Steel Plate Design — Engineering Reference

Steel plate design: gross yielding, net fracture, block shear rupture per AISC 360, plate buckling slenderness limits, and interactive block shear calculator.

Overview

Steel plates serve as the connecting elements in nearly every structural steel connection — gusset plates, splice plates, shear tabs, base plates, stiffeners, and doubler plates are all plate elements that must be designed for the specific forces they transfer. AISC 360-22 Chapter J (Design of Connections) provides the limit state checks for plates loaded in tension, compression, shear, and bending, or combinations thereof.

Plate design requires checking multiple limit states because the governing failure mode depends on the plate geometry, hole pattern, loading direction, and support conditions. A plate loaded in tension may fail by gross section yielding, net section rupture, or block shear rupture. A plate loaded in compression may fail by plate buckling before yielding. The designer must check all applicable limit states and use the lowest capacity as the design strength.

Tension limit states

Gross section yielding (AISC J4.1a)

phi x P_n = 0.90 x F_y x A_g

where A_g = plate width x thickness. This limit state ensures the plate can yield across its full cross-section without excessive elongation. Yielding is ductile and provides redistribution capacity.

Net section rupture (AISC J4.1b)

phi x P_n = 0.75 x F_u x A_e

where A_e = U x A_n. A_n is the net area (gross area minus hole deductions), and U is the shear lag factor (1.0 for plates connected at all elements, less for angles and channels). The effective hole width for deduction is the nominal hole diameter plus 1/16 in.

Block shear rupture (AISC J4.3)

phi x R_n = 0.75 x [0.6 x F_u x A_nv + U_bs x F_u x A_nt]

with an upper limit of 0.75 x [0.6 x F_y x A_gv + U_bs x F_u x A_nt]. A_nv is the net shear area, A_nt is the net tension area, and U_bs = 1.0 for uniform tension stress distribution. This limit state governs when the bolt pattern creates a distinct failure block that tears out of the plate.

Worked example — gusset plate in tension

Given: 1/2 in. thick A36 gusset plate (F_y = 36 ksi, F_u = 58 ksi), 12 in. wide, bolted with four 3/4 in. A325 bolts in two rows of two (gage = 4 in., pitch = 3 in.), edge distance = 1.5 in., standard holes (13/16 in.).

1. Gross yielding: A_g = 12 x 0.5 = 6.0 in^2. phi x P_n = 0.90 x 36 x 6.0 = 194.4 kip.
2. Net rupture: Two holes per row in tension. Effective hole = 13/16 + 1/16 = 7/8 in. A_n = (12 - 2 x 0.875) x 0.5 = 5.125 in^2. U = 1.0 (plate connected on all elements). phi x P_n = 0.75 x 58 x 5.125 = 222.9 kip.
3. Block shear: Shear planes: two vertical lines, each length = 1.5 + 3.0 = 4.5 in. A_gv = 2 x 4.5 x 0.5 = 4.5 in^2. A_nv = 4.5 - 2 x 1.5 x 0.875 x 0.5 = 3.19 in^2. Tension plane: width = 4.0 in. A_nt = (4.0 - 1 x 0.875) x 0.5 = 1.5625 in^2. phi x R_n = 0.75 x (0.6 x 58 x 3.19 + 1.0 x 58 x 1.5625) = 0.75 x (111.0 + 90.6) = 151.2 kip.
4. Controlling limit state: Block shear at 151.2 kip governs. This is 22% lower than the gross yielding capacity, demonstrating why block shear must always be checked.

Plates in compression — buckling

Plates loaded in compression (e.g., gusset plates in bracing connections, splice plates transferring compression) must be checked for plate buckling. The Thornton method treats the unbraced plate length as a column:

phi x P_n = 0.90 x F_cr x A_g

where F_cr is calculated using the AISC column equations (E3) with the slenderness ratio KL/r, and r = t / sqrt(12) for a rectangular plate (where t is the plate thickness). The effective length L is the average of the three distances from the Whitmore section corners and midpoint to the nearest plate edge.

For a 1/2 in. gusset plate with effective length L = 15 in. and K = 0.65 (fixed-free): KL/r = 0.65 x 15 / (0.5/sqrt(12)) = 9.75 / 0.1443 = 67.6. F_e = pi^2 x 29000 / 67.6^2 = 62.6 ksi. F_cr = 0.658^(36/62.6) x 36 = 28.4 ksi. phi x P_n = 0.90 x 28.4 x A_whitmore.

Code comparison — plate design checks

Key design considerations

• Plate grade — connection plates are commonly A36 (F_y = 36 ksi) or A572 Gr 50 (F_y = 50 ksi). Using higher-strength plate reduces thickness but does not help net section rupture proportionally because F_u for A572 Gr 50 (65 ksi) is only 12% higher than A36 (58 ksi), while F_y is 39% higher.
• Staggered bolt holes — when bolts are staggered, the net width must be checked along every possible failure path using the s^2/(4g) rule (AISC B4.3b), where s is the longitudinal pitch and g is the transverse gage. The path with the smallest net area governs.
• Whitmore section width — for gusset plates, the effective width for tension is defined by 30-degree lines from the first bolt to the last bolt: W_w = s x (n-1) x tan(30) x 2 + gage. This effective width, multiplied by the plate thickness, gives the area for gross yielding and net rupture checks.
• Minimum plate thickness — AISC does not specify a minimum connecting plate thickness, but practical considerations include weld heat distortion (minimum 1/4 in. for plates welded on both sides), bolt installation clearance, and durability. For exterior exposed connections, 3/8 in. minimum is common practice.

Common mistakes to avoid

• Not checking block shear — block shear frequently governs over both gross yielding and net rupture, especially for compact bolt patterns with small edge distances. It is the most commonly missed limit state in connection plate design.
• Using F_y instead of F_u for net rupture — net section rupture uses F_u (ultimate tensile strength), not F_y. Using F_y with phi = 0.75 dramatically underestimates the capacity and wastes material.
• Ignoring Whitmore section for gusset plates — the full plate width is not effective for transferring concentrated bolt group forces. The Whitmore effective width limits the area that can be used for yielding and rupture checks.
• Forgetting plate compression buckling — thin gusset plates in bracing connections can buckle under brace compression. The Thornton method check is mandatory for all gusset plates that carry compression.

Bearing Plates per AISC Chapter J

Bearing plates distribute concentrated loads from steel members to concrete or masonry supports, preventing local crushing of the weaker supporting material. AISC 360-22 Chapter J8 governs the design of bearing plates, which are a specialized application of steel plate design where the plate must be sufficiently thick to spread the load over the required concrete bearing area while remaining stiff enough to maintain uniform bearing pressure.

Bearing on Concrete (AISC J8)

The design bearing strength of concrete is:

phi x P_p = 0.65 x min(0.85 x f'_c x A_1, 0.85 x f'_c x A_1 x sqrt(A_2/A_1))

where f'_c is the concrete compressive strength, A_1 is the bearing area (plate area), and A_2 is the maximum area of the supporting surface that is geometrically similar to and concentric with the loaded area. The sqrt(A_2/A_1) term provides an increase in bearing strength when the supporting area is larger than the plate, reflecting the beneficial confinement from the surrounding concrete. This factor is limited to a maximum of 2.0.

The required plate area is determined from:

A_1,req = P_u / (0.65 x 0.85 x f'_c)

If the supporting surface area A_2 is larger than A_1, the required area can be reduced by the sqrt(A_2/A_1) factor.

Bearing Plate Thickness

Once the required plate area is determined, the plate dimensions (B x N) are selected to provide adequate bearing area and to fit within the concrete support. The plate thickness is then determined from the cantilevered portion of the plate that extends beyond the member flanges or web. The critical cantilever distance is:

n = (B - 2k) / 2  (projecting beyond the flange, transverse direction)
m = (N - d) / 2    (projecting beyond the beam depth, longitudinal direction)

where k is the distance from the outer face of the flange to the web toe of the fillet, and d is the beam depth. The larger of n and m governs the plate thickness:

t_p = sqrt(2 x P_u x l / (phi x F_yp x B))

where l is the cantilever distance (max of n, m), and B is the plate width. This equation derives from treating the projecting portion of the plate as a cantilever beam loaded by the uniform bearing pressure.

Bearing Plate Worked Example — W12x65 on Concrete

Given:

• Beam: W12x65 (A992), d = 12.1 in., b_f = 12.0 in., k = 1.12 in., t_w = 0.390 in.
• Factored reaction: P_u = 150 kip
• Concrete: f'_c = 4 ksi, supporting pedestal is 16 in. x 16 in.

Step 1 — Determine required bearing area:

A_1,req = P_u / (0.65 x 0.85 x f'_c) = 150 / (0.65 x 0.85 x 4) = 67.9 in.^2

Check A_2 factor: The pedestal is 16 x 16 = 256 in.^2. sqrt(A_2/A_1) = sqrt(256/67.9) = 1.94. This is less than 2.0, so the factor applies.

With the A_2 factor: A_1,req = 67.9 / 1.94 = 35.0 in.^2

Step 2 — Select plate dimensions:

The plate must be at least as wide as the beam flange (12.0 in.). Try B = 14 in. (2 in. projection beyond flange each side).

N = A_1 / B = 35.0 / 14 = 2.5 in. This is much smaller than the beam depth (12.1 in.), which is impractical.

The plate length N must be at least equal to the beam flange width to provide proper bearing. Set N = 14 in. (to match B for a square plate). Actual A_1 = 14 x 14 = 196 in.^2.

Check bearing pressure: p_u = P_u / A_1 = 150 / 196 = 0.765 ksi < 0.65 x 0.85 x 4 = 2.21 ksi. OK.

Step 3 — Determine plate thickness:

Cantilever in the transverse direction:

n = (B - 2 x k_des) / 2 = (14 - 2 x 1.12) / 2 = 5.88 in.

Cantilever in the longitudinal direction:

m = (N - d) / 2 = (14 - 12.1) / 2 = 0.95 in.

The transverse cantilever n = 5.88 in. governs.

t_p = sqrt(2 x P_u x n / (phi x F_yp x B))
    = sqrt(2 x 150 x 5.88 / (0.90 x 50 x 14))
    = sqrt(1764 / 630)
    = sqrt(2.80)
    = 1.67 in.

Use t_p = 1-3/4 in. (1.75 in. plate). Bearing plate: 14 x 14 x 1-3/4 in. A572 Gr 50.

Step 4 — Verify with web crippling check:

The beam web must also be checked for web crippling at the concentrated reaction per AISC J10.3:

phi x R_n = 0.75 x 0.80 x t_w^2 x [1 + 3 x (N/d) x (t_w/t_f)^1.5] x sqrt(E x F_yw x t_f / t_w)

With N/d = 14/12.1 = 1.16, t_w/t_f = 0.390/0.606 = 0.644:

phi x R_n = 0.75 x 0.80 x 0.390^2 x [1 + 3 x 1.16 x 0.644^1.5] x sqrt(29000 x 50 x 0.606/0.390)
          = 0.75 x 0.80 x 0.152 x [1 + 3 x 1.16 x 0.517] x sqrt(2,252,500)
          = 0.091 x 2.80 x 1501 = 382 kip > 150 kip. OK.

Base Plates per AISC Design Guide 1

Base plates are a specialized form of bearing plate that distribute column axial loads and moments to concrete foundations. AISC Design Guide 1 (DG1), "Base Plate and Anchor Rod Design" by Fisher and Kloiber, provides the standard design procedure used in North American practice.

Base Plate Design Procedure

The DG1 procedure for concentrically loaded columns (no moment) follows these steps:

1. Determine the required bearing area: A_1,req = P_u / (phi x 0.85 x f'_c), considering the A_2/A_1 factor.
2. Select plate dimensions B and N: The plate must extend at least m = (N - 0.95d)/2 and n = (B - 0.80b_f)/2 beyond the critical section. These projections are the "cantilever arms" used to compute plate bending.
3. Compute the bearing pressure: f_p = P_u / (B x N).
4. Determine plate thickness: The critical cantilever is the larger of m and n:

t_p = l x sqrt(2 x f_p / (phi x F_yp))

where l = max(m, n).

For columns with bending moments, the bearing pressure distribution becomes trapezoidal (or triangular for partial bearing), and the plate thickness is determined from the maximum bearing pressure at the plate edge. The uplift case (tension on one side) requires anchor rod design per ACI 318 Appendix D or Chapter 17.

Base Plate Worked Example

Given: W12x65 column, P_u = 400 kip, no moment, f'_c = 4 ksi, pedestal 20 x 20 in.

A_1,req = 400 / (0.65 x 0.85 x 4 x sqrt(400/78.5)) = 400 / (2.21 x 2.26) = 80.1 in.^2 (preliminary without A_2).

With A_2 = 400 in.^2: A_1,req = 400 / (0.65 x 0.85 x 4 x 2.0) = 400 / 4.42 = 90.5 in.^2. Wait, let us redo: A_1,req = P_u / (0.65 x 0.85 x f'_c x min(2, sqrt(A_2/A_1))).

Start with A_1 = B x N. Try B = 14 in., N = 14 in. A_1 = 196 in.^2.

m = (14 - 0.80 x 12.0) / 2 = (14 - 9.6) / 2 = 2.2 in.

n = (14 - 0.95 x 12.1) / 2 = (14 - 11.5) / 2 = 1.25 in.

l = max(2.2, 1.25) = 2.2 in.

f_p = 400 / 196 = 2.04 ksi.

t_p = 2.2 x sqrt(2 x 2.04 / (0.90 x 50)) = 2.2 x sqrt(0.0907) = 2.2 x 0.301 = 0.66 in.

Use t_p = 3/4 in. Base plate: 14 x 14 x 3/4 in. A36.

Gusset Plate Design — Buckling and Whitmore Section

Gusset plates transfer forces between diagonal braces and the beam-column joint in braced frames. They must be designed for tension, compression, shear, and block shear, with particular attention to plate buckling under brace compression.

Whitmore Section

The Whitmore section defines the effective width of a gusset plate at the point where forces from the bolt group spread out at 30-degree angles. Named after R.E. Whitmore (1962), the effective width is:

W_w = (n_bolts - 1) x s x tan(30) x 2 + gage

where n_bolts is the number of bolt rows, s is the bolt pitch (spacing between rows), and gage is the transverse bolt spacing. For a typical 4-bolt connection with s = 3 in. and gage = 4 in.:

W_w = 3 x 3 x 0.577 x 2 + 4 = 10.4 + 4 = 14.4 in.

The Whitmore area A_w = W_w x t_p is used for the gross yielding and net rupture checks. If the Whitmore section extends beyond the plate edge, the effective width is truncated at the plate boundary.

Plate Buckling Check (Thornton Method)

For gusset plates in compression (brace in compression), the Thornton method checks plate buckling using a column analogy:

phi x P_n = 0.90 x F_cr x A_w

where F_cr is determined from AISC Chapter E with the slenderness ratio:

KL/r = K x L_avg / (t_p / sqrt(12))

L_avg is the average of the three distances from the Whitmore section (two ends and midpoint) to the nearest unsupported edge of the gusset plate. The effective length factor K depends on the plate edge support conditions:

For the buckling check, F_cr is computed as:

• If KL/r <= 4.71 x sqrt(E/F_y): F_cr = 0.658^(F_y/F_e) x F_y
• If KL/r > 4.71 x sqrt(E/F_y): F_cr = 0.877 x F_e

where F_e = pi^2 x E / (KL/r)^2.

Block Shear in Gusset Plates

Block shear is particularly critical in gusset plates because the bolt group creates a well-defined failure block that can tear out of the plate. The check is identical to the general block shear check per AISC J4.3, but the failure path follows the bolt pattern:

• Shear planes run parallel to the brace force direction, from the lead bolt to the last bolt plus the edge distance
• Tension plane runs perpendicular to the brace force, between the two rows of bolts

The block shear capacity must exceed the brace force, and it frequently governs the gusset plate design when the edge distances are small or the bolt pattern is compact.

Gusset Plate Design Summary Table

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Related references

• Plate Weight Reference
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• gusset plate design
• structural capacity calculator
• bolt capacity calculator
• weld capacity calculator
• Anchor Bolts Reference

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This page is for educational and reference use only. It does not constitute professional engineering advice. All design values must be verified against the applicable standard and project specification before use. The site operator disclaims liability for any loss arising from the use of this information.