空間力系彎矩 - 土木

Valerie · 2009-05-31

Table of Contents

想問這一題謝謝

座標
A (2,7,2) B (4,0,5) C (-3,-3,0) D (1,3,5)
若自 D 點朝 A 點方向施一作用力大小為 20KN
求此力對 BC 線造成之彎矩大小為何?

答案： 70.98KN-m

...我只會算此力對"某一點"造成之力矩...囧

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All Comments

Tracy2009-06-04

rBA = (A-B) = [-2, 7, -3]

Daph Bay2009-06-08

eBC=(C-B)/(sum((C-B)**2))**0.5=[-0.7683,-0.3293,-0.548

Dora2009-06-10

eDA=(A-D)/(sum((A-D)**2))**0.5=[.19611 .78446 -.58834]

Noah2009-06-12

P =20*eDA=[ 3.9223227 , 15.68929081, -11.76696811]

Dora2009-06-15

mBC = dot(cross(rBA,P),eBC) = 71.037590070086907 kN-m

Ula2009-06-20

其中 rBA 也可以換成 rBD rCA rCD 答案一樣

Daph Bay2009-06-24

感謝

Rachel2009-06-29

原來-2,7,-3是醬出來的阿~