CELE Steel & Timber Design — Steel ConnectionsDetailed Explanation
If the summary was not enough, this is the deep dive. Detailed explanations for Steel Connections in the CELE Steel & Timber Design context, written to turn surface familiarity into genuine understanding. Professional Regulation Commission (PRC) — Board of Civil Engineering's toughest CELE questions on this chapter are answered by the reasoning built here.
Exam context
Professional Regulation Commission (PRC) — Board of Civil Engineering runs the Civil Engineer Licensure Examination on May and November 2026. Its Steel & Timber Design section sits under a "Core" weighting, and Steel Connections is the 4th chapter in the 5-chapter CELE Steel & Timber Design rotation. The CELE passing mark is 70% weighted average, no sub-test below 50%, and the most recent 2026 paper drew about a meaningful share of questions from Steel & Timber Design.
Steel Connections - Detailed Explanation
Steel connections are the critical junctions where individual structural members transfer forces to one another, and they represent one of the most heavily tested topics in the PRC Civil Engineer Licensure Examination under Steel and Timber Design. Unlike member design, where a single check often governs, connection design requires the engineer to evaluate multiple limit states simultaneously — bolt shear, bearing, block shear, weld throat rupture, and net-section fracture — and adopt the minimum capacity as the governing design strength. The Philippine practice references AISC 360-16 (adopted by NSCP 2015, Section 502) for structural steel, using the Load and Resistance Factor Design (LRFD) format with a uniform resistance factor φ = 0.75 for connection limit states. This chapter covers the two primary fastener systems — bolts and welds — along with the block shear failure mechanism, providing complete worked solutions and board-exam strategies.
Concepts
Bolted Connections — Bolt Shear Strength
A bolt in shear resists a force that tends to slide one connected plate relative to another across the bolt's cross-section. The failure plane passes through the bolt shank (or threads, depending on configuration). The nominal shear strength of a single bolt is: Rn = Fnv × Ab where: • Fnv = nominal shear stress of the bolt (MPa), tabulated in AISC 360-16 Table J3.2 • Ab = nominal (unthreaded) cross-sectional area of the bolt (mm²) = π/4 × db² The LRFD design (factored) strength is: φRn = 0.75 × Fnv × Ab For ASTM A325 bolts: • Threads IN shear plane (A325-N): Fnv = 372 MPa • Threads EXCLUDED from shear plane (A325-X): Fnv = 469 MPa For ASTM A490 bolts (higher-strength): • A490-N: Fnv = 457 MPa • A490-X: Fnv = 579 MPa SHEAR PLANES — THE MOST COMMON BOARD-EXAM TRAP: • Single shear: one shear plane passes through one bolt cross-section. Example: a lap splice plate — the bolt sees force from one side only. • Double shear: two shear planes pass through the bolt. Example: a bolt connecting a web plate (main member) sandwiched between two side plates (covers). The design strength doubles because two cross-sections resist the force. φRn (double shear) = 2 × 0.75 × Fnv × Ab For a group of n identical bolts, the total design shear strength is n × φRn (assuming equal load distribution, which is acceptable for most board problems; eccentric bolt groups require additional analysis).
Examples
The key is identifying 'lap-splice' as single shear (one shear plane) and using A325-N (threads in shear plane) with Fnv = 372 MPa. Double shear simply multiplies the result by 2 because two cross-sections of the bolt resist the applied force simultaneously.
Scenario
A single 20 mm diameter A325-N bolt is used in a lap-splice connection (single shear). Determine the LRFD design shear strength of the bolt.
Solution
Step 1 — Bolt area: Ab = π/4 × (20)² = π/4 × 400 = 314.16 mm² Step 2 — Design shear strength (φ = 0.75, Fnv = 372 MPa for A325-N): φRn = 0.75 × 372 × 314.16 φRn = 0.75 × 116,867.5 φRn = 87,651 N ≈ 87.65 kN per bolt If this were double shear: φRn = 2 × 87.65 = 175.3 kN per bolt
A325-X bolts are stronger than A325-N because excluding threads from the shear plane means the full unthreaded shank area resists shear with a higher allowable stress. The 'X' suffix signals this in exam problems. Double shear combined with 4 bolts gives a large total capacity — but remember, this must still be compared to bearing and block shear.
Scenario
Four 22 mm A325-X bolts connect a bracket to a column web in double shear. Find the total design bolt shear capacity of the connection.
Solution
Step 1 — Bolt area: Ab = π/4 × (22)² = π/4 × 484 = 380.13 mm² Step 2 — A325-X: Fnv = 469 MPa (threads excluded from shear plane) Step 3 — Design strength per bolt (double shear): φRn = 2 × 0.75 × 469 × 380.13 φRn = 2 × 133,346 = 266,693 N = 266.7 kN per bolt Step 4 — Total for 4 bolts: Total φRn = 4 × 266.7 = 1,066.8 kN
Applications
- Lap-splice connections in tension members (single shear)
- Shear tab and double-angle connections to beam webs (single or double shear)
- Gusset plate connections at trusses (typically single shear)
- Moment connections with bolted flanges (bolts in single shear per plate)
- Column splices using cover plates (bolts in double shear)
Misconceptions
- Using Fnv for A325-X when the problem says threads are in the shear plane (A325-N) — read the problem carefully
- Forgetting to multiply by 2 for double shear — the most common arithmetic error on board exams
- Using the root diameter of bolt threads instead of the nominal shank diameter for Ab
- Assuming all bolts in an eccentric group carry equal shear — only valid for concentric loading
- Confusing bolt shear strength with slip-critical strength — slip-critical uses a different formula based on clamping force
Related Concepts
- Bearing strength at bolt holes
- Slip-critical connections
- Net section fracture of connected member
- Block shear failure
- Bolt installation and pretensioning (snug-tight vs. fully pretensioned)
Common Exam Questions
Example
Given: 6 bolts, 24 mm A325-N, double shear. Find total bolt shear capacity. → Ab = π/4(24²) = 452.4 mm²; φRn per bolt = 2 × 0.75 × 372 × 452.4 = 252,840 N; Total = 6 × 252.84 = 1,517 kN
Approach
Identify bolt diameter, bolt grade (A325/A490) and thread condition (N or X), count shear planes, apply φRn = 0.75 × Fnv × Ab × (number of shear planes)
Question Type
Direct computation of bolt shear capacity
Example
Pu = 500 kN, 20 mm A325-N, single shear. φRn per bolt = 87.65 kN. n = 500/87.65 = 5.71 → use 6 bolts
Approach
Compute design factored load Pu; divide by single-bolt φRn to get minimum number of bolts; round UP to next integer
Question Type
Minimum number of bolts required
Key Points To Remember
- φ = 0.75 for all connection shear limit states (AISC J3.6)
- A325-N uses Fnv = 372 MPa; A325-X uses Fnv = 469 MPa — memorize both values
- Always count shear planes — single shear gives 1× capacity, double shear gives 2×
- Ab uses the NOMINAL (unthreaded) bolt diameter, NOT the root diameter of threads
- For a bolt group, multiply single-bolt capacity by the number of bolts (uniform-load assumption)
- A325 bolts are the most common in Philippine board problems; A490 bolts appear occasionally
Bearing Strength at Bolt Holes
When a bolt transfers load to a plate, the bolt bears against the edge of the hole, compressing the plate material and potentially causing either localized crushing of the plate (bearing failure) or tearing out of the plate material toward a free edge (tearout). AISC 360-16 Section J3.10 addresses both failure modes through a single unified formula that governs by the smaller of two checks: Rn = min(1.2 lc t Fu, 2.4 db t Fu) with φ = 0.75 where: • lc = clear distance in the direction of force, measured from the edge of the hole to the nearest edge of an adjacent hole or to the free edge of the material (mm) • t = thickness of the plate being checked (mm) • Fu = ultimate tensile strength of the plate material (MPa); for A36 steel Fu = 400 MPa; for A572 Gr.50, Fu = 450 MPa • db = nominal bolt diameter (mm) THE TWO TERMS EXPLAINED: 1. 1.2 lc t Fu — This is the TEAROUT (edge/between-holes) limit. It depends on lc, so it governs when bolts are placed close to an edge or close together (small lc). Increasing edge distance reduces tearout risk. 2. 2.4 db t Fu — This is the CRUSHING (bearing) limit. It depends on the bolt diameter and plate thickness. It caps the tearout term, preventing overestimation when lc is large. WHEN IS EACH TERM LIKELY TO GOVERN? • Small lc (short edge distance or closely spaced bolts) → 1.2 lc t Fu governs (tearout controls) • Large lc (generous edge distance, bolts well-spaced) → 2.4 db t Fu governs (crushing controls) ALTERNATE CASE (Deformation not a concern): When bolt-hole deformation at service load is NOT a design consideration (e.g., connections not subject to repeated loading), AISC permits the higher pair: Rn = min(1.5 lc t Fu, 3.0 db t Fu) Board problems usually specify which case to use. If silent, use the standard (deformation IS a concern) pair: 1.2 and 2.4. BEARING ON MULTIPLE PLATES: For a bolt passing through multiple plates, check bearing on each plate separately using its own t and Fu. The critical (minimum) bearing strength per bolt controls. CLEAR DISTANCE lc CALCULATION: • Edge bolt: lc = Le − (dh/2), where Le = edge distance, dh = hole diameter = db + 2 mm (standard oversize) • Interior bolt: lc = s − dh, where s = center-to-center bolt spacing
Examples
The tearout term governs because lc = 29 mm is relatively small. Notice that had lc been larger (say 60 mm), the tearout term would be 1.2 × 60 × 10 × 400 = 288,000 N > 192,000 N — then the crushing limit (192,000 N) would have governed instead.
Scenario
A 20 mm bolt is located at an edge distance Le = 40 mm from the free edge of a 10 mm plate (Fu = 400 MPa). The bolt hole is standard (dh = 22 mm). Check the bearing strength at this bolt.
Solution
Step 1 — Clear distance lc (edge bolt): lc = Le − dh/2 = 40 − 22/2 = 40 − 11 = 29 mm Step 2 — Two bearing limits: Tearout: 1.2 × lc × t × Fu = 1.2 × 29 × 10 × 400 = 139,200 N Crushing: 2.4 × db × t × Fu = 2.4 × 20 × 10 × 400 = 192,000 N Step 3 — Governing Rn (minimum): Rn = min(139,200; 192,000) = 139,200 N Step 4 — Design bearing strength: φRn = 0.75 × 139,200 = 104,400 N = 104.4 kN
With a generous bolt spacing (70 mm), lc = 48 mm is large enough that tearout > crushing, so the crushing limit (2.4 db t Fu) governs. This is the more common outcome for properly designed connections with adequate bolt spacing.
Scenario
An interior bolt (20 mm diameter) in a row has center-to-center spacing s = 70 mm on a 12 mm plate (Fu = 450 MPa). Find the design bearing strength at this bolt.
Solution
Step 1 — Hole diameter: dh = 20 + 2 = 22 mm Step 2 — Clear distance lc (interior bolt): lc = s − dh = 70 − 22 = 48 mm Step 3 — Two bearing limits: Tearout: 1.2 × 48 × 12 × 450 = 311,040 N Crushing: 2.4 × 20 × 12 × 450 = 259,200 N Step 4 — Governing Rn: Rn = min(311,040; 259,200) = 259,200 N Step 5 — Design strength: φRn = 0.75 × 259,200 = 194,400 N = 194.4 kN
Applications
- Checking plate thickness adequacy in gusset plate connections
- Sizing bolt edge distances and spacings for adequate bearing capacity
- Evaluating existing connections during building assessment or renovation projects
- Thin web connections where bearing on the web is often the critical limit state
- Combined shear-and-tension connections where bearing remains a separate check
Misconceptions
- Using center-to-center spacing instead of clear distance lc for the tearout check — always subtract hole diameter from spacing, and hole radius from edge distance
- Using Fy instead of Fu for bearing — this is an ultimate (rupture) limit state, always use Fu
- Forgetting the 2.4 db t Fu cap — some reviewees only compute the tearout term and report it without checking the cap
- Applying the same t for all bolts when they pass through plates of different thicknesses — check each plate individually
- Using db + 3 mm for hole diameter (used only for oversized holes) instead of the standard db + 2 mm
Related Concepts
- Bolt shear strength
- Net section fracture (reduced area at holes)
- Minimum edge distances and bolt spacings (AISC Table J3.4)
- Plate tearout and block shear
- Shear lag in connected elements
Common Exam Questions
Example
24 mm bolt, Le = 35 mm, t = 10 mm, Fu = 400 MPa. dh = 26 mm. lc = 35−13 = 22 mm. 1.2(22)(10)(400)=105,600; 2.4(24)(10)(400)=230,400. Rn = 105,600. φRn = 0.75(105,600) = 79,200 N = 79.2 kN
Approach
Calculate lc (clear distance), evaluate both terms (1.2 lc t Fu and 2.4 db t Fu), take the minimum, multiply by 0.75
Question Type
Compute bearing strength given edge or spacing distance
Example
φRn(shear) = 87.65 kN; φRn(bearing) = 104.4 kN → bolt shear governs at 87.65 kN per bolt
Approach
Compute φRn for bolt shear AND φRn for bearing; the smaller value governs the connection capacity per bolt
Question Type
Govern the connection by comparing bolt shear and bearing
Key Points To Remember
- The governing bearing strength is the MINIMUM of the tearout and crushing limits — never skip the min() check
- lc is the CLEAR distance, not center-to-center — subtract the hole radius from edge distance or the full hole diameter from spacing
- Standard bolt hole diameter = db + 2 mm (e.g., 20 mm bolt → 22 mm hole) per AISC J3.2
- φ = 0.75 for bearing, same as bolt shear
- Check bearing on all plates the bolt passes through; the weakest plate governs
- For a bolt group, sum the bearing capacities of all bolts if lc differs from bolt to bolt
- Fu of the PLATE is used, not Fy — this is an ultimate (rupture) limit state
Block Shear Failure
Block shear is a failure mode unique to connections where a bolt group tears out a block of material — the failure path involves SIMULTANEOUS shear along one or more lines parallel to the applied load AND tension rupture along a line perpendicular to the load. This most commonly occurs at: • Angle or gusset plate connections • Beam end connections (the beam web tears out) • Tension member end connections with staggered bolt patterns Think of it physically: imagine pulling a comb — the 'block' between the bolt holes and the plate edge tears along both shear surfaces (vertical) and a tension surface (horizontal) at the same time. AISC 360-16 Section J4.3 gives the nominal block shear strength as: Rn = 0.6 Fu Anv + Ubs Fu Ant ≤ 0.6 Fy Agv + Ubs Fu Ant Design strength: φRn, with φ = 0.75 where: • Agv = gross area subject to shear (mm²) — along the shear failure path, full area including holes • Anv = net area subject to shear (mm²) — gross shear area minus hole areas along shear path • Ant = net area subject to tension (mm²) — net area along the tension failure path • Ubs = 1.0 when tension stress is UNIFORM (most common case: gusset-to-member connections) • Ubs = 0.5 when tension stress is NON-UNIFORM (e.g., beam-end connections where only the web is connected — rarely tested directly) • Fy = yield stress of the plate material (MPa) • Fu = ultimate tensile strength of the plate material (MPa) UNDERSTANDING THE TWO-TERM FORMULA: • First term: 0.6 Fu Anv — shear RUPTURE along the shear path (ultimate failure) • Second term: Ubs Fu Ant — tension RUPTURE along the tension path (ultimate failure) • The upper bound: 0.6 Fy Agv + Ubs Fu Ant — replaces shear rupture with shear YIELD (gross area at Fy) when shear yield is less than shear rupture on the net area The inequality means: shear rupture (net) may be limited by shear yield (gross) on the shear planes. Take the SMALLER of the two expressions. HOW TO IDENTIFY THE AREAS: 1. Draw the failure path — the block being torn out 2. Shear path(s): parallel to the force, through bolt holes → compute Agv (all material), then subtract hole areas to get Anv 3. Tension path: perpendicular to force, through bolt holes → subtract half-holes at each end → compute Ant
Examples
The upper bound (shear yield on gross area) governs over shear rupture on net area in this case, giving a lower block shear capacity. Always check both expressions. Note the tension area uses the edge distance minus half the hole diameter — a very common calculation error on board exams.
Scenario
A gusset plate (t = 12 mm, Fy = 248 MPa, Fu = 400 MPa) has a single row of three 22 mm A325-N bolts in the direction of load. Bolt pitch = 75 mm, edge distance Le = 38 mm (in force direction), and the tension edge distance = 35 mm. Find the block shear design strength.
Solution
Step 1 — Hole diameter for net area: dh = 22 + 4 = 26 mm (using AISC J4 provision for punched holes, standard practice for board exams: add 2 mm to standard hole for damage) Step 2 — Shear areas (along the bolt line, parallel to force): Length of shear path = 2 × pitch + Le = 2(75) + 38 = 188 mm Agv = 188 × 12 = 2,256 mm² Anv = Agv − (2.5 holes × 26 × 12) [2.5 holes: at each bolt there is a half-hole on each side, but along the shear path from edge to last bolt, there are 2 full holes and 1 half-hole = 2.5 total] Anv = 2,256 − (2.5 × 26 × 12) = 2,256 − 780 = 1,476 mm² Step 3 — Tension area (perpendicular to force, at far end of bolt group): Ant = (tension edge distance − half hole) × t Ant = (35 − 13) × 12 = 22 × 12 = 264 mm² Step 4 — Block shear (Ubs = 1.0, uniform tension): Expression 1: 0.6 Fu Anv + Ubs Fu Ant = 0.6(400)(1,476) + 1.0(400)(264) = 354,240 + 105,600 = 459,840 N Expression 2 (upper bound): 0.6 Fy Agv + Ubs Fu Ant = 0.6(248)(2,256) + 1.0(400)(264) = 335,693 + 105,600 = 441,293 N Step 5 — Governing Rn = min(459,840; 441,293) = 441,293 N Step 6 — Design strength: φRn = 0.75 × 441,293 = 331,000 N = 331.0 kN
Applications
- Design of gusset plates at truss panel points
- Beam end connections where the web tears out
- Tension member connections with single or double angles
- Splice plate connections where tearout of the bolt group must be checked
- Column base plate anchor bolt arrangements (tension tearout check)
Misconceptions
- Using db + 2 mm (standard hole) instead of db + 4 mm for net area calculations in block shear — AISC J4 requires an additional 2 mm for punching damage
- Counting full holes along the shear path instead of correctly accounting for fractional holes at the tension path intersection
- Forgetting that BOTH expressions must be evaluated — some reviewees only compute one
- Setting Ubs = 0.5 for all problems — only use 0.5 for non-uniform tension stress distribution (partial web connections)
- Computing block shear for the bolt, not the plate — block shear is always a PLATE/MEMBER limit state, not a bolt limit state
Related Concepts
- Net section fracture of tension members
- Shear lag factor U
- Bolt bearing strength
- Gusset plate design
- Connection geometry: edge distances and bolt spacing
Common Exam Questions
Example
2-bolt row, 22 mm bolts, pitch = 70 mm, Le = 40 mm (force dir.), tension edge = 40 mm, t = 10 mm, Fy = 248, Fu = 400. Shear path length = 70+40=110mm. Agv=1100. Anv=1100−1.5(26)(10)=710. Ant=(40−13)(10)=270. Expr1=0.6(400)(710)+1.0(400)(270)=170,400+108,000=278,400. Expr2=0.6(248)(1100)+108,000=163,680+108,000=271,680. φRn=0.75(271,680)=203,760 N=203.8 kN
Approach
Identify shear path (parallel to force) and tension path (perpendicular), compute Agv, Anv, Ant using correct hole sizes, evaluate both block shear expressions, take the minimum, multiply by φ = 0.75
Question Type
Determine block shear strength of a bolt group in a gusset plate
Key Points To Remember
- Block shear involves SIMULTANEOUS shear + tension failure — both paths must be identified and their areas computed
- Ubs = 1.0 for uniform tension (standard gusset connections); Ubs = 0.5 for non-uniform tension (partial web connections)
- φ = 0.75 for block shear, same as all other connection limit states
- The two expressions in the inequality differ only in the SHEAR term (net area rupture vs. gross area yield); the tension term (Ubs Fu Ant) is identical in both
- Hole diameter for area calculation = db + 2 mm (standard) + 2 mm additional for damage during punching = db + 4 mm per AISC J4 for net area calculation
- The block shear limit state is always checked IN ADDITION TO bolt shear and bearing — all three must be evaluated
- Agv = (number of bolt spacings on shear path × pitch) + edge distance, multiplied by plate thickness
Welded Connections — Fillet Weld Strength
Welds provide a continuous force transfer between steel elements without requiring holes. The most common weld type in practice and on board exams is the FILLET WELD — a roughly triangular cross-section weld placed at the intersection of two plates meeting at a right angle (or close to it). FILLET WELD GEOMETRY: • Weld size a (or w) = the leg size of the fillet, measured from the root to the face of the weld along the plate surface • Throat = the perpendicular distance from the root to the hypotenuse face = 0.707a (for a 45°, equal-leg fillet weld) • Failure plane: fillet welds always fail through the THROAT, not along the leg DESIGN STRENGTH — FILLET WELD: The nominal shear stress on the weld throat is limited to 0.60 FEXX: Rn per unit length = 0.60 FEXX × 0.707a Design strength per unit length = φRn = 0.75 × 0.60 FEXX × 0.707a (N/mm) For total weld length L: φRn (total) = 0.75 × 0.60 × FEXX × 0.707a × L where: • FEXX = electrode classification strength (MPa) - E70 electrodes: FEXX = 482 MPa (most common in Philippine practice) - E60 electrodes: FEXX = 414 MPa - E80 electrodes: FEXX = 552 MPa • a = weld size (leg) in mm • L = effective weld length in mm • φ = 0.75 MINIMUM AND MAXIMUM WELD SIZES (AISC Table J2.4): • Minimum weld size depends on the THICKER of the two joined plates: - t ≤ 6 mm plate: min weld size = 3 mm - 6 < t ≤ 13 mm: min weld size = 5 mm - 13 < t ≤ 19 mm: min weld size = 6 mm - t > 19 mm: min weld size = 8 mm • Maximum weld size: - Along a plate edge ≥ 6 mm thick: weld size ≤ t − 2 mm (leave 2 mm gap) - Along a plate edge < 6 mm thick: weld size ≤ t MINIMUM EFFECTIVE WELD LENGTH: L ≥ 4a (otherwise use L = 4a for calculation) GROOVE (FULL-PENETRATION BUTT) WELDS: A complete joint penetration (CJP) groove weld has the SAME design strength as the base metal — it is considered as strong as the plate itself. No separate weld strength check is needed; the member capacity governs. PARTIAL JOINT PENETRATION (PJP) GROOVE WELDS: Throat = effective throat te (depends on process and groove angle); design as 0.60 FEXX on the throat, similar to a fillet weld.
Examples
The key insight is that 'both sides' means two separate welds, so total effective length is 400 mm. The throat factor 0.707 converts leg size to throat size. Using the leg size (6 mm) directly instead of the throat (4.24 mm) would overestimate the capacity by 41% — a critical error.
Scenario
A 6 mm fillet weld, 200 mm long, on each side of a plate (two welds total), uses E70 electrodes (FEXX = 482 MPa). Find the total design weld strength.
Solution
Step 1 — Weld throat: throat = 0.707 × 6 = 4.24 mm Step 2 — Design strength per unit length: φRn per mm = 0.75 × 0.60 × 482 × 4.24 = 0.75 × 0.60 × 2,043.7 = 0.75 × 1,226.2 = 919.6 N/mm Step 3 — Total weld length (both sides): L_total = 2 × 200 = 400 mm Step 4 — Total design strength: φRn = 919.6 × 400 = 367,840 N = 367.8 kN
This two-part problem tests both the tabulated weld size limits and the strength formula. In practice, the engineer selects a weld size between the minimum (6 mm) and maximum (13 mm) based on the required capacity. Minimum size is governed by preventing rapid cooling and cracking; maximum size prevents burning through the plate edge.
Scenario
Determine the minimum weld size and find the design capacity per mm for an E70 fillet weld on a 15 mm plate (F = 0.75, FEXX = 482 MPa).
Solution
Step 1 — Minimum weld size for 15 mm plate (13 < t ≤ 19 mm): Min a = 6 mm (from AISC Table J2.4) Step 2 — Design strength per unit length (a = 6 mm): φRn/L = 0.75 × 0.60 × 482 × 0.707 × 6 = 0.75 × 0.60 × 482 × 4.243 = 0.75 × 1,226.5 = 919.9 N/mm ≈ 920 N/mm Step 3 — Maximum weld size (plate edge ≥ 6 mm): Max a = 15 − 2 = 13 mm
This is the inverse design problem — finding L given a required capacity. Always check the minimum weld length requirement (L ≥ 4a) as a sanity check. In practice, the total 204 mm might be distributed over two weld lines (102 mm each) for a balanced connection.
Scenario
A double angle tension member is welded along its length to a gusset plate. The required weld capacity is 250 kN. Using 8 mm E70 fillet welds, find the required total weld length.
Solution
Step 1 — Design strength per unit length for 8 mm weld: φRn/L = 0.75 × 0.60 × 482 × 0.707 × 8 = 0.75 × 0.60 × 482 × 5.656 = 0.75 × 1,635.5 = 1,226.6 N/mm = 1.2266 kN/mm Step 2 — Required weld length: L_req = 250 kN ÷ 1.2266 kN/mm = 203.8 mm ≈ 204 mm Verify: L ≥ 4a = 4(8) = 32 mm ✓ (204 mm >> 32 mm)
Applications
- Welding shear tabs, double angles, and seat angles to beam webs and flanges
- Gusset plate connections at truss joints
- Welded moment connections (CJP groove welds at flanges, fillet welds at web)
- Column base plates welded to column flanges and webs
- Built-up girder fabrication (fillet welds connecting web to flanges)
Misconceptions
- Using the weld leg size (a) instead of the throat (0.707a) — inflates capacity by ~41%
- Forgetting to multiply by 2 for welds on both sides of a plate
- Using FEXX = 500 MPa — E70 is 482 MPa (70 ksi = 482 MPa); 500 MPa is not a standard electrode rating
- Applying CJP groove weld check as a weld strength — CJP welds are as strong as the base metal; check the member, not the weld
- Ignoring minimum weld size — a weld that is too small can crack during cooling due to high heat gradient
Related Concepts
- Groove (butt) welds — CJP and PJP
- Weld symbols and specifications on engineering drawings
- Heat-affected zone (HAZ) and weld quality
- Electrode selection and matching base metal (AISC prequalified weld procedures)
- Eccentric welds (weld groups with moment) — advanced topic
Common Exam Questions
Example
10 mm fillet weld, 300 mm long, E70. φRn = 0.75(0.60)(482)(0.707×10)(300) = 0.75(0.60)(482)(7.07)(300) = 459,730 N = 459.7 kN
Approach
Compute throat = 0.707a; apply φRn = 0.75 × 0.60 × FEXX × throat × L_total; ensure L ≥ 4a
Question Type
Compute total design weld capacity given size and length
Example
Need 300 kN with L = 250 mm, E70. Required: 0.75(0.60)(482)(0.707a)(250) ≥ 300,000. Solve: a ≥ 300,000/[0.75(0.60)(482)(0.707)(250)] = 300,000/38,318 = 7.83 mm → use 8 mm
Approach
Set φRn ≥ Pu; solve for a (given L) or L (given a); check min/max size limits
Question Type
Size a fillet weld to carry a given load
Key Points To Remember
- Fillet weld failure is always through the THROAT (0.707a), NOT the leg — using leg size instead of throat is the #1 weld error on board exams
- φ = 0.75 for fillet welds; nominal shear stress on throat = 0.60 FEXX
- E70 electrodes are standard Philippine practice: FEXX = 482 MPa
- Design strength per mm = 0.75 × 0.60 × 482 × 0.707 × a = 153.3a N/mm for E70
- CJP groove welds develop full base metal strength — no weld strength check needed
- Weld size limitations (min and max) are frequently tested as part of design problems
- When a weld is placed on 'both sides' (two welds), multiply total weld length L by 2
Connection Design Philosophy — Governing Limit State
The fundamental principle in connection design is that the SMALLEST capacity among all relevant limit states is the design capacity of the connection. This is the direct application of the LRFD philosophy: φRn ≥ Pu for every limit state. FOR A TYPICAL BOLTED CONNECTION, CHECK: 1. Bolt shear: φRn = 0.75 Fnv Ab × (shear planes) × n 2. Bearing on each plate: φRn = 0.75 min(1.2 lc t Fu, 2.4 db t Fu) × n 3. Block shear: φRn = 0.75 [min(0.6 Fu Anv, 0.6 Fy Agv) + Ubs Fu Ant] 4. Net section fracture of connected member: φRn = 0.75 Fu Ae (where Ae = U × An) 5. Gross section yielding of connected member: φRn = 0.90 Fy Ag The MINIMUM of all these is the connection's design strength. FOR A TYPICAL WELDED CONNECTION, CHECK: 1. Fillet weld strength: φRn = 0.75(0.60 FEXX)(0.707a)L 2. Base metal shear yielding: φRn = 1.00 × 0.60 Fy Agv 3. Base metal shear rupture: φRn = 0.75 × 0.60 Fu Anv 4. Net section fracture of connected member 5. Gross section yielding of connected member The weld strength and base metal strength must both be checked; the weld failure plane and the plate shear failure planes are different. DESIGN HIERARCHY IN PRACTICE: • Bolt shear and bearing are checked first (most frequently govern in standard connections) • Block shear becomes critical when bolt groups are arranged over short plate widths • Net section fracture is critical for tension members with many bolt holes • Weld base metal checks rarely govern when properly sized electrodes are used This multi-limit-state approach aligns with NSCP 2015 Section 502 requirements and AISC 360-16 Chapter J, which mandates that all applicable limit states be evaluated.
Examples
In this example, block shear governs over both bolt shear and bearing. This commonly occurs in connections with short shear paths or where the number of bolts is small relative to plate dimensions. Always check all three limit states before declaring the governing one.
Scenario
A lap connection uses 3 bolts (22 mm, A325-N, single shear) on a 10 mm plate (Fy = 248 MPa, Fu = 400 MPa). Edge distance Le = 38 mm, bolt pitch = 75 mm. Determine which limit state governs.
Solution
BOLT SHEAR (3 bolts, single shear, A325-N): Ab = π/4(22)² = 380.1 mm² φRn per bolt = 0.75 × 372 × 380.1 = 106,148 N Total bolt shear = 3 × 106,148 = 318,444 N = 318.4 kN BEARING (edge bolt, lc = 38 − 22/2 = 27 mm): 1.2(27)(10)(400) = 129,600; 2.4(22)(10)(400) = 211,200 → governs at 129,600 φRn per edge bolt = 0.75 × 129,600 = 97,200 N Interior bolts: lc = 75 − 22 = 53 mm → 1.2(53)(10)(400) = 254,400 > 211,200 → use 211,200 φRn per interior bolt = 0.75 × 211,200 = 158,400 N Total bearing = 97,200 + 2(158,400) = 97,200 + 316,800 = 414,000 N = 414.0 kN BLOCK SHEAR (estimate, Ubs = 1.0): Shear path length = 2(75) + 38 = 188 mm; Agv = 188(10) = 1,880 mm² Anv = 1,880 − 2.5(26)(10) = 1,880 − 650 = 1,230 mm² Ant = (35 − 13)(10) = 220 mm² [assuming tension edge = 35 mm] Expr1 = 0.6(400)(1,230) + 1.0(400)(220) = 295,200 + 88,000 = 383,200 Expr2 = 0.6(248)(1,880) + 88,000 = 279,936 + 88,000 = 367,936 φRn = 0.75(367,936) = 275,952 N = 276.0 kN GOVERNING LIMIT STATE: Block shear at 276.0 kN (lowest) → Connection design capacity = 276.0 kN
Applications
- Complete connection design packages for structural steel buildings
- Verification of existing connections under revised loading conditions
- Failure analysis of steel structures — identifying which limit state was exceeded
- Optimization of connection geometry to balance all limit states (ideal when all limits give equal capacity)
- Prequalified connection design using AISC Design Guide tables
Misconceptions
- Adding the capacities of different limit states — the connection can only develop the weakest link, not the sum
- Checking only bolt shear and ignoring block shear — block shear often governs and is a popular board exam topic
- Thinking that a larger bolt group is always stronger — more bolts may not help if block shear area is limited
- Applying φ = 0.90 to bearing or weld checks — φ = 0.75 for these limit states; φ = 0.90 is for member yielding only
Related Concepts
- LRFD load combinations (NSCP 2015 Section 203)
- Tension member design — gross yielding and net fracture
- Compression member connection requirements
- Seismic detailing requirements for connections (NSCP 2015 Section 506)
- Connection stiffness and moment-rotation behavior
Common Exam Questions
Example
Typical board problem: 'Which of the following governs?' with given capacities of 318 kN (shear), 414 kN (bearing), 276 kN (block shear) → Answer: block shear at 276 kN
Approach
Compute φRn for each limit state separately, tabulate results, report the minimum as the governing capacity
Question Type
Identify the governing limit state among bolt shear, bearing, and block shear
Key Points To Remember
- The connection capacity = MINIMUM of all limit state capacities — never sum them
- φ = 0.75 for all connection limit states (bolt shear, bearing, block shear, weld, net section fracture)
- φ = 0.90 for gross section yielding (a member limit state, not a connection limit state)
- Always check at least 3 limit states: bolt shear, bearing, and block shear for bolted connections
- The weld is NOT always the critical element — always check the base metal as well
- In board problems, if a limit state is not asked explicitly, check if it is implied by the problem statement
Practice Problems
A325-X (threads excluded) has the higher Fnv = 469 MPa vs. A325-N at 372 MPa. The double shear multiplier of 2 reflects two cross-sections of the bolt resisting the force. For a 6-bolt group under concentric loading, equal distribution is assumed, so total = 6 × individual capacity.
Problem
PROBLEM 1 (Bolt Shear — Double Shear) Six 24 mm A325-X bolts connect a W-section web to two side plates (double shear configuration). Given Fnv = 469 MPa for A325-X. Determine: a) The design shear strength per bolt b) The total design shear capacity of the bolt group
Solution
Step 1 — Bolt cross-sectional area: Ab = π/4 × (24)² = π/4 × 576 = 452.39 mm² Step 2 — Design shear strength per bolt (double shear, φ = 0.75): φRn = 2 × 0.75 × Fnv × Ab φRn = 2 × 0.75 × 469 × 452.39 φRn = 2 × 0.75 × 212,070.9 φRn = 2 × 159,053 = 318,106 N φRn ≈ 318.1 kN per bolt Step 3 — Total capacity (6 bolts): Total φRn = 6 × 318.1 = 1,908.6 kN ANSWER: a) 318.1 kN per bolt b) 1,908.6 kN total
The edge bolt has a smaller lc (24 mm < 48 mm) so its capacity is lower. The interior bolts all have the same lc and the same bearing capacity. The tearout limit (1.2 lc t Fu) governs the edge bolt; the crushing limit (2.4 db t Fu) governs the interior bolts. Always treat edge and interior bolts separately.
Problem
PROBLEM 2 (Bearing Strength — Interior and Edge Bolts) A row of four 20 mm bolts connects two A36 plates (Fu = 400 MPa, t = 10 mm). Edge distance Le = 35 mm; bolt pitch = 70 mm. Standard holes (dh = 22 mm). Find the total design bearing capacity of the connection (deformation is a design consideration).
Solution
Step 1 — Edge bolt (bolt 1): lc = Le − dh/2 = 35 − 11 = 24 mm 1.2 lc t Fu = 1.2(24)(10)(400) = 115,200 N 2.4 db t Fu = 2.4(20)(10)(400) = 192,000 N Rn (edge) = min(115,200; 192,000) = 115,200 N φRn (edge bolt) = 0.75 × 115,200 = 86,400 N = 86.4 kN Step 2 — Interior bolts (bolts 2, 3, 4): lc = s − dh = 70 − 22 = 48 mm 1.2 lc t Fu = 1.2(48)(10)(400) = 230,400 N 2.4 db t Fu = 2.4(20)(10)(400) = 192,000 N Rn (interior) = min(230,400; 192,000) = 192,000 N φRn (interior bolt) = 0.75 × 192,000 = 144,000 N = 144.0 kN Step 3 — Total bearing capacity: Total = 1(86.4) + 3(144.0) = 86.4 + 432.0 = 518.4 kN ANSWER: Total design bearing capacity = 518.4 kN
The total required weld length is divided by 2 because two weld lines resist the load simultaneously. Always verify that the selected weld length exceeds 4a to ensure proper weld performance. Round UP to the next millimeter to ensure capacity is not underestimated.
Problem
PROBLEM 3 (Fillet Weld — Design Length) A tension member must be welded to a gusset plate using 8 mm E70 fillet welds (FEXX = 482 MPa). The factored tensile force is Pu = 380 kN. Welds will be placed on both sides of the member. Find the required weld length per side.
Solution
Step 1 — Design strength per unit length for 8 mm E70 fillet weld: Throat = 0.707 × 8 = 5.656 mm φRn per mm = 0.75 × 0.60 × FEXX × throat = 0.75 × 0.60 × 482 × 5.656 = 0.75 × 0.60 × 2,726.2 = 0.75 × 1,635.7 = 1,226.8 N/mm Step 2 — Required TOTAL weld length: L_total = Pu ÷ (φRn per mm) L_total = 380,000 ÷ 1,226.8 = 309.8 mm Step 3 — Required weld length PER SIDE (welds on both sides): L per side = 309.8 ÷ 2 = 154.9 mm → Use 155 mm per side Step 4 — Minimum length check: Min L = 4a = 4(8) = 32 mm ≤ 155 mm ✓ ANSWER: Required weld length = 155 mm on each side
This comprehensive problem illustrates that the governing limit state is not always the most obvious one. Block shear with only 2 bolts gives a relatively short shear path length, resulting in small Agv and Anv — hence a lower block shear capacity. Increasing the bolt pitch or adding more bolts would improve block shear capacity significantly.
Problem
PROBLEM 4 (Complete Connection Check — Governing Limit State) A single-plate shear connection uses two 22 mm A325-N bolts (Fnv = 372 MPa) in single shear on a 10 mm plate (Fy = 248 MPa, Fu = 400 MPa). Edge distance Le = 38 mm (force direction), bolt pitch = 75 mm, tension edge = 32 mm. Standard holes (dh = 22 mm). Determine: a) Bolt shear capacity b) Bearing capacity (deformation is a concern) c) Block shear capacity d) Governing limit state
Solution
PART a — BOLT SHEAR: Ab = π/4(22)² = 380.13 mm² φRn per bolt = 0.75 × 372 × 380.13 = 106,176 N Total bolt shear = 2 × 106,176 = 212,352 N = 212.4 kN PART b — BEARING: Edge bolt: lc = 38 − 22/2 = 27 mm 1.2(27)(10)(400) = 129,600; 2.4(22)(10)(400) = 211,200 → governs: 129,600 φRn (edge) = 0.75(129,600) = 97,200 N Interior bolt (2nd bolt from edge): lc = 75 − 22 = 53 mm 1.2(53)(10)(400) = 254,400; 2.4(22)(10)(400) = 211,200 → governs: 211,200 φRn (interior) = 0.75(211,200) = 158,400 N Total bearing = 97,200 + 158,400 = 255,600 N = 255.6 kN PART c — BLOCK SHEAR: Hole for net area: dh_net = 22 + 4 = 26 mm (add 2mm for punching damage per AISC J4) Shear path length = 1(75) + 38 = 113 mm Agv = 113 × 10 = 1,130 mm² Anv = 1,130 − 1.5(26)(10) = 1,130 − 390 = 740 mm² [1.5 holes: 1 full interior hole + 0.5 hole at each end of shear path = 1.5 total] Ant = (32 − 13)(10) = 190 mm² [half hole = 26/2 = 13 mm] Expression 1: 0.6(400)(740) + 1.0(400)(190) = 177,600 + 76,000 = 253,600 N Expression 2: 0.6(248)(1,130) + 1.0(400)(190) = 168,144 + 76,000 = 244,144 N Rn = min(253,600; 244,144) = 244,144 N φRn = 0.75 × 244,144 = 183,108 N = 183.1 kN PART d — GOVERNING LIMIT STATE: Bolt shear: 212.4 kN Bearing: 255.6 kN Block shear: 183.1 kN ← MINIMUM ANSWER: Block shear governs at φRn = 183.1 kN
This comparison problem teaches that bolted and welded solutions can both satisfy a requirement but with different margins. Note that Option A only checked bolt shear — bearing and block shear would also need to be checked for a complete design. Option B weld check is complete as given, assuming adequate base metal.
Problem
PROBLEM 5 (Weld vs. Bolt — Comparison) A tension force of Pu = 450 kN must be transferred via a connection. Evaluate both options: Option A: Six 20 mm A325-N bolts in single shear Option B: 8 mm E70 fillet welds on both sides of a plate, 200 mm per side Which option provides adequate capacity?
Solution
OPTION A — BOLT SHEAR: Ab = π/4(20)² = 314.16 mm² φRn per bolt = 0.75 × 372 × 314.16 = 87,651 N Total (6 bolts) = 6 × 87,651 = 525,906 N = 525.9 kN > 450 kN ✓ ADEQUATE OPTION B — WELD: Throat = 0.707 × 8 = 5.656 mm φRn per mm = 0.75 × 0.60 × 482 × 5.656 = 1,226.8 N/mm Total length = 2 × 200 = 400 mm Total φRn = 1,226.8 × 400 = 490,720 N = 490.7 kN > 450 kN ✓ ADEQUATE MARGIN CHECK: Option A margin = 525.9 − 450 = 75.9 kN (16.9% excess) Option B margin = 490.7 − 450 = 40.7 kN (9.0% excess) ANSWER: BOTH options are adequate. Option A (bolts) has a larger safety margin. Option B (welds) provides a slimmer margin but eliminates hole-drilling and potential bearing/block shear issues.
Exam Preparation Tips
- MEMORIZE the four critical values: Fnv(A325-N) = 372 MPa, Fnv(A325-X) = 469 MPa, FEXX(E70) = 482 MPa, φ = 0.75 for ALL connection limit states. These appear in virtually every board problem.
- ALWAYS count shear planes before computing bolt shear. Draw a free body diagram of the bolt — the number of cuts through the bolt is the number of shear planes (1 for single shear, 2 for double shear).
- For bearing, compute lc (CLEAR distance, not center-to-center) and ALWAYS evaluate both the 1.2 lc t Fu and 2.4 db t Fu terms. Take the minimum — never skip the upper bound check.
- For block shear, add 4 mm (not 2 mm) to bolt diameter for hole size in net area calculations (AISC J4: standard hole db+2mm, plus 2mm damage = db+4mm for net area only).
- In fillet weld problems, the very first step should be computing the throat = 0.707 × weld size. Writing this prominently in your solution prevents the most common weld calculation error.
- When a board problem says 'determine the capacity of the connection,' check ALL applicable limit states and report the MINIMUM — not just bolt shear.
- For A36 steel: Fy = 248 MPa, Fu = 400 MPa. For A572 Grade 50: Fy = 345 MPa, Fu = 450 MPa. These are frequently given or implied in problems.
- Practice the block shear area calculations repeatedly — correctly identifying Agv, Anv, and Ant from a connection sketch is a skill that requires repeated exposure.
- E70 electrodes match A36 and A572 base metals (matching or overmatching). Remember: electrode must match or overmatch base metal tensile strength.
- Time management tip: In a 5-choice MCQ, eliminate options that violate φ = 0.75 (e.g., options using 0.90 or 0.85 for weld/bolt checks). This narrows choices before full computation.
- CJP groove welds develop full base metal strength — if a board exam asks for the capacity of a CJP weld, the answer is based on the member's tensile capacity (φ = 0.75 × Fu × Ae), NOT a separate weld calculation.
- For Ubs: use 1.0 in almost all standard exam problems unless the connection clearly shows a non-uniform tension distribution (e.g., partial web connection). When in doubt, use Ubs = 1.0.
In summary
Steel connections are the structural critical path in every steel building — they are where load transfer actually occurs and where failure most often initiates. For the PRC Civil Engineer Licensure Examination, mastery of this chapter demands both formula memorization and conceptual understanding of WHY each limit state exists physically. The three pillars of bolted connection design — bolt shear, bearing at holes, and block shear — must always be checked together; the minimum of all three governs. For welded connections, the throat (0.707a) is the critical dimension, and the design strength per unit length for E70 electrodes should become second nature: φRn/L = 0.75 × 0.60 × 482 × 0.707 × a ≈ 153.3a N/mm. The unifying thread across all connection limit states is φ = 0.75, reflecting the higher consequence of brittle failure modes (fracture, shear rupture) compared to yielding modes (where φ = 0.90). This difference in resistance factors is a direct application of structural reliability theory codified in AISC 360-16 and adopted by NSCP 2015 Section 502. Approach every board exam connection problem systematically: (1) identify all limit states that apply, (2) compute each one completely, (3) take the minimum as the governing capacity, and (4) compare to the factored demand. Never shortcut this process — the board exam is specifically designed to reward thorough, methodical engineers who check all limit states, not just the obvious one. With consistent practice of the worked examples in this chapter and the five practice problems provided, you will develop the speed and accuracy needed to solve any connection problem within the time constraints of the licensure examination. Mabuting swerte sa inyong board exam!
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