Moment of Inertia & Centroid Calculator
Find the centroid and second moment of area of a beam section. Rectangle, hollow box, circle, tube, I-beam, T, channel or angle, worked part by part with the parallel-axis theorem, plus the mass moment of inertia of common solids.
How to Use
- Choose Cross-section for a beam or column shape, or Solid body for the mass moment of inertia of something that spins.
- Pick the shape: rectangle, hollow rectangle, circle, round tube, I-beam, T-section, channel or angle.
- Enter its dimensions and the unit they are in (mm, cm, m or inches), and the unit you want the answers in.
- Read Ix, Iy, the area and the section modulus beside the drawing; the table under it lists the centroid, both moduli, the radii of gyration and, for an angle, the principal axes.
- Open Show Work to see each part’s area, its centroid, its own I and the A·d² term the parallel-axis theorem adds.
- Press a preset to load a worked example such as a 200 mm I-beam or a 3 × 3 × ¼ in angle.
Worked Example
A rectangle. A 100 mm wide, 200 mm deep section: A = 100 × 200 = 20,000 mm², the centroid is at mid-depth, and Ix = bh³/12 = 100 × 200³ ÷ 12 = 66,666,667 mm⁴ (6,666.7 cm⁴). The extreme fibre is c = 100 mm away, so S = Ix ÷ c = 666,667 mm³.
A T-section, part by part. Flange 150 × 12 mm (A = 1,800 mm², centre 144 mm up) on a stem 10 × 138 mm (A = 1,380 mm², centre 69 mm up). The centroid is ȳ = (1,800 × 144 + 1,380 × 69) ÷ 3,180 = 111.45 mm from the bottom. About that axis the flange gives 150 × 12³ ÷ 12 + 1,800 × 32.55² = 1,928,373 mm⁴ and the stem 10 × 138³ ÷ 12 + 1,380 × 42.45² = 4,677,155 mm⁴, so Ix = 6,605,528 mm⁴. The bottom of the stem is 111.45 mm from the axis and the top 38.55 mm, so the least section modulus is 6,605,528 ÷ 111.45 = 59,267 mm³.
The common mistake: adding each part’s own I and forgetting A·d². For a 200 × 100 I-beam (8.5 mm flanges, 5.6 mm web) the two flanges have only 5,118 mm⁴ each about their own centres, and the web 2,859,961 mm⁴. Adding those gives 2,870,196 mm⁴. The right answer moves each flange 95.75 mm to the neutral axis, adding 850 × 95.75² = 7,792,853 mm⁴ each, for Ix = 18,455,902 mm⁴, 6.4 times more. That distance term is the whole reason I-beams work.
Show Work
Formulas
From Galileo’s Cantilever to the I-Beam
Galileo asked how strong a beam is in Two New Sciences (1638). He saw that a beam’s strength grows with its depth faster than with its width, but he placed the turning point at the bottom edge of the section, so his numbers were too high. Edme Mariotte (1686) and Antoine Parent (1713) moved the axis inside the section, and Claude-Louis Navier settled it in his 1826 lectures: the neutral axis of an elastic beam passes through the centroid of the section, and stiffness depends on the integral of y² over the area, the quantity this calculator computes.
The rotating-body version is older in spirit. Christiaan Huygens worked out the swing of a compound pendulum in Horologium Oscillatorium (1673) using sums of mass times distance squared, and Leonhard Euler gave the quantity its name, the moment of inertia, in his 1765 book on the motion of rigid bodies. The rule for shifting it to a parallel axis is often called the Huygens–Steiner theorem, after Huygens and the Swiss geometer Jakob Steiner.
The rolled-iron I-beam that puts the idea into practice appeared in the mid-19th century: most of the metal sits in the flanges, far from the axis, where the A·d² term multiplies its value.
About This Tool
This calculator builds each section from rectangles and circles, adds holes as negative areas, and finds the centroid and the second moments about it with the parallel-axis theorem. Show Work lists every part with its area, its centre, its own moment and the A·d² term, so you can check it against a hand calculation. It also gives the section modulus to the top and bottom fibres separately for shapes that are not symmetric, the radii of gyration, the polar moment, and for angles the product of inertia and the principal axes. The drawing is to scale, with the centroid and both neutral axes marked.
Rolled steel sections have rounded root fillets and toes that a sharp-cornered model leaves out, so a catalogue figure for the same nominal size can differ by a few percent; use the catalogue value for design. The Solid body mode gives the mass moment of inertia of ten common shapes, the radius of gyration and the energy stored at 1,000 rpm. Everything runs in your browser; nothing you enter is sent anywhere.
Related tools: Beam Calculator, Torque Calculator, and Density, Mass & Volume Calculator.
Frequently Asked Questions
What is the moment of inertia of a rectangle?
About the axis through its centroid, parallel to the width, it is I = bh³/12. A 100 mm wide, 200 mm deep rectangle gives 100 × 200³ ÷ 12 = 66,666,667 mm⁴ (6,666.7 cm⁴). Turn it flat, 200 wide and 100 deep, and it drops to 16,666,667 mm⁴, a quarter, because the depth is cubed.
How do you find the centroid of a T-section?
Split it into rectangles and take the area-weighted average of their centres. A T 150 mm wide and 150 mm deep with a 12 mm flange and a 10 mm stem has a 1,800 mm² flange centred 144 mm up and a 1,380 mm² stem centred 69 mm up, so ȳ = (1,800 × 144 + 1,380 × 69) ÷ 3,180 = 111.45 mm from the bottom, or 38.55 mm below the top.
What is the parallel-axis theorem?
The moment of inertia about any axis equals the moment about a parallel axis through the part’s own centroid plus the area times the distance squared: I = I₀ + A·d². One flange of a 200 × 100 mm I-beam (100 × 8.5 mm) has I₀ = 5,118 mm⁴ but sits 95.75 mm from the neutral axis, so it adds 850 × 95.75² = 7,792,853 mm⁴ on top.
What is the difference between area and mass moment of inertia?
The area moment (second moment of area, mm⁴ or in⁴) measures how a cross-section resists bending. The mass moment (kg·m²) measures how a body resists being spun up. A 20 kg solid disc of 0.25 m radius has ½mr² = 0.625 kg·m² and stores 3,427 J at 1,000 rpm.
What is section modulus used for?
Section modulus S = I ÷ c turns a bending moment straight into the peak stress: σ = M ÷ S. A 100 × 200 mm rectangle has S = 66,666,667 ÷ 100 = 666,667 mm³, so a 10 kN·m bending moment gives 10,000,000 ÷ 666,667 = 15 MPa at the top and bottom faces.
How do I use the Moment of Inertia & Centroid Calculator?
Simply type your numbers and read the result, which refreshes the instant you change something. There is nothing to submit and nothing to wait for.
Is it free? Does it work without internet?
Yes to both. It is free with no sign-up, and once the page has loaded it keeps working even with no internet.
Where does my data go?
Nowhere — every calculation runs on your own device. Nothing you enter is uploaded, logged, or stored.
Common Use Cases
Timber joists
A 47 × 195 mm joist on edge has Ix = 29,041,594 mm⁴ (2,904 cm⁴) and S = 297,863 mm³; laid flat its Iy is only 1,687,124 mm⁴.
Steel beams
A 200 × 100 I-section with 8.5 mm flanges and a 5.6 mm web has Ix = 1,845.6 cm⁴ from 2,725 mm² of steel, 6.4 times what the parts give without their A·d² terms.
Tubes versus bars
A 60 × 5 mm round tube has Ix = 329,376 mm⁴; a solid bar with the same 864 mm² of metal (33.2 mm across) has 59,396 mm⁴, so the tube is 5.5 times stiffer for the same weight.
Angle brackets
A 3 × 3 × ¼ in angle has Ix = Iy = 1.244 in⁴, but it bends most easily about its 45° minor principal axis, where I = 0.504 in⁴.
Flywheels and rotors
A 20 kg disc of 0.25 m radius has I = 0.625 kg·m² (14.83 lb·ft²) and a radius of gyration of 176.8 mm.
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