内容发布更新时间 : 2024/4/27 16:24:53星期一 下面是文章的全部内容请认真阅读。

解：Iy??zdA??A2h0b31zdz?bh3 h4zbdz同理：

Iz??y2dA??Ab0h21y(h?y)dy?hb3 b12hhzyOIyz??zydA??A0?bzh01yzdydz?b2h2

8CⅠ.4、试求图示图形对形心轴的Iy和 IzC。 解：（a）建立如图坐标

zC?A1z1C?A2z2C?A3z3C40?180?(?130)?2???55.7mmA1?A2?A32?40?180?240?80zIyC?IyC1?IyC2?IyC311260?2[?40?1803?40?180?(130?55.7)2]??240?803?240?80?55.7212124?18818.2864cm40IzC?IzC1?IzC2?IzC311?2[?403?180?40?180?1002]??2403?80?23808cm4121280y240(a)40

（b）建立如图坐标

zC?A1z1C?A2z2C?A3z3C80?160?(?140)?360?60?130??18.68mm

A1?A2?A380?160?360?60?100?200180z18060yCIyC?IyC1?IyC2?IyC311??160?803?80?160?158.682??100?2003?200?100?18.68220012121??360?603?60?360?(130?18.68)2?11686.6246cm41210080160(b)

IzC?IzC1?IzC2?IzC3?111?3603?60??200?1003??1603?80?27725.33cm4121212

（c）建立坐标如图

zC?A1z1C?A2z2C60?100?(?20)???25.3mm 2A1?A260?100???20zIyC?IyC1?IyC21??60?1003?60?100?5.3212?(

1002030y?64?404???202?25.32)?424cm4IzC?IzC1?IzC2? 1??603?100??204?167cm4126460

第7章 弯曲应力

4.1、作图示结构的弯矩图和剪力图，并求最大弯矩Mmax和最大剪力FQ,max。（内力方程法）

PaaaP

q3a

7qa/6qaaqa11qa/6FQPM2Pa3PaFQ

qa2MFQmax?P49qa2/36；Mmax?2Pa FQmax?qa

49211qa； Mmax?qa 636

qa2

a5qa/22qa3qa/4FQFQqa/4qa/23qaa22qa2Mqaqa2/4qa2/323qa2/4M

FQmax?qa；Mmax?5233qa FQmax?qa；Mmax?qa2

424qLqLqL2

qaa3qa/4qa/4qaFQFQ

MqL2/2qL25qa/4M3qa2/435qa； Mmax?qa2 44FQmax?qL；Mmax?qL2 FQmax?qaqaqa

qaa

2q3aFQ11qa/62a

7qa/6FQ

7a/6qa2qaMqa/22Mqa2121qa2/72

FQmax?qa；Mmax?qa2 FQmax?121211qa ；Mmax?qa 672

4.2、作图示结构的弯矩图和剪力图，并求最大弯矩Mmax和最大剪力FSmax。（简

qaq易方法） a2a qa2qa FQ 3qa22

2qaM

FQmax?3qa；Mmax?5qa2 qqa a2a qa

FQ qa2qa

M qa2/2

F11Qmax?6qa；M2max?qa qa aq

FQ qa

qa2/2M

qa2qaaaqa/2FQqa/2Mqa2/225qa2/8qa/2F?15Qmax2qa；Mmax?8qa2 PPa2aaPFQPPPaMPaF?111212Qmaxq6qa ；Mmax?72qa qa2aaaFQqaqa/4qa2/23qa2/4qa2M/4

FQmax?qa；Mmax?qa2 FQmax?qa； Mmax?32qa 4qaqa/6qa2

qa

qa2aFQaaFQaqa/4qa2/6qa2/613qa2/725qa/6

qa2/4qa25qa/4M5qa/62MFQmax?555qa；Mmax?qa2 FQmax?qa；Mmax?qa2 646qaqa/4qaaqa

qaaqaq2aqa2aaFQ3qa/4qa2

FQqaqa2qaqa2qa2/4M

Mqa2/2qa/322FQmax?qa；Mmax?qa2 FSmax?qa；Mmax?qa2

4.3、截面为No24工字型的梁，受力如图所示。