过程设备设计第三版课后答案及重点(郑津洋) 下载本文

内容发布更新时间 : 2024/12/23 7:48:17星期一 下面是文章的全部内容请认真阅读。

○6边缘内力表达式

Nx?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?xEt○7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??????????esin?x?cos?x?esin?x?cos?xz??3ttt3Et???z?06Qx?x?3t?t224?3R2D?p?t22?2???x????z?ecos?x4?4?z?????Et???4?

○8综合应力表达式

pRNx12MxpR24?2R2D?p??x??x?2t?t?t3z?2t?Et4e?sin?x?cos?x?zpRN?12M?????t?t?t32???pR?24?RD????x??x?e?sin?x?cos?x?z???1?e??sin?x?cos?x??3t?Et?????z?0322??t2??24?RDpt22???x???z?ecos?x4?4?z??????Et???4?6Qx???xt3

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:

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转角:

?t?t2?r?w1?2?rw1?????1?t0?tc???1?t2?rrpr2p??t?2????r?1?tw1??2Et

???tw1??r?1?t

?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??2Et

p2?tw2??r?2?t温差引起的周向应变为:

???t转角:

?t?t2?r?w2?2?rw2?????2?t0?tc???2?t2?rrpr2p??t?2????r?2?tw2??2Et

??

?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

w1M0??12?M102?D?1??M0?D?M0Q0w1?13?Q10○4边缘力和边缘边矩作用下圆柱壳2的挠度和转角

2?D?1?Q022?D?

Q0w2M0??12?M202?D?1?M0?D?M00wQ??213?Q202?D?1?Q022?D?Q0○5变形协调条件

Q0Q0M0w1p??t?w1?w1M0?w2p??t?w2?w2?p??t1??Q01??M01??p??t2??Q02??M02

○6求解边缘力和边缘边矩

pr211pr21?2????r?1?t?2M0?3Q0???2????r?2?t?1?M?Q002Et2Et2?D?2?D?2?2D?2?3D?1111?M0?Q?M?Q00022????D??D2?D2?DM0?0Qo?r?3D??t0?tc???1??2? eq \\o\\ac(○,7)边缘内力表达式

Nx?0Et??xe?t0?tc???1??2?cos?x2Mx?r?2D?e??x?t0?tc???1??2?sin?xN??M???MxQx?r?3D?e??x?t0?tc???1??2??cos?x?sin?x?

○8边缘内力引起的应力表达式

Nx12Mx12z2??x??t0?tc???1??2?sin?x?x??z??r?De33tttN12M12z?E??????3?z?e??x?t0?tc???1??2??cos?x?3?r?2D?sin?x?ttt?2??z?06Qx?x?3t?t26r?t22?2?3??x??????t0?tc???1??2??cos?x?sin?x??z??z?De?4?t3?4?????

Nx12Mx?z?3ttN?12M??z?3ttpr12z2?3r?D?e??x?t0?tc???1??2?sin?x2ttpr12z?E??e??x?t0?tc???1??2??cos?x?3?r?2D?sin?x?tt?2?○9综合应力表达式

pr?2tpr????t???x???z?0?t26r?t22?2?3??x??????t0?tc???1??2??cos?x?sin?x??z??z?De?4?t3?4?????

7. 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=2×105MPa,μ=0.3。试求圆筒内外壁面应力值。 解:周向应变

6Qx???xt3???物理方程

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

???1???????r??z??Ew?r???仅承受内压时的Lamè公式

r???????r??z??E

22??piRi2?R0pi?R0?????r?21??1?2?2?22??R0?Ri?r?K?1?r?22??piRi2?R0pi?R0???????21??1?2?2?22??R0?Ri?r?K?1?r?piRi2pi?z?2?R0?Ri2K2?1在外壁面处的位移量及内径:

wr?R0?piR0?2????w0EK2?1??K?1?Ri?内壁面处的应力值:

piR0?2????1?36?4905??2?0.3??1.188Ew02?10?0.365

R0490??412.538mmK1.188?r??pi??36MPapi1.1882?12???21?K?36??211.036MPaK?11.1882?1p36?z?2i??87.518MPa2K?11.188?1

??外壁面处的应力值:

?r?02pi2?36??175.036MPaK2?11.1882?1p36?z?2i??87.518MPa2K?11.188?1

???8. 有一超高压管道,其外直径为78mm,内直径为34mm,承受内压力300MPa,操作温度下材料的

σb=1000MPa,σs=900MPa。此管道经自增强处理,试求出最佳自增强处理压力。

解:最佳自增强处理压力应该对应经自增强处理后的管道,在题给工作和结构条件下,其最大应力取最小值时对应的塑性区半径Rc情况下的自增强处理压力。对应该塑性区半径Rc的周向应力为最大拉伸应力,其值应为经自增强处理后的残余应力与内压力共同作用下的周向应力之和:

????s??R0?1???R3???c????2????Rc????R?????0?Ri2???R2?R2?0i2??Rc?1???R???02?Rc??piRi2????2lnR???R2?R2i???0i????R0?1???R???c????2令其一阶导数等于0,求其驻点

222????2?sR0Ri2??Rc?Rc????Rc???????????1??2ln???3?22????RcRRRRR?R3c??0?i??0i???0????2222?s??R0???Ri2?Rc1???Rc?2piRiR0??1????2?2??0???22?223??R3??R0R0?Ri?R0Rc???R0?RiRc??c????

????

解得:Rc=21.015mm。根据残余应力和拉美公式可知,该值对应周向应力取最大值时的塑性区半径。

由自增强内压pi与所对应塑性区与弹性区交界半径Rc的关系,最佳自增强处理压力为:

pi??S?R02?Rc2?3??2RoR?2lncRi???589.083MPa??

9. 承受横向均布载荷的圆平板,当其厚度为一定时,试证明板承受的总载荷为一与半径无关的定值。

证明:○1周边固支情况下的最大弯曲应力为

?max○2周边简支情况下的最大弯曲应力为:

3pR23p?R23P???4t24?t24?t2

???max10. 有一周边固支的圆板,半径R=500mm,板厚=38mm,板面上承受横向均布载荷p=3MPa,试求板的

5

最大挠度和应力(取板材的E=2×10MPa,μ=0.3) 解:板的最大挠度:

3?3???pR23?3???p?R23?3???P???8t28?t28?t2

??Et32?105?3839D????1.005?10121??212?1?0.32????w板的最大应力:

fmaxpR43?5004???2.915mm64D?64?1.005?109

?max11. 上题中的圆平板周边改为简支,试计算其最大挠度和应力,并将计算结果与上题作一分析比较。

解:板的最大挠度:

3pR23?3?5002???389.543MPa224t4?38

w板的最大应力:

smaxpR45??5?0.3???2.915?4.077?2.915?11.884mm?64D1??1?0.3

?max简支时的最大挠度是固支时的4.077倍;简支时的最大应力是固支时的1.65倍。

12. 一穿流式泡沫塔其内径为1500mm,塔板上最大液层为800mm(液体密度为ρ=1.5×103kg/m3),塔

5

板厚度为6mm,材料为低碳钢(E=2×10MPa,μ=0.3)。周边支承可视为简支,试求塔板中心处的挠度;若挠度必须控制在3mm以下,试问塔板的厚度应增加多少? 解:周边简支圆平板中心挠度

3?3???pR23?3?0.3??3?5002?3?0.3?????389.543?1.65?389.543?642.746MPa2228t8?38

Et32?105?635D????39.56?10121??212?1?0.32p?h?g?0.8?1500?9.81?11772Pa?0.012MPa????wsmaxpR45??0.012?75045?0.3???61.14mm64D?1??64?39.56?1051?0.3

挠度控制在3mm以下需要的塔板厚度

61.14?20.383需要的塔板刚度:D??20.38?39.56?105?806.2328?105塔板刚度需增加的倍数:225?121??D12?1?0.3?806.2328?10t?3?3?16.4mmE2?105

????需增加10.4mm以上的厚度。