Step 1: Member and Material Data

Step 2: Bolt Shear Strength (AISC 360 J3.6)

Single bolt shear capacity (threads included — N):

ϕRn = ϕ × Fnv × A_b

ϕ = 0.75 (LRFD shear) A_b = π × (0.75)² / 4 = 0.442 in²

ϕRn = 0.75 × 54 × 0.442 = 17.9 kips per bolt (single shear)

Double shear (two splice plates, one on each side of the web):

Since we will use two splice plates sandwiching the beam web, each bolt is in double shear:

ϕRn_double = 2 × 17.9 = 35.8 kips per bolt

Number of bolts required:

n = Vu / ϕRn_double = 65 / 35.8 = 1.82 → Use 3 bolts per side

Try two vertical rows of 3 bolts (6 total bolts connecting each splice plate to the web). But for a 20.2 in deep web, 3 bolts per row at 3 in spacing = 9 in total height — allows for 6 in end distance.

For simplicity, use 4 bolts in a single vertical row (one row each side of splice):

Effective double shear bolts = 4 bolts × 35.8 kips/bolt = 143.2 kips > 65 kips → OK

But check bolt spacing: 4 bolts at 3 in spacing = 9 in of bolt line. Allowable: min(12 in, 30×thickness) → OK.

Step 3: Bolt Bearing and Tear-Out at the Beam Web (AISC 360 J3.10)

Bearing strength per bolt:

ϕRn = ϕ × 2.4 × d × t × Fu (for deformation at service load considered)

ϕ = 0.75

For the beam web (t = tw = 0.350 in, Fu = 65 ksi):

ϕRn_bearing = 0.75 × 2.4 × 0.75 × 0.350 × 65 = 0.75 × 2.4 × 0.75 × 0.350 × 65 = 30.7 kips per bolt

Tear-out strength per bolt (end bolt):

For end bolts with Le = 1.5 in (minimum per AISC Table J3.4: Le_min = 1.125 in for 3/4 in bolts):

ϕRn_tearout_end = ϕ × 1.2 × Le × t × Fu

= 0.75 × 1.2 × 1.5 × 0.350 × 65 = 0.75 × 1.2 × 1.5 × 0.350 × 65 = 30.7 kips

Tear-out strength per bolt (interior bolts):

For interior bolts with spacing s = 3 in, using clear distance Lc = s - dh:

dh = 0.75 + 1/16 = 0.8125 in (standard hole) Lc_interior = 3.0 - 0.8125 = 2.1875 in

ϕRn_tearout_interior = ϕ × 1.2 × Lc × t × Fu = 0.75 × 1.2 × 2.1875 × 0.350 × 65 = 44.8 kips per bolt

Governing web bolt strength: Min(bearing = 30.7, tear-out end = 30.7, tear-out interior = 44.8) = 30.7 kips/bolt

Total web bolt strength = 4 × 30.7 = 122.8 kips > 65 kips → OK

Step 4: Splice Plate Design

Try 1/4 in thick A36 plates on each side of the web.

Plate dimensions:

• Width: 6 in (two vertical rows of bolts at 3 in gage + edges)
• Thickness: 1/4 in each plate
• Plate material: A36 (Fy = 36 ksi, Fu = 58 ksi)

Gross section yield (AISC 360 J4.1):

Ag_plate = 6 × 0.25 = 1.50 in² per plate Total Ag = 2 × 1.50 = 3.00 in²

ϕPn = ϕ × Fy × Ag = 0.90 × 36 × 3.00 = 97.2 kips > 65 kips → OK

Net section rupture (AISC 360 J4.2):

For each plate, standard holes: dh = 3/4 + 1/16 = 0.8125 in

Net section per plate: An = (6 - 2 × 0.8125) × 0.25 = (6 - 1.625) × 0.25 = 1.094 in²

U = 1.0 (tension on net section with two bolts per row in the connection)

Total Ae = 2 × 1.094 × 1.0 = 2.188 in²

ϕPn = ϕ × Fu × Ae = 0.75 × 58 × 2.188 = 95.2 kips > 65 kips → OK

Block shear in splice plate (AISC 360 J4.3):

Check the block shear failure path at the bolt group:

Gross shear area: Agv = 2 × (4 - 1 + 2 × 1.5 + (4-1) × 3) Wait — need to think about block shear path.

For the 4-bolt vertical row with end distance 1.5 in and spacing 3 in, the block shear path from the top (or bottom) bolt:

Agv = 2 plates × (1.5 + 3×3) × 0.25 = 2 × 10.5 × 0.25 = 5.25 in² (shear area along bolt line)

Net shear area: Anv = Agv - 2 × (3.5 × 0.8125 × 0.25) = 5.25 - 1.42 = 3.83 in²

Net tension area (across bottom bolt): Ant = 2 × (1.5 - 0.5 × 0.8125) × 0.25 = 2 × 1.094 × 0.25 = 0.547 in²

Fu × Ant = 58 × 0.547 = 31.7 kips 0.6 × Fu × Anv = 0.6 × 58 × 3.83 = 133.3 kips

Since 0.6 × Fu × Anv > Fu × Ant, use Equation J4-5: ϕRn = ϕ × min(0.6 × Fy × Agv + Fu × Ant, 0.6 × Fu × Anv + Fu × Ant)

ϕRn = 0.75 × (0.6 × 36 × 5.25 + 58 × 0.547) = 0.75 × (113.4 + 31.7) = 0.75 × 145.1 = 108.8 kips > 65 kips → OK

Step 5: Connection Geometry Summary

Step 6: Additional Checks

Minimum bolt spacing (AISC J3.3):

s_min = 2-2/3 d = 2.67 × 0.75 = 2.0 in (3.0 in provided → OK)

s_max = min(24 × t, 12 in) = min(24 × 0.25, 12) = 6 in (3.0 in provided → OK)

Minimum edge distance (AISC Table J3.4):

For 3/4 in bolts in standard holes at rolled edges: 1-1/8 in

Le = 1.5 in > 1.125 in → OK

Bearing at bolt holes (connection slip-critical if required):

This example uses bearing-type connection (N = threads included). If the splice must be slip-critical (e.g., fatigue loading), use SC (slip-critical category) per AISC J3.8:

ϕRn_slip = ϕ × μ × Du × hf × Tb × Ns

Where:

• μ = 0.30 (Class A surface, clean mill scale)
• Du = 1.13 (mean bolt pretension ratio)
• hf = 1.0 (standard holes)
• Tb = 28 kips (minimum pretension for 3/4 in A325)
• Ns = 2 (double shear)
• ϕ = 1.00 (slip-critical LRFD)

ϕRn_slip = 1.00 × 0.30 × 1.13 × 1.0 × 28 × 2 = 19.0 kips per bolt

For slip-critical: 4 × 19.0 = 76.0 kips > 65 kips → OK (slip-critical also works)

Step 7: Final Utilization Summary

The connection is adequate. The governing limit state for this design is splice plate net section rupture at 68% utilization.

Try the Calculator

Use the Bolted Connections Calculator to design bolted connections per AISC 360, EN 1993-1-8, AS 4100, or CSA S16 for your own member sizes, bolt grades, and loading.

Frequently Asked Questions

What is the difference between bearing-type and slip-critical connections? Bearing-type connections allow slip between plies until the bolt bears against the hole wall — the design assumes bearing and shear transfer. Slip-critical connections are designed to prevent slip under service loads through high bolt pretension and friction between faying surfaces. Per AISC 360 J3.8, slip-critical is required for connections subject to fatigue, oversized holes, or where slip would cause damage.

When should I use A325 versus A490 bolts? A490 bolts have higher tensile strength (Fy = 130 ksi, Fu = 150 ksi) compared to A325 (Fy = 92 ksi, Fu = 120 ksi). Use A490 when the connection is bolt-shear critical and space is limited. However, A490 bolts cannot be galvanized (hydrogen embrittlement risk) and have stricter pretensioning requirements. A325 bolts are sufficient for most building applications.

How does eccentric loading affect bolt group capacity? Eccentric loading creates both shear and torsion in the bolt group. The elastic (vector) method or the instantaneous center of rotation (ICR) method must be used. The ICR method in AISC Manual Table 7-7 to 7-14 gives higher capacities than the elastic method (typically 20-40% higher). This example uses concentric loading; for eccentric loading, bolt group strength can be significantly reduced depending on the eccentricity.

What is block shear and when does it govern? Block shear is a rupture failure along a path of least resistance — typically a combination of shear along the bolt line and tension across the last bolt row. It governs when the connection has a short end distance relative to the bolt spacing, or when the plate is thick relative to the edge distance. Per AISC 360 J4.3, the block shear strength is computed using both the gross shear area (yielding) and net shear area (rupture) paths.

See Also

• Beam Capacity Calculator
• Beam Displacement and Sag Tool
• Steel Beam Sizes Reference
• Beam Design Guide
• Beam Span Reference

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.