Solutions Manual For Advanced Mechanics Of Materials And .

Solutions Manual for Advanced Mechanics Of Materials And Applied Elasticity 5th Edition by UguralFull Download: ty-5th-editiSolutions Manual forAdvanced Mechanics of Materialsand Applied ElasticityFifth EditionAnsel C. UguralSaul K. FensterUpper Saddle River, NJ Boston Indianapolis San FranciscoNew York Toronto Montreal London Munich Paris MadridCapetown Sydney Tokyo Singapore Mexico CityFull download all chapters instantly please go to Solutions Manual, Test Bank site: TestBankLive.com

The author and publisher have taken care in the preparation of this book, but make noexpressed or implied warranty of any kind and assume no responsibility for errors oromissions No liability is assumed for incidental or consequential damages in connectionwith or arising out of the use of the information or programs contained herein.Visit us on the Web: InformIT.com/phCopyright 2012 Pearson Education, Inc.This work is protected by United States copyright laws and is provided solely for the useof instructors in teaching their courses and assessing student learning. Dissemination orsale of any part of this work (including on the World Wide Web) will destroy theintegrity of the work and is not permitted. The work and materials from it should neverbe made available to students except by instructors using the accompanying text in theirclasses. All recipients of this work are expected to abide by these restrictions and tohonor the intended pedagogical purposes and the needs of other instructors who rely onthese materials.ISBN-10:0-13-269049-7ISBN-13: 978-0-13-269049-2

CONTENTSChapter 1Analysis of Stress1–1Chapter 2Strain and Material Properties2–1Chapter 3Problem in Elasticity3–1Chapter 4Failure Criteria4–1Chapter 5Bending of Beams5–1Chapter 6Torsion of Prismatic Bars6–1Chapter 7Numerical Methods7–1Chapter 8Axisymmetrically Loaded Members8–1Chapter 9Beams on Elastic Foundations9–1Chapter 10Applications of Energy Methods10–1Chapter 11Stability of Columns11–1Chapter 12Plastic Behavior of Materials12–1Chapter 13Plates and Shells13–1

NOTES TO THE INSTRUCTORThe Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity,Fifth Edition supplements the study of stress and deformation analyses developed in thebook. The main objective of the manual is to provide efficient solutions for problemsdealing with variously loaded members. This manual can also serve to guide theinstructor in the assignments of problems, in grading these problems, and in preparinglecture materials as well as examination questions. Every effort has been made to have asolutions manual that can cut through the clutter and is as self-explanatory as possible,thus reducing the work on the instructor. It is written and class-tested by the author,Ansel Ugural.As indicated in the book’s Preface, the text is designed for the senior and/or firstyear graduate level courses in stress analysis. In order to accommodate courses of varyingemphasis, considerably more material has been presented in the book than can be coveredeffectively in a single three-credit course. The instructor has the choice of assigning avariety of problems in each chapter. Answers to selected problems are given at the end ofthe text. A description of the topics covered is given in the introduction of each chapterthroughout the text. It is hoped that the foregoing materials will help instructor inorganizing his or her course to best fit the needs of his or her students.Ansel C. UguralHolmdel, NJ

CHAPTER 1SOLUTION (1.1)We haveA 50 " 75 3.75(10 !3 ) m 2 , ! 90o " 40o 50o , and ! x P A .! 50o :! x ' 700(10 3 ) ! x cos 2 50o 0.413! x 110.18PEquations (1.8), withorP 6.35 kN" x ' y ' 560(10 3 )! x sin 50o cos 50o 0.492! x 131.2 PSolvingP 4.27 kN PallSOLUTION (1.2)Normal stress is"x PA 125(103 )0.05!0.05 50 MPa! 90o " 70o 20o :! x ' 50 cos 2 20o 44.15 MPa( a ) Equations (1.11), with" x ' y ' !50 sin 20o cos 20o !16.08 MPa! y ' 50 cos 2 ( 20o 90o ) 5.849 MPa5.849 MPay’44.15 MPa16.08 MPax’20 ox! 45o :! x ' 50 cos 2 45o 25 MPa( b ) Equations (1.11), with" x ' y ' !50 sin 45o cos 45o !25 MPa! y ' 50 cos 2 ( 45o 90o ) 25 MPa25 MPay’25 MPa25 MPax’45 oxSolutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, 2012 Pearson Education, Inc. 1–1

SOLUTION (1.3)From Eq. (1.11a),#x For# x'cos2 " !75cos2 30o !100 MPa! 50o , Eqs. (1.11) give then" x ' !100 cos 2 50o !41.32 MPa" x ' y ' !( !100) sin 50o cos 50o 49.24 MPaoSimilarly, for ! 140 :" x ' !100 cos 2 140o !58.68 MPa" x ' y ' !49.24 MPa58.68 MPa41.32 MPa50 o49.24 MPaSOLUTION (1.4)Refer to Fig. 1.6c. Equations (1.11) by substituting the double angle-trigonometric relations,or Eqs. (1.18) with ! y 0 and ! xy 0 , becomeor" x ' 12 " x 12 " x cos 2!and# x ' y ' 12 " x sin 2!20 and10 P2A(1 cos 2! )P2Asin 2!The foregoing lead to2 sin 2" ! cos 2" 1(a)By introducing trigonometric identities, Eq. (a) becomesThus,gives4 sin " cos " ! 2 cos 2 " 0 or tan ! 1 2 . Hence! 26.56o20 P2 (1300 ) (1 0.6)P 32.5 kNIt can be shown that use of Mohr’s circle yields readily the same result.SOLUTION (1.5)Equations (1.12):!1 ! maxP #150(103 ) #76.4 MPa"A2(50)4P 38.2 MPa2ASolutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, 2012 Pearson Education, Inc. 1–2

SOLUTION (1.6)Shaded transverse area:A 2at 2(10)(75) 1.5(103 ) mm 2Metal is capable of supporting the loadP ! A 90(106 )(1.5 #10"3 ) 135 kNApply Eqs. (1.11):P(cos 2 55o ) , P 114 kN"31.5(10 )P 12(106 ) "sin 55o cos 55o , P 38.3 kN"31.5(10 )! x ' 25(106 ) ! x' y'Thus,Pall 38.3 kNSOLUTION (1.7)Use Eqs. (1.11):P(cos 2 40o ) , P 51.1 kN31.5(10 )P 8(106 ) "sin 40o cos 40o , P 24.4 kN1.5(103 )! x ' 20(106 ) ! x' y'Thus,Pall 24.4 kNSOLUTION (1.8)A 15 ! 30 450 mm 2Apply Eqs. (1.11):120(103 )(cos 2 40o ) 156 MPa450 #10"6120(103 )sin 40o cos 40o "131 MPa "450 #10"6! x' ! x' y'SOLUTION (1.9)We have A 450(10!6) m2 .Use Eqs. (1.11):3"100(10 )(cos 2 60o ) "55.6 MPa"6450 #10"100(103 ) "sin 60o cos 60o 96.2 MPa"6450 #10! x' ! x' y'Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, 2012 Pearson Education, Inc. 1–3

SOLUTION (1.10)! 40o 90o 130o(103 )# x PA ! " ( 0150 !31.83 MPa.082 !0.07 2 )Equations (1.11):" x ' !31.83 cos 2 130o !13.15 MPa" x ' y ' 31.83 sin 130o cos 130o !15.67 MPax’13.15 MPa130 o15.67 MPaxPlane of weldy’SOLUTION (1.11)Use Eqs. (1.14),( 2 x ) ( !2 xy ) ( x ) Fx 0( ! y 2 ) ( !2 yz x ) (0) Fy 0( z ! 4 xy ) (0) ( !2 z ) Fz 0MN m 3 ):Fy ! x y 2 2 xzFx !3x 2 xySolving, we have (inFz 4 xy z(a)Substituting x -0.01 m, y 0.03 m, and z 0.06 m, Eqs. (a) yield the following valuesFx 29.4 kN m 3Fy 14.5 kN m 3Fz 58.8 kN m 3Resultant body force is thusF Fx2 Fy2 Fz2 67.32 kN m 3SOLUTION (1.12)Equations (1.14):" 2c1 y " 2c1 y 0 0 0,0 c3 z 0 0 0,0 0 0 0 04c1 y ! 0c3 z ! 0No. Eqs. (1.14) are not satisfied.Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, 2012 Pearson Education, Inc. 1–4

SOLUTION (1.13)( a ) No. Eqs. (1.14) are not satisfied.( b ) Yes. Eqs. (1.14) are satisfied.SOLUTION (1.14)Eqs. (1.14) for the given stress field yield:Fx Fy Fz 0SOLUTION (1.15)y! x ' y ' "A! x ' "Ay’40 !A cos20o50 !A cos20o20o20ox’x50 !A sin20o60 !A sin20o! x ' "A 40 cos 2 20o # 60"A sin 2 20o F 0: F 0 : ! x ' y ' "A # 40"A sin 20o cos 20ox'y'!2(50"A sin 20o cos 20o ) 0! x ' "35.32 7.02 32.14 3.84 MPa! x' y'!60"A sin 20o cos 20o ! 50"A cos 2 20o 50!A sin 2 20o 0 12.86 19.28 44.15 " 5.85 70.4 MPaSOLUTION (1.16)50 !A cos25oy’15 !A cos25o15 !A sin25o25o90 !A sin25o#Fx'! x ' y ' "A! x ' "A2o 0 : ! x ' "A 50"A cos 25x’!90"A sin 2 25o ! 2(15"A sin 25o cos 25o ) 0! x ' "41.7 16.07 11.49 "12.9 MPa(CONT.)Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, 2012 Pearson Education, Inc. 1–5

1.16 (CONT.) Fy' 0 : ! x ' y ' "A # 50"A sin 25o cos 25o!90"A sin 25o cos 25o ! 15"A cos 2 25o 15!A sin 2 25o 0! x ' y ' 19.15 34.47 12.32 " 2.68 63.3 MPaSOLUTION (1.17)yy’40 MPa! x' y'! x'20ox’! 20o! x50 MPa60 MPa11! x ' ("40 60) ("40 " 60) cos 40o 50sin 40o22 10 ! 38.3 32.1 !3.8 MPa1! x ' y ' " ("40 " 60) sin 40o 50 cos 40o2 32.14 38.3 70.4 MPaSOLUTION (1.18)! x'x’!y’! x' y'90 MPa25o15 MPax! 115o50 Mpa11! x ' (90 " 50) (90 50) cos 230o " 15sin 230o22 20 ! 45 11.5 !13.5 MPa1! x ' y ' " (90 50) sin 230o " 15cos 230o2 53.62 9.64 63.3 MPaSolutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, 2012 Pearson Education, Inc. 1–6

SOLUTION (1.19)! 40o to ! 0 . For convenience in computations, Let! x #160 MPa,! y #80 MPa," xy 40 MPa and ! "40oTransform fromThen11! x ' (! x ! y ) (! x ! y ) cos 2" # xy sin 2"2211 (!160 ! 80) (!160 80) cos(!80o ) 40sin(!80o )22 !138.6 MPa1! x ' y ' (" x " y ) sin 2# ! xy cos 2#21 ! (!160 80) sin(!80o ) 40 cos(!80o )2 !32.4 MPa! y ' ! x ! y " ! x ' "160 " 80 138.6 "101.4 MPaSoFor! 0o :y101.4 MPa32.4 MPa138.6 MPaxSOLUTION (1.20)4 53.1o345 90 45 " 90 cos106.2o! x' 22 67.5 6.28 73.8 MPa45 " 90sin106.2o 43.2 MPa! x' y' "2! tan "143.2 MPay’73.8 MPax’53.1oxSolutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, 2012 Pearson Education, Inc. 1–7

SOLUTION (1.21)" 70o! xy 0(a)" x ' y ' #30 #(b)! x ' 80 ! # 60sin140o2! 153.3 MPa! 60 ! " 60 cos140o22! 231 MPaSOLUTION (1.22)!(MPa)60 MPa20o(60, 50)1050 MParO#40 MPa(-40, -50)C" (MPa)40ox’50 39.8o6012r (60 502 ) 2 78.1! tan "1! x ' y ' sin 79.8o (78.1) 76.9 MPa! x ' cos 79.8o (78.1) "13.83 MPaSketch of results is as shown in solution of Prob. 1.15.SOLUTION (1.23)! (MPa)"’ 2050 MPar15 MPa25o(-50, -15)90 MPaxO62.1oy’#(90, 15)C" (MPa)("x’, ! x’y’)15! tan 12.1o7012r (15 702 ) 2 73.14! x ' y ' 73.14sin 62.1o 64.6 MPax’"1! x ' "73.14 cos 62.1o 20 !14.22 MPay’14.22 MPa25ox’64.6 MPa90 MPa15 MPax50 MPaSolutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, 2012 Pearson Education, Inc. 1–8

SOLUTION (1.24)! (MPa)! ' 67.5x’r 22.5r45 MPa904553.1oOxC90 MPa" (MPa)73.8ox’("x’, ! x’ y’ )106.2o! x ' y ' 22.5sin 73.8o 21.6 MPa! x ' 67.5 22.5cos 73.8o 73.8 MPaSketch of results is as shown in solution of Prob. 1.20.SOLUTION (1.25)! (MPa)(a)x’160 MPa140ox’O70o"60Crr "2(! " 60)" (MPa)x! # 60sin 40o ;! 153.3 MPa2! " 60! x ' 80 60 (1 " cos 40o )2! 231 MPa" x ' y ' #30 (b)SOLUTION (1.26)( a ) From Mohr’s circle, Fig. (a):# 1 121 MPao" p ' !19.3" s ' 25.7!(MPa) !max"# 2 !71 MPaA(100,60)2! p 'O2C" max 96 MPao"1" (MPa)B(CONT.)Figure (a)Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, 2012 Pearson Education, Inc. 1–9

1.26 (CONT.)By applying Eq. (1.20):502! 1, 2 or [22,4500 36000]2 25 961" 1 121 MPa" 2 !71 MPaUsing Eq. (1.19):tan 2" p ! 1215 !0.8" p ' !19.3o" s ' 25.7 o( b ) From Mohr’s circle, Fig. (b):# 1 200 MPa! p ' 26.55o!(MPa)"2O# 2 !50 MPa" max 125 MPa! s ' 71.55o!maxC2%p’"" (MPa)1A(150,-100)Figure (b)Through the use of Eq. (1.20),! 1, 2 75 [22,4500 10,000]2 75 1251or" 1 200 MPa" 2 !50 MPaUsing Eq. (1.19), tan 2! p 4 3 :! p ' 26.55o! s ' 71.55oSOLUTION (1.27)Referring to Mohr’s circle, Fig. 1.15:" x' " y' " x' y' " 1 " 22" 1 " 22# 1 # 22 " 1 #2" 2 cos 2!(a)# " 1 #2" 2 cos 2!sin 2!(b)From Eqs. (a),! x' ! y' ! 1 ! 2By usingcos 2 2! sin 2 2! 1 , and Eqs. (a) and (b), we have# x ' ! # y ' " x2' y ' # 1 ! # 2 const .Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, 2012 Pearson Education, Inc. 1–10