内科大材料力学习题解答

CC1AP2233BαlSSCXAYACA1P2233BαlΣMA=0SCcosα×l×tgα_P×l/2=0ΣX=0XA_SCcosα=0CSCC11N1XXAYAM2Q222N2AΣX=0XA+N2=03M3PXA3M3N33N33ΣX=0XA+N3=0 ABlcγdbaA’33332P24PRRR=50kNRRN11ΣX=0_R+N1=0N=50kN1RR2N2ΣX=0_R+40+N2=0N=10kN2N33ΣX=0_20_N3=0N3=_20kNxx-20N(KN)1N11N=012N24P2ΣX=0N2_4P=0N=4P2NN34PP33ΣX=0N+P_ΣX=03N=3P344P3PxN2.3.作用图示零件上的拉力P=382323PPPmax1CACAE2DBP=7.5kNG NC NCENE’AP=7.5kNGC—B—BNBENEDΣmD=0NE×1.5_NB×0.75=0N=NN=NBCEE∴N=6kNC22DC1BAXBBANN2DCN1BN=10KN1ΣY=0N2_N1_10=0N=20KN2A2.7.冷镦机的曲柄滑块机构如图所示。镦压工件时连杆接近水平位置，承受的镦压力P=1100kN。连杆的截面为矩形，高与宽之比为h/b=1.4。材料为45钢，许用应力为ABhBhbPPO。αCαABll工件DNNS1αS1BPS2PP=S1αααP[σ]1=7MPa；杆BC的横截面面积A2CC钢钢木BPAS1S2N1=Pctg30o=3PS1S2[P]=40.4kNDDCBAAA2A1A2A2/22A1k1N1=_20kNN2=40kN2.14在图示结构中，设CF为刚体(即CF的弯曲变形可以不计)，BC为铜杆，DF为钢杆，两BBA1A2PCxlFDNNBCPNNDFCFxlΣMC=0_P×x+NDF×l=0ΣMF=0P×(l_x)_NBC×l=0N(x)N(x)xqPqPPlN(x)=P+qx=P+ρgAxNmax=P+ρgAlaaCDaABPAPANABNADN’ABNBDBNBCNBDNBCN’AB2NBD=NA'B=P2BBBCAYACAYAChA解：(1)分析AB杆的受力，列平衡方程YB_YA_15=0Y∴NAC=0YNBC=15kN∴NAC=7kNYNBC=22kNAPAPBClNN1AP1N2B2N3C3Δl+Δl=2Δl2.31阶梯形钢杆的两端在T1=5oC时被固定，如图所示，杆件上、下两段的横截面面积分别AA1aaR2R1R1=R2)2aA1A1aA2aA2CACA’CDAB1lΣX=0(N2_N1)sinα=0ΣY=0(N1+N2)cosα_N3=0AA2A113δΔl1=(δ_Δl3)cosαPP/PP/2P/2dtPPbP拉杆拉杆2.41图示车床的传动光杆装有安全联轴器，过载时安全销即被剪断。已知安全销的平均直mm7.265×103)×0.02=145.3NmPPPbPPQ=PA=bhPbs=PAbs=bc2233T1T1x1:T1=2kN.mx22T2Σmx=0_2_T2=0:T2=_2kN.mTTxxmT1:T1=_4kN.m22T2T22xΣmx=03xx3=_2kN.mTxx_T21kN._T21kN.mDdT=m=57.29kN.mD2D2D1D1d2d1HAB解1）计算外力偶矩TC=701.9N.mCDBd2A1m1TTTL-xL-x传递的功率P=7.5kW，材料的许用剪应力[τ]=40MPa。试DD2d2d1T=m=716.2N.m πD(1_α4)≥TP2BP1AP3CTTd=84.6mmd=74.5mm(4)若AB和BC两段选用同一直径，则取d=d=84.6mmTOρdφdρφTOρdφdρφτρTOd解：(1)取微元dA，上面的切应力是τρ,则微力为τρdA：OOyMOαdRxROxlABT(x)=t(lx)m0.1txm0.1tx3.17.AB和CD两轴的B、C两端以凸缘相连接，A、D两端则都是固定的螺钉孔的中心线未能完全生命形成一个角度为的误差。当两个ψψmAbmAbABCDaMDmD_mA=0CCAB析EaaPDAB：L=LG=3GR=R'ABCDABCDEE(4)分配到AB和CD两杆上的受力分别为：3P/4BaBaM=qa2P=qaCaA1DB2211ACQ1P=qaqP=qaqBaCP=200N3D11ACB2M=qa2P=qa CD CDB八2A211Q1=P+qa=2qaCaCaBM=qa2P=qaM2RDARAP=200N3D11CB2RA=100NM1=_RA×0.2=_Q2=_RA=_100NM2=_RA×0.4=_40Nm=P=200NM2=_P×0.2=_40NmDB2RA200200200RB11ACΣMA=0ΣMB=0RB=2667NM2DBRBQ2=_RB+q×0.2=_0.667NM2=RB×0.2_(q×0.2)×0.1=333.4N.mM=M=qa2P=qaACDBRCRD221ΣMD=0RC×a+q×a×1.5a_M_P×a=0Q1=_qaM1=_qa×qa2AM2求内力RCM2=_RC×a_(q×a)(2)作剪力图和弯矩图；(3)判定lQlmax和lMlmax。22PM0=PaBC22PDPBaqqCBqqBqqCBqqqBCMM0A2M0BCPP6PCaB(f)qq=30kN/mP=20kNq=30kN/mCDEBAAAaCBqaaaAA/CaaA/ACaAAADaaAA/CaaA/ACaAAADaa22PCxMAARAM0=PaBΣY=0RA_2P=0RA=2PΣMA=0MA_2Pa+M0=0MA=PaQxMPa PaQmax=2PMmax=PaAACxBRBMBqΣY=0RB_qa=0RB=qaΣMB=0_MB+qa×1.5a=0xxQMMPP2PRBRB22RADABCaΣ030MRaPaPaBAΣ030MRaPaPaBAΣΣAB xxMxMxx22mCaABCaABRARBm5ΣMB=0_RA×2a+m+2m=0ΣY=0RA_RB=0xM xqqCARABRBΣMB=0_RA×aQxM2222x(f)RRCRBDaPCaBaAΣMB=0P×3a_RC×2a+6P×a=0ΣY=0RC+RB_P_6P=0PxQMMxPaC/C/RCAqqaRBaRBBΣMB=0_RC×axM2222xqq=30kN/mP=20kNq=30kN/mCDEBRCREARC=40kNΣY=0RE=40kNQxMxxQmax=30kNMmax=15kNmCCAqBMBRBqΣMB=0qa×a_qaa_MB=0BM=qa2BΣY=0RB_qa+qa=0RB=0QxQxMxMx2Qmax=qaMmax=qa2QxM2xQmax=qaMmax=qa2ddPPlARABRBxQxMMmaxMxxQxMMmaxMxxNNDBCAACACqBDDDRAHRAVRDABCMB=5kNΣX=01×4_RAH_1=0RAH=3kNΣY=0RAV_8+RD=0RAV=3kNNN’CM’CBCM’BN’BMCNCDDRDRAHARAVAAB杆ΣX=01×4_RAH_QB=0ΣY=0RAV_NB=0NB=3kNMB=4kN.mBC杆N'B=QB=1kNQ'B=NB=3kNM'B=MB=4kNmΣX=0N'B_NC=0NC=1kNQC=kNΣMC=0_M'B_Q'B×3+8×2_MC=0MC=3kN.mkMBMBNBBqqAMARADNCMCAMARAΣY=0RA_NC=0ΣMA=0MC_MA=0qqM’CDN’C2CC4CC4N=N'=qaMCC4CC4CCπ/4OPABANAANABRBHRBVπ/4OPΣX=0Pcosπ_RBH=0RBH=2PΣMB=0Psin×a_NA×2a=0NAPAC弧段BC弧段N(θ)Q(θ)AθM(AθAONAMPa_PasinooM(θ)=_0.437PRqqlMxxb≥277mmNo20aPNo20aPPCDBAMxhbhb[P]=57kNEBDAMxσ=σ=63.2MPamaxCmaxA-AAAP1=15.4kNMMxMA=308Nm6.12.图示横截面为⊥形的铸铁承受纯弯曲，材料的拉伸和压缩许用应力之比为MMMMy1zCCyCb6.13.⊥形截面铸铁梁如图所示。若铸铁的许用拉应力为[σt]=40MPa，许用压应力为荷P。PACB2PPACB2Ph1yzCMxx由弯矩图知：可能危险截面是A和C截面A截面的最大压应力A截面的最大拉应力[P]=44.2kN6.14.铸铁梁的载荷及截面尺寸如图所示。许用拉应力[σl]=40MPa，许用压应力[σc]=160zCu—zCu—yCABP=20kNDCMyxNo16CNo16CDBAMxxMxxQQmax=15kNMmax=20kNmAOAOBCDQPCOCOODRCQRDPDCxBRBARA由平衡方程求得A和B的约束反力R=50-6xR=10+6xAB(3)确定梁内发生最大弯矩时，起重机MC=98.27kNmMD=140.07kNmmaxMxQQ试求许可载荷PPlQmax=PMmax=PlBBRBlDaCRCAqaAPAPPBACqqBCPPBCPx2x1Axx1x2MAARABBaPaPllMMAARAPxRA(2)列AC段的弯矩方程M(x)=Px_Pax∈(0,a]EIv''=M(x)=Px_Pa(6)AC段的挠曲线方程和转角方程PPACJCJBBBC段保持为直线，则2EIB2EIBCPEIAPP2EIEIBCAx1x2MARARAM1(x1)=Px1_Plx∈(0,l/2]M2(x2)=Px2_Plx∈[l/2,l](x1)=Px1_PlEIv2'=M2(x2)=Px2_Pl=3Pl2D1=AABPM0=PLqP=qlBAP=P=qlBqAfA)PfA)ql3_2lPP=qaqBAAaqaBPBP=qaABABAAPlB解：(1)当起重机位于梁中央时，梁变形最qqACl/2PBl/2I=11100cm4q=52.717kg/m=516.6N/mPPAC5(3)计算AB变形引起B截面的位移(向下)7.14.滚轮沿等截面简支梁移动时，要求滚轮恰好走一水平路径，试问须将图示梁的轴线预yyPxlBxA解：(1)查表得水平简支梁受集中力作EECDBPAPEEACDBR’ERER’ECEAR’ECPa/2ECAECP/2P/2Pa/2CEACP/2AABmlmAmAxRABRBδδACBAAδCBBRCRARBCCBABCR’BBCR’BBRBAAARABRBMAxA截面上的弯矩Mmax=Mmax2=21.96kNmAA PPPBRBARB+0.00125=(P_RB)l3k3EIllAYARXBRYMAXA解：(1)对杆AB受力分析，可知AB杆发生压缩和弯曲变形，且XYR=RXY(2)AB杆的压缩变形大小(3)AB杆的弯曲变形大小Δlt_Δl=fB40(f)xyxytg2α0xyoσσ3σσ1τσ=_40MPaσ=_20MPaτ=_40MPaxyxyooσσ1σ3ττxyxyoσ3σ3σ1τ-84.74.7σxyxy00oo00σσ3σ1τσσxyτ=_20MPa0α=30oxyττ xyxyτxyxy2στσx=100MPaσy=50MPaτxy=0α=60oστσ2σ1px=2y=zMQ1234yx13=_120MPa11dδTodδToPPTAPσxyσσ1σ3解：(1)解：(1)A截面上的剪力和弯矩PhAh/4σσxW=875cm3I=15800czytg2α0__xy00τττyσ=80MPaτ=0σ=50MPaα=60oxxyαστyxyxzσ=30MPaσ=-20MPaτ=40MPaxyxyyyxAPPyyxz(2)x方向和y方向的线应变为零8.18.从钢构件内某一点的周围取出一部分如图所示。根据理论计算已经求得σ=30MPa，ττCAσxyxyσσrσtp1=500MPaσ2=420MPa=σδPPpPσσ’’σ’13=_13.64MPaσr1r21_μ(σ23)=22.7_0.25(0_13.64)=26.1MPa<]σt]aaM=P1aBxCIIa/2YCyAYAZCaaP1aZAIIP2IIazZZAYAP2P1QzIMyIMzITIΣY=0ΣZ=0ΣMx=0ΣMy=0ΣMz=0TI=_YA×a=_P1a/2MyI=ZA×2a=P2aMMyIIMxIIQyIIZCM=P1aNIIYCΣY=0ΣZ=0NII=ZC=P2/2ΣMx=0ΣMy=0MyII=ZC×a=P2a/2ΣMz=0AyzczCIIBYDYYαDαYαDαYMIEIINIzyNo18×2AzyNo18×2APBCYYCXCSAAPΣMC=0ΣX=0ΣMA=0SAsin30o×3.5_P×1.75=0SA=P=40kNXC_SAcos30o=0XC=SAcos30o=34.64kN2MxNN=34.64kNM=35kN.mmaxy9.5.单臂液压机架及其立柱的横截面尺寸如图所示。P=1600kN，材料的许用应力[σ]=160ycAycAbDBCPPMN900ycPN=P=1600kNM=P×(0.9+yc)=2256kNmPdPPPMNN=P=15kNM=0.4P=6kNm9.7.在力P和H联合作用下的短柱如图所示NBNBMzDQyMAyzMHBACDzPH=5kNP=25kNyM=25×103×0.025=625N.m(2)截面ABCD上的内力N=_P=_25kNM=M=625N.myMz=H×0.6=3kN.mA=0.15×0.1=0.015m2=_1.67×106+2.5×106+8×106=8.83MPaahbahbOP1 P2ybxz木梁在xy平面弯曲而引起的固定端截面上的最大木梁在xz平面弯曲而引起的固定端截面上的最[σ]3Mz3Mzmax+1.5Mymaxσ3=363PPPyzPzyP解：(1)将P力向y轴和z轴分解(2)画出梁在xz平面和xy平面内弯曲时的内力图xxxz平面内MyPzL/4MzPyL/4xy平面内xWy=141cm3Wz=21.2cm3PPPP0.180.18PPMMxTTxM=0.2PT=0.18PmaxP1=600NP2P1=600NP2DBACMM2DxP2sin80oCP2cos80oZCYCM1BYyAZAP1ΣMx=0ΣMz=0ΣY=0ΣMy=0ΣZ=0M1=M2120=0.295P2PYYCPZA_P1_ZC+P2sin80o=0ZA=4xx2.649N.m7xy平面内MzMMy71.28N.mxz平面内xxB截面是危险截面杆的强度足够.T1=2T2的啮合力Pn与齿轮节圆切线的夹角（压力T1=2T2AABQzz0.25T23T2xQPncos20oPnsin20oyTTxMMyxMzxBBzAM=M=802N.mBBzAB截面是危险截面MB=802.2NmTB=334.2Nmxz平面的弯矩图为Myxxy平面的弯矩图不变，B截面仍是危险截面T挺杆的横截面为圆形，两端可简化为铰支座，μ=12101024010E96λE96λππPllPPpABMPa。三杆均为两端铰支，长度分别为l1、l2和l3，且l1=2l2=4l3=5m。试求各杆的临界1crsddPlBAβCPP2BP2BP1 P2 P2PPtgθlBCABtgβEAB=EBCIAB=IBC欲使P为最大值，则两杆的压力同时达到各自的临tgθxzxzyIyI在xy平面和在xz平面内弯曲时的柔度值连杆容易在xz平面内失稳对于Q235钢sλa_σs1_3zyPDXAYRBDAA=32.837cm2i=2.09cmI=144cm4yyBD1BD杆为中长杆P杆AG的直径为d2=25mm。横梁ABC可视为刚体，规定的稳定安全系数nst=3，试求PDDABCGP解：(1)分析ABC受力ΣMB=0RBRAGRBRAGRADAPCB(RAD+RAG)×1-P×1=0RAD+RAG=PRAD×1.2RAG×1.2ADAGΔlADRAD×1.2RAG×1.2ADAGRAD=0.719PRAG=0.281P(2)杆AG受拉，考虑强度条件(3)杆AD受压，考虑稳定问题AD杆的柔度1P=82.3kN13.1.两根圆截面杆材料相同,尺寸dlPPdlPP2dd2dN(a)=PN(b)=PDDlBRBAYAXAPCΣX=0ΣMA=0P_XA=0XA=PRB×2l_P×l=0RB=P2MBCA/EIMBCA/EIdsEIdθARPBθOCCEIx1M/lx2AM/lMBPBM(θ)BRθROM(θ)=_PRsinθABCZABYACZA=ZB=0.5kNYA=YB=0.18kNMy(x)=ZAx=500xMz(x)=YAx=180xT(x)=80PPA。CBDPPACBPA。CBD