哈尔滨工程大学船舶结构力学课程习题集答案(2)

2.7图：（剪力弯矩图如２．６）

?v1?A1R1?v2?A2R2?0.05lEIl433??ql2ql24??ql440EIql450EI100EI5qlql?11??l?v????????2?384EI2EI?40100?ql?57?293ql?????EI?384400?9600EI44

图２．６ ??0??ql3

24EIql3?v1?v2l?ql?111?2ql??????EI?2440100?75EI333??l????v1?v2l24EI?ql?111??17ql??????EI?2440100?300EI3

图2.8（剪力弯矩图如２．７）

21?b???M???12A??1???24KA?l?????Qa由Q?qa,a?l,b?0,??1,A?1824 KA?1?1?1?1,代入得 82432M?R1?qlql2224??2?12?1ql8?ql3ql84?24?1??ql28,v0?AR1?64EI 图２．７

5qlqlMl5ql?l??v??????384EI?2?384EI128EI16EIql34424

ql?111??(0)????????24EIl6EIEI?246448??ql3v0Ml3

192EIl8EI?ql82?(l)???M??

??ql364EI6

2.6题

dv2??max.dx?v2??maxGdx??NGAsEIdx?GAN1dx?????N?EIv???sGAsv1???C132EI?v?v1?v2??f(x)?ax?bx?cx?d????62??GAs?f??(x)?ax?b??C11?f(x)?EIGAsqxf??(x)?ax436?bx2??EI??c?a?x?d2GAs??qx2式中由于由f(x)?24EIf??(x)?可得出2EId1?b?0v(0)?v1??(0)?0v(l)?v1??(l)?0得方程组：23?ql4??EIqlalEI???c?a???l?06GAs??24EIGAs2EI??2?ql?al?0??2EI解出：a=ql2EIqx4,c?qlx3ql324EIqx2 .

?qx3ql??v(x)???????x24EI12EI2GAs?24EI2GAs??v()??2384EI8GAsl5ql4

ql22.7.题

?12EI? 先推广到两端有位移?i,?i,?j,?j情形：?令???i??j,?? 2?GAsl???v?ax36?bx22?cx?d1?EIGAsax?d1?v(0)??i????由v1(0)??i?c??i??由v(l)??j而v0??i

?al??j?GAs?????EI?al63?2bl22??il??i?由v1?(l)??j?al2?bl??j 7

???????2??a?j2?l??il1?????解出????i31?b?j??i??j?2?l?ll?1????

???M(0)?EIv1??(0)?EIb???EI???6???2????4?????ji?l?1?????l?EI6?6???????4?????2?????ijij?l?1????l?l??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2.8题 已知：l?3?75?225cm, q??hs?102?51?0t?1.8cm,s?75cm?0?1050kgcm2

0?.75kg76.87 5cm面积 外板1.8?45 cm2距参考轴 cm 0 15.6 I?C?B2面积距 cm 3惯性矩 cm 4自惯性矩 cm 4 81 38.75 119.8 A e?BA?5.04cm0 604.5 B ?11662?604.520 9430.2 9430.2 C=11662 119.8?8610cm 24(21.87)略 2232 2253.9 球扁钢NO24a ? A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： 形心至球心表面y1?h?t2?e?24?0.9?5.04?19.86cm形心至最外板纤维

8

y2?e?tw2?I2?5.94cm?w1?Iy1?8610319.863?433.5cm

y2?8610?5.94?1449.4cm1050?174 u?l2?0AEI22522?10?86106?0.366

x?u??0.988,M?ql2?1(u)?0.98076.8752?225?1210.?988212x?u??2kg2cm0?4?324.) M??中ql24?1?u???M中 915?76.8?75?225??0.9（8kgcm015824158915?1416kg?中球头?固端?端球头板?2?cm?w1433.5?M320424?kgkg??0??1050??1271???max?141622cmcmw21450??M320424kg??0??1050??378?2cmw1433.5????0??1050? 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?1424kgcm21424?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。

2.9.题

?EIvIV?Tv???0,EIv????N?Tv?TEI2

?vIV?V???0,vIV?KV???0式中k?r??4k2T EI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) 9

??

??A3chkl?A4shkl?0??EIk3?A3shkl?A4chkl???p?Tk?A2?A3shkl?A4chkl?

解得：

A1??pkTthkl,A2?pkTpEIk3pkT,A?3pkTthklA,?4?pkT ?v(x)?????thkl?kx?thklchkx?shkx???thkl?1?chkx???shkx?kx???

2.10题

EIvvIVIV?Tv???0?2??EIv?22IV?N?Tv????kv???0式中k?4T?EI?特征方程：r?kr?0 特征根：r?0,1,2?r?3ik?r??4ik

??v?A1?Akx?Asinkx?Acoskx234???EIv??(0)?m?v(0)?0?A1?A4?0?A4k?2?????mEI???v??(l)?0????EIv???(l)??Tv?(l)???Asinkl?Acoskl?034????3???3???k?A3coskl?A4sinkl???k?A2?A3coskl?A4sinkl? ? ?解得：A3??mT?ctgkl,A2?0?m????????v?(0)??Ak?Akcosk?Aksink?Ak?343???2?x?0kEItgkl2.11题

图２．１２0

10