清空筛选条件
清空筛选条件

Solving simultaneous equations to give me forces at certain loads

4 次查看（过去 30 天）
Gian Paulo
Gian Paulo 2023-3-1
回答： Shushant 2023-3-14
I want to solve these simultaneous equations in a way that allows me to get the values of F (force) at any load applied (w); I want to be able to type in any load and the code gives me the forces F. The angle theta is 45 degrees. To illustrate this, I am building a cardboard bridge and I am trying to find the amount of load I can apply before any truss or member of that bridge fails. The force on the members is F and the load is w (the bridge has a certain breaking load that I am yet to determine but shouldn't be an issue in this code). How can I do this? I tried using syms but I did not get anywhere. Appreciate the help
  1 个评论
Star Strider
Star Strider 2023-3-1
They appear to be linear (unless you are solving for θ), so it should be straightforward.
The linsolve or lsqr functions or simply the mldivide, \ functions would likely be appropriate.

请先登录，再进行评论。

请先登录，再回答此问题。

回答（1 个）

Shushant
Shushant 2023-3-14
Ran in:
Hi @Gian Paulo,
According to my understanding, you are facing difficulty solving your set of simultaneous equation. You can use "syms" to solve the set of equations. Check out this documentation to get a deeper understanding on solving system of equations using "syms" Solve System of Linear Equations - MATLAB & Simulink (mathworks.com).
Here is a sample code on how you can use "syms" to solve your set of equations. I have made some random equation and solved them using "solve" function. Hope this gives you a better understanding.
syms F [2,1];
syms Rx [1,1];
theta = pi/4;
x(1) = F1*cos(theta)+F2+Rx1==0;
y(1) = F1*sin(theta)+Rx1==0;
x(2) = -F1*cos(theta)+F2*cos(theta)==0;
y(2) = -F1*sin(theta)+Rx1-F2*sin(theta)==0;
sol = solve([x(1:end), y(1:end)], [F1, F2, Rx1])
sol = struct with fields:
F1: 0 F2: 0 Rx1: 0

类别

在 Help Center 和 File Exchange 中查找有关 Symbolic Math Toolbox 的更多信息

标签

产品

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by