同济大学朱慈勉-结构力学-第9章超静定结构的实用计算方法与概念分析习题答案

9-11同济大学朱慈勉结构力学第9章超静定结构的实用计算方法与概念分析习题答案9-1试说出何为杆端转动刚度、弯矩分配系数和传递系数，为什么弯矩分配法一般只能用于无结点线位移的梁和刚架计算。9-2试用弯矩分配法计算图示梁和刚架，作出M图，并求刚结点B的转角φB。(a)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9067.545.940(b)解：设EI=9，则9m9m40kNC6mADBEEIEI2EI2EI3m3m20kN/m2m6m2m20kN/mABCEIEI40kN9-223,31,1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6061.27.23.65.473.8909-3试用弯矩分配法计算图示刚架，并作出M图。(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048-58124m4m8m2m6kN100kN·m8kN/m32kNBEIEIAC9-33分配传递050505512最后弯矩0103-31212.510331256.5(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-112.2215.57-15.4866.28-66.0560kN2m6m2m2m2m6m60kN40kN/mABCDE40kNEI=常数9-44112.2215.5766.2880(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15.6-15.6-62.4-31.272.48144.9636.24-36.2414.5-14.5-58-2911.623.25.8-5.82.32-2.32-9.28-4.643.70.93-0.9324kN/m4m5m5m3mABCDEIEIEI9-55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)EI1=∞3kN/m16m3EIk2EIEIEIABCDE4m6m4m2kN/mmkN20EI=常数ADCBE4m4m4m9-66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3.982.99-6.985-5-2.473.986.9852.992.47(f)解：截取对称结构为研究对象。0.5441/212/3323AAAAABABSEIEISEI同理可得：21,33BABB另AB6mA′B′4m2kN/mEI1.5EI1.5EIEI2kN/m9-77112AABBABBACCCCAB2/3041.33-0.890.15-0.104.491/3-62-0.44-0.05-4.49-3-2-0.440.05-4.51AAAAAB2/3022.67-0.440.29-0.050.034.501/3-61.330.150.02-4.50-3-1.33-0.15-0.02-4.50BABBBB4.494.514.504.504.50M图9-4试用弯矩分配法计算图示梁，并作出M图。设图a梁含无限刚性段；图b梁B支座处含转动弹簧，刚度系数为kθ=4i。(a)解：163i4iBBC4i6i283iBB43l4l4l43lEI1=∞MABCEIEI9-88733284411626411640,3163160)(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。CABkθiiM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4m2m2mC32kNABkθ4×2aii9-99328444244444BCiiiikikiθθθiθiθiθBABCBCBABCBC31241kikμμBCBA9-5试用弯矩分配法计算图示剪力静定刚架，并作出M图。(a)解：作出M图（在B处加刚臂）4.0,0,6.02,0,3BCBABDBCBABDiSSiS结点ABCE杆端ABBDBABCCBCEEC分配系数铰结0.600.4铰结固端弯矩0-2ql2-ql2/3-ql2/600分配传递021ql2/15014ql2/15-14ql2/150最后弯矩021ql2/15-2ql23ql2/5-33ql2/3000(b)解：提取左半部分分析10kN10kNABEFGC=+5kN5kN5kN5kN5kN5kN5kN5kN(a)(b)EI1=∞llllABCED2EI2EIEIq4m4m4m3m10kN10kNABCDEFGHEI=常数9-1010（a）图中结构不产生弯矩，（b）图中结构为反对称结构，因此可以取下半部分分析得：118111221241211112124141212/2419819124141414/25.1/3CBBAFBCBBACBEAFBABBAABEAABABAEEIEISEISSEISSEIEISEIEiS5kN5kNABEFC1/9-101.11-1.010.11-9.798/98.890.99.79ABAE1/11-10-1.111.01-0.110.01-10.22/112.020.022.048/118.080.088.16BABCBF9.7910.22.048.16M图9-6试回答：剪力分配法的适用范围如何？什么叫柱子的并联和串连？由并联和串连所构成的合成柱，其剪切刚度和剪切柔度应如何计算？9-7试用剪力分配法计算图示结构，并作出M图。(a)解：AB、CD、EF、GA均为并联结构。①首先转化结间荷载KNQKNqlQKNqlQFAGFBAFAB5.225.37835.628510kN/m10kN10m6kN/mEA=∞EA=∞EA=∞EIEI3EI3EIACEGBDFH9-1111固端弯矩：mKNqlMFAB1258223333243993lilEIlEIlEIlEIkkkkkGHEFCDAB并于是边柱和中柱的剪力分配系数为83,8121rr转化后的荷载为：37.5+22.5+10=70KN边柱和中柱的剪力分别为：KNrFKNrFQQ821070870702121边柱柱脚弯矩为：mKN5.21212510870中柱柱脚弯矩为：mKN5.262108210262.5262.5212.5MKNm图(b)解：同上题，边柱和中柱的剪力分配系数为83,8121rr转化结间荷载KNQFFE96.81041081032边柱和中柱的剪力分别为：mKNPMKNrFmKNMKNrFFFEQFEFQ8.1210028,36.396.82.31002810,12.196.8222121边柱柱脚弯矩为：mKN6.5512.1中柱CD柱脚弯矩为：mKN8.16536.3中柱EF柱脚弯矩为：mKN208.162.310kN8mEI1=∞EI1=∞EI1=∞10mEIEI3EI3EIACEGBDFH9-12125.65.616.816.829.635.7205.65.6MKNm图(c)解：R1530kN30kN151515151515(a)abcde当顶层横梁没有水平位移时，d、e、b、c并列R=45KNKNFFFFrrrrQeQdQcQbdecb5.74145KN3060(b)606030303030