朱慈勉结构力学课后习题答案 第9章VIP免费

习题9-2解：设EI=6，则5.1,1BCABii53.05.13145.1347.05.131414BCBA结点ABC杆端ABBABC分配系数固端0.470.53绞支固端弯矩-6060-300分配传递-7.05-14.1-15.90最后弯矩-67.0545.9-45.90逆时针方向215.216005.6721609.4522131mKNEIEImMmMiABABBABAB(b)解：设EI=9，则3,31,1BEBDBCABiiii12.0141333331316.0141333331436.01413333333BCBABEBD结点ABC杆端ABBABCBDBE分配系数固端0.160.120.360.36绞支固端弯矩00045-900分配传递3.67.25.416.216.20最后弯矩3.67.25.461.2-73.80顺时针方向22.1606.32102.732131mKNEIEImMmMiABABBABAB9-3(a)解：Ｂ为角位移节点设EI=8,则1BCABii，5.0BCBA固端弯矩mKNlblPabMBA4882124432222mKNlMBC582621892结点力偶直接分配时不变号结点ABC杆端ABBABC分配系数铰接0.50.5固端弯矩048-5812分配传递050505512最后弯矩0103-312(b)解：存在B、C角位移结点设EI=6，则1CDBCABiii73741413145.0141414BCCBBCBA固端弯矩：mKNMMMmKNMmKNMCDCBBCBAAB14021808640080802结点ABC杆端ABBABCCBCD分配系数固结0.50.54/73/7固端弯矩-808000-140分配传递-20-40-40-2047.591.468.6-11.4-22.8-22.8-11.43.256.54.9-0.82-1.63-1.63-0.820.60.45最后弯矩-112.2215.57-15.4866.28-66.05(c)解：B、C为角位移结点51411,5441454414,51411CDCBBCBA固端弯矩：mKNMmKNMmKNMmKNMmKNMmKNMDCCDCBBCBAAB10065242003524501252450125241283424646424222222结点ABCD杆端ABBABCCBCD滑动分配系数滑动0.20.80.80.2-100固端弯矩64128-5050-200分配传递15.6-15.6-62.4-31.272.48144.9636.24-36.2414.5-14.5-58-2911.623.25.8-5.82.32-2.32-9.28-4.643.70.93-0.93最后弯矩96.4295.58-95.6157.02-157.03-142.9796.4295.58157.02147.97(d)解：11313141413114131414145.0141414DBDEDCCDCA固端弯矩：mKNMmKNMEDDE383812422结点ACDE杆端ACCACDDCDBDEED分配系数固结0.50.54/113/114/11固结固端弯矩00000-2.672.67分配传递-5-10-10-546/3392/3369/3392/3346/33-0.35-23/33-23/33-0.350.1270.0960.1270.064最后弯矩-5.35-10.7-9.3-2.442.190.254.125.3510.79.32.092.440.254.12(e)解：当D发生单位转角时：2414mEIKYC则）假设12(441mEIEIMDC73,74,3716,379,371216,12,16,9,12EBEDDEDADCDEEBDEDADCSSSSS结点DEB杆端DCDADEEDEBBE分配系数12/379/3716/374/73/7固结固端弯矩00-9900分配传递-2.57-5.14-3.86-1.933.752.815-2.5-0.72-1.43-1.07-0.540.230.180.310.16最后弯矩3.982.99-6.985-5-2.47(f)解：截取对称结构为研究对象。0.5441/212/3323AAAAABABSEIEISEI同理可得：21,33BABB另112AABBABBACCCCAAAAABBABBBBM图9-4(a)解：163i4iBBC4i6i283iBB733284411626411640,3163160)(4413343434343BAABBABABAlABlABCBBCBCBCBCCBllCBMMCiMSiliiMiliiMMMCiSiMMEIiiliiM其中结点ABC杆端ABBABCCB分配系数固结7/114/11铰结固端弯矩00分配传递3M/117M/114M/110最后弯矩3M/117M/114M/110311M711M411MM图(b解：首先在B点偏右作用一力矩，如图所示。M根据杆BC端，可得①4BABCBCkiM根据杆BA端，可得②4BABABCik由②式得：③4BABCθkθkθiθ将②式代入①式得：④44BABCiθiθM328444244444BC...