Concrete Bearing — AS 3600 Clause 12.6

The design bearing strength of concrete under a base plate:

phi × 0.85 × f'_c × (A2/A1)^0.5 ≤ phi × 1.7 × f'_c

Where:

• A1 = loaded area (area of base plate)
• A2 = maximum area of the supporting surface that is geometrically similar to A1
• phi = 0.65 (AS 3600 for bearing)

For Grade 32 concrete (f'_c = 32 MPa), base plate 350 × 350 mm on 800 × 800 mm foundation: sqrt(A2/A1) = sqrt(640,000/122,500) = sqrt(5.22) = 2.29. Capped at 2.0 per practical limit matching ACI 318.

Bearing strength = 0.65 × 0.85 × 32 × 2.0 = 35.4 MPa, which coincides with the cap: 0.65 × 1.7 × 32 = 35.4 MPa.

Bearing capacity N_c = 35.4 × 350 × 350 / 1000 = 4337 kN.

Base Plate Bending — Yield Line Theory

Per Australian practice (based on ASI Design Guide 7 and AS 4100 Clause 5.2), the base plate is modelled as a cantilever plate from the column face:

Design moment in base plate: m* = N* × c² / (2 × A_eff)

Where c = maximum cantilever distance from column face to plate edge (or to bolt line).

For 250UC73 (b_f = 254 mm, d = 254 mm) with 400 × 400 mm plate: c_x = (400 - 254) / 2 = 73 mm, c_y = (400 - 254) / 2 = 73 mm.

Plate bending capacity per unit width: phi × m_s = phi × f_y × t_p² / 4 (plastic yield line, per AS 4100).

For 25 mm plate Grade 300: phi × m_s = 0.90 × 300 × 25² / 4 = 42,188 N·mm/mm.

Applied moment: m* = N* × c² / (2 × A1) = 1800 × 10³ × 73² / (2 × 400 × 400) = 1800 × 10³ × 5329 / 320,000 = 29,976 N·mm/mm.

Check: 29,976 < 42,188. OK (71% utilisation).

Anchor Bolt Tension — AS 4100 Clause 9.3

Tension capacity per bolt: phi × N_tf = phi × A_s × f_uf (similar to AS 4100 Clause 9.3.2.3 for preloaded bolts, adjusted for non-preloaded anchors).

For Grade 8.8/S M24 bolts (A_s = 353 mm², f_uf = 830 MPa): phi × N_tf = 0.80 × 353 × 830 / 1000 = 234.4 kN per bolt.

For cast-in holding-down bolts, Australian practice often uses AS 3850 (precast) or proprietary anchor systems with capacities verified by testing.

Grout Joint Design — AS 3600 / ASI Design Guide 7

The grout pad between the base plate and concrete foundation serves three purposes: (a) provides full bearing contact despite concrete surface irregularities, (b) protects anchor bolts from corrosion at the concrete interface, and (c) transfers shear through friction. Australian practice per ASI Design Guide 7:

• Minimum grout thickness: 25 mm for non-shrink cementitious grout (50 mm preferred for site-poured foundations)
• Grout strength: Minimum f'_c = 40 MPa at 28 days (one grade above typical foundation concrete of 32 MPa)
• Grout perimeter: Extend grout 25 mm beyond the base plate on all sides to prevent edge spalling
• Grout holes: Provide 50 mm diameter grout holes in the base plate for plates larger than 600 × 600 mm

Pinned vs Fixed Base Classification

Australian practice per AS 4100 Clause 4.4 distinguishes between nominally pinned and fixed column bases:

For pinned bases, the base plate thickness is primarily governed by bearing. For fixed bases, bending in the base plate and anchor bolt tension govern. Most Australian building columns are designed as nominally pinned bases unless the frame relies on base fixity for lateral stability.

Worked Example 1 — Australian Column Base (Compression)

Problem: 250UC73 column (Grade 300, AS/NZS 3679.1) with factored axial compression N* = 1800 kN and base moment M* = 65 kN·m. Design base plate and bolts on Grade 32 concrete.

Step 1 — Base Plate Size:

Base plate: 400 × 400 × 25 mm Grade 300. Four M24 Grade 8.8/S bolts at 310 mm gauge.

Step 2 — Concrete Bearing Check:

N* / N_c = 1800 / (35.4 × 400² / 1000) = 1800 / 5664 = 0.318. OK (32% utilisation).

Step 3 — Check with Moment (Partial Compression):

e = M*/N* = 65/1800 = 36 mm. Base is almost fully in compression. Triangular stress block: max bearing stress sigma_max < 35.4 MPa. By proportion: sigma_max ≈ 2 × 1800 × 10³ / (3 × 400 × (400/2 - 36)) ≈ 3600 / (3 × 400 × 164) = 3600 / 196,800 = 18.3 MPa < 35.4 MPa. OK.

No net tension in bolts for this gravity load case.

Step 4 — Wind Uplift Case:

N* = 250 kN tension, M* = 50 kN·m. Bolt tension: T_per_bolt_row = N*/n_b + M*/z = 250/4 + 50/(2 × 0.31) = 62.5 + 80.6 = 143.1 kN (two bolts, 71.6 kN each).

phi × N_tf = 234.4 kN > 71.6 kN. OK (31% utilisation).

Step 5 — Shear Transfer:

V* = 90 kN base shear. Friction capacity assuming 0.30 coefficient: phi × mu × N*_c = 0.90 × 0.30 × 1800 = 486 kN > 90 kN. OK — friction alone sufficient. No shear key required.

Selected: 400×400×25 mm base plate Grade 300, 4-M24 Grade 8.8/S bolts, 300 mm embedment, grout 25 mm minimum.

Worked Example 2 — Shear-Governed Light Column Base

Problem: A 150UC30 column (Grade 300) supports a canopy with N* = 85 kN compression and V* = 120 kN shear (wind on canopy). M* = 15 kN·m. Grade 25 concrete (f'_c = 25 MPa). Design a pinned base plate with shear key.

Step 1 — Base Plate Size (Bearing):

Try 250 × 250 × 16 mm plate Grade 250. Concrete bearing with foundation 500 × 500 mm: sqrt(A2/A1) = sqrt(250,000/62,500) = 2.0 (capped). Bearing strength = 0.65 × 0.85 × 25 × 2.0 = 27.6 MPa. N_c = 27.6 × 250² / 1000 = 1725 kN >> 85 kN. OK.

Step 2 — Base Plate Bending:

c = (250 - 148) / 2 = 51 mm (150UC30 b_f = 148 mm, d = 148 mm). m* = 85,000 × 51² / (2 × 250 × 250) = 85,000 × 2601 / 125,000 = 1769 N·mm/mm. phi × m_s = 0.90 × 250 × 16² / 4 = 14,400 N·mm/mm >> 1769. OK (12% utilisation — bearing governs plate thickness).

Step 3 — Shear Transfer:

Friction alone: phi × mu × N* = 0.90 × 0.30 × 85 = 23.0 kN < 120 kN — NOT OK. Shear key required.

Shear key: 100 × 20 mm flat bar Grade 300, 200 mm long, welded to base plate underside. Embed 75 mm into 125 mm deep grout pocket. Concrete bearing on shear key: V_c = phi × 0.85 × f'_c × A_key = 0.65 × 0.85 × 25 × (100 × 75) / 1000 = 103.6 kN < 120 kN — marginal.

Increase key depth to 100 mm: V_c = 0.65 × 0.85 × 25 × (100 × 100) / 1000 = 138.1 kN > 120 kN. OK.

Shear key weld: 8 mm FW all round, 100 + 2 × 20 = 140 mm length. Capacity approximately 0.8 × 140 × 8 × 410 / (√3 × 0.8) / 1000 ≈ 265 kN >> 120 kN. OK.

Step 4 — Anchor Bolts:

2-M20 Grade 4.6/S bolts (pinned base — bolts inside flanges). Bolt tension from M*: T = 15 / 0.2 = 75 kN (shared between 2 bolts, 37.5 kN each). phi × N_tf (M20 Gr 4.6, A_s = 245 mm²) = 0.80 × 245 × 400 / 1000 = 78.4 kN > 37.5 kN. OK.

Selected: 250×250×16 mm base plate Grade 250, 2-M20 Grade 4.6/S bolts, 100×20 mm shear key embedded 100 mm, 25 mm grout.

Frequently Asked Questions

How is the base plate thickness determined in Australian practice? The base plate thickness is governed by either concrete bearing (compression) or bending from the cantilevered portion outside the column footprint. For pinned bases, the thickness is typically 0.02-0.04 times the plate width. For fixed bases with significant moment, the yield line mechanism at the tension flange determines thickness. A 400 mm wide fixed base plate typically requires 25-32 mm thickness for Grade 300 steel.

When is a shear key required for Australian base plates? A shear key is required when the factored base shear V* exceeds the friction capacity phi × mu × N*_c, where mu = 0.30 for steel on grout (clean surface, no paint on underside). For columns with low axial load (canopy columns, bracing posts), friction is often insufficient and a shear key or anchor bolt shear governs the design. Embed the shear key into a grout pocket or cast-in recess — do not rely on anchor bolts for shear alone unless specifically designed for combined shear and tension per AS 4100 Clause 9.3.2.3.

What grout specification applies to Australian base plates? Non-shrink cementitious grout with minimum 28-day compressive strength of 40 MPa. Grout thickness 25-50 mm. For base plates larger than 600 × 600 mm, provide 50 mm diameter grout holes. The grout must extend at least 25 mm beyond the base plate perimeter on all sides. The grout surface must be finished level within ±3 mm across the plate footprint — air pockets under the plate concentrate bearing stress and can cause plate yielding at service loads.

How does AS 4100 base plate design differ from AISC 360? Both codes use the cantilever method for plate bending (effective cantilever distance c from column face). Key differences: AS 4100 uses phi = 0.90 for steel bending and phi = 0.65 for concrete bearing (AS 3600). AISC uses phi = 0.90 for bending and phi = 0.65 for bearing. The concrete bearing strength formulas differ: AS 3600 uses 0.85 × f'_c × sqrt(A2/A1) capped at 2.0, while ACI 318 uses 0.85 × f'_c × sqrt(A2/A1) capped at 2.0 with an additional limit of 1.7 × f'_c × A1. Australian practice caps sqrt(A2/A1) at 2.0 per ASI Design Guide 7.

Related Pages

• AS 4100 Connection Design — Bolted & Welded Overview
• AS 4100 Column Design — Worked Example
• AS 4100 Bolt Capacity — Bolt Tables
• AS 4100 Moment Connection — End Plate Design
• Australian Steel Grades — Grade 300 & 350
• AS 4100 Combined Loading — Interaction Checks
• Base Plate & Anchor Calculator — Free Tool
• Column Capacity Calculator — AS 4100

Educational reference only. Base plate design per AS 4100-2020, AS 3600-2018, and ASI Design Guide 7. Verify against current Australian Standards and local authority requirements. Results are PRELIMINARY — NOT FOR CONSTRUCTION without independent Chartered Professional Engineer verification per state building regulations.