Tag Archives: mechanics solutions

Expressing the Density of Water in SI Units

Water has a density of 1.94 slug/ft³. What is the density expressed in SI units? Express the answer to three significant figures.

Engineering Mechanics: Statics 13th Edition by RC Hibbeler, Problem 1-19
Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 1-17

Solution:

\begin{align*}
\rho _w & =\left(\frac{1.94\:\text{slug}}{1\:\text{ft}^3}\right)\left(\frac{14.59\:\text{kg}}{1\:\text{slug}}\right)\left(\frac{1\:\text{ft}^3}{0.3048^3\:\text{m}^3}\right) \\
& =\left(\frac{1.94\:\text{slug}}{1\:\text{ft}^3}\right)\left(\frac{14.59\:\text{kg}}{1\:\text{slug}}\right)\left(\frac{1\:\text{ft}^3}{0.3048^3\:\text{m}^3}\right) \\
& =999.6\:\frac{\text{kg}}{\text{m}^3}\\
& =1.00\:\text{Mg/m}^3\\
\end{align*}
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Mechanics of Materials: An Integrated Learning Approach 3rd Edition by Timothy A. Philpot Problem P1.2

A 2024-T4 aluminum tube with an outside diameter of 2.50 in. will be used to support a 27-kip load. If the axial normal stress in the member must be limited to 18 ksi, determine the wall thickness required for the tube.

Solution:

We are given the following values:

\begin{align*}
\text{Outside Diameter}, D & = 2.50\ \text{in} \\
\text{Axial Load}, P & = 27\ \text{kips} \\
\text{Maximum Axial Stress}, \sigma & = 18\ \text{ksi}\\
\text{Inside Diameter}, d & = D-2t
\end{align*}

We are solving for the unknown wall thickness of the tube, t.

From the definition of normal stress, solve for the minimum area required to support a 27-kip load without exceeding a stress of 18 ksi

\begin{align*}
\sigma & =\frac{P}{A} \\ 
A_{\text{min}} & =\frac{P}{\sigma } \\ 
A_{\text{min}} & = \frac{27\:\text{kips}}{18\:\text{ksi}} \\ 
A_{\text{min}} & = 1.500\:\text{in}^2 
\end{align*}

The cross-sectional area of the aluminum tube is given by

A=\frac{\pi }{4}\left(D^2-d^2\right)

Set this expression equal to the minimum area and solve for the maximum inside diameter, d

\begin{align*}
A & =\frac{\pi }{4}\left(D^2-d^2\right) \\
A & = \frac{\pi }{4}\left[\left(2.50\ \text{in}\right)^2-d^2\right] \\
1.500\ \text{in}^2 & = \frac{\pi }{4}\left[\left(2.50\ \text{in}\right)^2-d^2\right] \\
4 \left( 1.500\ \text{in}^2 \right) & = \pi \left[\left(2.50\ \text{in}\right)^2-d^2\right] \\
\frac{4 \left( 1.500\ \text{in}^2 \right)}{\pi} & = \left(2.50\ \text{in}\right)^2-d^2 \\
d^{2} & = \left( 2.50\ \text{in}^{2} \right) - \frac{4 \left( 1.500\ \text{in}^2 \right)}{\pi} \\
d & = \sqrt{\left( 2.50\ \text{in}^{2} \right) - \frac{4 \left( 1.500\ \text{in}^2 \right)}{\pi}} \\
d & = 2.08330\ \text{in}
\end{align*}

The outside diameter D, the inside diameter d, and the wall thickness t are related by

D = d+2t

Therefore, the minimum wall thickness required for the aluminum tube is

\begin{align*}
t & =\frac{D-d}{2} \\
t & = \frac{2.50\:\text{in}-2.08330\:\text{in}}{2} \\
t & = 0.208\ \text{in} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}