Engineering
Mohr's Circle 2D Principal Stresses (sigma_1, sigma_2) & Max Shear Stress formulas
Compute the two in-plane principal stresses and distinguish maximum in-plane shear from absolute three-dimensional maximum shear in plane stress.
Transform stress components at one point
All three inputs must use the same stress units, MPa in this example. Tension is positive and compression negative. These formulas transform a supplied symmetric in-plane stress tensor; they do not determine loads, stress concentrations, allowable stress or structural safety.
The circle radius is maximum in-plane shear. Under plane stress the third principal stress is zero, so absolute maximum shear is half the difference between the largest and smallest of the two in-plane principal stresses and zero. They need not equal the in-plane radius.
For (50,50,0) MPa the in-plane circle collapses and in-plane shear is zero, but the full principal set is (50,50,0), giving absolute maximum shear 25 MPa. This is equibiaxial plane stress, not three-dimensional hydrostatic stress. Changing the shear sign leaves these stress magnitudes unchanged but changes principal directions.
| Mohr's Circle Quantity | Formula | Inputs: sx=80, sy=-40, txy=35 | Transformed Stress (MPa) |
|---|---|---|---|
| Circle Center sigma_avg | (B2 + B3) / 2 | (80 + (-40)) / 2 | 20.00 MPa |
| Max In-Plane Shear (Radius R) | SQRT(((B2 - B3)/2)^2 + B4^2) | SQRT(60^2 + 35^2) = SQRT(4825) | 69.46 MPa |
| Major Principal Stress sigma_1 | sigma_avg + R | 20 + 69.4622 | +89.46 MPa (Peak tension!) |
| Minor Principal Stress sigma_2 | sigma_avg - R | 20 - 69.4622 | -49.46 MPa (Peak compression!) |
Major Principal Stress sigma_1 = (sx + sy)/2 + SQRT(((sx - sy)/2)^2 + txy^2)
Follow standard sign convention: tension positive (+), compression negative (-).
=IF(COUNT(B2:B4)<>3,"Invalid inputs",ROUND(((B2 + B3) / 2) + SQRT((((B2 - B3) / 2) ^ 2) + (B4 ^ 2)), 2))Maximum In-Plane Shear Stress tau_max (Mohr's Circle Radius)
Return the radius as a nonnegative magnitude. Equal in-plane normal components and zero in-plane shear collapse this circle, without necessarily eliminating out-of-plane shear.
=IF(COUNT(B2:B4)<>3,"Invalid inputs",ROUND(SQRT((((B2 - B3) / 2) ^ 2) + (B4 ^ 2)), 2))Smaller in-plane principal stress
Subtract the radius from the mean. This can still be positive; it is not necessarily the smallest of all three principal stresses.
=IF(COUNT(B2:B4)<>3,"Invalid inputs",ROUND(((B2 + B3) / 2) - SQRT((((B2 - B3) / 2) ^ 2) + (B4 ^ 2)), 2))Absolute maximum shear for the plane-stress assumption
Include the zero out-of-plane principal stress before taking half the full principal-stress range. Plane strain would require a different out-of-plane stress calculation.
=IF(COUNT(B2:B4)<>3,"Invalid inputs",LET(stressMean,(B2+B3)/2,circleRadius,SQRT(((B2-B3)/2)^2+B4^2),ROUND((MAX(stressMean+circleRadius,0)-MIN(stressMean-circleRadius,0))/2,2)))Try the example
Use the original values or change them to test your understanding.