path: /blog/base-plate-design-example/ canonical: https://steelcalculator.app/blog/base-plate-design-example/ metatitle: 'Base Plate Design Example -- AISC 360, AS 4100, EN 1993 & CSA S16' meta_description: 'Complete base plate design guide with worked examples for all four major steel codes. W200x52 on 400x400 base plate with anchor bolts. Concrete bearing, plate bending, and bolt checks.' robots: 'index,follow' lastmod: '2026-05-20' schema_file: 'schema/blog_base-plate-design-example.json' FAQPage: '@type': 'FAQPage' mainEntity: - '@type': 'Question' 'name': 'How do you calculate the required base plate thickness?' 'acceptedAnswer': '@type': 'Answer' 'text': 'Base plate thickness is calculated from the critical bending moment in the plate, treating the plate projection beyond the column footprint as a cantilever. Per AISC 360 Section 14 and the AISC Design Guide 1 method, the required thickness is t_req = L * sqrt(2 _ P_u / (0.9 _ Fy * B _ N)), where L is the larger of m and n (the cantilever distances), P_u is the factored axial load, F_y is the plate yield strength, and B and N are the base plate dimensions. For lightly loaded columns, the minimum practical thickness is typically 20 mm.' - '@type': 'Question' 'name': 'What is the concrete bearing capacity under a base plate?' 'acceptedAnswer': '@type': 'Answer' 'text': 'Concrete bearing capacity depends on the ratio of the concrete support area to the loaded area. Per AISC 360 J8, the nominal bearing strength is P_p = 0.85 _ f''c _ A1 _ sqrt(A2/A1) where A1 is the base plate area and A2 is the maximum concrete area that is geometrically similar and concentric with A1. The sqrt(A2/A1) term is capped at 2.0. AS 4100 uses a similar approach with phi=0.6 applied to 0.85f''cA1sqrt(A2/A1).' - '@type': 'Question' 'name': 'How are anchor bolts designed for base plates?' 'acceptedAnswer': '@type': 'Answer' 'text': 'Anchor bolts are designed for the tension, shear, and combined loading transferred from the column to the foundation. Per AISC 360 J9, bolt tension capacity uses the tensile stress area and nominal tensile strength (0.75F_u for headed anchors). Shear is transferred either by bolt bearing on the base plate, shear lugs, or friction. The concrete breakout strength in tension (ACI 318 Chapter 17) often governs for anchors near slab edges. Anchor bolt embedment depth, edge distance, and bolt spacing all affect the design. Typically, M20-M36 Grade 8.8 (or ASTM F1554 Grade 36/55/105) anchors are used.'

Base Plate Design Example -- AISC 360, AS 4100, EN 1993 & CSA S16

Base plate design is one of the most common tasks in structural steel engineering. Every column needs to transfer its load to the foundation, and the base plate -- a flat steel plate welded to the column base and anchored to the concrete footing -- is the component that makes that transfer possible. Despite its apparent simplicity, base plate design involves multiple checks: concrete bearing, plate bending, anchor bolt tension, anchor bolt shear, and the interaction of all four.

This post walks through a base plate design example under all four major structural steel codes (AISC 360, AS 4100, EN 1993, CSA S16) so you can compare the methodology across jurisdictions. The same column and base plate geometry is used throughout so you can see where the standards diverge and where they agree.

Disclaimer: All numeric examples are illustrative only and must not be treated as design-ready values. Always verify with the governing standard and a qualified Professional Engineer.

Why Base Plate Design Matters

When a steel column sits on a concrete footing, the concentrated load must be distributed over a large enough area to avoid crushing the concrete. The base plate spreads the column load across the footing, and anchor bolts keep the column from lifting or sliding. Getting the base plate thickness wrong means either an over-designed, expensive plate or -- worse -- an under-designed plate that yields in bending or punches through the concrete.

The typical design sequence is:

1. Determine the factored column loads (axial, shear, moment).
2. Estimate the required base plate area from concrete bearing strength.
3. Calculate the base plate bending moment (cantilever model beyond the column footprint).
4. Determine the required plate thickness from plate bending capacity.
5. Check anchor bolts for tension, shear, and combined loading.
6. Verify concrete breakout and pullout for tension anchors.

Worked Example: W200x52 Column on 400x400 Base Plate

For this comparison, we use the same base geometry across all four codes. This allows a direct side-by-side comparison of the methodology and any differences in the resulting design.

Design Inputs (Common to All Codes)

For quick lookup of column section properties used in base plate design (d, b_f, t_f, t_w), see the Section Properties Database.

AISC 360-22 Method (United States)

Step 1: Concrete bearing strength (AISC 360 J8)

The nominal bearing strength of the concrete is:

P*p = 0.85 * f''c _ A1 * sqrt(A2/A1)

where A1 = B _ N = 0.40 _ 0.40 = 0.16 m^2 and A2 is the area of the concrete support surface that is geometrically similar and concentric with the loaded area. For a footing that is 800 mm x 800 mm (A2 = 0.64 m^2):

sqrt(A2/A1) = sqrt(0.64/0.16) = 2.00

This is at the cap of 2.0, so:

P*p = 0.85 * 25 _ 160,000 * 2.0 = 6,800,000 N = 6,800 kN

Design bearing strength (phi_c = 0.65 per AISC 360 J8):

phic * Pp = 0.65 * 6,800 = 4,420 kN

Since 4,420 kN >> 850 kN, bearing is not critical. The base plate size is governed by plate bending, not concrete bearing.

Step 2: Base plate bending (AISC Design Guide 1)

The cantilever distances m and n are:

m = (N - 0.95d) / 2 = (400 - 0.95206) / 2 = (400 - 195.7) / 2 = 102.2 mm n = (B - 0.80b_f) / 2 = (400 - 0.80204) / 2 = (400 - 163.2) / 2 = 118.4 mm

The critical cantilever length L_crit = max(m, n, lambda*n'') = max(102.2, 118.4, ...). Using the larger of m and n:

L_crit = max(102.2, 118.4) = 118.4 mm

Base plate bending moment per unit width (treating the projection as a cantilever):

Mpl = (P_u / A1) * Lcrit^2 / 2 = (850,000 / 160,000) * (0.1184)^2 / 2

M_pl = 5.3125 * 0.01402 / 2 = 0.0372 MPa-m = 37.2 kN-mm/mm width

Step 3: Required plate thickness

Per AISC 360 Section 14 and DG1:

treq = sqrt(4 * Mpl / (phi_b * F_y))

where phi_b = 0.90 for plate bending:

t*req = sqrt(4 * 37,200 / (0.90 _ 250)) = sqrt(148,800 / 225) = sqrt(661.3) = 25.7 mm

Use a 26 mm base plate (or round up to 30 mm for practical procurement).

Step 4: Anchor bolt tension check

For the applied moment M_u = 12 kN-m with axial compression P_u = 850 kN, the column is in net compression. The moment produces a tension force in one pair of bolts, superimposed on the uniform compression:

T_b = M_u / (d_bolt_spacing) - P_u / 4

T_b = 12,000 / 0.320 - 850,000 / 4 = 37,500 - 212,500 = negative (net compression)

No tension governs in this case. The anchor bolts experience only shear.

Step 5: Anchor bolt shear

Shear per bolt = V_u / 4 = 45,000 / 4 = 11,250 N

M24 Grade 8.8 bolt tensile stress area A_t = 353 mm^2

Nominal shear strength per AISC 360 J3.6: Fnv = 0.563 * Fu for threads in shear plane = 0.563 * 830 = 467 MPa

phiv * Rnv = 0.75 * 467 _ 353 = 0.75 _ 164,851 = 123,638 N per bolt

Since 123.6 kN >> 11.25 kN, bolt shear is not critical.

AS 4100:2020 Method (Australia)

Step 1: Concrete bearing strength (AS 3600 / AS 4100 Section 9)

The Australian approach is conceptually identical to AISC but uses different phi factors. Per AS 4100:

P*n = 0.85 * f''c _ A1 * sqrt(A2/A1)

sqrt(A2/A1) = 2.0 (capped)

P*n = 0.85 * 25 _ 160,000 * 2.0 = 6,800 kN

phi = 0.60 (AS 4100 Table 3.4 for concrete bearing)

phi _ P_n = 0.60 _ 6,800 = 4,080 kN

4,080 kN >> 850 kN -- bearing is not critical.

Step 2: Base plate bending (AS 4100)

The Australian methodology uses the same cantilever model as AISC. The key difference is the capacity factor phi = 0.90 for plate bending.

M_pl = 37.2 kN-mm/mm (same as calculated above)

treq = sqrt(4 * Mpl / (phi * F*y)) = sqrt(4 * 37,200 / (0.90 _ 250)) = 25.7 mm

The result matches AISC for this case. Use 26 mm or 30 mm plate.

Step 3: Anchor bolts (AS 4100 Section 9)

M24 Grade 8.8 bolt shear capacity per AS 4100 Table 9.3.1:

Vf = phi * 0.62 _ f_uf _ (nn * A_c + n_x * A_o)

For threads in shear plane with one shear plane:

phi = 0.80 V*f = 0.80 * 0.62 _ 830 _ 1.0 _ 353 = 0.80 _ 0.62 _ 830 * 353 = 145,255 N per bolt

Bolt shear demand = 11.25 kN << 145.3 kN -- OK.

EN 1993-1-8 Method (Europe)

Step 1: Concrete bearing (EN 1992-1-1)

The Eurocode approach uses the design compressive strength of concrete:

FRdu = A_c0 * fcd * sqrt(A_c1/A_c0)

where fcd = alpha_cc * fck / gamma_c = 0.85 * 25 / 1.5 = 14.17 MPa

A_c0 = 160,000 mm^2 A_c1 = 640,000 mm^2 (assuming 800x800 footing) sqrt(A_c1/A_c0) = 2.0 (capped at 3.0 in EN 1992)

F*Rdu = 160,000 * 14.17 _ 2.0 = 4,534,400 N = 4,534 kN

Since the axial load is only 850 kN, bearing is not critical.

Step 2: Base plate bending (EN 1993-1-8 Section 6)

The Eurocode uses the T-stub model for base plates under moment, which is more detailed than the AISC cantilever model. For a purely axial case, the approach simplifies to the cantilever method.

Using the effective projection c:

c = tp * sqrt(fy / (3 * f_jd * gamma_M0))

where fjd = 2/3 * fcd = 2/3 * 14.17 = 9.44 MPa (joint bearing strength)

The required thickness is:

t*req = c * sqrt(3 _ f_jd * gamma_M0 / f_y)

For the cantilever projection L_crit = 118.4 mm:

t*req = 118.4 * sqrt(3 _ 9.44 _ 1.00 / 250) = 118.4 _ sqrt(0.1133) = 118.4 * 0.3366 = 39.9 mm

This is notably thicker than the AISC/AS 4100 result because the EN method uses a lower bearing stress and a different effective area assumption. In practice, the EN method often yields more conservative base plate thicknesses when the concrete bearing stress is low.

Using plate steel S275 (f_y = 275 MPa):

t*req = 118.4 * sqrt(3 _ 9.44 / 275) = 118.4 _ sqrt(0.103) = 118.4 _ 0.321 = 38.0 mm

Step 3: Anchor bolts (EN 1993-1-8)

M24 Grade 8.8 bolt shear resistance:

Fv,Rd = alpha_v * fub * A_s / gamma_M2

For Grade 8.8 with threads in shear plane (alpha_v = 0.6):

F*v,Rd = 0.6 * 830 _ 353 / 1.25 = 140,750 N per bolt

Bolt shear demand = 11.25 kN << 140.8 kN -- OK.

CSA S16:24 Method (Canada)

Step 1: Concrete bearing strength (CSA S16 Clause 25)

Br = 0.85 * phic * f''c _ A1 _ sqrt(A2/A1)

phi_c = 0.65 for concrete

B*r = 0.85 * 0.65 _ 25 _ 160,000 _ 2.0 = 4,420,000 N = 4,420 kN

4,420 kN >> 850 kN -- not critical.

Step 2: Base plate bending

The Canadian approach follows the same cantilever method as AISC:

M_pl = 37.2 kN-mm/mm

treq = sqrt(4 * Mpl / (phi * F*y)) = sqrt(4 * 37,200 / (0.90 _ 300)) (using 300W steel, F_y=300 MPa) = sqrt(148,800 / 270) = 23.5 mm

Step 3: Anchor bolts (CSA S16 Clause 13)

M24 Grade 8.8 bolt shear capacity:

Vr = 0.60 * phib * A_b * F_u

phi_b = 0.80 (bearing-type connections)

V*r = 0.60 * 0.80 _ 353 * 830 = 140,726 N per bolt

Bolt shear demand = 11.25 kN << 140.7 kN -- OK.

Code Comparison Summary

The key observation is that AISC, AS 4100, and CSA S16 produce similar base plate thicknesses (23-26 mm) when using the same cantilever method with code-specific phi factors and steel grades. The EN 1993 T-stub approach produces a thicker plate (38 mm), reflecting the more conservative European treatment of concrete bearing stress distribution.

Common Pitfalls in Base Plate Design

1. Forgetting the sqrt(A2/A1) cap. Both AISC (2.0) and EN 1992 (3.0) cap this ratio. If your footing is much larger than the base plate, do not use the actual area ratio -- apply the cap.

2. Ignoring moment. The worked example above had net compression from the combination of axial load and moment. When moment dominates, the anchor bolts on the tension side carry the uplift, and the base plate bending model changes from uniform pressure to a triangular distribution. Always check both pure compression and combined loading cases.

3. Assuming anchor bolts take shear automatically. Anchor bolts in oversized holes (common for construction tolerance) may not bear on the base plate until the column displaces several millimetres. For significant shear, provide a shear lug (a short section welded to the underside of the base plate, cast into a pocket in the footing) or rely on friction under the base plate (V <= mu * P_u, where mu = 0.30 to 0.55 for steel on grout).

4. Grout thickness. A 25-50 mm grout pad between the base plate and the concrete footing is standard. The grout fills irregularities and ensures uniform bearing. The AISC DG1 and EN 1993-1-8 both assume the grout is non-shrink and at least as strong as the concrete. If the grout thickness exceeds 50 mm, the grout pad should be checked as a separate element.

5. Weld between column and base plate. The fillet weld connecting the column to the base plate must transfer the full column load. For a W200x52, the perimeter weld length is approximately 2*(206+204) - 2*t_w = ~800 mm (minus the web thickness). With a 6 mm fillet weld (E70XX, capacity ~0.9 kN/mm at 45 degrees), the total weld capacity far exceeds the 850 kN column load. However, for columns subject to uplift, check the weld for the tension case.

When to Use a Stiffened Base Plate

For heavy columns or large moments, an unstiffened base plate may become impractically thick (t > 50 mm). In these cases, stiffeners (gusset plates) welded between the column flanges and the base plate reduce the cantilever span and allow a thinner plate.

The decision threshold varies by code:

• AISC DG1: consider stiffeners when the calculated t_req exceeds approximately 50 mm.
• AS 4100: similar guidance; stiffened plates reduce bending by supporting the plate between the column flange tips and the stiffener edges.
• EN 1993-1-8: the T-stub model naturally accommodates stiffened configurations by subdividing the effective bearing area.
• CSA S16: consistent with AISC practice.

For the 850 kN column in this example, the required 26 mm plate is reasonable without stiffeners. If the load doubled to 1,700 kN, the required thickness would be approximately 36 mm (AISC), still manageable without stiffeners, but approaching the range where stiffeners would reduce plate cost.

Practical Design Tips

• Base plate dimensions: Typically extend 100-150 mm beyond the column footprint in both directions. This provides room for anchor bolts and a reasonable bearing area.
• Anchor bolt diameter: M20-M30 for typical building columns. Four-bolt patterns are standard; eight-bolt patterns are used for heavy columns or high-moment bases.
• Hole size: Anchor bolt holes are typically oversized by 6-10 mm to allow for placement tolerance. Use plate washers (typically 75x75x10 mm) under the nut to provide bearing.
• Leveling nuts: A second nut below the base plate allows the column to be levelled before grouting. The lower nuts are hand-tight and carry no design load after grouting.

Interpreting the Results for Your Own Design

When you run a base plate design on the Steel Calculator, you will see the same sequence of checks described above. The calculator:

1. Computes the concrete bearing capacity from your input dimensions and concrete strength.
2. Calculates the cantilever dimensions m and n from the column section and plate geometry.
3. Determines the required plate thickness for the factored loads.
4. Checks each anchor bolt for tension (including prying effects where applicable), shear, and combined tension-shear interaction.
5. Flags any check that falls below the required capacity factor.

The output is preliminary only and must be verified by a licensed Professional Engineer. In particular, confirm the following items that are project-specific and not automatically assessed:

• Concrete edge distance and breakout (ACI 318 Ch. 17 or equivalent).
• Grout thickness and strength specification.
• Weld size and type between column and base plate.
• Base plate flatness tolerance (typically <= 1 mm / 300 mm).
• Corrosion protection requirements (galvanizing, paint system, or stainless steel).

Run This Calculation

Use the Steel Calculator base plate tool to verify the worked example above with your own inputs. The tool handles all four codes and provides per-code output with the same checks described in this post.

--> Base plate & anchors calculator -- axial and moment base plates, anchor bolt tension and shear per AISC 360, AS 4100, EN 1993, CSA S16.

--> Load combinations calculator -- ASCE 7-22, EN 1990, AS/NZS 1170, and NBC load combinations.

--> Bolted connections calculator -- bolt shear, tension, and combined checks.

Related Blog Posts

• AISC 360 Design Examples -- Beam, Column & Connection Worked Solutions
• AS 4100 Worked Examples -- Beam, Column & Bolt Group Design
• EN 1993-1-8 Steel Connection Design -- Bolt & Weld Checks
• CSA S16 Base Plate Design -- W250x73 Worked Example
• Steel Column Design Example -- AISC 360-22 LRFD Worked Solution
• Steel Connection Design Spreadsheet -- Free Online Alternative

Related Reference Pages

• Steel Fy & Fu Reference -- Yield and Tensile Strength by Grade
• Steel Base Plate Design Example -- AISC Design Guide 1 Method
• EN 1993 Base Plate Design -- Eurocode T-Stub Method
• Concrete on Steel Deck -- Composite Floor Design Guide
• Anchor Bolts -- Grades, Sizes, and Embedment Reference

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All real-world structural design depends on project-specific factors (loads, combinations, stability, detailing, fabrication, erection, tolerances, site conditions, and the governing standard and project specification). You are responsible for verifying inputs, validating results with an independent method, checking constructability and code compliance, and obtaining professional sign-off where required.

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