Australian Base Plate — AS 4100 Concrete and Anchor Design Guide
Quick access: [[Australian Steel Grades|/reference/australian-steel-grades/]] | [[Australian Steel Properties|/reference/australian-steel-properties/]] | [[Australian Beam Sizes|/reference/au-beam-sizes/]] | [[Australian Bolt Capacity|/reference/australian-bolt-capacity/]] | [[AS 4100 Beam Design|/reference/as4100-beam-design-example/]] | [[All Australian References|/reference/]]
Comprehensive guide for column base plate design per AS 4100:2020 Clause 9. Covers concrete bearing capacity (AS 3600), T-stub plate bending model, anchor bolt tension and shear, base plate welding, and stiffness considerations.
Design Workflow
| Step | Check | Formula | Reference |
|---|---|---|---|
| 1 | Base plate area | Ap ≥ N* / (φc × 0.85 × fc′) | AS 3600 Clause 12.3 |
| 2 | Concrete bearing | φc×0.85×fc′×A1√(A2/A1) ≤ 2×φc×0.85×fc′ | AS 3600 |
| 3 | Plate thickness | tp = m × √(2N*/(0.9×fy×B×D)) | T-stub model |
| 4 | Anchor bolts | Tension: φNtf ≥ T*, Shear: φVfn ≥ V* | AS 4100 Clause 9.3 |
| 5 | Weld (column to plate) | Fillet weld along column perimeter | AS 4100 Clause 9.7 |
Base Plate Sizing
| Column | Base Plate (mm) | tp (mm) | Bolts |
|---|---|---|---|
| 150UC37 | 300×300 | 16 | 4-M16 |
| 200UC52 | 350×350 | 20 | 4-M20 |
| 250UC89 | 400×400 | 25 | 4-M24 |
| 310UC137 | 450×450 | 32 | 4-M24 |
| 310UC158 | 500×500 | 36 | 4-M30 |
Typical for N* ≤ 0.6×φNc, fc′=32 MPa, Grade 300 plate.
Plate Bending — T-Stub Model
Critical section at m = (B − 0.95×dc)/2 (projection beyond column flange)
Required thickness: tp ≥ m × √(2 × fp / (φ × fy))
Where fp = actual concrete bearing stress = N* / (B × D)
Alternative (simplified): tp ≥ m × √(fp / (φ × fy / 2))
Worked Example
Problem: Design base plate for 250UC89 (dc=260 mm, bfc=256 mm). N*=1,800 kN (compression). fc′=32 MPa. Grade 300 plate.
Solution:
- Required area: Ap ≥ 1,800×10³/(0.65×0.85×32) = 101,700 mm² → 350×350=122,500 mm² OK
- Concrete bearing: fp = 1,800/122,500 = 14.7 MPa ≤ 0.85×0.65×32 = 17.7 MPa OK
- Plate projection: m = (350-0.95×260)/2 = 51.5 mm
- Required tp = 51.5 × √(2×14.7/(0.90×300)) = 51.5 × 0.33 = 17 mm → specify 20 mm
- Anchor bolts: 4-M20 (nominal) for erection, checked for any tension from uplift
- Weld: 8 mm fillet each side, full perimeter
Design Resources
- [[Australian Steel Grades|/reference/australian-steel-grades/]] | [[Australian Steel Properties|/reference/australian-steel-properties/]] | [[Australian Beam Sizes|/reference/au-beam-sizes/]] | [[Australian Bolt Capacity|/reference/australian-bolt-capacity/]] | [[AS 4100 Beam Design|/reference/as4100-beam-design-example/]] | [[All Australian References|/reference/]]
FAQ
What is the concrete bearing limit per AS 3600? φc × 0.85 × fc′ × √(A2/A1) ≤ 2 × φc × 0.85 × fc′. Where A2/A1 accounts for confinement from surrounding concrete (max 2x increase).
How is base plate thickness determined? Using the T-stub model: tp = m × √(2fp/(φ×fy)). The critical bending plane is at the column flange face, projecting distance m.
When are shear keys required? When lateral force exceeds anchor bolt shear capacity (φVfn) or when combined tension+shear governs. Shear keys (welded studs or plates) engage concrete bearing directly.
Educational Use Only — This reference is for educational and preliminary design purposes only. All structural designs must be independently verified by a licensed Professional Engineer (PE) or Structural Engineer (SE) in accordance with AS 4100:2020 and all applicable Australian Standards. Results are not for construction.