How is a steel base plate designed per AISC?

Per AISC 360-22 and AISC Design Guide 1: Step 1 — Determine factored column loads (axial P, moment M, shear V). Step 2 — Select trial base plate dimensions B x N (typically 4-8 in larger than column depth/flange width). Step 3 — Check concrete bearing: Pp = 0.85 x fc' x A1 x sqrt(A2/A1) <= 1.7 x fc' x A1 per AISC J8. Step 4 — Calculate plate thickness tp = L x sqrt(2Pu/(0.9 x Fy x B x N)) where L = max(m, n, lambda x n') for cantilever bending. Step 5 — Design anchor rods for tension (from moment) and shear. Step 6 — Design weld connecting column to base plate.

What controls base plate thickness?

Base plate thickness is governed by cantilever bending of the plate projection beyond the column footprint. Per AISC DG 1, the critical cantilever length L = max(m, n, lambda x n') where: m = (B - 0.95d)/2 (flange side projection), n = (N - 0.8bf)/2 (web side projection), and lambda x n' accounts for the yield line pattern at the column corners. For axially loaded plates: tp = L x sqrt(2Pu/(0.9 x Fy x B x N)). For plates with moment: the bearing stress is not uniform (rectangular or triangular block per DG 1 Section 3.3), and the plate thickness must account for the eccentric bearing pressure.

How many anchor rods does a base plate need?

Minimum 4 anchor rods per column (one at each corner) per AISC 360-22 J8. For lightly loaded columns (small axial force, no moment), 4 rods of 3/4 in diameter is typical. For columns with significant moment: the tension side rods resist uplift and must be designed for tension + shear interaction. Anchor rod design follows ACI 318 Appendix D for concrete breakout, pullout, and side-face blowout. The minimum embedment depth for headed anchors is 8 x rod diameter per ACI 318. For a typical W12 column base, 4 x 3/4 in or 4 x 7/8 in diameter anchor rods with 6-12 in embedment are common for building columns.

What concrete strength is used for base plate bearing?

Per AISC J8, the nominal concrete bearing strength is Pp = 0.85 x fc' x A1 x sqrt(A2/A1) but not more than 1.7 x fc' x A1. A1 is the base plate area. A2 is the maximum area of the supporting concrete that is geometrically similar to A1 and concentric with it. For a base plate on a pedestal the same size as the plate, A2/A1 = 1.0 and Pp = 0.85 x fc' x A1. For a plate on a large footing where A2 >> A1, the sqrt(A2/A1) factor approaches the cap of 2.0, giving Pp = 1.7 x fc' x A1. Typical concrete strengths: 3,000 psi for slabs on grade, 4,000 psi for building columns, 5,000-6,000 psi for heavy industrial. The phi factor for concrete bearing is 0.65 per AISC J8.

How do I account for moment on a base plate?

Per AISC DG 1 Section 3.3, moment on a base plate produces eccentric bearing pressure. For small eccentricity (e = M/P N/6): the pressure block is triangular and only a portion of the plate is in compression. The compressed length Y is solved from equilibrium: P + T = 0.85 x fc' x B x Y (with T from anchor rods). The plate thickness is designed for the maximum cantilever moment from the bearing pressure. Large eccentricity base plates require much thicker plates and larger anchor rods.

Steel Base Plate Design Example -- AISC LRFD Worked Problem

Q: What controls base plate thickness?

Base plate thickness is governed by cantilever bending of the plate projection beyond the column footprint. Per AISC DG 1, the critical cantilever length L = max(m, n, lambda x n') where: m = (B - 0.95d)/2 (flange side projection), n = (N - 0.8bf)/2 (web side projection), and lambda x n' accounts for the yield line pattern at the column corners. For axially loaded plates: tp = L x sqrt(2Pu/(0.9 x Fy x B x N)). For plates with moment: the bearing stress is not uniform (rectangular or triangular block per DG 1 Section 3.3), and the plate thickness must account for the eccentric bearing pressure.

Q: How many anchor rods does a base plate need?

Minimum 4 anchor rods per column (one at each corner) per AISC 360-22 J8. For lightly loaded columns (small axial force, no moment), 4 rods of 3/4 in diameter is typical. For columns with significant moment: the tension side rods resist uplift and must be designed for tension + shear interaction. Anchor rod design follows ACI 318 Appendix D for concrete breakout, pullout, and side-face blowout. The minimum embedment depth for headed anchors is 8 x rod diameter per ACI 318. For a typical W12 column base, 4 x 3/4 in or 4 x 7/8 in diameter anchor rods with 6-12 in embedment are common for building columns.

Q: What concrete strength is used for base plate bearing?

Per AISC J8, the nominal concrete bearing strength is Pp = 0.85 x fc' x A1 x sqrt(A2/A1) but not more than 1.7 x fc' x A1. A1 is the base plate area. A2 is the maximum area of the supporting concrete that is geometrically similar to A1 and concentric with it. For a base plate on a pedestal the same size as the plate, A2/A1 = 1.0 and Pp = 0.85 x fc' x A1. For a plate on a large footing where A2 >> A1, the sqrt(A2/A1) factor approaches the cap of 2.0, giving Pp = 1.7 x fc' x A1. Typical concrete strengths: 3,000 psi for slabs on grade, 4,000 psi for building columns, 5,000-6,000 psi for heavy industrial. The phi factor for concrete bearing is 0.65 per AISC J8.

This page presents a fully worked base plate design problem following AISC 360-22 LRFD provisions and AISC Design Guide 1. Every calculation step is shown explicitly. This is a pure worked example. For the underlying theory, bearing equations, and code comparison tables, refer to the Base Plate Design Reference. For column-specific considerations and stiffened base plates, see the Column Base Plate Reference.

Problem Statement

Design a base plate for an interior W12x96 column in a 6-story office building. The column supports 4 floors of composite steel framing plus roof. The foundation is a 32 in. x 32 in. reinforced concrete pier. Dead load is from composite slab (3-1/4 in. lightweight concrete on 3 in. metal deck, 20 ga), steel framing, MEP, ceiling, and partitions. Live load is office occupancy with a live load reduction per ASCE 7-22 Section 4.7.2.

Given Data

Parameter	Value	Source
Column	W12x96, ASTM A992	d = 12.7 in., bf = 12.2 in., tw = 0.550 in., tf = 0.900 in., k = 1.27 in.
Pu (LRFD)	540 kip	From frame analysis, load combo 1.2D + 1.6L
Vu (LRFD)	18 kip	From frame analysis
Mu (LRFD)	0 kip-in	Interior gravity column, negligible moment
f'c	5,000 psi	Normal weight concrete
Pier	32 x 32 in.	Below SOG, isolated footing
Plate Fy	36 ksi	ASTM A36 plate
Anchor rods	ASTM F1554 Gr 36	Threaded rods with heavy hex nuts
Grout	Non-shrink, f'c_grout = 6,000 psi	Prepackaged cementitious

Step 1: Determine Trial Plate Dimensions

Minimum bearing area from concrete strength:

Try a square plate. The minimum area A1 (assuming A2/A1 cap of 2.0):

A1_min = Pu / (phi_c x 0.85 x f'c x 2.0) A1_min = 540 / (0.65 x 0.85 x 5 x 2.0) A1_min = 540 / (0.65 x 8.5) A1_min = 540 / 5.525 A1_min = 97.7 in^2

This is the absolute minimum. For a square plate, side = sqrt(97.7) = 9.9 in. But this would not extend beyond the column footprint (bf = 12.2 in., d = 12.7 in.). The plate must project beyond the column for anchor rods and weld access.

Practical minimum dimensions:

B = bf + 2 x (2 in. min projection) = 12.2 + 4 = 16.2 in. Use B = 18 in. N = d + 2 x (2 in. min projection) = 12.7 + 4 = 16.7 in. Use N = 18 in.

Trial: B = 18 in., N = 18 in. Square plate.

A1 = 18 x 18 = 324 in^2.

Step 2: Check Concrete Bearing Capacity

Calculate A2:

The pier is 32 x 32 in. The projected area A2 is geometrically similar to A1 and concentric with it. The maximum A2 is the pier area = 32 x 32 = 1,024 in^2. For this geometry:

A2/A1 = 1,024 / 324 = 3.16

But the A2/A1 ratio is capped at 4.0 in the sqrt term, capped at 2.0 for the enhancement factor per ACI 318-19 Section 22.8.3. sqrt(A2/A1) = min(sqrt(3.16), 2.0) = min(1.778, 2.0) = 1.778.

Design bearing strength:

phi Pp = phi_c x 0.85 x f'c x A1 x sqrt(A2/A1) phi Pp = 0.65 x 0.85 x 5 x 324 x 1.778 phi Pp = 0.65 x 0.85 x 5 x 324 x 1.778 phi Pp = 0.65 x 2,448 phi Pp = 1,591 kip

Check: Pu = 540 kip < phi Pp = 1,591 kip. OK.

Bearing utilization: 540 / 1,591 = 0.34. Bearing is not the controlling limit state. The plate thickness will control.

Step 3: Compute Bearing Pressure

Under concentric axial load, the bearing pressure is uniform:

fp = Pu / A1 = 540 / 324 = 1.667 ksi

This is well below the grout compressive strength (6 ksi) and concrete bearing strength (0.85 x f'c x sqrt(A2/A1) = 0.85 x 5 x 1.778 = 7.56 ksi).

Step 4: Compute Cantilever Dimensions

The base plate is modeled as a cantilever projecting beyond the column bearing footprint. The three cantilever dimensions per AISC DG1 are:

m -- projection beyond column depth (parallel to web):

The column depth d = 12.7 in. The effective bearing depth is 0.95d = 0.95 x 12.7 = 12.07 in.

m = (N - 0.95 d) / 2 = (18 - 12.07) / 2 = 5.93 / 2 = 2.965 in.

n -- projection beyond column flange width:

bf = 12.2 in. The effective bearing width is 0.80 bf = 0.80 x 12.2 = 9.76 in.

n = (B - 0.80 bf) / 2 = (18 - 9.76) / 2 = 8.24 / 2 = 4.12 in.

lambda-n' -- effective projection for area inside flanges (Thornton model):

For W-shapes with bf/d >= 0.5, the parameter lambda = 1.0. Here bf/d = 12.2/12.7 = 0.961 > 0.5, so lambda = 1.0.

n' = sqrt(d x bf) / 4 = sqrt(12.7 x 12.2) / 4 = sqrt(154.9) / 4 = 12.45 / 4 = 3.112 in.

lambda x n' = 1.0 x 3.112 = 3.112 in.

Controlling cantilever dimension:

l = max(m, n, lambda-n') = max(2.965, 4.12, 3.112) = 4.12 in. (n governs)

This is consistent with the W12x96 section having a flange width slightly less than the depth -- the cantilever parallel to the flange (n) controls.

Step 5: Calculate Required Plate Thickness

The required thickness for the cantilever bending model:

tp_req = l x sqrt(2 x fp / (phi_b x Fy))

where:

l = 4.12 in. (controlling cantilever)
fp = 1.667 ksi (bearing pressure)
phi_b = 0.90 (flexure of the steel plate)
Fy = 36 ksi (A36 plate)

tp_req = 4.12 x sqrt(2 x 1.667 / (0.90 x 36)) tp_req = 4.12 x sqrt(3.334 / 32.40) tp_req = 4.12 x sqrt(0.10290) tp_req = 4.12 x 0.3207 tp_req = 1.32 in.

Select plate thickness: Use tp = 1-3/8 in. (1.375 in.). This is the next standard thickness increment above the required value.

Check with selected thickness:

phi Mn = 0.90 x Fy x (1 in. strip width) x tp^2 / 4 = 0.90 x 36 x 1 x 1.375^2 / 4 = 0.90 x 36 x 1.891 / 4 = 15.32 kip-in per inch width

Mu = fp x l^2 / 2 = 1.667 x 4.12^2 / 2 = 1.667 x 16.97 / 2 = 14.14 kip-in per inch width

phi Mn = 15.32 > Mu = 14.14. OK.

Step 6: Select Anchor Rods

For an interior gravity column with negligible moment and low shear, four anchor rods in a standard pattern are sufficient.

Rod diameter: Minimum practical rod = 3/4 in. diameter. The rod must engage enough concrete to develop the full steel tensile strength per ACI 318 Chapter 17. Use 4 x 7/8 in. diameter F1554 Gr 36 anchor rods.

Rod layout: Place rods at 14 in. x 14 in. square pattern (centered on the plate). Edge distance = (18 - 14) / 2 + 0.875/2 = 2.0 + 0.44 = 2.44 in. This exceeds the minimum edge distance of 6d = 6 x 0.875 = 5.25 in. ... wait, re-check.

Minimum edge distance for headed anchors per ACI 318 Table 17.9.2 is 6d for a headed anchor not limited by torque-controlled installation. For 7/8 in. rod: 6 x 0.875 = 5.25 in. Our edge distance of 2.44 in. is LESS than 5.25 in. This means we cannot develop full concrete breakout capacity.

Revised rod layout: Place rods at 10 in. x 10 in. square pattern. Edge distance = (18 - 10) / 2 = 4.0 in. center-to-edge. Still less than 5.25 in. The anchor group will have reduced breakout capacity.

Concrete breakout check (tension, single anchor):

hef = 8 in. (specified embedment for 7/8 in. rod).

For a single anchor influenced by edge distance:

ca1 = 4.0 in. (edge distance to plate edge; the pier edge is at (32-18)/2 + 4 = 11 in., so pier edge does not control).

ACI 318 Section 17.4.2.3, modification factor for edge distance:

phi_ed,N = 0.7 + 0.3 x (ca1 / 1.5 hef) = 0.7 + 0.3 x (4.0 / 12.0) = 0.7 + 0.3 x 0.333 = 0.80

Steel strength controls per rod: phi Nsa = 0.75 x 0.462 x 58 = 20.1 kip (tensile stress area Ase = 0.462 in^2 for 7/8 in. rod).

Since the column is in compression (no uplift), the anchor rods resist only nominal forces from erection and accidental eccentricity. The reduced concrete breakout capacity is acceptable.

Final rod specification: 4 x 7/8 in. diameter ASTM F1554 Gr 36 anchor rods, 8 in. minimum embedment, with 2-1/2 in. x 2-1/2 in. x 3/8 in. plate washers. Provide 2 in. projection above plate for nut and washer stack.

Step 7: Check Shear Transfer

Vu = 18 kip.

Friction capacity: mu x Pu = 0.55 x 540 = 297 kip >> 18 kip.

Friction alone provides ample shear resistance. No shear lug required. The anchor rods carry zero shear in the final condition.

Note on construction shear: During erection, before the grout sets and before gravity load is fully engaged, friction is not available. The anchor rods must resist erection shear. For 4 rods in shear: phi Vsa = 4 x 0.75 x 0.60 x 0.462 x 58 = 4 x 12.1 = 48.4 kip > 18 kip. OK.

Step 8: Weld Design (Column to Base Plate)

The column is shop-welded to the base plate. Use fillet welds all around the column profile.

Flange weld: The flange force is Pu x (Af / Ag). For W12x96, Af = bf x tf = 12.2 x 0.900 = 10.98 in^2 per flange. Ag = 28.2 in^2. Flange force = 540 x (10.98 / 28.2) = 210 kip per flange.

Weld length per flange (both sides) = 2 x 12.2 = 24.4 in. Required weld strength = 210 / 24.4 = 8.61 kip/in. For E70XX electrodes (0.60 x 70 = 42 ksi weld metal), minimum fillet size per Table J2.4 for 0.900 in. flange = 5/16 in. Capacity of 5/16 in. fillet: phi Rn = 0.75 x 0.60 x 70 x (5/16 x 0.707) = 0.75 x 42 x 0.221 = 6.96 kip/in < 8.61 kip/in. Insufficient.

Increase to 3/8 in. fillet: phi Rn = 0.75 x 0.60 x 70 x (3/8 x 0.707) = 0.75 x 42 x 0.265 = 8.35 kip/in < 8.61 kip/in. Still slightly insufficient.

Use 7/16 in. fillet weld at flanges: phi Rn = 0.75 x 0.60 x 70 x (7/16 x 0.707) = 0.75 x 42 x 0.309 = 9.74 kip/in > 8.61 kip/in. OK.

Web weld: Web force = Pu x (Aw / Ag) = 540 x ((12.7 - 2 x 0.900) x 0.550 / 28.2) = 540 x (10.90 x 0.550 / 28.2) = 540 x 0.213 = 115 kip. Weld length (both sides of web) = 2 x (12.7 - 2 x 1.27) = 2 x 10.16 = 20.32 in. Required strength = 115 / 20.32 = 5.66 kip/in. Minimum 5/16 in. fillet capacity = 6.96 kip/in > 5.66 kip/in. OK to use 5/16 in. fillet at the web.

Summary of Design

Item	Specification
Base plate	PL 1-3/8 x 18 x 18, ASTM A36
Anchor rods	4 x 7/8 in. dia. ASTM F1554 Gr 36, 8 in. embed
Rod pattern	10 in. x 10 in. square, centered on plate
Plate washers	2-1/2 x 2-1/2 x 3/8 in., ASTM F436
Grout	Non-shrink cementitious, f'c = 6,000 psi min., 1 in. thickness
Weld (flanges)	7/16 in. fillet weld, E70XX, both sides
Weld (web)	5/16 in. fillet weld, E70XX, both sides
Hole diameter	1-3/16 in. (1/16 over nut clearance, standard)

Design Verification

Independent check points an experienced engineer would verify:

Bearing: fp = 1.67 ksi << 0.85 f'c sqrt(A2/A1) = 7.56 ksi. OK.
Plate bending: tp_provided = 1.375 in. > tp_req = 1.32 in. OK.
Rod edge distance: Edge distance reduces concrete breakout but is acceptable for compression-only case.
Weld shear at flange: 9.74 kip/in capacity > 8.61 kip/in demand. OK.
Friction shear: 297 kip >> 18 kip. OK.

Related Tools and References

Disclaimer

This page is for educational and reference use only. This worked example illustrates a typical design approach but does not replace project-specific engineering analysis. All designs must be verified by a licensed Professional Engineer (PE) or Structural Engineer (SE) for the specific project conditions, loads, and governing building code. The site operator disclaims liability for any loss arising from the use of this information. Results are PRELIMINARY -- NOT FOR CONSTRUCTION.