Moment of Inertia and Section Modulus Calculator

What is the area moment of inertia?

I_x = ∫ y² dA: every small area dA of a cross-section times the square of its distance y from the x axis, summed over the section, in mm⁴ or in⁴. UK texts call it the second moment of area. Area far from the axis counts most, so depth adds stiffness fast: a rectangle has I_x = bh³/12.

Using the calculator

Pick one of the six shapes, choose millimetres or inches, and type the dimensions. The section redraws to scale with its centroid and its centroidal x and y axes, and the readouts give the area, the centroid, both second moments, the section moduli and the radii of gyration, with every line of the working underneath, ready to copy. The x axis is horizontal, so I_x is the one a beam loaded from above bends about. In millimetres each result also reads in the cm⁴ and cm³ that UK section tables print, and for an I or a T the flange width b and overall depth h are the b_f and d of AISC tables.

The integral is a sum of thin horizontal strips, each of area dA at height y, which is the same limit the Riemann sum explorer draws for an area under a curve. Every shape here has a closed form for it, from the standard tables in mechanics of materials texts such as Gere and Goodno.

Moment of inertia of an I-beam

An I-section is its bounding rectangle less the two voids beside the web, which is the quickest way to its I_x. The calculator opens on a 300 × 150 mm steel section with 10.7 mm flanges and a 7.1 mm web, the default section of the shear force and bending moment calculator, so the two pages give the same 7,999 cm⁴:

• A = 2 × b × tf + (h − 2tf) × tw = 2 × 150 × 10.7 + 278.6 × 7.1 = 5188 mm²
• Ix = [b × h³ − (b − tw) × (h − 2tf)³] / 12 = [150 × 300³ − 142.9 × 278.6³] / 12 = 79.99 × 10⁶ mm⁴
• Iy = [2 × tf × b³ + (h − 2tf) × tw³] / 12 = [2 × 10.7 × 150³ + 278.6 × 7.1³] / 12 = 6.027 × 10⁶ mm⁴
• Sx = Ix / (h / 2) = 79.99 × 10⁶ / 150 = 533.3 × 10³ mm³

The flanges carry the stiffness. The same 5,188 mm² of steel as a solid rectangle 300 mm deep would be 17.29 mm wide, with I_x = 38.91 × 10⁶ mm⁴, so the I-section is 2.06 times as stiff in bending for the same depth and weight. About the vertical axis no steel is more than 75 mm away, against 150 mm about the horizontal one, and I_y is only 7.5 percent of I_x, which is why an unbraced I-section column usually buckles sideways, about its weak axis.

Dividing a bending moment by S_x gives the largest bending stress, σ = M/S_x. The shear force and bending moment calculator’s default beam peaks at 40 kN·m over its roller, so the stress there is 40 × 10⁶ N·mm ÷ (533.3 × 10³ mm³) = 75.0 MPa, the figure that page gives.

Centroid of a T-section and the parallel axis theorem

A T is symmetric only about its vertical axis, so its centroid has to be found first. Split it into a flange and a web, take ȳ = ΣAy / ΣA from the bottom edge, then carry each part’s own second moment to the centroid with the parallel axis theorem, I = Σ(Ī + Ad²), where d is the distance from the part’s centroid to ȳ. Choose T-section and enter a 100 × 20 mm flange on a 20 mm web, 120 mm deep overall:

• Flange: Af = b × tf = 100 × 20 = 2000 mm², at yf = h − tf / 2 = 110 mm
• Web: Aw = tw × (h − tf) = 20 × 100 = 2000 mm², at yw = (h − tf) / 2 = 50 mm
• A = Af + Aw = 2000 + 2000 = 4000 mm²
• ȳ = (Af × yf + Aw × yw) / A = (2000 × 110 + 2000 × 50) / 4000 = 80 mm from the bottom
• Ix = b × tf³ / 12 + Af × (yf − ȳ)² + tw × (h − tf)³ / 12 + Aw × (ȳ − yw)²
• Ix = 66,670 + 2000 × 30² + 1.667 × 10⁶ + 2000 × 30² = 5.333 × 10⁶ mm⁴

Leaving out the Ad² terms is the classic mistake: the two parts’ own second moments add up to only 1.733 × 10⁶ mm⁴, 33 percent of the answer. Because the centroid sits 40 mm below the top and 80 mm above the bottom, the T has two section moduli about x, 133.3 × 10³ mm³ for the top of the flange and 66,670 mm³ for the foot of the web. The smaller one belongs to the farther face and gives the larger stress, so in a sagging T-beam the foot of the web, in tension, is the critical fibre.

Hollow tubes and box sections

Second moments of area subtract, because each is an integral over the area. A tube is a solid circle less its bore, I = π(D⁴ − d⁴)/64, and a box section is its outer rectangle less its void. Taking out the middle costs little stiffness, since that material sits closest to the axis. In inches, a tube 5 in across with a 0.25 in wall has A = 3.731 in² and I = 10.55 in⁴. A solid round bar with the same area is 2.179 in across and has I = 1.108 in⁴, so the tube is 9.53 times as stiff in bending for the same weight of metal.

That ratio is (D² + d²)/(D² − d²), and it keeps growing as the wall gets thinner. What stops it in practice is local buckling: a very thin wall crumples before the section as a whole is fully stressed, which this calculator does not check.

Section modulus and radius of gyration

The flexure formula gives the bending stress at a distance c from the neutral axis as σ = Mc/I, so the elastic section modulus S = I/c turns it into σ = M/S, one number that sizes a beam for strength, while I sizes it for stiffness through EI. The radius of gyration r = √(I/A) is the distance from the axis at which the whole area, concentrated, would give the same I. Columns use it: the Euler buckling stress is π²E/(KL/r)², so a column buckles about the axis with the smaller r. EI plays the part for a beam that the spring constant plays in the Hooke’s law calculator: double it and every deflection halves.

The letters swap between countries. In US texts and AISC tables, S is the elastic section modulus and Z the plastic one. UK tables to BS 5950 used Z for the elastic modulus and S for the plastic, and Eurocode 3 writes them W_el and W_pl. Eurocode also names the strong axis y-y and the weak axis z-z, so the I_y in a UK Eurocode section table is the I_x here. This calculator gives the elastic modulus only.

Converting mm⁴, cm⁴ and in⁴

A second moment of area is a length to the fourth power, so its units convert by the fourth power too. One inch is exactly 25.4 mm, so 1 in⁴ = 25.4⁴ mm⁴ = 416,231 mm⁴, or 41.62 cm⁴, and a section modulus goes by the cube, 1 in³ = 16,387 mm³. Going from mm⁴ to cm⁴ divides by 10⁴, not by 10: the default I-section’s 79.99 × 10⁶ mm⁴ is 7,999 cm⁴.

The calculator never converts partway through. Every formula is homogeneous in length, so millimetres in give mm², mm³ and mm⁴ out, and inches give in², in³ and in⁴. Switching the unit converts the dimensions you typed at 25.4 mm to the inch, and an untouched default switches to round numbers in the other unit instead.

Area moment of inertia versus mass moment of inertia

The name is shared with a different quantity. The mass moment of inertia, Σmr², in kg·m² or lb·ft², measures how hard a body is to spin up and appears in τ = Iα. The area moment of inertia has no mass in it: it is a property of a cross-section’s shape, in mm⁴ or in⁴, and it appears in beam bending, σ = My/I, and in beam deflection, EI d²v/dx² = M. Calling it the second moment of area, as UK texts do, keeps the two apart.

What this calculator leaves out

• Fillets and corner radii. A rolled I-section has curved root fillets where the web meets the flanges, and a hollow section has rounded corners. Both are left out, so a catalogue section differs a little: the default I-section gives 7,999 cm⁴ against the 8,356 cm⁴ a section table gives an IPE 300, fillets included.
• Unsymmetric sections. Every shape here is symmetric about its vertical axis, so the product of inertia is zero and x and y are the principal axes. An angle or a Z-section is not, and needs its principal axes found first.
• Plastic modulus and torsion. Only the elastic modulus is given. The polar moment J = I_x + I_y is shown only for the circle and the tube, where it is also the torsion constant. Any other section warps when it is twisted, and its torsion constant is smaller than its polar moment.
• Design checks. Nothing here classifies a section, checks the local buckling of a thin flange or wall, or applies AISC 360 or Eurocode 3. The figures are geometric properties, for coursework and for checking hand calculations.