As shown in the figure, a rigid slab CD of weight $W$ (distributed uniformly along its length) is hung from a ceiling using three vertical cables of identical length and cross-sectional area. The central cable is made of steel (Young's modulus = $3E$) and the other two outer cables are made of aluminium (Young's modulus = $E$). The percentage of the total weight taken by the central cable is _________ % (round off to the nearest integer).
Official Answer Key:
60.0
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2019 Q.59 / AE Q.49)**
**1. Intuitive Physical Setup & Given Parameters**:
- Rigid slab CD: does not bend (remains completely horizontal during displacement).
- Three vertical cables of equal length $L$ and equal cross-sectional area $A$.
- Two outer cables: made of aluminium with modulus $E_1 = E_3 = E$.
- Central cable: made of steel with modulus $E_2 = 3E$.
- Total downward weight: $W$.
**2. Step-by-Step Structural Mechanics Derivation**:
1. **Geometric Compatibility Condition**:
Because the slab is rigid and symmetric, it translates downward by a uniform vertical displacement $\delta$:
\delta_1 = \delta_2 = \delta_3 = \delta
2. **Force-Displacement Relations**:
Elongation of an axial member under tension $T_i$ is $\delta_i = \frac{T_i L}{A_i E_i}$.
Since length $L$ and area $A$ are identical:
T_i = \left( \frac{A E_i}{L} \right) \delta = k_i \delta
- For outer cables (1 and 3): $T_1 = T_3 = \left(\frac{A E}{L}\right) \delta = k \delta$
- For central cable (2): $T_2 = \left(\frac{A (3E)}{L}\right) \delta = 3k \delta = 3 T_1$
3. **Static Equilibrium**:
Sum of vertical forces:
T_1 + T_2 + T_3 = W
T_1 + 3T_1 + T_1 = W \implies 5T_1 = W \implies T_1 = \frac{W}{5}
4. **Load Taken by Central Cable ($T_2$)**:
T_2 = 3 T_1 = 3 \left(\frac{W}{5}\right) = \frac{3}{5} W = 0.60 W
5. **Percentage of Total Weight**:
\text{Percentage} = \frac{T_2}{W} \times 100\% = 60\%
**3. Comprehensive Options & Distractor Analysis**:
- **Question Type**: Numerical Answer Type (NAT).
- **Exact Analytical Result**: **60%** (Acceptable Range: [60, 60]).
**4. Official Master Key & Exam Takeaway**:
- **Official Answer**: **60.0**
- **Exam Rule**: In parallel indeterminate systems with identical deflection: load carried is directly proportional to axial stiffness $AE$ ($T_i \propto E_i$).