Question

A simply supported beam having length = L, width = B and depth = H, carries a concentrated load W at its centre and undergoes a deflection $\delta$ under the load. If the width and depth are interchanged, the deflection at the centre of the beam would attain a value:

• A. $\left(\frac{H}{B}\right) \delta$
• B. $\left(\frac{H}{B}\right)^2 \delta$
• C. $\left(\frac{H}{B}\right)^3 \delta$
• D. $\left(\frac{H}{B}\right)^{1.5} \delta$

Hint:

Deflection is highly sensitive to depth. [cite: 15] Because depth is cubed in the moment of inertia formula, changing which dimension acts as the depth has a squared effect on the overall deflection ratio. [cite: 37]

Answer 1

The Correct Option is B

To determine the new deflection when the dimensions are interchanged, we must examine the mathematical relationship between beam geometry and central deflection for a simply supported beam. 1. Initial Deflection Formula: [cite: 23, 25] For a simply supported beam with a central point load $W$, the maximum deflection $\delta$ at the center is given by: $$\delta = \frac{WL^3}{48EI}$$ Where $E$ is the Young's modulus and $I$ is the area moment of inertia. [cite: 23, 25]

2. Relation to Beam Dimensions: [cite: 23, 25] The moment of inertia for a rectangular cross-section is $I = \frac{BH^3}{12}$. [cite: 23, 25] Substituting this into the deflection formula: $$\delta \propto \frac{1}{I} \implies \delta \propto \frac{1}{BH^3}$$ [cite: 23, 25]

3. Interchanging Dimensions: [cite: 23, 25] If we interchange width ($B$) and depth ($H$), the new width becomes $B' = H$ and the new depth becomes $H' = B$. [cite: 23, 25] The new deflection $\delta'$ will be: $$\delta' \propto \frac{1}{H \cdot B^3}$$ [cite: 23, 25]

4. Calculating the Ratio: [cite: 23, 25] $$\frac{\delta'}{\delta} = \frac{BH^3}{HB^3} = \frac{H^2}{B^2} = \left(\frac{H}{B}\right)^2$$ [cite: 23, 25] Therefore, $\delta' = \left(\frac{H}{B}\right)^2 \delta$. [cite: 23, 25]