BIZOMAT MAKEOS · Strength

Three-Dimensional Principal Stresses

What are the three principal stresses, maximum shear and von Mises stress for this 3D stress state?

Principal stresses of a general three-dimensional stress state from its six components, by the stress invariants, with maximum shear, von Mises equivalent stress and the safety factor.

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How this tool works

Inputs, units and where to find them

InputUnitTypical source
σxMPaNormal stress in X (tension positive).
σyMPaNormal stress in Y.
σzMPaNormal stress in Z.
τxyMPaShear stress in the XY plane.
τyzMPaShear stress in the YZ plane.
τzxMPaShear stress in the ZX plane.
Yield strengthMPaMinimum yield strength from the material certificate or specification.

What each input means

Open an input to see what it is, its units, the worked example and the common mistakes, with a small diagram.

What this means: σx
Illustration: σxA test bar pulled from both ends: stress σx is the force carried by each square millimetre of its cross-section.σxstress = force ÷ areaExample: 80 MPaσx
What it is
The direct (normal) stress acting on the face square to the X axis at the point, from hand calculation, strain gauges or finite element results.
Units and format
Enter a number in MPa: megapascals (1 MPa = 1 N/mm²). The form accepts -100000 MPa to 100000 MPa. Required: the tool does not calculate until every input is valid.
Example
The form opens with 80 MPa as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
Get it right
Tension is positive and compression negative; keep the same sign rule for every input.
What this means: σy
Illustration: σyA test bar pulled from both ends: stress σy is the force carried by each square millimetre of its cross-section.σystress = force ÷ areaExample: -30 MPaσy
What it is
The direct (normal) stress acting on the face square to the Y axis at the same point.
Units and format
Enter a number in MPa: megapascals (1 MPa = 1 N/mm²). The form accepts -100000 MPa to 100000 MPa. Required: the tool does not calculate until every input is valid.
Example
The form opens with -30 MPa as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
Get it right
Tension is positive and compression negative.
What this means: σz
Illustration: σzA test bar pulled from both ends: stress σz is the force carried by each square millimetre of its cross-section.σzstress = force ÷ areaExample: 20 MPaσz
What it is
The direct (normal) stress acting on the face square to the Z axis at the same point. Use 0 for a plane-stress state.
Units and format
Enter a number in MPa: megapascals (1 MPa = 1 N/mm²). The form accepts -100000 MPa to 100000 MPa. Required: the tool does not calculate until every input is valid.
Example
The form opens with 20 MPa as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
Get it right
Tension is positive and compression negative.
What this means: τxy
Illustration: τxyA test bar pulled from both ends: stress τxy is the force carried by each square millimetre of its cross-section.τxystress = force ÷ areaExample: 40 MPaτxy
What it is
The shear stress acting in the XY plane: on the X face in the Y direction, equal to that on the Y face in the X direction.
Units and format
Enter a number in MPa: megapascals (1 MPa = 1 N/mm²). The form accepts -100000 MPa to 100000 MPa. Required: the tool does not calculate until every input is valid.
Example
The form opens with 40 MPa as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
Get it right
Shear signs must follow one consistent tensor convention, as in your analysis software.
What this means: τyz
Illustration: τyzA test bar pulled from both ends: stress τyz is the force carried by each square millimetre of its cross-section.τyzstress = force ÷ areaExample: 10 MPaτyz
What it is
The shear stress acting in the YZ plane: on the Y face in the Z direction, equal to that on the Z face in the Y direction.
Units and format
Enter a number in MPa: megapascals (1 MPa = 1 N/mm²). The form accepts -100000 MPa to 100000 MPa. Required: the tool does not calculate until every input is valid.
Example
The form opens with 10 MPa as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
Get it right
Shear signs must follow one consistent tensor convention.
What this means: τzx
Illustration: τzxA test bar pulled from both ends: stress τzx is the force carried by each square millimetre of its cross-section.τzxstress = force ÷ areaExample: -15 MPaτzx
What it is
The shear stress acting in the ZX plane: on the Z face in the X direction, equal to that on the X face in the Z direction.
Units and format
Enter a number in MPa: megapascals (1 MPa = 1 N/mm²). The form accepts -100000 MPa to 100000 MPa. Required: the tool does not calculate until every input is valid.
Example
The form opens with -15 MPa as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
Get it right
Shear signs must follow one consistent tensor convention.
What this means: Yield strength
Illustration: Yield strengthA test bar pulled from both ends: stress Yield strength is the force carried by each square millimetre of its cross-section.Systress = force ÷ areaExample: 350 MPaYield strength
What it is
The minimum yield strength of the material, used to give the safety factor against yield from the von Mises stress.
Units and format
Enter a number in MPa: megapascals (1 MPa = 1 N/mm²). The form accepts 0.000001 MPa to 100000 MPa. Required: the tool does not calculate until every input is valid.
Example
The form opens with 350 MPa as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
Effect on the result
The safety factor is directly proportional to the yield strength; the stresses do not change.
Get it right
Use the guaranteed minimum from the certificate or specification, not a typical value.

Method and formulas

  1. Invariants I₁ = σx + σy + σz; I₂ = σxσy + σyσz + σzσx − τxy² − τyz² − τzx²; I₃ = det of the stress tensor.
  2. Principal stresses are the roots of σ³ − I₁σ² + I₂σ − I₃ = 0 (trigonometric solution).
  3. τmax = (σ₁ − σ₃)/2; von Mises = √(½[(σ₁ − σ₂)² + (σ₂ − σ₃)² + (σ₃ − σ₁)²]).
  4. Example: the default state has principal stresses 94.711, 21.367 and -46.078 MPa.

Limitations

  • Static, isotropic material for the yield criteria.
  • Example values are illustrations. Replace them with values for your actual task.

Typical uses

  • Post-process FEA or strain gauge results
  • Check triaxial stress points in thick parts
  • Teach and check stress transformation

Who it is for

  • Design engineers
  • Structural and mechanical engineers
  • Engineering students

What you get

  • Principal stresses σ1 ≥ σ2 ≥ σ3
  • Maximum shear stress
  • von Mises stress and safety factor

Interpreting the result

  • Hydrostatic stress (equal principal stresses) does not cause yield in ductile metals.

References

  • Machinery’s Handbook, 29th edition (Industrial Press, 2012), printed pp. 212–218. Topic reference; no source tables or recommended settings are reproduced.

Learn the Concepts Behind This Tool

These BIZOMAT Library readings explain the method this tool uses, with worked examples.

  • Method explained here · BIZOMAT original book

    Three-dimensional stress

    In Stress Transformation by the Numbers

    What it covers: How to find the three principal stresses of a general stress state from its invariants, how three Mohr's circles show the true maximum shear, and when ignoring a direction misleads.

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Open the Three-Dimensional Principal Stresses calculator