Subtopic: Stress and Strain Tensors & Generalized Hooke's Law
Question Statement:
As shown in the figure, a beam of length 1 m is rigidly clamped at one end ($x = 0$) and simply supported at the other ($x = 1$ m). Under the action of a uniformly distributed downward load of $w = 10$ N/m, the magnitude of the vertical reaction force (in N) at the simply supported end is _________ (round off to two decimal places).
Official Answer Key:
3.75
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2020 Q.53)**
**1. Intuitive Physical Setup & Given Parameters**:
- Propped cantilever beam of length $L = 1\text{ m}$.
- Fixed support at $A$ ($x = 0$), simple roller support at $B$ ($x = L = 1\text{ m}$).
- Uniformly distributed load: $w = 10\text{ N/m}$.
- Target: Vertical reaction $R_B$ at the simple support.
**2. Step-by-Step Structural Mechanics Derivation**:
1. **Statically Indeterminate Beam Analysis (Method of Superposition)**:
- Remove the redundant reaction $R_B$ at $B$, leaving a cantilever beam subjected to $w$:
Downward tip deflection due to UDL $w$:
\delta_w = \frac{w L^4}{8 EI}
- Apply upward redundant reaction $R_B$ at tip $B$:
Upward tip deflection due to point force $R_B$:
\delta_{R} = \frac{R_B L^3}{3 EI}
2. **Compatibility Condition**:
Since the simple support at $B$ prevents vertical displacement (zero deflection):
\delta_{net} = \delta_w - \delta_R = 0 \implies \frac{R_B L^3}{3 EI} = \frac{w L^4}{8 EI}
3. **Solving for Reaction $R_B$**:
R_B = \frac{3}{8} w L
4. **Numerical Substitution**:
R_B = \frac{3}{8} \times (10\text{ N/m}) \times (1\text{ m}) = \frac{30}{8} = 3.75\text{ N}
The vertical reaction at the simple support is exactly **3.75 N**.
**3. Comprehensive Options & Distractor Analysis**:
- **Question Type**: Numerical Answer Type (NAT).
- **Exact Analytical Result**: **3.75 N** (Acceptable Range: [3.74, 3.76]).
**4. Official Master Key & Exam Takeaway**:
- **Official Answer**: **3.75**
- **Canonical Propped Cantilever Result**: $R_{prop} = \frac{3}{8} w L$; Fixed moment $M_A = \frac{1}{8} w L^2$.