中国石油大学工程力学答案合集1.ppt

中国石油大学工程力学答案合集1静力学第静力学第1章章作业作业/P23-1.1(g)*:2023/4/162KylinsoftKylinsoft静力学第静力学第2章章作业作业/P58-2.3*:Solution:由此得到如下方程组：解得：2023/4/166KylinsoftKylinsoft静力学第静力学第2章章作业作业/P60-2.14*:Solution:按定义有:So,expressions(a),(d),(e)are valid.2023/4/167KylinsoftKylinsoft静力学第静力学第2章章作业作业/P61-2.20:Solution:(a)In vector form:2023/4/168KylinsoftKylinsoft(b)In vector form:2023/4/169KylinsoftKylinsoft+Solution:静力学第静力学第2章章作作业/P63-2.28*:2023/4/1610KylinsoftKylinsoft+（a)Solution:静力学第静力学第2章章作作业/P63-2.30*:2023/4/1611KylinsoftKylinsoft(b)+2023/4/1612KylinsoftKylinsoftOOriginal forces and couple can be reduced to a force and a couple acting at point O:静力学第静力学第2章章作业作业/P64-2.32:Solution:于是得到合力的大小、方向和作用线位置（如图所示）：(或,或)2023/4/1613KylinsoftKylinsoft+Solution:静力学第静力学第2章章作作业/P64-2.34*:2023/4/1614KylinsoftKylinsoft静力学第静力学第3章章作业作业/P85-3.2*:Solution:RAyRByFrom the FBD of the beam in Fig:+RBx3kN2m+2023/4/1615KylinsoftKylinsoft静力学第静力学第3章章作业作业/P86-3.11:Solution:GPGLStudying person and ladder,draw FBD(a)x=1.5m+RBRAyRAx+2023/4/1616KylinsoftKylinsoft+GPGLRBRAyRAx(b)2023/4/1617KylinsoftKylinsoft静力学第静力学第3章章作业作业/P87-3.18:Solution:300N0.5mRDRE(1)Studying entire structureFrom the FBD of entire structure in Fig:+(2)Studying member ABFrom the FBD of member AB in Fig:RBxRByRAxRAy300N0.5m1.6m+2023/4/1618KylinsoftKylinsoft(3)Studying member ACEFrom the FBD of member ACE in Fig:0.8mRERAxRAyAECRCxRCy0.8m0.6m0.6m+2023/4/1619KylinsoftKylinsoft静力学第静力学第3章章作业作业/P90-3.28(f)*:Solution:(1)Studying member CDFrom the FBD of member CD in Fig:显然，本题中在约束A、B、C、D处没有水平方向的作用力RCRMRCR2qC3m1m+2023/4/1620KylinsoftKylinsoft(2)Studying member ABCRARCRB2qAC1m2mRARCRB1mFrom the FBD of member CD in Fig:+2023/4/1621KylinsoftKylinsoft静力学第静力学第3章章作业作业/P91-3.40:Solution:(1)From the FBD of joint A in Fig:APFABFAD12+2023/4/1622KylinsoftKylinsoft(2)From the FBD of joint B in Fig:BPFBCFBDFAB+2023/4/1623KylinsoftKylinsoft(3)From the FBD of joint D in Fig:DFBD+FDEFADFCD1212122023/4/1624KylinsoftKylinsoft(4)From the FBD of joint C in Fig:CRCFCEFBCFCD12+2023/4/1625KylinsoftKylinsoft静力学第静力学第3章章作业作业/P92-3.43*:Solution:(1)From the FBD of joint D in Fig:DPFCDFDH+(1)From the FBD of joint in Fig:HFDHFGFC2323+2023/4/1626KylinsoftKylinsoftSolution:+静力学第静力学第4章章作作业/P111-4.2*:2023/4/1627KylinsoftKylinsoft静力学第静力学第4章章作业作业/P113-4.12*:Solution:RAzRAyRByRBxRCxRCzFrom the FBD of bent pipe and cable in Fig:2023/4/1628KylinsoftKylinsoft无解！？2023/4/1629KylinsoftKylinsoftSolution:静力学第静力学第4章章作作业/P113-4.13*:2023/4/1630KylinsoftKylinsoft静力学第静力学第4章章作业作业/P114-4.14(b):Solution:S令形心坐标为(xc,yc),按定义有:2023/4/1631KylinsoftKylinsoft静力学第静力学第4章章作业作业/P115-4.24:Solution:已知crucible的密度rc=2000kg/m3,外径R=0.28m,内径r=0.2m;iron的密度ri=7200kg/m3OStudying crucible，Drawing FBD in Fig.WiWcGiGcFyFx+2023/4/1632KylinsoftKylinsoftOWiWcGiGcFyFx2023/4/1633KylinsoftKylinsoftSolution:x=0 静力学第静力学第4章章作作业/P116-4.27*:2023/4/1634KylinsoftKylinsoftxyO2xyO1静力学第静力学第5章作业章作业/P131-5.2:解：3036lb20lb18in假设系统是静力平衡的先研究圆柱，有P1G1G2N1N2f2f1P1得再研究方块，有得而因，所以系统不能保持静力平衡2023/4/1635KylinsoftKylinsoftDiscussion:1.方块会不会先翻倒?3036lb20lb18in2.如果圆柱和方块的接触面不光滑,会怎样?xyO2G2N2f2P1x2x2=?2023/4/1636KylinsoftKylinsoftSolution:总体分析:分析木条:圆盘分析:解之得：解之得：NfffCBA.=22.06=达到最大静摩擦力时:解之得：解之得：静力学第静力学第5章作章作业/P132-5.10*:2023/4/1637KylinsoftKylinsoft9000N/mBCDBCD静力学第静力学第5章作业章作业/P132-5.12:解：P假设系统是静力平衡的先研究桩子CBD，有NBNBNAfAfBRxRy再研究杆AB，有P9000N/m1.5m2m3m0.8mABCDABfBAB2023/4/1638KylinsoftKylinsoft只有当fAfAmax或fBfBmax时，杆AB才能移动，由以上计算结果可知，当P6300(N)，杆AB在A点处开始移动2023/4/1639KylinsoftKylinsoft静力学第静力学第5章作章作业/P133-5.18*:Solution:因此因此:2023/4/1640KylinsoftKylinsoftSolution:木块翻转临界时：滑动时：静力学第静力学第5章作章作业/P133-5.19*:2023/4/1641KylinsoftKylinsoft当P=600 lb时：当P=200 lb时：Solution:静力学第静力学第5章作章作业/P134-5.22*:2023/4/1642KylinsoftKylinsoft第六次作业第六次作业/P34-1.1(b):SolutionPPPP332211PPPPPPPPPPPPN1N2N3N1=PN2=0N3=PNPP2023/4/1643KylinsoftKylinsoft第六次作业第六次作业/P34-1.6:SolutionPpd1DPpp6Pb研究活塞杆，有研究右挡盖，有得2023/4/1644KylinsoftKylinsoft第六次作业第六次作业/P34-1.7*:SolutionPBPsPwBSteelWoodP30AC研究节点B，有研究杆BC，有研究杆AB，有取2023/4/1645KylinsoftKylinsoft第六次作业第六次作业/P35-1.13*:SolutionBPPCASteelAluminumL1L21m根据题意列出方程组：解得总伸长为2023/4/1646KylinsoftKylinsoft第六次作业第六次作业/P37-1.14*:SolutionBCA(1)A formula for the downward displacement dC of point CPP(2)The elongation dB of the entire bar2023/4/1647KylinsoftKylinsoft(3)The ratio r of the elongation of the upper half of the bar to the elongation of the lower half of the barThe elongation of the upper half of the barThe elongation of the lower half of the barSo2023/4/1648KylinsoftKylinsoft第六次作业第六次作业/P37-1.21:SolutionE1E2eebbPPP1P1P2P2(1)静力平衡方程：研究右挡板，有(2)变形协调方程：PeP1P2O(3)求解方程组：2023/4/1649KylinsoftKylinsoftP第六次作业第六次作业/P37-1.23*:SolutionPB1A2CD(1)Equation of equilibrium BACD(2)Equation of compatibility(3)Force-displacement relations 得补充方程2023/4/1650KylinsoftKylinsoft解得有2023/4/1651KylinsoftKylinsoft第六次作业第六次作业/P37-1.24*:Solution 变形过程中3根杆的长度始终满足下面的关系：而3根杆的长度变化满足下面的关系：(1)静力平衡方程：研究节点A，有132P3030APAP1P3P2(2)变形协调方程：2023/4/1652KylinsoftKylinsoft(3)求解方程组：2023/4/1653KylinsoftKylinsoftRBRA第六次作业第六次作业/P37-1.26:SolutionA1A2(1)Equation of equilibrium(2)Equation of compatibility A1A2RBA1A2A1A2RB(3)Force-displacement relations 得补充方程2023/4/1655KylinsoftKylinsoft解得2023/4/1656KylinsoftKylinsoftT2T1材料力学第材料力学第2章章作业作业/P58-2.1(b):Solutionm3mmm3mT2mm2023/4/1657KylinsoftKylinsoft材料力学第材料力学第2章章作业作业/P59-2.11:Solution123ABC400500N1N2N3(1)123ABCm1m2m3对于AB段，由强度条件得 由刚度条件得 2023/4/1658KylinsoftKylinsoft取由强度条件得 由刚度条件得 对于BC段，取(2)取(3)齿轮1安放在齿轮2、3中间。123ABC400500N1N2N3123ABCm1m2m32023/4/1659KylinsoftKylinsoft材料力学第材料力学第2章章作业作业/P59-2.14:Solution方法一（传统法）：L/4L/4L/2AB3T0T01.解除某些约束，施加相应约束反力，使之成为静定基；2.扭矩与变形协调方程：得TB=5T0/43.外扭矩3T0和T0作用处的转角分别为：3T03T0T0TBT0TB=2023/4/1660KylinsoftKylinsoft材料力学第材料力学第3章章作业作业/P72-3.1(c):BqaqACaaqa21122Solutionqaqqa2V1M1对于1-1截面，有qaqqa2V2M2对于2-2截面，有2023/4/1662KylinsoftKylinsoft补充：内力图BqaqACaaqa2V-2023/4/1663KylinsoftKylinsoft(1)Shear-Force and Bending-Moment Equations 材料力学第材料力学第3章章作业作业/P73-3.2(a):Solution2PBACaaPa2PBACPaVM2PBACPaVM2023/4/1664KylinsoftKylinsoft(2)Shear-Force and Bending-Moment Diagrams 2PBACaaPaV-(3)Maximum Shear Forces and Bending Moments2023/4/1665KylinsoftKylinsoft材料力学第材料力学第3章章作业作业/P73-3.3(b)*:qABCL/2L/2SolutionxVMVMxxM控制截面法？V2023/4/1666KylinsoftKylinsoftSolution材料力学第材料力学第4章章作业作业/P90-4.6*:PPACDB2m2m2mNo20a1.求约束反力2.求弯矩作出弯矩图RARB3.求许用载荷M由强度条件得危险截面在C或D，有查表得2023/4/1667KylinsoftKylinsoft材料力学第材料力学第4章章作业作业/P91-4.8:SolutionA T-beam made of cast iron in pure bending.Knowing the ratio of tensile allowable stress to compressive allowable stress is st/sc=1/4.Determine the reasonable width b of the flange.bzC3060400MM2023/4/1668KylinsoftKylinsoft材料力学第材料力学第4章章作业作业/P91-4.13:Solution1.求剪力和弯矩作出剪力图和弯矩图V2.根据胶合面强度条件求P1mPAB505050100G整个截面都是危险截面，考虑剪应力互等定理，有M2023/4/1669KylinsoftKylinsoftV1mPAB505050100GM3.根据截面剪应力强度条件求P整个截面都是危险截面，有4.根据截面正应力强度条件求PA截面是危险截面，有5.综上所述，取2023/4/1670KylinsoftKylinsoftq0LBAyx材料力学第材料力学第5章章作业作业/P110-5.3(c)*:Solution1.求约束反力RARB2.求弯矩方程OARAV Mx2023/4/1671KylinsoftKylinsoft3.求挠曲线方程：代入边界条件：得：最大挠度：两端转角：中间挠度：取合理解2023/4/1672KylinsoftKylinsoft材料力学第材料力学第5章章作业作业/P111-5.8:SolutionThe cantilever beam shown in the figure has moments of inertia I1 and I2 in parts AC and CB,respectively.(a)Using the method of superposition,determine the deflection dB at the free end due to the load P.(b)Determine the ratio r of the deflection dB to the deflection d1 at the free end of the prismatic cantilever beam with moment of inertia I1 carrying the same load.BAL/2L/2C(a)BCA2023/4/1673KylinsoftKylinsoftBAL/2L/2CBACBCBCA2023/4/1674KylinsoftKylinsoftBAL/2L/2C(b)BAL2023/4/1675KylinsoftKylinsoftBAL/2L/2CDiscussionUse energy method2023/4/1676KylinsoftKylinsoft材料力学第材料力学第5章章作业作业/P111-5.9:SolutionBAqL/2L/2BAq1.解除某些约束，施加相应约束反力，使之成为静定基；2.变形协调方程：RB=BAqBARBfBufBd查表得2023/4/1677KylinsoftKylinsoft3.求其它约束反力：BAqRBRAMAL/2L/22023/4/1678KylinsoftKylinsoft材料力学第材料力学第5章章作业作业/P112-5.13*:SolutionThe bars 1 and 2 have the same tensile rigidity EA.(1)If the beam AB is considered as rigid,determine the internal forces in the two bars.(2)If the beam AB is considered as deformable with bending rigidity EI,determine the internal forces in the two bars.(1)the beam AB is considered as rigidPAB21CPBAC(a)Equation of equilibrium(b)Equation of compatibility 2023/4/1679KylinsoftKylinsoft(1)the beam AB is considered as rigidPAB21CPBAC(a)Equation of equilibrium(b)Equation of compatibility(c)Force-displacement relations 得补充方程解得2023/4/1680KylinsoftKylinsoft(2)the beam AB is considered as deformablePAB21CPBAC(a)Equation of equilibrium(b)Equation of compatibility(c)Force-displacement relations 得补充方程解得2023/4/1681KylinsoftKylinsoft1.危险点位于圆轴固定端横截面的最上处和最下处（A点和B点）材料力学第材料力学第6章章作业作业/P135-6.1(d):mPLdTMSolution:忽略横力弯曲剪力的影响2.用单元体表示危险点的应力状态V3.用应力圆表示危险点的应力状态(,)(-,)(0,-,-)B点点A点点ABA点：或B点：或2023/4/1682KylinsoftKylinsoft材料力学第材料力学第6章章作业作业/P135-6.3(d):402040如右图所示单元体，画出对应的应力圆Solutions st tt tmaxmax(-40,-40)(-20,40)1.主应力和主方向s s3 3s s1 111.2371.232.如图所示在单元体上画出主平面方向52.023.最大剪应力2023/4/1683KylinsoftKylinsoft材料力学第材料力学第6章章作业作业/P135-6.4(b)*:30305020如右图所示单元体，画出对应的应力圆Solutions st ts s1 1s s2 2t tmaxmax(30,-20)(50,20)60(52.32,-18.66)a a=302023/4/1684KylinsoftKylinsoft材料力学第材料力学第6章章作业作业/P136-6.7*:Solution30如右下图所示，在A点取单元体并进行应力状态分析a a=120A test of thin walled tube is shown in the figure.If the loads are a concentrated force P=20kN and a torque T=600Nm.The diameter of the tube is d=50mm,thickness is d=2mm.(1)Determine the normal and shear stresses on the inclined section at point A.(2)Determine the magnitude and orientation of the principal stresses at point A.(Draw stress element.)2023/4/1685KylinsoftKylinsoft(2)s st t(61.21,-70.63)(0,70.63)s s3 3s s1 166.5733.29(1)2023/4/1686KylinsoftKylinsoft材料力学第材料力学第6章章作作业/P136-6.11:SolutionxxyyWhen a train pass through a steel bridge,the strains tested by instrument are ex=0.0004,ey=-0.00012 at point A.Find the normal stresses in directions x and y at point A.(E=200GPa,m=0.3)2023/4/1687KylinsoftKylinsoft材料力学第材料力学第6章章作业作业/P137-6.17*:Solution薄壁上任意一点都是危险点，在薄壁上任一点A取单元体并进行应力状态分析AA thin-walled tube shown in the figure is made of cast iron.Knowing its outside diameter is D=200mm,thickness of the wall is t=15mm,the inner pressure is p=4MPa,the compressive force is P=200kN.The allowable tensile and compressive stresses of cast iron are st=30MPa and sc=120MPa,respectively.Exam the strength of the thin-walled tube.2023/4/1688KylinsoftKylinsoft由上可知由第一强度理论可知强度满足由第二强度理论可知强度满足由莫尔强度理论可知强度满足2023/4/1689KylinsoftKylinsoft材料力学第材料力学第6章章作业作业/P137-6.18:Solution1.求A点所在横截面的内力PV M2.在点A处取单元体并进行应力状态分析AA cantilever beam of rectangular cross section is subjected to a concentrated load P=120kN,acting at the free end.The beam has width b=100mm,and height h=200mm.Point A is located at distance c=0.5m from the free end and distance d=50mm from the bottom of the beam.Calculate the principal stresses and the maximum shear stress tmax at point A,and the equivalent stresses of third and fourth strength condition.2023/4/1690KylinsoftKylinsoft3.点A处主应力A4.点A处等效应力2023/4/1691KylinsoftKylinsoft材料力学第材料力学第7章章作业作业/P145-7.2:Solution1.求约束反力2.求-横截面的内力RVMN3.-横截面性质RA construct is shown in the figure.Find the maximum normal stress on section I-I and the stress at point A.2023/4/1692KylinsoftKylinsoft4.应力RVMN（压应力，在横截面上端）（拉应力，在横截面下端）2023/4/1693KylinsoftKylinsoft材料力学第材料力学第7章章作业作业/P146-7.10:Solution1.简化模型，作内力图P0.18P0.18PTMV0.18P0.2P0.5P0.5P2.由内力图可见，中间横截面为危险截面，忽略横力弯曲剪力的影响，在危险截面处为弯扭组合变形In figure there is a manual winch.The diameter of the steel shaft is d=30mm and the allowable stress is s=80MPa.Find the maximum load for the winch,according to the third strength condition.2023/4/1694KylinsoftKylinsoftPyxzyx材料力学第材料力学第7章章作业作业/P146-7.12*:Solution如右下图所示，力P在a、b点所在横截面中心处产生的内力作用为一主失和一主矩：OabzF Fx xF Fy yF Fz zM Mz zM Mx xM My y2023/4/1695KylinsoftKylinsoftyxOabzF Fx xF Fy yF Fz zM Mz zM Mx xM My ya点：a、b点的应力状态如右下图所示aaaxyb点：bbbxy2023/4/1696KylinsoftKylinsoft此此课件下件下载可自行可自行编辑修改，修改，仅供参考！供参考！感感谢您的支持，我您的支持，我们努力做得更好！努力做得更好！谢谢!