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how to calculate scalar with matrix

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Kathleen
Kathleen 2024-2-8
编辑： VBBV 2024-2-8
Ran in:
I need to code the following: I got to part b and I am unsure how to get a scalar in the code.
Goal: (b)
% chapter 2-1
clear; clc; close all
%% normal stress
fprintf('\n ====== Exercise 2.1 a=======\n\n')
tau = [-30 -20; -20 -40]; %2D stress tensor (Mpa)
theta = 10;
fhat = [sind(theta) , cosd(theta)];
nhat = [ cosd(theta) , -sind(theta)];
tnhat = tau * nhat.';
tn = nhat * tnhat %normal stress
%% shear stress
ts = fhat * tnhat
fprintf('\n ====== Exercise 2.1 a end=======\n\n')
%%
fprintf('\n ====== Exercise 2.1 b =======\n\n')
I = [1 0; 0 1];
det[-30-x -20; -20 -40-x] = 0
Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for mismatched delimiters.
fprintf('\n ====== Exercise 2.1 b end=======\n\n')
  2 个评论
VBBV
VBBV 2024-2-8
编辑：VBBV 2024-2-8
Use the symbolic toolbox to declare the unknown variable. Then solve the determinant using solve function for the unknown
syms x
D = [-30-x -20; -20 -40-x];
S = det(D) % determinant
sol = solve(S==0,x)
VBBV
VBBV 2024-2-8
编辑：VBBV 2024-2-8
Ran in:
syms lambda % define lambda as symbolic variable (eigen value)
tau = [-30 -20; -20 -40]; % shear stress
I = [1 0; 0 1]; % identity matrix
S = det(tau - I*lambda) % determinant of characteristic equation
S = 
sol = solve(S==0,lambda) % solve for eigen values
sol = 
double(vpa(sol))
ans = 2×1
-55.6155 -14.3845

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回答（1 个）

Matt J
Matt J 2024-2-8
编辑：Matt J 2024-2-8
Ran in:
Don't see any reason why you'd have to reinvent eigendecomposition:
lambda=eig([-30,-20; -20 -40])
lambda = 2×1
-55.6155 -14.3845

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