Learn: Parallel-axis theorem
You may know the second moment of an area about its centre but need it about another parallel axis. The parallel-axis theorem adds the effect of moving the entire area away from its centroidal axis.
What the formula is saying
The extra contribution is area times the square of the perpendicular offset. Squaring the offset means shifting to either side of the centroid gives the same increase.
Read the symbols in plain language
- Ic
- Centroidal I
The area-weighted square of distance from the stated axis. It measures the spread of the cross-section, not its area or mass.
m⁴The fourth power of metres is used for a second moment of area; 1 m⁴ = 10¹² mm⁴.
- A
- Area
Area. Square the perpendicular distance and multiply by the entire area being transferred.
m²Square metres measure area; square the length conversion factor.
- d
- Axis offset
Perpendicular separation of two parallel axes, one of which passes through the section centroid.
mMetres measure length; 1 m = 1000 mm.
- I
- Result to find
Parallel-axis theorem. The shifted axis includes both the original spread and the added offset effect.
m⁴
Sort out the units first
Ic is in m⁴, area in m² and offset in m. The product A d² is m² × m² = m⁴, so it can be added to Ic.
Assumptions before calculating
Both axes are parallel and one of them is centroidal. Use a nonnegative centroidal second moment and positive area, with the perpendicular distance between axes.
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 I and explain the result in the stated output unit.
- Ic · Centroidal I
- 0.001 m⁴
- A · Area
- 0.02 m²
- d · Axis offset
- 0.4 m
Find the offset contribution
Square the perpendicular distance and multiply by the entire area being transferred.
(0.02) × (0.4)^2 = 0.0032 m⁴Add to the centroidal property
The shifted axis includes both the original spread and the added offset effect.
(0.001) + (0.0032) = 0.0042 m⁴
Does this worked answer make sense?
The transferred value cannot be smaller than Ic for a positive area. When d = 0 the axes coincide and I = Ic.
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.
- Ic · Centroidal I
- 0.0002 m⁴
- A · Area
- 0.01 m²
- d · Axis offset
- 0.15 m
Find the offset contribution
Square the perpendicular distance and multiply by the entire area being transferred.
(0.01) × (0.15)^2 = 0.000225 m⁴Add to the centroidal property
The shifted axis includes both the original spread and the added offset effect.
(0.0002) + (0.000225) = 0.000425 m⁴
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.
- Ic · Centroidal I
- 0.0005 m⁴
- A · Area
- 0.025 m²
- d · Axis offset
- 0.2 m
Find: Learn: Parallel-axis theorem
A hint, not the answer
The extra contribution is area times the square of the perpendicular offset. Squaring the offset means shifting to either side of the centroid gives the same increase.
Ic is in m⁴, area in m² and offset in m. The product A d² is m² × m² = m⁴, so it can be added to Ic.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Find the offset contribution
Square the perpendicular distance and multiply by the entire area being transferred.
(0.025) × (0.2)^2 = 0.001 m⁴Add to the centroidal property
The shifted axis includes both the original spread and the added offset effect.
(0.0005) + (0.001) = 0.0015 m⁴
Avoid the common trap
Do not square the area or measure the offset along an oblique line. The starting Ic must be centroidal, not a value about an arbitrary parallel axis.
When this method applies — and when it does not
This does not rotate an axis and does not find the centroid for you. For composite sections, first locate the common centroid and then sum each part’s Ic + A d²; holes are treated as removed areas.
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.
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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: Parallel-axis theorem. 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.