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— David Wagner 2009/09/22 12:35

√ME 5463-001 Fracture Mechanics, Homework #6

Verify that Z(z) = sigma/{sqrt{ 1-{{ sin^2({pi a}/W) }/{ sin^2({pi z}/W) }} }} satisfies the compatibility equation (biharmonic equation).

The biharmonic equation in two dimensions is

{partial^4phi}/{partial x^4} + {partial^4phi}/{partial y^4} + {partial^4phi}/{partial z^4} + 2{partial^4phi}/{partial x^2 partial y^2} + 2{partial^4phi}/{partial y^2 partial z^2} + 2{partial^4phi}/{partial x^2 partial z^2} = 0,
which simplifies in two dimensions x,y to {partial^4phi}/{partial x^4} + {partial^4phi}/{partial y^4} + 2{partial^4phi}/{partial x^2 partial y^2} = 0,
and in one dimension z to
{partial^4phi}/{partial z^4} = 0.

Mostly Unnecessary Setup

First, simplify the stress function it a little by factoring out the constant stress:
sigma Z(z) = sigma/{sqrt{ 1-{{ sin^2({pi a}/W) }/{ sin^2({pi z}/W) }} }}.

Define an Airy stress function
sigma F(z) = sigma( Re[hat{hat{Z}}(z)] + y(Im[hat{Z}(z)] + Im[hat{Y}(z)]) ), so
F(z) = Re[hat{hat{Z}}(z)] + y(Im[hat{Z}(z)] + Im[hat{Y}(z)]), where
{{d hat{hat{Z}}}/{dz}} =hat{Z}, {{d hat{Z}}/{dz}} =Z = 1/{sqrt{ 1-{{ sin^2({pi a}/W) }/{ sin^2({pi z}/W) }} }}, {{d hat{Y}}/{dz}} =Y, and z = x+iy.

A Failed Attempt at a Solution

Now, the proposition is that {partial^4 F(z)}/{partial z^4} = {partial^4}/{partial z^4} Re[hat{hat{Z}}(z)] + {partial^4}/{partial z^4} y(Im[hat{Z}(z)] + Im[hat{Y}(z)]) = 0.

{partial^4}/{partial z^4} Re[hat{hat{Z}}(z)] = Re[{partial^4}/{partial z^4} hat{hat{Z}}(z)] = Re[{partial^2}/{partial z^2} Z(z)] = a huge mess.

Solution

However, assuming Im[Y(z)] = 0 on y=0, to show Z(z) satisfies the boundary conditions, it is sufficient to show Re[Z(z)] = 0 on all traction-free crack faces. For this problem, that is when y = 0 and |x| < a.

Z(z) = sigma/{sqrt{ 1-{{ sin^2({pi a}/W) }/{ sin^2({pi z}/W) }} }} = sigma/{sqrt{ 1-{{ sin^2({pi a}/W) }/{ sin^2({pi (x+iy)}/W) }} }} = sigma/{sqrt{ 1-{{ sin^2({pi a}/W) }/{ sin^2({pi x}/W) }} }}

Now, Z(z) approaches zero as x approaches a, the crack tip. Also, since this problem defines a « W, then when |x| < a,
{{ sin^2({pi a}/W) }/{ sin^2({pi r}/W) }} approx 1,
which makes the denominator of Z(z) approach infinity and Z(z) approach zero.

So, Z(z) is approximately zero along the crack face and pretty much satisfies the biharmonic equations.


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