User:Eas4200c.f08.vqcrew.E/Homework3

Torsion Acting on a Uniform Bar with Circular Cross-Section
This case does not exhibit warping. The only deformation that occurs is contained within the y-z plane. The beam behaves as a series of rigid disks rotating entirely in the plane of the cross-section. Figure ## displays the reaction of the beam to a given torque.



The general formula for the torque is:


 * $$T = \iint_A r \tau dA $$

The shear stress is defined as the product of the shear modulus, G, and the angle of twist,  $$\gamma$$  , (Hooke's Law).


 * $$\tau = G \gamma  $$


 * $$\displaystyle \gamma = \displaystyle \frac{rd\alpha}{dx} = r\theta$$



Define $$\theta$$ as rate of twist:


 * $$\theta := \frac{d\alpha}{dx}$$

Substituting in, the torque is found to be:


 * $$\displaystyle T = \int_A rG(r\theta)dA = G\theta(\int_A r^{2} dA)$$

Where:   $$\displaystyle dA = dydz = rdrd\theta $$

The $$2^{nd}$$ polar moment of inertia for the beam is equal to the second term in the expression of the torque, $$\displaystyle J = (\int_A r^{2} dA)$$.



Thin-Walled Hollow Circular Cross-Section
Figure ## shows two circular cross-sections, one solid and the other thin-walled, with the same area. The two cross-sections have different $$2^{nd}$$ polar moments of inertia.

For the thin-walled cross-section,

Inner radius: $$\displaystyle r_i = a$$

Outer radius: $$\displaystyle r_o = b$$

The $$2^{nd}$$ polar moment of inertia for the thin-walled cross-section can be determined through the expression:


 * $$\displaystyle J = \frac{1}{2}\pi(b^{4}-a^{4}) = \frac{1}{2}\pi(b - a)(b + a)(b^{2} + a^{2})$$

Define $$\displaystyle \bar{r}$$ as the average radius:


 * $$\displaystyle \bar{r} = \frac{a+b}{2}$$

Define $$\displaystyle t$$ as the thickness:


 * $$\displaystyle t = (a - b)$$

Substitute into the $$2^{nd}$$ polar moment of inertia equation to get:


 * $$\displaystyle J = \frac{1}{2}\pi(t)(2\bar{r})(b^{2} + a^{2}) = \pi(t)(\bar{r})(b^{2} + a^{2})$$

Approximate and substitute:

$$\displaystyle a^{2} \cong \bar{r}^{2}$$

$$\displaystyle b^{2} \cong \bar{r}^{2}$$


 * $$\displaystyle \Rightarrow J = \pi t \bar{r}(\bar{r}^{2} + \bar{r}^{2})= 2\pi t \bar{r}^{3}$$

This result can be separated to allow for a proportional factor to be found between the $$2^{nd}$$ polar moment of inertia and the the average area, $$\displaystyle \bar{A}$$.

$$\displaystyle \bar{A} = \pi\bar{r}^{2}$$


 * $$\displaystyle \Rightarrow J = (2\pi^{-\frac{1}{2}}t)(\pi\bar{r}^{2})^{\frac{3}{2}} = (2\pi^{-\frac{1}{2}}t)(\bar{A})^{\frac{3}{2}}$$

In this form, it is clear that $$\displaystyle J$$ is proprtional to $$\displaystyle \bar{A}^{\frac{3}{2}}$$ by a factor of $$\displaystyle (2\pi^{-\frac{1}{2}}t)$$.