CELE Steel & Timber Design — Steel ConnectionsExam Answer Templates

Examiner tip

This is a one-mark recall question. Write exactly 'φ = 0.75' — no derivation needed. Any correct code citation (AISC 360 or NSCP 2015 Section 502) earns the mark.

Model answer

φ = 0.75 for bolt shear, bearing, and fillet weld limit states in LRFD (AISC 360-16, Section J).

Answer structure

• One sentence: state φ value and the limit states it applies to [1 mark]

Scoring breakdown

Common mark deductions

• Writing φ = 0.90 (which applies to member flexure, not connections)
• Writing φ = 0.85 (which applies to compression members)
• Leaving out the code reference when it is explicitly requested

Key phrases to include

• φ = 0.75
• LRFD
• bolt shear
• bearing
• AISC 360

Examiner tip

Memorize: N = threads iN the plane (lower strength), X = threads eXcluded (higher strength). Stating both Fnv values shows exam-ready knowledge and secures full marks.

Model answer

A325-N: threads included in the shear plane, Fnv = 372 MPa. A325-X: threads excluded from the shear plane, Fnv = 469 MPa. The 'X' condition gives a higher shear capacity.

Answer structure

• Define 'N' (threads in shear plane) and give Fnv [0.5 mark]
• Define 'X' (threads excluded) and give Fnv [0.5 mark]

Scoring breakdown

Common mark deductions

• Confusing N and X (stating that X has threads included — opposite is true)
• Omitting the Fnv values, which show you understand the practical consequence

Key phrases to include

• threads in shear plane
• threads excluded
• Fnv = 372 MPa
• Fnv = 469 MPa

Examiner tip

Examiners award one mark for the formula and one for variable identification. Even if you cannot recall Fnv values, defining what it represents earns the second mark.

Model answer

Design shear strength (single shear):

φRn = 0.75 × Fnv × Ab

Where:
φ = 0.75 (resistance factor for shear, AISC 360-16 §J3)
Fnv = nominal shear stress of the bolt (MPa); 372 MPa for A325-N, 469 MPa for A325-X
Ab = nominal (unthreaded) cross-sectional area of the bolt (mm²) = π/4 × db²

For double shear, multiply by 2: φRn = 2 × (0.75 Fnv Ab).

Answer structure

• Line 1: Write the formula φRn = 0.75 Fnv Ab [1 mark]
• Lines 2–4: Define φ, Fnv, and Ab with units and values [1 mark]

Scoring breakdown

Common mark deductions

• Writing φ = 0.90 instead of 0.75
• Using the threaded area instead of the nominal (gross) bolt area Ab
• Forgetting to mention the double-shear modifier

Key phrases to include

• φ = 0.75
• Fnv
• Ab = π/4 × db²
• single shear
• double shear — multiply by 2

Examiner tip

The key phrase examiners look for is 'minimum governs.' Write it explicitly. Also note: lc is the CLEAR distance (center-to-center minus hole diameter, or edge to hole edge) — not the bolt spacing.

Model answer

Two limit states in bearing:

1. Tear-out (clear-distance controlled): Rn = 1.2 lc t Fu
2. Crushing (bolt-diameter controlled): Rn = 2.4 db t Fu

The design bearing strength per bolt is:
φRn = 0.75 × min(1.2 lc t Fu , 2.4 db t Fu)

The SMALLER value governs — the plate fails at whichever limit state is reached first.

Answer structure

• Identify and name both limit states [1 mark]
• Write the φRn formula with the 'min' statement and φ = 0.75 [1 mark]

Scoring breakdown

Common mark deductions

• Choosing the larger (not smaller) of the two expressions
• Omitting lc and using bolt pitch or edge distance without subtracting hole diameter
• Using Fy instead of Fu for the plate material

Key phrases to include

• 1.2 lc t Fu
• 2.4 db t Fu
• minimum governs
• φ = 0.75
• lc = clear distance
• Fu = tensile strength

Examiner tip

Show your computation of Ab as a distinct step — this earns the first mark even if you multiply incorrectly. Always convert the final answer to kN (divide N by 1000).

Model answer

Given:
db = 22 mm, Fnv = 372 MPa (A325-N, threads in shear plane), single shear, φ = 0.75

Step 1 — Bolt area:
Ab = (π/4)(22)² = (π/4)(484) = 380.1 mm²

Step 2 — Design shear strength:
φRn = 0.75 × Fnv × Ab
= 0.75 × 372 × 380.1
= 106,148 N

∴ φRn ≈ 106.1 kN per bolt

Answer structure

• State given data including φ = 0.75 [0 marks — setup, but sets tone]
• Compute Ab = π/4 × (22)² = 380.1 mm² [1 mark]
• Compute φRn = 0.75 × 372 × 380.1 = 106.1 kN [1 mark]

Scoring breakdown

Common mark deductions

• Using Fnv = 469 MPa (wrong bolt condition — this is A325-X, not A325-N)
• Forgetting to apply φ = 0.75, giving Rn = 141.4 kN instead
• Using the threaded-shank area (≈ 303 mm² for 22 mm bolt) instead of gross area

Key phrases to include

• Ab = π/4 × db²
• φ = 0.75
• Fnv = 372 MPa
• single shear

Examiner tip

Label each limit state clearly ('Tear-out' and 'Bearing/Crushing'). Examiners follow your logic step by step. Applying φ to each before comparing is not wrong numerically, but labeling after comparison shows clearer AISC understanding.

Model answer

Given:
db = 20 mm, t = 10 mm, Fu = 400 MPa, lc = 35 mm, φ = 0.75

Step 1 — Tear-out limit state:
Rn1 = 1.2 × lc × t × Fu
= 1.2 × 35 × 10 × 400
= 168,000 N = 168 kN

Step 2 — Bearing (crushing) limit state:
Rn2 = 2.4 × db × t × Fu
= 2.4 × 20 × 10 × 400
= 192,000 N = 192 kN

Step 3 — Governing nominal strength (minimum):
Rn = min(168,000 ; 192,000) = 168,000 N

Step 4 — Design bearing strength:
φRn = 0.75 × 168,000 = 126,000 N

∴ φRn = 126 kN per bolt ← Tear-out governs

Answer structure

• Compute Rn1 = 1.2 lc t Fu = 168 kN [1 mark]
• Compute Rn2 = 2.4 db t Fu = 192 kN [1 mark]
• Apply minimum and φ = 0.75 → φRn = 126 kN [1 mark]

Scoring breakdown

Common mark deductions

• Taking the maximum instead of minimum of the two expressions
• Using Fy instead of Fu (e.g., using 250 MPa instead of 400 MPa)
• Applying φ = 0.75 to each expression before comparing — φ is applied only after the minimum is selected

Key phrases to include

• 1.2 lc t Fu
• 2.4 db t Fu
• minimum governs
• tear-out governs
• φ = 0.75
• 126 kN

Examiner tip

The statement 'weld fails in shear through the throat' is the key phrase. Write it. For 3-mark weld questions, examiners expect formula + computation + conceptual explanation — all three.

Model answer

Fillet Weld Design Strength (per unit length):

For an equal-leg fillet weld of size a = 6 mm, the effective throat is:
te = 0.707 × a = 0.707 × 6 = 4.24 mm

The weld fails by shear through this throat plane. The nominal shear strength of the weld metal:
fnw = 0.60 × FEXX = 0.60 × 482 = 289.2 MPa

Design strength per unit length:
φRn/L = φ × fnw × te
= 0.75 × 289.2 × 4.24
= 919.7 N/mm ≈ 920 N/mm

Significance of throat: The throat (0.707a) is the shortest distance across the weld cross-section and is the weakest plane through which failure occurs. Using the weld leg size 'a' instead of the throat OVERESTIMATES strength by 41% — a critical error.

Answer structure

• Compute throat te = 0.707 × 6 = 4.24 mm and explain its physical meaning [1 mark]
• Write and compute fnw = 0.60 × FEXX = 289.2 MPa [1 mark]
• Apply φ = 0.75 to get 920 N/mm and state significance of throat [1 mark]

Scoring breakdown

Common mark deductions

• Using weld size 'a = 6 mm' directly instead of 0.707a = 4.24 mm for the throat
• Using FEXX = 482 MPa directly instead of 0.60 × FEXX as the weld shear strength
• Omitting the physical explanation of throat significance in a 3-mark question

Key phrases to include

• throat = 0.707a
• 0.60 FEXX
• φ = 0.75
• weakest plane
• shear through throat
• 920 N/mm

Examiner tip

Structure your 5-mark answer with clear horizontal rules or headings for each limit state. Examiners mark limit state by limit state. Losing 'Bolt Shear Governs' statement costs the final comparison mark — always write it explicitly.

Model answer

Given:
db = 20 mm, Fnv = 372 MPa (A325-N), n = 4 bolts, single shear
t = 10 mm, Fu = 400 MPa, lc = 42 mm, φ = 0.75

━━━ LIMIT STATE 1: Bolt Shear ━━━

Bolt area:
Ab = (π/4)(20)² = 314.2 mm²

Design shear strength per bolt:
φRn,shear = 0.75 × 372 × 314.2 = 87,668 N ≈ 87.7 kN/bolt

For 4 bolts:
ΦRn,shear (total) = 4 × 87.7 = 350.8 kN

━━━ LIMIT STATE 2: Bearing on Plate ━━━

Tear-out check per bolt:
Rn1 = 1.2 × lc × t × Fu = 1.2 × 42 × 10 × 400 = 201,600 N

Crushing check per bolt:
Rn2 = 2.4 × db × t × Fu = 2.4 × 20 × 10 × 400 = 192,000 N

Governing Rn per bolt = min(201,600 ; 192,000) = 192,000 N ← Crushing governs

Design bearing strength per bolt:
φRn,bearing = 0.75 × 192,000 = 144,000 N = 144 kN/bolt

For 4 bolts:
ΦRn,bearing (total) = 4 × 144 = 576 kN

━━━ GOVERNING LIMIT STATE ━━━

Bolt Shear: 350.8 kN
Bearing: 576.0 kN

Design strength of connection = min(350.8, 576.0) = 350.8 kN

∴ Bolt shear governs. Design strength = 350.8 kN ≈ 351 kN

Answer structure

• Compute Ab = π/4 × (20)² = 314.2 mm² [0.5 mark]
• Compute φRn,shear per bolt = 87.7 kN → total = 4 × 87.7 = 350.8 kN [1 mark]
• Compute Rn1 (tear-out) = 201.6 kN and Rn2 (crushing) = 192 kN per bolt [1 mark]
• Select minimum bearing (192 kN/bolt) and apply φ → 144 kN/bolt → total 576 kN [1 mark]
• Compare both totals, state governing limit state, give final answer 351 kN [1.5 marks]

Scoring breakdown

Common mark deductions

• Not multiplying by 4 bolts at the end — computing per-bolt capacity only
• Taking maximum instead of minimum for bearing; taking maximum instead of minimum across limit states at the end
• Using double shear when the problem says single shear (lap connection = single shear)
• Forgetting to check both limit states and skipping bearing entirely
• Arithmetic error in Ab = π/4 × (20)² — use 314.16, not 400 (confusing diameter with area)

Key phrases to include

• Ab = π/4 × db²
• φ = 0.75
• Fnv = 372 MPa
• 1.2 lc t Fu
• 2.4 db t Fu
• minimum governs
• bolt shear governs
• 350.8 kN

Examiner tip

The formula has two parts — a rupture path and a yield cap. Write both. Examiners who are strict on AISC 360 will deduct if the cap is missing.

Model answer

Block shear is a connection failure mode where a block of material tears out along a combined failure path: shear yielding or rupture along the bolt line plus tension rupture perpendicular to the load direction.

Nominal block shear strength (AISC 360-16 §J4.3):

Rn = 0.6Fu Anv + Ubs Fu Ant ≤ 0.6Fy Agv + Ubs Fu Ant

Where:
Agv = gross shear area
Anv = net shear area
Ant = net tension area
Ubs = 1.0 (uniform tension) or 0.5 (non-uniform tension)
φ = 0.75

The first expression represents the rupture path; it is capped by the yielding expression.

Answer structure

• Define block shear conceptually (combined shear + tension failure path) [1 mark]
• Write the AISC formula with both the rupture expression and the yield cap; define Ubs [1 mark]

Scoring breakdown

Common mark deductions

• Omitting the upper bound (yield cap) expression — the formula requires both
• Using Fy instead of Fu in the tension rupture term (Ant uses Fu, not Fy)
• Confusing Agv and Anv (gross vs net shear areas)

Key phrases to include

• shear rupture + tension rupture
• 0.6 Fu Anv
• Ubs Fu Ant
• cap: 0.6 Fy Agv
• Ubs = 1.0 or 0.5
• φ = 0.75

Examiner tip

The Ant term (tension component) is the SAME in both expressions — Fu × Ant in both. Only the shear term changes (Fu × Anv vs Fy × Agv). This is the detail that separates top scorers from average ones.

Model answer

Given:
Anv = 1800 mm², Agv = 2100 mm², Ant = 600 mm²
Fy = 248 MPa, Fu = 400 MPa, Ubs = 1.0, φ = 0.75

Step 1 — Block shear rupture path:
Rn1 = 0.6 Fu Anv + Ubs Fu Ant
= 0.6(400)(1800) + 1.0(400)(600)
= 432,000 + 240,000
= 672,000 N

Step 2 — Upper bound (yield cap):
Rn2 = 0.6 Fy Agv + Ubs Fu Ant
= 0.6(248)(2100) + 1.0(400)(600)
= 312,480 + 240,000
= 552,480 N

Step 3 — Governing Rn = min(Rn1, Rn2) = min(672,000 ; 552,480) = 552,480 N
(Yield cap governs — shear yielding path is weaker)

Step 4 — Design block shear strength:
φRn = 0.75 × 552,480 = 414,360 N

∴ φRn ≈ 414 kN

Answer structure

• Compute rupture path Rn1 = 0.6 Fu Anv + Ubs Fu Ant = 672,000 N [1 mark]
• Compute yield cap Rn2 = 0.6 Fy Agv + Ubs Fu Ant = 552,480 N [1 mark]
• Take minimum and apply φ = 0.75 → φRn = 414 kN [1 mark]

Scoring breakdown

Common mark deductions

• Using Fy in the tension term (Ant × Fy instead of Ant × Fu)
• Taking maximum of the two expressions instead of minimum
• Omitting the Ubs factor (Ubs = 1.0 still needs to appear to show awareness)

Key phrases to include

• 0.6 Fu Anv
• 0.6 Fy Agv
• Ubs Fu Ant
• minimum governs
• yield cap governs
• φ = 0.75
• 414 kN

Examiner tip

Two key reductions on the weld: (1) 0.707a for throat, (2) 0.60 FEXX for weld shear strength. Both must appear in your solution. Apply φ = 0.75 at the end.

Model answer

Given:
a = 8 mm (weld size), L = 200 mm, FEXX = 482 MPa, φ = 0.75

Step 1 — Weld throat:
te = 0.707 × 8 = 5.656 mm

Step 2 — Design strength:
φRn = φ × (0.60 FEXX) × te × L
= 0.75 × (0.60 × 482) × 5.656 × 200
= 0.75 × 289.2 × 5.656 × 200
= 0.75 × 327,231
= 245,423 N

∴ φRn ≈ 245 kN

Answer structure

• Compute throat te = 0.707 × 8 = 5.656 mm [1 mark]
• Compute φRn = 0.75 × 0.60 × 482 × 5.656 × 200 = 245 kN [1 mark]

Scoring breakdown

Common mark deductions

• Using a = 8 mm directly (leg size) without converting to throat te = 5.656 mm
• Using FEXX = 482 MPa directly without the 0.60 factor
• Incorrect throat: using te = 0.5a (incorrect; 0.707 is for 45° equal-leg weld)

Key phrases to include

• te = 0.707a
• 0.60 FEXX
• φ = 0.75
• 245 kN

Examiner tip

This is a design (not analysis) problem. The key word 'both sides' doubles the weld length. Always round UP to the next standard weld size — never round down for capacity.

Model answer

Given:
Pu = 250 kN = 250,000 N (factored load)
L per weld = 150 mm; two sides → total L = 2 × 150 = 300 mm
FEXX = 482 MPa, φ = 0.75

Set design strength ≥ factored load:
φRn ≥ Pu
0.75 × (0.60 × 482) × (0.707 × a) × 300 ≥ 250,000

Simplify the left side:
0.75 × 289.2 × 0.707 × 300 × a ≥ 250,000
0.75 × 289.2 × 212.1 × a ≥ 250,000
46,000.7 × a ≥ 250,000 (approximately)

Solving for a:
a ≥ 250,000 / 46,000 ≈ 5.43 mm

Practical weld size (round up to next whole mm or standard size):
a_min = 6 mm

∴ Use a 6 mm fillet weld on both sides (300 mm total effective length).

Answer structure

• Identify total weld length = 2 × 150 = 300 mm and state φRn ≥ Pu [1 mark]
• Set up equation: 0.75 × 0.60 × FEXX × 0.707a × 300 = 250,000 [1 mark]
• Solve for a ≥ 5.43 mm → specify a = 6 mm [1 mark]

Scoring breakdown

Common mark deductions

• Using L = 150 mm instead of total 300 mm (forgetting the second weld side)
• Not rounding up from 5.43 to 6 mm — weld sizes must be practical/standard
• Setting up Rn ≥ Pu (unfactored) and forgetting φ on the left side

Key phrases to include

• total length = 300 mm
• 0.707a throat
• 0.60 FEXX
• φ = 0.75
• a_min = 6 mm
• round up

Examiner tip

Note that gross yielding uses φ = 0.90 while all the fracture/shear limit states use φ = 0.75. This distinction is a classic board trick — know which φ goes where.

Model answer

Five connection limit states and their resistance factors:

1. Bolt Shear: φRn = 0.75 Fnv Ab (per bolt, per shear plane)
2. Bearing on Plate: φRn = 0.75 min(1.2 lc t Fu, 2.4 db t Fu)
3. Net-Section Rupture of Connected Element: φRn = 0.75 Fu Ae
4. Gross-Section Yielding of Connected Element: φRn = 0.90 Fy Ag
5. Block Shear: φRn = 0.75 (0.6Fu Anv + Ubs Fu Ant)

The design strength of the connection is the MINIMUM of all applicable limit states.

Answer structure

• List limit states 1 and 2 with correct φ values [1 mark]
• List limit states 3, 4, and 5 with correct φ values [1 mark]
• State that connection capacity = minimum of all limit states [1 mark]

Scoring breakdown

Common mark deductions

• Using φ = 0.90 for bolt shear (wrong — 0.90 applies to gross yielding of member, not connections)
• Listing only bolt shear and bearing, omitting block shear and net rupture
• Not stating that the minimum of all limit states governs the connection

Key phrases to include

• bolt shear φ = 0.75
• bearing φ = 0.75
• net rupture φ = 0.75
• gross yielding φ = 0.90
• block shear φ = 0.75
• minimum governs

Examiner tip

This 3-part question is designed to test whether you know that the MINIMUM governs. Examiners often see students incorrectly use the bolt shear value (280 kN) — always take the minimum across ALL limit states.

Model answer

(a) Design strength of connection:
= min(280, 420, 310, 265) = 265 kN
(Block shear controls)

(b) Adequacy check:
Design strength (265 kN) > Factored load (250 kN) ✓
265 kN > 250 kN → Connection is ADEQUATE

(c) Governing limit state:
Block shear governs at 265 kN.

Answer structure

• Take minimum of all four values → 265 kN [1 mark]
• Compare 265 kN vs Pu = 250 kN and state adequacy [1 mark]
• Explicitly name block shear as governing limit state [1 mark]

Scoring breakdown

Common mark deductions

• Using the maximum (420 kN) or the bolt shear (280 kN) as the design strength
• Comparing 280 kN (bolt shear) with Pu instead of the true minimum of 265 kN
• Not stating adequacy explicitly (yes/no + reason)

Key phrases to include

• minimum governs
• 265 kN
• block shear governs
• adequate
• design strength > factored load

Examiner tip

For 5-mark problems, examiners expect headings for each limit state, an explicit comparison summary, and a clear final answer with the governing limit state named. Budget 8–10 minutes for this type of problem in the exam.

Model answer

Given:
db = 22 mm, Fnv = 469 MPa (A325-X, double shear), n = 6 bolts
t = 16 mm, Fu = 415 MPa, lc = 38 mm, φ = 0.75

━━━ (a) BOLT SHEAR — Double Shear ━━━

Bolt area:
Ab = (π/4)(22)² = 380.1 mm²

Design shear per bolt (double shear = 2 shear planes):
φRn,shear = 2 × 0.75 × Fnv × Ab
= 2 × 0.75 × 469 × 380.1
= 2 × 133,750
= 267,501 N = 267.5 kN/bolt

Total for 6 bolts:
ΦRn,shear = 6 × 267.5 = 1,605 kN

━━━ (b) BEARING ON PLATE ━━━

Tear-out per bolt:
Rn1 = 1.2 × 38 × 16 × 415 = 1.2 × 38 × 6,640 = 302,784 N

Crushing per bolt:
Rn2 = 2.4 × 22 × 16 × 415 = 2.4 × 22 × 6,640 = 350,592 N

Governing per bolt: min(302,784 ; 350,592) = 302,784 N (Tear-out governs)

Design bearing per bolt:
φRn,bearing = 0.75 × 302,784 = 227,088 N = 227.1 kN/bolt

Total for 6 bolts:
ΦRn,bearing = 6 × 227.1 = 1,362.5 kN

━━━ (c) GOVERNING LIMIT STATE ━━━

Bolt Shear: 1,605 kN
Bearing: 1,363 kN ← smaller

∴ Bearing (tear-out) governs.
Design connection strength = 1,362.5 kN ≈ 1,363 kN

Answer structure

• Compute Ab = 380.1 mm²; apply double shear (×2); φRn = 267.5 kN/bolt → 1,605 kN total [1.5 marks]
• Compute both bearing expressions per bolt (302,784 N and 350,592 N) [1 mark]
• Apply minimum bearing (tear-out) and φ → 227.1 kN/bolt → 1,363 kN total [1 mark]
• Compare 1,605 kN vs 1,363 kN; state bearing governs; give final answer [1.5 marks]

Scoring breakdown

Common mark deductions

• Omitting the ×2 factor for double shear — single shear gives half the correct shear capacity
• Using A325-N Fnv = 372 MPa instead of A325-X Fnv = 469 MPa
• Selecting maximum instead of minimum for bearing (tear-out vs crushing)
• Multiplying per-bolt bearing by 6 before applying φ vs applying φ first — both sequences give same answer but must be clearly shown
• Not comparing total bolt shear vs total bearing to determine governing limit state

Key phrases to include

• double shear — multiply by 2
• Ab = 380.1 mm²
• Fnv = 469 MPa (A325-X)
• 1.2 lc t Fu
• 2.4 db t Fu
• tear-out governs
• bearing governs
• 1,363 kN

Do

• Write the answer in one clear sentence or equation
• Include units if the answer is numerical (MPa, kN, mm²)
• Use the exact code terminology (e.g., 'nominal shear stress Fnv', not just 'shear stress')
• If asked for a formula, write it symbolically and label φ

Don't

• Do not derive or explain — it wastes time on a 1-mark recall item
• Do not write paragraphs — examiners scan for the keyword
• Do not leave units out of numerical answers

Do

• Write the formula explicitly before substituting numbers
• Show each computation step on a separate line
• State the governing condition (e.g., 'minimum governs', 'single shear')
• Convert the final answer to kN and label it clearly

Don't

• Do not skip the formula and go straight to numbers — the formula earns marks
• Do not use Fy where Fu is required (connection limit states use Fu for rupture)
• Do not mix up single and double shear without declaring which applies

Do

• Use headings (e.g., 'Step 1: Bolt Shear', 'Step 2: Bearing')
• Show both expressions in bearing problems (1.2 lc t Fu AND 2.4 db t Fu)
• Explicitly write the comparison and state which governs
• Box the final answer with its unit

Don't

• Do not omit the yield-cap (upper-bound) expression in block shear problems
• Do not forget to multiply per-bolt capacity by number of bolts
• Do not round intermediate values — carry at least 3 significant figures through to the end

Do

• Draw a clear horizontal rule or write 'LIMIT STATE 1:', 'LIMIT STATE 2:' headings
• Declare all given data at the top with units
• Write φ = 0.75 at the top once and reference it in each limit state computation
• Present a summary table: Limit State | φRn (kN) | Governs?
• State the governing limit state and final design strength in a conclusion line

Don't

• Do not omit any limit state — if the question provides data for bearing, compute bearing
• Do not take the maximum of limit states (rookie mistake — always take the minimum)
• Do not forget to check double shear (×2) for bolts through middle plate in double-shear configurations
• Do not present a wall of numbers without labels — examiners cannot follow unlabeled arithmetic