Airy stress function with body force

Airy stress function with body force potential

If a body force exists, the Airy stress function ($\phi$) has to be combined with a body force potential ($V$). Thus,

${\sigma }_{11}={\phi }_{,22}+V;{\sigma }_{22}={\phi }_{,11}+V;{\sigma }_{12}=-{\phi }_{,12}\text{(3)}$

or,

${\sigma }_{11}=\frac{{\partial }^2\phi }{\partial x_2^2}+V;{\sigma }_{22}=\frac{{\partial }^2\phi }{\partial x_1^2}+V;{\sigma }_{12}=-\frac{{\partial }^2\phi }{\partial x_1\partial x_2}\text{(4)}$

Is equilibrium still satisfied ?

Recall the equilibrium equation in two dimensions:

$\mathrm{\nabla }\bullet \mathit{\sigma }+𝐟=0;{\sigma }_{\beta \alpha ,\beta }+f_{\alpha }=0\text{(5)}$

or,

$\begin{array}{cc}{\sigma }_{11,1}+{\sigma }_{21,2}+f_1& =0\text{(6)}\\ {\sigma }_{12,1}+{\sigma }_{22,2}+f_2& =0\text{(7)}\end{array}$

In terms of $\phi$,

$\begin{array}{cc}{\phi }_{,122}+V_{,1}-{\phi }_{,212}+f_1& =0\text{(8)}\\ -{\phi }_{,112}+{\phi }_{,211}+V_{,2}+f_2& =0\text{(9)}\end{array}$

or,

$\begin{array}{cc}{V}_{,1}+f_1& =0\text{(10)}\\ V_{,2}+f_2& =0\text{(11)}\end{array}$

Therefore,

$f_1=-V_{,1};f_2=-V_{,2}\text{(12)}$

or,

$f_1=-\frac{\partial V}{\partial x_1};f_2=-\frac{\partial V}{\partial x_2}\text{(13)}$

Hence,

$𝐟=-\mathrm{\nabla }V\text{(14)}$

• Equilibrium is satisfied only if the body force field can be expressed as the gradient of a scalar potential.
• A force field that can be expressed as the gradient of a scalar potential is called conservative.

What condition is needed to satisfy compatibility ?

Recall that the compatibility condition in terms of the stresses can be written as

 Template:Center top${\mathrm{\nabla }}^2{\sigma }_{\gamma \gamma }=-\frac{1}{\alpha }{f}_{\gamma ,\gamma }\text{(15)}$Template:Center bottom

where,

 Template:Center top$\alpha =\{\begin{array}{cc}1-\nu & \mathrm{f}\mathrm{o}\mathrm{r}\mathrm{p}\mathrm{l}\mathrm{a}\mathrm{n}\mathrm{e}\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{i}\mathrm{n}\\ \frac{1}{1+\nu }& \mathrm{f}\mathrm{o}\mathrm{r}\mathrm{p}\mathrm{l}\mathrm{a}\mathrm{n}\mathrm{e}\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{s}\end{array}\text{(16)}$Template:Center bottom

or,

 Template:Center top${\mathrm{\nabla }}^2({\sigma }_{11}+{\sigma }_{22})+\frac{1}{\alpha }(f_{1,1}+f_{2,2})=0\text{(17)}$Template:Center bottom

or,

 Template:Center top${\sigma }_{11,11}+{\sigma }_{11,22}+{\sigma }_{22,11}+{\sigma }_{22,22}+\frac{1}{\alpha }(f_{1,1}+f_{2,2})=0\text{(18)}$Template:Center bottom

which is the same as,

 Template:Center top$\frac{{\partial }^2{\sigma }_{11}}{\partial x_1^2}+\frac{{\partial }^2{\sigma }_{11}}{\partial x_2^2}+\frac{{\partial }^2{\sigma }_{22}}{\partial x_1^2}+\frac{{\partial }^2{\sigma }_{22}}{\partial x_2^2}+\frac{1}{\alpha }(\frac{\partial f_1}{\partial x_1}+\frac{\partial f_2}{\partial x_2})=0\text{(19)}$Template:Center bottom

Plug in the stress potential and the body force potential in equation (18) to get

 Template:Center top${\phi }_{,2211}+V_{,11}+{\phi }_{,2222}+V_{,22}+{\phi }_{,1111}+V_{,11}+{\phi }_{,1122}+V_{,22}+\frac{1}{\alpha }(-V_{,11}-V_{,22})=0\text{(20)}$Template:Center bottom

or,

 Template:Center top$2{\phi }_{,2211}+2V_{,11}+{\phi }_{,2222}+2V_{,22}+{\phi }_{,1111}+\frac{1}{\alpha }(-V_{,11}-V_{,22})=0\text{(21)}$Template:Center bottom

Rearrange to get

 Template:Center top${\phi }_{,1111}+2{\phi }_{,2211}+{\phi }_{,2222}+(2-\frac{1}{\alpha })(V_{,11}+V_{,22})=0\text{(22)}$Template:Center bottom

which is the same as,

 Template:Center top$\frac{{\partial }^4\phi }{\partial x_1^4}+2\frac{{\partial }^4\phi }{\partial x_1^2\partial x_2^2}+\frac{{\partial }^4\phi }{\partial x_2^4}+(2-\frac{1}{\alpha })(\frac{{\partial }^{2}V}{\partial x_1^2}+\frac{{\partial }^{2}V}{\partial x_2^2})=0\text{(23)}$Template:Center bottom

Therefore,

 Template:Center top${\mathrm{\nabla }}^4\phi +(2-\frac{1}{\alpha }){\mathrm{\nabla }}^{2}V=0\text{(24)}$Template:Center bottom

Compatibility is satisfied only if equation (24) is satisfied.

Equations for Airy stress function with body force potential

The relation between the Cauchy stress and the Airy stress function is (in direct tensor notation)

$\mathit{\sigma }=\mathrm{\nabla }\times \mathrm{\nabla }\times \phi$

The relation between the body force and the body force potential is

$𝐟=-\mathrm{\nabla }V$

We also have to satisfy the compatibility condition for the Airy stress function to be a true stress potential, i.e.,

${\mathrm{\nabla }}^4\phi +(2-\frac{1}{\alpha }){\mathrm{\nabla }}^{2}V=0$

In rectangular Cartesian coordinates

In rectangular Cartesian coordinates, the relation between the Cauchy stress components and the Airy stress function + body force potential can be written as

${\sigma }_{11}={\phi }_{,22}+V;{\sigma }_{22}={\phi }_{,11}+V;{\sigma }_{12}=-{\phi }_{,12}$

or,

${\sigma }_{xx}=\frac{{\partial }^2\phi }{\partial y^2}+V;{\sigma }_{yy}=\frac{{\partial }^2\phi }{\partial x^2}+V;{\sigma }_{xy}=-\frac{{\partial }^2\phi }{\partial x\partial y}$

The relation between the body force components and the body force potential are:

$f_1=-V_{,1};f_2=-V_{,2}$

or,

$f_x=-\frac{\partial V}{\partial x};f_y=-\frac{\partial V}{\partial y}$

The compatibility condition is written as

${\phi }_{,1111}+2{\phi }_{,2211}+{\phi }_{,2222}+(2-\frac{1}{\alpha })(V_{,11}+V_{,22})=0$

or,

$\frac{{\partial }^4\phi }{\partial x^4}+2\frac{{\partial }^4\phi }{\partial x^2\partial y^2}+\frac{{\partial }^4\phi }{\partial y^4}+(2-\frac{1}{\alpha })(\frac{{\partial }^{2}V}{\partial x^2}+\frac{{\partial }^{2}V}{\partial y^2})=0$

In cylindrical coordinates

In cylindrical coordinates, the relation between the Cauchy stresses and the Airy stress function + body force potential can be written as

${\sigma }_{rr}=\frac{1}{r}\frac{\partial \phi }{\partial r}+\frac{1}{r^2}\frac{{\partial }^2\phi }{\partial {\theta }^2}+V;{\sigma }_{\theta \theta }=\frac{{\partial }^2\phi }{\partial r^2}+V;{\sigma }_{r\theta }=-\frac{\partial }{\partial r}\left(\frac{1}{r}\frac{\partial \phi }{\partial \theta }\right)\text{(25)}$

The components of the body force are related to the body force potential via

$f_r=-\frac{\partial V}{\partial r};f_{\theta }=-\frac{1}{r}\frac{\partial V}{\partial \theta }\text{(26)}$

The compatibility condition can be expressed as

$\begin{array}{cc}(\frac{{\partial }^2}{\partial r^2}+\frac{1}{r}\frac{\partial }{\partial r}+\frac{1}{r^2}\frac{{\partial }^2}{\partial {\theta }^2})& (\frac{{\partial }^2\phi }{\partial r^2}+\frac{1}{r}\frac{\partial \phi }{\partial r}+\frac{1}{r^2}\frac{{\partial }^2\phi }{\partial {\theta }^2})+\\ & (2-\frac{1}{\alpha })(\frac{{\partial }^{2}V}{\partial r^2}+\frac{1}{r}\frac{\partial V}{\partial r}+\frac{1}{r^2}\frac{{\partial }^{2}V}{\partial {\theta }^2})=0\text{(27)}\end{array}$

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