AISC 360 — Connections (Chapters J & K)

Bolt and weld design per AISC 360-16: F3125 bolts, fillet welds, bearing, slip-critical, block shear

Every structural steel member is only as strong as its connections. Chapters J and K of AISC 360-16 cover the design of bolted and welded connections for all member types, including special provisions for HSS connections. The fundamental principle: every connection must have a clearly defined load path from the member to the supporting element. Connection design targets a higher reliability index (β=4.0\beta = 4.0) than member design (β=2.6\beta = 2.6), reflected in the lower ϕ=0.75\phi = 0.75 for most connection limit states.

1. Bolt Types and Grades

ASTM F3125 is the unified bolt specification that replaced the individual standards A325, A490, F1852, and F2280. All structural bolts are now ordered under F3125 with a grade designation:

F3125 GradeOld NameFnvF_{nv} (N threads)FnvF_{nv} (X threads excl.)FntF_{nt}Notes
A325A32554 ksi68 ksi90 ksiStandard high-strength bolt. 120 ksi tensile. Most common.
A490A49068 ksi84 ksi113 ksiHigher strength. 150 ksi tensile. No galvanizing allowed.
F1852TC bolt (A325 equiv.)54 ksi68 ksi90 ksiTwist-off type tension-control bolt. Same strength as A325.
F2280TC bolt (A490 equiv.)68 ksi84 ksi113 ksiTwist-off type, A490 strength.
N vs X designation: A325-N means threads are in the shear plane (lower shear capacity). A325-X means threads are excluded from the shear plane (higher capacity but requires careful bolt length specification). In practice, always assume N (threads included) unless the connection detail explicitly shows thread exclusion and the installer verifies it.

2. Bolt Limit States

2.1. Bolt Shear (J3.6)

Rn=FnvAbR_n = F_{nv} \cdot A_b

where Ab=πd2/4A_b = \pi d^2/4 is the nominal bolt area (unthreaded body). For a single bolt:

LRFD: ϕRn=0.75FnvAb\phi R_n = 0.75 \cdot F_{nv} \cdot A_b

ASD: Rn/Ω=FnvAb/2.00R_n / \Omega = F_{nv} \cdot A_b / 2.00

For bolts in double shear, multiply by 2 (provided both shear planes have the same N or X condition).

2.2. Bolt Tension (J3.6)

Rn=FntAbϕ=0.75,  Ω=2.00R_n = F_{nt} \cdot A_b \quad \quad \phi = 0.75, \; \Omega = 2.00

2.3. Combined Tension and Shear (J3.7)

When a bolt is subjected to both shear and tension simultaneously, the tensile strength is reduced:

Fnt=1.3FntFntϕFnvfrvFnt(LRFD)F'_{nt} = 1.3 F_{nt} - \frac{F_{nt}}{\phi F_{nv}} f_{rv} \leq F_{nt} \quad \text{(LRFD)}
Fnt=1.3FntΩFntFnvfrvFnt(ASD)F'_{nt} = 1.3 F_{nt} - \frac{\Omega \, F_{nt}}{F_{nv}} f_{rv} \leq F_{nt} \quad \text{(ASD)}

where frvf_{rv} is the required shear stress. This is an elliptical interaction check linearized for simplicity.

2.4. Bearing and Tearout (J3.10)

The connected plate must also be checked for bearing (deformation of the hole) and tearout (bolt ripping through the plate edge):

Rn=2.4dtFu(bearing — deformation a consideration)R_n = 2.4 \, d \, t \, F_u \quad \text{(bearing — deformation a consideration)}
Rn=1.2lctFu(tearout)R_n = 1.2 \, l_c \, t \, F_u \quad \text{(tearout)}

where lcl_c is the clear distance from the bolt hole edge to the end of the plate or to the next bolt hole edge. The controlling value is the minimum of bearing and tearout for each bolt. ϕ=0.75\phi = 0.75.

2.5. Slip-Critical Connections (J3.8)

Slip-critical connections resist load through friction between the faying surfaces, not bearing. They are required when slip would be detrimental (e.g., oversized holes, connections subject to fatigue, or load reversal):

Rn=μDuhfTbnsR_n = \mu \, D_u \, h_f \, T_b \, n_s

where:

  • μ\mu = slip coefficient: 0.30 (Class A, clean mill scale) or 0.50 (Class B, blast-cleaned)
  • Du=1.13D_u = 1.13 = ratio of mean installed bolt pretension to specified minimum
  • hfh_f = hole factor: 1.00 (standard holes), 0.85 (oversized/short-slotted), 0.70 (long-slotted)
  • TbT_b = minimum bolt pretension from Table J3.1 (e.g., 28 kips for 3/4" A325)
  • nsn_s = number of slip planes

LRFD (strength level): ϕ=0.85\phi = 0.85

LRFD (serviceability level): ϕ=1.00\phi = 1.00

ASD (strength level): Ω=1.76\Omega = 1.76

ASD (serviceability level): Ω=1.50\Omega = 1.50

3. Bolt Spacing and Edge Distances

RequirementMinimumMaximum
Bolt spacing (center-to-center)223d2\frac{2}{3}d (preferred 3d)min(24t, 12 in) painted; min(14t, 7 in) unpainted
Edge distanceTable J3.4 (e.g., 1 in for 3/4" bolt)min(12t, 6 in)
End distanceSame as edge distance

4. Weld Types and Strength

4.1. Complete Joint Penetration (CJP) Groove Welds

CJP groove welds develop the full strength of the connected material. No separate weld check is needed — the base metal strength governs. CJP welds are required for moment connections in seismic applications (AISC 341).

4.2. Partial Joint Penetration (PJP) Groove Welds

PJP welds have an effective throat less than the full thickness. Strength is computed using Table J2.5 with the effective weld area AweA_{we} and the weld metal strength FnwF_{nw}.

4.3. Fillet Welds — The Workhorse of Steel Connections

Fillet welds are the most common weld type in structural steel, used in over 80% of all welded connections. The strength is based on the weld throat:

Rn=FnwAwewhere Fnw=0.60FEXX,  Awe=0.707aLR_n = F_{nw} \cdot A_{we} \quad \text{where } F_{nw} = 0.60 \, F_{EXX}, \; A_{we} = 0.707 \, a \, L

aa = weld leg size, LL = weld length, FEXXF_{EXX} = electrode classification strength. ϕ=0.75\phi = 0.75, Ω=2.00\Omega = 2.00.

Common Electrodes

ElectrodeFEXXF_{EXX}0.60FEXX0.60 F_{EXX}
E60XX60 ksi36.0 ksi
E70XX70 ksi42.0 ksi — most common
E80XX80 ksi48.0 ksi

Directional Strength Increase

Fillet welds loaded transversely (perpendicular to the weld axis) are stronger than those loaded longitudinally. Section J2.4 permits:

Fnw=0.60FEXX(1.0+0.50sin1.5θ)F_{nw} = 0.60 \, F_{EXX} \left(1.0 + 0.50 \sin^{1.5} \theta\right)

where θ\theta is the angle between the load direction and the weld axis. At θ=90°\theta = 90°(transverse), the strength increases by 50%. At θ=0°\theta = 0° (longitudinal), there is no increase.

Fillet Weld Capacity Table (E70XX, per inch of length)

Leg Size aThroat (0.707a)φRn per inch (LRFD)Rn/Ω per inch (ASD)
3/16 in0.133 in4.18 kip/in2.78 kip/in
1/4 in0.177 in5.57 kip/in3.71 kip/in
5/16 in0.221 in6.96 kip/in4.64 kip/in
3/8 in0.265 in8.35 kip/in5.57 kip/in
1/2 in0.354 in11.14 kip/in7.42 kip/in

5. Block Shear Rupture — Section J4.3

Rn=0.60FuAnv+UbsFuAnt0.60FyAgv+UbsFuAntR_n = 0.60 F_u A_{nv} + U_{bs} F_u A_{nt} \leq 0.60 F_y A_{gv} + U_{bs} F_u A_{nt}

ϕ=0.75\phi = 0.75, Ω=2.00\Omega = 2.00. The first expression assumes rupture on the shear plane and rupture on the tension plane. The cap (second expression) limits the shear plane to yielding strength. Ubs=1.0U_{bs} = 1.0 for uniform tension stress (symmetric connections), Ubs=0.5U_{bs} = 0.5 for non-uniform tension stress (e.g., coped beam with single row of bolts).

6. Common Connection Types

TypeCategoryKey Limit States
Shear tab (single plate)Simple shearBolt shear, bearing, plate shear yielding/rupture, block shear, weld
Double angleSimple shearBolt shear, bearing, angle leg rupture, block shear, coped beam
Single angleSimple shearSame as double angle + eccentricity in outstanding leg
Seated connectionSimple shearSeat angle bending, web crippling of beam, weld/bolt of seat
Extended end plateMomentBolt tension, plate bending, column flange bending, panel zone shear
Flange plateMomentPlate yielding/rupture, bolt shear, column panel zone
Directly welded flangeMomentCJP weld, column stiffeners, panel zone (J10.6)
Gusset plate (brace)BracingWhitmore section, Thornton method (buckling), bolt/weld to beam/column

Solved Example 1 — Shear Tab Connection

Given: Shear tab (single plate) connection. 3 bolts, 3/4" dia. F3125 Grade A325-N, standard holes. Plate: 1/4" × 9" A36 (Fy=36F_y = 36 ksi, Fu=58F_u = 58 ksi). Bolt spacing = 3 in, edge distance = 1.5 in (top and bottom). Factored reaction Vu=30V_u = 30 kips.

Step 1 — Bolt Shear (single shear)

Ab=π×0.7524=0.4418 in2A_b = \frac{\pi \times 0.75^2}{4} = 0.4418 \text{ in}^2
ϕRn=0.75×54×0.4418=17.89 kips/bolt\phi R_n = 0.75 \times 54 \times 0.4418 = 17.89 \text{ kips/bolt}
ϕRn(3 bolts)=3×17.89=53.7 kips30 kips\phi R_n (3 \text{ bolts}) = 3 \times 17.89 = 53.7 \text{ kips} \geq 30 \text{ kips} \quad \checkmark

Step 2 — Bearing/Tearout on Plate

Hole dia. = 3/4 + 1/16 = 13/16 in (standard hole)

Interior bolts (2 bolts): lc=3.013/16=2.188l_c = 3.0 - 13/16 = 2.188 in

Rn=min(2.4×0.75×0.25×58,  1.2×2.188×0.25×58)R_n = \min(2.4 \times 0.75 \times 0.25 \times 58, \; 1.2 \times 2.188 \times 0.25 \times 58)
=min(26.1,  38.1)=26.1 kips/bolt= \min(26.1, \; 38.1) = 26.1 \text{ kips/bolt}

Edge bolt (1 bolt): lc=1.513/32=1.094l_c = 1.5 - 13/32 = 1.094 in

Rn=min(26.1,  1.2×1.094×0.25×58)=min(26.1,  19.0)=19.0 kipsR_n = \min(26.1, \; 1.2 \times 1.094 \times 0.25 \times 58) = \min(26.1, \; 19.0) = 19.0 \text{ kips}
ϕRn=0.75×(2×26.1+19.0)=0.75×71.2=53.4 kips30\phi R_n = 0.75 \times (2 \times 26.1 + 19.0) = 0.75 \times 71.2 = 53.4 \text{ kips} \geq 30 \quad \checkmark

Step 3 — Plate Gross Shear Yielding

Agv=9.0×0.25=2.25 in2A_{gv} = 9.0 \times 0.25 = 2.25 \text{ in}^2
ϕRn=1.00×0.60×36×2.25=48.6 kips30\phi R_n = 1.00 \times 0.60 \times 36 \times 2.25 = 48.6 \text{ kips} \geq 30 \quad \checkmark

Step 4 — Plate Net Shear Rupture

Anv=(9.03×0.875)×0.25=6.375×0.25=1.594 in2A_{nv} = (9.0 - 3 \times 0.875) \times 0.25 = 6.375 \times 0.25 = 1.594 \text{ in}^2
ϕRn=0.75×0.60×58×1.594=41.6 kips30\phi R_n = 0.75 \times 0.60 \times 58 \times 1.594 = 41.6 \text{ kips} \geq 30 \quad \checkmark

(hole dia = 3/4 + 1/8 = 7/8 in for rupture — includes 1/16" damage allowance)

Step 5 — Block Shear

Failure path: shear along bolt line + tension across bottom edge

Agv=(1.5+2×3.0)×0.25=7.5×0.25=1.875 in2A_{gv} = (1.5 + 2 \times 3.0) \times 0.25 = 7.5 \times 0.25 = 1.875 \text{ in}^2
Anv=(7.52.5×0.875)×0.25=5.313×0.25=1.328 in2A_{nv} = (7.5 - 2.5 \times 0.875) \times 0.25 = 5.313 \times 0.25 = 1.328 \text{ in}^2
Ant=(1.50.5×0.875)×0.25=1.063×0.25=0.266 in2A_{nt} = (1.5 - 0.5 \times 0.875) \times 0.25 = 1.063 \times 0.25 = 0.266 \text{ in}^2

Ubs=1.0U_{bs} = 1.0 (symmetric, uniform tension)

Rn=0.60×58×1.328+1.0×58×0.266=46.2+15.4=61.6 kipsR_n = 0.60 \times 58 \times 1.328 + 1.0 \times 58 \times 0.266 = 46.2 + 15.4 = 61.6 \text{ kips}

Cap: 0.60×36×1.875+58×0.266=40.5+15.4=55.90.60 \times 36 \times 1.875 + 58 \times 0.266 = 40.5 + 15.4 = 55.9 kips (governs)

ϕRn=0.75×55.9=41.9 kips30\phi R_n = 0.75 \times 55.9 = 41.9 \text{ kips} \geq 30 \quad \checkmark

Summary of all limit states:

Bolt shear53.7 kips
Bearing/tearout53.4 kips
Gross shear yielding48.6 kips
Net shear rupture41.6 kips
Block shear (controls)41.9 kips

Controlling limit state: net shear rupture at 41.6 kips (72% utilization).

Verify automatically in CalcSteel — Open the editor with AISC 360 →

Solved Example 2 — Fillet Weld Connection

Given: Double angle connection (2L 4×3×1/4) welded to a W14×48 column flange. E70XX electrode. Required reaction Vu=55V_u = 55 kips (LRFD). Weld along the outstanding leg (4 in side), both toes and heel.

Step 1 — Required Weld Size

Each angle carries half: Vu/2=27.5V_u/2 = 27.5 kips per angle. Weld on both sides of the angle (two welds per angle), each weld length = 8.5 in (along the outstanding leg plus returns).

Total weld length per angle: Lw=2×8.5=17.0L_w = 2 \times 8.5 = 17.0 in

Required weld strength per inch:

qreq=27.517.0=1.62 kip/inq_{\text{req}} = \frac{27.5}{17.0} = 1.62 \text{ kip/in}

From the fillet weld table, a 3/16 in fillet weld provides ϕRn=4.18\phi R_n = 4.18 kip/in → 3/16 in fillet weld is adequate.

Step 2 — Base Metal Check

Column flange shear rupture along the weld:

ϕRn=0.75×0.60×65×0.595×17.0=297 kips27.5\phi R_n = 0.75 \times 0.60 \times 65 \times 0.595 \times 17.0 = 297 \text{ kips} \gg 27.5 \quad \checkmark

(A992 column flange: Fu=65F_u = 65 ksi, tf=0.595t_f = 0.595 in for W14×48)

Note on CalcSteel: CalcSteel provides the forces and moments at each node from FEM analysis. Connection design (selecting bolts, plate sizes, weld details) typically requires dedicated connection design tools such as IDEA StatiCa or the AISC connection design tables. CalcSteel outputs the demand; the connection designer provides the supply.
Verify automatically in CalcSteel — Open the editor with AISC 360 →

7. International Comparison — Connections

AspectAISC 360Eurocode 3NBR 8800
Bolt gradesF3125 (A325/A490)ISO 898-1 (8.8 / 10.9)ASTM A325/A490 (same as AISC)
Bolt shear φ0.751/γM2 = 0.801/γa2 ≈ 0.74
Hole deduction (rupture)db + 1/8 ind0 (actual hole dia)db + 2 mm
Fillet weld formula0.60FEXX × throatFu/(√3 × γM2) via directional method0.60Fw × throat (similar)
Directional strength increase1 + 0.50 sin1.5θDirectional method inherently capturesSame as AISC
Block shearJ4.3 (rupture + tension cap)EN 1993-1-8 §3.10.2Similar to AISC J4.3
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