Subtopic: Support Reactions & Equilibrium Equations
Question Statement:
For a 1 m long simply supported beam with a concentrated vertical load of 200 N and a concentrated bending moment of 100 N-m applied at mid-span, the vertical reaction at the left support is _________ N.
Options:
(A) 0
(B) 100
(C) 200
(D) 300
Official Answer Key:
A
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2019 Q.40 / AE Q.30)**
**1. Intuitive Physical Setup & Given Parameters**:
- Span of beam: $L = 1\text{ m}$.
- Left support $A$ at $x = 0$, Right support $B$ at $x = 1\text{ m}$.
- At mid-span ($x = 0.5\text{ m}$):
- Downward vertical concentrated load: $P = 200\text{ N}$.
- Concentrated clockwise bending moment: $M_0 = 100\text{ N m}$.
**2. Step-by-Step Statics Derivation**:
1. **Equations of Static Equilibrium**:
- Sum of vertical forces:
$R_A + R_B = P = 200\text{ N} \quad \text{--- (Equation 1)}$
- Moment equilibrium about right support $B$ (taking counter-clockwise as positive):
$\sum M_B = 0$
$(R_A \times 1\text{ m}) - (P \times 0.5\text{ m}) + M_0 = 0$
2. **Substituting Given Numerical Values**:
$R_A(1) - (200 \times 0.5) + 100 = 0$
$R_A - 100 + 100 = 0$
$R_A = 0\text{ N}$
3. Vertical reaction at the left support $R_A = 0\text{ N}$, while all 200 N is carried by support $B$ ($R_B = 200\text{ N}$).
**3. Comprehensive Options & Distractor Analysis**:
- **(A)** 0: ✅ **CORRECT**. Clockwise moment $M_0 = 100\text{ N m}$ exactly cancels the moment of the 200 N load about $B$.
- **(B)** 100: ❌ Incorrect. Would be the reaction if no moment were applied.
- **(C)** 200: ❌ Incorrect. This is the reaction at support B.
- **(D)** 300: ❌ Incorrect.
**4. Official Master Key & Exam Takeaway**:
- **Official Answer**: **A**
- **Exam Rule**: Always take moment about the opposite support to isolate the reaction directly.
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