Elasticity/Energy methods example 2

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Example 2

Given:

The potential energy functional for a membrane stretched over a simply connected region ๐’ฎ of the x1x2 plane can be expressed as

Π[w(x1,x2)]=12๐’ฎη[(w,1)2+(w,2)2]dA๐’ฎpwdA

where w(x1,x2) is the deflection of the membrane, p(x1,x2) is the prescribed transverse pressure distribution, and η is the membrane stiffness.

Find:

  1. The governing differential equation (Euler equation) for w(x1,x2) on ๐’ฎ.
  2. The permissible boundary conditions at the boundary ๐’ฎ of ๐’ฎ.

Solution

The principle of minimum potential energy requires that the functional Π be stationary for the actual displacement field w(x1,x2). Taking the first variation of Π, we get

δΠ=η2๐’ฎ[(2w,1)δw,1+(2w,2)δw,2]dA๐’ฎpδwdA

or,

δΠ=η๐’ฎ[w,1δw,1+w,2δw,2]dA๐’ฎpδwdA

Now,

(w,1δw),1=w,11δw+w,1δw,1(w,2δw),2=w,22δw+w,2δw,2

Therefore,

w,1δw,1+w,2δw,2=(w,1δw),1w,11δw+(w,2δw),2w,22δw

Plugging into the expression for δΠ,

δΠ=η๐’ฎ[(w,1δw),1+(w,2δw),2(w,11+w,22)δw]dA๐’ฎpδwdA

or,

δΠ=η๐’ฎ[(w,1δw),1+(w,2δw),2]dAη๐’ฎ2wδwdA๐’ฎpδwdA

Now, the Green-Riemann theorem states that

๐’ฎ(Q,1P,2)dA=๐’ฎ(Pdx1+Qdx2)

Therefore,

δΠ=η๐’ฎ[(w,1δw)dx2(w,2δw)dx1]๐’ฎ[η2w+p]δwdA

or,

δΠ=η๐’ฎ[w,1dx2dsw,2dx1ds]δwds๐’ฎ[η2w+p]δwdA

where s is the arc length around ๐’ฎ.


The potential energy function is rendered stationary if δΠ=0. Since δw is arbitrary, the condition of stationarity is satisfied only if the governing differential equation for w(x1,x2) on ๐’ฎ is

η2w+p=0(x1,x2)๐’ฎ

The associated boundary conditions are

w,1dx2dsw,2dx1ds=0(x1,x2)๐’ฎ

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