Plane Stress Transformation Equations

Find the normal and shear stresses on any rotated plane — the foundation of every stress-analysis tool that follows.

Mechanics of MaterialsMechanical Engineering Year 2Free preview
⏱️ About 18 min

A beam cracks along a 45° angle under torsion not because the material is weak at 45°, but because the stress state looks completely different on that slanted plane. Stress transformation is how you rotate your point of view mathematically — and discover where the real danger lies.

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The big idea: The stresses σx, σy, and τxy on the reference planes are just one way of looking at the same stress state. Rotate the viewing plane by an angle θ and the normal and shear stresses change in a perfectly predictable way, governed by equilibrium of a wedge-shaped element. The transformation equations are that equilibrium written in symbols.
🎯 By the end, you'll be able to
  • Apply the plane-stress transformation equations to find σx′ and τx′y′ on a plane rotated by θ
  • Compute principal stresses and the principal angle from a given stress state
  • Compute the maximum in-plane shear stress and the average normal stress on those planes
📎 Helpful to know first

Why stress changes with orientation

Stress is not a single number — it is a tensor, a complete description of how force is transmitted through a point in every possible direction. The three numbers you usually see (σx, σy, τxy) are just the components on the x and y faces of a conveniently oriented square element. If you slice that element at an angle, the same internal forces must still balance — but now they are resolved onto slanted faces, producing different normal and shear stresses.

The transformation equations come from cutting a tiny wedge at angle θ and enforcing equilibrium. They are not empirical fits; they are exact consequences of force balance, valid for any homogeneous, linear-elastic material and any plane-stress state.

A plane-stress element showing normal stresses sigma-x on vertical faces and sigma-y on horizontal faces, with complementary shear stresses tau-xy on all four faces x y σx σx σy σy τxy x′ θ = 30° CCW
The reference element (solid) and a rotated cut (dashed). The transformation equations resolve the same stress state onto the slanted faces.
\[ \sigma_{x'} = \frac{\sigma_x + \sigma_y}{2} + \frac{\sigma_x - \sigma_y}{2}\cos 2\theta + \tau_{xy}\sin 2\theta \]
Normal stress on a plane rotated by θ from the x-axis. The first term is the average normal stress; the remaining terms describe the oscillation as you rotate.
\[ \tau_{x'y'} = -\frac{\sigma_x - \sigma_y}{2}\sin 2\theta + \tau_{xy}\cos 2\theta \]
Shear stress on the same rotated plane. Note the negative sign on the first term — it ensures the shear vanishes on the principal planes.

Principal stresses and the principal angle

There is always some angle θ at which the shear stress τx′y′ drops to zero. Those planes — the principal planes — carry only normal stress, called the principal stresses σ1 and σ2. They are the maximum and minimum normal stresses attainable at that point under any rotation. Setting τx′y′ = 0 and solving gives the principal angle θp and the principal-stress formula below.

A quick sanity check: the sum of the principal stresses must equal the sum of the original normal stresses, because rotating the element cannot create or destroy normal force. That means σ1 + σ2 = σx + σy — a useful check after every calculation.

\[ \sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{\Bigl(\frac{\sigma_x - \sigma_y}{2}\Bigr)^2 + \tau_{xy}^2} \]
Principal stresses: the average normal stress plus or minus the radius of Mohr's circle.
🔑 Maximum in-plane shear stress

The largest shear stress in the x-y plane occurs on planes rotated 45° from the principal planes. Its magnitude equals the radius term in the principal-stress formula:

τmax(in-plane) = √[((σx − σy)/2)² + τxy²]

Crucially, the planes of maximum in-plane shear are not planes of zero normal stress. They carry the average normal stress σavg = (σx + σy)/2. This is one of the most common mistakes in introductory stress analysis — students often assume maximum-shear planes are 'pure shear' planes, but they only are when σavg happens to be zero.

🎮 Interactive: rotate a stress element LIVE
Predict first: Set σx = 80, σy = 40, τxy = 30. Rotate θ to about 32° — where does the shear on the rotated plane go to zero?

An interactive simulator showing a plane-stress element and its corresponding Mohr's circle, with sliders for sigma-x, sigma-y, tau-xy, and rotation angle theta.

Drag the rotation slider and watch the stress element and Mohr's circle update together. Positive shear on the reference face plots downward on the circle.
📝 Worked example: A stress element is loaded with σx = 80 MPa, σy = 40 MPa, and τxy = 30 MPa. Find the principal stresses, the principal angle, and the maximum in-plane shear stress.
  1. Average normal stress: σavg = (80 + 40)/2 = 60 MPa.
  2. Radius R = √[((80 − 40)/2)² + 30²] = √[20² + 30²] = √1300 ≈ 36.06 MPa.
  3. Principal stresses: σ1 = 60 + 36.06 = 96.06 MPa; σ2 = 60 − 36.06 = 23.94 MPa.
  4. Sanity check: σ1 + σ2 = 96.06 + 23.94 = 120 MPa = σx + σy — confirmed.
  5. Principal angle: tan(2θp) = τxy / ((σx − σy)/2) = 30/20 = 1.5 → 2θp = 56.31° → θp ≈ 28.2°.
  6. Maximum in-plane shear: τmax = R = 36.06 MPa, occurring on planes at θp + 45° ≈ 73.2°.
  7. Normal stress on those max-shear planes = σavg = 60 MPa (not zero).
✓ σ1 ≈ 96.1 MPa, σ2 ≈ 23.9 MPa, θp ≈ 28.2°, τmax(in-plane) ≈ 36.1 MPa, σavg = 60 MPa.
✏️ Practice: A stress element has σx = 80 MPa, σy = 40 MPa, τxy = 30 MPa. What is the normal stress on a plane rotated θ = 30° from the x-axis, in MPa?
MPa
Solution
  1. σx′ = (80+40)/2 + (80−40)/2·cos(60°) + 30·sin(60°) = 60 + 20·0.5 + 30·0.8660 = 60 + 10 + 25.98 = 95.98 MPa.
✏️ Practice: For the same stress state (σx = 80 MPa, σy = 40 MPa, τxy = 30 MPa), what is the maximum in-plane shear stress, in MPa?
MPa
Solution
  1. R = √[((80−40)/2)² + 30²] = √[400 + 900] = √1300 ≈ 36.06 MPa.
  2. τmax(in-plane) = R ≈ 36.1 MPa.

Check your understanding

1. What is always true about the principal planes?
By definition, principal planes are the orientations where shear stress vanishes — only normal (principal) stresses remain.
2. The planes of maximum in-plane shear stress carry what normal stress?
Maximum in-plane shear occurs at ±45° from the principal planes. On those planes the normal stress equals σavg, not zero (unless σavg = 0 by coincidence).
3. For a stress state with σx = 60 MPa, σy = 20 MPa, and τxy = 0, what are the principal stresses?
When τxy = 0, the x and y planes are already principal planes, so σ1 = σx = 60 MPa and σ2 = σy = 20 MPa.
🔎 Related lessons
Mohr's Circle for Stress What Is Stress? Normal & Shear Stress
✅ Key takeaways
  • Stress transformation resolves the same stress state onto rotated planes using equilibrium.
  • Principal stresses are the maximum and minimum normal stresses; they occur on planes where shear is zero.
  • Maximum in-plane shear equals the radius R and occurs on planes at 45° to the principal planes, carrying σavg.
  • Always check: σ1 + σ2 = σx + σy.
➡️ The transformation equations are exact but algebra-heavy. Mohr's circle turns the same math into a quick, visual construction that every engineer sketches by hand.
Want to test yourself on this? Try the Mechanical Aptitude test →
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