Computing Matrices with Multiple Variables that are Unknown
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Please utilized following code to optimize solutions for unknown variables. Only need assistance with %Solve Section
%% PS3 - 1A
clear all
%% Givens
L = 10 %in
E = 10 * 10^6 %PSI
w = .5 %in
I = (w^4)/12
%% k1 = k2
j = E*I/(L^3)
k = j * [12 6*L -12 6*L
6*L 4*L^2 -6*L 2*L^2
-12 -6*L 12 -6*L
6*L 2*L^2 -6*L 4*L^2]
%% K
K = [ k(1,1) k(1,2) k(1,3) k(1,4) 0 0
k(2,1) k(2,2) k(2,3) k(2,4) 0 0
k(3,1) k(3,2) (k(3,3) + k(1,1)) (k(3,4) + k(1,2)) k(1,3) k(1,4)
k(4,1) k(4,2) (k(4,3) + k(1,2)) (k(4,4) + k(2,2)) k(2,3) k(2,4)
0 0 k(1,3) k(2,3) k(3,3) k(4,3)
0 0 k(1,4) k(2,4) k(3,4) k(4,4)]
%% U
v1 = 0
%v2 =
v3 = 0
%t1 =
%t2 =
%t3 =
syms v2 t1 t2 t3
U = [v1 t1 v2 t2 v3 t3]'
%% F
q = 10 %lb/in
F1R = 75 %lb RESULTANT FORCE
F3R = 25 %lb RESULTANT FORCE
F1 = F1R - (q*L)/4
F2 = -(q*L)/4
F3 = F3R
M1 = -(q*L^2)/48
M2 = (q*L^2)/48
M3 = 0
F = [F1 M1 F2 M2 F3 M3]'
%%Solve
Solved=solve(K.*U==F,[t1 v2 t2 t3])
vpaSolved = vpa(struct2cell(Solved))
5 个评论
Walter Roberson
2024-3-11
There is no k2
Romualdo Cipro
2024-3-11
You must decide which of the variables you want to prescribe and which you want to solve for:
Solved=solve(K.*U==F,[f1 f3 F1 F3 v1 v2 v3 t1 t2 t3])
f1 and f3 are unknown variables in your code, F1 and F3 are set to numerical values - thus needn't be solved for, v1 and v3 are set to numerical values - thus needn't be solved for.
Romualdo Cipro
2024-3-12
Hi Romualdo,
I modified the code a bit to use all symbolic constants, real variables, and used transpose, .' instead of ctranspose, ' as it seemed that is the intent. The final equation to be solved used element-wise multiplication times, .* with singleton expansion instead of matrix mutliplication mtimes, *; the former results in an obviously inconsistent set of equations. Still can't find a solution.
L = sym(10); %in
E = sym(10) * 10^6; %PSI
%w = .5 %in
w = sym(1)/2;
I = (w^4)/12;
%% k1 = k2
j = E*I/(L^3);
k = j * [12 6*L -12 6*L
6*L 4*L^2 -6*L 2*L^2
-12 -6*L 12 -6*L
6*L 2*L^2 -6*L 4*L^2];
%% K
K = [ k(1,1) k(1,2) k(1,3) k(1,4) 0 0
k(2,1) k(2,2) k(2,3) k(2,4) 0 0
k(3,1) k(3,2) (k(3,3) + k(1,1)) (k(3,4) + k(1,2)) k(1,3) k(1,4)
k(4,1) k(4,2) (k(4,3) + k(1,2)) (k(4,4) + k(2,2)) k(2,3) k(2,4)
0 0 k(1,3) k(2,3) k(3,3) k(4,3)
0 0 k(1,4) k(2,4) k(3,4) k(4,4)];
%% U
v1 = 0;
v3 = 0;
% assume real
syms v2 t1 t2 t3 real
% use transpose
U = [v1 t1 v2 t2 v3 t3].' ;
%% F
q = sym(10); %lb/in
F1R = sym(75); %lb RESULTANT FORCE
F3R = sym(25); %lb RESULTANT FORCE
F1 = F1R - (q*L)/4;
F2 = -(q*L)/4;
F3 = F3R;
M1 = -(q*L^2)/48;
M2 = (q*L^2)/48;
M3 = 0;
% use transpose
F = [F1 M1 F2 M2 F3 M3].' ;
Here's the equation to solve. It appears to not be correctly formulated, just by looking at the first row.
%%Solve
K.*U == F
The preceding equation used element-wise multiplication. Maybe matrix multiplication was intended?
K*U == F
Though these six equations in four unknowns also don't have a solution.
Solved=solve(K*U==F,[t1 v2 t2 t3])
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2024-3-11
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