0; VD = 1.45 kip Ans - MD - 0.09(4) - 0.04(14)(7) - 0.8(14.25) = MD = - 15.7 kip # ft Ans 0.25 ft 01 Solutions 46060 5/6/10 2:43 PM Page © 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 1–15 Determine the resultant internal loading on the cross section through point C of the pliers There is a pin at A, and the jaws at B are smooth 20 N 40 mm 120 mm 15 mm C + c ©Fy = 0; + : ©Fx = 0; +d ©MC = 0; - VC + 60 = 0; VC = 60 N Ans NC = - MC + 60(0.015) = 0; MC = 0.9 N.m D Ans 80 mm 20 N *1–16 Determine the resultant internal loading on the cross section through point D of the pliers B A Ans 30Њ 20 N 40 mm 120 mm 15 mm R+ ©Fy = 0; VD - 20 cos 30° = 0; VD = 17.3 N Ans +b©Fx = 0; ND - 20 sin 30° = 0; ND = 10 N Ans +d ©MD = 0; MD - 20(0.08) = 0; MD = 1.60 N.m Ans C A D 80 mm 20 N 30Њ B 01 Solutions 46060 5/6/10 2:43 PM Page 10 © 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 •1–17 Determine resultant internal loadings acting on section a–a and section b–b Each section passes through the centerline at point C kN B b a Referring to the FBD of the entire beam, Fig a, a + ©MA = 0; NB sin 45°(6) - 5(4.5) = 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° = Na - a = - 3.75 kN Va - a = 1.25 kN Ans +a ©Fy¿ = 0; Va - a + 5.303 sin 45° - = a + ©MC = 0; 5.303 sin 45°(3) - 5(1.5) - Ma - a = Ma - a = 3.75 kN # m Ans Ans Referring to the FBD (section b–b) in Fig c, + ; ©Fx = 0; Nb - b - cos 45° + 5.303 = Nb - b = - 1.768 kN = - 1.77 kN + c ©Fy = 0; a + ©MC = 0; Vb - b - sin 45° = Vb - b = 3.536 kN = 3.54 kN Ans Ans 5.303 sin 45° (3) - 5(1.5) - Mb - b = Mb - b = 3.75 kN # m 10 Ans A 45Њ 3m 45Њ 1.5 m a 01 Solutions 46060 5/6/10 2:43 PM Page 58 © 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 1–86 The boom is supported by the winch cable that has an allowable normal stress of sallow = 24 ksi If it is required that it be able to slowly lift 5000 lb, from u = 20° to u = 50°, determine the smallest diameter of the cable to the nearest 16 in The boom AB has a length of 20 ft Neglect the size of the winch Set d = 12 ft B u 20 ft f A Maximum tension in cable occurs when u = 20° sin c sin 20° = 20 12 d c = 11.842° + : © Fx = 0; - T cos 20° + FAB cos 31.842° = + c ©Fy = 0; FAB sin 31.842° - T sin 20° - 5000 = T = 20 698.3 lb FAB = 22 896 lb P ; s = A Use 20 698.3 p (d) d = 1.048 in 24(103) = d = 1 in 16 Ans 1–87 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 58 01 Solutions 46060 5/6/10 2:43 PM Page 59 © 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 *1–88 The frame is subjected to the load of 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 kN 1m E 1.5 m C 45Њ D 1.5 m Referring to the FBD of member DCE, Fig a, a + ©ME = 0; Dy(2.5) - FBC sin 45° (1) = (1) + : ©Fx = FBC cos 45° - Dx = (2) B 1.5 m Referring to the FBD of member ABD, Fig b, a + ©MA = 0; cos 45° (3) + FBC sin 45° (1.5) - Dx (3) = (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 = Dx + Dy = 5.6572 + 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 = = 4.00 kN VC = VD = FD = 6.093 kN 2 For pin C, tallow = VC ; AC 40(106) = 4.00(103) p dC dC = 0.01128 m = 11.28 mm Use dC = 12 mm For pin D, VD ; tallow = AD 40(106) = Ans 6.093(103) p dD dD = 0.01393 m = 13.93 mm Use dD = 14 mm Ans 59 A 01 Solutions 46060 5/6/10 2:43 PM Page 60 © 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 •1–89 The eye bolt is used to support the load of kip Determine its diameter d to the nearest 18 in and the required thickness h to the nearest 18 in of the support so that the washer will not penetrate or shear through it The allowable normal stress for the bolt is sallow = 21 ksi and the allowable shear stress for the supporting material is tallow = ksi in h d kip Allowable Normal Stress: Design of bolt size 5(103) P sallow = ; 21.0(103) = p Ab d d = 0.5506 in Use d = in Ans Allowable Shear Stress: Design of support thickness 5(103) V ; 5(103) = tallow = A p(1)(h) Use h = in Ans 60 01 Solutions 46060 5/6/10 2:43 PM Page 61 © 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 1–90 The soft-ride suspension system of the mountain bike is pinned at C and supported by the shock absorber BD If it is designed to support a load P = 1500 N, determine the required minimum diameter of pins B and C Use a factor of safety of against failure The pins are made of material having a failure shear stress of tfail = 150 MPa, and each pin is subjected to double shear P A Internal Loadings: The forces acting on pins B and C can be determined by considering the equilibrium of the free-body diagram of the soft-ride suspension system shown in Fig a a + ©MC = 0; FBD = 5905.36 N + : ©Fx = 0; Cx - 5905.36 cos 60° = + c ©Fy = 0; 5905.36 sin 60° - 1500 - Cy = Cy = 3614.20 N Cx = 2952.68 N Thus, FC = Cx + Cy = 2952.682 + 3614.202 = 4666.98 N Since both pins are in double shear, FB 5905.36 VB = = = 2952.68 N 2 VC = FC 4666.98 = = 2333.49 N 2 Allowable Shear Stress: tfail 150 tallow = = = 75 MPa F.S Using this result, VB tallow = ; AB 75(106) = 2952.68 p d B dB = 0.007080 m = 7.08 mm tallow = VC ; AC 75(106) = Ans 2333.49 p d C dC = 0.006294 m = 6.29 mm Ans 61 30 mm B 60Њ 1500(0.4) - FBD sin 60°(0.1) - FBD cos 60°(0.03) = FB = FBD = 5905.36 N 100 mm 300 mm C D 01 Solutions 46060 5/6/10 2:43 PM Page 62 © 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 1–91 The soft-ride suspension system of the mountain bike is pinned at C and supported by the shock absorber BD If it is designed to support a load of P = 1500 N, determine the factor of safety of pins B and C against failure if they are made of a material having a shear failure stress of tfail = 150 MPa Pin B has a diameter of 7.5 mm, and pin C has a diameter of 6.5 mm Both pins are subjected to double shear P A Internal Loadings: The forces acting on pins B and C can be determined by considerning the equilibrium of the free-body diagram of the soft-ride suspension system shown in Fig a + ©MC = 0; 1500(0.4) - FBD sin 60°(0.1) - FBD cos 60°(0.03) = FBD = 5905.36 N + : ©Fx = 0; Cx - 5905.36 cos 60° = + c ©Fy = 0; 5905.36 sin 60° - 1500 - Cy = Cy = 3614.20 N Cx = 2952.68 N Thus, FB = FBD = 5905.36 N FC = Cx + Cy = 2952.682 + 3614.202 = 4666.98 N Since both pins are in double shear, VB = FB 5905.36 = = 2952.68N 2 VC = FC 4666.98 = = 2333.49 N 2 Allowable Shear Stress: The areas of the shear plane for pins B and C are p p A B = (0.00752) = 44.179(10 - 6)m2 and A C = (0.00652) = 33.183(10 - 6)m2 4 We obtain A tavg B B = VB 2952.68 = 66.84 MPa = AB 44.179(10 - 6) A tavg B C = VC 2333.49 = 70.32 MPa = AC 33.183(10 - 6) Using these results, tfail = (F.S.)B = A tavg B B tfail (F.S.)C = = A tavg B C 100 mm 300 mm 150 = 2.24 66.84 Ans 150 = 2.13 70.32 Ans 62 30 mm B 60Њ C D 01 Solutions 46060 5/6/10 2:43 PM Page 63 © 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 *1–92 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 kN 1.5 kN 1.5 m 1.5 m 1.5 m kN 2m 2m 1.5 m C D A B From FBD (a): a + ©MD = 0; FB(4.5) + 1.5(3) + 2(1.5) - FC(6) = 4.5 FB - FC = - 7.5 (1) From FBD (b): a + ©MD = 0; FB(5.5) - FC(4) - 3(2) = 5.5 FB - FC = (2) Solving Eqs (1) and (2) yields FB = 4.40 kN; FC = 4.55 kN For bolt: sallow = 150(106) = 4.40(103) p (dB) dB = 0.00611 m = 6.11 mm Ans For washer: sallow = 28 (104) = 4.40(103) p (d w - 0.006112) dw = 0.0154 m = 15.4 mm Ans •1–93 The assembly is used to support the distributed loading of w = 500 lb>ft Determine the factor of safety with respect to yielding for the steel rod BC and the pins at B and C if the yield stress for the steel in tension is sy = 36 ksi and in shear ty = 18 ksi The rod has a diameter of 0.40 in., and the pins each have a diameter of 0.30 in C ft For rod BC: s = A 1.667 P = 13.26 ksi = p A (0.4 ) F S = sy s = 36 = 2.71 13.26 Ans ft For pins B and C: w 0.8333 V = 11.79 ksi = p t = A (0.3 ) F S = ty t = B ft 18 = 1.53 11.79 Ans 63 01 Solutions 46060 5/6/10 2:43 PM Page 64 © 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 1–94 If the allowable shear stress for each of the 0.30in.-diameter steel pins at A, B, and C is tallow = 12.5 ksi, and the allowable normal stress for the 0.40-in.-diameter rod is sallow = 22 ksi, determine the largest intensity w of the uniform distributed load that can be suspended from the beam C ft Assume failure of pins B and C: tallow = 12.5 = 1.667w A p (0.3 ) w = 0.530 kip>ft Ans (controls) B Assume failure of pins A: ft FA = (2w)2 + (1.333w)2 = 2.404 w tallow w 1.202w = 12.5 = p (0.3 ) ft w = 0.735 kip>ft Assume failure of rod BC: sallow = 22 = 3.333w p (0.4 ) w = 0.829 kip>ft 1–95 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) = NB = 135 kN a + ©MB = 0; 40(1.5)(3.75) - 100(1.5) - NA(3) = NA = 25.0 kN For plate A¿ , NA (sb)allow = ; A A¿ 1.5(106) = 25.0(103) a2A¿ aA¿ = 0.1291 m = 130 mm Ans For plate B¿ , sallow = NB ; A B¿ 1.5(106) = 135(103) a2B¿ aB¿ = 0.300 m = 300 mm Ans 64 P A¿ B¿ 3m B 1.5 m 01 Solutions 46060 5/6/10 2:43 PM Page 65 © 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 40 kN/m *1–96 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 P A¿ B¿ 3m 1.5 m B 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) = NB = 1.5P - 15 a + ©MB = 0; 40(1.5)(3.75) - P(1.5) - NA(3) = NA = 75 - 0.5P For plate A¿ , NA (sb)allow = ; A A¿ 1.5(106) = (75 - 0.5P)(103) 0.15(0.15) P = 82.5 kN For plate B¿ , NB ; (sb)allow = A B¿ 1.5(106) = (1.5P - 15)(103) 0.25(0.25) P = 72.5 kN (Controls!) Ans •1–97 The rods AB and CD are made of steel having a failure tensile stress of sfail = 510 MPa Using a factor of safety of F.S = 1.75 for tension, determine their smallest diameter so that they can support the load shown The beam is assumed to be pin connected at A and C B D kN kN kN Support Reactions: a + ©MA = 0; FCD(10) - 5(7) - 6(4) - 4(2) = A FCD = 6.70 kN a + ©MC = 0; 4(8) + 6(6) + 5(3) - FAB(10) = FAB = 8.30 kN Allowable Normal Stress: Design of rod sizes For rod AB sallow = sfail FAB = ; F.S A AB 510(106) 8.30(103) = p 1.75 d AB dAB = 0.006022 m = 6.02 mm Ans For rod CD sallow = FCD sfail = ; F.S A CD C 2m 510(106) 6.70(103) = p 1.75 d CD dCD = 0.005410 m = 5.41 mm Ans 65 2m 3m 3m 01 Solutions 46060 5/6/10 2:43 PM Page 66 © 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 1–98 The aluminum bracket A is used to support the centrally applied load of 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 Equation of Equilibrium: + c ©Fy = 0; V - = 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) kip h = 1.74 in Ans 1–99 The hanger is supported using the rectangular pin Determine the magnitude of the allowable suspended load P if the allowable bearing stress is (sb)allow = 220 MPa, the allowable tensile stress is (st)allow = 150 MPa, and the allowable shear stress is tallow = 130 MPa Take t = mm, a = mm, and b = 25 mm 20 mm 75 mm 10 mm a a b Allowable Normal Stress: For the hanger (st)allow = P ; A 150 A 106 B = P (0.075)(0.006) P P = 67.5 kN 37.5 mm Allowable Shear Stress: The pin is subjected to double shear Therefore, V = tallow = 130 A 106 B = V ; A P P>2 (0.01)(0.025) P = 65.0 kN Allowable Bearing Stress: For the bearing area (sb)allow = P ; A 220 A 106 B = 37.5 mm t P>2 (0.005)(0.025) P = 55.0 kN (Controls!) Ans 66 h 01 Solutions 46060 5/6/10 2:43 PM Page 67 © 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 *1–100 The hanger is supported using the rectangular pin Determine the required thickness t of the hanger, and dimensions a and b if the suspended load is P = 60 kN The allowable tensile stress is (st)allow = 150 MPa, the allowable bearing stress is (sb)allow = 290 MPa, and the allowable shear stress is tallow = 125 MPa 20 mm 75 mm 10 mm a a 37.5 mm t P 37.5 mm Allowable Normal Stress: For the hanger (st)allow = P ; A 150 A 106 B = 60(103) (0.075)t t = 0.005333 m = 5.33 mm Ans Allowable Shear Stress: For the pin tallow = 125 A 106 B = V ; A 30(103) (0.01)b b = 0.0240 m = 24.0 mm Ans Allowable Bearing Stress: For the bearing area (sb)allow = P ; A 290 A 106 B = 30(103) (0.0240) a a = 0.00431 m = 4.31 mm Ans 67 b 01 Solutions 46060 5/6/10 2:43 PM Page 68 © 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 •1–101 The 200-mm-diameter aluminum cylinder supports a compressive load of 300 kN Determine the average normal and shear stress acting on section a–a Show the results on a differential element located on the section 300 kN a Referring to the FBD of the upper segment of the cylinder sectional through a–a shown in Fig a, +Q©Fx¿ = 0; Na - a - 300 cos 30° = +a©Fy¿ = 0; Va - a - 300 sin 30° = 30Њ Na - a = 259.81 kN a Va - a = 150 kN 0.1 Section a–a of the cylinder is an ellipse with a = 0.1 m and b = m Thus, cos 30° 0.1 b = 0.03628 m2 A a - a = pab = p(0.1)a cos 30° A sa - a B avg = 259.81(103) Na - a = = 7.162(106) Pa = 7.16 MPa Aa - a 0.03628 Ans A ta - a B avg = 150(103) Va - a = = 4.135(106) Pa = 4.13 MPa Aa - a 0.03628 Ans d The differential element representing the state of stress of a point on section a–a is shown in Fig b 1–102 The long bolt passes through the 30-mm-thick plate If the force in the bolt shank is 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 mm a mm b kN 18 mm b a P = ss = A (103) = 208 MPa Ans (tavg)a = (103) V = = 4.72 MPa A p (0.018)(0.030) Ans (tavg)b = (103) V = = 45.5 MPa A p (0.007)(0.008) Ans p (0.007)2 68 30 mm 01 Solutions 46060 5/6/10 2:43 PM Page 69 © 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 1–103 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 tallow = 10 ksi C 1.5 in Referring to the FBD of member AB, Fig a, a + ©MA = 0; 2(8)(4) - FBC sin 60° (8) = + : ©Fx = 0; 9.238 cos 60° - A x = + c ©Fy = 0; 9.238 sin 60° - 2(8) + A y = FBC = 9.238 kip 60Њ B A x = 4.619 kip ft A y = 8.00 kip kip/ft Thus, the force acting on pin A is FA = A 2x + A 2y = 4.6192 + 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 For member BC FBC ; sallow = A BC 29 = 9.238 1.5(t) t = 0.2124 in Use t = For pin A, VA ; tallow = AA 10 = 9.238 p dA in Ans dA = 1.085 in Use dA = in For pin B, VB ; tallow = AB 10 = 4.619 p dB Ans dB = 0.7669 in Use dB = 13 in 16 Ans 69 A 01 Solutions 46060 5/6/10 2:43 PM Page 70 © 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 *1–104 Determine the resultant internal loadings acting on the cross sections located through points D and E of the frame 150 lb/ft Segment AD: 1.5 ft + : ©Fx = 0; ND - 1.2 = 0; + T ©Fy = 0; VD + 0.225 + 0.4 = 0; a + ©MD = 0; ND = 1.20 kip Ans VD = - 0.625 kip B E Ans 2.5 ft MD + 0.225(0.75) + 0.4(1.5) = MD = A ft D C - 0.769 kip # ft Ans ft ft Segment CE: Q+ ©Fx = 0; NE + 2.0 = 0; R+ ©Fy = 0; VE = a + ©ME = 0; NE = - 2.00 kip Ans Ans ME = Ans •1–105 The pulley is held fixed to the 20-mm-diameter shaft using a key that fits within a groove cut into the pulley and shaft If the suspended load has a mass of 50 kg, determine the average shear stress in the key along section a–a The key is mm by mm square and 12 mm long a + ©MO = 0; a 75 mm F (10) - 490.5 (75) = F = 3678.75 N tavg = a 3678.75 V = = 61.3 MPa A (0.005)(0.012) Ans 70 01 Solutions 46060 5/6/10 2:43 PM Page 71 © 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 1–106 The bearing pad consists of a 150 mm by 150 mm block of aluminum that supports a compressive load of 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 kN a Equation of Equilibrium: +Q©Fx = 0; Va - a - cos 60° = Va - a = 3.00 kN a+ ©Fy = 0; Na - a - sin 60° = Na - a = 5.196 kN 30Њ a 150 mm Averge Normal Stress And Shear Stress: The cross sectional Area at section a–a is 0.15 b (0.15) = 0.02598 m2 A = a 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 1–107 The yoke-and-rod connection is subjected to a tensile force of kN Determine the average normal stress in each rod and the average shear stress in the pin A between the members kN 40 mm For the 40 – mm – dia rod: s40 30 mm (103) P = p = = 3.98 MPa A (0.04) Ans For the 30 – mm – dia rod: kN s30 = (10 ) V = p = 7.07 MPa A (0.03) Ans Average shear stress for pin A: tavg = A 25 mm 2.5 (103) P = p = 5.09 MPa A (0.025) Ans 71 01 Solutions 46060 5/6/10 2:43 PM Page 72 © 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 *1–108 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 = T = B gAL2 8s Average Normal Stress: gAL2 gL2 T 8s s = = = A A 8s Ans 72 L/2 ... - 0.02 2) = 13.3 MPa (C) Ans At E: sE = 8(103) (0 .012 2) B A = 70.7 MPa (T) Ans 22 kN kN E C kN 01 Solutions 46060 5/6/10 2:43 PM Page 23 © 2010 Pearson Education, Inc., Upper Saddle River, NJ... (0.6)(0.1) = 0. 2014 in2 0.3 in 0.1 in 0.5 in P 175 lb = s = = 869 psi A 0. 2014 in2 Ans Flat-heeled shoes: A = (p)(1.2)2 + 2.4(0.5) = 3.462 in2 s = P 175 lb = = 50.5 psi A 3.462 in2 Ans 30 ft E 01 Solutions... ME and VE act in the opposite direction to that shown on FBD E B ft ft ft C 01 Solutions 46060 5/6/10 2:43 PM Page © 2010 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This