Subtopic: Stress and Strain Tensors & Generalized Hooke's Law
Question Statement:
A simply supported horizontal beam is subjected to a distributed transverse load
varying linearly from q0 at A to zero at B, as shown in the figure. Which one of
the following options is correct?
Scientific schematic diagram reproduced from official IIT paper
Options:
(A) The magnitude of the vertical reaction force at A is larger than that at B.
(B) The magnitude of the vertical reaction force at B is larger than that at A.
(C) The magnitudes of the vertical reaction forces at A and B are equal.
(D) The reactions at points A and B are indeterminate.
Official Answer Key:
A
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2025 Q.27)**
**1. Intuitive Physical Setup & Given Parameters**:
- Simply supported horizontal beam of span $L$ resting on supports at $A$ (left) and $B$ (right).
- Subjected to a triangular distributed transverse load:
- Intensity at left support $A$: $q(0) = q_0$
- Intensity at right support $B$: $q(L) = 0$
- Load varies linearly: $q(x) = q_0 \left(1 - \frac{x}{L}\right)$
We need to compare the vertical reaction forces $R_A$ and $R_B$.
**2. Step-by-Step Static Equilibrium Derivation**:
1. **Total Downward Load ($W_{\text{total}}$)**:
The total downward force is the area of the load triangle:
W_{\text{total}} = \int_0^L q(x) dx = \frac{1}{2} q_0 L
2. **Centroid of the Triangular Load**:
- The centroid of a triangle with base at $x = 0$ is located at one-third of the span from the heavy end:
\bar{x} = \frac{L}{3} \text{ from support } A
\text{Distance from support } B = L - \frac{L}{3} = \frac{2L}{3}
3. **Moment Equilibrium About Support $B$**:
\sum M_B = 0 \implies R_A \times L - W_{\text{total}} \times \left(\frac{2L}{3}\right) = 0
R_A L = \left(\frac{1}{2} q_0 L\right) \left(\frac{2L}{3}\right) = \frac{1}{3} q_0 L^2
R_A = \frac{1}{3} q_0 L = \frac{2}{3} W_{\text{total}}
4. **Vertical Force Equilibrium**:
\sum F_y = 0 \implies R_A + R_B = W_{\text{total}}
R_B = W_{\text{total}} - R_A = \frac{1}{2} q_0 L - \frac{1}{3} q_0 L = \frac{1}{6} q_0 L = \frac{1}{3} W_{\text{total}}
5. **Comparison**:
R_A = \frac{q_0 L}{3} > R_B = \frac{q_0 L}{6}
- The vertical reaction force at $A$ is **twice** as large as that at $B$.
**3. Comprehensive Options & Distractor Analysis**:
- **(A) The magnitude of the vertical reaction force at A is larger than that at B**: ✅ **CORRECT**. $R_A = 2 R_B$ because the load resultant is located closer to $A$ ($\bar{x} = L/3$).
- **(B) The magnitude of the vertical reaction force at B is larger than that at A**: ❌ **INCORRECT**. Contradicts lever arm mechanics.
- **(C) The magnitudes of the vertical reaction forces at A and B are equal**: ❌ **INCORRECT**. Only true for symmetric loading.
- **(D) The reactions at points A and B are indeterminate**: ❌ **INCORRECT**. A simply supported beam is statically determinate.
**4. Official Master Key & Exam Takeaway**:
- **Official Answer Key**: **A**
- **Exam Takeaway**: A support closer to the resultant centroid of a transverse distributed load always carries the larger reaction share: $R_A = \frac{q_0 L}{3}$ vs $R_B = \frac{q_0 L}{6}$.
