工程力学教学第二部分

解:作轴力图如下:400kN340kN70kN130kN(a)400270340kNFF/3F/3F/3(b)F2F/3F/37-13FF2F(c)3F2Fll/2lFFF/lF(d)F2FFN题7-1图2F2FFF(f)2FF(e)2FF/lF/l7-3(a)2m2m4m1.2mABC1mDq=10kN/m12FAFBFCxFCyFAAFN2CDq=10kN/m1FN2FN1DN20404104236.42.2iCMFkN120,5/240.7,ixNNFFFkN3(1)11640.710/35.4,115010NFAMPa3(2)22636.410/31.7115010NFAMPa解：求反力,FA=FB=40kN;由结点D的平衡:取一半(如图)分析:7-3(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3max619109502010BCMPa解:(a)FN=F=14kN,.(b)轴力图如图,∴,8kN11kN17kNABCD11δb22b0FF(a)7-4(b)kN1982（b)336121078100.54003010BCMPa3336121078100.530/4078100.53004010ABMPamax400BCMPa（c）,12kN50cmABC(c)50cm解:由7-3式,11229696112230.20.3()20010100100107010200200100.410NNFlFllFEAEA解得,F=1932kN.7-6200F钢铝300解:222232()()2BBCgAllFgAlFlgAllEAEAEAEA方向向下7-7lFFl240,3120,360ABCDFFkNFFkNFFkN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.B7-8AGECDF1m1m1m0.5m123211022402HFkN2/311.143/5HABFFkNABFA33611.1103610/411103.14/4ABFdm7-9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.7-10max1122NFFgAgA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7-118-1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)8-2TT123d/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8-3600Nm500Nm200Nm600Nm300Nmd=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:8-4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8-5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8-6124525159.551.19.,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)1.191.911.910.488-7P1P1P3P4P51.5m1.75m2.5m1.5m作扭矩图如图若改用变截面15003009.559.559.55.,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;P2P1BP3AC8-8若采用同一直径,d=d1=92.0mm8-9600BA300T2T1C33612max1230.1,801015.7161616TTdTTkNd得122()ACCBABACCBPPTTlTlGIGI12249()0.60.30.0140.1801032TTTrad215.23,10.47TkNTkN解:Mxmax=T1+T2,由8-16式,解得:由强度条件（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A-1。解:(a)(b)2005010015050251005017591.750(200150100)Cymm*5315050(91.725)5.002510zSmm4005025250501752100502502134.150(40025021002)Cymm*53210050(250134.1)11.5910zSmm解:（a）(b)50zyc150(b)10050505015050A-21501501005050(a)zycy(a)442425,(),064644264yzyzdddddIII44158,2370yzIcmIcm3421.410158391.312yIcm324101.42370(101.410.72558012zIcm0yzIA-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A-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a(b)zy(a)zyaazyy’z’αC(a)a60°60°zyy’z’αC(b)4,012yzyzaIII40,12yzyzaIII43,096yzyzaIII430,96yzyzaIIIA-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A-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A-8解:(a)33228450200150502005053.615050(96.425)12121.0210zImmyyc150z2005050(a)12121124222,222233SSFFkNMMkNm12122,2,2SSFFFMFlMFlFlFl202(),(),33ABFkNFkN126212MMkNm解:(a)(c)(b)9-12m2m21(a)212kN/m2Fll21(b)21Fl122026,6,33SSFkNFkN2m2m21(c)218kNm6kN4mFAFB75,44ABFqlFql212732424lMMqllqlql(),()66ABqlqlFF(),()22ABqlqlFF(d)(e)(f)ll1(d)2122qqFAFBll1(e)1qqFAFBqll(f)1l12ql2qFAFB11;0623SqlqlqlFM113;2222SqlqlqllFqlMlqlql