V?2??rr?p??cos??cos??R?g?rdr?4?R3?g?103?h2?3R?h??g03?2??p?R?gcos?r???g0??rrdr?2?R3?g?cos2?sin?d??3?4R3?h2?3R?h??0?0??R2?p?R?gcos??20?sin2??sin2?0??3?R3?g?cos3??cos3?0????g3?4R3?h2?3R?h??V?2?R?2?tsin?R?p?R?gcos??22??3?0?sin??sin?20R?gcos??cos3?0???2tsin2??3tsin2???g?2h?6tsin2???4R?h2???3??R?????R?ptsin2???2?sin2??sin2???cos?022133??0?R?g??2?sin??sin?0??3?cos??cos?0???????g?6tsin2???4R2?h2???3?h??R?????pzR??????t??p??cos?0?cos??R?g?tR????p??cos?0?cos??R?gtR?R?tsin2??p?2?sin2??sin2?0??R?g??cos?0?2?sin2??sin2?0??13?cos3??cos3???0???????g?6tsin2???4R2?h2??h???3?R????11
???R?p1?cos?02222sin??sin??R?gsin??sin??cos3??cos3?0?00?23tsin??2?2?g?2h??2??4R?h3????R??6tsin2?????????????????100.2?106?sin2??0.5120.02?sin??????1???19656624?10?640?9.81??0.35?sin2??0.51?cos3??0.73???3sin2????50023?221974.4?sin??0.51?20928?cos??0.343?39313.2482sin?523?22.2?sin??0.51?2.1?cos??0.343?3.92sin?523?22.2?sin??2.1?cos??8.14MPasin2????????????????????p??cos?0?cos??R?g?gR?t6tsin2??2h??2?4R?h3??????R????R?p1?cos?0??222233sin??sin??R?gsin??sin??cos??cos??000???23tsin2??2???519.65662423 ?200?31.392??0.7?cos???22.2?sin??0.51?2.1?cos??0.343?22sin?sin?5?200?31.392??0.7?cos???22.2?sin2??2.1?cos3??8.142sin?5?221.974-31.392?cos??22.2?sin2??2.1?cos3??8.14MPa2sin?????????????????4. 有一锥形底的圆筒形密闭容器，如图所示，试用无力矩理论求出锥形底壳中的最大薄膜应力σθ与σφ的值及相应位置。已知圆筒形容器中面半径R，厚度t；锥形底的半锥角α，厚度t，内装有密度为ρ的液体，液面高度为H，液面上承受气体压力pc。
解：圆锥壳体：R1=∞，R2=r/cosα（α半锥顶角），pz=-[pc+ρg(H+x)]，φ=π/2-α,r?R?xtg?
12
31R2?pc?H?g??xR2?r2?Rr?g3???2rtcos??2x2tg2??2?R?pc?H?g??x??R?xRtg????g3???2?R?xtg??tcos?F??R2?pc?H?g???xR2?r2?Rr?g?2?r??tcos?????
r x ??R1???R2??pzttcos?????pc??H?x??g??R?xtg??d??1??g?R?xtg????pc??H?x??g?tg???dxtcos? d??pctg??d2??1?2?gtg??R?Htg???令：?0x????0?2dx2tg???gtcos?dx??在x处??有最大值。??的最大值在锥顶，其值为?。??pctg???pc??1??R??????????H??gR?Htg???pc??H??????????2tg??g?g???????????????2tcos???max5. 试用圆柱壳有力矩理论，求解列管式换热器管子与管板连接边缘处（如图所示）管子的不连续应力表达式（管板刚度很大，管子两端是开口的，不承受轴向拉力）。设管内压力为p，管外压力为零，管子中面半径为r，厚度为t。
1管板的转角与位移 解：○
Q0w1p?w1?w1M0?0???内压引起的周向应变为：
p1Q01??M01?0
2内压作用下管子的挠度和转角 ○
2?R?w2p?2?RpR????2?REtp??pR2w??Etp2转角：
?2p?0
3边缘力和边缘边矩作用下圆柱壳的挠度和转角 ○
13
w2M0??12?4变形协调条件 ○
2M02?D?1?M0?D?M00wQ??213?2Q02?D?1?Q022?D?Q0
Q0M0w2p?w2?w2?05求解边缘力和边缘边矩 ○
Q?2p??2??2M?0
00pR211??M?Q0?0023Et2?D?2?D?pR22M0?2?D?Et6边缘内力表达式 ○
11M0?Q0?02?D?2?D?
pR23Qo??4?D?EtNx?04?4R3D?p??xN???e?sin?x?cos?x???pRe??x?sin?x?cos?x?Et?2R2D?p??xMx??2e?sin?x?cos?x? EtM???Mx4?3R2D?p??xQx??ecos?xEt7边缘内力引起的应力表达式 ○
Nx12Mx24?2R2D?p??x?x??z??e?sin?x?cos?x?z34ttEt?N?12M?pR??x?24?2RD????x??????????e?sin?x?cos?x?esin?x?cos?xz?3ttt3Et???z?06Q?x?3xt322??t2??24?RDpt22???x???z?ecos?x4?4?z??????Et???4?
8综合应力表达式 ○
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pRNx12MxpR24?2R2D?p??x??x?2t?t?t3z?2t?Et4e?sin?x?cos?x?zpRN?12M?????t?t?t3??pR?24?2RD????x??x??e?sin?x?cos?x?z???1?e??sin?x?cos?x??3t?Et???
??z?0322??t2??24?RDpt22???x?????z???zecos?x4?4???Et???4?6Qx???xt36. 两根几何尺寸相同，材料不同的钢管对接焊如图所示。管道的操作压力为p，操作温度为0，环境温度为tc，而材料的弹性模量E相等，线膨胀系数分别α1和α2，管道半径为r，厚度为t，试求得焊接处的不连续应力（不计焊缝余高）。
1内压和温差作用下管子1的挠度和转角 解：○
内压引起的周向应变为：
2?r?w1p?2?r1?prpr?????????2?rE?t2t?p??pr2?2??? w??2Etp1温差引起的周向应变为：
???t?t?t2?r?w1?2?rw1?????1?t0?tc???1?t2?rr???tw1??r?1?t
w转角：
p??t1pr2?2????r?1?t ??2Et?1p??t?0
2内压和温差作用下管子2的挠度和转角 ○
内压引起的周向应变为：
2?r?w2p?2?r1?prpr?????????2?rE?t2t?p??pr2?2??? w??2Etp2 15