Mohr's Circle for Stress — Principal Stresses, Maximum Shear, and Worked Examples

Mohr's circle is one of the most elegant tools in the whole of solid mechanics. Devised by Otto Mohr in 1882, it turns the algebra of stress transformation into a single circle you can draw on graph paper — and from that one circle you read off the principal stresses, their directions, and the maximum shear stress at a point. This guide builds it from first principles and applies it to two complete numerical examples.

The Problem: Stress on an Inclined Plane

At a point in a loaded body, consider a small element in plane stress carrying a normal stress σ_x, a normal stress σ_y, and a shear stress τ_xy. If we cut the element on a plane inclined at angle θ, the normal stress σ_θ and shear stress τ_θ on that plane are given by the stress transformation equations:

σ<sub>θ</sub> = (σx + σy)/2 + (σx − σy)/2 · cos2θ + τxy · sin2θ
τ<sub>θ</sub> = − (σx − σy)/2 · sin2θ + τxy · cos2θ

These describe a circle when plotted with σ on the horizontal axis and τ on the vertical axis. That circle is Mohr's circle.

Key Results From the Circle

On the principal planes the shear stress is zero — that is the defining property of a principal plane.

Sign Convention (Get This Right First)

• Tensile normal stress is positive; compressive is negative.
• For the circle, a shear stress that tends to rotate the element clockwise is plotted upward (positive τ). Many textbooks use this "clockwise-up" rule so the geometry matches the physical rotation direction.
• Angles measured anticlockwise on the element are measured anticlockwise on the circle — but doubled.

Step-by-Step Construction

1. Draw the σ (horizontal) and τ (vertical) axes.
2. Plot point X = (σ_x, τ_xy) and point Y = (σ_y, −τ_xy).
3. Join X and Y; the line crosses the σ-axis at the centre C = (σ_avg, 0).
4. Draw the circle with centre C and radius R = CX.
5. The circle's intercepts on the σ-axis are σ₁ and σ₂; the top and bottom give ±τ_max.

Worked Example 1 — Combined Normal and Shear

At a point, σ_x = 50 MPa (tension), σ_y = −30 MPa (compression), τ_xy = 20 MPa. Find the principal stresses, maximum shear stress, and the orientation of the principal planes.

1. σ_avg = (50 + (−30))/2 = 10 MPa
2. R = √[ ((50 − (−30))/2)² + 20² ] = √[ 40² + 20² ] = √2000 = 44.72 MPa
3. σ₁ = 10 + 44.72 = 54.72 MPa (tension)
4. σ₂ = 10 − 44.72 = −34.72 MPa (compression)
5. τ_max = R = 44.72 MPa
6. tan 2θ_p = 2(20)/(50 − (−30)) = 40/80 = 0.5 → 2θ_p = 26.57° → θ_p = 13.28°

So the major principal stress of 54.72 MPa acts on a plane rotated 13.28° from the x-face.

Worked Example 2 — Pure Shear

A shaft surface is in pure shear: σ_x = σ_y = 0, τ_xy = 60 MPa. Find the principal stresses.

1. σ_avg = 0
2. R = √[0 + 60²] = 60 MPa
3. σ₁ = +60 MPa, σ₂ = −60 MPa
4. tan 2θ_p = 2(60)/0 → 2θ_p = 90° → θ_p = 45°

This is the famous result that pure shear is equivalent to equal tension and compression at 45° — which is exactly why a ductile shaft in torsion fails on a 45° helical surface, and a brittle one (like chalk) snaps along a 45° spiral.

Why Engineers Rely On It

• Failure theories (maximum shear stress / Tresca, maximum principal stress / Rankine) are stated in terms of σ₁, σ₂ and τ_max — all read directly off the circle.
• It reveals the worst-case plane, where cracks initiate, for welds, shafts and pressure vessels.
• The same construction works for plane strain and for moment-of-inertia transformation, making it a transferable skill.

Common Mistakes

• Forgetting that the circle angle is double the physical angle.
• Mixing up the shear sign convention, which flips the direction of θ_p.
• Reporting τ_max as σ₁ − σ₂ instead of (σ₁ − σ₂)/2.
• Ignoring the out-of-plane principal stress (zero in plane stress), which can govern the absolute maximum shear stress.