《材料力学》学习资料

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第一题（2010年3月5日前交该题作业）星期五1.21 Determine the smallest allowable cross-sectional areas of members BD, BE, and CE of the truss shown. The working stresses are 20 000 psi in tension and 12 000 psi in compression. (A reduced stress in compression is specified to reduce the danger of buckling.)SolutionThe free-body diagram of homogeneous BC in Fig.(b). The equilibrium equation are0243616,0=-⨯=∑Ay FP M, P=24(kips)=24000(lb)The free-body diagram of truss in Fig.(c). The equilibrium equation are0368)166488166448(316,0=⨯-+⨯++⨯+⨯=∑BD Ay E P P M ,P BD =－8.944(kips) (Compression)088,0=-=∑CE Ay BP P MP CE =24(kips) (Tension )01664436707.0,0=++--=∑BD BE Ay y P P P FP BE =-11.32(kN) (Compression)The normal stress of a member CE, DE and DF is)./(1200089442in lb A lbA P BDBD BD BD ≤==σ (Compression) A BD =0.745(in.2))./(12000113202in lb A lbA P BEBE BE BE ≤==σ (Compression)A BE =0.943(in.2))(2000024000psi A lbA P CECE CE CE ≤==σ (Tension ) A BE =1.2(in.2)1 .37 Compute the maximum force P that can be applied to the foot pedal. The l/4-in.-diameter pin at B is in single shear, and its working shear stress is 4000 psi. The cable attached at C has a diameter of 1/8 in. and a working normal stress of 20 000 psi.SolutionThe free-body diagram of bracket in Fig.(b). The equilibrium equation are010sin 26,0=-=∑οT P M B , P=0.05788T (a) 010cos ,0=+=∑οT P F Bx x , P Bx =-0.9848T 010sin ,0=--=∑οT P P FBy y, P By =0.2315TAccording to the normal stresses formula, we have )./(20000.125.0422in lb in T≤⨯=πσ T=245.4(lb) According to the shear stresses formula, we have)./(4000.25.02315.09848.04422222222in lb in T d P P A V By Bx ≤⨯+=+==ππτ T=194.1(lb)According to formula (a), we getP=0.05788T=0.05788×194.1=11.2(lb)第二题（2010年3月10日前交该题作业）星期三2.37 An initially rectangular element of a material is deformed into the shape shown in the figure. Find εx , εy , and γ for the element.SolutionAccording to the definition of the axial strain, we have:004.02.02.01992.0-=-=x ε 01.015.015.01515.0=-=y ε0157.09.07.156.16==-=οοογ2.69（g ） The bars AB, AC and AD are pinned together as shown in the figure. Calculate the axial force in the strut caused by the 10-kip load. For each steel bar. A = 0.3 in.2 and E = 29 x 106 psi. For the aluminum bar, A = 0.6 in.2 and E = 10 x 106 psi.Solution Equilibrium:)(101020cos 40cos 3lb P P P AD AC AB ⨯=⨯++⨯οοοο20sin 40sin ⨯=⨯AD AB P PCompatibilityAB x y δδδ=⨯-⨯οο40sin 40cosAD x y δδδ=⨯+⨯οο20sin 20cos y AC δδ=Hooke ’s law:)/.(1020.)(6.0)./(1010.)(12106226lb in P in in lb in P AC AC AC -⨯=⨯⨯⨯⨯=δ )/.(1001.18.)(3.0)./(102940cos /.)(12106226lb in P in in lb in P AB AB AB -⨯=⨯⨯⨯⨯=οδ )/.(1068.14.)(3.0)./(102920cos /.)(12106226lb in P in in lb in P AD AD AD-⨯=⨯⨯⨯⨯=οδ we getAB x AC P P 01.1840sin 40cos 20=⨯-⨯⨯οοδAD x AC P P 68.1420sin 20cos 20=⨯+⨯⨯οοδwe obtainAD AB AC P P P 436.9160.632.17+= and)(101020cos 40cos 3lb P P P AD AC AB ⨯=⨯++⨯οοοο20sin 40sin ⨯=⨯AD AB P P AD AB P P 5321.0=AD AD AC P P P 436.95321.0160.632.17+⨯=AD P 71.12=AC AD P P 362.1= 5321.0436.9160.632.17÷+=AB AB AC P P P AB P 89.23= AC AB P P 7249.0=So we obtainP AC =3.53×103 (lb) P AB =2.56×103 (lb) P AD =4.80×103 (lb)第三题（2010年3月15日前交该题作业）星期一3.27 The compound shaft, composed of steel, aluminum, and bronze segments, carries the two torques shown in the figure. If T C = 250 lb.ft, determine the maximum shear stress developed in each material. The moduli of rigidity for steel, aluminum, and bronze are 12 x 106 psi, 4 x 106 psi, and 6 x 106 psi, respectively.SolutionAccording to the angle of twist formula, we have032)500(32)750(32444=+-+-BroBro B DB AluAlu B CD SteSte B AC d G T L d G T L d G T L πππ016324)500(4112)750(6444=⨯⨯+⨯-⨯+⨯-⨯BB B T T T 0.5（T B -750）+0.0625(T B -500)+0.5T B =0 T B =382.4(lb.ft)According to the torsion formula, we have)(22466112)7504.382(163max psi Ste =⨯-⨯=πτ)(4.898212)5004.382(163max psi Alu =⨯-⨯=πτ )(233701124.382163max psi Ste =⨯⨯=πτ第四题（2010年3月19日前交该题作业的剪力图、弯矩图部分）星期五第四题（2010年3月24日前交该题作业的应力部分）星期三5.35 Determine the maximum tensile and compressive bending stresses in the beam shown.Solution ：1） FBD(support reactions at A and B)2） Shear-Moment Diagrams 3） Section Modulus Moment of inertiaI ＝100×106 (mm)44） Maximum Bending StressAt the top of section C it is in compression46610100130.105.12mmmmmm N I c M topC c ⨯⨯⨯==σ＝16.25 （Mpa ） At the bottom of section C it is in tension46610100200.105.12mmmmmm N I c M bot C t ⨯⨯⨯==σ＝25 （Mpa ） At the top of section B it is in tension46610100130.1012mmmmmm N I c M topB c ⨯⨯⨯==σ＝15.6 （Mpa ） At the bottom of section C it is in compression46610100200.1012mmmmmm N I c M bot B t ⨯⨯⨯==σ＝24 （Mpa ） So we get46610100130.105.12mmmmmm N I c M topC c ⨯⨯⨯==σ＝16.25 （Mpa ）46610100200.1012mmmmmm N I c M bot B t ⨯⨯⨯==σ＝24 （Mpa ）第五题（2010年3月29日前交该题作业）星期一5.68 For the beam shown, compute the shear stress at 1.0-in. vertical intervals on the cross section that carries the maximum shear force. Plot the results.Solution ：1)FBD(support reactions at B and C)2)Shear-Moment Diagrams 3)Section Modulus Moment of inertiaI ＝97.0in.4The first moment of this area at 1.0-in. vertical intervals on the cross section3112314in Q =⨯⨯=＝12 in.3 or3125.1525.115.2314in Q =⨯⨯+⨯⨯==15.125 in 3 4)Maximum shear Stresspsi in in in b I Q V 8.350.0.1.97.125.15225043max max =⨯⨯==τ=350.8psi Shear Stress at 1.0-in. vertical intervals on the cross sectionpsi in in in b I Q V 4.278.0.1.97.122250431max =⨯⨯==τ=278.4psi第六题（2010年4月12日前交该题作业）星期一6.72 Compute the value of EI δ at the overhanging end A of the beam, by superposition.According to deflection formulas for beams, we knowEIm N m EI m m m N L a LEI a w BC BwBC).(600)422()4(24)2()/(400)2(24222222=⨯-⨯⨯=-=θEIm N m BwBC AMBC ).(120023=⨯=θδ EI a M EI a M B B BMB 323)2(==θ EI m N m EI m m N m BMB AMB ).(213323)4().(80023=⨯⨯=⨯=θδ EIm N EI m m N EI a w AB AwAB ).(8008)2()/(4008344=⨯==δ EIm N AwABAMB AwBC A ).(17333=--=δδδδ7.44 The beam ABCD has four equally spaced supports. Find all the support reactions.SolutionAccording to slope formulas for beams, we know])())(3(2)3[(24)(3230L L L L EI L w B +⨯-⨯⨯-=δ ])2()()3[()3(6))(2(222L L L EI L L L R B --⨯⨯+0])()()3[()3(6))((222=--⨯⨯+L L L EIL L L R C])2()2)(3(2)3[(24)2(3230L L L L EI L w C +⨯-⨯⨯-=δ ])2(])2()3[(2)2(23[)3(6)2(3223L L L L L L LL EI L L R B --+-⨯⨯+ 0])()2()3[()3(6)2)((222=--⨯⨯+L L L EIL L L R C018718824220=++-C B R R L w 018818724220=++-C B R R L w30330Lw R R C B ==According to the equilibrium equation and according to symmetry ，A 、D supports have the same magnitude ， )3(0L w R R R R D C B A ⨯=+++ D A R R = we have30330L w R R C B ==, 520Lw R R D A ==第七题（2010年4月16日前交该题作业）星期五8.27 The cross sections of the members of the pin-jointed structure are 200-mm square. Find the maximum compressive stress in member BDE.SolutionThe maximum compressive stress in member BDE )(95.682.02.010001.582.02.010009062max MPa C =⨯⨯+⨯⨯⨯=σ第八题（2010年4月26日前交该题作业）星期一8.49 For the state of stress shown, determine the principal stresses and the principal directions. Show the results on a sketch of an element aligned with the principal directions.SolutionThe principal stresses222122xy yx y x τσσσσσσ+⎪⎪⎭⎫ ⎝⎛-±+=⎭⎬⎫ 228264264+⎪⎭⎫ ⎝⎛--±+-=which yields)(4.101ksi =σ )(4.82ksi -=σ The principal directions 6.158648222tan -=-=--⨯=-=yx xyσστθ The two solution areο99.572-=θ and οοο01.12218099.57=+- ο0.29-=θ and ο0.61Determine the angles 1θ (associated with 1σ) and the angles 2θ (associated with 2σ) θτθσσσσσ2sin 2cos 22'xy yx yx x +-++=)292sin(8)292cos(264264οο⨯-+⨯---++-=)(4.8ksi -=and θτθσσσσσ2sin 2cos 22'xy yx yx y ---+=)292sin(8)292cos(264264οο⨯--⨯----+-=)(4.10ksi =Therefore we conclude the angles 1θ=61.0o (associated with 1σ) and the angles ο292-=θ (associated with 2σ)第九题（2010年4月30日前交该题作业）星期五8.108 A shaft carries the loads shown in the figure. If the working shear stress is τw = 80 MPa, determine the smallest allowable diameter of the shaft. Neglect the weights of the pulleys and the shaft as well as the stress due to the transverse shear force.SolutionAccording the support, we know there is the largest bending moment occurs at C, the largest torque occurs in segment BC. Show in the figure.At C section, we haveM max =2.5kN ×0.6m=1.5(kN.m) T BC =0.3(kN.m)Therefore, the stress at the bottom of the section are3363max max )().(105.13232mm d mm N d M S M ⨯⨯⨯===ππσ 3363)().(103.01616mm d mm N d T J r T BC BC ⨯⨯⨯===ππτ Draw the Mohr ’s circleAppend(2)Figure (a) shows a reinforced concrete beam, where the cross-sectional area of the steel reinforcement is 19600 mm2. Using n = E st/E co= 8 and the working stresses of 12 MPa for concrete and 140 MPa for steel, determine the largest bending moment that the beam can carry.SolutionSuppose the tensile stress is zero and compressive stress is not zero in the concrete, the first moment of the transformed cross section about the neutral axis is zero. According to the maximum compressive stress in concrete and tensile stress in steel, the neutral axis is0)(2)()(221211=-⨯--⨯--⨯h d nA h h b b hhb st (b) 0)1670(1960082)280()9009820(2982022=-⨯⨯--⨯--⨯h h h 061152000026544004502=-+h h 013589337.58982=-+h h )(0.222mm h =The moment of inertia about the neutral axis221312131)(3))((3h h h nA h h b b h b I st -++---= (c) 23)2221670(19600832229820-⨯⨯+⨯=I49)(10)8.32881.35(mm I ⨯+=49)(106.364mm ⨯=The moment)(12106.3642229max MPa Mco ≤⨯⨯=σ ).(19708m kN M ≤)(140106.364)2221670(89MPa M st ≤⨯-⨯=σ).(4.4406m kN M ≤The beam carries a uniformly distributed load of intensity w o on a simply supported span 24 ft long. Determine the largest allowable value of w o2081l w M =20)30(814.4406⨯=w )/(16.390m kN w =)/(00.40m t w =第十题（2010年5月7日前交该题作业）星期五AppendThe 9-m-long concrete column is built in at its base and stayed by two beam at the top. Determine the largest axial load that can be carried. Use E =25 GPa and yp σ=20 MPa for concrete.SolutionDetermine the moment of inertia of the cross-sectional area about the z-axis and y-axis44123)(8432.1)(108432.112)2400(1600m mm mm mm I z =⨯=⨯=8192.0)(108192.012)1600(2400423=⨯=⨯=imm mm mm I yThe slenderness ratio of a 1600mm x 2400mm rectangle The least radius of gyration with z-axis)(48.0)(4.26.1)(8432.124m m m A I r z z =⨯== 5.3748.092=⨯=mmC zC The least radius of gyration with y-axis )(2133.0)(4.26.1)(8192.024m m m A I r yy =⨯==5.292133.097.0=⨯=mmC yCThe slenderness ratio157201000252222=⨯⨯==πσπypC ECFor the slenderness ratio zC yC C C , is less than C C , so that the concrete column is of intermediate length. These equations yield the factor of safety755.115785.3715785.37335883353333=⨯-⨯⨯+=-+=C zC C zC z C C C C N 736.115785.2915785.29335883353333=⨯-⨯⨯+=-+=C yC C yC y C C C C N and the working stress)(07.11755.12015725.371212222MPa N C C yp C zC zw =⨯⎥⎦⎤⎢⎣⎡⨯-=⎥⎥⎦⎤⎢⎢⎣⎡-=σσ )(32.11736.12015725.291212222MPa N C C yp C yC yw =⨯⎥⎦⎤⎢⎣⎡⨯-=⎥⎥⎦⎤⎢⎢⎣⎡-=σσ The largest allowable axial load thus becomes)(4338)(42512)(10512.424.26.11007.1166t kN N A P zw z ==⨯=⨯⨯⨯==σ )(4435)(43459)(10459.434.26.11032.1166t kN N A P yw y ==⨯=⨯⨯⨯==σ So we obtain axial load P )(4338t P P z ==。