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Principal stresses and Mohr circle

Intermediate Strength of Materials Reviewed 24 Oct 2026 2 min read

At any point in a loaded body, the stresses depend on the orientation of the plane you look at. A small square element with normal stresses σx, σy and shear τxy, if rotated, shows different values. There are always two perpendicular planes on which shear is zero; the normal stresses on those planes are the principal stresses σ₁ and σ₂, the largest and smallest normal stresses at that point. The maximum shear stress acts on planes at 45° to them and equals half their difference.

This matters because materials fail on particular planes. Concrete cracks perpendicular to the principal tension, which is why diagonal shear cracks appear near beam supports. Soils and rocks slip on planes where shear is high relative to normal stress. Welds and plates are checked against combined stress criteria built from principal values.

density k
Diagram: Indiacon original, not to scale.

Mohr's circle is a graphical way to see all of this. Plot σ on the horizontal axis and τ on the vertical, mark the points (σx, τxy) and (σy, −τxy), and draw a circle through them with its centre on the σ axis. The circle's extreme left and right points are the principal stresses, its radius is the maximum shear, and rotating the element by θ moves round the circle by 2θ. Soil mechanics uses the same circle with failure envelopes in triaxial testing.

Tip

Plot shear on the element face that turns the element clockwise as positive on the Mohr plot, and keep that convention throughout; then the circle rotation matches the element rotation in sense.

Key formulas
  • Centre C = (σx + σy)/2; radius R = √[((σx − σy)/2)² + τxy²]
  • σ₁,₂ = C ± R
  • τ_max = R
  • Principal angle: tan 2θp = 2 τxy / (σx − σy)
Common mistake

Forgetting to double the angle when moving between the element and the circle.

Solved example

Problem. At a point in a beam web, σx = 60 MPa (tension), σy = 0 and τxy = 40 MPa. Find the principal stresses, maximum shear and principal direction.

  1. C = 30 MPa; R = √(30² + 40²) = 50 MPa.
  2. σ₁ = 80 MPa tension; σ₂ = −20 MPa (compression).
  3. τ_max = 50 MPa.
  4. tan 2θp = 80 / 60 = 1.333, so 2θp = 53.1° and θp = 26.6° from the x-axis.

Answer. σ₁ = 80 MPa, σ₂ = −20 MPa, τ_max = 50 MPa, θp ≈ 26.6°

Quick quiz

1What is the shear stress on a principal plane?

Zero.

2How does a rotation θ of the element appear on Mohr's circle?

As a rotation of 2θ.

3Why do diagonal cracks form near RC beam supports?

Concrete cracks perpendicular to the principal tension, which is inclined where shear is high.

References and further reading

Standard texts, listed in no order of merit. We do not copy text, figures, tables or solved problems from any book; our explanations and examples are original.

  • J. M. Gere, B. J. Goodno, Mechanics of Materials, Cengage, latest edition.
  • F. P. Beer, E. R. Johnston, J. T. DeWolf, Mechanics of Materials, McGraw Hill, latest edition.
  • E. P. Popov, Engineering Mechanics of Solids, Prentice Hall, 2nd edition.

Free and official sources

All references for this subject →

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