Sections and stress

Loaded straight down, and it moves sideways

Every section drawn here so far had an axis of symmetry, and that symmetry has been doing silent work. Take it away and a vertical load produces a neutral axis that is not horizontal, a deflection that is not vertical, and on one ordinary section a sideways movement larger than the downward one.

Assumes The material far from the middle does nearly all the work, Bending is a pair of forces, pushing and pulling and The point that is not in the section.

The formula every beam in this collection has used is σ=My/I\sigma = My/I, and it contains a quiet assumption that has never been stated: that II is a single number, that yy 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.

Loaded straight down, and moving sideways. An equal angle with a moment applied about the horizontal axis. Its principal axes lie at 45.0° to the drawn ones, so the neutral axis runs at -30.6° rather than horizontally, and the section moves 59% as far sideways as it moves down. The product of inertia that causes it is -1.066 × 10⁶ mm⁴, and it is zero for every section drawn in this field until now.
Fig. 1 A 100 × 100 × 10 equal angle under a moment applied about the horizontal axis. Its principal axes lie at 45° to the drawn ones, so the neutral axis runs at −30.6° rather than horizontally, and the section moves 59% as far sideways as it moves down. The product of inertia that causes it is −1.066 × 10⁶ mm⁴, and it is zero for every section drawn in this field until now.

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, Ix=∫y2 dAI_x = \int y^2\,dA; about a vertical one, Iy=∫x2 dAI_y = \int x^2\,dA; and the third is the product of inertia,

Ixy=∫xy dAI_{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 xyxy term changes sign, so any section with an axis of symmetry has Ixy=0I_{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 second moment of area is a function of direction. Second moment of area of an equal angle against the angle of the axis it is taken about, with the product of inertia beneath it. The maximum is 2.866 × 10⁶ mm⁴ and the minimum 0.734 × 10⁶, a ratio of 3.90, and they occur where the product of inertia passes through zero — at 45.0° from the drawn axis. The value the drawing suggests, 1.800 × 10⁶, is neither of them.
Fig. 2 Second moment of area of the same angle against the angle of the axis it is taken about, with the product of inertia beneath it. The maximum is 2.866 × 10⁶ mm⁴ and the minimum 0.734 × 10⁶, a ratio of 3.90, and they occur exactly where the product of inertia crosses zero — at 45.0°. The value the drawing suggests, 1.800 × 10⁶, sits between them at an angle nothing physical happens at.

The two directions where Iuv=0I_{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 2θ2\theta and the algebra is Mohr’s circle in a different subject:

Iu=Ix+Iy2+Ix−Iy2cos⁡2θ−Ixysin⁡2θI_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

σ=M(Iy y−Ixy x)IxIy−Ixy2\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=(Ixy/Iy) xy = (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.

Four sections, and how far each moves sideways. The same vertical load on four profiles, with the neutral axis each produces drawn through its centroid. A rectangle and a channel are symmetric about a horizontal axis, their product of inertia is zero, and they deflect straight down. An angle and a zed have no such axis: their neutral axes are tilted, and they move sideways by a fraction of their vertical movement that is a property of the shape alone.
Fig. 3 Four profiles under the same vertical load, with the neutral axis each produces drawn through its centroid. A rectangle and a channel are symmetric about a horizontal axis, their product of inertia is zero, and their neutral axes are horizontal. The angle’s is tilted by 30.6° and the zed’s by 59.0°, which is what a product of inertia of 6.84 × 10⁶ mm⁴ does to a section whose Ix is 21.1.

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 Ix=21.12I_x = 21.12, Iy=4.11I_y = 4.11 and Ixy=6.84×106I_{xy} = 6.84 \times 10^6 mm⁴, and the ratio of sideways to downward movement is −Ixy/Iy=−1.66-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.

Loaded straight down, and moving sideways. An zed purlin with a moment applied about the horizontal axis. Its principal axes lie at -19.4° to the drawn ones, so the neutral axis runs at 59.0° rather than horizontally, and the section moves 166% as far sideways as it moves down. The product of inertia that causes it is 6.840 × 10⁶ mm⁴, and it is zero for every section drawn in this field until now.
Fig. 4 The zed alone, with its principal axes and neutral axis. The principal axes are at −19.4° and the neutral axis at 59.0°, so a vertical load bends the section about an axis nearer the vertical than the horizontal. Its minimum second moment is 1.70 × 10⁶ mm⁴ against a maximum of 23.5 — a ratio of 13.8, the largest in this figure set.

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:

σ=Mu vI1+Mv uI2\sigma = \frac{M_u\,v}{I_1} + \frac{M_v\,u}{I_2}

For the equal angle, a vertical moment MM resolves into M/2M/\sqrt2 about each principal axis, and the second of those acts on I2=0.734×106I_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 resolution is a statement about the load rather than about the section, which means it does not need an unsymmetric section to bite. Take a doubly symmetric I-section, whose product of inertia is exactly zero and whose principal axes are the two lines anybody would have drawn, and apply the moment five degrees away from the plane of its web. The rotation above is unnecessary, the ordinary formula applies to each component with no algebra at all, and the answer is still not a vertical deflection.

The neutral axis obeys neither the load nor the moment. A 305 × 102 mm I-section carrying a moment 5° out of the plane of its web. The moment vector is the short arrow; the neutral axis is the long line, at 69.7° to the strong axis. They do not line up, and the reason is that the neutral axis follows the moment ratio scaled by the stiffness ratio: tan α = (M_z/M_y)(I_y/I_z), and I_y ÷ I_z is 30.8 here. So a 5° tilt of the load puts the neutral axis 70° over, the corner that ends up furthest from it carries 489 N/mm² against the 258 the straight-down case would give, and the section has lost 47 per cent of its capacity to a misalignment nobody would draw on a detail.
Fig. 5 A 305 × 102 mm I-section carrying a moment 5° out of the plane of its web. The moment vector is the short arrow and the neutral axis the long line, at 69.7° to the strong axis. The neutral axis follows the moment ratio scaled by the stiffness ratio — tan α = (M_z/M_y)(I_y/I_z), and I_y ÷ I_z is 30.8 for this section — so the corner that ends up furthest from it carries 489 N/mm² against the 258 the straight-down case would give.

Nothing in that figure is a product of inertia. The neutral axis swings because the section answers the two components of one moment with two very different stiffnesses, and the tilt of the axis is the moment ratio multiplied by the stiffness ratio rather than the moment ratio alone. Five degrees of load becomes seventy degrees of neutral axis. The angle earlier in this essay had its axis tilted by its geometry; this one has its axis tilted by its loading, and the two arrive at the same place — a section bending about a line nobody drew.

What the misalignment costs is worth reading off a sweep rather than at a single angle, because the growth near zero is nothing like linear.

Five degrees is not five per cent. The corner stress on a 305 mm I-section, against how far the load is off the plane of the web, as a multiple of the stress with the load straight down. The two terms add — M cos θ ÷ Z_y and M sin θ ÷ Z_z — and Z_y ÷ Z_z is 10.3 for this section, so the weak-axis term catches the strong-axis one at only 5.5°. At the 5° drawn the stress is 1.90 times what a designer who ignored the tilt would have computed. A purlin on a roof pitch, a crane girder taking a lateral surge, a beam whose bearing is not quite level: none of them is five per cent worse than the calculation done for it.
Fig. 6 Corner stress on the same 305 mm I-section against how far the load lies off the plane of the web, as a multiple of the stress with the load straight down. The two terms add — M cos θ ÷ Z_y and M sin θ ÷ Z_z — and Z_y ÷ Z_z is 10.3 for this section, so the weak-axis term catches the strong-axis one at only 5.5°. At the 5° drawn the stress is 1.90 times what a calculation that ignored the tilt would have reported.

A purlin following a roof pitch, a crane girder taking a lateral surge, a beam whose bearing has been packed slightly out of level: none of them is five per cent worse than the calculation done for it. And the two mechanisms compound rather than compete, because the cold-formed sections that have a product of inertia are also the ones most often mounted on a slope.

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:

Ixy=∑(Ixy,own+A dxdy)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 AdxdyA d_x d_y terms. It is entirely a fact about where the pieces are, and not at all about what shape they are.

Moving the flanges apart. The second moment of area of an I-section against its depth, with the flange and web areas held constant. The growth is close to quadratic, because the parallel-axis term dominates everything the flanges contribute about their own centres.
Fig. 7 The parallel-axis theorem drawn for the second moment: each piece contributes its own value plus its area times the square of its offset. The product of inertia works the same way with the square replaced by the product of the two offsets — which can be negative, so pieces can cancel, and a section can have a large second moment and no product of inertia at all.

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

The sections that have no direction at all

The invariants have a degenerate case, and it is more useful than degenerate cases usually are.

If Ix=IyI_x = I_y and Ixy=0I_{xy} = 0, the Mohr circle has zero radius: it collapses to a point, and the second moment is the same about every axis through the centroid. Every axis is a principal axis, there is no strong direction and no weak one, and a load in any direction bends the section straight along itself.

A circle does it, obviously. A square does it, which surprises people: the second moment of a square about a diagonal is a4/12a^4/12, exactly the same as about a centre-line, and the shape that visibly has corners has no directional preference at all. So does an equal-legged cruciform, a regular hexagon, and in fact any regular polygon.

Two design consequences follow and both are worth having.

A column of such a section has no weak axis. A compression member buckles about the smallest second moment it has, and for these sections there is no smallest — every direction is equally resisted, so the section’s full stiffness is available whichever way the member decides to go. That is the real argument for a circular or square hollow section as a column, and it is stronger than the usual one about torsion: an I-section wastes most of its material buckling about an axis it is bad at, and a square tube has no such axis to waste it on.

And a beam of such a section deflects along the load, whatever direction the load arrives from. A member carrying an unpredictable load direction — a bracing member, a hanger, a member in a space frame, a mast — has no orientation to get wrong, which removes a whole class of erection error.

The handedness of a zed, and where its push goes

The product of inertia carries a sign, and the sign flips when the section is mirrored. A zed and its mirror image have equal and opposite IxyI_{xy}, identical second moments about both axes, and sideways deflections in opposite directions under the same vertical load.

That is a real decision on a roof rather than a curiosity. Purlins are laid all the same hand, because a zed nests with another zed of the same hand at a lap and does not nest with its mirror. So every purlin on a roof pushes the same way — 166% of its vertical deflection, all in one direction — and the sheeting, the sag rods and eventually the eaves beam have to absorb the accumulated push of every purlin on the slope.

Lay them in alternating hands and the pushes cancel in pairs, and the sheeting between two adjacent purlins carries the difference rather than the sum. That arrangement is structurally better and is not built, because the laps do not work and the erection sequence doubles in complexity.

So the standard detail is the one with the larger restraint demand, chosen for constructability, and the restraint is provided by elements — sheeting fixings, sag rods, an eaves tie — whose sizing is the residue of a sign in an integral that no purlin schedule mentions. It is a good example of a second-order consideration deciding a first-order component: the purlin is sized by bending and the roof is held together against a lateral force that exists because a shape has a handedness.

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 Ixy≠0I_{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.

The shear centre of a channel. A channel of 80 by 200, with the shear flow in its flanges drawn. Those flows form a couple, so the load has to be applied 31.7 outside the web to leave the section untwisted — a point in the air, outside the material entirely.
Fig. 8 The channel again, with the point a load must pass through if the section is not to twist. This has nothing to do with the product of inertia — the channel’s is zero — and everything to do with where the shear flows in the flanges put their resultant. A section can suffer either effect, both, or neither.

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 IuI_u, IvI_v and IuvI_{uv}, and it leaves two combinations of them alone. The trace Ix+IyI_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 IxIy−Ixy2I_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: Ix+Iy=∫(x2+y2) dAI_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 I1I_1 and I2I_2, add them, and compare with Ix+IyI_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 I2I_2 four times smaller than I1I_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?

What this makes readable

Essays that name this one as a prerequisite.

Named alongside this one

Essays reaching for the same objects. Nobody chose these; they are what the concept index makes visible.

What links here

The 8 essays that link to this one and share the most of its objects, of 11 that link here.

The objects this essay names

Each one links to every other essay that touches it.

Bending stressBiaxial bendingNeutral axisPrincipal axesProduct of inertiaSecond momentShear centreUnsymmetric bending