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    Calculates locally in your browser · AISC base plate design
    Steel / connections

    Steel base plate

    The plate has two jobs and each sets a dimension. The plan area has to spread the column load onto the concrete without exceeding the bearing strength of §J8 — which is allowed to be higher than f’c, because the concrete around the plate confines it. The thickness has to carry the cantilever bending between the column face and the free edge.

    BearingPassesConfinement2.00×Governsn
    Plate size
    12.0 × 15.0 in
    180 in² against 68 required
    Plate thickness
    1.00 in
    0.76 in required, rounded up
    Bearing pressure
    1667 psi
    38% of the 4420 psi capacity
    Plate weight
    51 lb
    steel at 0.2836 lb/in³
    Bearing, then bending
    The plan area comes from §J8; the thickness comes from the cantilever between the column face and the edge.
    1.00 in
    Bearing, then bending
    StepValueNote
    Factored load300 kipas entered
    A2/A1 available4.0capped at 4 by §J8
    Confinement √(A2/A1)2.00capped at 2
    Bearing capacity4420 psi0.65 × 0.85 f’c × √(A2/A1)
    Required area67.9 in²load ÷ capacity
    Plate B × N12.0 × 15.0 inrounded to half inches
    Bearing pressure1667 psi38% utilised
    Projection m0.85 in(N − 0.95d)/2
    Projection n2.80 in(B − 0.8bf)/2
    Governing projection2.80 inn governs
    Thickness required0.76 inl √(2Pu / 0.9 Fy B N)
    Thickness used1.00 innext quarter inch

    The bearing pressure is 1667 psi against a capacity of 4420 psi, and the n projection governs the thickness.

    Inspect the model
    Choose a quick result view, a scaled comparison, or the complete diagnostic set.
    Plate area against load
    Linear in the load, because the bearing capacity is fixed by the concrete.
    Plate area against load0.00040.5481.09121.63162.17202.71100 kip100 kip200 kip200 kip300 kip300 kip400 kip400 kip600 kip600 kip800 kip800 kipArea, in²
    Thickness against load
    Grows as the square root of the load — doubling the load adds about 40% to the plate.
    Thickness against load0.0000.3230.6460.9691.291.610.502100 kip100 kip0.656200 kip200 kip0.762300 kip300 kip0.839400 kip400 kip0.970600 kip600 kip1.37800 kip800 kipThickness, in
    Canonical equations
    The exact relationships implemented by this engine.

    φc 0.85 f’c A1 √(A2/A1) · φc = 0.65, root capped at 2.

    Pu / capacity per area · Plan size.

    m = (N − 0.95d)/2 · n = (B − 0.8bf)/2.

    tp = l √(2Pu / 0.9 Fy B N) · l is the larger of m and n.

    engineering methodology

    A clear path from inputs to results.

    The plate has two jobs and each sets a dimension. The plan area has to spread the column load onto the concrete without exceeding the bearing strength of §J8 — which is allowed to be higher than f’c, because the concrete around the plate confines it. The thickness has to carry the cantilever bending between the column face and the free edge.

    Step-by-step methodWorked exampleAssumptions explained

    How to use

    1. Enter the factored axial load and the column dimensions.
    2. Set the concrete and plate strengths.
    3. Choose the A2/A1 ratio the pier can provide.
    4. Read the plate size, the bearing pressure and the required thickness.

    The formula

    A1=Pu0.65×0.85fc′A2/A1A_1 = \frac{P_u}{0.65 \times 0.85 f'_c \sqrt{A_2/A_1}}

    Reference: AISC 360 §J8 for bearing; AISC Design Guide 1 for the plate thickness.

    Worked example

    300 kip on a W14×8 column

    Inputs
    • 300 kip factored load, d = 14 in, bf = 8 in.
    • 4,000 psi concrete, A572 Gr 50 plate.
    • A2/A1 = 4, the maximum §J8 allows.
    Calculation path
    1. 1

      The confinement factor is √4 = 2, so the bearing capacity is 0.65 × 0.85 × 4 × 2 = 4.42 ksi.

    2. 2

      The required area is 300 / 4.42 = 67.9 in².

    3. 3

      A 12 × 15 in plate provides 180 in², comfortably above the requirement.

    4. 4

      The cantilever projections are m = 0.85 in and n = 2.80 in, so n governs.

    5. 5

      The required thickness is 2.80 × √(2 × 300 / (0.9 × 50 × 12 × 15)) = 0.76 in, rounded to 1 in.

    Result

    12 × 15 in · 0.76 in required · 1 in plate.

    Interpreting the result

    The confinement credit is the part most often missed. A plate set back from the pier edges can bear at twice the pressure of one that covers the whole pier, which halves the plate area — and the plate is usually the expensive item, not the concrete.

    Assumptions & limitations

    • A concentrically loaded column on a grouted plate.
    • A2/A1 capped at 4 by §J8 and the root capped at 2.
    • Plate thickness from the DG1 cantilever model with l = max(m, n).
    • Anchor bolt tension and moment are separate checks.

    Frequently asked questions

    What is A2/A1?

    The ratio of the concrete area that can confine the plate to the plate area itself. It is capped at 4 because beyond that the surrounding concrete stops helping.

    How thick should a base plate be?

    Usually 0.75 to 2 inches for building columns. Thicker plates are a sign the load or the concrete is unusual.

    Does grout count?

    The grout is assumed to spread the load but not to add bearing capacity; the plate bears on the concrete through it.

    Why does n govern more often than m?

    Because n uses 0.8 times the flange width while m uses 0.95 times the depth. A wide-flange column is deeper than it is wide, so the transverse cantilever is usually longer.

    Do I need stiffeners?

    Only when the plate becomes impractically thick, typically above 2.5 inches. Stiffeners change the yield-line pattern and are a different calculation.

    Methodology last reviewed 2026-10-22. Results are informational and not financial or engineering advice — see our disclaimer.