过程设备设计第三版答案（郑津洋 董其伍 桑芝富主编）最新最全版

Nx?04?RD?pEt43N???Mee??x?sin?x?cos?sin?x?cos?x???pRe?x???x?sin?x?cos?x?x??2?RD?pEtx3222??xM???MQx??4?RD?pEte??xcos?x

○7边缘内力引起的应力表达式

?x?Ntx??12Mtt3xz????24?RD?pEtte??x422e??x?sin?x?cos?x?z???N?t12M?3pR2??24?RD????x????esin?x?cos?xz??sin?x?cos?x?3Et???z?0?x2322????x6Qx?t24?RD?p?t22????e??z???zcos?x34????t?4Et??4?

○8综合应力表达式

??x??pR2tpRt??Ntx??12Mtt3xz?pR2t?24?RD?pEt422e??x?sin?x?cos?x?z???N?t12M?32???pR?24?RD????x??x?sin?x?cos?x?z???e?1?e??sin?x?cos?x??3t?Et???????z?0232?6Qx?t24?RD?p2???z???34??t?4Et?x?t2???x2??cos?x?4?z?e??

6.两根几何尺寸相同，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）。 解：○1内压和温差作用下管子1的挠度和转角 内压引起的周向应变为：

???p2?r?w1?p??2?r2?r1?prpr??????E?t2t?wp1??pr22Et?2???

温差引起的周向应变为：

????t2?r?w1??t??2?r2?rpr2??w1r?t??1?t0?tc???1?tw1??r?1?t?t

wp??t1??2Et?2????r?1?t

转角： ?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：

???p2?r?w2?2?r2?r?p?1?prpr??????E?t2t?wp2??pr22Et?2???

温差引起的周向应变为：

????t2?r?w2??t??2?r2?rpr2??w2r?t??2?t0?tc???2?tw2??r?2?t?t

wp??t2??2Et?2????r?2?t

转角： ?2p??t?0

○3边缘力和边缘边矩作用下圆柱壳1的挠度和转角

wM01????12?D?12M0w1Q0?Q012?D??123Q0Q0

?M01?D?M0?12?D?○4边缘力和边缘边矩作用下圆柱壳2的挠度和转角

w2M0???12?D?M02M0w2Q0???12?D?123Q0?M021?D??Q022?D?Q0

○5变形协调条件

wp??t1p??t?w??1Q01?w??1M01?wp??t2?wQ0Q02?wM0M02?1??MQ0M0??2Mp??t??21??2pr2

○6求解边缘力和边缘边矩

pr122Et?2????r?1?t?M012?D?120?2?D?123Q0??2Et?2????r?2?t?12?D?2M0?12?D?3Q0?D?0?12?D?2Q0??D?M0?2?D?Q0?03Qo?r?D??t0?tc???1??2?○7边缘内力表达式

Nx?0Et2e2??xN??Mx?t0??x?tc???1??2?cos?x?r?D?ex?t0?tc???1??2?sin?x?tc???1??2??cos?x?sin?x?M???M3Qx?r?D?e??x?t0x

○8边缘内力引起的应力表达式

?x?Ntx??12Mtt3z??z?e12zt??x3r?D?e2??x?t0?tc???1??2?sin?x?????zN?t12M?3?t012z?E?2?tc???1??2??cos?x?3?r?D?sin?x?t?2??0?6r6Qx?t22?????z33??t?4t??t2?32???D?e??x?t0?tc???1??2??cos?x?sin?x??z?4???x

○9综合应力表达式

??x?pr2tprt??Ntx??12Mtt3xz?z?pr2tprt?12zt3r?D?e2??x?t0?tc???1??2?sin?xE?2cos?x?12zt3????????

zN?t12M?3?e??x?t0?tc???1??2????r?D?sin?x??2??022???36Qx?t6rt22?????D?e??x?t0?tc???1??2??cos?x?sin?x???z??z33?t?4?t??4???x7. 一单层厚壁圆筒，承受内压力pi=36MPa时，测得（用千分表）筒体外表面的径向位移w0=0.365mm，圆筒外直径D0=980mm，E=2×10MPa，μ=0.3。试求圆筒内外壁面应力值。 解：周向应变

5

????r?w?d?rd??rd??wrw?r??

物理方程

???1E???????r????zw?r???rE???????r??z??

仅承受内压时的Lamè公式

?r??22?R0?pi?R0??1?2???1?2??22?2?R0?Ri?r?K?1?r???piRi222?R0?pi?R0??1?2???1?2??22?2?R0?Ri?r?K?1?r???piRi2?z?piRi222R0?Ri?piK2?1

在外壁面处的位移量及内径： wr?R0?K?E?KpiR02?1??2????w036?490??2?0.3?2?10?0.36551?R0KpiR0Ew?0?2????1??1.188Ri?4901.188?412.538mm

内壁面处的应力值：

?r??pi??36MPapiKK2??????1?1?1?K??36?21.1881.18822?1?1?211.036MPapi2z?361.1882?1?87.518MPa