---------------------- | ------------------------- | ------------------------- | ----------------- | ----------------- | | Bearing on concrete | J8 (phi=0.65) | AS 3600 Cl 12.6 (phi=0.6) | Cl 6.2.5 | Cl 25.3.1 | | Plate bending | Design Guide 1 | Cl 5.12 (phi=0.9) | Cl 6.2.5 | Cl 25.3 | | Anchor tension (steel) | ACI 318 Ch.17 (phi=0.75) | AS 4100 Cl 9.3.2.2 | EN 1992-4 | CSA A23.3 Annex D | | Anchor tension (breakout) | ACI 318 17.6.2 (phi=0.70) | AS 5216 | EN 1992-4 | CSA A23.3 D.5 | | Combined T+V interaction | ACI 318 17.6.3 | AS 4100 Cl 9.3.3 | EN 1992-4 | CSA A23.3 D.7 | | Shear lug design | Design Guide 1, Ch. 5 | Not covered | EN 1993-1-8 Annex | CSA S16 Cl 25.3 |

Key difference: AISC/ACI uses phi = 0.65 for concrete bearing, while AS 3600 uses phi = 0.6. EN 1993-1-8 uses the T-stub analogy for base plate bending, which can give different plate thicknesses than the AISC cantilever method.

Step-by-Step Example

Problem: Design a base plate for a W12x58 column carrying Pu = 250 kips axial compression (no moment) on 4000 psi concrete. Pedestal is 24x24 inches.

Step 1 -- Trial plate size: Column: d = 12.2 in, bf = 10.0 in. Try B = 16 in, N = 16 in. A1 = 16 * 16 = 256 in^2.

Step 2 -- Bearing capacity: A2/A1 = (2424)/(1616) = 576/256 = 2.25. sqrt(A2/A1) = 1.50 (capped at 2.0, OK). phi*Pp = 0.65 * 0.85 _ 4.0 _ 256 _ 1.50 = 0.65 _ 0.85 _ 4.0 _ 384 = 849 kips. Demand = 250 kips. Utilization = 250/849 = 0.29. OK.

Step 3 -- Bearing pressure: q = 250 / 256 = 0.977 ksi.

Step 4 -- Cantilever projections: m = (16 - 0.9512.2)/2 = (16 - 11.59)/2 = 2.205 in. n = (16 - 0.8010.0)/2 = (16 - 8.0)/2 = 4.0 in. n' = sqrt(12.210.0)/4 = sqrt(122)/4 = 2.76 in. lambdan' requires X = (4dbf/(d+bf)^2) * (Pu/phiPp) = (412.210.0/(22.2)^2)(250/849) = (488/492.84)0.294 = 0.291. lambda = 2sqrt(0.291)/(1+sqrt(1-0.291)) = 1.079/1.841 = 0.586. lambda*n' = 0.586*2.76 = 1.62 in. Controlling l = max(2.205, 4.0, 1.62) = 4.0 in (n controls).

Step 5 -- Required plate thickness: tp = 4.0 _ sqrt(2 _ 0.977 / (0.90 _ 36)) = 4.0 _ sqrt(1.954/32.4) = 4.0 _ sqrt(0.0603) = 4.0 _ 0.2455 = 0.98 in. Use 1-inch plate, A36. (For Grade 50 plate: tp = 4.0 _ sqrt(1.954/45) = 4.0 _ 0.2084 = 0.83 in, use 7/8-inch plate.)

Result: 16x16x1" A36 base plate. Bearing utilization = 0.29. Plate bending controls the thickness.

Common Design Mistakes

• Forgetting the A2/A1 confinement factor: When the concrete pedestal is larger than the base plate, the bearing capacity increases by sqrt(A2/A1) up to a factor of 2.0. Ignoring this factor leads to oversized plates.
• Using phi = 0.90 for concrete bearing: The resistance factor for concrete bearing is phi = 0.65 per AISC (or 0.60 per AS 3600), not 0.90. Using the steel bending phi for the concrete check is unconservative by 38%.
• Not checking plate bending at the web: For wide-flange columns, the plate bending between the column flanges (the "web yield line") can control when the column depth is significantly less than the plate length. AISC Design Guide 1 provides the lambda*n' check for this case.
• Ignoring grout effects on bearing: The grout layer under the base plate must have compressive strength at least equal to f'c of the concrete. Weak grout creates a bearing capacity bottleneck that the standard bearing equation does not capture.
• Undersizing anchor bolts for erection loads: Even gravity-only columns need anchor bolts sized for erection stability -- typically a minimum of four 3/4-inch rods. OSHA 1926.755 requires a minimum of four anchors for column erection safety.
• Not considering base plate flexibility in moment connections: For moment-loaded base plates, the plate must be thick enough to develop the anchor bolt tension without excessive bending. A flexible plate redistributes force away from the bolts and can invalidate the rigid plate assumption.

Frequently Asked Questions

What plate thickness is required for a W10×49 column on a 14×14 base plate with 300 kips axial load on 3,000 psi concrete? For a W10×49 (d = 10.0 in, bf = 10.0 in) on a 14×14 plate, the cantilever projection is n = (14 − 10.0) / 2 = 2.0 in. Bearing pressure q = 300,000 / (14 × 14) = 1,531 psi = 1.53 ksi. Required thickness: tp = n × √(2q / (φ × Fy)) = 2.0 × √(2 × 1.53 / (0.9 × 36)) = 2.0 × √(0.0944) = 2.0 × 0.307 = 0.61 in. Use 3/4 in plate (A36). For Grade 50 plate, tp drops to about 0.52 in — use 5/8 in. These are cantilever bending checks only; concrete bearing and anchor bolt design require separate checks.

How do anchor bolts resist combined tension and shear simultaneously? When an anchor bolt carries both tension (from overturning moment or uplift) and shear (from lateral loads), the interaction is checked using a combined demand equation. ACI 318-19 Section 17.8 provides a trilinear interaction check: if Nua/(φNn) ≤ 0.2, full shear capacity governs; if Vua/(φVn) ≤ 0.2, full tension capacity governs; otherwise Nua/(φNn) + Vua/(φVn) ≤ 1.2. For ductile steel failure modes specifically, ACI 318-19 R17.6.1.3 permits the less conservative (5/3) power interaction: (Nua/φNn)^(5/3) + (Vua/φVn)^(5/3) ≤ 1.0. This calculator uses the trilinear method (Section 17.8) which is applicable to all failure modes. The steel capacity φNn and φVn must account for the applicable failure modes — steel fracture, concrete breakout, pullout, and side-face blowout for tension; steel fracture and concrete pryout for shear. All applicable modes must be checked independently and the controlling one governs.

What is the concrete breakout capacity for a single 3/4-inch anchor bolt with 6-inch embedment in 4,000 psi concrete? Using ACI 318-19 Chapter 17, the basic breakout strength in tension for a single anchor is Nb = kc × √f’c × hef^1.5 = 24 × √4,000 × 6^1.5 = 24 × 63.2 × 14.7 = 22,250 lb ≈ 22.3 kips (normal-weight concrete, kc = 24). Applying φ = 0.70 for tension breakout gives φNb = 15.6 kips. This single-anchor value must be modified for group effects, edge distance, and eccentricity using the AN/ANo ratio and modification factors ψ — the tabulated value assumes the full projected cone area is available.

What are typical minimum base plate dimensions relative to the column size? Base plate dimensions must extend sufficiently beyond the column flanges to spread the bearing load to the concrete at a reasonable stress. As a starting point, add at least 2–3 inches per side beyond the column flange width and depth to allow for anchor bolt placement and grout. Minimum plate dimensions are also constrained by the anchor bolt pattern — bolts typically need at least 1.5 to 2 inches of edge distance from the plate edge, and standard anchor bolt patterns have minimum center-to-center spacings of 3–4 inches depending on rod diameter. Anchor bolt edge distances into the concrete (not just the plate) govern the ACI breakout capacity and often control plate size indirectly.

What happens to the bearing pressure distribution when the moment is large relative to the axial load? Under small eccentricity (e = M/P less than about N/6), the entire base plate stays in compression and bearing pressure varies linearly from a maximum on one side to a minimum on the other. As eccentricity increases, the minimum stress approaches zero and then the plate begins to lift off on the tension side — part of the plate loses contact with the grout and anchor bolts on that side must carry the resulting tensile reaction. For large eccentricity, the bearing stress distributes over a reduced contact length and can be several times the value under concentric load. This non-linear regime requires an iterative solution to locate the neutral axis of the contact zone.

How much grout should be specified under a steel base plate, and what strength is required? Non-shrink cementitious grout under base plates is typically 1–2 inches thick, providing clearance for leveling nuts during erection and allowing grout to flow completely under the plate. Grouting should be specified after the column is plumbed and final elevation is set. Thin grout pads (under 1/2 inch) can be difficult to place without voids. Thicker pads (over 3 inches) may require investigation of the grout shear and bearing capacity as a structural layer. The grout compressive strength should meet or exceed the concrete strength, typically f’c ≥ 5,000 psi (35 MPa) for standard structural applications — this ensures the bearing capacity equation using f’c is not reduced by a weaker grout layer.

Related pages

• Anchor bolts reference
• Concrete footing calculator
• Steel Column Buckling Calculator
• Steel grades reference
• Unit converter
• Tools directory
• Reference tables directory
• Guides and checklists
• How to verify calculator results
• Disclaimer (educational use only)
• concrete footing design calculator
• weld capacity for base plate connections
• Steel Calculator
• Base Plate Checklist

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AS 4100 Base Plate Design Procedure

What is AS 4100 Base Plate Design?

A steel column base plate transfers axial compression, shear, and moment from the column to the concrete foundation. Under AS 4100:2020, base plates are designed using the limit state method — checking bearing capacity, plate bending, and anchor bolt tension.

Design Flowchart

1. Determine Design Actions — axial force N*, shear V*, and moment M* at the column base
2. Size the Base Plate — select plate dimensions (B × D) based on bearing pressure
3. Check Bearing Capacity — verify concrete bearing stress ≤ φ·0.85·f'c·A2/A1 (AS 3600 Cl 12.6)
4. Check Plate Bending — cantilever projection method for uniform bearing
5. Design Anchor Bolts — check tension, shear, and combined actions
6. Check Raised Base Plate — if anchor bolts are in tension, check bolt bending

Bearing Capacity Calculation

The design bearing capacity of the concrete foundation:

φ·Nc = φ · 0.85 · f'c · A1 · √(A2/A1) ≤ 1.7 · φ · f'c · A1

Where:

• φ = 0.6 (concrete in bearing, AS 3600 Table 2.2.2)
• f'c = characteristic compressive strength (MPa)
• A1 = bearing area of the plate (mm²)
• A2 = maximum area of the supporting surface that is geometrically similar to A1 (mm²)

The bearing pressure under the plate:

q = N* / (B × D)

This must satisfy: q ≤ φ·0.85·f'c·√(A2/A1)

Base Plate Bending (Cantilever Projection)

For uniform bearing (no net uplift), the plate bends as a cantilever from the column flange edges.

The cantilever projection (n) is typically the larger of:

n = max[(B - bf) / 2, (D - d) / 2]

Where bf = column flange width, d = column depth.

The required plate thickness:

tp,min = n · √(2·q / (φ·fy))

Where:

• φ = 0.9 (steel in bending)
• fy = plate yield strength (MPa)
• q = bearing pressure (MPa)

Anchor Bolt Tension (Combined Actions)

When the base plate has both axial force and moment, one side may experience net uplift. In this case, anchor bolts resist tension.

The design tensile force per bolt:

N*tf = (M* - N*·e) / (n_bolts · s)

Where e = eccentricity from plate centroid to compression resultant, s = bolt spacing.

Combined shear and tension check per AS 4100 Cl 9.3.3:

(N*tf / φ·Ntf)^2 + (V*f / φ·Vf)^2 ≤ 1.0

Raised Base Plate (Bolt Bending)

If anchor bolt chairs or levelling plates are used, the bolts must be checked for bending over the grout height. The bolt acts as a cantilever with the tension force at the top of the grout pad.

Worked Example — 400×400×20 Base Plate, W250×73 Column

Given:

• Column: W250×73 (d = 254mm, bf = 254mm)
• Base plate: 400×400×20 PL, Grade 250 (fy = 250 MPa)
• Design axial: N* = 800 kN (compression)
• Concrete: f'c = 32 MPa
• No shear or moment

Step 1: Bearing pressure

q = N* / (B×D) = 800,000 / (400×400) = 5.0 MPa

Step 2: Check bearing capacity

φ·Nc = 0.6 × 0.85 × 32 × (400×400) × √(A2/A1)

Assuming A2/A1 = 1.5: φ·Nc = 0.6 × 0.85 × 32 × 160,000 × 1.225 = 3.19 MN

3.19 MN > 0.800 MN ✓ OK

Step 3: Cantilever projection

n = (400 - 254) / 2 = 73 mm

Step 4: Required plate thickness

tp = 73 × √(2 × 5.0 / (0.9 × 250)) = 73 × √(0.0444) = 73 × 0.211 = 15.4 mm

Use 20mm plate ✓

Result: 400×400×20 PL Grade 250 base plate satisfies bearing and bending checks.

Key AS 4100 Clauses

→ Use the Base Plate & Anchor Calculator to automate these checks for your specific configuration.