Question Statement:
In the given figure, plate ABCD in its undeformed configuration (solid line) is a
rhombus with all the internal angles being 90o. The lengths of the undeformed
diagonals are 20 cm. ABCD deforms as shown by the dotted lines. Upon
deformation, diagonal AC reduces to 19.96 cm and BD increases to 20.04 cm. In
the given x-y coordinate system, the engineering shear strain $\gamma$xy is equal to ____.
Scientific schematic diagram reproduced from official IIT paper
Options:
-
(A) 0
-
(B) 0.002
-
(C) 0.004
-
(D) -0.004
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2025 Q.30)**
**1. Intuitive Physical Setup & Given Parameters**:
- Undeformed plate $ABCD$ is a rhombus with all internal angles $90^\circ$ (i.e., a square rotated by $45^\circ$).
- Coordinate axes: $x$-axis lies along diagonal $AC$, and $y$-axis lies along diagonal $BD$.
- Initial diagonal lengths:
L_{AC} = 20\text{ cm}, \quad L_{BD} = 20\text{ cm}
- Deformed configuration:
L'_{AC} = 19.96\text{ cm}, \quad L'_{BD} = 20.04\text{ cm}
- We need to find the **engineering shear strain** $\gamma_{xy}$ in the specified $x$-$y$ coordinate system.
**2. Step-by-Step Continuum Mechanics Derivation**:
1. **Definition of Engineering Shear Strain $\gamma_{xy}$**:
- Engineering shear strain $\gamma_{xy}$ measures the change in the right angle ($90^\circ$) between two line segments originally oriented along the $x$ and $y$ coordinate axes:
\gamma_{xy} = \frac{\pi}{2} - \theta'_{xy}
2. **Analysis of the Diagonals**:
- The diagonals $AC$ and $BD$ lie precisely along the $x$-axis and $y$-axis, respectively.
- Throughout the deformation shown:
- Diagonal $AC$ shortens symmetrically along the $x$-axis ($\Delta L_{AC} = 19.96 - 20 = -0.04\text{ cm}$).
- Diagonal $BD$ lengthens symmetrically along the $y$-axis ($\Delta L_{BD} = 20.04 - 20 = +0.04\text{ cm}$).
- Because the deformation is symmetric about both the $x$ and $y$ axes, the diagonals remain strictly perpendicular to each other ($\theta'_{xy} = 90^\circ$).
3. **Evaluate $\gamma_{xy}$**:
- Since the angle between the $x$-axis (line $AC$) and $y$-axis (line $BD$) does not change:
\gamma_{xy} = 90^\circ - 90^\circ = 0
- The axes $x$ and $y$ are the **principal strain axes**, meaning the shear strain referred to these axes is identically zero:
\gamma_{xy} = 0
- (Note: The sides of the square rotate relative to each other, so shear strain referred to edge axes $x'-y'$ is non-zero, but along the specified diagonal axes $x-y$, $\gamma_{xy} = 0$).
**3. Comprehensive Options & Distractor Analysis**:
- **(A) 0**: ✅ **CORRECT**. $x$ and $y$ axes lie along the diagonals, which remain orthogonal by symmetry, so $\gamma_{xy} = 0$.
- **(B) 0.002**: ❌ **INCORRECT**. Represents normal strain magnitude $|\varepsilon_{xx}| = \frac{0.04}{20} = 0.002$, not shear strain.
- **(C) 0.004**: ❌ **INCORRECT**. Represents shear strain referred to the rotated edge axes ($45^\circ$).
- **(D) -0.004**: ❌ **INCORRECT**. Sign variation for edge axes.
**4. Official Master Key & Exam Takeaway**:
- **Official Answer Key**: **A**
- **Exam Takeaway**: Lines of symmetry are always principal strain axes. Along principal axes, shear strain is identically zero ($\gamma_{xy} = 0$).