How to use this calculator
- Map the tear-out block. Identify the shear plane(s) parallel to the load and the tension plane across the end of the bolt group — that L-shaped (or U-shaped) block is what tears out.
- Compute the three areas. Gross shear area Agv (plane length × thickness), net shear area Anv (deduct the holes the shear plane crosses), and net tension area Ant (tension plane length × thickness minus its holes). Standard hole = bolt + 2 mm + 2 mm.
- Enter Fu, Fy and Ubs. Use the connected material’s strengths and Ubs = 1.0 (uniform) or 0.5 (nonuniform tension stress).
- Read Rn, then φRn or Rn/Ω. Compare φRn (LRFD) to the factored load or Rn/Ω (ASD) to the service load, and confirm block shear is not the smallest connection limit state.
How it works
Block shear is a tear-out failure: a chunk of the connected base metal pulls free, shearing along the plane(s) parallel to the load and rupturing across the tension plane at the end of the bolt group — both at once. AISC 360 Section J4.3 adds the two contributions:
Rₙ = 0.6·Fu·Anv + Ubs·Fu·Ant ≤ 0.6·Fy·Agv + Ubs·Fu·Ant
The tension term Ubs·Fu·Ant is common to both expressions, so
in practice the shear term is the lesser of net-section rupture
0.6·Fu·Anv and gross-section yield 0.6·Fy·Agv.
The yield cap stops the rupture equation from over-predicting when the
shear plane has few holes. Areas come from your connection geometry; the
calculator runs the equation and names which shear limit governs.
Block shear is one of several limit states that share the same bolt group. The bolt shear strength calculator checks the fasteners themselves, the bolt bearing & tearout calculator checks the material around each hole, and the bolt pattern force calculator distributes an eccentric load to the bolts. The connection capacity is the smallest of all of them.
Worked example
Verified against the live calculator
A coped beam web, A992 steel
(Fu = 450 MPa / 65 ksi, Fy = 345 MPa / 50 ksi),
with a tear-out block of Agv = 1800 mm²,
Anv = 1200 mm², Ant = 400 mm² and uniform
tension (Ubs = 1.0):
shear rupture 0.6·450·1200 = 324 kN · shear yield 0.6·345·1800 = 372.6 kN → rupture governs
Rₙ = 324 + (1.0·450·400 = 180) = 504 kN · φRₙ = 0.75·504 = 378 kN · Rₙ/Ω = 504/2.00 = 252 kN
Net-section shear rupture (324 kN) beats gross-section yield
(372.6 kN), so it sets the shear term; the tension plane adds
180 kN. The connection can take a 378 kN
factored load (LRFD) or a 252 kN service load (ASD) in
block shear — provided no other limit state is smaller.
Frequently asked questions
How do you calculate block shear capacity?
With AISC 360 Eq. J4-5: Rn = 0.6·Fu·Anv + Ubs·Fu·Ant, capped at 0.6·Fy·Agv + Ubs·Fu·Ant. The shear term is the lesser of net-section rupture (0.6·Fu·Anv) and gross-section yield (0.6·Fy·Agv); the tension term (Ubs·Fu·Ant) adds on top. For Fu = 450 MPa, Fy = 345 MPa, Agv = 1800 mm², Anv = 1200 mm², Ant = 400 mm², Ubs = 1.0: shear rupture 324 kN governs over shear yield 372.6 kN, plus 180 kN tension, so Rn = 504 kN.
What is the difference between shear rupture and shear yield in block shear?
Shear rupture (0.6·Fu·Anv) acts on the NET shear area — the material left after the bolt holes are removed — and uses the ultimate strength Fu. Shear yield (0.6·Fy·Agv) acts on the GROSS shear area and uses the yield strength Fy; it is an upper bound that prevents the rupture equation from over-predicting on lightly-perforated planes. AISC takes the smaller of the two as the shear contribution. On normal connections rupture governs.
When is Ubs 0.5 instead of 1.0?
Ubs = 1.0 when the tension stress on the tension plane is uniform — the great majority of connections: single line of bolts, angles, most gusset plates. Ubs = 0.5 applies where the tension stress is nonuniform, the classic case being a coped beam connected with two rows of bolts. When in doubt 1.0 is unconservative, so confirm the tension-stress distribution against the AISC commentary figures.
Is block shear an LRFD or ASD check?
Either — AISC 360 gives both. The same nominal Rn is factored by φ = 0.75 for LRFD (compare φRn to the factored load) or divided by Ω = 2.00 for ASD (compare Rn/Ω to the service load). For Rn = 504 kN that is φRn = 378 kN and Rn/Ω = 252 kN. Use whichever design philosophy the rest of your connection uses.
Method & assumptions
- AISC 360 Section J4.3 block shear rupture, Eq. J4-5. The shear term is the lesser of net-section rupture (0.6·Fu·Anv) and gross-section yield (0.6·Fy·Agv); the tension term Ubs·Fu·Ant adds to either.
- You supply the areas. Net areas deduct the standard hole = bolt diameter + 2 mm (1/16 in) clearance + 2 mm (1/16 in) for hole damage; gross areas use the full plane length. No steel-grade or hole-dimension tables are embedded — Fu, Fy and the geometry are your inputs.
- Ubs = 1.0 for uniform tension stress (most bolted connections); 0.5 only where the tension stress is nonuniform, such as coped beams with two bolt lines.
- φ = 0.75 (LRFD) and Ω = 2.00 (ASD) per AISC J4.3, stable across the 360-10/16/22 editions. Block shear is one limit state among several on the same connection — also verify bolt shear, bearing/tearout, member rupture/yield and any welds. A licensed engineer owns the final connection design.