Question Statement:
Consider a unit square body as shown in the figure below. The body is subjected
to the deformation field u = -ay and v = ax, where 'a' is a constant. Due to the
application of this deformation field, the body undergoes _______ in the x-y plane.
Scientific schematic diagram reproduced from official IIT paper
Options:
-
(A) biaxial deformation
-
(B) pure shear
-
(C) pure bending
-
(D) rigid body rotation
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2026 Q.40)**
**1. Intuitive Physical Setup & Given Parameters**:
A unit square body in the $x$-$y$ plane is subjected to the 2D displacement field:
u = -ay, \quad v = ax
where $a$ is a non-zero constant ($a \ll 1$ in linear elasticity).
We need to determine the physical nature of the deformation experienced by the body.
**2. Step-by-Step Continuum Mechanics Derivation**:
1. **Evaluate the Strain Tensor Components**:
In linear elasticity, the components of the infinitesimal strain tensor are:
\varepsilon_{xx} = \frac{\partial u}{\partial x} = \frac{\partial}{\partial x}(-ay) = 0
\varepsilon_{yy} = \frac{\partial v}{\partial y} = \frac{\partial}{\partial y}(ax) = 0
\gamma_{xy} = \frac{\partial u}{\partial y} + \frac{\partial v}{\partial x} = (-a) + (a) = 0
2. **Evaluate the Rigid-Body Rotation Tensor**:
The infinitesimal rotation (spin) tensor component $\omega_{xy}$ about the $z$-axis is:
\omega_{xy} = \frac{1}{2}\left(\frac{\partial v}{\partial x} - \frac{\partial u}{\partial y}\right) = \frac{1}{2}(a - (-a)) = \frac{1}{2}(2a) = a \neq 0
3. **Physical Interpretation**:
- Because all normal strains ($\varepsilon_{xx} = \varepsilon_{yy} = 0$) and the shear strain ($\gamma_{xy} = 0$) vanish identically, no element of the body changes length, area, or internal angle.
- The body experiences zero internal deformation or stress.
- The non-zero rotation $\omega_z = a$ corresponds to a pure **rigid body rotation** by an angle $\theta \approx a$ radians in the counter-clockwise direction around the origin ($x' = x\cos a - y\sin a \approx x - ay$, $y' = x\sin a + y\cos a \approx ax + y$).
**3. Comprehensive Options & Distractor Analysis**:
- **(A) biaxial deformation**: ❌ **INCORRECT**. Requires non-zero normal strains $\varepsilon_{xx}, \varepsilon_{yy} \neq 0$.
- **(B) pure shear**: ❌ **INCORRECT**. Requires non-zero engineering shear strain $\gamma_{xy} = \frac{\partial u}{\partial y} + \frac{\partial v}{\partial x} \neq 0$. Here $(-a) + (a) = 0$.
- **(C) pure bending**: ❌ **INCORRECT**. Requires linear variation of axial strain $\varepsilon_{xx} \propto y$.
- **(D) rigid body rotation**: ✅ **CORRECT**. All strain components are zero while the rotation tensor is non-zero ($\omega_{xy} = a$).
**4. Official Master Key & Exam Takeaway**:
- **Official Answer Key**: **D**
- **Exam Takeaway**: When examining displacement fields, always evaluate $\varepsilon_{xx}, \varepsilon_{yy}$ and $\gamma_{xy} = \frac{\partial u}{\partial y} + \frac{\partial v}{\partial x}$. If all strains vanish, the field is purely rigid-body translation/rotation.