Loaded straight down, and it moves sideways

The formula every beam in this collection has used is $\sigma = My/I$, and it contains a quiet assumption that has never been stated: that $I$ is a single number, that $y$ is measured from a horizontal axis, and that a vertical load bends the beam vertically.

All three are true for a section with an axis of symmetry, which is every section drawn so far — the I-section, the tee, the box, the rectangle. For an angle, a channel loaded about its weak axis, a Z-purlin or a cranked bracket, none of them are.

The load in that figure is vertical. The beam moves down and to the left, by amounts in the ratio 1 : 0.59, and nothing has been applied sideways at all.

The quantity symmetry was hiding #

Second moment of area comes in three flavours, and the collection has so far needed two. About a horizontal axis, $I_x = \int y^2\,dA$; about a vertical one, $I_y = \int x^2\,dA$; and the third is the product of inertia,

$$I_{xy} = \int xy\,dA$$

which is not a second moment about anything. It measures how much of the section sits in the diagonally opposite quadrants — positive if the material clusters in the first and third quadrants, negative in the second and fourth. Reflect the section about either axis and every $xy$ term changes sign, so any section with an axis of symmetry has $I_{xy} = 0$, which is why it has never appeared.

The three together form a tensor, and the practical consequence of that word is the one the next figure makes: the second moment of area is a function of direction.

The two directions where $I_{uv} = 0$ are the principal axes, and they are the ones the section really has. For an equal angle they are at 45°, which is the angle of the axis of symmetry the shape does have — and that is the general rule: an axis of symmetry is always a principal axis, and a section with no symmetry has two principal axes anyway, at some angle that has to be computed.

The curve is a sinusoid in $2\theta$ and the algebra is Mohr’s circle in a different subject:

$$I_u = \frac{I_x + I_y}{2} + \frac{I_x - I_y}{2}\cos 2\theta - I_{xy}\sin 2\theta$$

with the same construction, the same invariants and the same trap. This site has met the circle once already, for stress, and the shared mathematics is not a coincidence: both quantities are symmetric second-order tensors, and everything true of one transformation is true of the other.

Where the neutral axis actually goes #

The stress formula for a moment applied about the horizontal axis, with no symmetry available, is

$$\sigma = \frac{M(I_y\,y - I_{xy}\,x)}{I_xI_y - I_{xy}^2}$$

and the neutral axis is where that vanishes: the line $y = (I_{xy}/I_y)\,x$, which passes through the centroid and is tilted. For the angle it is at −30.6° to the horizontal.

That tilt is the whole phenomenon. The neutral axis is by definition the axis the section bends about, and a section bending about an axis at −30.6° moves perpendicular to it — down and sideways. The deflection direction and the neutral axis are always at right angles, and neither of them is aligned with the load.

The zed moves further sideways than down #

The last cell in that figure is the one worth pausing on, because it is not a curiosity — it is the section most roof purlins in the world are made from.

A Z-purlin loaded vertically has $I_x = 21.12$, $I_y = 4.11$ and $I_{xy} = 6.84 \times 10^6$ mm⁴, and the ratio of sideways to downward movement is $-I_{xy}/I_y = -1.66$. It moves 166% as far across as it moves down. A purlin sagging 20 mm under snow has moved 33 mm sideways, out of the roof plane, in a direction nobody designed for.

This is the reason a Z-purlin is never used bare. The roof sheeting is fixed to its top flange and restrains it laterally, which converts the problem into one the section is good at — and the design of that connection is then load-bearing in a way that looks cosmetic on a drawing — a brace whose job is stiffness rather than strength, in a field that does not usually call it one. Take the sheeting off during a re-roofing and the purlins are suddenly members with a 13.8 : 1 stiffness ratio being loaded 59° away from their strong axis.

The general lesson generalises past purlins: a section chosen for how efficiently it can be rolled or folded is not a section chosen for the axes it will be loaded about, and the two considerations have no reason to agree. Cold-formed sections are the extreme case because their shapes are dictated by what a rolling line can do to a coil of steel.

The resolution that makes it tractable #

There is a way to make an unsymmetric section behave, and it is the reason principal axes are worth finding rather than merely knowing about.

Resolve the applied moment into components about the principal axes. About those axes the product of inertia is zero by construction, so the ordinary formula works separately for each, and the answer is the sum of two well-behaved bending problems:

$$\sigma = \frac{M_u\,v}{I_1} + \frac{M_v\,u}{I_2}$$

For the equal angle, a vertical moment $M$ resolves into $M/\sqrt2$ about each principal axis, and the second of those acts on $I_2 = 0.734 \times 10^6$ mm⁴ — a quarter of the first’s. That is where the large stresses come from: a section loaded off its principal axes puts a share of the moment onto its weakest direction, and the weak direction of an unsymmetric section is very weak indeed.

The parallel-axis theorem has a third line #

The theorem this collection has used a dozen times to shift a second moment from a piece’s own centroid to the section’s has a companion for the product of inertia, and it is the one that makes an unsymmetric section computable at all:

$$I_{xy} = \sum \left( I_{xy,\text{own}} + A\,d_x d_y \right)$$

For a rectangle with its sides parallel to the axes, the first term is zero — a rectangle is symmetric about both of its own centre-lines — so the whole product of inertia of a built-up section comes from the $A d_x d_y$ terms. It is entirely a fact about where the pieces are, and not at all about what shape they are.

That the terms can cancel is what makes symmetry so effective. The two flanges of a channel sit at $+d_y$ and $-d_y$ with the same $d_x$, their contributions are equal and opposite, and the total is exactly zero. The two flanges of a zed sit at $+d_y$ and $-d_y$ with opposite $d_x$, their contributions add, and the total is the 6.84 × 10⁶ that tilts everything.

The neighbouring effect it is not #

Two distinct things happen to a channel loaded vertically through its web, and they are constantly confused, so it is worth separating them precisely.

Unsymmetric bending is what this essay is about: it needs $I_{xy} \ne 0$, it produces a tilted neutral axis, and a channel loaded about its strong axis does not have it, because a channel is symmetric about its horizontal axis.

Twisting from a load off the shear centre is a different effect entirely: it needs the load’s line of action to miss a particular point, it produces a rotation rather than a tilt, and a channel loaded through its web does have it.

An angle suffers both, which is why it is the worst common section to load carelessly and why an angle used as a beam is nearly always either restrained or paired with a second one back to back — an arrangement that restores symmetry, zeroes the product of inertia, and puts the shear centre back on the axis in one move.

Two numbers that do not move #

There is a check on all of this that costs nothing and catches the commonest arithmetic mistake, and it is worth carrying because it applies to every rotation of every symmetric tensor.

Rotating the axes changes $I_u$, $I_v$ and $I_{uv}$, and it leaves two combinations of them alone. The trace $I_x + I_y$ is the same at every angle — for the equal angle it is 3.600 × 10⁶ mm⁴ at 0° and at 45° and at every angle between. So is the determinant $I_xI_y - I_{xy}^2$, which for the same section is 2.104 × 10¹² mm⁸ whichever axes it is computed in.

The first of those has a physical reading worth having: $I_x + I_y = \int (x^2 + y^2)\,dA$, which is the polar second moment about the centroid — a quantity with no direction in it at all, being about a point rather than an axis. It cannot change when the axes turn because it never referred to them.

The invariants are how a computed rotation is checked without repeating it. Compute $I_1$ and $I_2$, add them, and compare with $I_x + I_y$: for the zed that is 23.529 + 1.701 against 21.120 + 4.110, and both come to 25.230 × 10⁶. An error in the rotation angle breaks that sum immediately, and an error in a sign — which is the mistake everybody makes with the product of inertia — breaks the determinant.

This is the same apparatus that Mohr’s circle provides graphically: the centre of the circle is the trace over two and its radius comes from the determinant, so the two invariants are the two numbers that fix the circle, and every angle is a point on it.

Why it was so late to arrive #

Bending theory was complete for symmetric sections by the 1820s, and the unsymmetric case waited nearly a century for a reason that is entirely practical: nobody built with unsymmetric sections. A masonry arch, a timber beam and a cast-iron girder are all symmetric about a vertical plane, and the first structural members that are not are rolled and cold-formed steel shapes — angles, channels and zeds — which arrive with the industrialisation of steel sections at the end of the nineteenth century.

The theory that answered it is the same rotation Cauchy had written for stress in 1822, applied to a different tensor. That is the ordinary way this subject advances: the mathematics is finished long before the objects that need it exist, and the delay is in the rolling mill rather than in the analysis. The same is true of the shear centre, which is a 1920s idea about sections that had been in use for thirty years.

What the picture cannot show #

Everything above is elastic and small-displacement. Once the section has moved sideways the load is no longer where the analysis put it, and for a slender member the sideways movement feeds back — which is the second-order effect that turns unsymmetric bending into lateral-torsional buckling for members long enough.

The moment is applied and the load is not drawn. A real load arrives at some point on the section and produces both a moment about the centroid and a torque about the shear centre; the figures resolve only the first, which is the standard treatment and is exact only when the load happens to act through the shear centre. Where it does not, the leftover is a torque and the member twists.

Restraint is absent everywhere. Almost every unsymmetric section in a real structure is restrained by something — sheeting, a slab, a paired member — and the restrained problem is a completely different one, in which the sideways movement is prevented and the restraint force becomes the design quantity. The figures show what the section would do left alone, which is a case that exists mainly during construction.

Where the ladder goes #

The first rung is the buckling case. A section with $I_2$ four times smaller than $I_1$ is a section with a weak axis, and a compression member’s critical load is set by the smallest second moment there is — which for an unsymmetric section is about a principal axis nobody drew, at an angle nobody quoted.

The second is what happens when the two effects of the last section combine on one member: bending about a tilted neutral axis while twisting about a shear centre outside the metal. That is the general behaviour of a thin-walled open section and it is the reason cold-formed design has a code of its own.

The third is the practical one. The whole apparatus above exists to compute the stress in a section loaded off its axes, and the ordinary engineering answer is to avoid being in that position — restrain the member, pair it, or use a section whose principal axes are the ones the load arrives on. The most useful thing to know about unsymmetric bending is which sections have it, and the check is a single number: is the product of inertia zero?