Shear Lag — Definition, Reduction Factor U & Tension Member Design
Shear lag is the phenomenon in structural steel tension members where tensile stress is not uniformly distributed across the cross-section at the connection. When only part of a member's cross-section is directly attached to the gusset plate or connected member (e.g., one leg of an angle, one flange of a channel), the unconnected elements "lag behind" in carrying their share of the load. The shear transfer from the connected elements to the unconnected elements requires distance along the member, and near the connection, stress is concentrated in the directly attached portion.
The consequence is that the full net area An cannot be fully utilized. AISC 360 accounts for this by introducing the shear lag factor U, which reduces the effective net area:
Ae = An * U (effective net area in tension)
Physical Mechanism
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Consider an angle tension member bolted through one leg only:
- At the connection, most load enters through the connected leg
- The outstanding leg is unloaded at the connection
- Shear must transfer load from the connected leg to the outstanding leg
- This shear transfer occurs gradually along the member length
- Near the connection, the outstanding leg carries less stress than the connected leg — this is shear lag
- Farther from the connection (beyond the transition length), stress becomes uniform
The shear lag factor U accounts for this non-uniform stress distribution at the critical section (through the bolt holes).
AISC 360 Table D3.1 — Shear Lag Factors
| Connection Type | U Factor | Notes |
|---|---|---|
| W, M, S shapes — flange connected, >= 3 bolts per line | U = 0.90 (bfd >= 2/3d) or U = 0.85 (bfd < 2/3d) | Both flanges connected: U = 1.0 |
| W, M, S shapes — web connected, >= 4 bolts per line | U = 0.70 | Conservative; see Commentary for refinement |
| Structural tees — flange connected, >= 3 bolts per line | U = 0.90 | Similar to W-shape flange case |
| Single angle — one leg connected, >= 4 bolts | U = 0.80 | For 2 or 3 bolts: U = 0.60 |
| Double angle — connected to each side of plate | U = 0.80 | >= 3 bolts per line |
| Channel — web connected, >= 4 bolts | U = 0.70 | Similar to W-shape web case |
| HSS Rectangular — concentric connection to plate | U = 1.00 | All walls attached (slot through both sides) |
| HSS Round — concentric, slotted, fillet welded | U = 0.90 | Reduced if slot width inadequate |
General Case — U = 1 - x_bar / L
For most bolted connections where the connection length L can be defined, a more precise U factor is:
U = 1 - x_bar / L <= 0.90 (and >= 0.60)
Where:
- x_bar = distance from the plane of the connection to the centroid of the connected element (in)
- L = length of the connection (in), taken as the distance between the first and last bolt in the line
Key observation: Longer connections (larger L) reduce shear lag (U approaches 1.0). Short, stubby connections with concentrated bolts produce more shear lag. This is why AISC specifies minimum numbers of bolts per line — fewer bolts means shorter L, larger x_bar/L, and smaller U.
Effect on Tension Member Design
The tension member yield and fracture checks are:
Yielding on gross area (AISC 360 D2):
phi * Rn = 0.90 * Fy * Ag
Fracture on effective net area (AISC 360 D2):
phi * Rn = 0.75 * Fu * Ae = 0.75 * Fu * An * U
Where An = Ag - (number of holes _ (dh + 1/16") _ t) for a straight-line failure path, or accounting for stagger if bolts are staggered.
The fracture capacity on Ae often governs for tension members because:
- Fu is only ~20-30% higher than Fy (58 vs. 36 ksi for A36), but phi_fracture = 0.75 vs. phi_yield = 0.90
- Net area reduction from bolt holes further reduces capacity
- Shear lag U further reduces effective area
Worked Example — Single Angle in Tension
Problem: A 4"x4"x3/8" single angle (A36: Fy = 36 ksi, Fu = 58 ksi) is used as a tension diagonal brace. It is connected by one leg with four 3/4" A325 bolts in a single line. Bolt spacing = 3" c/c. Determine the design tensile strength.
Angle properties:
- Ag = 2.86 in^2
- x_bar = 1.12 in (centroid distance from back of connected leg)
- t = 3/8" = 0.375 in
Step 1: Gross section yielding
phi*Pn_yield = 0.90 * 36 * 2.86 = 92.7 kips
Step 2: Net area
dh = 3/4" + 1/16" = 13/16" = 0.8125" (hole diameter)
An = Ag - 1 hole * 0.8125 * 0.375 = 2.86 - 0.305 = 2.555 in^2
Step 3: Shear lag factor
L = 3 * 3 = 9.0" (distance between first and last bolt)
U = 1 - x_bar/L = 1 - 1.12/9.0 = 1 - 0.124 = 0.876
Check: 0.876 <= 0.90 and >= 0.60. OK. (AISC Table D3.1 gives U = 0.80 for >= 4 bolts — the general formula gives a more favorable value.)
Step 4: Effective net area fracture
Ae = An * U = 2.555 * 0.876 = 2.238 in^2
phi*Pn_fracture = 0.75 * 58 * 2.238 = 97.4 kips
Step 5: Controlling capacity phi*Pn = min(92.7, 97.4) = 92.7 kips (gross yielding governs for this case with a long connection).
If the connection had only 2 bolts (L = 3"), then U = 1 - 1.12/3 = 0.627, Ae = 2.555 * 0.627 = 1.60 in^2, and phiPnfracture = 0.75 * 58 _ 1.60 = 69.6 kips — fracture would govern at 25% lower capacity.
Code Comparison
| Code | Section | Shear Lag Factor U | Approach |
|---|---|---|---|
| AISC 360 | Table D3.1 | 0.60 to 1.00 | Pre-calculated or U = 1 - x_bar/L |
| AS 4100 | 7.3 | Similar reduction factor kt | Equivalent to U, connection eccentricity factor |
| EN 1993-1-1 | 6.2.3 | Net section reduction for staggered holes only | Shear lag via partial safety factors, less explicit |
| CSA S16 | 12.3.3 | U = 0.60 to 1.00 | Similar to AISC, Table 2 values |
Design Strategies to Minimize Shear Lag
| Strategy | Effect |
|---|---|
| Increase connection length L | Reduces x_bar/L, increases U |
| Connect more elements directly | Connect both angle legs, both flanges, full perimeter |
| Use concentric connections | HSS with slotted gusset through both walls: U = 1.0 |
| Weld instead of bolt | Weld along the entire connected edge, U = 1.0 (if weld >= member width) |
| Use heavier member section | Capacity increase may offset U reduction |
| Provide transition length | Longer grip lengths allow stress redistribution |
Frequently Asked Questions
What is shear lag in tension members? Shear lag is the non-uniform stress distribution in a tension member near its connection, caused by only part of the cross-section being directly attached. The unconnected elements lag in carrying load until shear transfer occurs, reducing the effective net area to Ae = An * U (U <= 1.0).
How do I calculate the shear lag factor U? Use AISC 360 Table D3.1 for standard cases, or the general formula U = 1 - x_bar/L where x_bar is the connection eccentricity and L is the connection length. For welds, U depends on weld length relative to member width. U typically ranges from 0.60 to 1.00.
Does shear lag affect compression members? Shear lag is primarily a tension member phenomenon because tension fracture is sensitive to stress concentration at net sections. Compression members are checked for buckling, not fracture, and shear lag effects are less critical. However, connection eccentricity from partial connection can produce bending moments that must be accounted for in compression.
When does shear lag govern over gross yielding? Shear lag + net section fracture governs when the connection is short (small L), only part of the section is connected (high x_bar), there are many bolt holes removing significant area, or the Fu/Fy ratio is low. For A36 steel (Fu/Fy = 1.61), shear lag often governs. For A572 Gr. 50 (Fu/Fy = 1.30), gross yielding typically governs.
Related Terms and Pages
- Block Shear — Definition & Failure Mode
- Prying Action — Definition & Bolt Force
- Web Crippling — Definition & Design Check
- Compact Section — Definition & Limits
- Tension Member Design — Bolted & Welded
- Connection Design Workflow
- Bolt Patterns & Gage Distances
- Angle Section Properties
Educational reference only. Shear lag factors must be determined per the governing design code (AISC 360 Table D3.1, AS 4100 Section 7.3, CSA S16 Clause 12.3.3) by a licensed Professional Engineer for all construction applications.
Disclaimer: This content is for educational purposes only. Results must be verified by a licensed professional engineer. Steel Calculator provides preliminary design tools — NOT a substitute for professional engineering judgment.