Structural Analysis Fundamentals / Complex stress and strain

Strain transformation and Mohr's circle of strain

The same circle again, with one factor of two that catches everybody.

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Why this matters

Strain gauges measure direct strain in whatever direction they were stuck down, which is almost never the direction you want. Getting from those readings to the principal strains — and then to the stresses that matter — is strain transformation. The algebra is identical to the stress case you already know, apart from one factor of two, and that factor is where nearly every error in this topic comes from.

By the end of this lesson you should be able to

Write the strain transformation equations
Explain why the shear term carries a factor of one half
Construct Mohr's circle of strain and read the principal strains from it
Relate the maximum shear strain to the circle's diameter

What you should already know

Stress transformation and Mohr's circle (this module)
Shear strain as an angle change (Module 7)
Strain rosettes (this module)

The transformation equations for strain are:

ε_θ = (ε_x + ε_y)/2 + ((ε_x − ε_y)/2) cos 2θ + (γ_xy/2) sin 2θ
γ_θ/2 = −((ε_x − ε_y)/2) sin 2θ + (γ_xy/2) cos 2θ

Put them beside the stress equations and they are the same expressions, with ε in place of σ — except that wherever τ appeared, γ/2 appears rather than γ.

That factor of one half is not a convention someone chose to be awkward. It is a consequence of how engineering shear strain is defined.

Direct strain ε measures the extension of a line. Engineering shear strain γ measures the change in a right angle — which is the sum of two rotations, one of each of the two lines forming that angle. The off-diagonal component of the symmetric small-strain tensor is γ/2. It is not generally the rotation of either individual line, because rigid-body rotation can also be present. Engineering shear strain is twice the quantity that actually belongs in the tensor.

So the rule is simple and absolute: whenever strain enters a transformation, use γ/2, never γ.

Mohr's circle of strain is then built exactly as for stress:

Horizontal axis: direct strain ε
Vertical axis: half the engineering shear strain, γ/2
Centre at (ε_x + ε_y)/2, on the horizontal axis
Radius R = √( ((ε_x − ε_y)/2)² + (γ_xy/2)² )

And everything follows as before:

Principal strains ε₁, ε₂ = centre ± R, where the circle crosses the horizontal axis — the directions in which there is no shear strain at all
Maximum shear strain: the top and bottom of the circle are at γ/2 = R, so the maximum engineering shear strain is γ_max = 2R = ε₁ − ε₂, the full diameter
Angles on the circle are twice the physical angles, as always

For isotropic linear elasticity without additional imposed strain, the principal strain directions coincide with the principal stress directions, because there is no preferred orientation to separate them. That is what lets you go from rosette readings straight to principal stresses.

Predict first

Why does Mohr's circle of strain plot γ/2 rather than γ on its vertical axis?

Worked example

Principal strains from a measured strain state

Given

At a point: ε_x = 400 µ, ε_y = −100 µ, γ_xy = 200 µ (all in microstrain)

Find

The principal strains, the maximum shear strain, the principal direction, and the strain on a plane at 30°.

Practice

At a point ε_x = 400 µ, ε_y = −100 µ and γ_xy = 200 µ. What is the radius of Mohr's circle of strain, in microstrain?

Practice

For that same state, what is the major principal strain, in microstrain?

Practice

For the same state, what is the maximum engineering shear strain, in microstrain?

Summary

Strain transforms exactly like stress, with γ/2 in place of τ
The half arises because γ is the change in a right angle — two line rotations, not one
Mohr's circle of strain: ε horizontal, γ/2 vertical
Centre = (ε_x + ε_y)/2, radius = √(((ε_x − ε_y)/2)² + (γ_xy/2)²)
Principal strains = centre ± R; maximum engineering shear strain = 2R = ε₁ − ε₂
For an isotropic material, principal strain and principal stress directions coincide