Chapter 10, Problem 117RP

Determine (a) the critical load for the steel strut, (b) the dimension d for which the aluminum strut will have the same critical load. (c) Express the weight of the aluminum strut as a percent of the weight of the steel strut.

Fig. P10.117

To determine

Find the critical load for the steel strut.

Answer to Problem 117RP

The critical load of the steel strut is $\underset{\_}{647\text{lb}}$.

Explanation of Solution

The modulus of elasticity of the steel strut is $E_{st}=29\times {10}^6\text{psi}$.

The size of the steel strut is $a_{st}=\frac{1}{2}\text{in}\text{.}$.

The length of the steel strut is $L_{st}=4\text{ft}$.

Determine the moment of inertia of the steel strut $\left(I_{st}\right)$ using the relation.

$I_{st}=\frac{{\left(a_{st}\right)}^4}{12}$

Here, the size of the steel strut is $a_{st}$.

Substitute $\frac{1}{2}\text{in}\text{.}$ for $a_{st}$.

$\begin{array}{c}{I}_{st}=\frac{{\left(\frac{1}{2}\right)}^4}{12}\\ =5.2083\times {10}^{-3}\text{in}{\text{.}}^4\end{array}$

The effective length $\left(L_e\right)$ of the column is equal to the length of the column.

$L_e=L_{st}=4\text{ft}$

Determine the critical load $\left(P_{\text{cr,st}}\right)$ of the steel strut using the equation.

$P_{\text{cr},st}=\frac{{\pi }^2{E}_{st}{I}_{st}}{{\left(L_e\right)}^2}$

Here, the modulus of elasticity of the steel strut is $E_{st}$ and the effective length of the steel strut is $L_e$.

Substitute $29\times {10}^3\text{psi}$ for $E_{st}$, $5.2083\times {10}^{-3}\text{in}{\text{.}}^4$ for $I_{st}$, and 4 ft for $L_e$.

$\begin{array}{c}{P}_{\text{cr},st}=\frac{{\pi }^2\times 29\times {10}^6\times 5.2083\times {10}^{-3}}{{\left(4\text{ft}\times \frac{\text{12}\text{in}\text{.}}{\text{1}\text{ft}}\right)}^2}\\ =647\text{lb}\end{array}$

Therefore, the critical load of the steel strut is $\underset{\_}{647\text{lb}}$.

To determine

Find the dimension d for the condition that the critical load is same for steel strut and aluminum strut.

Answer to Problem 117RP

The dimension d of the aluminum strut for the given condition is $\underset{\_}{0.651\text{in}\text{.}}$.

Explanation of Solution

The modulus of elasticity of the aluminium strut is $E_{al}=10.1\times {10}^6\text{psi}$.

The length of the aluminium strut is $L_{st}=4\text{ft}$.

The critical load of the steel strut and the aluminum strut is equal.

$P_{\text{cr},st}=P_{\text{cr},al}=647\text{lb}$.

The effective length $\left(L_e\right)$ of the column is equal to the length of the column.

$L_e=L_{al}=4\text{ft}$

Determine the moment of inertia of the aluminium strut $\left(I_{al}\right)$ using the equation.

$P_{\text{cr},al}=\frac{{\pi }^2{E}_{al}{I}_{al}}{{\left(L_e\right)}^2}$

Here, the critical load in the aluminium strut is $P_{\text{cr},al}$, the modulus of elasticity of the aluminium strut is $E_{al}$, and the effective length of the aluminium strut is $L_e$.

Substitute 647 lb for $P_{\text{cr},al}$, $10.1\times {10}^6\text{psi}$ for $E_{al}$, and 4 ft for $L_e$.

$\begin{array}{c}647=\frac{{\pi }^2\times 10.1\times {10}^6\times I_{al}}{{\left(4\text{ft}\times \frac{\text{12}\text{in}\text{.}}{\text{1}\text{ft}}\right)}^2}\\ I_{al}=149.543\times {10}^{-4}\text{in}{\text{.}}^4\end{array}$

Determine the dimension of the aluminum strut (d) using the relation.

$I_{al}=\frac{d^4}{12}$

Substitute $149.543\times {10}^{-4}\text{in}{\text{.}}^4$ for $I_{al}$.

$\begin{array}{l}149.543\times {10}^{-4}=\frac{d^4}{12}\\ d=0.651\text{in}\text{.}\end{array}$

Therefore, the dimension d of the aluminum strut for the given condition is $\underset{\_}{0.651\text{in}\text{.}}$.

To determine

Find the percentage weight of aluminum strut to the steel strut.

Answer to Problem 117RP

The percentage weight of aluminum strut to the steel strut is $\underset{\_}{58.8\%}$.

Explanation of Solution

The weight density of the aluminium strut is ${\gamma }_{al}=170{\text{lb/ft}}^3$.

The weight density of the steel strut is ${\gamma }_{st}=490{\text{lb/ft}}^3$.

The length of the steel strut is $L_{st}=4\text{ft}$.

The length of the aluminium strut is $L_{al}=4\text{ft}$.

The dimension of the aluminium strut is $d=0.651\text{in}\text{.}$.

The size of the steel strut is $a_{st}=\frac{1}{2}\text{in}\text{.}$.

In general,

The weight density $\left(\gamma \right)$ of a material is defined as follows;

$\begin{array}{c}\gamma =\frac{W}{V}\\ =\frac{W}{AL}\\ W=\gamma AL\end{array}$

Here, the weight of the material is W, the volume of the material is V, the cross sectional area of the material is A, and the length of the material is L.

Write the expression as percentage weight of aluminium strut $\left(W_{al}\right)$ to the steel strut $\left(W_{st}\right)$ as follows;

$\begin{array}{c}\frac{W_{al}}{W_{st}}\times 100\%=\frac{{\gamma }_{al}{A}_{al}{L}_{al}}{{\gamma }_{st}{A}_{st}{L}_{st}}\times 100\%\\ =\frac{{\gamma }_{al}{d}^2{L}_{al}}{{\gamma }_{st}{\left(a_{st}\right)}^2{L}_{st}}\times 100\%\end{array}$

Substitute $170{\text{lb/ft}}^3$ for ${\gamma }_{al}$, 0.651 in. for d, 4 ft for $L_{al}$, $490{\text{lb/ft}}^3$ for ${\gamma }_{st}$, $\frac{1}{2}\text{in}\text{.}$ for $a_{st}$, 4 ft for $L_{st}$.

$\begin{array}{c}\frac{W_{al}}{W_{st}}\times 100\%=\frac{170\times {0.651}^2\times 4}{490\times {\left(\frac{1}{2}\right)}^2\times 4}\times 100\%\\ =58.8\%\end{array}$

Therefore, the percentage weight of aluminum strut to the steel strut is $\underset{\_}{58.8\%}$.