Base Plate Design Worked Example — Column per AISC Design Guide 1

Problem: Design a base plate for a W10x49 column (A992 steel, Fy = 50 ksi) supporting a factored axial load of Pu = 400 kips (LRFD). The column bears on a concrete pier with f'c = 4,000 psi. The pier is 24 in × 24 in. Use A36 plate steel (Fy = 36 ksi, Fu = 58 ksi) and design per AISC Design Guide 1 (3rd Edition) and AISC 360-22.

Step 1: Column and Material Data

Step 2: Required Bearing Area (AISC DG1 Section 3.1)

The concrete bearing strength per AISC 360-22 J8:

ϕc × Pp = ϕc × 0.85 × f'c × A1 × √(A2/A1)

Where:

• ϕc = 0.65 (bearing on concrete)
• A1 = base plate area (in²)
• A2 = concrete support area (pier area = 24 × 24 = 576 in²)
• √(A2/A1) ≤ 2.0

For the initial trial, assume √(A2/A1) = 2.0 (full confinement):

Required A1 = Pu / (ϕc × 0.85 × f'c × 2.0) A1 = 400,000 / (0.65 × 0.85 × 4,000 × 2.0) A1 = 400,000 / 4,420 A1 = 90.5 in²

Try N × B = 14 × 14 = 196 in² (A1 = 196 in²)

Check √(A2/A1) = √(576/196) = √2.94 = 1.71 ≤ 2.0 → OK

Check bearing capacity:

ϕc × Pp = 0.65 × 0.85 × 4,000 × 196 × 1.71 ϕc × Pp = 0.65 × 0.85 × 4,000 × 335 ϕc × Pp = 0.65 × 1,139,000 ϕc × Pp = 740 kips > 400 kips → OK

Step 3: Plate Dimensions (DG1 Section 3.2)

Required plate dimensions:

N_min = √(A1_req) + Δ, where Δ = (0.95d - 0.8bf)/2

Δ = (0.95 × 10.0 - 0.8 × 10.0)/2 = (9.5 - 8.0)/2 = 0.75 in

N_min = √90.5 + 0.75 = 9.51 + 0.75 = 10.26 in

B_min = A1_req / N_min = 90.5 / 10.26 = 8.82 in

Use N = 16 in, B = 14 in (A1 = 224 in² > 90.5 in² → OK)

Bearing check with actual dimensions:

√(A2/A1) = √(576/224) = √2.57 = 1.60 ≤ 2.0 → OK

ϕcPp = 0.65 × 0.85 × 4,000 × 224 × 1.60 ϕcPp = 0.65 × 0.85 × 4,000 × 358.4 ϕcPp = 0.65 × 1,218,560 ϕcPp = 792 kips > 400 kips → OK

Step 4: Plate Thickness (DG1 Section 3.3)

Cantilever dimensions (m and n):

m = (N - 0.95d)/2 = (16 - 0.95 × 10.0)/2 = (16 - 9.5)/2 = 3.25 in

n = (B - 0.80bf)/2 = (14 - 0.80 × 10.0)/2 = (14 - 8.0)/2 = 3.00 in

Required bearing pressure on plate:

fp = Pu / (B × N) = 400 / (14 × 16) = 400 / 224 = 1.79 ksi

Plate thickness by bending:

For A36 plate (Fy = 36 ksi), the required thickness:

t_min = m × √(2 × fp / (0.90 × Fy))

t_min (from m direction) = 3.25 × √(2 × 1.79 / 32.4) = 3.25 × √(0.110) = 3.25 × 0.332 = 1.08 in

t_min (from n direction) = 3.00 × √(2 × 1.79 / 32.4) = 3.00 × 0.332 = 1.00 in

Check n' (DG1 alternative method):

n' = √(d × bf)/4 = √(10.0 × 10.0)/4 = √100/4 = 10/4 = 2.50 in

λ = min(1.0, 2 × √(n') / (1 + √(1 - bearing_factor)) in tension/compression)

For simplicity with concentric load, λ = min(1.0, 2 × n' / (m + n)):

λ = min(1.0, 2 × 2.50 / (3.25 + 3.00)) = min(1.0, 5.0/6.25) = min(1.0, 0.80)

t_min (DG1) = n' × √(2 × fp / (0.90 × Fy)) × (m × n / (m + n)) factor...

Using the simplified DG1 approach:

l = max(m, n, λn') = max(3.25, 3.00, 0.80 × 2.50) = max(3.25, 3.00, 2.00) = 3.25 in

t_req = l × √(2 × fp / (0.90 × Fy)) = 3.25 × √(3.58 / 32.4) = 3.25 × 0.332 = 1.08 in

Use 1-1/4 in thick plate (next standard 1/4-in increment above 1.08 in).

Step 5: Anchor Rod Design (AISC DG1 Section 3.5)

For a pinned base (no moment), use four 3/4-in diameter anchor rods for stability and erection:

Rod material: ASTM F1554 Grade 36 (Fy = 36 ksi, Fu = 58 ksi)

Tension check: No net uplift at service loads (assume compression only). However, anchor rods must resist handling and construction loads (minimum 5 kips/rod per DG1).

Shear check: If lateral loads are small (say Vu = 20 kips at base):

Shear per rod = 20 / 4 = 5.0 kips

ϕvVn = ϕv × 0.60 × Fy × A_b × C (where C accounts for threads in shear plane)

For 3/4 in rod, A_b = 0.442 in² Assume threads are not excluded from shear plane: ϕvVn = 0.75 × 0.60 × 36 × 0.442 × 0.80 = 5.73 kips per rod → 5.0 < 5.73 → OK

Step 6: Weld Design (Column-to-Base Plate)

Minimum weld per AISC 360 J2.2b: For column flange tf = 0.560 in (W10x49), minimum fillet weld = 3/16 in.

Weld at column flanges (load transfer):

Required weld strength = Pu / 2 = 200 kips per flange (assuming web carries minimal compression)

For E70XX electrodes (FEXX = 70 ksi), 3/16 in fillet weld strength:

ϕRn = 0.75 × 0.60 × 70 × (0.707 × 3/16) per inch = 0.75 × 42 × 0.1326 = 4.18 kip/in

Flange width available = bf = 10.0 in. Use two return welds of 1 in each on each flange.

Total weld length per flange = 10 + 2 × 1 = 12 in (at two flanges = 24 in total at flanges)

Add web welds: W10x49 web depth ≈ 8.88 in, use 3/16 in fillet both sides:

Web weld capacity = 2 × 8.88 × 4.18 = 74.2 kips

Total weld capacity = (2 flanges × 12 in × 4.18) + 74.2 = 100.3 + 74.2 = 174.5 kips

This is insufficient — need larger weld or more weld length.

Increase to 5/16 in fillet:

ϕRn = 0.75 × 0.60 × 70 × (0.707 × 5/16) = 0.75 × 42 × 0.221 = 6.96 kip/in

Total capacity = 24 × 6.96 + 2 × 8.88 × 6.96 = 167 + 123.6 = 290.6 kips < 400 kips

Add stiffener plates or use CJP groove weld at flanges.

Alternative: Use complete-joint-penetration (CJP) groove weld at flanges (standard for fully-loaded columns). CJP weld capacity equals the base metal strength of the column flange. This is the typical detail for columns subject to heavy loads.

Step 7: Base Plate Stiffeners (Optional)

For a 1-1/4 in plate on a 16 × 14 bearing area, stiffeners are not required. The cantilever distance m = 3.25 in with t = 1.25 in gives a reasonable m/t ratio of 2.6.

If plate thickness needed to be reduced (e.g., to 1 in or less), stiffeners between the column flanges and the base plate would be required. Each stiffener would be designed as a compression element (AISC E) with:

• Stiffener length = m = 3.25 in
• Minimum thickness = 3/8 in (practical minimum)
• Weld: 3/16 in fillet to column flange and base plate

Step 8: Summary

Try the Calculator

Use the Base Plate & Anchors Calculator to design base plates for your own column sections, loads, and concrete strengths. Supports AISC DG1, AS 4100, EN 1993-1-8, and CSA S16.

Frequently Asked Questions

What is the concrete confinement factor √(A2/A1) and when does it apply? The confinement factor accounts for the increased bearing capacity of concrete when the loaded area is smaller than the total support area. The confining concrete surrounds the bearing area and resists lateral expansion, triaxially confining the concrete under the plate. It applies only when the base plate is not at the edge of the concrete pier (minimum 4 in edge distance on all sides for full development). For edge conditions, use √(A2/A1) = 1.0.

When should I use stiffeners on base plates? Stiffeners are typically needed when:

• Plate thickness exceeds 2 in (economy)
• Cantilever dimension m or n exceeds 6-8 in
• Large moment is present (moment-resisting base plate)
• Column size is small relative to the base plate area The DG1 alternative method (n' and λ) helps determine whether stiffeners are actually needed or if a thicker plate is more economical.

How do I design a base plate for moment (not just axial load)? Moment-resisting base plates follow DG1 Section 3.4. The approach distributes the moment to the flanges as a tension-compression couple. The tension side requires anchor rods designed per AISC 360 J3 (tension + shear interaction), while the compression side follows the same bearing design as above. The plate thickness is governed by the maximum moment from either the compression cantilever (m) or the tension anchor rod cantilever distance.

What is the minimum edge distance for anchor rods? Per AISC 360 Table J3.4M, the minimum edge distance for a 3/4 in anchor rod in a rolled edge is 1-1/8 in. For concrete edge distances, ACI 318-19 Chapter 17 requires minimum edge distance of 1.5 × anchor diameter (1.125 in for 3/4 in) but recommends 4-6 in for proper development of the concrete breakout cone.

See Also

• Column Capacity Calculator
• Column Buckling Guide
• Column Design Guide
• Effective Length Reference