Australian Base Plate Design — AS 4100 and AS 3600

Complete reference for column base plate design in accordance with AS 4100:2020 Clause 9 and AS 3600:2018 concrete provisions. Covers all limit states: concrete bearing, plate bending, anchor bolt tension and shear, shear keys, weld design, and grouting requirements.

Related pages: AS 4100 Column Design | AS 4100 Bolt Group Capacity | Australian Steel Grades | Base Plate Calculator

Design Limit States

Columns transmit factored axial load N*, shear V*, and overturning moment M* to the foundation through the base plate assembly. Four limit states must be satisfied:

Concrete Bearing Resistance — AS 3600 Clause 12.3.4

The nominal concrete bearing capacity beneath a confined base plate is:

Nc = 0.85 x f'c x A1 x sqrt(A2/A1) <= 2 x 0.85 x f'c x A1

Where A1 is the base plate area and A2 is the maximum area of the supporting concrete surface that is geometrically similar to and concentric with A1. The confinement factor sqrt(A2/A1) accounts for the beneficial effect of surrounding concrete restraining lateral expansion.

For a base plate on a pier of the same plan dimensions, A2/A1 = 1.0 and the confinement benefit is lost. For a plate on a significantly larger pedestal, the factor can reach 2.0, effectively doubling the bearing resistance. In practice, most base plates sit on pedestals that extend 100-150 mm beyond the plate on all sides, yielding A2/A1 in the range of 1.3 to 1.8.

Design bearing capacity: phi x Nc = 0.65 x 0.85 x f'c x A1 x sqrt(A2/A1)

For 32 MPa concrete (the most common foundation concrete in Australian commercial construction):

• 300x300 plate, A2/A1 = 1.0: phi_Nc = 0.65 x 0.85 x 32 x 90000 x 1.0 / 1000 = 1,591 kN
• 400x400 plate, A2/A1 = 1.56 (assuming 500x500 pedestal): phi_Nc = 0.65 x 0.85 x 32 x 160000 x 1.25 / 1000 = 3,536 kN
• 500x500 plate, A2/A1 = 2.0: phi_Nc = 0.65 x 0.85 x 32 x 250000 x 2.0 / 1000 = 8,840 kN

Plate Bending — T-Stub Model per AS 4100 Clause 9.4

The base plate acts as a cantilever projecting beyond the column footprint. The critical bending plane is at the face of the column flange or web.

Bearing pressure: fp = N* / (B_plate x D_plate)

Cantilever distance: m = (B_plate - 0.95 x bfc) / 2 for projection parallel to column web; n = (D_plate - 0.80 x dc) / 2 for projection parallel to column flange. The factor 0.95 accounts for weld root; 0.80 accounts for the typical effective bearing width of the column web and root radius.

Required plate thickness: tp >= m x sqrt(2 x fp / (phi x fy_plate))

This is derived from the bending moment in a unit-width strip of plate: M* = fp x m^2 / 2, and the moment capacity per unit width: phi_Ms = phi x fy_plate x tp^2 / 4.

Alternative method using yield-line analysis: For base plates with four bolts, yield-line analysis gives: tp >= sqrt(4 x M_plate / (phi x fy_plate x L_eff)), where M_plate is the plate bending moment and L_eff is the effective yield-line length.

For Grade 300 plate (fy = 280 MPa for t > 12 mm), phi = 0.90: tp_required = m x sqrt(2 x fp / (0.90 x 280)) = m x sqrt(fp / 126)

Anchor Bolt Design — AS 4100 Clause 9.3

Anchor bolts resist tension from uplift, overturning moment, and shear from lateral loads.

Tension capacity per bolt (Clause 9.3.2.1): phi_Ntf = phi x 0.85 x fuf x As

Where fuf = bolt minimum tensile strength (830 MPa for Grade 8.8 bolts), As = tensile stress area of bolt. The factor 0.85 accounts for the reduced thread area and stress concentration.

Shear capacity per bolt: phi_Vfn = phi x 0.62 x fuf x Ac (threads in shear plane) or phi_Vfn = phi x 0.80 x fuf x As (shank in shear plane).

For bolts subject to combined tension and shear (Clause 9.3.2.4): (V*/phi_Vfn)^2 + (T*/phi_Ntf)^2 <= 1.0

Anchor bolt capacities for common Grade 8.8 bolts (threads excluded from shear plane):

Shear Key Design

When V* exceeds the shear capacity of the anchor bolts (typically 4 x phi_Vfn) or when combined tension and shear interaction exceeds unity, a shear key should be provided. The shear key is a welded steel block or section (typically a short length of UC or flat bar) that engages the concrete in bearing.

Shear key bearing capacity on concrete: phi x 1.3 x 0.85 x f'c x A_key, where A_key is the bearing area of the key against the concrete and 1.3 accounts for the triaxial confinement of the key within the grout pocket.

Shear key weld: Full-strength fillet weld around the perimeter of the shear key to the base plate. The weld must transfer the factored shear V*.

Grout Requirements

Per the ASI and industry practice (AS 4100 does not directly specify grouting, but references AS 3600 for concrete bearing):

• Grout thickness: 20-50 mm nominal between base plate and concrete
• Grout strength: Minimum fc'_grout >= fc'_concrete. Specify 40 MPa grout for 32 MPa concrete.
• Levelling nuts: Required on all anchor bolts to set plate elevation before grouting. Torque nuts after grout cures (7 days minimum).
• Grout holes: For base plates exceeding 600 x 600 mm, provide 50 mm diameter grout holes at plate centre to ensure complete filling.
• Edge form: Form around plate perimeter with timber or foam to contain grout during placement.

Worked Example — 310UC158 Compression Base Plate

Problem: Design the base plate for a 310UC158 column (Grade 300) subject to N* = 2,800 kN axial compression and V* = 120 kN shear. Concrete pedestal is 600 x 600 mm with f'c = 32 MPa. Base plate steel is Grade 300.

Section properties — 310UC158: dc = 326 mm, bfc = 316 mm, tf = 25.0 mm, tw = 15.4 mm

Step 1 — Determine plate size: Try 500 x 500 mm plate: A1 = 250,000 mm^2 Pedestal area = 600 x 600 = 360,000 mm^2 A2/A1 = 360,000 / 250,000 = 1.44, sqrt(1.44) = 1.20 phi_Nc = 0.65 x 0.85 x 32 x 250,000 x 1.20 / 1000 = 5,304 kN >= 2,800 kN. OK.

Step 2 — Bearing pressure: fp = 2,800,000 / 250,000 = 11.2 MPa

Step 3 — Cantilever distances: m = (500 - 0.95 x 316) / 2 = (500 - 300) / 2 = 100 mm (flange projection) n = (500 - 0.80 x 326) / 2 = (500 - 261) / 2 = 119.5 mm (web projection) c = max(m, n) = 119.5 mm

Step 4 — Required plate thickness: tp_req = 119.5 x sqrt(2 x 11.2 / (0.90 x 280)) = 119.5 x sqrt(22.4 / 252) = 119.5 x sqrt(0.0889) = 119.5 x 0.298 = 35.7 mm Specify 40 mm plate (standard thickness).

Step 5 — Anchor bolt selection: Use 4-M24 Grade 8.8 bolts. Minimum practical embedment = 300 mm per AS 3600. Check shear: V* = 120 kN, 4 bolts at 180.7 kN/bolt = 722.8 kN >> 120 kN. OK. Use 4-M24 with shear key for 120 kN (not required but good practice).

Step 6 — Weld design: Column to base plate: 8 mm fillet weld each side of flange (2 x 316 mm) = 632 mm length. Per AS 4100 Clause 9.7, phi_vw = 0.80 x 0.60 x 490 x 0.707 x 8 / 1000 = 1.33 kN/mm. Weld capacity = 632 x 1.33 = 840 kN >> 100 kN (tension from prying/nominal). OK. Also weld web: 6 mm fillet both sides of web = 2 x 326 x 1.00 kN/mm = 652 kN. OK.

Final specification: 500 x 500 x 40 mm base plate, Grade 300, 4-M24 Grade 8.8 anchor bolts at 400 mm embedment, 8 mm fillet welds all around column section, levelling nuts and 40 MPa grout.

Moment Base Plates

When the column carries significant moment in addition to axial load, the base plate must be designed for both compression and tension. The plate is divided into a compression zone (concrete bearing) and a tension zone (anchor bolts).

Neutral axis depth: xn = N* / (0.85 x f'c x B_plate) — iterative, as plate dimensions affect the concrete stress block.

Tension per bolt row: Ti = (M* - N* x (D/2 - 0.4 x xn)) / (d_tension)

Where d_tension is the distance from bolt centre to the compression centroid. The lever arm d_tension must be balanced against the bolt tension capacity, and the concrete bearing stress must remain below phi x 0.85 x f'c.

For large moments relative to axial load, the neutral axis may be shallow, placing the compression zone close to the edge. Under these conditions:

• Anchor bolts must develop full tension capacity
• The plate thickness may be governed by the tension side (prying action)
• Base plate stiffeners between bolt rows can significantly reduce required plate thickness

Construction and Erection Considerations

Base plates are fabricated off-site and must allow for site tolerances. Key practical points:

• Oversize anchor bolt holes: Use holes 6-10 mm larger than bolt diameter per AS 4100 Clause 14.3.5 to accommodate setting-out tolerances.
• Plate washers: 10 mm thick plate washers under nuts to distribute bolt pretension across oversized holes.
• Bolt projection: Minimum 3 threads above fully engaged nut for field inspection.
• Corrosion protection: Hot-dip galvanize base plates for exterior or corrosive environments. Unprotected steel in contact with concrete must have minimum 50 mm cover on all sides.
• Level mark on plate: Steel fabricator to mark a level datum on the plate for the site crew.

Design Checklist

Before finalising a base plate design, verify:

1. Concrete bearing stress <= phi_0.85_f'c_sqrt(A2/A1)
2. Plate bending at both flange and web projections
3. Anchor bolt tension from uplift or overturning
4. Anchor bolt shear (or provide shear key)
5. Combined tension + shear interaction on anchors
6. Weld capacity between column and plate
7. Grout thickness and strength specification on drawing
8. Anchor embedment depth adequate for concrete breakout
9. Oversize holes and plate washers specified for erection tolerance
10. Corrosion protection for exposure condition

Frequently Asked Questions

When should a shear key be provided on a base plate? A shear key is required when the factored shear V* exceeds the total shear capacity of the anchor bolts, or when combined tension and shear interaction on the bolts exceeds unity. In practice, shear keys become economic when V* exceeds about 250 kN (the shear capacity of 4-M20 bolts). The key engages the concrete in bearing and eliminates shear demand on bolts.

What governs base plate thickness — the T-stub model or the cantilever projection method? Both are equivalent formulations. The T-stub model (AS 4100 Clause 9.4) treats the plate as a T-stub flange in bending, while the cantilever projection method treats it as a cantilever strip. For concentrically loaded compression-only plates, they give identical results. For moment plates, the tension side must be checked for prying per Clause 9.3.4, which may govern plate thickness.

What is the difference between base plates and column splices in AS 4100? Base plates transfer column forces to concrete (different material, different stiffness), while splices transfer forces between steel sections of similar stiffness. Base plates must account for concrete bearing, grout, and anchor embedment. Splices are designed per Clause 9.5 for bearing or end-plate connections, and generally follow the same bolt and weld provisions.

How does grout thickness affect base plate design? Grout thicknesses up to 50 mm do not affect the concrete bearing capacity calculation. For thicknesses exceeding 50 mm, the grout must be designed as a structural element and may require reinforcement. The standard 25-40 mm grout bed is treated as a load-transfer medium only — the bearing check is based on the underlying concrete strength, not the grout.

What is the practical minimum plate thickness for Australian base plates? For lightly loaded columns (N* < 500 kN), 12 mm plate is common. For typical commercial building columns (N* = 1,000-3,000 kN), 20-32 mm is standard. For heavy industrial columns, 40-50 mm may be required. Plates thicker than 50 mm should be avoided where possible due to cost and welding preheat requirements.

This page is for educational reference. Base plate design per AS 4100:2020 Clause 9 and AS 3600:2018 Clause 12. All structural designs must be independently verified by a licensed Professional Engineer or Structural Engineer registered with Engineers Australia or the relevant state registration board. Results are PRELIMINARY — NOT FOR CONSTRUCTION.