Matlab:如何解决多个变量出现多次的方程组

时间:2015-08-06 10:51:55

标签: matlab wolfram-mathematica wolframalpha

我的方程式如下:

w1   = (1/EI)*(heaviside(x - a1).*(((x - a1).^3)*F1(1))/6 + (C(1, 1)*x.^3)/6 + (C(2, 1)*x.^2)/2 + C(3, 1)*x + C(4, 1)) + (1/GAs).*(- heaviside(x - a1).*(x - a1)*F1(1) - C(1, 1).*x) - C(1, 1)/c_wa;

我使用测量中的七对值来求解这个等式。我设法用cftool-toolbox解决了这个问题。但我真的不想知道Cs,但是c1,c2,k1和k2。 我有四个Cs方程式。现在我想解决c1,c2,k1和k2的这些方程。到目前为止,我不知道如何做到这一点,因为我想解决的变量在每个等式中出现多次。

我尝试将四个方程式插入到我的第一个方程式中并使用cftool来解决它,但这样做不正常,因为它太复杂了。

 C(1, 1) = ((c1*F*(-2*a^3*c2*(EI*(k1 + k2) + k1*k2*l) + 3*a^2*c2*k1*l*(2*EI + k2*l) + 6*a*c2*EI*l*(2*EI + k2*l) - ...
     12*EI^2*(c2*l^2 + k1 + k2) - 4*c2*EI*l^3*(k1 + k2) - c2*k1*k2*l^4 - 12*EI*k1*k2*l))/ ...
     (12*EI^2*(c1 + c2)*(k1 + k2) + 12*c1*c2*EI^2*l^2 + 4*c1*c2*EI*l^3*(k1 + k2) + ...
     12*EI*k1*k2*l*(c1 + c2) + c1*c2*k1*k2*l^4));

C(2, 1) = ((F*k1*(2*EI*l*(a^3*c1*c2 + 6*a*k2*(c1 + c2) - 6*c2*EI) + c2*l^2*(a^3*c1*k2 - 6*a^2*c1*EI - 6*EI*k2) - ...
     2*a*c1*c2*l^3*(a*k2 - 2*EI) - 6*a*EI*(c1 + c2)*(a*k2 - 2*EI) + a*c1*c2*k2*l^4))/ ...
     (12*EI^2*(c1 + c2)*(k1 + k2) + 12*c1*c2*EI^2*l^2 + 4*c1*c2*EI*l^3*(k1 + k2) + ...
     12*EI*k1*k2*l*(c1 + c2) + c1*c2*k1*k2*l^4));

C(3, 1) = ((EI*F*(2*EI*l*(a^3*c1*c2 + 6*a*k2*(c1 + c2) - 6*c2*EI) + c2*l^2*(a^3*c1*k2 - 6*a^2*c1*EI - 6*EI*k2) - ...
     2*a*c1*c2*l^3*(a*k2 - 2*EI) - 6*a*EI*(c1 + c2)*(a*k2 - 2*EI) + a*c1*c2*k2*l^4))/ ...
     (12*EI^2*(c1 + c2)*(k1 + k2) + 12*c1*c2*EI^2*l^2 + 4*c1*c2*EI*l^3*(k1 + k2) + ...
     12*EI*k1*k2*l*(c1 + c2) + c1*c2*k1*k2*l^4));

C(4, 1) = ((EI*F*(2*a^3*c2*(EI*(k1 + k2) + k1*k2*l) - 3*a^2*c2*k1*l*(2*EI + k2*l) - 6*a*c2*EI*l*(2*EI + k2*l) + ...
     12*EI^2*(c2*l^2 + k1 + k2) + 4*EI*l*(c2*l^2*(k1 + k2) + 3*k1*k2) + c2*k1*k2*l^4))/ ...
     (12*EI^2*(c1 + c2)*(k1 + k2) + 12*c1*c2*EI^2*l^2 + 4*c1*c2*EI*l^3*(k1 + k2) + ...
     12*EI*k1*k2*l*(c1 + c2) + c1*c2*k1*k2*l^4));

我会非常感谢一些提示!如果这种问题只能通过Mathematica或Derive解决,我也很乐意为您提供解决方案!

编辑(致Bill 8月7日23:08):

感谢您的回复!

  • “......”表示新行
  • EI,F1(1),GAs,F,c_wa给出变量(你在哪里看到F?)

因此:

C(1, 1) = ((c1*F*(-2*a^3*c2*(EI*(k1 + k2) + k1*k2*l) + 3*a^2*c2*k1*l*(2*EI + k2*l) + 6*a*c2*EI*l*(2*EI + k2*l) - 12*EI^2*(c2*l^2 + k1 + k2) - 4*c2*EI*l^3*(k1 + k2) - c2*k1*k2*l^4 - 12*EI*k1*k2*l))/ (12*EI^2*(c1 + c2)*(k1 + k2) + 12*c1*c2*EI^2*l^2 + 4*c1*c2*EI*l^3*(k1 + k2) + 12*EI*k1*k2*l*(c1 + c2) + c1*c2*k1*k2*l^4));

C(2, 1) = ((F*k1*(2*EI*l*(a^3*c1*c2 + 6*a*k2*(c1 + c2) - 6*c2*EI) + c2*l^2*(a^3*c1*k2 - 6*a^2*c1*EI - 6*EI*k2) - ...
     2*a*c1*c2*l^3*(a*k2 - 2*EI) - 6*a*EI*(c1 + c2)*(a*k2 - 2*EI) + a*c1*c2*k2*l^4))/ (12*EI^2*(c1 + c2)*(k1 + k2) + 12*c1*c2*EI^2*l^2 + 4*c1*c2*EI*l^3*(k1 + k2) + 12*EI*k1*k2*l*(c1 + c2) + c1*c2*k1*k2*l^4));

C(3, 1) = ((EI*F*(2*EI*l*(a^3*c1*c2 + 6*a*k2*(c1 + c2) - 6*c2*EI) + c2*l^2*(a^3*c1*k2 - 6*a^2*c1*EI - 6*EI*k2) - 2*a*c1*c2*l^3*(a*k2 - 2*EI) - 6*a*EI*(c1 + c2)*(a*k2 - 2*EI) + a*c1*c2*k2*l^4))/ (12*EI^2*(c1 + c2)*(k1 + k2) + 12*c1*c2*EI^2*l^2 + 4*c1*c2*EI*l^3*(k1 + k2) + 12*EI*k1*k2*l*(c1 + c2) + c1*c2*k1*k2*l^4));

C(4, 1) = ((EI*F*(2*a^3*c2*(EI*(k1 + k2) + k1*k2*l) - 3*a^2*c2*k1*l*(2*EI + k2*l) - 6*a*c2*EI*l*(2*EI + k2*l) + 12*EI^2*(c2*l^2 + k1 + k2) + 4*EI*l*(c2*l^2*(k1 + k2) + 3*k1*k2) + c2*k1*k2*l^4))/ (12*EI^2*(c1 + c2)*(k1 + k2) + 12*c1*c2*EI^2*l^2 + 4*c1*c2*EI*l^3*(k1 + k2) + 12*EI*k1*k2*l*(c1 + c2) + c1*c2*k1*k2*l^4));

即使您不需要它们来解决问题的理解:

EI = 8.0934e+11
F1 = [10000 345] 
=> F = F1(1) = 10000 
=> a = F1(2) = 345
GAs = 7.0852e+08 
c_wa = 2.115e+05; 

亲切的问候,

AMK

0 个答案:

没有答案