Problem 5.3
The deflection of a beam on an elastic foundation is governed by the equation
(d4w/ dx4 ) +w=1, where x and w are dimensionless quantities. The boundary
conditions for a simply supported beam are given by transverse deflection =w=0
and bending moment = (d2w/ dx2 ) =0
By taking a two term trial solution W(x) = C1f1(x) +C2f2(x) with f1(x) =sinπx f2(x)
=sin3πx, find the solution of the problem using the Galerkin method.
SOLUTION
(d4w/ dx4 ) +w-1=0
Boundary conditions
W=0 d2w/dx2=0
At supports x=0, x=1
2 term trial solution is
W(x)= C1f1(x) +C2f2(x) with f1(x)= sinπx
F2(x) =sin3πx
Use of Glarkin method :
In Galerkin method ∫v fiRdv=0
So w(x) = C1sinπx+ C2 sin3πx
D4w/dx4= (π)4C1 sinπx+(3π)4C2sin3πx
Substitute values in governing equation, we obtain
R = (π)4 C1 sinπx +(3π)4C2sin3πx+C1 sinπx+C2 sin3πx-1
R= (π4+1)C1 sinπx+(81π4+1)C2 sin3πx-1
∫1f1(x)Rdx=∫01(sinπx)[ (π4+1)C1 sinπx+(81π4+1)C2 sin3πx]
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