OpenSpecEngineering

Finds the principal stresses, the maximum in-plane shear stress and the von Mises stress at a point in plane stress, and draws Mohr’s circle for it. The angle slider rotates the stress element and shows the stresses on the rotated faces. Designers use it to turn the stresses from a bending, torsion or pressure calculation into the values a failure criterion needs.

Equations

σ_avg = (σ_x + σ_y)/2     R = √[((σ_x − σ_y)/2)² + τ_xy²]

σ_1,2 = σ_avg ± R     τ_max = R     σ′ = √(σ₁² − σ₁σ₂ + σ₂²)

σ_x and σ_y are the normal stresses on the x and y faces of the element, tension positive. τ_xy is the shear stress on the x face, positive when it acts in the +y direction. σ_avg is the circle’s center and R its radius. σ₁ and σ₂ are the largest and smallest normal stresses, on faces that carry no shear, and σ′ is the von Mises stress [1].

For a rotation θ (counterclockwise positive) the Stress at Rotation Angle θ panel gives

σ_x′ = σ_avg + ((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

and σ_y′ = 2σ_avg − σ_x′. On the circle the point for the x face moves through 2θ as the element turns through θ. The diagram plots it at (σ_x′, −τ_x′y′), as in [1].

Limits and common mistakes

• The page treats plane stress only: the third principal stress, normal to the plane, is zero. When σ₁ and σ₂ have the same sign, the absolute maximum shear stress is the larger of |σ₁|/2 and |σ₂|/2, not R.
• Keep the sign of τ_xy. Flipping it mirrors the circle, so the principal directions change even though the principal stresses don’t.
• Enter compression as a negative number.

Worked example

A point on the surface of a rotating shaft sees a bending stress of 120 MPa along the axis and a torsional shear stress of 50 MPa. Find the principal stresses and the von Mises stress for a yield check.

1. Enter Normal Stress (σ_x) 120 MPa and Normal Stress (σ_y) 0 MPa.
2. Enter Shear Stress (τ_xy) 50 MPa and press Solve. The Solution shows: Max Principal Stress (σ1): 138.102 MPa
   Min Principal Stress (σ2): −18.1025 MPa
   In-Plane Shear (τmax): ±78.1025 MPa
   Average Stress (σavg): 60 MPa
   von Mises Stress (σ′): 147.986 MPa
3. Set θ to 20°. The Stress at Rotation Angle panel shows: Normal Stress (σx′): 138.102 MPa
   Normal Stress (σy′): −18.102 MPa
   Shear Stress (τx′y′): −0.265034 MPa The shear is nearly zero, so the principal direction is just under 20° from the shaft axis.

Check: for σ_y = 0 the von Mises stress is √(σ_x² + 3τ_xy²) = √(14,400 + 7,500) = 148.0 MPa, and the principal angle is ½·atan(2·50/120) = 19.9°.