CELE Steel & Timber Design — Steel ConnectionsSummary

The nominal shear strength of a bolt is Rn = Fnv × Ab, where Fnv is the nominal shear stress (typically 372 MPa for A325-N with threads in the shear plane, or 469 MPa for A325-X with threads excluded per AISC 360 Table J3.2) and Ab is the nominal bolt area (π/4 × d² in mm²). Design strength is φRn = 0.75 Rn. For **single shear** (one shear plane, e.g., a simple lap splice), the bolt crosses one plane. For **double shear** (two planes, e.g., a bolt through two plates and a central gusset), multiply the single-shear capacity by 2. Example: A 20 mm A325-N bolt in single shear has Ab = 314.2 mm², Rn = 372 × 314.2 = 116,864 N, and φRn = 0.75 × 116,864 = 87.6 kN.

When a bolt bears on a plate, the plate can tear out (rupture) along a path through the hole. The nominal bearing strength per bolt is Rn = min(1.2 lc t Fu, 2.4 db t Fu), where lc = clear edge distance (mm) in the direction of load, t = plate thickness (mm), db = bolt diameter (mm), and Fu = ultimate tensile strength of the plate (MPa). The 1.2 and 2.4 coefficients apply **when bolt-hole deformation at service load IS a design consideration** (the standard case). If deformation is NOT a concern (rare, explicitly stated), the higher limits 1.5 lc t Fu ≤ 3.0 db t Fu may be used. Design strength is φRn = 0.75 Rn. The 2.4 db t Fu term acts as a cap; the 1.2 lc t Fu term depends on edge distance. Example: A 20 mm bolt on a 10 mm plate (Fu = 400 MPa) with lc = 35 mm gives: 1.2(35)(10)(400) = 168,000 N and 2.4(20)(10)(400) = 192,000 N. The smaller, 168 kN, governs, so φRn = 0.75(168) = 126 kN.

A bolt group (especially in angle connections or gussets) can fail by tearing out a rectangular 'block' of the connected element. The failure path combines **shear rupture** along one edge (perpendicular to load) and **tension rupture** along the opposite edge (parallel to load). Per AISC J4.3, the nominal strength is Rn = 0.6 Fu Anv + Ubs Fu Ant ≤ 0.6 Fy Agv + Ubs Fu Ant (both terms capped by the shear-yield path), where Agv = gross shear area, Anv = net shear area (shear plane with bolt holes removed), Ant = net tension area (perpendicular to load with bolt holes removed), and Ubs = 1.0 for uniform tension or 0.5 for non-uniform. Design strength is φRn = 0.75 Rn. The first form uses rupture (Fu Anv); it is limited by the second form (yield Fy Agv). Example: A 2-bolt angle with Agv = 2000 mm², Anv = 1600 mm² (one hole removed), Ant = 800 mm², Fu = 400 MPa, Fy = 250 MPa, Ubs = 1.0 gives: 0.6(400)(1600) + 1.0(400)(800) = 384,000 + 320,000 = 704,000 N (first form), capped by 0.6(250)(2000) + 1.0(400)(800) = 300,000 + 320,000 = 620,000 N (second form). Thus Rn = 620 kN and φRn = 465 kN.

A fillet weld (the most common weld type) is a triangular weld deposited in the corner between two plates. It fails through its **throat**, a plane perpendicular to the weld face. For an equal-leg fillet weld of size a (leg length, mm), the throat thickness is 0.707a (the perpendicular distance). The design strength per unit weld length is φRn = 0.75 × 0.6 × FExx × (0.707a) × L, where FExx is the electrode strength (E70 ≈ 482 MPa per AISC; lower-strength E60 ≈ 414 MPa is also common) and L is the effective weld length (mm). A common shorthand: φRn ≈ 0.75(0.6)(482)(0.707a) = 154.5a (MPa·mm) for E70. Example: A 6 mm E70 fillet weld, 200 mm long, has φRn = 0.75(0.6)(482)(0.707 × 6)(200) = 0.75(1,226.8)(200) ≈ 184 kN. Minimum weld size depends on plate thickness (e.g., 3 mm for plates < 6 mm, 5 mm for 6–13 mm, 6 mm for 13–25 mm); maximum is limited by melt-through or heat distortion.

A **groove weld** (also called a full-penetration or complete-joint-penetration weld) is deposited in a prepared groove between two members and, when properly executed, develops the full strength of the base metal (Fy or Fu as applicable). A **fillet weld** is a partial-strength weld limited by its throat. In design: groove welds are used for member continuity (e.g., splice plates) where full strength is required; fillet welds are used for ancillary connections (e.g., gusset plates to flanges, stiffeners). A groove weld's capacity is limited by the base metal, not the weld itself, so the design strength equals 0.75 × Fy × A (or 0.75 × Fu × A for rupture). Fillet welds require explicit throat calculation.

A connection typically involves multiple potential failure modes: bolt shear, bearing, block shear, weld throat, and member net-section rupture. The **design capacity of the connection** is the **minimum** capacity across all these checks, not the sum. Example: A bolted connection with four bolts in single shear has individual bolt shear capacity 87.6 kN each (total 350 kN), bearing capacity 126 kN per bolt (total 504 kN), and block shear capacity 465 kN. The governing limit is bolt shear at 350 kN; the connection capacity cannot exceed this, even though the other paths are stronger. Failure will occur by bolt shear, not bearing or block shear.

Connection design uses LRFD with φ = 0.75 (compared to φ = 0.90 for member tension or φ = 0.85 for member compression). The lower factor reflects the higher uncertainty and consequence of connection failure. Per AISC 360-16 and NSCP 2015, all connection limit states (bolt shear, bearing, weld throat) are designed with φ = 0.75. Some sources (ACI 318 for concrete anchors) use φ = 0.65 or 0.70, but for steel, φ = 0.75 is standard. Design strength = 0.75 × Nominal strength.

ASTM A325 high-strength bolts come in two grades for shear: **A325-N** (standard, threads allowed in the shear plane) and **A325-X** (excluded, threads excluded from shear plane). The nominal shear stress is Fnv = 372 MPa (N grade, ~54 ksi) or 469 MPa (X grade, ~68 ksi). The X-grade bolts are stronger because the threads (a stress-concentration region) are not in the plane where shear occurs. In practice, A325-N is more common and cheaper; X is used in high-shear applications or when thread-in-shear is problematic (e.g., fatigue). Board exams often specify the grade to test whether students know the corresponding Fnv value.

Some connections are designed to resist slip at service load (not ultimate load) by friction between bolt faces. Slip-critical design is used in fatigue-prone or oscillating-load applications (e.g., bridge expansion joints, machinery mounts) where bolt slip must be prevented. The nominal slip resistance is Rn = μ × Du × hf × Tb × ns, where μ is the slip coefficient (~0.3–0.4 for clean steel), Du is a factor depending on bolt type and hole condition, hf is the number of friction surfaces, Tb is the bolt tension, and ns is the number of bolts. Design uses LRFD with φ = 1.0 (no reduction, because the limit state is service slip, not ultimate rupture). Bolt tension is typically 70% of its ultimate tensile capacity. Slip-critical is less common in building design than in bridge design but appears on board exams.

Bolts can also carry tension (e.g., in moment connections, anchor bolts, or tie connections). Nominal tension strength is Rn = Fnt × Ab, where Fnt ≈ 621 MPa for A325 (per AISC Table J3.2). When both shear fv and tension ft act on the same bolt, an interaction equation applies: fv/Fnv + ft/Fnt ≤ 1.0 (linear interaction). Example: If fv = 100 MPa and ft = 200 MPa, and Fnv = 372 and Fnt = 621, then 100/372 + 200/621 = 0.269 + 0.322 = 0.591 < 1.0, so the bolt is adequate. This is less common on initial-level board exams but appears in advanced questions.

Bolt shear strength is calculated first (straightforward: Fnv × Ab). Bearing is then checked (min of 1.2 lc t Fu and 2.4 db t Fu). The connection capacity cannot exceed the lesser of these two. If shear < bearing, bolts slip or shear; if bearing < shear, the plate tears at the hole. In a typical connection, bearing is larger, so shear governs; but in thick plates with large edge distances, bearing can be the limit.

Block shear depends on the layout of the bolt group. A single row of bolts perpendicular to load has one block shear plane; angled or staggered rows create multiple potential planes. The failure path is the weakest rectangular path combining shear and tension rupture. As bolts are moved further apart (larger spacing), the block shear area increases and its risk decreases. Conversely, closely spaced bolts in a small gusset are prone to block shear.

The throat is always 0.707 times the leg size for equal-leg welds (a common simplification). Doubling the weld size a doubles the throat area, doubling the capacity. The relationship is linear: capacity ∝ a. This allows easy weld sizing: to carry 2× the load, size the weld 2× larger.

When designing a connection for a high-strength steel member, bolt shear may become the controlling limit state because the plate is strong but the bolts are not. Low-strength bolts on high-strength plates lead to unbalanced design.

The lower φ = 0.75 (vs. φ = 0.90 for members) reflects greater uncertainty in connection fabrication, bolt tension variability, and the severity of connection failure. This lower factor means connection designs are more conservative; a connection is 'less efficient' than a member because of this. Understanding why φ is lower helps students recognize that connections must be checked carefully—they are not just 'automatic' once members are sized.

Slip-critical connections use friction (controlled by bolt preload) to resist movement at service (unfactored) load. The design strength is φRn with φ = 1.0 (no reduction) because the limit state is service slip, not ultimate rupture. In contrast, strength-based (non-slip-critical) connections use φ = 0.75 and are checked at factored (ultimate) load. This dual system (service-level check for slip; ultimate-level check for strength) is an important conceptual distinction that appears on advanced board questions.

A common steel connection: two lap plates bolted with typically 4–8 bolts in two rows. Steps: (1) Count shear planes—usually single shear if each bolt crosses one interface. (2) Calculate bolt shear capacity: Rn = Fnv Ab, design = 0.75 Rn per bolt, multiply by number of bolts. (3) Check bearing: 1.2 lc t Fu and 2.4 db t Fu, minimum governs. (4) Check block shear if the bolt group is near an edge. (5) Check member net-section rupture (An Fu). (6) The connection capacity is the minimum of all these; it must be ≥ design load Pu. Example: Four 20 mm A325-N bolts, single shear, 10 mm plates (Fu = 400 MPa), lc = 40 mm. Bolt shear: 4 × 0.75(372)(314.2) ≈ 350 kN. Bearing per bolt: min(1.2 × 40 × 10 × 400, 2.4 × 20 × 10 × 400) = min(192, 192) = 192 kN; total 4 × 0.75 × 192 ≈ 576 kN. Bolt shear governs at 350 kN.

A small angle (e.g., L76×76×6, length 100 mm) is bolted to a girder flange with 2–4 bolts. This is prone to block shear because the force concentrates in a small area. Steps: (1) Calculate bolt shear and bearing as before. (2) **Critical**: Calculate block shear carefully. The shear rupture path runs along the angle leg perpendicular to load; the tension path runs along the angle leg parallel to load. Agv = leg width × thickness; Anv = Agv minus bolt-hole area on shear plane; Ant = perpendicular leg area minus hole. (3) Compare all three limit states. Example: A 2-bolt angle, Agv = 76 × 6 = 456 mm², Ant = 456 mm² (other leg). With 16 mm bolts, Anv ≈ 456 − 16 × 6 = 360 mm². Fu = 400 MPa, Fy = 250 MPa. Block shear: Rn = min(0.6 × 400 × 360 + 1.0 × 400 × 456, 0.6 × 250 × 456 + 1.0 × 400 × 456) = min(268,800, 432,000) = 268.8 kN. Bolt shear per bolt: 0.75 × 372 × 201 ≈ 56 kN, total ≈ 112 kN. Block shear governs here; the angle capacity is limited to ~270 kN despite adequate bolt shear.

A small gusset or bracket is welded to a column flange using continuous fillet welds on two sides. Steps: (1) Determine the load to be carried (shear or moment). (2) Size the fillet weld(s). For shear: φRn = 0.75 × 0.6 × FExx × (0.707a) × (2L), where 2L is the total weld length (both sides of the gusset). (3) Check weld size constraints (min, max). (4) Verify base-metal strength is not exceeded. Example: A 100 mm tall gusset, welded on both sides, carries V = 200 kN shear. Total weld length 2L = 2 × 100 = 200 mm. Using E70 (FExx = 482 MPa), φRn = 0.75 × 0.6 × 482 × 0.707 × a × 200. Solving for a: 0.75 × 0.6 × 482 × 0.707 × a × 200 = 200,000 ⇒ a ≈ 6.2 mm. Specify a 6 mm fillet (slightly under, so iterate upward) or a 7 mm fillet (safe margin).

A beam is spliced mid-span using two cover plates (one top, one bottom flange) bolted through the web and flanges. Each bolt sees **double shear** because it passes through the web and the cover plates on both sides. Steps: (1) Recognize double shear: multiply single-shear capacity by 2. (2) Calculate bolt shear: 0.75 × Fnv × Ab × (number of bolts) × 2. (3) Check bearing on the web (thin, so lc and 2.4 db t Fu may be tight). (4) Check bearing on cover plates (usually larger, less restrictive). (5) Check member rupture through the web and flanges. Example: Six 24 mm A325-N bolts, double shear, web thickness 12 mm. Bolt shear per bolt, single: 0.75 × 372 × (π/4 × 24²) ≈ 0.75 × 372 × 452.4 ≈ 126 kN. Double shear: 126 × 2 = 252 kN per bolt, total 6 × 252 = 1,512 kN. Bearing on thin web (lc = 35 mm): min(1.2 × 35 × 12 × 400, 2.4 × 24 × 12 × 400) = min(201.6, 276.5) = 201.6 kN per bolt (governs at higher load), total 6 × 0.75 × 201.6 ≈ 910 kN. Member rupture and plate bearing are then checked. The connection capacity is the minimum across all checks.

Anchor bolts embed in a concrete foundation and resist shear and tension from the superstructure (column base or equipment mount). The connection must resist both vertical tension (from moment or lifting) and horizontal shear. Steps: (1) Check bolt shear (if horizontal load applies). (2) Check bolt tension (if uplift or moment applies). (3) If both shear and tension coexist, apply interaction: fv/Fnv + ft/Fnt ≤ 1.0. (4) Check concrete bearing and pullout (per ACI 318 for anchorage). Example: A 24 mm A325 anchor bolt in tension (uplift from overturning moment). Tension capacity: Rn = 621 × 452.4 ≈ 280.5 kN per bolt. Design: φRn = 0.75 × 280.5 ≈ 210 kN. If also 100 kN shear: fv = 100/452.4 ≈ 221 MPa (estimated), ft = 210 × 0.75 / 0.75 / 452.4 ≈ 310 MPa. Interaction: 221/372 + 310/621 ≈ 0.59 + 0.50 = 1.09 > 1.0, so the bolt is overstressed in combined loading; a larger bolt or fewer bolts subjected to tension would be needed.

A 6-bolt connection (3 rows, 2 bolts per row) is designed for shear. The engineer must check: (1) Individual bolt shear (usually straightforward). (2) Bearing on the main plate (may vary if holes are near an edge). (3) Bearing on the gusset or splice plate. (4) Block shear (especially if the bolt group is near a corner or edge). (5) Member net-section rupture through the main plate or member. The connection capacity is the minimum of all these checks. A common scenario: bolt shear looks adequate, bearing looks adequate, but block shear through the gusset is small because the gusset is small, or the bolt group is very close to an edge. The engineer must then either enlarge the gusset, move the bolts further from the edge, or increase bolt size to spread the load over a larger area. This iterative process is a realistic design activity.

A vertical member (tee stem or stiffener) is welded to a column flange using fillet welds on two sides of the stem. The stem carries a vertical load and a horizontal shear. Steps: (1) Determine the demand: shear force V and/or axial force P. (2) Assume a fillet size a (mm). (3) Calculate capacity: φRn = 0.75 × 0.6 × FExx × (0.707a) × L, where L is the effective weld length on one side. If two sides are welded, use 2L. (4) Size a such that φRn ≥ demand. (5) Check that a is within min/max constraints. (6) Verify that the base metal (column flange and stem) is not overstressed. Example: A stem 80 mm tall, 10 mm thick, carries V = 250 kN. Weld on both sides, 2L = 2 × 80 = 160 mm, E70. φRn = 0.75 × 0.6 × 482 × 0.707 × a × 160 = 38.7a (kN). To carry 250 kN: a ≥ 250 / 38.7 ≈ 6.5 mm. Specify a = 7 mm fillet (round up). Check base metal: shear stress in weld = 250 / (0.707 × 7 × 160) ≈ 316 MPa < 0.6 × FExx ≈ 289 MPa... slight overstress, so iterate to a = 8 mm.