Construction Tolerances

The Survey Says the Column Is 25 mm Out of Plumb. Now What?

Published: August 23, 2026  |  By: RHCES Engineering Team  |  17 min read

The forms are gone and the column looks fine from the ground. Then the as-built survey lands: the top of column C-7 leans 25 mm out of plumb over its own lift, all of it one direction.

What follows is an argument, not an engineering exercise. The contractor says it is only 25 mm over a 3.60 m storey. QA/QC says the specification says what it says. Three weeks later the floor is still not accepted, because nobody has written the thing that settles it: an arithmetic path from the survey number to a structural quantity.

Tolerance is an acceptance criterion, not a capacity check

A construction tolerance is a contractual and workmanship limit: it keeps the built work close enough to the drawing for everything downstream to fit. It lives in your specification, usually by reference to a published tolerance document — commonly ACI 117 for concrete, the AISC Code of Standard Practice for steel. The numeric limits sit in those documents, not in a blog post, and they differ by project, by member type, and per-storey versus total-height case.

Structural adequacy is a different question: does the member as built still satisfy the governing code? A member can be outside tolerance and still adequate — which is why blanket demolition orders are so often wrong — or inside tolerance and still a problem once the deviation combines with a second one going the same way. A breach is a non-conformance to raise and disposition, not automatically a defect.

From a survey number to a moment

This stalls on site because the tolerance is a length and the design check is a moment, and nobody converts one to the other out loud. A plumb deviation is an eccentricity, and an eccentricity on an axially loaded column is a moment: M = P × e. Keep e in metres, so kN × m gives kN.m.

The second half is knowing which eccentricity belongs to which member, and this is where most screenings go wrong. The moment induced inside a given lift is driven by the offset of that lift's top relative to that lift's own base — its relative lean. The offset measured against the true grid line is a different quantity: it is what a total-height limit is written against, and it is what the lower lift, the pedestal and the foundation actually carry, because the accumulation translates everything above them sideways without bending the upper lift. If the lower column already leaned 12 mm north and the upper lift leans a further 25 mm north, the top of the upper lift stands 37 mm off station against the grid, but the upper lift's own eccentricity is 25 mm. Put 37 mm into the upper lift and you overstate it by half while letting the members that do carry the inherited 12 mm walk out of the disposition.

Accumulation also has to be earned. Deviations can run one way when they share a cause — one badly set template will do it — but they just as often partially cancel, and same-direction drift is one possible pattern rather than the default. Do not add them because adding is conservative; make the survey demonstrate that they act the same way, resolved into two orthogonal grid directions, before you add anything.

A workflow that ends the argument

Worked example: two columns, the same 25 mm

A 400 mm × 400 mm interior column, 3.60 m storey, factored axial load 1,850 kN. The top of the lift is 25 mm out of plumb relative to its own base, north; the column below had already put that base 12 mm north of the grid. Everything that follows is screening arithmetic on stated assumptions, and the assumptions are set out at the end of the example because several of them can overturn it.

1. The two eccentricities

Storey rate: 25 mm ÷ 3,600 mm = 0.00694, which is 1 in 144, or 0.694 percent of storey height. Quote it as 0.694, or round up to 0.70, and never down to 0.69: this figure gets compared against a percentage-of-height plumb limit, and rounding down at that comparison turns a marginal fail into an apparent pass. The rate is what goes against the per-storey plumb tolerance.

Offset from the grid at the top of the lift: 12 mm + 25 mm = 37 mm = 0.037 m. That is the number for any total-height limit, and it is the eccentricity the lower lift, the pedestal and the foundation see.

The eccentricity that bends this lift is its lean relative to its own base: 25 mm = 0.025 m.

2. The induced moment

M = P × e = 1,850 kN × 0.025 m = 46.25 kN.m in the upper lift.

The inherited 12 mm is not lost; it lands somewhere else. The load arriving at the top of the lift below is 37 mm off station, so the screening moment to carry into the lower lift, the pedestal and the footing is 1,850 × 0.037 = 68.45 kN.m, of which 1,850 × 0.012 = 22.20 kN.m is inherited from a column already accepted on a survey that did not know this lift was coming. Those members belong in the same disposition. The commonest failure of these write-ups is to load the whole 37 mm onto the new lift — overstating it by 68.45 ÷ 46.25 = 1.48 — and then never mention the members that actually carry the accumulation.

3. The screening comparison

Take the design output as showing 95 kN.m about that axis. Then 46.25 ÷ 95 = 0.4868, so the induced moment is 48.7 percent of the design moment; crudely combined, 95 + 46.25 = 141.25 kN.m, a factor of 141.25 ÷ 95 = 1.49.

That sum is only legitimate if the 1,850 kN and the 95 kN.m come from the same factored load combination and act about the same axis in the same sense. Nothing in the inputs establishes it, and it is the single assumption most likely to make the whole comparison meaningless: if the 95 kN.m came out of a lateral case with a much lower axial load, adding a moment computed from the gravity axial load to it compares nothing with nothing. Ask for the combination before you print a factor.

Even when it survives that test, the screening does not settle the disposition, and a reviewer will say so:

4. The same deviation, a different column

Put the identical 25 mm relative lean on a column carrying 420 kN factored, holding the design moment at 95 kN.m so only the axial load changes. M = 420 kN × 0.025 m = 10.50 kN.m, and 10.50 ÷ 95 = 0.1105 → 11.1 percent, against 48.7 percent before. The ratio of induced moments, 46.25 ÷ 10.50 = 4.40, is exactly the ratio of axial loads, 1,850 ÷ 420 = 4.40.

That is the point: same 25 mm, same breach — one needs a full recheck, the other a note. Hold the second case lightly, though. A column carrying 420 kN would not normally have the same section or the same design moment, so it shows how P scales the answer rather than standing as a real second column.

5. What this example assumes

None of these is established by a survey, and each can move or overturn the result above. Put them in the disposition instead of leaving a reviewer to guess:

6. The anchor bolt

A 24 mm anchor rod is found 18 mm off position. Base plate 400 mm × 400 mm, bolt gauge 300 mm × 300 mm, holes detailed at 30 mm. Two things to state before the arithmetic. First, a 30 mm hole for a 24 mm anchor rod is a structural-bolt clearance, not an anchor-rod clearance: anchor rod holes are normally detailed well beyond rod diameter plus 6 mm precisely to absorb setting-out error, so this project has detailed an unusually tight hole and the misplacement will look more fatal here than it usually is. Second, the 18 mm is taken as a single-axis offset toward one plate edge; if it is instead the resultant of two components, an 18 mm resultant at 45 degrees costs about 12.7 mm on each of two edges rather than 18 mm on one, and every figure below changes.

Radial clearance = (30 - 24) ÷ 2 = 3 mm. The rod centre sits 18 mm off, which is 15 mm outside the 3 mm the hole can accept, so 15 mm of additional radial relief is required and the plate will not drop over the rod. It is not a clean miss: a 12 mm rod radius centred 18 mm off spans 6 mm to 30 mm from the hole centre while the 30 mm hole spans ±15 mm, so rod and hole overlap over 9 mm and the rod fouls the edge of the hole. The distinction is not pedantry — "misses" invites the reader to picture no engagement at all, and that picture quietly rules out every remedy that works by relieving the hole.

Theoretical bolt centre to plate edge = (400 - 300) ÷ 2 = 50 mm, so as detailed the clear distance from hole edge to plate edge is 50 - 15 = 35 mm. With the rod 18 mm off toward the near edge, the actual centre sits 50 - 18 = 32 mm from that edge. What each fix leaves as clear distance:

Check the washer too: a 100 mm square washer centred on theory reaches 50 mm each way, so its outer edge lands exactly on the plate edge — nothing to weld to on that side. Bearing and tear-out in the plate follow from these clear distances in the steel standard your project adopted, judged against that standard's own edge-distance basis rather than a remembered rule of thumb.

Then check the concrete, which none of that arithmetic has touched. Moving a rod 18 mm also moved its edge distance in the pedestal, and on the concrete side breakout and pryout usually govern anchor shear and are far more edge-sensitive than steel bearing. On the 600 mm pedestal priced below, a 300 mm gauge gives 150 mm nominal edge distance in the concrete, cut to 132 mm by the misplacement. Neither the pedestal size nor the embedment is something a base plate drawing will tell you, and both belong in the disposition next to the plate geometry.

Anchor bolt remedies, in rough order of preference

Two of those refusals turn on base fixity, so the disposition has to say whether the base was designed pinned or fixed. The worked example does not, and that is the first gap a reviewer will find. Take it as pinned for the arithmetic above; if it is fixed, the bolt group geometry is part of a design assumption and the slot and oversize options go back to the designer before anyone cuts steel.

Illustrative rates only, and not reproducible without the fillet weld size behind the rate and the grout and plate-washer thicknesses, none of which are stated here. Oversize plus washer: call-out ₱6,500, washer ₱450, and because the washer's outer edge lands on the plate edge, only three sides can be welded — 0.30 m of fillet weld at ₱1,200 per metre = ₱360, so 6,500 + 450 + 360 = ₱7,310. Pricing the full 0.40 m perimeter, as these estimates usually do, buys a weld along an edge with nothing behind it; if you want four sides, specify a washer small enough to sit clear of the plate edge and price that instead. Recasting a 600 mm × 600 mm × 900 mm pedestal: 0.60 × 0.60 × 0.90 = 0.324 m³ at ₱5,200 per m³ = ₱1,684.80, plus ₱13,204 for demolition, dowels, formwork and re-survey — ₱14,888.80, or 14,888.80 ÷ 7,310 = 2.04 times the first option.

Writing a disposition a reviewer will accept

"It is only 25 mm" is not a disposition; it is an opinion with a number attached. A signable one contains:

The test for a disposition: could a reviewer redo it a year from now from the paper alone?

Every item in the list below exists so that the answer is yes. The screening arithmetic is worthless without the inputs it was run on, and the inputs are worthless without the datum they were measured from.

The sketch and photographs matter because a year later the arithmetic still reads but the site is gone.

Common pitfalls

P × e tells you how hard to look; the interaction diagram tells you what to do. If you want section properties, unit conversions and quick geometry checks in one place while you write it up, the RHCES web tools page collects the ones that come up most often here.

FAQ

The column is outside tolerance but the screening moment is small. Can we just accept it?

Probably, but not silently. Outside tolerance is a non-conformance whatever the structural outcome, so it is raised, dispositioned and closed out on paper by whoever the contract says signs. A small screening moment supports accept-as-is instead of remedial work; it does not replace the designer's confirmation.

Do I measure against the column below or against the grid?

Both, and they are not interchangeable. Against the lift's own base, divided by storey height, gives the rate you compare with the per-storey plumb tolerance, and that same relative lean is the eccentricity that bends the lift. Against the grid gives the accumulated offset, which goes against any total-height limit and is the eccentricity seen by the lift below, the pedestal and the foundation, because the accumulation carries everything above them sideways.

The bolt is 18 mm off but the connection carries no tension. Does it still matter?

Probably, but check the design basis rather than asserting it. In many base plates shear is transferred by friction under the plate or through a shear lug, and rods sitting in oversized holes are often explicitly assumed not to carry shear until a plate washer is welded. Where the rods are relied on for shear, that shear is governed by the bearing and tear-out geometry the misplacement changed — in the plate, and more sensitively at the concrete edge of the pedestal. The bolt group may also be providing base fixity the frame analysis assumed. And it flags the template crew before the next base plate.