User:Eas4200c.f08.vqcrew.d/hw6

Boundary Conditions for Phi
For solid cross sections, the constant value of $$\phi \;$$ is arbitrary, therefore it can be set to zero. Thus, we have the boundary condition $$\phi =0 \;$$ on the lateral surface of a solid bar. This boundary condition is depicted in Figure #, where $$\phi \;$$ is equal to zero on $$S_0$$. In the case of a thin-walled cross section, $$\phi \;$$ has a different constant value on each surface. This is depicted in Figure #, where $$\phi =C_0\;$$ on $$S_0$$ and $$\phi =C_1\;$$ on $$S_1$$.

For a solid bar with circular cross section, as shown in Figure #, the Prandtl stress function is assumed to be
 * $$\phi = C\left(\frac{y^2}{a^2} + \frac{z^2}{a^2}-1 \right)$$
 * {| class="collapsible collapsed"

!Proof that this equation satisfies the boundary condition We can prove that this satisfies the surface boundary condition of $$\phi =0\;$$ as follows:
 * 
 * 
 * $$\phi = C\left(\frac{y^2}{a^2} + \frac{z^2}{a^2}-1 \right) = C\left(\frac{y^2+z^2}{a^2}-1 \right)$$

Substituting in the Pythagorean theorem, $$a^2=y^2+z^2$$, we get:
 * $$\phi = C\left(\frac{a^2}{a^2}-1 \right)=C(1-1)=0$$

To find C, we can plug this function into the compatibility equation:
 * }
 * $$\frac{\partial^2 \phi }{\partial y^2} + \frac{\partial^2 \phi }{\partial z^2}=-2G \theta$$

which results in:
 * $$\frac{2C}{a^2} + \frac{2C}{a^2} = -2G \theta \Rightarrow C=-\frac{1}{2}a^2G\theta$$

This can also be expressed in terms of torque $$T$$ and polar moment of inertia $$J$$:
 * $$C=-\frac{1}{2}a^2TJ$$