Learn: Fatigue stress range
A fatigue stress cycle moves between a minimum and maximum stress. Stress range is the full difference between those extremes, not half the difference.
What the formula is saying
Subtract σmin from σmax while retaining their algebraic signs. A cycle from −50 MPa to +150 MPa has range 200 MPa because subtracting a negative adds its magnitude.
Read the symbols in plain language
- σmax
- Maximum stress
Signed extreme stress in the same loading cycle; use the same tension/compression convention for both values.
MPaOne megapascal equals one N/mm² and 1000 kPa.
- σmin
- Minimum stress
Signed extreme stress in the same loading cycle; use the same tension/compression convention for both values.
MPaOne megapascal equals one N/mm² and 1000 kPa.
- Δσ
- Result to find
Fatigue stress range. Subtracting the signed minimum gives full range, while half of this would be amplitude.
MPa
Sort out the units first
Both stresses and the range use MPa. Use one convention, typically tension positive and compression negative. The range is nonnegative because σmax must be at least σmin.
Assumptions before calculating
Assume the extremes describe the same stress component at the same point during one defined cycle. The nominal, structural or local stress definition must match the fatigue method being used.
Let’s solve one together
Read the given values, follow each operation, then check what the result means.
Read the supplied values as one complete study case. Find Δσ and explain the result in the stated output unit.
- σmax · Maximum stress
- 200 MPa
- σmin · Minimum stress
- 50 MPa
Identify the algebraic maximum stress
Use the maximum of the same signed stress component over the selected cycle.
(200) = 200 MPaSubtract the algebraic minimum
Subtracting the signed minimum gives full range, while half of this would be amplitude.
(200)-(50) = 150 MPa
Does this worked answer make sense?
Equal maximum and minimum stresses give zero range. Adding the same offset to both extremes changes mean stress but leaves range unchanged.
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.
- σmax · Maximum stress
- 150 MPa
- σmin · Minimum stress
- -50 MPa
Identify the algebraic maximum stress
Use the maximum of the same signed stress component over the selected cycle.
(150) = 150 MPaSubtract the algebraic minimum
Subtracting the signed minimum gives full range, while half of this would be amplitude.
(150)-(-50) = 200 MPa
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.
- σmax · Maximum stress
- 180 MPa
- σmin · Minimum stress
- 30 MPa
Find: Learn: Fatigue stress range
A hint, not the answer
Subtract σmin from σmax while retaining their algebraic signs. A cycle from −50 MPa to +150 MPa has range 200 MPa because subtracting a negative adds its magnitude.
Both stresses and the range use MPa. Use one convention, typically tension positive and compression negative. The range is nonnegative because σmax must be at least σmin.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Identify the algebraic maximum stress
Use the maximum of the same signed stress component over the selected cycle.
(180) = 180 MPaSubtract the algebraic minimum
Subtracting the signed minimum gives full range, while half of this would be amplitude.
(180)-(30) = 150 MPa
Avoid the common trap
Do not divide by two unless asked for stress amplitude. Do not discard the negative sign of compression, or take absolute values of both extremes before subtracting.
When this method applies — and when it does not
Stress range alone does not give fatigue life. Cycle counting, mean stress, detail classification, stress concentrations, residual stress and the applicable S–N curve must also be considered. Nonproportional multiaxial cycles require more than subtracting two unrelated extrema.
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.
One idea understood. Keep going.
This optional checkmark is saved only in this browser. It is your own progress note, not a certificate.
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: Fatigue stress range. Stress cycles, ranges and life under repeated loading. Linear cumulative damage is an idealization and does not account for load-sequence effects.
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.