On this page
How to use the calculator
- Choose the unit first: mm, cm, m or in. Every input and result uses it, so a section typed in mm gives I in mm⁴. Changing the unit relabels your numbers; it does not convert them.
- Single shape: tap a shape and type its dimensions. Boxes accept decimals and fractions such as
3/4or1 1/2. Leave the bottom flange of an I-beam blank for a symmetric beam, and give an angle two different legs for an unequal angle. - Built-up section: add parts and give each one an x and y, the lower-left corner of that part measured from an origin you choose. Rotate and Mirror turn angles, channels and triangles. Tick Hole to subtract a part, such as a bolt hole. A part can also be a rolled shape copied from a steel table (A, Ix, Iy, depth and width).
- Read the drawing, results and working table. Copy link saves the exact section; Print gives a clean sheet of the working for checking an FE or GATE answer.
Links can open a section directly. For example, ?shape=ibeam&b=200&h=300&tf=15&tw=10 opens a 200 × 300 mm I-beam.
| Link parameter | Meaning | Example |
|---|---|---|
shape | rect, box, circle, tube, semicircle, triangle, ibeam, tee, channel, angle, zsection | shape=tee |
b h tf tw t d di r a bb tb | The dimensions named on the input labels | b=150&tf=20 |
parts | A built-up section: type:dimensions@x,y, dimensions in the order of the input boxes, then r90, m or hole | rect:240,120@0,0 |
u | mm, cm, m or in | u=in |
Second moment of area and the parallel axis theorem
The second moment of area, which beam courses call the moment of inertia of a section, measures how far the area sits from an axis: Ix = ∫y² dA. Its unit is length to the fourth power. Bending stress is σ = M·y/I, and a larger I means less deflection. Doubling the depth of a rectangle multiplies b·h³/12 by 8, which is why beams are deep and why an I-beam puts its steel in the flanges, far from the centroid.
The parallel axis theorem
A part of area A whose own centroid sits a distance d from an axis has I = Iown + A·d² about that axis. The own values come from standard formulas: b·h³/12 for a rectangle, π·d⁴/64 for a circle, b·h³/36 for a triangle and 0.1098·r⁴ for a semicircle about its own centroid.
The table method
- Split the section into simple parts. A hole is a part with negative area.
- For each part write A, its centroid x and y, and the products A·x and A·y.
- Find the centroid: x̄ = ΣA·x / ΣA and ȳ = ΣA·y / ΣA.
- For each part find dy = y − ȳ, then Ix = Σ(Iown + A·dy²). Repeat with dx for Iy.
The tool prints this table with its sums.
Principal axes
The product of inertia Ixy = ∫x·y dA is zero when a section is symmetric about either axis. Angles and Z-sections are not, so their largest and smallest moments act about tilted principal axes. Mohr's circle finds them: centre C = (Ix + Iy)/2, radius R = √[((Ix − Iy)/2)² + Ixy²], I1,2 = C ± R and tan 2θp = −2Ixy/(Ix − Iy). The tool measures θp counterclockwise from the x axis to the axis of I1.
Section modulus versus moment of inertia
The elastic section modulus S = I/c, with c the distance from the centroid to the extreme fibre, turns a moment straight into the peak stress: σmax = M/S. An unsymmetric section has a top and a bottom value, and the smaller one gives the larger stress. The plastic modulus Z is the sum of the first moments of the two halves of the area about the axis that splits it equally, used in plastic and LRFD design. The radius of gyration r = √(I/A) goes into column buckling. For units, 1 cm⁴ = 10 000 mm⁴ and 1 in⁴ = 25.4⁴ mm⁴ = 416 231 mm⁴.
Three worked examples
1. T-section: centroid and Ix by the table
A 150 × 20 mm flange sits on a 20 × 180 mm web, 200 mm deep in total. Measure y from the bottom of the web.
| Part | A (mm²) | y (mm) | A·y (mm³) | Own Ix (mm⁴) | dy (mm) | A·dy² (mm⁴) |
|---|---|---|---|---|---|---|
| Flange | 3000 | 190 | 570 000 | 100 000 | 54.545 | 8.9256 × 10⁶ |
| Web | 3600 | 90 | 324 000 | 9.72 × 10⁶ | −45.455 | 7.438 × 10⁶ |
| Σ | 6600 | 894 000 | 9.82 × 10⁶ | 16.364 × 10⁶ |
So ȳ = 894 000 / 6600 = 135.45 mm from the bottom, and Ix = 9.82 × 10⁶ + 16.364 × 10⁶ = 26.184 × 10⁶ mm⁴. The section moduli are Stop = 405.66 × 10³ mm³ and Sbottom = 193.3 × 10³ mm³, so for any moment the bottom of the web is stressed about twice as hard as the top of the flange. Half the area lies above y = 165 mm, 15 mm into the web, which gives Zx = 349.5 × 10³ mm³.
2. A plate with a semicircular cut-out
This is Beer and Johnston's Sample Problem 9.5: a 240 × 120 mm rectangle with a half circle of radius 90 mm cut from the middle of its top edge, with Ix wanted about the bottom edge. In the tool the half circle is a semicircle part turned 180° (flat side up), placed at (30, 30) and marked Hole.
- Rectangle: A = 28 800 mm² at y = 60 mm.
- Half circle: A = −12 723 mm². Its centroid is 4r/(3π) = 38.197 mm below the top edge, at y = 81.803 mm.
- Net area 16 077 mm², so ȳ = 687 190 / 16 077 = 42.745 mm.
- About the centroid: Ix = 27.359 × 10⁶ − 10.835 × 10⁶ = 16.524 × 10⁶ mm⁴.
- About the bottom edge: Ix + A·ȳ² = 45.897 × 10⁶ mm⁴.
The textbook answer is 45.9 × 10⁶ mm⁴. In the hole row the area, the own I and the A·d² term all carry a minus sign.
3. Principal axes of a Z-shaped member
Hibbeler's Statics has a member built from three 100 mm thick rectangles: a 600 mm web centred on the origin, a 300 mm leg rising from its left end and another hanging from its right end. The table gives Ix = 2.9 × 10⁹, Iy = 5.6 × 10⁹ and Ixy = −3 × 10⁹ mm⁴, because each leg adds A·dx·dy = 30 000 × 250 × (−200) = −1.5 × 10⁹ mm⁴.
Mohr's circle then has C = 4.25 × 10⁹ and R = 3.2898 × 10⁹ mm⁴, so I1 = 7.5398 × 10⁹ mm⁴ and I2 = 960.24 × 10⁶ mm⁴. The major axis is at θp = 57.11° and the minor one at −32.89°. The book gives 7.54 × 10⁹, 0.960 × 10⁹, 57.1° and −32.9°.
Reading the result and common mistakes
- Forgetting A·d². Adding only the own values gives 9.82 × 10⁶ mm⁴ for the T-section above instead of 26.184 × 10⁶. The A·d² terms are often the larger part.
- Measuring d to the wrong line. d runs from each part's centroid to the centroid of the whole section, not to the base. For I about the base, read Ix about y = 0.
- Adding a hole. A hole has a negative area, a negative own I and a negative A·d² row.
- Unit slips. Lengths go to the fourth power: 26.184 × 10⁶ mm⁴ is 2618.4 cm⁴ or 62.906 in⁴.
- Treating Ix + Iy as the torsion constant. It is the polar moment about the centroid. It equals the torsion constant J only for a solid or hollow circle; for an I-beam or an angle the real J is far smaller.
- Steel tables read higher. Rolled shapes have root fillets that this tool does not draw. An IPE 300 drawn here (300 × 150 mm, web 7.1 mm, flanges 10.7 mm) has A = 51.881 cm², and the published 53.81 cm² includes four 15 mm fillets. For a rolled shape in a built-up section, use the table part with the published A, Ix and Iy.
- Angles bend about tilted axes. A 100 × 100 × 10 mm angle has rx = 30.78 mm, but its minimum radius of gyration about the 45° axis is only 19.658 mm, and a buckling check needs the smaller one.
Embed the Moment of Inertia Calculator on your website
Teachers, bloggers and course sites can add this tool to any web page for free. Paste the code below into an HTML or embed block (WordPress, Google Sites, Canvas, Moodle, Blogger, Wix). It loads the same tool you see above, without our menus or ads. Please keep the credit line under the frame. Sizes, a live preview and step-by-step guides for each platform are on the calculator widget page.
Frequently asked questions
Is moment of inertia the same as second moment of area?
In beam and column work, yes. The moment of inertia of a section means the second moment of area, ∫y² dA, in mm⁴ or in⁴. The mass moment of inertia of dynamics, in kg·m², is a different quantity that this tool does not compute.
How do I find the moment of inertia of a T-section?
Split it into the flange and the web, find the centroid with ΣA·y / ΣA, then add Iown + A·d² for both parts. For a 150 × 20 mm flange on a 20 × 180 mm web, ȳ = 135.45 mm from the bottom and Ix = 26.184 × 10⁶ mm⁴. Choose T-section above to see every row of the table.
What is the difference between section modulus and moment of inertia?
I, in length⁴, measures stiffness and goes into deflection formulas. S = I/c, in length³, measures bending strength, because the peak stress is M/S. Two sections with the same I can have different S if their extreme fibres sit at different distances from the centroid.
Why does my answer differ from the steel table?
The tool draws rolled shapes with sharp corners, while tables include the root fillets between web and flange. For an IPE 300 the four fillets add 1.9314 cm² of area. For design work, enter the rolled shape as a table part with the published A, Ix and Iy.
How do I convert mm⁴ to cm⁴ or in⁴?
Raise the length factor to the fourth power: 1 cm⁴ = 10⁴ mm⁴, 1 m⁴ = 10¹² mm⁴ and 1 in⁴ = 25.4⁴ mm⁴ = 416 231 mm⁴. So 26.184 × 10⁶ mm⁴ is 2618.4 cm⁴ or 62.906 in⁴. The tool lists the main results in all four units.
Does the calculator give the torsion constant J?
Only for circles and tubes, where J equals Ix + Iy exactly (π·d⁴/32 for a solid circle). For other shapes it labels Ix + Iy as the polar moment, which overstates the twisting stiffness of I-beams and angles.