Mohr's Circle: Method

Mohr's Circle: Conventions, Equations and Reading the Figure

What the tool does

Given the stress (or strain) components at a point on one set of axes, the tool finds the components on every other set of axes through the point: the principal values and their directions, the maximum shear and its planes, the components on an element rotated through an entered angle, and the equivalent stresses used in yield criteria. The algebra is the transformation equations of Mohr (1882); the figure is their graphical form, which makes the result easy to check by eye. Three states are offered: plane stress (σ_z = τ_xz = τ_yz = 0, a free surface), plane strain (ε_z = γ_xz = γ_yz = 0) and the general three-dimensional stress state.

Sign conventions

• Normal stress is positive in tension. Shear stress τ_xy is positive when it acts toward +y on the face whose outward normal is +x (and, by moment equilibrium, toward +x on the +y face). This is the convention of Hibbeler, Beer and Johnston, Gere, Timoshenko and of the site's strain tools; it is the tensor convention, so a positive τ_xy is a counter-clockwise couple on the x faces.
• The rotation θ carries the x axis to the x′ axis counter-clockwise (seen from +z); a clockwise rotation is entered as a negative angle. Principal and maximum-shear directions are reported in (−90°, 90°]; adding or subtracting 180° names the same plane.
• Strain: ε is positive in extension; γ_xy is the engineering shear strain, positive when the right angle between the +x and +y faces closes, so a strain state transforms exactly like a stress state once γ/2 stands in for τ.
• In three dimensions τ_ij acts toward +j on the +i face; the rotation θ is about z.

Transformation equations (plane stress)

Equilibrium of a wedge cut from the element by the plane whose normal is x′ gives

σ_x′ = (σ_x + σ_y)/2 + (σ_x − σ_y)/2 cos 2θ + τ_xy sin 2θ
σ_y′ = (σ_x + σ_y)/2 − (σ_x − σ_y)/2 cos 2θ − τ_xy sin 2θ
τ_x′y′ = −(σ_x − σ_y)/2 sin 2θ + τ_xy cos 2θ

The sum σ_x′ + σ_y′ = σ_x + σ_y for every θ (the first invariant); the tool prints it as a check. Writing σ_avg = (σ_x + σ_y)/2 and R = √[((σ_x − σ_y)/2)² + τ_xy²], the pair (σ_x′, τ_x′y′) traces a circle of centre (σ_avg, 0) and radius R as θ varies: Mohr's circle. For strain the same equations hold with ε for σ and γ/2 for τ.

Principal stresses and maximum shear

The normal stress is stationary, and the shear zero, where dσ_x′/dθ = 0: tan 2θ_p = 2τ_xy/(σ_x − σ_y). The two roots 90° apart are the principal directions; σ₁ = σ_avg + R and σ₂ = σ_avg − R. The tool resolves the quadrant of 2θ_p from the signs of the numerator and denominator (the two-argument arctangent), so θ_p1 is always the direction of σ₁, and it confirms this by substituting θ_p1 back into the transformation equation. When σ_x = σ_y and τ_xy = 0 the state is isotropic (every direction principal) and the angle is reported as 0.

The shear stress is largest where dτ_x′y′/dθ = 0: tan 2θ_s = −(σ_x − σ_y)/(2τ_xy), which places the maximum-shear planes 45° from the principal planes, θ_s = θ_p ∓ 45°. The maximum in-plane shear is τ_max = R = (σ₁ − σ₂)/2, and the normal stress on both of those planes is σ_avg. The tool reports θ_s1 as the plane where τ_x′y′ = +R and θ_s2 where it is −R.

Absolute maximum shear: the three circles

Plane stress is a three-dimensional state with σ₃ = 0 on the free surface. Three Mohr's circles exist, on the diameters σ₁–σ₂, σ₁–σ₃ and σ₂–σ₃, and every plane through the point plots on or between them. The largest radius is the absolute maximum shear stress, τ_{abs max} = (σ_max − σ_min)/2 of the three principal stresses. When σ₁ and σ₂ have opposite signs the in-plane circle is the largest and τ_{abs max} = R; when they have the same sign an out-of-plane plane, inclined 45° to the surface, carries the larger shear |σ₁|/2 or |σ₂|/2 (Hibbeler Section 9.7). The figure draws the out-of-plane circles dashed. In plane strain the third principal strain is ε_z = 0 and the same construction gives the absolute maximum shear strain.

Equivalent stresses

The von Mises (distortion-energy) stress is σ_vm = √[((σ₁ − σ₂)² + (σ₂ − σ₃)² + (σ₃ − σ₁)²)/2], which for plane stress reduces to √(σ₁² − σ₁σ₂ + σ₂²); yielding is predicted when it reaches the tensile yield strength. The Tresca (maximum-shear) stress is σ₁ − σ₃ = 2τ_{abs max}, compared with the same yield strength. Tresca is the more conservative of the two by at most 15 % (pure shear). No equivalent is reported for a strain state, since it needs the material constants.

Reading Mohr's circle

The figure follows Beer and Johnston: point X, the x face, is plotted at (σ_x, −τ_xy) and point Y, the y face, at (σ_y, +τ_xy), so a shear that tends to rotate the element clockwise plots above the σ axis. With this choice a rotation θ of the element is a rotation 2θ of the diameter XY about the centre C in the same sense, and the new end X′ has coordinates (σ_x′, −τ_x′y′). The circle crosses the σ axis at the principal stresses (no shear), 2θ_p away from X; its top and bottom are the maximum in-plane shear planes, 2θ_s = 2θ_p ∓ 90° away, at the normal stress σ_avg. Other books plot X at (σ_x, τ_xy) with the τ axis positive downward (Hibbeler, Gere); the picture is the same, mirrored, and every number agrees. The element drawing shows the given faces and the rotated faces with the computed components as arrows: tension points out of the face, and a positive shear acts toward +y on the +x face.

Three-dimensional stress

The principal stresses are the eigenvalues of the symmetric stress tensor, the roots of σ³ − I₁σ² + I₂σ − I₃ = 0 with the invariants I₁ = σ_x + σ_y + σ_z, I₂ = σ_xσ_y + σ_yσ_z + σ_zσ_x − τ_xy² − τ_yz² − τ_xz², I₃ = det[σ]. The tool solves the cubic in closed form (the trigonometric solution, since a symmetric tensor has three real roots) and also diagonalises the tensor by Jacobi rotations, which gives the principal directions as unit eigenvectors with their direction cosines and angles to the axes; the two agree to round-off. The three circles on the diameters σ₁–σ₂, σ₂–σ₃, σ₁–σ₃ bound every plane's (σ, τ) point; the given x, y and z faces are marked at their normal stress and the resultant shear on the face, and lie on a circle only when that axis is principal. The octahedral plane, equally inclined to the three principal axes, carries σ_oct = I₁/3 and τ_oct = ⅓√[(σ₁ − σ₂)² + (σ₂ − σ₃)² + (σ₃ − σ₁)²] = (√2/3) σ_vm. The rotation θ about z transforms σ_x, σ_y, τ_xy as in plane stress and turns τ_xz, τ_yz as a vector: τ_x′z = τ_xz cos θ + τ_yz sin θ, τ_y′z = −τ_xz sin θ + τ_yz cos θ.

Limits

The analysis is purely geometric: it needs no material properties and applies to any stress or strain tensor, elastic or not. It does not relate stress to strain (use the Stress to Strain and Strain to Stress tools), does not combine loads into the stress components, and the plane strain option assumes ε_z = 0 exactly; the strain state on a free surface under plane stress has ε_z = −ν(σ_x + σ_y)/E, not zero, and its absolute maximum shear strain must be found from that value. Angles are printed in degrees in both unit systems; stresses follow the unit chosen on the form.

Worked examples checked

The default case is Example 9.3 of Hibbeler's Mechanics of Materials (σ_x = −20 MPa, σ_y = 90 MPa, τ_xy = 60 MPa): σ₁ = 116 MPa at θ_p1 = 66.3°, σ₂ = −46.4 MPa at −23.7°, and (Example 9.4) τ_max = 81.4 MPa at 21.3° with σ_avg = 35 MPa. The library also reproduces Hibbeler's Examples 9.1, 9.10, 9.13, 10.1, 10.2 and 10.3, Beer and Johnston's Concept Application 7.1 and Sample Problems 7.1 and 7.2, Gere and Goodno's Examples 7-1 to 7-6, and the classic three-dimensional tensor with eigenvalues 4, 1, −2, to the digits those books print.