Six months after handover, the client calls. The warehouse floor in Laguna — 2,000 m² of it, poured to the "standard" 150 mm slab on grade with 10 mm bars at 300 mm both ways — has cracked along the wheel paths, where the owner's new 3.5-tonne counterbalance forklift swings in and out of the racking and crosses the sawcut joints on the run to the dock.
The site engineer opens the drawing set looking for the calculation behind 150 mm. There is none. The thickness came from the last project, which took it from the one before. Nobody ever saw a forklift data sheet.
This is probably the most commonly uncalculated number in Philippine light-industrial work — and one of the easiest to check by hand.
The reason is structural, not lazy. The reinforced concrete provisions Philippine designers work to — NSCP 2015, whose concrete chapter tracks ACI 318-14 — say in their scope that they do not govern slabs on ground unless the slab transmits vertical loads or lateral forces from other portions of the structure to the soil. Both limbs matter, and the wording is edition dependent. A warehouse floor on compacted fill carrying only forklift and storage loads generally falls outside; a floor that restrains column bases, acts as a diaphragm or resists portal-frame thrust does not, and is back inside the concrete code. Otherwise the slab is nobody's "code member", and the concrete chapters give the reviewer no thickness or steel table to check it against.
The second trap is reaching for the wrong tool. The minimum-thickness and deflection rules from suspended slabs are derived for a member spanning between supports. A slab on grade is supported continuously by soil; it does not span. Its thickness is governed not by the moment capacity of its reinforcement, but by whether the flexural tensile stress in the plain concrete under a wheel load on an elastic subgrade stays below a safe fraction of the modulus of rupture. The slab is designed to stay uncracked. That is why more rebar does not rescue a thin floor. It is also why the reference used here is recommended practice, not law: ACI 360R sets out the Portland Cement Association method followed below alongside alternatives including Wire Reinforcement Institute and Corps of Engineers methods and post-tensioned and fibre approaches. This article walks one item on that menu.
All figures are illustrative. Everything the answer depends on is listed, including the two declarations that decide the headline.
The two declarations that decide the headline. Those two are not measurements but declarations, and between them they set the allowable every result below is compared against. This article takes the conservative pair. Move either inside the ranges quoted and the verdict moves: across the four corners of those ranges the 150 mm demand-to-capacity ratio runs 0.86, 1.01, 1.04 and 1.22, so at 0.75 √f'c with a factor of 1.7 — both inside the article's own stated ranges — 150 mm passes at 0.86. The answer is a function of the declared pair, not a property of the slab. The slab-on-ground literature also conventionally pairs a higher tested modulus of rupture, around 0.75 √f'c in megapascals, with that factor range, so taking the low coefficient with the top of the range compounds two conservatisms.
Axle force = 7,200 × 9.81 = 70,632 N = 70.6 kN. Two front wheels, so P = 70,632 ÷ 2 = 35,316 N (35.3 kN). Carry 35,316: stress is linear in P, so rounding it down to 35,300 N is the one rounding that goes the unconservative way. Tire pressure converts as 700 kPa = 700,000 N/m², and 700,000 ÷ the 1,000,000 mm² in a square metre = 0.70 N/mm², or 0.70 MPa. Contact area A = 35,316 ÷ 0.70 = 50,451 mm². Contact radius a = √(50,451 ÷ 3.1416) = √16,059 = 126.7 mm. Products and roots below are rounded for display; figures downstream of them come from the unrounded values.
√28 = 5.2915. Ec = 4700 × 5.2915 = 24,870 MPa. Modulus of rupture fr = 0.62 × 5.2915 = 3.281 MPa. Both expressions are the ACI 318 and NSCP normal-weight concrete relations, borrowed deliberately: this page has just argued the concrete code does not govern the member, then reached into it for material properties, for want of beam tests. Say so on the sheet. With the safety factor of 2.0 for effectively unlimited repetitions (common practice, not a code value), allowable stress = 3.281 ÷ 2.0 = 1.64 MPa. Constant in the stiffness term: 12 × (1 - 0.15 × 0.15) = 12 × 0.9775 = 11.73, and 11.73 × 0.04 = 0.4692.
Ec h³ = 24,870 × 3,375,000 = 83,936,250,000; ÷ 0.4692 = 178,892,263,000. The radius of relative stiffness l is the fourth root: √178,892,263,000 = 422,956.6, and √422,956.6 = 650.35 mm.
a = 126.7 mm is less than 1.724 × 150 = 258.6 mm, so use the adjusted radius. a² = 16,059; 1.6 × 16,059 = 25,694; h² = 22,500; sum = 48,194; √48,194 = 219.53; 0.675 × 150 = 101.25; b = 219.53 - 101.25 = 118.28 mm.
l ÷ b = 650.35 ÷ 118.28 = 5.498; log 5.498 = 0.7402; 4 × 0.7402 = 2.9608; plus 1.069 = 4.0298. P ÷ h² = 35,316 ÷ 22,500 = 1.5696. Stress = 0.316 × 1.5696 × 4.0298 = 2.00 MPa. Ratio = 2.00 ÷ 1.64 = 1.22. Not adequate — over by 22 percent at mid-panel on one wheel, before any second wheel, joint, impact or tolerance allowance.
Ec h³ = 24,870 × 5,359,375 = 133,287,656,250; ÷ 0.4692 = 284,074,289,000; √284,074,289,000 = 532,986.2; √532,986.2 = 730.06 mm. Then 25,694 + 30,625 = 56,319; √56,319 = 237.32; 0.675 × 175 = 118.13; b = 237.32 - 118.13 = 119.19 mm.
l ÷ b = 730.06 ÷ 119.19 = 6.125; log 6.125 = 0.7871; 4 × 0.7871 = 3.1484; plus 1.069 = 4.2174. P ÷ h² = 35,316 ÷ 30,625 = 1.1532. Stress = 0.316 × 1.1532 × 4.2174 = 1.54 MPa. Ratio = 1.54 ÷ 1.64 = 0.94 — clearing mid-panel by six percent on one wheel alone. Hold that verdict until Step F.
Ec h³ = 24,870 × 8,000,000 = 198,960,000,000; ÷ 0.4692 = 424,040,921,000; √424,040,921,000 = 651,184.2; √651,184.2 = 806.96 mm. Then 25,694 + 40,000 = 65,694; √65,694 = 256.31; 0.675 × 200 = 135.00; b = 256.31 - 135.00 = 121.31 mm.
l ÷ b = 806.96 ÷ 121.31 = 6.652; log 6.652 = 0.8230; 4 × 0.8230 = 3.2920; plus 1.069 = 4.3610. P ÷ h² = 35,316 ÷ 40,000 = 0.8829. Stress = 0.316 × 0.8829 × 4.3610 = 1.22 MPa. Ratio = 1.22 ÷ 1.64 = 0.74. Adequate at mid-panel on one wheel, 26 percent in hand. Step F spends most of it.
The pitfall list says the wheels interact; the trials above load one wheel and stop. That gap decides 175 mm. At a track of 0.9 to 1.2 m the second wheel stands between about one and two times the radius of relative stiffness away — l runs 650 to 807 mm here — so the deflection bowls overlap and superposing it on the same Winkler model adds real stress. The addition grows with thickness because l grows with it, putting the same track relatively closer.
At 175 mm the second wheel adds roughly 10 to 18 percent, taking 1.54 MPa to about 1.69 to 1.81 MPa against the 1.64 MPa allowable, a ratio of about 1.03 to 1.10. 175 mm fails. Its six percent margin was never a margin; it was the second wheel, unaccounted for. At 200 mm the addition is roughly 12 to 20 percent, taking 1.22 MPa to about 1.37 to 1.46 MPa and the ratio to 0.83 to 0.89. 200 mm still passes, but the margin falls from 26 percent to between 11 and 17 percent.
Free edge, and the edges that cannot be dowelled. The pitfall list gives free-edge stress as one and a half to two times the interior value, so run both ends. At 1.5, 200 mm gives 1.2162 × 1.5 = 1.82 MPa — 1.82 unrounded, not the 1.83 a rounded 1.22 produces. At 2.0 it gives 2.43 MPa against the 1.64 MPa allowable, a ratio of 1.48: a true free edge fails by 48 percent, not marginally. Note what this is, a practice-based multiplier on an interior result rather than a computed free-edge solution, and on that basis load transfer at the joints is what makes 200 mm work. It does not answer the perimeter: at dock thresholds and around the building there is no adjacent panel to dowel into, the free-edge stress stands there as computed, and those edges need thickening, which has not been sized here.
Impact, early age and the rupture value. No dynamic allowance has been applied above; take a modest 1.15 and the single-wheel ratios become 1.40, 1.08 and 0.85 at 150, 175 and 200 mm, so 175 mm fails on that basis too, independently of the second wheel. The safety factor of 2.0 was justified as covering effectively unlimited repetitions, not impact, and if it is being asked to cover both, say so on the sheet. At 10 days, with flexural strength there taken as an illustrative 80 percent of the 28-day value — an assumption, not a test result — the allowable drops to 3.281 × 0.8 ÷ 2.0 = 1.31 MPa, so 200 mm at 1.22 MPa still holds on one wheel and 175 mm at 1.54 MPa does not; though reapplying the full factor of 2.0, justified as covering unlimited repetitions over the life of the slab, to a short early-age exposure is a different check that arguably warrants a lower factor. And practice often uses a tested rupture value nearer 0.75 √f'c, giving 3.97 MPa and an allowable of 1.98 MPa, which would change several verdicts above; the 0.62 coefficient is the conservative default, so adopt more only with beam test data and say which safety factor you paired it with.
The recommendation. Apply the tolerance pitfall before adopting anything: at an as-placed 190 mm the single-wheel ratio is 0.81 rather than 0.74, so 200 mm absorbs a 10 mm shortfall and stays adequate at mid-panel. 200 mm is therefore the forklift-governed minimum, to be confirmed against rack post reactions before adoption, with dowelled construction joints and a thickened edge wherever there is no adjacent panel.
The decision in pesos (illustrative rates). Going from 150 mm to 200 mm over 2,000 m² adds 0.05 × 2,000 = 100 m³ of concrete: 300 m³ becomes 400 m³. At an illustrative ₱5,500 per m³ delivered and placed, that is 100 × 5,500 = ₱550,000, plus dowels. Demolishing and replacing even a quarter of a cracked floor, with operations halted, costs several times that.
That comparison is what settles the argument with an owner. If you are pricing 150 mm against 200 mm across a whole floor, the RHCES Estimator will carry concrete, base course and steel for both thicknesses so the difference lands on one page.
Not for this check. Thickness is set by tensile stress in plain concrete under a wheel, and light mesh or bars do not raise the load at which that concrete cracks. Shrinkage and temperature steel in a slab on ground is often sized by the subgrade-drag approach, which is one accepted method among several rather than the normal one: equivalent-strength and minimum-percentage methods are also used, and jointless, fibre-reinforced and post-tensioned floors are sized on different bases entirely. That steel holds cracks tight and keeps joints closed — serviceability, not capacity.
Ask for a plate load test on the prepared subgrade. Failing that, use a published CBR-to-k or soil-classification-to-k correlation and say on the calculation that it is one. Do not substitute allowable bearing pressure: one is a strength limit, the other a stiffness, and they are not convertible. The result is forgiving on this input, though not in the way the sensitivity is usually described. On this model, halving k from 0.04 to 0.02 N/mm³ raises the interior stress by 7.5 percent at 150 mm and 6.9 percent at 200 mm, while doubling the contact pressure from 0.70 to 1.40 MPa raises it by 13.4 percent and 9.8 percent. Halving k raises the stress only modestly, and so, on the same scale, does getting the contact pressure badly wrong: both are second-order inputs, worth single digits to the mid teens in percent. The first-order input is the joint and edge condition, worth 50 to 100 percent.
Yes, and on many floors the racking governs rather than the forklift. Base plates are small, loads static and permanent, back-to-back frames put two plates close together, and posts often sit near an aisle joint. The concentrated-load check uses the same subgrade model with different geometry, and frequently drives a thicker slab or a local pad. Uniform storage load in the racks is a third case again, driven by rack loading and aisle width rather than wheel geometry, and not performed here. Get the rack layout and post reactions before you finalise the floor.