AISC 360-22 Table D3.1 — Shear Lag Factor U for Tension Members
AISC 360-22 Table D3.1 defines the shear lag factor U used to calculate the effective net area A_e of tension members. When a tension member is connected through only some of its cross-section elements (e.g., a W-shape bolted through its flanges only, or a single angle connected through one leg), the stress in the unconnected elements lags behind the stress at the connection. This uneven stress distribution reduces the member's fracture capacity. Table D3.1 provides eight cases for determining U without performing a detailed elastic analysis.
This page reproduces all eight cases with values, examples, and practical guidance. For a complete treatment of shear lag theory and multi-code comparison, see our Shear Lag Factor reference.
Quick access:
- How to Use Table D3.1
- Case 1 — All Elements Connected
- Case 2 — General Formula
- Case 3 — W-Shapes Connected Through Flanges
- Case 4 — W-Shapes Connected Through Web
- Case 5 — Single and Double Angles
- Case 6 — Round and Rectangular HSS
- Case 7 — Plates with Longitudinal Welds
- Case 8 — HSS with Non-Concentric Gusset
- U Values for Common Shapes
- Worked Examples
- Common Mistakes
- Frequently Asked Questions
How to Use Table D3.1
The shear lag factor U is applied in the tensile rupture limit state:
phi P_n = 0.75 x F_u x A_e = 0.75 x F_u x A_n x U
Where:
- phi = 0.75 (LRFD resistance factor for rupture)
- F_u = specified minimum tensile strength (58 ksi for A36, 65 ksi for A992)
- A_n = net area (gross area minus bolt hole deductions)
- U = shear lag factor from Table D3.1 (0 < U <= 1.0)
Selection logic:
- Identify the member type and connection configuration
- Match to the appropriate Table D3.1 case (1 through 8)
- If multiple cases could apply, use the case that gives the larger U value
- If no prescriptive case applies, use the general formula U = 1 - x_bar/L (Case 2)
Case 1 — All Elements Connected
When it applies: Tension load is transmitted to all elements of the cross-section by transverse welds.
U = 1.0
This is the ideal case — the entire cross-section is engaged at the connection, so there is no shear lag. Examples:
- A W-shape with transverse welds across both flanges and the web
- A round HSS with a full-perimeter weld to a gusset plate
- A plate with transverse welds across its full width
Case 2 — General Formula
When it applies: Any configuration not specifically covered by Cases 1 or 3-8. This is the most general case and can be used as a fallback.
U = 1 - x_bar / L
Where:
- x_bar = distance from the centroid of the connected elements to the connection plane (perpendicular to the direction of loading)
- L = length of the connection, measured between the outermost fasteners (for bolted) or the length of longitudinal weld (for welded)
Practical interpretation: Longer connections (larger L) produce U closer to 1.0. A connection length of at least 3 times the eccentricity (L >= 3 x_bar) typically gives U >= 0.67.
Case 3 — W-Shapes, M-Shapes, S-Shapes, HP-Shapes (Flange Connections, 3+ Fasteners)
When it applies: Tension load transmitted only through the flanges of W, M, S, or HP shapes using bolts. Must have 3 or more fasteners per line in the direction of loading.
| Condition | U |
|---|---|
| b_f >= (2/3) d | 0.90 |
| b_f < (2/3) d | 0.85 |
Where b_f = flange width, d = overall depth.
Most standard W-shapes qualify for U = 0.90. For example:
- W14x43: b_f = 7.995", d = 13.66" → b_f/d = 0.585 < 0.667 → U = 0.85
- W12x50: b_f = 8.080", d = 12.19" → b_f/d = 0.663 < 0.667 → U = 0.85
- W10x49: b_f = 10.00", d = 9.98" → b_f/d = 1.002 >= 0.667 → U = 0.90
- W8x31: b_f = 7.995", d = 8.00" → b_f/d = 0.999 >= 0.667 → U = 0.90
With fewer than 3 fasteners per line, use Case 2 (general formula) instead.
Case 4 — W-Shapes Connected Through Web (4+ Fasteners)
When it applies: Tension load transmitted only through the web of W, M, S, or HP shapes, or through the stem of tees. Requires 4 or more fasteners per line in the direction of loading.
U = 0.90
With fewer than 4 fasteners per line, use Case 2 (general formula).
Case 5 — Single and Double Angles
When it applies: Single angles connected through one leg, or double angles (back-to-back) each connected through one leg.
| Fasteners per line | U |
|---|---|
| 4 or more | 0.80 |
| 3 | 0.60 |
| 2 or fewer | Use Case 2 formula |
Note: The designer may use the larger of the Case 2 formula value and the tabulated value from Case 5. For a single L4x4x3/8 with 4 bolts and a typical bolt pattern, the Case 2 formula often gives U > 0.80, so always check both.
Case 6 — Round and Rectangular HSS (Concentric Gusset)
When it applies: Round or rectangular HSS connected by a single concentric gusset plate (slotted through the tube wall and welded along both sides).
| Condition | U |
|---|---|
| L >= 1.3D | 1.0 |
| L < 1.3D | Use Case 2 |
Where D = outside diameter (round HSS) or the loaded dimension (rectangular HSS). L = length of the gusset connection.
Practical note: For a 6-inch diameter round HSS, the gusset must be at least 7.8 inches long to achieve U = 1.0.
Case 7 — Plates with Longitudinal Welds
When it applies: Flat plates or bars connected by longitudinal welds along both edges, with no transverse weld.
| Weld Length / Width (l/w) | U |
|---|---|
| l >= 2w | 1.00 |
| 1.5w <= l < 2w | 0.87 |
| w <= l < 1.5w | 0.75 |
Where l = length of longitudinal weld, w = plate width (distance between welds).
Example: A 6-inch wide plate with 14-inch long longitudinal welds: l/w = 14/6 = 2.33 >= 2.0, so U = 1.0. If the welds are only 8 inches long: l/w = 8/6 = 1.33, which is between 1.0 and 1.5, so U = 0.75.
Case 8 — HSS with Non-Concentric Gusset
When it applies: Rectangular HSS or box sections connected through a single side wall or with a non-concentric gusset plate.
U = 1 - x_bar / L (same as Case 2)
The eccentricity x_bar is measured from the centroid of the full HSS section to the connected wall. This is distinct from Case 6, which applies only to concentric gusset connections.
U Values for Common Shapes
Standard W-Shapes (Flange-Connected, 3+ Bolts per Line)
| Shape | b_f/d | U (Case 3) |
|---|---|---|
| W14x43 | 0.585 | 0.85 |
| W12x50 | 0.663 | 0.85 |
| W12x65 | 0.752 | 0.90 |
| W10x49 | 1.002 | 0.90 |
| W10x22 | 0.607 | 0.85 |
| W8x31 | 0.999 | 0.90 |
| W8x18 | 0.688 | 0.90 |
| W6x15 | 0.757 | 0.90 |
| W4x13 | 0.862 | 0.90 |
Common Angles (One Leg Connected, 4+ Bolts)
| Shape | x_bar (in) | Typical U (Case 5) |
|---|---|---|
| L4x4x3/8 | 1.13 | 0.80 |
| L6x4x3/8 | 1.03 | 0.80 |
| L3-1/2x3-1/2x5/16 | 0.978 | 0.80 |
| L5x3-1/2x5/16 | 0.872 | 0.80 |
Tip: Always check the Case 2 formula (1 - x_bar/L) in addition to the tabulated value. For long connections with many bolts, the Case 2 value may exceed 0.80.
Worked Examples
Example 1 — W10x49 Tension Member (Flange-Connected)
A W10x49 (A992, F_y = 50 ksi, F_u = 65 ksi) is connected through its flanges with four 3/4" A325 bolts per line (2 per flange) in standard holes. The connection has two lines of bolts spaced at 3 inches apart.
Given:
- A_g = 14.4 in^2
- d = 9.98", b_f = 10.00"
- Bolt holes: 4 holes x (13/16" + 1/16") = 4 x 7/8" diameter deduction per line
Step 1 — Select U:
- Case 3 applies: flange-connected with 4 bolts per line (>= 3)
- b_f/d = 10.00/9.98 = 1.002 >= 2/3 → U = 0.90
Step 2 — Net area:
- Flange thickness t_f = 0.560"
- Hole deduction per line = 2 x (13/16 + 1/16) x 0.560 = 2 x 0.875 x 0.560 = 0.980 in^2
- Both flanges: A_n = 14.4 - 2(0.980) = 12.44 in^2 (approximate; web holes if any would also deduct)
Step 3 — Rupture capacity:
- A_e = A_n x U = 12.44 x 0.90 = 11.20 in^2
- phi P_n = 0.75 x 65 x 11.20 = 546 kips
Step 4 — Compare to yielding:
- phi P_n (yielding) = 0.90 x 50 x 14.4 = 648 kips
- Rupture (546 kips) governs over yielding (648 kips)
Example 2 — Single L4x4x3/8 Tension Member
A single L4x4x3/8 (A36, F_y = 36 ksi, F_u = 58 ksi) is connected through one leg with 3 bolts at 3" spacing.
Given:
- A_g = 2.86 in^2
- x_bar = 1.13" (for the connected leg)
- Bolt holes: 3 holes x (13/16 + 1/16) x 3/8 = 3 x 0.875 x 0.375 = 0.984 in^2
Step 1 — U from two cases:
- Case 5: 3 bolts → U = 0.60
- Case 2: L = 2 x 3 = 6" (distance from first to last bolt), x_bar = 1.13", U = 1 - 1.13/6 = 0.812
- Use the larger: U = 0.812
Step 2 — Net area:
- A_n = 2.86 - 0.984 = 1.876 in^2
Step 3 — Rupture capacity:
- A_e = 1.876 x 0.812 = 1.523 in^2
- phi P_n = 0.75 x 58 x 1.523 = 66.3 kips
Step 4 — Compare to yielding:
- phi P_n (yielding) = 0.90 x 36 x 2.86 = 92.7 kips
- Rupture (66.3 kips) governs
Common Mistakes
Using Case 5 tabulated value without checking Case 2. Always check the general formula — it often gives a larger U for long connections.
Confusing connection length with member length. L is the distance between the outermost fasteners (or the length of longitudinal weld), not the overall member length.
Applying U to the yielding limit state. Shear lag only affects the rupture limit state. Yielding uses the full gross area (phi P_n = 0.90 x F_y x A_g).
Using Case 3 with fewer than 3 bolts per line. Case 3's tabulated values (0.85 or 0.90) require at least 3 fasteners per line. With 2 bolts, use Case 2.
Forgetting that the net area deduction uses hole diameter + 1/16". The standard hole for a 3/4" bolt is 13/16", but the net area deduction is 13/16 + 1/16 = 7/8".
Not using the maximum of Cases 2 and 5 for angles. AISC 360 permits the designer to use the larger value. Always compute both.
Calculator
Check any tension member against AISC 360-22 with automatic shear lag factor selection:
- Bolted Connections Calculator — Tension rupture with Table D3.1 U values
- Section Properties Lookup — b_f, d, t_f, and x_bar for all AISC shapes
FAQ
Q: What does the shear lag factor U represent? A: U represents the ratio of the maximum stress in the connected element to the average stress across the entire cross-section. When U = 1.0, all elements carry equal stress (no lag). When U < 1.0, the unconnected elements carry less stress than the connected elements, reducing the effective area available for fracture.
Q: When does U = 1.0? A: When all elements of the cross-section are connected by transverse welds (Case 1), when a round HSS has a concentric gusset at least 1.3D long (Case 6), or when a plate has longitudinal welds at least 2 times the plate width (Case 7).
Q: Can I use U = 1.0 for a W-shape connected through both flanges? A: Only if transverse welds connect both flanges and the web. If the connection is bolted through the flanges only, U is governed by Case 3 (typically 0.85 or 0.90) regardless of how many bolts are used.
Q: Does the shear lag factor apply to bolt shear or bearing checks? A: No. U applies only to the tensile rupture limit state of the tension member itself. Bolt shear, bearing, and tearout are checked independently at each bolt location.
Q: What if my connection doesn't match any of the 8 cases? A: Use the general formula from Case 2: U = 1 - x_bar/L. Alternatively, AISC 360 permits the use of a U value determined by rational analysis (finite element or experimental).
Related: AISC 360-22 Steel Design Overview | AISC Block Shear — Section J4.3 | AISC Standard Hole Sizes | Shear Lag Factor — Complete Reference | Bolt Bearing and Tearout