Question

For the same cross section area and for a given load, the ratio of depression for the beam of a square cross section and circular cross section is:
(A) $3:\pi $
(B) $\pi :3$
(C) $1:\pi $
(D) $\pi :1$

Answer 1

Hint: The ratio of the depression can be determined by dividing the deflection values of the square cross section to the circular cross section. And by using the given information in the question the ratio of depression can be determined.

Useful formula:
The deflection of the beam is given by,
$\delta = \dfrac{{W{l^3}}}{{3EI}}$
Where, $\delta $ is the deflection of the beam, $W$ is the load given to the beam, $l$ is the length of the beam, $E$ is the Young’s modulus of the beam and $I$ is the moment of inertia.
Moment of inertia for rectangular beam is given by,
$I = \dfrac{{b{d^3}}}{{12}}$
Where, $I$ is the moment of inertia, $b$ is the breadth of the rectangle and $d$ is the depth of the rectangle.
Moment of inertia for square beam is given by,
$I = \dfrac{{{b^4}}}{{12}}$ (For square both the breadth and depth are same and in other words all sides are equal)
Where, $I$ is the moment of inertia and $b$ is the breadth of the square.
Moment of inertia for circular beam is given by,
$I = \dfrac{{\pi {r^4}}}{4}$
Where, $I$ is the moment of inertia and $r$ is the radius of the circular beam.

Complete step by step solution:
Given that,
Both the rectangular beam and the circular beam have the same cross section.
Now,
The deflection of the square beam is given by,
${\delta _1} = \dfrac{{W{l^3}}}{{3EI}}\,..................\left( 1 \right)$
Here, ${\delta _1}$ is the deflection in the square cross section beam.
By substituting the moment of inertia value for square cross section in the equation (1), then
${\delta _1} = \dfrac{{W{l^3}}}{{3E \times \left( {\dfrac{{{b^4}}}{{12}}} \right)}}$
By rearranging the terms in the above equation, then
${\delta _1} = \dfrac{{W{l^3} \times 12}}{{3E \times {b^4}}}\,....................\left( 2 \right)$
The deflection of the circular beam is given by,
${\delta _2} = \dfrac{{W{l^3}}}{{3EI}}\,..................\left( 1 \right)$
Here, ${\delta _2}$ is the deflection in the circular cross section beam.
By substituting the moment of inertia value for circular beam in the equation (1), then
${\delta _2} = \dfrac{{W{l^3}}}{{3E \times \left( {\dfrac{{\pi {r^4}}}{4}} \right)}}$
By rearranging the terms in the above equation, then
${\delta _2} = \dfrac{{W{l^3} \times 4}}{{3E \times \pi {r^4}}}\,....................\left( 3 \right)$
On dividing the equation (2) and equation (3), then
\[\dfrac{{{\delta _1}}}{{{\delta _2}}} = \dfrac{{\left( {\dfrac{{W{l^3} \times 12}}{{3E \times {b^4}}}} \right)}}{{\left( {\dfrac{{W{l^3} \times 4}}{{3E \times \pi {r^4}}}} \right)}}\]
By cancelling the same terms, then
\[\dfrac{{{\delta _1}}}{{{\delta _2}}} = \dfrac{{\left( {\dfrac{{12}}{{{b^4}}}} \right)}}{{\left( {\dfrac{4}{{\pi {r^4}}}} \right)}}\]
By rearranging the terms, then
\[\dfrac{{{\delta _1}}}{{{\delta _2}}} = \dfrac{{12}}{{{b^4}}} \times \dfrac{{\pi {r^4}}}{4}\]
On further simplification, then
\[\dfrac{{{\delta _1}}}{{{\delta _2}}} = \dfrac{{3\pi {r^4}}}{{{b^4}}}\,..................\left( 4 \right)\]
Both the beam having same cross section, then $\left( {{b^2} = \pi {r^2}} \right)$, substituting the term in the above equation, then
\[\dfrac{{{\delta _1}}}{{{\delta _2}}} = \dfrac{{3\pi {r^4}}}{{{\pi ^2}{r^4}}}\]
By cancelling the same terms, then
\[\dfrac{{{\delta _1}}}{{{\delta _2}}} = \dfrac{3}{\pi }\]
Then the above equation is written as,
\[{\delta _1}:{\delta _2} = 3:\pi \]

Hence, the option (A) is the correct answer.

Note: Moment of inertia for the square cross section does not have a direct formula, it has been derived from the moment of inertia for the square cross section formula. In equation (4), both the beams have the same cross section area, so the assumption is taken for the simple calculation.