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Chapter02 Chapter02 Document Transcript

  • CHAPTER 2
  • PROBLEM 2.1An 80-m-long wire of 5-mm diameter is made of a steel with E = 200 GPa and an ultimate tensile strength of400 MPa. If a factor of safety of 3.2 is desired, determine (a) the largest allowable tension in the wire, (b) thecorresponding elongation of the wire.SOLUTION π π(a) σ U = 400 × 106 Pa A= d2 = (5) 2 = 19.635 mm 2 = 19.635 × 10−6 m 2 4 4 P = σ U A = (400 × 106 ) (19.635 × 10−6 ) = 7854 N U PU 7854 Pall = = = 2454 N Pall = 2.45 kN  F .S 3.2 PL (2454) (80)(b) δ= = = 50.0 × 10−3 m δ = 50.0 mm  AE (19.635 × 10−6 )(200 × 109 )PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.2A steel control rod is 5.5 ft long and must not stretch more than 0.04 in. when a 2-kip tensile load is applied to it.Knowing that E = 29 × 106 psi, determine (a) the smallest diameter rod that should be used, (b) thecorresponding normal stress caused by the load.SOLUTION PL (2000 lb) (5.5 × 12 in.)(a) δ= : 0.04 in. = 6 AE A (29 × 10 psi) 1 A = π d 2 = 0.11379 in 2 4 d = 0.38063 in. d = 0.381 in.  P 2000 lb(b) σ= = 2 = 17580 psi σ = 17.58 ksi  A 0.11379 inPROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.3Two gage marks are placed exactly 10 in. apart on a 1 -in.-diameter aluminum rod with E = 10.1 × 106 psi and 2an ultimate strength of 16 ksi. Knowing that the distance between the gage marks is 10.009 in. after a load isapplied, determine (a) the stress in the rod, (b) the factor of safety.SOLUTION(a) δ = 10.009 − 10.000 = 0.009 in. δ σ Eδ (10.1 × 106 ) (0.009) ε= = σ= = = 9.09 × 103 psi σ = 9.09 ksi  L E L 10 σU 16(b) F. S . = = F. S . = 1.760  σ 9.09PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.4An 18-m-long steel wire of 5-mm diameter is to be used in the manufacture of a prestressed concrete beam. It is observed that the wire stretches 45 mm when a tensile force P is applied. Knowing that E = 200 GPa,determine (a) the magnitude of the force P, (b) the corresponding normal stress in the wire.SOLUTION PL δ AE(a) δ= , or P= AE L 1 1 with A = π d 2 = π (0.005) 2 = 19.6350 × 10−6 m 2 4 4 (0.045 m)(19.6350 × 10−6 m 2 )(200 × 109 N/m 2 ) P= = 9817.5 N 18 m P = 9.82 kN  P 9817.5 N 6(b) σ= = −6 2 = 500 × 10 Pa σ = 500 MPa  A 19.6350 × 10 mPROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.5A polystyrene rod of length 12 in. and diameter 0.5 in. is subjected to an 800-lb tensile load. Knowing thatE = 0.45 × 106 psi, determine (a) the elongation of the rod, (b) the normal stress in the rod.SOLUTION π π A= d2 = (0.5) 2 = 0.19635 in 2 4 4 PL (800) (12)(a) δ= = 6 = 0.1086 δ = 0.1086 in.  AE (0.19635) (0.45 ×10 ) P 800(b) σ= = = 4074 psi σ = 4.07 ksi  A 0.19635PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.6A nylon thread is subjected to a 8.5-N tension force. Knowing that E = 3.3 GPa and that the length of thethread increases by 1.1%, determine (a) the diameter of the thread, (b) the stress in the thread.SOLUTION δ 1.1(a) Strain: ε= = = 0.011 L 100 Stress: σ = Eε = (3.3 × 109 )(0.011) = 36.3 × 106 Pa P σ = A P 8.5 Area: A= = = 234.16 × 10−9 m 2 σ 36.3 × 106 4A (4)(234.16 × 10−9 ) Diameter: d = = = 546 × 10−6 m d = 0.546 mm  π π(b) Stress: σ = 36.3 MPa PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.7Two gage marks are placed exactly 250 mm apart on a 12-mm-diameter aluminum rod. Knowing that, with anaxial load of 6000 N acting on the rod, the distance between the gage marks is 250.18 mm, determine themodulus of elasticity of the aluminum used in the rod.SOLUTION δ = Δ L = L − L0 = 250.18 − 250.00 = 0.18 mm δ 0.18 mm ε= = = 0.00072 L0 250 mm π π A= (12)2 = 113.097 mm 2 = 113.097 ×10−6 m 2 d2 = 4 4 P 6000 σ= = = 53.052 × 106 Pa  A 113.097 × 10−6 σ 53.052 × 106 E= = = 73.683 × 109 Pa E = 73.7 GPa  ε 0.00072PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.8An aluminum pipe must not stretch more than 0.05 in. when it is subjected to a tensile load. Knowing thatE = 10.1 × 106 psi and that the maximum allowable normal stress is 14 ksi, determine (a) the maximumallowable length of the pipe, (b) the required area of the pipe if the tensile load is 127.5 kips.SOLUTION PL(a) δ= ; AE EAδ Eδ (10.1 × 106 ) (0.05) Thus, L= = = P σ 14 × 103 L = 36.1 in.  P(b) σ = ; A P 127.5 × 103 Thus, A= = A = 9.11 in 2  σ 14 × 103PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.9An aluminum control rod must stretch 0.08 in. when a 500-lb tensile load is applied to it. Knowing thatσ all = 22 ksi and E = 10.1 × 106 psi, determine the smallest diameter and shortest length that can be selected forthe rod.SOLUTION P = 500 lb, δ = 0.08 in. σ all = 22 × 103 psi P P 500 σ= < σ all A> = = 0.022727 in 2 A σ all 22 × 103 π 4A (4)(0.022727) A= d2 d= = d min = 0.1701 in.  4 π π Eδ σ = Eε = < σ all L Eδ (10.1 × 106 )(0.08) L> = = 36.7 in. Lmin = 36.7 in.  σ all 22 × 103PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.10A square yellow-brass bar must not stretch more than 2.5 mm when it is subjected to a tensile load. Knowingthat E = 105 GPa and that the allowable tensile strength is 180 MPa, determine (a) the maximum allowablelength of the bar, (b) the required dimensions of the cross section if the tensile load is 40 kN.SOLUTIONσ = 180 × 106 Pa P = 40 × 103 NE = 105 × 109 Pa δ = 2.5 × 10−3 m PL σ L(a) δ= = AE E Eδ (105 × 109 )(2.5 × 10−3 ) L= = = 1.45833 m σ 180 × 106 L = 1.458 m  P(b) σ = A P 40 × 103 A= = = 222.22 × 10−6 m 2 = 222.22 mm 2 σ 180 × 106 A = a2 a = A = 222.22 a = 14.91 mm PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.11A 4-m-long steel rod must not stretch more than 3 mm and the normal stress must not exceed 150 MPa whenthe rod is subjected to a 10-kN axial load. Knowing that E = 200 GPa, determine the required diameter ofthe rod.SOLUTIONL =4mδ = 3 × 10−3 m, σ = 150 × 106 PaE = 200 × 109 Pa, P = 10 × 103 N PStress: σ = A P 10 × 103 A= = 6 = 66.667 × 10−6 m 2 = 66.667 mm 2 σ 150 × 10 PLDeformation: δ = AE PL (10 × 103 ) (4) A= = = 66.667 × 10−6 m 2 = 66.667 mm 2 Eδ (200 × 109 ) (3 × 10−3 )The larger value of A governs: A = 66.667 mm 2 π 4A 4(66.667) A= d2 d= = d = 9.21 mm  4 π πPROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.12A nylon thread is to be subjected to a 10-N tension. Knowing that E = 3.2 GPa, that the maximum allowablenormal stress is 40 MPa, and that the length of the thread must not increase by more than 1%, determine therequired diameter of the thread.SOLUTIONStress criterion: σ = 40 MPa = 40 × 106 Pa P = 10 N P P 10 N σ = : A= = = 250 × 10−9 m 2 A σ 40 × 106 Pa π A 250 × 10−9 A= d 2: d = 2 =2 = 564.19 × 10−6 m 4 π π d = 0.564 mmElongation criterion: δ = 1% = 0.01 L PL δ = : AE P /E 10 N/3.2 × 109 Pa A= = = 312.5 × 10−9 m 2 δ /L 0.01 A 312.5 × 10−9 d =2 =2 = 630.78 × 10−6 m 2 π π d = 0.631 mmThe required diameter is the larger value: d = 0.631 mm PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.13 The 4-mm-diameter cable BC is made of a steel with E = 200 GPa. Knowing that the maximum stress in the cable must not exceed 190 MPa and that the elongation of the cable must not exceed 6 mm, find the maximum load P that can be applied as shown.SOLUTION LBC = 62 + 42 = 7.2111 mUse bar AB as a free body.  4  Σ M A = 0: 3.5 P − (6)  FBC  = 0  7.2111  P = 0.9509 FBCConsidering allowable stress: σ = 190 × 106 Pa π π A= d2 = (0.004) 2 = 12.566 × 10−6 m 2 4 4 FBC σ= ∴ FBC = σ A = (190 × 106 ) (12.566 × 10−6 ) = 2.388 × 103 N AConsidering allowable elongation: δ = 6 × 10−3 m FBC LBC AEδ (12.566 × 10−6 )(200 × 109 )(6 × 10−3 ) δ= ∴ FBC = = = 2.091 × 103 N AE LBC 7.2111Smaller value governs. FBC = 2.091 × 103 N P = 0.9509 FBC = (0.9509)(2.091 × 103 ) = 1.988 × 103 N P = 1.988 kN PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.14 The aluminum rod ABC ( E = 10.1 × 106 psi), which consists of two cylindrical portions AB and BC, is to be replaced with a cylindrical steel rod DE ( E = 29 × 106 psi) of the same overall length. Determine the minimum required diameter d of the steel rod if its vertical deformation is not to exceed the deformation of the aluminum rod under the same load and if the allowable stress in the steel rod is not to exceed 24 ksi.SOLUTIONDeformation of aluminum rod. PLAB PLBC δA = + AAB E ABC E P  LAB LBC  =  +  E  AAB ABC   12 28 × 103 18  = π +   (1.5)2 10.1 × 106 π (2.25) 2  4 4  = 0.031376 in.Steel rod. δ = 0.031376 in. PL PL (28 × 103 )(30) δ= ∴ A= = = 0.92317 in 2 EA Eδ (29 × 10 )(0.031376) 6 P P 28 × 103 σ= ∴ A= = = 1.1667 in 2 A σ 24 × 103Required area is the larger value. A = 1.1667 in 2 4A (4)(1.6667)Diameter: d= = d = 1.219 in.  π πPROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.15 A 4-ft section of aluminum pipe of cross-sectional area 1.75 in2 rests on a fixed support at A. The 5 -in.-diameter steel rod BC hangs from a rigid 8 bar that rests on the top of the pipe at B. Knowing that the modulus of elasticity is 29 × 106 psi for steel, and 10.4 × 106 psi for aluminum, determine the deflection of point C when a 15-kip force is applied at C.SOLUTIONRod BC: LBC = 7 ft = 84 in. EBC = 29 × 106 psi π π ABC = d2 = (0.625) 2 = 0.30680 in 2 4 4 PLBC (15 × 103 )(84) δ C/B = = = 0.141618 in. EBC ABC (29 × 106 )(0.30680)Pipe AB: LAB = 4 ft = 48 in. E AB = 10.4 × 106 psi AAB = 1.75 in 2 PLAB (15 × 103 ) (48) δ B/A = = 6 = 39.560 × 10−3 in. E AB AAB (10.4 × 10 ) (1.75)Total: δ C = δ B/A + δ C/B = 39.560 × 10−3 + 0.141618 = 0.181178 in. δ C = 0.1812 in. PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.16 The brass tube AB ( E = 105 GPa) has a cross-sectional area of 140 mm2 and is fitted with a plug at A. The tube is attached at B to a rigid plate that is itself attached at C to the bottom of an aluminum cylinder ( E = 72 GPa) with a cross-sectional area of 250 mm2. The cylinder is then hung from a support at D. In order to close the cylinder, the plug must move down through 1 mm. Determine the force P that must be applied to the cylinder.SOLUTIONShortening of brass tube AB: LAB = 375 + 1 = 376 mm = 0.376 m AAB = 140 mm 2 = 140 × 10−6 m 2 E AB = 105 × 109 Pa PLAB P(0.376) δ AB = = 9 −6 = 25.578 × 10−9 P E AB AAB (105 × 10 )(140 × 10 )Lengthening of aluminum cylinder CD: LCD = 0.375 m ACD = 250 mm 2 = 250 × 10−6 m 2 ECD = 72 × 109 Pa PLCD P(0.375) δ CD = = 9 −6 = 20.833 × 10−9 P ECD ACD (72 × 10 )(250 × 10 )Total deflection: δ A = δ AB + δ CD where δ A = 0.001 m 0.001 = (25.578 × 10−9 + 20.833 × 10−9 ) P P = 21.547 × 103 N P = 21.5 kN PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.17 A 250-mm-long aluminum tube ( E = 70 GPa) of 36-mm outer diameter and 28-mm inner diameter can be closed at both ends by means of single-threaded screw-on covers of 1.5-mm pitch. With one cover screwed on tight, a solid brass rod ( E = 105 GPa) of 25-mm diameter is placed inside the tube and the second cover is screwed on. Since the rod is slightly longer than the tube, it is observed that the cover must be forced against the rod by rotating it one-quarter of a turn before it can be tightly closed. Determine (a) the average normal stress in the tube and in the rod, (b) the deformations of the tube and of the rod.SOLUTION π π Atube = (d o − di2 ) = 2 (362 − 282 ) = 402.12 mm 2 = 402.12 × 10−6 m 2 4 4 π π A rod = d2 = (25)2 = 490.87 mm 2 = 490.87 × 10−6 m 2 4 4 PL P(0.250) δ tube = = 9 −6 = 8.8815 × 10−9 P Etube Atube (70 × 10 )(402.12 × 10 ) PL P(0.250) δ rod =− = = −4.8505 × 10−9 P Erod Arod (105 × 106 )(490.87 × 10−6 ) 1  δ * =  turn  × 1.5 mm = 0.375 mm = 375 × 10−6 m 4  δ tube = δ * + δ rod or δ tube − δ rod = δ * 8.8815 × 10−9 P + 4.8505 × 10−9 P = 375 × 10−6 0.375 × 10−3 P= −9 = 27.308 × 103 N (8.8815 + 4.8505)(10 ) P 27.308 × 103(a) σ tube = = = 67.9 × 106 Pa σ tube = 67.9 MPa  Atube 402.12 × 10−6 P 27.308 × 103 σ rod = − =− = −55.6 × 106 Pa σ rod = −55.6 MPa  Arod 490.87 × 10−6(b) δ tube = (8.8815 × 10−9 )(27.308 × 103 ) = 242.5 × 10−6 m δ tube = 0.2425 mm  δ rod = −(4.8505 × 10−9 )(27.308 × 103 ) = −132.5 × 10−6 m δ rod = −0.1325 mm PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.18 The specimen shown is made from a 1-in.-diameter cylindrical steel rod with two 1.5-in.-outer-diameter sleeves bonded to the rod as shown. Knowing that E = 29 × 106 psi, determine (a) the load P so that the total deformation is 0.002 in., (b) the corresponding deformation of the central portion BC.SOLUTION Pi Li P Li(a) δ =Σ = Σ Ai Ei E Ai −1  L  π P = Eδ  Σ i  Ai = di2  Ai  4 L, in. d, in. A, in2 L/A, in–1 AB 2 1.5 1.7671 1.1318 BC 3 1.0 0.7854 3.8197 CD 2 1.5 1.7671 1.1318 6.083 ← sum P = (29 × 106 )(0.002)(6.083) −1 = 9.353 × 103 lb P = 9.53 kips  PLBC P LBC 9.535 × 103(b) δ BC = = = (3.8197) δ = 1.254 × 10−3 in.  ABC E E ABC 29 × 106PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.19 Both portions of the rod ABC are made of an aluminum for which E = 70 GPa. Knowing that the magnitude of P is 4 kN, determine (a) the value of Q so that the deflection at A is zero, (b) the corresponding deflection of B.SOLUTION π π(a) AAB = 2 d AB = (0.020)2 = 314.16 × 10−6 m 2 4 4 π π ABC = 2 d BC = (0.060)2 = 2.8274 × 10−3 m 2 4 4 Force in member AB is P tension. Elongation: PLAB (4 × 103 )(0.4) δ AB = = 9 −6 = 72.756 × 10−6 m EAAB (70 × 10 )(314.16 × 10 ) Force in member BC is Q − P compression. Shortening: (Q − P) LBC (Q − P)(0.5) δ BC = = = 2.5263 × 10−9 (Q − P ) EABC (70 × 109 )(2.8274 × 10−3 ) For zero deflection at A, δ BC = δ AB 2.5263 × 10−9 (Q − P ) = 72.756 × 10−6 ∴ Q − P = 28.8 × 103 N Q = 28.3 × 103 + 4 × 103 = 32.8 × 103 N Q = 32.8 kN (b) δ AB = δ BC = δ B = 72.756 × 10−6 m δ AB = 0.0728 mm ↓ PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.20 The rod ABC is made of an aluminum for which E = 70 GPa. Knowing that P = 6 kN and Q = 42 kN, determine the deflection of (a) point A, (b) point B.SOLUTION π π AAB = 2 d AB = (0.020) 2 = 314.16 × 10−6 m 2 4 4 π π ABC = 2 d BC = (0.060)2 = 2.8274 × 10−3 m 2 4 4 PAB = P = 6 × 103 N PBC = P − Q = 6 × 103 − 42 × 103 = −36 × 103 N LAB = 0.4 m LBC = 0.5 m PAB LAB (6 × 103 )(0.4) δ AB = = = 109.135 × 10−6 m AAB E A (314.16 × 10−6 )(70 × 109 ) PBC LBC (−36 × 103 )(0.5) δ BC = = = −90.947 × 10−6 m ABC E (2.8274 × 10−3 )(70 × 109 )(a) δ A = δ AB + δ BC = 109.135 × 10−6 − 90.947 × 10−6 m = 18.19 × 10−6 m δ A = 0.01819 mm ↑ (b) δ B = δ BC = −90.9 × 10−6 m = −0.0909 mm or δ B = 0.0919 mm ↓ PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.21 Members AB and BC are made of steel ( E = 29 × 106 psi) with cross- sectional areas of 0.80 in2 and 0.64 in2, respectively. For the loading shown, determine the elongation of (a) member AB, (b) member BC.SOLUTION(a) LAB = 62 + 52 = 7.810 ft = 93.72 in. Use joint A as a free body. 5 ΣFy = 0: FAB − 28 = 0 7.810 FAB = 43.74 kip = 43.74 × 103 lb FAB LAB (43.74 × 103 ) (93.72) δ AB = = δ AB = 0.1767 in.  EAAB (29 × 106 ) (0.80)(b) Use joint B as a free body. 6 ΣFx = 0: FBC − FAB = 0 7.810 (6) (43.74) FBC = = 33.60 kip = 33.60 × 103 lb. 7.810 FBC LBC (33.60 × 103 ) (72) δ BC = = δ BC = 0.1304 in.  EABC (29 × 106 ) (0.64)PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.22 The steel frame ( E = 200 GPa) shown has a diagonal brace BD with an area of 1920 mm2. Determine the largest allowable load P if the change in length of member BD is not to exceed 1.6 mm.SOLUTION δ BD = 1.6 × 10−3 m, ABD = 1920 mm 2 = 1920 × 10−6 m 2 LBD = 52 + 62 = 7.810 m, EBD = 200 × 109 Pa FBD LBD δ BD = EBD ABD EBD ABDδ BD (200 × 109 ) (1920 × 10−6 )(1.6 × 10−3 ) FBD = = LBD 7.81 = 78.67 × 103 N Use joint B as a free body. ΣFx = 0: 5 FBD − P = 0 7.810 5 (5)(78.67 × 103 ) P= FBD = 7.810 7.810 = 50.4 × 103 N P = 50.4 kN PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.23 For the steel truss ( E = 200 GPa) and loading shown, determine the deformations of the members AB and AD, knowing that their cross-sectional areas are 2400 mm2 and 1800 mm2, respectively.SOLUTIONStatics: Reactions are 114 kN upward at A and C.Member BD is a zero force member. LAB = 4.02 + 2.52 = 4.717 mUse joint A as a free body. 2.5 ΣFy = 0 : 114 + FAB = 0 4.717 FAB = −215.10 kN 4 ΣFx = 0 : FAD + FAB = 0 4.717 (4)(−215.10) FAD = − = 182.4 kN 4.717Member AB: FAB LAB (−215.10 × 103 )(4.717) δ AB = = EAAB (200 × 109 )(2400 × 10−6 ) = −2.11 × 10−3 m δ AB − 2.11 mm  FAD LAD (182.4 × 103 )(4.0)Member AD: δ AD = = EAAD (200 × 109 )(1800 × 10−6 ) = 2.03 × 10−3 m δ AD = 2.03 mm PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.24 For the steel truss ( E = 29 × 106 psi) and loading shown, determine the deformations of the members BD and DE, knowing that their cross- sectional areas are 2 in2 and 3 in2, respectively.SOLUTIONFree body: Portion ABC of truss ΣM E = 0 : FBD (15 ft) − (30 kips)(8 ft) − (30 kips)(16 ft) = 0 FBD = + 48.0 kipsFree body: Portion ABEC of truss ΣFx = 0 : 30 kips + 30 kips − FDE = 0  FDE = + 60.0 kips PL (+48.0 × 103 lb)(8 × 12 in.) δ BD = = δ BD = +79.4 × 10−3 in.  AE (2 in 2 )(29 × 106 psi) PL (+60.0 × 103 lb)(15 × 12 in.) δ DE = = δ DE + 124.1 × 10−3 in.  AE (3 in 2 )(29 × 106 psi)PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.25 Each of the links AB and CD is made of aluminum ( E = 10.9 × 106 psi) and has a cross-sectional area of 0.2 in.2. Knowing that they support the rigid member BC, determine the deflection of point E.SOLUTIONFree body BC: ΣM C = 0 : − (32) FAB + (22) (1 × 103 ) = 0 FAB = 687.5 lb ΣFy = 0 : 687.5 − 1 × 103 + FCD = 0 FCD = 312.5 lb FAB LAB (687.5) (18)δ AB = = 6 = 5.6766 × 10−3 in = δ B EA (10.9 × 10 ) (0.2) F L (312.5) (18)δ CD = CD CD = 6 = 2.5803 × 10−3 in = δ C EA (10.9 × 10 ) (0.2)Deformation diagram: δ B − δC 3.0963 × 10−3 Slope θ = = LBC 32 = 96.759 × 10−6 rad δ E = δ C + LECθ = 2.5803 × 10−3 + (22) (96.759 × 10−6 ) = 4.7090 × 10−3 in δ E = 4.71 × 10−3 in ↓ PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.26 3 The length of the 32 -in.-diameter steel wire CD has been adjusted so that 1 with no load applied, a gap of 16 in. exists between the end B of the rigid beam ACB and a contact point E. Knowing that E = 29 × 106 psi, determine where a 50-lb block should be placed on the beam in order to cause contact between B and E.SOLUTIONRigid beam ACB rotates through angle θ to close gap. 1/16 θ= = 3.125 × 10−3 rad 20Point C moves downward. δ C = 4θ = 4(3.125 × 10−3 ) = 12.5 × 10−3 in. δ CD = δ C = 12.5 × 10−3 in. 2 π π 3  −3 ACD = d2 =  32  = 6.9029 × 10 in 2 d 4  F L δ CD = CD CD EACD EACDδ CD (29 × 106 ) (6.9029 × 10−3 ) (12.5 × 10−3 ) FCD = = LCD 12.5 = 200.18 lbFree body ACB: ΣM A = 0: 4 FCD − (50) (20 − x) = 0 (4) (200.18) 20 − x = = 16.0144 50 x = 3.9856 in.For contact, x < 3.99 in. PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.27 Link BD is made of brass ( E = 105 GPa) and has a cross-sectional area of 240 mm2. Link CE is made of aluminum ( E = 72 GPa) and has a cross- sectional area of 300 mm2. Knowing that they support rigid member ABC determine the maximum force P that can be applied vertically at point A if the deflection of A is not to exceed 0.35 mm.SOLUTIONFree body member AC: ΣM C = 0 : 0.350 P − 0.225 FBD = 0 FBD = 1.55556 P ΣM B = 0 : 0.125 P − 0.225 FCE = 0 FCE = 0.55556 P FBD LBD (1.55556 P) (0.225)δ B = δ BD = = 9 −6 = 13.8889 × 10−9 P EBD ABD (105 × 10 ) (240 × 10 ) F L (0.55556 P) (0.150)δ C = δ CE = CE CE = = 3.8581 × 10−9 P ECE ACE (72 × 109 ) (300 × 10−6 )Deformation Diagram:From the deformation diagram, δ B + δC 17.7470 × 10−9 P Slope, θ= = = 78.876 × 10−9 P LBC 0.225 δ A = δ B + LABθ = 13.8889 × 10−9 P + (0.125) (78.876 × 10−9 P) = 23.748 × 10−9 PApply displacement limit. δ A = 0.35 × 10−3 m = 23.748 × 10−9 P P = 14.7381 × 103 N P = 14.74 kN PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.28 Each of the four vertical links connecting the two rigid horizontal members is made of aluminum ( E = 70 GPa) and has a uniform rectangular cross section of 10 × 40 mm. For the loading shown, determine the deflection of (a) point E, (b) point F, (c) point G.SOLUTIONStatics. Free body EFG. ΣM F = 0 : − (400)(2 FBE ) − (250)(24) = 0 FBE = −7.5 kN = −7.5 × 103 N ΣM E = 0 : (400)(2 FCF ) − (650)(24) = 0 FCF = 19.5 kN = 19.5 × 103 NArea of one link: A = (10)(40) = 400 mm 2 = 400 × 10−6 m 2Length: L = 300 mm = 0.300 mDeformations. FBE L (−7.5 × 103 )(0.300) δ BE = = 9 −6 = −80.357 × 10−6 m EA (70 × 10 )(400 × 10 ) FCF L (19.5 × 103 )(0.300) δ CF = = = 208.93 × 10−6 m EA (70 × 109 )(400 × 10−6 )PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.28 (Continued)(a) Deflection of Point E. δ E = |δ BF | δ E = 80.4 μ m ↑ (b) Deflection of Point F. δ F = δ CF δ F = 209 μ m ↓  Geometry change. Let θ be the small change in slope angle. δE + δF 80.357 × 10−6 + 208.93 × 10−6 θ= = = 723.22 × 10−6 radians LEF 0.400(c) Deflection of Point G. δ G = δ F + LFG θ δ G = δ F + LFG θ = 208.93 × 10−6 + (0.250)(723.22 × 10−6 ) = 389.73 × 10−6 m δ G = 390 μ m ↓ PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.29 A vertical load P is applied at the center A of the upper section of a homogeneous frustum of a circular cone of height h, minimum radius a, and maximum radius b. Denoting by E the modulus of elasticity of the material and neglecting the effect of its weight, determine the deflection of point A.SOLUTIONExtend the slant sides of the cone to meet at a point O and place the origin of the coordinate system there. b−aFrom geometry, tan α = h a b a1 = , b1 = , r = y tan α tan α tan αAt coordinate point y, A = π r 2 PdyDeformation of element of height dy: dδ = AE P dy P dy dδ = = Eπ r 2 π E tan α y 2 2Total deformation. b1 P b1 dy P  1 P 1 1 δA = π E tan 2 α  a1 y 2 = 2 −  π E tan α  y  a1 = 2  −  π E tan α  a1 b1  P b1 − a1 P(b1 − a1 ) Ph = = δA = ↓  2 π E tan α a1b1 π Eab π EabPROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.30A homogeneous cable of length L and uniform cross section is suspended from one end. (a) Denoting by ρ thedensity (mass per unit volume) of the cable and by E its modulus of elasticity, determine the elongation of thecable due to its own weight. (b) Show that the same elongation would be obtained if the cable were horizontaland if a force equal to half of its weight were applied at each end.SOLUTION(a) For element at point identified by coordinate y, P = weight of portion below the point = ρ g A(L − y ) Pdy ρ gA( L − y )dy ρ g ( L − y ) dδ = = = dy EA EA E L L ρ g (L − y) ρg  1 2 δ=  0 E dy =  Ly − y  E  2 0 ρg  2 L  2 1 ρ gL2 = L −  δ=  E   2  2 E(b) Total weight: W = ρ gAL EAδ EA 1 ρ gL2 1 1 F= = ⋅ = ρ gAL F= W  L L 2 E 2 2PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.31The volume of a tensile specimen is essentially constant while plastic deformation occurs. If the initialdiameter of the specimen is d1, show that when the diameter is d, the true strain is ε t = 2 ln(d1 /d ).SOLUTION π πIf the volume is constant, d 2L = d12 L0 4 4 2 L d12  d1  = = L0 d 2  d    2 L d  d1 ε t = ln = ln  1  ε t = 2ln  L0 d  dPROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.32Denoting by ε the “engineering strain” in a tensile specimen, show that the true strain is ε t = ln (1 + ε ).SOLUTION L L +δ  δ  ε t = ln = ln 0 = ln 1 +  = ln (1 + ε ) L0 L0  L0 Thus ε t = ln (1 + ε ) PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.33 An axial force of 200 kN is applied to the assembly shown by means of rigid end plates. Determine (a) the normal stress in the aluminum shell, (b) the corresponding deformation of the assembly.SOLUTIONLet Pa = Portion of axial force carried by shell Pb = Portion of axial force carried by core. Pa L Ea Aa δ= , or Pa = δ Ea Aa L Pb L Eb Ab δ= , or Pb = δ Eb Ab L δThus, P = Pa + Pb = ( Ea Aa + Eb Ab ) L πwith Aa = [(0.060)2 − (0.025)2 ] = 2.3366 × 10−3 m 2 4 π Ab = (0.025) 2 = 0.49087 × 10−3 m 2 4 δ P = [(70 × 109 ) (2.3366 × 10−3 ) + (105 × 109 ) (0.49087 × 10−3 )] L δ P = 215.10 × 106 L δ P 200 × 103Strain: ε = = = = 0.92980 × 10−3 L 215.10 × 106 215.10 × 106(a) σ a = Ea ε = (70 × 109 ) (0.92980 × 10−3 ) = 65.1 × 106 Pa σ a = 65.1 MPa (b) δ = ε L = (0.92980 × 10−3 ) (300 mm) δ = 0.279 mm PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.34 The length of the assembly shown decreases by 0.40 mm when an axial force is applied by means of rigid end plates. Determine (a) the magnitude of the applied force, (b) the corresponding stress in the brass core.SOLUTIONLet Pa = Portion of axial force carried by shell and Pb = Portion of axial force carried by core. Pa L Ea Aa δ= , or Pa = δ Ea Aa L Pb L Eb Ab δ= , or Pb = δ Eb Ab L δThus, P = Pa + Pb = ( Ea Aa + Eb Ab ) L πwith Aa = [(0.060)2 − (0.025)2 ] = 2.3366 × 10−3 m 2 4 π Ab = (0.025)2 = 0.49087 × 10−3 m 2 4 δ δ P = [(70 × 109 ) (2.3366 × 10−3 ) + (105 × 109 ) (0.49087 × 10−3 )] = 215.10 × 106 L Lwith δ = 0.40 mm, L = 300 mm 0.40(a) P = (215.10 × 106 ) = 286.8 × 103 N P = 287 kN  300 Pb Eδ (105 × 109 )(0.40 × 10−3 )(b) σb = = b = −3 = 140 × 106 Pa σ b = 140.0 MPa  Ab L 300 × 10PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.35 A 4-ft concrete post is reinforced with four steel bars, each with a 3 -in. diameter. 4 Knowing that Es = 29 × 106 psi and Ec = 3.6 × 106 psi, determine the normal stresses in the steel and in the concrete when a 150-kip axial centric force P is applied to the post.SOLUTION  π  3 2  As = 4     = 1.76715 in 2 44    Ac = 82 − As = 62.233 in 2 Ps L Ps (48) δs = = = 0.93663 × 10−6 Ps As Es (1.76715)(29 × 106 ) Pc L Pc (48) δc = = = 0.21425 × 10−6 Pc Ac Ec (62.233)(3.6 × 106 )But δ s = δ c : 0.93663 × 10−6 Ps = 0.21425 × 10−6 Pc Ps = 0.22875 Pc (1)Also: Ps + Pc = P = 150 kips (2)Substituting (1) into (2): 1.22875Pc = 150 kips Pc = 122.075 kipsFrom (1): Ps = 0.22875(122.075) = 27.925 kips Ps 27.925 σs = − =− σ s = −15.80 ksi  As 1.76715 Pc 122.075 σc = − =− σ c = −1.962 ksi  Ac 62.233PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.36 A 250-mm bar of 15 × 30-mm rectangular cross section consists of two aluminum layers, 5-mm thick, brazed to a center brass layer of the same thickness. If it is subjected to centric forces of magnitude P = 30 kN, and knowing that Ea = 70 GPa and Eb = 105 GPa, determine the normal stress (a) in the aluminum layers, (b) in the brass layer.SOLUTIONFor each layer, A = (30)(5) = 150 mm 2 = 150 × 10−6 m 2Let Pa = load on each aluminum layer Pb = load on brass layer Pa L Pb L Eb 105Deformation. δ= = Pb = Pa = Pa = 1.5 Pa Ea A Eb A Ea 70Total force. P = 2 Pa + Pb = 3.5 Pa 2 3Solving for Pa and Pb, Pa = P Pb = P 7 7 Pa 2P 2 30 × 103(a) σa = − =− =− −6 = −57.1 × 106 Pa σ a = −57.1 MPa  A 7 A 7 150 × 10 Pb 3P 3 30 × 103(b) σb = − =− =− = −85.7 × 106 Pa σ b = −85.7 MPa  A 7 A 7 150 × 10−6PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.37 Determine the deformation of the composite bar of Prob. 2.36 if it is subjected to centric forces of magnitude P = 45 kN. PROBLEM 2.36 A 250-mm bar of 15 × 30-mm rectangular cross section consists of two aluminum layers, 5-mm thick, brazed to a center brass layer of the same thickness. If it is subjected to centric forces of magnitude P = 30 kN, and knowing that Ea = 70 GPa and Eb = 105 GPa, determine the normal stress (a) in the aluminum layers, (b) in the brass layer.SOLUTIONFor each layer, A = (30)(5) = 150 mm 2 = 150 × 10−6 m 2Let Pa = load on each aluminum layer Pb = load on brass layer Pa L PL Eb 105Deformation. δ =− =− b Pb = Pa = Pa = 1.5 Pa Ea A Eb A Ea 70 2Total force. P = 2 Pa + Pb = 3.5 Pa Pa = P 7 Pa L 2 PL δ =− =− Ea A 7 Ea A 2 (45 × 103 )(250 × 10−3 ) =− 7 (70 × 109 )(150 × 10−6 ) = −306 × 10−6 m δ = −0.306 mm PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.38 Compressive centric forces of 40 kips are applied at both ends of the assembly shown by means of rigid plates. Knowing that Es = 29 × 106 psi and Ea = 10.1 × 106 psi, determine (a) the normal stresses in the steel core and the aluminum shell, (b) the deformation of the assembly.SOLUTIONLet Pa = portion of axial force carried by shell Ps = portion of axial force carried by core Pa L Ea Aa δ= Pa = δ Ea Aa L Ps L Es As δ= Ps = δ Es As L δTotal force. P = Pa + Ps = ( Ea Aa + Es As ) L δ P =ε = L Ea Aa + Es AsData: P = 40 × 103 lb π π Aa = (d 0 − di2 ) = 2 (2.52 − 1.0)2 = 4.1233 in 2 4 4 π 2 π As = d = (1) = 0.7854 in 2 2 4 4 − 40 × 103 ε= 6 6 = −620.91 × 10−6 (10.1 × 10 )(4.1233) + (29 × 10 )(0.7854)(a) σ s = Es ε = (29 × 106 )(−620.91 × 10−6 ) = −18.01 × 103 psi −18.01 ksi  σ a = Ea ε = (10.1 × 106 )(620.91 × 10−6 ) = −6.27 × 103 psi  −6.27 ksi (b) δ = Lε = (10)(620.91 × 10−6 ) = −6.21 × 10−3 δ = −6.21 × 10−3 in. PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.39 Three wires are used to suspend the plate shown. Aluminum wires of 1 -in.8 1 diameter are used at A and B while a steel wire of 12 -in. diameter is used at C. Knowing that the allowable stress for aluminum ( Ea = 10.4 × 106 psi) is 14 ksi and that the allowable stress for steel ( Es = 29 × 106 psi) is 18 ksi, determine the maximum load P that can be applied.SOLUTIONBy symmetry, PA = PB , and δ A = δ BAlso, δC = δ A = δ B = δStrain in each wire: δ δ ε A = εB = , εC = = 2ε A 2L LDetermine allowable strain. σA 14 × 103Wires A&B: εA = = = 1.3462 × 10−3 EA 10.4 × 106 ε C = 2 ε A = 2.6924 × 10−4 σC 18 × 103Wire C: εC = = = 0.6207 × 10−3 EC 29 × 106 1 ε A = ε B = ε C = 0.3103 × 10−6 2Allowable strain for wire C governs, ∴ σ C = 18 × 103 psi σ A = E Aε A PA = AA E Aε A 2 π 1 = (10.4 × 106 )(0.3103 × 10−6 ) = 39.61 lb 4 8   PB = 39.61 lb 2 π1 σ C = EC ε C PC = ACσ C = (18 × 103 ) = 98.17 lb 4  12   For equilibrium of the plate, P = PA + PB + PC = 177.4 lb P = 177.4 lb PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.40 A polystyrene rod consisting of two cylindrical portions AB and BC is restrained at both ends and supports two 6-kip loads as shown. Knowing that E = 0.45 × 106 psi, determine (a) the reactions at A and C, (b) the normal stress in each portion of the rod.SOLUTION(a) We express that the elongation of the rod is zero: PAB LAB PBC LBC δ= + =0 π d2 E 4 AB π 4 2 d BC E But PAB = + RA PBC = − RC Substituting and simplifying: RA LAB RC LBC 2 − 2 =0 d AB d BC 2 2 LAB  d BC  25  2  RC =   RA =  RA LBC  d AB  15  1.25   RC = 4.2667 RA (1) From the free body diagram: RA + RC = 12 kips (2) Substituting (1) into (2): 5.2667 RA = 12 RA = 2.2785 kips RA = 2.28 kips ↑  From (1): RC = 4.2667 (2.2785) = 9.7217 kips RC = 9.72 kips ↑  PAB + RA 2.2785(b) σ AB = = = σ AB = +1.857 ksi  AAB AAB π (1.25)2 4 PBC − RC −9.7217 σ BC = = = σ BC = −3.09 ksi  ABC ABC π 4 (2)2PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.41 Two cylindrical rods, one of steel and the other of brass, are joined at C and restrained by rigid supports at A and E. For the loading shown and knowing that Es = 200 GPa and Eb = 105 GPa, determine (a) the reactions at A and E, (b) the deflection of point C.SOLUTIONA to C: E = 200 × 109 Pa π A= (40) 2 = 1.25664 × 103 mm 2 = 1.25664 × 10−3 m 2 4 EA = 251.327 × 106 NC to E: E = 105 × 109 Pa π A= (30) 2 = 706.86 mm 2 = 706.86 × 10−6 m 2 4 EA = 74.220 × 106 NA to B: P = RA L = 180 mm = 0.180 m PL RA (0.180) δ AB = = EA 251.327 × 106 = 716.20 × 10−12 RAB to C: P = RA − 60 × 103 L = 120 mm = 0.120 m PL ( RA − 60 × 103 )(0.120) δ BC = = EA 251.327 × 106 = 447.47 × 10−12 RA − 26.848 × 10−6PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.41 (Continued)C to D: P = RA − 60 × 103 L = 100 mm = 0.100 m PL ( RA − 60 × 103 )(0.100) δ BC = = EA 74.220 × 106 = 1.34735 × 10−9 RA − 80.841 × 10−6D to E: P = RA − 100 × 103 L = 100 mm = 0.100 m PL ( RA − 100 × 103 )(0.100) δ DE = = EA 74.220 × 106 = 1.34735 × 10−9 RA − 134.735 × 10−6A to E: δ AE = δ AB + δ BC + δ CD + δ DE = 3.85837 × 10−9 RA − 242.424 × 10−6Since point E cannot move relative to A, δ AE = 0(a) 3.85837 × 10−9 RA − 242.424 × 10−6 = 0 RA = 62.831 × 103 N RA = 62.8 kN ←  RE = RA − 100 × 103 = 62.8 × 103 − 100 × 103 = −37.2 × 103 N RE = 37.2 kN ← (b) δ C = δ AB + δ BC = 1.16367 × 10−9 RA − 26.848 × 10−6 = (1.16369 × 10−9 )(62.831 × 103 ) − 26.848 × 10−6 = 46.3 × 10−6 m δ C = 46.3 μ m → PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.42 Solve Prob. 2.41, assuming that rod AC is made of brass and rod CE is made of steel. PROBLEM 2.41 Two cylindrical rods, one of steel and the other of brass, are joined at C and restrained by rigid supports at A and E. For the loading shown and knowing that Es = 200 GPa and Eb = 105 GPa, determine (a) the reactions at A and E, (b) the deflection of point C.SOLUTIONA to C: E = 105 × 109 Pa π A= (40) 2 = 1.25664 × 103 mm 2 = 1.25664 × 10−3 m 2 4 EA = 131.947 × 106 NC to E: E = 200 × 109 Pa π A= (30) 2 = 706.86 mm 2 = 706.86 × 10−6 m 2 4 EA = 141.372 × 106 NA to B: P = RA L = 180 mm = 0.180 m PL RA (0.180) δ AB = = EA 131.947 × 106 = 1.36418 × 10−9 RAB to C: P = RA − 60 × 103 L = 120 mm = 0.120 m PL ( RA − 60 × 103 )(0.120) δ BC = = EA 131.947 × 106 = 909.456 × 10−12 RA − 54.567 × 10−6C to D: P = RA − 60 × 103 L = 100 mm = 0.100 m PL ( RA − 60 × 103 )(0.100) δ CD = = EA 141.372 × 106 = 707.354 × 10−12 RA − 42.441 × 10−6PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.42 (Continued)D to E: P = RA − 100 × 103 L = 100 mm = 0.100 m PL ( RA − 100 × 103 )(0.100) δ DE = = EA 141.372 × 106 = 707.354 × 10−12 RA − 70.735 × 10−6A to E: δ AE = δ AB + δ BC + δ CD + δ DE = 3.68834 × 10−9 RA − 167.743 × 10−6Since point E cannot move relative to A, δ AE = 0(a) 3.68834 × 10−9 RA − 167.743 × 10−6 = 0 RA = 45.479 × 103 N R A = 45.5 kN ←  RE = RA − 100 × 103 = 45.479 × 103 − 100 × 103 = −54.521 × 103 RE = 54.5 kN ← (b) δ C = δ AB + δ BC = 2.27364 × 10−9 RA − 54.567 × 10−6 = (2.27364 × 10−9 )(45.479 × 103 ) − 54.567 × 10−6 = 48.8 × 10−6 m δ C = 48.8 μ m → PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.43 The rigid bar ABCD is suspended from four identical wires. Determine the tension in each wire caused by the load P shown.SOLUTIONDeformations Let θ be the rotation of bar ABCD and δ A , δ B , δ C and δ D be the deformations of wires A, B,C, and D. δB − δ AFrom geometry, θ= L δ B = δ A + Lθ δ C = δ A + 2 Lθ = 2δ B − δ A (1) δ D = δ A + 3Lθ = 3δ B − 2δ A (2)Since all wires are identical, the forces in the wires are proportional to the deformations. TC = 2TB − TA (1′) TD = 3TB − 2TA (2′)PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.43 (Continued)Use bar ABCD as a free body. ΣM C = 0 : − 2 LTA − LTB + LTD = 0 (3) ΣFy = 0 : TA + TB + TC + TD − P = 0 (4)Substituting (2′) into (3) and dividing by L, −4TA + 2TB = 0 TB = 2TA (3′)Substituting (1′), (2′), and (3′) into (4), TA + 2TA + 3TA + 4TA − P = 0 10TA = P 1 TA = P  10  1  1 TB = 2TA = (2)   P TB = P   10  5 1   1  3 TC = (2)  P  −  P  TC = P   5   10  10 1   1  2 TD = (3)  P  − (2)  P  TD = P   5   10  5PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.44 1 The rigid bar AD is supported by two steel wires of 16 -in. diameter 6 ( E = 29 × 10 psi) and a pin and bracket at D. Knowing that the wires were initially taut, determine (a) the additional tension in each wire when a 120-lb load P is applied at B, (b) the corresponding deflection of point B.SOLUTIONLet θ be the rotation of bar ABCD.Then δ A = 24θ δ C = 8θ  PAE LAE δA = AE 6 1 2 EAδ A (29 × 10 ) π ( 16 ) (24θ ) 4 PAE = = LAE 15 = 142.353 × 103θ PCF LCF δC = AE EAδ C (29 × 10 ) 4 ( 16 ) (8θ ) 6 π 1 2 PCF = = LCF 8 = 88.971 × 103θUsing free body ABCD, ΣM D = 0 : −24PAE + 16P − 8PCF = 0 −24(142.353 × 103θ ) + 16(120) − 8(88.971 × 103θ ) = 0 θ = 0.46510 × 10−3 radۗ(a) PAE = (142.353 × 103 ) (0.46510 × 10−3 ) PAE = 66.2 lb  PCF = (88.971 × 103 ) (0.46510 × 10−3 ) PCF = 41.4 lb (b) δ B = 16θ = 16(0.46510 × 10−3 ) δ B = 7.44 × 10−3 in. ↓ PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.45 The steel rods BE and CD each have a 16-mm diameter ( E = 200 GPa); the ends of the rods are single-threaded with a pitch of 2.5 mm. Knowing that after being snugly fitted, the nut at C is tightened one full turn, determine (a) the tension in rod CD, (b) the deflection of point C of the rigid member ABC.SOLUTIONLet θ be the rotation of bar ABC as shown. Then δ B = 0.15θ δ C = 0.25θ PCD LCD But δ C = δ turn − ECD ACD ECD ACD PCD = (δ turn − δ C ) LCD (200 × 109 Pa) π (0.016 m) 2 = 4 (0.0025m − 0.25θ ) 2m = 50.265 × 103 − 5.0265 × 106 θ PBE LBE EBE ABE δB = or PBE = δB EBE ABE LBE (200 × 109 Pa) π (0.016 m) 2 PBE = 4 (0.15θ ) 3m = 2.0106 × 106 θFrom free body of member ABC: ΣM A = 0 : 0.15 PBE − 0.25PCD = 0 0.15(2.0106 × 106 θ ) − 0.25(50.265 × 103 − 5.0265 × 106 θ ) = 0 θ = 8.0645 × 10−3 rad(a) PCD = 50.265 × 103 − 5.0265 × 106 (8.0645 × 10−3 ) = 9.7288 × 103 N PCD = 9.73 kN (b) δ C = 0.25θ = 0.25(8.0645 × 10−3 ) = 2.0161 × 10−3 m δ C = 2.02 mm ← PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.46 Links BC and DE are both made of steel ( E = 29 × 106 psi) and are 1 2 in. wide and 1 in. thick. Determine (a) the force in each link 4 when a 600-lb force P is applied to the rigid member AF shown, (b) the corresponding deflection of point A.SOLUTIONLet the rigid member ACDF rotate through small angle θ clockwise about point F.Then δ C = δ BC = 4θ in. → δ D = −δ DE = 2θ in. → FL EAδ δ= or F= EA L  1  1 For links: A =    = 0.125 in 2  2  4  LBC = 4 in. LDE = 5 in. EA δ BC (29 × 106 )(0.125)(4θ ) FBC = = = 3.625 × 106 θ LBC 4 EAδ DE (29 × 106 )(0.125)( −2θ ) FDE = = = −1.45 × 106 θ LDE 5Use member ACDF as a free body. ΣM F = 0 : 8 P − 4 FBC + 2 FDE = 0 1 1 P= FBC − FDE 2 4 1 1 600 = (3.625 × 106 )θ − ( −1.45 × 106 )θ = 2.175 × 106 θ 2 4 −3 θ = 0.27586 × 10 rad ۖ +(a) FBC = (3.625 × 106 )(0.27586 × 10−3 ) FBC = 1000 lb  FDE = −(1.45 × 106 )(0.27586 × 10−3 ) FDE = −400 lb (b) Deflection at Point A. δ A = 8θ = (8)(0.27586 × 10−3 ) δ A = 2.21 × 10−3 in → PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.47 The concrete post ( Ec = 3.6 × 106 psi and α c = 5.5 × 10−6 / °F) is reinforced with six steel bars, each of 7 -in. diameter ( Es = 29 × 106 psi and 8 α s = 6.5 × 10−6 / °F). Determine the normal stresses induced in the steel and in the concrete by a temperature rise of 65°F.SOLUTION 2 π π 7 As = 6 d2 = 6 = 3.6079 in 2 4 48   Ac = 102 − As = 102 − 3.6079 = 96.392 in 2Let Pc = tensile force developed in the concrete.For equilibrium with zero total force, the compressive force in the six steel rods equals Pc . Pc PcStrains: εs = − + α s (ΔT ) εc = + α c ( ΔT ) Es As Ec Ac Pc PMatching: ε c = ε s + α c (ΔT ) = − c + α s (ΔT ) Ec Ac Es As  1 1   +  Pc = (α s − α c )(ΔT )  Ec Ac Es As   1 1  −6  6 + 6  Pc = (1.0 × 10 )(65)  (3.6 × 10 )(96.392) (29 × 10 )(3.6079)  Pc = 5.2254 × 103 lb Pc 5.2254 × 103 σc = = = 54.210 psi σ c = 54.2 psi Ac 96.392 Pc 5.2254 × 103 σs = − =− = −1448.32 psi  σ s = −1.448 ksi  As 3.6079PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.48 The assembly shown consists of an aluminum shell ( Ea = 10.6 × 106 psi, αa = 12.9 × 10–6/°F) fully bonded to a steel core ( Es = 29 × 106 psi, αs = 6.5 × 10–6/°F) and is unstressed. Determine (a) the largest allowable change in temperature if the stress in the aluminum shell is not to exceed 6 ksi, (b) the corresponding change in length of the assembly.SOLUTIONSince α a > α s , the shell is in compression for a positive temperature rise.Let σ a = −6 ksi = −6 × 103 psi π π Aa = 4 (d 2 o ) − di2 = 4 (1.252 − 0.752 ) = 0.78540 in 2 π π As = d2 = (0.75) 2 = 0.44179 in 2 4 4 P = −σ a Aa = σ s Aswhere P is the tensile force in the steel core. σ a Aa (6 × 103 )(0.78540) σs = − = = 10.667 × 103 psi As 0.44179 σs σa ε= + α s ( ΔT ) = + α a ( ΔT ) Es Ea σs σa (α a − α s )(ΔT ) = − Es Ea 10.667 × 103 6 × 103 (6.4 × 10−6 )(ΔT ) = + = 0.93385 × 10−3 29 × 106 10.6 × 106(a) ΔT = 145.91°F ΔT = 145.9°F  10.667 × 103(b) ε= + (6.5 × 10−6 )(145.91) = 1.3163 × 10−3 29 × 106 or −6 × 103 ε= + (12.9 × 10−6 )(145.91) = 1.3163 × 10−3 10.6 × 106 δ = Lε = (8.0)(1.3163 × 10−3 ) = 0.01053 in.  δ = 0.01053 in. PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.49 The aluminum shell is fully bonded to the brass core and the assembly is unstressed at a temperature of 15 °C. Considering only axial deformations, determine the stress in the aluminum when the temperature reaches 195 °C.SOLUTIONBrass core: E = 105 GPa α = 20.9 × 10−6/ °CAluminum shell: E = 70 GPa α = 23.6 × 10−6 / °CLet L be the length of the assembly.Free thermal expansion: ΔT = 195 − 15 = 180 °C Brass core: (δT )b = Lα b ( ΔT ) Aluminum shell: (δT ) = Lα a (ΔT )Net expansion of shell with respect to the core: δ = L(α a − α b )(ΔT )Let P be the tensile force in the core and the compressive force in the shell.Brass core: Eb = 105 × 109 Pa π Ab = (25) 2 = 490.87 mm 2 4 = 490.87 × 10−6 m 2 PL (δ P )b = Eb AbPROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.49 (Continued)Aluminum shell: Ea = 70 × 109 Pa π Aa = (602 − 252 ) 4 = 2.3366 × 103 mm 2 = 2.3366 × 10−3 m 2 δ = (δ P )b + (δ P ) a PL PL L(α b − α a )(ΔT ) = + = KPL Eb Ab Ea Aawhere 1 1 K= + Eb Ab Ea Aa 1 1 = + (105 × 10 )(490.87 × 10 ) (70 × 10 )(2.3366 × 10−3 ) 9 −6 9 = 25.516 × 10−9 N −1Then (α b − α a )(ΔT ) P= K (23.6 × 10−6 − 20.9 × 10−6 )(180) = 25.516 × 10−9 = 19.047 × 103 N P 19.047 × 103Stress in aluminum: σa = − =− −3 = −8.15 × 106 Pa σ a = −8.15 MPa  Aa 2.3366 × 10PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.50 Solve Prob. 2.49, assuming that the core is made of steel ( Es = 200 GPa, α s = 11.7 × 10−6 / °C) instead of brass. PROBLEM 2.49 The aluminum shell is fully bonded to the brass core and the assembly is unstressed at a temperature of 15 °C. Considering only axial deformations, determine the stress in the aluminum when the temperature reaches 195 °C.SOLUTIONAluminum shell: E = 70 GPa α = 23.6 × 10−6 / °CLet L be the length of the assembly.Free thermal expansion: ΔT = 195 − 15 = 180 °C Steel core: (δT ) s = Lα s (ΔT ) Aluminum shell: (δT ) a = Lα a (ΔT )Net expansion of shell with respect to the core: δ = L(α a − α s )( ΔT )Let P be the tensile force in the core and the compressive force in the shell. πSteel core: Es = 200 × 109 Pa, As = (25) 2 = 490.87 mm 2 = 490.87 × 10−6 m 2 4 PL (δ P ) s = Es AsAluminum shell: Ea = 70 × 109 Pa PL (δ P ) a = Ea Aa π Aa = (602 − 25) 2 = 2.3366 × 103 mm 2 = 2.3366 × 10−3 m 2 4 δ = (δ P ) s + (δ P )a PL PL L(α a − α s )(ΔT ) = + = KPL Es As Ea Aawhere 1 1 K= + Es As Ea Aa 1 1 = + (200 × 10 )(490.87 × 10 ) (70 × 10 )(2.3366 × 10−3 ) 9 −6 9 = 16.2999 × 10−9 N −1PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.50 (Continued)Then (α a − α s )(ΔT ) (23.6 × 10−6 − 11.7 × 10−6 )(180) P= = = 131.41 × 103 N K 16.2999 × 10−9 P 131.19 × 103Stress in aluminum: σ a = − =− −3 = −56.2 × 106 Pa σ a = −56.2 MPa  Aa 2.3366 × 10PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.51 A rod consisting of two cylindrical portions AB and BC is restrained at both ends. Portion AB is made of steel ( Es = 200 GPa, α s = 11.7 × 10−6 / °C) and portion BC is made of brass ( Eb = 105 GPa, α b = 20.9 × 10−6 / °C). Knowing that the rod is initially unstressed, determine the compressive force induced in ABC when there is a temperature rise of 50 °C.SOLUTION π π AAB = 2 d AB = (30)2 = 706.86 mm 2 = 706.86 × 10−6 m 2 4 4 π π ABC = 2 d BC = (50)2 = 1.9635 × 103 mm 2 = 1.9635 × 10−3 m 2 4 4Free thermal expansion: δT = LABα s (ΔT ) + LBC α b (ΔT ) = (0.250)(11.7 × 10−6 )(50) + (0.300)(20.9 × 10−6 )(50) = 459.75 × 10−6 mShortening due to induced compressive force P: PL PL δP = + Es AAB Eb ABC 0.250 P 0.300 P = + (200 × 10 )(706.86 × 10 ) (105 × 10 )(1.9635 × 10−3 ) 9 −6 9 = 3.2235 × 10−9 PFor zero net deflection, δ P = δ T 3.2235 × 10−9 P = 459.75 × 10−6 P = 142.62 × 103 N P = 142.6 kN PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.52A steel railroad track ( Es = 200 GPa, α s = 11.7 × 10−6 /°C) was laid out at a temperature of 6°C. Determinethe normal stress in the rails when the temperature reaches 48°C, assuming that the rails (a) are welded toform a continuous track, (b) are 10 m long with 3-mm gaps between them.SOLUTION(a) δT = α ( ΔT ) L = (11.7 × 10−6 )(48 − 6)(10) = 4.914 × 10−3 m PL Lσ (10)σ δP = = = = 50 × 10−12 σ AE E 200 × 109 δ = δT + δ P = 4.914 × 10−3 + 50 × 10−12 σ = 0 σ = −98.3 × 106 Pa σ = −98.3 MPa(b) δ = δT + δ P = 4.914 × 10−3 + 50 × 10−12 σ = 3 × 10−3 3 × 10−3 − 4.914 × 10−3 σ= 50 × 10−12 = −38.3 × 106 Pa σ = −38.3 MPa PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.53 A rod consisting of two cylindrical portions AB and BC is restrained at both ends. Portion AB is made of steel ( Es = 29 × 106 psi, α s = 6.5 × 10−6 /°F) and portion BC is made of aluminum ( Ea = 10.4 × 106 psi, α a = 13.3 × 10−6 /°F). Knowing that the rod is initially unstressed, determine (a) the normal stresses induced in portions AB and BC by a temperature rise of 70°F, (b) the corresponding deflection of point B.SOLUTION π π AAB = (2.25) 2 = 3.9761 in 2 ABC = (1.5) 2 = 1.76715 in 2 4 4Free thermal expansion. ΔT = 70°F (δ T ) AB = LABα s (ΔT ) + (24)(6.5 × 10−6 )(70) = 10.92 × 10−3 in (δT ) BC = LBC α a (ΔT ) = (32)(13.3 × 10−6 (70) = 29.792 × 10−3 in.Total: δT = (δT ) AB + (δT ) BC = 40.712 × 10−3 in.Shortening due to induced compressive force P. PLAB 24 P (δ P ) AB = = = 208.14 × 10−9 P Es AAB (29 × 106 )(3.9761) PLBC 32 P (δ P ) BC = = = 1741.18 × 10−9 P Ea ABC (10.4 × 106 )(1.76715)Total: δ P = (δ P ) AB + (δ P ) BC = 1949.32 × 10−9 PFor zero net deflection, δ P = δ T 1949.32 × 10−9 P = 40.712 × 10−3 P = 20.885 × 103 lb P 20.885 × 103(a) σ AB = − =− = −5.25 × 103 psi σ AB = −5.25 ksi  AAB 3.9761 P 20.885 × 103 σ BC = − =− = −11.82 × 103 psi σ BC = −11.82 ksi  ABC 1.76715(b) (δ P ) AB = (208.14 × 10−9 )(20.885 × 103 ) = 4.3470 × 10−3 in. δ B = (δT ) AB → + (δ P ) AB ← = 10.92 × 10−3 → + 4.3470 × 10−3 ← δ B = 6.57 × 10−3 in. →  or (δ P ) BC = (1741.18 × 10−9 )(20.885 × 103 ) = 36.365 × 10−3 in. δ B = (δT ) BC ← + (δ P ) BC → = 29.792 × 10−3 ← + 36.365 × 10−3 → = 6.57 × 10−3 in. → (checks)PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.54 Solve Prob. 2.53, assuming that portion AB of the composite rod is made of aluminum and portion BC is made of steel. PROBLEM 2.53 A rod consisting of two cylindrical portions AB and BC is restrained at both ends. Portion AB is made of steel ( Es = 29 × 106 psi, α s = 6.5 × 10−6 /°F) and portion BC is made of aluminum ( Ea = 10.4 × 106 psi, α a = 13.3 × 10−6 /°F). Knowing that the rod is initially unstressed, determine (a) the normal stresses induced in portions AB and BC by a temperature rise of 70°F, (b) the corresponding deflection of point B.SOLUTION π π AAB = (2.25) 2 = 3.9761 in 2 ABC = (1.5) 2 = 1.76715 in 2 4 4Free thermal expansion. ΔT = 70°F (δ T ) AB = LABα a ( ΔT ) = (24)(13.3 × 10−6 )(70) = 22.344 × 10−3 in. (δT ) BC = LBC α s (ΔT ) = (32)(6.5 × 10−6 )(70) = 14.56 × 10−3 in.Total: δT = (δT ) AB + (δT ) BC = 36.904 × 10−3 in.Shortening due to induced compressive force P. PLAB 24 P (δ P ) AB = = = 580.39 × 10−9 P Ea AAB (10.4 × 106 )(3.9761) PLBC 32 P (δ P ) BC = = 6 = 624.42 × 10−9 P Es ABC (29 × 10 )(1.76715)Total: δ P = (δ P ) AB + (δ P ) BC = 1204.81 × 10−9 PFor zero net deflection, δ P = δ T 1204.81 × 10−9 P = 36.904 × 10−3 P = 30.631 × 103 lb P 30.631 × 103(a) σ AB = − =− = −7.70 × 103 psi σ AB = −7.70 ksi  AAB 3.9761 P 30.631 × 103 σ BC = − =− = −17.33 × 103 psi σ BC = −17.33 ksi  ABC 1.76715PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.54 (Continued)(b) (δ P ) AB = (580.39 × 10−9 ) (30.631 × 103 ) = 17.7779 × 10−3 in. δ B = (δT ) AB → + (δ P ) AB ← = 22.344 × 10−3 → + 17.7779 × 10−3 ← δ B = 4.57 × 10−3 in. →  or (δ P ) BC = (624.42 × 10−9 )(30.631 × 103 ) = 19.1266 × 10−3 in. δ B = (δT ) BC ← + (δ P ) BC → = 14.56 × 10−3 ← + 19.1266 × 10−3 → = 4.57 × 10−3 in. → (checks)PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.55 A brass link ( Eb = 105 GPa, α b = 20.9 × 10−6 /°C) and a steel rod (Es = 200 GPa, α s = 11.7 × 10−6 / °C) have the dimensions shown at a temperature of 20 °C. The steel rod is cooled until it fits freely into the link. The temperature of the whole assembly is then raised to 45 °C. Determine (a) the final stress in the steel rod, (b) the final length of the steel rod.SOLUTIONInitial dimensions at T = 20 °C.Final dimensions at T = 45 °C. ΔT = 45 − 20 = 25 °CFree thermal expansion of each part: Brass link: (δT )b = α b (ΔT ) L = (20.9 × 10−6 )(25)(0.250) = 130.625 × 10−6 m Steel rod: (δT ) s = α s (ΔT ) L = (11.7 × 10−6 )(25)(0.250) = 73.125 × 10−6 mAt the final temperature, the difference between the free length of the steel rod and the brass link is δ = 120 × 10−6 + 73.125 × 10−6 − 130.625 × 10−6 = 62.5 × 10−6 mAdd equal but opposite forces P to elongate the brass link and contract the steel rod. Brass link: E = 105 × 109 Pa Ab = (2)(50)(37.5) = 3750 mm 2 = 3.750 × 10−3 m 2 PL P(0.250) (δ P ) = = = 634.92 × 10−12 P EA (105 × 109 )(3.750 × 10−3 ) π Steel rod: E = 200 × 109 Pa As = (30) 2 = 706.86 mm 2 = 706.86 × 10−6 m 2 4 PL P(0.250) (δ P ) s = = = 1.76838 × 10−9 P Es As (200 × 109 )(706.86 × 10−6 ) (δ P )b + (δ P ) s = δ : 2.4033 × 10−9 P = 62.5 × 10−6 P = 26.006 × 103 N P (26.006 × 103 )(a) Stress in steel rod: σs = − =− −6 = −36.8 × 106 Pa σ s = −36.8 MPa  As 706.86 × 10(b) Final length of steel rod: L f = L0 + (δ T ) s − (δ P ) s L f = 0.250 + 120 × 10−6 + 73.125 × 10−6 − (1.76838 × 10−9 )(26.003 × 103 ) = 0.250147 m L f = 250.147 mm PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.56 Two steel bars ( Es = 200 GPa and α s = 11.7 × 10−6 /°C) are used to reinforce a brass bar ( Eb = 105 GPa, α b = 20.9 × 10−6 / °C) that is subjected to a load P = 25 kN. When the steel bars were fabricated, the distance between the centers of the holes that were to fit on the pins was made 0.5 mm smaller than the 2 m needed. The steel bars were then placed in an oven to increase their length so that they would just fit on the pins. Following fabrication, the temperature in the steel bars dropped back to room temperature. Determine (a) the increase in temperature that was required to fit the steel bars on the pins, (b) the stress in the brass bar after the load is applied to it.SOLUTION(a) Required temperature change for fabrication: δT = 0.5 mm = 0.5 × 10−3 m Temperature change required to expand steel bar by this amount: δT = Lα s ΔT , 0.5 × 10−3 = (2.00)(11.7 × 10−6 )(ΔT ), ΔT = 0.5 × 10−3 = (2)(11.7 × 10−6 )(ΔT ) ΔT = 21.368 °C 21.4 °C (b) Once assembled, a tensile force P* develops in the steel, and a compressive force P* develops in the brass, in order to elongate the steel and contract the brass. Elongation of steel: As = (2)(5)(40) = 400 mm2 = 400 × 10−6 m 2 F *L P* (2.00) (δ P ) s = = = 25 × 10−9 P* As Es (400 × 10−6 )(200 × 109 ) Contraction of brass: Ab = (40)(15) = 600 mm 2 = 600 × 10−6 m 2 P* L P* (2.00) (δ P )b = = = 31.746 × 10−9 P* Ab Eb (600 × 10−6 )(105 × 109 ) But (δ P ) s + (δ P )b is equal to the initial amount of misfit: (δ P ) s + (δ P )b = 0.5 × 10−3 , 56.746 × 10−9 P* = 0.5 × 10−3 P* = 8.811 × 103 NStresses due to fabrication: * P* 8.811 × 103 Steel: σs = = = 22.03 × 106 Pa = 22.03 MPa As 400 × 10−6PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.56 (Continued) * P* 8.811 × 103 Brass: σb = − =− −6 = −14.68 × 106 Pa = −14.68 MPa Ab 600 × 10To these stresses must be added the stresses due to the 25 kN load.For the added load, the additional deformation is the same for both the steel and the brass. Let δ ′ be theadditional displacement. Also, let Ps and Pb be the additional forces developed in the steel and brass,respectively. Ps L PL δ′ = = b As Es Ab Eb As Es (400 × 10−6 )(200 × 109 ) Ps = δ′ = δ ′ = 40 × 106 δ ′ L 2.00 Ab Eb (600 × 10−6 )(105 × 109 ) Pb = δ′ = δ ′ = 31.5 × 106 δ ′ L 2.00Total P = Ps + Pb = 25 × 103 N 40 × 106 δ ′ + 31.5 × 106 δ ′ = 25 × 103 δ ′ = 349.65 × 10−6 m Ps = (40 × 106 )(349.65 × 10−6 ) = 13.986 × 103 N Pb = (31.5 × 106 )(349.65 × 10−6 ) = 11.140 × 103 N Ps 13.986 × 103 σs = = = 34.97 × 106 Pa As 400 × 10−6 Pb 11.140 × 103 σb = = = 18.36 × 106 Pa Ab 600 × 10−6Add stress due to fabrication.Total stresses: σ s = 34.97 × 106 + 22.03 × 106 = 57.0 × 106 Pa σ s = 57.0 MPa σ b = 18.36 × 106 − 14.68 × 106 = 3.68 × 106 Pa σ b = 3.68 MPa PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.57 Determine the maximum load P that may be applied to the brass bar of Prob. 2.56 if the allowable stress in the steel bars is 30 MPa and the allowable stress in the brass bar is 25 MPa. PROBLEM 2.56 Two steel bars ( Es = 200 GPa and α s = 11.7 × 10–6/°C) are used to reinforce a brass bar ( Eb = 105 GPa, α b = 20.9 × 10–6/°C) that is subjected to a load P = 25 kN. When the steel bars were fabricated, the distance between the centers of the holes that were to fit on the pins was made 0.5 mm smaller than the 2 m needed. The steel bars were then placed in an oven to increase their length so that they would just fit on the pins. Following fabrication, the temperature in the steel bars dropped back to room temperature. Determine (a) the increase in temperature that was required to fit the steel bars on the pins, (b) the stress in the brass bar after the load is applied to it.SOLUTIONSee solution to Problem 2.56 to obtain the fabrication stresses. * σ s = 22.03 MPa * σ b = 14.68 MPaAllowable stresses: σ s ,all = 30 MPa, σ b ,all = 25 MPaAvailable stress increase from load. σ s = 30 − 22.03 = 7.97 MPa σ b = 25 + 14.68 = 39.68 MPaCorresponding available strains. σs 7.97 × 106 εs = = = 39.85 × 10−6 Es 200 × 109 σb 39.68 × 106 εb = = = 377.9 × 10−6 Eb 105 × 109Smaller value governs ∴ ε = 39.85 × 10−6Areas: As = (2)(5)(40) = 400 mm2 = 400 × 10−6 m 2 Ab = (15)(40) = 600 mm 2 = 600 × 10−6 m 2Forces Ps = Es As ε = (200 × 109 )(400 × 10−6 )(39.85 × 10−6 ) = 3.188 × 103 N Pb = Eb Abε = (105 × 109 )(600 × 10−6 )(39.85 × 10−6 ) = 2.511 × 10−3 NTotal allowable additional force: P = Ps + Pb = 3.188 × 103 + 2.511 × 103 = 5.70 × 103 N P = 5.70 kN PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.58 Knowing that a 0.02-in. gap exists when the temperature is 75 °F, determine (a) the temperature at which the normal stress in the aluminum bar will be equal to −11 ksi, (b) the corresponding exact length of the aluminum bar.SOLUTION σ a = −11 ksi = −11 × 103 psi P = −σ a Aa = (11 × 103 )(2.8) = 30.8 × 103 lbShortening due to P: PLb PLa δP = + Eb Ab Ea Aa (30.8 × 103 )(14) (30.8 × 103 )(18) = + (15 × 106 )(2.4) (10.6 × 106 )(2.8) = 30.657 × 10−3 in.Available elongation for thermal expansion: δT = 0.02 + 30.657 × 10−3 = 50.657 × 10−3 in.But δ T = Lbα b (ΔT ) + Laα a (ΔT ) = (14)(12 × 10−6 )( ΔT ) + (18)(12.9 × 10−6 )(ΔT ) = 400.2 × 10−6 ) ΔTEquating, (400.2 × 10−6 )ΔT = 50.657 × 10−3 ΔT = 126.6 °F(a) Thot = Tcold + ΔT = 75 + 126.6 = 201.6 °F Thot = 201.6°F  PLa(b) δ a = Laα a (ΔT ) − Ea Aa (30.8 × 103 )(18) = (18)(12.9 × 10−6 )(26.6) − 6 = 10.712 × 10−3 in. (10.6 × 10 )(2.8) Lexact = 18 + 10.712 × 10−3 = 18.0107 in. L = 18.0107 in. PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.59 Determine (a) the compressive force in the bars shown after a temperature rise of 180 °F, (b) the corresponding change in length of the bronze bar.SOLUTIONThermal expansion if free of constraint: δT = Lbα b (ΔT ) + Laα a (ΔT ) = (14)(12 × 10−6 )(180) + (18)(12.9 × 10−6 )(180) = 72.036 × 10−3 in.Constrained expansion: δ = 0.02in.Shortening due to induced compressive force P: δ P = 72.036 × 10−3 − 0.02 = 52.036 × 10−3 in. PLb PLa  Lb L But δP = + = + a P Eb Ab Ea Aa  Eb Ab Ea Aa   14 18  −9 = 6 + 6  P = 995.36 × 10 P  (15 × 10 )(2.4) (10.6 × 10 )(2.8) Equating, 995.36 × 10−9 P = 52.036 × 10−3 P = 52.279 × 103 lb(a) P = 52.3 kips  PLb(b) δ b = Lbα b (ΔT ) − Eb Ab (52.279 × 103 )(14) = (14)(12 × 10−6 )(180) − = 9.91 × 10−3 in. δ b = 9.91 × 10−3 in.  (15 × 106 )(2.4)PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.60 At room temperature (20 °C) a 0.5-mm gap exists between the ends of the rods shown. At a later time when the temperature has reached 140°C, determine (a) the normal stress in the aluminum rod, (b) the change in length of the aluminum rod.SOLUTION ΔT = 140 − 20 = 120 °CFree thermal expansion: δT = Laα a (ΔT ) + Lsα s ( ΔT ) = (0.300)(23 × 10−6 )(120) + (0.250)(17.3 × 10−6 )(120) = 1.347 × 10−3 mShortening due to P to meet constraint: δ P = 1.347 × 10−3 − 0.5 × 10−3 = 0.847 × 10−3 m PLa PLs  La L  δP = + = + s P Ea Aa Es As  Ea Aa Es As   0.300 0.250  = 9 −6 + 9 −6  P  (75 × 10 )(2000 × 10 ) (190 × 10 )(800 × 10 )  = 3.6447 × 10−9 PEquating, 3.6447 × 10−9 P = 0.847 × 10−3 P = 232.39 × 103 N P 232.39 × 103(a) σa = − =− = −116.2 × 106 Pa σ a = −116.2 MPa  Aa 2000 × 10−6 PLa(b) δ a = Laα a (ΔT ) − Ea Aa (232.39 × 103 )(0.300) = (0.300)(23 × 10−6 )(120) − 9 −6 = 363 × 10−6 m δ a = 0.363 mm  (75 × 10 )(2000 × 10 )PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.61 A 600-lb tensile load is applied to a test 1 coupon made from 16 -in. flat steel plate 6 (E = 29 × 10 psi and v = 0.30). Determine the resulting change (a) in the 2-in. gage length, (b) in the width of portion AB of the test coupon, (c) in the thickness of portion AB, (d) in the cross-sectional area of portion AB.SOLUTION  1  1  A =    = 0.03125 in 2  2  16  P 600 σ= = = 19.2 × 103 psi A 0.03125 σ 19.2 × 103 εx = = = 662.07 × 10−6 E 29 × 106(a) δ x = L0ε x = (2.0)(662.07 × 10−6 ) δ l = 1.324 × 10−3 in.  ε y = ε z = −vε x = −(0.30)(662.07 × 10−6 ) = −198.62 × 10−6 1(b) δ width = w0ε y =   ( −198.62 × 10−6 ) δ w = −99.3 × 10−6 in.  2  1 (c) δ thickness = t0ε z =   (−198.62 × 10−6 ) δ t = −12.41 × 10−6 in.   16 (d) A = wt = w0 (1 + ε y )t0 (1 + ε z ) = w0t0 (1 + ε y + ε z + ε y ε z ) ΔA = A − A0 = w0t0 (ε y + ε z + ε y ε z )  1  1  =    ( −198.62 × 10−6 − 198.62 × 10−6 + negligible term)  2  16  = −12.41 × 10−6 in 2 ΔA = −12.41 × 10−6 in 2 PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.62 In a standard tensile test, a steel rod of 22-mm diameter is subjected to a tension force of 75 kN. Knowing that v = 0.3 and E = 200 GPa, determine (a) the elongation of the rod in a 200-mm gage length, (b) the change in diameter of the rod.SOLUTION π π P = 75 kN = 75 × 103 N A= d2 = (0.022)2 = 380.13 × 10−6 m 2 4 4 P 75 × 103 σ = = = 197.301 × 106 Pa A 380.13 × 10−6 σ 197.301 × 106 εx = = = 986.51 × 10−6 E 200 × 109 δ x = Lε x = (200 mm)(986.51 × 10−6 ) (a) δ x = 0.1973 mm  ε y = −vε x = −(0.3)(986.51 × 10−6 ) = −295.95 × 10−6  δ y = d ε y = (22 mm)(−295.95 × 10−6 )  (b) δ y = −0.00651 mm PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.63A 20-mm-diameter rod made of an experimental plastic is subjected to a tensile force of magnitude P = 6 kN.Knowing that an elongation of 14 mm and a decrease in diameter of 0.85 mm are observed in a 150-mm length,determine the modulus of elasticity, the modulus of rigidity, and Poisson’s ratio for the material.SOLUTIONLet the y-axis be along the length of the rod and the x-axis be transverse. π A= (20)2 = 314.16 mm 2 = 314.16 × 10−6 m 2 P = 6 × 103 N 4 P 6 × 103 σy = = −6 = 19.0985 × 106 Pa A 314.16 × 10 δ y 14 mm εy = = = 0.093333 L 150 mm σy 19.0985 × 106Modulus of elasticity: E = = = 204.63 × 106 Pa E = 205 MPa  εy 0.093333 δx 0.85 εx = =− = −0.0425 d 20 εx −0.0425Poisson’s ratio: v=− =− v = 0.455  εy 0.093333 E 204.63 × 106Modulus of rigidity: G= = = 70.31 × 106 Pa G = 70.3 MPa  2(1 + v) (2)(1.455)PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.64 The change in diameter of a large steel bolt is carefully measured as the nut is tightened. Knowing that E = 29 × 106 psi and v = 0.30, determine the internal force in the bolt, if the diameter is observed to decrease by 0.5 × 10−3 in.SOLUTION δ y = −0.5 × 10−3 in. d = 2.5 in. εy 0.5 × 10−3 εy = =− = −0.2 × 10−3 d 2.5 εy −ε y 0.2 × 10−3 v=− : εx = = = 0.66667 × 10−3 εx v 0.3 σ x = Eε x = (29 × 106 )(0.66667 × 10−3 ) = 19.3334 × 103 psi π π A= d2 = (2.5) 2 = 4.9087 in 2 4 4 F = σ x A = (19.3334 × 103 )(4.9087) = 94.902 × 103 lb F = 94.9 kips PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.65 A 2.5-m length of a steel pipe of 300-mm outer diameter and 15-mm wall thickness is used as a column to carry a 700-kN centric axial load. Knowing that E = 200 GPa and v = 0.30, determine (a) the change in length of the pipe, (b) the change in its outer diameter, (c) the change in its wall thickness.SOLUTION d o = 0.3 m t = 0.015 m L = 2.5 m di = do − 2t = 0.3 − 2(0.015) = 0.27 m P = 700 × 103 N π π A= ( do2 − di2 ) = (0.32 − 0.272 ) = 13.4303 × 10−3 m 2 4 4 PL (700 × 103 )(2.5)(a) δ =− =− EA (200 × 109 )(13.4303 × 10−3 ) = −651.51 × 10−6 m δ = −0.652 mm  δ −651.51 × 10−6 ε= = = −260.60 × 10−6 L 2.5 ε LAT = −vε = −(0.30)(−260.60 × 10−6 ) = 78.180 × 10−6(b) Δdo = do ε LAT = (300 mm)(78.180 × 10−6 ) Δd o = 0.0235 mm (c) Δt = tε LAT = (15 mm)(78.180 × 10−6 ) Δt = 0.001173 mm PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.66 An aluminum plate ( E = 74 GPa and v = 0.33) is subjected to a centric axial load that causes a normal stress σ. Knowing that, before loading, a line of slope 2:1 is scribed on the plate, determine the slope of the line when σ = 125 MPa.SOLUTION 2(1 + ε y )The slope after deformation is tan θ = 1+ εx σx 125 × 106 εx = = = 1.6892 × 10−3 E 74 × 109 ε y = −vε x = −(0.33)(1.6892 × 10−3 ) = −0.5574 × 10−3 2(1 − 0.0005574) tan θ = = 1.99551 tan θ = 1.99551  1 + 0.0016892PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.67 The block shown is made of a magnesium alloy, for which E = 45 GPa and v = 0.35. Knowing that σ x = −180 MPa, determine (a) the magnitude of σ y for which the change in the height of the block will be zero, (b) the corresponding change in the area of the face ABCD, (c) the corresponding change in the volume of the block.SOLUTION(a) δy = 0 εy = 0 σz = 0 1 εy = (σ x − vσ y − vσ z ) E σ y = vσ x = (0.35)(−180 × 106 ) = −63 × 106 Pa σ y = −63 MPa  1 v (0.35)( −243 × 106 ) εz = (σ z − vσ x − vσ y ) = − (σ x + σ y ) = = −1.89 × 10−3 E E 45 × 109 1 σ x − vσ y 157.95 × 106 ε x = (σ x − vσ y − vσ Z ) = =− = −3.51 × 10−3 E E 45 × 109(b) A0 = Lx Lz A = Lx (1 + ε x ) Lz (1 + ε z ) = Lx Lz (1 + ε x + ε z + ε xε z ) Δ A = A − A0 = Lx Lz (ε x + ε z + ε x ε z ) ≈ Lx Lz (ε x + ε z ) Δ A = (100 mm)(25 mm)(−3.51 × 10−3 − 1.89 × 10−3 ) Δ A = −13.50 mm 2 (c) V0 = Lx Ly Lz V = Lx (1 + ε x ) Ly (1 + ε y ) Lz (1 + ε z ) = Lx Ly Lz (1 + ε x + ε y + ε z + ε xε y + ε y ε z + ε z ε x + ε x ε y ε z ) ΔV = V − V0 = Lx Ly Lz (ε x + ε y + ε z + small terms) ΔV = (100)(40)(25)(−3.51 × 10−3 + 0 − 1.89 × 10−3 ) ΔV = −540 mm3 PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.68 A 30-mm square was scribed on the side of a large steel pressure vessel. After pressurization, the biaxial stress condition at the square is as shown. For E = 200 GPa and v = 0.30, determine the change in length of (a) side AB, (b) side BC, (c) diagnonal AC.SOLUTIONGiven: σ x = 80 MPa σ y = 40 MPa 1 80 − 0.3(40)Using Eq’s (2.28): εx = (σ x − vσ y ) = = 340 × 10−6 E 200 × 103 1 40 − 0.3(80) εy = (σ y − vσ x ) = = 80 × 10−6 E 200 × 103(a) Change in length of AB. δ AB = ( AB)ε x = (30 mm)(340 × 10−6 ) = 10.20 × 10−3 mm δ AB = 10.20 μ m (b) Change in length of BC. δ BC = ( BC )ε y = (30 mm)(80 × 10−6 ) = 2.40 × 10−3 mm δ BC = 2.40 μ m (c) Change in length of diagonal AC. From geometry, ( AC )2 = ( AB)2 + ( BC ) 2 Differentiate: 2( AC ) Δ( AC ) = 2( AB)Δ( AB) + 2( BC )Δ( BC ) But Δ( AC ) = δ AC Δ( AB) = δ AB Δ( BC ) = δ BC Thus, 2( AC )δ AC = 2( AB)δ AB + 2( BC )δ BC AB BC 1 1 δ AC = δ AB + δ BC = (10.20 μ m)+ (2.40 μ m) AC AC 2 2 δ AC = 8.91 μ m PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.69 The aluminum rod AD is fitted with a jacket that is used to apply a hydrostatic pressure of 6000 psi to the 12-in. portion BC of the rod. Knowing that E = 10.1 × 106 psi and v = 0.36, determine (a) the change in the total length AD, (b) the change in diameter at the middle of the rod.SOLUTION σ x = σ z = − P = −6000 psi σy = 0 1 εx = (σ x − vσ y − vσ z ) E 1 = [−6000 − (0.36)(0) − (0.36)( −6000)] 10.1 × 106 = −380.198 × 10−6 1 ε y = (−vσ x + σ y − vσ z ) E 1 = [−(0.36)(−6000) + 0 − (0.36)( −6000)] 10.1 × 106 = 427.72 × 10−6Length subjected to strain ε x: L = 12 in.(a) δ y = Lε y = (12)(427.72 × 10−6 ) δ l = 5.13 × 10−3 in. (b) δ x = d ε x = (1.5)(−380.198 × 10−6 ) δ d = −0.570 × 10−3 in. PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.70 For the rod of Prob. 2.69, determine the forces that should be applied to the ends A and D of the rod (a) if the axial strain in portion BC of the rod is to remain zero as the hydrostatic pressure is applied, (b) if the total length AD of the rod is to remain unchanged. PROBLEM 2.69 The aluminum rod AD is fitted with a jacket that is used to apply a hydrostatic pressure of 6000 psi to the 12-in. portion BC of the rod. Knowing that E = 10.1 × 106 psi and v = 0.36, determine (a) the change in the total length AD, (b) the change in diameter at the middle of the rod.SOLUTIONOver the pressurized portion BC, σx =σz = −p σy =σy 1 (ε y ) BC = (−vσ x + σ y − vσ z ) E 1 = (2vp + σ y ) E(a) (ε y ) BC = 0 2vp + σ y = 0 σ y = −2vp = −(2)(0.36)(6000) = −4320 psi π π A= d2 = (1.5) 2 = 1.76715 in 2 4 4 F = Aσ y = (1.76715)(−4320) = −7630 lb i.e., 7630 lb compression (b) Over unpressurized portions AB and CD, σx =σz = 0 σy (ε y ) AB = (ε y )CD = E For no change in length, δ = LAB (ε y ) AB + LBC (ε y ) BC + LCD (ε y )CD = 0 ( LAB + LCD )(ε y ) AB + LBC (ε y ) BC = 0 σy 12 (20 − 12) (2vp + σ y ) = 0 + E E 24vp (24)(0.36)(6000) σy = − =− = −2592 psi 20 20 P = Aσ y = (1.76715)(−2592) = −4580 lb P = 4580 lb compression PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.71 In many situations, physical constraints prevent strain from occurring in a given direction. For example, ε z = 0 in the case shown, where longitudinal movement of the long prism is prevented at every point. Plane sections perpendicular to the longitudinal axis remain plane and the same distance apart. Show that for this situation, which is known as plane strain, we can express σ z , ε x , and ε y as follows: σ z = v(σ x + σ y ) 1 εx = [(1 − v 2 )σ x − v(1 + v)σ y ] E 1 ε y = [(1 − v 2 )σ y − v(1 + v)σ x ] ESOLUTION 1 εz = 0 = (−vσ x − vσ y + σ z ) or σ z = v(σ x + σ y )  E 1 εx = (σ x − vσ y − vσ z ) E 1 = [σ x − vσ y − v 2 (σ x + σ y )] E 1 = [(1 − v 2 )σ x − v(1 + v)σ y ]  E 1 εy = ( −vσ x + σ y − vσ z ) E 1 = [−vσ x + σ y − v 2 (σ x + σ y )] E 1 = [(1 − v 2 )σ y − v(1 + v)σ x ]  EPROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.72 In many situations, it is known that the normal stress in a given direction is zero, for example, σ z = 0 in the case of the thin plate shown. For this case, which is known as plane stress, show that if the strains εx and εy have been determined experimentally, we can express σ x , σ y , and ε z as follows: ε x + vε y ε y + vε x v σx = E 2 σy = E 2 εz = − (ε x + ε y ) 1− v 1− v 1− vSOLUTION σz = 0 1 εx = (σ x − vσ y ) (1) E 1 εy = ( −vσ x + σ y ) (2) EMultiplying (2) by v and adding to (1), 1 − v2 E ε x + vε y = σx or σx = (ε x + vε y )  E 1 − v2Multiplying (1) by v and adding to (2), 1 − v2 E ε y + vε x = σy or σy = (ε y + vε x )  E 1 − v2 1 v E εz = (−vσ x − vσ y ) = − ⋅ (ε x + vε y + ε y + vε x ) E E 1 − v2 v(1 + v) v =− 2 (ε x + ε y ) = − (ε x + ε y )  1− v 1− vPROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.73 For a member under axial loading, express the normal strain ε′ in a direction forming an angle of 45° with the axis of the load in terms of the axial strain εx by (a) comparing the hypotenuses of the triangles shown in Fig. 2.49, which represent, respectively, an element before and after deformation, (b) using the values of the corresponding stresses of σ ′ and σx shown in Fig. 1.38, and the generalized Hooke’s law.SOLUTION Figure 2.49(a) [ 2(1 + ε ′)]2 = (1 + ε x ) 2 + (1 − vε x ) 2 2(1 + 2ε ′ + ε ′2 ) = 1 + 2ε x + ε x + 1 − 2vε x + v 2ε x 2 2 4ε ′ + 2ε ′2 = 2ε x + ε x − 2vε x + v 2ε x 2 2 1− v Neglect squares as small 4ε ′ = 2ε x − 2vε x ε′ = εx  2 (A) (B)PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.73 (Continued) σ′ vσ ′(b) ε′ = − E E 1− v P = ⋅ E 2A 1− v = σx 2E 1− v = εx  2PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.74 The homogeneous plate ABCD is subjected to a biaxial loading as shown. It is known that σ z = σ 0 and that the change in length of the plate in the x direction must be zero, that is, ε x = 0. Denoting by E the modulus of elasticity and by v Poisson’s ratio, determine (a) the required magnitude of σ x , (b) the ratio σ 0 / ε z .SOLUTION σ z = σ 0 , σ y = 0, ε x = 0 1 1 εx = (σ x − vσ y − vσ z ) = (σ x − vσ 0 ) E E(a) σ x = vσ 0  1 1 1 − v2 σ0 E(b) εz = (−vσ x − vσ y + σ z ) = (−v 2σ 0 − 0 + σ 0 ) = σ0 =  E E E ε z 1 − v2PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.75 A vibration isolation unit consists of two blocks of hard rubber bonded to a plate AB and to rigid supports as shown. Knowing that a force of magnitude P = 25 kN causes a deflection δ = 1.5 mm of plate AB, determine the modulus of rigidity of the rubber used.SOLUTION 1 1 F = P = (25 × 103 N) = 12.5 × 103 N 2 2 F 12.5 × 103 N) τ = = = 833.33 × 103 Pa A (0.15 m)(0.1 m) δ = 1.5 × 10−3 m h = 0.03 m 1.5 × 10−3 δ γ= = = 0.05 h 0.03 τ 833.33 × 103 G = = = 16.67 × 106 Pa γ 0.05 G = 16.67 MPa PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.76 A vibration isolation unit consists of two blocks of hard rubber with a modulus of rigidity G = 19 MPa bonded to a plate AB and to rigid supports as shown. Denoting by P the magnitude of the force applied to the plate and by δ the corresponding deflection, determine the effective spring constant, k = P/δ , of the system.SOLUTION δ Shearing strain: γ = h Gδ Shearing stress: τ = Gγ = h 1 GAδ 2GAδ Force: P = Aτ = or P= 2 h h P 2GA Effective spring constant: k = = δ h with A = (0.15)(0.1) = 0.015 m 2 h = 0.03 m 2(19 × 106 Pa)(0.015 m 2 ) k = = 19.00 × 106 N/m 0.03 m k = 19.00 × 103 kN/m PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.77 The plastic block shown is bonded to a fixed base and to a horizontal rigid plate to which a force P is applied. Knowing that for the plastic used G = 55 ksi, determine the deflection of the plate when P = 9 kips.SOLUTIONConsider the plastic block. The shearing force carried is P = 9 × 103 lbThe area is A = (3.5)(5.5) = 19.25 in 2 P 9 × 103Shearing stress: τ = = = 467.52 psi A 19.25 τ 467.52Shearing strain: γ = = = 0.0085006 G 55 × 103 δBut γ = ∴ δ = hγ = (2.2)(0.0085006) δ = 0.0187 in.  hPROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.78 A vibration isolation unit consists of two blocks of hard rubber bonded to plate AB and to rigid supports as shown. For the type and grade of rubber used τ all = 220 psi and G = 1800 psi. Knowing that a centric vertical force of magnitude P = 3.2 kips must cause a 0.1-in. vertical deflection of the plate AB, determine the smallest allowable dimensions a and b of the block.SOLUTION 1Consider the rubber block on the right. It carries a shearing force equal to 2 P. 1 PThe shearing stress is τ= 2 A P 3.2 × 103or required area A= = = 7.2727 in 2 2τ (2)(220)But A = (3.0)b AHence, b= = 2.42 in. bmin = 2.42 in.  3.0Use b = 2.42 in and τ = 220 psi τ 220Shearing strain. γ= = = 0.12222 G 1800 δBut γ= a δ 0.1Hence, a= = = 0.818 in. amin = 0.818 in.  γ 0.12222PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.79 The plastic block shown is bonded to a rigid support and to a vertical plate to which a 55-kip load P is applied. Knowing that for the plastic used G = 150 ksi, determine the deflection of the plate. SOLUTION A = (3.2)(4.8) = 15.36 in 2 P = 55 × 103 lb P 55 × 103 τ = = = 3580.7 psi A 15.36 G = 150 × 103 psi τ 3580.7 γ = = = 23.871 × 10−3 G 150 × 103 h = 2 in.  δ = hγ = (2)(23.871 × 10−3 ) δ = 0.0477 in. ↓  = 47.7 × 10−3 in.PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.80 1What load P should be applied to the plate of Prob. 2.79 to produce a 16 -in. deflection?PROBLEM 2.79 The plastic block shown is bonded to a rigid support and to a vertical plate to which a55-kip load P is applied. Knowing that for the plastic used G = 150 ksi, determine the deflection of the plate.SOLUTION 1 δ = in. = 0.0625 in. 16 h = 2 in. δ 0.0625 γ = = = 0.03125 h 2 G = 150 × 103 psi τ = Gγ = (150 × 103 )(0.03125) = 4687.5 psi A = (3.2)(4.8) = 15.36 in 2 P = τ A = (4687.5)(15.36) = 72 × 103 lb 72 kips PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.81 Two blocks of rubber with a modulus of rigidity G = 12 MPa are bonded to rigid supports and to a plate AB. Knowing that c = 100 mm and P = 45 kN, determine the smallest allowable dimensions a and b of the blocks if the shearing stress in the rubber is not to exceed 1.4 MPa and the deflection of the plate is to be at least 5 mm.SOLUTION δ τShearing strain: γ= = a G Gδ (12 × 106 Pa)(0.005 m) a= = = 0.0429 m a = 42.9 mm  τ 1.4 × 106 Pa 1 P PShearing stress: τ= 2 = A 2bc P 45 × 103 N b= = = 0.1607 m b = 160.7 mm  2cτ 2(0.1 m) (1.4 × 106 Pa)PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.82 Two blocks of rubber with a modulus of rigidity G = 10 MPa are bonded to rigid supports and to a plate AB. Knowing that b = 200 mm and c = 125 mm, determine the largest allowable load P and the smallest allowable thickness a of the blocks if the shearing stress in the rubber is not to exceed 1.5 MPa and the deflection of the plate is to be at least 6 mm.SOLUTION 1 P PShearing stress: τ= 2 = A 2bc P = 2bcτ = 2(0.2 m)(0.125 m)(1.5 × 103 kPa) P = 75.0 kN  δ τShearing strain: γ= = a G Gδ (10 × 106 Pa)(0.006 m) a= = = 0.04 m a = 40.0 mm  τ 1.5 × 106 PaPROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.83∗ Determine the dilatation e and the change in volume of the 200-mm length of the rod shown if (a) the rod is made of steel with E = 200 GPa and v = 0.30, (b) the rod is made of aluminum with E = 70 GPa and v = 0.35.SOLUTION π π A= d 2 = (22) 2 = 380.13 mm 2 = 380.13 × 10−6 m 2 4 4 3 P = 46 × 10 N P σ x = = 121.01 × 106 Pa A σy =σz = 0 1 σ εx = (σ x − vσ y − vσ z ) = x E E σx ε y = ε z = −vε x = −v E 1 (1 − 2v)σ x e = ε x + ε y + ε z = (σ x − vσ x − vσ x ) = E EVolume: V = AL = (380.13 mm 2 )(200 mm) = 76.026 × 103 mm3 ΔV = Ve (1 − 0.60)(121.01 × 106 )(a) Steel: e= 9 = 242 × 10−6 e = 242 × 10−6  200 × 10 ΔV = (76.026 × 103 )(242 × 10−6 ) = 18.40 mm3 ΔV = 18.40 mm3  (1 − 0.70)(121.01 × 106 )(b) Aluminum: e= 9 = 519 × 10−6 e = 519 × 10−6  70 × 10 ΔV = (76.026 × 103 )(519 × 10−6 ) = 39.4 mm3 ΔV = 39.4 mm3 PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.84∗ Determine the change in volume of the 2-in. gage length segment AB in Prob. 2.61 (a) by computing the dilatation of the material, (b) by subtracting the original volume of portion AB from its final volume.SOLUTIONFrom Problem 2.61, thickness = 1 16 in., E = 29 × 106 psi, v = 0.30.  1  1 (a) A =    = 0.03125 in 2  2  16  Volume: V0 = AL 0 = (0.03125)(2.00) = 0.0625 in 3 P 600 σx = = = 19.2 × 103 psi σy =σz = 0 A 0.03125 1 σ 19.2 × 103 ε x = (σ x − vσ y − vσ z ) = x = = 662.07 × 10−6 E E 29 × 106 ε y = ε z = −v ε x = −(0.30)(662.07 × 10−6 ) = −198.62 × 10−6 e = ε x + ε y + ε z = 264.83 × 10−6 ΔV = V0 e = (0.0625) (264.83 × 10−6 ) = 16.55 × 10−6 in 3 (b) From the solution to Problem 2.61, δ x = 1.324 × 10−3 in., δ y = −99.3 × 10−6 in., δ z = −12.41 × 10−6 in. The dimensions when under a 600-lb tensile load are: Length: L = L0 + δ x = 2 + 1.324 × 10−3 = 2.001324 in. 1 Width: w = w0 + δ y = − 99.3 × 10−6 = 0.4999007 in. 2 1 Thickness: t = t0 + δ z = − 12.41 × 10−6 = 0.06248759 in. 16 Volume: V = Lwt = 0.062516539 in 3 ΔV = V − V0 = 0.062516539 − 0.0625 = 16.54 × 10−6 in 3 PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.85∗A 6-in.-diameter solid steel sphere is lowered into the ocean to a point where the pressure is 7.1 ksi (about3 miles below the surface). Knowing that E = 29 × 106 psi and v = 0.30, determine (a) the decrease indiameter of the sphere, (b) the decrease in volume of the sphere, (c) the percent increase in the density of thesphere.SOLUTION π 3For a solid sphere, V0 = d0 6 π = (6.00)3 6 = 113.097 in.3 σ x = σ y = σ z = −p = −7.1 × 103 psi 1 εx = (σ x − vσ y − vσ z ) E (1 − 2v) p (0.4)(7.1 × 103 ) =− =− E 29 × 106 = −97.93 × 10−6Likewise, ε y = ε z = −97.93 × 10−6 e = ε x + ε y + ε z = −293.79 × 10−6(a) −Δd = −d 0ε x = −(6.00)(−97.93 × 10−6 ) = 588 × 10−6 in. −Δd = 588 × 10−6 in. (b) −ΔV = −V0 e = −(113.097)(−293.79 × 10−6 ) = 33.2 × 10−3 in 3 − ΔV = 33.2 × 10−3 in 3 (c) Let m = mass of sphere. m = constant. m = ρ0V0 = ρV = ρV0 (1 + e) ρ − ρ0 ρ m V 1 = −1 = × 0 −1 = −1 ρ0 ρ0 V0 (1 + e) m 1+ e = (1 − e + e 2 − e3 + ) − 1 = −e + e 2 − e3 +  ≈ −e = 293.79 × 10−6 ρ − ρ0 × 100% = (293.79 × 10−6 )(100%) 0.0294%  ρ0PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.86∗ (a) For the axial loading shown, determine the change in height and the change in volume of the brass cylinder shown. (b) Solve part a, assuming that the loading is hydrostatic with σ x = σ y = σ z = −70 MPa.SOLUTION h0 = 135 mm = 0.135 m π π A 0= 2 (85) 2 = 5.6745 × 103 mm 2 = 5.6745 × 10−3 m 2 d0 = 4 4 V0 = A0 h0 = 766.06 × 103 mm3 = 766.06 × 10−6 m3(a) σ x = 0, σ y = −58 × 106 Pa, σ z = 0 1 σy 58 × 106 εy = (−vσ x + σ y − vσ z ) = =− = −552.38 × 10−6 E E 105 × 109 Δh = h 0ε y = (135 mm)( − 552.38 × 10−6 ) Δh = −0.0746 mm  1 − 2v (1 − 2v)σ y (0.34)(−58 × 106 ) e= (σ x + σ y + σ z ) = = 9 = −187.81 × 10−6 E E 105 × 10 ΔV = V0 e = (766.06 × 103 mm3 )(−187.81 × 10−6 ) ΔV = −143.9 mm3 (b) σ x = σ y = σ z = −70 × 106 Pa σ x + σ y + σ z = −210 × 106 Pa 1 1 − 2v (0.34)(−70 × 106 ) εy = (−vσ x + σ y − vσ z ) = σy = = −226.67 × 10−6 E E 105 × 109 Δh = h 0ε y = (135 mm)( −226.67 × 10−6 ) Δh = −0.0306 mm  1 − 2v (0.34)( −210 × 106 ) e= (σ x + σ y + σ z ) = 9 = −680 × 10−6 E 105 × 10 ΔV = V0 e = (766.06 × 103 mm3 )(−680 × 10−6 ) ΔV = −521 mm3 PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.87∗ A vibration isolation support consists of a rod A of radius R1 = 10 mm and a tube B of inner radius R 2 = 25 mm bonded to an 80-mm-long hollow rubber cylinder with a modulus of rigidity G = 12 MPa. Determine the largest allowable force P that can be applied to rod A if its deflection is not to exceed 2.50 mm.SOLUTIONLet r be a radial coordinate. Over the hollow rubber cylinder, R1 ≤ r ≤ R2 .Shearing stress τ acting on a cylindrical surface of radius r is P P τ= = A 2π rhThe shearing strain is τ P γ= = G 2π GhrShearing deformation over radial length dr, dδ =γ dr P dr d δ = γ dr = 2π Gh rTotal deformation. R2 P dr R2 δ= R1 dδ = 2π Gh  R1r P R2 P = ln r = (ln R2 − ln R1 ) 2π Gh R1 2π Gh P R 2π Ghδ = ln 2 or P = 2π Gh R1 ln( R2 / R1 )Data: R1 = 10 mm = 0.010 m, R2 = 25 mm = 0.025 m, h = 80 mm = 0.080 m G = 12 × 106 Pa δ = 2.50 × 10−3 m (2π )(12 × 106 ) (0.080) (2.50 × 10−3 ) P= = 16.46 × 103 N 16.46 kN  ln (0.025/0.010)PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.88 A vibration isolation support consists of a rod A of radius R1 and a tube B of inner radius R2 bonded to a 80-mm-long hollow rubber cylinder with a modulus of rigidity G = 10.93 MPa. Determine the required value of the ratio R2/R1 if a 10-kN force P is to cause a 2-mm deflection of rod A.SOLUTIONLet r be a radial coordinate. Over the hollow rubber cylinder, R1 ≤ r ≤ R2 .Shearing stress τ acting on a cylindrical surface of radius r is P P τ= = A 2π rhThe shearing strain is τ P γ= = G 2π GhrShearing deformation over radial length dr, dδ =γ dr d δ = γ dr P dr drδ = 2π Gh rTotal deformation. R2 P dr R2 δ=  R1 dδ = 2π Gh R1 r P R2 P = ln r = (ln R2 − ln R1 ) 2π Gh R1 2π Gh P R = ln 2 2π Gh R1 R2 2π Ghδ (2π ) (10.93 × 106 ) (0.080) (0.002) ln = = = 1.0988 R1 P 10.103 R2 = exp (1.0988) = 3.00 R2 /R1 = 3.00  R1PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.89∗The material constants E, G, k, and v are related by Eqs. (2.33) and (2.43). Show that any one of theseconstants may be expressed in terms of any other two constants. For example, show that(a) k = GE/(9G − 3E) and (b) v = (3k – 2G)/(6k + 2G).SOLUTION E E k= and G= 3(1 − 2v) 2(1 + v) E E(a) 1+ v = or v = −1 2G 2G E 2 EG 2 EG EG k= = = k=    E   3[2G − 2 E + 4G ] 18G − 6 E 9G − 6 E 3 1 − 2  − 1    2G   k 2(1 + v)(b) = G 3(1 − 2v) 3k − 6kv = 2G + 2Gv 3k − 2G 3k − 2G = 2G + 6k v=  6k + 2GPROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.90∗Show that for any given material, the ratio G/E of the modulus of rigidity over the modulus of elasticity isalways less than 1 but more than 1 . [Hint: Refer to Eq. (2.43) and to Sec. 2.13.] 2 3SOLUTION E E G= or = 2(1 + v) 2(1 + v) GAssume v > 0 for almost all materials, and v < 1 2 for a positive bulk modulus. E  1Applying the bounds, 2≤ < 2 1 +  = 3 G  2 1 G 1Taking the reciprocals, > > 2 E 3 1 G 1or < <  3 E 2PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.91∗ A composite cube with 40-mm sides and the properties shown is made with glass polymer fibers aligned in the x direction. The cube is constrained against deformations in the y and z directions and is subjected to a tensile load of 65 kN in the x direction. Determine (a) the change in the length of the cube in the x direction, (b) the stresses σ x , σ y , and σ z . Ex = 50 GPa vxz = 0.254 E y = 15.2 GPa vxy = 0.254 Ez = 15.2 GPa vzy = 0.428SOLUTIONStress-to-strain equations are σx v yxσ y vzxσ z εx = − − (1) Ex Ey Ez vxyσ x σy vzyσ z εy = − + − (2) Ex Ey Ez vxzσ x v yzσ y σ z εz = − − + (3) Ex Ey Ez vxy v yx = (4) Ex Ey v yz vzy = (5) Ey Ez vzx vxz = (6) Ez ExThe constraint conditions are ε y = 0 and ε z = 0.Using (2) and (3) with the constraint conditions gives 1 vzy vxy σy − σz = σx (7) Ey Ez Ex v yz 1 V − σy + σ z = xz σ x (8) Ey Ez Ex 1 0.428 0.254 σy − σz = σ x or σ y − 0.428σ z = 0.077216 σ x 15.2 15.2 50 0.428 1 0.254 − σy + σz = σ x or − 0.428 σ y + σ z = 0.077216 σ x 15.2 15.2 50PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.91∗ (Continued) Solving simultaneously, σ y = σ z = 0.134993 σ x 1 vxy v Using (4) and (5) in (1), εx = σx − σ y − xz σ z Ex Ex E 1 Ex = [1 − (0.254) (0.134993) − (0.254)(0.134993)]σ x Ex 0.93142 σ x = Ex A = (40)(40) = 1600 mm 2 = 1600 × 10−6 m 2 P 65 × 103 σx = = = 40.625 × 106 Pa A 1600 × 10−6 (0.93142) (40.625 × 103 ) εx = 9 = 756.78 × 10−6 50 × 10(a) δ x = Lx ε x = (40 mm) (756.78 × 10−6 ) δ x = 0.0303 mm (b) σ x = 40.625 × 106 Pa σ x = 40.6 MPa  σ y = σ z = (0.134993) (40.625 × 106 ) = 5.48 × 106 Pa  σ y = σ z = 5.48 MPa PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.92∗ The composite cube of Prob. 2.91 is constrained against deformation in the z direction and elongated in the x direction by 0.035 mm due to a tensile load in the x direction. Determine (a) the stresses σ x , σ y , and σ z , (b) the change in the dimension in the y direction. Ex = 50 GPa vxz = 0.254 E y = 15.2 GPa vxy = 0.254 Ez = 15.2 GPa vzy = 0.428SOLUTION σx v yxσ y vzxσ z εx = − − (1) Ex Ey Ez vxyσ x σy vzyσ z εy = − + − (2) Ex Ey Ez vxzσ x v yzσ y σ z εz = − − + (3) Ex Ey Ez vxy v yx = (4) Ex Ey v yz vzy = (5) Ey Ez vzx vxz = (6) Ez ExConstraint condition: εz = 0Load condition : σy = 0 vxz 1From Equation (3), 0=− σx + σz Ex Ez vxz Ez (0.254) (15.2) σz = σx = = 0.077216 σ x Ex 50PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.92∗ (Continued)From Equation (1) with σ y = 0, 1 v 1 v εx = σ x − zx σ z = σ x − xz σ z Ex Ez Ex Ex 1 1 = [σ x − 0.254 σ z ] = [1 − (0.254) (0.077216)]σ x Ex Ex 0.98039 = σx Ex Exε x σx = 0.98039 δx 0.035 mmBut, εx = = = 875 × 10−6 Lx 40 mm (50 × 109 ) (875 × 10−6 )(a) σx = = 44.625 × 103 Pa σ x = 44.6 MPa  0.98039 σy = 0 σ z = (0.077216) (44.625 × 106 ) = 3.446 × 106 Pa σ z = 3.45 MPa  vxy 1 vzy From (2), εy = σx + σy − σz Ex Ey Ez (0.254) (44.625 × 106 ) (0.428) (3.446 × 106 ) =− +0− 50 × 109 15.2 × 109 −6 = −323.73 × 10(b) δ y = Ly ε y = (40 mm) ( − 323.73 × 10−6 ) δ y = −0.0129 mm PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.93 Two holes have been drilled through a long steel bar that is subjected to a centric axial load as shown. For P = 6.5 kips, determine the maximum value of the stress (a) at A, (b) at B.SOLUTION 1 1 1(a) At hole A: r = ⋅ = in. 2 2 4 1 d =3− = 2.50 in. 2 1 Anet = dt = (2.50)   = 1.25 in 2 2 P 6.5 σ non = = = 5.2 ksi Anet 1.25 2r 2( 1 ) 4 = = 0.1667 D 3 From Fig. 2.60a, K = 2.56 σ max = Kσ non = (2.56)(5.2) σ max = 13.31 ksi  1(b) At hole B: r = (1.5) = 0.75, d = 3 − 1.5 = 1.5 in. 2 1 P 6.5 Anet = dt = (1.5)   = 0.75 in 2 , σ non = = = 8.667 ksi 2 Anet 0.75 2r 2(0.75) = = 0.5 D 3 From Fig. 2.60a, K = 2.16 σ max = Kσ non = (2.16)(8.667) σ max = 18.72 ksi PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.94 Knowing that σ all = 16 ksi, determine the maximum allowable value of the centric axial load P.SOLUTION 1 1 1At hole A: r = ⋅ = in. 2 2 4 1 d =3− = 2.50 in. 2 1 Anet = dt = (2.50)   = 1.25 in 2 2 2r 2( 1 ) 4 = = 0.1667 D 3From Fig. 2.60a, K = 2.56 KP Anetσ max (1.25)(16) σ max = ∴ P= = = 7.81 kips Anet K 2.56 1At hole B: r = (1.5) = 0.75 in, d = 3 − 1.5 = 1.5 in. 2 1 Anet = dt = (1.5)   = 0.75 in 2 , 2 2r 2(0.75) = = 0.5 D 3From Fig. 2.60a, K = 2.16 Anetσ max (0.75)(16) P= = = 5.56 kips K 2.16Smaller value for P controls. P = 5.56 kips PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.95 Knowing that the hole has a diameter of 9 mm, determine (a) the radius rf of the fillets for which the same maximum stress occurs at the hole A and at the fillets, (b) the corresponding maximum allowable load P if the allowable stress is 100 MPa.SOLUTION 1For the circular hole, r =   (9) = 4.5 mm 2 2r 2(4.5) d = 96 − 9 = 87 mm = = 0.09375 D 96 Anet = dt = (0.087 m)(0.009 m) = 783 × 10−6 m 2From Fig. 2.60a, K hole = 2.72 K hole P σ max = Anet Anetσ max (783 × 10−6 )(100 × 106 ) P= = = 28.787 × 103 N K hole 2.72(a) For fillet, D = 96 mm, d = 60 mm D 96 = = 1.60 d 60 Amin = dt = (0.060 m)(0.009 m) = 540 × 10−6 m 2 K fillet P A σ (5.40 × 10−6 )(100 × 106 ) σ max = ∴ K fillet = min max = Amin P 28.787 × 103 = 1.876 rf From Fig. 2.60b, ≈ 0.19 ∴ rf ≈ 0.19d = 0.19(60) r f = 11.4 mm  d (b) P = 28.8 kN PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.96 For P = 100 kN, determine the minimum plate thickness t required if the allowable stress is 125 MPa.SOLUTIONAt the hole: rA = 20 mm d A = 88 − 40 = 48 mm 2rA 2(20) = = 0.455 DA 88From Fig. 2.60a, K = 2.20 KP KP KP σ max = = ∴ t = Anet d At d Aσ max (2.20)(100 × 103 N) t = 6 = 36.7 × 10−3 m = 36.7 mm (0.048 m)(125 × 10 Pa) D 88At the fillet: D = 88 mm, d B = 64 mm = = 1.375 dB 64 rB 15 rB = 15 mm = = 0.2344 dB 64From Fig. 2.60b, K = 1.70 KP KP σ max = = Amin d Bt KP (1.70)(100 × 103 N) t = = = 21.25 × 10−3 m = 21.25 mm d Bσ max (0.064 m)(125 × 106 Pa)The larger value is the required minimum plate thickness. t = 36.7 mm PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.97 The aluminum test specimen shown is subjected to two equal and opposite centric axial forces of magnitude P. (a) Knowing that E = 70 GPa and σ all = 200 MPa, determine the maximum allowable value of P and the corresponding total elongation of the specimen. (b) Solve part a, assuming that the specimen has been replaced by an aluminum bar of the same length and a uniform 60 × 15-mm rectangular cross section.SOLUTION σ all = 200 × 106 Pa E = 70 × 109 Pa Amin = (60 mm) (15 mm) = 900 mm 2 = 900 × 10−6 m 2(a) Test specimen. D = 75 mm, d = 60 mm, r = 6 mm D 75 r 6 = = 1.25 = = 0.10 d 60 d 60 P From Fig. 2.60b K = 1.95 σ max = K A Aσ max (900 × 10−6 ) (200 × 106 ) P= = = 92.308 × 103 N P = 92.3 kN  K 1.95 Wide area A* = (75 mm) (15 mm) = 1125 mm 2 = 1.125 × 10−3 m 2 Pi Li P L 92.308 × 103  0.150 0.300 0.150  δ =Σ = Σ i = 9 1.125 × 10−3 + 900 × 10−6 + 1.125 × 10−3  Ai Ei E Ai 70 × 10   = 7.91 × 10−6 m δ = 0.791 mm (b) Uniform bar. P = Aσ all = (900 × 10−6 )(200 × 106 ) = 180 × 103 N P = 180.0 kN  PL (180 × 103 )(0.600) δ= = = 1.714 × 10−3 m  δ = 1.714 mm  AE (900 × 10−6 )(70 × 109 )PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.98 For the test specimen of Prob. 2.97, determine the maximum value of the normal stress corresponding to a total elongation of 0.75 mm. PROBLEM 2.97 The aluminum test specimen shown is subjected to two equal and opposite centric axial forces of magnitude P. (a) Knowing that E = 70 GPa and σ all = 200 MPa, determine the maximum allowable value of P and the corresponding total elongation of the specimen. (b) Solve part a, assuming that the specimen has been replaced by an aluminum bar of the same length and a uniform 60 × 15-mm rectangular cross section.SOLUTION Pi Li P Li δ =Σ = Σ δ = 0.75 × 10−3 m Ei Ai E Ai L1 = L3 = 150 mm = 0.150 m, L2 = 300 mm = 0.300 m A1 = A3 = (75 mm) (15 mm) = 1125 mm 2 = 1.125 × 10−3 m 2 A2 = (60 mm) (15 mm) = 900 mm2 = 900 × 10−6 m 2 Li 0.150 0.300 0.150 Σ = −3 + −6 + −3 = 600 m −1 Ai 1.125 × 10 900 × 10 1.125 × 10 Eδ (70 × 109 ) (0.75 × 10−3 ) P= = = 87.5 × 103 N Li 600 Σ AiStress concentration. D = 75 mm, d = 60 mm, r = 6 mm D 75 r 6 = = 1.25 = = 0.10 d 60 d 60From Fig. 2.60b K = 1.95 P (1.95) (87.5 × 103 ) σ max = K = −6 = 189.6 × 106 Pa σ max = 189.6 MPa  Amin 900 × 10Note that σ max < σ all . PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.99 A hole is to be drilled in the plate at A. The diameters of the bits available to drill the hole range from 1 to 11/2 in. in 1 -in. 2 4 increments. If the allowable stress in the plate is 21 ksi, determine (a) the diameter d of the largest bit that can be used if the allowable load P at the hole is to exceed that at the fillets, (b) the corresponding allowable load P.SOLUTION D 4.6875 r 0.375At the fillets: = = 1.5 = = 0.12 d 3.125 d 3.125From Fig. 2.60b, K = 2.10 Amin = (3.125)(0.5) = 1.5625 in 2 Pall σ max = K = σ all Amin Aminσ all (1.5625)(21) Pall = = = 15.625 kips K 2.10At the hole: Anet = ( D − 2r )t , K from Fig. 2.60a P Anetσ all σ max = K = σ all ∴ Pall = Anet Kwith D = 4.6875 in. t = 0.5 in. σ all = 21 ksi Hole diam. r d = D − 2r 2r/D K Anet Pall 0.5 in. 0.25 in. 4.1875 in. 0.107 2.68 2.0938 in2 16.41 kips 2 0.75 in. 0.375 in. 3.9375 in. 0.16 2.58 1.96875 in 16.02 kips 1 in. 0.5 in. 3.6875 in. 0.213 2.49 1.84375 in2 15.55 kips 2 1.25 in. 0.625 in. 3.4375 in. 0.267 2.41 1.71875 in 14.98 kips 1.5 in. 0.75 in. 3.1875 in. 0.32 2.34 1.59375 in2 14.30 kips 3(a) Largest hole with Pall > 15.625 kips is the 4 -in. diameter hole. (b) Allowable load Pall = 15.63 kips PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.100 (a) For P = 13 kips and d = 1 in., determine the maximum stress 2 in the plate shown. (b) Solve part a, assuming that the hole at A is not drilled.SOLUTIONMaximum stress at hole: Use Fig. 2.60a for values of K. 2r 0.5 = = 0.017, K = 2.68 D 4.6875 Anet = (0.5)(4.6875 − 0.5) = 2.0938 in 2 P (2.68)(13) σ max = K = = 16.64 ksi, Anet 2.0938Maximum stress at fillets: Use Fig. 2.60b for values of K. r 0.375 D 4.6875 = = 0.12 = = 1.5 K = 2.10 d 3.125 d 3.125 Amin = (0.5)(3.125) = 1.5625 in 2 P (2.10)(13) σ max = K = = 17.47 ksi Amin 1.5625(a) With hole and fillets: 17.47 ksi (b) Without hole: 17.47 ksi PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.101 Rod ABC consists of two cylindrical portions AB and BC; it is made of a mild steel that is assumed to be elastoplastic with E = 200 GPa and σ Y = 250 MPa. A force P is applied to the rod and then removed to give it a permanent set δ p = 2 mm. Determine the maximum value of the force P and the maximum amount δ m by which the rod should be stretched to give it the desired permanent set.SOLUTION π AAB = (30)2 = 706.86 mm 2 = 706.86 × 10−6 m 2 4 π ABC = (40) 2 = 1.25664 × 103 mm 2 = 1.25664 × 10−3 m 2 4 Pmax = Aminσ Y = (706.86 × 10−6 )(250 × 106 ) = 176.715 × 103 N Pmax = 176.7 kN  P′LAB P ′LBC (176.715 × 103 )(0.8) (176.715 × 103 )(1.2) δ′ = + = + EAAB EABC (200 × 109 )(706.86 × 10−6 ) (200 × 109 )(1.25664 × 10−3 ) = 1.84375 × 10−3 m = 1.84375 mm δ p = δ m − δ ′ or δ m = δ p + δ ′ = 2 + 1.84375 δ m = 3.84 mm PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.102 Rod ABC consists of two cylindrical portions AB and BC; it is made of a mild steel that is assumed to be elastoplastic with E = 200 GPa and σ Y = 250 MPa. A force P is applied to the rod until its end A has moved down by an amount δ m = 5 mm. Determine the maximum value of the force P and the permanent set of the rod after the force has been removed.SOLUTION π AAB = (30)2 = 706.86 mm 2 = 706.86 × 10−6 m 2 4 π ABC = (40) 2 = 1.25664 × 103 mm 2 = 1.25644 × 10−3 m 2 4 Pmax = Aminσ Y = (706.86 × 10−6 )(250 × 106 ) = 176.715 × 103 N Pmax = 176.7 kN  P′LAB P ′LBC (176.715 × 103 )(0.8) (176.715 × 103 )(1.2) δ′ = + = + EAAB EABC (200 × 109 )(706.68 × 10−6 ) (200 × 109 )(1.25664 × 10−3 ) = 1.84375 × 10−3 m = 1.84375 mm δ p = δ m − δ ′ = 5 − 1.84375 = 3.16 mm δ p = 3.16 mm PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.103 The 30-mm square bar AB has a length L = 2.2 m; it is made of a mild steel that is assumed to be elastoplastic with E = 200 GPa and σ Y = 345 MPa. A force P is applied to the bar until end A has moved down by an amount δ m . Determine the maximum value of the force P and the permanent set of the bar after the force has been removed, knowing that (a) δ m = 4.5 mm, (b) δ m = 8 mm.SOLUTION A = (30)(30) = 900 mm2 = 900 × 10−6 m 2 Lσ Y (2.2)(345 × 106 ) δY = Lε Y = = = 3.795 × 10−3 = 3.795 mm E 200 × 109If δ m ≥ δ Y , Pm = Aσ Y = (900 × 10−6 )(345 × 106 ) = 310.5 × 103 N Pm L σ Y LUnloading: δ ′ = = = δY = 3.795 mm AE E δ P = δm − δ ′(a) δ m = 4.5 mm > δY Pm = 310.5 × 103 N δ m = 310.5 kN  δ perm = 4.5 mm − 3.795 mm δ perm = 0.705 mm (b) δ m = 8 mm > δY Pm = 310.5 × 103 N δ m = 310.5 kN  δ perm = 8.0 mm − 3.795 mm δ perm = 4.205 mm PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.104 The 30-mm square bar AB has a length L = 2.5 m; it is made of mild steel that is assumed to be elastoplastic with E = 200 GPa and σ Y = 345 MPa. A force P is applied to the bar and then removed to give it a permanent set δ p . Determine the maximum value of the force P and the maximum amount δ m by which the bar should be stretched if the desired value of δ p is (a) 3.5 mm, (b) 6.5 mm.SOLUTION A = (30)(30) = 900 mm 2 = 900 × 10−6 m 2 Lσ Y (2.5)(345 × 106 ) δY = Lε Y = = 9 = 4.3125 × 103 m = 4.3125 mm E 200 × 10When δ m exceeds δ Y , thus producing a permanent stretch of δ p , the maximum force is Pm = Aσ Y = (900 × 10−6 )(345 × 106 ) = 310.5 × 103 N = 310.5 kN  δ p = δ m − δ ′ = δ m − δY ∴ δ m = δ p + δY(a) δ p = 3.5 mm δ m = 3.5 mm + 4.3125 mm = 7.81 mm (b) δ p = 6.5 mm δ m = 6.5 mm + 4.3125 mm = 10.81 mm PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.105 Rod AB is made of a mild steel that is assumed to be elastoplastic with E = 29 × 106 psi and σ Y = 36 ksi. After the rod has been attached 3 to the rigid lever CD, it is found that end C is 8 in. too high. A vertical force Q is then applied at C until this point has moved to position C ′. Determine the required magnitude of Q and the deflection δ1 if the lever is to snap back to a horizontal position after Q is removed.SOLUTIONSince the rod AB is to be stretched permanently, the peak force in the rod is P = PY , where 2 π 3 PY = Aσ Y = (36) = 3.976 kips 4 8  Referring to the free body diagram of lever CD, ΣM D = 0: 33 Q − 22 P = 0 22 (22)(3.976) Q= P= = 2.65 kips Q = 2.65 kips  33 33During unloading, the spring back at B is LABσ Y (60)(36 × 103 ) δ B = LABε Y = = = 0.0745 in. E 29 × 106From the deformation diagram, δB δC 33Slope: θ= = ∴ δC = δ B = 0.1117 in. δ C = 0.1117 in.  22 33 −22PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.106 Solve Prob. 2.105, assuming that the yield point of the mild steel is 50 ksi. PROBLEM 2.105 Rod AB is made of a mild steel that is assumed to be elastoplastic with E = 29 × 106 psi and σ Y = 36 ksi. After the rod 3 has been attached to the rigid lever CD, it is found that end C is 8 in. too high. A vertical force Q is then applied at C until this point has moved to position C′. Determine the required magnitude of Q and the deflection δ1 if the lever is to snap back to a horizontal position after Q is removed.SOLUTIONSince the rod AB is to be stretched permanently, the peak force in the rod is P = PY , where 2 π 3 PY = Aσ Y = (50) = 5.522 kips 4 8  Referring to the free body diagram of lever CD, ΣM D = 0: 33Q − 22 P = 0 22 (22)(5.522) Q= P= = 3.68 kips Q = 3.68 kips  33 33During unloading, the spring back at B is LABσ Y (60)(50 × 103 ) δ B = LAB ε Y = = = 0.1034 in. E 29 × 106From the deformation diagram, δB δC 33Slope: θ= = ∴ δC = δB δ C = 0.1552 in.  22 33 22PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.107 Each cable has a cross-sectional area of 100 mm2 and is made of an elastoplastic material for which σ Y = 345 MPa and E = 200 GPa. A force Q is applied at C to the rigid bar ABC and is gradually increased from 0 to 50 kN and then reduced to zero. Knowing that the cables were initially taut, determine (a) the maximum stress that occurs in cable BD, (b) the maximum deflection of point C, (c) the final displacement of point C. (Hint: In Part c, cable CE is not taut.)SOLUTIONElongation constraints for taut cables.Let θ = rotation angle of rigid bar ABC. δ BD δ CE θ= = LAB LAC LAB 1 δ BD = δ CE = δ CE (1) LAC 2Equilibrium of bar ABC. M A = 0 : LAB FBD + LAC FCE − LAC Q = 0 LAB 1 Q = FCE + FBD = FCE + FBD (2) LAC 2Assume cable CE is yielded. FCE = Aσ Y = (100 × 10−6 )(345 × 106 ) = 34.5 × 103 NFrom (2), FBD = 2(Q − FCE ) = (2)(50 × 103 − 34.5 × 103 ) = 31.0 × 103 NSince FBD < Aσ Y = 34.5 × 103 N, cable BD is elastic when Q = 50 kN.PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.107 (Continued)(a) Maximum stresses. σ CE = σ Y = 345 MPa FBD 31.0 × 103 σ BD = = −6 = 310 × 106 Pa σ BD = 310 MPa  A 100 × 10(b) Maximum of deflection of point C. FBD LBD (31.0 × 103 )(2) δ BD = = = 3.1 × 10−3 m EA (200 × 109 )(100 × 10−6 )From (1), δ C = δ CE = 2δ BD = 6.2 × 10−3 m 6.20 mm ↓  σ Y LCEPermanent elongation of cable CE: (δ CE ) p = (δ CE ) − E FCE LCE σ L (δ CE ) P = (δ CE ) max − = (δ CE ) max − Y CE EA E 6 (345 × 10 )(2) = 6.20 × 10−3 − 9 = 2.75 × 10−3 m 200 × 10(c) Unloading. Cable CE is slack ( FCE = 0) at Q = 0. From (2), FBD = 2(Q − FCE ) = 2(0 − 0) = 0 FBD LBD Since cable BD remained elastic, δ BD = = 0.  EAPROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.108 Solve Prob. 2.107, assuming that the cables are replaced by rods of the same cross-sectional area and material. Further assume that the rods are braced so that they can carry compressive forces. PROBLEM 2.107 Each cable has a cross-sectional area of 100 mm2 and is made of an elastoplastic material for which σ Y = 345 MPa and E = 200 GPa. A force Q is applied at C to the rigid bar ABC and is gradually increased from 0 to 50 kN and then reduced to zero. Knowing that the cables were initially taut, determine (a) the maximum stress that occurs in cable BD, (b) the maximum deflection of point C, (c) the final displacement of point C. (Hint: In Part c, cable CE is not taut.)SOLUTIONElongation constraints.Let θ = rotation angle of rigid bar ABC. δ BC δ CE θ= = LAB LAC LAB 1 δ BD = δ CE = δ CE (1) LAC 2Equilibrium of bar ABC. M A = 0: LAB FBD + LAC FCE − LAC Q = 0 LAB 1 Q = FCE + FBD = FCE + FBD (2) LAC 2Assume cable CE is yielded. FCE = Aσ Y = (100 × 10−6 )(345 × 106 ) = 34.5 × 103 NFrom (2), FBD = 2(Q − FCE ) = (2)(50 × 103 − 34.5 × 103 ) = 31.0 × 103 NSince FBD < Aσ Y = 34.5 × 103 N, cable BD is elastic when Q = 50 kN.PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.108 (Continued)(a) Maximum stresses. σ CE = σ Y = 345 MPa FBD 31.0 × 103 σ BD = = −6 = 310 × 106 Pa σ BD = 310 MPa  A 100 × 10(b) Maximum of deflection of point C. FBD LBD (31.0 × 103 )(2) δ BD = = = 3.1 × 10−3 m EA (200 × 109 )(100 × 10−6 ) From (1), δ C = δ CE = 2δ BD = 6.2 × 10−3 m 6.20 mm ↓  Unloading. Q′ = 50 × 103 N, δ CE = δ C ′ ′ From (1), ′ ′ δ BD = 1 δ C 2 9 −6 ′ EAδ BD (200 × 10 )(100 × 10 )( 1 δ C ) ′ ′′ Elastic FBD = = 2 = 5 × 106 δ C ′ LBD 2 EAδ CE (200 × 109 )(100 × 10−6 )(δ C ) ′ ′ ′ FCE = = = 10 × 106 δ C ′ LCE 2 From (2), Q′ = FCE + 1 FBD = 12.5 × 106 δ C ′ 2 ′ ′ Equating expressions for Q′, 12.5 × 106 δ C = 50 × 103 ′ δ C = 4 × 10−3 m ′(c) Final displacement. δ C = (δ C ) m − δ C = 6.2 × 10−3 − 4 × 10−3 = 2.2 × 10−3 m ′ 2.20 mm ↓ PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.109 Rod AB consists of two cylindrical portions AC and BC, each with a cross-sectional area of 1750 mm2. Portion AC is made of a mild steel with E = 200 GPa and σ Y = 250 MPa, and portion CB is made of a high-strength steel with E = 200 GPa and σ Y = 345 MPa. A load P is applied at C as shown. Assuming both steels to be elastoplastic, determine (a) the maximum deflection of C if P is gradually increased from zero to 975 kN and then reduced back to zero, (b) the maximum stress in each portion of the rod, (c) the permanent deflection of C.SOLUTIONDisplacement at C to cause yielding of AC. L σ (0.190)(250 × 106 ) δ C ,Y = LAC ε Y , AC = AC Y , AC = 9 = 0.2375 × 10−3 m E 200 × 10Corresponding force. FAC = Aσ Y , AC = (1750 × 10−6 )(250 × 106 ) = 437.5 × 103 N EAδ C (200 × 109 )(1750 × 10−6 )(0.2375 × 10−3 ) FCB = − =− = −437.5 × 103 N LCB 0.190For equilibrium of element at C, FAC − ( FCB + PY ) = 0 PY = FAC − FCB = 875 × 103 NSince applied load P = 975 × 103 N > 875 × 103 N, portion AC yields. FCB = FAC − P = 437.5 × 103 − 975 × 103 N = −537.5 × 103 N FCB LCD (537.5 × 103 )(0.190)(a) δC = − = = 0.29179 × 10−3 m EA (200 × 109 )(1750 × 10−6 ) 0.292 mm (b) Maximum stresses: σ AC = σ Y , AC = 250 MPa 250 MPa  FBC 537.5 × 103 σ BC = =− −6 = −307.14 × 106 Pa = −307 MPa −307 MPa  A 1750 × 10(c) Deflection and forces for unloading. P′ L P′ L L δ ′ = AC AC = − CB CB ∴ PCB = − PAC AC = − PAC ′ ′ ′ EA EA LAB P′ = 975 × 103 = PAC − PCB = 2 PAC PAC = 487.5 × 10−3 N ′ ′ ′ ′ (487.5 × 103 )(0.190) δ′ = 9 −6 = 0.26464 × 103 m (200 × 10 )(1750 × 10 ) δ p = δ m − δ ′ = 0.29179 × 10−3 − 0.26464 × 10−3 = 0.02715 × 10−3 m 0.0272 mm PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.110 For the composite rod of Prob. 2.109, if P is gradually increased from zero until the deflection of point C reaches a maximum value of δ m = 0.3 mm and then decreased back to zero, determine (a) the maximum value of P, (b) the maximum stress in each portion of the rod, (c) the permanent deflection of C after the load is removed. PROBLEM 2.109 Rod AB consists of two cylindrical portions AC and BC, each with a cross-sectional area of 1750 mm2. Portion AC is made of a mild steel with E = 200 GPa and σ Y = 250 MPa, and portion CB is made of a high-strength steel with E = 200 GPa and σ Y = 345 MPa. A load P is applied at C as shown. Assuming both steels to be elastoplastic, determine (a) the maximum deflection of C if P is gradually increased from zero to 975 kN and then reduced back to zero, (b) the maximum stress in each portion of the rod, (c) the permanent deflection of C.SOLUTIONDisplacement at C is δ m = 0.30 mm. The corresponding strains are δm 0.30 mm ε AC = = = 1.5789 × 10−3 LAC 190 mm δm 0.30 mm ε CB = − =− = −1.5789 × 10−3 LCB 190 mmStrains at initial yielding: σ Y, AC 250 × 106 ε Y, AC = = = 1.25 × 10−3 (yielding) E 200 × 109 σ Y, BC 345 × 106 ε Y, CB = =− = −1.725 × 10−3 (elastic) E 200 × 109(a) Forces: FAC = Aσ Y = (1750 × 10−6 )(250 × 106 ) = 437.5 × 10−3 N FCB = EAε CB = (200 × 109 )(1750 × 10−6 )(−1.5789 × 10−3 ) = −552.6 × 10−3 N For equilibrium of element at C, FAC − FCB − P = 0 P = FAC − FCD = 437.5 × 103 + 552.6 × 103 = 990.1 × 103 N = 990 kN (b) Stresses: AC : σ AC = σ Y, AC = 250 MPa  FCB 552.6 × 103 CB : σ CB = =− −6 = −316 × 106 Pa −316 MPa  A 1750 × 10PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.110 (Continued)(c) Deflection and forces for unloading. ′ PAC LAC P′ L L δ′ = ′ ′ = − CB CB ∴ PCB = − PAC AC = − PAC EA EA LAB P′ = PAC − PCB = 2 PAC = 990.1 × 103 N ∴ PAC = 495.05 × 103 N ′ ′ ′ ′ (495.05 × 103 )(0.190) δ′ = = 0.26874 × 10−3 m = 0.26874 mm (200 × 109 )(1750 × 10−6 ) δ p = δ m − δ ′ = 0.30 mm − 0.26874 mm 0.031mm PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.111 3 Two tempered-steel bars, each 16 -in. thick, are bonded to a 1 -in. mild-steel 2 bar. This composite bar is subjected as shown to a centric axial load of magnitude P. Both steels are elastoplastic with E = 29 × 106 and with yield strengths equal to 100 ksi and 50 ksi, respectively, for the tempered and mild steel. The load P is gradually increased from zero until the deformation of the bar reaches a maximum value δ m = 0.04 in. and then decreased back to zero. Determine (a) the maximum value of P, (b) the maximum stress in the tempered-steel bars, (c) the permanent set after the load is removed.SOLUTION 1 Lσ Y 1 (14)(50 × 103 )For the mild steel, A1 =   (2) = 1.00 in 2 δY1 = = = 0.024138 in. 2 E 29 × 106  3 Lσ Y 2 (14)(100 × 103 )For the tempered steel, A2 = 2   (2) = 0.75 in 2 δY 2 = = = 0.048276 in.  16  E 29 × 103Total area: A = A1 + A2 = 1.75 in 2δ Y 1 < δ m < δY 2 . The mild steel yields. Tempered steel is elastic.(a) Forces: P = A1σ Y 1 = (1.00)(50 × 103 ) = 50 × 103 lb 1 EA2δ m (29 × 103 )(0.75)(0.04) P2 = = = 62.14 × 103 lb L 14 P = P + P2 = 112.14 × 103 lb = 112.1kips 1 P = 112.1 kips  P(b) Stresses: σ1 = 1 = σ Y 1 = 50 × 103 psi = 50 ksi A1 P2 62.14 × 103 σ2 = = = 82.86 × 103 psi = 82.86 ksi 82.86 ksi  A2 0.75 PL (112.14 × 103 )(14) Unloading: δ′ = = = 0.03094 in. EA (29 × 106 )(1.75)(c) Permanent set: δ p = δ m − δ ′ = 0.04 − 0.03094 = 0.00906 in. 0.00906in. PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.112 For the composite bar of Prob. 2.111, if P is gradually increased from zero to 98 kips and then decreased back to zero, determine (a) the maximum deformation of the bar, (b) the maximum stress in the tempered-steel bars, (c) the permanent set after the load is removed. 3 PROBLEM 2.111 Two tempered-steel bars, each 16 -in. thick, are bonded to a 1 -in. mild-steel bar. This composite bar is subjected as shown to a centric 2 axial load of magnitude P. Both steels are elastoplastic with E = 29 × 106 psi and with yield strengths equal to 100 ksi and 50 ksi, respectively, for the tempered and mild steel.SOLUTION 1Areas: Mild steel: A1=   (2) = 1.00 in 2 2  3 Tempered steel: A2 = 2   (2) = 0.75 in 2  16  Total: A = A1+ A2 = 1.75 in 2Total force to yield the mild steel: PY σY1 = ∴ PY = Aσ Y 1 = (1.75)(50 × 103 ) = 87.50 × 103 lb AP > PY , therefore, mild steel yields.Let P = force carried by mild steel. 1 P2 = force carried by tempered steel. P = A1σ1 = (1.00)(50 × 103 ) = 50 × 103 lb 1 P + P2 = P, P2 = P − P = 98 × 103 − 50 × 103 = 48 × 103 lb 1 1 P2 L (48 × 103 )(14)(a) δm = = = 0.03090in.  EA2 (29 × 106 )(0.75) P2 48 × 103(b) σ2 = = = 64 × 103 psi = 64 ksi  A2 0.75 PL (98 × 103 )(14) Unloading: δ ′ = = = 0.02703 in. EA (29 × 106 )(1.75)(c) δ P = δ m − δ ′ = 0.03090 − 0.02703 = 0.00387 in. PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.113 The rigid bar ABC is supported by two links, AD and BE, of uniform 37.5 × 6-mm rectangular cross section and made of a mild steel that is assumed to be elastoplastic with E = 200 GPa and σ Y = 250 MPa. The magnitude of the force Q applied at B is gradually increased from zero to 260 kN. Knowing that a = 0.640 m, determine (a) the value of the normal stress in each link, (b) the maximum deflection of point B.SOLUTIONStatics: ΣM C = 0 : 0.640(Q − PBE ) − 2.64 PAD = 0Deformation: δ A = 2.64θ , δ B = aθ = 0.640θElastic analysis: A = (37.5)(6) = 225 mm 2 = 225 × 10−6 m 2 EA (200 × 109 )(225 × 10−6 ) PAD = δA = δ A = 26.47 × 106 δ A LAD 1.7 = (26.47 × 106 )(2.64θ ) = 69.88 × 106 θ P σ AD = AD = 310.6 × 109 θ A EA (200 × 109 )(225 × 10−6 ) PBE = δB = δ B = 45 × 106 δ B LBE 1.0 = (45 × 106 )(0.640θ ) = 28.80 × 106 θ P σ BE = BE = 128 × 109 θ A 2.64From Statics, Q = PBE + PAD = PBE + 4.125PAD 0.640 = [28.80 × 106 + (4.125)(69.88 × 106 ] θ = 317.06 × 106 θθY at yielding of link AD: σ AD = σ Y = 250 × 106 = 310.6 × 109 θ θY = 804.89 × 10−6 QY = (317.06 × 106 )(804.89 × 10−6 ) = 255.2 × 103 NPROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.113 (Continued)(a) Since Q = 260 × 103 > QY , link AD yields. σ AD = 250 MPa  PAD = Aσ Y = (225 × 10−6 )(250 × 10−6 ) = 56.25 × 103 N From Statics, PBE = Q − 4.125PAD = 260 × 103 − (4.125)(56.25 × 103 ) PBE = 27.97 × 103 N PBE 27.97 × 103 σ BE = = −6 = 124.3 × 106 Pa σ BE = 124.3 MPa  A 225 × 10 PBE LBE (27.97 × 103 )(1.0)(b) δB = = = 621.53 × 10−6 m δ B = 0.622 mm ↓  EA (200 × 109 )(225 × 10−6 )PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.114 Solve Prob. 2.113, knowing that a = 1.76 m and that the magnitude of the force Q applied at B is gradually increased from zero to 135 kN. PROBLEM 2.113 The rigid bar ABC is supported by two links, AD and BE, of uniform 37.5 × 6-mm rectangular cross section and made of a mild steel that is assumed to be elastoplastic with E = 200 GPa and σ Y = 250 MPa. The magnitude of the force Q applied at B is gradually increased from zero to 260 kN. Knowing that a = 0.640 m, determine (a) the value of the normal stress in each link, (b) the maximum deflection of point B.SOLUTIONStatics: ΣM C = 0 : 1.76(Q − PBE ) − 2.64 PAD = 0Deformation: δ A = 2.64θ , δ B = 1.76θElastic Analysis: A = (37.5)(6) = 225 mm 2 = 225 × 10−6 m 2 EA (200 × 109 )(225 × 10−6 ) PAD = δA = δ A = 26.47 × 106 δ A LAD 1.7 = (26.47 × 106 )(2.64θ ) = 69.88 × 106 θ P σ AD = AD = 310.6 × 109 θ A EA (200 × 109 )(225 × 10−6 ) PBE = δB = δ B = 45 × 106 δ B LBE 1.0 = (45 × 106 )(1.76θ ) = 79.2 × 106 θ P σ BE = BE = 352 × 109 θ A 2.64From Statics, Q = PBE + PAD = PBE + 1.500 PAD 1.76 = [73.8 × 106 + (1.500)(69.88 × 106 ] θ = 178.62 × 106 θθY at yielding of link BE: σ BE = σ Y = 250 × 106 = 352 × 109 θY θY = 710.23 × 10−6 QY = (178.62 × 106 )(710.23 × 10−6 ) = 126.86 × 103 NSince Q = 135 × 103 N > QY , link BE yields. σ BE = σ Y = 250 MPa  PBE = Aσ Y = (225 × 10−6 )(250 × 106 ) = 56.25 × 103 NPROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.114 (Continued) 1From Statics, PAD = (Q − PBE ) = 52.5 × 103 N 1.500 PAD 52.5 × 103(a) σ AD = = −6 = 233.3 × 106 σ AD = 233 MPa  A 225 × 10 PAD From elastic analysis of AD, θ= = 751.29 × 10−3 rad 69.88 × 106(b) δ B = 1.76θ = 1.322 × 10−3 m δ B = 1.322 mm ↓ PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.115∗ Solve Prob. 2.113, assuming that the magnitude of the force Q applied at B is gradually increased from zero to 260 kN and then decreased back to zero. Knowing that a = 0.640 m, determine (a) the residual stress in each link, (b) the final deflection of point B. Assume that the links are braced so that they can carry compressive forces without buckling. PROBLEM 2.113 The rigid bar ABC is supported by two links, AD and BE, of uniform 37.5 × 6-mm rectangular cross section and made of a mild steel that is assumed to be elastoplastic with E = 200 GPa and σ Y = 250 MPa. The magnitude of the force Q applied at B is gradually increased from zero to 260 kN. Knowing that a = 0.640 m, determine (a) the value of the normal stress in each link, (b) the maximum deflection of point B.SOLUTIONSee solution to Problem 2.113 for the normal stresses in each link and the deflection of Point B after loading. σ AD = 250 × 106 Pa σ BE = 124.3 × 106 Pa δ B = 621.53 × 10−6 mThe elastic analysis given in the solution to Problem 2.113 applies to the unloading. Q′ = 317.06 × 106 θ ′ Q 260 × 103 Q′ = = = 820.03 × 10−6 317.06 × 106 317.06 × 106 σ ′ = 310.6 × 109 θ = (310.6 × 109 )(820.03 × 10−6 ) = 254.70 × 106 Pa AD σ BE = 128 × 109 θ = (128 × 109 )(820.03 × 10−6 ) = 104.96 × 106 Pa ′ δ B = 0.640θ ′ = 524.82 × 10−6 m ′(a) Residual stresses. σ AD , res = σ AD − σ AD = 250 × 106 − 254.70 × 10−6 = −4.70 × 106 Pa ′ = −4.70 MPa  σ BE , res = σ BE − σ BE = 124.3 × 106 − 104.96 × 106 = 19.34 × 106 Pa ′ = 19.34 MPa (b) δ B , P = δ B − δ B = 621.53 × 10−6 − 524.82 × 10−6 = 96.71 × 10−6 m ′ = 0.0967 mm ↓ PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.116 A uniform steel rod of cross-sectional area A is attached to rigid supports and is unstressed at a temperature of 45 °F. The steel is assumed to be elastoplastic with σ Y = 36 ksi and E = 29 × 106 psi. Knowing that α = 6.5 × 10−6 /°F, determine the stress in the bar (a) when the temperature is raised to 320 °F, (b) after the temperature has returned to 45 °F .SOLUTIONLet P be the compressive force in the rod.Determine temperature change to cause yielding. PL σ L δ =− + L α (ΔT ) = − Y + Lα ( ΔT )Y = 0 AE E 3 σ 36 × 10 (ΔT )Y = Y = = 190.98 °F Eα (29 × 106 )(6.5 × 10−6 )But ΔT = 320 − 45 = 275 °F > (ΔTY )(a) Yielding occurs. σ = −σ Y = −36 ksi  Cooling: (ΔT)′ = 275 °F P′L δ ′ = δ P = δT = − ′ ′ + Lα (ΔT )′ = 0 AE P′ σ′= = − Eα (ΔT )′ A = −(29 × 106 )(6.5 × 10−6 )(275) = −51.8375 × 103 psi(b) Residual stress: σ res = −σ Y − σ ′ = −36 × 103 + 51.8375 × 103 = 15.84 × 10 psi 15.84 ksi PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.117 The steel rod ABC is attached to rigid supports and is unstressed at a temperature of 25 °C. The steel is assumed elastoplastic, with E = 200 GPa and σ y = 250 MPa. The temperature of both portions of the rod is then raised to 150 °C . Knowing that α = 11.7 × 10−6 / °C, determine (a) the stress in both portions of the rod, (b) the deflection of point C.SOLUTION AAC = 500 × 10−6 m 2 LAC = 0.150 m ACB = 300 × 10−6 m 2 LCB = 0.250 mConstraint: δ P + δT = 0Determine ΔT to cause yielding in portion CB. PLAC PLCB − − = LABα ( ΔT ) EAAC EACB P  LAC LCB  ΔT =  +  LAB Eα  AAC ACB At yielding, P = PY = ACBσ Y = (300 × 10−6 )(2.50 × 106 ) = 75 × 103 N PY  LAC LCB  (ΔT )Y =  +  LAB Eα  AAC ACB  75 × 103  0.150 0.250  = −6  −6 +  = 90.812 °C 9 (0.400)(200 × 10 )(11.7 × 10 )  500 × 10 300 × 10−6 Actual ΔT: 150 °C − 25 °C = 125 °C > (ΔT )YYielding occurs. For ΔT > (ΔT )Y , P = PY = 75 × 103 N PY 75 × 103(a) σ AC = − =− = −150 × 10−6 Pa σ AC = −150 MPa  AAC 500 × 10−6 PY σ CB = − = −σ Y σ CB = −250 MPa  ACB(b) For ΔT > (ΔT )Y , portion AC remains elastic. PY LAC δ C /A = − + LACα (ΔT ) EAAC (75 × 103 )(0.150) =− + (0.150)(11.7 × 10−6 )(125) = 106.9 × 10−6 m (200 × 109 )(500 × 10−6 )Since Point A is stationary, δ C = δ C /A = 106.9 × 10−6 m δ C = 0.1069 mm → PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.118∗ Solve Prob. 2.117, assuming that the temperature of the rod is raised to 150°C and then returned to 25 °C. PROBLEM 2.117 The steel rod ABC is attached to rigid supports and is unstressed at a temperature of 25 °C. The steel is assumed elastoplastic, with E = 200 GPa and σ Y = 250 MPa. The temperature of both portions of the rod is then raised to 150 °C. Knowing that α = 11.7 × 10−6 / °C, determine (a) the stress in both portions of the rod, (b) the deflection of point C.SOLUTION AAC = 500 × 10−6 m 2 LAC = 0.150 m ACB = 300 × 10−6 m 2 LCB = 0.250 mConstraint: δ P + δT = 0Determine ΔT to cause yielding in portion CB. PLAC PLCB − − = LABα ( ΔT ) EAAC EACB P  LAC LCB  ΔT =  +  LAB Eα  AAC ACB At yielding, P = PY = ACBσ Y = (300 × 10−6 )(250 × 106 ) = 75 × 103 N PY  LAC LCB  75 × 103  0.150 0.250 (ΔT )Y =  + = −6  −6 +  LAB Eα  AAC ACB  (0.400)(200 × 10 )(11.7 × 10 )  500 × 10 9 300 × 10−6  = 90.812 °CActual ΔT : 150 °C − 25 °C = 125 °C > (ΔT )YYielding occurs. For ΔT > (ΔT )Y P = PY = 75 × 103 N ELABα (ΔT )′ (200 × 109 )(0.400)(11.7 × 10−6 )(125)Cooling: (ΔT )′ = 125 °C P′ = = = 103.235 × 103 N ( LAC AAC + LCB ACB ) 0.150 500 × 10 −6 + 0.250 −6 300 × 10Residual force: Pres = P′ − PY = 103.235 × 103 − 75 × 103 = 28.235 × 103 N (tension)PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.118∗ (Continued) Pres 28.235 × 103(a) Residual stresses. σ AC = = σ AC = 56.5 MPa  AAC 500 × 10−6 Pres 28.235 × 103 σ CB = = σ CB = 9.41 MPa  ACB 300 × 10−6 Pres LAC(b) Permanent deflection of point C. δ C = δ C = 0.0424 mm →  EAACPROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.119∗ For the composite bar of Prob. 2.111, determine the residual stresses in the tempered-steel bars if P is gradually increased from zero to 98 kips and then decreased back to zero. 3 PROBLEM 2.111 Two tempered-steel bars, each 16 -in. thick, are 1 bonded to a 2 -in. mild-steel bar. This composite bar is subjected as shown to a centric axial load of magnitude P. Both steels are elastoplastic with E = 29 × 106 psi and with yield strengths equal to 100 ksi and 50 ksi, respectively, for the tempered and mild steel. The load P is gradually increased from zero until the deformation of the bar reaches a maximum value δ m = 0.04 in. and then decreased back to zero. Determine (a) the maximum value of P, (b) the maximum stress in the tempered-steel bars, (c) the permanent set after the load is removed.SOLUTION 1Areas: Mild steel: A1 =   (2) = 1.00 in 2 2  3 Tempered steel: A2 = (2)   (2) = 0.75 in 2  16  Total: A = A1+ A2 = 1.75 in 2 PYTotal force to yield the mild steel: σ Y 1 = ∴ PY = Aσ Y 1 = (1.75)(50 × 103 ) = 87.50 × 103 lb AP > PY; therefore, mild steel yields.Let P = force carried by mild steel. 1 P2 = force carried by tempered steel. P = A1σ Y 1 = (1.00)(50 × 103 ) = 50 × 103 lb 1 P + P2 = P, P2 = P − P = 98 × 103 − 50 × 103 = 48 × 103 lb 1 1 P2 48 × 103 σ2 = = = 64 × 103 psi A2 0.75 P 98 × 103Unloading: σ ′ = = = 56 × 103 psi A 1.75Residual stresses. Mild steel: σ1,res = σ 1 − σ ′ = 50 × 103 − 56 × 103 = −6 × 10−3 psi = −6 ksi Tempered steel: σ 2,res = σ 2 − σ 1 = 64 × 103 − 56 × 103 = 8 × 103 psi 8.00 ksi PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.120∗ For the composite bar in Prob. 2.111, determine the residual stresses in the tempered-steel bars if P is gradually increased from zero until the deformation of the bar reaches a maximum value δ m = 0.04 in. and is then decreased back to zero. 3 PROBLEM 2.111 Two tempered-steel bars, each 16 -in. thick, are bonded 1 to a 2 -in. mild-steel bar. This composite bar is subjected as shown to a centric axial load of magnitude P. Both steels are elastoplastic with E = 29 × 106 psi and with yield strengths equal to 100 ksi and 50 ksi, respectively, for the tempered and mild steel. The load P is gradually increased from zero until the deformation of the bar reaches a maximum value δ m = 0.04 in. and then decreased back to zero. Determine (a) the maximum value of P, (b) the maximum stress in the tempered-steel bars, (c) the permanent set after the load is removed.SOLUTION 1 Lδ (14)(50 × 103 )For the mild steel, A1 =   (2) = 1.00 in 2 δ Y 1 = Y 1 = = 0.024138 in 2 E 29 × 106  3 Lδ (14)(100 × 103 )For the tempered steel, A2 = 2   (2) = 0.75 in 2 δ Y 2 = Y 2 = = 0.048276 in.  16  E 29 × 106Total area: A = A1 + A2 = 1.75 in 2 δY 1 < δ m < δ Y 2The mild steel yields. Tempered steel is elastic.Forces: P = A1δY 1 = (1.00)(50 × 103 ) = 50 × 103 lb 1 EA2δ m (29 × 106 )(0.75)(0.04) P2 = = = 62.14 × 103 lb L 14 P P 62.14 × 103Stresses: σ1 = 1 = δY 1 = 50 × 103 psi σ 2 = 2 = = 82.86 × 103 psi A1 A2 0.75 P 112.14Unloading: σ′ = = = 64.08 × 103 psi A 1.75Residual stresses. σ1,res = σ 1 − σ ′ = 50 × 103 − 64.08 × 103 = −14.08 × 103 psi = −14.08 ksi σ 2,res = σ 2 − σ ′ = 82.86 × 103 − 64.08 × 103 = 18.78 × 103 psi = 18.78 ksi PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.121∗ Narrow bars of aluminum are bonded to the two sides of a thick steel plate as shown. Initially, at T1 = 70°F, all stresses are zero. Knowing that the temperature will be slowly raised to T2 and then reduced to T1, determine (a) the highest temperature T2 that does not result in residual stresses, (b) the temperature T2 that will result in a residual stress in the aluminum equal to 58 ksi. Assume α a = 12.8 × 10−6 / °F for the aluminum and α s = 6.5 × 10−6 / °F for the steel. Further assume that the aluminum is elastoplastic, with E = 10.9 × 106 psi and σ Y = 58 ksi. (Hint: Neglect the small stresses in the plate.)SOLUTIONDetermine temperature change to cause yielding. PL δ= + Lα a ( ΔT )Y = Lα s ( ΔT )Y EA P = σ = − E (α a − α s )(ΔT )Y = −σ Y A σY 58 × 103 (ΔT )Y = = = 844.62 °F E (α a − α s ) (10.9 × 10 )(12.8 − 6.5)(10−6 ) 6(a) T2Y = T1 + (ΔT )Y = 70 + 844.62 = 915 °F 915 °F  After yielding, σY L δ= + Lα a (ΔT ) = Lα s (ΔT ) E Cooling: P′L δ′ = + Lα a (ΔT )′ = Lα s (ΔT )′ AE The residual stress is P′ σ res = σ Y − = σ Y − E (α a − α s )(ΔT ) A Set σ res = −σ Y −σ Y = σ Y − E (α a − α s )(ΔT ) 2σ Y (2)(58 × 103 ) ΔT = = = 1689 °F E (α a − α s ) (10.9 × 106 )(12.8 − 6.5)(10−6 )(b) T2 = T1 + ΔT = 70 + 1689 = 1759 °F 1759 °F  If T2 > 1759 °F, the aluminum bar will most likely yield in compression.PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.122∗ Bar AB has a cross-sectional area of 1200 mm2 and is made of a steel that is assumed to be elastoplastic with E = 200 GPa and σ Y = 250 MPa. Knowing that the force F increases from 0 to 520 kN and then decreases to zero, determine (a) the permanent deflection of point C, (b) the residual stress in the bar.SOLUTION A = 1200 mm 2 = 1200 × 10−6 m 2Force to yield portion AC: PAC = Aσ Y = (1200 × 10−6 )(250 × 106 ) = 300 × 103 NFor equilibrium, F + PCB − PAC = 0. PCB = PAC − F = 300 × 103 − 520 × 103 = −220 × 103 N PCB LCB (220 × 103 )(0.440 − 0.120) δC = − = EA (200 × 109 )(1200 × 10−6 ) = 0.293333 × 10−3 m PCB 220 × 103 σ CB = = A 1200 × 10−6 = −183.333 × 106 PaPROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.122∗ (Continued)Unloading: ′ PAC LAC P′ L ′ ( F − PAC ) LCB ′ δC = = − CB CB = EA EA EA L L  FL ′ PAC  AC + BC  = CB  EA EA  EA FLCB (520 × 103 )(0.440 − 0.120) ′ PAC = = = 378.182 × 103 N LAC + LCB 0.440 PCB = PAC − F = 378.182 × 103 − 520 × 103 = −141.818 × 103 N ′ ′ ′ PAC 378.182 × 103 ′ σ AC = = −6 = 315.152 × 106 Pa A 1200 × 10 P′ 141.818 × 103 ′ σ BC = BC = − −6 = −118.182 × 106 Pa A 1200 × 10 (378.182)(0.120) ′ δC = 9 6 = 0.189091 × 10−3 m (200 × 10 )(1200 × 10 )(a) δ C , p = δ C − δ C = 0.293333 × 10−3 − 0.189091 × 10−3 = 0.1042 × 10−3 m ′ = 0.1042 mm (b) σ AC , res = σ Y − σ ′ = 250 × 106 − 315.152 × 106 = −65.2 × 106 Pa AC = −65.2 MPa  σ CB , res = σ CB − σ CB = −183.333 × 106 + 118.182 × 106 = −65.2 × 106 Pa ′ = −65.2 MPa PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.123∗ Solve Prob. 2.122, assuming that a = 180 mm. PROBLEM 2.122 Bar AB has a cross-sectional area of 1200 mm2 and is made of a steel that is assumed to be elastoplastic with E = 200 GPa and σ Y = 250 MPa. Knowing that the force F increases from 0 to 520 kN and then decreases to zero, determine (a) the permanent deflection of point C, (b) the residual stress in the bar.SOLUTION A = 1200 mm 2 = 1200 × 10−6 m 2Force to yield portion AC: PAC = Aσ Y = (1200 × 10−6 )(250 × 106 ) = 300 × 103 NFor equilibrium, F + PCB − PAC = 0. PCB = PAC − F = 300 × 103 − 520 × 103 = −220 × 103 N PCB LCB (220 × 103 )(0.440 − 0.180) δC = − = EA (200 × 109 )(1200 × 10−6 ) = 0.238333 × 10−3 m PCB 220 × 103 σ CB = =− A 1200 × 10−6 = −183.333 × 106 PaPROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.123∗ (Continued)Unloading: ′ PAC LAC P′ L ′ ( F − PAC ) LCB ′ δC = = − CB CB = EA EA EA L L  FL ′ = PAC  AC + BC  = CB  EA EA  EA FLCB (520 × 103 )(0.440 − 0.180) ′ PAC = = = 307.273 × 103 N LAC + LCB 0.440 PCB = PAC − F = 307.273 × 103 − 520 × 103 = −212.727 × 103 N ′ ′ (307.273 × 103 )(0.180) ′ δC = = 0.230455 × 10−3 m (200 × 109 )(1200 × 10−6 ) ′ PAC 307.273 × 103 ′ σ AC = = −6 = 256.061 × 106 Pa A 1200 × 10 PCB −212.727 × 103 ′ ′ σ CB = = = −177.273 × 106 P A 1200 × 10−6(a) δ C , p = δ C − δ C = 0.238333 × 10−3 − 0.230455 × 10−3 = 0.00788 × 10−3 m ′ = 0.00788 mm (b) σ AC ,res = σ AC − σ ′ = 250 × 106 − 256.061 × 106 = −6.06 × 106 Pa AC = −6.06 MPa  σ CB ,res = σ CB − σ CB = −183.333 × 106 + 177.273 × 106 = −6.06 × 106 Pa ′ = −6.06 MPa  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.124 Rod BD is made of steel ( E = 29 × 106 psi) and is used to brace the axially compressed member ABC. The maximum force that can be developed in member BD is 0.02P. If the stress must not exceed 18 ksi and the maximum change in length of BD must not exceed 0.001 times the length of ABC, determine the smallest-diameter rod that can be used for member BD.SOLUTION FBD = 0.02 P = (0.02) (130) = 2.6 kips = 2.6 × 103lbConsidering stress: σ = 18 ksi = 18 × 103 psi FBD FBD 2.6 σ= ∴ A= = = 0.14444 in 2 A σ 18Considering deformation: δ = (0.001)(144) = 0.144 in. FBD LBD FBD LBD (2.6 × 103 ) (54) δ= ∴ A= = = 0.03362 in 2 AE Eδ 6 (29 × 10 ) (0.144)Larger area governs. A = 0.14444 in 2 π 4A (4) (0.14444) A= d2 ∴ d= = d = 0.429in.  4 π πPROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.125 Two solid cylindrical rods are joined at B and loaded as shown. Rod AB is made of steel ( E = 200 GPa) and rod BC of brass ( E = 105 GPa). Determine (a) the total deformation of the composite rod ABC, (b) the deflection of point B.SOLUTIONRod AB: FAB = − P = −30 × 103 N LAB = 0.250 m E AB = 200 × 109 GPa π AAB = (30)2 = 706.85 mm 2 = 706.85 × 10−6 m 2 4 F L (30 × 103 )(0.250) δ AB = AB AB = − 9 −6 = −53.052 × 10−6 m E AB AAB (200 × 10 )(706.85 × 10 )Rod BC: FBC = 30 + 40 = 70 kN = 70 × 103 N LBC = 0.300 m EBC = 105 × 109 Pa π ABC = (50) 2 = 1.9635 × 103 mm 2 = 1.9635 × 10−3 m 2 4 F L (70 × 103 )(0.300) δ BC = BC BC = − 9 −3 = −101.859 × 10−6 m EBC ABC (105 × 10 )(1.9635 × 10 )(a) Total deformation: δ tot = δ AB + δ BC = −154.9 × 10−6 m = −0.1549 mm (b) Deflection of Point B. δ B = δ BC δ B = 0.1019 mm ↓ PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.126 Two solid cylindrical rods are joined at B and loaded as shown. Rod AB is made of steel ( E = 29 × 106 psi), and rod BC of brass ( E = 15 × 106 psi). Determine (a) the total deformation of the composite rod ABC, (b) the deflection of point B.SOLUTIONPortion AB: PAB = 40 × 103 lb LAB = 40 in. d = 2 in. π π AAB = d2 = (2) 2 = 3.1416 in 2 4 4 E AB = 29 × 106 psi PAB LAB (40 × 103 )(40) δ AB = = = 17.5619 × 10−3 in. E AB AAB (29 × 106 )(3.1416)Portion BC: PBC = −20 × 103 lb LBC = 30 in. d = 3 in. π π ABC = d2 = (3) 2 = 7.0686 in 2 4 4 EBC = 15 × 106 psi PBC LBC (−20 × 103 )(30) δ BC = = 6 = −5.6588 × 10−3 in. EBC ABC (15 × 10 )(7.0686)(a) δ = δ AB + δ BC = 17.5619 × 10−6 − 5.6588 × 10−6 δ = 11.90 × 10−3 in. ↓ (b) δ B = −δ BC δ B = 5.66 × 10−3 in. ↑ PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.127 The uniform wire ABC, of unstretched length 2l, is attached to the supports shown and a vertical load P is applied at the midpoint B. Denoting by A the cross-sectional area of the wire and by E the modulus of elasticity, show that, for δ  l , the deflection at the midpoint B is P δ =l3 AESOLUTIONUse approximation. δ sin θ ≈ tan θ ≈ lStatics: ΣFY = 0 : 2 PAB sin θ − P = 0 P Pl PAB = ≈ 2sin θ 2δ PAB l Pl 2Elongation: δ AB = = AE 2 AEδDeflection:From the right triangle, (l + δ AB ) 2 = l 2 + δ 2 δ 2 = l 2 + 2l δ AB + δ AB − l 2 2  1 δ AB  = 2l δ AB 1 +  ≈ 2l δ AB  2 l  Pl 3 ≈ AEδ Pl 3 P δ3 ≈ ∴ δ ≈ l3  AE AEPROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.128 The brass strip AB has been attached to a fixed support at A and rests on a rough support at B. Knowing that the coefficient of friction is 0.60 between the strip and the support at B, determine the decrease in temperature for which slipping will impend.SOLUTIONBrass strip: E = 105 GPa α = 20 × 10−6 / °C ΣFy = 0 : N − W = 0 N =W ΣFx = 0 : P − μ N = 0 P = μW = μ mg PL P μ mg δ =− + Lα ( ΔT ) = 0 ΔT = = EA EAα EAαData: μ = 0.60 A = (20)(3) = 60 mm 2 = 60 × 10−6 m 2 m = 100 kg g = 9.81 m/s 2 E = 105 × 109 Pa (0.60)(100)(9.81) ΔT = ΔT = 4.67 °C  (105 × 109 )(60 × 10−6 )(20 × 106 )PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.129 Members AB and CD are 1 1 -in.-diameter steel rods, and members BC and AD 8 are 7 -in.-diameter steel rods. When the turnbuckle is tightened, the diagonal 8 member AC is put in tension. Knowing that E = 29 × 106 psi and h = 4 ft, determine the largest allowable tension in AC so that the deformations in members AB and CD do not exceed 0.04 in.SOLUTION δ AB = δ CD = 0.04 in. h = 4ft = 48 in. = LCD π π ACD = d2 = (1.125) 2 = 0.99402 in 2 4 4 FCD LCD δ CD = EACD EACDδ CD (29 × 106 )(0.99402)(0.04) FCD = = = 24.022 × 103 lb LCD 48Use joint C as a free body. 4 5 ΣFy = 0 : FCD − FAC = 0 ∴ FAC = FCD 5 4 5 FAC = (24.022 × 103 ) = 30.0 × 103 lb FAC = 30.0 kips  4PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.130 The 1.5-m concrete post is reinforced with six steel bars, each with a 28-mm diameter. Knowing that Es = 200 GPa and Ec = 25 GPa, determine the normal stresses in the steel and in the concrete when a 1550-kN axial centric force P is applied to the post.SOLUTIONLet Pc = portion of axial force carried by concrete. Ps = portion carried by the six steel rods. Pc L Ec Acδ δ= Pc = Ec Ac L Ps L Es Asδ δ= Ps = Es As L δ P = Pc + Ps = ( Ec Ac + Es As ) L δ −P ε= = L Ec Ac + Es As π 6π As = 6 d s2 = (28) 2 = 3.6945 × 103 mm 2 4 4 = 3.6945 × 10−3 m 2 π π Ac = d c2 − As = (450) 2 − 3.6945 × 103 4 4 = 155.349 × 103 mm 2 = 153.349 × 10−3 m 2 L = 1.5 m 1550 × 103 ε= = 335.31 × 10−6 (25 × 109 )(155.349 × 10−3 ) + (200 × 109 )(3.6945 × 10−3 ) σ s = Es ε = (200 × 109 )(335.31 × 10−6 ) = 67.1 × 106 Pa σ s = 67.1 MPa  σ c = Ecε = (25 × 109 )( −335.31 × 10−6 ) = 8.38 × 106 Pa σ c = 8.38 MPa PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.131 The brass shell (α b = 11.6 × 10−6 /°F) is fully bonded to the steel core (α s = 6.5 × 10−6 /°F). Determine the largest allowable increase in temperature if the stress in the steel core is not to exceed 8 ksi.SOLUTIONLet Ps = axial force developed in the steel core.For equilibrium with zero total force, the compressive force in the brass shell is Ps . PsStrains: εs = + α s (ΔT ) Es As Ps εb = − + α b (ΔT ) Eb AbMatching: ε s = εb Ps P + α s (ΔT ) = − s + α b (ΔT ) Es As Eb Ab  1 1   +  Ps = (α b − α s )(ΔT ) (1)  Es As Eb Ab  Ab = (1.5) (1.5) − (1.0) (1.0) = 1.25 in 2 As = (1.0) (1.0) = 1.0 in 2 αb − α s = 5.1 × 10−6 /°F Ps = σ s As = (8 × 103 ) (1.0) = 8 × 103 lb 1 1 1 1 + = + = 87.816 × 10−9 lb −1 Es As Eb Ab (29 × 106 ) (1.0) (15 × 106 ) (1.25)From (1), (87.816 × 10−9 ) (8 × 103 ) = (5.1 × 10−6 ) (ΔT ) ΔT = 137.8 °F PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.132 A fabric used in air-inflated structures is subjected to a biaxial loading that results in normal stresses σ x = 120 MPa and σ z = 160 MPa. Knowing that the properties of the fabric can be approximated as E = 87 GPa and v = 0.34, determine the change in length of (a) side AB, (b) side BC, (c) diagonal AC.SOLUTION σ x = 120 × 106 Pa, σ y = 0, σ z = 160 × 106 Pa 1 1 εx = (σ x − vσ y − vσ z ) = [120 × 106 − (0.34)(160 × 106 )] = 754.02 × 10−6 E 87 × 109 1 1 ε z = ( −vσ x − vσ y + σ z ) = [−(0.34)(120 × 106 ) + 160 × 106 ] = 1.3701 × 10−3 E 87 × 109(a) δ AB = ( AB)ε x = (100 mm)(754.02 × 10−6 ) = 0.0754 mm (b) δ BC = ( BC )ε z = (75 mm)(1.3701 × 10−6 ) = 0.1028 mm  Label sides of right triangle ABC as a, b, and c. c2 = a 2 + b2 Obtain differentials by calculus. 2c dc = 2a da + 2b db a b dc = da + db c c But a = 100 mm, b = 75 mm, c = (1002 + 752 ) = 125 mm da = δ AB = 0.0754 mm db = δ BC = 0.1370 mm 100 75(c) δ AC = dc = (0.0754) + (0.1028) = 0.1220 mm  125 125PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.133 An elastomeric bearing (G = 0.9 MPa) is used to support a bridge girder as shown to provide flexibility during earthquakes. The beam must not displace more than 10 mm when a 22-kN lateral load is applied as shown. Knowing that the maximum allowable shearing stress is 420 kPa, determine (a) the smallest allowable dimension b, (b) the smallest required thickness a.SOLUTIONShearing force: P = 22 × 103 NShearing stress: τ = 420 × 103 Pa P P τ= ∴ A= A τ 22 × 103 = = 52.381 × 10−3 m 2 420 × 103 = 52.381 × 103 mm 2 A = (200 mm)(b) A 52.381 × 103(a) b= = = 262 mm b = 262 mm  200 200 τ 420 × 103 γ= = 6 = 466.67 × 10−3 G 0.9 × 10 δ δ 10 mm(b) But γ = ∴ a= = = 21.4 mm a = 21.4 mm  a γ 466.67 × 10−3PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.134 Knowing that P = 10 kips, determine the maximum stress when (a) r = 0.50 in., (b) r = 0.625 in.SOLUTION P = 10 × 103 lb D = 5.0 in. d = 2.50 in. D 5.0 = = 2.00 d 2.50 3 Amin = dt = (2.50)   = 1.875 in 2 4 r 0.50(a) r = 0.50 in. = = 0.20 d 2.50 From Fig. 2.60b, K = 1.94 KP (1.94) (10 × 103 ) σ max = = = 10.35 × 103 psi Amin 1.875 10.35 ksi  r 0.625(b) r = 0.625 in. = = 0.25 K = 1.82 d 2.50 KP (1.82) (10 × 103 ) σ max = = = 9.71 × 103 psi Amin 1.875 9.71 ksi PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.
  • PROBLEM 2.135 The uniform rod BC has a cross-sectional area A and is made of a mild steel that can be assumed to be elastoplastic with a modulus of elasticity E and a yield strength σ y . Using the block-and-spring system shown, it is desired to simulate the deflection of end C of the rod as the axial force P is gradually applied and removed, that is, the deflection of points C and C′ should be the same for all values of P. Denoting by μ the coefficient of friction between the block and the horizontal surface, derive an expression for (a) the required mass m of the block, (b) the required constant k of the spring.SOLUTIONForce-deflection diagram for Point C or rod BC.For P < P = Aσ Y Y PL EA δC = P= δC EA L Pmax = P = Aσ Y YForce-deflection diagram for Point C′ of block-and-spring system. ΣFy = 0 : N − mg = 0 N = mg ΣFx = 0 : P − F f = 0 P = FfIf block does not move, i.e., F f < μ N = μ mg or P < μ mg , Pthen δ c′ = or ′ P = kδ c KIf P = μ mg, then slip at P = Fm = μ mg occurs.If the force P is the removed, the spring returns to its initial length. Aσ Y(a) Equating PY and Fmax, Aσ Y = μ mg m=  μg EA(b) Equating slopes, k=  LPROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using itwithout permission.