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How to calculate elastic section modulus

The outermost fibres of a bending section are furthest from its neutral axis. Section modulus combines the area distribution I with this outer-fibre distance.

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01

What the formula is saying

Divide the second moment I by y, the distance from the neutral axis to the extreme fibre. For a symmetric rectangle about its centroid, y is half the depth, not the full depth.

W = I / ymax

Read the symbols in plain language

I
Second moment of aream⁴
ymax
Extreme-fibre distancem

Sort out the units first

I in m⁴ divided by y in m gives W in m³. The example takes I = 0.0054 m⁴ and y = 0.3 m from a 0.6 m deep rectangle.

02

Let’s solve one together

Read the given values, follow each operation, then check what the result means.

I · Second moment of area
0.0054 m⁴
ymax · Extreme-fibre distance
0.3 m
  1. Locate the extreme fibre

    Measure from the relevant neutral axis, not from the bottom edge.

    (0.3) = 0.3 m
  2. Divide the area property by that distance

    m⁴ divided by m leaves m³.

    (0.0054) ÷ (0.3) = 0.018 m³
Answer0.018 m³

Does this worked answer make sense?

For the worked rectangle, bh²/6 = 0.3 × 0.6² / 6 = 0.018 m³ gives the same result independently.

03

Now try your own values

Change a value or its unit. The same method will show your calculation, step by step.

English, Arabic and Persian digits are supported. The steps convert inputs to the formula’s base units.

Results update only when you calculate. The lesson example above stays unchanged.

04

Your turn — check your understanding

Use these new values. Work it out first, then check your answer.

I · Second moment of area
0.00045 m⁴
ymax · Extreme-fibre distance
0.15 m

Find: elastic section modulus

For repeating decimals, use at least four significant figures. Accepted rounding tolerance: 0.05% of the expected value; zero uses an absolute tolerance of 10⁻¹².

A hint, not the answer

Divide the second moment I by y, the distance from the neutral axis to the extreme fibre. For a symmetric rectangle about its centroid, y is half the depth, not the full depth.

I in m⁴ divided by y in m gives W in m³. The example takes I = 0.0054 m⁴ and y = 0.3 m from a 0.6 m deep rectangle.

Show the full practice solution

Compare the steps with your work; revealing a solution does not mark the lesson complete.

  1. Locate the extreme fibre

    Measure from the relevant neutral axis, not from the bottom edge.

    (0.15) = 0.15 m
  2. Divide the area property by that distance

    m⁴ divided by m leaves m³.

    (0.00045) ÷ (0.15) = 0.003 m³
Answer0.003 m³

Avoid the common trap

Using full depth instead of half depth for a symmetric rectangle halves W incorrectly. Elastic W is not the plastic section modulus.

When this method applies — and when it does not

Elastic section modulus for the stated bending axis. An asymmetric section may have different top and bottom values.

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.

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