How to calculate maximum elastic bending stress
Bending stretches one side of a beam and compresses the other. For simple elastic bending, the largest stress magnitude is found at an extreme fibre.
What the formula is saying
The flexure relation is stress = My/I. Since W = I/y, the outer-fibre magnitude can be written M/W. Use the section modulus for the same axis as the bending moment.
Read the symbols in plain language
- M
- Bending momentN·m
- W
- Section modulusm³
Sort out the units first
Use N·m for moment and m³ for W to obtain Pa. 90 kN·m = 90,000 N·m. Do not combine kN·m with mm³ without conversion.
Let’s solve one together
Read the given values, follow each operation, then check what the result means.
- M · Bending moment
- 90000 N·m
- W · Section modulus
- 0.018 m³
Identify the moment at the section
Use the bending moment, not the shear force or total applied load.
(90000) = 90000 N·mDivide by the matching section modulus
For a given moment, a larger W gives a smaller elastic stress.
(90000) ÷ (0.018) = 5000000 Pa
Does this worked answer make sense?
The example gives 5 MPa. Doubling M doubles the stress; doubling W halves it.
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
Use these new values. Work it out first, then check your answer.
- M · Bending moment
- 12000 N·m
- W · Section modulus
- 0.003 m³
Find: maximum elastic bending stress
A hint, not the answer
The flexure relation is stress = My/I. Since W = I/y, the outer-fibre magnitude can be written M/W. Use the section modulus for the same axis as the bending moment.
Use N·m for moment and m³ for W to obtain Pa. 90 kN·m = 90,000 N·m. Do not combine kN·m with mm³ without conversion.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Identify the moment at the section
Use the bending moment, not the shear force or total applied load.
(12000) = 12000 N·mDivide by the matching section modulus
For a given moment, a larger W gives a smaller elastic stress.
(12000) ÷ (0.003) = 4000000 Pa
Avoid the common trap
Mixing moment and force units is a common error. A stress below a remembered material number is not a complete structural safety check.
When this method applies — and when it does not
Small, linear-elastic bending about a principal axis with the appropriate section model. This is not a cracked-concrete or plastic-capacity calculation.
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.
Lesson updated: · Worked examples checked against the implemented formula; not an independent engineering certification.