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CELE Steel & Timber DesignSteel ConnectionsRevision Notes

Condensed revision notes for Steel Connections, built for the final weeks before the CELE 2026. These are the distilled key points you need when there is no time left for full study notes — just the concepts, formulas, and traps Professional Regulation Commission (PRC) — Board of Civil Engineering tests.

Exam context

The Civil Engineer Licensure Examination is conducted by Professional Regulation Commission (PRC) — Board of Civil Engineering and is scheduled for May and November 2026. The Steel & Timber Design subtest is marked as "Core" in the official pattern, and Steel Connections appears in position 4th of 5 in the CELE Steel & Timber Design review rotation. Passing mark: 70% weighted average, no sub-test below 50%. Recent CELE 2026 papers have drawn roughly a meaningful share of questions from this subject.

Steel Connections - Revision Notes

Steel connections are the critical junctions where structural members transfer forces to one another. Because connections are where many steel structures actually fail, the PRC board examination consistently tests this topic. This chapter covers the two primary connection systems — bolted connections and welded connections — along with their associated limit states: bolt shear, bearing, block shear, slip-critical behavior, and fillet/groove weld strength. All design in this chapter uses the LRFD philosophy per AISC 360-16 (adopted by NSCP 2015 Section 502), with the resistance factor φ = 0.75 governing all connection limit states. Mastery of these concepts, formulas, and their proper application is essential for the licensure examination.

Sections

Exam Tips

  • Memorize φ = 0.75 for ALL connection limit states: shear, bearing, block shear, weld rupture.
  • The design strength of a connection = minimum of all limit state capacities — always state which limit state governs.
  • Read the problem carefully for whether bolts are in single or double shear — this doubles the capacity.
  • When asked for 'number of bolts required,' divide the factored load Pu by the governing per-bolt φRn, then round UP.

Key Points

  • Connections must transfer all applied forces and moments between structural members without premature failure.
  • LRFD design philosophy: φRn ≥ Pu, where φ = 0.75 for connection limit states (bolt shear, bearing, block shear, weld rupture).
  • The two primary fastener types are high-strength bolts (ASTM A325, A490) and welds (fillet, groove).
  • The governing (controlling) design strength of a connection is always the MINIMUM value across all applicable limit states.
  • Limit states for bolted connections: bolt shear, bearing, block shear, net-section tension, gross-section yielding.
  • Limit states for welded connections: weld throat shear rupture, base metal shear yielding, base metal shear rupture.
  • NSCP 2015 Section 502.3 and AISC 360-16 Chapter J govern connection design in Philippine practice.
  • RA 544 (Philippine Civil Engineering Law) requires the Registered Civil Engineer to ensure all connections comply with applicable codes.

Definitions

Term

LRFD (Load and Resistance Factor Design)

Definition

A design method where factored loads (Pu) must not exceed the reduced nominal strength (φRn). For connections, φ = 0.75.

Importance

All board problems on connections use LRFD. Ensure you apply φ = 0.75, not 0.90 (which is for members in tension/compression).

Term

Limit State

Definition

A condition at which a structural element ceases to fulfill its intended function. Connection design checks multiple limit states and uses the most critical (lowest strength).

Importance

The most common board-exam error is checking only bolt shear and ignoring bearing, block shear, or weld throat — always check ALL applicable limit states.

Term

Nominal Strength (Rn)

Definition

The theoretical maximum strength of a connection element based on material properties and geometry, before applying the resistance factor φ.

Importance

Always distinguish Rn (nominal) from φRn (design strength). Board problems ask for design strength.

Term

ASTM A325 / A490 Bolts

Definition

High-strength structural bolts. A325: Fu = 825 MPa (120 ksi); A490: Fu = 1040 MPa (150 ksi). Both are used in bearing-type and slip-critical connections.

Importance

A325 is more common in Philippine practice. Bolt classification (A325-N, A325-X) determines Fnv — a frequent board-exam variable.

Section Title

1. Fundamentals of Steel Connection Design

Common Mistakes

  • Using φ = 0.90 (tension member factor) instead of φ = 0.75 for connection elements — always φ = 0.75 for connections.
  • Forgetting to check all limit states and reporting only the bolt shear capacity as the connection capacity.
  • Confusing nominal bolt area (based on nominal diameter) with the stress area (used only for tension rupture of bolts, not shear).
  • Applying the wrong resistance factor for slip-critical connections (φ = 1.00 for serviceability-based slip check, φ = 0.85 for strength-level slip).

Formulas

Example

20 mm A325-N bolt, single shear: Ab = π/4(20²) = 314.2 mm²; φRn = 0.75(372)(314.2) = 87,650 N = 87.7 kN per bolt.

Formula

φRn = 0.75 × Fnv × Ab

Variables

φ = 0.75 (resistance factor for bolt shear); Fnv = nominal shear stress of bolt (MPa), from AISC Table J3.2; Ab = nominal (unthreaded) cross-sectional area of bolt = π/4 × db² (mm²); db = nominal bolt diameter (mm)

Application

Use for SINGLE SHEAR. Multiply by 2 for double shear. Multiply by number of bolts n for a bolt group. This gives the design bolt shear strength per bolt.

Example

Same 20 mm A325-N bolt in double shear: φRn = 2 × 87.7 = 175.3 kN per bolt.

Formula

φRn = 0.75 × Fnv × Ab × 2

Variables

Factor of 2 accounts for two shear planes in double shear configuration (e.g., bolts in a lap-splice with cover plates on both sides).

Application

Use for DOUBLE SHEAR only. This is the most common board-exam variant where a gusset plate is sandwiched between two member plates.

Exam Tips

  • Memorize key Fnv values: A325-N = 372 MPa, A325-X = 469 MPa, A490-N = 457 MPa, A490-X = 579 MPa.
  • Memorize common bolt areas: 16 mm → 201 mm², 20 mm → 314 mm², 22 mm → 380 mm², 24 mm → 452 mm², 25 mm → 491 mm².
  • Always write out Ab = π/4 × db² in your solution — board examiners award partial credit for correct formulas.
  • For a bolt group, the answer = n × φRn (shear) vs n × φRn (bearing) — pick the smaller as governing.

Key Points

  • Bolt shear is the most commonly tested connection limit state in the PRC board exam.
  • A bolt in shear fails by sliding of the connected plates across each other, shearing the bolt shaft.
  • Single shear: one shear plane through the bolt; double shear: two shear planes — double the strength.
  • Nominal bolt shear strength: Rn = Fnv × Ab per bolt per shear plane.
  • Design bolt shear strength: φRn = 0.75 × Fnv × Ab (single shear); φRn = 0.75 × Fnv × Ab × 2 (double shear).
  • Fnv depends on bolt grade AND thread condition: A325-N (threads IN shear plane) = 372 MPa; A325-X (threads EXCLUDED from shear plane) = 469 MPa per AISC 360-16 Table J3.2.
  • A490-N: 457 MPa; A490-X: 579 MPa.
  • For a group of n identical bolts, total capacity = n × φRn per bolt.

Definitions

Term

A325-N vs A325-X

Definition

N = threads included in the shear plane (lower Fnv = 372 MPa, conservative, most common assumption). X = threads excluded from the shear plane (higher Fnv = 469 MPa). The problem statement or drawing must specify.

Importance

If the problem does not specify N or X, assume N (threads in shear plane) — the conservative, safe assumption for the board exam.

Term

Nominal Bolt Area (Ab)

Definition

The unthreaded cross-sectional area of the bolt shank: Ab = π/4 × db². This is used for SHEAR calculations. Do NOT use the tensile stress area (smaller) for shear checks.

Importance

Using the tensile stress area instead of the full shank area for shear is a common and costly computational error.

Term

Single Shear vs Double Shear

Definition

Single shear: bolt crosses ONE shear plane (e.g., two plates lapped together). Double shear: bolt crosses TWO shear planes (e.g., a middle plate sandwiched by two outer plates — common in gusset connections).

Importance

Misidentifying single vs double shear will halve or double your answer — always sketch the connection to count shear planes.

Section Title

2. Bolt Shear Strength

Common Mistakes

  • Using Fnv = 469 MPa (A325-X) when the problem specifies or implies threads are in the shear plane — default to N (372 MPa).
  • Computing bolt shear strength for double shear without multiplying by 2.
  • Using bolt tensile stress area (≈ 245 mm² for 20 mm bolt) instead of the full shank area (314.2 mm²) for shear calculations.
  • Forgetting to multiply per-bolt capacity by the number of bolts n when finding total connection shear capacity.

Formulas

Example

20 mm bolt, 10 mm plate (Fu = 400 MPa), lc = 35 mm: 1.2(35)(10)(400) = 168,000 N; 2.4(20)(10)(400) = 192,000 N. Min = 168,000 N. φRn = 0.75(168,000) = 126,000 N = 126 kN.

Formula

φRn = 0.75 × min(1.2 lc t Fu, 2.4 db t Fu)

Variables

φ = 0.75; lc = clear distance in direction of force (mm) — see below; t = thickness of the plate being checked (mm); Fu = ultimate tensile strength of the plate material (MPa); db = nominal bolt diameter (mm)

Application

Standard bearing check (hole deformation IS a design consideration — the default board-exam assumption). Apply PER BOLT. For a group of bolts, sum individual bolt bearing capacities (edge bolt has smaller lc than interior bolts).

Example

20 mm bolt (dh = 22 mm), edge distance Le = 40 mm: lc = 40 - 22/2 = 40 - 11 = 29 mm.

Formula

lc(edge) = Le - dh/2

Variables

Le = edge distance from center of bolt to edge of plate (mm); dh = bolt hole diameter = db + 2 mm (standard hole) or db + 3 mm (oversized)

Application

Clear distance for the edge (outermost) bolt in the direction of applied force. This is usually the controlling (smallest lc) bolt.

Example

20 mm bolt (dh = 22 mm), spacing s = 75 mm: lc = 75 - 22 = 53 mm.

Formula

lc(interior) = s - dh

Variables

s = center-to-center spacing of bolts in the direction of force (mm); dh = hole diameter (mm)

Application

Clear distance between holes for interior bolts. Standard minimum spacing is 2.67db, preferred 3db per AISC.

Exam Tips

  • The 2.4 db t Fu term is the UPPER LIMIT (crushing cap). If 1.2 lc t Fu > 2.4 db t Fu, use 2.4 db t Fu.
  • If the problem gives only one bolt row, the edge bolt controls bearing — compute lc for the edge bolt first.
  • In multi-bolt problems, the total bearing capacity = sum of individual bolt capacities (each computed with its own lc).
  • Quick check: if lc > 2db, then 1.2 lc t Fu > 2.4 db t Fu and crushing governs → φRn = 0.75 × 2.4 db t Fu.

Key Points

  • Bearing is the localized compression (crushing) of the plate material around the bolt hole.
  • Two sub-limit states exist: tearout (low clear distance) governed by 1.2 lc t Fu, and crushing governed by 2.4 db t Fu.
  • Bearing capacity per bolt = smaller of the two sub-limits, then multiplied by φ = 0.75.
  • lc is the CLEAR distance in the direction of force: from the edge of the hole to the edge of the plate (edge bolt) or to the edge of the adjacent hole (interior bolt).
  • lc = L - dh/2 for edge bolts (where dh = hole diameter = db + 2 mm for standard holes).
  • lc = s - dh for interior bolts (where s = center-to-center bolt spacing).
  • Bearing applies to EVERY plate the bolt passes through — check the thinner/weaker plate.
  • When hole deformation IS a design consideration (standard condition): use 1.2 lc t Fu ≤ 2.4 db t Fu.
  • When hole deformation is NOT a design consideration: use 1.5 lc t Fu ≤ 3.0 db t Fu (higher, less common in board problems).

Definitions

Term

Clear Distance (lc)

Definition

The distance from the edge of the bolt hole to the nearest edge of the adjacent hole or to the edge of the plate, measured in the direction of the applied force. It is the critical geometric parameter for tearout.

Importance

Incorrect computation of lc is the most common error in bearing problems. Remember: lc is measured from the HOLE EDGE, not the bolt center.

Term

Tearout vs Crushing

Definition

Tearout: the plate tears along two shear planes radiating from the hole toward the plate edge (controlled by 1.2 lc t Fu). Crushing: the plate material directly beneath the bolt compresses and yields/ruptures (controlled by 2.4 db t Fu).

Importance

The minimum of these two values is the nominal bearing strength. A small lc (short edge distance) means tearout governs; a large lc (long edge distance) means crushing governs.

Section Title

3. Bearing Strength at Bolt Holes

Common Mistakes

  • Computing lc from the BOLT CENTER instead of from the HOLE EDGE — subtract dh/2 for edge bolts, subtract dh for interior bolt-to-bolt.
  • Using hole deformation NOT a design consideration (1.5/3.0 coefficients) when the standard (1.2/2.4) applies — read the problem carefully.
  • Forgetting that bearing must be checked for EVERY plate (connected plate AND connecting plate) — the thinner/weaker plate governs.
  • Summing bearing capacities incorrectly when bolts have different lc values (edge vs interior) — compute each bolt separately then sum.

Formulas

Example

Plate: Fy = 250 MPa, Fu = 400 MPa, t = 10 mm. Block: Agv = 1500 mm², Anv = 1200 mm², Ant = 300 mm², Ubs = 1.0. Rupture: 0.60(400)(1200) + 1.0(400)(300) = 288,000 + 120,000 = 408,000 N. Cap: 0.60(250)(1500) + 1.0(400)(300) = 225,000 + 120,000 = 345,000 N. Min = 345,000 N. φRn = 0.75(345,000) = 258,750 N = 258.8 kN.

Formula

φRn = 0.75 × [min(0.60 Fu Anv + Ubs Fu Ant, 0.60 Fy Agv + Ubs Fu Ant)]

Variables

Fu = ultimate tensile strength of plate (MPa); Fy = yield strength of plate (MPa); Anv = net shear area = gross shear area minus hole areas on shear plane (mm²); Agv = gross shear area on shear plane (mm²); Ant = net tension area = gross tension area minus hole areas on tension plane (mm²); Ubs = 1.0 (uniform tension) or 0.5 (non-uniform tension)

Application

Apply to the critical block shear failure path. Identify the shear planes (parallel to load) and tension plane (perpendicular to load). Compute areas carefully using net areas (deducting hole areas = (db + 2 mm) × t per hole).

Example

3 bolts on shear plane, 20 mm bolts (dh = 22 mm), plate t = 10 mm, shear length = 200 mm: Agv = 200(10) = 2000 mm²; Anv = 2000 - 3(22)(10) = 2000 - 660 = 1340 mm².

Formula

Anv = Agv - n_shear × (dh + 2) × t

Variables

Agv = gross shear area (length of shear plane × thickness); n_shear = number of bolt holes on the shear plane; dh = bolt hole diameter (db + 2 mm for standard holes); t = plate thickness

Application

Computing net shear area for the block shear formula. The '+ 2 mm' accounts for damage to plate material around the hole per AISC.

Exam Tips

  • Always check BOTH expressions: rupture-rupture path and the shear-yield cap. Report whichever is smaller.
  • In a simple gusset plate with a single row of bolts: Ubs = 1.0 (standard board-exam assumption).
  • Block shear often appears in combination with bolt shear and bearing — always compare all three and report the governing limit state.
  • Draw the failure block outline: shade it, then compute shear plane area (parallel to force) and tension plane area (perpendicular to force).

Key Points

  • Block shear is a tearing-out failure of a block of plate material, combining shear yielding or rupture on one plane with tension rupture on a perpendicular plane.
  • It is critical at gusset plates, clip angles, beam webs, and any element with a bolt-group layout.
  • AISC 360-16 Section J4.3 (adopted in NSCP 2015): Rn = 0.60 Fu Anv + Ubs Fu Ant ≤ 0.60 Fy Agv + Ubs Fu Ant.
  • The first expression (with Fu Anv) is the rupture-rupture path; the cap (with Fy Agv) prevents it from exceeding the shear-yield path.
  • φ = 0.75 applies to block shear.
  • Ubs = 1.0 when tension stress is uniform (single line of bolts, symmetric connection).
  • Ubs = 0.5 when tension stress is non-uniform (two or more lines of bolts, eccentric connections).
  • Block shear frequently governs when edge distances are small or when few bolts are used.

Definitions

Term

Block Shear Failure Path

Definition

The combined shear-tension tear-out path. The shear planes are parallel to the applied load direction; the tension plane is perpendicular to the load direction. The block 'tears out' as a unit.

Importance

Correctly identifying the failure block geometry (which plane is shear, which is tension) is prerequisite to any calculation. Sketch the connection first.

Term

Ubs (Uniform Shear Factor)

Definition

A modifier that accounts for non-uniform tension stress distribution in the net tension area. Ubs = 1.0 for connections with uniform tension stress (single bolt line). Ubs = 0.5 for connections where tension is non-uniform (multiple bolt lines, beam-end connections to one flange only).

Importance

Using Ubs = 1.0 when 0.5 is required will overestimate block shear capacity by up to 25-30% — a critical error in design.

Term

Net Area vs Gross Area

Definition

Gross area: full cross-section without deducting holes. Net area: gross area minus the area of bolt holes (using hole diameter = db + 2 mm to account for punching damage). Anv and Ant use net areas; Agv uses gross area.

Importance

The hole deduction uses dh = db + 2 mm (not db) — a 2 mm damage allowance per AISC 360 Section B4.3b.

Section Title

4. Block Shear

Common Mistakes

  • Deducting hole areas from BOTH the shear and tension areas without checking whether the cap (shear yield + tension rupture) governs.
  • Using Ubs = 1.0 for a two-row bolt pattern — check symmetry and load path first.
  • Forgetting to add the tension rupture term (Ubs Fu Ant) to the shear term — some students compute only the shear component.
  • Using db instead of (db + 2) mm for the hole width deduction in net area calculations.

Formulas

Example

20 mm A325, Class A (μ = 0.35), single shear (ns = 1), serviceability check (φ = 1.00): φRn = 1.00(0.35)(1.13)(1.0)(142)(1) = 56.2 kN per bolt.

Formula

φRn = φ × μ × Du × hf × Tb × ns

Variables

φ = 1.00 (serviceability slip) or 0.85 (strength slip); μ = slip coefficient (0.35 Class A, 0.50 Class B); Du = 1.13; hf = hole factor (1.0 standard, 0.85 oversized); Tb = bolt pretension (kN); ns = number of slip planes

Application

Used ONLY for slip-critical connections. Apply per bolt. The problem must specify Class A or Class B surface condition.

Example

20 mm A325 bolt in tension: φRn = 0.75(620)(314.2) = 146,100 N = 146.1 kN.

Formula

φRn = 0.75 × Fnt × Ab

Variables

Fnt = nominal tensile stress of bolt: 620 MPa (A325), 780 MPa (A490); Ab = nominal bolt area (mm²)

Application

Bolt in pure tension (e.g., hanger connections, prying action situations). When combined with shear, use the interaction formula.

Exam Tips

  • If the problem says 'slip-critical,' use the μ Du hf Tb ns formula. If it says 'bearing-type,' use Fnv Ab.
  • Class A (μ = 0.35) is the default assumption for unpainted steel — most board problems specify Class A.
  • Bolt pretension Tb values are given in AISC Table J3.1 — the most common are for 20 mm and 22 mm A325 bolts (142 kN and 176 kN).

Key Points

  • Bearing-type connections allow slip; slip-critical connections develop friction to prevent slip entirely.
  • Slip-critical connections are required for: connections subject to fatigue, oversized holes, connections where slip is unacceptable (precision machinery, bridges).
  • Design slip resistance (serviceability): φRn = φ μ Du hf Tb ns, with φ = 1.00 (serviceability) or 0.85 (strength).
  • μ = mean slip coefficient: 0.35 (Class A, unpainted clean mill scale), 0.50 (Class B, blast-cleaned).
  • Tb = minimum bolt pretension (from AISC Table J3.1): 142 kN for 20 mm A325; 176 kN for 22 mm A325.
  • Du = 1.13 (ratio of mean installed pretension to specified minimum).
  • hf = 1.0 for standard-hole connections (no filler plates).
  • ns = number of slip planes.
  • Bolts in tension: φRn = 0.75 × Fnt × Ab, where Fnt = 620 MPa (A325), 780 MPa (A490).
  • Combined shear + tension (bearing type): use reduced Fnt' = 1.3 Fnt - (Fnt/φFnv) × frv ≤ Fnt.

Definitions

Term

Slip-Critical Connection

Definition

A bolted connection designed to resist applied shear loads through friction between the mating surfaces (clamping force from pretensioned bolts), with no allowable slip at service load levels.

Importance

Distinguish from bearing-type connections on the exam. Slip-critical uses μ, Du, hf, Tb; bearing-type uses Fnv. The problem context (fatigue, oversized holes) signals slip-critical.

Term

Bolt Pretension (Tb)

Definition

The minimum clamping force induced in the bolt by tightening, specified in AISC Table J3.1. For A325 bolts: Tb = 91 kN (16 mm), 142 kN (20 mm), 176 kN (22 mm), 213 kN (24 mm).

Importance

Tb must be memorized for common bolt sizes for slip-critical calculations. It is not computed from first principles in board problems.

Section Title

5. Slip-Critical and Bolt Tension

Common Mistakes

  • Applying the slip-critical formula to ordinary bearing-type connections — read the problem context carefully.
  • Using φ = 0.75 for the serviceability slip check instead of φ = 1.00 (serviceability) or 0.85 (strength-level).
  • Forgetting to check bolt shear and bearing EVEN FOR slip-critical connections — slip-critical classification governs service behavior but strength-level checks still apply.

Formulas

Example

6 mm fillet weld, E70 (FEXX = 482 MPa), L = 200 mm: throat = 0.707(6) = 4.24 mm; φRn = 0.75(0.60)(482)(4.24)(200) = 0.75(122,850) = 92,138 N ≈ 184 kN. Note: per unit length φrn = 0.75(0.60)(482)(4.24) = 920 N/mm.

Formula

φRn = 0.75 × 0.60 × FEXX × 0.707a × L

Variables

φ = 0.75; FEXX = electrode strength (MPa): E70 = 482 MPa, E60 = 414 MPa; 0.60 = shear strength factor on electrode; 0.707 = throat-to-leg ratio for 45° equal-leg fillet weld; a = weld leg size (mm); L = effective weld length (mm)

Application

Use for all fillet weld strength calculations. Can be rearranged to find required weld length L or required leg size a when design load is known.

Example

Carry 250 kN with E70, weld both sides, L_total = 600 mm: a = 250,000 / [0.75(0.60)(482)(0.707)(600)] = 250,000 / 92,138 ≈ 2.7 mm → round up, check min size (say 5 mm for typical plate thickness).

Formula

Required weld leg size: a_req = Pu / [0.75 × 0.60 × FEXX × 0.707 × L]

Variables

Pu = factored applied load (N); other variables same as above; L = total effective weld length (both sides if applicable)

Application

When designing a weld (finding required size), rearrange the basic formula. Always check min/max weld size limits after computing a_req.

Example

8 mm fillet weld: te = 0.707(8) = 5.66 mm effective throat.

Formula

Effective throat: te = 0.707a

Variables

te = effective throat size (mm); a = fillet weld leg size (mm); 0.707 = sin(45°) for equal-leg fillet weld

Application

Always use 0.707a for the throat — never use the full leg size a as the throat in strength calculations.

Exam Tips

  • Memorize φrn per mm for E70, 6 mm fillet weld: φrn = 0.75(0.60)(482)(0.707)(6) = 0.75(0.60)(482)(4.242) ≈ 0.75 × 122,850/200 ≈ 920 N/mm — confirm each time.
  • Quick formula memory: φRn = 0.75 × 0.60 × FEXX × 0.707 × a × L = 0.3182 × FEXX × a × L.
  • If asked to 'size the weld,' find a_req, round up to the next 1 mm increment, then verify a_req ≥ a_min and a_req ≤ a_max.
  • When weld is on both sides of a plate: L_total = 2 × weld length per side — explicitly state this in your solution.

Key Points

  • Fillet welds are the most common weld type in structural steel connections (gussets, brackets, beam-to-column).
  • Fillet weld strength is based on shear rupture through the weld THROAT, not the leg.
  • Effective throat for a 45° equal-leg fillet weld: te = 0.707a, where a = weld leg size (mm).
  • Design fillet weld strength: φRn = 0.75 × 0.60 FEXX × 0.707a × L per weld.
  • FEXX = electrode classification strength: E70 → 482 MPa (70 ksi), E60 → 414 MPa, E80 → 552 MPa.
  • Per unit length: φrn = 0.75 × 0.60 × FEXX × 0.707a (N/mm).
  • Minimum weld size: controlled by thicker plate (AISC Table J2.4): plate t ≤ 6 mm → a_min = 3 mm; 6 < t ≤ 13 mm → 5 mm; 13 < t ≤ 19 mm → 6 mm; t > 19 mm → 8 mm.
  • Maximum weld size: a_max = t - 2 mm for plate t ≥ 6 mm; a_max = t for plate t < 6 mm.
  • Minimum effective weld length: L_eff ≥ 4a (otherwise, use the actual leg as the effective leg).
  • Groove (full-penetration) welds: develop full base-metal strength — no separate weld strength check needed.

Definitions

Term

Fillet Weld Leg Size (a)

Definition

The length of each leg of a right-triangle cross-section fillet weld. For a 45° equal-leg fillet weld, both legs are equal to a. The actual strength is based on the throat (0.707a), not the leg.

Importance

Board examiners specifically test whether you use 0.707a (correct) vs a (incorrect) as the throat dimension. Using a instead of 0.707a overestimates strength by 41%.

Term

FEXX (Electrode Classification Strength)

Definition

The minimum tensile strength of the weld electrode (filler metal). E70XX electrodes have FEXX = 482 MPa (70 ksi). The 'E' prefix and two-digit number give the ksi value (E70 → 70 ksi → 482 MPa).

Importance

E70 (482 MPa) is the most common Philippine board-exam electrode. Memorize the conversion: 70 ksi × 6.895 = 482 MPa.

Term

Groove (Complete Joint Penetration, CJP) Weld

Definition

A weld that fuses through the entire thickness of the connected member. Develops the full strength of the base metal in tension, compression, and shear. No separate weld capacity check is required — base metal governs.

Importance

When a problem specifies a full-penetration groove weld, do NOT apply the fillet weld formula — use the base metal cross-section strength directly.

Section Title

6. Welded Connections — Fillet Welds

Common Mistakes

  • Using the leg size a as the throat instead of 0.707a — ALWAYS use 0.707a for equal-leg fillet welds.
  • Forgetting to multiply by the total weld length L (in mm) when computing total capacity — φrn per mm × L mm = total φRn.
  • Not checking minimum and maximum weld size limits after computing the required size.
  • Counting weld length as the member length when only portions are welded — use effective weld length only.

Formulas

Example

Pu = 500 kN; per-bolt shear = 87.7 kN; per-bolt bearing = 126 kN. Governing = 87.7 kN. n = ⌈500/87.7⌉ = ⌈5.70⌉ = 6 bolts.

Formula

Number of bolts required: n = ⌈Pu / (min φRn per bolt)⌉

Variables

Pu = factored design load (N or kN); min φRn = governing per-bolt design strength (minimum of shear and bearing); ⌈ ⌉ = ceiling function (round up to next integer)

Application

After computing per-bolt shear and bearing capacities, use the smaller to find n. Always round UP — never use a fractional number of bolts.

Exam Tips

  • In a 5-item connection problem, the last sub-question often asks 'which limit state governs?' — this tests whether you checked all states.
  • State your governing limit state explicitly: 'Bolt shear governs: φRn = 87.7 kN per bolt.' This earns full solution credit.
  • If block shear and bolt shear give similar values, double-check your net area calculations — they're likely the discriminating factor.

Key Points

  • Step 1 — Identify connection type: bolted (bearing-type or slip-critical) or welded (fillet or groove).
  • Step 2 — Identify ALL applicable limit states based on connection geometry and loading.
  • Step 3 — Compute φRn for each applicable limit state.
  • Step 4 — The governing design strength = MINIMUM of all φRn values.
  • Step 5 — For design (finding number of bolts or weld size): n = Pu / (governing φRn per bolt), round UP.
  • For bolted connections, always check: (a) bolt shear, (b) bearing on connected plate, (c) bearing on connecting plate, (d) block shear, (e) net-section tension of member.
  • For welded connections, always check: (a) weld throat shear rupture, (b) base metal shear yielding, (c) base metal shear rupture.
  • The connection capacity is always the MINIMUM — never average or sum the limit states.
  • φ = 0.75 for ALL connection limit states listed in AISC 360-16 Chapter J.

Definitions

Term

Governing Limit State

Definition

The limit state that yields the LOWEST design strength (φRn) and therefore controls the connection capacity. The connection must be designed so that its governing capacity ≥ factored load Pu.

Importance

The single most important concept in connection design — always identify and explicitly state the governing limit state in your solution.

Section Title

7. Summary of Limit States and Design Procedure

Common Mistakes

  • Using the largest φRn (optimistic) instead of the smallest (governing) as the connection capacity.
  • Rounding the number of bolts DOWN instead of UP — a fractional bolt is impossible, and rounding down is unconservative.
  • Checking only bolt shear and calling it the connection capacity without checking bearing and block shear.

Connections

  • Steel Connections → Member Design: Connection capacity must be compatible with the member's tensile (net section and gross section) and compressive capacities; a well-designed member can be rendered inadequate by an undersized connection.
  • Block Shear → Net Section Tension: Both limit states require computing net areas (deducting hole areas using dh = db + 2 mm); the computational procedure is identical — mastering one helps with the other.
  • Bolt Bearing → Plate Geometry: Clear distance lc directly controls tearout; this links back to detailing requirements (minimum edge distances and bolt spacing) specified in AISC 360 Table J3.4.
  • Fillet Weld Size → Plate Thickness: Minimum and maximum weld sizes are controlled by plate thickness, linking connection design to member selection and plate sizing.
  • Slip-Critical Connections → Serviceability Design: Slip resistance uses service-level loads in some formulations — connecting this chapter to the broader LRFD serviceability versus strength design philosophy.
  • AISC 360-16 Chapter J → NSCP 2015 Section 502: Philippine practice adopts AISC 360 by reference; RA 544 requires the Registered Civil Engineer to comply with the current edition of applicable codes.
  • Weld Strength → Electrode Selection: FEXX links weld design to materials science — matching electrode strength to base metal strength (matching or under-matching) is a practical design decision.
  • Connection Limit States → Structural Redundancy: The 'minimum governs' principle reflects the load path — forces route through the weakest link, which is why all limit states must be checked.

Exam Strategy

Steel Connections is a consistently high-yield topic in the PRC Civil Engineer Licensure Exam, typically appearing as 3–6 items per exam. Follow this strategy: (1) MEMORIZE key formulas — φRn = 0.75 Fnv Ab (bolt shear), φRn = 0.75 min(1.2 lc t Fu, 2.4 db t Fu) (bearing), φRn = 0.75(0.60 FEXX)(0.707a)L (fillet weld). Write these on scratch paper at the start of the exam. (2) KNOW key constants — A325-N: Fnv = 372 MPa; A325-X: Fnv = 469 MPa; E70: FEXX = 482 MPa; φ = 0.75 for all connection limit states. (3) SKETCH the connection before computing — counting shear planes (single vs double shear) and identifying the block shear failure path from a sketch prevents systematic errors. (4) CHECK ALL LIMIT STATES — bolt shear, bearing (edge and interior bolts separately), and block shear for bolted connections; throat rupture, base metal shear for welded connections. Report the MINIMUM as the governing capacity. (5) ROUND BOLTS UP — when finding required number of bolts, always round the decimal answer upward to the next whole number. (6) WELD THROAT — never forget 0.707a; using a instead of 0.707a is the single most common weld calculation error. (7) ALLOCATE TIME WISELY — a complete 5-limit-state connection problem takes 8–12 minutes; if time is tight, prioritize bolt shear and bearing (most commonly tested) before block shear.

Quick Review Questions

A 22 mm A325-N bolt (Fnv = 372 MPa) is in double shear. What is its design shear strength in kN?

Ab = π/4(22²) = 380.1 mm². For double shear, multiply by 2. φRn = 0.75(372)(380.1)(2) = 212,700 N = 212.7 kN. Note: A325-N means threads are IN the shear plane → Fnv = 372 MPa (conservative).

A 20 mm bolt bears on a 12 mm plate (Fu = 410 MPa) with clear edge distance lc = 28 mm. Compute the design bearing strength. Deformation at hole IS a design consideration.

Since lc = 28 mm < 2db = 40 mm, tearout governs (1.2 lc t Fu < 2.4 db t Fu). Apply φ = 0.75 to the smaller value. Design bearing = 124.0 kN per bolt.

A 8 mm fillet weld using E70 electrodes (FEXX = 482 MPa) has an effective length of 150 mm. Find the design strength of the weld.

Throat = 0.707(8) = 5.656 mm. Shear stress on throat = 0.60 FEXX = 0.60(482) = 289.2 MPa. φRn = 0.75 × 289.2 × 5.656 × 150 = 0.75 × 245,600 = 184,200 N. Re-check: 0.75(0.60)(482)(0.707)(8)(150) = 0.75(1,640 N/mm)(150 mm)... Let me recompute: 0.60(482) = 289.2; 289.2 × 0.707 × 8 = 289.2 × 5.656 = 1,635.8 N/mm; × 150 mm = 245,370 N; × 0.75 = 184,028 N ≈ 184.0 kN.

In block shear, what is Ubs and when is Ubs = 0.5 used?

The Ubs factor reduces the tension component when the tension stress is not uniformly distributed across the net tension area. In standard gusset plate connections with a single bolt row, Ubs = 1.0. For beam-end connections with bolts in the web only (wide-flange sections), Ubs = 0.5 per AISC 360-16 Section J4.3.

What is the minimum and maximum fillet weld size for a 16 mm thick plate connected to a 10 mm thick plate?

Per AISC 360-16 Table J2.4: minimum weld size is determined by the THICKER plate to ensure adequate heat input for fusion. Maximum weld size is determined by the THINNER plate to prevent burning through or excessive weld. The weld leg must satisfy both limits simultaneously.

A connection has the following computed design strengths: Bolt shear = 320 kN, Bearing = 280 kN, Block shear = 350 kN, Net section tension = 420 kN. What is the design strength of the connection and which limit state governs?

The connection design strength is ALWAYS the MINIMUM of all applicable limit states. Here: min(320, 280, 350, 420) = 280 kN → Bearing governs. The connection can carry a maximum factored load Pu ≤ 280 kN.

How many 22 mm A490-N bolts (Fnv = 457 MPa) in single shear are needed to carry a factored load of 900 kN? Use bolt shear only.

Single shear: no factor of 2. φRn = 0.75 × 457 × 380.1 = 130,290 N ≈ 130.3 kN per bolt. Required bolts: 900/130.3 = 6.91 → round UP to 7 bolts (never round down for safety).

What electrode classification strength (FEXX) corresponds to E70XX electrodes in SI units?

The classification number in the electrode designation (E70) gives the tensile strength in ksi. Converting: 70 ksi × 6.895 = 482.7 MPa, rounded to 482 MPa. E60 → 414 MPa; E80 → 552 MPa. E70 (482 MPa) is the most common electrode for structural steel in Philippine practice.

For a slip-critical connection with 20 mm A325 bolts in single shear on a Class B surface (μ = 0.50), find the design slip resistance per bolt at the serviceability level. Use Du = 1.13, hf = 1.0, Tb = 142 kN.

For serviceability-level slip check, φ = 1.00. Class B surface: μ = 0.50. Single shear: ns = 1. φRn = 1.00(0.50)(1.13)(1.0)(142)(1) = 80.23 kN per bolt. This is significantly lower than the bearing-type bolt shear capacity (87.7 kN for A325-N), illustrating that slip-critical does not always give higher capacity.

In block shear, the computed rupture path value is 0.60 Fu Anv + Ubs Fu Ant = 480 kN, but the yield cap gives 0.60 Fy Agv + Ubs Fu Ant = 390 kN. Which value is used for Rn?

AISC 360 Section J4.3 states that block shear Rn = 0.60 Fu Anv + Ubs Fu Ant BUT NOT GREATER THAN 0.60 Fy Agv + Ubs Fu Ant. The 'not greater than' clause is a cap on the rupture path. Since 480 kN > 390 kN, the cap governs: Rn = 390 kN, and φRn = 0.75(390) = 292.5 kN.

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