2kNm+2单位:kN·m1单位:kN·m[单题3-1]画扭矩图1kNm2kNm1kNm21kNm2kNm3kNm312–+2mm3m[孙题3-6]等直圆杆,已知d=40mm,a=400mm,G=80GPa,DB=1°。试求:(1)最大切应力;(2)截面A相对于截面C的扭转角。解:1PCBCBPDCDCBDGIlTGIlTm2m3mABCDaa2am2m3m+2mm3mABCD1PCBCBPDCDCBDGIlTGIlT18012PPGIamGIam)mN(4.292PBABAPCBCBACGIlTGIlTaGImP3180∴33.27PGImam2m3mA[错误解法]BCD12PDBPDCBDGImlGIml122PPGIamGIma13PGIma[孙题3-16]EB段的强度和刚度可以不计算阶梯形圆轴,AE段为空心,外径D=140mm,内径d=100mm;BC段为实心,d=100mm。外力偶MA=18kNm,MB=32kNm,MC=14kN·m,已知[]=80MPa,[′]=1.2()/m,G=80GPa。试校核该轴的强度和刚度。MAMBMCAEBCdDd+18kNm–14kNm阶梯形圆轴,d1=40mm,d2=70mm,皮带轮3的输入功率P3=30kW,轮1的输出功率P1=13kW,转速n=200r/min,已知[]=60MPa,[′]=2()/m,G=80GPa。试校核轴的强度和刚度。[刘题3.8]BMe1Me2Me3ACD–6211432单位:Nm解:求出外力偶矩,画扭矩图。][MPa4.4911maxtWT[AC段]][m/77.118011pGITBMe1Me2Me3ACD–6211432CD段的强度和刚度可以不计算单位:Nm[DB段]][m/44.018022pGIT][MPa3.2122maxtWT[例1]MBMCABCd1d2已知:MB=5kN·m,MC=1.2kN·m,l1=0.8m,l2=0.5m,d1=60mm,d2=40mm,G=80GPa。求:1、画扭矩图;2、全杆的最大切应力;3、截面C的扭转角。l1l2解:MBMCABCd1d2l1l2xT3.81.2(kN·m)–解:(1)画扭矩图111maxtWTMPa6.89(2)AB段和BC段的最大切应力1660108.336222maxtWTMPa5.951640102.136全杆的最大切应力τmax=95.5MPaAB222111ppIGlTIGlT10271.1008800108.363610.5121008500102.153610)88.299.29(3(rad)1025(rad)32411dIP3260446mm1027.1(3)截面C的扭转角。32422dIP3240445mm1051.2已知变截面钢轴上的外力偶矩MB=1.8kN·m,MC=1.2kN·m,l1=0.75m,l2=0.5m,d1=75mm,d2=50mm,G=80GPa。试求最大切应力和最大相对扭转角。[北科大题3-6]MBMCABCd1d2l1l2解:(1)画扭矩图xT31.2(kN·m)–111maxtWTMPa36(2)AB段和BC段的最大切应力167510336222maxtWTMPa491650102.136全杆的最大切应力τmax=49MPaCB222111ppIGlTIGlT103.110087501036361014.61008500102.153610)2.129(3(rad)0212.0(rad)32411dIP3275446mm101.3(3)最大相对扭转角。32422dIP3250445mm1014.6216.1一钢轴,转速n=240r/min,输入功率Pk=44.1kW,已知[]=40MPa,[]=1()/m,G=80GPa。试按强度和刚度条件计算轴的直径。[北科大题3-6]解:求出外力偶矩m)(N6.75419549nPMktmaxWT][163dT3][16Td)mm(6.60][180pGITmm8.59d