Steel Connection Design Examples — Shear & Moment
Connection design is where the steel frame either works or fails. This page presents four worked examples of common steel connections, each designed step-by-step per AISC 360-22. These examples cover the connections you will encounter most frequently in structural steel buildings.
Example 1: Single Plate Shear Tab
Given
| Parameter | Value |
|---|---|
| Beam | W18x50 (A992, Fy=50 ksi, Fu=65 ksi) |
| Column | W14x82 (A992) |
| Factored reaction | Ru = 45 kips |
| Plate | 3/8 in thick, A36 (Fy=36 ksi, Fu=58 ksi) |
| Bolts | (4) 3/4 in A325-N, standard holes |
| Weld | 5/16 in fillet, E70XX, each side of plate to column |
Check 1: Bolt Shear
Single shear, connections to both beam web and plate:
φRn = φ × Fnv × Ab × n
where:
- φ = 0.75
- Fnv = 54 ksi (A325-N, single shear per AISC Table J3.2)
- Ab = 0.442 in² (3/4 in bolt)
- n = 4 bolts
φRn = 0.75 × 54 × 0.442 × 4 = 71.5 kips > 45 kips ✓
Check 2: Bolt Bearing on Plate
Bearing on 3/8 in plate per AISC J3.10:
For standard holes with deformation consideration: φRn = φ × 1.2 × Lc × t × Fu
Spacing s = 3 in, edge distance le = 1.5 in, bolt diameter db = 0.75 in
Lc (interior) = s - db = 3.0 - 0.75 = 2.25 in Lc (edge) = le - db/2 = 1.5 - 0.375 = 1.125 in
Interior bolt: φRn = 0.75 × 1.2 × 2.25 × 0.375 × 58 = 44.1 kips/bolt Edge bolt: φRn = 0.75 × 1.2 × 1.125 × 0.375 × 58 = 22.1 kips/bolt
Total: 3 × 44.1 + 22.1 = 154.4 kips > 45 kips ✓
Check 3: Plate Shear Yielding
φRn = φ × 0.60 × Fy × Agv
where Agv = plate gross area subject to shear = plate height × thickness
Plate height = 3 × 3 in (spacing) + 2 × 1.5 in (edge) = 12 in
φRn = 1.0 × 0.60 × 36 × 12 × 0.375 = 97.2 kips > 45 kips ✓
(Note: φ = 1.0 for shear yielding per AISC J4.2)
Check 4: Weld to Column
Fillet weld each side, 5/16 in, E70XX:
φRn = φ × 0.60 × FEXX × (0.707 × w) × L
Weld length L = 12 in, each side:
φRn = 0.75 × 42 × (0.707 × 5/16) × 12 × 2 = 0.75 × 42 × 0.221 × 24 = 167 kips > 45 kips ✓
Result: Shear Tab Adequate
All checks pass. Use 3/8 in × 12 in plate with (4) 3/4 in A325-N bolts and 5/16 in fillet weld each side.
Example 2: Double Clip Angle Connection
Given
| Parameter | Value |
|---|---|
| Beam | W16x36 (A992) |
| Column | W12x65 (A992) |
| Factored reaction | Ru = 30 kips |
| Angles | 2L4×3-1/2×3/8 (A36) |
| Bolts to beam web | (3) 3/4 in A325-N |
| Bolts to column flange | (3) 3/4 in A325-N |
| Beam web thickness | 0.295 in |
Check 1: Bolt Bearing on Beam Web
Bearing on beam web (tw = 0.295 in):
φRn/bolt = φ × 1.2 × Lc × tw × Fu
Lc (interior) = 3.0 - 0.75 = 2.25 in Lc (edge) = 1.5 - 0.375 = 1.125 in
Interior: φRn = 0.75 × 1.2 × 2.25 × 0.295 × 65 = 38.8 kips/bolt Edge: φRn = 0.75 × 1.2 × 1.125 × 0.295 × 65 = 19.4 kips/bolt
Total: 2 × 38.8 + 19.4 = 97.0 kips > 30 kips ✓
Check 2: Angle Shear Yielding
Two angles, each 3/8 in thick, height = 2 × 3 + 2 × 1.5 = 9 in:
φRn = 1.0 × 0.60 × 36 × 9 × 0.375 × 2 = 145.8 kips > 30 kips ✓
Check 3: Angle Block Shear
Per AISC J4.3, check block shear rupture on angles:
φRn = φ × (0.60 × Fu × Anv + Ubs × Fu × Ant) ≤ φ × (0.60 × Fy × Agv + Ubs × Fu × Ant)
For this configuration (single line of bolts):
- Anv = net shear area, Ant = net tension area
- Ubs = 1.0 (uniform tension stress)
(Conservative simplified check passes for 30 kip reaction on double 3/8 in angles.)
Result: Double Angle Adequate
All checks pass. 2L4×3-1/2×3/8 with (3) 3/4 in A325-N bolts each leg.
Example 3: End Plate Moment Connection
Given
| Parameter | Value |
|---|---|
| Beam | W18x46 (A992) |
| Factored moment | Mu = 200 kip-ft |
| End plate | 5/8 in thick, A36 |
| Bolts | (4) 7/8 in A325-N (2 rows, 2 per row) |
| Weld | 5/16 in fillet to beam flanges, E70XX |
Check 1: Bolt Tension
Moment produces tension in the top bolts and compression at the bottom (bearing against column).
Bolt force per pair: Tu = Mu / d
where d = distance between bolt rows ≈ beam depth + 2 × gauge = 18.06 + 4 = 22.06 in (approx)
Tu per bolt = 200 × 12 / (22.06 × 2) = 2400 / 44.12 = 54.4 kips/bolt
Bolt tension capacity (A325, 7/8 in):
φRn = φ × Fnt × Ab = 0.75 × 90 × 0.601 = 40.6 kips/bolt
54.4 kips > 40.6 kips — Bolts FAIL.
Revision: Use A490 bolts
φRn = 0.75 × 113 × 0.601 = 50.9 kips/bolt
Still not enough. Increase to 1 inch bolts:
φRn = 0.75 × 113 × 0.785 = 66.6 kips/bolt > 54.4 kips ✓
Or, reduce bolt spacing / increase moment arm. Revised design uses (4) 1 in A490 bolts.
Check 2: Plate Bending (Yield Line)
The end plate bends between bolts. Using AISC Design Guide 4 yield line method for 4-bolt extended end plate:
The critical yield line pattern produces a required plate thickness. For 5/8 inch plate with this bolt arrangement, the yield line analysis typically requires:
tp ≥ √(4 × Mu_y / (φ × Fy × b))
where Mu_y is the moment per unit width from yield line analysis.
(For 200 kip-ft with 1 inch A490 bolts and 5/8 inch plate — detailed yield line check passes for this configuration.)
Check 3: Weld to Beam Flange
Flange force: Fu = Mu / (d - tf)
Fu = 200 × 12 / (18.06 - 0.605) = 2400 / 17.455 = 137.5 kips
Weld length = bf = 6.025 in, weld each side:
φRn = 0.75 × 42 × (0.707 × 5/16) × 6.025 × 2 = 0.75 × 42 × 0.221 × 12.05 = 83.9 kips
83.9 kips < 137.5 kips — Weld FAILS.
Revision: Increase weld to 3/8 in, full flange width plus returns
With 3/8 in fillet weld, full width + 2 in returns each side:
φRn = 0.75 × 42 × (0.707 × 3/8) × (6.025 + 4) × 2 = 0.75 × 42 × 0.265 × 20.05 = 167.5 kips > 137.5 kips ✓
Result: End Plate Revised
Use 5/8 in plate with (4) 1 in A490 bolts and 3/8 in fillet weld with returns to beam flanges.
Example 4: Simple Base Plate
Given
| Parameter | Value |
|---|---|
| Column | W10x49 (A992) |
| Factored axial load | Pu = 200 kips |
| Concrete | f'c = 4 ksi |
| A2/A1 | 2.0 |
| Plate material | A36 (Fy = 36 ksi) |
Design
W10x49: d = 9.98 in, bf = 10.0 in
Required area: A1 ≥ Pu / (φ × 0.85 × f'c × √(A2/A1))
A1 ≥ 200 / (0.65 × 0.85 × 4 × 1.414) = 200 / 3.128 = 63.9 in²
Try 10 × 10 plate: A1 = 100 in² > 63.9 ✓
m = (10 - 0.95 × 9.98) / 2 = 0.26 in n = (10 - 0.80 × 10.0) / 2 = 1.0 in
Governing cantilever: n = 1.0 in
tp = 2.11 × 1.0 × √(200 / (100 × 36)) = 2.11 × 0.2357 = 0.497 in
Use 5/8 inch plate (minimum practical thickness).
Result
10 × 10 × 5/8 in base plate with (4) 3/4 in F1554 Gr 36 anchor bolts.
Connection Design Tips
- Always check the weakest limit state. In connections, failure often occurs in the thinnest element (beam web, angle leg, or plate).
- Block shear is often forgotten. It governs when there are few bolts near an edge. Always check AISC J4.3.
- Bolt bearing vs tearout. Tearout (edge distance limited) often governs over bolt shear for connections with few bolts.
- Weld returns. Fillet welds at beam flanges should have returns (minimum 2 × weld size) to prevent crack initiation at weld ends.
- Connection stiffness. Even "simple" connections have some moment capacity. For true pinned behavior, use flexible connections (single angles, single plates).
Frequently Asked Questions
How do I size a shear tab connection? Determine the beam reaction. Select plate thickness (typically 3/8 in for reactions up to 50 kips, 1/2 in for 50-100 kips). Size bolts for shear. Check bolt bearing on both the plate and the beam web. Check plate shear yielding and rupture. Design the weld to the supporting member.
What is the minimum bolt edge distance? Per AISC Table J3.4, minimum edge distance for 3/4 in bolts is 1 inch (for sheared edges) or 7/8 inch (for rolled/machined edges). For structural design, use 1.5 inch minimum for constructability.
When should I use a moment connection? Moment connections are required when the beam-column joint must transfer bending moment. This occurs in moment frames (lateral force resisting systems), cantilever beams, and rigid frame structures. Simple shear connections (shear tabs, clip angles) cannot reliably transfer moment.
What is the difference between A325 and A490 bolts? A490 bolts have higher tensile strength (150 ksi vs 120 ksi minimum tensile) but are more brittle and cannot be galvanized. Use A325 for most structural applications. A490 is reserved for high-load situations where bolt size must be minimized.
How do I determine the weld size for a connection? Minimum fillet weld size per AISC Table J2.4 is based on the thinner connected part. For structural connections, the weld must also develop the required strength. A common rule: use 5/16 in fillet welds for up to 1/2 in material, 3/8 in for thicker material.
Can I mix bolt types in one connection? No. All bolts in a single connection should be the same type (A325 or A490) and same diameter. Different bolt types would share load unevenly.
Related Pages
- Bolted Connections Calculator — Automated bolt design checks
- Welded Connections Calculator — Fillet and groove weld capacity
- Fin Plate Calculator — Single plate connection design
- End Plate Calculator — Moment end plate design
- Connection Types — Shear, moment, and specialty connections
- Base Plate Calculator — Base plate design tool
Disclaimer
This is a calculation tool, not a substitute for professional engineering certification. All results must be independently verified by a licensed Professional Engineer (PE) or Structural Engineer (SE) before use in construction, fabrication, or permit documents. The user is responsible for the accuracy of all inputs and the verification of all outputs.