I get "empty syms" solution when solving inequalities

1 view (last 30 days)
I am trying to solve the following inequality. I have done similar inequality and they all worked out fine, could anyone tell me what is wrong with my setup?
syms s
ox = 2*s;
oy = -3*s;
txy = 4*s;
oxy = [ox; oy; txy];
%ultimate stresses, 1 is fiber direction 2 is matrix direction
ut1stress = 1500;
uc1stress = -1500;
ut2stress = 40;
uc2stress = -246;
ut12 = 68;
%layup angle
theta = deg2rad(60); %angle from x to 1, input degree value ,change to rad
%o1 = ox * cos(theta)^2 + oy * sin(theta)^2 + 2 * txy * sin(theta) * cos (theta)
%o2 = ox * sin(theta)^2 + oy * cos(theta)^2 - 2 * txy * sin(theta) * cos (theta)
c = cos(theta);
s = sin(theta);
%T is global to local stress
T = [c^2 s^2 2*s*c; s^2 c^2 -2*s*c; -1*s*c s*c c^2 - s^2];
o12 = T*oxy;
o1 = o12(1, 1);
o2 = o12(2, 1);
t12 = o12(3, 1);
%tsaihill needs to be less than 1
tsaihill = (o1/ut1stress)^2-((o1*o2)/(ut1stress^2))+(o2/ut2stress)^2+(t12/ut12)^2;
%check how to get less than 1
checkvalue = 1;
eqn1 = tsaihill < checkvalue;
solvetsaihill = solve(eqn1, s,'ReturnConditions',true)
solvetsaihill =
struct with fields:
s: [0×1 sym]
parameters: [1×0 sym]
conditions: [0×1 sym]

Accepted Answer

Walter Roberson
Walter Roberson on 17 Dec 2021
Edited: Walter Roberson on 17 Dec 2021
syms s
ox = 2*s;
oy = -3*s;
txy = 4*s;
oxy = [ox; oy; txy];
%ultimate stresses, 1 is fiber direction 2 is matrix direction
ut1stress = 1500;
uc1stress = -1500;
ut2stress = 40;
uc2stress = -246;
ut12 = 68;
%layup angle
theta = deg2rad(60); %angle from x to 1, input degree value ,change to rad
%o1 = ox * cos(theta)^2 + oy * sin(theta)^2 + 2 * txy * sin(theta) * cos (theta)
%o2 = ox * sin(theta)^2 + oy * cos(theta)^2 - 2 * txy * sin(theta) * cos (theta)
c = cos(theta);
s = sin(theta);
%T is global to local stress
T = [c^2 s^2 2*s*c; s^2 c^2 -2*s*c; -1*s*c s*c c^2 - s^2];
o12 = T*oxy;
o1 = o12(1, 1);
o2 = o12(2, 1);
t12 = o12(3, 1);
%tsaihill needs to be less than 1
tsaihill = (o1/ut1stress)^2-((o1*o2)/(ut1stress^2))+(o2/ut2stress)^2+(t12/ut12)^2;
%check how to get less than 1
checkvalue = 1;
eqn1 = tsaihill < checkvalue;
string(eqn1(:))
ans = "(((3*s)/4 - 2*3^(1/2)*s)*((7*s)/4 - 2*3^(1/2)*s))/2250000 + ((3*s)/160 - (3^(1/2)*s)/20)^2 + (s/34 + (5*3^(1/2)*s)/272)^2 + ((7*s)/6000 - (3^(1/2)*s)/750)^2 < 1"
%solvetsaihill = solve(eqn1, s,'ReturnConditions',true)
F = lhs(eqn1) - rhs(eqn1);
bounds = vpasolve(F, -10)
bounds = 
  2 Comments
JingChong Ning
JingChong Ning on 17 Dec 2021
Thank you for answering my question. But this requires me to know that -10 is a value close to my guess, right? Could you help me find a way to solve this if the value of all input parameter changes?

Sign in to comment.

More Answers (0)

Products


Release

R2020b

Community Treasure Hunt

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

Start Hunting!