Bending is a pair of forces, pushing and pulling

A bending moment arrives at a cross-section as a single number, in units of force times length, with no obvious physical picture attached. What the section does with it is much more concrete: it pushes on one side and pulls on the other, and the two forces form a couple that matches the moment.

That is all bending is. The push and the pull are equal, so there is no net force; they are separated, so there is a moment. Everything else — the linear stress distribution, the section modulus, why depth matters — follows from working out how a section arranges that push and pull.

Where the stress comes from #

The chain has three links and no gaps.

Geometry. Bending a beam curves it. Fibres on the outside of the curve get longer, fibres on the inside get shorter, and one surface keeps its length. If cross-sections stay flat as the beam bends — the central assumption — then the change in length is proportional to distance from that surface, so the strain is linear in $y$.

Material. Elastic material has stress proportional to strain. So the stress is also linear in $y$: zero at the neutral axis, greatest at the extreme fibres, opposite in sign above and below.

Equilibrium. The stresses have to add up to the applied moment and to no net force. Zero net force places the neutral axis at the centroid of the area. The moment condition then gives

$$\sigma = \frac{My}{I}.$$

Three sentences, and the only assumption doing real work is the first one.

The couple, and where its arm is #

The stress block is triangular on each side, so the resultant of each half acts at the centroid of a triangle — two-thirds of the way out from the neutral axis.

For a rectangle of depth $d$, each resultant sits at $d/3$ from the axis, so the lever arm between push and pull is $2d/3$. The compressive resultant is the average stress times the area of the compressed half, which is $\sigma_{\max}/2 \times bd/2$.

Multiplying gives $M = \sigma_{\max} bd^2/6$, so $Z = bd^2/6$, which is the familiar result obtained without any integration. It is a useful way to hold the section modulus in mind: a fraction of the peak stress, over a fraction of the area, at a fraction of the depth.

The same reasoning explains the I-section immediately. Move nearly all the area to the extreme fibres and two things improve at once — the average stress in the material approaches the peak instead of being half of it, and the lever arm approaches the full depth instead of two-thirds. An ideal I-section with all its area in two thin flanges has $Z = A d/2$, which is fifty percent better than a rectangle of the same area and depth.

The tee is the case that catches people. Its stem reaches much further from the neutral axis than its flange does, so the stem’s extreme fibre governs. Turn the load the other way and the flange becomes the tension face, the governing distance changes, and the capacity is quite different. A section with one axis of symmetry has one capacity per direction of bending.

Tension, compression, and which material cares #

The two sides of the stress block are not equivalent, because materials are not.

Steel is nearly identical in tension and compression, so a symmetric section makes sense and the extreme fibre on either face governs equally.

Concrete is strong in compression and useless in tension, so the tension side of the block is assumed to have cracked and to carry nothing at all. Reinforcement is placed where the tension would have been, and the couple becomes a concrete compression block against a steel tension force. The lever arm survives; the material on one side of it is replaced.

Timber is stronger in tension than compression along the grain — the opposite of most expectations — so a timber beam usually crushes on its compression face before it tears on its tension face, unless a knot intervenes.

Masonry has no tensile capacity at all, which is a material fact with a geometric consequence, which is why an arch has to keep its thrust line inside the section: the moment a stress block would require tension anywhere, the joint opens instead.

Past the elastic limit #

The linear stress block holds only while the material is elastic, and steel is not, past yield.

When the extreme fibre reaches yield, the section has reached its elastic capacity $M_y = \sigma_y Z$. Nothing dramatic happens. The outer fibres stop taking more stress and start taking more strain, while the material further in continues to load up. The stress block spreads from a triangle toward a rectangle.

When the whole section has yielded, the block is fully rectangular and the section can take no more. That moment is the plastic moment, and for a rectangle it is exactly 1.5 times the elastic one — the shape factor. For an I-section, where the material was already near the extremes, the factor is only about 1.14, because there was less under-used material in the middle to recruit.

There is a second change hiding inside that ratio, and it is not a change of magnitude. The elastic block bends about the centroid, which is the axis that makes the first moment of area vanish. The fully rectangular block bends about the axis that makes the two areas equal, because a uniform stress on each side gives a force proportional to area alone. On a symmetric section those two axes are the same line and the distinction never has to be made. On a section with one axis of symmetry they are not the same line, and the shape factor is then a ratio of two moduli taken about two different axes.

The tee in that figure carries the point further. Its two elastic section moduli differ by a factor of 2.78, because one extreme fibre is far further from the centroid than the other; its plastic modulus is a single number, because a fully yielded section has no memory of which fibre reached yield first. An asymmetric section has two elastic moduli and one plastic one, and that asymmetry is where the largest shape factors in the set come from.

The consequences are large. A steel beam does not fail when its extreme fibre yields; it forms a plastic hinge at that section and continues to carry load while rotating. In a redundant structure that hinge lets moment redistribute to less-stressed regions, and collapse requires enough hinges to form a mechanism. Plastic design counts hinges rather than stresses, and it exists because the elastic stress block understates what a ductile section can do.

Reading the diagram back to the beam #

The stress block is a property of a section; the moment that fills it comes from the whole beam.

Combining the two — the diagram along the beam and the block across the section — gives the design statement in one sentence: the section modulus must be at least the peak moment divided by the permitted stress, everywhere along the beam. Where the moment is small, a smaller section would do — which is what a haunch, a plate curtailment or a tapered rafter is exploiting.

The lever arm, when half the material has gone #

The couple picture earns its keep most in reinforced concrete, where the elastic stress block is abandoned entirely and the couple survives.

Concrete cracks in tension at a low stress and is then assumed to carry none. So the free body of the cut face has, on the tension side, no concrete at all — only the reinforcing bars — and on the compression side a block of concrete near its crushing strength. Two forces, one lever arm, exactly as before.

The force equation locates the neutral axis. With a compressive stress $f_c$ acting over a depth $a$ of a section $b$ wide, and steel of area $A_s$ at a stress $f_y$,

$$f_c\,b\,a = A_s f_y \quad\Longrightarrow\quad a = \frac{A_s f_y}{f_c\,b}.$$

Note what has just happened: the depth of concrete in compression is set by the amount of steel in tension. Adding reinforcement pushes the neutral axis down, because more tension force requires more concrete to balance it.

The moment equation gives the capacity. The compression resultant acts at the centroid of the block, $a/2$ from the top, so the lever arm to the steel at effective depth $d$ is $z = d - a/2$, and

$$M = A_s f_y\left(d - \frac{a}{2}\right).$$

The useful fact hiding in that expression is how little the lever arm moves. For ordinary proportions, $a$ comes out between about a tenth and a quarter of $d$, so $z$ lies between $0.95d$ and $0.87d$ — a range of under ten per cent across the whole practical range of reinforcement. Capacity is therefore very nearly proportional to the steel area alone, and $M \approx 0.9\,A_s f_y d$ is accurate enough to size a beam in one line on the back of a drawing.

That insensitivity is why reinforced concrete design remained a hand calculation long after the analysis it fed had been computerised, and it is a direct consequence of the couple: the lever arm is a geometric quantity, bounded above by the section depth and below by the requirement that the concrete not be crushed, and there is not much room between the two.

What an axial force does to the couple #

A beam has a moment. A column has an axial force. Almost every real member has both, and the couple picture handles the combination in a way that produces one genuinely counter-intuitive result.

Adding a compressive axial force shifts the whole stress distribution downward: the compression side grows, the tension side shrinks, and the neutral axis moves toward the tension face. For a steel section that is straightforwardly bad news, because the extreme compression fibre reaches yield sooner, and the capacity in bending falls as the axial force rises. The interaction is monotonic, and a member carrying half its squash load has appreciably less than half its bending capacity left.

For a concrete section it is not monotonic, and the reason is that the tension zone was contributing nothing anyway. Compression from an axial load partly closes the crack, restoring concrete to the tension face where it can now resist. Up to a point — the balance point, where the concrete crushes and the steel yields simultaneously — adding axial compression increases the moment the section can carry. Above it, the concrete runs out and the capacity falls away steeply.

The resulting interaction diagram is therefore a curve that bulges outward, with its widest point at some intermediate axial load rather than at zero. A concrete column with a modest axial load is a better beam than the same section with none, which is a sentence that reads as an error and is not.

Two consequences follow that shape into practice. Prestressing is this effect used deliberately: applying compression to a concrete member before the load arrives moves the whole section into a state where far more of it participates in bending. And a concrete column that loses its axial load — during demolition, or when the storeys above are removed — can be less able to carry moment than it was when more heavily loaded, which is a genuine hazard in refurbishment and one that no ordinary intuition about load anticipates.

The couple has to be delivered #

There is one more thing the stress block does not show, and it becomes visible only when the moment is allowed to change along the beam.

In pure bending — the constant-moment region between two symmetric point loads — the compression resultant in the top of the section is the same at every station. Nothing has to move. Away from that region the moment varies, so the compression force at one station differs from the compression force a short distance along, and the difference has to be delivered into the flange from somewhere.

The somewhere is the web, and the mechanism is horizontal shear. A flange carrying $600$ at one station and $620$ a little further along has had $20$ handed to it through the plane where it joins the web, and that plane is under shear stress for exactly that reason.

This is why a timber beam built up from planks laid loose on top of each other is enormously weaker than the same planks glued: unglued, each plank forms its own small couple with its own small lever arm; glued, the whole depth participates and the glue lines carry the horizontal shear that makes it one member. Sliding at the interfaces is visible at the ends of a loaded stack of planks, and it is the failure the glue prevents.

The distribution of that shear across the section is a separate calculation with a shape of its own — peaking where the bending stress is zero, which is the neutral axis, and vanishing at the extreme fibres where the bending stress peaks. The two quantities a cut reveals are complementary in exactly that sense: neither is largest where the other is.

Where the stress block comes from #

The linear distribution is not an assumption about materials. It is a consequence of an assumption about geometry.

Cut the section into thin strips and the chain is visible one link at a time. Plane sections staying plane makes the strain in each strip proportional to its distance from the axis; an elastic material makes the stress proportional to the strain, and so linear in distance too; and summing each strip’s force times its distance is the operation whose result is called the second moment of area. Nothing in that sequence is a statement about steel or concrete or timber except the middle link, and the middle link is the one most easily replaced.

The whole derivation rests on cross-sections remaining flat as the beam bends, which is very accurate for a slender member and progressively wrong as it gets deeper. A beam shorter than about twice its depth does not obey it at all.

That is the argument for the I-section stated in terms of the block rather than the integral: an efficient section is one where the average stress is close to the peak, because the material near the neutral axis is being wasted.

The shape factor, split into its two halves #

Reading a moment as a force times an arm makes the plastic reserve decompose into two effects that pull in opposite directions, which no single ratio shows.

Take a rectangle of depth $d$ and width $b$ at a yield stress $\sigma$.

Elastic. Each half carries a triangular stress block, so the resultant is $\tfrac12\sigma(d/2)b = \sigma bd/4$ and it acts at the triangle’s centroid, two thirds of the way out — so the two resultants are $2d/3$ apart. The moment is $\sigma bd^2/6$.

Fully plastic. Each half carries a rectangular block, so the resultant has doubled to $\sigma bd/2$, and it acts at the centroid of a rectangle, which is only $d/4$ from the axis — so the arm has fallen to $d/2$. The moment is $\sigma bd^2/4$.

The force went up by a factor of two and the lever arm went down by a quarter, and the product went up by 1.5.

That is the shape factor, and reading it this way says where it comes from: the gain is entirely in recruiting material and it is partly given back by recruiting material that is close to the axis. A section with little material near the neutral axis — an I-section — has less to recruit and loses less arm, which is why its shape factor is 1.09 rather than 1.5, and why the two effects nearly cancel there.

Where the model stops #

Plane sections stay plane. The assumption behind the linear strain distribution. It fails for deep beams, near concentrated loads, and where shear is large relative to bending.

Linear elastic material. Covered above; the plastic reserve is real and is deliberately excluded from the elastic calculation.

Bending about a principal axis. Load a section about any other axis and it bends in two directions at once. For an angle, whose principal axes are diagonal, the everyday loading direction is not principal, and an angle loaded vertically deflects sideways.

No axial force. A beam-column has a stress block that is the sum of a uniform stress and a bending one, and the neutral axis moves off the centroid — possibly off the section entirely.

No twisting. If the load does not pass through the shear centre, the section twists as well as bends, and a channel loaded through its web is the standard example.

The figures share a specific limitation: the stress block is drawn as a shape beside the section, which suggests the stress lives outside the material. It does not — the block is a graph of stress against height, plotted sideways, and its horizontal axis is a stress rather than a distance. Nothing about the section is that wide.

The ladder from here #

Later rungs: plane sections and what the assumption costs. Elastic and plastic section moduli, and shape factors. The plastic hinge and collapse mechanisms. Unsymmetric bending and principal axes. Combined bending and axial force, and the interaction diagram. Composite sections. Cracked reinforced concrete. Prestress, which puts the stress block in backwards before the load arrives. And the shear stress distribution, which peaks exactly where the bending stress does not.

Galileo’s 1638 analysis of a cantilever assumed the whole depth was in tension about the bottom edge, giving a capacity three times too high. The error stood for seventy-five years, and it is the reason the neutral axis is worth naming.