Module 14 · Lesson 14.2

Principal stresses and Mohr's circle

The planes with no shear, the largest shear, and the picture that ties it all together.

Why this matters

Materials do not fail because of σx. They fail because of the largest tension, or the largest shear, somewhere at that point. The principal stresses tell you what those are, and Mohr's circle makes the whole family of planes visible at once.

By the end of this lesson you should be able to

Find the principal stresses and their directions
Calculate the maximum shear stress
Read and construct Mohr's circle
Relate every part of the circle to a physical plane

As θ sweeps round, the shear stress on the plane rises and falls. There are two perpendicular planes where it passes through zero. Those are the principal planes, and the direct stresses on them — the largest and smallest normal stresses anywhere at that point — are the principal stresses.

Principal stresses and their direction

σ_{1, 2} = \frac{σ _{x} + σ _{y}}{2} \pm (\frac{σ _{x} - σ _{y}}{2})^{2} + τ_{x y}^{2}, tan 2 θ_{p} = \frac{2 τ _{x y}}{σ _{x} - σ _{y}}

What it calculates: The greatest and least normal stress at a point, and the orientation of the planes on which they act.

σ1: Major principal stress (algebraically largest) (N/mm²)
σ2: Minor principal stress (N/mm²)
θp: Angle from the x axis to a principal plane normal (degrees)

This assumes

Plane stress
The three in-plane components are known

In plain terms: The first term is the mean stress and the square root is the radius of Mohr's circle. Setting the shear expression to zero is exactly how the angle formula is obtained — the principal planes are defined by having no shear.

Maximum in-plane shear stress

τ_{ma x} = (\frac{σ _{x} - σ _{y}}{2})^{2} + τ_{x y}^{2} = \frac{σ _{1} - σ _{2}}{2}

What it calculates: The largest shear stress at the point, and hence what governs yielding in ductile materials.

τmax: Maximum in-plane shear stress (N/mm²)
σ1, σ2: Principal stresses (N/mm²)

This assumes

Plane stress; a full 3D check may give a larger value if the third principal stress matters

In plain terms: It is the radius of Mohr's circle, and it always occurs on planes at 45° to the principal planes. Ductile metals slip on planes of maximum shear, which is why this quantity predicts yielding.

Mohr's circle is not a trick to be memorised. It is simply a graph of the two transformation equations, plotting σ horizontally and τ vertically as θ varies. Because both equations are sinusoids of the same amplitude about the mean stress, the point traces out a circle.

Every point on that circle is a real plane through the material. The two crossings of the horizontal axis are the principal planes, where τ = 0. The top and bottom are the planes of maximum shear. And because the algebra uses 2θ, rotating the physical plane by θ moves you 2θ around the circle.

Try it

Mohr's circle

Change the stress state, then rotate the plane. Every point on the circle is a real plane through the same point in the material.

σx

N/mm²

σy

N/mm²

τxy

N/mm²

Plane angle θ

degrees

σ1: 87.1 N/mm²
σ2: -47.1 N/mm²
τmax: 67.1 N/mm²
θp: 13.3°
σ at 0°: 80.0 N/mm²
τ at 0°: 30.0 N/mm²

Rotating the physical plane by θ moves you 2θ around the circle. The two points where the circle crosses the horizontal axis have no shear stress at all — those are the principal planes. The top and bottom of the circle are the planes of maximum shear, always 45° from the principal planes.

Worked example

Principal stresses and maximum shear

Given

σx = 80 N/mm²
σy = −40 N/mm²
τxy = 30 N/mm²

Find

The principal stresses, the maximum shear stress, and the principal directions.

Assumptions

Plane stress

Practice

At a point, σx = 60 N/mm², σy = 20 N/mm² and τxy = 25 N/mm². What is the major principal stress σ1?

Practice

For that same stress state, what is the maximum in-plane shear stress?

Practice

An element is in pure shear with τxy = 50 N/mm² and no direct stresses. What is the major principal stress?

Check yourself

What is special about the principal planes?

Summary

Principal planes carry no shear; principal stresses are the extreme normal stresses
σ1,2 = mean ± R, where R = √[((σx−σy)/2)² + τxy²]
τmax = R = (σ1 − σ2)/2, on planes 45° from the principal planes
Mohr's circle is a graph of the transformation equations: physical θ is 2θ on the circle

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This is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions. Module overview and checkpoint