Question

A cantilever beam of rectangular cross section is subjected to a load W at its free end. If the depth of the beam is doubled and the load is halved, the deflection of the free end as compared to original deflection will be _______.

• A. 1/2
• B. 1/8
• C. 1/16
• D. 2.0

Hint:

Bending deflection is highly sensitive to the depth of the beam because depth is cubed in the denominator of the stiffness term.
Doubling depth alone reduces deflection to $1/8$, and halving the load further reduces it to $1/16$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
This question asks for the ratio of the end deflection of a cantilever beam after specific changes are made to its depth and the applied load.

Step 2: Key Formula or Approach:
The deflection ($\delta$) at the free end of a cantilever beam of length $L$ carrying a point load $W$ at its free end is:
\[ \delta = \frac{W \cdot L^3}{3 E \cdot I} \]
where $I$ is the area moment of inertia. For a rectangular cross-section of width $b$ and depth $d$:
\[ I = \frac{b \cdot d^3}{12} \]
Substituting this gives:
\[ \delta \propto \frac{W}{I} \propto \frac{W}{d^3} \]

Step 3: Detailed Explanation:

• Let the original parameters be load $W_1$, depth $d_1$, and deflection $\delta_1$.
- The original deflection is proportional to:
\[ \delta_1 \propto \frac{W_1}{d_1^3} \]

• Let the modified parameters be load $W_2$, depth $d_2$, and deflection $\delta_2$. We are given:
- The load is halved: $W_2 = \frac{W_1}{2}$
- The depth is doubled: $d_2 = 2 \cdot d_1$

• Express the new deflection ($\delta_2$):
\[ \delta_2 \propto \frac{W_2}{d_2^3} = \frac{\left(\frac{W_1}{2}\right)}{(2 \cdot d_1)^3} \]
\[ \delta_2 \propto \frac{W_1}{2 \cdot (8 \cdot d_1^3)} = \frac{1}{16} \left( \frac{W_1}{d_1^3} \right) \]

• Comparing the two deflections:
\[ \delta_2 = \frac{1}{16} \cdot \delta_1 \]

Step 4: Final Answer:
The deflection of the free end will be 1/16 of the original deflection.