Steel Base Plate Design

A column base plate distributes the column load over a larger concrete bearing area, keeping bearing stresses within allowable limits. The design follows AISC Design Guide 1, 3rd Edition for plate proportioning and thickness, combined with ACI 318-25 §17 for anchor rod design.

For concentrically loaded bases the design sequence is: (1) select plate area to limit bearing pressure, (2) compute cantilever projections to find the governing moment arm, (3) size the plate thickness for the bending demand, (4) check shear friction, and (5) design anchor rods.

Notation

Symbol	Meaning
d, bf	Column depth and flange width
B, N	Plate width (∥ bf) and plate length (∥ d)
Pu, Vu	Factored axial and shear forces (LRFD)
fp	Actual bearing pressure = Pu/(B×N)
Fy, f′c	Plate yield strength; concrete compressive strength
m, n, n′	Cantilever projection distances
λ	Yield-line theory modification factor
ℓ	Governing cantilever length = max(m, n, λn′)
tp	Required plate thickness

The plate area must keep bearing stress within the design bearing capacity of the concrete (ACI 318-25 §22.8.3.2):

A₁ = B × N is the plate footprint. A₂ is the largest area of the bearing surface that is geometrically similar to A₁, centered on A₁, and bounded by the actual concrete cross-section. The √(A₂/A₁) term reflects the confinement benefit of surrounding concrete mass pressing against the bearing zone.

A₂/A₁ by Column Location

The ratio changes significantly with column position. A common design error is using A₂/A₁ = 4.0 for edge or corner columns when the geometry does not support it.

How to Compute A₂ Geometrically

Expand the plate outline (maintaining the B/N aspect ratio) symmetrically outward from each side until it hits the concrete boundary. For common cases:

Enter the computed ratio in the inputs. Do not default to 4.0 for edge or corner columns — using the maximum where the geometry does not allow it is unconservative.

Eccentric Loading — Rectangular Bearing Block (AISC DG1 §3.2)

Under eccentric axial load (moment M = P_u×e), part of the plate lifts off the concrete. AISC DG1 3rd Ed. (Drake & Elkin 1999) replaces the triangular stress distribution with a uniform (rectangular) block of intensity equal to the design bearing capacity φF_p, acting over a contact length Y:

When T > 0 the bearing block is at its capacity limit and anchor tension is the binding design check. The bearing DCR shows "At cap." in the results — the anchor tension DCR is the number that governs.

The plate acts as a cantilever projecting from the column footprint. Three cantilever lengths are computed and the largest governs:

This ensures the plate cross-section at the critical cantilever can develop the plastic moment capacity φM_p = φ F_y t_p²/4 under the uniform bearing pressure f_p.

Shear Friction (ACI 318-25 §22.9)

Horizontal shear at the base is first resisted by friction between the base plate and grout/concrete:

If V_u > φV_n,fric, the excess shear must be resisted by anchor rods acting in bearing. The calculator also checks ACI 318-25 §17.7 concrete failure modes when shear is applied.

Pryout Failure — Anchor Group (ACI 318-25 §17.7.3)

A shear load can "pry" the anchor out of the concrete in the direction opposite to the applied shear. The pryout capacity is proportional to the group's tensile breakout strength:

Note: φ = 0.70 is already embedded in φN_cbg, so φV_cpg = k_cp × φN_cbg directly (no additional φ).

Concrete Breakout in Shear — Anchor Group (ACI 318-25 §17.7.2)

When anchors are near a free concrete edge, shear can cause a concrete wedge to break out toward that edge. The basic single-anchor breakout strength in shear:

A_Vc0 = 4.5c_a1² (projected area for single anchor). The group projected area A_Vc adds the spacing between bolts parallel to the shear direction:

The thickness modifier ψ_h,V = max(1.0, √(1.5c_a1/h_a)) amplifies capacity when h_a > 1.5c_a1 (thick member). When h_a is not entered, h_a = ∞ is assumed (ψ_h,V = 1.0).

Tension-Shear Interaction (ACI 318-25 §17.8)

When anchors carry simultaneous tension N_ua and shear V_ua, ACI 318-25 §17.8 requires an interaction check. Let n_rat = N_ua/φN_n and v_rat = V_ua/φV_n:

Cast-in headed anchor rods are designed per ACI 318-25 Chapter 17. For typical base plates the critical tensile failure modes are steel fracture and concrete breakout. Near a free concrete face both capacities drop substantially — edge and corner columns require special attention.

Steel Fracture — Tension (§17.6.1.2)

Concrete Breakout — Tension Group (§17.6.2)

A breakout cone projects from the anchor head at ~35° to the surface. The plan area of the cone (A_Nc) relative to a single-anchor reference area (A_Nc0 = 9h_ef²) scales the group capacity. Basic single-anchor strength for cast-in anchors:

Concrete Edge Distance — A_Nc Truncation and ψ_ed,N (§17.6.2.1 & §17.6.2.4)

Two independent mechanisms reduce φN_cbg when anchors are placed close to a free concrete face. They act simultaneously and their effects compound — together they can cut φN_cbg by 50–70% compared to an interior group of the same size.

Mechanism 1 — Breakout cone truncated at the edge (§17.6.2.1).
Without a nearby edge, each anchor's cone projects 1.5h_ef outward on all sides. A free face physically cuts the cone, reducing the projected plan area A_Nc. The effective projection toward the edge is limited by the available cover:

Mechanism 2 — Edge distance modification factor ψ_ed,N (§17.6.2.4).
Edge proximity reduces concrete confinement in the breakout zone. This is captured as a separate multiplier on top of the A_Nc reduction:

The critical distance is 1.5h_ef. Closer than this, both A_Nc and ψ_ed,N decrease together. At the extreme (c_a,min → 0): ψ_ed,N → 0.70 and h_ef,e → 0 (A_Nc collapses to nearly zero).

Typical Impact by Column Location

Column location	ca,min vs. 1.5hef	Typical φNcbg reduction
Interior — anchors well away from any face	≥ 1.5hef	None — baseline capacity
Edge column — one anchor line near a face	0.5 – 1.0 × 1.5hef	15 – 45%
Corner column — anchors near two adjacent faces	< 0.5 × 1.5hef	50 – 75%+

Note: This calculator applies c_a,min to all sides of the bolt group (conservative). The physical truncation occurs on the near side only, but since a single c_a,min is entered without specifying which face is exposed, the all-sides assumption is safe.

Pullout Failure — Headed Anchor (ACI 318-25 §17.6.3)

A headed anchor can fail by the head pulling through the concrete without forming a breakout cone — the bearing area of the head directly crushes and displaces the concrete below it. Pullout is checked per individual anchor (not as a group mode):

When A_brg is not entered, the calculator estimates it as 1.5d_a² — derived from a heavy hex nut head (across-flats ≈ 1.6d_a, giving A_brg = A_hex − π/4⋅d_a² ≈ 1.4–1.5d_a²). For precision, obtain A_brg from the bolt manufacturer's data or ASME B18.2.2 heavy hex nut tables.

Steel Failure — Shear (§17.7.1.2)

Given: W10×60 column (d = 10.2 in, b_f = 10.1 in, A = 17.6 in²). P_u = 400 kip, V_u = 20 kip. Plate: B = N = 16 in, F_y = 36 ksi. Concrete: f′_c = 4 ksi, A₂/A₁ = 1.0. 4 × 1" F1554 Gr 36 rods, h_ef = 15 in, s_a = 12 in.

Step 1 — Bearing check

A₁ = 16 × 16 = 256 in². f_p = 400/256 = 1.56 ksi.

φ_cP_p/A₁ = 0.65 × 0.85 × 4 × 1.0 = 2.21 ksi. DCR = 1.56/2.21 = 0.71 — OK.

Step 2 — Cantilever lengths

m = (16 − 0.95×10.2)/2 = 3.15 in.
n = (16 − 0.80×10.1)/2 = 3.96 in.
n′ = √(10.2×10.1)/4 = 2.54 in.

X = [4×10.2×10.1/(10.2+10.1)²] × 400/(0.65×0.85×4×256) = 0.994×400/567 = 0.701.
λ = 2√(0.701)/(1+√(0.299)) = 1.673/1.547 = 1.08 → capped at 1.0.
ℓ = max(3.15, 3.96, 1.0×2.54) = 3.96 in.

Step 3 — Plate thickness

t_p = ℓ × √(2f_p/(φF_y)) = 3.96 × √(2×1.56/(0.90×36)) = 3.96 × √(0.0965) = 3.96 × 0.311 = 1.23 in → use 1.25 in plate.

Step 4 — Shear friction

φV_n = 0.90×0.55×400 = 198 kip > 20 kip — friction governs, OK.

Step 5 — Anchor steel fracture (tension)

A_se = 0.606 in² (1" rod, UNC). f_uta = 58 ksi.
φN_sa/bolt = 0.75 × 0.606 × 58 = 26.4 kip. Group = 4 × 26.4 = 105.5 kip.

Step 6 — Concrete breakout (tension)

N_b = 24×√4000×15^1.5 = 24×63.2×58.1 = 88,180 lb = 88.2 kip.
h_ef,e = min(1.5×15, ∞) = 22.5 in (no edge constraint). A_Nc = (12+2×22.5)² = 57² = 3249 in² → capped at 4×2025 = 8100 in², so A_Nc = 3249 in².
A_Nc0 = 9×225 = 2025 in². ψ_ed,N = 1.0 (no edge; c_a,min ≥ 1.5h_ef).
φN_cbg = 0.70×(3249/2025)×1.0×1.0×1.25×1.0×88.2 = 123 kip.
Controls steel fracture vs. breakout: min(105.5, 123) = 105.5 kip.

Step 7 — Pullout (§17.6.3)

A_brg ≈ 1.5×1² = 1.50 in² (heavy hex nut approx.).
N_p = 8×1.50×4 = 48 kip/bolt. φN_pn = 0.70×1.4×48 = 47.0 kip/bolt.
Group: 4×47.0 = 188 kip. φN_t,min = min(105.5, 123, 188) = 105.5 kip (steel fracture governs).