請問反矩陣解法 - 土木
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By Edward Lewis
at 2012-08-03T16:57
at 2012-08-03T16:57
Table of Contents
K= { 6EI/L -9EI/2L^2 }
{-9EI/2L^2 27EI/2L^3 }
然後求反矩陣後解為
F= 2L/81EI [ 9 3L ]
[ 3L 4L^2]
----------------------------------------
假設:
外力為[R] 變位為[r]
解題過程通常為[r]=[K]^-1 [R]
其實可以不用套公式去求反矩陣 用你平常熟悉得解聯立就OK!
而且可以省去解K的反矩陣 直接求變位
比如外力[R]=[0]
[P]
可以想成[R]=[K][r]
列聯立
[0] = EI/L [6 -9/2L ] [r1]
[P] [-9/2L 27/2L^2] [r2]
乘出來為
0=EI/L(6*r1-9/2L*r2)--1式
P=EI/L(-9/2L*r1+27/2L^2*r2)--2式
解1 . 2 式 就可以求得r1 r2
其實這過程中都沒有解反矩陣的影子
直接解聯立求變位 優點就是可以少算反矩陣
除非題目刻意要求 不然求反矩陣這步是可以忽略的!
--
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| | | | | | | |
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--
{-9EI/2L^2 27EI/2L^3 }
然後求反矩陣後解為
F= 2L/81EI [ 9 3L ]
[ 3L 4L^2]
----------------------------------------
假設:
外力為[R] 變位為[r]
解題過程通常為[r]=[K]^-1 [R]
其實可以不用套公式去求反矩陣 用你平常熟悉得解聯立就OK!
而且可以省去解K的反矩陣 直接求變位
比如外力[R]=[0]
[P]
可以想成[R]=[K][r]
列聯立
[0] = EI/L [6 -9/2L ] [r1]
[P] [-9/2L 27/2L^2] [r2]
乘出來為
0=EI/L(6*r1-9/2L*r2)--1式
P=EI/L(-9/2L*r1+27/2L^2*r2)--2式
解1 . 2 式 就可以求得r1 r2
其實這過程中都沒有解反矩陣的影子
直接解聯立求變位 優點就是可以少算反矩陣
除非題目刻意要求 不然求反矩陣這步是可以忽略的!
--
|◣ ◢| |◣ ◢| |◣ ◢| ◣ ◤ ◣ ◢ ζ
| | | | | | | |
◢ Bryant ◣ ◢ Bryant ◣ ◢ Bryant ◣ ◣Bryant◢ ◢ Bryant ◣Forever
| | ▌ ▌ ◣ | | ◢ ◣ ◢ Superstar
| 33 | 8 ▌ 24 ▌ ▌|10| ▌ 24 ▌
| | ▌ ▌ | | ▌ ▌ KobeBryant
--
Tags:
土木
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