Hello Matthew,
for me it is by no means clear what you want to do.
What is the w[alpha] you are talking about? In your last post I see only w[ x, y ].
If the "Boundary Conditions" are the equations you want to be fulfilled with a certain alpha that seems to be impossible.
With (giving your w [ x, y ] from above, but as function)
q = 15.904;
v = 0.12157;
E1 = 18904;
b = 1830;
a = 1*b;
h = .04*b;
D1 = (E1 h^3)/(12 (1 - v^2));
w[x_, y_] :=
q/(24*D1)*(y^4 - 2*b*y^3 + b^3*y) + (2*q)/(D1*b)*
Evaluate[
Sum[(((n Pi)/b*x*Sinh[(n Pi)/b*x] -
Cosh[(n Pi)/b*
x]*(2 + (\[Alpha] n Pi)/(2 b)*
Tanh[(\[Alpha] n Pi)/(2 b)])) Sinh[(n Pi)/b*
y])/(((n Pi)/b)^5*Cosh[(\[Alpha] n Pi)/(2 b)]), {n, 1,
5}]];
some equations are true for any alpha:
w[0, 0]
w[.5 a, 0]
Out[170]= 0
Out[171]= 0
For dealing with other equations I define a new function
f[x_, y_, a_] := w[x, y] /. \[Alpha] -> a
and find
f[-.5 a, b, 1340]
f[-.5 a, b, 1360]
Out[155]= 1.01387*10^6
Out[156]= -459226.
Here we have different signs, therefore I can do a bisection
eps = 10^-6;
xxx = 10;
a1 = 1340;
a2 = 1360;
While[
xxx > eps,
a3 = (a1 + a2)/2;
If[
Sign[f[.5 a, b, a3]] == Sign[f[.5 a, b, a2]],
a2 = a3,
a1 = a3];
xxx = Abs[a2 - a1]
]
a3 // N
f[.5 a, b, a3]
1353.4
0.0386934
0.0386 is not zero, ok, but decreasing eps should give better values.
So w[ .5 a , .5 b ] == 0 is almost fulfilled.
But w [ -.5 a, .5 b ] is not
f[-.5 a, .5 b, a3]
Out[26]= -9041.8
In fact I doubt that you can achieve all your "Boundary Conditions" with a single alpha
