Question

An isotropic linear elastic material point under plane strain condition in the x–y plane always obeys:

• A. out-of-plane normal strain, \(\varepsilon_{zz} = 0\)
• B. out-of-plane normal stress, \(\sigma_{zz} = 0\)
• C. out-of-plane shear stress, \(\tau_{xz} = 0\)
• D. out-of-plane shear strain, \(\gamma_{xz} = 0\)

Hint:

In plane strain, all out-of-plane strains are zero. But stresses may develop due to constraints and Poisson’s effect.

Answer 1

The Correct Option is A, C, D

Step 1: Understand plane strain condition.
Under plane strain, deformation is allowed only in the x–y plane. There is no strain in the z-direction: \(\varepsilon_{zz} = 0\), and out-of-plane shear strains are zero: \(\gamma_{xz} = 0\), \(\gamma_{yz} = 0\).
Step 2: Address out-of-plane stress.
The out-of-plane stress \(\sigma_{zz}\) is generally \textit{not zero} because it adjusts to maintain \(\varepsilon_{zz} = 0\) via Poisson’s ratio effect.
Step 3: Evaluate options.
Option (A): True — \(\varepsilon_{zz} = 0\) by definition of plane strain.
Option (B): False — \(\sigma_{zz}\) can be non-zero to enforce \(\varepsilon_{zz} = 0\).
Option (C): True — \(\tau_{xz} = 0\) in plane strain.
Option (D): True — \(\gamma_{xz} = 0\) in plane strain.