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Physics

Principal Stress Calculator

Enter the three independent stress components at a point on a loaded element (normal stress in X, normal stress in Y, and the XY shear stress) and the calculator instantly finds both principal stresses, the maximum in-plane shear stress, the principal-plane angle, the average normal stress, and the Von Mises equivalent stress. A signed bar chart compares sigma1 and sigma2, and the step panel shows every arithmetic step with your actual numbers substituted in.

Your details

Stress unit for all inputs and outputs. The formulas are dimensionless; pick whichever unit your problem uses.
Normal stress on the X face (positive = tension, negative = compression).
MPa
Normal stress on the Y face (positive = tension, negative = compression).
MPa
Shear stress on the XY plane. Sign convention: positive when the stress on the +X face points in the +Y direction.
MPa
Angle to rotate the element (degrees). Outputs for the rotated element appear as extra fields. Leave at 0 for the standard principal-stress result only.
deg
Maximum principal stress (sigma1)Combined tension-compression
94.031

Largest normal stress; shear stress is zero on this plane.

Minimum principal stress (sigma2)-34.031
Max in-plane shear stress (tau_max)64.031
Average normal stress (sigma_avg)30
Von Mises stress114.891
Principal angle (theta_p)19.33deg
Max-shear angle (theta_s)64.33deg
Rotated normal stress X80
Rotated normal stress Y-20
Rotated shear stress40
sigma1 (max principal)94.031
sigma2 (min principal)-34.031
tau_max (in-plane shear)64.031
Von Mises114.891

Principal stresses: 94.03 MPa (max) and -34.03 MPa (min).

  • sigma1 = 94.03 MPa and sigma2 = -34.03 MPa. These are the extreme normal stresses at this point; shear stress vanishes on the planes they act on.
  • The maximum in-plane shear stress is 64.03 MPa, acting on planes rotated 45 degrees from the principal planes. This is the radius of Mohr's circle.
  • Von Mises equivalent stress is 114.89 MPa. Compare this to your material's tensile yield strength to check for yielding under this loading.
  • The principal plane is rotated 19.3 degrees counter-clockwise from the X axis.

Next stepTo check whether the material will yield, divide the Von Mises stress by the material yield strength. A ratio above 1.0 predicts yielding under von Mises (distortion-energy) theory.

Formula

σ1,2=σx+σy2±(σxσy2)2+τxy2,τmax=(σxσy2)2+τxy2,θp=12arctan2τxyσxσy\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}, \quad \tau_{\max}=\sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}, \quad \theta_p=\tfrac{1}{2}\arctan\frac{2\tau_{xy}}{\sigma_x-\sigma_y}

Worked example

For sigma_x = 80 MPa, sigma_y = -20 MPa, tau_xy = 40 MPa: avg = (80 + (-20)) / 2 = 30 MPa, R = sqrt(50^2 + 40^2) = sqrt(4100) = 64.03 MPa. sigma1 = 30 + 64.03 = 94.03 MPa, sigma2 = 30 - 64.03 = -34.03 MPa, tau_max = 64.03 MPa, theta_p = 0.5 * atan2(80, 100) = 19.3 degrees.

What are principal stresses?

At every point in a stressed body the stress state depends on how you orient your reference axes. For most orientations there is both a normal stress and a shear stress on each face of an imaginary cube. There is, however, always one special pair of perpendicular axes for which the shear stress on every face is exactly zero. The normal stresses on those faces are called the principal stresses. They are the algebraically largest (sigma1) and smallest (sigma2) normal stresses that the point experiences in any direction. Knowing them is essential for strength analysis because most failure theories (von Mises, Tresca, Rankine) are expressed in terms of principal stresses rather than the arbitrary sigma_x and sigma_y you read from your free-body diagram.

How the formulas work: Mohr's circle in algebra

The stress-transformation equations for a plane stress state (sigma_z = tau_zx = tau_zy = 0) form a circle when you plot normal stress on the horizontal axis and shear stress on the vertical axis. The centre of that circle sits at sigma_avg = (sigma_x + sigma_y) / 2, and the radius is R = sqrt( ((sigma_x - sigma_y) / 2)^2 + tau_xy^2 ). The two intersections of the circle with the horizontal axis give you sigma1 = sigma_avg + R and sigma2 = sigma_avg - R, i.e. the points where shear stress is zero. The radius R is also the maximum in-plane shear stress. The angle between the X axis and the principal plane is theta_p = 0.5 * atan2(2 * tau_xy, sigma_x - sigma_y), and the maximum-shear plane is always 45 degrees away from the principal plane.

Von Mises stress and yield prediction

For ductile materials the distortion-energy (von Mises) theory is the most widely used yield criterion. It compares the actual multi-axial stress state to a single equivalent stress: sigma_vm = sqrt(sigma1^2 - sigma1 * sigma2 + sigma2^2) for plane stress. If sigma_vm is less than the material's uniaxial tensile yield strength (Sy), the material stays elastic. If sigma_vm exceeds Sy, yielding is predicted. This calculator outputs sigma_vm directly so you can compare it straight to a material data sheet without extra arithmetic.

Stress transformation at an arbitrary angle

Sometimes you need the stresses on a plane that is neither the principal plane nor the X-Y plane, for instance to check a weld bead oriented at a specific angle. The general transformation equations are: sigma_x' = sigma_avg + ((sigma_x - sigma_y) / 2) * cos(2*theta) + tau_xy * sin(2*theta), sigma_y' = sigma_avg - ((sigma_x - sigma_y) / 2) * cos(2*theta) - tau_xy * sin(2*theta), tau_xy' = -((sigma_x - sigma_y) / 2) * sin(2*theta) + tau_xy * cos(2*theta), where theta is the angle from the original X axis. Enter that angle in the "Rotation angle" field and the calculator shows all three rotated stress components.

Stress state classification

Stress statesigma1sigma2Typical source
Uniaxial tension> 00Axially loaded bar, cable
Uniaxial compression0< 0Column, strut
Pure shear> 0< 0 (equal magnitude)Torsion shaft, weld
Biaxial tension> 0> 0Pressure vessel (hoop + axial)
Biaxial compression< 0< 0Deep underground rock
Combined bending + torsion> 0< 0 (unequal)Drive shaft, crank

Typical stress states in engineering components and their principal stress signatures.

Frequently asked questions

What is the difference between sigma1, sigma2, and tau_max?

Sigma1 and sigma2 are the two principal stresses: the maximum and minimum normal stresses a point experiences when you rotate the reference element to eliminate shear stress. Tau_max is the maximum in-plane shear stress, equal to the radius of Mohr's circle: (sigma1 - sigma2) / 2. It acts on planes oriented 45 degrees from the principal planes. Note that for a 3D body the absolute maximum shear stress may be larger than the in-plane value if sigma_z differs from both sigma1 and sigma2.

What is plane stress?

Plane stress is the simplification that all stress components acting on one face of the element (the Z face) are zero: sigma_z = tau_zx = tau_zy = 0. This applies to thin plates and shells loaded in their own plane, to the surface of any three-dimensional body (because there is no material pressing on the outer surface), and to many practical engineering cross-sections under bending or torsion. If sigma_z is not zero you need a 3x3 stress tensor and a more involved eigenvalue calculation.

How do I find the principal angle?

The principal angle theta_p is the rotation from the X axis to the plane carrying sigma1, measured counter-clockwise. The formula is theta_p = 0.5 * arctan(2 * tau_xy / (sigma_x - sigma_y)). The arctan function returns a value between -90 and 90 degrees, so theta_p lies between -45 and 45 degrees. The second principal plane is perpendicular: theta_p + 90 degrees. The maximum-shear planes are at theta_p + 45 degrees and theta_p - 45 degrees.

When should I use Von Mises stress instead of principal stresses?

Use Von Mises (distortion-energy) stress when checking whether a ductile material will yield, because yielding in ductile materials is driven by shear energy, which the Von Mises criterion captures well. Use principal stresses when checking brittle fracture (maximum-normal-stress or Rankine criterion) or when designing against fatigue with a specific SN curve referenced to a uniaxial stress state. Both checks are often needed in a full design review.

What units should I use?

All three inputs must share the same unit. Use MPa for most mechanical engineering calculations (steel yield strengths are commonly listed in MPa). Use psi for American engineering standards and many material data sheets written in US units. kPa suits geotechnical or low-pressure analyses. GPa is useful when working with very stiff materials such as ceramics or when stresses are expressed per gigapascal in research literature. All outputs appear in the same unit you choose.

Why is shear stress zero on the principal planes?

Mathematically, the principal planes are the eigenvectors of the stress tensor: rotating to those directions diagonalises the tensor, which means off-diagonal (shear) entries become zero. Physically, it means there is no tendency for the material to slide parallel to those planes; all the force is pushing or pulling perpendicular to the face. This is why principal stresses appear in failure criteria: they represent the pure tension or compression a point experiences, free from the complicating effect of shear.

Sources

Written by Dr. Tomás Okafor, PhD Physicist · Lagos, Nigeria

Physicist specializing in classical mechanics, bringing 17 years of research and applied dynamics expertise to every calculator he reviews.

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