Equivalent Stress Calculator (von Mises)
Reduce a multiaxial stress state to a single equivalent stress and check it against the yield strength. Choose between the plane stress state (σ_x, σ_y, τ_xy) and the combined shaft/beam case (normal stress from tension/bending plus shear stress from torsion) - the calculator returns the von Mises, Tresca and Rankine values, the principal stresses, Mohr's circle and the available safety factor with a traffic-light rating, live with every input.
Calculation
Model: static yield check via a strength theory (von Mises, Tresca or Rankine) in the plane stress state (σ_3 = 0). The reference is the freely selectable yield strength R_p. No fatigue, notch or stability check. For a detailed fatigue verification of rotating shafts use the shaft calculator to DIN 743.
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Formulas and fundamentals
Equivalent stress for shafts and beams
The equivalent stress reduces a multiaxial stress state to a fictitious uniaxial stress σ_v that produces the same material loading and can be compared directly with the yield strength R_p. For a combined shaft or beam cross section with normal stress σ (from tension or bending) and shear stress τ (from torsion or transverse force), the closed forms are:
The three theories differ only in the shear factor and in how they treat the principal stresses.
Plane stress state and principal stresses
For the general plane stress state with σ_x, σ_y and τ_xy, the distortion energy theory reads:
The principal stresses follow from Mohr's circle with centre and radius:
The maximum shear stress τ_max equals the radius of the circle. In plane stress the third principal stress is σ_3 = 0 - this zero matters for the maximum shear theory when σ_1 and σ_2 share the same sign.
Maximum shear theory, Rankine theory and safety
The maximum shear theory (Tresca) sets σ_v equal to the largest difference of the three principal stresses:
The maximum principal stress theory (Rankine) sets σ_v equal to the principal stress of largest magnitude. From the governing equivalent stress the available safety against yielding follows as:
It is compared with the required safety S_req and rated as a traffic light.
Worked example
A shaft shoulder transmits a bending moment and a torque at the same time. At the governing point of the outer fibre a normal stress σ = 100 N/mm² from bending and a shear stress τ = 50 N/mm² from torsion act. The material has a yield strength R_p = 355 N/mm² and the required safety is S_req = 1.5.
By the von Mises theory σ_v = √(100² + 3·50²) = √17500 = 132.3 N/mm². Tresca gives a slightly higher σ_v = √(100² + 4·50²) = 141.4 N/mm², Rankine σ_v = 50 + 0.5·√(100² + 4·50²) = 120.7 N/mm². The principal stresses are σ_1 = 120.7 N/mm² and σ_2 = −20.7 N/mm², the maximum shear stress τ_max = 70.7 N/mm².
With von Mises as the governing theory the available safety is S = 355/132.3 = 2.68, well above S_req = 1.5 - the check passes (green). Choosing the more conservative Tresca theory lowers the safety to S = 355/141.4 = 2.51; the rating stays green. The example shows the typical difference: for combined loading Tresca sits roughly 7 to 15 percent on the safe side compared with von Mises.
Frequently asked questions
When should I use von Mises (GEH), Tresca (SH) or Rankine (NH)?
The von Mises theory describes ductile materials such as structural or heat-treated steel most accurately and is the machine-design standard. Tresca yields a slightly higher equivalent stress under combined loading and is therefore conservative. Rankine applies to brittle materials such as grey cast iron or hardened steel, which fail through the largest tensile principal stress.
How do von Mises and Tresca differ numerically?
Under pure shear von Mises gives σ_v = √3·τ ≈ 1.73·τ and Tresca σ_v = 2·τ, so Tresca is about 15 percent higher and thus conservative. Under uniaxial tension both are identical. For combined tension/bending and torsion the difference between the two theories is typically 7 to 15 percent.
What is the plane stress state and when does it occur?
Plane stress occurs when stresses act in one plane and none perpendicular to it (σ_3 = 0) - typical for thin sheets, vessel walls or free surfaces. It is described by σ_x, σ_y and τ_xy. The shaft/beam case is the special case σ_y = 0 of this state.
How do I compute the principal stresses?
The principal stresses are the extreme normal stresses, in whose section direction no shear stress acts. They follow from Mohr's circle: σ_1,2 = (σ_x + σ_y)/2 ± √(((σ_x − σ_y)/2)² + τ_xy²). The first term is the centre, the root term the radius and simultaneously the maximum shear stress τ_max.
Do I relate the safety to R_p or R_m?
The onset of yielding is referenced to the yield strength R_p (or R_p0.2 for materials without a pronounced yield point): S = R_p/σ_v. For a fracture check on brittle materials or against the static ultimate load the tensile strength R_m is used instead. The calculator works with a freely entered reference stress, by default the yield strength.
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