User:Eas4200c.f08.gator.edwards/Homework Problems

Area of a Triangle
Prove $$A = \frac{1}{2}bh$$

First begin by considering a rectangle with base of BD and height of AD. The area of this rectangle would be (BD)(AD). Now, if we just consider half of this rectangle (triangle ABD) we would have an area of $$\frac{1}{2}$$(BD)(AD). This is the area of the triangle pictured including the area enclosed by the dotted lines. If we subtract the area enclosed by the dotted lines from the area we just found, we will get the area of the triangle we are interested in. The area of the dotted line would be $$\frac{1}{2}$$(AD)(CD) by the same logic as before. Performing this, we find that $$A = \frac{1}{2}\left(AD \right)\left(BD \right)- \frac{1}{2}\left(AD \right)\left(CD \right) = \frac{1}{2}\left(AD \right)\left[ (BD)-(CD)\right]$$

Looking at the picture, we can see that by definition BD - CD = b and AD = h. Plugging these relationships into the above equation yields $$A = \frac{1}{2}bh$$

Polar Moment of Inertia of Circular Cross-Section
For a circle of radius a, the polar moment of inertia ($$J_{0}$$) can be calculated as follows:

$$J_{0} = \int_{A}\rho^{2}dA = \int_{0}^{a}\int_{0}^{2\pi}\rho^{2}(\rho d\theta)d\rho $$
 * $$= \int_{0}^{a} \rho^{2}\left(2\pi\rho d\rho \right) $$
 * $$= 2\pi\int_{0}^{a}\rho^{3}d\rho = \frac{1}{2}\pi a^{4}$$
 * $$\Rightarrow J_{0} = \frac{1}{2}\pi a^{4}$$

Solid vs. Hollow Cross-Section
Compare a solid circular cross section to a hollow thin walled circular cross section.

$$r_{o}^{(a)}=1cm$$

$$r_{i}^{(b)}=5cm$$

$$t^{(b)}=0.1cm$$

$$A^{(a)}=\pi r^{2}=\pi (1cm)^{2}=3.1415cm^{2}$$

$$A^{(b)}=\pi (r_{o}^{2}-(r_{o}-t_{b})^{2})=\pi (5cm^{2}-(5cm-0.1cm)^{2})=3.1102cm^{2}$$

$$J^{(a)}=\frac{\pi }{2}r^{4}=\frac{\pi }{2}(1cm)^{4}=1.5708cm^{4}$$

$$J^{(b)}=2\pi t\bar{r}^{3}=2\pi t(\frac{2r_{o}-t^{(b)}}{2})^{3}=2\pi (0.1cm)(\frac{2(5cm)-(0.1cm)}{2})^{3}=76.207cm^{4}$$

$$\frac{J^{(a)}}{J^{(b)}}=\frac{1.5708cm^{4}}{76.207cm^{4}}=0.0206$$

Find $$r_{i}^{(c)}$$ if $$t^{(c)}=0.02r_{o}^{(c)}$$ and $$J^{(b)}=J^{(c)}$$.

$$J^{(b)}=76.207cm^{4}=2\pi (0.02r_{i}^{(c)})(\frac{2(r_{i}^{(c)}+(0.02r_{i}^{(c)}))}{2})^{3}=1.06684r_{o}^{(c)}=J^{(c)}$$

$$r_{i}^{(c)}=2.907$$

Single Cell Section Area


To calculate the area of this figure, we break it down into two smaller parts: a semicircle and a triangle. The area of a semicircle is $$\frac{1}{2}\pi R^{2}\quad$$ with R being $$\frac{b}{2}$$ = 1 m. The area of a triangle is $$\frac{1}{2}bh\quad$$ with the base being a (4 m) and the height being b (2 m).

Therefore the overall area is:

$$\bar{A} = \frac{1}{2}\pi \left( \frac{b}{2}\right)^{2} + \frac{1}{2}ab $$
 * $$= \frac{1}{2}\pi \left( \frac{1}{2}\right)^{2} + \frac{1}{2}\left(4 \right)\left(2 \right)$$
 * $$=\frac{\pi}{8} + 4 \quad m.$$
 * $$= \frac{1}{2}\pi \left( \frac{1}{2}\right)^{2} + \frac{1}{2}\left(4 \right)\left(2 \right)$$
 * $$=\frac{\pi}{8} + 4 \quad m.$$
 * $$=\frac{\pi}{8} + 4 \quad m.$$
 * $$=\frac{\pi}{8} + 4 \quad m.$$