PROBLEM 2.45 (Continued) PA  5 P  1  P   21 P PA  0.525P  8 2  5  40 PC  0.275P  PC  3 P  1 P  11 P 8 2  5  40Check:PA  PB  PC  1.000P Ok PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 141

• Mechanics Of Materials Beer 6th Edition Solution Manual Pdf

MECHANICS OF MATERIALS 6TH EDITION GERE SOLUTION MANUAL PDF INTRODUCTION This eBook discuss about the subject of MECHANICS OF MATERIALS 6TH EDITION GERE SOLUTION MANUAL PDF, along with the whole set of accommodating tips plus details about that subject. Mechanics of Materials 6th Edition Beer and Johnston's Mechanics of Materials is the uncontested leader for the teaching of solid mechanics. Used by thousands of students around the globe since its publication in 1981, Mechanics of Materials, provides a precise presentation of the subject illustrated with numerous engineering examples that students both understand and relate to theory and application.

E PROBLEM 2.46 15 in. F The rigid bar AD aisndsuapppoinrteadndbybratwckoetstaetelDw. Kirensowoifng116t-hiant. diameterAB 8 in. (E  29  106 psi) the wires C 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. D 8 in. 8 in. 8 in. PSOLUTIONLet  be the rotation of bar ABCD.Then  A  24 C  8  A  PAE LAE AE PAE  EA A  (29 106 )  (116 )2 (24 ) LAE 4 15  142.353103 C  PCF LCF AE 106 2 (8 )  PCF (29 )  1  EAC  4 16 LCF 8  88.971103Using free body ABCD, M D  0 : 24PAE  16P  8PCF  0 24 (142.353  103 )  16(120)  8(88.971  103 )  0   0.46510  103 radۗ(a) PAE  (142.353103 )(0.46510 103 ) PAE  66.2 lb  PCF  (88.971103 )(0.46510 103 ) PCF  41.4 lb  B  7.44 103 in.  (b) B  16  16(0.46510 103 )PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 142

25 mm Brass core PROBLEM 2.4760 mm E ϭ 105 GPa ␣ ϭ 20.9 ϫ 10–6/ЊC The aluminum shell is fully bonded to the brass core and the assembly is unstressed at a temperature of 15C. Considering only Aluminum shell axial deformations, determine the stress in the aluminum when the E ϭ 70 GPa temperature reaches 195C. ␣ ϭ 23.6 ϫ 10–6/ЊCSOLUTIONBrass core: E  105 GPa   20.9 106/ CAluminum shell: E  70 GPaLet L be the length of the assembly.   23.6 106/ CFree thermal expansion: T  195 15  180 CBrass core: (T )b  Lb (T )Aluminum shell: (T )a  La (T )Net expansion of shell with respect to the core:   L(a  b )(T )Let P be the tensile force in the core and the compressive force in the shell.Brass core: Eb  105 109 Pa Ab   (25)2  490.87 mm2 4  490.87 106 m2 (P )b  PL Eb AbPROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 143

PROBLEM 2.47 (Continued)Aluminum shell: ( p )a  PL Ea AawhereThen Ea  70 109 PaStress in aluminum: Aa   (602  252 ) 4  2.3366 103 mm2  2.3366 103 m2   (P )b  (P )a L(b  a )(T )  PL  PL  KPL Eb Ab Ea Aa K  1  1 Eb Ab Ea Aa 11  (105 109 )(490.87 106 )  (70 109 )(2.3366 103 )  25.516 109 N1 P  (b  a )(T ) K  (23.6 106  20.9 106 )(180) 25.516 109  19.047 103 N a  P   19.047 103  8.15 106 Pa  a  8.15 MPa  Aa 2.3366 103PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 144

25 mm PROBLEM 2.4860 mm Brass core Solve Prob. 2.47, assuming that the core is made of steel (Es  200 GPa, E ϭ 105 GPa s  11.7 106/ C) instead of brass. ␣ ϭ 20.9 ϫ 10–6/ЊC PROBLEM 2.47 The aluminum shell is fully bonded to the brass core Aluminum shell and the assembly is unstressed at a temperature of 15C. Considering E ϭ 70 GPa only axial deformations, determine the stress in the aluminum when the ␣ ϭ 23.6 ϫ 10–6/ЊC temperature reaches 195C.SOLUTIONAluminum shell: E  70 GPa   23.6 106/ CLet L be the length of the assembly.Free thermal expansion: T  195 15  180CSteel core: (T )s  Ls (T )Aluminum shell: (T )a  La (T )Net expansion of shell with respect to the core:   L(a  s )(T )Let P be the tensile force in the core and the compressive force in the shell.Steel core: Es  200 109 Pa, As   (25)2  490.87 mm2  490.87 106 m2 4 (P )s  PL Es AsAluminum shell: Ea  70 109 Pa (P )a  PL Ea Aa Aa   (602  25)2  2.3366 103 mm2  2.3366 103 m2 4   (P )s  (P )a L(a  s )(T )  PL  PL  KPL Es As Ea Aawhere K  1  1 Es As Ea Aa  1  1 (200 109 )(490.87 106 ) (70 109 )(2.3366 103 )  16.2999 109 N1PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 145

PROBLEM 2.48 (Continued)Then P  (a  s )(T )  (23.6  106  11.7  106 )(180)  131.412  103 N K 16.2999  109Stress in aluminum: a   P   131.412 103  56.241106 Pa  a  56.2 MPa  Aa 2.3366 103PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 146

1 in. 1 in. 1 in. PROBLEM 2.49 4 4 1 in.1 14 in. 4 in. The brass s1h0el6l/(F).b  11.6  106 /F) is fully bonded to the steel core ( s  6.5  Determine the largest allowable increase inSteel core temperature if the stress in the steel core is not to exceed 8 ksi.E ϭ 29 ϫ 106 psiBrass shell 12 in.E ϭ 15 ϫ 106 psiSOLUTIONLet Ps  axial force developed in the steel core.For equilibrium with zero total force, the compressive force in the brass shell is Ps.Strains: s  Ps  s (T ) Es As b   Ps  b (T ) Eb AbMatching: s  b Ps  s(T )   Ps  b(T ) Es As Eb Ab  1  1  Ps  (b  s )(T ) (1)  Es As Eb Ab  T  137.8F    Ab  (1.5)(1.5)  (1.0)(1.0)  1.25 in2 As  (1.0)(1.0)  1.0 in2 b  s  5.1106 /F Ps   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)(T )PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 147

PROBLEM 2.50 nTbaohrresm,caeolansccthrreeotsefseps78o-isnitnd.(uEdcceiadmi3ne.t6tehre1(s0Et6eseplsa2in9adnid1n0t6hc epcs5oi.n5acnrd1et0es6b/yF6a).5tiesmr1pe0ienr6fa/oturFcr)ee.drDiwseetietohrfms6iix5n°esFtte.heel 6 ft10 in. 10 in.SOLUTION As  6  d2  6   7 2  3.6079 in2 4 4  8  Ac  102  As  102  3.6079  96.392 in2Let Pc  tensile force developed in the concrete.For equilibrium with zero total force, the compressive force in the six steel rods equals Pc.Strains: s   Pc  s (T ) c  Pc  c (T ) Es As Ec AcMatching: c  s Pc  c (T )   Pc  s (T ) Ec Ac Es As 1  1  Pc  (s  c )(T )  Es As   Ec Ac  1  1 Pc  (1.0 106 )(65)    (3.6  106 )(96.392) (29  106 )(3.6079)  Pc  5.2254 103 lb c  Pc  5.2254 103  54.210 psi  c  54.2 psi Ac 96.392 s  Pc   5.2254 103  1448.32 psi   s  1.448 ksi  As 3.6079PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 148

A PROBLEM 2.51 30-mm diameter250 mm A rod consisting of two cylindrical portions AB and BC is restrained at both300 mm B GbPa,20s.91110.76/ 1C0).6/KCno)wainndg 50-mm diameter ends. Portion AB is made of steel (Es  200 portion BC is made of brass ( Eb  105 GPa, that the rod is initially unstressed, determine the compressive force induced in ABC when there is a temperature rise of 50C. CSOLUTION AAB   d 2   (30)2  706.86 mm2  706.86 106 m2 4 AB 4 ABC   d 2   (50)2  1.9635 103 mm2  1.9635 103 m2 4 BC 4Free thermal expansion: T  LABs (T )  LBCb (T )  (0.250)(11.7 106 )(50)  (0.300)(20.9 106 )(50)  459.75 106 mShortening due to induced compressive force P: P  PL  PL Es AAB Eb ABC  0.250P  0.300P (200 109 )(706.86 106 ) (105 109 )(1.9635 103 )  3.2235 109PFor zero net deflection, P  T P  142.6 kN  3.2235 109P  459.75 106 P  142.624 103 NPROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 149

24 in. 32 in. PROBLEM 2.52 AB C A rod consisting of two cylindrical portions AB and BC is restrained at both ends. Portion AB is made of steel (E(Es a2190.410610p6si,psi,s  6.5  106/F) and portion BC is made of aluminum a  13.3  106/°F).2 1 -in. diameter 1 1 -in. diameter Knowing that the rod is initially unstressed, determine (a) the normal stresses 4 2 induced in portions AB and BC by a temperature rise of 70°F, (b) the corresponding deflection of point B.SOLUTION AAB   (2.25)2  3.9761 in 2 ABC   (1.5)2  1.76715 in 2 4 4Free thermal expansion. T  70FTotal: (T )AB  LABs (T )  (24)(6.5 106 )(70)  10.92 103 in. (T )BC  LBCa (T )  (32)(13.3106 )(70)  29.792 103 in. T  (T )AB  (T )BC  40.712 103 in.Shortening due to induced compressive force P. ( P ) AB  PLAB  24P  208.14 109P Es AAB (29 106 )(3.9761) ( P )BC  PLBC  32P  1741.18 109P Ea ABC (10.4 106 )(1.76715)Total: P  (P )AB  (P )BC  1949.32 109P P  20.885 103 lbFor zero net deflection, P  T 1949.32 109P  40.712 103(a)  AB  P   20.885 103  5.25 103 psi  AB  5.25 ksi  AAB 3.9761  BC   P   20.885 103  11.82 103 psi  BC  11.82 ksi  ABC 1.76715(b) (P )AB  (208.14 109 )(20.885 103 )  4.3470 103in. B  (T )AB   (P )AB   10.92 103   4.3470 103  B  6.57 103 in.   or (P )BC  (1741.18 109 )(20.885 103 )  36.365 103in. B  (T )BC   (P )BC   29.792 103   36.365 103   6.57 103in.  (checks)PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 150

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24 in. 32 in. PROBLEM 2.53 AB C Solve Prob. 2.52, assuming that portion AB of the composite rod is made of aluminum and portion BC is made of steel.2 1 -in. diameter 1 1 -in. diameter 4 2 PROBLEM 2.52 A rod consisting of two cylindrical portions AB and BC is reasstra16i3.n5.e3d1a10t06b/6o/F°tFh) )a. enKnddnpsoo.wrtPiinoognrtiBtohCnatiAsthBme aridsoedmoifsaadlieunmitoiiafnlulsymteue(nlEsta(rEess 1se0d.24,9de1t10e06rm6 ppisnsiie, (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   (2.25)2  3.9761 in2 ABC   (1.5)2  1.76715 in2 4 4Free thermal expansion. T  70F (T )AB  LABa (T )  (24)(13.3 106 )(70)  22.344 103in. (T )BC  LBCs (T )  (32)(6.5 106 )(70)  14.56 103in.Total: T  (T )AB  (T )BC  36.904 103in.Shortening due to induced compressive force P. ( P ) AB  PLAB  24P  580.39 109P Ea AAB (10.4 106 )(3.9761) ( P )BC  PLBC  32P  624.42 109P Es ABC (29 106 )(1.76715)Total: P  (P )AB  (P )BC  1204.81109PFor zero net deflection, P  T 1204.81109 P  36.904 103 P  30.631103 lb(a)  AB  P   30.631103  7.70 103 psi  AB  7.70 ksi  AAB 3.9761  BC P   30.631103  17.33103 psi  BC  17.33 ksi    ABC 1.76715PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 151

PROBLEM 2.53 (Continued)(b) (P )AB  (580.39 109 )(30.631103 )  17.7779 103 in.B  (T )AB   (P )AB   22.344 103   17.7779 103  B  4.57 103 in.   or (P )BC  (624.42 109 )(30.631103 )  19.1266 103 in.B  (T )BC   (P )BC   14.56 103   19.1266 103   4.57 103 in.  (checks)PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 152

PROBLEM 2.54The steel rails of a railroad track (Es  200 GPa, αs  11.7 × 102–6/C) were laid at a temperature of 6C.Determine the normal stress in the rails when the temperature reaches 48C, assuming that the rails (a) arewelded to form a continuous track, (b) are 10 m long with 3-mm gaps between them.SOLUTION(a) T   (T )L  (11.7 106 )(48  6)(10)  4.914 103 mP  PL  L  (10)  50 1012 AE E 200 109  T  P  4.914 103  50 1012  0  98.3106 Pa   98.3 MPa   38.3 MPa (b)   T  P  4.914 103  50 1012  3 103  3 103  4.914 103 50 1012  38.3106 PaPROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 153

PЈ PROBLEM 2.55 2m 15 mm Two steel bars (Es  200 GPa and s  11.7 106/C) are used toSteel reinforce a brass bar (Eb  105 GPa, b  20.9 106/C) that is subjected Brass 5 mm Steel P to a load P  25 kN. When the steel bars were fabricated, the distance 40 mm 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: T  0.5 mm  0.5 103 mTemperature change required to expand steel bar by this amount: T  LsT , 0.5 103  (2.00)(11.7 106 )(T ), 21.4 C  T  0.5 103  (2)(11.7 106 )(T ) T  21.368C(b) Once assembled, a tensile force P* develops in the steel, and a compressive force P* develops in the brass, in order to elongate the steel and contract the brass.Elongation of steel: As  (2)(5)(40)  400 mm2  400 106 m2 ( P )s  F*L  P* (2.00)  25 109P* As Es (400 106 )(200 109 )Contraction of brass: Ab  (40)(15)  600 mm2  600 106 m2 ( P )b  P*L  P* (2.00)  31.746 109P* Ab Eb (600 106 )(105 109 )But (P )s  (P )b is equal to the initial amount of misfit: (P )s  (P )b  0.5 103, 56.746 109P*  0.5 103 P*  8.8112 103 NStresses due to fabrication:Steel:  *  P*  8.8112  103  22.028  106 Pa  22.028 MPa s As 400  106PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 154

PROBLEM 2.55 (Continued)Brass:  *   P*   8.8112  103  14.6853  106 Pa  14.685 MPa b Ab 600  106To 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   be theadditional displacement. Also, let Ps and Pb be the additional forces developed in the steel and brass,respectively.  Ps L  Pb L As Es Ab Eb Ps  As Es   (400  106 )(200 109 )    40 106  L 2.00 Pb  Ab Eb  (600  106 )(105  109 )    31.5 106  L 2.00Total: P  Ps  Pb  25 103 N 40 106   31.5 106   25 103    349.65 106 m Ps  (40  106 )(349.65  106 )  13.9860  103 N Pb  (31.5  106 )(349.65  106 )  11.0140  103 N s  Ps  13.9860  103  34.965  106 Pa As 400  106 b  Pb  11.0140  103  18.3566  106 Pa Ab 600  106Add stress due to fabrication.Total stresses:  s  34.965  106  22.028  106  56.991  106 Pa  s  57.0 MPa b  18.3566 106 14.6853106  3.6713106 Pa b  3.67 MPa PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 155

PЈ PROBLEM 2.56 2m 15 mm Determine the maximum load P that may be applied to the brass bar ofSteel Prob. 2.55 if the allowable stress in the steel bars is 30 MPa and the Brass allowable stress in the brass bar is 25 MPa. Steel 5 mm PROBLEM 2.55 Two steel bars (Es  200 GPa and s  11.7  10–6/C) P are used to reinforce a brass bar (Eb  105 GPa, b  20.9  10–6/C) 40 mm that is subjected to a load P  25 kN. When the steel bars were fabricated, the distance between the centers of the holes that were to fit on the pins was made 0.5 mm smaller than the 2 m needed. The steel bars were then placed in an oven to increase their length so that they would just fit on the pins. Following fabrication, the temperature in the steel bars dropped back to room temperature. Determine (a) the increase in temperature that was required to fit the steel bars on the pins, (b) the stress in the brass bar after the load is applied to it.SOLUTIONSee solution to Problem 2.55 to obtain the fabrication stresses.  *  22.028 MPa s  *  14.6853 MPa bAllowable stresses:  s,all  30 MPa,  b,all  25 MPaAvailable stress increase from load.  s  30  22.028  7.9720 MPa b  25  14.6853  39.685 MPaCorresponding available strains. s  s  7.9720  106  39.860  106 Es 200  109 b  b  39.685  106  377.95  106 Eb 105  109Smaller value governs    39.860  106Areas: As  (2)(5)(40)  400 mm2  400 106 m2 Ab  (15)(40)  600 mm2  600 106 m2Forces Ps  Es As  (200  109 )(400  106 )(39.860  106 )  3.1888  103 N Pb  Eb Ab  (105  109 )(600  106 )(39.860  106 )  2.5112  103 NTotal allowable additional force:P  Ps  Pb  3.1888 103  2.5112 103  5.70 103 N P  5.70 kN PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 156

Dimensions in mm PROBLEM 2.57 0.15 An aluminum rod (Ea  70 GPa, αa  23.6 × 106/C) and a steel link20 20 200 30 (Es × 200 GPa, αa  11.7 × 106/C) have the dimensions shown at a temperature of 20C. The steel link is heated until the aluminum rod can be fitted freely into the link. The temperature of the whole assembly is then raised to 150C. Determine the final normal stress (a) in the rod, (b) in the link.AA20 Section A-ASOLUTIONT  Tf  Ti  150C  20C  130CUnrestrained thermal expansion of each part:Aluminum rod: (T )a  La (T ) (T )a  (0.200 m)(23.6 106/C)(130C)  6.1360  104 mSteel link: (T )s  Ls (T ) (T )s  (0.200 m)(11.7 106/C)(130C)  3.0420  104 mLet P be the compressive force developed in the aluminum rod. It is also the tensile force in the steel link.Aluminum rod: ( P )a  PL Ea Aa P(0.200 m)  (70  109 Pa)( /4)(0.03 m)2  4.0420 109PSteel link: ( P )s  PL Es As P(0.200)  (200  109 Pa)(2)(0.02 m)2  1.250 109PSetting the total deformed lengths in the link and rod equal gives(0.200)  (T )s  (P )s  (0.200)  (0.15 103 )  (T )a  (P )aPROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 157

PROBLEM 2.57 (Continued) (P)s  (P )a  0.15  103  (T )a  (T )s1.25  109P  4.0420  109P  0.15  103  6.1360  104  3.0420  104 P  8.6810  104 N(a) Stress in rod:   P A R  8.6810 104 N  1.22811108 Pa ( /4)(0.030 m)2(b) Stress in link:  R  122.8 MPa   L  108.5 MPa  L  8.6810  104 N  1.08513  108 Pa (2)(0.020 m)2PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 158

Mechanics Of Materials Beer 6th Edition Solution Manual Pdf

0.02 in. 18 in. PROBLEM 2.58 14 in. Knowing that a 0.02-in. gap exists when the temperature is 75F,Bronze Aluminum determine (a) the temperature at which the normal stress in theA ϭ 2.4 in2 A ϭ 2.8 in2 aluminum bar will be equal to 11 ksi, (b) the corresponding exactE ϭ 15 ϫ 106 psi E ϭ 10.6 ϫ 106 psi length of the aluminum bar.␣ ϭ 12 ϫ 10–6/ЊF ␣ ϭ 12.9 ϫ 10–6/ЊFSOLUTION  a  11 ksi  11103 psiShortening due to P: P   a Aa  (11103 )(2.8)  30.8 103 lb P  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3in.Available elongation for thermal expansion: T  0.02  30.657 103  50.657 103 in.But T  Lbb (T )  Laa (T )  (14)(12 106 )(T )  (18)(12.9 106 )(T )  (400.2 106 )TEquating, (400.2 106 )T  50.657 103 T  126.6F(a) Thot  Tcold  T  75  126.6  201.6F Thot  201.6F  L  18.0107 in. (b) a  Laa (T )  PLa Ea Aa  (18)(12.9 106 )(26.6)  (30.8 103 )(18)  10.712 103 in. (10.6 106 )(2.8) Lexact  18  10.712 103  18.0107 in.PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 159

0.02 in. 18 in. PROBLEM 2.59 14 in. 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.Bronze AluminumA ϭ 2.4 in2 A ϭ 2.8 in2E ϭ 15 ϫ 106 psi E ϭ 10.6 ϫ 106 psi␣ ϭ 12 ϫ 10–6/ЊF ␣ ϭ 12.9 ϫ 10–6/ЊFSOLUTIONThermal expansion if free of constraint: T  Lbb (T )  Laa (T )  (14)(12 106 )(180)  (18)(12.9 106 )(180)  72.036 103 in.Constrained expansion:   0.02 in.Shortening due to induced compressive force P: P  72.036 103  0.02  52.036 103in.But P  PLb  PLa   Lb  La  PEquating, Eb Ab Ea Aa  Eb Ab Ea Aa      14 )(2.4)  (10.6 18 )(2.8)  P  995.36 109P  (15 106 106    995.36 109P  52.036 103 P  52.279 103 lb(a) P  52.3 kips  b  9.91  103 in. (b) b  Lbb (T )  PLb Eb Ab  (14)(12 106 )(180)  (52.279 103 )(14)  9.91103 in. (15 106 )(2.4)PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 160

0.5 mm PROBLEM 2.60 300 mm 250 mm 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 reachedAB 140C, determine (a) the normal stress in the aluminum rod, (b) the change in length of the aluminum rod.Aluminum Stainless steelA 5 2000 mm2 A 5 800 mm2E 5 75 GPa E 5 190 GPaa 5 23 3 10–6/8C a 5 17.3 3 10–6/8CSOLUTION T  140  20  120CFree thermal expansion: T  Laa (T )  Lss (T )  (0.300)(23106 )(120)  (0.250)(17.3 106 )(120)  1.347 103 mShortening due to P to meet constraint: P  1.347 103  0.5 103  0.847 103 m P  PLa  PLs   La  Ls  P Ea Aa Es As  Ea Aa Es As      (75 0.300 106 )  0.250 106 )  P  109 )(2000 (190 109 )(800     3.6447 109PEquating, 3.6447 109P  0.847 103 P  232.39 103 N(a) a  P   232.39 103  116.2 106 Pa  a  116.2 MPa  Aa 2000 106 a  0.363 mm (b) a  Laa (T )  PLa Ea Aa  (0.300)(23 106 )(120)  (232.39 103 )(0.300)  363  106 m (75 109 )(2000 106 )PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 161

P PROBLEM 2.61 A standard tension test is used to determine the properties of an experimental plastic. el85o-ning.a-tdioianmoefte0r.4r5odina. nadndita The test specimen is a is subjected to an 800-lb tensile force. Knowing that an decrease in diameter of 0.025 in.5.0 in. 5 in. diameter are observed in a 5-in. gage length, determine the modulus of elasticity, the modulus 8 of rigidity, and Poisson’s ratio for the material. P'SOLUTION A   d2    5 2  0.306796 in 2 4 4  8  P  800 lb y  P  800  2.6076 103 psi A 0.306796 y  y  0.45  0.090 L 5.0 x  x  0.025  0.040 d 0.625 E  y  2.6076 103  28.973103 psi E  29.0 103 psi  y 0.090 v  0.444  v  x  0.040  0.44444   10.03103 psi  y 0.090   E v)  28.973  103  10.0291103 psi 2(1  (2)(1  0.44444)PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 162

640 kN PROBLEM 2.62 A 2-m length of an aluminum pipe of 240-mm outer diameter and 10-mm wall thickness is used as a short column to carry a 640-kN centric axial load. Knowing that E  73 GPa and v  0.33, determine (a) the change in length of the pipe, (b) the change in its outer diameter, (c) the change in its wall thickness. 2mSOLUTION do  0.240 t  0.010 L  2.0 di  do  2t  0.240  2(0.010)  0.220 m P  640 103 N  A  103 m2  4 do2  di2  4 (0.240  0.220)  7.2257(a)    PL   (640 103 )(2.0) EA (73109 )(7.2257 103 )  2.4267 103 m   2.43 mm      2.4267  1.21335 103 L 2.0  LAT  v  (0.33)(1.21335 103 )  4.0041104(b) do  do LAT  (240 mm)(4.0041104 )  9.6098 102 mm do  0.0961 mm  t  t LAT  (10 mm)(4.0041104 )  4.0041103 mm t  0.00400 mm PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 163

200 kN 4 200 kN PROBLEM 2.63 10 150 mm A line of slope 4:10 has been scribed on a cold-rolled yellow-brass 200 mm plate, 150 mm wide and 6 mm thick. Knowing that E  105 GPa and v  0.34, determine the slope of the line when the plate is subjected to a 200-kN centric axial load as shown.SOLUTION A  (0.150)(0.006)  0.9 103 m2x  P  200 103  222.22 106 Pa A 0.9 103 x  x  222.22 106  2.1164 103 E 105 109  y   x  (0.34)(2.1164 103 )  0.71958 103  tan   4(1   y ) 10(1   x )  4(1  0.71958 103 ) 10(1  2.1164 103 )  0.39887 tan   0.399 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 164

50 mm PROBLEM 2.642.75 kN 2.75 kN A 2.75-kN tensile load is applied to a test coupon A B made from 1.6-mm flat 12 mm steel plate (E  200 GPa, v  0.30). Determine the resulting change (a) in the 50-mm 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.SOLUTIONA  (1.6)(12)  19.20 mm2  19.20 106 m2P  2.75 103 Nx  P  2.75 103 A 19.20 106  143.229 106 Pax  x  143.229 106  716.15 106 E 200 109 y   z   x  (0.30)(716.15 106 )  214.84 106(a) L  0.050 m x  L x  (0.50)(716.15 106 )  35.808 106 m(b) w  0.012 m  y  w y  (0.012)(214.84 106 )  2.5781106 m 0.0358 mm  0.00258 mm (c) t  0.0016 m  z  t z  (0.0016)(214.84 106 )  343.74 109 m 0.000344 mm (d) A  w0 (1   y )t0 (1   z )  w0t0 (1   y   z   y z ) A0  w0t0 0.00825 mm2  A  A  A0  w0t0 ( y   z  negligible term)  2w0t0 y  (2)(0.012)(0.0016)(214.84 106 )  8.25 109 m2PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 165

PROBLEM 2.6575 kN 22-mm diameter 75 kN 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, 200 mm determine (a) the elongation of the rod in a 200-mm gage length, (b) the change in diameter of the rod.SOLUTION A   d2   (0.022)2  380.13  106 m2 P  75 kN  75  103 N 4 4   P  75  103  197.301  106 Pa A 380.13  106 x    197.301  106  986.51  106 E 200  109 x  L x  (200 mm)(986.51  106) (a) x  0.1973 mm   y  v x  (0.3)(986.51  106)  295.95  106   y  d y  (22 mm)(295.95  106) (b)  y  0.00651 mm PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 166

2.5 in. PROBLEM 2.66SOLUTION 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.  y  0.5  103 in. d  2.5 in. y  y   0.5  103  0.2  103 d 2.5 v  y : x   y  0.2  103  0.66667  103 x v 0.3  x  E x  (29  106)(0.66667  103)  19.3334  103 psi A   d2   (2.5)2  4.9087 in2 4 4 F   x A  (19.3334  103)(4.9087)  94.902  103 lb F  94.9 kips PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 167

A PROBLEM 2.67 B The brass rod AD is fitted with a jacket that is used to apply a 600 mm hydrostatic pressure of 48 MPa to the 240-mm portion BC of the rod. Knowing that E  105 GPa and v  0.33, determine (a) the change in 240 mm the total length AD, (b) the change in diameter at the middle of the rod. CD50 mmSOLUTION  x   z   p  48 106 Pa,  y  0 x  1 ( x  y  z ) E  1 48 106  (0.33)(0)  (0.33)(48 106 ) 105 109  306.29 106 y  1 ( x y  z ) E  105 1 (0.33)(48 106 )  0  (0.33)(48 106 ) 109  301.71106(a) Change in length: only portion BC is strained. L  240 mm  y  L y  (240)(301.71106 )  0.0724 mm  (b) Change in diameter: d  50 mm  x  z  d x  (50)(306.29 106 )  0.01531 mmPROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 168

y PROBLEM 2.68 3 in. A 4 in. A fabric used in air-inflated structures is subjected to a biaxial C loading that results in normal stresses  x  18 ksi and  z  24 ksi. D B Knowing that the properties of the fabric can be approximated asz ␴z ␴x x E  12.6 × 106 psi and v  0.34, determine the change in length of (a) side AB, (b) side BC, (c) diagonal AC.SOLUTION  x  18 ksi y 0  z  24 ksi x 1 ( x   y  z )  1 18, 000  (0.34)(24, 000)  780.95 106 E 12.6 106 z 1 ( x  y z)  1  (0.34)(18, 000)  24, 000  1.41905 103 E 12.6 106(a)  AB  ( AB) x  (4 in.)(780.95 106 )  0.0031238 in.(b) BC  (BC) z  (3 in.)(1.41905 103 )  0.0042572 in. 0.00312 in.  0.00426 in.  Label sides of right triangle ABC as a, b, c. Then c2  a2  b2 Obtain differentials by calculus. 2cdc  2ada  2bdb dc  a da  b db c c But a  4 in. b  3 in. c  42  32  5 in. da   AB  0.0031238 in. db  BC  0.0042572 in.(c)  AC  dc  4 (0.0031238)  3 (0.0042572) 5 5 0.00505 in. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 169

␴y ϭ 6 ksi PROBLEM 2.69 A B A 1-in. square was scribed on the side of a large steel pressure vessel.1 in. After pressurization the biaxial stress condition at the square is as shown. Knowing that E  29 × 106 psi and v  0.30, determine the ␴x ϭ 12 ksi change in length of (a) side AB, (b) side BC, (c) diagonal AC.DC 1 in.SOLUTION x  1 ( x  y )  1 12 103  (0.30)(6 103 ) E 29 106  351.72 106 y  1 ( y  x )  1 6 103  (0.30)(12 103) E 29 106  82.759 106(a)  AB  ( AB)0 x  (1.00)(351.72 106 )  352 106 in. (b) BC  (BC)0 y  (1.00)(82.759 106 )  82.8 106 in. (c) ( AC)  ( AB)2  (BC)2  ( AB0   x )2  (BC0   y )2   (1  351.72 106 )2  (1  82.759 106 )2  1.41452( AC)0  2 AC  ( AC)0  307 106or use calculus as follows: Label sides using a, b, and c as shown. c2  a2  b2 Obtain differentials. 2cdc  2ada  2bdcfrom which dc  a da  b dc c cBut a  100 in., b  1.00 in., c  2 in. da   AB  351.72 106 in., db  BC  82.8 106 in. AC  dc  1.00 (351.7 106 )  1.00 (82.8 106 ) 22  307 106 in.PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 170

PROBLEM 2.70 The block shown is made of a magnesium alloy, for which E  45 GPa and v  0.35. Knowing that  x  180 MPa, determine (a) the magnitude of  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)  y  0  y  0  z  0y  1 ( x  v y  v z ) E y  v x  (0.35)(180 106 )  63106 Pa  y  63.0 MPa  z  1 ( z  v x  v y )   v ( x y)   (0.35)(243  106 )  1.890 103 E E 45 109 x  1 ( x  v y  v Z )   x  v y   157.95 106  3.510 103 E E 45 109(b) A0  Lx Lz A  Lx (1   x )Lz (1   z )  Lx Lz (1   x   z   x z )  A  A  A0  Lx Lz ( x   z   x z )  Lx Lz ( x   z ) A  (100 mm)(25 mm)(3.510 103  1.890 103 )  A  4.05 mm2 (c) V0  Lx Ly Lz V  162.0 mm3  V  Lx (1   x )Ly (1   y )Lz (1   z )  Lx Ly Lz (1   x   y   z   x y   y z   z x   x y z ) V  V  V0  Lx Ly Lz ( x   y   z  small terms) V  (100)(40)(25)(3.510 103  0  1.890 103 )PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 171

y PROBLEM 2.71 A The homogeneous plate ABCD is subjected to a biaxial loading as shown. It is known that  z   0 and that the change in length of DB the plate in the x direction must be zero, that is,  x  0. Denoting x by E the modulus of elasticity and by v Poisson’s ratio, determinez ␴z C ␴x (a) the required magnitude of  x , (b) the ratio  0 /  z .SOLUTION  z   0 ,  y  0,  x  0 x  1 ( x  v y  v z )  1 ( x  v0 ) E E(a)  x  v 0 (b) z  1 (v x  v y z)  1 (v2 0  0  0)  1 v2 0 0 E  E E E z  1 v2PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 172

y PROBLEM 2.72P' ␴x P x For a member under axial loading, express the normal strain  in a z direction forming an angle of 45 with the axis of the load in terms of the ␴x ϭ P A axial strain x by (a) comparing the hypotenuses of the triangles shown in Fig. 2.43, which represent, respectively, an element before and after (a) deformation, (b) using the values of the corresponding stresses of  and x shown in Fig. 1.38, and the generalized Hooke’s law.P' ␴' ␴' P 45Њ ␶m ϭ P 2A ␶m ␴' ␴' ϭ P 2A (b)SOLUTION Figure 2.49(a) [ 2(1   )]2  (1   x )2  (1  v x )2 2(1  2    2 ) 1 2 x   2 1 2v x  v2 2 x x 4   2 2  2 x   2  2v x  v 2 2 x x Neglect squares as small. 4   2 x  2v x    1  v   2 x (A) (B)PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 173

PROBLEM 2.72 (Continued)(b)      v  E E  1  v  P E 2A  1 v  x 2E  1  v   2 xPROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 174

␴y PROBLEM 2.73 In many situations, it is known that the normal stress in a given direction is zero. For example,  z  0 in the case of the thin plate shown. For this case, which is ␴x known as plane stress, show that if the strains x and y have been determined experimentally, we can express  x ,  y , and  z as follows: x  E  x  v y y  E  y  v x z   1 v v ( x y) 1 v2 1 v2 SOLUTION z 0 x  1 ( x  v y ) (1) E (2)  y  1 (v x y)  E Multiplying (2) by v and adding to (1), x  v y  1 v2 x or x  E ( x  v y ) E 1 v2Multiplying (1) by v and adding to (2), y  v x  1 v2 y or y  E ( y  v x ) E 1 v2 z  1 (v x  v y )   v  1 E ( x  v y y  v x ) E E  v2   v(1  v) ( x  y)   1 v v ( x   y) 1 v2 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 175

PROBLEM 2.74 y In many situations, physical constraintsz (a) prevent strain from occurring in a given direction. For example,  z  0 in the case shown, where longitudinal movement of ␴y 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 x ␴x situation, which is known as plane strain, ␴z we can express  z ,  x , and  y as follows: (b)  z  v( x   y ) x  1 [(1  v2 ) x  v(1  v) y ] E y  1 [(1  v2 ) y  v(1  v) x ] ESOLUTION z  0  1 (v x  v y z) or  z  v( x   y )  E   x  1 ( x  v y  v z ) E  1 [ x  v y  v2 ( x   y )] E  1 [(1  v2 ) x  v(1  v) y] E y  1 (v x y  v z ) E  1 [v x y  v2 ( x   y )] E  1 [(1  v2 ) y  v(1  v) x ] EPROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 176

PROBLEM 2.753.2 in. 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.4.8 in.  2 in. P A  (3.2)(4.8)  15.36 in2SOLUTION P  55  103 lb    P  55  103  3580.7 psi A 15.36 G  150  103 psi     3580.7  23.871  103 G 150  103 h  2 in.   h  (2)(23.871  103)   0.0477 in.    47.7  103 in.PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 177

PROBLEM 2.76 3.2 in. What load P should be applied to the plate of Prob. 2.75 to produce a 1 -in.4.8 in. deflection? 16 PROBLEM 2.75 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.2 in. PSOLUTION   1 in.  0.0625 in. 16 h  2 in.     0.0625  0.03125 h 2 G  150  103 psi   G  (150  103)(0.03125)  4687.5 psi A  (3.2)(4.8)  15.36 in2 P   A  (4687.5)(15.36)  72.0  103 lb 72.0 kips PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 178

b aa PROBLEM 2.77 B A Two blocks of rubber with a modulus of rigidity G  12 MPa areP c 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.SOLUTIONShearing strain:     Shearing stress: a G a G  (12 106 Pa)(0.005 m)  0.0429 m a  42.9 mm   1.4 106 Pa b  160.7 mm    1 P  P 2 2bc A b P  45 103 N  0.1607 m 2c 2(0.1 m)(1.4 106 Pa)PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 179

b aa PROBLEM 2.78 B A Two blocks of rubber with a modulus of rigidity G  10 MPa are bondedP c 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.SOLUTIONShearing stress:   1 P  PShearing strain: 2 2bc A P  2bc  2(0.2 m)(0.125 m)(1.5 103 kPa) P  75.0 kN  a  40.0 mm       a G a G  (10 106 Pa)(0.006 m)  0.04 m  1.5 106 PaPROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 180

PROBLEM 2.79 An elastomeric bearing (G  130 psi) is used to support a bridge girder as shown to provide flexibility during earthquakes. The beam must not 83mina.xwimhuemn displace more than a 5-kip lateral load is applied as shown. Knowing that the allowable shearing stress is 60 psi, determine (a) the smallest allowable dimension b, (b) the smallest P required thickness a.a b 8 in.SOLUTION P  5 kips  5000 lbShearing force:Shearing stress:   60 psi   P, or A  P  5000  83.333 in2 A  60 and A  (8 in.)(b)(a) b  A  83.333  10.4166 in. b  10.42 in.  8 8 a  0.813 in.     60  461.54 103 rad  130(b) But    , or a  0.375 in. a  461.54 103PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 181

PROBLEM 2.80 For the elastomeric bearing in Prob. 2.79 with b  10 in. and a  1 in., determine the shearing modulus G and the shear stress  for a maximum lateral load P  5 kips and a maximum displacement   0.4 in. P ba 8 in.SOLUTION Shearing force: P  5 kips  5000 lb  Area: A  (8 in.)(10 in.)  80 in2   62.5 psi  G  156.3 psi  Shearing stress:   P  5000 A 80Shearing strain:     0.4 in.  0.400 radShearing modulus: a 1 in. G   62.5  0.400PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 182

P PROBLEM 2.81150 mm A 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 100 mm magnitude P  25 kN causes a deflection   1.5 mm of plate AB, determine the modulus of rigidity of the rubber used. B30 mm 30 mmSOLUTION F  1 P  1 (25  103 N)  12.5  103 N 2 2   F  (12.5  103 N)  833.33  103 Pa A (0.15 m)(0.1 m)   1.5  103 m h  0.03 m     1.5  103  0.05 h 0.03 G  833.33  103  16.67  106 Pa  0.05 G  16.67 MPa PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 183

P PROBLEM 2.82150 mm A 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 100 mm supports as shown. Denoting by P the magnitude of the force applied to the plate and by  the corresponding deflection, determine the effective spring constant, k  P/ , of the system. B30 mm 30 mmSOLUTION Shearing strain:    h Shearing stress:   G  G h Force: 1 P  A  GA or P  2GA 2 h h Effective spring constant: k  P  2GA  h with A  (0.15)(0.1)  0.015 m2 h  0.03 m k  2(19  106 Pa)(0.015 m2)  19.00  106 N/m 0.03 m k  19.00  103 kN/m PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 184

PROBLEM 2.83A 6-in.-diameter solid steel sphere is lowered into the ocean to a point where the pressure is 7.1 ksi (about3 miles below the surface). Knowing that E  29 106 psi and v  0.30, determine (a) the decrease indiameter of the sphere, (b) the decrease in volume of the sphere, (c) the percent increase in the density of thesphere.SOLUTIONFor a solid sphere, V0   d03Likewise, 6   (6.00)3 6  113.097 in3 x y z  p  7.1103 psi x  1 ( x  v y  v z ) E   (1  2v) p   (0.4)(7.1103 ) E 29 106  97.93 106  y   z  97.93 106 e   x   y   z  293.79 106 d  588 106 in.  V  33.2 103 in3 (a) d  d0 x  (6.00)(97.93106 )  588 106 in.(b) V  V0e  (113.097)(293.79 106 )  33.2 103 in3 0.0294% (c) Let m  mass of sphere. m  constant. m  0V0  V  V0 (1  e)   0    1  V0 m e)  V0  1  1 1 e 1 0 0 (1  m   (1  e  e2  e3  ) 1  e  e2  e3    e  293.79 106   0 100%  (293.79 106 )(100%) 0PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 185

85 mm PROBLEM 2.84 sy 5 258 MPa (a) For the axial loading shown, determine the change in height and the E 5 105 GPa change in volume of the brass cylinder shown. (b) Solve part a, n 5 0.33 assuming that the loading is hydrostatic with  x   y   z  70 MPa.135 mmSOLUTION h0  135 mm  0.135 m A0  d02   (85)2  5.6745 103 mm2  5.6745 103 m2 4 4 V0  A0h0  766.06 103 mm3  766.06 106 m3(a)  x  0,  y  58 106 Pa,  z  0 y  1 (v x y  v z )  y   58 106  552.38 106 E E 105 109 h  h 0 y  (135 mm)(  552.38 106 ) h  0.0746 mm  V  143.9 mm3  e  1  2v ( y z)  (1  2v) y  (0.34)(58 106 )  187.81106 h  0.0306 mm  E E 105 109 x V  521 mm3 V  V0e  (766.06 103 mm3 )(187.81106 )(b)  x   y   z  70 106 Pa  x   y   z  210 106 Pa y  1 (v x y  v z )  1  2v   (0.34)(70 106 )  226.67 106 E E 105 109 y h  h 0 y  (135 mm)( 226.67 106 ) e  1  2v ( x   y  z)  (0.34)(210 106 )  680 106 E 105 109V  V0e  (766.06 103 mm3 )(680 106 )PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 186

11 kips 1 in. diameter PROBLEM 2.85* 11 kips Determine the dilatation e and the change in volume of the 8-in. length 8 in. of the rod shown if (a) the rod is made of steel with E  29 × 106 psi and v  0.30, (b) the rod is made of aluminum with E  10.6 × 106 psi and v  0.35.SOLUTION A   d2   (1)2  0.78540 in 2 4 4 P  11103 lb Stresses : x  P  11103  14.0056 103 psi A 0.78540 y z 0(a) Steel. E  29 106 psi v  0.30 x  1 ( x  v y  v z )  x  14.0056 103  482.95 106 E E 29 106 y  1 (v x y  v z )   v x  v x  (0.30)(482.95 106 ) E E  144.885 106 z  1 (v x  v y   z )   v x y  144.885 106 E E e   x   y   z  193.2 106   v  ve  Le  (0.78540)(8)(193.2 106 )  1.214 103 in3  (b) Aluminum. E  10.6 106 psi v  0.35 x  x  14.0056 103  1.32128 103 E 10.6 106  y  v x  (0.35)(1.32128 103 )  462.45 106  z   y  462.45 106 e   x   y   z  396 106 v  ve  Le  (0.78540)(8)(396 106 )  2.49 103 in3PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 187

PROBLEM 2.86Determine the change in volume of the 50-mm gage length segment AB in Prob. 2.64 (a) by computing thedilatation of the material, (b) by subtracting the original volume of portion AB from its final volume.PROBLEM 2.64 A 2.75-kN tensile load is applied to a test coupon made from 1.6-mm flat steel plate(E  200 GPa, v  0.30). Determine the resulting change (a) in the 50-mm gage length, (b) in the width ofportion AB of the test coupon, (c) in the thickness of portion AB, (d) in the cross-sectional area of portion AB. 2.75 kN 50 mm 2.75 kN AB 12 mmSOLUTION(a) A0  (12)(1.6)  19.2 mm2  19.2 106 m2Volume V0  L0 A0  (50)(19.2)  960 mm3x  P  2.75 103  143.229 106 Pa y z 0 A0 19.2 106x  1 ( x  y   z )  x  143.229 106  716.15 106 E E 200 109 y   z   x  (0.30)(716.15 103 )  214.84 106e   x   y   z  286.46 106  v  v0e  (960)(286.46 106 )  0.275 mm3 (b) From the solution to problem 2.64,  x  0.035808 mm  y  0.0025781  z  0.00034374 mmThe dimensions when under the 2.75-kN load areLength: L  L0  x  50  0.035808  50.035808 mmWidth: w  w0   y  12  0.0025781  11.997422 mmThickness: t  t0   z  1.6  0.00034374  1.599656 mmVolume: V  Lwt  (50.03581)(11.997422)(1.599656)  960.275 mm3  V  V  V0  960.275  960  0.275 mm3PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 188

P PROBLEM 2.87 R1 A vibration isolation support consists of a rod A of radius R1  10 mm and A R2 a tube B of inner radius R2  25 mm bonded to an 80-mm-long hollow rubber cylinder with a modulus of rigidity G  12 MPa. Determine theB 80 mm largest allowable force P that can be applied to rod A if its deflection is not to exceed 2.50 mm.SOLUTIONLet r be a radial coordinate. Over the hollow rubber cylinder, R1  r  R2.Shearing stress  acting on a cylindrical surface of radius r is   P  P A 2 rhThe shearing strain is     P G 2 GhrShearing deformation over radial length dr: d  dr d  dr  P dr 2 Gh rTotal deformation. R2 d P R2 dr R1 2 Gh R1 r      P ln r R2  P (ln R2  ln R1 ) 2 Gh R1 2 Gh  P ln R2 or P  2 Gh 2 Gh R1 ln(R2 / R1)Data: R1  10 mm  0.010 m, R2  25 mm  0.025 m, h  80 mm  0.080 m G  12 106 Pa   2.50 103 m P  (2 )(12 106 ) (0.080) (2.50 103 )  16.46 103 N 16.46 kN  ln (0.025/0.010)PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 189

A P PROBLEM 2.88B R1 R2 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 80 mm modulus of rigidity G  10.93 MPa. Determine the required value of the ratio R2/R1 if a 10-kN force P is to cause a 2-mm deflection of rod A.SOLUTIONLet r be a radial coordinate. Over the hollow rubber cylinder, R1  r  R2.Shearing stress  acting on a cylindrical surface of radius r is   P  P A 2 rhThe shearing strain is     P G 2 GhrShearing deformation over radial length dr: d  dr d   dr dr  P dr 2 Gh rTotal deformation. R2 d P R2 dr R1 2 Gh R1 r      P ln r R2  P (ln R2  ln R1) 2 Gh R1 2 Gh  P ln R2 2 Gh R1 ln R2  2 Gh  (2 ) (10.93106 ) (0.080) (0.002)  1.0988 R1 P 10.103 R2  exp (1.0988)  3.00 R2/R1  3.00  R1PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or postedon a website, in whole or part. 190