AISC Block Shear — Section J4.3 Design Guide

Block shear is a limit state where a block of material at a connection tears out from the member along a perimeter defined by the bolt pattern. It combines tension rupture on one face with shear yielding or rupture on one or two other faces. AISC 360-22 Section J4.3 provides two design equations that account for the interaction of these failure modes. Block shear frequently governs the design of connections in angles, gusset plates, and beam web copes — and it is one of the most commonly searched AISC provisions.

Quick access:

• What is Block Shear?
• AISC 360-22 Section J4.3 Equations
• How to Identify the Failure Path
• U_bs — Eccentricity Factor
• Worked Example 1 — Single Angle
• Worked Example 2 — Gusset Plate
• Worked Example 3 — Beam Web Cope
• Block Shear vs. Bearing/Tearout
• Common Mistakes
• Frequently Asked Questions

What is Block Shear?

Block shear failure occurs when a section of connected material tears out along a perimeter defined by the bolt pattern. The failure path consists of:

• Shear surfaces: One or two planes of shear rupture parallel to the applied load, running along the bolt line(s)
• Tension surface: One plane of tension rupture perpendicular to the applied load, typically at the last row of bolts

The block "tears out" as a unit — the material between the bolts and the free edge shears along the bolt line, while the material at the end of the bolt group fractures in tension.

When Block Shear Governs

Block shear is most critical when:

• Bolts are near a free edge — shorter shear planes mean lower shear resistance
• Few bolts in the shear plane — 2 or 3 bolts produce short shear lengths
• Thin material — lower plate thickness reduces both shear and tension areas
• High-strength bolts in thin plates — bolt strength exceeds the plate's block shear capacity
• Angle connections — single angles with one-leg connections are particularly susceptible

In many practical connections, block shear governs over bolt shear, bearing, or tearout. Always check it.

AISC 360-22 Section J4.3 Equations

AISC 360-22 provides two equations for the nominal block shear strength R_n. The design strength is the lesser of the two:

Equation J4-5 (Shear yielding + Tension rupture):

R_n = 0.60 x F_y x A_gv + U_bs x F_u x A_nt

Equation J4-6 (Shear rupture + Tension rupture):

R_n = 0.60 x F_u x A_nv + U_bs x F_u x A_nt

Design strength (LRFD):

phi R_n = 0.75 x R_n     (where R_n = lesser of J4-5 and J4-6)

Where:

How to Compute Each Area

A_gv (Gross shear area):

A_gv = t x L_v

Where t = plate thickness, L_v = total length of the shear path (sum of all shear segments).

A_nv (Net shear area):

A_nv = t x (L_v - n_holes x d_h)

Where n_holes = number of bolt holes in the shear path, d_h = standard hole diameter + 1/16".

A_nt (Net tension area):

A_nt = t x (L_t - 0.5 x n_holes_t x d_h)

Where L_t = length of the tension path, n_holes_t = number of bolt holes in the tension path. The factor of 0.5 on hole deductions in the tension path accounts for the stagger effect per AISC Section B4.3.

Net Area Hole Deductions

For standard holes, the deduction per hole is:

d_h = bolt diameter + 1/16" (for punched holes)
d_h = bolt diameter + 1/16" (for drilled holes — same rule)

For a 3/4" bolt: d_h = 13/16" + 1/16" = 7/8" = 0.875"

For an A325 3/4" bolt in a standard hole (hole diameter = 13/16"):

d_h = 13/16" + 1/16" = 7/8"

How to Identify the Failure Path

The block shear failure path must be traced as a continuous perimeter around the bolt group. The path follows:

1. Along the bolt line(s) in the direction of load (shear surfaces)
2. Between or around bolt holes perpendicular to the load (tension surface)

Single Row of Bolts — One Shear Plane

The most common case. One row of bolts near a free edge:

Free edge
    |
    |---- bolt 1 ----|
    |                |
    |---- bolt 2 ----|   <-- tension rupture
    |                |
    |---- bolt 3 ----|
    |
Shear plane (along bolt line, parallel to load)

• Shear path: along the bolt line from the free edge to the last bolt
• Tension path: across the last bolt row, perpendicular to the load

Double Row of Bolts — Two Shear Planes

When bolts are arranged in two rows:

Row 1    Row 2
bolt 1   bolt 1
  |        |
bolt 2   bolt 2    <-- tension rupture across both rows
  |        |
bolt 3   bolt 3

• Two shear planes (one along each row)
• One tension plane across the end

U_bs — Eccentricity Factor

The eccentricity factor U_bs accounts for how uniformly the tension stress is distributed across the tension surface:

When U_bs = 1.0:

• Connection is symmetric about the line of load transfer
• Both shear planes are equidistant from the load path
• Examples: gusset plates with centered bolt groups, wide plates with bolts on both sides

When U_bs = 0.5:

• Connection has eccentricity (single shear plane offset from the load path)
• Single angle connections connected through one leg
• Tee sections connected through the flange with bolts on one side of the web

Practical rule: When in doubt, use U_bs = 0.5. It is always conservative.

Worked Example 1 — Single Angle

Problem: A single L4x4x3/8 (A36 steel, F_y = 36 ksi, F_u = 58 ksi) is connected to a gusset plate through one leg with three 3/4" A325 bolts at 3" spacing. Check block shear.

Given:

• t = 3/8" = 0.375"
• Bolt diameter = 3/4", standard hole diameter = 13/16"
• d_h = 13/16 + 1/16 = 7/8" = 0.875"
• Edge distance from the end bolt to the free end = 1.5"
• Spacing between bolts = 3"

Identify the failure path:

• Shear plane: along the 3 bolts in the direction of load
  • L_v = edge distance + 2 x spacing = 1.5 + 2(3) = 7.5"
• Tension plane: across the last bolt, perpendicular to load
  • L_t = 4" (leg width, approximately)

Compute areas:

• A_gv = t x L_v = 0.375 x 7.5 = 2.813 in^2
• A_nv = t x (L_v - 3 x d_h) = 0.375 x (7.5 - 3 x 0.875) = 0.375 x 4.875 = 1.828 in^2
• A_nt = t x (L_t - 0.5 x 1 x d_h) = 0.375 x (4 - 0.5 x 0.875) = 0.375 x 3.563 = 1.336 in^2

Determine U_bs:

• Single angle, one leg connected → U_bs = 0.5

Check both equations:

J4-5 (shear yielding + tension rupture):

R_n = 0.60 x 36 x 2.813 + 0.5 x 58 x 1.336
R_n = 60.76 + 38.74 = 99.50 kips

J4-6 (shear rupture + tension rupture):

R_n = 0.60 x 58 x 1.828 + 0.5 x 58 x 1.336
R_n = 63.61 + 38.74 = 102.35 kips

Governor: J4-5 = 99.50 kips (lesser)

Design strength:

phi R_n = 0.75 x 99.50 = 74.6 kips

Compare to bolt shear: 3 bolts x 0.75 x 54 x (pi/4 x 0.75^2) = 3 x 17.9 = 53.6 kips. Block shear (74.6 kips) does not govern in this case — bolt shear governs. However, for thinner material or more bolts, block shear could govern.

Worked Example 2 — Gusset Plate

Problem: A 1/2" gusset plate (A572 Grade 50, F_y = 50 ksi, F_u = 65 ksi) is connected with six 7/8" A490 bolts in two rows of 3, as shown. The bolt spacing is 3" and the edge distance is 1.5". Check block shear.

Given:

• t = 1/2" = 0.50"
• Bolt diameter = 7/8", d_h = 1" (standard hole: 15/16" + 1/16")
• 3 bolts per row, 3" spacing, 1.5" edge distance

Identify the failure path (two shear planes):

• Shear path each row: L_v = 1.5 + 2(3) = 7.5"
• Total A_gv = 2 rows x (0.50 x 7.5) = 7.50 in^2
• Total A_nv = 2 rows x [0.50 x (7.5 - 3 x 1.0)] = 2 x [0.50 x 4.5] = 4.50 in^2

Tension path (across the 3 bolts at the end):

• A_nt = 0.50 x (3" - 2 x 0.5 x 1.0) = 0.50 x 2.0 = 1.00 in^2 (2 holes in the tension path, staggered)

Determine U_bs:

• Symmetric connection with two shear planes → U_bs = 1.0

Check both equations:

J4-5:

R_n = 0.60 x 50 x 7.50 + 1.0 x 65 x 1.00
R_n = 225.0 + 65.0 = 290.0 kips

J4-6:

R_n = 0.60 x 65 x 4.50 + 1.0 x 65 x 1.00
R_n = 175.5 + 65.0 = 240.5 kips

Governor: J4-6 = 240.5 kips (lesser)

Design strength:

phi R_n = 0.75 x 240.5 = 180.4 kips

Worked Example 3 — Beam Web Cope

Problem: A W18x35 beam (A992, F_y = 50 ksi, F_u = 65 ksi) has a top cope at the connection to a column. The cope removes the top flange and part of the web. Two 3/4" bolts connect the remaining web to a clip angle. Web thickness t_w = 0.300". The bolt pattern is 2 bolts in a single row, 3" spacing, with 1.5" edge distance from the cope to the first bolt.

Identify the failure path:

• Shear plane: along the bolt line, from the cope edge down
  • L_v = 1.5 + 3 = 4.5" (edge distance + one spacing)
• Tension plane: across the bottom bolt
  • L_t = distance from bolt to free edge (web height below the bolt)

Compute areas:

• A_gv = 0.300 x 4.5 = 1.350 in^2
• A_nv = 0.300 x (4.5 - 2 x 0.875) = 0.300 x 2.750 = 0.825 in^2
• A_nt = 0.300 x (L_t - 0.5 x 1 x 0.875) = depends on geometry below the bolt

For this example, assume L_t = 3.0":

• A_nt = 0.300 x (3.0 - 0.438) = 0.300 x 2.563 = 0.769 in^2

U_bs = 1.0 (symmetric, bolts centered in the cope)

J4-5:

R_n = 0.60 x 50 x 1.350 + 1.0 x 65 x 0.769
R_n = 40.50 + 49.99 = 90.49 kips

J4-6:

R_n = 0.60 x 65 x 0.825 + 1.0 x 65 x 0.769
R_n = 32.18 + 49.99 = 82.17 kips

Governor: J4-6 = 82.17 kips

Design strength:

phi R_n = 0.75 x 82.17 = 61.6 kips

Beam web copes are a common location for block shear failure. The loss of the flange removes the primary tension element, leaving the thin web to resist all forces through the bolt connection.

Block Shear vs. Bearing/Tearout

Block shear and bolt tearout are related but distinct limit states:

When to check both: Always. Block shear checks the entire bolt group; bearing/tearout checks individual bolts. Neither is a substitute for the other.

In many connections, block shear governs for the connection as a whole, while individual bolts near edges may have reduced bearing/tearout capacity. Check both and use the lesser.

Common Mistakes

1. Using U_bs = 1.0 for single angles. Single angle connections connected through one leg have eccentricity. Use U_bs = 0.5.

2. Forgetting the 0.5 factor on tension hole deductions. In the tension path, the hole deduction is 0.5 x d_h per hole (to account for stagger), not the full d_h.

3. Confusing A_nv with A_gv. A_nv includes bolt hole deductions in the shear path; A_gv does not. Both are needed (J4-5 uses A_gv, J4-6 uses A_nv).

4. Not checking both equations. J4-5 (shear yielding governs) applies when the shear path is long and the tension path is short. J4-6 (shear rupture governs) applies when the shear path is short. Always check both.

5. Missing the correct failure path. The failure path must be continuous and physically realistic. Trace it around the bolt group — shear planes follow the bolt line, tension planes are perpendicular to the load.

6. Using block shear instead of tearout (or vice versa). Block shear (J4.3) applies to the bolt group as a whole. Bearing and tearout (J3.10) apply to individual bolts. Both must be checked.

7. Using wrong F_y or F_u. Block shear uses the material properties of the connected part (the plate or angle), not the bolt.

Calculator

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FAQ

Q: What is the difference between block shear and tearout? A: Block shear (AISC J4.3) checks the entire bolt group as a unit — a block of material tears out along a continuous perimeter. Tearout (AISC J3.10) checks individual bolts — the material between a single bolt hole and the nearest edge shears out. Both must be checked. Block shear typically governs for bolt groups; tearout governs for individual bolts near edges.

Q: When does block shear govern over bolt shear? A: Block shear governs when the connected material is thin, the edge distance is short, or the bolt group is small (2-3 bolts). High-strength bolts (A490) in thin plates are particularly susceptible because the bolt shear strength exceeds the plate's block shear capacity.

Q: Can I use U_bs = 1.0 for a gusset plate connected on both sides? A: Yes, if the bolt pattern is symmetric about the line of load transfer. If the load is eccentric to the bolt group (e.g., bolts on one side of the gusset centerline), use U_bs = 0.5.

Q: Do I need to check block shear for bolted web connections on beams? A: Yes. Beam web connections with copes or short edge distances are common locations for block shear failure. The web is thin (typically 0.3-0.5 inches), making it susceptible even with high-strength bolts.

Q: What if the tension surface has staggered bolts? A: For staggered bolts in the tension path, use the stagger formula from AISC Section B4.3 to compute A_nt. The net width is: gross width minus sum of d_h for each hole, plus s^2/(4g) for each stagger (where s = stagger pitch, g = gage distance). The 0.5 factor on hole deductions in the A_nt expression already accounts for this in block shear calculations.

Q: Does block shear apply to welded connections? A: Block shear (J4.3) is primarily for bolted connections. For welded connections, the equivalent limit state is shear rupture along the weld, checked per J2 or J4 provisions. However, if a connection has a combination of bolts and welds, block shear may apply to the bolted portion.

Related: AISC 360-22 Steel Design Overview | AISC Standard Hole Sizes | AISC Bolt Spacing | Bolt Bearing and Tearout | AISC Table D3.1 — Shear Lag Factor | Shear Tab Design