曲線面積的二次矩(慣性矩) - 土木
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By Franklin
at 2013-12-16T14:57
at 2013-12-16T14:57
Table of Contents
題目:求x軸的慣性矩
http://farm8.staticflickr.com/7420/11398182013_374eae3198_c.jpg
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答案:18
小弟的算法:
x = k*y^3
4 = k*27
k = 4/27 = 0.15
x = 0.15*y^3
y = 1.88*x^(1/3)
4
Ix =∫{[(1/2)*y]^2} * dA = ∫{[(1/2)*y]^2}*y*dx = ∫ [(1/4)*y^3]*dx
0
4 4
=∫ 1.66*x*dx = 0.83x^2 | = 13.28
0 0
小弟積分出來的結果為Ix = 13.28,但是答案為Ix = 18
請問版上前輩們,後學哪個地方計算錯誤,還是哪邊的觀念有問題呢?
麻煩不吝嗇指導與告知,謝謝!
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