Learn: Principal stresses — plane stress
The stresses on a tiny element change when you rotate the axes, even though its physical loading stays the same. Mohr’s circle turns the three plane-stress components into a centre and a radius so you can identify the important extremes.
What the formula is saying
Principal directions have zero in-plane shear; their normal stresses are the mean plus and minus the radius. The maximum in-plane shear equals that radius, obtained from the normal-stress half-difference and shear component.
Read the symbols in plain language
- σx
- Normal stress x
Normal stress x. The average is the centre of the in-plane Mohr circle.
PaOne pascal is one newton per square metre. 1 MPa = 10⁶ Pa.
- σy
- Normal stress y
Normal stress y. The average is the centre of the in-plane Mohr circle.
PaOne pascal is one newton per square metre. 1 MPa = 10⁶ Pa.
- τxy
- Shear stress
Shear stress. Use the square root of the sum of squares; the radius is nonnegative.
PaOne pascal is one newton per square metre. 1 MPa = 10⁶ Pa.
- σ₁
- Result to find
Larger in-plane principal stress.
Pa - σ₂
- Result to find
Smaller in-plane principal stress.
Pa
Sort out the units first
Use the same stress unit for σx, σy and τxy; tension is positive and compression negative. Every intermediate and final stress is in Pa after conversion, and squared stresses are square-rooted back to stress units.
Assumptions before calculating
The inputs represent a symmetric plane-stress tensor at one point, with out-of-plane normal stress and shear taken as zero. This is a stress transformation, not a material failure criterion.
Let’s solve one together
Read the given values, follow each operation, then check what the result means.
Use the following study data and find the requested result. Follow the calculation before trying the second case.
- σx · Normal stress x
- 10000000 Pa
- σy · Normal stress y
- 3000000 Pa
- τxy · Shear stress
- 2000000 Pa
Find the mean normal stress
The average is the centre of the in-plane Mohr circle.
((10000000) + (3000000)) ÷ 2 = 6500000 PaFind half the normal-stress difference
The departure from the mean is combined with the original shear stress.
((10000000)-(3000000)) ÷ 2 = 3500000 PaFind the Mohr-circle radius
Use the square root of the sum of squares; the radius is nonnegative.
√((3500000)^2 + (2000000)^2) ≈ 4031128.874 PaFind the smaller in-plane principal stress
Subtracting the radius gives the left-hand end of the in-plane stress circle.
(6500000)-(4031128.874) ≈ 2468871.126 PaFind the larger in-plane principal stress
Adding the radius gives the other principal stress; report both values together.
(6500000) + (4031128.874) ≈ 10531128.87 Pa
Both in-plane principal stresses, larger then smaller. The third plane-stress principal value is zero.
Does this worked answer make sense?
For pure shear, the in-plane principal stresses are +|τ| and −|τ|. Changing only the sign of τxy changes orientation but not the two principal values or the radius.
A second worked example — different values
A second case uses different data. Predict which way the answer will change, then calculate it without reusing the first answer.
- σx · Normal stress x
- 0 Pa
- σy · Normal stress y
- 0 Pa
- τxy · Shear stress
- 5000000 Pa
Find the mean normal stress
The average is the centre of the in-plane Mohr circle.
((0) + (0)) ÷ 2 = 0 PaFind half the normal-stress difference
The departure from the mean is combined with the original shear stress.
((0)-(0)) ÷ 2 = 0 PaFind the Mohr-circle radius
Use the square root of the sum of squares; the radius is nonnegative.
√((0)^2 + (5000000)^2) = 5000000 PaFind the smaller in-plane principal stress
Subtracting the radius gives the left-hand end of the in-plane stress circle.
(0)-(5000000) = -5000000 PaFind the larger in-plane principal stress
Adding the radius gives the other principal stress; report both values together.
(0) + (5000000) = 5000000 Pa
Both in-plane principal stresses, larger then smaller. The third plane-stress principal value is zero.
Now try your own values
Change a value or its unit. The same method will show your calculation, step by step.
Results update only when you calculate. The lesson example above stays unchanged.
Your turn — check your understanding
Solve this separate case yourself. Use only the values below; the two worked examples use different data. Give the requested result in the selected unit.
- σx · Normal stress x
- 8000000 Pa
- σy · Normal stress y
- -2000000 Pa
- τxy · Shear stress
- 3000000 Pa
Find: Learn: Principal stresses — plane stress
A hint, not the answer
Principal directions have zero in-plane shear; their normal stresses are the mean plus and minus the radius. The maximum in-plane shear equals that radius, obtained from the normal-stress half-difference and shear component.
Use the same stress unit for σx, σy and τxy; tension is positive and compression negative. Every intermediate and final stress is in Pa after conversion, and squared stresses are square-rooted back to stress units.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Find the mean normal stress
The average is the centre of the in-plane Mohr circle.
((8000000) + (-2000000)) ÷ 2 = 3000000 PaFind half the normal-stress difference
The departure from the mean is combined with the original shear stress.
((8000000)-(-2000000)) ÷ 2 = 5000000 PaFind the Mohr-circle radius
Use the square root of the sum of squares; the radius is nonnegative.
√((5000000)^2 + (3000000)^2) ≈ 5830951.895 PaFind the smaller in-plane principal stress
Subtracting the radius gives the left-hand end of the in-plane stress circle.
(3000000)-(5830951.895) ≈ -2830951.895 PaFind the larger in-plane principal stress
Adding the radius gives the other principal stress; report both values together.
(3000000) + (5830951.895) ≈ 8830951.895 Pa
Both in-plane principal stresses, larger then smaller. The third plane-stress principal value is zero.
Avoid the common trap
Do not omit the factor 1/2 from the normal-stress difference. Square shear before adding it. Do not confuse maximum in-plane shear with the absolute maximum shear in three dimensions.
When this method applies — and when it does not
The two returned principal values are in-plane values. For absolute three-dimensional maximum shear, include the third principal stress 0 and use half the difference between the largest and smallest of all three; the in-plane radius need not equal that value.
For study and understanding, not approval of a real structure, site operation or design. Apply the correct standard, National Annex and professional review to actual engineering work.
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Sources & further reading
References open in a new tab and explain the underlying principles. The teaching text and examples here are SimpleFlick’s own; the source organisations have not endorsed this calculator.
Reading focus: Principal stresses — plane stress. Use the relevant elastic-response, equilibrium, constitutive-relations or shear-and-torsion module; the lesson states its particular sign convention.
Lesson updated: · Both examples and the separate practice case are checked against an independent high-precision numerical implementation. This verifies arithmetic for the stated model, not engineering certification.