The element is subjected to two equal and like stress 'σ' on two mutually perpendicular planes. The shape of the Mohr's circle will be

Question

Question

This question was previously asked in

UPPSC AE Civil 2022 Official Paper I (Held on 29 May 2022)

Answer 1

Concept:

When the normal stresses on the two mutually perpendicular planes are equal and like then the radius of the Mohr circle will be zero.

The radius of Mohr's circle,

R = \(\sqrt { (\frac{σ_x-σ_y}{2})^2 + τ_xy^2}\)

Here,

σ_x = σ_y and τ_xy = 0

∴ R = 0 means Mohr's circle reduces to a point.