Two strengths, depending which way up

A symmetric section has one section modulus, and the word “one” in that sentence is doing more work than it looks. $Z = I/c$ needs a distance $c$ from the neutral axis to the extreme fibre, and a symmetric section has the same distance to both faces, so the ambiguity never arises and the habit of thinking of $Z$ as a property of the shape takes hold.

An asymmetric section has two of them, and they can differ by a great deal.

More second moment, and less strength #

Start with the fact that makes the subject worth an essay rather than a footnote.

The tee above has an area of 3,000 mm² and a depth of 200 mm, exactly like the rectangle beside it. Its second moment of area is $11.23 \times 10^6$ mm⁴ against the rectangle’s $10.00 \times 10^6$ — twelve per cent more. Its elastic section modulus is $75.3 \times 10^3$ mm³ against $100.0 \times 10^3$ — a quarter less.

The two facts are the same fact seen twice. Concentrating material near one face moves the centroid toward it, and moving the centroid toward one face moves it away from the other. Second moment of area rewards the concentration; section modulus divides by the distance to the furthest fibre, and that distance has just grown.

$$I = 11.23 \times 10^6 \;\uparrow\,, \qquad c_{\max} = 149\;\text{mm} \;\uparrow\,, \qquad Z = \frac{I}{c_{\max}} \;\downarrow$$

So the tee is the stiffer member and the weaker one, at the same weight and the same depth. Anything that reads a section catalogue looking for the largest $I$ has just chosen the beam that will yield first.

The factor of forty between four shapes of equal area is a real ranking and it is a ranking on $I$; the ranking on $Z$ is not the same ranking, and for asymmetric shapes it is not even close.

Which free body produced the number #

The two moduli are one arithmetic performed twice, and it is worth doing once explicitly.

Cut the beam. On the cut face the direct stress varies linearly with distance from the neutral axis, because plane sections stay plane and nothing else. The resultant of that stress distribution has to be zero force and a moment $M$, which is the push and the pull the moment is made of.

Zero net force puts the neutral axis at the centroid, which is a statement about the first moment of area, and it puts it at 149 mm from the web tip for this section. The stress at any fibre is then $\sigma = Mc/I$, and the two extreme fibres are at $c = 149$ and $c = 51$.

Set the larger of those two stresses to the yield stress and the moment that does it is $M = f_y I / 149 = f_y Z_{bot}$. Set the smaller and the moment is $f_y Z_{top}$, which is 2.92 times larger and completely unavailable — the section will have yielded at the other face long before.

Only the smaller modulus is a strength. The larger one is a number about a fibre that never governs, for a moment of that sign.

It is worth noticing what the construction of $I$ itself does not contain. The second moment is assembled strip by strip, each strip’s area times the square of its distance from the axis, and the parallel-axis theorem that assembles it for a built-up shape rewards distance from the centroid symmetrically. Neither operation knows where the extreme fibre is, and neither can tell the two sides of the section apart, because both square the distance before they add it. That is precisely why the number they produce is a stiffness and not a strength, and why an argument that stops at the second moment has stopped one division too early.

Turn it over and the answer changes #

The sign of the moment decides which face is in compression, and for an asymmetric section that decides which modulus governs.

A tee used as a simply supported beam sags: the flange is on top, in compression, and the web tip at the bottom is the extreme tension fibre at $c = 149$. The governing modulus is 75.3.

The same tee over a support hogs: the flange is now in tension and the web tip is in compression, but the distances have not moved — $c = 149$ is still the larger. The governing modulus is still 75.3.

So the sign of the moment does not change the elastic strength of a section made of one material. What changes it is turning the section over, which exchanges which physical face is far from the centroid. A tee with its flange at the bottom of a sagging beam has its flange as the extreme tension fibre at $c = 149$ and its web tip in compression at $c = 51$ — the same two numbers attached to different pieces of steel.

That distinction is academic in a steel beam, where both faces yield at the same stress, and it is the whole design in three cases where they do not:

Concrete, whose compressive strength is an order of magnitude above its tensile one, so the arrangement that puts the small $c$ on the tension face is disastrous rather than merely inefficient.

Cast iron, whose tensile strength is about a quarter of its compressive, which is why every cast-iron beam ever made has an enormous bottom flange and a small top one — a tee upside down. The proportions were arrived at empirically before anybody could compute them, and they are the correct answer to this arithmetic.

Any composite arrangement where the two faces are different materials, where the transformed section does the same job with an extra step.

The plastic neutral axis is somewhere else entirely #

Push the section past first yield and the neutral axis moves — not gradually toward some limit, but to a completely different geometric definition.

At 60% of its plastic moment the outer fibres of the web have yielded, the flange is still elastic, and the axis has already begun to leave the centroid. It does not stop there.

Elastically, zero net force means the first moment of area about the axis vanishes, which puts it at the centroid. Fully plastic, the stress is $\pm f_y$ everywhere, so zero net force means the areas above and below are equal. Those are different conditions and they give different answers for any section that is not symmetric.

For this tee: the flange is 1,680 mm² and half the total area is 1,500 mm², so the equal-area axis lies inside the flange, 12.5 mm down from the top — at 187.5 mm from the web tip, against the centroid’s 149.0.

The neutral axis moves 38.5 mm, which is 19% of the section’s depth, entirely because of how hard the section is being pushed. It is not a property of the shape.

The plastic modulus that follows is $S = 134.3 \times 10^3$ mm³, and putting it beside the elastic one gives a shape factor of

$$\frac{S}{Z_{\min}} = \frac{134.3}{75.3} = 1.78$$

against 1.50 for the rectangle and 1.09 for the I-section.

The two sections’ journeys from first yield to full plasticity differ in exactly that way: the tee starts lower and finishes closer, because the reserve between the two is what a poor elastic section has and a good one does not.

The bonus is the size of the earlier penalty #

A shape factor of 1.78 reads as a merit. It is arithmetically the opposite.

A shape factor is a price quoted as a bonus in general, and for an asymmetric section the quotation is at its most misleading. The tee’s plastic modulus, 134.3, is 89% of the rectangle’s 150.0. Its elastic modulus, 75.3, is 75% of the rectangle’s 100.0. So the tee is 11% weaker than the rectangle plastically and 25% weaker elastically — and the whole of its impressive shape factor is the difference between those two shortfalls.

Plastic capacity very nearly does not care about asymmetry. That is the useful form of the result: at full plasticity the far fibre has no special status, every fibre is at $f_y$, and the only thing that matters is how the area is distributed about the equal-area axis. The elastic calculation is the one that punishes a lopsided shape, and it punishes it for a reason that disappears the moment the material yields.

Which is a genuine argument for plastic design on asymmetric sections and a genuine trap: the reserve is only available if the section can reach it, and a tee’s slender outstanding web in compression frequently cannot.

Both calculations rest on plane sections staying plane, which is what makes the strain distribution linear in the first place. Neither the elastic argument nor the plastic one on this page survives without it, and for a stubby member neither is true.

Add an axial force and both axes move again #

The interaction between axial force and moment makes the asymmetry visible in a second dimension. A rectangle’s interaction curve is symmetric about the moment axis because the section is; a tee’s is not, and the compression it can carry alongside a sagging moment is not the compression it can carry alongside a hogging one.

An axial force adds a uniform stress to the block, so the elastic zero moves by $N/A$ divided by the stress gradient — and the plastic axis moves so that the area difference above and below carries $N$. Both moves happen from different starting points and at different rates.

The consequence for a tee is that its interaction diagram is lopsided, and that a member carrying compression and a moment has four combinations to check rather than two. Nothing about this is exotic: a tee cut from a universal beam is one of the commonest members in a truss, it carries axial force and a moment from the loads applied between its nodes, and both signs of moment occur along its length.

The same asymmetry appears in the axial problem with no moment applied at all. The kern is the region within which a compressive force can be applied without producing tension anywhere, and its boundary sits at $Z/A$ from the centroid in each direction — which is the two moduli again, divided by one area. For a symmetric section the kern is centred; for a tee it is not, and its two halves are as different in size as its two moduli are, in the ratio 2.92 to one.

Where an asymmetric section is the right answer #

Nothing above is an argument against lopsided sections. It is an argument that they are chosen for a reason, and that the reason is never the second moment of area.

Where the two faces are made of different things. A composite floor beam is a steel section with a concrete slab on top of it, and the whole point of the arrangement is that the concrete is in compression and the steel in tension. The transformed section is violently asymmetric — its neutral axis often sits inside the slab — and that is the design working rather than failing.

Where the two faces meet different demands. A crane girder carries a vertical load from the wheel and a horizontal surge from the crab, and the horizontal load acts at the top flange. So a channel is welded along the top of an I-section: the result is asymmetric about the horizontal axis, badly so, and it is the correct member because the two loads arrive at different places.

Where a member is continuous and the moment reverses. A plate girder over several spans has a hogging region and a sagging one, and the flange that governs is different in each. Unequal flanges, or a flange that changes thickness along the span, are the honest response — and they turn a single member into two sections joined by a butt weld, each asymmetric, each right where it is.

And where the material forces it. Cast iron’s tension-to-compression ratio of about one to four asks for a section whose small $c$ is on the compression face, which is a tee upside down. That the ratio of the two moduli should approximately match the ratio of the two strengths is the design rule, and it says that the right asymmetry is not a compromise but a match — the point at which both faces reach their own limit at the same moment, which is the only arrangement in which no material is wasted.

The stiffness does not care which way up it is #

There is one quantity in the whole argument that is indifferent to the asymmetry, and noticing it separates two checks that are usually made together.

A section has one second moment of area about a given axis. Turn the tee over and $I$ is unchanged, because it is an integral of $y^2$ and squaring destroys the sign. So the deflection under a given moment, the natural frequency, the buckling load about that axis and the share of load the member attracts in a redundant structure are all identical either way up.

Only the stress differs, because the stress needs $y$ rather than $y^2$, and $y$ has a sign.

Which means an asymmetric section splits its two checks cleanly. Serviceability does not know the section is asymmetric at all. Strength knows about nothing else. A beam that fails its deflection check will fail it whichever way it is installed, and a beam that fails its stress check may pass simply by being turned over — a remedy available in no other part of this collection.

It also says which mistake is recoverable. Installing an asymmetric member upside down is a stress problem and not a stiffness one, so a member found the wrong way up on site is either fine or is a strength failure with the deflection unchanged, and the diagnosis is a section modulus rather than a survey.

And it is only worth having if the moment has one sign #

The whole benefit of an asymmetric section is that it puts more material where the material is needed, which presupposes knowing which face that is. Reverse the moment and the arrangement is exactly wrong.

For the tee here the two elastic moduli differ by a factor of 2.92. A section arranged so that both fibres reach their limits together under a sagging moment reaches neither under a hogging one: the fibre that was comfortably inside its limit is now the extreme one, at 2.92 times the stress it had, and the capacity in the reverse direction is a third of the capacity in the intended one.

And the penalty grows with the very proportion that makes the section worth choosing.

So an asymmetric section is a member for statically determinate, one-directional bending — a simply supported beam under gravity, a cast-iron girder in a Victorian mill, a precast lintel. It is a poor member for anything continuous, anything carrying uplift, anything seismic, and anything where a construction stage reverses the moment before the permanent condition establishes it.

That last case is the one that catches people, because it is temporary. A precast tee-beam lifted by two points inboard of its ends is hogging over the lifting points while the finished member sags between its supports, and the section that was arranged for the second is upside down for the first. The handling condition and the service condition load an asymmetric section in opposite directions, and only one of them was designed.

Where the model stops #

Everything here is bending about one axis. A tee has a second axis, about which it is symmetric, and a load applied in neither principal direction gives a beam that moves sideways when it is pushed down. An angle is the section where that matters most and it is asymmetric about both axes.

The plastic argument assumes the whole section can yield. A tee’s web is an outstand supported along one edge only, and if the web tip is in compression it is very likely to buckle locally first — at which point the section cannot reach its own strength and the shape factor is a number about a member that does not exist.

The two moduli are elastic quantities, so they carry the elastic assumptions: no residual stress, no yielding anywhere, and a material with the same modulus in tension and compression. Cast iron fails the last of those, which makes the very case the asymmetry was invented for the case where the arithmetic above is least exact.

And the shear centre has been ignored throughout. A tee’s shear centre is at the junction of the flange and the web, not at the centroid, so a load applied through the centroid twists it — which is a point that is not in the section doing its usual damage to an argument that only looked at bending.

What the pictures cannot show #

The stress blocks are drawn at a scale chosen to make both triangles visible. In the elastic figure the two peak stresses differ by the factor 2.92, and drawing them to a common scale makes the small one nearly invisible — which is the honest picture and a poor illustration.

The plastic figure shows the neutral axis at 182.6 mm, which is where it has reached at 99% of the plastic moment and not where it ends. It cannot show the path: the axis migrates from 149.0 to 187.5 mm as the moment rises, so a drawing at any single moment ratio is one frame of a movement that has no single answer, and the frame chosen is the one nearest the end.

Nor can the two ranking figures show why they disagree, only that they do. Both are computed from one second moment of area, assembled the same way from the same strips; the disagreement is entirely in what each of them divides that number by, and a division is the one thing a bar chart has no way to draw.

The ladder from here #

Later rungs on this anchor: the angle, asymmetric about both axes, where the principal axes are inclined and nothing decomposes conveniently. The cast-iron beam as a designed object, and the ratios its makers arrived at without the arithmetic. Composite and reinforced sections, where the asymmetry is in the material rather than in the geometry and the transformed section restores it. The tee in a truss, carrying axial force and both signs of moment. The plastic section modulus of a section with a hole in it, where the equal-area axis moves and the elastic one barely does. And the deep question underneath all of it: why the neutral axis is defined by two different conditions at two different load levels, and what that says about treating it as a property of a shape at all.