Question:

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.

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Compute the strain tensor from the displacement field first; if every strain component is zero but the antisymmetric rotation term is not, the motion is a pure rotation.
Updated On: Jul 16, 2026
  • biaxial deformation
  • pure shear
  • pure bending
  • rigid body rotation
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The Correct Option is D

Solution and Explanation

Step 1: Write down the displacement gradient tensor.
The displacement field is \(u=-ay\), \(v=ax\). Differentiate each component with respect to \(x\) and \(y\):
\[ \frac{\partial u}{\partial x}=0,\quad \frac{\partial u}{\partial y}=-a,\quad \frac{\partial v}{\partial x}=a,\quad \frac{\partial v}{\partial y}=0 \]

Step 2: Split the displacement gradient into its symmetric (strain) and antisymmetric (rotation) parts.
The normal strains are
\[ \varepsilon_{xx}=\frac{\partial u}{\partial x}=0,\qquad \varepsilon_{yy}=\frac{\partial v}{\partial y}=0 \]
The engineering shear strain is
\[ \gamma_{xy}=\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}=-a+a=0 \]
So the shear strain \(\varepsilon_{xy}=\tfrac{1}{2}\gamma_{xy}=0\) as well.

Step 3: Find the rotation part.
The (small) rotation about the z-axis is
\[ \omega_z=\frac{1}{2}\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right)=\frac{1}{2}(a-(-a))=a \]
This is nonzero, so the field has a rotation but no strain of any kind.

Step 4: Match this to the given options.
Since every strain component (\(\varepsilon_{xx}\), \(\varepsilon_{yy}\), \(\varepsilon_{xy}\)) is exactly zero, the square does not stretch along either axis (rules out biaxial deformation), does not distort its right angles (rules out pure shear), and does not bend or curve (rules out pure bending, which would need a strain varying through the body). All that remains is a plain rotation of the whole square by angle \(a\) about the origin, which changes its orientation but not its size or shape.

Final Answer:
The body undergoes rigid body rotation in the x-y plane. \[ \boxed{\text{Rigid body rotation}} \]
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