Mohr's Circle for Stress Analysis

Mohr’s circle is the most compact tool in stress analysis. It tells you the principal stresses, maximum shear stress, and the orientation of the failure plane — all from a simple 2D plot. Once you understand the geometry, you’ll never forget it.

When You Need It

Mohr’s circle applies when a point in a structure sees combined loading — bending and torsion, axial and shear, or any combination that produces both normal and shear stress on the same plane.

A shaft under simultaneous bending and torque is the classic example: the outer surface sees bending stress σ (normal) and torsional shear stress τ at the same time. Neither stress alone tells you the failure risk. Mohr’s circle combines them into the actual maximum and minimum stresses at that point.

The 2D Stress State

Any point in a stressed body can be described by three values on the x–y faces:

• σ_x — normal stress on the x-face (positive = tension)
• σ_y — normal stress on the y-face (positive = tension)
• τ_xy — shear stress (positive convention: shear on +x face pointing in +y direction)

These three numbers fully define the 2D stress state. Mohr’s circle converts them to principal stresses and max shear.

Building the Circle

Step 1 — Plot two points:

Point A: (σ_x, τ_xy) — the stress state on the x-face Point B: (σ_y, −τ_xy) — the stress state on the y-face

Note: Point B uses negative τ_xy because the shear on the y-face is opposite to the x-face shear.

Step 2 — Connect A and B. The line AB is a diameter of the circle.

Step 3 — Find the center:

C = ((σ_x + σ_y) / 2, 0)

The center is always on the horizontal (normal stress) axis.

Step 4 — Find the radius:

R = √[((σ_x − σ_y) / 2)² + τ_xy²]

Step 5 — Draw the circle centered at C with radius R.

That’s it. The circle is fully defined.

Reading the Results

Principal stresses (maximum and minimum normal stress, zero shear):

σ_1 = C + R = (σ_x + σ_y)/2 + R σ_2 = C − R = (σ_x + σ_y)/2 − R

σ_1 ≥ σ_2 always. σ_1 is the maximum normal stress; σ_2 is the minimum.

Maximum in-plane shear stress:

τ_max = R

This is the radius of the circle — which is why the graphical method is so powerful.

Maximum absolute shear stress (for 3D failure criteria):

For plane stress (σ_3 = 0), the absolute maximum shear depends on the signs of σ_1 and σ_2:

• If σ_1 and σ_2 are both the same sign: τ_abs_max = σ_1 / 2 (the larger principal stress dominates)
• If σ_1 and σ_2 are opposite signs: τ_abs_max = (σ_1 − σ_2) / 2 = R

Von Mises yield criterion uses these principal stresses directly:

σ_VM = √(σ_1² − σ_1·σ_2 + σ_2²)

Worked Example: Shaft Under Bending and Torsion

A 1-inch diameter steel shaft carries:

• Bending moment M = 500 lb·in → bending stress at outer fiber
• Torque T = 300 lb·in → torsional shear stress

Section properties for 1” solid circle:

• I = π·(1)⁴ / 64 = 0.0491 in⁴
• c = 0.5 in
• J = 2I = 0.0982 in⁴

Step 1 — Stresses at the critical point (outer fiber, max bending):

• σ_x = M·c / I = 500 × 0.5 / 0.0491 = 5,093 psi (bending stress)
• σ_y = 0 (no axial load)
• τ_xy = T·c / J = 300 × 0.5 / 0.0982 = 1,527 psi (torsional shear)

Step 2 — Mohr’s circle parameters:

• Center: C = (5,093 + 0) / 2 = 2,547 psi
• Radius: R = √[(5,093/2)² + 1,527²] = √[6,490,209 + 2,331,729] = √8,821,938 = 2,970 psi

Step 3 — Results:

• σ_1 = 2,547 + 2,970 = 5,517 psi
• σ_2 = 2,547 − 2,970 = −423 psi
• τ_max = 2,970 psi
• Von Mises: σ_VM = √(5,517² − 5,517 × (−423) + 423²) = 5,736 psi

If this is A36 steel (36 ksi yield), FoS = 36,000 / 5,736 = 6.3. Very comfortable for a static load.

Orientation of the Principal Planes

The angle from the x-axis to the σ_1 plane is:

2θ_p = arctan(2τ_xy / (σ_x − σ_y))

So θ_p = half that value. This is the angle you’d need to rotate the element to get pure normal stress with no shear.

The principal stress planes and the maximum shear planes are always 45° apart in physical space (90° on the Mohr’s circle).

The Key Insight

Mohr’s circle isn’t magic — it’s a visualization of a rotation transformation. Rotating a stress element in physical space by angle θ corresponds to rotating 2θ around the Mohr’s circle. The principal stresses are just the rotation that eliminates shear stress entirely.

This is why failure planes in ductile materials under tension (45° fractures) and in brittle materials (0° or 90° fractures) make physical sense — they correspond directly to the maximum shear and maximum normal stress planes.

Use PartCalc for Combined Loading

PartCalc computes principal stresses, Von Mises stress, and FoS from a real McMaster part URL. Paste in a shaft or structural member, enter your moment and torque, and get the full Mohr’s circle results — labeled by source so you know which numbers came from the McMaster data and which were computed.