I get "empty syms" solution when solving inequalities

1 vue (au cours des 30 derniers jours)
JingChong Ning
JingChong Ning le 17 Déc 2021
Commenté : Walter Roberson le 17 Déc 2021
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]

Réponse acceptée

Walter Roberson
Walter Roberson le 17 Déc 2021
Modifié(e) : Walter Roberson le 17 Déc 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 commentaires
JingChong Ning
JingChong Ning le 17 Déc 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?
Walter Roberson
Walter Roberson le 17 Déc 2021
bounds = vpasolve(F)

Connectez-vous pour commenter.

Plus de réponses (0)

Produits


Version

R2020b

Community Treasure Hunt

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

Start Hunting!

Translated by