Elasticity/Fourier series solutions

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Using the Airy Stress Function : Fourier Series Solutions

Useful for more general boundary conditions.

Suppose Template:Center topφ=f(x2)cos(λx1)=or=φ=f(x2)sin(λx1)Template:Center bottom Substitute into the biharmonic equation. Then, Template:Center topf(x2)=(A+Bx2)eλx2+(C+Dx2)eλx2Template:Center bottom or, equivalently, Template:Center topf(x2)=(A+Bx2)coshλx2+(C+Dx2)sinhλx2Template:Center bottom The hyperbolic form allows us to take advantage of symmetry about the x2=0 plane.

If φ=f(x2)cos(λx1), Template:Center topσ11=λ2f(x2)cos(λx1);σ22=f'(x2)cos(λx1);σ12=λf'(x2)sin(λx1)Template:Center bottom


Example of Fourier Series Technique

File:Elastic beam on foundation.png
Bending of an elastic beam on a foundation

The traction boundary conditions are

 Template:Center topσ12=0;x2=±bσ22=p1(x1);x2=bσ22=p2(x1);x2=bσ11=0;x1=±aTemplate:Center bottom

The problem is broken up into four subproblems which are superposed. The subproblems are chosen so that the even/odd properties of hyperbolic functions can be exploited.

The loads for the four subproblems are chosen to be

 Template:Center topf1(x1)=f1(x1)=14[p1(x1)+p1(x1)+p2(x1)+p2(x1)]f2(x1)=f2(x1)=14[p1(x1)p1(x1)+p2(x1)p2(x1)]f3(x1)=f3(x1)=14[p1(x1)+p1(x1)p2(x1)p2(x1)]f4(x1)=f4(x1)=14[p1(x1)p1(x1)p2(x1)+p2(x1)]Template:Center bottom

The new boundary conditions are

 Template:Center topσ12=0;x2=±bσ22=f1(x1)f2(x1)f3(x1)f4(x1);x2=bσ22=f1(x1)f2(x1)+f3(x1)+f4(x1);x2=bσ11=0;x1=±aTemplate:Center bottom

Let us look at the subproblem with loads ±f3(x1) applied on the top and bottom of the beam. The problem is even in x1 and odd in x2. So we use,

 Template:Center topφ=n=1fn(x2)cos(λnx1)=n=1[Anx2cosh(λnx2)+Bnsinh(λnx2)]cos(λnx1)Template:Center bottom

At x1=a,

 Template:Center topσ11=n=1λn2fn(x2)cos(λna)Template:Center bottom

Hence σ11=0 if λn=(2n1)π/2a.

We can substitute φ and express the stresses in terms of Fourier series.

Applying the boundary conditions of x2=±b we get

 Template:Center topn=1[Anλncosh(λnb)+Anλn2bsinh(λnb)+Bnλn2cosh(λnb)]sin(λnx1)=0n=1[Anλn2bcosh(λnb)+Bnλn2sinh(λnb)]cos(λnx1)=f3(x1)Template:Center bottom

The first equation is satisfied if

 Template:Center topAmλmcosh(λmb)+Amλm2bsinh(λmb)+Bmλm2cosh(λmb)=0(1) Template:Center bottom

Integrate the second equation from a to a after multiplying by cos(λmx1).

All the odd functions are zero, except the case where n=m.

Therefore, all that remains is

 Template:Center top[Amλm2bcosh(λmb)+Bmλm2sinh(λmb)]a=aaf3(x1)cos(λmx1)dx1(2)Template:Center bottom

We can calculate Am and Bm from equations (1) and (2), substitute them into the expressions for stress to get the solution.

We do the same thing for the other subproblems.

The Fourier series approach is particularly useful if we have discontinuous or point loads.


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