Compare The Loads Required To Produce Equal Depression For Two Beams Made Of The Same Material And Having The Same Length And Weight With Only Different That One Has Circular Cross Section While The Other Has Square Cross Section.

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Let us consider a bar of length l and density \( \rho \). This bar has a circular cross-sectional area of radius r. So the mass of this bar is \( \pi{r^2}l\rho \).
Now let us consider another bar of the same length and density but it has a square cross-sectional area with each side of length a. So the mass of this bar is \( a^2{l}\rho \).

Since the weight of these two bars is equal, then
\( \pi{r^2}l\rho=a^2{l}\rho\\or,\ \pi{r^2}=a^2\tag{1} \)

Let, \( I_1 \) and \( I_2 \) be the geometrical moment of inertia of the circular bar and the square bar respectively. So we can write,

\( I_1=\frac{\pi{r^4}}{4}\tag{2} \)

and, \( I_2=\frac{a^4}{12}\tag{3} \)

Let, \( W_1 \) and \( W_2 \) be the loads of the circular bar and the square bar respectively. So we can write,
the depression of the circular bar is \( y_1=\frac{W_1{l^3}}{3YI_1} \)
and the depression of the square bar is \( y_2=\frac{W_2{l^3}}{3YI_2} \).
[Read In Detail] Where Y is Young’s modulus of the material of the bars.

Using equation (2) we can write, \( y_1=\frac{4W_1{l^3}}{3\pi{Y}r^4} \)
Using equation (3) we can write, \( y_2=\frac{12W_2{l^3}}{3Y{a^4}}=\frac{4W_2{l^3}}{Ya^4} \)

Since the loads produce the equal depression for the two beams,
therefore, \( \displaystyle{\frac{4W_1{l^3}}{3\pi{Y}r^4}=\frac{4W_2{l^3}}{Ya^4}} \)
or, \( \displaystyle{\frac{W_1}{W_2}=\frac{3\pi{r^4}}{a^4}} \)
Using equation (1) we can write
\( \displaystyle{\frac{W_1}{W_2}=\frac{3\pi{r^4}}{{\pi}^2{r^4}}} \)
or, \( \displaystyle{\frac{W_1}{W_2}=\frac{3}{\pi}} \)

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