Moments of inertia

Start with the name, because it misleads almost everybody. This quantity is a second moment of area. It has units of length to the fourth power, it belongs to a cross-section drawn on paper, and it has nothing whatever to do with how hard a body is to spin up. The thing that resists angular acceleration is the mass moment of inertia, measured in kilogram metres squared, and it is a different quantity that unfortunately shares a name. Everything below is about area.

Where the integral comes from

Nobody invented \(\int y^2\,dA\) because it looked interesting. It arrives, uninvited, the first time anyone tries to add up the moment of a bending stress across a section.

Bend a beam and its fibres do not all stretch equally. One surface shortens, the opposite one lengthens, and somewhere between them lies a layer that does neither — the neutral axis. Because plane sections stay plane, the strain in a fibre is proportional to its distance from that axis, and while the material behaves elastically the stress follows:

Now take a thin strip of the cross-section, of area \(dA\), sitting a distance \(y\) from the axis. The force it carries is the stress times its area, \(dF = k y\,dA\). That force acts a distance \(y\) from the axis, so it contributes a moment \(dM = y\,dF = k y^2\,dA\). Add every strip in the section and the constant comes out of the integral:

Whatever else is going on — how stiff the steel is, how sharply the beam is curved — has been packed into \(k\). What is left is a single number that depends on the shape of the section and on nothing else. That number is the second moment of area:

Read the subscripts carefully, because they trip people up. \(I_x\) is taken about the \(x\) axis, and the distance from that axis is \(y\). Writing \(I_x = \int x^2 dA\) is the single commonest slip in the whole topic, and it is silent: the answer is a perfectly plausible number for the wrong axis. Note also that the integrand is never negative, so a second moment of area about any axis is always positive. It cannot cancel itself out the way a first moment can.

Why distance counts twice

The square is the whole story. Everything surprising about beam design follows from the fact that the weighting is quadratic rather than linear, so it is worth seeing why it could not have been linear.

The linear version already exists. It is the first moment of area, \(\int y\,dA\), and about a centroidal axis it is exactly zero — that is the definition of the centroid. Material above the axis cancels material below it. A quantity that always vanishes cannot possibly measure stiffness, and the square is what stops the cancellation: every contribution counts positively, and distant area counts enormously.

Take 4700 square millimetres of steel — a fixed weight per metre, a fixed cost. Rolled into a square 68.6 mm on a side, it gives \(I = 1.84 \times 10^{6}\) mm\(^4\). Drawn out into a strip 23.5 mm wide and 200 mm deep, standing on end, the same steel gives \(15.7 \times 10^{6}\) — eight and a half times as much, for nothing. Roll it instead into an I-section 200 mm deep with 15 mm flanges and a 10 mm web and it reaches \(29.8 \times 10^{6}\), sixteen times the square.

For a rectangle the arithmetic is visible in the formula. About the axis through its centroid parallel to the base,

Depth is cubed while width is not, which is why a floor joist is stood on edge and never laid flat. A 50 by 150 joist on edge is nine times stiffer than the same joist lying down, because tripling the depth multiplies \(h^3\) by twenty-seven while dividing \(b\) by three.

The I-beam is this observation taken to its conclusion. In the section above, the two flanges hold two thirds of the area and supply \(25.7\) of the \(29.8\) — eighty-six per cent of the stiffness. The web contributes almost nothing to bending; it is there to carry shear and, more importantly, to hold the flanges apart. Material near the neutral axis is nearly idle, and the entire art of section design is getting rid of it without letting the section buckle or shear.

The parallel-axis theorem

Tables give \(\bar{I}\) for simple shapes about their own centroidal axes. Real problems almost never ask about those axes. The bridge between them is one line:

where \(d\) is the distance between the two parallel axes. The derivation is short and it contains the warning. Measuring \(y\) from the centroidal axis, a point on the new axis is at \(y + d\), so

The first term is \(\bar{I}\) and the third is \(A d^2\). The middle term is \(2d\) times the first moment of area, and it disappears only because that first moment is zero about the centroid. Start anywhere else and the cross term survives, unaccounted for, and the theorem is simply false.

So the rule that matters is not the formula but the restriction: you may only transfer \(I\) from a centroidal axis. Given \(I\) about one off-centre axis, you cannot hop to another off-centre axis by adding \(A(d_2 - d_1)^2\). The 60 by 40 rectangle in the figure has \(I = 4.16 \times 10^{6}\) at 40 mm and \(15.68 \times 10^{6}\) at 80 mm — a difference of \(11.52 \times 10^{6}\), where the illegal shortcut predicts \(3.84 \times 10^{6}\). You must come back to the centroid first, then go out again.

Two things follow. Because \(A d^2\) can never be negative, the centroidal axis gives the smallest second moment of any axis in its direction — the shape is always closest to its own centroid. And the transfer term usually dominates: that same rectangle has \(\bar{I} = 0.32 \times 10^{6}\) against a transfer of \(15.36 \times 10^{6}\), so its own shape accounts for two per cent of the answer. This is exactly why the flanges of an I-beam do so much work, and it is also why a rough sketch of where the area sits will get you closer than careful arithmetic on \(\bar{I}\) alone.

Composite shapes

Almost no real section is a shape in a table, and almost every real section is two or three of them stuck together. Because integration is additive, so is \(I\): cut the section into rectangles, triangles and circles, work out each part, and add.

The order of operations is fixed and skipping a step is the usual cause of a wrong answer. First find the centroid of the whole section, because that is the axis everything must eventually be referred to. Second, for each part, look up its own \(\bar{I}_i\) about its own centroid. Third, transfer each part to the composite axis and sum:

with \(d_i\) measured from the composite centroidal axis to the centroid of part \(i\). Never add \(I\) values that were taken about different axes. They are not the same kind of number and the sum means nothing.

Holes are handled by giving them negative area. A drilled or cut-away region enters with a minus sign in every term it touches: negative \(A\) in the centroid calculation, negative \(\bar{I}\), and negative \(A d^2\). The T-section in the figure has a 140 by 25 flange, a 40 by 95 web and a 30 mm hole through the web. Its signed areas give \(\bar{y} = 39.8\) mm below the top, and the three transferred terms are \(2.80\), \(6.91\) and \(-1.18\), for \(I = 8.53 \times 10^{6}\) mm\(^4\).

Get that last sign wrong and the arithmetic still works, which is what makes it dangerous. Adding the hole rather than subtracting it gives \(10.89 \times 10^{6}\) — larger than the \(9.60 \times 10^{6}\) the section would have with no hole in it at all. Drilling a beam has apparently made it stiffer. Any composite answer that exceeds the answer for the same section with its holes filled in has a sign error in it, and that is the cheapest check available.

The same trick works in reverse, and it is often quicker. Rather than adding up three rectangles to make a channel, take one big rectangle and subtract the two rectangles that are missing. Subtraction usually means fewer parts and fewer transfer distances to measure, and fewer chances to mismeasure one.

Radius of gyration and the polar moment

Two pieces of book-keeping remain, both of which are just \(I\) in a different dress.

The radius of gyration answers a hypothetical question: if the whole area were squeezed into a thin line, how far from the axis would that line have to be to give the same second moment? Setting \(A k^2 = I\),

For the 120 by 60 rectangle in the figure, \(A = 7200\) mm\(^2\), \(I_x = 2.16 \times 10^{6}\) and \(I_y = 8.64 \times 10^{6}\), giving \(k_x = 17.3\) mm and \(k_y = 34.6\) mm. For any rectangle these come out as \(h/\sqrt{12}\) and \(b/\sqrt{12}\). Note that \(k\) is not the distance to anything — not to the centroid, not to the edge, not to any material that exists. It is a length that summarises a distribution, and its value is that it lets sections of different size be compared on equal terms. It is the length that appears in the slenderness ratio \(L/k\) that governs whether a column buckles, which is where you will meet it next.

The polar moment measures resistance to twisting rather than bending, so distance is taken from a point rather than a line:

The two rectangular second moments about any pair of perpendicular axes through a point add to give the polar moment about that point — for the rectangle above, \(J = 10.8 \times 10^{6}\). It governs the torsion of circular shafts, where \(J = \pi d^4 / 32\) for a solid section, and it explains why a hollow shaft is so nearly as strong as a solid one of the same diameter: the material near the centre was carrying almost no torque. The polar radius of gyration follows the same pattern, and because \(J = I_x + I_y\) it obeys \(k_O^2 = k_x^2 + k_y^2\) — for the rectangle, \(38.7^2 = 17.3^2 + 34.6^2\).