12

Bolted Connections

Why Structural Steel?

Bolt shear, bearing, tearout, slip-critical (Chapter J).

⏱ 100 min★ Core✎ 3 objectives
§01Section 01

Engineering story

Engineering story
Chapter 12 · Bolted Connections

Bolt shear, bearing, tearout, slip-critical (Chapter J).

A real project narrative for this chapter will be authored as this chapter migrates to the v3.0 structured schema.

§02Section 02

Learning objectives

After this chapter you will be able to
  • Compute bolt shear, bearing, tearout
  • Distinguish bearing vs slip-critical
  • Detail hole types and spacing
§03Section 03

Engineering motivation

§04Section 04

Failure mechanisms

Failure mechanisms & lessons learned

Photographs and lessons-learned case studies for this topic will be added during chapter migration.

AISC Reference Box
  • AISC 360-22 §J3 — Bolts and Threaded Parts

Why This Chapter Matters

Connections transfer forces between members. The Hyatt Regency (1981) and countless smaller failures traced back to bolt shear, bearing, tearout, or slip-critical assumptions that the designer got wrong.

Learning Objectives

  • Distinguish A325 vs A490, snug-tight vs pretensioned vs slip-critical.
  • Compute bolt shear rupture (φRn = φ · Fnv · Ab · Ns).
  • Compute bearing and tearout at each bolt (AISC J3.10).
  • Design slip-critical connections (AISC J3.8).
  • Detail minimum edge distance, pitch, and gage per AISC Table J3.4.

Where This Chapter Is Used

Every bolted joint in Chapters 14 (shear), 15 (moment), 16 (base plates), 19 (framing), and the capstone.

ANSI / AISC 360-22Specification for Structural Steel Buildings16.1-127 to 16.1-140 (§J3)
Chapter
J
AISC 360-22

Chapter J. Design of Connections — Bolts (§J3)

Use this reference to flip directly to the correct page of the AISC 360-22 Specification while solving problems in this course chapter.

§Section titlePage
J3.1High-Strength Bolts (A325, A490, F3125)16.1-127
J3.3Minimum Spacing16.1-129
J3.4Minimum Edge Distance (Table J3.4)16.1-130
J3.6Tensile and Shear Strength of Bolts16.1-131
J3.7Combined Tension and Shear in Bearing-Type16.1-133
J3.8Slip-Critical Connections16.1-133
J3.10Bearing and Tearout at Bolt Holes16.1-136

Companion reference: AISC Manual Part 7 — Design Considerations for Bolts

Lecture Notes

Chapter 12 — Bolted Connections (AISC 360-22 Chapter J)

Chapter focus. Members are sized in Chapters 4–11; this chapter is where those members are actually joined. A bolted joint is a system of limit states: the bolt can shear, the plate can crush or tear out at the hole, the plate can yield or rupture on its net section, and a block of plate can tear free entirely. AISC Chapter J gives one equation per mode, and the connection strength is the smallest of them. Learn the ladder, work it in the same order every time, and you will never miss a check.

1. Why Connections Deserve More Care Than Members

Structural failures overwhelmingly begin at connections, not at midspan. Three reasons:

  • Stress concentration. A member carries stress smoothly; a connection funnels the entire member force through a handful of small holes. Local stresses at a hole edge are 2–3× the average.
  • Multiple materials and modes. Bolt steel, plate steel, weld metal and base metal all interact — and each has a different ductility.
  • Fabrication and erection tolerance. Members are rolled to tight tolerance in a mill. Connections are drilled, punched, fitted and bolted by people, often at height in bad weather.

This is why AISC assigns rupture-mode resistance factors (φ = 0.75) that are stingier than member yielding (φ = 0.90), and why the entire philosophy of connection design is to make the ductile mode govern where possible.

The six limit states of a bolted tension connection 1. Bolt shear rupture shank shears on the faying plane 2. Bearing (hole ovalizes) plate crushes ahead of the bolt 3. Tearout bolt plows out through the edge 4. Tensile yielding (A_g) gross section stretches, φ = 0.90 5. Tensile rupture (A_e) fracture across the net section, φ = 0.75 6. Block shear (§J4.3) shear + tension block tears free Connection strength = MINIMUM of all applicable limit states — never a sum Modes 1–3 are checked bolt-by-bolt (§J3.7, §J3.11). Modes 4–6 are checked on the connected element (§J4). Slip-critical joints add a seventh check (§J3.9) — and if the bolts also carry tension, the k_sc reduction of §J3.10. Every check must be satisfied. The engineer reports the governing one.
Figure 12.1 — The complete limit-state ladder for a bolted tension connection. Work every rung; report the minimum.

2. Bolt Types, Grades and Installation

AISC 360-22 organizes high-strength bolts by strength group rather than by the historical A325/A490 designations. All are covered by ASTM F3125/F3125M.

GroupLegacy nameFuFntFnv (N)Fnv (X)Notes
Group 120A325, F1852120 ksi90 ksi54 ksi68 ksiWorkhorse; reusable if not pretensioned
Group 144A354 Gr. BC144 ksi108 ksi65 ksi82 ksiIntermediate grade
Group 150A490, F2280150 ksi113 ksi68 ksi84 ksiNever galvanized; not reusable
A307"common" bolt60 ksi45 ksi27 ksi27 ksiSnug-tight only; light framing, anchor rods

Read the "N" and "X" carefully. N means threads are iNcluded in the shear plane; X means eXcluded. The 26% difference between 54 and 68 ksi is entirely about whether the shear plane cuts the threaded root area or the full shank. In practice, always assume N unless you can guarantee the geometry — a shop that substitutes a longer bolt will put threads back in the plane and you will never know.

Installation classes (RCSC Specification, §J3.1)

  • Snug-tight. Plies in firm contact from the full effort of an ironworker with a spud wrench. Adequate for most bearing-type connections, which is the majority of building steel.
  • Pretensioned. Installed to at least Tb from Table J3.1 (turn-of-nut, calibrated wrench, twist-off, or DTI washer). Required for: joints subject to fatigue or significant load reversal, joints with A490 bolts in tension, and column splices in tall/slender frames.
  • Slip-critical. Pretensioned and with a specified faying-surface condition. Required where slip cannot be tolerated: oversized holes, slots parallel to load, fatigue with reversal, and connections to existing structures.
Cost note. A slip-critical joint costs roughly 2–3× a snug-tight joint once you account for surface preparation, controlled installation and inspection. Specify it when the behaviour requires it — not as a blanket "extra safety" default.

3. Bolt Shear Rupture (§J3.7)

Spec. Eq. J3-1Rn = Fnv · Ab (φ = 0.75, Ω = 2.00)

Ab = πdb²/4 is the nominal shank area — always, even for thread condition N. The thread reduction is already baked into the lower Fnv value, so never reduce Ab as well (a classic double-count).

dbAb (in²)Std. hole dhφrn, Grp 120-N, single shearTb, Grp 120 (kips)
⅝ in.0.307¹¹⁄₁₆ in.12.4 k19
¾ in.0.442¹³⁄₁₆ in.17.9 k28
⅞ in.0.601¹⁵⁄₁₆ in.24.3 k39
1 in.0.7851⅛ in.31.8 k51
1⅛ in.0.9941¼ in.40.3 k56

Counting shear planes

Multiply Rn by the number of shear planes the bolt crosses. A lap splice (two plies) → 1 plane. A butt/double-lap splice (three plies) → 2 planes. A double-angle framing connection → the web bolts are in double shear, the support bolts are usually single shear. Getting this wrong is a factor-of-two error, and it is the single most common student mistake in this chapter.

Long joints (§J3.7 User Note)

In connections longer than 38 in. between extreme fasteners in the line of force, the end bolts pick up more load than the interior ones (shear lag in the fastener group). The Specification's Fnv values already contain a modest reduction for this; for very long splices, refer to the Commentary.

4. Bearing and Tearout at Bolt Holes (§J3.11)

These two modes describe the plate, not the bolt, and they are checked at every bolt individually because each bolt has a different clear distance.

Spec. Eq. J3-6a — bearing, deformation IS a design considerationRn = 2.4·db·t·Fu
Spec. Eq. J3-6b — bearing, deformation is NOT a considerationRn = 3.0·db·t·Fu
Spec. Eq. J3-6c — tearout, deformation IS a design considerationRn = 1.2·lc·t·Fu
Spec. Eq. J3-6d — tearout, deformation is NOT a considerationRn = 1.5·lc·t·Fu

"Deformation at the bolt hole at service load is a design consideration" means you care whether the hole ovalizes by more than about ¼ in. For nearly all building work the answer is yes, so use the 2.4/1.2 forms. The 3.0/1.5 forms apply only where hole elongation genuinely does not matter (e.g. a redundant, non-architectural, non-fatigue connection).

Computing the clear distance lc

  • Edge bolt: lc = le − dh/2, where le is the centre-to-edge distance in the direction of force.
  • Interior bolt: lc = s − dh, where s is the centre-to-centre spacing in the direction of force.

At each bolt, the plate strength is the lesser of bearing and tearout. Then the bolt's contribution to the joint is the lesser of that plate value and the bolt shear value — this "effective strength per bolt" logic is stated in the User Note to §J3.7 and is exactly how AISC Manual Part 10 builds its tables.

Additional Design Aids & Stratified Equations

Bolt tearout — clear distance Lc Lc,edge Lc,int = s − dh Load →
Clear distance l_c is measured hole-edge to plate-edge (edge bolt) or hole-edge to next hole-edge (interior bolt). It is never the centre-to-centre dimension.
Bearing per bolt (φ = 0.75)φR_n = 0.75 · 2.4 · d_b · t · F_u
Tearout per boltφR_n = 0.75 · 1.2 · l_c · t · F_u
Effective strength per boltr_n = min(bolt shear, bearing, tearout)

5. Bolts in Tension (§J3.7) and Prying Action

Spec. Eq. J3-1 (tension)Rn = Fnt · Ab (φ = 0.75, Ω = 2.00)

Where a bolt loads a flexible flange or plate (hanger connections, tee stubs, end-plate moment connections), the fitting bends and bears back on the bolt at its toe. This prying action adds a force q to the applied tension:

Manual Part 9Ttotal = Tapplied + q, where q depends on b′, a′, tfitting and the bolt pitch

Prying can add 20–60% to the bolt force in a thin fitting. The design cure is thickness: a fitting thick enough that it does not bend has no prying. AISC Manual Part 9 gives the closed-form tmin to eliminate prying entirely.

6. Combined Tension and Shear in Bearing-Type Joints (§J3.8)

A bolt resisting both tension and shear has a reduced tensile strength. AISC uses a straight-line clip of the true elliptical interaction:

Spec. Eq. J3-3a (LRFD)F′nt = 1.3Fnt − (Fnt / φFnv)·frv ≤ Fnt
Spec. Eq. J3-3b (ASD)F′nt = 1.3Fnt − (Ω Fnt / Fnv)·frv ≤ Fnt

where frv = required shear stress = Vr/Ab. Exemption: per the User Note, the interaction may be ignored if either the required shear stress or the required tensile stress is ≤ 30% of the corresponding available stress. Note that the equation reduces the tensile capacity; the Commentary gives the mirror form if you prefer to reduce shear instead.

7. Slip-Critical Connections (§J3.9)

Spec. Eq. J3-4Rn = μ · Du · hf · Tb · ns
SymbolMeaningTypical value
μMean slip coefficient of the faying surface0.30 Class A (mill scale / blast-cleaned Class A coating); 0.50 Class B (blast-cleaned, unpainted or Class B coating)
DuRatio of mean actual pretension to specified minimum1.13 (a different value may be used with approval)
hfFiller factor1.0 with no fillers; reduced per §J3.9 with fillers
TbMinimum fastener pretensionTable J3.1 (28 k for ¾ in. Grp 120; 39 k for ⅞ in.)
nsNumber of slip planes1 (lap) or 2 (double lap)

Resistance factors depend on hole type — this is where slip-critical design bites:

Hole condition§J3.9 clauseφΩEffective penalty
Standard holes; short slots ⟂ to load(a)1.001.50none
Oversized holes; short slots ∥ to load(b)0.851.76−15%
Long-slotted holes (any orientation)(c)0.702.14−30%
Slip is not a strength limit state — it is a serviceability-flavoured strength check. A slip-critical connection must also satisfy every bearing-type limit state (bolt shear, bearing, tearout, plate yield/rupture, block shear). Passing the slip check does not finish the design; it adds a check.

8. Combined Tension and Shear in Slip-Critical Joints (§J3.10)

Applied tension partially un-clamps the joint, so there is less normal force available to generate friction. The nominal slip resistance is multiplied by ksc:

Spec. Eq. J3-5a (LRFD)ksc = 1 − Tu / (Du·Tb·nb) ≥ 0
Spec. Eq. J3-5b (ASD)ksc = 1 − 1.5·Ta / (Du·Tb·nb) ≥ 0

Tu and Ta are the total tension on the bolt group and nb is the number of bolts carrying that tension — the per-bolt division is already inside the formula. Omitting ksc is the most common slip-critical design error, and Worked Example 12.5 shows a case where including it moves the joint from 71% to 99% utilization.

Load–slip behaviour: what "slip-critical" actually buys you Joint slip / deformation → Load → bearing-type major slip into bearing (up to hole clearance ⅟₁₆ in.) slip-critical friction limit (Eq. J3-4) — design stops here Both joints have the same ultimate strength; the slip-critical joint simply refuses to move at service load. Use it for fatigue, load reversal, oversized/slotted holes, and where movement is unacceptable.
Figure 12.2 — Slip-critical and bearing-type joints reach the same ultimate load; they differ entirely in how much the joint moves on the way there.

9. Hole Types, Spacing and Edge Distance (§J3.2–J3.5)

Hole typeSize (¾ in. bolt)Where permitted
Standard (STD)¹³⁄₁₆ in. (db + ⅟₁₆)Default everywhere
Oversized (OVS)¹⁵⁄₁₆ in.Slip-critical only; washers required
Short slot (SSL)¹³⁄₁₆ × 1 in.Any; ⟂ to load if bearing-type
Long slot (LSL)¹³⁄₁₆ × 1⅞ in.One ply only; washers/plate washers required
  • Minimum spacing (§J3.3): s ≥ 2⅔db; 3db preferred (2¼ in. for a ¾ in. bolt, 3 in. is the shop standard).
  • Minimum edge distance (§J3.4): Table J3.4 — 1 in. for a ¾ in. bolt, 1⅛ in. for ⅞ in., 1¼ in. for 1 in. Detailers routinely use 1¼ in. as a universal minimum.
  • Maximum spacing (§J3.5): 24t or 12 in. for painted members; tighter for weathering steel in the elements, to keep moisture out of the faying surface.
  • Maximum edge distance (§J3.5): 12t or 6 in.

Note that spacing and edge distance are not merely detailing rules — they feed directly into lc, and hence into tearout strength. Tightening an edge distance to save an inch of plate can cost you 25% of the connection.

10. Affected Elements: Yielding, Rupture and Block Shear (§J4)

Eq. J4-1 tensile yieldingRn = FyAg (φ = 0.90, Ω = 1.67)
Eq. J4-2 tensile ruptureRn = FuAe (φ = 0.75, Ω = 2.00)
Eq. J4-3 shear yieldingRn = 0.60FyAgv (φ = 1.00, Ω = 1.50)
Eq. J4-4 shear ruptureRn = 0.60FuAnv (φ = 0.75, Ω = 2.00)
Eq. J4-5 block shearRn = 0.60FuAnv + UbsFuAnt ≤ 0.60FyAgv + UbsFuAnt (φ = 0.75)

Ubs = 1.0 when tension stress is uniform (most gusset plates, angles, coped beams); Ubs = 0.5 when it is non-uniform (e.g. a coped beam with two bolt lines). In net-area calculations, remember §B4.3b: add ⅟₁₆ in. to the hole diameter to allow for damage from punching, so a ¾ in. bolt in a standard hole removes ⅞ in. of material.

11. Design Workflow — Work It in This Order

  1. Compute required strength Ru / Ra from ASCE/SEI 7 load combinations.
  2. Choose bolt diameter, grade, thread condition and hole type. Choose bearing-type vs slip-critical and justify it.
  3. Bolt shear (Eq. J3-1) × number of shear planes → strength per bolt.
  4. Bearing and tearout at every bolt, in every connected ply (Eqs. J3-6). Take the minimum at each bolt.
  5. Effective strength per bolt = min(shear, bearing, tearout). Sum over the group.
  6. Connected-element checks: tensile yield (J4-1), tensile rupture (J4-2), shear yield (J4-3), shear rupture (J4-4), block shear (J4-5).
  7. If slip-critical: slip resistance (J3-4) with the correct φ for the hole type; apply ksc (J3-5) if tension is present.
  8. If bolts carry tension: bolt tension (J3-1), combined interaction (J3-3), and prying action (Manual Part 9).
  9. Verify spacing, edge distances and maximum spacing (J3.3–J3.5). Report the governing limit state and D/C.

12. Practice and Detailing Judgment

  • Standardize. One bolt diameter and one grade per project wherever possible. Mixed ¾ in. and ⅞ in. bolts on the same job guarantees a field error.
  • Shop weld, field bolt. Welding is cheapest downhand in a controlled shop; bolting is fastest and safest in the field.
  • Leave room for the wrench. An impact wrench needs roughly 1¾ in. of clearance around the nut. A connection that calculates perfectly and cannot be installed is a failed design.
  • Design for ductility. Prefer configurations where bolt shear or plate yielding governs over brittle tearout or block shear at low utilization.
  • Do not over-specify slip-critical. It is a real cost with a real inspection burden. Use it where the code or the behaviour demands it.

⚠ Common mistakes

  • Using Fnv = 68 ksi (X) when the threads are actually in the shear plane (N = 54 ksi).
  • Reducing Ab to the thread root area and using the N value — the reduction is already in Fnv.
  • Miscounting shear planes: single vs. double shear is a factor-of-two error.
  • Using the nominal edge distance le instead of the clear distance lc in tearout.
  • Applying the edge-bolt tearout value to all bolts in the group.
  • Forgetting the ⅟₁₆ in. damage allowance on hole diameter for net-area calculations (§B4.3b).
  • Applying φ = 1.00 for slip in oversized or long-slotted holes (should be 0.85 or 0.70).
  • Omitting the ksc reduction when a slip-critical joint also carries tension.
  • Stopping after the bolt checks and never looking at plate yielding, rupture or block shear.
  • Adding limit-state capacities together. The connection strength is the minimum, always.

📖 Using the AISC Manual — Bolted Connections

Companion reference: Manual Part 7 — Tables 7-1 (φrn shear), 7-2 (tension), 7-4/7-5 (slip-critical), 7-6 (bearing & tearout), 7-11 (eccentric bolt groups). In practice, engineers rarely compute every quantity from first principles — the AISC Steel Construction Manual (16th Ed.) tabulates φRn (or Rn/Ω) for every rolled shape so you can pick a member in seconds. Formulas remain essential for understanding, verifying, and for anything the tables do not cover.

TableGives youHow to read
7-1φrn per bolt in shear (single & double, thread cases N/X)Read directly by bolt grade × diameter → multiply by number of bolts. Case N (threads Not excluded) is the safe default.
7-6Bearing φrn and tearout φrn per bolt per 1″ of ply thicknessφrn = smaller of bearing and tearout, times ply thickness. Governs when edge distance or spacing is small.
7-11Instantaneous-center coefficient C for eccentric bolt groupsLook up C by bolt pattern (rows × columns) and eccentricity → φRn,group = C · φrn,bolt. Replaces the elastic-vector approximation.

Example — 6 bolts, 3/4″ A325-N, single shear, 1/2″ ply: Table 7-1 → φrn = 17.9 kip/bolt; Table 7-6 → bearing 39.2 kip/in × 0.5″ = 19.6 kip, tearout 40.8 × 0.5 = 20.4 kip. Bolt shear governs → φRn = 6 × 17.9 = 107 kip. Two look-ups, done.

Table vs. formula — which to use? Use tables to select a shape quickly. Use formulas to verify odd geometry, non-standard grades (Fy≠50 ksi), unusual K-factors or Lb, and to answer exam problems that hand you a section not in the current Manual.
Project case study — Cardinal Square — 4-story braced-frame office

Every chapter's worked example is one step in the design of the same building: Plan: 4 bays N–S × 3 bays E–W, each 30 ft × 30 ft. Stories: 4 @ 13 ft (52 ft roof). Composite floor: 4.5 in NW concrete on 3 VLI20 deck. Roof: 1.5 in B-deck + insulation + membrane. Materials: Wide-flange members A992 (Fy = 50 ksi, Fu = 65 ksi). Plates A572 Gr. 50. HSS bracing A500 Gr. C. Bolts A325-N 7/8 in dia. Welds E70XX. Concrete f'c = 4 ksi. Anchor rods F1554 Gr. 36.

Chapter 12 — Bolted shear-tab connection
Filler-to-girder shear tab — 3 bolts, A325-N 7/8 in
Demand carried forward
From Chapter 7: Vu = 25.5 k per bolt group.
This chapter contributes
Checks bolt shear (φRn = 0.75 · Fnv · Ab), plate bearing and tearout per §J3.7 / §J3.11. Verifies plate edge distance and bolt spacing.
PuLap splice in single shear
Bolted lap splice — limit states: bolt shear (§J3.7), bearing (§J3.11), tearout, slip (§J3.9).

Formula Sheet

NameEquationAISC Ref
Bolt shearφRn = 0.75 · Fnv · AbAISC §J3.7
BearingφRn = 0.75 · 2.4 · db · t · FuAISC §J3.11
TearoutφRn = 0.75 · 1.5 · lc · t · FuAISC §J3.11

Worked Example

Worked Example 12.1 — Bolt Shear, Bearing and Tearout in a Lap Splice

Problem statement. A tension splice joins two ½ in. thick ASTM A572 Grade 50 plates (Fy = 50 ksi, Fu = 65 ksi) in a single-shear lap with four ¾ in. diameter Group 120 (A325) bolts in standard holes, thread condition N. Bolts are in two rows of two, with edge distance le = 1.5 in. in the direction of force and spacing s = 3.0 in. Determine the available strength of the bolt group considering bolt shear, bearing and tearout. Deformation at the bolt hole at service load is a design consideration.

Four-bolt single-shear lap splice — ½ in. A572 Gr. 50 plates, ¾ in. Group 120 bolts, standard holes l_e = 1.5 in s = 3.0 in gage = 3.0 in P Standard hole d_h = ¾ + ⅟₁₆ = ¹³⁄₁₆ in. Single shear plane (one plate over the other). Bearing/tearout must be checked at BOTH the edge bolts (short l_c) and the interior bolts.
Figure 12.1a — Lap splice geometry with edge distance, pitch and gage dimensioned
Step 1 — Bolt shear rupture (§J3.7).
Spec. Eq. J3-1Rn = Fnv·Ab
From Table J3.2, Group 120 with threads not excluded (N): Fnv = 54 ksi. From Manual Table 7-2, Ab = 0.442 in².
Rn = 54(0.442) = 23.9 kips/bolt (single shear)
LRFD: φRn = 0.75(23.9) = 17.9 kips/bolt; ASD: Rn/Ω = 23.9/2.00 = 11.9 kips/bolt
Step 2 — Bearing at bolt holes (§J3.11).
Spec. Eq. J3-6aRn = 2.4·db·t·Fu (deformation is a design consideration)
Rn = 2.4(0.75)(0.500)(65) = 58.5 kips/bolt → φRn = 0.75(58.5) = 43.9 kips/bolt
Step 3 — Tearout at each bolt (§J3.11). Standard hole diameter: dh = db + ⅟₁₆ = 0.750 + 0.0625 = 0.8125 in.
Spec. Eq. J3-6cRn = 1.2·lc·t·Fu
Edge bolts: lc = le − dh/2 = 1.50 − 0.406 = 1.094 in.
Rn = 1.2(1.094)(0.500)(65) = 42.7 kips → φRn = 32.0 kips/bolt
Interior bolts: lc = s − dh = 3.00 − 0.8125 = 2.19 in.
Rn = 1.2(2.19)(0.500)(65) = 85.4 kips → φRn = 64.1 kips/bolt
Step 4 — Governing strength per bolt, then the group. §J3.11 requires the lesser of bearing and tearout at each bolt, compared with bolt shear:
BoltShearBearingTearoutGoverns (LRFD)
2 edge bolts17.9 k43.9 k32.0 k17.9 k (shear)
2 interior bolts17.9 k43.9 k64.1 k17.9 k (shear)
Group strength: φRn = 4(17.9) = 71.6 kips (LRFD); ASD: 4(11.9) = 47.6 kips
Result: The connection can deliver φRn = 71.6 kips (LRFD) / 47.6 kips (ASD), governed by bolt shear rupture.

Engineering interpretation: shear governs because the plate is thick relative to the bolt (t = 0.5 in. of 65 ksi steel is stronger in bearing than the 54 ksi bolt is in shear). Flip that around — a ¼ in. plate — and bearing/tearout would govern instead. Also note the edge bolt is only 1.8× stronger in tearout than the bolt is in shear; shortening le to 1¼ in. would drop tearout to 26.7 kips and it would still not govern, but at 1 in. it would. This is why AISC Table J3.4 minimum edge distances exist.

Common mistakes:
  • Using Fnv = 68 ksi (thread condition X) when the specification says N.
  • Doubling Rn for "double shear" in a lap splice — a lap splice has only one shear plane.
  • Using lc = le instead of le − dh/2. lc is the clear distance to the hole edge.
  • Applying the edge-bolt tearout value to all bolts, or the interior value to all bolts. Each bolt has its own lc.
  • Stopping here. Plate tensile yielding, tensile rupture on the net section and block shear rupture (§J4) still have to be checked.

Worked Example 12.2 — Combined Tension and Shear in a Bearing-Type Connection

Adapted from AISC Design Examples V16.0, Example J.3

Problem statement. A ¾ in. diameter Group 120 bolt with threads not excluded from the shear plane (condition N) is subjected to a tension force of 3.5 kips dead load + 12 kips live load, and a shear force of 1.33 kips dead + 4 kips live. Check the combined stresses using AISC Specification Equations J3-3a/J3-3b.

¾ in. Group 120 (A325) bolt, thread condition N — combined tension and shear T_u = 23.4 k (tension in bolt) V_u = 8.00 k (shear across shank) Service loads: T_D = 3.5 k, T_L = 12 k; V_D = 1.33 k, V_L = 4 k A_b = 0.442 in²; F_nt = 90 ksi; F_nv = 54 ksi (Table J3.2)
Figure 12.2a — A single bolt carrying tension along its axis and shear across its shank at the same time
Step 1 — Required strengths (ASCE/SEI 7, Chapter 2). Tu = 1.2(3.5) + 1.6(12) = 4.2 + 19.2 = 23.4 kips; Ta = 3.5 + 12 = 15.5 kips
Vu = 1.2(1.33) + 1.6(4) = 1.60 + 6.40 = 8.00 kips; Va = 1.33 + 4 = 5.33 kips
Step 2 — Nominal stresses and bolt area. From Table J3.2 (Group 120): Fnt = 90 ksi, Fnv = 54 ksi. From Manual Table 7-2: Ab = 0.442 in².
Step 3 — Required shear stress vs. available shear stress. LRFD: φFnv = 0.75(54) = 40.5 ksi; frv = Vu/Ab = 8.00/0.442 = 18.1 ksi ≤ 40.5 ksi o.k.
ASD: Fnv/Ω = 54/2.00 = 27.0 ksi; frv = 5.33/0.442 = 12.1 ksi ≤ 27.0 ksi o.k.
Step 4 — Is the combined check even required? Per the User Note at the end of §J3.8, combined stresses need not be investigated if either the required shear or the required tensile stress is ≤ 30% of the corresponding available stress.
Shear: 18.1/40.5 = 44.7% > 30%. Tension: frt = 23.4/0.442 = 52.9 ksi vs φFnt = 0.75(90) = 67.5 ksi → 78.4% > 30%.
Both exceed 30% → the combined check is required.
Step 5 — Reduced nominal tensile stress.
Spec. Eq. J3-3a (LRFD)F'nt = 1.3Fnt − (Fnt/(φFnv))·frv ≤ Fnt
F'nt = 1.3(90) − (90/40.5)(18.1) = 117 − 2.222(18.1) = 117 − 40.2 = 76.8 ksi ≤ 90 ksi ✓
Spec. Eq. J3-3b (ASD)F'nt = 1.3Fnt − (Ω·Fnt/Fnv)·frv ≤ Fnt
F'nt = 117 − (90/27.0)(12.1) = 117 − 3.333(12.1) = 117 − 40.3 = 76.7 ksi
Step 6 — Available tensile strength of the bolt.
Spec. Eq. J3-2Rn = F'nt·Ab , with φ = 0.75 / Ω = 2.00 (§J3.8)
LRFD: φRn = 0.75(76.8)(0.442) = 25.5 kips ≥ Tu = 23.4 kips o.k. (D/C = 0.92)
ASD: Rn/Ω = 76.7(0.442)/2.00 = 17.0 kips ≥ Ta = 15.5 kips o.k.
CheckRequiredAvailable (LRFD)Ratio
Shear stress18.1 ksi40.5 ksi0.447 ✓
Tension (unreduced)23.4 k29.8 k0.785
Tension (reduced, Eq. J3-3a)23.4 k25.5 k0.918 ✓
Result: The ¾ in. Group 120 bolt is adequate — φRn = 25.5 kips ≥ Tu = 23.4 kips with the shear interaction included.

Engineering interpretation: the shear present here costs the bolt 14% of its tensile capacity (29.8 → 25.5 kips). The Eq. J3-3a form is a straight-line clip of the true elliptical interaction, and the 1.3 coefficient is what makes the straight line tangent to that ellipse. Practically: if a bolt is doing two jobs, size it for the pair, not for each job separately.

Common mistakes:
  • Forgetting the F'nt ≤ Fnt cap — with small shear, Eq. J3-3a returns a value above 90 ksi.
  • Using the ASD form (Ω·Fnt/Fnv) with LRFD stresses, or vice versa.
  • Applying the 30% exemption when only one of the two happens to be small but the other is near 100% — read it carefully: the exemption applies if either ratio ≤ 30%.
  • Reducing the shear strength instead of the tensile strength. Eqs. J3-3a/b reduce Fnt; the Commentary gives the mirror-image form (C-J3-6a/b) if you need F'nv.

Worked Example 12.3 — Slip-Critical Connection with Short-Slotted Holes

Adapted from AISC Design Examples V16.0, Example J.4A

Problem statement. A double-lap splice (two slip planes) must carry PD = 17 kips and PL = 51 kips. The connection plates have short slots transverse to the direction of load, no fillers, Class A faying surfaces, and washers per RCSC Specification Section 6. Using ¾ in. Group 120 bolts, determine the number of bolts required for slip resistance only.

Slip-critical double-lap splice — 2 slip planes, Class A faying surfaces, ¾ in. Group 120 bolts splice plate main plate splice plate slip plane 1 slip plane 2 P_D = 17 kips, P_L = 51 kips → P_u = 102 kips Short slots ⟂ load (Ex. 12.3): φ = 1.00, Ω = 1.50 Long slots ∥ load (Ex. 12.4): φ = 0.70, Ω = 2.14
Figure 12.3a — Double-lap splice: two faying surfaces means two slip planes per bolt
Step 1 — Required strength. Pu = 1.2(17) + 1.6(51) = 20.4 + 81.6 = 102 kips; Pa = 17 + 51 = 68.0 kips
Step 2 — Slip resistance per bolt (§J3.9).
Spec. Eq. J3-4Rn = μ · Du · hf · Tb · ns
TermValueSource
μ (slip coefficient)0.30 (Class A)§J3.9
Du1.13§J3.9
hf (filler factor)1.0 (no fillers)§J3.9
Tb (min. pretension)28 kipsTable J3.1, Group 120, ¾ in.
ns (slip planes)2Double lap
Rn = 0.30(1.13)(1.0)(28)(2) = 19.0 kips/bolt
Step 3 — Resistance/safety factors for short slots perpendicular to the load. Per §J3.9(a), for standard holes and short-slotted holes perpendicular to the direction of load: φ = 1.00, Ω = 1.50.
LRFD: φRn = 1.00(19.0) = 19.0 kips/bolt; ASD: Rn/Ω = 19.0/1.50 = 12.7 kips/bolt
(These values are also tabulated directly in Manual Table 7-3.)
Step 4 — Number of bolts.
§J3.9nb = Pr / (available slip resistance per bolt)
LRFD: nb = 102/19.0 = 5.37 → use 6 bolts
ASD: nb = 68.0/12.7 = 5.35 → use 6 bolts
Result: Use 6 bolts for slip resistance. Slip-critical connections must also satisfy every bearing-type limit state — bolt shear, bearing, tearout, tensile yielding, tensile rupture and block shear — before the design is complete.

Engineering interpretation: slip resistance is a serviceability-flavoured strength check: it prevents the connection from ratcheting into bearing. Because it is friction, it scales with pretension Tb, not with bolt shear area — which is why doubling bolt diameter does not double slip resistance the way it does bolt shear. Note also that ns = 2 doubles the answer: a double-lap joint is dramatically more slip-efficient than a single lap.

Common mistakes:
  • Using ns = 1 in a double-lap splice.
  • Using Ab anywhere in Eq. J3-4. Slip strength has nothing to do with bolt cross-sectional area.
  • Using Class B μ = 0.50 without specifying and inspecting a blast-cleaned Class B surface on the drawings.
  • Reporting the connection "complete" after the slip check. It is one of six or more limit states.
  • Forgetting hf < 1.0 when fillers are present (§J3.9).

Worked Example 12.4 — The Same Splice with Long-Slotted Holes

Adapted from AISC Design Examples V16.0, Example J.4B

Problem statement. Repeat Example 12.3 with the same loads, but with long-slotted holes parallel to the direction of load.

Step 1 — Nominal slip resistance is unchanged. All of μ, Du, hf, Tb and ns are identical, so Rn = 0.30(1.13)(1.0)(28)(2) = 19.0 kips/bolt exactly as before.
Step 2 — Only the resistance factor changes (§J3.9(c)). For long-slotted holes: φ = 0.70, Ω = 2.14 — a 30% penalty, because a slip in a long slot lets the joint move a long way before it picks up bearing.
LRFD: φRn = 0.70(19.0) = 13.3 kips/bolt; ASD: Rn/Ω = 19.0/2.14 = 8.88 kips/bolt
Step 3 — Number of bolts. LRFD: nb = 102/13.3 = 7.67 → use 8 bolts
ASD: nb = 68.0/8.88 = 7.66 → use 8 bolts
Hole type§J3.9 clauseφ / ΩφRn per boltBolts required
Standard / short slot ⟂ loadJ3.9(a)1.00 / 1.5019.0 k6
Oversized / short slot ∥ loadJ3.9(b)0.85 / 1.7616.2 k7
Long slot ∥ loadJ3.9(c)0.70 / 2.1413.3 k8
Result: Use 8 bolts — two more than the short-slot case for exactly the same load, purely because of the hole geometry.

Engineering interpretation: hole type is not a detailing afterthought — it is a strength decision that can cost you 33% more bolts. Long slots buy erection tolerance; the code charges for it. If the fabricator asks for long slots late in the job, that request has to come back through the engineer, because the connection was almost certainly designed for standard holes.

Common mistakes:
  • Reducing Rn instead of φ. The nominal slip resistance does not change with hole type — the resistance factor does.
  • Confusing slots perpendicular to the load (mild penalty) with slots parallel to the load (severe penalty).
  • Omitting the washers required by RCSC §6 for slotted holes.
  • Assuming a long-slotted slip-critical joint still delivers full bearing strength — the bearing check must use the slot geometry too.

Worked Example 12.5 — Combined Tension and Shear in a Slip-Critical Connection

Adapted from AISC Design Examples V16.0, Example J.5

Problem statement. The slip-critical bolt group shown has 8 bolts, ¾ in. diameter, Group 120, standard holes, Class A surfaces, one slip plane per bolt. An inclined brace delivers PD = 15 kips and PL = 45 kips along a 3-4-5 slope, so the connection sees both tension and shear. Determine whether the bolts are adequate for tension and for slip.

Inclined load on a slip-critical bolt group (8 bolts) — tension reduces the clamping force 8 bolts brace, 3-4-5 slope T_u = (4/5)P_u = 72.0 k V_u = (3/5)P_u = 54.0 k P_D = 15 k, P_L = 45 k → P_u = 90.0 k
Figure 12.5a — The inclined brace resolves into a tension component that partly relieves the bolt clamping force and a shear component that must be resisted by friction
Step 1 — Required strengths and their components. Pu = 1.2(15) + 1.6(45) = 18 + 72 = 90.0 kips; Pa = 15 + 45 = 60.0 kips
Tu = (4/5)(90.0) = 72.0 kips; Vu = (3/5)(90.0) = 54.0 kips
Ta = (4/5)(60.0) = 48.0 kips; Va = (3/5)(60.0) = 36.0 kips
Step 2 — Bolt tensile strength (§J3.7).
from Spec. Eq. J3-1Rn = Fnt·Ab = 90(0.442) = 39.8 kips/bolt
LRFD: φRn = 0.75(39.8) = 29.9 kips/bolt vs required 72.0/8 = 9.00 kips/bolt o.k.
ASD: Rn/Ω = 39.8/2.00 = 19.9 kips/bolt vs 48.0/8 = 6.00 kips/bolt o.k.
Step 3 — Slip resistance with no tension present.
Spec. Eq. J3-4Rn = μ·Du·hf·Tb·ns = 0.30(1.13)(1.0)(28)(1) = 9.49 kips/bolt
Standard holes → φ = 1.00, Ω = 1.50.
Step 4 — Reduce the slip resistance for the applied tension (§J3.10).
Spec. Eq. J3-5a (LRFD)ksc = 1 − Tu/(Du·Tb·nb) ≥ 0
ksc = 1 − 72.0/[1.13(28)(8)] = 1 − 72.0/253 = 1 − 0.285 = 0.715
Spec. Eq. J3-5b (ASD)ksc = 1 − 1.5·Ta/(Du·Tb·nb) ≥ 0
ksc = 1 − 1.5(48.0)/253 = 1 − 0.285 = 0.715
Step 5 — Final slip check. LRFD: φRn·ksc = 1.00(9.49)(0.715) = 6.79 kips/bolt vs required Vu/nb = 54.0/8 = 6.75 kips/bolt o.k. (D/C = 0.994)
ASD: (Rn/Ω)·ksc = (9.49/1.50)(0.715) = 4.52 kips/bolt vs 36.0/8 = 4.50 kips/bolt o.k.
CheckReferenceRequired/boltAvailable/bolt (LRFD)Ratio
Bolt tension§J3.7, Eq. J3-19.00 k29.9 k0.301 ✓
Slip (no tension)§J3.9, Eq. J3-46.75 k9.49 k0.711
Slip (ksc = 0.715)§J3.10, Eq. J3-5a6.75 k6.79 k0.994 ✓
Result: The 8-bolt slip-critical group is adequate, but only just — the governing slip check runs at 99.4% utilization once the tension reduction is applied.

Engineering interpretation: this example is the whole point of §J3.10. Ignore ksc and the connection looks 71% utilized and comfortable; include it and it is at 99.4%. The physics is direct — externally applied tension partially un-clamps the joint, so there is less normal force available to generate friction. Notice that bolt tension rupture is nowhere near critical (0.301). In slip-critical joints with axial tension, slip almost always governs long before bolt strength does.

Common mistakes:
  • Omitting ksc entirely — the single most common slip-critical error.
  • Using Tu per bolt in Eq. J3-5a. The denominator already contains nb, so Tu must be the total tension on the group.
  • Dropping the 1.5 factor in the ASD form (Eq. J3-5b).
  • Using Eq. J3-3a (bearing-type combined stress) instead of ksc for a slip-critical joint. They are different provisions for different behaviours — and a slip-critical connection with tension must satisfy both its slip check and its bearing-type checks.
  • Resolving the brace force with the wrong slope components (3-4-5: shear = 3/5, tension = 4/5 here).
Chapter 12 takeaway. A bolted connection is never one number. Work the ladder every time: bolt shear → bearing → tearout → plate yielding → plate rupture → block shear, plus slip if the joint is slip-critical, plus the J3-3/J3-5 interaction if the bolts carry tension too. The smallest number on that ladder is the connection.

Additional Worked Examples

Textbook — Aghayere & Vigil (2009)

Chapter 9 develops bolt mechanics. A bolt may fail in (a) shear of the bolt body, (b) bearing of the plate against the bolt, (c) tearout of plate material to the edge, (d) tension of the bolt body, or (e) slip in a slip-critical joint. AISC §J3 governs.

Example 9-1 — Shear capacity of A325-N bolt

Example 9-1 — Shear capacity of A325-N bolt

Setup. 3/4" A325-N bolt (Fnv = 54 ksi, Ab = 0.442 in²) in single shear.

AISC Reference: AISC §J3.7

Numerical practice

φrn per bolt?

  1. A. 12.0 k
  2. B. 15.9 k
  3. C. 17.9 k (Answer)
  4. D. 21.6 k

Step-by-step solution

φrn = 0.75·54·0.442 = 17.9 k.

Example 9-2 — Bolt group capacity

Example 9-2 — Bolt group capacity

Setup. 6 bolts (3/4" A325-N) in double shear connecting a tension splice.

AISC Reference: AISC §J3.7

Numerical practice

Total φRn?

  1. A. 108 k
  2. B. 144 k
  3. C. 180 k
  4. D. 215 k (Answer)

Step-by-step solution

Per bolt double shear = 2·17.9 = 35.8 k; 6·35.8 = 215 k.

Example 9-4 — Bearing on plate

Example 9-4 — Bearing on plate

Setup. 3/4" bolt through PL 3/8", A36 (Fu = 58 ksi), 3-in spacing, 1.5-in edge.

AISC Reference: AISC §J3.11

Numerical practice

φRn (bearing, deformation OK)?

  1. A. 29 k
  2. B. 39 k (Answer)
  3. C. 47 k
  4. D. 55 k

Step-by-step solution

φRn = 0.75·2.4·0.75·0.375·58 = 29.4 k. (Tearout is also OK: lc = 1.5 − 0.5·(13/16) = 1.09 in; φRn,t = 0.75·1.5·1.09·0.375·58 = 26.7 k → tearout governs ≈ 27 k; pick bearing/tearout min.)

Example 9-5 — Eccentric bolt group — elastic method

Example 9-5 — Eccentric bolt group — elastic method

Setup. Vertical bracket bolted with 6 bolts (3 rows × 2 cols) at 3-in spacing, load P at 6-in eccentricity.

AISC Reference: AISC Manual Table 7-7

Numerical practice

Direct shear component per bolt (P = 60 k)?

  1. A. 6 k
  2. B. 8 k
  3. C. 10 k (Answer)
  4. D. 12 k

Step-by-step solution

rv = P/n = 60/6 = 10 k. Add torsional rt,x and rt,y components vectorially to find resultant.

Example 9-7 — Slip-critical capacity

Example 9-7 — Slip-critical capacity

Setup. 3/4" A325 SC, Class A surface, single shear, Tb = 28 k pretension, single slip plane.

AISC Reference: AISC §J3.8

Numerical practice

φRn (slip)?

  1. A. 6.3 k
  2. B. 8.4 k (Answer)
  3. C. 10.5 k
  4. D. 13.4 k

Step-by-step solution

φRn = 1.0·0.30·1.0·1.0·28·1 = 8.4 k per bolt (Class A).

FE-Style Worked Examples(7)

Each example mirrors the NCEES FE Civil Reference Handbook style: brief givens, a labeled figure, AISC section reference, step-by-step numeric solution, and a single boxed answer.

Given
3/4" A325-N bolt, Fnv=54 ksi, Ab=0.442 in².
AISC Reference
AISC §J3.7 Table J3.2
Step-by-step solution
  1. φRn
    0.75 × 54 × 0.442 = 17.9 k per shear plane
Answer φRn = 17.9 k.
Single bolt shear (Group A, threads in plane)
Problem statement image
PuLap splice in single shear
DIMDimensions from the problem statement
Fnv = 54kAb = 0.442in²
Bolted lap — plate + bolt pattern
  • Pitch s along the load, gage g perpendicular
  • Edge distances L_ev (parallel), L_eh (perpendicular)
  • Hole diameter d_h = d_b + 1/8 in (fabrication + damage)
  • Net area A_n = A_g − Σ(d_h · t) + Σ(s²/4g)·t

Interactive Calculator

Bolt Shear / Bearing / Tearout

AISC §J3.7, §J3.11
Ab0.601 in²
φ Rn (shear, all bolts)97.4 kips
φ Rn (bearing, all bolts)182.7 kips
φ Rn (tearout, all bolts)195.8 kips
Governing φ Rn97.4 kips

Graded Chapter Quiz(13 FE-style questions · AISC Manual required)

These questions reference AISC Steel Construction Manual (16th ed.) — sections, equations, and tables are cited explicitly. Use a calculator. Each question offers a clue you may reveal before answering. Submissions are recorded to your account once signed in.

C12-01AISC 360-22 Table J3.2 & Eq. J3-1
1. Single 7/8-in Ø A325-N bolt (Group A, threads in shear plane, Ab = 0.601 in²). Compute nominal shear Rn per shear plane (Fnv = 54 ksi).
4-bolt lap splice
C12-02AISC 360-22 Eq. J3-1
2. Same bolt in SINGLE shear. Compute design shear per bolt φRn.
4-bolt lap splice
C12-03AISC 360-22 Eq. J3-1
3. 4-bolt lap splice, single shear: design shear strength of the BOLT GROUP.
4-bolt lap splice
C12-04AISC 360-22 §J3.11
4. Same 3/8-in PL 6 wide (Fu = 65). For an INTERIOR bolt with s = 3 in, dh = 15/16, compute the clear distance Lc used for bearing/tearout.
4-bolt lap splice
C12-05AISC 360-22 §J3.11
5. Same bolt (interior, deformation not a design consideration). Compute φRn for BEARING / TEAROUT and identify which governs.
4-bolt lap splice
C12-06AISC 360-22 §J3.11
6. At the EDGE bolt with Le = 1.5 in, dh = 15/16: compute Lc and φRn (tearout).
4-bolt lap splice
C12-07AISC 360-22 §J3
7. Assemble the group strength for the 4-bolt lap (1 edge bolt + 3 interior bolts). Take the MIN of bolt shear, bearing, and tearout for each bolt, then sum.
4-bolt lap splice
C12-08AISC 360-22 §J3.3
8. Minimum spacing between bolt centers per §J3.3:
C12-09AISC 360-22 Eq. J3-4
9. Slip-critical strength (STD holes, Class A faying surface, 1 slip plane, μ = 0.30, Du = 1.13, hf = 1.0, Tb = 39 k):
C12-10AISC Manual Part 9
10. T-stub bolted hanger in tension: what phenomenon amplifies the bolt tension beyond the applied load / n?
C12-11AISC 360-22 Table J3.1
11. Minimum pretension Tb for a 7/8-in A325 (Table J3.1):
C12-12AISC 360-22 Eq. J3-3a
12. Combined bolt shear + tension (bearing bolt): reduced tension F'nt =
C12-13AISC 360-22 §J3.2 Table J3.3
13. Standard hole diameter for a ¾-in bolt (Table J3.3):

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§11Section 11

Chapter summary

Formula sheet
  • Bolt shear
    φRn = 0.75 · Fnv · Ab
    AISC §J3.7
  • Bearing
    φRn = 0.75 · 2.4 · db · t · Fu
    AISC §J3.11
  • Tearout
    φRn = 0.75 · 1.5 · lc · t · Fu
    AISC §J3.11
Engineering checklist
  • Module 12: Bolted Connections
  • Key limit states and AISC references are listed in the reference box.
  • Use φRn ≥ Ru for every check.
  • Verify section properties with the official AISC Manual.
Professional tips
  • • Mixing ASD and LRFD load combinations in the same problem.
  • • Using nominal strength Rn instead of design strength φRn.
  • • Forgetting to check every limit state listed in the AISC chapter.
§13Section 13

FE exam preparation

FE exam preparation
Concept review
Concept summary coming soon.
Calculator tips

Calculator tips coming soon.

Common exam traps

Traps coming soon.

Time management

Aim for ~3 minutes per FE problem; skip and return to any item that takes longer than 5 minutes.