Mohr’s construction: The locus of the end point of the resultant of normal and tangential stresses on an inclined plane is a ____.

Introduction:This question targets your understanding of Mohr’s circle, a powerful graphical tool for plane stress transformation. It asks about the geometric path traced by the resultant stress vector as the physical plane rotates through different inclinations.

Given Data / Assumptions:

• Biaxial (plane) stress state defined by σx, σy, and τxy at a point.
• We examine stresses on all possible plane orientations passing through that point.
• Linear elasticity and standard sign conventions apply.

Concept / Approach:When the plane is rotated by an angle θ in the body, the normal and shear stresses on that plane (σn, τn) satisfy parametric equations that, in σ–τ coordinates, describe a circle. This construction is Mohr’s circle, with center at (σavg, 0), where σavg = (σx + σy)/2, and radius R = √[( (σx − σy)/2 )^2 + τxy^2 ].

Step-by-Step Solution:Write transformation equations: σn = σavg + ((σx − σy)/2) * cos(2θ) + τxy * sin(2θ).Also τn = −((σx − σy)/2) * sin(2θ) + τxy * cos(2θ).Eliminate θ to obtain (σn − σavg)^2 + τn^2 = R^2, which is the equation of a circle.Therefore, the locus of the stress resultant endpoint is a circle.

Verification / Alternative check:Special cases: pure uniaxial stress gives a circle passing through (σx, 0) and (0, 0); pure shear gives a circle centered at the origin.

Why Other Options Are Wrong:Parabola / ellipse / straight line / hyperbola: None satisfy the derived relation (σn − σavg)^2 + τn^2 = R^2.

Common Pitfalls:Confusing stress transformation with strain transformation or inertia ellipse; forgetting the angle-doubling (2θ) relation in Mohr’s construction.

Final Answer:circle.