AS 4100 Base Plate Design -- Complete Procedure with Worked Examples

A column base plate transfers axial compression, bending moment, and shear from the steel column into the concrete foundation. Under AS 4100 (Steel Structures), the design must satisfy three distinct checks: concrete bearing per AS 3600, plate bending capacity per AS 4100 Section 5, and anchor bolt tension and shear per AS 4100 Section 9. The interaction among these three mechanisms determines the plate dimensions, bolt layout, and weld specification.

This reference covers the full design sequence, from determining design actions through to weld detailing and grout specification. Two worked examples are included: a pure compression base and a combined axial-plus-moment base. The levelling nut (yield line) method and seismic detailing provisions are addressed separately.

Governing Standards and Design Philosophy

Base plate design in Australia draws on two primary standards:

The design philosophy is limit state design with capacity factors from AS 4100 Table 3.4 and AS 3600 Table 2.2.2. The connection must be designed for the more severe of (a) the actions determined from structural analysis and (b) the minimum notional capacity per the standard.

Capacity Factors

Limit State Reference phi Application
Plate bending AS 4100 Tbl 3.4 0.90 Yield of plate in flexure
Bolt tension AS 4100 Tbl 3.4 0.80 Anchor bolt tensile rupture
Bolt shear AS 4100 Tbl 3.4 0.80 Anchor bolt shear rupture
Concrete bearing AS 3600 Tbl 2.2.2 0.60 Crushing of concrete under plate
Concrete cone breakout AS 3600 Sec 17 0.60 Tensile pullout cone in concrete
Fillet weld AS 4100 Tbl 3.4 0.80 Column-to-plate weld

Step-by-Step Design Procedure

Step 1: Determine Design Actions

The base plate design starts with the factored actions at the column base from the structural analysis. These come from the governing load combinations per AS/NZS 1170.0:

For moment-resisting frames, also determine the plastic moment capacity M_p of the column section -- the connection may need to be capacity-designed for 1.25 to 1.40 times the column plastic capacity.

Step 2: Select Preliminary Plate Geometry

Base plate plan dimensions are established from the column profile plus an edge margin:

For a typical UC section, plates are square or nearly square. The initial plate thickness estimate can be based on past experience (20-25 mm for UC200-300 series columns under moderate load) and verified by the bending check in Step 4.

Step 3: Concrete Bearing Check (AS 3600 Clause 12.6)

The bearing capacity of the concrete under the base plate is:

phi Nb = phi_b * 0.85 _ f'_c _ A1 * sqrt(A_2 / A_1)

subject to the upper bound: sqrt(A2 / A_1) <= 2.0, giving a maximum bearing stress of phi_b * 0.85 _ f'_c _ 2.0 = phib * 1.70 * f'_c.

Where:

The sqrt(A*2 / A_1) factor accounts for confinement from the surrounding concrete mass. For a base plate on a large footing or slab, A_2 >> A_1 and the full factor of 2.0 can be used, yielding a maximum usable bearing stress of approximately 1.02 * f'_c (phi reduced) or, in pre-factored terms, 1.70 _ f'_c.

Required plate area: A_1 >= N* / (phi N_b / A_1), i.e., N* / f_b where f_b is the factored bearing strength per unit area.

Step 4: Plate Bending Check

The concrete bearing pressure acts upward on the plate underside while the column applies concentrated reactions downward at the flange tips and web. This produces bending in the plate in two regions:

4a. Cantilever bending (plate projection beyond column face):

The plate cantilevers from the column face to the free edge, subject to the uniform bearing pressure q:

M*_cant = q * m^2 / 2 (bending moment per unit width)

Where m = (plate width B - column flange width b_f) / 2 is the cantilever projection.

Required thickness from the plastic section modulus: t*p >= sqrt(4 * M__cant / (phi * f_y))

Where phi = 0.90 for plate bending and f_y is the plate yield stress (typically 250, 300, or 350 MPa).

4b. Two-way bending (plate interior region):

For plates significantly larger than the column footprint, also check the bending of the interior plate region spanning between the column flanges. This is a plate strip loaded by bearing pressure and supported at the flange tips. For a rectangular plate under uniform load:

M*_interior = q * (d_i - 2 * k_s) ^ 2 / 8 (approximately, where k_s approximates the column flange outstand)

The controlling thickness is the larger of the cantilever and interior bending requirements. In most practical cases, cantilever bending governs.

Step 5: Anchor Bolt Tension Design

Anchor bolts resist tension from uplift, overturning moments, and nominal erection loads. For each bolt:

Tensile capacity: phi Ntf = phi * As * f_uf

Where:

Nominal tensile strength f_uf by bolt grade per AS 4100 Table 9.3.1:

Bolt Grade f_uf (MPa) f_yf (MPa) phi N_tf M24 (kN)
4.6 400 240 113
8.8 830 660 234

For seismic applications, Grade 4.6 bolts are preferred over Grade 8.8 because of their higher elongation capacity (minimum 20% vs 12%), which accommodates inelastic deformation without fracture.

Step 6: Bolt Tension Distribution (Moment-Resisting Bases)

For a base plate subject to combined axial load N* and moment M*:

  1. Compute the eccentricity e = M* / N*
  2. If e <= d_i/6: The entire plate is in compression. No bolt tension. The bearing pressure varies linearly from q_max at one edge to q_min at the other.
  3. If e > d_i/6: A portion of the plate develops tension. The tension force is resolved by the anchor bolts on the tension side.

For a rectangular plate with bolts symmetrically placed at distance d_b (centre-to-centre):

T_total = (M* / d_b) - (N* / 2) [subject to T_total >= 0]

T_per_bolt = T_total / n_bolts_tension_side

Where n_bolts_tension_side is the number of bolts on the tension face (typically 2 for a 4-bolt pattern).

Step 7: Shear Transfer

Shear at the column base transfers through three mechanisms in decreasing order of preference:

  1. Friction: V*f = mu * N_ where mu = 0.30 for steel on grout (unpainted, clean contact surface). Friction alone is often sufficient for gravity columns.
  2. Anchor bolts in shear: The bolt shear capacity per AS 4100 Clause 9.3.2.1: phi Vf = phi * 0.62 _ f_uf _ As (threads excluded from shear plane) or phi * 0.80 _ f_uf _ A_core (threads included).
  3. Shear key: A structural steel lug welded to the underside of the base plate, cast into a recess in the concrete. Required if friction plus bolt shear is insufficient. The key bears against the concrete in a formed pocket and transfers horizontal reaction directly.

Combined tension + shear interaction: Per AS 4100 Clause 9.3.2.3, for bolts with threads in the shear plane:

(T* / phi N_tf)^1.5 + (V* / phi V_f)^1.5 <= 1.0

Step 8: Anchor Bolt Embedment

The bolt must be embedded deeply enough into the concrete to develop its full tensile capacity without pullout or cone breakout. Check three failure modes:

  1. Concrete cone breakout (AS 3600 Section 17): The pullout cone angle is approximately 35 degrees from the bolt axis. The projected area of the cone at the concrete surface determines capacity. Edge distance and group effects reduce capacity when bolts are closely spaced.
  2. Bond failure along the bolt shaft (plain bars and headed anchors).
  3. Steel tensile failure of the bolt itself (the intended failure mode for ductile design).

Minimum practical embedment: 12 _ d_b for hooked anchors, 15 _ d_b for straight headed anchors. For M24 bolts: 288-360 mm. For M30 bolts: 360-450 mm.

Worked Example 1: Axial Compression Only

Problem

Design the base plate for a 250UC72.9 column carrying a design axial compression N* = 1,800 kN. The column bears on a mass concrete pad footing with f'_c = 32 MPa. Use Grade 250 plate and M24 Grade 8.8 anchor bolts. The footing plan dimensions are 2,400 mm x 2,400 mm -- large enough that A_2/A_1 will reach the 2.0 limit.

Column Properties (250UC72.9)

Step 1: Preliminary Plate Dimensions

Edge distance a_e = 55 mm (comfortably exceeds 1.5 * 24 = 36 mm and the 40 mm minimum).

B = 254 + 2 _ 55 = 364 mm -- round up to 370 mm N = 254 + 2 _ 55 = 364 mm -- round up to 370 mm

A_1 = 370 * 370 = 136,900 mm^2 = 0.1369 m^2

Step 2: Concrete Bearing

With A_2/A_1 taken as 2.0 (large footing relative to plate):

phi N*b = 0.60 * 0.85 _ 32 _ 136,900 _ sqrt(2.0) / 1000 = 0.60 _ 0.85 _ 32 _ 136,900 _ 1.414 / 1000 = 3,162 kN

Bearing capacity check: phi N_b = 3,162 kN > N* = 1,800 kN. OK. Substantial reserve.

Bearing stress under the plate: q = N* / A_1 = 1,800,000 / 136,900 = 13.15 MPa

Step 3: Plate Bending

Cantilever projection: m = (370 - 254) / 2 = 58 mm

Bending moment per unit width: M* = q * m^2 / 2 = 13.15 _ 58^2 / 2 = 13.15 _ 3,364 / 2 = 22,112 Nmm/mm

Required thickness: t*p = sqrt(4 * 22,112 / (0.90 _ 250)) = sqrt(88,448 / 225) = sqrt(393.1) = 19.8 mm

Adopt t_p = 20 mm Grade 250 plate.

Step 4: Anchor Bolt Check

For a pure compression connection, bolts resist nominal erection loads only. Minimum 4 * M24 Grade 8.8.

Tension capacity per bolt: phi N*tf = 0.80 * 353 _ 830 / 1000 = 234.4 kN

Four bolts provide 938 kN total -- more than adequate for any reasonable erection load.

Step 5: Weld Design

For full bearing on grout, the column-to-plate weld transfers only enough load to hold the column in place. Per AS 4100 minimum fillet weld sizes:

Plate thickness 20 mm, column flange thickness 14.2 mm: minimum 6 mm fillet weld.

Specify: 6 mm fillet weld, all around column profile, E48XX electrodes.

Result

Component Specification
Base plate 370 x 370 x 20 mm, Grade 250
Anchor bolts 4 x M24 Grade 8.8, 300 mm embedment
Weld 6 mm fillet all around, E48XX
Grout 40 mm non-shrink cementitious, f_c >= 40 MPa

Worked Example 2: Combined Axial Compression and Major-Axis Moment

Problem

Design a base plate for a 310UC118 column carrying N* = 900 kN (compression) and M*_x = 220 kNm. The footing is 32 MPa concrete on a 2,800 mm x 2,800 mm pad. Use Grade 300 plate and M30 Grade 8.8 anchor bolts.

Column Properties (310UC118)

Step 1: Preliminary Plate Dimensions

Because the overturning moment is significant, the plate is extended in the moment direction to engage a wider bolt couple.

B (width, parallel to moment axis): 314 + 2 _ 70 = 454 mm -- round up to 460 mm N (depth, perpendicular): 307 + 2 _ 55 = 417 mm -- round up to 420 mm

A_1 = 460 * 420 = 193,200 mm^2 = 0.1932 m^2

Step 2: Eccentricity Check

e = M* / N* = 220 / 900 = 0.244 m = 244 mm

Plate dimension in the moment direction B = 460 mm. Kern distance = 460/6 = 76.7 mm.

e = 244 mm > 76.7 mm -- significant tension develops on one side. The compression zone is a rectangular stress block at the bearing face.

Step 3: Tension Force in Anchor Bolts

Bolt layout: 4 x M30 bolts at 70 mm edge distance, giving a bolt couple lever arm d_bolt = 460 - 2 * 70 = 320 mm centre-to-centre.

Taking moments about the compression face centroid (assume the resultant compression acts at the centre of the bearing stress block, approximately d_comp = B * 0.42 = 193 mm from the compression edge for full bearing, or at mid-depth of the compression zone):

Using the simplified plastic stress distribution:

C = N* + T_total (equilibrium) C * (B/2 - a/2) = M* + T_total * d_bolt/2 (moment equilibrium about column centroid)

where a = depth of compression stress block. Iterating on a:

Assume a = 180 mm initially. Compression centroid from plate edge ~ a/2 = 90 mm.

Ttotal = (M - N* * (B/2 - a/2)) / d*bolt = (220 - 900 * (0.230 - 0.090)) / 0.320 = (220 - 900 * 0.140) / 0.320 = (220 - 126) / 0.320 = 94 / 0.320 = 293.8 kN

Check compression: C = 900 + 293.8 = 1,193.8 kN at bearing stress q = 1,193.8 _ 1000 / (420 _ 180) = 1,193,800 / 75,600 = 15.8 MPa < 0.60 _ 0.85 _ 32 * 2.0 = 32.6 MPa. OK.

Step 4: Bolt Tension Capacity

M30 Grade 8.8: A_s = 561 mm^2, f_uf = 830 MPa.

phi N*tf = 0.80 * 561 _ 830 / 1000 = 372.5 kN per bolt

Tension per bolt (2 bolts on tension face): T*_bolt = 293.8 / 2 = 146.9 kN

phi N_tf = 372.5 kN > T*_bolt = 146.9 kN. OK. Reserve is more than 2.5x -- adequate for prying allowance.

Step 5: Plate Bending (Cantilever)

Cantilever projection in moment direction: m = 70 - t_f/2 = 70 - 18.7/2 = 60.7 mm

Bearing pressure (worst case, compression side): q = 15.8 MPa

M*_cant = 15.8 * 60.7^2 / 2 = 15.8 * 3,685 / 2 = 29,107 Nmm/mm

Required thickness: t*p = sqrt(4 * 29,107 / (0.90 _ 300)) = sqrt(116,428 / 270) = sqrt(431.2) = 20.8 mm

This is for the compression side cantilever. The tension side also requires checking for prying (plate bending under bolt pull). For tension-side bending, the yield line from the bolt centre to the plate edge governs.

Step 6: Plate Bending (Tension Side -- Prying Check)

Bolt centre to plate edge (at 70 mm edge distance): the plate yields between the bolt and the plate edge under the upward bolt pull.

Using the simplified prying model: the plate must be sufficiently thick to develop the bolt tension without excessive deformation. For M30 bolts, a 25 mm thick plate with 70 mm edge distance provides adequate stiffness to limit prying to below 10% of bolt tension. (Detailed yield-line verification available in the full design guide.)

Adopt t_p = 25 mm Grade 300 plate.

Step 7: Combined Shear + Tension Interaction

For completeness, check the bolt group for the case where lateral shear coexists with tension. If V* = 80 kN (from wind or seismic lateral load):

Shear per bolt: V*_bolt = 80 / 4 = 20 kN

phi V*f (M30, threads in shear plane) = 0.80 * 0.62 _ 830 * 561 / 1000 = 230.6 kN

Interaction: (146.9/372.5)^1.5 + (20/230.6)^1.5 = (0.394)^1.5 + (0.087)^1.5 = 0.247 + 0.026 = 0.273 < 1.0. OK.

Step 8: Embedment

Minimum embedment for M30 hooked anchors: 12 * 30 = 360 mm. Use 400 mm to the hook tangent point plus 150 mm hook tail.

Result

Component Specification
Base plate 460 x 420 x 25 mm, Grade 300
Anchor bolts 4 x M30 Grade 8.8, 400 mm embedment + hook
Weld 8 mm fillet all around column profile, E48XX
Grout 40 mm non-shrink cementitious, f_c >= 40 MPa
Shear key Not required (friction: 0.30 _ 900 = 270 kN > V_ = 80 kN)

The Levelling Nut Method -- Yield Line Analysis

When the base plate is supported on levelling nuts rather than a continuous grout bed, the load path changes fundamentally: the column load passes through discrete contact points (the nuts) to the anchor bolts and then into the foundation. The plate spans between the nuts and must be designed for the associated bending.

When Levelling Nuts Govern

Yield Line Mechanism

The yield line method models the plate as a plastic mechanism at failure. For a rectangular base plate with bolts at the corners:

  1. Yield line pattern: Diagonal yield lines radiate from the column corners to the plate edges, subdividing the plate into triangular rigid regions that rotate about the column face.
  2. External work: The bearing pressure q acting through the virtual displacement of each triangular region.
  3. Internal work: The plastic moment capacity M_p = f_y * t_p^2 / 4 per unit width, times the rotation angle, integrated along each yield line.
  4. Equilibrium: The internal work dissipated in the hinges equals the external work done by the bearing pressure:

SUM (Mp * thetai * Lhinge_i) = SUM (q * Aregion_j * delta_j)

Solving for t_p gives the required plate thickness for the yield line pattern. This typically produces a thickness 20-40% higher than the cantilever bending method because the mechanism involves two-way bending with longer effective spans.

Practical Guidance

Weld Design: Column to Base Plate

Minimum Fillet Weld Sizes (AS 4100 Clause 9.7.3.10)

Thicker part thickness t (mm) Minimum fillet leg size (mm)
t <= 10 4
10 < t <= 20 6
20 < t <= 30 8
30 < t <= 40 10
40 < t <= 50 12

For a column in full compression bearing on grout, the weld is nominally loaded: it transfers enough force to secure the column during erection but does not need to develop the full column axial capacity. A 6 mm fillet weld around the profile is the standard minimum for UC200-UC310 columns.

For columns with tension or moment (the tension flange in Example 2), the weld on the tension side must develop the full tension flange force. For the 310UC118 example, the tension flange force at the moment M*_x = 220 kNm is approximately M* / (d - t*f) = 220,000 / (0.314 - 0.019) = 220,000 / 0.295 = 746 kN. An 8 mm fillet weld (capacity ~ 1.03 kN/mm for E48XX electrodes on Grade 300 plate, design capacity per AS 4100 Clause 9.7.3.10) around the flange (perimeter ~ 307 * 2 + 18.7 + 307 = 940 mm) provides 940 _ 1.03 = 968 kN > 746 kN. OK.

Weld Electrode Selection

Electrode Nominal tensile strength (MPa) Matching plate steel
E41XX 410 Grade 250
E48XX 480 Grade 300, 350
E55XX 550 Grade 400, 450 (quenched & tempered)

Match the electrode to the weaker of the two joined parts. For Grade 250 and 300 plate, E48XX is standard.

Grout Specification

Material Requirements

Structural non-shrink cementitious grout for base plates:

Placement Procedure

  1. Concrete surface preparation: Roughen to minimum 3 mm amplitude (scabbling or needle-gunning). Clean of laitance, oil, and debris. Saturate surface-dry (SSD) immediately before grouting -- dry concrete absorbs mix water and causes localized shrinkage.
  2. Form construction: Build a tight form around the plate perimeter with grout inlet holes at one or two sides and vent holes at the opposite edges. The form must extend 10-15 mm above the plate underside to provide hydrostatic head.
  3. Grout mixing: Use a mechanical paddle mixer. Add the powder to the water (not water to powder). Mix for 3-5 minutes until uniform. Do not over-water -- each 1% excess water reduces compressive strength by approximately 3-5%.
  4. Placement: Pump or pour from one side only to push air out the opposite vents. Maintain continuous flow until all vents show fresh grout. Do not vibrate the plate during grouting -- vibration can cause segregation and entrapment.
  5. Curing: Wet cure for minimum 7 days (wet burlap, polyethylene sheet, or curing compound). Ambient temperature must be above 5 deg C and below 35 deg C during curing.

Quality Control

Seismic Detailing Provisions

For columns in seismic-resisting frames designed to AS 4100 Supplement 1 and AS/NZS 1170.5:

Capacity Design

The base plate connection must be designed for 1.25 to 1.40 times the column plastic moment capacity M_p, not just the actions from linear analysis. This ensures the column yields in flexure before the base plate, welds, or anchor bolts fracture.

Bolt Grade Selection

Property Grade 4.6 Grade 8.8
f_uf (MPa) 400 830
f_yf (MPa) 240 660
Elongation (min) 20% 12%
Seismic use Preferred Acceptable if verified

Grade 4.6 bolts are preferred in seismic applications because of higher elongation. The lower strength is compensated by using more or larger bolts.

Detailing Requirements

Detail Non-seismic Seismic
Design basis Factored actions from analysis 1.25 * column M_p capacity
Minimum weld 6 mm fillet 8 mm fillet or CJP groove at tension flange
Shear transfer Friction permitted Shear key required
Weld inspection Visual Visual + MPI or UT at tension region
Bolt embedment 12 * d_b 15 * d_b minimum
Grout strength 30 MPa 40 MPa minimum
Base plate stiffeners Not typically required Required for M* > 300 kNm

Comparison with AISC Design Guide 1

For engineers familiar with the AISC method, the AS 4100 approach differs in several details:

Item AISC DG 1 (US) AS 4100 (Australia)
Bearing stress phi*c * 0.85 _ f'_c * sqrt(A2/A1) phi _ 0.85 _ f'_c * sqrt(A2/A1)
Bearing phi 0.65 (concrete) 0.60 (AS 3600)
Plate bending phi 0.90 0.90
Bolt tension phi 0.75 0.80
Cantilever bending model M* = q * (m - 0.95*d)^2 / 2 M* = q * m^2 / 2
Yield line method AISC Manual Part 14, bracket approach First principles per AS 4100 Commentary
Minimum edge distance 1.5 * d_b (preferred) 1.5 * d_b or 40 mm (whichever greater)
Grout requirements 1-2 in typical, 4,000 psi min 25-50 mm typical, 30-40 MPa min

The AISC DG 1 method accounts for the column profile area when computing the cantilever moment (deducting 0.95 * column depth from the projection), which results in slightly thinner plates for a given geometry. The AS 4100 method (using the full projection m without deduction) is more conservative. When base plate designs from both standards are compared, the AS 4100 plates are typically 2-4 mm thicker for the same column and load combination.

Frequently Asked Questions

What is the governing AS 4100 clause for base plate design? Base plates are governed by AS 4100 Clause 9 (Connections) for the bolted interface and Clause 5 (Members in Bending) for plate flexure. The concrete bearing check is per AS 3600 Clause 12.6. There is no single clause titled "base plate design" -- the design is assembled from the relevant provisions in both standards.

How is the A1/A2 ratio determined for bearing? A_1 is the plan area of the base plate (B * N). A_2 is the maximum area of the supporting concrete surface that is geometrically similar to and concentric with A_1. For a base plate on a footing much larger than the plate, A_2 is limited by the footing plan dimensions. The ratio sqrt(A_2/A_1) is capped at 2.0, representing the maximum beneficial confinement from surrounding concrete.

When do I need to check prying action on anchor bolts? Prying should be checked when the base plate thickness is less than t_min for a given bolt tension -- see the simplified prying model in Step 6 of Example 2. Plates 25 mm and thicker with standard edge distances (>= 1.5 * d_b) typically develop minimal prying for M24 and M30 bolts. For thinner plates or larger bolts, calculate the required thickness or increase the bolt design tension by 20-30%.

What is the minimum base plate thickness per AS 4100? AS 4100 does not specify a numerical minimum for base plates. Engineering practice uses t_p >= max(16 mm, 2 * t_f_column). For most structural columns (UC200-UC310), 20-25 mm is typical. Plates thinner than 16 mm are susceptible to handling damage, welding distortion, and are difficult to grout under without warping.

How do I design for large overturning moments? For M* > 500 kNm on a single column, consider: (a) extending the plate in the moment direction to increase the bolt couple lever arm, (b) using 6 or 8 bolts instead of 4, (c) adding vertical stiffener plates (gussets) between the column flanges and the base plate, (d) specifying a deeper embedment for the tension bolts with supplementary hairpin reinforcement across the pullout cone, and (e) verifying the concrete footing itself for the uplift reaction.

What grout strength is required under a base plate? AS 4100 does not specify grout strength directly, but AS 3600 requires the grout to be at least as strong as the foundation concrete. For typical foundations with f'_c = 32 MPa, specify grout with a minimum 28-day strength of 40 MPa to provide margin. Non-shrink cementitious grout meeting the requirements of AS 1478.2 is the standard product.

Related Pages

Disclaimer

This is a calculation reference, not a substitute for professional engineering certification. All results must be independently verified by a licensed Professional Engineer (PE), Chartered Professional Engineer (CPEng), or Registered Structural Engineer before use in construction, fabrication, or permit documents. The user bears full responsibility for the accuracy of all inputs and the independent verification of all outputs against the applicable edition of AS 4100, AS 3600, and the project-specific design brief.