AS 4100 Base Plate Design — Complete Worked Example

Base plate design is one of the most scrutinized calculations in Australian structural engineering practice. The base plate transfers the column axial load, shear, and moment into the concrete footing or pedestal, and a failure at this interface can be catastrophic. AS 4100:2020 Section 9 provides the steel-side checks, while AS 3600 governs the concrete bearing capacity. This complete worked example walks through every limit state for a typical column base plate in an Australian industrial building.

Design parameters

The column is a 250UC89.5 supporting a portal frame rafter. The base plate is shop-welded to the column and site-bolted to the concrete footing with cast-in anchor bolts. The column carries axial compression, a small shear from wind bracing, and a nominal base moment from frame action.

Parameter Value
Column section 250UC89.5 (d=260 mm, bf=256 mm)
Steel grade (column) AS/NZS 3679.1 Grade 300 (Fy=300 MPa, Fu=440 MPa)
Base plate dimensions 400 x 400 x 25 mm
Plate steel grade AS/NZS 3678 Grade 300 (Fy=300 MPa, Fu=440 MPa)
Concrete strength f'c = 32 MPa (pedestal 600 x 600 mm)
Anchor bolts 4 x M24 4.6/S, cast-in, 125 mm embedment
Factored axial load (N*) 1,200 kN (compression)
Factored shear (V*) 45 kN
Factored moment (M*) 15 kN-m (minor axis)

Step 1: Concrete bearing check per AS 3600

AS 4100:2020 Clause 9.6.1 references AS 3600 for the concrete bearing strength under the base plate. The bearing capacity depends on the concrete compressive strength, the loaded area, and the supporting area:

Nc = phi * 0.85 * f'c * A1 * sqrt(A2/A1)

where:
  phi   = 0.6 (AS 3600 Table 2.2.2 — bearing)
  f'c   = 32 MPa
  A1    = base plate area = 400 x 400 = 160,000 mm^2
  A2    = pedestal area = 600 x 600 = 360,000 mm^2
  sqrt(A2/A1) = sqrt(360000/160000) = 1.50 (capped at 2.0 per AS 3600 Cl 12.6)

Nc = 0.6 x 0.85 x 32 x 160,000 x 1.50 / 1000
   = 3,917 kN

Check: N* = 1,200 kN < Nc = 3,917 kN  -->  OK (ratio 0.31)

The concrete bearing capacity is well above the demand. The sqrt(A2/A1) factor provides a significant increase — without it, the capacity would be 2,611 kN, still adequate but with less margin. This factor accounts for the confining effect of the surrounding concrete in the pedestal.

Note the phi factor of 0.6 for concrete bearing in AS 3600 compared to phi = 0.65 in CSA A23.3 and phi = 0.65 in ACI 318. Australian engineers working across codes should verify the applicable phi factor for each jurisdiction.

Step 2: Bearing pressure distribution

For a concentrically loaded base plate with small moment, the bearing pressure is assumed uniform. The pressure must be checked against the allowable bearing stress:

fp = N* / A1 = 1,200,000 / 160,000 = 7.50 MPa

Allowable bearing stress:
fp,max = phi * 0.85 * f'c * sqrt(A2/A1) = 0.6 x 0.85 x 32 x 1.50 = 24.5 MPa

7.50 MPa < 24.5 MPa  -->  OK

Check for moment effect (minor axis, M* = 15 kN-m):
  P/A = 7.50 MPa
  M/Z = 15e6 / (400 x 400^2 / 6) = 15e6 / 10,666,667 = 1.41 MPa
  Max pressure = 7.50 + 1.41 = 8.91 MPa  < 24.5 MPa  -->  OK
  Min pressure = 7.50 - 1.41 = 6.09 MPa  > 0  -->  No uplift

The small moment does not cause uplift, so the full base plate area remains effective in bearing. The pressure remains uniform enough for the simplified cantilever method to be valid.

Step 3: Plate bending thickness check

The base plate bends as a cantilever beyond the column footprint. The critical cantilever dimensions are defined by the projection beyond the column face:

Column dimensions: d = 260 mm, bf = 256 mm
Plate dimensions: N = 400 mm (depth direction), B = 400 mm (width direction)

Cantilever in depth direction:
  m = (N - 0.95*d) / 2 = (400 - 0.95*260) / 2 = (400 - 247) / 2 = 76.5 mm

Cantilever in width direction:
  n = (B - 0.80*bf) / 2 = (400 - 0.80*256) / 2 = (400 - 204.8) / 2 = 97.6 mm

Critical cantilever = max(m, n) = 97.6 mm

The 0.95d and 0.80bf factors in the effective bearing width account for the fact that the column distributes load into the plate over a slightly reduced footprint. These factors are consistent with AISC Design Guide 1 and are widely adopted in Australian practice in the absence of a specific AS 4100 clause.

Required plate thickness per unit width (AS 4100:2020 Cl 5.2):

  tp,req = n_crit * sqrt(2 * fp / (phi * Fy))

  where:
    phi = 0.90 (AS 4100:2020 Table 3.4 — steel in bending)
    Fy  = 300 MPa
    fp  = 8.91 MPa (including moment effect for conservatism)

  tp,req = 97.6 * sqrt(2 * 8.91 / (0.90 * 300))
         = 97.6 * sqrt(17.82 / 270)
         = 97.6 * sqrt(0.0660)
         = 97.6 * 0.257
         = 25.1 mm

  Provided: 25 mm vs Required: 25.1 mm  -->  essentially OK (ratio 1.00)

The 25 mm plate is right at the calculated limit. In practice, an engineer might increase to 30 mm for construction tolerance and corrosion allowance, or enlarge the plate to reduce the cantilever projection and therefore the required thickness. A 450 x 450 mm plate would reduce the required thickness to approximately 19 mm. The trade-off between plate area and plate thickness is a common optimisation in base plate design.

Step 4: Anchor bolt shear check per AS 4100:2020

Anchor bolts transfer shear from the column base into the concrete footing. The shear capacity per bolt depends on the bolt grade, diameter, and whether the shear plane passes through the threaded or unthreaded portion:

Bolt shear capacity per AS 4100:2020 Clause 9.2.2.1:

  Vf = phi * 0.62 * fuf * (n_n * A_n + n_x * A_x)

For M24 4.6/S bolts with threads in shear plane:
  phi = 0.8 (AS 4100:2020 Table 3.4 — bolts)
  fuf = 400 MPa (minimum tensile strength, Grade 4.6 per AS/NZS 1252)
  Ac  = 353 mm^2 (tensile stress area, M24)
  A_n = Ac = 353 mm^2 (core area in shear plane)
  n_n = 1.0 (single shear plane)

  Vf = 0.8 x 0.62 x 400 x 353 / 1000
     = 70.0 kN per bolt

4 bolts: total shear capacity = 4 x 70.0 = 280 kN

Check: V* = 45 kN < 280 kN  -->  OK (ratio 0.16)

Shear is not the governing limit state for this base plate. The 4 x M24 bolts provide substantial reserve capacity. However, the connection detail must ensure the shear is reliably transferred — if the base plate holes are oversized (common for construction tolerance), shear transfer through bolt bearing may require additional measures such as shear lugs or welded plate washers after alignment.

Step 5: Weld design — column to base plate

The column is fillet-welded to the base plate around the full column profile. The weld must transfer the axial compression and, more critically, the tension that develops on the windward side of the column under uplift load combinations. AS 4100:2020 Clause 9.7.3.10 governs fillet weld capacity:

Weld perimeter around column:
  Lw = 2 x (bf + d) = 2 x (256 + 260) = 1,032 mm
  (welded both sides of each flange + both sides of web)

Fillet weld capacity per mm (SP category, E48XX electrode):
  phi_w = 0.8 (SP)
  fuw   = 480 MPa (E48XX)
  Try 8 mm fillet weld:
    tt  = 0.707 x 8 = 5.66 mm
    vw  = phi_w x 0.6 x fuw x tt = 0.8 x 0.6 x 480 x 5.66 = 1,304 N/mm

Total weld capacity:
  phi_w * Vw = 1,032 x 1.304 / 1000 = 1,346 kN

Check compression transfer:
  N* = 1,200 kN < 1,346 kN  -->  OK (ratio 0.89)

The weld is adequate for compression transfer. The minimum weld size per AS 4100:2020 Table 9.7.3.2 for the thicker part (base plate, 25 mm) is 6 mm, so the 8 mm fillet exceeds the minimum. A 6 mm weld would provide a capacity of 978 N/mm, or 1,009 kN total, which would be insufficient (ratio 1.19). This illustrates why weld size checks, not just member checks, can govern base plate design.

Step 6: Minimum plate dimensions and detailing

Beyond strength, AS 4100:2020 imposes minimum dimensional requirements for base plates. These ensure constructability, adequate grout coverage, and bolt clearance:

Step 7: Check minimum axial capacity (uplift case)

Under wind uplift, the column may experience net tension. The anchor bolts must resist this tension through a combination of bolt tensile strength and concrete breakout capacity. AS 4100:2020 Clause 9.2.2.2 gives the bolt tension capacity:

Bolt tension capacity per AS 4100:2020 Clause 9.2.2.2:

  Ntf = phi * fuf * As

For M24 4.6/S bolts:
  phi = 0.8 (AS 4100:2020 Table 3.4)
  fuf = 400 MPa
  As  = 353 mm^2 (tensile stress area)

  Ntf = 0.8 x 400 x 353 / 1000 = 113.0 kN per bolt

4 bolts: total tension capacity = 4 x 113.0 = 451.8 kN

For a typical wind uplift of N*uplift = 180 kN:
  180 kN < 451.8 kN  -->  OK (ratio 0.40)

The bolt tension capacity governs over concrete breakout for well-embedded anchors in competent concrete. The concrete breakout cone capacity per AS 3600 should also be verified, particularly for shallow embedments or edge-proximate anchors.

Summary of checks

Check Demand Capacity Ratio Status
Concrete bearing 1,200 kN 3,917 kN 0.31 PASS
Bearing pressure 8.91 MPa 24.5 MPa 0.36 PASS
Plate bending 25.1 mm req'd 25 mm 1.00 MARGINAL
Anchor bolt shear 45 kN 280 kN 0.16 PASS
Weld (compression) 1,200 kN 1,346 kN 0.89 PASS
Bolt tension (uplift) 180 kN 452 kN 0.40 PASS

The plate bending check is marginally acceptable. In practice, the engineer would either increase the plate thickness to 30 mm or enlarge the plate to 450 x 450 mm. The choice depends on whether plate material cost or concrete pedestal size is the driving economic constraint.

Key differences: AS 4100:2020 vs other codes for base plates

Try the calculator

The Steel Calculator Base Plate and Anchor Calculator performs all the above checks per AS 4100:2020, AS 3600, EN 1993-1-8, CSA S16:24, and AISC 360-22. Select the Australian code, enter your column size, plate dimensions, concrete strength, and anchor bolt configuration. Every intermediate value, formula, and clause reference is displayed.

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