Question

assume p = 20,000 lb and l = 30 in. the aluminum rod shown below has a circular cross - section with a diameter of 1.5 in. determine the tensile stress of the rod.
8,888.9 psi
11,317.7 psi
13,215.4 psi
6,542.6 psi
16,976.5 psi

Explanation:

Step1: Calculate cross - sectional area

The formula for the area of a circle is $A=\pi(\frac{d}{2})^2$, where $d = 1.5$ in. So $A=\pi(\frac{1.5}{2})^2=\frac{\pi\times1.5^2}{4}\text{ in}^2$.

Step2: Calculate tensile stress

The formula for tensile stress $\sigma=\frac{P}{A}$, where $P = 20000$ lb. Substitute $A=\frac{\pi\times1.5^2}{4}$ into the stress formula: $\sigma=\frac{20000}{\frac{\pi\times1.5^2}{4}}=\frac{20000\times4}{\pi\times1.5^2}$.
\[

$$\begin{align*} \sigma&=\frac{80000}{\pi\times2.25}\\ &\approx\frac{80000}{3.14\times2.25}\\ &=\frac{80000}{7.065}\\ &\approx11323.4\text{ psi} \end{align*}$$

\]
The closest value to our result is $11317.7$ psi.

Answer:

11,317.7 psi