Connection Design Workflow — Engineering Reference

AISC 360 steel connection design workflow: load path, bolt shear/bearing, block shear, weld sizing, plate checks, and interactive bolt capacity calculator.

Overview

Designing a steel connection is a multi-step process that starts with identifying the forces to be transferred, selecting a connection type, sizing the fasteners and plates, and then verifying all applicable limit states. A systematic workflow prevents missed checks — a common source of connection failures. The AISC Manual Part 9 (Design of Connecting Elements) and Part 10 (Design of Simple Shear Connections) provide the framework used by most practicing engineers in North America.

The workflow differs depending on whether the connection is a simple shear connection (transferring only vertical reaction), a moment connection (transferring shear plus moment), or a bracing connection (transferring axial force). Each type has a different set of governing limit states, but the overall approach follows the same logical sequence.

Step-by-step workflow for a bolted shear connection

Step 1: Determine design forces

Identify the factored beam end reaction R_u from the structural analysis. Include gravity, wind, and seismic load combinations. For connections that also serve as collectors or drag struts, include the axial force from the lateral analysis.

Step 2: Select connection type

Choose from shear tab, double angle, single angle, or seated connection based on reaction magnitude, beam depth, column orientation, and erection requirements. Use the AISC Manual selection tables as a starting point.

Step 3: Size bolts for shear

Determine the number of bolts required: n >= R_u / (phi x r_n), where phi x r_n is the single-bolt design shear strength. For 3/4 in. A325-N in single shear: phi x r_n = 0.75 x 54 x 0.4418 = 17.9 kip.

Step 4: Check bolt bearing and tearout

For each bolt, check bearing on both the connected plate and the beam web. The bearing strength is: phi x R_n = 0.75 x 1.2 x L_c x t x F_u (per bolt), capped at 0.75 x 1.5 x d x t x F_u. L_c is the clear distance to the next hole or to the edge.

Step 5: Check block shear rupture

Block shear (AISC J4.3) checks the combined failure of a tension rupture plane and a shear yielding/rupture plane: R_n = 0.6 x F_u x A_nv + U_bs x F_u x A_nt, capped at 0.6 x F_y x A_gv + U_bs x F_u x A_nt. This check applies to the beam web, the connection plate, and the gusset plate if applicable.

Step 6: Check plate yielding and rupture

Step 7: Size weld (if welded to support)

For a shear tab welded to a column flange: weld length = plate length, fillet weld size per AISC Table J2.4 minimum. Weld strength: phi x R_n = 0.75 x 0.60 x F_EXX x t_e x L, where t_e = 0.707 x weld leg size.

Step 8: Check supporting member

Verify the column web or flange can resist the concentrated beam reaction — check web yielding (J10.2), web crippling (J10.3), and web sidesway buckling (J10.4).

Limit state checklist

Limit State AISC Section Applies To phi
Bolt shear J3.6 Bolt group 0.75
Bolt bearing/tearout J3.10 Plate, beam web 0.75
Block shear rupture J4.3 Beam web, plate, gusset 0.75
Gross section yielding J4.1 Connecting plate 1.00
Net section rupture J4.1 Connecting plate 0.75
Weld shear J2.4 Fillet or CJP weld 0.75
Web local yielding J10.2 Column or beam web 1.00
Web crippling J10.3 Column or beam web 0.75
Flexural yielding F11/J4 Connecting plate 0.90
Buckling of gusset J4 + E3 Gusset in compression 0.90

Worked example — shear tab limit state summary

Given: W21x44 beam, R_u = 60 kip, 3/8 in. x 9 in. shear tab plate (A36, F_y = 36 ksi, F_u = 58 ksi), three 3/4 in. A325-N bolts at 3 in. spacing, 1.5 in. edge distance, 5/16 in. fillet weld to W14 column.

  1. Bolt shear: 3 bolts x 17.9 kip = 53.7 kip. 53.7 < 60 — need 4 bolts.** Revise to 4 bolts at 3 in. spacing (plate becomes 12 in. long). 4 x 17.9 = **71.6 kip > 60. OK.
  2. Bearing on plate (t = 3/8 in.): L_c at edge bolt = 1.5 - 13/32 = 1.094 in. R_n = 0.75 x 1.2 x 1.094 x 0.375 x 58 = 21.4 kip. Interior bolts: L_c = 3.0 - 13/16 = 2.188 in. R_n = 0.75 x min(1.2 x 2.188 x 0.375 x 58, 1.5 x 0.75 x 0.375 x 58) = 0.75 x min(57.1, 24.5) = 18.4 kip. Total = 21.4 + 3 x 18.4 = 76.6 kip > 60. OK.
  3. Block shear on plate: A_gv = 2 x (10.5 x 0.375) = 7.875 in^2. A_nv = 7.875 - 2 x 3.5 x (13/16+1/16) x 0.375 = 5.578 in^2. A_nt = (1.5 x 0.375) - 0.5 x (13/16+1/16) x 0.375 = 0.398 in^2. R_n = 0.75 x (0.6 x 58 x 5.578 + 1.0 x 58 x 0.398) = 0.75 x (194.1 + 23.1) = 162.9 kip > 60. OK.
  4. Weld: 5/16 in. fillet, two lines x 12 in. = 24 in. total. phi x R_n = 0.75 x 0.6 x 70 x 0.707 x 5/16 x 24 = 0.75 x 0.6 x 70 x 0.221 x 24 = 167 kip > 60. OK.

Common mistakes to avoid

Run this calculation

Related references

Disclaimer

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.