solving three non linear equation simultaneously??

2018 1 30
2 个回答
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更新时间：2018 2 17
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i have following set of equations-
ex = (1/Ec)*(sigx - v*(sigy + sigz); ---(1)
ey = (1/Ec)*(sigy - v*(sigx + sigz); ---(2)
ez= (1/Ec)*(sigz - v*(sigy + sigx); ---(3)
where,
sigx= -(px*ex*Es + Pfx*Ef*ex + lx);
sigy= -(py*ey*Es + Pfy*Ef*ey + ly);
sigz= -(px*ex*Es + lz);
For equations (1) (2) and(3), known term is ex and hence sigx
Unknowns are ey and ez
Values of other terms are-
Ef = 180000;
Es= 2*10^5;
Ec= 25000;
fx=fy=420;
lx=0;
ly=5;
lz=0;
px=0.016;
py=0.024;
pz= 0.016;
v=0.15;
pfx=0.0125;
pfy=0.0125;
PROBLEM - when i am solving (2) and (3) simultaneously for a given value of ex and hence sigx , I am getting values of ey and ez which are not satisfying (1) equation. What is the reason for this contradiction? Please help!!

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Birdman
Birdman 2018-1-30
What are your expected results for ey and ez?
udita pant
udita pant 2018-1-31
there are no such expected results for ey and ez . But the values of ey and ez should be such that when kept in eq(1) it should give the same value of ex which was initially assumed.

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 采纳的回答

Birdman
Birdman 2018-1-30

0 个投票

I believe what you are looking for is as follows:
syms ex ey ez
Ef=180000;Es=2*10^5;Ec=25000;fx=420;fy=420;Ix=0;Iy=5;Iz=0;px=0.016;py=0.024;pz=0.016;v=0.15;pfx=0.0125;pfy=0.0125;
ex = (1/Ec)*(-(px*ex*Es + pfx*Ef*ex + Ix) - v*(-(py*ey*Es + pfy*Ef*ey + Iy) + -(pz*ez*Es + Iz)));
ey = (1/Ec)*(-(py*ey*Es + pfy*Ef*ey + Iy) - v*(-(px*ex*Es + pfx*Ef*ex + Ix) + -(pz*ez*Es + Iz)));
ez= (1/Ec)*(-(pz*ez*Es + Iz) - v*(-(py*ey*Es + pfy*Ef*ey + Iy) + -(px*ex*Es + pfx*Ef*ex + Ix)));
sol=solve([ex,ey,ez])
See the solutions by typing
sol.ex
sol.ey
sol.ez

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udita pant
udita pant 2018-1-31
Thanks for your help.... but actually i want to find the value of ey and ez for a given value of ex.
Birdman
Birdman 2018-1-31
Then set the valıe of ex to something, then you will have two equations for two unknowns.
udita pant
udita pant 2018-1-31
编辑：Walter Roberson 2018-1-31
syms ey ez
Ef=180000;Es=2*10^5;Ec=25000;fx=420;fy=420;Ix=0;Iy=5;Iz=0;px=0.016;py=0.024;pz=0.016;v=0.15;pfx=0.0125;pfy=0.0125;
ex= 0.00003;
sigx= -(px*ex*Es + pfx*Ef*ex + Ix);
ey = (1/Ec)*(-(py*ey*Es + pfy*Ef*ey + Iy) - v*(sigx + -(pz*ez*Es + Iz)));
ez= (1/Ec)*(-(pz*ez*Es + Iz) - v*(-(py*ey*Es + pfy*Ef*ey + Iy) + sigx));
sol=solve([ey,ez]);
y=sol.ey
z=sol.ez
excheck = (1/Ec)*(sigx - v*(-(py*y*Es + pfy*Ef*y + Iy) + -(pz*z*Es + Iz)))
As u said, i fixed the value of ex and got the value of ey and ez .
But when these values of ey and ez are not satisfying the first equation then...i.e excheck is not equal to ex..
this is the problem!!
Walter Roberson
Walter Roberson 2018-1-31
How are you deriving your equation for excheck ?
Birdman
Birdman 2018-1-31
You can not expect it to be the same. Actually, when you check their difference:
err=abs(ex)-abs(vpa(excheck))
>> 0.00002895
which means that their difference can be easily thought as zero.
udita pant
udita pant 2018-2-17
but this value is very large if in terms of strain so for my project i cannot take it as zero. Is there any method through which i can solve this problem.? Kindly answer.
Walter Roberson
Walter Roberson 2018-2-17
How are you deriving your equation for excheck ?
udita pant
udita pant 2018-2-17
By assuming ex we get the value of ex and ey and consequently sigx ,sigy and sigz. Therefore by back substituting these values in eq(1) i.e sigx= -(px*ex*Es + pfx*Ef*ex + Ix) , I am trying to check the value of ex.
Walter Roberson
Walter Roberson 2018-2-17
I notice that your lines
ey = (1/Ec)*(-(py*ey*Es + pfy*Ef*ey + Iy) - v*(sigx + -(pz*ez*Es + Iz)));
ez= (1/Ec)*(-(pz*ez*Es + Iz) - v*(-(py*ey*Es + pfy*Ef*ey + Iy) + sigx));
assign to symbols on the left that are also used on the right. I wonder if those should be
eqns = [ey == (1/Ec)*(-(py*ey*Es + pfy*Ef*ey + Iy) - v*(sigx + -(pz*ez*Es + Iz))),
ez == (1/Ec)*(-(pz*ez*Es + Iz) - v*(-(py*ey*Es + pfy*Ef*ey + Iy) + sigx))];
sol = solve(eqns,[ey, ez]);
This does not bring ex and excheck into agreement, but might hint at further difficulties.

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更多回答（1 个）

Torsten
Torsten 2018-1-30

0 个投票

As far as I can see, the equations are linear in the unknowns. So you only have to solve a system of linear equations using the "backslash" operator.
Best wishes
Torsten.

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