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Strain gauge rosette calculator

Enter the three readings of a strain gauge rosette, or the strains εx, εy and γxy. Get the principal strains and their directions, the maximum shear strain, Mohr’s circle of strain and, with E and ν, the stresses.

Also see: Mohr’s circle calculator. This is a learning and checking tool for plane stress and plane strain. Check your own working and your course’s sign convention.

What do you know?

degrees
με
degrees
με
degrees
με

Stresses from the strains (optional, plane stress)

Result

600 μεεx−100 μεεy500 μεγxy680 μεε1−180 μεε217.8°θp of ε1860 μεγmax in plane138 MPaσ15.26 MPaσ266.2 MPaτmax135 MPavon Mises
εγ/2 (positive downward)C 250ε1 680ε2 −180γmax/2 430XY2θp
Mohr’s circle of strain. The vertical axis is half the engineering shear strain, γ/2, positive downward. Values on the axes are in microstrain (με).
Gauge directions
xabc
Solid lines: gauges a, b, c. Dashed: direction of ε1.

Working

  1. Strains from the rosette
    Each gauge reads εθ = εx cos2θ + εy sin2θ + γxy sin θ cos θ. Three readings give three equations, solved for εx, εy and γxy.
    For a 45° rectangular rosette this simplifies to εx = εa = 600 με, εy = εc = (−100) με, γxy = 2εb − εa − εc = 2 × 500 − 600 − (−100) = 500 με.
    Result: εx = 600 με, εy = −100 με, γxy = 500 με.
  2. Centre and radius
    εavg = (εx + εy) / 2 = (600 + (−100)) / 2 = 250 με
    R = √[ ((εx − εy) / 2)2 + (γxy / 2)2 ] = 430 με
  3. Principal strains
    ε1 = εavg + R = 680 με
    ε2 = εavg − R = −180 με
    ε1 acts along the direction θp = 17.8° from the x axis (anticlockwise positive), and ε2 is 90° from that.
  4. Maximum in-plane shear strain
    γmax = 2R = ε1 − ε2 = 860 με, on planes at −27.2° where the normal strain is εavg = 250 με.
  5. Stresses from Hooke’s law (plane stress)
    Assume the surface is in plane stress (σz = 0) with E = 200 GPa and ν = 0.3.
    σx = E (εx + ν εy) / (1 − ν2) = 125 MPa
    σy = E (εy + ν εx) / (1 − ν2) = 17.6 MPa
    τxy = G γxy with G = E / (2(1 + ν)) = 76.9 GPa, so τxy = 38.5 MPa
    Principal stresses: σ1 = 138 MPa, σ2 = 5.26 MPa, maximum shear τmax = 66.2 MPa. Von Mises stress: 135 MPa.
    Out-of-plane strain εz = −ν (εx + εy) / (1 − ν) = −214 με.

Practise strain rosette questions

Press the button for a question with random numbers. Work it out on paper first, then type your answer.

How a strain gauge rosette is read

A single gauge measures strain along one line. Mounted at three different angles, three gauges give three equations, and these are enough to find the strain in any direction at that point: the normal strains εx and εy and the shear strain γxy. From those, Mohr’s circle of strain gives the principal strains and their directions.

  1. Each gauge at angle θ reads εθ = εx cos2θ + εy sin2θ + γxy sin θ cos θ.
  2. Solve the three equations for εx, εy, γxy.
  3. Principal strains: ε1,2 = εavg ± R, with R = √[((εx − εy) / 2)2 + (γxy / 2)2].
  4. With E and ν, use Hooke’s law for plane stress to get the stresses.

Worked example

A 45° rectangular rosette on a steel surface reads εa = 600 με at 0°, εb = 500 με at 45° and εc = −100 με at 90°. Take E = 200 GPa and ν = 0.3.

  1. Strains from the rosette
    Each gauge reads εθ = εx cos2θ + εy sin2θ + γxy sin θ cos θ. Three readings give three equations, solved for εx, εy and γxy.
    For a 45° rectangular rosette this simplifies to εx = εa = 600 με, εy = εc = (−100) με, γxy = 2εb − εa − εc = 2 × 500 − 600 − (−100) = 500 με.
    Result: εx = 600 με, εy = −100 με, γxy = 500 με.
  2. Centre and radius
    εavg = (εx + εy) / 2 = (600 + (−100)) / 2 = 250 με
    R = √[ ((εx − εy) / 2)2 + (γxy / 2)2 ] = 430 με
  3. Principal strains
    ε1 = εavg + R = 680 με
    ε2 = εavg − R = −180 με
    ε1 acts along the direction θp = 17.8° from the x axis (anticlockwise positive), and ε2 is 90° from that.
  4. Maximum in-plane shear strain
    γmax = 2R = ε1 − ε2 = 860 με, on planes at −27.2° where the normal strain is εavg = 250 με.
  5. Stresses from Hooke’s law (plane stress)
    Assume the surface is in plane stress (σz = 0) with E = 200 GPa and ν = 0.3.
    σx = E (εx + ν εy) / (1 − ν2) = 125 MPa
    σy = E (εy + ν εx) / (1 − ν2) = 17.6 MPa
    τxy = G γxy with G = E / (2(1 + ν)) = 76.9 GPa, so τxy = 38.5 MPa
    Principal stresses: σ1 = 138 MPa, σ2 = 5.26 MPa, maximum shear τmax = 66.2 MPa. Von Mises stress: 135 MPa.
    Out-of-plane strain εz = −ν (εx + εy) / (1 − ν) = −214 με.

Related

Library formulas: Hooke’s law, Young’s modulus from a test, Poisson’s ratio from measured strains, Shear strain from shear stress, Shear modulus from Young’s modulus and Poisson’s ratio. If you already have stresses rather than strains, use the Mohr’s circle calculator.

Common questions

What is a strain gauge rosette?

A rosette is a group of three strain gauges at known angles on the same spot. One gauge measures strain in only one direction, so on a surface where the directions of the principal strains are not known, three readings at different angles are needed to find the full strain state.

How do I get εx, εy and γxy from the readings?

Each gauge at angle θ reads εθ = εx cos²θ + εy sin²θ + γxy sinθ cosθ. Three readings give three equations, solved for the three unknowns. For a 45° rectangular rosette (0°, 45°, 90°) this gives εx = εa, εy = εc and γxy = 2εb − εa − εc. For a 60° delta rosette (0°, 60°, 120°), εx = εa, εy = (2(εb + εc) − εa) / 3 and γxy = 2(εb − εc) / √3.

How are the principal strains found?

The principal strains are εavg ± R, where εavg = (εx + εy) / 2 and R = √[((εx − εy) / 2)² + (γxy / 2)²]. The direction of ε₁ comes from tan 2θp = γxy / (εx − εy), taken in the correct quadrant.

Why is the shear strain halved on Mohr’s circle?

The rotation formulas for strain match those for stress only when the tensor shear strain γ / 2 is used. The engineering shear strain γ is twice that, so the circle uses γ / 2 on its vertical axis and the maximum shear strain is twice the radius.

How do I get the stresses from the strains?

For a free surface in plane stress with a linear elastic, isotropic material, σx = E(εx + νεy) / (1 − ν²), σy = E(εy + νεx) / (1 − ν²) and τxy = Gγxy with G = E / (2(1 + ν)). Enter E in GPa and ν to get the stresses in MPa. The result is only as good as the material values and the plane-stress assumption.

Can I use my own gauge angles?

Yes. Choose custom angles and type three different angles in degrees from the x axis. The calculator solves the three equations directly and tells you if the angles cannot separate the strains.

Is it free, and does it keep my numbers?

It is free to use, with no sign-up, no adverts and no personal tracking. The numbers you type stay in your browser.