哈尔滨工程大学船舶结构力学课程习题集答案(5)

???l2AG??ftds?

????8b?tsin??????22AGt??82?b?t??2sicn8lf?4?10(弪）5?410?09?.5?b6?t?8?8sinos?8

100?9.56?84?29.8?cos?8?8?10?0.23.4题

.将剪流对内部任一点取矩

?215667f1rds??(f62731?f2)rds??f3)rds??32f2rds??f2rds??(f2?7843f3rds ???21562If1rds?f2??32673rds?f3??78437rds

?f1??rds?f2??rds?f3??rdsIIIII?2A1f1?2A2f2?2A3f3?Mt.........(1) 由于I区与II区，II区与III区扭率相等可得两补充方程

?f1ds???2GA1?t?11???f2tf21??ds??ds????2GAt?2?II?ds???26f1tds??73f3?ds?t??f3????ds?2GA3?IIIt?37?f2t2f2?f1?f3f?f23f?f2即：1??3.....(2)A1A2A3(1)(2)联立（注意到A1?A3，2A1?A2?a）??2A1?f1?2f2?f3??Mt?解得?3f1?f2?3f3?f2?1?3f1?f2?(?f1?4f2?f3)?2?????1???f1ds????2GA1?t1知J0?3Mt??f1?f3?214a??f?2M7a2t?22

?62f2?2Mt?5Mta?9Mtds?????222?3t7a?14atG ?2Gat?14a23????Mt145J0Gat

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第４章 力法

4.1题

令l?l0?2.75cm由对称性考虑一半I?I0I2?26I0 2.5??q??1???0.8?1.025?1.845吨/米2??对0，1节点列力法方程3?M0l0M1l0ql0???0??3EI6EI24EI?000?33?M0l0?M1l0?ql0??M1(0.8)l0?M2(0.8)l0?q0(0.8l0)?6EI3EI024EI03E(26I0)6E(26I0)24E(26I0)0?2??M0?M1/2?ql8即：?2??M0?2.09M1?0.2549ql2??M1?0.0817ql?1.139?t?m???2M?0.0842ql?1.175?t?m???0

4.2.题

将第一跨载荷向c支座简化M1?Q1l12,p?Q1由2节点转轴连续条件：?Q1l12?l6EI2解得?2M2l23EI2?Q2l2224EI2??M2l33EI3?I2l3?1???Il32???Ql?8?2?Ql16M2??Q1l1?Q2l2?2??8?Q1l1?若不计各跨载荷与尺度的区别则简化为M2??RA??M2l?Q16???QM2?M1?MR????B?2?ll???

2??Q84.3题

由于折曲连续梁足够长且多跨在a, b周期重复。可知各支座断面弯矩且为M 对2节点列角变形连续方程

Ma3EI?Ma6EI?qa324EI??Mb3EI?Mb6EI?qb324EI解得

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2332??q?a?b?qqbaa??22 M?1????? ?a?ab?b??12?????12?a?b?12b?b????

４．４题

图4.4，对2，1节点角连续方程：

2?M?l?Ml7QlM?l/1?20????10200????I0?E3?I4E1?8I0?04EI3??6E?400?2M?l8Q?l?0??M1?l02?0???0 ??3E?4 IE6I4E18I04?????000??41?Ql?0.1242Ql?M1?解得：330??M?Ql/55?0.0182Ql?

?

4.5图令I12?I34?4I0，I23?3I0l12?l23?l34?l0,由对称考虑一半

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2?M?l?Ml2Ql????10200???0??3E4IE6I4E45I?0??0???04??2M?l7Ql????M2l0Ml20?M1?l0200????? ?6E?4I0?E3?I4E1?8I0?04E（I）3E（3I） 06?00?3?解出：??M411?330Ql?0.1242Ql??M2?Ql/55?0.0182Ql

4.5题

对图4.4?刚架?1??l02?2?3EI?l006EI0对图4.5?所示刚架考虑2，杆3，由对称性?M2l02l0M2l2?M3E（3I?6E（3I?00）0）6EI0??2?l06EI0?均可按右图示单跨梁计算。由附录表A-6（5）?l0E（4I0）21?0?2?6EI?0l?03K??1?0???21?111?3???3?????3?3636??M?2Ql0?1??2?9?41Ql0?0.1242Ql?145?1136??316??0?????330? ?M7Ql0?1??1?Ql02?0??0.0182Ql0?180??1136????7???55

4.6题

??2为刚节点，转角唯一（不考虑23杆） ?M2l1??M243EI2?l3EI

?M2节点平衡21?M?24?????M22

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??2??M22?l3EI?M2l6EI,??2?1??2M2l?l6EIK?6EIl13EIl 若21杆单独作用，K21?3EI?21,若24杆单独作用，K24?6EIl?24?

?两杆同时作用，K?K21?K24?4.7.题

已知：受有对称载荷Q的对称弹性固定端单跨梁（EIl）， 证明：相应固定系数?与?关

??2?EI?? l?系为：??1?1?

证：梁端转角?i??M??l??M???Q????2EI?Ml3EI?Ml6EI???Q???.............................?1??令??0则相应M?M?固端弯矩?即M???Q?l2EIMM?........................................?2?l2EI?1?12?EIl???EIl

?1??2?得????l2EI????11?2?或：???1?1?1??2??? 讨论：

1）只要载荷与支撑对称，上述结论总成立

2）当载荷与支撑不对称时，重复上述推导可得

?i??2?ij?1??ij?ijor2?ij?ij?1?3?i??1j?i?6?1??j???i13?13式中?ij?MiM??外荷不对称系数 ?ij??i?j??支撑不对称系数 仅当?ij??ij?1即外荷与支撑都对称时有?i?11?2?i

否则会出现同一个固定程度为?i的梁端会由载荷不对称或支撑不对称而影响该端的柔度?i，这与?i对梁端的约束一定时为唯一的前提矛盾，所以适合?i??Mi定义的?i~?i普遍关系式是不存在的。

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