结构力学答案部分(6)

? v?x??Apcl2x?2plEI?34??n?11n4sinn?clsinn?xl

3）6.50如图所示 令v?x??ax2?l?x? 所以， V??EI2?l0vdx''2

2EI22?0?2al?6ax?3ldx

5192?2aEIl ??由

?

?l/20qU?x?dx??l0/2qax52?l?x?dxql4?qal4

??V????a?0 得 4aEIl?3192 所以，a?5ql768EI v?x??5ql768EIx2?l?x?

4）6.60所示如图， 设v?x??a1x2?a2x3，v?x??2?a1?3a2x?

V?EI2''?l0vdx?''2EI2?l04?a1?3a2x?dx222

2 ?2EI?l1a?31a2a?l3?a

2l3???ll/2qv?x?dx?q?ll/2?ax12?a2x?dx

ql?7a115a2l?? ???

8?38?3由

??V????a1??V????a2?0 得 2EIl?2a1?3a2l??7ql3/24

由

?0 得 6EIl?a1l?2a2l2??15ql4/64

2?67qla???1384EI解上述两式得 ?

?13ql?a???2192EI? v?x??0.1745ql2EIx?0.06772qlEI3x

6.6题

如图所示

设 v?x??a1sin?xl

?EIV?2??2?l/40vdx?''2E?2I?2?l/2l/4?''2vdx?

? ?EI?

l/40l/2?????x??x?????2a???sindx?2EIasin1?????dx ?l/4l?l??l???l??2144242l?31?2????EIa1?????l4???2??

?xldx?2qla1/?4???l0qv?x?dx?q?a1sin0l

由

??V????a1???l?31?2ql?0 得 EIa1?? ????l22??????4ql4所以， a1?1?EI5?3?????2??4?0.00718ql4EI

ql U?x??0.00718EI?xsi nl6.7题

如图所示

?设 v?x???an?1nsin?2n?1??x2l

2V?EI2?2l0?v?l????v?x?''?dx??

??2A224??????2n?1???2?EI??2n?1????3????? ?an???ansin?2l?n?1?22?????n?1???其中，A??Vl3EI

4EI?2n?1?EI???2n?1?3???an?3??ansin???anl?2l?n?1??2????2n?1??????sin?? ??2??????2l0qv?x?dx?q?2l0??n?1ansin?2n?1??x2ldx

?2l?q?an???2n?1??n?1??4ql1?cos2n?1??????????????n?1an2n?1

所以，

???an?4ql?2n?1??

44?VEI?3??EIEI???EI取前两项得 ?3?a??3??a1?3?a1?a2?，?a1?a2? ?23?a2l?2?l?a1l?2?l?V 由

??V????a1??V????a2?0 得

??EI?3?l?????4??EI4ql?????1??a1?3a2?l?????2????

由

?0 得

??EI?3l????3??4??EI4ql??1a?a?????21?32l3?????????

4?4ql?7.088a1?a2???EI即： ? 4?a?494.133a??4ql12?3?EI?4?ql?a1?0.1798?EI解得 ? 4?a?0.001ql182??EI?x3?x?ql??0.0012sin ?v?x???0.180sin?2l2lEI??ql?l? ?中点挠度v???0.1786EI?2?446.8题 取v1(x)??ansinn?xl,v2(x)??bnsinn?xl

V?EI2?l0v1dx?12GAs2?l0v2dx2'22GAsEIl?n?x??n?? ??asindx??????n?20?ll2??????n?x??n??bcos?n?l??dx?0?l????l2 ?EI2EIl4???lGAs2?n??an???l22??GAsl?n??2a???n4?l?44??l2?n??bn???l?222

??V?an?n??2??bn?l?GAsln?2EIln?4?V()an,?()bn 2l?bn2ll???0qv1dx?l0?l0qv2dxn?xl?1 ?q??ansindx?q?l0?bnsinn?xldx?1

(1?cosn?) ?q??n??an???l?(1?cosn?)?q??n??bn???l?∴???an?q??(V??)?an?(V??)?an?l??l?(1?cosn?),???b?qn???(1?cosn?) ?n???n??2ql(1?cosn?)(n?)EI2ql(1?cosn?)(n?)GAs3254n为奇数由

?0得an??n为奇数4ql54(n?)EI4ql32

由

?0得bn??(n?)GAs

∴U?x??U1?x??U2(x)

?4ql54?EI?nn15sinn?xl?4ql32?GAS?nn13sinn?xl

(N?1,3,5,? ? ? ? ? )6.13题

由对称性可知，对称断面处剪力为零，转角?0?0，静不定内力T0和M0可最小功原理求出：

2?qs1—(OA段)?M?2M(s)??0 22??(M0?qr/2)?2qrsin??T0r(1?cos?)—(AB段)?M(s)?M0?1 (OA段)?M(s)?0 —(OA段)?? ??

?T1 (AB段)r(1?cos?) —(AB段)??0最小功原理：

?V?M0?1EI?r0?sM(s)?M(s)EI?M0ds??21??qs1M?ds?1?02?EI??M????20022?qr/2??2qrsin??T0r?1?cos???rd???0

?V?T?1EI??202??qr2M0??2qrsin??T0r(1?cos?)r?1?cos???rd??0 ??2??????1??qr2?2?1???M1??Tr0???02264???分别得：?

1??2??M1???T0r2?3??qr???024??24???????????M0??0.5388qr2?解得：??M(s)表达式正确??T0??2.7452qr

由

?M?s1?M?s2?0

得极值点在st?0点,该处极值为M1?M0

2qr?0.7285,??0.6296 得tg???T0由?0

??1?2?22M2????0.5388??qr?2qrsin0.6296???2.7452qr??1?cos???2?极值为???

?0.61qr2

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