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时间:2020-03-18
《【精品】材料力学习题解答.doc》由会员上传分享,免费在线阅读,更多相关内容在工程资料-天天文库。
1、F1NAB8kN_80cm2_F1NBC19kN^BC20cm2F1NCD2kN^CD~120cm2950MPa16.7MPa2-2(b)lOOMPa^CD习题解答(b)FbcWP—dx二%=950MPa2・3答:以3点为研究对象,山平而汇交力系的平衡条件Fab=97.14kNFbc=-12.l2kN%=137.5MPacrl{C=一12.IMP"2・7答:BC段轴力为Fn(x)=-2F-pgA(l+x)BC段变形为A/二f[-2F-/7gA(/+x)](irbc_AEA3-2FI__p^Al2_EA-4Fl-3pgAl?2EA2・9答:任一截面上轴力为F,由x_2bId—d.rA⑴牛
2、02+2疔得而积为心)工a+2疔工円一〃2)龙+竹244I伸长量为△/=f仃张bEA(x)AEtt[(£_d2)x+d2l]2_4FlE7idxd22・10答:对水塔XMa=o,100x1+400x1+^x2=0F3=-25MN工比=0,100+r2xV2/2=0F3=-100应=-141.4册工化=0,F}+F2xV2/2+F3+400=0F、=—5MNFn/A】<[crc],£>500〃伽2FN2/A2<[b」,A2>1414mm2FN3/A3<[cre]A3>2500mm22・12答:对CD杆,HMc=°Fabsin30"x2-Fx2.5=0,F=Fw/2.5对杆,Fnab/^A
3、B-[b]Fnab5[b]比B=I60x106x^-x0.032/4=113.04kNF<113.04jUV/2.5=45.216EN2・14答:山变形协调,Ac/A«=1/3=,/3,Fn=2FnJ3>[t]=70MPa2・17答:(一)剪切强度:r=—<[r]5A%挤压强度:200①OOO£_o®_160EAFn2•2dEA对AB杆,工Ma=O,片xd+化x3a—Fx2d=0即2F2/3xa+F2x3a-Fx2a=0F2=6F/11,F[=4F/11=FV1/A=4F/11Aer#?—Fn?/A=6F/1A2-16答:F120xl03t=——==84.9MPa2A兀nQ22x—x
4、0.034剪切强度不够。F120xl03”“6v=——==70xl062A令得d=0.033m①板处FW5^[%]=420kN6=字5[a]拉伸强度:F5、-F<(160-2r/)xl6xl0-6[cr]2=246.4kN故拉力[F2240kN(二)排列顺序相反时,剪切强度与挤压强度同前相同。拉伸强度:Fr.CTt=<[6]①板1-1截面(200-2t/)xl2xl0-6F<(200-26/)x12x10-6[6、=288kN3F/5bt=7①板2-2截面(200-3r/)xl2xl0-6F<5/3x(200-3J)xl2xl0^[a]I=400kN2F/5rnS=<[7、16x10'FW(160—3〃)x16x10一6Q]2=190.4kN故拉力[F]=190.4kN3-1答:1点处:rpi贬呼卓J.43MPa/P加厂/323册2点处:•••p=0/.rp=03点处:T八俨=47.15MPap3/P知4/32最大切应变了=曽=5.89x10"3-3答:M〃/2_16M加4/32加'M•d/216MfmaxJ?龙(/—(0.5d)°)0.9375加彳32.•.最大应力增加了rmax~fmaxxl00%=6.67%3-4答:x60兀max=35.56MPa1640兀=24MPa•••—=35.56MPa0=BA+®CB+0DC+Wed=Mab(ab+Mcb8、I(:b+Md」dc+MedIedGIpABGIpenGIpDCG]pED=0.004444+0.001481+0.0015+0.004=0.011433・5答:C和对B的扭转角©肚_MlBC_100N•m•a_a%=莎二=373.06m80GPamm32B相对A的扭转角久p_MlAB_100N-m-(0.9-6/)_0.9-aGIPQ2D^-5044490.94m80GPamm32.・.总扭转角(P=(Pbc+
5、-F<(160-2r/)xl6xl0-6[cr]2=246.4kN故拉力[F2240kN(二)排列顺序相反时,剪切强度与挤压强度同前相同。拉伸强度:Fr.CTt=<[6]①板1-1截面(200-2t/)xl2xl0-6F<(200-26/)x12x10-6[6、=288kN3F/5bt=7①板2-2截面(200-3r/)xl2xl0-6F<5/3x(200-3J)xl2xl0^[a]I=400kN2F/5rnS=<[7、16x10'FW(160—3〃)x16x10一6Q]2=190.4kN故拉力[F]=190.4kN3-1答:1点处:rpi贬呼卓J.43MPa/P加厂/323册2点处:•••p=0/.rp=03点处:T八俨=47.15MPap3/P知4/32最大切应变了=曽=5.89x10"3-3答:M〃/2_16M加4/32加'M•d/216MfmaxJ?龙(/—(0.5d)°)0.9375加彳32.•.最大应力增加了rmax~fmaxxl00%=6.67%3-4答:x60兀max=35.56MPa1640兀=24MPa•••—=35.56MPa0=BA+®CB+0DC+Wed=Mab(ab+Mcb8、I(:b+Md」dc+MedIedGIpABGIpenGIpDCG]pED=0.004444+0.001481+0.0015+0.004=0.011433・5答:C和对B的扭转角©肚_MlBC_100N•m•a_a%=莎二=373.06m80GPamm32B相对A的扭转角久p_MlAB_100N-m-(0.9-6/)_0.9-aGIPQ2D^-5044490.94m80GPamm32.・.总扭转角(P=(Pbc+
6、=288kN3F/5bt=7①板2-2截面(200-3r/)xl2xl0-6F<5/3x(200-3J)xl2xl0^[a]I=400kN2F/5rnS=<[7、16x10'FW(160—3〃)x16x10一6Q]2=190.4kN故拉力[F]=190.4kN3-1答:1点处:rpi贬呼卓J.43MPa/P加厂/323册2点处:•••p=0/.rp=03点处:T八俨=47.15MPap3/P知4/32最大切应变了=曽=5.89x10"3-3答:M〃/2_16M加4/32加'M•d/216MfmaxJ?龙(/—(0.5d)°)0.9375加彳32.•.最大应力增加了rmax~fmaxxl00%=6.67%3-4答:x60兀max=35.56MPa1640兀=24MPa•••—=35.56MPa0=BA+®CB+0DC+Wed=Mab(ab+Mcb8、I(:b+Md」dc+MedIedGIpABGIpenGIpDCG]pED=0.004444+0.001481+0.0015+0.004=0.011433・5答:C和对B的扭转角©肚_MlBC_100N•m•a_a%=莎二=373.06m80GPamm32B相对A的扭转角久p_MlAB_100N-m-(0.9-6/)_0.9-aGIPQ2D^-5044490.94m80GPamm32.・.总扭转角(P=(Pbc+
7、16x10'FW(160—3〃)x16x10一6Q]2=190.4kN故拉力[F]=190.4kN3-1答:1点处:rpi贬呼卓J.43MPa/P加厂/323册2点处:•••p=0/.rp=03点处:T八俨=47.15MPap3/P知4/32最大切应变了=曽=5.89x10"3-3答:M〃/2_16M加4/32加'M•d/216MfmaxJ?龙(/—(0.5d)°)0.9375加彳32.•.最大应力增加了rmax~fmaxxl00%=6.67%3-4答:x60兀max=35.56MPa1640兀=24MPa•••—=35.56MPa0=BA+®CB+0DC+Wed=Mab(ab+Mcb
8、I(:b+Md」dc+MedIedGIpABGIpenGIpDCG]pED=0.004444+0.001481+0.0015+0.004=0.011433・5答:C和对B的扭转角©肚_MlBC_100N•m•a_a%=莎二=373.06m80GPamm32B相对A的扭转角久p_MlAB_100N-m-(0.9-6/)_0.9-aGIPQ2D^-5044490.94m80GPamm32.・.总扭转角(P=(Pbc+
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