Plane stress and transformation equations
Stress is force per unit area. Under plane stress, components involving the z direction vanish: σz = τxz = τyz = 0. The components σx, σy and τxy describe the stress state at a point; rotating the axes changes the components while the physical state stays the same.
The traction on a face with normal n is t = σn. For an infinitesimal wedge in equilibrium, the areas of the faces parallel to the axes are projections of the inclined area. Projecting the forces onto the normal and tangent of the inclined face gives the transformation equations:
θ is the rotation of the face normal from +x toward +y, not the angle of the face itself. Normal stress is positive in tension. Positive τxy acts toward +y on the +x face and toward +x on the +y face; the arrows reverse on opposite faces.
Constructing and orienting the circle
Points X = (σx, τxy) and Y = (σy, −τxy) are opposite ends of a diameter. The center is (C, 0). Eliminating θ from the transformation equations gives:
To draw it by hand:
- Plot X and Y using the same scale on both axes.
- Join X and Y: the diameter crosses the σ axis at the center C.
- Measure the radius R, the distance from C to X.
- Read σ₁ and σ₂ where the circle crosses the σ axis, and ±τmax at the points farthest from that axis.
- For a plane rotated by θ, rotate the diameter XY by 2θ; its ends are X′ and Y′.
With positive τ downward, the diameter rotates in the same direction as the element; with positive τ upward, in the opposite direction. Textbooks use either convention: it changes the drawing, not the values or the physical shear sign.
In the diagram, the solid diameter connects X′ and Y′ and the dashed one connects X and Y; the arcs mark 2θ and 2θp₁. Because θ and θ + 180° give the same components, half a turn of the element is a full turn on the circle. The unit circle helps explain this double angle.
Principal stresses and maximum in-plane shear
A principal plane carries no shear stress. The circle crosses the σ axis at σ₁ = C + R and σ₂ = C − R. The equation tan 2θp = 2τxy/(σx − σy) has two solutions 90° apart; the atan2 function selects the one for σ₁:
Angles are given in (−90°, 90°]. Maximum in-plane shear equals R and is accompanied by the average normal stress C. θs+ produces +R and θs− produces −R. If R = 0, every in-plane direction is principal and the angles are not unique.
The state (60, 60, 0) is equibiaxial, with R = 0. Under plane stress it is not a three-dimensional hydrostatic state: the third principal stress is still zero.
Why absolute maximum shear can be larger
The three principal stresses are σ₁, σ₂ and σz = 0. σ₁ and σ₂ are ordered within the plane (σ₁ ≥ σ₂); zero can lie above, below or between them. A traction-free surface has zero normal and shear stress on it, which is why a thin plate loaded in its plane is approximated as plane stress.
If both in-plane principal stresses are nonzero and have the same sign, the outer circle involves the z direction. The planes where that maximum occurs have normals tilted out of xy, so no in-plane rotation θ reaches τabs.
For example, (100, 50, 0) MPa gives R = 25 MPa, but τabs = 50 MPa. This is why additional circles are drawn between each principal stress and zero. If the in-plane principals have opposite signs or one is zero, τabs = R.
Equivalent stresses and eigenvalues
Von Mises and Tresca express a multiaxial state as a single equivalent stress. For plane stress:
To assess failure, the equivalent stress is compared with the material strength under the chosen criterion. Tresca is the more conservative of the two: in any state, σTresca ≥ σvM.
The in-plane principal stresses are eigenvalues of the symmetric stress matrix, and its eigenvectors give the principal directions; the connection can be explored with the matrix analyzer. Trace and determinant are preserved when the axes rotate.
Example, scope and references
For σx = 80, σy = −40 and τxy = 25 MPa: C = 20 MPa, R = 65 MPa, σ₁ = 85 MPa and σ₂ = −45 MPa. θp₁ ≈ 11.30993°. At a rotation of 30°, σx′ ≈ 71.65064 MPa, σy′ ≈ −31.65064 MPa and τx′y′ ≈ −39.46152 MPa.
The calculation covers stress, not strain or strain rosettes, and is limited to plane stress: it does not accept a general 3D tensor. Geomechanics often treats compression as positive; such data must be converted to the tension-positive convention used here.
MIT OCW, Raúl Radovitzky: Transformation of stress components. Tensor derivation, principal stresses and Mohr's circle.
David Roylance, MIT: Transformation of stresses and strains. Transformation and the three-dimensional extension.
R. C. Hibbeler, Mechanics of Materials, chapter 9. Beer and Johnston, Statics and Mechanics of Materials: reference textbooks on stress transformation.