Module 14 · Lesson 14.5
Strain transformation and Mohr's circle of strain
The same circle again, with one factor of two that catches everybody.
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:
εθ = (εx + εy)/2 + ((εx − εy)/2) cos 2θ + (γxy/2) sin 2θ
γθ/2 = −((εx − εy)/2) sin 2θ + (γxy/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 quantity that transforms in the same way as a direct strain is the rotation of one line, which is γ/2. 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 (εx + εy)/2, on the horizontal axis
- Radius R = √( ((εx − εy)/2)² + (γxy/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 γmax = 2R = ε₁ − ε₂, the full diameter
- Angles on the circle are twice the physical angles, as always
For an isotropic material 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: εx = 400 µ, εy = −100 µ, γxy = 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 εx = 400 µ, εy = −100 µ and γxy = 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 = (εx + εy)/2, radius = √(((εx − εy)/2)² + (γxy/2)²)
- Principal strains = centre ± R; maximum engineering shear strain = 2R = ε₁ − ε₂
- For an isotropic material, principal strain and principal stress directions coincide
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