Chapter 10.1, Problem 14P

Determine (a) the critical load for the square strut, (b) the radius of the round strut for which both struts have the same critical load. (c) Express the cross-sectional area of the square strut as a percentage of the cross-sectional area of the round strut. Use E = 200 GPa.

Fig. P10.13 and P10.14

To determine

Find the critical load of the square strut.

Answer to Problem 14P

The critical load of the square strut is $\underset{\_}{64.3\text{kN}}$.

Explanation of Solution

The modulus of elasticity of the strut is $E=200\text{GPa}$.

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

$I_{\text{square}}=\frac{a^4}{12}$

Substitute 25 mm for a.

$\begin{array}{c}{I}_{\text{square}}=\frac{{\left(25\text{mm}\times \frac{\text{1}\text{m}}{\text{1000}\text{mm}}\right)}^4}{12}\\ =3.2552\times {10}^{-8}{\text{m}}^4\end{array}$

Determine the critical load $\left(P_{\text{cr}}\right)$ for both the end pinned condition using the equation.

$P_{\text{cr}\left(\text{square}\right)}=\frac{{\pi }^{2}EI}{L^2}$

Here, the modulus of elasticity is E and the length of the strut is L.

Substitute 200 GPa for E, $3.2552\times {10}^{-8}{\text{m}}^4$ for I, and 1 m for L.

$\begin{array}{c}{P}_{\text{cr}\left(\text{square}\right)}=\frac{{\pi }^2\times \left(200\text{GPa}\times \frac{{\text{10}}^{\text{9}}{\text{N/m}}^{\text{2}}}{\text{1}\text{GPa}}\right)\times 3.2552\times {10}^{-8}}{1^2}\\ =64.3\times {10}^3\text{N}\times \frac{\text{1}\text{kN}}{\text{1000}\text{N}}\\ =64.3\text{kN}\end{array}$

Therefore, the critical load of the square strut is $\underset{\_}{64.3\text{kN}}$.

To determine

Find the radius of the round strut when the critical load is same for square strut and round strut.

Answer to Problem 14P

The radius of the round strut is $\underset{\_}{14.27\text{mm}}$.

Explanation of Solution

The modulus of elasticity of the strut is $E=200\text{GPa}$.

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

$I_{square}=\frac{a^4}{12}$

Here, the size of the square strut is a.

Substitute 25 mm for a.

$\begin{array}{c}{I}_{square}=\frac{{\left(25\text{mm}\times \frac{\text{1}\text{m}}{\text{1000}\text{mm}}\right)}^4}{12}\\ =3.2552\times {10}^{-8}{\text{m}}^4\end{array}$

Determine the critical load $\left(P_{\text{cr}}\right)$ for both the end pinned condition using the equation.

$P_{\text{cr}\left(square\right)}=\frac{{\pi }^{2}EI}{L^2}$

Substitute 200 GPa for E, $3.2552\times {10}^{-8}{\text{m}}^4$ for I, and 1 m for L.

$\begin{array}{c}{P}_{\text{cr}\left(square\right)}=\frac{{\pi }^2\times \left(200\text{GPa}\times \frac{{\text{10}}^{\text{9}}{\text{N/m}}^{\text{2}}}{\text{1}\text{GPa}}\right)\times 3.2552\times {10}^{-8}}{1^2}\\ =64.3\times {10}^3\text{N}\times \frac{\text{1}\text{kN}}{\text{1000}\text{N}}\\ =64.3\text{kN}\end{array}$

The critical load of the square strut and the round strut is equal.

$P_{\text{cr}\left(square\right)}=P_{\text{cr}\left(round\right)}=64.3\text{kN}$

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

$P_{\text{cr}\left(round\right)}=\frac{{\pi }^{2}E{I}_{round}}{L^2}$

Substitute 64.3 kN for $P_{\text{cr}\left(round\right)}$, 200 GPa for E, and 1 m for L.

$\begin{array}{c}64.3\text{kN}\times \frac{\text{1000}\text{N}}{\text{1}\text{kN}}=\frac{{\pi }^2\times \left(200\text{GPa}\times \frac{{\text{10}}^{\text{9}}{\text{N/m}}^{\text{2}}}{\text{1}\text{GPa}}\right)\times I_{round}}{1^2}\\ I_{round}=3.25748\times {10}^{-8}{\text{m}}^4\times {\left(\frac{1000\text{mm}}{1\text{m}}\right)}^4\\ =3.25748\times {10}^4{\text{mm}}^4\end{array}$

Determine the radius of the round strut (c) using the relation.

$I_{round}=\frac{\pi c^4}{4}$

Substitute $3.25748\times {10}^4{\text{mm}}^4$ for $I_{round}$.

$\begin{array}{c}3.25748\times {10}^4=\frac{\pi \times c^4}{4}\\ c=14.27\text{mm}\end{array}$

Therefore, the radius of the round strut is $\underset{\_}{14.27\text{mm}}$.

To determine

Find the percentage of cross-sectional area of square strut to the cross-sectional area of round strut.

Answer to Problem 14P

The percentage of cross-sectional area of square strut to the cross-sectional area of round strut is $\underset{\_}{97.7\%}$.

Explanation of Solution

The modulus of elasticity of the strut is $E=200\text{GPa}$.

Find the cross sectional area of the square strut $\left(A_{square}\right)$  using the equation.

$A_{square}=a^2$

Here, the size of the square strut is a.

Substitute 25 mm for a.

$\begin{array}{c}{A}_{square}={25}^2\\ =625{\text{mm}}^2\end{array}$

Find the cross sectional area of the round strut $\left(A_{round}\right)$ using the equation.

$A_{round}=\pi c^2$

Here, the radius of the round strut is c.

Substitute 14.27 mm for c.

$\begin{array}{c}{A}_{round}=\pi \times {14.27}^2\\ =639.73{\text{mm}}^2\end{array}$

Find the percentage of area of square strut to the area of round strut as follows;

$\text{Percentage}=\frac{A_{square}}{A_{round}}\times 100\%$

Substitute $625{\text{mm}}^2$ for $A_{square}$ and $639.73{\text{mm}}^2$ for $A_{round}$.

$\begin{array}{c}\text{Percentage}=\frac{625}{639.73}\times 100\%\\ =97.7\%\end{array}$

Therefore, the percentage of cross-sectional area of square strut to the cross-sectional area of round strut is $\underset{\_}{97.7\%}$.