09年理工大作业材料力学

５．５当力Ｐ直接作用在跨长ｌ＝６ｍ的梁ＡＢ的中点时，梁内的最大正应力σ超过了许用值３０％。为了消除这种过载现象，配置了如图所示的辅助梁ＣＤ，试求此辅助梁应有的跨长ａ。2Pl412alP41zmaxmaxWMPl41301alP41％l133a５．６如图５．２９所示某施工支架上钢梁的计算简图。若钢梁是由两个槽钢组成，材料为３号钢，其许用应力［σ］＝１７０ＭＰａ，［τ］＝１００ＭＰａ。问在不计梁的自重时应选用多大的槽钢？73.02.11zmaxmaxWMMPa1703zcm05.24MPa170273.02.11W号槽钢选82５．９已知图示铸铁简支梁的Ｅ＝１２０ＧＰａ，许用拉应力［σｔ］＝３０ＭＰａ，许用压应力［σｃ］＝９０ＭＰａ。试求：（１）许可荷载Ｐ。（２）在许可荷载作用下，梁下边缘的总伸长量。2P4144222222zm1055.210050200502001212520050200501211505010050100121IzmaxmaxtI125MMPa30tkN4.122PzmaxmaxCI175MMPa90CkN262PkN4.122P５．９已知图示铸铁简支梁的Ｅ＝１２０ＧＰａ，许用拉应力［σｔ］＝３０ＭＰａ，许用压应力［σｃ］＝９０ＭＰａ。试求：（１）许可荷载Ｐ。（２）在许可荷载作用下，梁下边缘的总伸长量。2P41kN4.122PztEI125xMEx2kN4.122xMdxEI125xMdxdzmm25.010z10dxEI125xM2dx2５．１２一铸铁梁如图所示。已知材料的抗拉强度极限（σｂ）ｔ＝１５０ＭＰａ，抗压强度极限（σｂ）ｃ＝６３０ＭＰａ。试求此梁的安全系数。kNm8kNm1253601402001023060140100200102yc2323z2360140601401214720010200101212I47mm109.2z1cmaxcIyMC截面：533109.2101471012MPa8.60z2cmaxtIyM533109.210531012MPa9.21５．１２一铸铁梁如图所示。已知材料的抗拉强度极限（σｂ）ｔ＝１５０ＭＰａ，抗压强度极限（σｂ）ｃ＝６３０ＭＰａ。试求此梁的安全系数。kNm8kNm1253601402001023060140100200102yc2323z2360140601401214720010200101212I47mm109.2z2BmaxcIyMB截面：533109.21053108MPa6.14z1BmaxtIyM533109.210147108MPa5.40５．１２一铸铁梁如图所示。已知材料的抗拉强度极限（σｂ）ｔ＝１５０ＭＰａ，抗压强度极限（σｂ）ｃ＝６３０ＭＰａ。试求此梁的安全系数。z2BmaxcIyMB截面：533109.21053108MPa6.14z1BmaxtIyM533109.210147108MPa5.40z1cmaxcIyMC截面：533109.2101471012MPa8.60z2cmaxtIyM533109.210531012MPa9.215.40150nmaxttbt7.38.60630nmaxccbc36.1040kN040kN40kNmmaxmaxzM170MPaW3zW235.29cm3zNo.20a,W237cm选用*3Smaxz23zFS401033.2MPaId17.210710６．４试分别用积分法和叠加法求图示变截面梁的最大挠度和转角。６．５用叠加法求图示各梁截面Ａ的挠度和截面Ｂ的转角。设ＥＩ＝常量。3A1lF2y3EI2A1B1lP22EI2A2lPl2y2EIB2PllEIAA1A2yyyBB1B2xdx22Aqdxxdy3l4x48EIll2222AA00qdxxydy3l4x48EIBqdxlxxlxd6lEIl2B0qdxlxxlx6lEI0.50.50.5q0.5q６．６试按叠加法计算图示梁的θＣ和ｙＣ。设ＥＩ＝常量。C1yC12qaC2yC2６．１１试求图所示各超静定梁的支座反力，并作Ｑ、Ｍ图。BFB0BF12BP2af9a2a6EI23BBF3af3EIB14FP27BF13P2714Pa274Pa9BFBFB012BM2af2EI23BBF2af3EIB3MF4a3M4aM1M20601005075cos60262.5MPa06010050sin60221.65MPa06080508050cos602217.5MPa0608050sin60256.29MPa(0,40(-100,40(-50,01322101005040214.03MPa223010050402114MPa02040tan2010020019.3310122max010040264.03MPa(-20,45(45,4513221452012.545268MPa(12.5,0223452012.545243MPa02045tan2452020027.08110(-20,45(-45,4513221204532.5452(-32.5,0(-20,10(-50,10(-35,022220503510216.97MPa22320503510253MPa232020010tan2205020016.852022max205010218.03MPa(-80，30160,-30(40,02211608040302163.69MPa223160804030283.69MPa22max16080302123.69MPa02030tan2160802007.02110(100,-40(160,40(130,0221160100130402180MPa22316010013040280MPa22max16010040250MPa02040tan216010020026.57110(0,60(120,-60(60,022max120060284.85MPa221120060602144.85MPa22312006060224.85MPa02060tan2120020022.5110(38,28(114,4812(38,28(114,481222382811448cc86c2382856cR(38,28(114,4812c1cR142MPa2cR30MPa2075.3020029.7012７．４某点处的应力如图所示，设σα、τα、σｙ值为已知，试考虑如何根据已知数据直接作出应力圆。(σy,0(σα,τα(70,40(30,40(50,013222max703040244.72MPa22170305040294.72MPa35044.725.28MPa７．７对图示的梁进行试验时，测得梁上点Ａ处的应变为εｘ＝０．５×１０－３，εｙ＝１．６５×１０－４。若梁材料的弹性模量Ｅ＝２１０ＧＰａ，泊松比ν＝０．３，试求梁上点Ａ处的正应力σｘ和σｙ。xxy2E126.8MPa1yyx2E72.7MPa1７．９如图所示为一体积为１０ｍｍ×１０ｍｍ×１０ｍｍ的立方体铝块，放入宽度正好是１０ｍｍ的钢槽中。设在立方体顶面施加的压力Ｐ＝６ｋＮ，铝的横向变形系数ν＝０．３３，钢槽的变形可以忽略不计，试求铝块的三个主应力。xzyz0y2P600060MPaA0.01xxyz10Ex19.8MPa1z02x19.8MPa3y60MPa如图８．５所示的简支梁，已知其材料的许用应力［σ］＝１７０ＭＰａ，［τ］＝１００ＭＰａ。试为此梁选择工字钢的型号，并按第四强度理论进行强度校核。210kN210kN210kN208kN208kN210kN8kN8kN41.8kNm41.8kNm45kNmmaxmaxzMW*3maxzzQS21010Id16.9cm7.5mm165.7MPa3zW264.7cm选择22a工字钢*maxzzQSId试算法*zzI:S24.6cm取：选择28a工字钢*maxzzQS100.4MPaId如图８．５所示的简支梁，已知其材料的许用应力［σ］＝１７０ＭＰａ，［τ］＝１００ＭＰａ。试为此梁选择工字钢的型号，并按第四强度理论进行强度校核。210kN210kN210kN208kN208kN210kN8kN8kN41.8kNm41.8kNm45kNm*zz200kNS76.59MPaId考虑C、D截面腹板与翼板交界处：CDz41.8kNmy71.14MPaI22r43151MPa170MPa８．３如图８．７所示为钢轨与火车车轮接触点处的应力状态。已知σ１＝－６５０ＭＰａ，σ２＝－７００ＭＰａ，σ３＝－９００ＭＰａ，钢轨材料的许用应力［σ］＝２５０ＭＰａ。试用第三强度理论和第四强度理论分别校核接触点处材料的强度。r313650900250MPa250MPa222r41223311229MPa2250MPa８．５如图所示两端封闭的铸铁薄壁容器，已知其内径Ｄ＝１００ｍｍ，壁厚ｔ＝１０ｍｍ，承受的荷载为内压力ｐ＝５ＭＰａ和两端的轴向压力Ｐ＝１００ｋＮ，材料的许用拉应力［σｔ］＝４０ＭＰａ，横向变形系数ν＝０．２５。试用第二强度理论校核其强度。pDppDp19.3MPa4tA4tDtpD25MPa2tp5MPa125MPa25MPa319.3MPar212331MPat40MPa８．１０如图所示，在船舶螺旋桨轴的Ｆ—Ｆ截面上，由主机扭矩引起的剪应力τ＝１４．９ＭＰａ，由推力引起的压应力σｘ′＝－４．２ＭＰａ，由螺旋桨等重力引起的最大弯曲正应力σｘ″＝±２２ＭＰａ。试求