电力出版社--材料力学课后习题答案

解:作轴力图如下:400kN340kN70kN130kN(a)400270340kNFF/3F/3F/3(b)F2F/3F/32-13FF2F(c)3F2Fll/2lFFF/lF(d)F2FFN题7-1图2F2FFF(f)2FF(e)2FF/lF/l2-3(a)2m2m4m1.2mABC1mDq=10kN/m12FAFBFCxFCyFAAFN2CDq=10kN/m1FN2FN1DN10404104236.42.2iCMFkN120,5/240.7,ixFFFkN3(1)11636.410/35.4,115010FAMPa3(2)22640.710/31.7115010FAMPa解：求反力,FA=FB=40kN;由结点D的平衡:取一半(如图)分析:2-3(b)1m1m1m1mABECDq=3kN/m123AF1ECDq=3kN/mF2F1CF310,331.5/113.5AMFkN310,219.1ixFFFkN230,1/213.5iyFFFkN31(1)613(2)63(3)613.51015.98501013.51022.56001019.11038.250010FMPaAMPaMPa解：去掉(1)杆的约束(如图),用约束力F1代之,由由结点C的平衡3max6min1410350(2010)410FMPaA368102673010BCMPa解:(a)FN=F=10kN,.(b)轴力图如图,∴,AB17kN9kN16kNCD11δb22b0FF(a)2-4(b)kN81724（b)36241020012010CDMPa3617102138010ABMPa361210403010BCMPa3618104540010ABMPamax45BCMPa（c）,12kN50cmABC(c)50cm解:由7-3式,1122969611220.20.3()0.42001010010010701020020010NNFlFllFEAEA解得,F=1932kN.2-6200F钢铝300解:222232()()2BBCgAllFgAlFlgAllEAEAEAEA方向向下2-7lFFl240,3120,360ABCDFFkNFFkNFFkN39611401010.42001050010AADADFllmmEA39660100.50.210010150010CGlmm396201010.671010300010BElmm0.670.40.40.20.693GCGCClmm解:由平衡条件可求得ΔlADABΔlBECC,G点的竖直位移.B2-11AGECDF1m1m1m0.5m123F1F3F1F1F1FBFA211022402HFkN2/311.143/5HABFFkNABFA33611.1103610/411103.14/4ABFdm2-12解:水压力:AB杆的内力由强度条件7-7式故2m2m2mA3m4mB2mΔFABFHAF30°CDB0.5m2mFN2.52.52sin30NFFFNFA260.0317010448.02.52.5AFkN解:AB杆的轴力:由强度条件7-7式得所以容许荷载[F]=48kN.2-13max1122NFFgAgAmax2NFA33211632161016100.430.5730.081020100.4FgAlamgh解:最大轴力:由强度条件7-7式,式中A2=a2Fa3m0.4m所以2-14FDCABaa2-16aa12Δl1Δl2F1F2A12122,FaFallEAEA123ll1232FF解:变形协调的几何关系:物理关系得补充方程1252FFF1246,1111FFFF由平衡条件解得1246,1111FFAA所以3-1解:作扭矩图如图:2TTT6T2T1.5aaaa(a)T3T4T2T10Nm90Nm500Nm/m100100200(b)10100（Nm）aaaa3kNm2kNm4kNm1kNm(c)3512(kN.m)(d)aa2a1.5kNm1kNm1.52.5(kN.m)3-2TT123d/3331242106010/331.5,0,0.0632xPMMPaI33321047.20.0616xPMMPaW3maxmax347.20.59108010radG解:由8-5式,8-7式和8-2式,,maxmax3min500163.00.02516xPMMPaW500100300300Mx(N.m)解:作扭矩图如图;由3-7式3-3600Nm500Nm200Nm600Nm300Nmd=25d=100d=75d=751max316xMD2max34(1)16xMD42max1max1max1111001006.71空心圆轴:增加的百分比为解:原实心圆轴:3-4从直径为300mm的实心轴中镗出一个直径为150mm的通孔而成为空心轴，问最大切应力增大了百分之几?444()0.12180(1)3232ACABBCTalaDDG3-6解：由3-16式,把100N.m,G=80GPa,D=50mm,d=35mm,l=900mm代入上式,900CAaB100Nm可解得a=402mmlmx解:Mx=mxx,由3-15式,2440dd1632lxBAPlMxtxxtldGIGdG3-7124525159.551.19.,9.550.72.,20020030109.551.43.,9.550.48.200200TkNmTkNmTkNmTkNm33361.91107920101616xmaxMdmm331223344533661.19100.481067,79,50201020101616dmmddmmdmm解:求外力偶矩:由强度条件：Mx(kN.m)1.191.911.910.483-8P1P1P3P4P51.5m1.75m2.5m1.5m作扭矩图如图若改用变截面15002009.559.559.55.,9.555.73.500500xABxBCPMkNmMkNmn13max139.5510,89.016xABABPMdmmdW3max2325.7310,75.016BCdmmd314919.55101,92.0180801032xABABPMdmmdGI324925.7310,80.0801032BCdmmd解:计算扭矩由强度条件由刚度条件取d1=92.0mm,d2=80mm;P2P1BP3AC3-10若采用同一直径,d=d1=92.0mm3-11600CA300T2T1B33612max1230.1,801015.7161616TTdTTkNd得122()ACCBABACCBPPTTlTlGIGI12249()0.60.30.0140.1801032TTTrad125.23,10.47TkNTkN解:Mxmax=T1+T2,由8-16式,解得:由强度条件ACCB解：几何关系,代入几何关系，得由已知条件，TA=TB，得(),ABACCBPAPBTlaTaGIGI物理关系3-16lCAaBTd1d2PBABPAIaTTIla4211()dlad（b)(b)757525zyOz3050y(a)O222250(50)30(10030)0,38.75(5030)CCzymm1757575/3257575/2230,7575/22575Czmm17575(75/325)7525(25/2)2357575/22575)CymmI-1。解:(a)(b)20zy200(c)O2020801508012040010220150(15020)20,46.420400220150CCzymm(c)2005010015050251005017591.750(200150100)Cymm*5315050(91.725)5.002510zSmm4005025250501752100502502134.150(40025021002)Cymm*53210050(250134.1)11.5910zSmmI-2解:（a）(b)1501501505050(a)zyc50zyc150(b)10050505015050(a)442425,(),064644264yzyzdddddIII44158,2370yzIcmIcm3421.410158391.312yIcm324101.42370(101.410.72558012zIcm0yzII-3解:(a)(b)20a号工字钢:所以,zcd/2y(a)O(b)INO:20a100×14100×14yz442000553.7,53.2,5.8,1.71,18.51zyIcmIcmbcmzcmAcm2253.218.51(5.81.71)2553.72yzaII2.32acm44200245,33,6.8,14.3zyIcmIcmbcmAcm26.8233()14.3224522yzaII0.9acmI-4解:(a)14a号槽钢:,(b)10号工字钢:azyzay4200048.17,2.03,7.0,10.667zyIIcmzcmbcmAcm2414(48.172.0310.667)368.51zIcm242448.17(7.02.03)10.6671246.62zIcm12368.510.29561246.62zzIIA-5解:70×70×8等边角钢:(a)(b)所以(b)zy(a)zyaazyy’z’αC(a)a60°60°zyy’z’αC(b)4,012yzyzaIII40,12yzyzaIII43,096yzyzaIII430,96yzyzaIIII-6证明:(a)代入转轴公式,对于任意的角度,得(b)代入转轴公式,对于任意的角度,得I-7(d)zy(a)zy(b)zy(c)zy解：形心主轴的大致位置如图；对z轴的惯性矩最大。20050150150502596.450(200150)Cymm338420050501500.161101212yImmI-8解:(a)33228450200150502005053.615050(96.425)12121.0210zImmyyc150z2005050(a)12121124222,222233SSFFkNMMkNm12122,2,2SSFFFMFlMF