Bolted Shear Tab Connection

A shear tab (single-plate) connection transfers shear from a beam web to a support via a vertical row of bolts through a plate. This calculator checks all bolt and plate failure modes per AISC 360-22 Chapter J. The plate-to-support weld or the support itself are outside the scope of this check.

The ten limit states checked, in order of typical criticality:

#	Limit State	Section	φ
1	Bolt shear rupture	§J3.6	0.75
2	Bolt group eccentricity (elastic method)	§J3.6	0.75
3	Bearing on plate	§J3.10	0.75
4	Bearing on beam web	§J3.10	0.75
5	Web gross shear yielding	§J4.2	1.00
6	Web net shear rupture	§J4.2	0.75
7	Plate gross shear yielding	§J4.2	1.00
8	Plate net shear rupture	§J4.2	0.75
9	Block shear on plate	§J4.3	0.75
10	Plate bending at weld (cantilever)	§F11	0.90

Each bolt resists a vertical shear of V_u/n (assuming uniform distribution for a concentric connection).

φR_n = n × φ × F_nv × A_b

where φ = 0.75, A_b = πd_b²/4, and F_nv is from AISC Table J3.2:

Grade	Condition	Fnv (ksi)	Fnv (MPa)
Group A (A325, F1852)	-N (threads in shear plane)	48	330
Group A	-X (threads excluded)	54	372
Group B (A490, F2280)	-N	60	414
Group B	-X	68	469

A shear tab bolt group is offset a horizontal distance a from the support face. This eccentricity creates an in-plane moment M = V_u × a on the bolt group, so each bolt resists a combination of direct shear and torsional shear.

For a single vertical row of n bolts at spacing s, the polar moment of inertia of the bolt group (treating bolt areas as unit areas) is:

I_p = Σy_i² = s² × n(n²−1)/12

The critical (outermost) bolt at distance c_max = (n−1)s/2 from the centroid carries:

• Direct vertical shear: V' = V_u/n
• Eccentric horizontal shear: V″ = (V_u×a×c_max) / I_p

r_crit = √(V'² + V″²)

The equivalent group capacity is the max V_u such that r_crit ≤ φR_n,1bolt:

φR_n,ecc = φR_n,1bolt × V_u / r_crit

When a = 0 the result equals the direct bolt shear capacity (no penalty). Eccentricity can reduce group capacity by 20–50% for typical shear tab geometries.

Bolt bearing is checked on both the plate and the beam web independently. For standard holes when deformation is a design consideration (φ = 0.75):

R_n = 1.2 l_c t F_u ≤ 2.4 d_b t F_u

where l_c is the clear distance from hole edge to next hole edge (interior bolts: l_c = s − d_h) or to plate edge (end bolts: l_c = e_v − d_h/2). Standard hole diameter d_h = d_b + 1/16″ (US) or d_b + 2 mm (SI).

The plate height h_p = (n−1)·s + 2·e_v.

Gross shear yielding (φ = 1.00):

φR_n = 1.00 × 0.60 F_y × h_p × t_p

Net shear rupture (φ = 0.75, deduct n hole diameters):

φR_n = 0.75 × 0.60 F_u × (h_p − n·d_h) × t_p

For a single vertical bolt row with uniform shear (U_bs = 1.0), the block tears along the bolt line (shear) and across the top or bottom (tension).

R_n = 0.60 F_u A_nv + U_bs F_u A_nt ≤ 0.60 F_y A_gv + U_bs F_u A_nt

where φ = 0.75 and:

Area	Expression	Description
Agv	[(n−1)s + ev] × tp	Gross shear area (vertical path)
Anv	Agv − (n−0.5)dhtp	Net shear area
Ant	(eh − dh/2) × tp	Net tension area (horizontal)

e_h is the horizontal edge distance from bolt centerline to the free (far) edge of the plate.

The shear tab plate acts as a short cantilever from the support weld to the bolt group. The applied shear V_u at eccentricity a creates a moment at the weld line:

M_u = V_u × a

The plate cross-section at the weld (h_p × t_p, no holes) is checked for flexural yielding per §F11. Using the plastic section modulus:

Z_x = t_p × h_p² / 4 φM_n = 0.90 × F_y × Z_x

The calculator converts this moment capacity to an equivalent shear capacity at distance a for direct comparison in the DCR table:

φR_n,bending = φM_n / a

When a = 0 this check is not applicable (shown as —). Plate bending typically governs for long eccentricities combined with thin or short plates.

Given: W18×35 beam (t_w = 0.300″, A992: F_u = 65 ksi), V_u = 40 kip.
Bolts: 3/4″ A325-N (n = 3, s = 3″, e_v = 1.25″, e_h = 1.75″), eccentricity a = 3″.
Plate: t_p = 3/8″, b_p = 3.5″, A36 (F_y = 36 ksi, F_u = 58 ksi).

Geometry: d_h = 3/4 + 1/16 = 13/16″ = 0.8125″. h_p = 2×3 + 2×1.25 = 8.5″.
A_b = π(0.75)²/4 = 0.4418 in².

Step 1 — Bolt shear

F_nv = 48 ksi. φR_n = 0.75×48×0.4418×3 = 47.7 kip > 40 kip ✔

Step 2 — Bearing on plate

l_c,end = 1.25 − 0.8125/2 = 0.844″; l_c,int = 3.0 − 0.8125 = 2.188″.
End bolt: R_n = min(1.2×0.844×0.375×58, 2.4×0.75×0.375×58) = min(22.1, 39.2) = 22.1 kip.
Interior bolt: R_n = min(1.2×2.188×0.375×58, 39.2) = min(57.3, 39.2) = 39.2 kip.
φR_n = 0.75×(2×22.1 + 1×39.2) = 0.75×83.4 = 62.5 kip ✔

Step 3 — Bearing on web

t_w = 0.300″, F_u,web = 65 ksi.
End bolt: R_n = min(1.2×0.844×0.300×65, 2.4×0.75×0.300×65) = min(19.7, 35.1) = 19.7 kip.
Interior bolt: R_n = min(1.2×2.188×0.300×65, 35.1) = min(51.2, 35.1) = 35.1 kip.
φR_n = 0.75×(2×19.7 + 35.1) = 0.75×74.5 = 55.9 kip ✔

Step 4 — Plate shear yielding

φR_n = 1.00×0.60×36×8.5×0.375 = 68.9 kip ✔

Step 5 — Plate shear rupture

A_nv = (8.5 − 3×0.8125)×0.375 = 6.063×0.375 = 2.273 in².
φR_n = 0.75×0.60×58×2.273 = 59.4 kip ✔

Step 6 — Block shear

A_gv = (2×3 + 1.25)×0.375 = 7.25×0.375 = 2.719 in².
A_nv = (7.25 − 2.5×0.8125)×0.375 = 5.219×0.375 = 1.957 in².
A_nt = (1.5 − 0.8125/2)×0.375 = 1.094×0.375 = 0.410 in².
R_n = min(0.60×58×1.957 + 58×0.410, 0.60×36×2.719 + 58×0.410) = min(91.9, 82.6) = 82.6 kip.
φR_n = 0.75×82.6 = 62.0 kip ✔

Step 7 — Bolt group eccentricity

a = 3″. I_p = 2×3² = 18 in². c_max = 3″.
V' = 40/3 = 13.33 kip. V″ = 40×3×3/18 = 20.0 kip.
r_crit = √(13.33² + 20.0²) = 24.04 kip.
φR_n,1bolt = 47.7/3 = 15.9 kip. φR_n,ecc = 15.9×40/24.04 = 26.5 kip < 40 kip ⚠

Step 8 — Plate bending

Z_x = 0.375×8.5²/4 = 6.76 in³.
φM_n = 0.90×36×6.76 = 219 kip·in.
φR_n,bending = 219/3 = 73.0 kip ✔

Governing: Bolt group eccentricity, φR_n,min = 26.5 kip. DCR = 40/26.5 = 1.51 — FAIL. Increase n or reduce a.