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Mechanical

Von Mises equivalent stress from stress components (three dimensions)

Von Mises equivalent stress straight from the six stress components, with no need to find the principal stresses first.

Ref
MEC-66
Rev
v1
Field
Mechanical
Result
σVM [MPa]

σVM = √{[(σx − σy)² + (σy − σz)² + (σz − σx)²]/2 + 3(τxy² + τyz² + τxz²)} gives the von Mises stress directly from the six components, without first finding the principal stresses. It gives exactly the same value as the principal-stress form, and reduces to the plane stress expression when σz and the out-of-plane shears are zero. The shears carry a weight of 3 and only their squares appear, so their sign does not matter.

Von Mises equivalent stress straight from the six stress components, with no need to find the principal stresses first. It gives exactly the same value as the principal-stress form.

Calculator

Solve for
MPa
MPa
−200300
MPa
−200300
MPa
−200300
MPa
−150150
MPa
−150150
MPa
−150150
  • From the principal stresses (should be the same)79.37 MPa
  • Tresca equivalent stress σ1 − σ390 MPa
  • Factor of safety for Sy = 250 MPa3.15

What drives the result

Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.

Von Mises equivalent stress (MPa) current value 79.37 MPa
σzNormal stress in zKey driver
52.1285
τxyShear stress xy
60267
τyzShear stress yz
272
τxzShear stress xz
272
σxNormal stress in x
73.5261
σyNormal stress in y
73.5261
  • The lecture state gives √6300 = 79.373 MPa, the same as from the principal stresses 110, 50 and 20 MPa, which is a good check on both calculations. The factor of safety against a 250 MPa yield strength is 3.15, against 2.778 for Tresca.
  • Set σz and the out-of-plane shears to zero and the result is 95.394 MPa, the same as √(σx² − σxσy + σy² + 3τxy²) = √9100. A rotation of axes leaves the result unchanged, which is a useful test of any program.
Chart data as tables

Ranked variables

VariableTypical rangeLowest resultHighest resultSwing (MPa)Local stepChange in result
σz Normal stress in z−200 to 300 MPa52.07284.8232.7+10 MPa−9.15 %
τxy Shear stress xy−150 to 150 MPa60266.6206.6+6 MPa+9.02 %
τyz Shear stress yz−150 to 150 MPa79.37271.7192.3+6 MPa+0.854 %
τxz Shear stress xz−150 to 150 MPa79.37271.7192.3+6 MPa+0.854 %
σx Normal stress in x−200 to 300 MPa73.48260.6187.1+10 MPa+5.41 %
σy Normal stress in y−200 to 300 MPa73.48260.6187.1+10 MPa+5.41 %

Sweep of Shear stress xy

Shear stress xy (MPa)Von Mises equivalent stress (MPa)
−150266.6
−112.5203.9
−75143.1
−37.588.42
060
37.588.42
75143.1
112.5203.9
150266.6

Sweep one variable

Von Mises equivalent stress (MPa)Hover, touch, or use the arrow keys on the chart
50100150200250 −150−100−50050100150 Shear stress xy (MPa) now 79.37

Variables and how each one contributes

σVM

Von Mises equivalent stress

Result · MPa

What the equation gives you. Everything else feeds it.

σx

Normal stress in x

MPa · typical −200 to 300

Normal stress on the x face, in MPa. Tension is positive and compression negative, so the slider runs from −200 to +300 MPa. Raises σVM when σx is further from the other two normal stresses, and lowers it when it moves towards them. Raising it by 10 MPa (to 90 MPa) moves the result by +5.41 %, from 79.37 to 83.67 MPa.

σy

Normal stress in y

MPa · typical −200 to 300

Normal stress on the y face, in MPa. Tension is positive and compression negative, so the slider runs from −200 to +300 MPa. Raises σVM when σy is further from the other two normal stresses, and lowers it when it moves towards them. Raising it by 10 MPa (to 90 MPa) moves the result by +5.41 %, from 79.37 to 83.67 MPa.

σz

Normal stress in z

MPa · typical −200 to 300

Normal stress on the z face, in MPa. Tension is positive and compression negative, so the slider runs from −200 to +300 MPa. Raises σVM when σz moves further from the other two normal stresses, and lowers it when it moves towards them (as it does at the default, where σz is below σx and σy). Raising it by 10 MPa (to 30 MPa) moves the result by −9.15 %, from 79.37 to 72.11 MPa.

τxy

Shear stress xy

MPa · typical −150 to 150

Shear stress τxy, in MPa, acting on the x face in the +y direction. Shear is signed here, so the slider runs from −150 to +150 MPa. Raises σVM with weight 3 inside the root. The sign does not matter. Raising it by 10 MPa (to 40 MPa) moves the result by +15.5 %, from 79.37 to 91.65 MPa.

τyz

Shear stress yz

MPa · typical −150 to 150

Shear stress τyz, in MPa, acting on the y face in the +z direction. Shear is signed here, so the slider runs from −150 to +150 MPa. Raises σVM with weight 3 inside the root. The sign does not matter. Raising it by 10 MPa (to 10 MPa) moves the result by +2.35 %, from 79.37 to 81.24 MPa.

τxz

Shear stress xz

MPa · typical −150 to 150

Shear stress τxz, in MPa, acting on the x face in the +z direction. Shear is signed here, so the slider runs from −150 to +150 MPa. Raises σVM with weight 3 inside the root. The sign does not matter. Raising it by 10 MPa (to 10 MPa) moves the result by +2.35 %, from 79.37 to 81.24 MPa.

Worked example with the values above

  1. Normal-stress part: [(σx − σy)² + (σy − σz)² + (σz − σx)²]/2 = 3,600 MPa².
  2. Shear part: 3(τxy² + τyz² + τxz²) = 2,700 MPa².
  3. σVM = √(3,600 + 2,700) = √6,300 = 79.373 MPa.

Assumptions and limits

  • A symmetric stress tensor with six components, tension positive.
  • An isotropic ductile material with equal tensile and compressive yield strength, under static load.

Common mistakes

  • Forgetting the factor 3 on the shear terms.
  • Using the plane stress expression when σz or the out-of-plane shears are not zero.

Statistics and uncertainty

Finite-element von Mises stress is often reported at integration points, with mesh error of a few per cent at a stress concentration. The formula itself adds no error, but a stress state close to hydrostatic involves a tiny difference of large numbers, so round-off matters more than in the principal-stress form.

Where it is used

  • Yield check on finite-element output and on a measured stress tensor in three dimensions.
  • Testing a program: the result must not change when the axes are rotated.

Maths you need

  • The stress tensor: six independent components
  • Von Mises from principal stresses: the same value in different axes

Related formulas

Practise this formula