Connections

Where the structure meets the ground, and when the bolts start working

Push a base plate off its middle third and it lifts off the foundation. The holding-down bolts then carry exactly nothing, and go on carrying nothing until the plate has crushed the concrete underneath it.

Assumes The connection is not a point, and every diagram on this site says it is and The hinge put in on purpose.

A column arrives at the ground and has to be joined to it. The joint is a steel plate welded to the column, sitting on grout, with bolts cast into the concrete.

Everything about that arrangement is unlike the connections above it, and the reason is one sentence: the interface can push and cannot pull.

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. 1 A 500 × 400 plate carrying 600 kN and 90 kN·m, so the resultant sits 150 mm from the plate’s centre against a middle third of ±83.3 mm. The plate has lifted — bearing over 300 mm rather than 500 — and the holding-down bolt is carrying nothing at all. What is climbing is the pressure under the compressed end.

The one sentence, and what it makes determinate

A base plate on a foundation is a rigid body resting on a surface that resists compression and offers no tension. That is a unilateral contact problem, and it is the reason base plates behave unlike anything else in this field.

The consequence is that the plate cannot be analysed by assuming a stress distribution. The distribution has to be found, subject to being non-negative everywhere, and it is the requirement of non-negativity that makes the answer determinate: given NN and MM, there is exactly one distribution of bearing pressure over exactly one contact length that satisfies both equilibrium equations without going negative anywhere.

Three regimes come out of that, and the middle one is the result.

It is worth noticing how unusual that determinacy is. Every other joint in this field is statically indeterminate at the level of the connection: the division of load between bolt rows, between weld segments, between friction and bearing, all require an assumption about stiffness or a plastic redistribution to settle. Here, two equilibrium equations and one inequality settle the whole distribution with nothing else supplied. The base plate is the only connection on this site whose internal force distribution can be found from statics alone, and the reason is that its material has been forbidden from doing one of the two things a material can do.

Regime one: the middle third

While the resultant of NN and MM falls within the middle third of the plate, the whole plate bears and the stress distribution is linear:

σ=NA±MZ\sigma = \frac{N}{A} \pm \frac{M}{Z}

The minimum stress is zero exactly when e=M/N=L/6e = M/N = L/6, which is the middle-third rule — and it is the same rule, arrived at from the same argument, that keeps a thrust line inside the middle third of a masonry arch’s ring. Masonry cannot take tension; nor can a foundation interface. The identical constraint produces the identical kern.

For the plate above, L/6=83.3L/6 = 83.3 mm and the moment at which the plate first lifts is N×L/6=50N \times L/6 = 50 kN·m.

Regime two, which is the interesting one

Push past e=L/6e = L/6 and the plate lifts off at one edge. The instinct is that the holding-down bolts now do something.

They do not. Not at ee slightly past L/6L/6, and not at ee twice that.

What happens instead is that the contact length shrinks. The bearing pressure becomes a triangle over a length yy at the compressed end, and equilibrium is satisfied by putting that triangle’s resultant directly under the applied resultant — which requires

y=3(L2e)y = 3\left(\frac{L}{2} - e\right)

with no tension anywhere. A compression-only distribution exists, so no tension is needed, so none develops. The bolts are along for the ride.

A base plate, and when the bolts start workingA 500 × 400 mm plate carrying 600 kN and 40 kN·m, so the resultant sits 66.67 mm from the centre against a kern of 83.33 mm. The plate is in full contact: bearing over 500 mm at a peak of 5.4 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 kN40 kN·mresultant at e = 66.67middle thirdbolt carries nothingFull contactbearing over 500 mm at 5.4 N/mm² · 27% of 20
Fig. 2 The same plate at 40 kN·m, inside the kern. The whole plate bears, the pressure distribution is a trapezoid from 5.4 down to 0.6 N/mm², and nothing is lifting. This is the only one of the three regimes in which the familiar N/A±M/ZN/A \pm M/Z applies.

What is happening in regime two is that the peak pressure climbs, and it climbs fast:

σmax=2NBy\sigma_{\max} = \frac{2N}{By}

with yy shrinking towards zero as eL/2e \to L/2. So the pressure goes to infinity while the bolt force stays at exactly zero — which is the clearest possible statement that the two questions are unrelated.

Where regime two actually ends

The formal end of regime two is e=L/2e = L/2, where the resultant leaves the plate entirely and no compression-only distribution can balance it. For this plate that is e=250e = 250 mm and M=150M = 150 kN·m.

But the plate does not get there. The pressure reaches the concrete’s bearing limit first, at

e=L22N3Bfjde = \frac{L}{2} - \frac{2N}{3Bf_{jd}}

which for N=600N = 600 kN, B=400B = 400 mm and fjd=20f_{jd} = 20 N/mm² is e=200e = 200 mm, or M=120M = 120 kN·m.

So the three boundaries for this plate are:

event ee MM
plate first lifts (middle third) 83.3 mm 50 kN·m
concrete reaches its bearing limit 200 mm 120 kN·m
bolts would first be needed 250 mm 150 kN·m

The governing boundary is the middle one, and it is the one that has no rule of thumb attached. The middle-third rule is famous and marks an event with no structural consequence. The bolt-tension threshold is what everybody expects to govern and never gets reached. The limit that actually decides is a bearing check on the grout.

A base plate, and when the bolts start workingA 500 × 400 mm plate carrying 600 kN and 150 kN·m, so the resultant sits 250 mm from the centre against a kern of 83.33 mm. The plate is in bolts engaged: bearing over 172.56 mm at a peak of 20 N/mm², with the holding-down bolts carrying 90.23 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 kN150 kN·mresultant at e = 250middle third90.23 kNBolts engagedbearing over 172.56 mm at 20 N/mm² · 100% of 20
Fig. 3 Regime three, at 150 kN·m. Now the resultant is outside the plate, a compression-only distribution is impossible, and the holding-down bolt is genuinely in tension — 90.2 kN of it, with the compression block 172.6 mm long at the limit pressure. Everything about the calculation has changed: it is now two equations in two unknowns rather than a geometric construction.

What “rigid plate” is doing, and when it fails

Every result above assumes the plate is rigid — that it rotates as a body and delivers whatever pressure distribution the geometry requires. That assumption is doing a great deal of work and it is worth knowing when it stops.

A base plate is a cantilever spanning out from the column’s flanges over a bearing pressure. If it is thick relative to that projection it distributes the pressure the way the analysis says. If it is thin, it bends: the pressure concentrates under the column’s own footprint and falls away towards the plate’s edges, and the effective bearing area is smaller than the plate.

The design response is the effective area method, which abandons the whole-plate calculation and instead takes an area consisting of the column section plus a fringe of width cc around it, where cc is chosen so that the plate at that projection is exactly at its own bending capacity:

c=tfy3fjdc = t\sqrt{\frac{f_y}{3 f_{jd}}}

That formula is the plate’s cantilever moment set equal to its plastic moment, solved for the projection, and it is a tidy inversion: instead of checking whether a plate of given thickness can distribute a pressure, it computes how much plate is participating and uses only that.

Two consequences worth carrying. A thicker plate genuinely spreads the load further, so plate thickness buys bearing area rather than merely bending capacity. And for a plate on the thin side, the middle-third and lift-off analysis above is being applied to an area that is not the drawn one — which is a reminder that the three regimes are exact for a rigid plate and approximate for a real one.

The grout, which is the weakest thing in the load path

fjdf_{jd} has appeared as a number throughout and it deserves a paragraph, because it is not the concrete’s strength and the difference matters.

Under the plate there is a bedding layer of grout, typically 25 to 50 mm thick, whose job is to take up the gap between a plate set at a nominal level and a foundation cast to a construction tolerance. Structurally it is the weakest material anywhere in the column’s load path — weaker than the steel by a factor of forty, and generally weaker than the concrete below it.

What saves it is confinement. A block of grout squeezed between a steel plate above and a large concrete mass below cannot spread sideways, and a material prevented from spreading can carry far more than its uniaxial strength. The bearing value used in design is therefore larger than the grout’s cylinder strength, sometimes substantially, and the enhancement is a function of the ratio between the loaded area and the foundation’s area.

That is the same mechanism that lets a masonry joint carry more than its mortar’s strength, and it has the same weakness: it depends on the surrounding material being there. A base plate near the edge of a foundation has confinement on three sides rather than four, and the enhancement it is designed with may not exist.

Why this is the lower-bound theorem again

The whole analysis above is an application of the same theorem that runs through the masonry arch and block shear, and seeing it as one thing is worth more than the individual results.

The lower-bound theorem says: if a distribution of internal forces can be found that is in equilibrium with the applied load and nowhere exceeds the material’s strength, the structure carries that load.

For the base plate, the distribution has to satisfy two equilibrium equations, be non-negative everywhere, and stay below fjdf_{jd}. The construction above finds one. It is not necessarily the real one — the plate is not perfectly rigid, the grout is not perfectly plastic, and the real pressure distribution is something else — but it does not have to be, and that is exactly the theorem’s point.

Which is why the same question that is unanswerable for the arch — what is this structure actually doing? — is unanswerable here and equally unnecessary.

Reading the three regimes as one curve

Put the three regimes on one axis — bolt tension against applied moment, at constant NN — and the shape is a flat line at zero followed by a rising one, with the corner at e=L/2e = L/2.

That is an unusual shape for a structural response and it is worth naming why. Most quantities in this subject are proportional, or at worst piecewise proportional: double the load, double the moment, roughly double the stress. Bolt tension in a base plate is exactly zero over a finite range and then rises steeply, which means it has no useful sensitivity anywhere. A designer cannot reason about it by scaling, and a small change in NN moves the corner rather than the slope.

The reason is that the bolt force is the residual of two much larger quantities. It is CNC - N, the difference between a compression block and an axial load, and both are of order hundreds of kilonewtons where the difference is zero or tens. A residual of two large numbers is exactly the quantity that behaves discontinuously and that is worst to estimate — the same property that makes a moment computed as the difference between two large end moments the thing worth being careful about.

Which leads to the practical instruction. Do not interpolate bolt tension. Compute which regime the plate is in, and then compute within that regime. A designer who has the bolt force at 150 kN·m and wants it at 120 knows nothing useful, because at 120 the answer is not proportionally smaller — it is zero.

A base plate, and when the bolts start workingA 500 × 400 mm plate carrying 600 kN and 120 kN·m, so the resultant sits 200 mm from the centre against a kern of 83.33 mm. The plate is in partial contact: bearing over 150 mm at a peak of 20 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 kN120 kN·mresultant at e = 200middle thirdbolt carries nothingPartial contactbearing over 150 mm at 20 N/mm² · 100% of 20
Fig. 4 The boundary that actually governs, at 120 kN·m. The bearing block has shrunk to 150 mm and the peak pressure has reached the grout’s 20 N/mm² limit, so the plate is at its capacity — with the resultant still inside the plate and the holding-down bolts still carrying nothing whatever.

What the bolts are for, since it is not moment

If holding-down bolts spend most of their life carrying nothing, it is worth asking why they are there. There are four answers and none of them is the one the calculation above tests.

Erection. A column has to stand up before anything is connected to it, and four bolts and a plate are what holds it plumb while the frame is assembled. This is the reason there are at least four, always, whatever the analysis says.

Uplift. A light structure under wind can have net tension at its base — no NN at all, or a negative one — and then the bolts are the entire load path. The analysis above assumes N>0N > 0 and does not apply.

Shear. Base shear is carried by friction under the plate, by the bolts in shear, or by a shear key. Friction needs NN, so a column in uplift loses its friction at exactly the moment its shear demand is highest.

Robustness. A base with no tension capacity has no reserve against anything the analysis did not include — an impact, a settlement, an accidental load case, or a load reversal the design combinations happened not to contain.

So the bolts are structural for reasons that are mostly not about the moment they are checked against, and the check that dominates their sizing in an ordinary building is a load case in which NN is small rather than large.

Nominally pinned bases, which are not

A base detailed with four bolts inside the column’s flanges is called a pinned base and drawn as a pin. It is not one, for exactly the reasons the joint-stiffness essay gave — and the consequence at a base is larger than elsewhere, because a frame’s sway is very sensitive to base fixity.

The conventional allowance is to analyse a nominally pinned base as pinned for member forces and as having some stiffness for sway — commonly 10% of the column’s EI/LEI/L — and there is no pretence that this is anything but a calibrated fudge. What it recognises is that a real base plate on real grout has a moment–rotation curve like every other joint, and that the two idealisations bracket it.

The base is also where the unintended redistribution has the largest reach, because a frame’s sway stiffness enters its second-order amplification, which multiplies everything. A base assumed pinned that is partly fixed makes the frame stiffer than modelled, which is safe; a base assumed fixed that is partly pinned does the reverse, and that error propagates into every column in the frame.

What to take from it

A base plate rests on something that cannot pull, and that single constraint makes the problem determinate and produces three regimes.

The middle-third rule marks a lift-off with no structural consequence. Past it, the plate lifts and the bolts still carry nothing, because a compression-only distribution still exists.

The governing boundary is a bearing check nobody names. 120 kN·m here, against 50 for the kern and 150 for the bolts — the two famous numbers bracket the one that decides.

And it is the same theorem as the masonry arch. Any admissible distribution proves the load can be carried, so the question of what the plate is really doing does not have to be answered.

A base plate, and when the bolts start workingA 500 × 400 mm plate carrying 600 kN and 240 kN·m, so the resultant sits 400 mm from the centre against a kern of 83.33 mm. The plate is in bolts engaged: bearing over 247.57 mm at a peak of 20 N/mm², with the holding-down bolts carrying 390.27 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 kN240 kN·mresultant at e = 400, outside the platemiddle third390.27 kNBolts engagedbearing over 247.57 mm at 20 N/mm² · 100% of 20
Fig. 5 Well into regime three, at 240 kN·m. The resultant is 400 mm from the plate’s centre and well outside it, so the bolts are carrying 390 kN of tension against a compression block of 990. Both numbers are large and both are the difference of larger ones, which is why this regime cannot be interpolated into from the one below it.
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. 6 The same rule, three fields away. A masonry arch stands if some line of compression can be drawn inside its stonework; a base plate bears if some non-negative pressure distribution can be found under it. Both are the lower-bound theorem, both refuse to answer what the structure is really doing, and both have a middle third in them for the same reason.

Where the field ends

This is the last of the fifteen, and the base plate is a fitting place to stop because it is the connection at which the structure stops being a structure.

Every joint above it transfers force between two members that were designed by the same person out of the same material with the same theory. This one hands the load to concrete, and then to soil, neither of which obeys any equation in the seven fields that precede this one on this site. The plate is the boundary of the model.

What the field has done is take one sentence — the connection is not a point — and follow it through fifteen distances on drawings. The eccentricity from a bracket to a bolt group. The distance from a web face to a bolt, which generates prying. The end distance that decides between tear-out and bearing. The stagger that makes a tear diagonal. The connection length that decides how much of a section is working. The lever arm between bolt rows. The kern of a base plate.

None of those appears in a frame analysis. Every one of them decided an answer. And the recurring result is the one the sections field established at the beginning of this site and that has now appeared in six settings: where the material sits decides more than how much of it there is — with the connections field’s own addition, which is that in a connection, unlike in a rolled section, somebody chose where it sits.

Named alongside this one

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

The objects this essay names

Each one links to every other essay that touches it.

Base plateBearingConnectionEccentricityHolding down boltLower bound theoremMiddle thirdNo tension