结构力学-朱慈勉-第6章课后答案全解

结构力学第6章习题答案6-1试确定图示结构的超静定次数。(a)(b)(c)(d)(e)(f)(g)所有结点均为全铰结点2次超静定6次超静定4次超静定3次超静定II去掉复铰，可减去2（4-1）=6个约束，沿I-I截面断开，减去三个约束，故为9次超静定沿图示各截面断开，为21次超静定III刚片I与大地组成静定结构，刚片II只需通过一根链杆和一个铰与I连接即可，故为4次超静定(h)6-2试回答：结构的超静定次数与力法基本结构的选择是否有关？力法方程有何物理意义？6-3试用力法计算图示超静定梁，并绘出M、FQ图。(a)解：上图=l1MpM01111pX其中：EIlllllllEIllllEI8114232332623232333211311EIlFllFllFEIlpppp81733232226323108178114313EIlFXEIlppFX211pMXMM11lFp61lFp61FP4×2aA2l3l3B2EIEIC题目有错误，为可变体系。+pFplF32X1=1M图pQXQQ11pF21pF21(b)解：基本结构为：l1M3ll2MlFp21pMlFp310022221211212111ppXXXXpMXMXMM2211pQXQXQQ22116-4试用力法计算图示结构，并绘其内力图。(a)l2l2l2lABCDEI=常数FP4×2al2EFQ图FP4×2aX1X2FP4×2a解：基本结构为：1MpM01111pXpMXMM11(b)解：基本结构为：EI=常数qACEDB4a2a4a4a20kN/m3m6m6mAEI1.75EIBCD20kN/mX1166810810计算1M，由对称性知，可考虑半结构。a211M计算pM：荷载分为对称和反对称。对称荷载时：aq22q2qa26qa26qa26qa反对称荷载时：aq22qa2q2qa2qa28qa28qa28qa22qa214qa22qa2qaX11122pM01111pXpMXMM116-5试用力法计算图示结构，并绘出M图。(a)解：基本结构为：1M2MpM用图乘法求出pp21221211,,,,0022221211212111ppXXXX(b)6m6m3m3mABCEI2EIEID11kNX1X211KN1121663311KN33解：基本结构为：1M2MpMMEIEI108662332332661103323326612EIEIEI1086623323326622EIEIp27003231806212362081632323180621121EIEIp5403231806212362081632323180621122EI=常数6m6m6mEDACB20kN/mX1X120kN/mX2X236336111118090150301505250540108027001082111XXEIXEIEIXEImKNMCA9035253180mKNMCB12035253180mKNMCD3056(c)解：基本结构为：1N1MpMEIIEEI5558293299233256633263111EIIEp144210310910923102566101111pX29.11XmKNMAC61.11029.196m3m5III10kN·m10kN·mEA=∞CABD5I12m10kN·m10kN·mX110kN·m11933910kN·m10kN·m1010mKNMDA13.61029.13mKNMDC87.329.13M(d)解：基本结构为：1M2MpMEIIEEI6.1112932992332566233263116m3m5IIIEA=∞DABE2I5ICEA=∞10kN/mFG3.873.876.136.131.611.6110kN/mX1X21339966140545EIIE2.256396256612EIIEIE4.5066226666256622EIEIIEEIp25.1721645632519454053405924532566433453311102p69.839.1704.502.25025.17212.256.111212121XXXEIXEIEIXEIXEImKNMAD49.24839.179405mKNMBF37.10439.17969.86mKNMFE17.5239.173mKNMCG14.5269.8617.52M49.24837.10414.526-6试用力法求解图示超静定桁架，并计算1、2杆的内力。设各杆的EA均相同。(a)(b)题6-6图6-7试用力法计算图示组合结构，求出链杆轴力并绘出M图。(a)aFPFPaaa121.5m2m2m1230kN解：基本结构为：PFl2lFpPF1MpMEIllklllEIlEAl272222262311EIlFlklFllFllFEIlppppp222263101111pXpFX721lFlFlFMpppA73272lFp73lFp72M(b)lllEIABCFP4×2akθ==12EIlEA==2EIl2aaaaABCDEFGqqaEAEI=常数EA=EI/a2116-8试利用对称性计算图示结构，并绘出M图。(a)解：原结构=+①②①中无弯矩。②取半结构：基本结构为：1MpMEIEI22433299921211pppFEIFEI22433292992111ppFXX410111116m6m9mABCEA=∞FP4×2a2EIEIEIDEFEA=∞2pF2pF2pF2pF2pF2pF2pFX112pFpF2999M图整体结构M图(b)(c)解：根据对称性，考虑1/4结构：q基本结构为：q1X128lq11MpMEIllEI2121111EIqlqllqllEIp121821823112221llABCDEI=常数qq3m4m5m4m60kNABCDEI=常数pF49pF49pF49pF49pF2901111pX1221qlXpMXMM11242ql242ql242ql122ql242ql242qlM(d)解：取1/4结构：q基本结构为：qX2X11l111M12MpMlllDEABEI=常数qqCF122ql122ql22lq22lq22lqEIlllEI332213211EIllEI212112212EIlllEI2311112122EIqllqllEIp8432311421EIqlqllEIp612311322221321242213361125062320823qlXqlXEIqlXEIlXEIlEIqlXEIlXEIl362ql362ql362ql362ql362qlM(e)9mABC50kN4×2aIDEF6m6mI2I2III92ql92ql92ql92ql92ql(f)(BEH杆弯曲刚度为2EI，其余各杆为EI)取1/2结构：=+①②②中弯矩为0。考虑①：反对称荷载作用下，取半结构如下：=+③④④中无弯矩。考虑③：弯矩图如下：aaa2a2aa4FPGDEFABCHIpF2pF2pFpFpFpFpFpFpFpFpFpFpFpFpFpFpFpFpF2pF2pF2pF2pF2pF2pF2pF2pF2pF2pF2pF2pFaFp2aFp2aFp2aFp2aFp2aFp2aFp2aFp2aFp2aFp2aFp2aFp2aFp2aFp2aFp2aFp2aFp2aFp2aFp2aFp2aFpaFpaFp(g)解：原结构=+①②①弯矩为0。反对称荷载下：基本结构为：X112a1MpMEIaaaaEI3832222211311FP4×2aaaaaEI=常数ADk=3EI4a3kBGCEF2pF2pF2pF2pF2pF2pF2pFaFp2aFp2EIaFaFaaFaEIapppp1252222631pppFXXEIaaEIFXEIakXX485341253811331311111M图如下：(h)6-9试回答：用力法求解超静定结构时应如何恰当地选取基本结构？6-10试绘出图示结构因支座移动产生的弯矩图。设各杆EI相同。(a)(b)题6-10图6-11试绘出图示结构因温度变化产生的M图。已知各杆截面为矩形，EI=常数，截面高度h=l/10，材料线膨胀系数为α。(a)(b)lllABCD+15℃-15℃-10℃+15℃+15℃+5℃4a4a4a3aABB′EI=常数CDllABC+25℃-15℃-10℃2l2l2lABDElCEI=常数4FP4×2alhllllACEBDFI2I2I2IIIIIIaFp485aFp485aFp247aFp247题6-11图6-12图示平面链杆系各杆l及EA均相同，杆AB的制作长度短了，现将其拉伸（在弹性范围内）拼装就位，试求该杆轴力和长度。题6-12图题6-13图6-13刚架各杆正交于结点，荷载垂直于结构平面，各杆为相同圆形截面，G=0.4E，试作弯矩图和扭矩图。6-14试求题6-11a所示结构铰B处两截面间的相对转角BΔ。6-15试判断下列超静定结构的弯矩图形是否正确，并说明理由。(a)(b)(c)(d)题6-15图6-16试求图示等截面半圆形两铰拱的支座水平推力，并画出M图。设EI=常数，并只考虑弯曲变形对位移的影响。题6-16图ABlBACDFPFPqFPqFPRRFPRABC