结构力学答案部分(7)

区间端点B处

22 MB????0.538?8?qr?qr22?????1??s?in?2?22qr.7?4??5?2??1qr2 0.79?Mmax?max?M0,M1,M?Mmax?M2BB??MB

??0.79qr(发生在支撑处)6.14题

由左右对称,∴对陈断面01上无剪力。

有垂向静力平衡条件：?qrsin?d??P

0?2解得：q?P/4r

任意断面弯矩为：

M(s)?M0?Pr2sin??T0r(1?cos?)?Pr2??0qr???1?cos???????d?22 ?M0?T0r(1?cos?)??M?M0?1,?M?T0?r?1?cos?sin??qr??sin????

?有最小功原理确定T0和M0

?V?M0?1EI??0Pr??2M?Tr(1?cos?)?sin??qr(?sin???)rd??0 00??2??2即：M0??T0?r?Pr?qr2(?2??2)?0

?V?T0?1EI??0Pr??2M?Tr(1?cos?)?sin??qr(?sin???)r(1?cos?)rd??0 0?0?2??即?M(s)(1?cos?)d??0??M(s)cos?d??0

00????Pr?22???(M0?T0r)cos??T0rcos??sin??qr(?sin?cos???cos?)?d??002??

得：?T0r?2?2qr?0?T0??4qr/???P22?(与图中假设T0方向相反)

?M0?Pr(4??)8?

Pr?M(s)?Pr8?(4??)?2?(1?cos?)?Pr4sin??Pr4?

?4??21cos?sin??? ???????Pr

??44??8?第7章 矩阵法

7.1题

解：由ch2/2.4题/2.6图计算结果

v??1x?2?1??2lx?2?1??2l2x

3v(x)??1?'2?1??2lx?3?1??2l2x，v(x)??2''2?1??2l?6?1??2l2x

???46x???26x????i??2???2???? ∵??yv?y??l??ll????j???l''∴?B??2y??3x??3x???2?1????,?D??E ??l??l??l???Ke??????B??T?D??B?d?????3x??22?4y?l??3x?E?2???????2l?3x???l??1???l???l??l??2?3x???1???d??l??2??3x??2???4EIl??l??2?0?l?对称??2?3x?3x??2????3ll????4EI22?l?3x???1????l??

l?2?2EI??ll????2??11??2?7.2题

解：如图示离散为 3个节点，2个单元

?122?l???????2??6?l?????E?2I???2????l??12???2?2???l?????2??6???l??????2?K23??2??K33??2?6?l????2?4?12?l????2?62??K??1??6?l????2?2?l????2?122?l????2?6??l????2????l?????2???2???1???K11?6??K?1????21?l?????2???4????6??2K?1??K22?K12?1???2???K22?2????2?K?32

?K11?1???1?形成?K??K21??0??K12K22?1??1??2??K22?2?K32???1???2???K23???2? ?2?K33???3?????0?将各子块代入得：

?24?2??l/2?2?6x??l/2????24EI??l/2?2?l/2??12???l/2???????6x2?241212?l/2??l/2?24?2?12?l/2?4?624?6?l/2?36?6?l/2??l/2?24?l/2??l/2?212?6?l/2?24?l/2?24?6?l/2?2?6?l/2??l/2?22?l/2??l/2???????v1??Ry1???????6???z1??MR1??v2????0???l/2???? ?????0??2??z2????v3??P?????6???????0???z3???l/2??4???划去1、2行列，（∵v1??z1?0）约束处理后得： ?1442?l??12?2EI?ll??482?l?12??l?12l12?12l2?48l?12l48l?12l2212?l???v2??0?2?????0???z2???????? ?12?v3???P?l?????0???z3???4??图7.3 离散如图

e?不变，离散方式一样，组装成的整∵杆元尺寸图7.2（以2l代l），∴?K??体刚度矩一样?K?

?P?T??R1y??v1MR1P0R3yv30?T

???T?z1v2?z2?z3?

T约束条件 v1??z1?v3?0，划去1、2、5行列得（注意用上题结果时要以2l代l）

?36?l2?EI??6l?l?6??l?16l1226?l?v??2??P?????2???z2???0? ????????z3??0?4???图7.4，由对称计算一半，注意到?z2?0,v3?0

?12?l2??6EI?l??l?12?2?l?6???l(2)6l4?6l2?12l2?6l12l2?K?(1)?6l6?l??2?(1)??K11???(1)?6K?21?l??4???K23?(2)?K33?(2)?(1)?K22? K12(1)????????K以2l代l，4I代I?K22(2)???(2)??K32?K11(1)?(1)K21???0??K12K22(1)(1)(2)?K22(2)K32???1??P1????(2)??K23???2???P2?，将各子块代入得 (2)????P?K33???3??3??0??12?l2??6?l??12?EI?l2l?6??l??????6l4?6l2?12l26l2012?6l4?6l2?6l18l20?6l2?6l6l26l?6l??ql??R?y1??2????2v??1ql???M???R1?12?6??z1???????v3ql?l??2???kv???22 ?????2?4??z2??2ql???v3??????6??4??z3??????qll??????MR3??8??

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