Section Properties Calculator

Area, both second moments, both section moduli, the centroid and the radii of gyration for eight machine-part shapes — with the polar second moment and the torsion constant printed separately, because they are not the same quantity for anything but a round bar.

Last updated 24 Sep 2026

Section properties

Shape and four dimensions → A, I, Z, r and the centroid

Section shapeWhich four of the dimensions below get read depends on this. A round bar uses only the first. A solid rectangle uses the first two. A tube or a box uses the wall thickness. An I, a T and a channel use all four; an angle uses three, with the flange thickness serving as the leg thickness.

Overall DEPTH h (mm) — the diameter for a round sectionThe dimension in the direction of the bending you care about. Depth is the powerful one: the second moment goes as h³ and the section modulus as h², so putting the material where the depth is buys stiffness faster than anything else you can do.

Overall WIDTH b (mm) — flange width, or the horizontal legIgnored for a round section, where the depth is the diameter. For an I, a T or a channel this is the flange width; for an angle it is the length of the horizontal leg.

FLANGE or WALL thickness t_f (mm)The wall of a tube or box, the flange of an I, T or channel, the thickness of an angle’s legs. Capped at just under half the smaller overall dimension, because a wall thicker than that is a solid section.

WEB thickness t_w (mm) — the upright of an I, T or channelRead only by the I, T and channel. For an I it is the vertical member joining the two flanges; for a T it is the stem; for a channel it is the flat back. It carries the shear and it contributes almost nothing to the strong-axis second moment, which is the point of an I-section.

Not a circuit: the section itself. The outline is a SCHEMATIC — it shows which shape you have selected rather than its proportions — but the crosshair is the real centroid, at the height and offset computed from your own dimensions, and the two lines through it are the two centroidal axes that every second moment on this page is taken about. Watch the horizontal line as you switch between a rectangle and a T: for anything with a horizontal axis of symmetry it sits at mid-depth, and for a T or an angle it does not, which is the single thing that makes those shapes harder. The bar pair underneath is the consequence — the two section moduli, one to each face. They are equal for a symmetric section and differ by more than two to one for this page's default T, and the SHORTER bar is the one that governs, because the fibre further from the centroid reaches the allowable stress first. On the right, note the last two rows: I_p is I_x + I_y for every shape as a matter of arithmetic, and the torsion constant J is a different quantity that happens to equal it only for a round bar.

28.749cm³Example

A 120 mm deep T-section, 80 mm flange, 10 mm flange thickness, 8 mm stem

Assemble about any axis, then translate once

A = ΣA_i · ȳ = ΣA_i y_i / A · I_x = Σ(I_i + A_i d_i²) · Z = I/c · r = √(I/A) · I_p = I_x + I_y · I_xy = ΣA_i x̄_i ȳ_i

A: area. Add the parts, subtract the holes
ȳ: centroid height: the area-weighted mean of the parts’ own centroids. For anything without a horizontal axis of symmetry this is NOT mid-depth, and it is the step people skip
I: second moment of area about a CENTROIDAL axis. bh³/12 for a rectangle about its own centre; πd⁴/64 for a circle
A d²: the parallel-axis (Steiner) term. d is the distance from the part’s centroid to the section’s. It is always positive, so any axis other than the centroid gives too much I
c: distance from the centroidal axis to the fibre you care about. An asymmetric section has two of them and therefore two section moduli
r: radius of gyration. The radius at which all the area could be concentrated without changing I. It is the length that turns a member’s length into a slenderness
I_p: POLAR second moment, ∫r²dA. It is I_x + I_y for every shape, as an identity. It is NOT the torsion constant except for a round bar
I_xy: product of inertia. Zero if the section has any axis of symmetry. Non-zero for an angle, and then x and y are not the principal axes

Worked example

A 120 mm deep T-section, 80 mm flange, 10 mm flange thickness, 8 mm stem
Break it into two rectangles and find each one's own area and centroid height. Flange: 80 × 10 = 800 mm², centred at 120 − 5 = 115 mm. Stem: 8 × 110 = 880 mm², centred at 110/2 = 55 mm. Total area 1,680 mm²
Centroid, area-weighted: ȳ = (800 × 115 + 880 × 55) / 1,680 = (92,000 + 48,400) / 1,680 = 83.571 mm. Not 60 mm. That single number is what makes a T harder than a rectangle, and every line below depends on it
Now each rectangle's I about its OWN centre, plus its Steiner term. Flange: 80 × 10³/12 = 6,667, plus 800 × (31.429)² = 790,204. Stem: 8 × 110³/12 = 887,333, plus 880 × (28.571)² = 718,367
Add the four: I_x = 2,402,571 mm⁴, which is 240.26 cm⁴ in the units a section table uses. Note that the stem contributes most of it, even though it is the thinner part — because it is the tall part, and I goes as the cube of height
The weak axis needs no Steiner term at all, because both rectangles are centred on the same vertical line: I_y = 10 × 80³/12 + 110 × 8³/12 = 426,667 + 4,693 = 431,360 mm⁴. This section is 5.6 times stiffer one way than the other
TWO section moduli, because the centroid is not central. To the flange face: Z = I/c = 2,402,571/36.429 = 65,953 mm³. To the stem tip: 2,402,571/83.571 = 28,749 mm³. The second is 56 per cent smaller, so the stem tip reaches the allowable stress first and the GOVERNING modulus is 28.749 cm³
Radius of gyration, which is what a slenderness check wants: r_x = √(I/A) = √(2,402,571/1,680) = 37.82 mm and r_y = 16.02 mm. The smaller one governs anything to do with stability
AND THE ONE EVERYBODY GETS WRONG. The polar second moment is I_x + I_y = 2,833,931 mm⁴ — that is an identity and it is correct. The TORSION CONSTANT of this section is 45,440 mm⁴, which is 62 times smaller. Putting the polar value into θ = TL/GJ would say this T-section is 62 times stiffer in torsion than it is. They are different quantities that share a letter

The eight shapes, all at 120 × 80 so the comparison is fair

Read the last three columns together and the point of this page arrives. I_p is I_x + I_y for every single row, because the polar second moment is DEFINED as ∫r²dA and r² = x² + y² — that is arithmetic, not engineering, and it holds for any shape whatever. J, the torsion constant, is the thing that actually appears in θ = TL/GJ, and the last column is the ratio between them. It is exactly 1.0 for the round bar and the round tube and for nothing else on the list. For the open sections it is in the hundreds. Put I_p where J belongs on an I-section and you will calculate a shaft three hundred times stiffer in torsion than it is. The torsion of non-circular sections page is where J comes from and why. Notice also that the angle is the only row with a non-zero I_xy: it has no axis of symmetry, so x and y are not its principal axes, and its genuinely weakest axis is rotated off the legs. This is a first-pass calculation on an idealised part — uniform section, static load, no stress raisers beyond those stated, and room-temperature material properties. Real parts have fillets, keyways, surface finish and duty cycles that a closed-form answer cannot see.
Shape	Size	A (mm²)	ȳ (mm)	I_x (cm⁴)	I_y (cm⁴)	Z_x (cm³)	r_x (mm)	I_xy (cm⁴)	I_p (cm⁴)	J (cm⁴)	I_p ÷ J
Solid round bar	Ø50 bar	1,963	25.00	30.68	30.68	12.27	12.50	0.00	61.36	61.359	1.0×
Hollow round (tube)	Ø50 × 4 tube	578	25.00	15.41	15.41	6.16	16.32	0.00	30.81	30.810	1.0×
Solid rectangle	120 × 80 solid	9,600	60.00	1,152.00	512.00	192.00	34.64	0.00	1,664.00	1,201.999	1.4×
Rectangular box (RHS)	120 × 80 × 5 box	1,900	60.00	375.58	197.58	62.60	44.46	0.00	573.17	391.530	1.5×
I-section (symmetric)	120 × 80 I, 10/8	2,400	60.00	552.00	85.76	92.00	47.96	0.00	637.76	7.040	90.6×
T-section	120 × 80 T, 10/8	1,680	83.57	240.26	43.14	28.75	37.82	0.00	283.39	4.544	62.4×
Channel (C-section)	120 × 80 channel, 10/8	2,400	60.00	552.00	154.88	92.00	47.96	0.00	706.88	7.115	99.4×
Angle (L-section)	120 × 80 × 10 angle	1,900	39.74	278.32	100.32	34.68	38.27	-97.26	378.64	6.333	59.8×

Z matters for strength and I matters for stiffness, and they do not rank sections the same way

Stress is M/Z and deflection is proportional to 1/EI, so a part that must not YIELD is sized by Z and a part that must not MOVE is sized by I. The two are related by Z = I/c, which means that making a section deeper raises I faster than Z — I as h³, Z as h² — so a deep section is disproportionately better at stiffness than at strength. That is why a machine frame that has to hold a tolerance ends up deeper than one that only has to hold a load, and it is why a designer who sizes everything on stress ends up with a machine that is strong enough and still flexes. The last two columns are the honest efficiency measures: how much I and how much Z you got per square millimetre of material you paid for. This is a first-pass calculation on an idealised part — uniform section, static load, no stress raisers beyond those stated, and room-temperature material properties. Real parts have fillets, keyways, surface finish and duty cycles that a closed-form answer cannot see.
Section	A (mm²)	I_x (cm⁴)	Z_x (cm³)	I per unit area (cm²)	Z per unit area (mm)	Material used
Solid rectangle	9,600	1,152.00	192.00	12.000	20.000	100 %
Box, 5 mm wall	1,900	375.58	62.60	19.768	32.946	20 %
I-section, 10/8	2,400	552.00	92.00	23.000	38.333	25 %
T-section, 10/8	1,680	240.26	28.75	14.301	17.112	18 %

Getting the centroid of a T wrong, and what it costs

For this page’s 120 × 80 T-section with a 10 mm flange and an 8 mm stem, the centroid is 83.57 mm above the bottom of the stem, not 60 mm. The area-weighted average is the only definition: ȳ = ΣA_iy_i / ΣA_i, taking each rectangle’s own centre. Assume mid-depth and the second moment you compute is bigger than the truth, because you have added an A·d² term about an axis that is not the centroid — the parallel-axis theorem only ever increases I, so ANY wrong axis gives too much stiffness. Then the section moduli split 2.29 to one between the two faces, and the stem tip, which is the far fibre, is the one that yields first in sagging. A T-section is strong one way up and weak the other, and which way up it goes is a drawing note, not a calculation. This is a first-pass calculation on an idealised part — uniform section, static load, no stress raisers beyond those stated, and room-temperature material properties. Real parts have fillets, keyways, surface finish and duty cycles that a closed-form answer cannot see.
Assumed neutral axis	ȳ (mm)	I about that axis (cm⁴)	Z to the top (cm³)	Z to the bottom (cm³)
Mid-depth, which is the wrong answer	60.00	333.60	55.60	55.60	the usual guess
Centroid of the flange alone	115.00	406.20	812.40	35.32	flange only
Correct centroid, area-weighted	83.57	240.26	65.95	28.75	what this page computes

The centroid, the two moduli, and why the polar second moment is not the torsion constant

The centroid is the whole difficulty, and it is the step that gets skipped. Every second moment on this page is taken about an axis through the centroid, because that is the axis a beam actually bends about — the neutral axis of a section in pure bending passes through its centroid, and nowhere else. Finding it is an area-weighted average and nothing more: break the section into rectangles, multiply each area by the height of its own centre, add, divide by the total area. For a rectangle, a round bar, a tube, a box or a symmetric I the answer is mid-depth and nobody has to think. For a T, a channel loaded about its weak axis, or an angle it is not, and the error propagates into everything: the second moment comes out too large, because the parallel-axis theorem only ever adds, and the two section moduli come out equal when they should differ by a factor of two.

Z is for strength and I is for stiffness, and they do not rank sections the same way. Bending stress is M/Z, so a part that must not yield is sized by its section modulus. Deflection is proportional to 1/EI, so a part that must not move is sized by its second moment. The two are related by Z = I/c, and since c grows with depth, making a section deeper raises I faster than it raises Z — I as the cube of depth, Z as the square. That asymmetry is why a machine frame that has to hold a micron ends up deeper than one that only has to hold a tonne, and it is why a design checked only on stress can pass every calculation and still flex visibly. The chart on this page scales the depth and shows both curves separating.

The polar second moment is not the torsion constant, and this is the most common error in the whole subject. The polar second moment I_p = ∫r²dA is I_x + I_y for every shape that exists; that is arithmetic, since r² = x² + y². The torsion constant J is the thing that appears in θ = TL/(GJ) and τ = T/(…), and it comes from solving the St Venant torsion problem for the actual cross-section. For a circle — and only for a circle — the two are equal, because the cross-section of a twisted round bar stays plane and circular. For a solid square the torsion constant is already only about half the polar value; for a 120 × 80 I-section it is a few hundred times smaller. The table above prints the ratio for all eight shapes. The non-circular torsion page owns the calculation and the reason.

The radius of gyration is a length, and it is the one that turns a member into a slenderness. r = √(I/A) is the radius at which all of a section’s area could be concentrated without changing its second moment. On its own it says nothing; divided into a member’s effective length it gives the slenderness ratio L/r, which is what every column formula is written in. This page gives r about both axes and, separately, the smallest r there is — computed from the minor PRINCIPAL second moment rather than from the smaller of I_x and I_y, because for an angle those are different numbers and the principal one is smaller. Buckling of a structural member is out of scope here; the radius of gyration that a buckling check needs is not.

What this page is not. It is geometry. It knows nothing about the material, the load, the support conditions or the manufacturing route. It does not compute beam bending stress or deflection — feed its Z and its I into whichever beam case you have. It does not know about root fillets, so a built-up I-section here will have a slightly smaller area and second moment than a hot-rolled beam of the same nominal dimensions, whose fillets add material at the web-flange junction; use the maker’s own section table for a rolled profile and this page for anything fabricated or machined. It does not compute shear centres, which for a channel and an angle are outside the section and are why those shapes twist when you load them through the centroid. And it does not compute local or lateral buckling, which is what actually limits a thin deep section.

Frequently asked questions

Is the polar second moment J the same as I_x + I_y?

The POLAR SECOND MOMENT is I_x + I_y for every shape, always, as a matter of arithmetic: it is ∫r²dA and r² = x² + y². What is not I_x + I_y is the TORSION CONSTANT — the J in θ = TL/(GJ) — and the two quantities unfortunately share a letter. They are equal only for a circular section, solid or hollow. For a solid square the torsion constant is about 52 per cent of the polar value; for a thin-walled I-section it can be a few hundred times smaller. This page prints both and the ratio between them, and the non-circular torsion page computes the torsion constant properly.

Why does my T-section have two different section moduli?

Because its centroid is not half way up, so the two extreme fibres are at different distances from the neutral axis, and Z = I/c gives a different answer for each. The smaller Z belongs to the fibre further from the centroid, and that fibre reaches the allowable stress first, so the smaller Z governs. It matters which way up the part goes: a T with its flange in tension and a T with its flange in compression are two different parts made from the same extrusion. If the load can reverse, the smaller modulus governs both directions.

Which radius of gyration should I use for a slenderness check?

The smallest one, and for a shape without an axis of symmetry that is not simply the smaller of r_x and r_y. An angle’s second moment is smallest about a principal axis rotated off its legs, and the principal value I₂ is below BOTH I_x and I_y. This page computes it and gives the radius of gyration from it, which is the number a column check needs. Published angle section tables print this as the u–v axis pair for exactly this reason.

Why does an angle twist when I load it straight down?

Two separate reasons and both are geometry. First, the load plane is not a principal plane: because I_xy is not zero, a vertical load produces curvature about both principal axes, so the section deflects sideways as well as down. Second, the shear centre of an angle is at the intersection of its legs, not at its centroid, so a load applied through the centroid applies a torque about the shear centre. This page computes the first effect — the principal axes and their rotation. It does not compute the shear centre.

Are these the same numbers as a steel section table?

Close but not identical for a rolled profile, and identical for anything fabricated or machined. A hot-rolled beam has root fillets where the web meets the flanges, and a rolled angle has a root radius and toe radii; those add a little material near the neutral axis, which raises the area by a per cent or two and the second moment rather less. This page builds sections from sharp-cornered rectangles, which is exactly right for a welded box, a machined part or a plate assembly. For a catalogue profile use the maker’s own table — and for an extruded profile, use the extruder’s, because extrusion tolerances on wall thickness are wider than you may expect.

Where does the plastic section modulus fit in?

It is the modulus that applies once a ductile section has yielded all the way through rather than just at the extreme fibre, and it is larger than the elastic one — 1.5 times for a solid rectangle and 1.70 times for a round bar. This page computes it for those two shapes, where it is a closed form, and refuses it for the rest, where it needs the equal-area axis rather than the centroid and is a different calculation. The ratio between them is the shape factor, and it is the margin between first yield and a fully plastic hinge. Machine design generally stays elastic and uses the elastic modulus; it is worth knowing the margin exists.

How do I get the section properties of a shape that is not on this list?

Break it into rectangles, circles and holes, and use the same three steps this page uses. Add the areas, taking holes as negative. Find the area-weighted centroid. Then sum each part’s own second moment plus its area times the square of its distance from the section centroid. That last step — the parallel-axis theorem — is the whole method, and it works for any shape you can cut up. Add the products of inertia the same way if the result has no axis of symmetry.

Related calculators

References

E. J. Hearn, Mechanics of Materials, chapter 5, “Torsion of non-circular and thin-walled sections”. The source for Timoshenko’s tabulated torsion coefficients k₁ and k₂ against the aspect ratio d/b (0.208 and 0.1406 at a square, both tending to 1/3 as the section thins), for the Bredt–Batho thin-walled closed-section result τ = T/(2A_mt), and for the narrow-strip limit τ = 3T/(dt²). Every entry in that table was reproduced here from the exact Fourier series of the St Venant torsion problem to within 0.0012, which is the last printed digit.
W. C. Young, R. G. Budynas and A. M. Sadegh, Roark’s Formulas for Stress and Strain, Table 10.1 case 4 (solid rectangular section in torsion), as reproduced by AmesWeb’s rectangular-torsion calculator. The two polynomial fits used on this page — K = ab³[16/3 − 3.36(b/a)(1 − b⁴/12a⁴)] for the torsion constant and the four-term bracket for the peak shear stress — agree with the exact series to better than 0.8 per cent over aspect ratios from 1 to 50, and with Timoshenko’s printed table to its last digit. They are used in preference to interpolating the table because they are continuous and because a fit that reproduces the table is a better object than the table it reproduces.
R. C. Hibbeler, Mechanics of Materials, on the superposition of axial and bending stress, on Mohr’s circle and the stress-transformation equations, and on the absolute maximum shear stress in plane stress — the point that the third principal stress is zero rather than absent, so the governing shear may be σ₁/2 and not (σ₁ − σ₂)/2.
EN 1993-1-1 (Eurocode 3) and AISC 360 are named here only to say what this page is NOT. Structural member design — beam reactions, shear and moment diagrams, column buckling of building members, weld groups, plate design — belongs to those codes and to a structural calculator, not to a machine-part stress page. A cantilever bracket on a machine frame is in scope here; a floor beam is not.