材料力学(柴国钟、梁利华)第6章

材料力学(柴国钟、梁利华)第6章6.1 用积分法求图示各梁的挠曲线方程及自由端的挠度和转角，设梁的刚度EI为常数。解：（a）Mx1Fx1,x10,ax1x22FABMx2F2ax2,x2a,2aaCaF由EIwxMx，可得当x10,a时，EIwx1Fx1，EIx1Fx12Fx13D1C1，EIwx1C1x126当x2a,2a时，EIwx2F2ax2，F4ax2x22C2，EIx2F6ax22x232C2x2D2EIwx26边界条件：当x10时，0w00，代入，即得C1D10连续性条件：当x1x2a时，1a2a，w1aw2a，代入Fa23Fa2C2C2Fa2；Fa35Fa3C2aD2D2Fa322663因此，梁的挠曲线方程为：wx1Fx13,x10,a6EIx236ax226a2x22a3wx2Fa,2a6EI,x2梁的转角为：x1Fx12,x10,a2EIx2Fx224ax22a2,x2a,2a2EI自由端的挠度和转角为：wBFa3Fa2EI，BEIMx1122ax1,x10,aqq3aAB（b）2Mx21q2ax22,x2a,2ax1x22aa由EIwxMx，可得当x10,a时，2EIwx11qa3a2x1，EIx11qax13ax1C1，22EIwx1129a2x1C1x1D112qax1当x2a,2a时，EIwx21q2ax22，EIx2qx212a26ax2x22C2，26EIwx2qx2224a28ax2x22C2x2D224边界条件：当x10时，0w00，代入，即得C1D10连续性条件：当x1x2a时，1a2a，w1aw2a，代入3317qa44qa37qaC2C2qa；7qa4C2x2D2D2qa24661224因此，梁的挠曲线方程为：wx1qax129a2x1,x10,a12EIwx2qx248ax2324a2x224a3x2a4a,2a24EI,x2梁的转角为：x1qax13ax1,x10,a2EIx2q12a2x26ax22x23a3,x2a,2a6EI自由端的挠度和转角为：wB41qa4，B7qa324EI6EI（c）Mx3q01q0lx6l由EIwxMx，可得AxlB1lx3，EIwxq0l61l4xC，EIxq024l5EIwx1q0llxCxD120边界条件：当x0时，0w00，代入，即得Cq0l3，Dq0l424120因此，梁的挠曲线方程为：3q0lx54xl55lEIwx120lEI梁的转角为：q0lx44lEIx24lEI自由端的挠度和转角为：q0l4， q0l3wB B30EI 24EId）Mx11qx12,x10,a2Mx2qax23qa2,x2a,2a2由EIwxMx，可得当x10,a时，

x2q qa2A x1 Ba aEIwx11qx12，EIx11qx13C1，EIwx11qx14C1x1D12624当x2a,2a时，EIwx2qax23qa2，EIx21qax223qa2x2C2222EIwx21qax233qa2x22C2x2D264边界条件：当x22a时，2aw2a0，代入，即得C2qa3，D21qa43连续性条件：当x1x2a时，1a2a，w1aw2a，代入，即得C11qa3，D15qa4624因此，梁的挠曲线方程为：qx144a3x15a4,x10,awx124EIqa2x239ax2212a2x24a3a,2awx212EI,x2梁的转角为：4qa3x13,x10,ax16EIqax22ax2a,x2a,2ax22EI自由端的挠度和转角为：5qa4，qa3wBB24EI6EI6.2 用积分法求图示各梁的挠曲线方程、截面A和截面B的转角、跨度中点C的挠度以及外伸端的挠度与转角，设梁的刚度EI为常数。

x2x1ABqa/4C3qa/4aa(a)解：（a）Mx1,x10,aqax14q2x225ax22a2a,2aMx24,x2由EIwxMx，可得当x10,a时，EIwx1qax1，EIx1qax12C1，EIwx1qax13C1x1D14824当x2a,2a时，22，EI322，EIwx2q2x25ax22ax2q4x215ax212ax2C2424EIwx2qx245ax236a2x22C2x2D224边界条件：当x10时，w00；当x22a时，w2a0，代入，即得D10，2aC2D20连续性条件：当x1x2a时，1a2a，w1aw2a，代入qa3C1C2，qa4aC1aC2D268334联立求解，可得：C17qa48，C2qa48，D2qa24。因此，梁的挠曲线方程为：5wx1qa2x137a2x1，wx2q2x2410ax2312a2x22a3x22a448EI48EI梁的转角为：qa6x127a2x148EI

q8x2330ax2224a2x2a3，x248EI截面A和截面B的转角为：A7qa3，B3qa3，Cqa348EI16EI48EI跨度中点C的挠度为：wC5qa448EIq02l2x3lx2x3x（b）Mx6lq0BA由EIwxMx，可得Cq0l/3lq0l/6EIwxq02l2x3lx2x3，(b)6lEIxq04l2x24lx3x4C，24lEIwxq020l2x315lx43x5D360lCx边界条件：当x10时，w00；当x2l时，wl0，代入，即得Cq0l3，D045因此，梁的挠曲线方程为：wxq020l2x315lx43x58l4x360EIl梁的转角为：xq04l2x24lx3x4q0l324EIl45EI截面A和截面B的转角为：Aq0l3，B7q0l345EI360EI跨度中点C的挠度为：5q0l4wC（c）2x2Mx1qax1x1,x10,2ax1qa2qq3a2ABMx2ax2,x22a,3aCDqa/22a5qa/2a6(c)由EIwxMx，可得当x10,2a时，EIwx1qax1x12，EIx1q3ax122x13C1，21234EIwx1q2ax1x1C1x1D124当x22a,3a时，EIwx2q3a2ax2，EIx2q6a2x2ax22C2，q9a2x22ax232EIwx2D26C2x2边界条件：当x10时，w00；当x22a时，w2a0，代入，即得D10，C10，14qa42aC2D203连续性条件：当x1x22a时，12a22a，代入联立求解，可得：C10，D10，C211qa3，D28qa4。33因此，梁的挠曲线方程为：wx1q2ax13x14，wx2q9a2x22ax2311qa3x28qa424EI6EI3EI3EI梁的转角为：x1q3ax122x13，x2q6a2x2ax2211qa312EI2EI3EI截面A和截面B的转角为：A0，B5qa36EI跨度中点qa4C的挠度为：wC24EI为：2qa4wB

；外伸端B的挠度（d）Mqax14a2,x10,2ax2qx1qa2x142ABq3ax2M,x22a,3aCx22qa/43qa/42aa(d)7由EIwxMx，可得当x10,2a时，EIwqax14a2，x14322qax112ax1C1x1D1EIwx124当x22a,3a时，q3ax22EIwx2，2q3a4x2C2x2D2EIwx224边界条件：当x10时，w0代入，即得

qax128a2x1C1，EIx18q3ax23，C2EIx260；当 x22a时，w2a0，D10，C15qa3，2aC2D2qa4624连续性条件：当x1x22a时，12a22a，代入C25qa36联立求解，可得：C15qa3，D10，C25qa3，D241qa4。6624因此，梁的挠曲线方程为：qax1312ax1220a2x1，wx2q3ax245qa3x241qa4wx124EI24EI6EI24EI梁的转角为：qax128a2x15qa3，x235qa3x1q3ax28EI6EI6EI6EI截面A和截面B的转角为：5qa35qa3A6EI，B6EI跨度中点C的挠度为：43qa8EI为：wB19qa4。24EI

；外伸端B的挠度6.3用叠加法求图示各梁截面A和截面B的挠度，以及截面C的转角，设梁的刚度EI为常数。82FFaqqa2BAAaaCaBaC(a)(b)qqa2qaqBCABACaaaaaa(c)(d)解：（a）梁截面A的挠度为2Fa32232FaFa2awAFBMAFBFBMAFawAwAwBBawA3EI2EIa2EI3EI梁截面B的挠度为wBwBFBwBMA2Fa3Faa2Fa33EI2EI6EI梁截面C的转角为2Fa2Fa22Fa2Fa2FBMAFBMAFa2aCCCBC2EIEIEIEIEI（b）梁截面A的挠度为42wAwAqwAMAq2aqa22a08EI2EI梁截面B的挠度为wBqMAqa2a28a224a2qa45qa4wBwB24EI2EI24EI梁截面C的转角为qMA8qa3qa22a2qa3CCC6EIEI3EI（c）梁截面A的挠度为5q2a4122wA2qa2aqa4384EI16EI12EI梁截面B的挠度为q2a31qa22aqa4qa4wBa2a3EI8EI8EI24EI梁截面C的转角为q2a31qa22aqa3qa3C224EI3EI6EI6EI9（d）梁截面A的挠度为2a4a2a231qa22a25qa4qaqa2a2wA12EIa48EI16EI24EI梁截面B的挠度为qa22aaqa2a21qa22aaqa4qa4qa413qa4wBa6EI16EI23EI8EI3EI8EI24EI梁截面C的转角为qa22aqa2a2qa37qa3C16EI6EI12EI3EI6.4用叠加法求图示梁截面A的挠度，设梁的刚度为常数。解：因为wAwA(a)wA(b)wA(c)，又wAwA(c)，q故AEIwA1wA(b)因为A2A(a)15q2a4(b)wAwA2384EI2

5q45qa42a2384EI48EI，故AA，又AA(b)(c)(c)5qa448EI

a a2EI EIa a

FA6.5用叠加法求图示变截面梁自由端A的挠度和转角。解：wAFa3Fa2Faa2FaaFa33Fa332EI22EIa2EIa2EI22EI3EIAFa2FaaFa25Fa222EI2EI2EI4EI6.6 图示结构中，拉杆AC的横截面面积为A2500mm2，弹性模量E1100GPa。梁BC的横截面为200mm 200mm的正方形，弹性模量E2200GPa。试用叠加法求梁的中间截面D的铅垂位移。

q=40kN/mBD1m 1m

A2mC10解：wD5qa4qal384E2I4E1A125402000440200020003842001032004410010325000.31250.160.4725mmqa6.7用叠加法求图示刚架截面A的水平位移和铅垂位移。设各个梁段的刚度EI为常数，拉压变形忽略不计。

EI aqaA解：qa3qa2aqaa33qa4uAEIa2EI6EI3EI4qavA8EIy

F6.8 滚轮沿简支梁移动时，要求滚 A轮恰好走一水平路径。试问需将l梁的轴线预先应弯成怎样的曲线？设梁的刚度常数。解：wxFlxxl2x2lx2Flxx26EIl3EI6.9桥式起重机的最大载