CELE Steel & Timber Design — Steel Beams: Flexure and ShearExam Answer Templates

Marks

1

Topic

Plastic Moment Capacity

Difficulty

easy

Template Id

T1

Examiner Tip

Even at 1 mark, examiners want to see Z_x, not S_x. The single most common error on this question in PRC board exams is confusing the plastic modulus with the elastic section modulus.

Model Answer

The plastic moment M_p is the internal bending moment at which the entire cross-section has yielded in tension and compression, given by M_p = F_y Z_x, where Z_x is the plastic section modulus. It represents the maximum moment capacity of a compact steel section.

Question Type

very_short_answer

Answer Structure

• Line 1: State that M_p is the moment when the full cross-section yields AND give the formula M_p = F_y Z_x [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing M_p = F_y S_x (using elastic modulus instead of plastic modulus) — zero marks
• Defining it as 'yield moment' M_y = F_y S_x, which is a different quantity
• Omitting the formula and giving only a vague verbal description

Key Phrases To Include

• entire cross-section yielded
• F_y Z_x
• plastic section modulus
• maximum moment capacity
• compact section

Marks

2

Topic

Lateral-Torsional Buckling — Limiting Lengths

Difficulty

easy

Template Id

T2

Examiner Tip

The coefficient 1.76 is specific to doubly-symmetric I-shapes. Examiners look for the √(E/F_y) term; if you write the formula structurally correct but cannot explain its implication for brace spacing, you earn only 1 of the 2 marks.

Model Answer

The limiting unbraced length L_p is given by: L_p = 1.76 r_y √(E / F_y) where r_y = radius of gyration about the weak axis (mm), E = modulus of elasticity (200,000 MPa for steel), and F_y = yield stress (MPa). Significance: When the compression-flange brace spacing L_b ≤ L_p, lateral-torsional buckling (LTB) does not occur and the beam can develop its full plastic moment M_p = F_y Z_x. If L_b exceeds L_p, the nominal flexural strength M_n is reduced below M_p.

Question Type

short_answer

Answer Structure

• Line 1: State the formula L_p = 1.76 r_y √(E/F_y) with variables defined [1 mark]
• Line 2: Explain significance — L_b ≤ L_p means full M_p is available; L_b > L_p means LTB reduces M_n [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using coefficient 1.76 incorrectly (e.g., writing 1.67 or 1.73)
• Omitting the significance part and only giving the formula
• Mixing up L_p with L_r (the limit for elastic LTB)

Key Phrases To Include

• 1.76 r_y
• √(E/F_y)
• lateral-torsional buckling
• full plastic moment
• compression flange bracing

Marks

2

Topic

Compact vs. Non-Compact Sections

Difficulty

easy

Template Id

T3

Examiner Tip

Two-mark concept questions follow a definition + consequence structure. State what the section satisfies (λ ≤ λ_p) and what that allows (full M_p). Both parts must appear for full credit.

Model Answer

A compact section is a steel cross-section whose flange and web width-to-thickness ratios (λ) are both below the plastic limit λ_p: Flange: b_f / (2t_f) ≤ 0.38√(E/F_y) Web: h / t_w ≤ 3.76√(E/F_y) Importance: A compact section can reach the full plastic moment M_p = F_y Z_x without premature local buckling. A non-compact or slender section buckles locally before full yielding, so M_n must be reduced below M_p.

Question Type

short_answer

Answer Structure

• Line 1: Define compact section — all plate elements satisfy λ ≤ λ_p [1 mark]
• Line 2: State the consequence — compact section reaches M_p; non-compact/slender sections have reduced M_n [1 mark]

Scoring Breakdown

Common Mark Deductions

• Describing 'compact' only verbally without referencing width-to-thickness ratios
• Confusing compactness with lateral bracing requirements
• Not stating the consequence for M_n

Key Phrases To Include

• width-to-thickness ratio
• plastic limit λ_p
• local buckling
• M_p = F_y Z_x
• non-compact
• slender

Marks

2

Topic

Plastic Moment Capacity

Difficulty

easy

Template Id

T4

Examiner Tip

For 2-mark numerical problems, the formula line and the arithmetic line each carry a mark. Write both clearly. Never skip the formula.

Model Answer

Given: Z_x = 1.50 × 10⁶ mm³ F_y = 248 MPa Section is compact and fully braced → M_n = M_p Formula (AISC 360-10 / NSCP 2015 Sec. 502.3): M_p = F_y Z_x φ_b = 0.90 Solution: M_p = 248 N/mm² × 1.50 × 10⁶ mm³ = 3.720 × 10⁸ N·mm = 372.0 kN·m φ_b M_n = 0.90 × 372.0 = 334.8 kN·m ∴ Design flexural strength = 334.8 kN·m

Question Type

numerical

Answer Structure

• Line 1: State Given data and note that M_n = M_p applies (compact + braced) [0.5 mark]
• Line 2: Write formula M_p = F_y Z_x and φ_b = 0.90 [0.5 mark]
• Line 3: Substitute and compute M_p in kN·m [0.5 mark]
• Line 4: Multiply by φ_b and state final answer with units [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using S_x in place of Z_x (wrong modulus)
• Using φ_b = 1.0 instead of 0.90
• Forgetting to convert N·mm to kN·m (answer left in N·mm without conversion note)
• Not stating the condition M_n = M_p and why it applies

Key Phrases To Include

• M_n = M_p
• F_y Z_x
• φ_b = 0.90
• compact and fully braced
• 334.8 kN·m

Marks

3

Topic

Shear Strength of Steel Beams

Difficulty

medium

Template Id

T5

Examiner Tip

The most common shear error is using clear height h instead of overall depth d for A_w. AISC 360 Section G2.1 explicitly uses d (total depth). State this explicitly to show code awareness.

Model Answer

Given: d = 500 mm t_w = 11 mm F_y = 248 MPa C_v = 1.0 (stocky web, h/t_w ≤ 2.24√(E/F_y)) φ_v = 1.0 (rolled I-shape with compact web per AISC 360-10 Sec. G2.1) Step 1 — Web shear area: A_w = d × t_w = 500 × 11 = 5,500 mm² Step 2 — Nominal shear strength: V_n = 0.6 F_y A_w C_v = 0.6 × 248 N/mm² × 5,500 mm² × 1.0 = 818,400 N = 818.4 kN Step 3 — Design shear strength: φ_v V_n = 1.0 × 818.4 = 818.4 kN ∴ φ_v V_n = 818.4 kN

Question Type

numerical

Answer Structure

• Line 1: Write Given data, state C_v = 1.0 and φ_v = 1.0 with justification [0.5 mark]
• Line 2: Compute A_w = d × t_w with units [1 mark]
• Line 3: Apply V_n = 0.6 F_y A_w C_v and substitute [1 mark]
• Line 4: Apply φ_v and state final answer in kN [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using A_w = h × t_w (clear height) instead of d × t_w — AISC 360 uses full depth d
• Using φ_v = 0.90 instead of 1.0 for compact-web rolled I-shapes
• Omitting C_v and its basis (C_v = 1.0 for h/t_w ≤ 2.24√(E/F_y))
• Not converting N to kN in the final answer

Key Phrases To Include

• A_w = d × t_w
• V_n = 0.6 F_y A_w C_v
• C_v = 1.0
• φ_v = 1.0
• stocky web
• 818.4 kN

Marks

3

Topic

Lateral-Torsional Buckling — Limiting Lengths

Difficulty

medium

Template Id

T6

Examiner Tip

Always show the square root step explicitly — examiners cannot award the intermediate mark if they cannot see whether you computed √(E/F_y) or just (E/F_y).

Model Answer

Given: r_y = 35 mm F_y = 345 MPa E = 200,000 MPa Formula (AISC 360-10 / NSCP 2015): L_p = 1.76 r_y √(E / F_y) Step 1 — Compute √(E / F_y): E/F_y = 200,000 / 345 = 579.71 √579.71 = 24.08 Step 2 — Compute L_p: L_p = 1.76 × 35 mm × 24.08 = 61.6 × 24.08 = 1,483 mm ≈ 1.48 m ∴ L_p = 1.48 m Interpretation: The compression flange must be braced at intervals ≤ 1.48 m for the beam to develop its full plastic moment M_p.

Question Type

numerical

Answer Structure

• Line 1: Write Given data [0 marks — organizational, but required]
• Line 2: State correct formula L_p = 1.76 r_y √(E/F_y) [1 mark]
• Line 3: Compute √(E/F_y) = 24.08 [1 mark]
• Line 4: Compute L_p = 1,483 mm = 1.48 m and state interpretation [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using wrong coefficient (e.g., 1.67 or 1.73) instead of 1.76
• Forgetting to take the square root of (E/F_y) — computing E/F_y instead of √(E/F_y)
• Not converting mm to m in the final answer
• No interpretation of what L_p means for brace spacing

Key Phrases To Include

• 1.76 r_y √(E/F_y)
• r_y = 35 mm
• F_y = 345 MPa
• 1.48 m
• full plastic moment
• brace spacing

Marks

3

Topic

Lateral-Torsional Buckling — Limiting Lengths

Difficulty

medium

Template Id

T7

Examiner Tip

In a 3-mark question covering three zones, structure your answer into three clearly labeled parts. One zone = one mark. If the examiner cannot identify your three zones, they cannot award the marks.

Model Answer

Steel I-beams exhibit three LTB zones depending on the compression-flange unbraced length L_b (AISC 360-10 / NSCP 2015 Sec. 502.3.2): Zone 1 — No LTB (L_b ≤ L_p): The beam reaches full plastic moment. M_n = M_p = F_y Z_x Zone 2 — Inelastic LTB (L_p < L_b ≤ L_r): Yielding and LTB interact; M_n decreases linearly from M_p to 0.7 F_y S_x: M_n = C_b [M_p − (M_p − 0.7 F_y S_x)((L_b − L_p)/(L_r − L_p))] ≤ M_p Zone 3 — Elastic LTB (L_b > L_r): Beam buckles elastically before yielding. M_n = F_cr S_x ≤ M_p where F_cr is the elastic LTB stress (function of L_b, r_ts, J, c, h_o). Note: C_b is the moment-gradient factor (C_b = 1.0 for uniform moment; C_b > 1.0 for non-uniform moment, which increases M_n, capped at M_p).

Question Type

short_answer

Answer Structure

• Part 1: Zone 1 — L_b ≤ L_p, M_n = M_p = F_y Z_x [1 mark]
• Part 2: Zone 2 — L_p < L_b ≤ L_r, linear interpolation formula with C_b [1 mark]
• Part 3: Zone 3 — L_b > L_r, elastic LTB, M_n = F_cr S_x [1 mark]

Scoring Breakdown

Common Mark Deductions

• Describing only two zones instead of three
• Interchanging Zone 2 and Zone 3 behavior
• Omitting the M_n formula for each zone
• Not mentioning C_b at all for Zone 2

Key Phrases To Include

• L_b ≤ L_p
• M_n = M_p
• inelastic LTB
• L_p < L_b ≤ L_r
• elastic LTB
• L_b > L_r
• F_cr S_x
• C_b

Marks

1

Topic

Plastic Moment Capacity

Difficulty

easy

Template Id

T8

Examiner Tip

One-mark definition questions must contain both the formula and the value. A verbal description alone earns zero marks in numerical-intensive subjects like Steel Design.

Model Answer

The shape factor f is the ratio of the plastic section modulus Z_x to the elastic section modulus S_x: f = Z_x / S_x For doubly-symmetric wide-flange I-shapes, f ≈ 1.12 (range: 1.10–1.15). It quantifies the additional moment capacity beyond first yield that a compact section can develop through plasticity.

Question Type

very_short_answer

Answer Structure

• Line 1: Define shape factor f = Z_x / S_x and state typical value ≈ 1.12 for I-shapes [1 mark]

Scoring Breakdown

Common Mark Deductions

• Inverting the ratio (writing S_x / Z_x)
• Giving the shape factor for a rectangular section (1.5) instead of an I-shape
• Omitting the numerical value

Key Phrases To Include

• Z_x / S_x
• shape factor
• 1.12
• I-shape
• plastic section modulus
• elastic section modulus

Marks

1

Topic

Shear Strength of Steel Beams

Difficulty

easy

Template Id

T9

Examiner Tip

The coefficient 2.24 is distinct from 2.46 used in earlier AISC editions. Board exam writers test this specific coefficient frequently. Memorize it.

Model Answer

C_v = 1.0 when the web is stocky, specifically when: h / t_w ≤ 2.24 √(E / F_y) For this condition, shear yielding governs and the full shear capacity V_n = 0.6 F_y A_w is developed without shear buckling. For most standard rolled W-shapes, this criterion is satisfied.

Question Type

very_short_answer

Answer Structure

• Line 1: State the criterion h/t_w ≤ 2.24√(E/F_y) for C_v = 1.0 [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using 2.46 (the limit for φ_v = 1.0 in older editions) instead of 2.24
• Writing the criterion as an equality rather than inequality
• Not including √(E/F_y)

Key Phrases To Include

• h/t_w ≤ 2.24√(E/F_y)
• C_v = 1.0
• shear yielding
• stocky web
• 2.24

Marks

5

Topic

Plastic Moment Capacity

Difficulty

medium

Template Id

T10

Examiner Tip

Five-mark problems always have at least three sub-parts. Structure your answer with clearly labeled (a), (b), (c). Examiners grade sub-parts independently — a wrong (a) does not disqualify you from marks in (b) and (c) if your method is correct.

Model Answer

Given: Span L = 6 m = 6,000 mm w_u = 40 kN/m (factored uniform load) F_y = 248 MPa Z_x = 3.0 × 10⁶ mm³ (proposed section) Section: compact, fully braced → M_n = M_p φ_b = 0.90 (AISC 360-10 / NSCP 2015) (a) Maximum Factored Moment: M_u = w_u L² / 8 = (40 kN/m)(6 m)² / 8 = 40 × 36 / 8 = 180 kN·m (b) Required Plastic Section Modulus: Demand: M_u ≤ φ_b M_n = φ_b F_y Z_x Required Z_x = M_u / (φ_b F_y) = 180 × 10⁶ N·mm / (0.90 × 248 N/mm²) = 180 × 10⁶ / 223.2 = 806,624 mm³ ≈ 8.07 × 10⁵ mm³ (c) Adequacy Check for Z_x = 3.0 × 10⁶ mm³: φ_b M_n = φ_b F_y Z_x = 0.90 × 248 N/mm² × 3.0 × 10⁶ mm³ = 6.696 × 10⁸ N·mm = 669.6 kN·m Check: φ_b M_n = 669.6 kN·m > M_u = 180 kN·m ✓ ∴ The section is ADEQUATE. (Demand-to-capacity ratio = 180/669.6 = 0.27; the section is over-designed.)

Question Type

numerical

Answer Structure

• Part (a): State formula M_u = w_u L²/8 and compute M_u = 180 kN·m [1 mark]
• Part (b): Set up M_u = φ_b F_y Z_x, rearrange for required Z_x, compute Z_x = 8.07 × 10⁵ mm³ [2 marks]
• Part (c): Compute φ_b M_n for given Z_x = 3.0 × 10⁶ mm³, compare with M_u, state adequacy conclusion [2 marks]

Scoring Breakdown

Common Mark Deductions

• Using M_u = w_u L²/4 (point load formula instead of UDL formula) — loses part (a) mark
• Using φ_b = 1.0 instead of 0.90 for flexure
• Not explicitly stating the comparison φ_b M_n vs. M_u — the conclusion mark requires the comparison statement
• Mixing units: using kN/m for w_u but mm for L without conversion
• Using S_x (elastic modulus) instead of Z_x for M_p

Key Phrases To Include

• M_u = w_u L²/8
• φ_b = 0.90
• M_u ≤ φ_b M_n
• required Z_x
• 669.6 kN·m
• ADEQUATE
• compact and fully braced

Marks

3

Topic

Lateral-Torsional Buckling — Limiting Lengths

Difficulty

medium

Template Id

T11

Examiner Tip

In LTB zone classification problems, the comparison statement is mandatory — do not just compute L_p and stop. The examiner wants to see you apply the result.

Model Answer

Given: r_y = 38.6 mm F_y = 248 MPa E = 200,000 MPa L_b = 3,000 mm Step 1 — Compute L_p: L_p = 1.76 r_y √(E/F_y) = 1.76 × 38.6 × √(200,000 / 248) = 1.76 × 38.6 × √806.45 = 1.76 × 38.6 × 28.40 = 1,929 mm ≈ 1.93 m Step 2 — Classify LTB Zone: L_b = 3.0 m vs. L_p = 1.93 m L_b > L_p ∴ Zone: L_b > L_p — at minimum, Inelastic LTB governs (Zone 2 if L_b ≤ L_r; elastic LTB if L_b > L_r). Full M_p is NOT available; M_n is reduced below M_p. A check against L_r (requiring J, c, r_ts, h_o section properties) is needed to confirm whether Zone 2 or Zone 3 governs.

Question Type

numerical

Answer Structure

• Line 1: Write formula L_p = 1.76 r_y √(E/F_y) [0.5 mark]
• Line 2: Compute √(E/F_y) = 28.40 [0.5 mark]
• Line 3: Compute L_p = 1,929 mm = 1.93 m [1 mark]
• Line 4: Compare L_b = 3.0 m > L_p = 1.93 m → LTB reduces M_n, Zone 2 minimum [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating M_n = M_p despite L_b > L_p — fundamental conceptual error
• Not comparing L_b with L_p after computing L_p
• Rounding E/F_y before taking square root, leading to significant rounding error

Key Phrases To Include

• L_p = 1.93 m
• L_b = 3.0 m > L_p
• inelastic LTB
• M_n < M_p
• L_r check required
• Zone 2

Marks

2

Topic

Lateral-Torsional Buckling — Limiting Lengths

Difficulty

medium

Template Id

T12

Examiner Tip

The M_p cap on C_b is critical — without it, the answer is incomplete. Examiners test whether you know that C_b can never push M_n above M_p.

Model Answer

C_b is the lateral-torsional buckling modification factor that accounts for the beneficial effect of non-uniform bending moment along the unbraced length. A non-uniform moment diagram is less severe than uniform moment because only part of the length is subjected to the maximum compressive stress; hence LTB is resisted more effectively. Effect: C_b multiplies M_n in the LTB equations, potentially increasing the design flexural strength — but M_n is always capped at M_p: M_n(with C_b) ≤ M_p C_b = 1.0 when the bending moment is uniform (constant) along the entire unbraced length — this is the most critical (worst-case) condition for LTB. Using C_b = 1.0 for all cases is conservative.

Question Type

short_answer

Answer Structure

• Line 1: Define C_b as moment-gradient factor that increases M_n for non-uniform moment, capped at M_p [1 mark]
• Line 2: State C_b = 1.0 for uniform (constant) moment along unbraced length [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating C_b < 1.0 for uniform moment (it equals 1.0, not less than 1.0)
• Saying C_b can increase M_n without noting the M_p cap
• Confusing C_b with Cb used in column design (different context)

Key Phrases To Include

• moment-gradient factor
• non-uniform moment
• capped at M_p
• C_b = 1.0
• uniform moment
• conservative

Marks

5

Topic

Shear Strength of Steel Beams

Difficulty

hard

Template Id

T13

Examiner Tip

Part (a) sets up part (b). Even if you get C_v wrong, show all your V_n steps — partial marks are awarded for correct methodology even with a wrong C_v value.

Model Answer

Given: d = 600 mm t_w = 12 mm F_y = 248 MPa E = 200,000 MPa (Assume h ≈ d = 600 mm for the stocky-web check, conservative) (a) Web Slenderness Check (C_v criterion): Limit = 2.24 √(E/F_y) = 2.24 × √(200,000/248) = 2.24 × √806.45 = 2.24 × 28.40 = 63.6 Actual h/t_w = 600 / 12 = 50.0 50.0 ≤ 63.6 ✓ → C_v = 1.0 and φ_v = 1.0 (b) Design Shear Strength: Step 1 — Shear area: A_w = d × t_w = 600 × 12 = 7,200 mm² Step 2 — Nominal shear strength: V_n = 0.6 F_y A_w C_v = 0.6 × 248 N/mm² × 7,200 mm² × 1.0 = 1,071,360 N = 1,071.4 kN Step 3 — Design shear strength: φ_v V_n = 1.0 × 1,071.4 = 1,071.4 kN ∴ φ_v V_n = 1,071.4 kN

Question Type

numerical

Answer Structure

• Part (a) Line 1: Compute limit 2.24√(E/F_y) = 63.6 [1 mark]
• Part (a) Line 2: Compute h/t_w = 50.0, compare with 63.6, state C_v = 1.0 and φ_v = 1.0 [1 mark]
• Part (b) Line 1: Compute A_w = d × t_w = 7,200 mm² [1 mark]
• Part (b) Line 2: Apply V_n = 0.6 F_y A_w C_v, substitute and compute [1 mark]
• Part (b) Line 3: Apply φ_v = 1.0, state final φ_v V_n = 1,071.4 kN [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using 2.46 instead of 2.24 for the C_v criterion
• Using t_w² instead of d × t_w for A_w
• Using φ_v = 0.90 instead of 1.0 after correctly confirming C_v = 1.0
• Not showing the comparison h/t_w ≤ 2.24√(E/F_y) explicitly

Key Phrases To Include

• 2.24√(E/F_y) = 63.6
• h/t_w = 50.0
• C_v = 1.0
• φ_v = 1.0
• A_w = d × t_w
• 1,071.4 kN

Marks

2

Topic

Plastic Moment Capacity

Difficulty

easy

Template Id

T14

Examiner Tip

The conceptual differentiation must include both the formula AND the physical meaning (linear vs. rectangular stress block). Formulae alone without physical interpretation score 1 of 2 marks.

Model Answer

Elastic section modulus S_x: S_x = I_x / c It represents the moment capacity at first yielding (when the extreme fiber just reaches F_y): M_y = F_y S_x Based on linear (triangular) stress distribution across the cross-section. Plastic section modulus Z_x: Z_x = A_top × ȳ_top + A_bot × ȳ_bot (first moment of area of half-sections about the plastic neutral axis) It represents the moment capacity when the entire cross-section has yielded: M_p = F_y Z_x Based on rectangular (fully plastic) stress blocks. Why Z_x > S_x: The plastic neutral axis divides the section into equal areas (not equal distances), so material farther from the centroid contributes more than in the elastic distribution, resulting in Z_x > S_x. Their ratio f = Z_x/S_x ≈ 1.12 for I-shapes (1.5 for rectangles).

Question Type

short_answer

Answer Structure

• Line 1: Define S_x = I_x/c → M_y = F_y S_x, linear stress distribution [1 mark]
• Line 2: Define Z_x (first moment of half-areas) → M_p = F_y Z_x, rectangular stress blocks; explain why Z_x > S_x [1 mark]

Scoring Breakdown

Common Mark Deductions

• Defining Z_x as I_x/c (that is S_x, not Z_x)
• Not explaining WHY Z_x > S_x (missing the equal-area / stress distribution explanation)
• Reversing the two definitions

Key Phrases To Include

• S_x = I_x/c
• M_y = F_y S_x
• first yielding
• Z_x
• M_p = F_y Z_x
• equal area
• plastic neutral axis
• Z_x > S_x

Marks

5

Topic

Plastic Moment Capacity

Difficulty

medium

Template Id

T15

Examiner Tip

Five-part sub-questions allow partial credit recovery. Even if you err on M_p, the shape factor f = Z_x/S_x uses given data directly — always answer every sub-part.

Model Answer

Given: Z_x = 2.5 × 10⁶ mm³ S_x = 2.2 × 10⁶ mm³ F_y = 345 MPa Beam: compact, fully braced → M_n = M_p φ_b = 0.90 (a) Plastic Moment M_p: M_p = F_y Z_x = 345 N/mm² × 2.5 × 10⁶ mm³ = 862.5 × 10⁶ N·mm = 862.5 kN·m (b) Yield Moment M_y: M_y = F_y S_x = 345 N/mm² × 2.2 × 10⁶ mm³ = 759.0 × 10⁶ N·mm = 759.0 kN·m (c) Design Flexural Strength φ_b M_n: Since beam is compact and fully braced: M_n = M_p φ_b M_n = 0.90 × 862.5 = 776.3 kN·m (d) Shape Factor f: f = Z_x / S_x = 2.5 × 10⁶ / 2.2 × 10⁶ = 1.136 ∴ Summary: M_p = 862.5 kN·m M_y = 759.0 kN·m φ_b M_n = 776.3 kN·m f = 1.136 (within typical range 1.10–1.15 for I-shapes ✓)

Question Type

numerical

Answer Structure

• Part (a): M_p = F_y Z_x = 862.5 kN·m [1 mark]
• Part (b): M_y = F_y S_x = 759.0 kN·m [1 mark]
• Part (c): φ_b M_n = 0.90 × M_p = 776.3 kN·m, state M_n = M_p because compact + braced [2 marks]
• Part (d): f = Z_x/S_x = 1.136 [1 mark]

Scoring Breakdown

Common Mark Deductions

• Swapping Z_x and S_x in the formulas for M_p and M_y
• Using φ_b = 1.0 for the design strength
• Not justifying why M_n = M_p (must say compact + fully braced)
• Computing f = S_x/Z_x (inverted ratio)

Key Phrases To Include

• M_p = F_y Z_x
• M_y = F_y S_x
• compact and fully braced
• M_n = M_p
• φ_b = 0.90
• 776.3 kN·m
• f = Z_x/S_x = 1.136