Shear Tab Design Worked Example — Single Plate Bolted to Web per AISC DG21

Complete step-by-step shear tab (single plate) connection design per AISC Design Guide 21 and AISC 360-22. W18x55 beam connected to W14x90 column web with 4-bolt shear tab. All limit states checked with actual numbers.

Problem Statement

PRELIMINARY — NOT FOR CONSTRUCTION. All results are for educational and reference use only. Must be independently verified by a licensed Professional Engineer (PE) or Structural Engineer (SE) before use in any project.

Design a shear tab connection for a W18x55 beam framing into the web of a W14x90 column. The connection transfers only shear (simple connection). The beam end reaction is from a 40 ft simple span with uniform dead and live loads.

Beam: W18x55, ASTM A992 (Fy = 50 ksi, Fu = 65 ksi) Column: W14x90, ASTM A992 (Fy = 50 ksi, Fu = 65 ksi) Shear tab plate: ASTM A36 (Fy = 36 ksi, Fu = 58 ksi) Bolts: 3/4 in. diameter ASTM A325N, standard holes

Beam Properties (W18x55):

Property Value Units
d 18.1 in.
tw 0.390 in.
T (flange-to-flange) 18.1 in.

Column Properties (W14x90):

Property Value Units
tw 0.440 in.
T (clear between fillets) 10.2 in.

Design shear (factored, LRFD):

Bolt layout: 4 bolts in single vertical row at 3 in. spacing. Top edge distance = 1.5 in. Bottom edge distance = 1.5 in.

Step 1: Plate Sizing

Plate height based on bolt layout:

h_p = (4 - 1) × 3.0 + 2 × 1.5 = 9.0 + 3.0 = 12.0 in.

Plate width (standard practice): 4.5 in. (provides 2.5 in. from bolt line to weld line: a = 2.5 in.)

Plate thickness per ductility requirement (AISC J2.3):

t_p ≤ d_b/2 + 1/16 = 0.75/2 + 0.0625 = 0.375 + 0.0625 = 0.4375 in.

Use t_p = 3/8 in. = 0.375 in. < 0.4375 in. Ductility requirement satisfied.

Step 2: Bolt Shear Check (AISC J3.6)

Bolts are in single shear (one shear plane between plate and beam web):

A_b = pi × (3/4)² / 4 = 0.442 in.²
r_n = Fnv × A_b = 54 × 0.442 = 23.9 kips per bolt (threads included — N condition)
phi r_n = 0.75 × 23.9 = 17.9 kips per bolt

Total: phi R_n = 4 × 17.9 = 71.6 kips

Check: V_u = 25.3 kips < 71.6 kips. Bolt shear OK. Utilization = 0.353.

Step 3: Bolt Bearing on Shear Tab Plate (AISC J3.10)

Plate: t_p = 0.375 in., Fu = 58 ksi. Hole: dh_eff = 13/16 + 1/16 = 7/8 in.

Edge bolts (top and bottom):

L_e = 1.5 in.
L_c = L_e - dh_eff/2 = 1.5 - 0.875/2 = 1.5 - 0.438 = 1.063 in.
r_n_edge = 1.2 × L_c × t × Fu = 1.2 × 1.063 × 0.375 × 58 = 27.7 kips
≤ 2.4 × d × t × Fu = 2.4 × 0.75 × 0.375 × 58 = 39.2 kips
r_n_edge = 27.7 kips (tearout governs)

Interior bolts (2 middle bolts):

L_c = s - dh_eff = 3.0 - 0.875 = 2.125 in.
r_n_int = 1.2 × 2.125 × 0.375 × 58 = 55.5 kips
≤ 39.2 kips (bearing upper bound)
r_n_int = 39.2 kips (bearing governs for interior bolts)

Total bearing strength:

R_n_brg = 2 × 27.7 + 2 × 39.2 = 55.4 + 78.4 = 133.8 kips
phi R_n_brg = 0.75 × 133.8 = 100.4 kips

Check: 25.3 kips < 100.4 kips. Bolt bearing on plate OK. Utilization = 0.252.

Step 4: Bolt Bearing on Beam Web (AISC J3.10)

Beam web: tw = 0.390 in., Fu = 65 ksi.

Edge bolts:

r_n_edge_web = 1.2 × 1.063 × 0.390 × 65 = 32.3 kips
≤ 2.4 × 0.75 × 0.390 × 65 = 45.6 kips
r_n_edge_web = 32.3 kips

Interior bolts:

r_n_int_web = 2.4 × 0.75 × 0.390 × 65 = 45.6 kips
(interior L_c = 2.125 in., tearout = 1.2 × 2.125 × 0.390 × 65 = 64.6 kips > 45.6 kips)

R_n_brg_web = 2 × 32.3 + 2 × 45.6 = 64.6 + 91.2 = 155.8 kips
phi R_n = 0.75 × 155.8 = 116.9 kips > 25.3 kips. **Web bearing OK.**

Step 5: Plate Shear Yielding (AISC J4.2)

Gross shear area of plate:

A_gv = h_p × t_p = 12.0 × 0.375 = 4.500 in.²
R_n = 0.60 × Fy × A_gv = 0.60 × 36 × 4.500 = 97.2 kips

For plates meeting h/tw limits, phi = 1.00 per AISC G1:

phi R_n = 1.00 × 97.2 = 97.2 kips > 25.3 kips. **Shear yielding OK.** Utilization = 0.260.

Step 6: Plate Shear Rupture (AISC J4.2)

Net shear area (4 bolts, dh_eff = 7/8 in.):

A_nv = (h_p - 4 × dh_eff) × t_p = (12.0 - 4 × 0.875) × 0.375
     = (12.0 - 3.5) × 0.375 = 8.5 × 0.375 = 3.188 in.²

R_n = 0.60 × Fu × A_nv = 0.60 × 58 × 3.188 = 110.9 kips
phi R_n = 0.75 × 110.9 = 83.2 kips > 25.3 kips. **Shear rupture OK.**

Step 7: Plate Block Shear (AISC J4.3)

Shear path along bolt line (vertical):

L_gv = (4 - 1) × 3.0 + 1.5 + 1.5 = 9.0 + 3.0 = 12.0 in. (but this is just the full plate height; the actual shear path length from first to last bolt plus edge distances)

More precisely: L_gv = 12.0 in. (full plate height including edge distances)

A_gv = 12.0 × 0.375 = 4.500 in.²
A_nv = (12.0 - 3.5 × 0.875) × 0.375 = (12.0 - 3.063) × 0.375 = 8.938 × 0.375 = 3.352 in.²

(3.5 holes deducted: 3 full holes + 2 half holes at the edges = 3 + 2×0.5 = 4, adjusted — the more correct way is 4 bolts so deduct 3.5 holes for the shear plane as done above)

Tension path perpendicular to load (horizontal edge from bolt line to plate edge):

L_gt = a - dh_eff/2 = 2.5 - 0.438 = 2.063 in. (distance from bolt center to plate free edge)
A_gt = 2.063 × 0.375 = 0.773 in.²
A_nt = (2.063 - 0.5 × 0.875) × 0.375 = (2.063 - 0.438) × 0.375 = 1.625 × 0.375 = 0.609 in.²

For shear tabs with one row of bolts, U_bs = 1.0:

R_n = 0.60 × Fu × A_nv + U_bs × Fu × A_nt
    = 0.60 × 58 × 3.352 + 1.0 × 58 × 0.609
    = 116.6 + 35.3 = 151.9 kips

Upper bound: 0.60 × Fy × A_gv + U_bs × Fu × A_nt
    = 0.60 × 36 × 4.500 + 1.0 × 58 × 0.609
    = 97.2 + 35.3 = 132.5 kips

Governing: R_n = 132.5 kips (gross shear yielding + net tension rupture controls)

phi R_n = 0.75 × 132.5 = 99.4 kips > 25.3 kips. **Block shear OK.**

Step 8: Weld Design — Plate to Column Web

The weld connects the shear tab plate to the column web. Eccentricity e = a = 2.5 in. (distance from bolt line to weld line).

Direct shear per weld (2 vertical welds, length L_weld = h_p = 12.0 in.):

f_v = V_u / (2 × L) = 25.3 / (2 × 12.0) = 25.3 / 24.0 = 1.054 kip/in.

Moment from eccentricity:

M_e = V_u × a = 25.3 × 2.5 = 63.3 kip-in.

For two vertical welds, the polar moment of inertia for out-of-plane loading:

I_weld = 2 × L³ / 12 = 2 × 12.0³ / 12 = 2 × 1,728 / 12 = 2 × 144 = 288 in.³

Equivalent force per unit length from moment (at the extreme fiber c = L/2 = 6.0 in.):

f_m = M_e × c / I_weld = 63.3 × 6.0 / 288 = 379.8 / 288 = 1.319 kip/in.

Vector resultant:

f_r = sqrt(f_v² + f_m²) = sqrt(1.054² + 1.319²) = sqrt(1.111 + 1.740) = sqrt(2.851) = 1.688 kip/in.

Required weld strength per inch = 1.688 kip/in.

Select weld size: 3/16 in. fillet weld (E70XX electrode, F_EXX = 70 ksi).

Per AISC Table J2.5, weld capacity (LRFD, loading angle theta = tan⁻¹(f_m/f_v) = 51.4 degrees):

For theta ≠ 0, use AISC Eq. J2-5:

phi r_n = 0.75 × (0.60 × F_EXX) × (1.0 + 0.50 × sin^1.5(theta)) × (0.707 × D)

Where D = 3 (sixteenths of an inch for 3/16 in. weld):

phi r_n = 0.75 × 42 × (1.0 + 0.50 × sin^1.5(51.4°)) × (0.707 × 3)
sin(51.4°) = 0.781
sin^1.5(51.4°) = 0.781^1.5 = 0.690

phi r_n = 0.75 × 42 × (1.0 + 0.50 × 0.690) × 2.121
       = 0.75 × 42 × 1.345 × 2.121
       = 0.75 × 42 × 2.853
       = 89.9 kips per in.??? No — that's total. Per inch:

Actually, let me use the simpler AISC Manual Table 8-4 approach. For E70 electrode, 3/16 in. fillet weld:

phi r_n = 1.392 × D × (1.0 + 0.5 × sin^1.5(theta))
       = 1.392 × 3 × 1.345
       = 4.176 × 1.345
       = 5.62 kip/in.

Check: 5.62 kip/in. > 1.688 kip/in. Weld OK. Utilization = 0.300.

Alternatively, for theta = 0 (conservative):

phi r_n = 1.392 × 3 = 4.176 kip/in. > 1.688 kip/in. **Still OK.**

Use 3/16 in. fillet weld both sides, E70XX electrode.

Step 9: Column Web Check

Column web thickness tw = 0.440 in. > beam web 0.390 in., so bearing and block shear checks on column web are more favorable. Additionally, the weld is on the column web side and does not create additional stress concentrations.

Column web local yielding (AISC J10.2):

The shear tab plate applies a concentrated force to the column web. For a W14x90 column:

R_n_web_local = Fy × tw × (5k + lb) where lb = plate length = 12.0 in.
k = kdes + tf = 1.25 + 0.710 = 1.96 in. (W14x90)

R_n = 50 × 0.440 × (5 × 1.96 + 12.0) = 22 × (9.8 + 12.0) = 22 × 21.8 = 479.6 kips
phi R_n = 1.00 × 479.6 = 479.6 kips > 25.3 kips. **Column web local yielding OK.**

Step 10: Summary — Pass/Fail

Limit State Reference Demand Capacity D/C Ratio Status
Bolt shear (4 bolts, single) AISC J3.6 25.3 k 71.6 k 0.353 PASS
Bolt bearing on plate AISC J3.10 25.3 k 100.4 k 0.252 PASS
Bolt bearing on beam web AISC J3.10 25.3 k 116.9 k 0.217 PASS
Plate shear yielding AISC J4.2 25.3 k 97.2 k 0.260 PASS
Plate shear rupture AISC J4.2 25.3 k 83.2 k 0.304 PASS
Plate block shear AISC J4.3 25.3 k 99.4 k 0.255 PASS
Ductility (t_p ≤ d_b/2 + 1/16) AISC J2.3 0.375 in. 0.438 in. — PASS
Weld (3/16 in. fillet) AISC J2.4 1.69 k/in. 5.62 k/in. 0.300 PASS
Column web local yielding AISC J10.2 25.3 k 479.6 k 0.053 PASS

All checks pass. The governing limit state is bolt shear at D/C = 0.353. This is an efficient, conventional shear tab design with significant reserve capacity.

Final Shear Tab Details:

This connection satisfies the AISC definition of a simple shear connection: it transfers only shear and permits end rotation of the beam under load.

Related Calculators

Use the Bolted Connection Calculator to check bolt shear, bearing, and block shear. For plate and weld design, see the Welded Connections Calculator. Pre-designed shear tab capacity tables are available on the Shear Tab Reference page.