Filter löschen
Filter löschen

I get "empty syms" solution when solving inequalities

2 Ansichten (letzte 30 Tage)
JingChong Ning
JingChong Ning am 17 Dez. 2021
Kommentiert: Walter Roberson am 17 Dez. 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]

Akzeptierte Antwort

Walter Roberson
Walter Roberson am 17 Dez. 2021
Bearbeitet: Walter Roberson am 17 Dez. 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 Kommentare
JingChong Ning
JingChong Ning am 17 Dez. 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?

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Produkte


Version

R2020b

Community Treasure Hunt

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

Start Hunting!

Translated by