How do you solve for constants in a system of equations?

Good Morning,

I'm trying to determine how to solve for constants in a system of equations with specified conditions. Please see the below forumlaic explanation. I've explored using "solve", but am unsure how to set the condition values as seed for the solution. FYI, this is for solving a cantilever beam deflection with varying cross section (changing area moment of inertia). This is a simplistic representation of the problem by solving for only two distinct sections, but in practice there would be many more depending on the considered sampling length.

Thank you for any feedback or hints for functions to explore.

P, L, E1 & E2 are all known values

s1=(P/(2*E*I1))*(2*L*x-x^2)+C1*x;

s2=(P/(2*E*I2))*(2*L*x-x^2)+C2*x;

v1=(P/(6*E*I1))*(3*L*x^2-x^3)+C1*x+C3;

v2=(P/(6*E*I2))*(3*L*x^2-x^3)+C2*x+C4;

at x=0 s1=0 and v1=0

at x=L/2 s1=s2, and v1=v2

[Sam Chak]

Hi Daniel Wilkinson

What is when ? If you can get , then you can find all of them in this order , , , .

, are found when , while , can be found when .

Answer 1

[Paul]

syms P  L  E1  E2 % are all known values, E1 and E2 not used?
syms I1 I2 E      % also known values?
syms C1 C2 C3 C4
syms x
s1(x) = (P/(2*E*I1))*(2*L*x-x^2)+C1*x;
s2(x) = (P/(2*E*I2))*(2*L*x-x^2)+C2*x;
v1(x) = (P/(6*E*I1))*(3*L*x^2-x^3)+C1*x+C3;
v2(x) = (P/(6*E*I2))*(3*L*x^2-x^3)+C2*x+C4;

% at x=0 s1=0  and  v1=0
eqn1 = s1(0) == 0;
eqn2 = v1(0) == 0;

% at x=L/2   s1=s2,   and  v1=v2
eqn3 = s1(L/2) == s2(L/2);
eqn4 = v1(L/2) == v2(L/2);

sol = solve([eqn1 eqn2 eqn3 eqn4],[C1 C2 C3 C4],'ReturnConditions',true)

sol = struct with fields:
            C1: (3*I1*L*P - 3*I2*L*P + 4*E*I1*I2*z)/(4*E*I1*I2)
            C2: z
            C3: 0
            C4: ((- 5*P*L^3 + 18*P*L^2)*(I1 - I2))/(48*E*I1*I2)
    parameters: z
    conditions: L ~= 0

[sol.C1 sol.C2 sol.C3 sol.C4]

ans = 

It looks like C3 must be zero, C2 can be anything, and C1 and C4 have the values shown, as long as

sol.conditions

ans = 

[Sam Chak]

Kudos to Paul.

Yes, , and this can be obtained from the 3rd equation when .

By rights, is obtainable as well when .

When , we know the relationship from , and we can express from the 3rd equation.

Since , and making the substitution in , we can possibly solve for as a function of , which is .

If , then logically, we can find , followed by .

Since MATLAB returns the value for as a function of the known constants, isn't it true that we can fully find as well?

[Paul]

It appears that the contraints are not sufficient to specify C2.

syms P  L  E1  E2 % are all known values, E1 and E2 not used?
syms I1 I2 E      % also known values?
syms C1 C2 C3 C4
syms x
s1(x) = (P/(2*E*I1))*(2*L*x-x^2)+C1*x;
s2(x) = (P/(2*E*I2))*(2*L*x-x^2)+C2*x;
v1(x) = (P/(6*E*I1))*(3*L*x^2-x^3)+C1*x+C3;
v2(x) = (P/(6*E*I2))*(3*L*x^2-x^3)+C2*x+C4;
% at x=0 s1=0  and  v1=0
eqn1 = s1(0) == 0;
eqn2 = v1(0) == 0;
% at x=L/2   s1=s2,   and  v1=v2
eqn3 = s1(L/2) == s2(L/2);
eqn4 = v1(L/2) == v2(L/2);
sol = solve([eqn1 eqn2 eqn3 eqn4],[C1 C2 C3 C4],'ReturnConditions',true)

sol = struct with fields:
            C1: (3*I1*L*P - 3*I2*L*P + 4*E*I1*I2*z)/(4*E*I1*I2)
            C2: z
            C3: 0
            C4: ((- 5*P*L^3 + 18*P*L^2)*(I1 - I2))/(48*E*I1*I2)
    parameters: z
    conditions: L ~= 0

Now substitute the solution back into the functions:

s1(x) = subs(s1(x),[C1 C2 C3 C4],[sol.C1 sol.C2 sol.C3 sol.C4]);
s2(x) = subs(s2(x),[C1 C2 C3 C4],[sol.C1 sol.C2 sol.C3 sol.C4]);
v1(x) = subs(v1(x),[C1 C2 C3 C4],[sol.C1 sol.C2 sol.C3 sol.C4]);
v2(x) = subs(v2(x),[C1 C2 C3 C4],[sol.C1 sol.C2 sol.C3 sol.C4]);

Evaluate at the constraint conditions. The first two at x =0:

[s1(0) v1(0)]

ans = 

Now we see that s1(L/2) and s2(L/2) are both dependent on the free parameter z = C2

[s1(L/2)  s2(L/2)]

ans = 

But apparently they are equal to each other for any z

simplify(s1(L/2)-s2(L/2))

ans = 

0

And similarly for the contraints on v1(L/2) and v2(L/2)

[v1(L/2) v2(L/2)]

ans = 

simplify(v1(L/2)-v2(L/2))

ans = 

0

It appears that the given conditions are insufficient to uniquely specify C2.

[Sam Chak]

Hi @Paul,

Thank you for the analysis. I learned something new on using the subs function to validate the results.