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 13^th Edition by RC Hibbeler, Problem 1-19}
_{Engineering Mechanics: Statics 14^th 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 3rd Edition by Timothy A. Philpot, 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:

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

$\displaystyle \sigma =\frac{P}{A}$

$\displaystyle A_{min}=\frac{P}{\sigma }$

$\displaystyle A_{min}=\frac{27\:kips}{18\:ksi}$

$\displaystyle A_{min}=1.500\:in.^2$

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

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

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

$\displaystyle \frac{\pi }{4}\left[\left(2.50\:in\right)^2-d^2\right]=1.500\:in^2$

$\displaystyle \left(2.50\:in\right)^2-d^2=\frac{4}{\pi }\left(1.500\:in^2\right)$

$\displaystyle \left(2.50\:in\right)^2-\frac{4}{\pi }\left(1.500\:in^2\right)=d^2$

$\displaystyle d_{max}=2.08330\:in$

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

$\displaystyle t_{min}=\frac{D-d}{2}$

$\displaystyle t_{min}=\frac{2.50\:in-2.08330\:in}{2}$

$\displaystyle t_{min}=0.20835\:in$

$\displaystyle t_{min}=0.208\:in$

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