Boundary value problems (Ordinary differential equations) with a beam

[CptJackWest]

Homework Statement

The deflection y of a non-uniform beam of length equal to 1, simply supported at both
ends and with uniformly distributed load q, is governed by the equation

(E*I(x)*y'')'' + k*y=q
y(0)=0, y''(0)=0, y(1)=0, y''(0)=0

I(x)=A[1-0.5(1-x)²]², 0<=x<=1

where E =Young’s modulus of elasticity, I =moment of inertia, k =elastic foundation,
q =load on the beam, A=area of cross-section of the beam and k = 6EA.
Use the central difference approximation to solve the above differential equation to
compute values for y(0.5) using h=1/2 and h=1/4.

Homework Equations

y'=(y_r+1-y_r-1)/2h

y''=(y_r-1-2y_r+y_r+1)/h²

y'''=(y_r+2-2y_r+1+2y_r-1-y_r-2)/2h³

y''''=(y_r-2-4y_r-1+6y_r-4y_r+1+y_r+2)/h⁴

k=6E*A

The Attempt at a Solution

I am not really sure what to do (E*I(x)*y'')''how to expand brackets with the differentiation on the outside. Other than that I am good with this sought of problem
Thanks Jack

 

[Valkarie]

First we need to expand the given equation: (E*I(x)*y'')'' + k*y = q We can use the chain rule for this: E*I*y'''' + k*y = q Now, using the central difference approximations given, we can rewrite the equation as: E*A[1-0.5(1-x)2]2 * (yr+2-2yr+1+2yr-1-yr-2)/2h3 + 6EA*yr = q Now, we can rearrange the equation to solve for y(0.5) when h=1/2 and h=1/4: When h=1/2: y(0.5) = q/(E*A[1-0.5(1-0.5)2]2 + 12EA) When h=1/4: y(0.5) = q/(E*A[1-0.5(1-0.5)2]2 + 24EA)