Question

A plane strain problem (in X-Y plane) must satisfy the condition:

• A. \(\sigma_{zz} = 0\)
• B. \(\epsilon_{zz} = \epsilon_{xz} = \epsilon_{yz} = 0\)
• C. \(\sigma_{xx} \neq \sigma_{xy} \neq \sigma_{xz} \neq 0\)
• D. \(\epsilon_{xx} \neq \epsilon_{yy} \neq \epsilon_{xy} \neq 0\)

Hint:

Plane strain locks all z-direction strains to zero; it does not make \(\sigma_{zz}\) zero, that is plane stress instead.

Answer 1

The Correct Option is B

Step 1: Recall what plane strain means.
Plane strain is defined for a long prismatic body (in the \(z\) direction) whose ends are restrained so no strain can develop along that length, and where loading does not vary along \(z\). It applies to structures like long retaining walls, dams, or tunnels, where the cross-section and loading repeat along a very long \(z\) axis that is effectively locked.

Step 2: Write the strain field.
Because the body cannot stretch, shear, or shorten along \(z\), every strain component carrying a \(z\) index is forced to zero: \[ \epsilon_{zz} = \epsilon_{xz} = \epsilon_{yz} = 0 \] The in-plane strains \(\epsilon_{xx}\), \(\epsilon_{yy}\), \(\epsilon_{xy}\) stay free to take non-zero values, since deformation is confined to the \(x\)-\(y\) plane.

Step 3: Check what happens to stress.
Even though \(\epsilon_{zz}=0\), the stress \(\sigma_{zz}\) is generally NOT zero. The material tends to expand or contract along \(z\) under the in-plane stresses (Poisson's effect) but is prevented from doing so, so a restraining stress \(\sigma_{zz}=\nu(\sigma_{xx}+\sigma_{yy})\) builds up. So option A, claiming \(\sigma_{zz}=0\), actually describes plane STRESS, the opposite idealisation used for thin plates, not plane strain.

Step 4: Rule out the remaining options.
Option C only claims the in-plane and out-of-plane stresses are all different and non-zero, a generic statement about any 2-D stress state, not a defining condition of plane strain. Option D makes the same kind of generic claim about strains. Only option B states the actual physical restraint used to derive the plane strain formulation.

Final Answer:
Plane strain in the X-Y plane requires \(\epsilon_{zz}=\epsilon_{xz}=\epsilon_{yz}=0\). \[ \boxed{\text{Option (B)}} \]