User:EAS4200C.F08.WIKI.A/My Contribution: 6

=Torsional Analysis Continued=

Two Cases of Torsional Analysis
There are two specific cases that we will concern ourselves with. The first being the closed thin walled cross-section and the second being the solid cross section. the two are depicted below.

Case 1: Closed Thin Walled Cross-Section (compare equations)
This case can be compared to the NACA Airfoil that is continuously being analyzed. It is important to note that the Prandtl stress function $$\displaystyle \Phi$$ is constant on the outer surface $$\displaystyle S_o$$ and the inner surface $$\displaystyle S_i$$. $$\displaystyle S_i$$ can also be represented by $$\displaystyle S_1$$ as well. The image below represents this case. It is comparable to Figure 3.11 in the text (Sun 2006).



Case 2: Solid Cross-Section
The one thing that is important to note about this case is that the Prandtl stress function $$\displaystyle \Phi$$ is zero on the outter surface $$\displaystyle S_o$$. This case is depicted below.



Uniform Bar With Solid Circular Cross-Section
This section coincides with section 3.3 in the text (Sun 2006). The following figure depicts the cross section of a solid circular bar and we will be determining the expression for the torque and polar moment of inertia in terms of the the Prandtl stress function.



Torque In Terms of $$\displaystyle \phi$$
The axis of the bar coincides with the origin of the coordinate system. The boundary of the bar can be expressed as the following equation;

$$\displaystyle x^2 +y^2 = a^2$$ where $$\displaystyle a^2$$ is the radius of the cross section

It is important to note that this is equivalent to the equation of a circle. We will assume that the stress function equals the following;

 $$\displaystyle \phi = C \left(\frac{x^2}{a^2} + \frac{y^2}{a^2} -1 \right)$$

With this information we are able to compute the torque. This is done below.

$$\displaystyle T = 2\int \int_{A}^{}{\phi dxdy} =2 C \int \int_{A}^{}{\left(\frac{x^2}{a^2} + \frac{y^2}{a^2} -1 \right) dxdy} = 2C\int \int_{A}^{}{\left(\frac{r^2}{a^2} -1 \right) dA} = 2C \left(\frac{{\pi} a^2}{2} - ({\pi}a^2) \right) = -C \pi a^2$$

We must now solve for the variable $$\displaystyle C$$. We know that the Prandtl stress function is equal to zero on the outer surface. We also that the gradient of the Prandtl stress function is equal to the following;

$$\displaystyle \nabla^2 \phi = -2G\theta = \frac{\partial^2 \phi }{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2} = C\frac{2}{a^2} + C\frac{2}{a^2} = C\frac{4}{a^2}$$

Solving for $$\displaystyle C$$ yields;

 $$\displaystyle C = -\frac{a^2G\theta}{2}$$

Plugging this back into the equation for torque we can obtain the final expression.

$$\displaystyle T = \frac {\theta G \pi a^4}{2}$$

Polar Moment of Inertia In Terms of $$\displaystyle \phi$$
We know that the polar moment of inertia $$\displaystyle J$$ equals the following;

 $$\displaystyle J = \frac {T}{G\theta}$$

where

$$\displaystyle T = 2\int \int_{A}^{}{\phi dxdy}= -C\pi a^2$$  $$\displaystyle G \theta = -\frac{\nabla^2 \phi}{2} = \frac{2C}{a^2}$$

The polar moment of inertia then equals the following;

$$\displaystyle J = \frac {T}{G\theta} = --\frac{C\pi a^2}{\frac{2C}{a^2}} = \frac{\pi a^4}{2}$$

This is the exact equation that has been derived multiple time before using other methods.