Principal stresses, principal strains and strain gauge rosettes

Why it matters

You cannot measure stress directly; you measure strain. Every aircraft structural test, from a coupon to a full-scale wing on a test rig, is instrumented with strain gauges, often hundreds of rosettes. Turning three gauge readings into principal strains, principal stresses and their directions is a core skill of structural test and flight-loads engineers.

Key ideas

Principal stresses (recap). At any point there are planes free of shear; the normal stresses on them are the principal stresses. In plane stress σ₁,₂ = (σx + σy)/2 ± √[((σx − σy)/2)² + τxy²], with the third principal stress σ₃ = 0 normal to the surface. In three dimensions they are the eigenvalues of the stress tensor (roots of a cubic).

Strain transformation. Plane strain components εx, εy and γxy transform exactly like σx, σy and τxy, with γxy/2 in place of τxy. This is because the tensor shear strain is γ/2. So Mohr's circle for strain has centre (εx + εy)/2 and radius √[((εx − εy)/2)² + (γxy/2)²], with γ/2 on the vertical axis.

Principal strains ε₁ and ε₂ are the extreme normal strains, on directions with zero shear strain. For an isotropic linear-elastic material the principal strain directions coincide with the principal stress directions. That is what lets a rosette give stress directions.

What a single gauge measures. A gauge measures only the normal strain along its own axis. Shear strain cannot be measured directly, so three gauges at different angles are needed to find the three unknowns εx, εy, γxy.

Rosettes.

• Rectangular (0°/45°/90°): εx = εa, εy = εc, γxy = 2εb − εa − εc.
• Delta or equiangular (0°/60°/120°): εx = εa, εy = [2(εb + εc) − εa]/3, γxy = 2(εb − εc)/√3. Each comes from writing εθ = (εx + εy)/2 + (εx − εy)/2·cos 2θ + (γxy/2)·sin 2θ for each gauge angle and solving.

From strain to stress. A gauge is bonded to a free surface, so the material is in plane stress (σz = 0) even though εz ≠ 0. Use the plane-stress Hooke's law with the principal strains: σ₁ = E(ε₁ + νε₂)/(1 − ν²) and σ₂ = E(ε₂ + νε₁)/(1 − ν²). Do not simply multiply each strain by E: the Poisson coupling matters.

Out-of-plane strain. Under plane stress εz = −ν(σx + σy)/E = −ν(εx + εy)/(1 − ν). It is a principal strain too, and it matters for the absolute maximum shear strain.

Practical points: gauge readings are in microstrain (µε = 10⁻⁶); temperature-compensated gauges or a dummy gauge remove thermal output; misalignment by a few degrees changes the results noticeably where strain gradients are steep.

Formulas

εθ = (εx + εy)/2 + (εx − εy)/2·cos 2θ + (γxy/2)·sin 2θ Normal strain (–) along a direction at θ from x.

ε₁,₂ = (εx + εy)/2 ± √[((εx − εy)/2)² + (γxy/2)²] Principal strains (–).

tan 2θp = γxy / (εx − εy) Direction of principal strain (and stress, for isotropic materials).

γmax,in-plane = ε₁ − ε₂ Maximum in-plane engineering shear strain (rad).

εx = εa, εy = εc, γxy = 2·εb − εa − εc Rectangular rosette with gauges a, b, c at 0°, 45°, 90°.

εx = εa, εy = [2·(εb + εc) − εa] / 3, γxy = 2·(εb − εc) / √3 Delta rosette at 0°, 60°, 120°.

σ₁ = E·(ε₁ + ν·ε₂) / (1 − ν²), σ₂ = E·(ε₂ + ν·ε₁) / (1 − ν²) Principal stresses (Pa) from principal strains, plane stress; E in Pa, ν = Poisson's ratio.

Worked examples

Example 1 (standard). A rectangular rosette on an aluminium skin (E = 70 GPa, ν = 0.33) reads εa = 600 µε (0°), εb = 200 µε (45°), εc = −100 µε (90°). Find the principal strains, their direction and the principal stresses.

1. εx = 600 µε, εy = −100 µε, γxy = 2εb − εa − εc = 400 − 600 + 100 = −100 µε.
2. Centre = (600 − 100)/2 = 250 µε; radius = √(350² + 50²) = 353.6 µε.
3. ε₁ = 250 + 353.6 = 603.6 µε; ε₂ = 250 − 353.6 = −103.6 µε.
4. tan 2θp = γxy/(εx − εy) = −100/700, so 2θp = −8.13° and θp = −4.07° (ε₁ is 4.07° clockwise from gauge a; substituting back confirms it gives the larger value).
5. E/(1 − ν²) = 70 000/(1 − 0.1089) = 78 555 MPa.
6. σ₁ = 78 555 × (603.6 − 0.33 × 103.6) × 10⁻⁶ = 78 555 × 569.4 × 10⁻⁶ = 44.7 MPa.
7. σ₂ = 78 555 × (−103.6 + 0.33 × 603.6) × 10⁻⁶ = 78 555 × 95.6 × 10⁻⁶ = 7.5 MPa.

Answer: ε₁ ≈ 604 µε, ε₂ ≈ −104 µε at θp ≈ −4.1°; σ₁ ≈ 44.7 MPa, σ₂ ≈ 7.5 MPa. Note that σ₂ is tensile although ε₂ is compressive: the Poisson effect of σ₁ causes that contraction.

Example 2 (GATE level). A delta rosette on a steel part (E = 200 GPa, ν = 0.3) reads εa = 400 µε (0°), εb = 100 µε (60°), εc = −200 µε (120°). Find the principal strains, their direction and the principal stresses.

1. εx = 400 µε.
2. εy = [2(εb + εc) − εa]/3 = [2(100 − 200) − 400]/3 = −600/3 = −200 µε.
3. γxy = 2(εb − εc)/√3 = 2 × 300/1.732 = 346.4 µε.
4. Check with εθ at 60°: 100 + 300 × cos 120° + 173.2 × sin 120° = 100 − 150 + 150 = 100 µε. Correct.
5. Centre = 100 µε; radius = √(300² + 173.2²) = 346.4 µε.
6. ε₁ = 446.4 µε, ε₂ = −246.4 µε; tan 2θp = 346.4/600 = 0.577, so θp = 15.0° from gauge a.
7. E/(1 − ν²) = 200 000/0.91 = 219 780 MPa.
8. σ₁ = 219 780 × (446.4 − 0.3 × 246.4) × 10⁻⁶ = 81.9 MPa; σ₂ = 219 780 × (−246.4 + 0.3 × 446.4) × 10⁻⁶ = −24.7 MPa.

Answer: ε₁ ≈ 446 µε, ε₂ ≈ −246 µε at 15° from gauge a; σ₁ ≈ 81.9 MPa, σ₂ ≈ −24.7 MPa.

Common mistakes

• Using γxy instead of γxy/2 in the strain transformation and the Mohr's circle radius.
• Converting strains to stresses with σ = E·ε gauge by gauge, ignoring Poisson coupling.
• Forgetting that the rosette relations depend on the gauge angles; the 45° and 60° formulas are not interchangeable.
• Treating the surface as plane strain. A bonded gauge sits on a free surface, which is plane stress.
• Losing the 10⁻⁶ of microstrain when computing stresses.

For GATE AE

Expect numericals that give rosette readings and ask for principal strains, shear strain, the principal angle or the principal stresses, and short questions on why three gauges are needed and how γ/2 enters Mohr's circle for strain. Practise the rectangular and delta rosette equations and checking a result by substituting a gauge angle back into the transformation equation.

Quick check

1. Why does a rosette need at least three gauges?
2. A rectangular rosette reads εa = 300, εb = 300, εc = 300 µε. What is γxy?
3. In Mohr's circle for strain, what is plotted on the vertical axis?
4. Do principal stress and principal strain directions coincide in an isotropic elastic material?

Answers: 1. There are three unknowns, εx, εy and γxy, and each gauge gives one equation. 2. Zero. 3. γ/2. 4. Yes.