船舶结构力学课程习题集答案

1

第1章 绪 论 .................................................................................1

第2章 单跨梁的弯曲理论 ............................................................2

第３章 杆件的扭转理论 ............................................................15

第４章 力法 ...............................................................................17

第5章 位移法 .............................................................................29

第6章 能量法 .............................................................................41

第7章 矩阵法 .............................................................................56

第9章 矩形板的弯曲理论 ..........................................................69

第10章 杆和板的稳定性 ............................................................75

第1章 绪 论

１．１ 题

１）承受总纵弯曲构件：

连续上甲板，船底板，甲板及船底纵骨，连续纵桁， 龙骨等远离中和轴的纵向连续构件（舷侧列板等） ２）承受横弯曲构件：甲板强横梁，船底肋板，肋骨 ３）承受局部弯曲构件：甲板板，平台甲板，船底板，

目 录

1

2

纵骨等

４）承受局部弯曲和总纵弯曲构件：甲板，船底板，纵

骨，递纵桁，龙骨等

１．２ 题

甲板板：纵横力（总纵弯曲应力沿纵向，横向货物或上

浪水压力，横向作用）

舷侧外板：横向水压力等骨架限制力沿中面

内底板：主要承受横向力货物重量，骨架限制力沿中面

为纵向力

舱壁板：主要为横向力如水，货压力也有中面力

?v(l)?0v'(l)?012?12 ?p'?v1(0)?0N1(0)??2２）图2.2?v(x)??0x?Mx22EI?N?x36EIx0?l33p(x?l)3 6EI3p(x?l)2 6EI３）图2.3?v(xx)??0xx?N?x36EI??qxdx6EI3?l2２.２题

a) v1?vpppl?1131?pl?1131??vp?(3???)??(?2?)? ?6EI?41626EI?164444????pl3第2章 单跨梁的弯曲理论

２.1题

设坐标原点在左跨时与在跨中时的挠曲线分别为v(x)与v(x1)

１

）M0x2EI233 =

?512EIpl3

31?13?pl3?9pl???()??? ?192EI96EI1624???图

N?x33p(x?l)4?6EI2.13p(x?l)2?6EI

V2?3v(x)??6EI?l4l23l4

p(x?3l)42?MlMl2Pl9'6EI b) v(0)???(1?2)

33EI6EI6EI0.1Pl6EI2?16EI?4?原点在跨中：v1(x1)?v0?M0x12EI2?N?x16EI3?l43p(x?l)46EI，

=??5Pl23?27EI?73Pl21620EI

2

?(l)??MlMl23EI?6EI?2Pl96EI(1?13)

24Pl2 =?0.1Pl26EI?3?27EI??107Pl1620EI

22p??l?v?l???3??2l?2???3??l?3?1?1??3?3??3EIl6EI??m?2?13??m?1?13???

2 =

37pl2430EI

7ql4?3ql4 c) vl图2.1

2??ql4192EI?5768EI?2304EI v'(0)?ql3?pl2?ql2l3124EI16EI?16?6EI?ql?11?ql38EI????3612???96EI d)2.1?图、2.2?图和2.3?图的弯矩图与剪力图如图2.1、图2.2和图2.3

3

3

4

1）

??右??M?Ml6EI13q1l?2ql324EI?Ml?l?q?q??0??21?45EI?23EI??

2l2120 2）?0??Ml3EI3?q1l324EI?Ml?l? lq?1?180EI?26EI??37l2ql?1q1l1713? =1?? ??????EI?18243606?120?80EI2.4 题

图2.2

Nx ? 图2.5?v(x)?v0??0x?036EI，

v0?A?p?N0?

?x3??v(x)?Ap??0x???A?N0

?6EI?如图2.4， 由v?l??v??l??0得

图2.3

2.3题

4

5

??l3?Ap??0l???A?N0?0?6EI????2l??0?N0?0?2EI??pl?Ap?????0l6EI??N?p3?022.5题

解出 图2.5：（剪力弯矩图如２．５）

?

?R1?pl?Mll3?p??2p32p3??pl322p39EI33v0?AR?6EI33pl?3xx?3 ?v(x)??1?9EI?2l2lMlplpl5pl?l?vv???0????216EI18EI48EI144EI?2??? 图2.4 ?v??0???0??v0l?Ml6EI??pl29EI?pl218EI??pl26EI 图2.6

v?x???1x?由v?l??0,Mx2? M?pa?bb?A?1?l?KA?6l???? ，

02EI?N0x6EI3图2.5

将a?l,b?0A?l6,KA?16?13?12代入得v??l???2?N0l3得：

4EI2EI?M?????2M?pl1?pl 01??ll6312解得??N?6EI?????122?0?l? 2.7图：（剪力弯矩图如２．６）

23?2?1??2?x??1??2?x?v?x???1x??2ll??0??2EI6EI?2M0lN0l?1????2??EI2EI??1l?M0l2??

5

v0.05l31?A1R1?EI?ql2?ql440EIM?Qa24?1?K?12A??b2?1?????A???l???vl3ql42?A2R2?50EI?ql2?100EI由Q?qa,a?l,b?0,44

v?l5qlql??1??8,A?124?2????11??384EI2EI???40100?? K44A?18?124?13?12,代入得

?ql?57?2932EI???384400??ql?9600EIM?ql?1ql2 图

24?2?12?24?1??8

R1?ql2?ql3ql,3v338?8??0??ql2ql4v

24EI?v1?l?ql?1EI??24?140?1?2ql100???75EI0?AR1?64EIql333

??l????v1?v2ql? 图２．７24EIl?EI???111??17ql24?40?100???300EI4424?v?l图2.8?（剪力弯矩图如２．７）

??5qlqlMl5ql??EI?128EI?16EI??2?384384EI?(0)?ql3?v0Mlql3?111?

24EIl?6EI?EI????246448??3?ql192EI?M??l2?(l)??8EI?ql8??ql364EI

2.6题

6

6

２．６

7

.

7

8

dv2??max.dx??maxGdx??NGAsEIdx

2.7.题

,?,情形：先推广到两端有位移?,??NN?EIv???v2????GAdx????1??C1sGAv1sv?v1?v2???ax3bx2EI?f(x)?6?2?cx?d????GA?f??(x)?ax?b??C1s?f(x)?EI32GAf??(x)?axs6?bx2???c?EIa??GA?x?d1s?中f(x)?qx4qx2式24EIf??(x)?2EI于v(0)?v1??(0)?0可得出d1?b?0v(l)?v1??(l)?0得方程组：?4?qlEIql2al3?EI?????24EIGAs2EI6??c??GAa?l?0s??ql2??2EI?al?0解出：a=ql2EI,c?ql324EIv(x)?qx424EI?qlx312EI?qx22GA???qx3?ql??xs?24EI2GAs?v(l5ql4ql22)?384EI?8GAs8

iijj??令???12EI??i??j,??GAsl2? ??v?ax36?bx22?cx?d1?EIGAaxs而v0??i?d1?v(0)??i??

由v?1?(0)??i?c??i??2由v(l)??j?al3EI6?bl2??il??i?GAal???j?s?2?由v?(l)???1j?al2?bl??j?????a??l21???i??j?2?l??解出????????

?i31?b?j??i??j?2??ll?1????l??由由?? 9

球扁钢NO24a ??????M(0)?EIv1(0)?EIb?6????2??j????4??i?l?1?????l?EI6?6?????????4?????2?????ijij?l?1????l?l?EI38.75 119.8 A e?BA?5.04cmI?C?B2 15.6 A 604.5 B ?11662?604.529430.2 9430.2 C=11662 119.8?8610cm 2422? 225 ?6EI2????N(0)?EIv(0)?EIa??????????1ijji??2l?1????l?????N(l)?N(0)?EI????M(l)?EIv1??(l)?EI?b?al????4??2????6????ji??l1??l??????令上述结果中?i?0，即???j同书中特例计算外力时面积A?75?1.8?38.75?174cm?l?l计算I时，带板be?min?,s???45cm?5?5 1）.计算组合剖面要素：

2.8

l?3??题

cm7t?形心至球心表面y1?h?t?e?24?0.9?5.04?19.86cm形心

cm5s?已cm知：52?20?kg76.87 5cmkg22 cm至最外板纤维

,y2?e?t21?5.94cm?w1?Iy1.?861019.8683?433.5cm

,7 q??hs?102?5面积 外板1.8?45 cm21?00?.75

惯性矩 cm 4 81 距参考轴 cm 0 面积距 cm 33w2?I自惯?8610?1449.4cmy25.94性矩 l?0A42251050?174u??0.366 cm? 62EI22?10?86100 0 x?u??0.988,?1(u)?0.980(21.87) 略 9

10

M?Mql212??x?u??ql276.87512?225?0.988?320424?kg.cm?2?EIvIV?Tv???0,EIv????N?Tv?TEI2

?vIV中24?1?u???M中124?76.875?225?0.980??158915（kgcm)158915kg2?V???0,vIV?KV???0式中k?r??4k2T EI?中球头?固端?端球头板???0??1050??14162?cm?w1433.5?M320424?kgkg??0??1050??1271???max?141622cmcmw21450??M320424kg??0??1050??378?2cmw1433.5??特征根：r1,?0,2r?3k?v?A1?Akx?Achkx?Ashkx234 ?v(0)?0??v?(0)?0??A1?A3?0??A2?A4?0?

?v??(l)?0???EIv???(l)?N(l)?Tv?(l)??A3chkl?A4shkl?0 ?? 3EIkAshkl?Achkl??p?TkA?Ashkl?Achkl?3??2??434?

若不计轴向力影响，则令u=0重复上述计算：

ql2?max??中球头??0?24w1?1050?76.875?22524?433.52?1424kgcm2

解得：

A1??pkTthkl,A2?pkTpEIk31424?1416相对误差：?0.5624结论：轴向力对弯曲应力的影响可忽略不及计。结果是偏安全的。pkT,A?3pkTthklA,?4?pkT ?v(x)?????thkl?kx?thklchkx?shkx???thkl?1?chkx???shkx?kx???

2.9.题

2.10题

10

11

EIvvIVIV由v?(0)?0?Tv???0?2?协调条件查附录图：l24kEI?l2464EI4EIl?1?EIv?22IV?N?Tv???令 A=0 ?B=0 u=?kv???0式中k?4T?EI?

ql3特征方程：r?kr?0特征根：r1,?0,2?24EI??2?u??Ml3EI??0?u??0ql2

r?3ikr??4ik?v?A1?Akx?Asinkx?Acoskx234???EIv??(0)?m?v(0)?0?A1?A4?0?A4k?2??M?2ql?2?u?.6098?0?u?80.752?0.101ql2????mEI??

??l??l??v(2u)v??v(2u)v????1??133??u??M?l?q0???2??2??? v????1???2221?B?2?EI?v1(2u)?v3(2u)?2?k??????

?v??(l)?0????EIv???(l)??Tv?(l)???Asinkl?Acoskl?034????3???3???k?A3coskl?A4sinkl???k?A2?A3coskl?A4sinkl? ??

?1.91150.66354.93011.9335???44?ql0.101ql??22222l?u???????1?0.448??22?1B?01.911564EI8EI??4.930122???0.0049ql4解得：A3??mT?ctgkl,A2?0???????EIm????????v?(0)??Ak?Akcosk?Aksink?Ak?343???2?x?0kEItgkl

2.13图

?2.11题

图２．１２0

11

??0??pl216EIx0?u??Ml3EI?0?u???Mpl2M??? 16EIx0?u?l???0?u???3EI??

将u?1,??l12EI代入得：M?pl16?0.591??1?12?0.72?3???0.111pl

?v(2u)v??l??v(2u)?l??v?lpl3?1?3?3v??1????M?2????2??2???48EI?2?u??2?2EI??v21(2u)?v3(22u)?????????23?lu?1??pl?0.6090.1110.9115?0.6635?4.8301?1.9335EI?????4881.91152?4.93012???0.0086Pl3EI2.12题

1）先计算剖面参数：

12

W?bh26?2?10236?1003?cm?Wp??Aiyii?2?Ah???2?24??bh3?4?50?cm??形状系数?f?WpWbh2?4bh2?362

图２．８ａ

2）求弹性阶段最大承载能力Pma如x(图2.8a) 令M4ma?xW?y?1003?2400?8?10kgcm2

即516P16W?16?8?104malx?W?y解出P?yma5xl5?500?512?kg? 12

13

3)求Pu?极限载荷?则边界条件：v0?0得N0?3ml2l?6EI2?0?0v(l)?0EIv??(l)?m3ml2?用机动法?此结构

达到极限状态时将出现三个塑性铰，其上作用有塑性力矩Mp?Wp?y如图由虚功原理：??Pu????? Pu???2??4Mp2????l?2?? 图２．８??M0?m?GAs

22l?6EIGAs23?m??3?lx6??llx2?v(x)??x??2?22EI?2l?6?2l?6?2(2l?6?)???EIGAs

2.14. 补充题

试用静力法及破坏机构法求右图示机构的极限载荷 p，已知梁的极限弯矩为Mp（20分） （1983年华中研究生入学试题） 解： 1）用静力法：（如图２．９）

由对称性知首先固端和中间支座达到塑性铰，再加力

p?puｂ

?Pu?4Mpl?4Wp?yl00?4?24?5?0960?kg?

5002.13补充题

剪切对弯曲影响补充题，求图示结构剪切影响下的v(x)

解：可直接利用

v(x)?v0??0x?

M0x2EI2，当p

?N0?36EI?x? ?x?6EI?GAs? 作用点处也形成塑性铰时结构达到极限

状态。即：

pul4?Mp?Mp?pu?l8Mpl8Mp

2）用机动法： 2p??8Mp?2??pu?l2.15.补充题

求右图所示结构的极限载荷其中??l3EI,船工研究生入学试题）

p?ql（1985

年哈

13

14

解：由对称性只需考虑一半，用机动法。当此连续梁中任意一个跨度的两

端及中间发生三个塑性铰时，梁将达到极限状态。考虑a) 、b)两种可能：

对a)解得?2??2?qu???xdx?4M0l??2l

图２．９ 图２．１０

2?pl?0

对b)?qu?qu?16Mplp2pu???4M16Mp2?l

?0l2（如图２．１０）取小者为极限载荷为qu?荷p的跨度是破坏。

8Mpl2即承受集中载

14

15

对于a)示闭室其扭转惯性矩为J0?4A2??dst?4?a?t?4t4?t?a?t?

3?a?t?第３章 杆件的扭转理论

3.1题

a) 由狭长矩形组合断面扭转惯性矩公式：

J?13 对于b)开口断面有J?

1ht?33ii?t33??4?a?t???

?hti3ii?1?650?10?200?8?80?8??26.4cm ?3?3334?两者扭转之比为?b??a??MtGJMtGJ0dst?J0 b) J??13?a?t?J??(）??271倍4?t?2?70?1.23?35?13?15?1.23??60.6cm4 ?3? c) 由环流方程

本题易将??的积分路径取为截面外缘使答案为300倍，误差为10%，可用但概念不对。若采用s为外缘的话，J大，?小偏于危险。??????ds2AG??M????tf?2ABredt公式Mt4AG22??dst?????材力MtGJ0?J0?4A2

ds ??t题 3.3

8本题A?40?41.6???20?0.8??3023.2?cm2???

dst?11.6Mt??n?1pb2?8?b2?p?4pb?2?40?41.6???131.682 A?131.68?2.775?10cm54?J0?4??3023.2?8????1???2b?tsinb?tcos?b?tsin???????2?8?8?4???2?Mt2A?4bp2?b?t?sin2

3.2题

?f??4?2?100?302(300?0.2)2?9.555kg/cm 15

?09?.5?b6?t?8????l??flf?108sin2AGtds??2AGt?8??b?t?sin????82?b?t?2?2s??8icn8os?100?9.56?8?4?10?4(弪）4?29.8?cos?58?8?10?0.2

3.4题

.将剪流对内部任一点取矩

?f1rds??(f1?f2)rds??f2rds21566232??f2rds??(f2?f3)rds??f3rds67737843 ???f1rds?f2??rds?f3??rds

215623267378437?f1??rds?f2??rds?f3??rdsIIIIII?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

16

16

1?ff32GA???1ds???f2ds?1?f??21?t?t2GAds??????f12?IIt26tds??tds??73??1?f?f22GA????3ds?ds??3IIIt?37t?即：3f1?f22f2?f1?f3f?f2A??3A.....(2)1A23(1)(2)联立（注意到A21?A3，2A1?A2?a）??2A?1?f1?2f2?f3??Mt??f?3Mt?3f1?f2?3f3?f2解得?1f3?14a2????f?2Ma22t7?3ff11?2?2(?f1?4f2?f3)????1?f11??2GA???ds??f21?ttds???a?9Mt2Mt?5Mt62?2Ga2??14a2?7a2???14a3tG2t???Mt知J143J0?0G5at??17

第４章 力法

4.1题

令l?l0?2.75cm由对称性考虑一半2.5??q??1???0.8?1.025?1.845吨/米2??对0，1节点列力法方程I?I0I2?26I0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???I3l2???Ql?8?2?Ql16M2??Q1l1?Q2l2?2??8?Q1l1?

3?M0l0M1l0ql0???0???3EI06EI024EI0?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?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??Q8

4.3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?qa34.2.题

24EI??Mb3EI?Mb6EI?qb324EI解得

17

M?q?a3?b3?q12?a22qb?2a?a?2??a?b???12??ab?b??12??1??b???b?????

图4.?4，对2，1节点角连续方程：

??M21?l?0?M?l2?0?7Q?l?0?M?l/1?20?6E??4I0?E3?I4?0E1?8I0?04EI03?M?1?l?0M?22l?08Q?l?0 ????0?3E?4I? 0?E6?I40E1?8I0?04?41解得：??M1?330Ql?0.1242Ql??M?Ql/55?0.0182Ql18

?

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

４．４题

18

???M1?l?0M?l2?02Q?l?2?3E?4I???E6?I4??00E4?5I??0004?M1?l?0M?7Q?l?22l?00M ?2l0Ml2?6E?4I?4??0E1?8I0????0?E3?I04E（I）30E（3I）?41解出：??M1?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?M2l0M2l0M2l02?3E（3I?E（3I?0）60）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???3??21????1?11??33?063?3636??M2Ql0?1??1?45?1136??2?3?9?41Ql016???0.1242Ql0?????330??M7Ql0?1??1?Ql02??180?0??0.0182Ql?1136????7???550

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?M2?4?????M2219

19

0 20

证：梁端转角?Mlli??M??3EI?M6EI???Q??M???Q????l???2EI??.............................?1?令??0则相应M?M?固端弯矩?即M???Q?l2EI........................................?2??1??

2?得??Ml2EI1???EI??l1

或：M???l2EI???1?2??1?2?EIM22?l?l??26EI2?3EI?M2l6EI,??2?M?l26EIK?l

1 若21杆单独作用，K?3EI121??l,若24杆单独作用，K24?? 3EI 讨论：

1）只要载荷与支撑对称，上述结论总成立 21?24 l 2）当载荷与支撑不对称时，重复上述推导可得 ?两杆同时作用，K?KK6EI21?24?

l

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 系为：??1??2?EI??1?l? ?

20

?1?12??1??????

21

?i??2?ij?1??ij?ijor2?ij?ij?1?3?i??1j?i?6?1????3j1?i?13A1??2l?348EI?l36EI列出1节点的角变形连续方程：2?Mlvp2l??M(2l)v11?1????1??3EIl3EI2l16EI??v?AR?A??M1?2p???M1?p??

111???????12????2l??l?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联立解出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3323pl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，M所1??11pl,v1?36EI以适合?i??Mi定义的?i~?i普遍关系式是不存在的。画弯矩图见右图

4.9题

１）如图所示刚架提供的

支撑柔度为A1?A2?V而由5节点?5?0得p?14.8 题

??pl?l3E?7I?6E?7I?M5l???pl???pl2?l?0

?M5?pl2,F???3p2 21

由卡瓦定理：

A?VM?M1p??1?EI?Pdsp?1

?1?ll?3p??3E?7I????0?ps1?sds1?1?0?pl??2s????2l?2s??ds??2??p?2

?1??l?E?7I??l3?0?l?3?2?1?l3l3?l3??3?2s2?ds2????7EI???3???4?12EI ２）由对称性只需对0，1节点列出方程组求解

???M30l?M1l?v1?ql?0?3EI6EIl24EI33 ??M0lM1lv1qlM1lM1lql?6EI?3EI?l?24EI??3EI?6EI?24EI ?3?v1?Al??M1?M0ql?ql?1R1??12EI?????l2???2?? 联立解得：M20?11ql36,M1??ql236,v1?2v2?ql418EI４．１０题

22

a)? =13?8?4,Q1?qal192,q??Q?a?ql2k?192Eial3

b)Q?Q1?Qqal3qal2?qal?2?2?Q21?3Q,Q2?13Q5Q3?v1l7中??Q2l3?310?384Ei180Ei???225?8??5Q31l5Q3?2lQl35?25??i??3384Ei?Ei??3?384E?384384???5Ql3384Ei???5384,??148q??Q?5?48?a384?3qal2a?1516qlk?Ei?al3?48Eial3

22

1

23

c)??148,??148,p?p,Q?p

pl3?1?m????1???1??????48Ei?k?Ei768Ei2?23?3?????2?alal

768?131m3m24768Ei??p=-???????k?37al7?122?222?l7lf)令???1???34?1???4?11p?,k87pl31?1???1?????m?26Ei2?2???l2pl?2l1?4d)令48Ei48p4?3?1??p=??26Ei?44?6g)p同348?Ei(同ac图l)a)即p???Q?qal2?q?pa?ql2

?192Eil3?k0?x?6a?1? k???192EiA?192E(2i)l3?2??2k03l??k?k0a?192Eial3?x?6a?

e)??5pl33843,??1,Q?qal2?481?4??1????9?p=?q?Q?a?5?48qal??384a2123???36p6?51623qlpl?11?令???1??48Ei6Ei?32?4?8p??6?

27

4.11题

?支柱处v???0,可简化为刚性固定约束?仅考虑右半边板架

23

??148??1?11?116??21????11?4??164?????48?16p=??p?1116p

k?48Eial3?48EI30l0?6l?0?2EI09l40???2EI0u??6l0??9l4?40?2?4E9I02??111?M?p?6l0???6l0?8?1?1??16p8?0.874?0.4507pl0

N??p?p1121?1???2?0.852?32P?0.852??0.2929pvma??l?xv???v??3l?4?0p?6l311?63?0.889?0??pl3301?1??16192E??9?96?9?EI?0.1528pl00EI0?2I0??24

4.12题

设a?l540?1mi?I0?5.833?10cmL1?l?10l0b?2.5l0I?1.857I0,Q?q0al,q2620?1kgcmE?2?10kgcm '求：中纵桁跨中及端部弯曲应力及v中

解：因主向梁两端简支受均布载荷Q故其形状可设为sin?ylc1?c3?sin?y1l?sin?4?0.707c2?sin?2?1

24

25

?11??12????11?1?6?411???3?????0.02083?44????按对称跨中求??端?M?h?I??t??2?q2L242?q2L122??2?1.?2I51?3.q1l0?608l1220?20?100.8?1310.833?10551?1010kgcm1?11??6?24211???1??????0.01432164???10.707???i?1?11iciIic1I1??1?0.02083?0.707?0.01432?1??0.0411?2?45?中?

??1?u?I51?3.168ql0?10l024??0.774?5110.833?105?481kgcm?2??I1I2???23?0.0411,6384?0.0130223k2?Ei?q2??2al?2?10?5.833?10?i0.01302?10q0l00.0411l0?20.0411?10??10?34.13?283kgcm补充题 写出下列构件的边界条件：（15分）

?2Q?L24l0?3.168q0l03u?4alI1?13?10l024I04l0?10l0?1.857I0?0.0411?1.2?1?1.2??0.728,v中?q2k21?2?u??0.813,3.168?1?102832?1?u??0.774(1?0.728)?0.304?cm??1???u???

1）

25

31

???M??M'25'32'23?M?M'34'21??(M32?M34)???M23?M21?M25??M

M2334EI0??12q0l0?2EI0????l0?1045EI0?l03?16q0l0?11222??ql??0.0357ql??00003135?3?1045EI0?则有

2EI04EI0EI?0??????0233?ll2l?000?28EIEI4EI2EI2ql0000??2?????2???32?l2l0ll1500?0

，得

00M3232EI0??12q0l0?4EI0????l0?1045EI0?l03?16q0l0?822??q0l0??0.0026q0l0??3135?3?1045EI0?

其余由对称性可知（各差一负号）：M65??M12，M56??M21，

M52??M253??12q0l0??2?1045EI0? ?3q0l016????33?1045EI0?，M54??M23，M45??M32，M43??M34?M32；

4）

M12?M'12弯矩图如图5.1 5.3 题

3（4EI0??12q0l0?1257222M14?M25?0）M12??pl8，M21?pl8，其余固端弯矩??(?ql)??ql??0.246ql??000000l0?1045EI0?51045都为0

'? M41?M12

M2138EI0??12q0l0?226222??ql?q0l0?0.0415q0l0 ??00l0?1045EI0?156273EI0??12q0l0?622???ql??0.0057ql??0000 2l0?1045EI0?10452EIM63?M12?M'23''l2EIl4EI?1，M14?''4EI?3，M36??1?2EIl2EIll4EIl?1，M52?'2EIl?2，M25?'4EIl?2

?3

'M25?l4EIl?2， M21?2EIl2EIl?1??2?4EI?2??3， M32?'l4EIl?2 ?3

由1、2、3节点的平衡条件

31

32

M14?M12?0???M21?M25?M23?0 即

M3624EI?5pl?5??pl?0.0142pl??M32 ??l?22?64EI?352??M32?M36?0??M'M'14?12???M14?M12???M''25?M23?M'21???M23?M21?M25?

???M'M'32?36???M32?M36??4EI4EI2EIpl??l?1?l?1?l?2?8??2EI?4EI4EI4EI1??2???2EIpl?lll2?l2?l?3??8

?2EI???4EI4EI0?l2l?3?l?3?22解得：?27pl1?22?64EI，?2??5pl22?16EI，?3?5pl222?64EIM?4EI?27pl2?14??M12l?64EI?27pl?0.0767?22??pl ?3522M41?2EI?27pl?27l??22?64EI??pl?0.0383pl ?704M63?2EI?5pl2?5l?22?64EI?pl?0.007pl ???704M25?4EI?2l??5pl?22?16EI??5pl??0.0568pl ???88M23?4EI?22l??5pl??2EI?5pl???35pl??0.0497?22?16EI??l??22?64EI?704pl?

M21??M25?M23?75704pl?0.1065pl

2M52?2EI?l??5pl??5?22?16EI??pl??0.0284pl ?176弯矩图如图5.2

32

33

Q0?12q2l12?12q0l0，q4?4q0，

12(3q0)3l0?6Q0?9Q0 Q24?Q矩24?Q三角24?q（3l0）?01）求固端弯矩

M21?Q0l010

，M12??Q0l015，M32?M23?0

??(6Q0)(3l0)12(6Q)(3l)0012???33Q0l010 M24?? M42?(9Q0)(3l0)15(9Q0)(3l0)10

21Q0l05

2）转角弯矩

M12?'4E(2I0)l0'

5.4题

图5.2（单位：ql） ?1?2E?2I0?l0?2，

M21?2E(2I0)l0?1?4E?2I0?l0?2

'? M234E(3I0)2?(2l0)已知l12?l0?3m，l23?2.2l0?6.6m，l24?3l0?9m I0?0.3?104cm4，I12?2I0，I23?3I0，I24?8I0

?2?2E(3I0)2?(2l0)?3，

M32?'2E(3I0)2?(2l0)?2?4E(3I0)2?(2l0)?3

33

34

M'24?4E(8I0)(3l0)??2，

?3??209Q0l02740EI02??0.03814q0l02EI0

M'422E(8I0)(3l0)?2

图5.3（单位：Q0l0） 4）求出节点弯矩 M21?????4?223432880?8?2091370?1??Q0l0?1.0487Q0l0 10???Q0l0?0.6241Q0l0?3）对1、2、3节点列平衡方程

M12?0?? ?M21?M24?M23?0?M32?0?即：

2096209?12M23??????1.213702.22740

8EI04EI0????2?Q0l0151?l0l0??796EI030EI0?4EI0?16??1??2??3????Q0l0??33l011l0?5??l0?30EI060EI0?2??3?0?11l11l?00?33??32209M24?????Q0l0??1.6727Q0l03137010??

21??14209M24?????Q0l0?5.0136Q0l05??31370

解得：

?1??22234Q0l032880EI02??0.03397q0l02弯矩图如图5.3。

5.5 题

由对称性只考虑一半； ，

节点号 杆件号ij 1 12 —— —— —— 21 4 1 4 2 23 3 1 3 EI0?2?209Q0l01370EI02?0.07628q0l0EI0，

Iij/I0 lij/l0 kij 34

35

Cij Cijkij —— —— —— —— —— -1/10 -4/165 -41/330 1 4 （1/2）对称 3/2 11/2 Iij/I0 lij/l0 kij —— —— —— —— —— —— —— —— -1/10 -1/45 -11/90 1 1 1 1 1 3/2 2/3 1/2 1/15 -2/45 1/45 1 1.5 2/3 3/4 1/2 —— —— —— —— —— ?Cijkij ?ij nij 8/11 1/2 1/15 -8/165 1/55 3/11 —— 0 -1/55 -1/55 Cij Cijkij Mij/Ql0 ?Cijkij mij/Ql0m'ij ?ij nij 1/3 0 0 -1/45 -1/45 —— —— 0 —— 0 /Ql0Mij/Ql0 Mij/Ql0 所以：

M12??MM??M32??Ql05543mij/Ql041Ql0330??，

M21??M34?Ql055m'ij /Ql0，

Mij/Ql0 23

由表格解出 M01??0.1222QlM10?0.0222Ql5.6题

1.图5.40：令I10?I0?I12,l10?l0,l12?1.5l0

节点号 杆件号ij 0 01 10 1 12 2 21

M12??0.0222Ql 35

36

M21?0

mij/qlm'ij2/ql2 -5/512 -0.0931 -5/256 0.0638 -5/768 -0.0638 -5/1536 0.0228 2.图5.50

令I10?3I0，I0?I12，

l10?l0，l12?l0

Mij/ql 2由表格解出： M01??0.0931ql2，M10??M12?0.0638ql2，M21?0.0228ql2

q?q0，Q10?q0l0，Q12?节点号 杆件号ij Iij/I0 lij/l0 kij Cij Cijkij q0l0若将图5.5中的中间支座去掉，用位移法解之，可有：

2

1 2 12 1 1 1 1 1 4 21 —— —— —— —— —— 1/4 1/2 -11/192 —— —— 5/192 0 01 —— —— —— —— —— —— —— —— 3/4 1/2 1/12 10 3 1 3 1 3 4?5ql?16l?2?12?2???192EI ?4??12l??48??29ql22?32EI?解得：

?2?77ql396?52EI227ql4?0.0514ql3EIql，

4?2?256?39EI2?0.0227EI

?Cijkij M12??0.140ql，

?ij nij M23?0.14ql2 ，

Mij/ql 2-1/12 N21?0.040ql 36

37

N23??0.040ql

1）不计45杆的轴向变形，由对称性知，4、5节点可视为刚5.7题

性固定端

计算如表所示 2) Q??3节点号 1 2 3 4 23?12q0?3l02q0l0,Q34?0.6q0?3l0??1.8q0l0 杆件号ij 12 21 23 24 32 42 MQ323?23(3l0)/15?10q20l0I—— 2 3 8 ——— ij/I0 M?Q9232?23(3l0)/10??20q0l0

l/l—— 1 2.2 3 —— —— ij0 — M34?Q34(3l0)/12?9k20q0l20

ij —— 2 15/11 8/3 — —— —3) 计算由下表进行： C—— 3/4 3/4 1 — —— ij — M2C—— 3/2 45/44 8/3 — —— 18??M12?0.0039q0l0，

ijkij ?—— 198/685 297/1507 1056/2055 —— —— M2ij —21??0.0786q0l0

nij —— 0 0 1/2 — —— — M?0.518q232??M34?0l0，

Mij/Ql0 0 2/15 0 -3.3 0 21/5 Mm25??0.0341q20l0

ij/Ql0 0 0.9153 0.6241 1.6273 0 0.8136 m'ij/Ql0 M43??0.4159q0l20，M23?0.1127q20l0

Mij/Ql0 0 1.0487 0.6241 -1.6273 0 5.0136 M52??0.0170q20l0， 其它均可由对称条件得出。 5.8题

37

,

38

节点号 杆件号ij Iij/I0 lij/l0 kij Cij Cijkij 1 18 1 6 1/6 1/2 1/12 13/12 1/13 —— 12/13 1/2 0 -.045 .04154 -.00537 0.3 1/2 0 -.009 .02077 -.01073 12 1 1 1 1 1 21 1 1 1 1 1 2 25 1 3 1/3 1 1/3 10/3 0.1 1/2 0 -.003 -.00358 0.6 1/2 0.3 -.018 .015 -.02146 1/3 1/2 -0.45 -.009 .003 -.01073 23 6 3 2 1 2 32 6 3 2 1 2 3 34 12 3 4 1 4 2/3 1/2 0.45 .06 4 43 -0.45 .03 5 52 0 -.015 -.00179 ?Cijkij ?ij nij Mij/ql 20 mij/qlm'ij20.00346 /ql2 .00041 .00496 -.00064 .00248 -.00128 -.00043 .00179 -.00256 .00358 -.00128 .00715 .00358 .00022 38

39

Mij/q0l0 2.00005 -0.0039 .00059 -.00008 0.0039 .00030 -.00016 -.00001 -0.0786 -.00005 -.00000 -0.0341 .00022 -.00031 .00003 -.00002 0.1127 .00043 -.00016 .00005 -.00001 -0.5181 .00085 .00011 0.5181 .00043 .00006 -0.4159 -.00003 -0.0170

图5.4a 图5.4b

39

40

5.9 题

任一点i的不平衡力矩为

Mi??sMis?ql12?ql12?0（i=1，2，…,h,i,j,…n-1. s=i-1,i+1）

所以任一中间节点的分配弯矩mij与传导弯矩m'ij?njimji均为0。 任一杆端力矩：Mij?Mij?mij?m'ij

?Mij??ij?Mis?nji???ji?Mjs??Mij?0?i?n?

s????s对两端i?0,n，由于只吸收传导弯矩m'ij?0 Mij?Mij?m'ij?Mij

两端所以对于每个节都有杆端力矩Mij?Mij

说明：对图5.4b所示载荷由于也能使?Mi?0，也可以看作两端刚固的单

跨梁。

40

41

第6章 能量法

6.1题

1）方法一 虚位移法

考虑b),c)所示单位载荷平衡系统， 分别给予a)示的虚变形 ：

M(x)EIdx??d?

外力虚功为 ?W??虚应变能为

?V=?1??i?? ?1??j??M(x)MEI01l0(x)dx

?1?? =?EI?1??EI?0?Ril0ilx?Mi??0R?x1?idx

??Rx?M??Rx?dx0ii

?l???EI??=??l??EI???Mi3Mj3?Mj?l?1??M?M..........b)?j??i6?3EI?2?Mi?l?1?M?M...........c)??i??j6?3EI?2?

?由虚功原理：?W=?V 得：

?1??i?l? ?????3EIj??1????2?1?2??Mi???? Mj?1????2）方法二 虚力法（单位虚力法） ?　梁弯曲应力：???＝M?x?Iy

??＝?E＝M?x?EIy

Mx??Mi??Mi?Mlj?x

41

42

?M?x??1?(1?0)

lx给Mi以虚变化?Mi?1 虚应力为 ????＝虚余功：?W?＝?i?1

?M?x?Iy

虚余能：?V*＝?（真实应变）?（虚应力）d?

?????1EI1EIM?x??M?x?yydxdydz EII??2?l0l0M?x??M?x?dx?Ay2dA

j??Mi??Mi?M?1?x/l?dx ?x/l????

Qi?1??M?Mj??i3EI?2?l

同理：给Mj以虚变化?Mj?1，??Mi?0?可得（将i换为j）

?j??Mi???Mj??3EI?2?l

3）方法三 矩阵法（柔度法） ??i??Mi???Mi?设??????,?p????，虚力??p????,?????p????

?M?Mj??j??j?? ??????M?x?Iy??Mi??Mi?MI?yy?x/l??j?Ix?1??l??x??Mi?????c??p? ?l??Mj?式中?c??y??x??x??1??,??????I??l??l??（不妨称为物理矩阵以便与刚度法中几何矩阵

?B?对应）

??Mi?? ?Mj???1虚应力??????c???p???c???1实应变?????D??????D??C??p?

虚余功 ?W*??????p????p???????i?Mi??j?MTTj?

42

43

虚余能 ?V*?????????d??????????d?

??TT

????p??C??D??C??P?d????p??TT?1T??T?1???C??D??C?d???p? ???于虚力原理：?W*??V*考虑到虚力??p?的任意性。得： ?????p???C??D???A??? p??C?d?T?1式中 ?A????C??D??C?d?——柔度矩阵（以上推导具有普遍意义）

?T?1对本题：

x??1??y??y?x??l?1??????I?x?EI??l???l???2??x???1??ll?x?1????d???l?EI0?x?x???1???l??l???A????x?x????1???l?l??dx2??x??????l??

?由

1?l/3?l/?6l???EI??l/6l/3?3EI????1?1/21?/2? ?1????A??p?展开得： ??i?l ?????j?3EI?1/?2?Mi??1???M? ?1/21???j?6.2题

方法一 单位位移法

???uj?ui?/l ， ??E??E?uj?ui?/l

ui/l??1/l设 ?ui?1，则 ?????

Ti?1???l?uEj?ui???1/l?d???EAl2??u0lj?ui?dx?EAl?ui?uj?

同理，令?uj? 可得

Tj?1???l?uEj?ui??1/l?d??EAl?uj?ui?

?Ti?EA?1即：????Tlj??1???1??ui???? 可记为 1??uj??p???K????

ijij?K?为刚度矩阵。

43

44

方法二 矩阵虚位移法 设?pij????TiTj??T ??i???j?u1iu??jT

? {?}??uj?ui?1l?ui?/l???1??1???l?uj??B??????j i?式中 ?B????11?——几何矩阵

? ?????D??????D??B???ij?

设虚位移

??ij?????uiT?uj??T ， 虚应变 ??????B????ij?

T外力虚功 ?W??pij?????ij?????ij???p?

ij虚应变能 ?V?????????d??????????d? ?TT????i???????B??D??BjTTij d?d??? ????i??????B??D??Bj???T?T??

ij ????i?K???j??

ij由 ?W??V 得： ?pi???K???jT?

ijd——刚度矩阵 式中 ?K????B??D???B??1??1?1EA?1对拉压杆元 ?K??EA?????11?dx??l?1?ll??1l?1?? 详细见方法一。 1?方法三 矩阵虚力法 设 ?pij??Ti??ui???? ， ??ij???? ， ?????D???? ?Tj??uj?Tj?TiA1A?1? ?????Ti????11?????C??pij? A?Tj?11?——物理矩阵（指联系杆端力与应力的系数矩阵）

?1 式中 ?C???

??1 ?????D??????D??C??pij? 虚应力 ??????C???pij?

??Ti??1??? ， 则 ??????D??C???pij? ??Tj? 设虚力 ??pij? 44

45

虚余功 ?W*???ij?T??pij????pij??TT?1T???

ij 虚余能 ?V*?????????d??????????d?

?TT ?

p??C??????p??C??Dij?ij?d

?T???pi?C??j?????D???1?Cd??????p

ij ???pij??A??pij? 式中

A????C???C??D?T?1?d ——柔度矩阵

?1?? 1?对拉压杆： ?K???

AE?l1??1?1l?1?11dx??????A?1?AEA??1 ??i???A??pj?

ij?ui?l?1 即 ????uEAj??1???1??Ti???? 1??Tj?讨论： 比较方法二、三。

结论： ?pij???K???ij?， ??i???A??pj?

ij?1若 ?K?与?A?的逆矩阵存在（遗憾的是并非总是存在），则，?K?实际上是一个柔度矩阵，?A?实际上是一个刚度矩阵

6.3题

1）6.30如图所示 设vx??2n?x??a1?cos?n??

l??n?1??1

显然满足x?0,x?l处的

变形约束条件

v?0??v?l??0v?0??v?l??0''

EI?2n?x??2n??acos??n??dx ??02l???n?1?l??l?22变形能 V?EI2?l0(v)dx?''2 45

46

?2n??l ? a???2n?1?l?2EI2n?4力函数 ??pv?c??pv?l?c??2pv?c?（对称） ?2p?an??1?cosn?1??2n?c??l?

EIl22n?l2n?c??4)?2p?1?cos?

l??由

??V????an?0 ，所以

?V?an????an 。即

an(所以， an?pl43pl344EI??2n?c???1?cos?l???4n

v?x??4?EI?n?11?2n?c??2n?x?1?cos1?cos??? 4?n?l??l? 2）6.40如图所示

?设v?x??a0x??ansinn?1n?xl

2?EI??n?x?EI?n???v?l???V??asindx????n????20?ll?2A2??n?1??l?22l?a0l?2?n???an?l?2?2A

??n?1?42 ??pU?c??p?ansinn?1n?cl?a0pc

由

??V????a0?0

得 a0l2/A?pc ， 所以，a0?Apc/l2 由

??V????an4?0， 得

32plEIl?n??n?ca? 所以，a?psinn??n2?l?lEI?n??4sinn?cl

46

47

? 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 47

48

? v?x??0.1745ql2EIx?0.06772qlEI3x

6.4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???0.00718ql4EI

ql U?x??0.00718EI4?xsi nl6.5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??? 48

49

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?44

49

50

6.6题 取v1(x)?V?EI2?ansinn?xl,v2(x)?'2?bnsinn?xl

?l0v12dx?1GAs2?l0v2dx22GAsEIl?n?n?x??? ??asindx??????n?20?ll2??????n?x??n??bcos?n?l??dx?0?l????l2 ?EI2EIl4???lGAs2?n??an???2?l?2GAsl?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,? ? ? ? ? )6.7题

1）图6.9 对于等断面轴向力沿梁长不变时，复杂弯曲方程为：

EIVIV?TV?q?0''

能满足梁段全部边界条件

取v(x)??nansinn?xl 50

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