FM_TOC 46060 6/22/10 11:26 AM Page iii CONTENTS To the Instructor iv Stress Strain 73 Mechanical Properties of Materials 92 Axial Load 122 Torsion 214 Bending 329 Transverse Shear 472 Combined Loadings 532 Stress Transformation 619 10 Strain Transformation 738 11 Design of Beams and Shafts 830 12 Deflection of Beams and Shafts 883 13 Buckling of Columns 1038 14 Energy Methods 1159 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 Nm 300 mm TC - 250 = TC = 250 N # m ©Mx = 0; (b) 400 Nm 200 mm Ans TD = 150 Nm 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 lbft B 350 lbft TB + 350 - 500 = Ans TC - 500 = TC = ft C 500 lbft 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 30 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 45 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 80 lb 30 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 45 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 30 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 30 B 03 Solutions 46060 5/7/10 8:45 AM Page 16 © 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 *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-mmlong 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 e = 0.12 120(103) = 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 - e)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 = eL = 0.12(200) = 24 mm Ans 16 P 03 Solutions 46060 5/7/10 8:45 AM Page 17 © 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 •3–21 The stress–strain diagram for a polyester resin is given in the figure If the rigid beam is supported by a strut AB and post CD, both made from this material, and subjected to a load of P = 80 kN, determine the angle of tilt of the beam when the load is applied The diameter of the strut is 40 mm and the diameter of the post is 80 mm B 2m P A C 0.75 m 0.75 m D 0.5 m From the stress–strain diagram, E = 32.2(10)6 = 3.22(109) Pa 0.01 s (MPa) 100 95 Thus, 70 60 40(10 ) FAB = p = 31.83 MPa AAB (0.04) sAB = eAB 50 31.83(106) sAB = 0.009885 mm>mm = = E 3.22(109) 20 7.958(106) sCD = 0.002471 mm>mm = E 3.22(109) dAB = eABLAB = 0.009885(2000) = 19.771 mm dCD = eCDLCD = 0.002471(500) = 1.236 mm Angle of tilt a: tan a = 18.535 ; 1500 tension 40 32.2 40(103) FCD = p = 7.958 MPa ACD (0.08) sCD = eCD = compression 80 a = 0.708° Ans 17 0.01 0.02 0.03 0.04 P (mm/mm) 03 Solutions 46060 5/7/10 8:45 AM Page 18 © 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 3–22 The stress–strain diagram for a polyester resin is given in the figure If the rigid beam is supported by a strut AB and post CD made from this material, determine the largest load P that can be applied to the beam before it ruptures The diameter of the strut is 12 mm and the diameter of the post is 40 mm B 2m P Rupture of strut AB: sR = FAB ; AAB 50(106) = P>2 A ; p (0.012) 0.75 m 0.75 m P = 11.3 kN (controls) D 0.5 m Ans s (MPa) Rupture of post CD: FCD sR = ; ACD C 95(10 ) = 100 95 P>2 p (0.04) compression 80 70 60 P = 239 kN 50 tension 40 32.2 20 0 0.01 0.02 0.03 0.04 P (mm/mm) s (ksi) 3–23 By adding plasticizers to polyvinyl chloride, it is possible to reduce its stiffness The stress–strain diagrams for three types of this material showing this effect are given below Specify the type that should be used in the manufacture of a rod having a length of in and a diameter of in., that is required to support at least an axial load of 20 kip and also be able to stretch at most 14 in 15 P unplasticized 10 copolymer flexible (plasticized) Normal Stress: P P s = = A 20 p = 6.366 ksi (2 ) 0 Normal Strain: e = 0.25 = 0.0500 in.>in From the stress–strain diagram, the copolymer will satisfy both stress and strain requirements Ans 18 0.10 0.20 0.30 P (in./in.) 03 Solutions 46060 5/7/10 8:45 AM Page 19 © 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 *3–24 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 = 3011032 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 •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 elong = 300 p (0.015) 300 N 300 N 200 mm = 1.697 MPa 1.697(106) s = 0.0006288 = E 2.70(109) d = elong L = 0.0006288 (200) = 0.126 mm Ans elat = - Velong = - 0.4(0.0006288) = - 0.0002515 ¢d = elatd = - 0.0002515 (15) = - 0.00377 mm Ans 19 03 Solutions 46060 5/7/10 8:45 AM Page 20 © 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 3–26 The short cylindrical block of 2014-T6 aluminum, having an original diameter of 0.5 in and a length of 1.5 in., is placed in the smooth jaws of a vise and squeezed until the axial load applied is 800 lb Determine (a) the decrease in its length and (b) its new diameter 800 lb 800 lb a) s = P = A elong = p 800 = 4074.37 psi (0.5)2 s - 4074.37 = - 0.0003844 = E 10.6(106) d = elong L = - 0.0003844 (1.5) = - 0.577 (10 - 3) in Ans b) V = -elat = 0.35 elong elat = - 0.35 ( -0.0003844) = 0.00013453 ¢d = elat d = 0.00013453 (0.5) = 0.00006727 d¿ = d + ¢d = 0.5000673 in Ans s(MPa) 3–27 The elastic portion of the stress–strain diagram for a steel alloy is shown in the figure The specimen from which it was obtained had an original diameter of 13 mm and a gauge length of 50 mm When the applied load on the specimen is 50 kN, the diameter is 12.99265 mm Determine Poisson’s ratio for the material 400 Normal Stress: s = P = A 50(103) p (0.0132) = 376.70 Mpa 0.002 Normal Strain: From the stress–strain diagram, the modulus of elasticity 400(106) = 200 GPa Applying Hooke’s law E = 0.002 elong = elat = 376.70(106) s = 1.8835 A 10 - B mm>mm = E 200(104) d - d0 12.99265 - 13 = = - 0.56538 A 10 - B mm>mm d0 13 Poisson’s Ratio: The lateral and longitudinal strain can be related using Poisson’s ratio V = - - 0.56538(10 - 3) elat = 0.300 = elong 1.8835(10 - 3) Ans 20 P(mm/mm) 03 Solutions 46060 5/7/10 8:45 AM Page 21 © 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 s(MPa) *3–28 The elastic portion of the stress–strain diagram for a steel alloy is shown in the figure The specimen from which it was obtained had an original diameter of 13 mm and a gauge length of 50 mm If a load of P = 20 kN is applied to the specimen, determine its diameter and gauge length Take n = 0.4 400 Normal Stress: s = P = A 20(103) p (0.0132) = 150.68Mpa 0.002 P(mm/mm) Normal Strain: From the Stress–Strain diagram, the modulus of elasticity 400(106) E = = 200 GPa Applying Hooke’s Law 0.002 elong = 150.68(106) s = 0.7534 A 10 - B mm>mm = E 200(109) Thus, dL = elong L0 = 0.7534 A 10 - B (50) = 0.03767 mm L = L0 + dL = 50 + 0.03767 = 50.0377 mm Ans Poisson’s Ratio: The lateral and longitudinal can be related using poisson’s ratio elat = - velong = - 0.4(0.7534) A 10 - B = - 0.3014 A 10 - B mm>mm dd = elat d = - 0.3014 A 10 - B (13) = - 0.003918 mm d = d0 + dd = 13 + ( - 0.003918) = 12.99608 mm Ans •3–29 The aluminum block has a rectangular cross section and is subjected to an axial compressive force of 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 = elat = in kip kip in P = = 2.667 ksi A (2)(1.5) elong = v = 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¿ = + 0.0000880(2) = 2.000176 in Ans 21 03 Solutions 46060 5/7/10 8:45 AM Page 22 © 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 3–30 The block is made of titanium Ti-6A1-4V and is subjected to a compression of 0.06 in along the y axis, and its shape is given a tilt of u = 89.7° Determine Px, Py, and gxy y Normal Strain: ey = dLy Ly = in u -0.06 = - 0.0150 in.>in Ans Poisson’s Ratio: The lateral and longitudinal strain can be related using Poisson’s ratio x in ex = - vey = - 0.36( -0.0150) = 0.00540 in >in Ans Shear Strain: b = 180° - 89.7° = 90.3° = 1.576032 rad gxy = p p - b = - 1.576032 = - 0.00524 rad 2 Ans 3–31 The shear stress–strain diagram for a steel alloy is shown in the figure If a bolt having a diameter of 0.75 in is made of this material and used in the double lap joint, determine the modulus of elasticity E and the force P required to cause the material to yield Take n = 0.3 P/2 P/2 P t(ksi) 60 The shear force developed on the shear planes of the bolt can be determined by considering the equilibrium of the FBD shown in Fig a + ©F = 0; : x V + V - P = V = = P 0.00545 From the shear stress–strain diagram, the yield stress is ty = 60 ksi Thus, ty = Vy A ; 60 = P>2 p A 0.752 B P = 53.01 kip = 53.0 kip Ans From the shear stress–strain diagram, the shear modulus is G = 60 ksi = 11.01(103) ksi 0.00545 Thus, the modulus of elasticity is G = E ; 2(1 + y) 11.01(103) = E 2(1 + 0.3) E = 28.6(103) ksi Ans 22 g(rad) 03 Solutions 46060 5/7/10 8:45 AM Page 23 © 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 *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 = 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 = - P ln ro + C 2p h G y = 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 23 03 Solutions 46060 5/7/10 8:45 AM Page 24 © 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 •3–33 The support consists of three rigid plates, which are connected together using two symmetrically placed rubber pads If a vertical force of N is applied to plate A, determine the approximate vertical displacement of this plate due to shear strains in the rubber Each pad has cross-sectional dimensions of 30 mm and 20 mm Gr = 0.20 MPa C B 40 mm 40 mm A tavg = g = V 2.5 = = 4166.7 Pa A (0.03)(0.02) 5N 4166.7 t = 0.02083 rad = G 0.2(106) d = 40(0.02083) = 0.833 mm Ans 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 P t = V P = = 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 24 a a 03 Solutions 46060 5/7/10 8:45 AM Page 25 © 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 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 in and a diameter of 0.5 in When the applied load is 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.) 70 s = = 11400.65 ksi e 0.00614 Eal = When specimen is loaded with a - kip load, s = P = A p = 45.84 ksi (0.5)2 s 45.84 = = 0.0040208 in.>in E 11400.65 elong = 0.49935 - 0.5 d¿ - d = = - 0.0013 in.>in d 0.5 elat = V = - Gal = elat - 0.0013 = 0.32332 = elong 0.0040208 11.4(103) Eat = = 4.31(103) ksi 2(1 + v) 2(1 + 0.32332) Ans 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 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 0.00614 10 = 50.9296 ksi p (0.5) From the stress–strain diagram E = 70 = 11400.65 ksi 0.00614 elong = G = s 50.9296 = = 0.0044673 in.>in E 11400.65 E ; 2(1 + v) 3.8(103) = 11400.65 ; 2(1 + v) v = 0.500 elat = - velong = - 0.500(0.0044673) = - 0.002234 in.>in ¢d = elat d = - 0.002234(0.5) = - 0.001117 in d¿ = d + ¢d = 0.5 - 0.001117 = 0.4989 in Ans 25 P (in./in.) 03 Solutions 46060 5/7/10 8:45 AM Page 26 © 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 s(psi) 3–37 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 E = 11 = 5.5 psi Ans ut = 1 (2)(11) + (55 + 11)(2.25 - 2) = 19.25 psi 2 Ans ut = (2)(11) = 11 psi Ans 3–38 A short cylindrical block of 6061-T6 aluminum, having an original diameter of 20 mm and a length of 75 mm, is placed in a compression machine and squeezed until the axial load applied is kN Determine (a) the decrease in its length and (b) its new diameter a) s = - 5(103) P = p = - 15.915 MPa A (0.02) s = E elong ; - 15.915(106) = 68.9(109) elong elong = - 0.0002310 mm>mm d = elong L = - 0.0002310(75) = - 0.0173 mm b) v = - elat ; elong 0.35 = - Ans elat - 0.0002310 elat = 0.00008085 mm>mm ¢d = elat d = 0.00008085(20) = 0.0016 mm d¿ = d + ¢d = 20 + 0.0016 = 20.0016 mm Ans 26 2.25 P(in./in.) 03 Solutions 46060 5/7/10 8:45 AM Page 27 © 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 3–39 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 a + ©MA = 0; FB(3) - 80(x) = 0; a + ©MB = 0; - FA(3) + 80(3 - x) = 0; FB = 80 kN x A 80x FA = B 210 mm 220 mm (1) 80(3 - x) 3m (2) Since the beam is held horizontally, dA = dB s = P ; A d = eL = a P e = P A E dA = dB ; s A = E E bL = PL AE 80(3 - x) (220) 80x (210) = AE 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 (0.03 ) elong = 55.27(106) sA = - 0.000756 = E 73.1(109) elat = - velong = - 0.35( - 0.000756) = 0.0002646 œ dA = dA + d elat = 30 + 30(0.0002646) = 30.008 mm Ans *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 bolts Each bolt has a diameter of 16 in If sY = 40 ksi and Est = 29110 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 16 = 28.97 ksi sg = 40 ksi Normal Strain: Since s sg, Hooke’s law is still valid e = 28.97 s = 0.000999 in.>in = E 29(103) Ans If the nut is unscrewed, the load is zero Therefore, the strain e = 27 Ans 03 Solutions 46060 5/7/10 8:45 AM Page 28 © 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 •3–41 The stone has a mass of 800 kg and center of gravity at G It rests on a pad at A and a roller at B The pad is fixed to the ground and has a compressed height of 30 mm, a width of 140 mm, and a length of 150 mm If the coefficient of static friction between the pad and the stone is ms = 0.8, determine the approximate horizontal displacement of the stone, caused by the shear strains in the pad, before the stone begins to slip Assume the normal force at A acts 1.5 m from G as shown The pad is made from a material having E = MPa and n = 0.35 0.4 m B Equations of Equilibrium: a + ©MB = 0; + ©F = 0; : x FA(2.75) - 7848(1.25) - P(0.3) = [1] P - F = [2] Note: The normal force at A does not act exactly at A It has to shift due to friction Friction Equation: F = ms FA = 0.8 FA [3] Solving Eqs [1], [2] and [3] yields: FA = 3908.37 N F = P = 3126.69 N Average Shear Stress: The pad is subjected to a shear force of V = F = 3126.69 N t = V 3126.69 = = 148.89 kPa A (0.14)(0.15) Modulus of Rigidity: G = E = = 1.481 MPa 2(1 + v) 2(1 + 0.35) Shear Strain: Applying Hooke’s law for shear g = 148.89(103) t = 0.1005 rad = G 1.481(106) Thus, dh = hg = 30(0.1005) = 3.02 mm Ans 28 P G 1.25 m 0.3 m 1.5 m A 03 Solutions 46060 5/7/10 8:45 AM Page 29 © 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 3–42 The bar DA is rigid and is originally held in the horizontal position when the weight W is supported from C If the weight causes B to be displaced downward 0.025 in., determine the strain in wires DE and BC Also, if the wires are made of A-36 steel and have a cross-sectional area of 0.002 in2, determine the weight W E ft ft D ft B = 0.025 d A ft d = 0.0417 in eDE = C 0.0417 d = = 0.00116 in.>in L 3(12) Ans W sDE = EeDE = 29(10 )(0.00116) = 33.56 ksi FDE = sDEADE = 33.56 (0.002) = 0.0672 kip a + ©MA = 0; - (0.0672) (5) + 3(W) = W = 0.112 kip = 112 lb Ans sBC = W 0.112 = = 55.94 ksi ABC 0.002 eBC = sBC 55.94 = 0.00193 in.>in = E 29 (103) Ans 3–43 The 8-mm-diameter bolt is made of an aluminum alloy It fits through a magnesium sleeve that has an inner diameter of 12 mm and an outer diameter of 20 mm If the original lengths of the bolt and sleeve are 80 mm and 50 mm, respectively, determine the strains in the sleeve and the bolt if the nut on the bolt is tightened so that the tension in the bolt is kN Assume the material at A is rigid Eal = 70 GPa, Emg = 45 GPa 50 mm A 30 mm Normal Stress: 8(103) sb = P = Ab p (0.008 ) ss = P = As p (0.02 = 159.15 MPa 8(103) - 0.0122) = 39.79 MPa Normal Strain: Applying Hooke’s Law eb = 159.15(106) sb = 0.00227 mm>mm = Eal 70(109) Ans es = 39.79(106) ss = 0.000884 mm>mm = Emg 45(109) Ans 29 03 Solutions 46060 5/7/10 8:45 AM Page 30 © 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 *3–44 The A-36 steel wire AB has a cross-sectional area of 10 mm2 and is unstretched when u = 45.0° Determine the applied load P needed to cause u = 44.9° A 400 mm u 400 m m B P ¿ LAB 400 = sin 90.2° sin 44.9° ¿ = 566.67 mm LAB LAB = e = 400 = 565.69 sin 45° ¿ - LAB LAB 566.67 - 565.69 = = 0.001744 LAB 565.69 s = Ee = 200(109) (0.001744) = 348.76 MPa a + ©MA = P(400 cos 0.2°) - FAB sin 44.9° (400) = (1) However, FAB = sA = 348.76(106)(10)(10 - 6) = 3.488 kN From Eq (1), P = 2.46 kN Ans 30