Mohr’s Circle: Plane Stress Equations and Example
Introduction
Mohr’s circle turns plane stress transformation into a clear geometric construction, helping engineers find principal stresses, maximum shear stress, and stresses on any rotated plane. This guide develops the governing equations, solves a worked example, and highlights the sign conventions that commonly cost students marks.
Mohr’s Circle and Plane Stress Transformation
A plane stress element is described by normal stresses σx and σy plus in-plane shear stress τxy. For an element rotated counterclockwise through angle θ, the transformed normal stress is σθ = (σx + σy)/2 + [(σx − σy)/2]cos(2θ) + τxy sin(2θ). The transformed shear stress is τθ = −[(σx − σy)/2]sin(2θ) + τxy cos(2θ).
These equations define a circle in σ–τ coordinates. Its centre lies on the normal-stress axis at C = (σx + σy)/2, and its radius is R = √{[(σx − σy)/2]² + τxy²}. The two points representing perpendicular x and y faces lie at opposite ends of a diameter.
The circle uses an angular movement of 2θ because the transformation equations contain cos(2θ) and sin(2θ). Therefore, a physical rotation of 20° corresponds to 40° around the circle. Depending on the textbook’s shear-axis convention, the graphical direction may oppose the physical rotation, so always state the convention before plotting.
Mohr’s Circle Worked Example
Consider σx = 80 MPa tension, σy = 20 MPa tension, and τxy = 30 MPa. The centre is C = (80 + 20)/2 = 50 MPa. The radius is R = √{[(80 − 20)/2]² + 30²} = √(30² + 30²) = 42.43 MPa.
The principal stresses occur where shear stress is zero, so σ1 = C + R = 92.43 MPa and σ2 = C − R = 7.57 MPa. The maximum in-plane shear stress equals the radius, τmax = 42.43 MPa, while the associated normal stress is 50 MPa. These results also provide a fast numerical check on a hand-drawn circle.
The principal-plane angle follows tan(2θp) = 2τxy/(σx − σy) = 60/60 = 1. Thus 2θp = 45° and θp = 22.5°, with the perpendicular principal plane at 112.5°. Use the original transformation equations to identify which orientation carries σ1 when the shear-sign convention is uncertain.
Principal Stresses and Maximum Shear Stress in Design
Mechanical designers use principal stresses when a brittle material is assessed with maximum-normal-stress criteria and when stress directions guide strain-gauge placement. Maximum shear stress supports the Tresca yield criterion for ductile metals, where yielding begins when the largest three-dimensional shear stress reaches the material limit.
Applications include shafts under combined bending and torsion, pressure-vessel walls, welded joints, aircraft structures, and finite element analysis result checks. For a thin plate, plane stress often assumes σz = τxz = τyz = 0, but σz = 0 still acts as the third principal stress. This detail matters when comparing the absolute maximum shear stress across all three principal stresses.
Common Mohr’s Circle Exam Mistakes
The most frequent mistake is mixing sign conventions for shear stress. Some mechanics texts plot positive shear downward, while standard Cartesian graphs plot it upward; either can work if point coordinates, rotation direction, and equations remain consistent. Students also forget that angles measured on the circle are twice the physical angles on the stress element.
Keep stress units consistent and distinguish tensile normal stress, usually positive, from compressive stress, usually negative. Do not read maximum shear at σ = 0; it occurs at the top and bottom of the circle, directly above and below C. Finally, verify that σ1 + σ2 = σx + σy and that (σ1 − σ2)/2 = R.
Conclusion
Mohr’s circle packages plane stress transformation into one diagram: the centre gives average normal stress, the radius gives maximum in-plane shear, and the axis intersections give principal stresses. Build the circle from consistent signs, remember the 2θ relationship, and check the stress invariants after every calculation. Explore more mechanical engineering topics on Mechtics, or share a stress-analysis question in the comments.


