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Question 62
Question Statement:
The components of stress in a body under plane stress condition, in the absence of body forces, is given by: $\sigma_{xx} = Ax^2; \sigma_{yy} = 12x^2 - 6y^2$ and $\sigma_{xy} = 12xy$. The coefficient, $A$, such that the body is in static equilibrium is _________.
Official Answer Key:
-6.0
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2018 Q.62 / AE Q.52)**
**1. Intuitive Physical Setup & Given Parameters**:
- Stress tensor components under plane stress without body forces:
$\sigma_{xx} = A x^2$
$\sigma_{yy} = 12x^2 - 6y^2$
$\tau_{xy} = \sigma_{xy} = 12xy$
**2. Step-by-Step Mathematical Derivation**:
1. **Differential Equations of Equilibrium (Plane Stress)**:
In the absence of body forces ($F_x = 0, F_y = 0$):
$\frac{\partial \sigma_{xx}}{\partial x} + \frac{\partial \tau_{xy}}{\partial y} = 0 \quad \text{--- (Equation 1)}$
$\frac{\partial \tau_{xy}}{\partial x} + \frac{\partial \sigma_{yy}}{\partial y} = 0 \quad \text{--- (Equation 2)}$
2. **Checking Equation 2**:
$\frac{\partial \tau_{xy}}{\partial x} = \frac{\partial}{\partial x}(12xy) = 12y$
$\frac{\partial \sigma_{yy}}{\partial y} = \frac{\partial}{\partial y}(12x^2 - 6y^2) = -12y$
$12y + (-12y) = 0 \implies \text{Equation 2 is identically satisfied!}$
3. **Evaluating Equation 1 for Coefficient $A$**:
$\frac{\partial \sigma_{xx}}{\partial x} = \frac{\partial}{\partial x}(A x^2) = 2A x$
$\frac{\partial \tau_{xy}}{\partial y} = \frac{\partial}{\partial y}(12xy) = 12x$
Substituting into Equation 1:
$2A x + 12x = 0$
$(2A + 12) x = 0 \quad \forall x$
$2A + 12 = 0 \implies 2A = -12 \implies A = -6.0$
**3. Comprehensive Options & Distractor Analysis**:
- **Question Type**: Numerical Answer Type (NAT).
- **Exact Analytical Result**: **-6.0** (Acceptable Range: [-6.0, -6.0]).
**4. Official Master Key & Exam Takeaway**:
- **Official Answer**: **-6.0**
- **Exam Rule**: In plane elasticity equilibrium, differentiate directly and set the divergence $\nabla \cdot \boldsymbol{\sigma} = 0$.
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