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CHAPTER 2 PROBLEM 2.1 An 80-m-long wire of 5-mm diameter is made of a steel with E = 200 GPa and an ultimate tensile strength of 400 MPa. If a factor of safety of 3.2 is desired, determine (a) the largest allowable tension in the wire, (b) the corresponding elongation of the wire. SOLUTION (a) σ U = 400 × 106 Pa A= π 4 d2 = π 4 (5) 2 = 19.635 mm 2 = 19.635 × 10−6 m 2 PU = σ U A = (400 × 106 ) (19.635 × 10−6 ) = 7854 N Pall = (b) δ= PU 7854 = = 2454 N F .S 3.2 PL (2454) (80) = = 50.0 × 10−3 m AE (19.635 × 10−6 )(200 × 109 ) Pall = 2.45 kN δ = 50.0 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.2 A 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) the corresponding normal stress caused by the load. SOLUTION (a) δ= PL AE : 0.04 in. = (2000 lb) (5.5 × 12 in.) 6 A (29 × 10 psi) 1 A = π d 2 = 0.11379 in 2 4 d = 0.38063 in. (b) σ= P A = 2000 lb 0.11379 in 2 = 17580 psi d = 0.381 in. σ = 17.58 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.3 Two gage marks are placed exactly 10 in. apart on a 12 -in.-diameter aluminum rod with E = 10.1 × 106 psi and an ultimate strength of 16 ksi. Knowing that the distance between the gage marks is 10.009 in. after a load is applied, determine (a) the stress in the rod, (b) the factor of safety. SOLUTION (a) δ = 10.009 − 10.000 = 0.009 in. ε= (b) F. S . = σU 16 = 9.09 σ δ L = σ E σ= Eδ (10.1 × 106 ) (0.009) = = 9.09 × 103 psi L 10 σ = 9.09 ksi F. S . = 1.760 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.4 An 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 (a) δ= PL , or AE P= δ AE L 1 1 with A = π d 2 = π (0.005) 2 = 19.6350 × 10−6 m 2 4 4 P= (0.045 m)(19.6350 × 10−6 m 2 )(200 × 109 N/m 2 ) = 9817.5 N 18 m P = 9.82 kN (b) σ= P A = 9817.5 N 19.6350 × 10 −6 6 m 2 = 500 × 10 Pa σ = 500 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.5 A polystyrene rod of length 12 in. and diameter 0.5 in. is subjected to an 800-lb tensile load. Knowing that E = 0.45 × 106 psi, determine (a) the elongation of the rod, (b) the normal stress in the rod. SOLUTION A= (a) δ= (b) σ= PL AE P A = = (800) (12) 6 (0.19635) (0.45 ×10 ) 800 0.19635 = 4074 psi π 4 d2 = π 4 = 0.1086 (0.5) 2 = 0.19635 in 2 δ = 0.1086 in. σ = 4.07 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.6 A nylon thread is subjected to a 8.5-N tension force. Knowing that E = 3.3 GPa and that the length of the thread increases by 1.1%, determine (a) the diameter of the thread, (b) the stress in the thread. SOLUTION (a) 1.1 = 0.011 100 ε= Stress: σ = Eε = (3.3 × 109 )(0.011) = 36.3 × 106 Pa σ = (b) δ Strain: Area: A= Diameter: d = Stress: = L P A P σ = 4A π 8.5 = 234.16 × 10−9 m 2 36.3 × 106 = (4)(234.16 × 10−9 ) π = 546 × 10−6 m d = 0.546 mm 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.7 Two gage marks are placed exactly 250 mm apart on a 12-mm-diameter aluminum rod. Knowing that, with an axial load of 6000 N acting on the rod, the distance between the gage marks is 250.18 mm, determine the modulus 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 π 250 mm π (12)2 = 113.097 mm 2 = 113.097 ×10−6 m 2 4 4 6000 P σ= = = 53.052 × 106 Pa A 113.097 × 10−6 A= E= d2 = σ 53.052 × 106 = = 73.683 × 109 Pa ε 0.00072 E = 73.7 GPa 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.8 An aluminum pipe must not stretch more than 0.05 in. when it is subjected to a tensile load. Knowing that E = 10.1 × 106 psi and that the maximum allowable normal stress is 14 ksi, determine (a) the maximum allowable length of the pipe, (b) the required area of the pipe if the tensile load is 127.5 kips. SOLUTION (a) δ= PL ; AE L= Thus, EAδ Eδ (10.1 × 106 ) (0.05) = = σ P 14 × 103 L = 36.1 in. (b) σ = Thus, P ; A A= P σ = 127.5 × 103 14 × 103 A = 9.11 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.9 An aluminum control rod must stretch 0.08 in. when a 500-lb tensile load is applied to it. Knowing that σ all = 22 ksi and E = 10.1 × 106 psi, determine the smallest diameter and shortest length that can be selected for the rod. SOLUTION σ all = 22 × 103 psi P = 500 lb, δ = 0.08 in. σ= A= P < σ all A π 4 d2 A> d= P σ all = 4A π 500 = 0.022727 in 2 22 × 103 = (4)(0.022727) π Eδ < σ all L Eδ (10.1 × 106 )(0.08) L> = = 36.7 in. σ all 22 × 103 d min = 0.1701 in. σ = Eε = Lmin = 36.7 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.10 A square yellow-brass bar must not stretch more than 2.5 mm when it is subjected to a tensile load. Knowing that E = 105 GPa and that the allowable tensile strength is 180 MPa, determine (a) the maximum allowable length of the bar, (b) the required dimensions of the cross section if the tensile load is 40 kN. SOLUTION σ = 180 × 106 Pa P = 40 × 103 N E = 105 × 109 Pa δ = 2.5 × 10−3 m (a) PL σ L = AE E Eδ (105 × 109 )(2.5 × 10−3 ) = = 1.45833 m L= σ 180 × 106 δ= L = 1.458 m (b) σ = A= P A P σ A = a2 = 40 × 103 = 222.22 × 10−6 m 2 = 222.22 mm 2 180 × 106 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.11 A 4-m-long steel rod must not stretch more than 3 mm and the normal stress must not exceed 150 MPa when the rod is subjected to a 10-kN axial load. Knowing that E = 200 GPa, determine the required diameter of the rod. SOLUTION L =4m δ = 3 × 10−3 m, σ = 150 × 106 Pa E = 200 × 109 Pa, P = 10 × 103 N Stress: σ = A= Deformation: P A P σ = 10 × 103 = 66.667 × 10−6 m 2 = 66.667 mm 2 6 150 × 10 δ = PL AE A= PL (10 × 103 ) (4) = = 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 A= π 4 d2 d= 4A π = 4(66.667) π d = 9.21 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.12 A nylon thread is to be subjected to a 10-N tension. Knowing that E = 3.2 GPa, that the maximum allowable normal stress is 40 MPa, and that the length of the thread must not increase by more than 1%, determine the required diameter of the thread. SOLUTION Stress criterion: σ = 40 MPa = 40 × 106 Pa P = 10 N σ = A= P P 10 N : A= = = 250 × 10−9 m 2 A σ 40 × 106 Pa π 4 d 2: d = 2 A π 250 × 10−9 =2 π = 564.19 × 10−6 m d = 0.564 mm Elongation criterion: δ L = 1% = 0.01 δ = PL : AE A= P /E 10 N/3.2 × 109 Pa = = 312.5 × 10−9 m 2 0.01 δ /L d =2 A π =2 312.5 × 10−9 π = 630.78 × 10−6 m 2 d = 0.631 mm The 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 m Use bar AB as a free body. 4 3.5 P − (6) FBC = 0 7.2111 P = 0.9509 FBC Σ M A = 0: Considering allowable stress: σ = 190 × 106 Pa A= π d2 = 4 FBC σ= A π 4 (0.004) 2 = 12.566 × 10−6 m 2 ∴ FBC = σ A = (190 × 106 ) (12.566 × 10−6 ) = 2.388 × 103 N Considering allowable elongation: δ = 6 × 10−3 m δ= FBC LBC AE ∴ FBC = AEδ (12.566 × 10−6 )(200 × 109 )(6 × 10−3 ) = = 2.091 × 103 N LBC 7.2111 Smaller 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. SOLUTION Deformation of aluminum rod. δA = = PLAB PLBC + AAB E ABC E P LAB LBC + E AAB ABC 12 + π (1.5)2 4 = 0.031376 in. = Steel rod. 28 × 103 10.1 × 106 18 (2.25) 2 4 π δ = 0.031376 in. δ= PL PL (28 × 103 )(30) ∴ A= = = 0.92317 in 2 6 EA Eδ (29 × 10 )(0.031376) σ= P A ∴ A= P σ = 28 × 103 = 1.1667 in 2 24 × 103 Required area is the larger value. A = 1.1667 in 2 Diameter: d= 4A π = (4)(1.6667) π d = 1.219 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 58 -in.-diameter steel rod BC hangs from a rigid 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. SOLUTION Rod BC: LBC = 7 ft = 84 in. EBC = 29 × 106 psi ABC = δ C/B = Pipe AB: π 4 d2 = π 4 (0.625) 2 = 0.30680 in 2 PLBC (15 × 103 )(84) = = 0.141618 in. EBC ABC (29 × 106 )(0.30680) LAB = 4 ft = 48 in. E AB = 10.4 × 106 psi AAB = 1.75 in 2 δ B/A = Total: PLAB (15 × 103 ) (48) = = 39.560 × 10−3 in. 6 E AB AAB (10.4 × 10 ) (1.75) 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. SOLUTION Shortening 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 δ AB = PLAB P(0.376) = = 25.578 × 10−9 P −6 9 E AB AAB (105 × 10 )(140 × 10 ) Lengthening of aluminum cylinder CD: LCD = 0.375 m δ CD = Total deflection: ACD = 250 mm 2 = 250 × 10−6 m 2 ECD = 72 × 109 Pa PLCD P(0.375) = = 20.833 × 10−9 P 9 −6 ECD ACD (72 × 10 )(250 × 10 ) 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 Atube = A rod = π 4 π 4 (d o2 − di2 ) = d2 = π 4 π 4 (362 − 282 ) = 402.12 mm 2 = 402.12 × 10−6 m 2 (25)2 = 490.87 mm 2 = 490.87 × 10−6 m 2 PL P(0.250) = = 8.8815 × 10−9 P 9 −6 Etube Atube (70 × 10 )(402.12 × 10 ) PL P(0.250) = −4.8505 × 10−9 P =− = Erod Arod (105 × 106 )(490.87 × 10−6 ) δ tube = δ rod 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 P= (a) σ tube = P 27.308 × 103 = = 67.9 × 106 Pa Atube 402.12 × 10−6 σ rod = − (b) 0.375 × 10−3 = 27.308 × 103 N −9 (8.8815 + 4.8505)(10 ) P 27.308 × 103 =− = −55.6 × 106 Pa Arod 490.87 × 10−6 δ tube = (8.8815 × 10−9 )(27.308 × 103 ) = 242.5 × 10−6 m δ rod = −(4.8505 × 10−9 )(27.308 × 103 ) = −132.5 × 10−6 m σ tube = 67.9 MPa σ rod = −55.6 MPa δ tube = 0.2425 mm 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 (a) δ =Σ Pi Li P Li = Σ Ai Ei E Ai −1 L π P = Eδ Σ i Ai = di2 4 Ai 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 P = (29 × 106 )(0.002)(6.083) −1 = 9.353 × 103 lb (b) δ BC = PLBC P LBC 9.535 × 103 = = (3.8197) ABC E E ABC 29 × 106 ← sum P = 9.53 kips δ = 1.254 × 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 (a) AAB = ABC = π 4 π 4 2 d AB = 2 d BC = π 4 (0.020)2 = 314.16 × 10−6 m 2 π (0.060)2 = 2.8274 × 10−3 m 2 4 Force in member AB is P tension. Elongation: δ AB = PLAB (4 × 103 )(0.4) = = 72.756 × 10−6 m −6 9 EAAB (70 × 10 )(314.16 × 10 ) Force in member BC is Q − P compression. Shortening: δ BC = (Q − P) LBC (Q − P)(0.5) = = 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 (b) δ AB = δ BC = δ B = 72.756 × 10−6 m Q = 32.8 kN 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 AAB = ABC = π 4 π 4 2 d AB = 2 d BC = π 4 π 4 (0.020) 2 = 314.16 × 10−6 m 2 (0.060)2 = 2.8274 × 10−3 m 2 PAB = P = 6 × 103 N PBC = P − Q = 6 × 103 − 42 × 103 = −36 × 103 N LAB = 0.4 m LBC = 0.5 m δ AB = PAB LAB (6 × 103 )(0.4) = = 109.135 × 10−6 m AAB E A (314.16 × 10−6 )(70 × 109 ) δ BC = PBC LBC (−36 × 103 )(0.5) = = −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 (b) δ B = δ BC = −90.9 × 10−6 m = −0.0909 mm or δ A = 0.01819 mm ↑ 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.21 Members AB and BC are made of steel ( E = 29 × 106 psi) with crosssectional 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 FAB − 28 = 0 7.810 = 43.74 kip = 43.74 × 103 lb ΣFy = 0: FAB δ AB = (b) FAB LAB (43.74 × 103 ) (93.72) = EAAB (29 × 106 ) (0.80) δ AB = 0.1767 in. Use joint B as a free body. ΣFx = 0: FBC − FBC = δ BC = 6 FAB = 0 7.810 (6) (43.74) = 33.60 kip = 33.60 × 103 lb. 7.810 FBC LBC (33.60 × 103 ) (72) = EABC (29 × 106 ) (0.64) δ BC = 0.1304 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 δ BD = FBD LBD EBD ABD FBD = (200 × 109 ) (1920 × 10−6 )(1.6 × 10−3 ) EBD ABDδ BD = 7.81 LBD = 78.67 × 103 N Use joint B as a free body. ΣFx = 0: 5 FBD − P = 0 7.810 P= 5 (5)(78.67 × 103 ) 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. SOLUTION Statics: Reactions are 114 kN upward at A and C. Member BD is a zero force member. LAB = 4.02 + 2.52 = 4.717 m Use joint A as a free body. ΣFy = 0 : 114 + 2.5 FAB = 0 4.717 FAB = −215.10 kN ΣFx = 0 : FAD + FAD = − 4 FAB = 0 4.717 (4)(−215.10) = 182.4 kN 4.717 Member AB: δ AB = FAB LAB (−215.10 × 103 )(4.717) = EAAB (200 × 109 )(2400 × 10−6 ) = −2.11 × 10−3 m Member AD: δ AD = δ AB − 2.11 mm FAD LAD (182.4 × 103 )(4.0) = 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 crosssectional areas are 2 in2 and 3 in2, respectively. SOLUTION Free body: Portion ABC of truss ΣM E = 0 : FBD (15 ft) − (30 kips)(8 ft) − (30 kips)(16 ft) = 0 FBD = + 48.0 kips Free body: Portion ABEC of truss ΣFx = 0 : 30 kips + 30 kips − FDE = 0 FDE = + 60.0 kips δ BD = PL (+48.0 × 103 lb)(8 × 12 in.) = AE (2 in 2 )(29 × 106 psi) δ BD = +79.4 × 10−3 in. δ DE = PL (+60.0 × 103 lb)(15 × 12 in.) = AE (3 in 2 )(29 × 106 psi) δ DE + 124.1 × 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. SOLUTION Free 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) = = 5.6766 × 10−3 in = δ B 6 EA (10.9 × 10 ) (0.2) F L (312.5) (18) = CD CD = = 2.5803 × 10−3 in = δ C 6 EA (10.9 × 10 ) (0.2) δ AB = δ CD Deformation diagram: Slope θ = δ B − δC LBC = 3.0963 × 10−3 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.26 3 The length of the 32 -in.-diameter steel wire CD has been adjusted so that with no load applied, a gap of 161 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. SOLUTION Rigid beam ACB rotates through angle θ to close gap. θ= 1/16 = 3.125 × 10−3 rad 20 Point 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. ACD = δ CD π d2 = π 3 d F L = CD CD EACD FCD = 2 = 6.9029 × 10−3 in 2 4 32 EACDδ CD (29 × 106 ) (6.9029 × 10−3 ) (12.5 × 10−3 ) = 12.5 LCD = 200.18 lb Free body ACB: ΣM A = 0: 4 FCD − (50) (20 − x) = 0 (4) (200.18) = 16.0144 50 x = 3.9856 in. 20 − x = 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 crosssectional 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. SOLUTION Free 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) = = 13.8889 × 10−9 P 9 −6 EBD ABD (105 × 10 ) (240 × 10 ) F L (0.55556 P) (0.150) = CE CE = = 3.8581 × 10−9 P ECE ACE (72 × 109 ) (300 × 10−6 ) δ B = δ BD = δ C = δ CE Deformation Diagram: From the deformation diagram, Slope, θ= δ B + δC LBC = 17.7470 × 10−9 P = 78.876 × 10−9 P 0.225 δ A = δ B + LABθ = 13.8889 × 10−9 P + (0.125) (78.876 × 10−9 P) = 23.748 × 10−9 P Apply 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. SOLUTION Statics. 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 N Area of one link: A = (10)(40) = 400 mm 2 = 400 × 10−6 m 2 Length: L = 300 mm = 0.300 m Deformations. δ BE = FBE L (−7.5 × 103 )(0.300) = = −80.357 × 10−6 m 9 −6 EA (70 × 10 )(400 × 10 ) δ CF = FCF L (19.5 × 103 )(0.300) = = 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. θ= (c) δE + δF LEF Deflection of Point G. = 80.357 × 10−6 + 208.93 × 10−6 = 723.22 × 10−6 radians 0.400 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. SOLUTION Extend the slant sides of the cone to meet at a point O and place the origin of the coordinate system there. tan α = From geometry, a1 = b−a h a b , b1 = , tan α tan α r = y tan α At coordinate point y, A = π r 2 Deformation of element of height dy: dδ = Pdy AE dδ = P dy P dy = 2 2 Eπ r π E tan α y 2 Total deformation. P δA = π E tan 2 α = b1 a1 1 dy P = − 2 2 π E tan α y y b1 − a1 P(b1 − a1 ) P = 2 π Eab π E tan α a1b1 b1 = a1 1 1 P − 2 π E tan α a1 b1 δA = Ph ↓ π Eab 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.30 A homogeneous cable of length L and uniform cross section is suspended from one end. (a) Denoting by ρ the density (mass per unit volume) of the cable and by E its modulus of elasticity, determine the elongation of the cable due to its own weight. (b) Show that the same elongation would be obtained if the cable were horizontal and 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 δ= = (b) Total weight: L ρ g (L − y) 0 E dy = ρg L 1 2 Ly − y E 2 0 ρg 2 L 2 L − 2 E δ= 1 ρ gL2 2 E W = ρ gAL F= EAδ EA 1 ρ gL2 1 = ⋅ = ρ gAL L L 2 E 2 1 F= W 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.31 The volume of a tensile specimen is essentially constant while plastic deformation occurs. If the initial diameter of the specimen is d1, show that when the diameter is d, the true strain is ε t = 2 ln(d1 /d ). SOLUTION If the volume is constant, π 4 d 2L = π 4 d12 L0 L d12 d1 = = L0 d 2 d ε t = ln 2 L d = ln 1 L0 d 2 ε t = 2ln d1 d 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.32 Denoting by ε the “engineering strain” in a tensile specimen, show that the true strain is ε t = ln (1 + ε ). SOLUTION ε t = ln Thus L +δ L δ = ln 0 = ln 1 + = ln (1 + ε ) L0 L0 L0 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. SOLUTION Let Pa = Portion of axial force carried by shell Pb = Portion of axial force carried by core. Thus, with δ= Pa L , or Ea Aa Pa = Ea Aa δ L δ= Pb L , or Eb Ab Pb = Eb Ab δ L P = Pa + Pb = ( Ea Aa + Eb Ab ) Aa = Ab = π 4 π 4 δ L [(0.060)2 − (0.025)2 ] = 2.3366 × 10−3 m 2 (0.025) 2 = 0.49087 × 10−3 m 2 P = [(70 × 109 ) (2.3366 × 10−3 ) + (105 × 109 ) (0.49087 × 10−3 )] P = 215.10 × 106 Strain: ε = δ L = δ L δ L 200 × 103 P = = 0.92980 × 10−3 215.10 × 106 215.10 × 106 (a) σ a = Ea ε = (70 × 109 ) (0.92980 × 10−3 ) = 65.1 × 106 Pa (b) δ = ε L = (0.92980 × 10−3 ) (300 mm) σ a = 65.1 MPa 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. SOLUTION Let Pa = Portion of axial force carried by shell and Pb = Portion of axial force carried by core. Thus, with δ= Pa L , or Ea Aa Pa = Ea Aa δ L δ= Pb L , or Eb Ab Pb = Eb Ab δ L P = Pa + Pb = ( Ea Aa + Eb Ab ) Aa = Ab = π 4 π 4 δ L [(0.060)2 − (0.025)2 ] = 2.3366 × 10−3 m 2 (0.025)2 = 0.49087 × 10−3 m 2 P = [(70 × 109 ) (2.3366 × 10−3 ) + (105 × 109 ) (0.49087 × 10−3 )] with δ L = 215.10 × 106 δ L δ = 0.40 mm, L = 300 mm (a) P = (215.10 × 106 ) (b) σb = 0.40 = 286.8 × 103 N 300 Pb Eδ (105 × 109 )(0.40 × 10−3 ) = b = = 140 × 106 Pa −3 Ab L 300 × 10 P = 287 kN σ b = 140.0 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.35 A 4-ft concrete post is reinforced with four steel bars, each with a 34 -in. diameter. 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 δs = Ps L Ps (48) = = 0.93663 × 10−6 Ps As Es (1.76715)(29 × 106 ) δc = Pc L Pc (48) = = 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 Also: Substituting (1) into (2): Ps + Pc = P = 150 kips (1) (2) 1.22875Pc = 150 kips Pc = 122.075 kips From (1): Ps = 0.22875(122.075) = 27.925 kips σs = − Ps 27.925 =− As 1.76715 σ s = −15.80 ksi σc = − Pc 122.075 =− Ac 62.233 σ c = −1.962 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. SOLUTION For each layer, A = (30)(5) = 150 mm 2 = 150 × 10−6 m 2 Let Pa = load on each aluminum layer Pb = load on brass layer Pa L Pb L = Ea A Eb A Deformation. δ= Total force. P = 2 Pa + Pb = 3.5 Pa Solving for Pa and Pb, Pa = 2 P 7 Pb = Pb = Eb 105 Pa = Pa = 1.5 Pa 70 Ea 3 P 7 (a) σa = − Pa 2P 2 30 × 103 =− =− = −57.1 × 106 Pa −6 A 7 A 7 150 × 10 σ a = −57.1 MPa (b) σb = − Pb 3P 3 30 × 103 =− =− = −85.7 × 106 Pa A 7 A 7 150 × 10−6 σ b = −85.7 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. SOLUTION For each layer, A = (30)(5) = 150 mm 2 = 150 × 10−6 m 2 Let Pa = load on each aluminum layer Pb = load on brass layer Pa L PL =− b Ea A Eb A Deformation. δ =− Total force. P = 2 Pa + Pb = 3.5 Pa δ =− =− Pb = Eb 105 Pa = Pa = 1.5 Pa 70 Ea Pa = 2 P 7 Pa L 2 PL =− 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. SOLUTION Let Pa = portion of axial force carried by shell Ps = portion of axial force carried by core Total force. δ= Pa L Ea Aa Pa = Ea Aa δ L δ= Ps L Es As Ps = Es As δ L P = Pa + Ps = ( Ea Aa + Es As ) δ L Data: δ =ε = L P Ea Aa + Es As P = 40 × 103 lb Aa = As = π 4 π 4 (d 02 − di2 ) = 2 d = π 4 π 4 (2.52 − 1.0)2 = 4.1233 in 2 2 (1) = 0.7854 in 2 ε= − 40 × 103 = −620.91 × 10−6 6 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 σ a = Ea ε = (10.1 × 106 )(620.91 × 10−6 ) = −6.27 × 103 psi (b) δ = Lε = (10)(620.91 × 10−6 ) = −6.21 × 10−3 −18.01 ksi −6.27 ksi 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.39 Three wires are used to suspend the plate shown. Aluminum wires of 18 -in. diameter are used at A and B while a steel wire of 121 -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. SOLUTION By symmetry, PA = PB , and δ A = δ B Also, δC = δ A = δ B = δ Strain in each wire: δ ε A = εB = 2L , εC = δ L = 2ε A Determine allowable strain. Wires A&B: εA = σA EA = 14 × 103 = 1.3462 × 10−3 10.4 × 106 ε C = 2 ε A = 2.6924 × 10−4 Wire C: 18 × 103 = 0.6207 × 10−3 EC 29 × 106 1 ε A = ε B = ε C = 0.3103 × 10−6 2 εC = σC = σ A = E Aε A σ C = 18 × 103 psi ∴ Allowable strain for wire C governs, PA = AA E Aε A π 1 2 (10.4 × 106 )(0.3103 × 10−6 ) = 39.61 lb 4 8 PB = 39.61 lb = σ C = EC ε C PC = ACσ C = π1 2 (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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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: δ= But PAB LAB d2 E 4 AB π + PAB = + RA PBC LBC π 4 2 d BC E =0 PBC = − RC Substituting and simplifying: RA LAB RC LBC − 2 =0 2 d AB d BC RC = LAB LBC 2 2 d BC 25 2 RA = RA 15 d 1.25 AB RC = 4.2667 RA From the free body diagram: Substituting (1) into (2): RA + RC = 12 kips (2) 5.2667 RA = 12 RA = 2.2785 kips From (1): (1) RA = 2.28 kips ↑ RC = 4.2667 (2.2785) = 9.7217 kips RC = 9.72 kips ↑ (b) σ AB = PAB + RA 2.2785 = = AAB AAB π4 (1.25)2 σ AB = +1.857 ksi σ BC = PBC − RC −9.7217 = = π ABC ABC (2)2 4 σ BC = −3.09 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. SOLUTION A to C: E = 200 × 109 Pa π (40) 2 = 1.25664 × 103 mm 2 = 1.25664 × 10−3 m 2 4 EA = 251.327 × 106 N A= C to E: E = 105 × 109 Pa π (30) 2 = 706.86 mm 2 = 706.86 × 10−6 m 2 4 EA = 74.220 × 106 N A= A to B: P = RA L = 180 mm = 0.180 m RA (0.180) PL = δ AB = EA 251.327 × 106 = 716.20 × 10−12 RA B to C: P = RA − 60 × 103 L = 120 mm = 0.120 m δ BC = PL ( RA − 60 × 103 )(0.120) = EA 251.327 × 106 = 447.47 × 10−12 RA − 26.848 × 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.41 (Continued) C to D: P = RA − 60 × 103 L = 100 mm = 0.100 m PL ( RA − 60 × 103 )(0.100) = EA 74.220 × 106 = 1.34735 × 10−9 RA − 80.841 × 10−6 δ BC = D to E: P = RA − 100 × 103 L = 100 mm = 0.100 m PL ( RA − 100 × 103 )(0.100) = EA 74.220 × 106 = 1.34735 × 10−9 RA − 134.735 × 10−6 δ DE = A to E: δ AE = δ AB + δ BC + δ CD + δ DE = 3.85837 × 10−9 RA − 242.424 × 10−6 Since point E cannot move relative to A, (a) (b) δ AE = 0 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 ← 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. SOLUTION A to C: E = 105 × 109 Pa π (40) 2 = 1.25664 × 103 mm 2 = 1.25664 × 10−3 m 2 4 EA = 131.947 × 106 N A= C to E: E = 200 × 109 Pa π (30) 2 = 706.86 mm 2 = 706.86 × 10−6 m 2 4 EA = 141.372 × 106 N A= A to B: P = RA L = 180 mm = 0.180 m RA (0.180) PL = δ AB = EA 131.947 × 106 = 1.36418 × 10−9 RA B to C: P = RA − 60 × 103 L = 120 mm = 0.120 m δ BC = PL ( RA − 60 × 103 )(0.120) = EA 131.947 × 106 = 909.456 × 10−12 RA − 54.567 × 10−6 C to D: P = RA − 60 × 103 L = 100 mm = 0.100 m δ CD = PL ( RA − 60 × 103 )(0.100) = EA 141.372 × 106 = 707.354 × 10−12 RA − 42.441 × 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.42 (Continued) D to E: P = RA − 100 × 103 L = 100 mm = 0.100 m PL ( RA − 100 × 103 )(0.100) = EA 141.372 × 106 = 707.354 × 10−12 RA − 70.735 × 10−6 δ DE = A to E: δ AE = δ AB + δ BC + δ CD + δ DE = 3.68834 × 10−9 RA − 167.743 × 10−6 Since point E cannot move relative to A, (a) (b) δ AE = 0 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 ← 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. SOLUTION Deformations 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. From geometry, θ= δB − δ A 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 TA = 1 TB = 2TA = (2) P 10 1 P 10 TB = 1 1 TC = (2) P − P 5 10 TC = 1 1 TD = (3) P − (2) P 5 10 TD = 1 P 5 3 P 10 2 P 5 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.44 The rigid bar AD is supported by two steel wires of 161 -in. diameter ( E = 29 × 106 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. SOLUTION Let θ be the rotation of bar ABCD. Then δ A = 24θ δ C = 8θ δA = PAE LAE AE PAE = 6 2 EAδ A (29 × 10 ) π4 ( 161 ) (24θ ) = LAE 15 = 142.353 × 103θ δC = PCF LCF AE 6 π 1 EAδ C (29 × 10 ) 4 ( 16 ) (8θ ) = = 8 LCF 2 PCF = 88.971 × 103θ Using 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 rad哷 (a) (b) PAE = (142.353 × 103 ) (0.46510 × 10−3 ) PAE = 66.2 lb PCF = (88.971 × 103 ) (0.46510 × 10−3 ) PCF = 41.4 lb 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. SOLUTION Let θ be the rotation of bar ABC as shown. Then δ B = 0.15θ But δ C = δ turn − PCD = = δ C = 0.25θ PCD LCD ECD ACD ECD ACD (δ turn − δ C ) LCD (200 × 109 Pa) π4 (0.016 m) 2 2m (0.0025m − 0.25θ ) = 50.265 × 103 − 5.0265 × 106 θ δB = PBE = PBE LBE EBE ABE or PBE = EBE ABE δB LBE (200 × 109 Pa) π4 (0.016 m) 2 3m (0.15θ ) = 2.0106 × 106 θ From 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 (b) PCD = 9.73 kN 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.46 Links BC and DE are both made of steel ( E = 29 × 106 psi) and are 1 in. wide and 14 in. thick. Determine (a) the force in each link 2 when a 600-lb force P is applied to the rigid member AF shown, (b) the corresponding deflection of point A. SOLUTION Let the rigid member ACDF rotate through small angle θ clockwise about point F. δ C = δ BC = 4θ in. → δ D = −δ DE = 2θ in. → Then δ= For links: FL EA or F= EAδ L 1 1 A = = 0.125 in 2 2 4 LBC = 4 in. LDE = 5 in. FBC = EA δ BC (29 × 106 )(0.125)(4θ ) = = 3.625 × 106 θ 4 LBC FDE = EAδ DE (29 × 106 )(0.125)( −2θ ) = = −1.45 × 106 θ 5 LDE Use member ACDF as a free body. ΣM F = 0 : 8 P − 4 FBC + 2 FDE = 0 1 1 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 哶 P= (a) (b) FBC = (3.625 × 106 )(0.27586 × 10−3 ) FBC = 1000 lb FDE = −(1.45 × 106 )(0.27586 × 10−3 ) FDE = −400 lb 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 78 -in. diameter ( Es = 29 × 106 psi and 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 As = 6 π 4 d2 = 6 π 7 2 = 3.6079 in 2 4 8 Ac = 102 − As = 102 − 3.6079 = 96.392 in 2 Let Pc = tensile force developed in the concrete. For equilibrium with zero total force, the compressive force in the six steel rods equals Pc . Strains: εs = − Matching: ε c = ε s Pc + α s (ΔT ) Es As εc = Pc + α c ( ΔT ) Ec Ac Pc P + 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 + Pc = (1.0 × 10 )(65) 6 6 (3.6 × 10 )(96.392) (29 × 10 )(3.6079) Pc = 5.2254 × 103 lb σc = Pc 5.2254 × 103 = = 54.210 psi Ac 96.392 σs = − Pc 5.2254 × 103 =− = −1448.32 psi As 3.6079 σ c = 54.2 psi σ s = −1.448 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. SOLUTION Since α a > α s , the shell is in compression for a positive temperature rise. σ a = −6 ksi = −6 × 103 psi Let Aa = As = π 4 π 4 (d ) 2 o − di2 = d2 = π 4 π 4 (1.252 − 0.752 ) = 0.78540 in 2 (0.75) 2 = 0.44179 in 2 P = −σ a Aa = σ s As where P is the tensile force in the steel core. σs = − ε= (α a − α s )(ΔT ) = (6.4 × 10−6 )(ΔT ) = (a) ΔT = 145.91°F (b) ε= σ a Aa σs Es σs Es As = (6 × 103 )(0.78540) = 10.667 × 103 psi 0.44179 + α s ( ΔT ) = − σa Ea + α a ( ΔT ) σa Ea 10.667 × 103 6 × 103 + = 0.93385 × 10−3 29 × 106 10.6 × 106 ΔT = 145.9°F 10.667 × 103 + (6.5 × 10−6 )(145.91) = 1.3163 × 10−3 29 × 106 or ε= −6 × 103 + (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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. SOLUTION Brass core: E = 105 GPa α = 20.9 × 10−6/ °C Aluminum shell: E = 70 GPa α = 23.6 × 10−6 / °C Let L be the length of the assembly. Free thermal expansion: ΔT = 195 − 15 = 180 °C Brass core: Aluminum shell: (δT )b = Lα b ( ΔT ) (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 (25) 2 = 490.87 mm 2 4 = 490.87 × 10−6 m 2 PL (δ P )b = Eb Ab Ab = 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.49 (Continued) Aluminum shell: Ea = 70 × 109 Pa π (602 − 252 ) 4 = 2.3366 × 103 mm 2 Aa = = 2.3366 × 10−3 m 2 δ = (δ P )b + (δ P ) a L(α b − α a )(ΔT ) = PL PL + = KPL Eb Ab Ea Aa where K= = 1 1 + Eb Ab Ea Aa 1 1 + −6 9 (105 × 10 )(490.87 × 10 ) (70 × 10 )(2.3366 × 10−3 ) 9 = 25.516 × 10−9 N −1 Then (α b − α a )(ΔT ) K (23.6 × 10−6 − 20.9 × 10−6 )(180) = 25.516 × 10−9 = 19.047 × 103 N P= Stress in aluminum: σa = − P 19.047 × 103 =− = −8.15 × 106 Pa −3 Aa 2.3366 × 10 σ a = −8.15 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. SOLUTION E = 70 GPa α = 23.6 × 10−6 / °C Aluminum shell: Let L be the length of the assembly. ΔT = 195 − 15 = 180 °C Free thermal expansion: Steel core: (δT ) s = Lα s (ΔT ) Aluminum shell: (δT ) a = Lα a (ΔT ) δ = L(α a − α s )( ΔT ) Net expansion of shell with respect to the core: Let P be the tensile force in the core and the compressive force in the shell. Es = 200 × 109 Pa, As = Steel core: (δ P ) s = Aluminum shell: π 4 (25) 2 = 490.87 mm 2 = 490.87 × 10−6 m 2 PL Es As Ea = 70 × 109 Pa (δ P ) a = PL Ea Aa π (602 − 25) 2 = 2.3366 × 103 mm 2 = 2.3366 × 10−3 m 2 4 δ = (δ P ) s + (δ P )a Aa = L(α a − α s )(ΔT ) = PL PL + = KPL Es As Ea Aa where K= = 1 1 + Es As Ea Aa 1 1 + −6 9 (200 × 10 )(490.87 × 10 ) (70 × 10 )(2.3366 × 10−3 ) 9 = 16.2999 × 10−9 N −1 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.50 (Continued) Then P= (α a − α s )(ΔT ) (23.6 × 10−6 − 11.7 × 10−6 )(180) = = 131.41 × 103 N K 16.2999 × 10−9 Stress in aluminum: σ a = − P 131.19 × 103 =− = −56.2 × 106 Pa −3 Aa 2.3366 × 10 σ a = −56.2 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 AAB = ABC = π 4 π 4 2 d AB = 2 d BC = π 4 π 4 (30)2 = 706.86 mm 2 = 706.86 × 10−6 m 2 (50)2 = 1.9635 × 103 mm 2 = 1.9635 × 10−3 m 2 Free 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 m Shortening due to induced compressive force P: δP = = PL PL + Es AAB Eb ABC 0.250 P 0.300 P + −6 9 (200 × 10 )(706.86 × 10 ) (105 × 10 )(1.9635 × 10−3 ) 9 = 3.2235 × 10−9 P For 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.52 A steel railroad track ( Es = 200 GPa, α s = 11.7 × 10−6 /°C) was laid out at a temperature of 6°C. Determine the normal stress in the rails when the temperature reaches 48°C, assuming that the rails (a) are welded to form 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 δP = PL Lσ (10)σ = = = 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 (b) σ = −98.3 MPa δ = δT + δ P = 4.914 × 10−3 + 50 × 10−12 σ = 3 × 10−3 3 × 10−3 − 4.914 × 10−3 50 × 10−12 = −38.3 × 106 Pa σ= 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 AAB = Free thermal expansion. π 4 (2.25) 2 = 3.9761 in 2 ABC = π 4 (1.5) 2 = 1.76715 in 2 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. δT = (δT ) AB + (δT ) BC = 40.712 × 10−3 in. Total: Shortening due to induced compressive force P. PLAB 24 P = = 208.14 × 10−9 P Es AAB (29 × 106 )(3.9761) PLBC 32 P = = = 1741.18 × 10−9 P Ea ABC (10.4 × 106 )(1.76715) (δ P ) AB = (δ P ) BC δ P = (δ P ) AB + (δ P ) BC = 1949.32 × 10−9 P Total: For zero net deflection, δ P = δ T (a) (b) 1949.32 × 10−9 P = 40.712 × 10−3 P = 20.885 × 103 lb σ AB = − P 20.885 × 103 =− = −5.25 × 103 psi 3.9761 AAB σ AB = −5.25 ksi σ BC = − P 20.885 × 103 =− = −11.82 × 103 psi 1.76715 ABC σ BC = −11.82 ksi (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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 AAB = π 4 (2.25) 2 = 3.9761 in 2 Free thermal expansion. ABC = π 4 (1.5) 2 = 1.76715 in 2 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. δT = (δT ) AB + (δT ) BC = 36.904 × 10−3 in. Total: Shortening due to induced compressive force P. PLAB 24 P = = 580.39 × 10−9 P Ea AAB (10.4 × 106 )(3.9761) PLBC 32 P = = = 624.42 × 10−9 P 6 Es ABC (29 × 10 )(1.76715) (δ P ) AB = (δ P ) BC δ P = (δ P ) AB + (δ P ) BC = 1204.81 × 10−9 P Total: For zero net deflection, δ P = δ T (a) 1204.81 × 10−9 P = 36.904 × 10−3 P = 30.631 × 103 lb σ AB = − P 30.631 × 103 =− = −7.70 × 103 psi 3.9761 AAB σ AB = −7.70 ksi σ BC = − P 30.631 × 103 =− = −17.33 × 103 psi 1.76715 ABC σ BC = −17.33 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 ← or δ B = 4.57 × 10−3 in. → (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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. SOLUTION Initial dimensions at T = 20 °C. Final dimensions at T = 45 °C. ΔT = 45 − 20 = 25 °C Free 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 m At 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 m Add 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 (δ P ) = Steel rod: PL P(0.250) = = 634.92 × 10−12 P EA (105 × 109 )(3.750 × 10−3 ) E = 200 × 109 Pa (δ P ) s = As = π 4 (30) 2 = 706.86 mm 2 = 706.86 × 10−6 m 2 PL P(0.250) = = 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 σs = − P (26.006 × 103 ) =− = −36.8 × 106 Pa −6 As 706.86 × 10 (a) Stress in steel rod: (b) Final length of steel rod: L f = L0 + (δ T ) s − (δ P ) s σ s = −36.8 MPa 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 (b) 21.4 °C 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. As = (2)(5)(40) = 400 mm2 = 400 × 10−6 m 2 Elongation of steel: (δ P ) s = F *L P* (2.00) = = 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 (δ P )b = P* L P* (2.00) = = 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 N Stresses due to fabrication: Steel: σ s* = P* 8.811 × 103 = = 22.03 × 106 Pa = 22.03 MPa As 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.56 (Continued) Brass: σ b* = − P* 8.811 × 103 =− = −14.68 × 106 Pa = −14.68 MPa −6 Ab 600 × 10 To 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 the additional displacement. Also, let Ps and Pb be the additional forces developed in the steel and brass, respectively. δ′ = Ps L PL = b As Es Ab Eb As Es (400 × 10−6 )(200 × 109 ) δ′ = δ ′ = 40 × 106 δ ′ 2.00 L Ab Eb (600 × 10−6 )(105 × 109 ) δ′ = δ ′ = 31.5 × 106 δ ′ Pb = 2.00 L Ps = P = Ps + Pb = 25 × 103 N Total 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 σs = Ps 13.986 × 103 = = 34.97 × 106 Pa As 400 × 10−6 σb = Pb 11.140 × 103 = = 18.36 × 106 Pa Ab 600 × 10−6 Add 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. SOLUTION See solution to Problem 2.56 to obtain the fabrication stresses. σ s* = 22.03 MPa σ b* = 14.68 MPa Allowable stresses: σ s ,all = 30 MPa, σ b ,all = 25 MPa Available stress increase from load. σ s = 30 − 22.03 = 7.97 MPa σ b = 25 + 14.68 = 39.68 MPa Corresponding available strains. εs = εb = σs Es σb Eb = 7.97 × 106 = 39.85 × 10−6 200 × 109 = 39.68 × 106 = 377.9 × 10−6 105 × 109 Smaller value governs ∴ ε = 39.85 × 10−6 Areas: As = (2)(5)(40) = 400 mm2 = 400 × 10−6 m 2 Ab = (15)(40) = 600 mm 2 = 600 × 10−6 m 2 Forces 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 N Total 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 lb Shortening due to P: δP = = PLb PLa + 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 ) ΔT Equating, (400.2 × 10−6 )ΔT = 50.657 × 10−3 (a) (b) ΔT = 126.6 °F Thot = Tcold + ΔT = 75 + 126.6 = 201.6 °F δ a = Laα a (ΔT ) − Thot = 201.6°F PLa Ea Aa = (18)(12.9 × 10−6 )(26.6) − (30.8 × 103 )(18) = 10.712 × 10−3 in. 6 (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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. SOLUTION Thermal 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. δP = But PLb PLa Lb L + = + a Eb Ab Ea Aa Eb Ab Ea Aa P 14 18 −9 = + P = 995.36 × 10 P 6 6 (15 × 10 )(2.4) (10.6 × 10 )(2.8) 995.36 × 10−9 P = 52.036 × 10−3 Equating, P = 52.279 × 103 lb P = 52.3 kips (a) (b) δ b = Lbα b (ΔT ) − PLb Eb Ab = (14)(12 × 10−6 )(180) − (52.279 × 103 )(14) = 9.91 × 10−3 in. (15 × 106 )(2.4) δ b = 9.91 × 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 °C Free 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 m Shortening 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 + = + s P Ea Aa Es As Ea Aa Es As 0.300 0.250 = + P 9 −6 9 −6 (75 × 10 )(2000 × 10 ) (190 × 10 )(800 × 10 ) = 3.6447 × 10−9 P δP = 3.6447 × 10−9 P = 0.847 × 10−3 Equating, P = 232.39 × 103 N P 232.39 × 103 =− = −116.2 × 106 Pa Aa 2000 × 10−6 (a) σa = − (b) δ a = Laα a (ΔT ) − σ a = −116.2 MPa PLa Ea Aa = (0.300)(23 × 10−6 )(120) − (232.39 × 103 )(0.300) = 363 × 10−6 m 9 −6 (75 × 10 )(2000 × 10 ) δ a = 0.363 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.61 A 600-lb tensile load is applied to a test coupon made from 161 -in. flat steel plate (E = 29 × 106 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 16 600 P 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 ) 2 (c) δ thickness = t0ε z = (−198.62 × 10−6 ) 16 (d) 1 δ w = −99.3 × 10−6 in. δ t = −12.41 × 10−6 in. 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 P = 75 kN = 75 × 103 N A= σ = εx = π 4 d2 = π 4 (0.022)2 = 380.13 × 10−6 m 2 P 75 × 103 = = 197.301 × 106 Pa A 380.13 × 10−6 σ E = 197.301 × 106 = 986.51 × 10−6 200 × 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.63 A 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. SOLUTION Let the y-axis be along the length of the rod and the x-axis be transverse. π (20)2 = 314.16 mm 2 = 314.16 × 10−6 m 2 4 P 6 × 103 σy = = = 19.0985 × 106 Pa −6 A 314.16 × 10 δ y 14 mm εy = = = 0.093333 L 150 mm A= Modulus of elasticity: E = εx = σy εy δx d Poisson’s ratio: v=− Modulus of rigidity: G= = 19.0985 × 106 = 204.63 × 106 Pa 0.093333 =− P = 6 × 103 N E = 205 MPa 0.85 = −0.0425 20 εx −0.0425 =− 0.093333 εy E 204.63 × 106 = = 70.31 × 106 Pa 2(1 + v) (2)(1.455) v = 0.455 G = 70.3 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. εy = εy d v=− =− d = 2.5 in. 0.5 × 10−3 = −0.2 × 10−3 2.5 εy : εx εx = −ε y v = 0.2 × 10−3 = 0.66667 × 10−3 0.3 σ x = Eε x = (29 × 106 )(0.66667 × 10−3 ) = 19.3334 × 103 psi A= π 4 d2 = π 4 (2.5) 2 = 4.9087 in 2 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 A= (a) δ =− π 4 ( do2 − di2 ) = δ L 4 (0.32 − 0.272 ) = 13.4303 × 10−3 m 2 PL (700 × 103 )(2.5) =− EA (200 × 109 )(13.4303 × 10−3 ) = −651.51 × 10−6 m ε= π P = 700 × 103 N = δ = −0.652 mm −651.51 × 10−6 = −260.60 × 10−6 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 The slope after deformation is tan θ = 2(1 + ε y ) 1+ εx 125 × 106 = 1.6892 × 10−3 E 74 × 109 ε y = −vε x = −(0.33)(1.6892 × 10−3 ) = −0.5574 × 10−3 εx = tan θ = σx = 2(1 − 0.0005574) = 1.99551 1 + 0.0016892 tan θ = 1.99551 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 (σ x − vσ y − vσ z ) E σ y = vσ x = (0.35)(−180 × 106 ) εy = = −63 × 106 Pa σ y = −63 MPa 1 v (0.35)( −243 × 106 ) (σ z − vσ x − vσ y ) = − (σ x + σ y ) = = −1.89 × 10−3 E E 45 × 109 σ x − vσ y 1 157.95 × 106 ε x = (σ x − vσ y − vσ Z ) = =− = −3.51 × 10−3 E E 45 × 109 εz = (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 ) (c) Δ A = −13.50 mm 2 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. SOLUTION Given: σ x = 80 MPa σ y = 40 MPa Using Eq’s (2.28): εx = 1 80 − 0.3(40) (σ x − vσ y ) = = 340 × 10−6 200 × 103 E εy = 1 40 − 0.3(80) (σ y − vσ x ) = = 80 × 10−6 200 × 103 E (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, Differentiate: But Thus, ( AC )2 = ( AB)2 + ( BC ) 2 2( AC ) Δ( AC ) = 2( AB)Δ( AB) + 2( BC )Δ( BC ) Δ( AC ) = δ AC Δ( AB) = δ AB Δ( BC ) = δ BC 2( AC )δ AC = 2( AB)δ AB + 2( BC )δ BC δ AC = AB BC 1 1 (10.20 μ m)+ (2.40 μ m) δ AB + δ BC = 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 (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 × 10−6 εx = Length subjected to strain ε x: L = 12 in. (a) δ y = Lε y = (12)(427.72 × 10−6 ) (b) δ x = d ε x = (1.5)(−380.198 × 10−6 ) δ l = 5.13 × 10−3 in. 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. SOLUTION Over the pressurized portion BC, σx =σz = −p σy =σy 1 (−vσ x + σ y − vσ z ) E 1 = (2vp + σ y ) E (ε y ) BC = (a) (ε y ) BC = 0 2vp + σ y = 0 σ y = −2vp = −(2)(0.36)(6000) = −4320 psi A= π d2 = π (1.5) 2 = 1.76715 in 2 4 4 F = Aσ y = (1.76715)(−4320) = −7630 lb i.e., (b) Over unpressurized portions AB and CD, (ε y ) AB 7630 lb compression σx =σz = 0 σy = (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 (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 (20 − 12) + 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 [(1 − v 2 )σ x − v(1 + v)σ y ] E 1 ε y = [(1 − v 2 )σ y − v(1 + v)σ x ] E εx = SOLUTION εz = 0 = 1 (−vσ x − vσ y + σ z ) or σ z = v(σ x + σ y ) E 1 (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 εx = 1 ( −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 ] E εy = 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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: σx = E ε x + vε y 1− v 2 σy = E ε y + vε x 1− v 2 εz = − v (ε x + ε y ) 1− v SOLUTION σz = 0 εx = 1 (σ x − vσ y ) E (1) εy = 1 ( −vσ x + σ y ) E (2) Multiplying (2) by v and adding to (1), ε x + vε y = 1 − v2 σx E or σx = E (ε x + vε y ) 1 − v2 or σy = E (ε y + vε x ) 1 − v2 Multiplying (1) by v and adding to (2), ε y + vε x = 1 − v2 σy E 1 v E (−vσ x − vσ y ) = − ⋅ (ε x + vε y + ε y + vε x ) E E 1 − v2 v(1 + v) v =− (ε x + ε y ) = − (ε x + ε y ) 2 1− v 1− v εz = 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 x2 + 1 − 2vε x + v 2ε x2 4ε ′ + 2ε ′2 = 2ε x + ε x2 − 2vε x + v 2ε x2 4ε ′ = 2ε x − 2vε x Neglect squares as small (A) ε′ = 1− v εx 2 (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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.73 (Continued) (b) vσ ′ E E 1− v P = ⋅ E 2A 1− v σx = 2E ε′ = = σ′ − 1− v εx 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 εx = 1 1 (σ x − vσ y − vσ z ) = (σ x − vσ 0 ) E E σ x = vσ 0 (a) (b) εz = 1 1 1 − v2 (−vσ x − vσ y + σ z ) = (−v 2σ 0 − 0 + σ 0 ) = σ0 E E E σ0 E = ε z 1 − v2 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 F = 1 1 P = (25 × 103 N) = 12.5 × 103 N 2 2 τ = F 12.5 × 103 N) = = 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 = 0.05 h 0.03 τ 833.33 × 103 G = = = 16.67 × 106 Pa γ 0.05 γ= δ = 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 h Shearing strain: γ = Shearing stress: τ = Gγ = Force: Gδ h 1 GAδ P = Aτ = 2 h P P= k = with A = (0.15)(0.1) = 0.015 m 2 k = 2GAδ h 2GA h Effective spring constant: δ = or h = 0.03 m 2(19 × 106 Pa)(0.015 m 2 ) = 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. SOLUTION Consider the plastic block. The shearing force carried is P = 9 × 103 lb The area is A = (3.5)(5.5) = 19.25 in 2 Shearing stress: τ = Shearing strain: γ = But γ = P 9 × 103 = = 467.52 psi A 19.25 τ G = 467.52 = 0.0085006 55 × 103 δ ∴ δ = hγ = (2.2)(0.0085006) h δ = 0.0187 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 Consider the rubber block on the right. It carries a shearing force equal to 1 2 1 2 P. P The shearing stress is τ= or required area A= But A = (3.0)b Hence, b= Use b = 2.42 in Shearing strain. γ= But γ= δ a Hence, a= δ 0.1 = = 0.818 in. γ 0.12222 A P 3.2 × 103 = = 7.2727 in 2 2τ (2)(220) A = 2.42 in. 3.0 τ G = and bmin = 2.42 in. τ = 220 psi 220 = 0.12222 1800 amin = 0.818 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 τ = P 55 × 103 = = 3580.7 psi A 15.36 G = 150 × 103 psi γ = τ = G h = 2 in. 3580.7 = 23.871 × 10−3 150 × 103 δ = hγ = (2)(23.871 × 10−3 ) = 47.7 × 10−3 in. δ = 0.0477 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.80 What load P should be applied to the plate of Prob. 2.79 to produce a 1 16 -in. deflection? 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 1 in. = 0.0625 in. 16 h = 2 in. δ = γ = δ h = 0.0625 = 0.03125 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 Shearing strain: γ= a= Shearing stress: τ= b= δ = a Gδ τ 1 2 P A τ G = (12 × 106 Pa)(0.005 m) = 0.0429 m 1.4 × 106 Pa = P 2bc P 45 × 103 N = = 0.1607 m 2cτ 2(0.1 m) (1.4 × 106 Pa) a = 42.9 mm b = 160.7 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 Shearing stress: τ= 1 2 P A = P 2bc P = 2bcτ = 2(0.2 m)(0.125 m)(1.5 × 103 kPa) Shearing strain: γ= a= δ a = Gδ τ P = 75.0 kN τ G = (10 × 106 Pa)(0.006 m) = 0.04 m 1.5 × 106 Pa a = 40.0 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 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 A= εx = σ 1 (σ x − vσ y − vσ z ) = x E E ε y = ε z = −vε x = −v σx E (1 − 2v)σ x 1 e = ε x + ε y + ε z = (σ x − vσ x − vσ x ) = E E Volume: (a) V = AL = (380.13 mm 2 )(200 mm) = 76.026 × 103 mm3 ΔV = Ve Steel: e= (1 − 0.60)(121.01 × 106 ) = 242 × 10−6 9 200 × 10 ΔV = (76.026 × 103 )(242 × 10−6 ) = 18.40 mm3 (b) Aluminum: e= (1 − 0.70)(121.01 × 106 ) = 519 × 10−6 9 70 × 10 ΔV = (76.026 × 103 )(519 × 10−6 ) = 39.4 mm3 e = 242 × 10−6 ΔV = 18.40 mm3 e = 519 × 10−6 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. SOLUTION From Problem 2.61, thickness = (a) 1 in., 16 E = 29 × 106 psi, v = 0.30. 1 1 A = = 0.03125 in 2 2 16 Volume: V0 = AL 0 = (0.03125)(2.00) = 0.0625 in 3 P 600 = = 19.2 × 103 psi σy =σz = 0 A 0.03125 σ 1 19.2 × 103 = 662.07 × 10−6 ε x = (σ x − vσ y − vσ z ) = x = E E 29 × 106 σx = 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. Width: w = w0 + δ y = Thickness: Volume: 1 − 99.3 × 10−6 = 0.4999007 in. 2 1 t = t0 + δ z = − 12.41 × 10−6 = 0.06248759 in. 16 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 (about 3 miles below the surface). Knowing that E = 29 × 106 psi and v = 0.30, determine (a) the decrease in diameter of the sphere, (b) the decrease in volume of the sphere, (c) the percent increase in the density of the sphere. SOLUTION For a solid sphere, V0 = π 6 d 03 π (6.00)3 6 = 113.097 in.3 = σ x = σ y = σ z = −p = −7.1 × 103 psi 1 (σ x − vσ y − vσ z ) E (1 − 2v) p (0.4)(7.1 × 103 ) =− =− E 29 × 106 εx = = −97.93 × 10−6 Likewise, 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. (b) −ΔV = −V0 e = −(113.097)(−293.79 × 10−6 ) = 33.2 × 10−3 in 3 (c) Let m = mass of sphere. −Δd = 588 × 10−6 in. − ΔV = 33.2 × 10−3 in 3 m = constant. m = ρ0V0 = ρV = ρV0 (1 + e) ρ − ρ0 ρ V m 1 = −1 = × 0 −1 = −1 V0 (1 + e) m 1+ e ρ0 ρ0 = (1 − e + e 2 − e3 + ) − 1 = −e + e 2 − e3 + ≈ −e = 293.79 × 10−6 ρ − ρ0 × 100% = (293.79 × 10−6 )(100%) ρ0 0.0294% 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 (85) 2 = 5.6745 × 103 mm 2 = 5.6745 × 10−3 m 2 4 4 V0 = A0 h0 = 766.06 × 103 mm3 = 766.06 × 10−6 m3 A 0= (a) d 02 = σ x = 0, σ y = −58 × 106 Pa, σ z = 0 σy 1 58 × 106 (−vσ x + σ y − vσ z ) = =− = −552.38 × 10−6 E E 105 × 109 εy = Δh = h 0ε y = (135 mm)( − 552.38 × 10−6 ) e= (1 − 2v)σ y (0.34)(−58 × 106 ) 1 − 2v (σ x + σ y + σ z ) = = = −187.81 × 10−6 9 E E 105 × 10 ΔV = V0 e = (766.06 × 103 mm3 )(−187.81 × 10−6 ) (b) Δh = −0.0746 mm σ x = σ y = σ z = −70 × 106 Pa εy = σ x + σ y + σ z = −210 × 106 Pa 1 1 − 2v (0.34)(−70 × 106 ) (−vσ x + σ y − vσ z ) = = −226.67 × 10−6 σy = E E 105 × 109 Δh = h 0ε y = (135 mm)( −226.67 × 10−6 ) e= ΔV = −143.9 mm3 Δh = −0.0306 mm 1 − 2v (0.34)( −210 × 106 ) (σ x + σ y + σ z ) = = −680 × 10−6 9 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. SOLUTION Let 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 τ= P P = A 2π rh The shearing strain is γ= τ = G P 2π Ghr Shearing deformation over radial length dr, dδ =γ dr d δ = γ dr = P dr 2π Gh r Total deformation. δ= R2 R1 dδ = P 2π Gh R2 P ln r R1 2π Gh R P ln 2 = R1 2π Gh = Data: R1 = 10 mm = 0.010 m, R1 R2 = 25 mm = 0.025 m, h = 80 mm = 0.080 m G = 12 × 106 Pa P= dr r P (ln R2 − ln R1 ) = 2π Gh 2π Ghδ or P = ln( R2 / R1 ) R2 δ = 2.50 × 10−3 m (2π )(12 × 106 ) (0.080) (2.50 × 10−3 ) = 16.46 × 103 N ln (0.025/0.010) 16.46 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. SOLUTION Let 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 τ= P P = A 2π rh The shearing strain is γ= τ G = P 2π Ghr Shearing deformation over radial length dr, dδ =γ dr d δ = γ dr P dr drδ = 2π Gh r Total deformation. δ= R2 R1 dδ = P 2π Gh R2 P ln r R1 2π Gh R P = ln 2 R1 2π Gh = ln dr R1 r P (ln R2 − ln R1 ) = 2π Gh R2 R2 2π Ghδ (2π ) (10.93 × 106 ) (0.080) (0.002) = = = 1.0988 R1 P 10.103 R2 = exp (1.0988) = 3.00 R1 R2 /R1 = 3.00 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 these constants 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 k= (a) 1+ v = and G= E 2(1 + v) E E or v = −1 2G 2G k= (b) E 3(1 − 2v) E 2 EG 2 EG = = E 3[2G − 2 E + 4G ] 18G − 6 E − 1 3 1 − 2 2G k= EG 9G − 6 E v= 3k − 2G 6k + 2G k 2(1 + v) = G 3(1 − 2v) 3k − 6kv = 2G + 2Gv 3k − 2G = 2G + 6k 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.90∗ Show that for any given material, the ratio G/E of the modulus of rigidity over the modulus of elasticity is always less than 12 but more than 13 . [Hint: Refer to Eq. (2.43) and to Sec. 2.13.] SOLUTION G= E 2(1 + v) Assume v > 0 for almost all materials, and v < or 1 2 E = 2(1 + v) G for a positive bulk modulus. E 1 < 2 1 + = 3 G 2 Applying the bounds, 2≤ Taking the reciprocals, 1 G 1 > > 2 E 3 or 1 G 1 < < 3 E 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 = 0.428 SOLUTION Stress-to-strain equations are εx = σx Ex εy = − εz = − − v yxσ y Ey vxyσ x Ex + − σy Ey vzxσ z Ez − vzyσ z Ez vxzσ x v yzσ y σ z − + Ex Ey Ez vxy Ex v yz Ey = = v yx Ey vzy Ez vzx vxz = Ez Ex (1) (2) (3) (4) (5) (6) ε y = 0 and ε z = 0. The constraint conditions are Using (2) and (3) with the constraint conditions gives vzy vxy 1 σy − σz = σx Ey Ez Ex − v yz Ey σy + V 1 σ z = xz σ x Ez Ex (7) (8) 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 50 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.91∗ (Continued) Solving simultaneously, σ y = σ z = 0.134993 σ x Using (4) and (5) in (1), εx = Ex = = vxy v 1 σx − σ y − xz σ z Ex Ex E 1 [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 = = 40.625 × 106 Pa A 1600 × 10−6 (0.93142) (40.625 × 103 ) εx = = 756.78 × 10−6 9 50 × 10 σx = (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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 = 0.428 SOLUTION εx = σx Ex εy = − εz = − − v yxσ y vxyσ x Ex Ey + − σy Ey vzxσ z Ez − vzyσ z Ez vxzσ x v yzσ y σ z − + Ex Ey Ez vxy Ex v yz Ey = = v yx Ey vzy Ez vzx vxz = Ez Ex Constraint condition: Load condition : (1) (2) (3) (4) (5) (6) εz = 0 σy = 0 0=− From Equation (3), σz = vxz 1 σx + σz Ex Ez vxz Ez (0.254) (15.2) σx = = 0.077216 σ x Ex 50 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.92∗ (Continued) From Equation (1) with σ y = 0, εx = = 1 1 [σ x − 0.254 σ z ] = [1 − (0.254) (0.077216)]σ x Ex Ex = 0.98039 σx Ex σx = But, εx = (a) σx = δx Lx = v v 1 1 σ x − zx σ z = σ x − xz σ z Ex Ez Ex Ex Exε x 0.98039 0.035 mm = 875 × 10−6 40 mm (50 × 109 ) (875 × 10−6 ) = 44.625 × 103 Pa 0.98039 σ x = 44.6 MPa σy = 0 σ z = (0.077216) (44.625 × 106 ) = 3.446 × 106 Pa From (2), εy = vxy Ex σx + σ z = 3.45 MPa vzy 1 σy − σz 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 (a) At hole A: r = 1 1 1 ⋅ = in. 2 2 4 d =3− 1 = 2.50 in. 2 1 Anet = dt = (2.50) = 1.25 in 2 2 σ non = P 6.5 = = 5.2 ksi Anet 1.25 2 ( 14 ) 2r = = 0.1667 D 3 From Fig. 2.60a, K = 2.56 σ max = Kσ non = (2.56)(5.2) (b) r = At hole B: 1 (1.5) = 0.75, 2 σ max = 13.31 ksi d = 3 − 1.5 = 1.5 in. 1 Anet = dt = (1.5) = 0.75 in 2 , 2 σ non = P 6.5 = = 8.667 ksi 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.94 Knowing that σ all = 16 ksi, determine the maximum allowable value of the centric axial load P. SOLUTION At hole A: r = 1 1 1 ⋅ = in. 2 2 4 d =3− 1 = 2.50 in. 2 1 Anet = dt = (2.50) = 1.25 in 2 2 2 ( 14 ) 2r = = 0.1667 D 3 From Fig. 2.60a, K = 2.56 σ max = At hole B: r = KP Anet ∴ P= Anetσ max (1.25)(16) = = 7.81 kips K 2.56 1 (1.5) = 0.75 in, 2 d = 3 − 1.5 = 1.5 in. 1 Anet = dt = (1.5) = 0.75 in 2 , 2 2r 2(0.75) = = 0.5 D 3 From Fig. 2.60a, K = 2.16 P= Smaller value for P controls. Anetσ max (0.75)(16) = = 5.56 kips K 2.16 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 r = (9) = 4.5 mm 2 For the circular hole, d = 96 − 9 = 87 mm 2r 2(4.5) = = 0.09375 D 96 Anet = dt = (0.087 m)(0.009 m) = 783 × 10−6 m 2 K hole = 2.72 From Fig. 2.60a, σ max = P= (a) For fillet, K hole P Anet Anetσ max (783 × 10−6 )(100 × 106 ) = = 28.787 × 103 N K hole 2.72 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 σ max = (5.40 × 10−6 )(100 × 106 ) K fillet P A σ ∴ K fillet = min max = Amin P 28.787 × 103 = 1.876 From Fig. 2.60b, rf d ≈ 0.19 ∴ rf ≈ 0.19d = 0.19(60) r f = 11.4 mm (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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.96 For P = 100 kN, determine the minimum plate thickness t required if the allowable stress is 125 MPa. SOLUTION At the hole: rA = 20 mm d A = 88 − 40 = 48 mm 2rA 2(20) = = 0.455 DA 88 From Fig. 2.60a, K = 2.20 σ max = t = At the fillet: From Fig. 2.60b, KP KP KP = ∴ t = Anet d At d Aσ max (2.20)(100 × 103 N) = 36.7 × 10−3 m = 36.7 mm 6 (0.048 m)(125 × 10 Pa) D 88 = = 1.375 dB 64 D = 88 mm, d B = 64 mm rB = 15 mm rB 15 = = 0.2344 dB 64 K = 1.70 σ max = t = KP KP = Amin d Bt KP d Bσ max = (1.70)(100 × 103 N) = 21.25 × 10−3 m = 21.25 mm (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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 = = 1.25 d 60 From Fig. 2.60b K = 1.95 P= r 6 = = 0.10 d 60 σ max = K P A Aσ max (900 × 10−6 ) (200 × 106 ) = = 92.308 × 103 N K 1.95 P = 92.3 kN Wide area A* = (75 mm) (15 mm) = 1125 mm 2 = 1.125 × 10−3 m 2 δ =Σ Pi Li P L 92.308 × 103 0.150 0.300 0.150 = Σ i = + + −3 −6 9 Ai Ei E Ai 70 × 10 900 × 10 1.125 × 10−3 1.125 × 10 = 7.91 × 10−6 m (b) Uniform bar. P = Aσ all = (900 × 10−6 )(200 × 106 ) = 180 × 103 N δ = 0.791 mm δ= PL (180 × 103 )(0.600) = = 1.714 × 10−3 m AE (900 × 10−6 )(70 × 109 ) P = 180.0 kN δ = 1.714 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 δ =Σ Pi Li P Li = Σ Ei Ai E Ai δ = 0.75 × 10−3 m 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 Σ Li 0.150 0.300 0.150 = + + = 600 m −1 −3 −6 −3 Ai 1.125 × 10 900 × 10 1.125 × 10 P= Stress concentration. (70 × 109 ) (0.75 × 10−3 ) Eδ = = 87.5 × 103 N Li 600 Σ Ai D = 75 mm, d = 60 mm, r = 6 mm D 75 = = 1.25 d 60 From Fig. 2.60b 6 r = = 0.10 d 60 K = 1.95 σ max = K P (1.95) (87.5 × 103 ) = = 189.6 × 106 Pa −6 Amin 900 × 10 σ max = 189.6 MPa Note 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 12 to 11/2 in. in 14 -in. 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 At the fillets: D 4.6875 = = 1.5 d 3.125 From Fig. 2.60b, K = 2.10 r 0.375 = = 0.12 d 3.125 Amin = (3.125)(0.5) = 1.5625 in 2 σ max = K Pall = Aminσ all (1.5625)(21) = = 15.625 kips K 2.10 Anet = ( D − 2r )t , K from Fig. 2.60a At the hole: σ max = K with Hole diam. 0.5 in. Pall = σ all Amin r 0.25 in. P = σ all Anet ∴ Pall = D = 4.6875 in. t = 0.5 in. d = D − 2r 2r/D 4.1875 in. 0.107 Anetσ all K σ all = 21 ksi K 2.68 Anet Pall 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 (a) Largest hole with Pall > 15.625 kips is the (b) Allowable load Pall = 15.63 kips 3 4 -in. diameter hole. 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.100 (a) For P = 13 kips and d = 12 in., determine the maximum stress in the plate shown. (b) Solve part a, assuming that the hole at A is not drilled. SOLUTION Maximum stress at hole: Use Fig. 2.60a for values of K. 2r 0.5 = = 0.017, D 4.6875 K = 2.68 Anet = (0.5)(4.6875 − 0.5) = 2.0938 in 2 σ max = K P (2.68)(13) = = 16.64 ksi, Anet 2.0938 Maximum stress at fillets: Use Fig. 2.60b for values of K. r 0.375 = = 0.12 d 3.125 D 4.6875 = = 1.5 d 3.125 K = 2.10 Amin = (0.5)(3.125) = 1.5625 in 2 σ max = K (2.10)(13) P = = 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 AAB = ABC = π 4 π (30)2 = 706.86 mm 2 = 706.86 × 10−6 m 2 (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 δ′ = P′LAB P ′LBC (176.715 × 103 )(0.8) (176.715 × 103 )(1.2) + = + 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 AAB = 4 (30)2 = 706.86 mm 2 = 706.86 × 10−6 m 2 π (40) 2 = 1.25664 × 103 mm 2 = 1.25644 × 10−3 m 2 4 = Aminσ Y = (706.86 × 10−6 )(250 × 106 ) = 176.715 × 103 N ABC = Pmax π Pmax = 176.7 kN δ′ = P′LAB P ′LBC (176.715 × 103 )(0.8) (176.715 × 103 )(1.2) + = + 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 δY = Lε Y = Lσ Y (2.2)(345 × 106 ) = = 3.795 × 10−3 = 3.795 mm E 200 × 109 If δ m ≥ δ Y , Pm = Aσ Y = (900 × 10−6 )(345 × 106 ) = 310.5 × 103 N Unloading: δ ′ = Pm L σ Y L = = δY = 3.795 mm AE E δ P = δm − δ ′ (a) δ m = 4.5 mm > δY Pm = 310.5 × 103 N δ perm = 4.5 mm − 3.795 mm (b) δ m = 8 mm > δY Pm = 310.5 × 103 N δ perm = 8.0 mm − 3.795 mm δ m = 310.5 kN δ perm = 0.705 mm δ m = 310.5 kN 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 δY = Lε Y = Lσ Y (2.5)(345 × 106 ) = = 4.3125 × 103 m = 4.3125 mm 9 E 200 × 10 When 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 to the rigid lever CD, it is found that end C is 83 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. SOLUTION Since the rod AB is to be stretched permanently, the peak force in the rod is P = PY , where PY = Aσ Y = π 3 2 (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 33 33 Q = 2.65 kips During unloading, the spring back at B is δ B = LABε Y = LABσ Y (60)(36 × 103 ) = = 0.0745 in. E 29 × 106 From the deformation diagram, Slope: θ= δB 22 = δC 33 ∴ δC = 33 δ B = 0.1117 in. −22 δ C = 0.1117 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 has been attached to the rigid lever CD, it is found that end C is 83 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. SOLUTION Since the rod AB is to be stretched permanently, the peak force in the rod is P = PY , where PY = Aσ Y = π 3 2 (50) = 5.522 kips 4 8 Referring to the free body diagram of lever CD, ΣM D = 0: 33Q − 22 P = 0 Q= 22 (22)(5.522) P= = 3.68 kips 33 33 Q = 3.68 kips During unloading, the spring back at B is δ B = LAB ε Y = LABσ Y (60)(50 × 103 ) = = 0.1034 in. E 29 × 106 From the deformation diagram, Slope: θ= δB 22 = δC 33 ∴ δC = 33 δB 22 δ C = 0.1552 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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.) SOLUTION Elongation constraints for taut cables. Let θ = rotation angle of rigid bar ABC. θ= δ BD LAB δ BD = = δ CE LAC LAB 1 δ CE = δ CE LAC 2 (1) Equilibrium of bar ABC. M A = 0 : LAB FBD + LAC FCE − LAC Q = 0 Q = FCE + LAB 1 FBD = FCE + FBD 2 LAC Assume cable CE is yielded. FCE = Aσ Y = (100 × 10−6 )(345 × 106 ) = 34.5 × 103 N From (2), FBD = 2(Q − FCE ) = (2)(50 × 103 − 34.5 × 103 ) = 31.0 × 103 N (2) Since 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.107 (Continued) (a) σ CE = σ Y = 345 MPa Maximum stresses. σ BD = (b) FBD 31.0 × 103 = = 310 × 106 Pa −6 A 100 × 10 σ BD = 310 MPa Maximum of deflection of point C. δ BD = From (1), FBD LBD (31.0 × 103 )(2) = = 3.1 × 10−3 m EA (200 × 109 )(100 × 10−6 ) δ C = δ CE = 2δ BD = 6.2 × 10−3 m Permanent elongation of cable CE: (δ CE ) p = (δ CE ) − 6.20 mm ↓ σ Y LCE E FCE LCE σ L = (δ CE ) max − Y CE EA E 6 (345 × 10 )(2) = 6.20 × 10−3 − = 2.75 × 10−3 m 9 200 × 10 (δ CE ) P = (δ CE ) max − (c) Unloading. Cable CE is slack ( FCE = 0) at Q = 0. From (2), FBD = 2(Q − FCE ) = 2(0 − 0) = 0 Since cable BD remained elastic, δ BD = FBD LBD = 0. EA 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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.) SOLUTION Elongation constraints. Let θ = rotation angle of rigid bar ABC. θ= δ BD = δ BC LAB = δ CE LAC LAB 1 δ CE = δ CE LAC 2 (1) Equilibrium of bar ABC. M A = 0: LAB FBD + LAC FCE − LAC Q = 0 Q = FCE + LAB 1 FBD = FCE + FBD LAC 2 (2) Assume cable CE is yielded. FCE = Aσ Y = (100 × 10−6 )(345 × 106 ) = 34.5 × 103 N From (2), FBD = 2(Q − FCE ) = (2)(50 × 103 − 34.5 × 103 ) = 31.0 × 103 N Since 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.108 (Continued) (a) σ CE = σ Y = 345 MPa Maximum stresses. σ BD = (b) FBD 31.0 × 103 = = 310 × 106 Pa −6 A 100 × 10 σ BD = 310 MPa Maximum of deflection of point C. δ BD = FBD LBD (31.0 × 103 )(2) = = 3.1 × 10−3 m EA (200 × 109 )(100 × 10−6 ) δ C = δ CE = 2δ BD = 6.2 × 10−3 m From (1), 6.20 mm ↓ ′ = δ C′ Unloading. Q′ = 50 × 103 N, δ CE ′ = 12 δ C′ δ BD From (1), ′′ = Elastic FBD ′ = FCE (200 × 109 )(100 × 10−6 )( 12 δ C′ ) ′ EAδ BD = = 5 × 106 δ C′ LBD 2 ′ EAδ CE (200 × 109 )(100 × 10−6 )(δ C′ ) = = 10 × 106 δ C′ LCE 2 From (2), ′ + 12 FBD ′ = 12.5 × 106 δ C′ Q′ = FCE 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. SOLUTION Displacement at C to cause yielding of AC. L σ (0.190)(250 × 106 ) δ C ,Y = LAC ε Y , AC = AC Y , AC = = 0.2375 × 10−3 m 9 E 200 × 10 FAC = Aσ Y , AC = (1750 × 10−6 )(250 × 106 ) = 437.5 × 103 N Corresponding force. FCB = − EAδ C (200 × 109 )(1750 × 10−6 )(0.2375 × 10−3 ) =− = −437.5 × 103 N 0.190 LCB For equilibrium of element at C, FAC − ( FCB + PY ) = 0 PY = FAC − FCB = 875 × 103 N Since 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 (a) δC = − FCB LCD (537.5 × 103 )(0.190) = = 0.29179 × 10−3 m EA (200 × 109 )(1750 × 10−6 ) 0.292 mm (b) Maximum stresses: σ AC = σ Y , AC = 250 MPa (c) FBC 537.5 × 103 =− = −307.14 × 106 Pa = −307 MPa −6 A 1750 × 10 Deflection and forces for unloading. P′ L P′ L L ′ = − PAC ′ AC = − PAC ′ δ ′ = AC AC = − CB CB ∴ PCB EA EA LAB σ BC = 250 MPa −307 MPa ′ − PCB ′ = 2 PAC ′ PAC ′ = 487.5 × 10−3 N P′ = 975 × 103 = PAC δ′ = (487.5 × 103 )(0.190) = 0.26464 × 103 m −6 9 (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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. SOLUTION Displacement at C is δ m = 0.30 mm. The corresponding strains are ε AC = δm LAC ε CB = − = δm LCB 0.30 mm = 1.5789 × 10−3 190 mm =− 0.30 mm = −1.5789 × 10−3 190 mm Strains at initial yielding: ε Y, AC = ε Y, CB = (a) σ Y, AC E σ Y, BC E 250 × 106 = 1.25 × 10−3 (yielding) 200 × 109 345 × 106 =− = −1.725 × 10−3 (elastic) 200 × 109 = 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 (b) Stresses: AC : σ AC = σ Y, AC CB : σ CB = FCB 552.6 × 103 =− = −316 × 106 Pa −6 A 1750 × 10 = 990 kN = 250 MPa −316 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.110 (Continued) (c) Deflection and forces for unloading. δ′ = ′ LAC PAC P′ L L ′ = − PAC ′ AC = − PAC = − CB CB ∴ PCB EA EA LAB ′ − PCB ′ = 2 PAC ′ = 990.1 × 103 N ∴ PAC ′ = 495.05 × 103 N P′ = PAC δ′ = (495.05 × 103 )(0.190) = 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.111 Two tempered-steel bars, each 163 -in. thick, are bonded to a 12 -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 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 For the mild steel, A1 = (2) = 1.00 in 2 2 δY1 = Lσ Y 1 (14)(50 × 103 ) = = 0.024138 in. E 29 × 106 3 For the tempered steel, A2 = 2 (2) = 0.75 in 2 16 δY 2 = Lσ Y 2 (14)(100 × 103 ) = = 0.048276 in. E 29 × 103 Total area: A = A1 + A2 = 1.75 in 2 δ Y 1 < δ m < δY 2 . The mild steel yields. Tempered steel is elastic. (a) Forces: P1 = A1σ Y 1 = (1.00)(50 × 103 ) = 50 × 103 lb P2 = EA2δ m (29 × 103 )(0.75)(0.04) = = 62.14 × 103 lb 14 L P = P1 + P2 = 112.14 × 103 lb = 112.1kips (b) Stresses: σ1 = σ2 = Unloading: (c) P = 112.1 kips P1 = σ Y 1 = 50 × 103 psi = 50 ksi A1 P2 62.14 × 103 = = 82.86 × 103 psi = 82.86 ksi A2 0.75 δ′ = 82.86 ksi PL (112.14 × 103 )(14) = = 0.03094 in. EA (29 × 106 )(1.75) 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. PROBLEM 2.111 Two tempered-steel bars, each 163 -in. thick, are bonded to a 12 -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. SOLUTION Areas: Mild steel: 1 A1= (2) = 1.00 in 2 2 Tempered steel: 3 A2 = 2 (2) = 0.75 in 2 16 A = A1+ A2 = 1.75 in 2 Total: Total force to yield the mild steel: σY1 = PY ∴ PY = Aσ Y 1 = (1.75)(50 × 103 ) = 87.50 × 103 lb A P > PY , therefore, mild steel yields. Let P1 = force carried by mild steel. P2 = force carried by tempered steel. P1 = A1σ1 = (1.00)(50 × 103 ) = 50 × 103 lb P1 + P2 = P, P2 = P − P1 = 98 × 103 − 50 × 103 = 48 × 103 lb (a) δm = P2 L (48 × 103 )(14) = EA2 (29 × 106 )(0.75) (b) σ2 = P2 48 × 103 = = 64 × 103 psi A2 0.75 Unloading: δ ′ = (c) = 0.03090in. = 64 ksi PL (98 × 103 )(14) = = 0.02703 in. EA (29 × 106 )(1.75) 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. SOLUTION ΣM C = 0 : 0.640(Q − PBE ) − 2.64 PAD = 0 Statics: δ A = 2.64θ , δ B = aθ = 0.640θ Deformation: Elastic analysis: A = (37.5)(6) = 225 mm 2 = 225 × 10−6 m 2 PAD = σ AD PBE σ BE EA (200 × 109 )(225 × 10−6 ) δA = δ A = 26.47 × 106 δ A LAD 1.7 = (26.47 × 106 )(2.64θ ) = 69.88 × 106 θ P = AD = 310.6 × 109 θ A (200 × 109 )(225 × 10−6 ) EA = δB = δ B = 45 × 106 δ B 1.0 LBE = (45 × 106 )(0.640θ ) = 28.80 × 106 θ P = BE = 128 × 109 θ A From Statics, Q = PBE + 2.64 PAD = PBE + 4.125PAD 0.640 = [28.80 × 106 + (4.125)(69.88 × 106 ] θ = 317.06 × 106 θ θY 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 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 σ BE = (b) δB = PBE 27.97 × 103 = = 124.3 × 106 Pa −6 A 225 × 10 PBE LBE (27.97 × 103 )(1.0) = = 621.53 × 10−6 m EA (200 × 109 )(225 × 10−6 ) σ BE = 124.3 MPa δ B = 0.622 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. SOLUTION ΣM C = 0 : 1.76(Q − PBE ) − 2.64 PAD = 0 Statics: δ A = 2.64θ , δ B = 1.76θ Deformation: Elastic Analysis: A = (37.5)(6) = 225 mm 2 = 225 × 10−6 m 2 PAD = σ AD PBE σ BE EA (200 × 109 )(225 × 10−6 ) δA = δ A = 26.47 × 106 δ A LAD 1.7 = (26.47 × 106 )(2.64θ ) = 69.88 × 106 θ P = AD = 310.6 × 109 θ A (200 × 109 )(225 × 10−6 ) EA = δB = δ B = 45 × 106 δ B 1.0 LBE = (45 × 106 )(1.76θ ) = 79.2 × 106 θ P = BE = 352 × 109 θ A From Statics, Q = PBE + 2.64 PAD = PBE + 1.500 PAD 1.76 = [73.8 × 106 + (1.500)(69.88 × 106 ] θ = 178.62 × 106 θ θY 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 N Since 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.114 (Continued) From Statics, PAD = 1 (Q − PBE ) = 52.5 × 103 N 1.500 (a) From elastic analysis of AD, (b) σ AD = PAD 52.5 × 103 = = 233.3 × 106 −6 A 225 × 10 θ= PAD = 751.29 × 10−3 rad 69.88 × 106 δ B = 1.76θ = 1.322 × 10−3 m σ AD = 233 MPa 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. SOLUTION See 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 m The elastic analysis given in the solution to Problem 2.113 applies to the unloading. Q′ = 317.06 × 106 θ ′ Q′ = Q 260 × 103 = = 820.03 × 10−6 317.06 × 106 317.06 × 106 σ ′AD = 310.6 × 109 θ = (310.6 × 109 )(820.03 × 10−6 ) = 254.70 × 106 Pa ′ = 128 × 109 θ = (128 × 109 )(820.03 × 10−6 ) = 104.96 × 106 Pa σ BE δ B′ = 0.640θ ′ = 524.82 × 10−6 m (a) (b) Residual stresses. ′ = 250 × 106 − 254.70 × 10−6 = −4.70 × 106 Pa σ AD , res = σ AD − σ AD = −4.70 MPa ′ = 124.3 × 106 − 104.96 × 106 = 19.34 × 106 Pa σ BE , res = σ BE − σ BE = 19.34 MPa 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 . SOLUTION Let P be the compressive force in the rod. Determine temperature change to cause yielding. σ L PL + 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 ) δ =− But ΔT = 320 − 45 = 275 °F > (ΔTY ) (a) σ = −σ Y = −36 ksi Yielding occurs. Cooling: (ΔT)′ = 275 °F δ ′ = δ P′ = δ T′ = − P′L + Lα (ΔT )′ = 0 AE P′ = − Eα (ΔT )′ A = −(29 × 106 )(6.5 × 10−6 )(275) = −51.8375 × 103 psi σ′= (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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 m δ P + δT = 0 Constraint: Determine ΔT to cause yielding in portion CB. − PLAC PLCB − = LABα ( ΔT ) EAAC EACB ΔT = P LAC LCB + LAB Eα AAC ACB At yielding, P = PY = ACBσ Y = (300 × 10−6 )(2.50 × 106 ) = 75 × 103 N (ΔT )Y = = PY LAB Eα LAC LCB + AAC ACB 75 × 103 0.250 0.150 + 9 −6 −6 (0.400)(200 × 10 )(11.7 × 10 ) 500 × 10 300 × 10−6 Actual ΔT: 150 °C − 25 °C = 125 °C > (ΔT )Y Yielding occurs. For ΔT > (ΔT )Y , (a) (b) = 90.812 °C P = PY = 75 × 103 N σ AC = − PY 75 × 103 =− = −150 × 10−6 Pa AAC 500 × 10−6 σ AC = −150 MPa σ CB = − PY = −σ Y ACB σ CB = −250 MPa For ΔT > (ΔT )Y , portion AC remains elastic. δ C /A = − =− PY LAC + 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 Constraint: LAC = 0.150 m ACB = 300 × 10−6 m 2 LCB = 0.250 m δ P + δT = 0 Determine ΔT to cause yielding in portion CB. − PLAC PLCB − = LABα ( ΔT ) EAAC EACB ΔT = P LAB Eα LAC LCB + AAC ACB At yielding, P = PY = ACBσ Y = (300 × 10−6 )(250 × 106 ) = 75 × 103 N PY LAC LCB 75 × 103 0.250 0.150 + + = 9 −6 −6 LAB Eα AAC ACB (0.400)(200 × 10 )(11.7 × 10 ) 500 × 10 300 × 10−6 = 90.812 °C (ΔT )Y = Actual ΔT : 150 °C − 25 °C = 125 °C > (ΔT )Y Yielding occurs. For ΔT > (ΔT )Y Cooling: (ΔT )′ = 125 °C P′ = P = PY = 75 × 103 N ELABα (ΔT )′ ( LAC AAC + LCB ACB ) = (200 × 109 )(0.400)(11.7 × 10−6 )(125) = 103.235 × 103 N 0.150 0.250 + −6 −6 500 × 10 300 × 10 Residual 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.118∗ (Continued) (a) (b) Residual stresses. σ AC = Pres 28.235 × 103 = AAC 500 × 10−6 σ AC = 56.5 MPa σ CB = Pres 28.235 × 103 = ACB 300 × 10−6 σ CB = 9.41 MPa Permanent deflection of point C. δ C = Pres LAC EAAC δ C = 0.0424 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. PROBLEM 2.111 Two tempered-steel bars, each 163 -in. thick, are bonded to a 12 -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 Areas: Mild steel: 1 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 Total force to yield the mild steel: σ Y 1 = PY A ∴ PY = Aσ Y 1 = (1.75)(50 × 103 ) = 87.50 × 103 lb P > PY; therefore, mild steel yields. P1 = force carried by mild steel. Let P2 = force carried by tempered steel. P1 = A1σ Y 1 = (1.00)(50 × 103 ) = 50 × 103 lb P1 + P2 = P, P2 = P − P1 = 98 × 103 − 50 × 103 = 48 × 103 lb σ2 = Unloading: σ ′ = P2 48 × 103 = = 64 × 103 psi A2 0.75 P 98 × 103 = = 56 × 103 psi A 1.75 Residual 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. PROBLEM 2.111 Two tempered-steel bars, each 163 -in. thick, are bonded to a 12 -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 Lδ (14)(50 × 103 ) 1 = 0.024138 in A1 = (2) = 1.00 in 2 δ Y 1 = Y 1 = E 29 × 106 2 For the mild steel, Lδ (14)(100 × 103 ) 3 For the tempered steel, A2 = 2 (2) = 0.75 in 2 δ Y 2 = Y 2 = = 0.048276 in. E 29 × 106 16 A = A1 + A2 = 1.75 in 2 Total area: δY 1 < δ m < δ Y 2 The mild steel yields. Tempered steel is elastic. P1 = A1δY 1 = (1.00)(50 × 103 ) = 50 × 103 lb Forces: P2 = Stresses: Unloading: σ1 = EA2δ m (29 × 106 )(0.75)(0.04) = = 62.14 × 103 lb 14 L P1 P 62.14 × 103 = δY 1 = 50 × 103 psi σ 2 = 2 = = 82.86 × 103 psi 0.75 A1 A2 σ′ = P 112.14 = = 64.08 × 103 psi A 1.75 Residual 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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.) SOLUTION Determine temperature change to cause yielding. δ= PL + 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 6 E (α a − α s ) (10.9 × 10 )(12.8 − 6.5)(10−6 ) (a) T2Y = T1 + (ΔT )Y = 70 + 844.62 = 915 °F 915 °F After yielding, δ= σY L E + Lα a (ΔT ) = Lα s (ΔT ) Cooling: δ′ = P′L + Lα a (ΔT )′ = Lα s (ΔT )′ AE The residual stress is σ res = σ Y − Set σ res = −σ Y −σ Y = σ Y − E (α a − α s )(ΔT ) ΔT = (b) P′ = σ Y − E (α a − α s )(ΔT ) A 2σ Y (2)(58 × 103 ) = = 1689 °F E (α a − α s ) (10.9 × 106 )(12.8 − 6.5)(10−6 ) 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 2 Force to yield portion AC: PAC = Aσ Y = (1200 × 10−6 )(250 × 106 ) = 300 × 103 N For equilibrium, F + PCB − PAC = 0. PCB = PAC − F = 300 × 103 − 520 × 103 = −220 × 103 N δC = − PCB LCB (220 × 103 )(0.440 − 0.120) = EA (200 × 109 )(1200 × 10−6 ) = 0.293333 × 10−3 m PCB 220 × 103 = A 1200 × 10−6 = −183.333 × 106 Pa σ CB = 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.122∗ (Continued) Unloading: δ C′ = ′ = PAC ′ LAC ′ ) LCB PAC P′ L ( F − PAC = − CB CB = EA EA EA L FL L ′ AC + BC = CB PAC EA EA EA FLCB (520 × 103 )(0.440 − 0.120) = = 378.182 × 103 N LAC + LCB 0.440 ′ = PAC ′ − F = 378.182 × 103 − 520 × 103 = −141.818 × 103 N PCB ′ PAC 378.182 × 103 = = 315.152 × 106 Pa −6 A 1200 × 10 P′ 141.818 × 103 ′ = BC = − σ BC = −118.182 × 106 Pa −6 A 1200 × 10 (378.182)(0.120) = 0.189091 × 10−3 m δ C′ = 9 6 (200 × 10 )(1200 × 10 ) ′ = σ AC (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 ′AC = 250 × 106 − 315.152 × 106 = −65.2 × 106 Pa = −65.2 MPa ′ = −183.333 × 106 + 118.182 × 106 = −65.2 × 106 Pa σ CB , res = σ CB − σ CB = −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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 2 Force to yield portion AC: PAC = Aσ Y = (1200 × 10−6 )(250 × 106 ) = 300 × 103 N For equilibrium, F + PCB − PAC = 0. PCB = PAC − F = 300 × 103 − 520 × 103 = −220 × 103 N δC = − PCB LCB (220 × 103 )(0.440 − 0.180) = EA (200 × 109 )(1200 × 10−6 ) = 0.238333 × 10−3 m PCB 220 × 103 =− A 1200 × 10−6 = −183.333 × 106 Pa σ CB = 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.123∗ (Continued) Unloading: ′ LAC ′ ) LCB PAC P′ L ( F − PAC = − CB CB = EA EA EA L FL L ′ AC + BC = CB = PAC EA EA EA δ C′ = ′ = PAC FLCB (520 × 103 )(0.440 − 0.180) = = 307.273 × 103 N 0.440 LAC + LCB ′ = PAC ′ − F = 307.273 × 103 − 520 × 103 = −212.727 × 103 N PCB δ C′ = (307.273 × 103 )(0.180) = 0.230455 × 10−3 m (200 × 109 )(1200 × 10−6 ) ′ PAC 307.273 × 103 = = 256.061 × 106 Pa −6 A 1200 × 10 ′ PCB −212.727 × 103 ′ = σ CB = = −177.273 × 106 P A 1200 × 10−6 ′ = σ AC (a) δ C , p = δ C − δ C′ = 0.238333 × 10−3 − 0.230455 × 10−3 = 0.00788 × 10−3 m (b) σ AC ,res = σ AC − σ ′AC = 250 × 106 − 256.061 × 106 = −6.06 × 106 Pa = −6.06 MPa ′ = −183.333 × 106 + 177.273 × 106 = −6.06 × 106 Pa σ CB ,res = σ CB − σ CB = −6.06 MPa = 0.00788 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 × 103lb Considering stress: σ = 18 ksi = 18 × 103 psi σ= FBD A ∴ A= FBD σ = 2.6 = 0.14444 in 2 18 Considering deformation: δ = (0.001)(144) = 0.144 in. δ= FBD LBD AE ∴ A= FBD LBD (2.6 × 103 ) (54) = = 0.03362 in 2 6 Eδ (29 × 10 ) (0.144) Larger area governs. A = 0.14444 in 2 A= π 4 d2 ∴ d= 4A π = (4) (0.14444) π d = 0.429in. 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. SOLUTION Rod AB: FAB = − P = −30 × 103 N LAB = 0.250 m E AB = 200 × 109 GPa δ AB Rod BC: π (30)2 = 706.85 mm 2 = 706.85 × 10−6 m 2 4 F L (30 × 103 )(0.250) = AB AB = − = −53.052 × 10−6 m −6 9 E AB AAB (200 × 10 )(706.85 × 10 ) AAB = FBC = 30 + 40 = 70 kN = 70 × 103 N LBC = 0.300 m EBC = 105 × 109 Pa δ BC π (50) 2 = 1.9635 × 103 mm 2 = 1.9635 × 10−3 m 2 4 F L (70 × 103 )(0.300) = BC BC = − = −101.859 × 10−6 m 9 −3 EBC ABC (105 × 10 )(1.9635 × 10 ) ABC = δ tot = δ AB + δ BC = −154.9 × 10−6 m (a) Total deformation: (b) Deflection of Point B. δ B = δ BC = −0.1549 mm 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. SOLUTION Portion AB: PAB = 40 × 103 lb LAB = 40 in. d = 2 in. E AB δ AB = Portion BC: π d2 = π (2) 2 = 3.1416 in 2 4 4 = 29 × 106 psi AAB = PAB LAB (40 × 103 )(40) = = 17.5619 × 10−3 in. E AB AAB (29 × 106 )(3.1416) PBC = −20 × 103 lb LBC = 30 in. d = 3 in. ABC = π d2 = π (3) 2 = 7.0686 in 2 4 4 EBC = 15 × 106 psi δ BC = PBC LBC (−20 × 103 )(30) = = −5.6588 × 10−3 in. 6 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 δ =l3 P AE SOLUTION Use approximation. sin θ ≈ tan θ ≈ Statics: l ΣFY = 0 : 2 PAB sin θ − P = 0 PAB = Elongation: δ P Pl ≈ 2sin θ 2δ δ AB = PAB l Pl 2 = AE 2 AEδ Deflection: From the right triangle, (l + δ AB ) 2 = l 2 + δ 2 2 δ 2 = l 2 + 2l δ AB + δ AB − l2 1 δ AB = 2l δ AB 1 + 2 l ≈ δ3 ≈ ≈ 2l δ AB Pl 3 AEδ Pl 3 P ∴ δ ≈ l3 AE AE 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. SOLUTION Brass strip: E = 105 GPa α = 20 × 10−6 / °C ΣFy = 0 : N − W = 0 N =W ΣFx = 0 : P − μ N = 0 δ =− Data: PL + Lα ( ΔT ) = 0 EA P = μW = μ mg ΔT = μ mg P = EAα EAα 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 ΔT = (0.60)(100)(9.81) (105 × 109 )(60 × 10−6 )(20 × 106 ) ΔT = 4.67 °C 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PROBLEM 2.129 Members AB and CD are 1 18 -in.-diameter steel rods, and members BC and AD are 78 -in.-diameter steel rods. When the turnbuckle is tightened, the diagonal 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 ACD = δ CD π d2 = π 4 4 FCD LCD = EACD FCD = (1.125) 2 = 0.99402 in 2 EACDδ CD (29 × 106 )(0.99402)(0.04) = = 24.022 × 103 lb 48 LCD Use joint C as a free body. ΣFy = 0 : FCD − FAC = 4 5 FAC = 0 ∴ FAC = FCD 5 4 5 (24.022 × 103 ) = 30.0 × 103 lb 4 FAC = 30.0 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. SOLUTION Let Pc = portion of axial force carried by concrete. Ps = portion carried by the six steel rods. δ= Pc L Ec Ac Pc = Ec Acδ L δ= Ps L Es As Ps = Es Asδ L P = Pc + Ps = ( Ec Ac + Es As ) δ L δ −P ε= = L Ec Ac + Es As 6π (28) 2 = 3.6945 × 103 mm 2 4 4 = 3.6945 × 10−3 m 2 As = 6 Ac = π π d s2 = d c2 − As = π (450) 2 − 3.6945 × 103 4 4 = 155.349 × 103 mm 2 = 153.349 × 10−3 m 2 L = 1.5 m ε= 1550 × 103 = 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. SOLUTION Let Ps = axial force developed in the steel core. For equilibrium with zero total force, the compressive force in the brass shell is Ps . Strains: εs = Ps + α s (ΔT ) Es As εb = − Matching: Ps + α b (ΔT ) Eb Ab ε s = εb Ps P + α s (ΔT ) = − s + α b (ΔT ) Es As Eb Ab 1 1 + Ps = (α b − α s )(ΔT ) Es As Eb Ab (1) 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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 (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 εx = (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 dc = a b 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 (c) δ AC = dc = 100 75 (0.0754) + (0.1028) 125 125 = 0.1220 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. SOLUTION Shearing force: P = 22 × 103 N Shearing stress: τ = 420 × 103 Pa P P ∴ A= τ A 22 × 103 = = 52.381 × 10−3 m 2 420 × 103 = 52.381 × 103 mm 2 A = (200 mm)(b) τ= (a) b= γ= (b) A 52.381 × 103 = = 262 mm 200 200 τ G = But γ = b = 262 mm 420 × 103 = 466.67 × 10−3 6 0.9 × 10 δ a ∴ a= δ 10 mm = = 21.4 mm γ 466.67 × 10−3 a = 21.4 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. 5.0 D = = 2.00 2.50 d 3 Amin = dt = (2.50) = 1.875 in 2 4 (a) r = 0.50 in. From Fig. 2.60b, σ max = 0.50 r = = 0.20 2.50 d K = 1.94 KP (1.94) (10 × 103 ) = = 10.35 × 103 psi Amin 1.875 10.35 ksi (b) r = 0.625 in. σ max = r 0.625 = = 0.25 K = 1.82 d 2.50 KP (1.82) (10 × 103 ) = = 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without 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. SOLUTION Force-deflection diagram for Point C or rod BC. P < PY = Aσ Y For PL EA = PY = Aσ Y δC = Pmax P= EA δC L Force-deflection diagram for Point C′ of block-and-spring system. ΣFy = 0 : N − mg = 0 N = mg ΣFx = 0 : P − F f = 0 P = Ff If block does not move, i.e., F f < μ N = μ mg or P < μ mg , δ c′ = then P K or P = kδ c′ If 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. (a) Equating PY and Fmax, (b) Equating slopes, Aσ Y = μ mg k= EA L m= Aσ Y μg 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.