User:Eas4200c.f08.aeris.guan/hw3

Torque, Shear Flow, and Polar Moment of Inertia
In order to calculate torque on a member induced by shear flow, it is necessary to analyze the average area in which that shear flow acts. In the collapsible box below, it is proved that a triangle's area is half the product of its base length and its height even if the height is measured outside of the triangle.

The torque on an arbitrarily shaped shear panel can be expressed in terms of shear flow, and the average area. The shear flow is the force per unit length the shear panel experiences and the average area is the area bounded by the contour of the shear panel relative to the designated origin of the coordinate system in which the shear panel is described. The math process of obtaining this equation is shown below. $$\displaystyle T= \oint{\rho qds}= \int\int_{ \bar{A}}{2qdA}=2q \bar{A}$$ The below picture describes the forces, shear flow, average area and coordinate system of an open thin-walled cross section.  PQ is the shear panel contour, A is the average area enclosed by the panel relative to the origin, e is the perpendicular vector arm from the origin to the force F (which is calculated by multiplying the shear flow with the length in with the shear flow acts). F is the essentially the equivalent force of the shear flow and is used to calculate the torque. Another equation can be written to accommodate for F and the vector e. $$\displaystyle Fe=T=2Q\bar{A}$$ Now, let's consider a uniform bar with circular cross section subject to torque and resulting torsion. This is not a case in which warping of the member is a problem and thus the cross section behaves like rigid disks with only angular displacement rather than lateral displacement (along the length of the bar). The picture below represents an infinitesimally small part of the circlar bar where dA is the infinitesimally small area. This area will be displaced in the (y,z) plane an angel gamma when the object is torqued.  The derivation of an equation to find the torque in terms of the bar's shear modulus, the polar moment of inertia of the cross section, and the rate of twist (gamma). $$\displaystyle T=\int\int_{A}{r \tau dA}=G \gamma \int\int_{A}{rdA}$$ where $$\tau =G \gamma $$ is obtained from Hooke's Law The below picture describes the deformation after torque when the vector r (vector from the origin to dA) is displaced by a small angle alpha to a new vector (r') in the (y,z) plane. $$\displaystyle \gamma= \frac{rd \alpha}{dx}$$ where $$\displaystyle \frac{d \alpha}{dx}=: \theta$$ (rate of twist) so $$\displaystyle T= \int_A{rG(r \theta)dA}$$ where dA= dydz, and G and $$\theta$$ are independent of y and z. $$\displaystyle = G \theta \int_A{r^2dA}=G \theta J$$ G is the shear modulus, theta is the rate of twist, and J is the polar moment of inertia. If a is the radius of the circular cross section then $$ J=\frac{1}{2} \pi a^4$$.

If it is a hollow, thin-walled (t<$$J=\frac{1}{2}\pi(b^4-a^4)=\frac{1}{2}\pi(b-a)(b+a)(b^2+a^2)=\frac{1}{2}\pi t(2 \bar{r})(b^2+a^2)$$ where $$\bar{r}$$ is the average radius ((a+b)/2) $$\displaystyle b^2 \cong \bar{r}^2$$ and $$a^2 \cong \bar{r}^2$$ so $$J=2\pi t \bar{r}^3$$

$$\displaystyle J=(2\pi^{1/2}t)(\pi \bar{r}^2)^{3/2}$$ where $$\displaystyle \bar{a}^=\pi \bar{r}^2$$ This means that J is proportional to the average area to the 1.5 power with a proportional factor of $$2\pi^{1/2}t$$