CELE Steel & Timber Design — Steel Tension MembersExam Answer Templates

Marks

1

Topic

Two Limit States

Difficulty

easy

Template Id

T1

Examiner Tip

This is a 1-mark definition question; examiners want a complete, paired answer. A one-word response like 'yielding and rupture' earns zero — you must include the φ factors and area terms.

Model Answer

The two limit states are: (1) Tensile yielding on the gross area: φPn = 0.90 Fy Ag, and (2) Tensile rupture on the effective net area: φPn = 0.75 Fu Ae. The lower value governs.

Question Type

very_short_answer

Answer Structure

• Line 1: Name both limit states with their resistance factors and area references [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating only one limit state (-1 mark, i.e., zero)
• Swapping the φ factors (0.75 for yield, 0.90 for rupture) — a fatal error
• Using 'net area An' instead of 'effective net area Ae' for the rupture limit state

Key Phrases To Include

• tensile yielding
• gross area Ag
• φ = 0.90
• tensile rupture
• effective net area Ae
• φ = 0.75
• lower value governs

Marks

2

Topic

Shear-Lag Factor

Difficulty

easy

Template Id

T2

Examiner Tip

Give a concrete example (plate bolted full-width vs. angle bolted one leg) — examiners reward practical application of the concept.

Model Answer

The shear-lag factor U accounts for the non-uniform stress distribution across a cross-section when not all elements are directly connected at a joint. It reduces the net area to give the effective net area: Ae = U·An, where U ≤ 1.0. U = 1.0 is applicable when the tensile load is transmitted through ALL cross-sectional elements (e.g., a plate bolted across its entire width, or a W-section with both flanges and web connected). When only one leg of an angle or one flange of a tee is connected, U < 1.0.

Question Type

short_answer

Answer Structure

• Line 1–2: Define U and write the formula Ae = U·An [1 mark]
• Line 3–4: State the condition for U = 1.0 and give one example where U < 1.0 [1 mark]

Scoring Breakdown

Common Mark Deductions

• Defining U without the formula Ae = U·An
• Stating U = 1.0 'always' without specifying the connection condition
• Confusing An (net area) with Ae (effective net area)

Key Phrases To Include

• shear lag
• non-uniform stress
• Ae = U·An
• U ≤ 1.0
• all cross-sectional elements connected
• U = 1.0

Marks

2

Topic

Resistance Factors

Difficulty

medium

Template Id

T3

Examiner Tip

Use the words 'ductile' and 'brittle' explicitly — these are the engineering-precise terms examiners want to see.

Model Answer

The difference in φ reflects the relative predictability and consequence of each failure mode: φ = 0.90 for yielding: Yielding is ductile and gradual — there is visible deformation and warning before failure. The gross-area and material yield strength Fy are well-defined, so a higher φ (less conservatism) is justified. φ = 0.75 for rupture: Fracture at the net section is sudden and brittle — there is little or no warning. Additionally, Fu (ultimate strength) has greater variability than Fy, and hole quality introduces uncertainty. A lower φ (more conservatism) is therefore required.

Question Type

short_answer

Answer Structure

• Line 1–2: Explain why φ = 0.90 is used for yielding (ductile, predictable, Fy well-defined) [1 mark]
• Line 3–4: Explain why φ = 0.75 is used for rupture (brittle, sudden, Fu more variable, hole uncertainty) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Simply stating 'rupture is more dangerous' without linking to ductility or variability
• Reversing the explanation (calling yielding brittle or rupture ductile)
• Not mentioning the consequence of hole-related uncertainty

Key Phrases To Include

• ductile
• gradual
• visible deformation
• Fy well-defined
• brittle
• sudden
• no warning
• Fu variability
• conservatism

Marks

1

Topic

Net Area Calculation

Difficulty

easy

Template Id

T4

Examiner Tip

Write the formula dh = db + 2 mm explicitly before substituting — this shows the examiner you know the rule, not just the number.

Model Answer

Use dh = 20 + 2 = 22 mm. The 2 mm allowance accounts for the punching/drilling damage to the surrounding material beyond the nominal hole diameter. (Some references use +3 mm total, giving dh = 23 mm; either value is acceptable if the basis is stated.)

Question Type

very_short_answer

Answer Structure

• Line 1: State dh = db + 2 mm = 22 mm with brief justification [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using the bolt diameter (20 mm) directly without adding the clearance — most common error
• Adding 3 mm but not stating the basis
• Using 25 mm (standard bolt clearance hole) which applies to hole fabrication, not the design deduction

Key Phrases To Include

• dh = db + 2 mm
• 22 mm
• punching damage
• clearance allowance
• net area

Marks

5

Topic

Two Limit States — Full Design Check

Difficulty

medium

Template Id

T5

Examiner Tip

Write 'Limit State 1' and 'Limit State 2' explicitly as headings — it signals to the examiner that you are systematically checking both, which is the NSCP 2015 requirement.

Model Answer

GIVEN: Plate: 250 mm × 12 mm Fy = 248 MPa, Fu = 400 MPa Bolt diameter = 22 mm → dh = 24 mm Number of bolts in critical section = 2 U = 1.0 STEP 1 — Gross Area: Ag = 250 × 12 = 3 000 mm² STEP 2 — Net Area: An = Ag − n(dh × t) = 3 000 − 2(24)(12) An = 3 000 − 576 = 2 424 mm² STEP 3 — Effective Net Area: Ae = U·An = 1.0 × 2 424 = 2 424 mm² STEP 4 — Limit State 1: Tensile Yielding (φ = 0.90) φPn,yield = 0.90 × Fy × Ag = 0.90 × 248 × 3 000 φPn,yield = 669 600 N = 669.6 kN STEP 5 — Limit State 2: Tensile Rupture (φ = 0.75) φPn,rupture = 0.75 × Fu × Ae = 0.75 × 400 × 2 424 φPn,rupture = 727 200 N = 727.2 kN STEP 6 — Governing Design Strength: φPn = min(669.6, 727.2) = 669.6 kN CONCLUSION: The design tensile strength is φPn = 669.6 kN. Yielding on the gross section governs.

Question Type

numerical

Answer Structure

• Section 1: List all given data clearly [0.5 mark implicit — shows organized thinking]
• Step 1: Compute Ag correctly [0.5 mark]
• Step 2: Compute An with correct dh and subtraction [1 mark]
• Step 3: Compute Ae = U·An [0.5 mark]
• Step 4: Compute φPn,yield = 0.90 Fy Ag [1 mark]
• Step 5: Compute φPn,rupture = 0.75 Fu Ae [1 mark]
• Step 6: Select the lower value and state which limit state governs [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using bolt diameter (22 mm) instead of hole diameter (24 mm) for net-area subtraction — loses the An and all subsequent marks
• Skipping either the yielding or rupture limit state
• Using φ = 0.75 for yielding or φ = 0.90 for rupture
• Omitting the conclusion statement about which limit state governs
• Missing units (N vs. kN) causing incorrect final answer

Key Phrases To Include

• Ag = 250 × 12
• dh = 24 mm
• An = Ag − n(dh)(t)
• Ae = U·An
• φPn,yield = 0.90 Fy Ag
• φPn,rupture = 0.75 Fu Ae
• yielding governs
• minimum

Marks

3

Topic

Staggered Holes — Net Area

Difficulty

medium

Template Id

T6

Examiner Tip

Always compare ALL failure paths (straight and zigzag) and state which gives the smallest net area — this comparison earns an often-overlooked mark.

Model Answer

GIVEN: Plate width Wg = 200 mm, thickness t = 10 mm dh = 22 mm, s = 60 mm, g = 80 mm Failure path: 2 holes + 1 stagger segment STEP 1 — Net Width Formula (NSCP 2015 / AISC 360-D3): wnet = Wg − Σdh + Σ(s²/4g) STEP 2 — Substitute: wnet = 200 − 2(22) + (60²)/(4 × 80) wnet = 200 − 44 + 3 600/320 wnet = 200 − 44 + 11.25 wnet = 167.25 mm STEP 3 — Net Area: An = wnet × t = 167.25 × 10 = 1 672.5 mm² CONCLUSION: The net width is 167.25 mm and the net area for this failure path is An = 1 672.5 mm². (Note: The straight path across 1 hole gives An = (200 − 22) × 10 = 1 780 mm². The staggered path (1 672.5 mm²) is smaller and therefore governs.)

Question Type

numerical

Answer Structure

• Step 1: Write the staggered net-width formula [1 mark]
• Step 2: Substitute values correctly (dh, s, g) and compute wnet [1 mark]
• Step 3: Compute An = wnet × t and state which path governs [1 mark]

Scoring Breakdown

Common Mark Deductions

• Forgetting the s²/4g term entirely — most common error in staggered-hole problems
• Using bolt diameter (20 mm) instead of hole diameter (22 mm)
• Not checking the straight-path net area for comparison
• Computing s²/4g incorrectly (e.g., dividing by 4g² instead of 4g)

Key Phrases To Include

• wnet = Wg − Σdh + Σ(s²/4g)
• stagger
• s = 60 mm
• g = 80 mm
• s²/4g = 11.25 mm
• governing net area

Marks

1

Topic

Slenderness Recommendation

Difficulty

easy

Template Id

T7

Examiner Tip

The word 'recommended' is a key differentiator from compression member limits. Use it explicitly.

Model Answer

Per NSCP 2015 (AISC 360 Section D1), the recommended slenderness limit for tension members is L/r ≤ 300. This limit is a recommendation, not a mandatory requirement (except for rods, which are exempt). It is provided to prevent excessive sag and vibration under service conditions.

Question Type

very_short_answer

Answer Structure

• Line 1: State L/r ≤ 300 with code reference [0.5 mark]
• Line 2: Clarify that it is recommended, not mandatory [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Stating L/r ≤ 300 is mandatory (it is only recommended)
• Citing the compression member limit (L/r ≤ 200) instead
• Omitting the reason (sag/vibration) when the question asks for an explanation

Key Phrases To Include

• L/r ≤ 300
• recommended
• not mandatory
• sag
• vibration

Marks

5

Topic

Angle Tension Member with Shear Lag

Difficulty

hard

Template Id

T8

Examiner Tip

In angle problems, rupture often governs because U < 1.0 substantially reduces Ae. Warn yourself at the start: 'angle + one leg = shear lag = U < 1.0.'

Model Answer

GIVEN: Angle: L75×75×8, Ag = 1 150 mm² Fy = 248 MPa, Fu = 400 MPa Bolt: 20 mm diameter → dh = 22 mm Connected leg thickness = 8 mm U = 0.85 (shear lag, one leg connected) STEP 1 — Gross Area: Ag = 1 150 mm² (given) STEP 2 — Net Area: An = Ag − dh × t = 1 150 − 22 × 8 An = 1 150 − 176 = 974 mm² STEP 3 — Effective Net Area: Ae = U × An = 0.85 × 974 = 827.9 mm² STEP 4 — Limit State 1: Tensile Yielding (φ = 0.90) φPn,yield = 0.90 × 248 × 1 150 φPn,yield = 256 680 N = 256.7 kN STEP 5 — Limit State 2: Tensile Rupture (φ = 0.75) φPn,rupture = 0.75 × 400 × 827.9 φPn,rupture = 248 370 N = 248.4 kN STEP 6 — Governing Design Strength: φPn = min(256.7, 248.4) = 248.4 kN CONCLUSION: The design tensile strength is φPn = 248.4 kN. Tensile rupture on the effective net section governs.

Question Type

numerical

Answer Structure

• Step 1: Identify Ag (given) [implicit]
• Step 2: Compute An = Ag − dh × t (one hole, one leg) [1 mark]
• Step 3: Compute Ae = U × An with U = 0.85 [1 mark]
• Step 4: Compute φPn,yield = 0.90 × 248 × 1 150 [1 mark]
• Step 5: Compute φPn,rupture = 0.75 × 400 × 827.9 [1 mark]
• Step 6: Select minimum and state governing limit state [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using U = 1.0 (ignoring shear lag for an angle bolted through one leg)
• Applying U to Ag instead of An
• Using the wrong leg thickness (using 75 mm instead of 8 mm for hole deduction)
• Not comparing both limit states

Key Phrases To Include

• An = Ag − dh × t
• Ae = U × An = 0.85 × 974
• shear lag
• one leg connected
• φPn,yield = 0.90 Fy Ag
• φPn,rupture = 0.75 Fu Ae
• rupture governs

Marks

2

Topic

Slenderness Check

Difficulty

easy

Template Id

T9

Examiner Tip

Always show the unit conversion (5.0 m → 5 000 mm) explicitly — examiners give credit for dimensional consistency.

Model Answer

GIVEN: Length L = 5.0 m = 5 000 mm Least radius of gyration r = 15 mm STEP 1 — Compute Slenderness Ratio: L/r = 5 000 / 15 = 333.3 STEP 2 — Compare with Recommendation: NSCP 2015 (AISC 360 D1) recommends L/r ≤ 300 for tension members. 333.3 > 300 → NOT SATISFIED. CONCLUSION: The slenderness ratio L/r = 333.3 exceeds the recommended limit of 300. Although this is not a mandatory failure, the member may experience unacceptable sag or vibration and the designer should select a section with larger r.

Question Type

numerical

Answer Structure

• Step 1: Compute L/r = 5 000 / 15 = 333.3 [1 mark]
• Step 2: Compare with limit 300 and state pass/fail + recommendation [1 mark]

Scoring Breakdown

Common Mark Deductions

• Forgetting to convert L to mm (using L = 5.0 directly gives L/r = 0.33 — nonsensical)
• Stating the member 'fails' without clarifying that the limit is only a recommendation
• Not providing a conclusion or remediation comment

Key Phrases To Include

• L/r = 333.3
• recommended limit 300
• not satisfied
• recommendation not mandatory
• sag or vibration

Marks

3

Topic

Staggered Holes — Formula Derivation

Difficulty

medium

Template Id

T10

Examiner Tip

If you sketch the bolt pattern showing the diagonal failure path, you earn the explanation mark visually — a 30-second sketch saves 3 lines of written explanation.

Model Answer

The staggered net-width formula accounts for a diagonal failure path through staggered holes: wnet = Wg − Σdh + Σ(s²/4g) Where: wnet = net width of the failure path (mm) Wg = gross width of the plate (mm) Σdh = sum of all hole diameters on the failure path (mm); each dh = bolt diameter + 2 mm allowance s = longitudinal center-to-center spacing (pitch) between adjacent holes on the diagonal path (mm) g = transverse center-to-center spacing (gage) between adjacent bolt lines (mm) s²/4g = empirical correction added for each diagonal segment to account for the longer path Physical Basis: A diagonal path through staggered holes is longer than a straight-line path. The term s²/4g (derived from the Cochrane formula) partially compensates for this longer path by adding back some effective width. However, the net deduction is still larger than for a straight path through fewer holes when multiple staggered holes exist. Note: All possible failure paths (straight and zigzag) must be evaluated; the path giving the smallest wnet (hence smallest An) governs.

Question Type

short_answer

Answer Structure

• Line 1–2: Write the formula with correct symbols [1 mark]
• Lines 3–8: Define each symbol (Wg, dh, s, g, s²/4g) with physical meaning [1 mark]
• Lines 9–12: Explain the physical basis of the s²/4g term and the need to check all paths [1 mark]

Scoring Breakdown

Common Mark Deductions

• Subtracting s²/4g instead of adding it
• Confusing pitch (s, longitudinal) with gage (g, transverse)
• Omitting the note about checking all failure paths

Key Phrases To Include

• wnet = Wg − Σdh + Σ(s²/4g)
• pitch s
• gage g
• diagonal failure path
• Cochrane formula
• all failure paths must be checked
• smallest net area governs

Marks

3

Topic

Sizing a Tension Member

Difficulty

medium

Template Id

T11

Examiner Tip

Always round UP when sizing members — a 33.78 mm rod means you must use 35 mm (next commercial size), never 33 mm. Write 'round up' explicitly.

Model Answer

GIVEN: Pu = 200 kN = 200 000 N Fy = 248 MPa, Fu = 400 MPa Round solid rod → no bolt holes in the typical sizing (threaded ends may reduce area, but problem specifies yielding limit state) STEP 1 — Required Gross Area (Yielding): φPn ≥ Pu → 0.90 × Fy × Ag ≥ Pu Ag,req = Pu / (0.90 × Fy) Ag,req = 200 000 / (0.90 × 248) Ag,req = 200 000 / 223.2 = 896.1 mm² STEP 2 — Required Diameter: Ag = π d² / 4 → d = √(4 Ag / π) d = √(4 × 896.1 / π) d = √(1 140.9) = 33.78 mm STEP 3 — Select Standard Size: Use d = 35 mm (round up to next standard size) Check: Ag = π(35)²/4 = 962.1 mm² φPn = 0.90 × 248 × 962.1 = 214 740 N = 214.7 kN > 200 kN ✓ CONCLUSION: Use a 35 mm diameter solid round tie rod. φPn = 214.7 kN > Pu = 200 kN — yielding limit state satisfied.

Question Type

numerical

Answer Structure

• Step 1: Apply φPn ≥ Pu for yielding to find Ag,req [1 mark]
• Step 2: Convert Ag to required diameter using Ag = πd²/4 [1 mark]
• Step 3: Round up to practical size and verify φPn > Pu [1 mark]

Scoring Breakdown

Common Mark Deductions

• Forgetting to convert Pu from kN to N (200 vs. 200 000)
• Rounding down the diameter (33 mm instead of 35 mm) — unsafe design
• Omitting the final verification check

Key Phrases To Include

• φPn ≥ Pu
• Ag,req = Pu / (0.90 Fy)
• Ag = πd²/4
• round up
• verify
• yielding limit state

Marks

2

Topic

Net Area vs. Effective Net Area

Difficulty

easy

Template Id

T12

Examiner Tip

Use a contrast structure ('An accounts for... whereas Ae accounts for...') — it signals clear understanding of the two-step area calculation.

Model Answer

Net Area (An): The gross cross-sectional area minus the material removed by bolt holes at the critical section. Formula: An = Ag − Σ(dh × t). It accounts only for the geometric reduction due to holes. Effective Net Area (Ae): The net area further reduced by the shear-lag factor U to reflect the non-uniform stress distribution when not all cross-sectional elements are connected at the joint. Formula: Ae = U × An, where U ≤ 1.0. An = Ae when U = 1.0, which occurs when the tensile force is transmitted through ALL elements of the cross-section (e.g., a plate fully bolted across its entire width, or a W-section with both flanges and web connected at the joint). In this case, shear lag is absent and no further reduction is needed.

Question Type

short_answer

Answer Structure

• Lines 1–2: Define An with formula [0.5 mark]
• Lines 3–4: Define Ae with formula [0.5 mark]
• Lines 5–6: State and justify when An = Ae (U = 1.0, all elements connected) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Defining An and Ae identically (not distinguishing shear lag)
• Stating An = Ae without justifying why (U = 1.0 condition not stated)
• Omitting examples or conditions

Key Phrases To Include

• An = Ag − Σ(dh × t)
• Ae = U × An
• shear lag
• U = 1.0
• all elements connected
• non-uniform stress

Marks

5

Topic

Full Adequacy Check — Tension Plate

Difficulty

medium

Template Id

T13

Examiner Tip

The adequacy check (Step 7) is often the easiest mark but is forgotten under time pressure. Always write 'φPn ___ Pu → ADEQUATE / INADEQUATE' as your last line.

Model Answer

GIVEN: Plate: 150 mm × 10 mm Fy = 250 MPa, Fu = 410 MPa Pu = 280 kN = 280 000 N Bolts: 2 × 16 mm → dh = 18 mm U = 1.0 STEP 1 — Gross Area: Ag = 150 × 10 = 1 500 mm² STEP 2 — Net Area: An = Ag − n(dh × t) = 1 500 − 2(18)(10) An = 1 500 − 360 = 1 140 mm² STEP 3 — Effective Net Area: Ae = U × An = 1.0 × 1 140 = 1 140 mm² STEP 4 — Yielding Capacity: φPn,yield = 0.90 × 250 × 1 500 = 337 500 N = 337.5 kN STEP 5 — Rupture Capacity: φPn,rupture = 0.75 × 410 × 1 140 = 350 550 N = 350.6 kN STEP 6 — Governing Capacity: φPn = min(337.5, 350.6) = 337.5 kN (yielding governs) STEP 7 — Adequacy Check: φPn = 337.5 kN > Pu = 280 kN ✓ ADEQUATE CONCLUSION: The 150×10 plate is adequate. Design strength φPn = 337.5 kN exceeds the factored load Pu = 280 kN. Yielding on the gross section governs.

Question Type

numerical

Answer Structure

• Steps 1–3: Area computations (Ag, An, Ae) [1 mark]
• Step 4: Yielding capacity [1 mark]
• Step 5: Rupture capacity [1 mark]
• Step 6: Governing value identified [1 mark]
• Step 7: Adequacy check (φPn vs. Pu) with conclusion [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using 16 mm instead of 18 mm for hole diameter
• Stopping after calculating one limit state
• Not writing the adequacy conclusion statement
• Errors in unit conversion (kN vs. N)

Key Phrases To Include

• φPn ≥ Pu
• ADEQUATE
• yielding governs
• minimum
• dh = 18 mm (not 16 mm)

Marks

3

Topic

Block Shear

Difficulty

hard

Template Id

T14

Examiner Tip

Board exams frequently ask for the formula with all symbols defined. Write the formula first, then define symbols below — this ordered approach avoids omissions.

Model Answer

Block Shear: A failure mode at bolt connections where a block of material tears out, combining SHEAR yielding or rupture along one plane and TENSION rupture along a perpendicular plane. It is a third limit state (in addition to gross yielding and net section rupture) that must be checked at the connection region. NSCP 2015 / AISC 360 Section J4.3 — Block Shear Rupture Strength: Rn = 0.60 Fu Anv + Ubs Fu Ant ≤ 0.60 Fy Agv + Ubs Fu Ant Where: Agv = gross area subject to shear Anv = net area subject to shear Ant = net area subject to tension Ubs = 1.0 for uniform tension stress; 0.50 for non-uniform tension stress φ = 0.75 (LRFD), Ω = 2.00 (ASD) The two expressions represent: (1) shear rupture + tension rupture [first term], and (2) shear yielding + tension rupture [second term]. The LOWER of the two governs (upper-bound limit on the shear term).

Question Type

short_answer

Answer Structure

• Lines 1–2: Define block shear (combined shear + tension failure at connection) [1 mark]
• Lines 3–5: Write the complete NSCP 2015 formula with both terms [1 mark]
• Lines 6–8: Define all symbols and identify which expression governs [1 mark]

Scoring Breakdown

Common Mark Deductions

• Confusing block shear with net-section rupture (they occur in different planes)
• Writing only one of the two expressions (the inequality is part of the formula)
• Omitting Ubs (and hence the distinction between uniform and non-uniform tension)

Key Phrases To Include

• block shear
• tearout
• shear rupture
• tension rupture
• Rn = 0.60 Fu Anv + Ubs Fu Ant
• Agv, Anv, Ant
• Ubs = 1.0 or 0.50
• φ = 0.75

Marks

3

Topic

Common Pitfalls

Difficulty

easy

Template Id

T15

Examiner Tip

For 'list and explain' questions, always match each item with an explicit avoidance action — passive lists earn 50% of available marks; action-based answers earn full marks.

Model Answer

Four Common Pitfalls and How to Avoid Them: 1. Using bolt diameter (db) instead of hole diameter (dh) for net area: Avoid by always writing dh = db + 2 mm (or +3 mm) before computing An. 2. Skipping a limit state (checking only yield or only rupture): Avoid by using a two-column layout: compute both φPn,yield and φPn,rupture, then take the minimum. 3. Applying U = 1.0 to angles/tees bolted through one leg: Avoid by flagging: 'angle + one leg connected → shear lag → U < 1.0' at the start of the problem. 4. Using the wrong resistance factor (φ = 0.90 for rupture or φ = 0.75 for yield): Avoid by writing a memory rule: 'Yield = 0.90 (higher φ, lower F); Rupture = 0.75 (lower φ, higher F).'

Question Type

short_answer

Answer Structure

• Pitfall 1: Hole diameter error + avoidance strategy [0.75 mark]
• Pitfall 2: Missing limit state + avoidance strategy [0.75 mark]
• Pitfall 3: Shear lag assumption + avoidance strategy [0.75 mark]
• Pitfall 4: Wrong φ factor + memory aid [0.75 mark]

Scoring Breakdown

Common Mark Deductions

• Listing pitfalls without avoidance strategies
• Repeating the same pitfall in different words
• Vague answers like 'be careful' without specific guidance

Key Phrases To Include

• dh = db + 2 mm
• both limit states
• shear lag
• U < 1.0
• φ = 0.90 for yield
• φ = 0.75 for rupture