Question

In case of bending of a beam, depression $\delta$ depends on Young modulus of elasticity Y as
A. $\propto Y$
B. $\propto {Y^2}$
C. $\propto {Y^{ - 1}}$
D. $\propto {Y^{ - 2}}$

Answer 1

Beam is a structural member of the horizontal axis in which loads applied on it are perpendicular to its axis. Due to the application of load, the beam bends into positive or negative direction depending upon the nature of load and the value of bending from the neutral axis is called deflection of the beam. We will use the expression for deflection of a simply supported beam when a load is applied at its centre.

Complete step by step answer:
We can find the answer to this question by considering any beam, but for the sake of our simplicity, we will consider a simply supported beam subjected to a point load at its centre.

We know that the deflection of a simply supported beam of length L which is subjected to a point load at its midpoint is given by:
$d = \dfrac{{P{L^3}}}{{48YI}}$
Here Y is Young’s modulus for beam material, and I is the moment of inertia of beam.

From the above expression, we can see that deflection is inversely proportional to Young’s modulus of elasticity.
$d \propto \dfrac{1}{Y}$

This deflection is the depression of the beam so that we can write:

\delta \propto \dfrac{1}{Y}\\\ \delta \propto {Y^{ - 1}} \end{array}$$ Therefore, in case of bending of a beam depression $$\delta $$ is inversely proportional to the Young modulus of elasticity of the beam material. **So, the correct answer is “Option C”.** **Note:** Young’s modulus of elasticity of the beam is equal to the ratio of stress and strain within the limit of proportion. It is the property of the material which does not depend on the external factors like force, moment, etc.