Sections and stress

The middle third

A material that cannot be pulled imposes a condition on where the load may land, and the condition is a region rather than a point. For a rectangle it is the famous middle third; for every other section it is a shape nobody quotes, and one ordinary section's is nearly twice as generous as the rule allows.

Assumes The material far from the middle does nearly all the work, Bending is a pair of forces, pushing and pulling and Weight is the only thing resisting it.

Most of this collection is about materials that can be pulled. Steel is as good in tension as in compression, reinforcement exists precisely so that concrete can be, and every bending calculation drawn so far has a tension side that carries stress.

Take that away — masonry, unreinforced concrete, a plate bearing on grout, a footing on soil — and a new question appears that has no analogue in the rest of the subject. The material cannot pull, so the resultant of everything pressing on the section has to land somewhere that leaves no part of it in tension. That somewhere is a region, and it has a name.

The middle third, computedThe kern of a 400 × 600 mm rectangle, computed by asking, for every direction, how far the resultant can move before the far face would be pulled. It reaches ±100.0 mm vertically and ±66.7 mm horizontally, which are h/6 and b/6 exactly, and the region between is a rhombus rather than the ellipse the two numbers suggest — a resultant on a diagonal has less room than either axis allows.rectangle±100 of 600 mm33.3% of the depth
Fig. 1 The kern of a 400 × 600 mm rectangle, computed by asking, for every direction, how far the resultant can move before the far face would be pulled. It reaches ±100.0 mm vertically and ±66.7 mm horizontally — h/6 and b/6 exactly — and the region between is a rhombus rather than the ellipse those two numbers suggest. On the diagonal it reaches only 56.7 mm.

The vertical half-width is the middle third, arriving as a piece of section arithmetic rather than as a rule:

emax=ZA=bh2/6bh=h6e_{\text{max}} = \frac{Z}{A} = \frac{bh^2/6}{bh} = \frac{h}{6}

which is a third of the depth, centred. The same ZZ that decides how much moment a section can take decides how much eccentricity it can tolerate, and the second use is the one nobody teaches. Everything else in this essay follows from noticing that Z/AZ/A is a property of a section, so it has a different value for every shape, and that the argument runs in every direction and not only the two the drawing has axes for.

Why it is a rhombus and not an ellipse

The condition to be satisfied is that the stress at every point of the section stays compressive. For a resultant at (ex,ey)(e_x, e_y) that stress is

σ(x,y)1+exxry2+eyyrx2\sigma(x,y) \propto 1 + \frac{e_x x}{r_y^2} + \frac{e_y y}{r_x^2}

with r2=I/Ar^2 = I/A the radius of gyration squared. Each point of the boundary supplies one inequality, so the kern is the intersection of a half-plane per boundary point — and the intersection of half-planes is a polygon.

For a rectangle only the four corners bind, so the kern has four sides. Its vertices are on the axes at b/6b/6 and h/6h/6, and the sides run straight between them, which is why the diagonal is the tight direction: a resultant on the diagonal is being watched by a corner, and a corner is further from the centroid than a face is.

The practical consequence is the one the refutation states. A check that reads “the eccentricity is within b/6 across and within h/6 along, so it is inside the middle third” is checking a rectangle when it should be checking a rhombus, and the two differ by up to 38% on the diagonal. Biaxial eccentricity is not two uniaxial checks, which is exactly the same statement an interaction diagram makes about combined actions and is missed for exactly the same reason.

Every section has one and they are not alike

Every section has one, and they are not alikeThe kern of three sections, shaded: the region a compressive resultant has to land in if no part of the section is to go into tension. A rectangle's is a rhombus reaching a sixth of the depth, 16.7% of it; a circle's is a disc of a quarter of its radius; an I-section's is 1.77 times the rectangle's in the strong direction and much smaller across it. The shape follows from the section's own radii of gyration and nothing else — no material property enters anywhere.rectangle±100 of 600 mm33.3% of the depthcircle±75 of 600 mm25.0% of the depthI-section±177 of 600 mm58.9% of the depth
Fig. 2 The kern of three sections, shaded. A rectangle’s is a rhombus reaching a sixth of the depth — 16.7% of it either way, a third in total. A circle’s is a disc of a quarter of its radius, so its kern is 25% of the diameter, narrower than the rectangle’s. An I-section’s is 1.77 times the rectangle’s in the strong direction and much smaller across it.

The I-section is the surprise, and the reason is instructive. Its kern half-depth is Z/AZ/A, and an I-section is precisely the shape that maximises ZZ for a given AA — putting the material at the extremes is the whole point of it. The kern is therefore enormous: ±176.6 mm on a 600 mm section, which is 58.9% of the depth against a rectangle’s 33.3%.

An I-section on a bearing that cannot pull is far more tolerant of an eccentric load than a solid one of the same depth. That is the opposite of what the phrase “middle third” leads people to expect, and it is worth stating as a rule: the kern is generous exactly where the section is efficient in bending, because both are measured by Z/AZ/A.

Across the section the same efficiency works against it. The I’s horizontal kern is ±39.0 mm against the rectangle’s ±66.7, because almost all its area is at the middle of its width. A section that is excellent in one direction is poor in the other, and here that shows up not as a stiffness but as a tolerance.

Outside the kern, the arithmetic changes kind

The kern is not a strength limit. Nothing fails when the resultant crosses it. What happens is that part of the section stops working, the section that remains is smaller than the one that was designed, and every stress in it is computed from a different set of equations.

The pressure runs away outside the middle thirdPeak bearing pressure under a 4 × 3 m base carrying 900 kN, against the eccentricity of the load. Inside the middle third the line is straight and the pressure has doubled by the time it reaches the edge of it: 75 kPa at the centre, 150 kPa at e = B/6. Beyond that the base lifts, the contact length shortens, and the curve turns upward without limit — at e = 1.54 m the peak is 433 kPa on 1.38 m of base.00.20.40.60.811.21.40100200300400eccentricity of the resultant (m)peak pressure (kPa)B/6: the base is on the point of liftinguniform: 75 kPa
Fig. 3 Peak pressure against eccentricity for a 4 × 3 m base carrying 900 kN. Inside the middle third the relation is a straight line and the pressure has exactly doubled by the edge of it — 75 kPa at the centre, 150 at e = B/6. Beyond that the contact length is 3(B/2 − e) and shrinks toward zero, so the curve turns upward and has a vertical asymptote at the edge of the base.

The factor of two at the kern boundary is worth remembering, because it is the same for every section: at the kern the stress block is a triangle from zero to σmax\sigma_{\max}, whose average must be N/AN/A, so the peak is 2N/A2N/A whatever the shape. Reaching the edge of the kern doubles the peak stress, and the design decision about whether to allow it is usually made on that basis rather than on the uplift itself.

Past it, the unknown is the contact length, and it is found by requiring the pressure block’s centroid to sit under the resultant. For a rectangle that gives c=3(B/2e)c = 3(B/2 - e) and a peak of 2N/Dc2N/Dc — an expression that goes to infinity as the resultant approaches the edge, which is the mathematics saying that a rigid body on a rigid surface balanced on its corner carries infinite stress.

The pressure under a base that cannot pullBearing pressure under a 4 × 3 m base carrying 900 kN, at four eccentricities. Inside the middle third — ±0.67 m here — the pressure is a trapezoid and the whole base is working. Beyond it the base lifts: at e = 1.00 m only 3.00 m of the 4 m is in contact and the peak pressure is 200 kPa against 75 kPa at no eccentricity.e = 0.00 m75 kPawholly in bearinge = 0.33 m113 kPawholly in bearinge = 0.67 m150 kPawholly in bearingthe middle third, exactlye = 1.00 m200 kPa1.00 m lifted
Fig. 4 The same four states drawn as pressure blocks under the base. The first three are trapezoids on the whole width; the fourth has lifted over a quarter of it. The transition is not gradual in the quantity that matters — the moment the far edge reaches zero pressure, the length of base doing the work starts to shrink.

Where the same idea keeps arriving

The kern is one of the few pieces of arithmetic in this collection that appears identically in four fields, and it is worth collecting the appearances because the vocabulary hides the identity.

A whole body being blown over. Uplift at the windward edge begins exactly when the resultant leaves the base’s middle third, and the ratio between that height and the toppling height is √3.

A masonry arch. The line of thrust must stay inside the middle third of the ring for the joints not to open, which is the same condition applied section by section along a curve.

A line of thrust, and the masonry it has to stay insideAn arch ring of 9% of the span in thickness, rising 28% of the span, under its own weight as a uniform load. Any horizontal thrust between 3.85 and 5.23 puts a line of compression entirely inside the masonry, so the arch stands — and which of them it actually takes is not decided by statics. The two extremes are drawn: the minimum-thrust line, which rides high at the crown and low at the haunches, and the maximum-thrust line, which does the opposite.thrust anywhere from 3.85 to 5.23 fitsH = 3.85, leastH = 5.23, most
Fig. 5 The thrust line in an arch ring: the locus of the resultant’s position at every section, searched for over the range of horizontal thrusts that keep it inside the stonework. Each section is a kern problem, and a thrust line touching the extrados is a joint that has opened at the intrados — a hinge, not a failure.

A base plate on grout. The plate is a section that cannot pull, the holding-down bolts are what happens when it does, and the moment at which they start working is the moment the resultant leaves the plate’s kern.

A base plate, and when the bolts start workingA 500 × 400 mm plate carrying 600 kN and 90 kN·m, so the resultant sits 150 mm from the centre against a kern of 83.33 mm. The plate is in partial contact: bearing over 300 mm at a peak of 10 N/mm², with the holding-down bolts carrying 0 kN. The plate lifts at 50 kN·m and crushes at 120 kN·m, and the bolts are not needed until 150 kN·m.600 kN90 kN·mresultant at e = 150middle thirdbolt carries nothingPartial contactbearing over 300 mm at 10 N/mm² · 50% of 20
Fig. 6 A column base under 600 kN and 90 kNm. The eccentricity is 150 mm against a kern of 83.3 mm for a 500 mm plate, so the resultant is outside and the bolts on the tension side are carrying load — the arrangement changes from a bearing problem into an anchorage one, at a threshold the plate’s own geometry sets.

A prestressed beam near its ends. With no applied moment to help, the requirement that the top fibre stay in compression is exactly the requirement that the prestressing force land inside the kern.

The zone the tendon has to stay insideThe eccentricities that keep the top fibre out of tension at transfer and the bottom fibre out of tension in service, along a 12 m beam. The two limits cross the section at different rates, and the parabolic profile drawn between them is the tendon: 220 mm at midspan, where the zone is 148 mm deep, and on the centroid at the ends, where a tendon left low would crack the top of a beam carrying nothing but itself.024681012-300-200-1000100200300distance along the span (m)eccentricity below the centroid (mm)above this line the top cracks at transferbelow this line the bottom cracks in service
Fig. 7 The zone a tendon has to lie in along a beam. At midspan the applied moment lets the force sit far outside the kern — 220 mm against a kern of 116.7 — and the zone is 148 mm deep. At the support the applied moment is zero, the kern is all there is, and the tendon has to come back to the centroid.

Four fields, four vocabularies, one inequality. That is worth stating because it is how a subject is actually organised: the useful unit is not the topic but the condition, and the resultant must land inside Z/A is a condition that reaches from a Roman arch to a post-tensioned floor.

Reading it backwards, as a way of sizing things

The kern is usually met as a check: compute the eccentricity, compare, worry. Read the other way it is a sizing rule, and in that direction it is one of the fastest pieces of arithmetic in structural engineering.

A footing under a column carrying NN with a moment MM has an eccentricity e=M/Ne = M/N, fixed before anything is chosen. Requiring the resultant to stay inside the kern then gives the base length directly:

MNB6B6MN\frac{M}{N} \le \frac{B}{6} \quad\Longrightarrow\quad B \ge \frac{6M}{N}

so a column with 900 kN and 300 kNm needs a base at least 2.0 m long if the whole of it is to stay in contact. That is a dimension arrived at in one line, with no soil property, no concrete grade and no reinforcement in it — and it is usually within a size or two of the final answer, because the eccentricity is the thing that decides base plan dimensions and the bearing pressure is the thing that decides them only when the moment is small.

The same reading sizes a wall thickness for a given thrust eccentricity, a base plate for a given column moment, and the prestress a beam needs at a section where the tendon must be near the centroid. In each case the quantity that goes in is a ratio of two actions, which is often known long before either action is, and that is why the rule survives as a design instinct: eccentricity over depth is a dimensionless number an engineer can carry in their head, and a sixth is the value at which the arithmetic changes.

The rule is older than the mechanics

The middle third predates every piece of theory in this essay. It appears in eighteenth- and nineteenth-century masonry practice as a rule about joints — keep the thrust within the middle third and the joint will not open — and it was stated, argued over and applied long before Navier’s bending theory gave it the derivation above in the 1820s.

That order of events explains the rule’s peculiar status. It was arrived at empirically, from the observation that arch and wall joints open on the side away from the thrust, and it was then found to be exactly Z/AZ/A for a rectangle — which is a much stronger statement, because it generalises to every other shape and to sections nobody had built with. The empirical rule and the derived one agree for the case that generated it and diverge everywhere else, which is the usual fate of a good rule of thumb.

There is a second reason it stuck. Masonry design in the nineteenth century had no way to compute stresses in a curved structure, and the middle third gave a geometrical criterion — draw the thrust line, see whether it stays in the middle third of the ring — that could be executed with a straightedge. That is drawing as calculation in its purest form, and it is why the rule outlived several generations of the theory that justified it.

Heyman’s twentieth-century reworking made the modern position clear: the middle third is sufficient for a masonry structure to be safe and it is not necessary, because a joint that opens is a hinge and a structure with hinges can still be perfectly stable. The rule is conservative, it was known to be conservative, and it was kept because the alternative required knowing how much hinge rotation the structure could take — which is the same trade the safe theorem makes everywhere it is used.

The one section whose kern contains nothing

There is a case that reads as a paradox and is not. A thin ring — a chimney, a silo, a tubular pile — has almost all of its area at the extreme fibre, so its Z/AZ/A is close to R/2R/2, and the kern is a disc of half the radius. That is by far the most generous kern of any shape, and it is why a chimney can take an enormous wind moment without uplift.

The opposite extreme is a section whose material is concentrated at its centroid, for which Z/AZ/A goes to zero. A cable is the limiting case: its kern is a point, it cannot take any eccentricity at all, and that is precisely the statement that a cable can carry no bending. The kern is a measure of how much bending a no-tension section can absorb before it stops being a section, and it runs from a point for a cable to half the radius for a thin tube.

Between them, the ordinary shapes: 33% of the depth for a rectangle, 25% of the diameter for a solid circle, 59% for an I-section, and the limits of both extremes reached by things that are barely sections at all.

What the picture cannot show

Every figure here assumes a rigid section on a rigid support. A real base plate bends, a real footing is flexible relative to the soil under it, and the linear pressure distribution the whole calculation rests on is an idealisation whose error grows with the flexibility. The kern boundary is robust — it depends only on where the resultant is — but the peak pressure past it is not.

The material is assumed to have unlimited compressive strength. In practice the peak pressure runs into a bearing limit long before the contact length approaches zero, and what actually happens beyond the kern is a plastic block of limited stress and finite length rather than the triangle drawn. The base-plate figure uses that treatment; the others use the elastic one, and they disagree once the eccentricity is large.

Nothing here has a safety factor in it. The kern is a geometrical boundary and not a limit state: designs routinely and deliberately operate outside it, with the consequences computed rather than avoided. What the boundary is good for is knowing which calculation applies, and that is a question with a yes-or-no answer.

Where the ladder goes

The first rung is the arch, where the same condition applied continuously along a curve becomes a thrust line and a collapse mechanism — and where the interesting question is not whether a line exists but how many can.

The second is the section that has left its kern and is working on a reduced area: the cracked no-tension section, whose depth of compression is itself an unknown and which behaves nonlinearly under a load that is applied linearly.

The third is the one this essay has kept close to the surface. Everything above is about a material property — the inability to be pulled — expressed as a geometric region. That translation is what makes the rule usable by somebody with a drawing and no calculation, and it is why the middle third has survived as a rule of thumb for four hundred years while the mechanics behind it was being rewritten twice.

What this makes readable

Essays that name this one as a prerequisite.

What links here

Every essay whose body links to this one.

The objects this essay names

Each one links to every other essay that touches it.

Bearing pressureEccentricityKernNo tension materialSecond momentSection modulusThrust lineUplift