Reinforced Concrete

Concrete Slab on Grade — A Structural Engineer's Design Handbook

Structural Notes · No. 06 · 07 Aug 2026 · 20 min read

1. What is a concrete slab on grade?

  • A concrete slab placed directly on the ground (subgrade / subbase).
  • Not a structural element of the building — it is non-structural per ACI 360.
  • Its purpose is to provide a flat surface carrying operational loads such as forklifts, racks and partition walls.
  • Design focuses on serviceability: crack control, curling and flatness.

2. The Winkler subgrade model

  • The subgrade is modelled as a system of independent springs (dense liquid foundation).
  • Each point reacts linearly with settlement: q = k × δ.
  • The model is simple yet accurate enough for practical design.

Modulus of subgrade reaction, k

k is the pressure required to cause one unit of deformation, in pci (lb/in³) or MN/m³. It is measured by a field plate load test (Ø750 mm plate) or estimated from a CBR correlation. Remember that k is not a constant — it depends on soil type, moisture content and plate size.

Soil type (USCS)k (pci)k (MN/m³)Notes
Organic soils (OL, OH, Pt)25–1007–27Very weak, improvement needed
High-plasticity clays/silts (CH, MH)50–15014–41Moisture-sensitive, k varies widely
Low-plasticity clays/silts (CL, ML)50–20014–54Most common
Silty/clayey sands (SM, SC)50–25014–68
Sands (SW, SP)150–40041–109Good
Silty/clayey gravels (GC, GM)200–50054–136
Gravels (GW, GP)300–50082–136Very good

3. Radius of relative stiffness — the key parameter

Lr = ⁴√[ E·t³ / ( 12·k·(1 − μ²) ) ]
SymbolMeaningUnit
LrRadius of relative stiffnessmm (in)
EConcrete modulus of elasticityMPa (psi)
tSlab thicknessmm (in)
kModulus of subgrade reactionMN/m³ (pci)
μPoisson's ratio (≈ 0.15–0.20)

Lr represents the zone of influence of a load on the slab. A large Lr means the slab is stiff relative to the subgrade, the load spreads further and stresses are lower. A small Lr means the slab is flexible relative to the subgrade, the load concentrates and stresses are higher.

4. Westergaard theory — three loading positions

Figure 1. Westergaard's three loading positions: interior, edge and corner.
Figure 1. Westergaard's three loading positions: interior, edge and corner.

Interior loading

  • The load sits far from any edge or corner.
  • Maximum tensile stress is at the bottom of the slab, directly under the load.
  • This is the most common case in industrial buildings — forklifts and rack posts.
fb = [ 3P(1+μ) / (2π·t²) ] × [ ln(2Lr/b) + 0.5 − γ ]

P = concentrated load      γ ≈ 0.5772 (Euler's constant)
b = equivalent radius of the resisting section
  when a < 1.724t :  b = √(1.6a² + t²) − 0.675t
  when a ≥ 1.724t :  b = a

Edge and corner loading

  • Edge loading: on the slab edge but away from corners. Tensile stress is higher than interior loading because support is lost on one side — check it separately.
  • Corner loading: the maximum tensile stress is at the top of the slab — the opposite of the other two cases.
Corner:    fc = [ 3P / t² ] × [ 1 − (a√2 / Lr)^0.6 ]

5. Slab thickness design — the PCA / ACI 360 method

This is an allowable stress approach: compute the applied stress and compare it with the allowable stress, which is the modulus of rupture divided by a factor of safety.

MR = 9 × √f'c      (f'c in psi)
MR = 0.75 × √f'c   (f'c in MPa)

Allowable stress   WS = MR / FoS
Loading conditionRecommended FoS
Non-repetitive, light traffic1.7
Moderate repetition1.7–2.0
Heavy loads, dense traffic2.0
Special or critical loads> 2.0
Figure 2. The three design load cases: concentrated, continuous wall and uniform load.
Figure 2. The three design load cases: concentrated, continuous wall and uniform load.

Case 1 — concentrated or wheel load

  • Sources: rack posts, forklift wheels, equipment legs.
  • Checks: Westergaard flexure, punching shear, bearing, and dowel bearing if near a joint.
  • Inputs: P, contact area Ac, f'c, t, k and FoS — where Ac = P / p, with p the tyre or post pressure.

Case 2 — continuous wall load

The load comes from masonry or partition walls built on the slab. It is analysed as a beam on elastic foundation per TM 5-809-12, checked at two positions: near the centre (or a joint) and near a free edge.

Case 3 — uniform load

The load comes from stacked goods or stored materials. The critical condition is the aisle width between loaded areas: the maximum stress occurs when the aisle is narrow, where the slab bends in the opposite direction.

Punching shear and finding the minimum thickness

fv = P / (b₀ × t)

b₀ = perimeter of the critical section, taken at d/2 from the load edge
Allowable stress  Fv = 4√f'c (psi)  or  0.33√f'c (MPa)
  • Assume a trial thickness t.
  • Compute Lr, fb and fv for each load case.
  • Check fb ≤ MR/FoS and fv ≤ the allowable Fv.
  • If a check fails, increase t and repeat; t_min is the smallest thickness at which every check passes.

6. Joint design

Figure 3. Joint types in a slab on grade.
Figure 3. Joint types in a slab on grade.
Joint typePurposeLocation
Contraction jointCreates a weakened plane to control crackingSaw cut 1/4–1/3 of the slab depth
Construction jointBoundary of a day's concrete pourEnd of the pour or a planned break
Isolation jointSeparates the slab from fixed structuresAround columns, walls and equipment bases

Joint spacing

  • Rule of thumb: L = (24 to 36) × slab thickness. A 150 mm slab gives L = 3.6–5.4 m; a 200 mm slab gives L = 4.8–7.2 m.
  • Prefer square panels; if rectangular, keep the length-to-width ratio below 1.5.
  • Align joints with the column grid where possible.

Dowel bars

  • Purpose: transfer vertical load across the joint and prevent faulting between panels.
  • Use a smooth round bar; one half must be debonded so the joint can open and close.
  • The diameter is typically ≈ t/8 (about 20–25 mm for a 150–200 mm slab).
  • Length 400–500 mm with 200–250 mm projecting each side; spacing 300 mm.
Dowel bearing stress:   fb,dowel = P_dowel / (d × t_eff)
Allowable (Friberg/PCA): Fb,allow = (4/3 − d/3t) × f'c

7. Reinforcement and crack control

Reinforcement becomes necessary when joint spacing is large, when crack width must be controlled, or when the temperature differential is significant.

Shrinkage and temperature reinforcement

As = ( f × L × W × t × γ ) / ( 2 × fs )
SymbolMeaning
fSlab-to-subgrade friction coefficient (1.0–2.5, typically ≈ 1.5)
LDistance between joints
WConcrete unit weight (≈ 2400 kg/m³)
tSlab thickness
γFactor, equal to 1 for a single layer at mid-depth
fsAllowable steel stress (typically 0.67fy)
  • Crack-control reinforcement is placed in the upper third of the slab depth.
  • Do not run deformed bars through a contraction joint — it stops the joint working and moves the crack elsewhere.
  • Crack width is estimated as w = ε × L_joint / 2, where ε combines drying shrinkage and thermal contraction.

8. Subgrade preparation and curling

Figure 4. The layer system beneath a concrete slab on grade.
Figure 4. The layer system beneath a concrete slab on grade.
  • Subgrade: must be compacted and uniform; remove weak and organic soils.
  • Subbase: 100–150 mm of sand or crushed stone to spread load, drain water and provide a working platform.
  • Vapour barrier: a PE film ≥ 0.15 mm beneath the slab to stop moisture rising from the ground.
  • Slip membrane: reduces slab-to-subgrade friction and therefore shrinkage stress.

Curling and how to reduce it

Slab edges curl upward because of moisture and temperature gradients between top and bottom: the top dries and shrinks faster while the underside stays damp. The slab then loses contact with the subgrade at edges and corners, load concentrates there and cracking follows.

  • Reduce the water–cement ratio to reduce shrinkage.
  • Use the largest practical aggregate size to reduce water demand.
  • Cure properly for at least seven days to reduce the moisture gradient.
  • Keep joint spacing sensible so panels stay small.
  • Provide top reinforcement to control curling cracks.

9. Quick reference tables

Concrete properties

PropertyFormulaExample, f'c = 35 MPa
Modulus of rupture (MR)0.75√f'c (MPa)4.44 MPa
Elastic modulus (Ec)4700√f'c (MPa)27,800 MPa
Poisson's ratio (μ)0.15–0.200.15

Factor of safety and working stress

ScenarioFoSWS = MR/FoS
Light racking, few forklifts1.72.61 MPa
Standard warehouse1.82.47 MPa
Heavy warehouse, continuous forklift traffic2.02.22 MPa

Typical slab thickness

ApplicationThickness (mm)Thickness (in)
Residential, light use100–1254–5
Commercial, office125–1505–6
Light industrial150–1756–7
Heavy industrial175–2507–10
Warehouse, heavy forklifts200–3008–12

10. Design checklist from A to Z

Survey and input data

  • Geotechnical report available: soil type, k-value, groundwater level?
  • Operational loads defined: forklifts (P, Ac), rack posts, walls?
  • Flatness requirements (FF/FL) specified?
  • Environmental conditions considered: temperature differential, moisture?

Design phase

  • f'c selected and MR, Ec calculated?
  • k-value determined by test or lookup table?
  • Radius of relative stiffness Lr calculated?
  • Flexure checked for all three load cases?
  • Punching shear and bearing checked?
  • Minimum thickness t_min determined?
  • Joints designed: location, spacing, type?
  • Dowel bars designed where needed?
  • Shrinkage/temperature reinforcement calculated and crack width checked?

Construction phase

  • Subgrade compacted and uniform?
  • Subbase placed to the correct thickness?
  • Vapour barrier installed?
  • Reinforcement or mesh placed correctly in the upper third?
  • Dowels placed correctly and debonded on one side?
  • Concrete placed at the correct slump and grade?
  • Contraction joints cut at the right time (4–12 hours after placement)?
  • Cured for at least seven days?

11. Related standards

StandardScopeWhat engineers need to know
ACI 360RSlab on grade designThe primary reference — methods, formulas and detailing
ACI 318Structural concrete designApplies only when the slab is a structural element
ACI 302.1RConcrete floor constructionConstruction, curing and flatness requirements
PCA IS195Industrial floor thickness designPCA design charts and tables
TM 5-809-12Heavy-load concrete floor slabsWall load, uniform load and lookup tables
Westergaard (1926)Elastic plate on elastic foundationThe original equations for interior, edge and corner loading

Supporting design tools

Closing — ten things to remember

ContentKeyword
The subgrade is a Winkler spring system, q = k × δWinkler k
Lr is the key parameter behind every formulaLr
Three loading positions: interior, edge, corner3 Positions
Interior loading usually governsInterior Governs
Working stress is MR divided by a factor of 1.7–2.0Working Stress
MR = 9√f'c (psi) or 0.75√f'c (MPa)MR
Joints at 24–36 times the thickness, square panelsJoint Spacing
Dowels must be round, smooth and debonded on one sideDowel Bar
Reinforcement does not prevent cracks, it only keeps them tightCrack Control
A uniform subgrade makes a good slab — investing there always paysSubgrade First

Part of the series "Structural design for industrial facilities" — Roberto Structural. The content is technical guidance; the engineer remains responsible for checking and adapting it to the conditions of each project and the requirements of the governing code.

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