CELE Strength of Materials — Stresses in BeamsExam Answer Templates

Marks

1

Topic

Section Modulus

Difficulty

easy

Template Id

T1

Examiner Tip

One mark means one fact. Give the formula AND the unit — both are required for full credit. Do not elaborate unnecessarily.

Model Answer

The section modulus S is the ratio of the moment of inertia I of a cross-section about its neutral (centroidal) axis to the distance c from the neutral axis to the extreme fiber: S = I/c. Its SI unit is mm³.

Question Type

very_short_answer

Answer Structure

• Line 1: State the formula S = I/c with identification of terms [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating S = bh²/6 without the general formula S = I/c (formula is section-specific, not the definition)
• Omitting the SI unit
• Confusing S (mm³) with I (mm⁴)

Key Phrases To Include

• S = I/c
• moment of inertia
• neutral axis
• extreme fiber
• mm³

Marks

1

Topic

Flexure Formula

Difficulty

easy

Template Id

T2

Examiner Tip

The examiner wants to see that you know the formula is σ = My/I (general) and that the maximum occurs at y = c. Citing both forms distinguishes a well-prepared answer.

Model Answer

The flexure formula is: σ = My/I, where σ = bending stress (MPa), M = bending moment (N·mm), y = distance from neutral axis to the point of interest (mm), and I = moment of inertia about the neutral axis (mm⁴). Maximum bending stress occurs at y = c (extreme fiber): σ_max = Mc/I = M/S.

Question Type

very_short_answer

Answer Structure

• Line 1: Write the formula σ = My/I [0.5 mark]
• Line 2: Define at least two terms with correct units [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Writing σ = M/I without the y term
• Incorrect units (e.g., stating I in m⁴ without converting)
• Not distinguishing between y (general) and c (extreme fiber)

Key Phrases To Include

• σ = My/I
• bending moment M
• moment of inertia I
• neutral axis
• MPa
• σ_max = Mc/I

Marks

2

Topic

Horizontal Shear Stress

Difficulty

easy

Template Id

T3

Examiner Tip

This is a 2-mark question. Allocate one mark per major item: the formula, then Q. Write Q = A'ȳ' explicitly — this phrase separates prepared students from those guessing.

Model Answer

The horizontal shear stress formula is: τ = VQ / (Ib) Where: • V = transverse shear force at the section (N) • Q = first moment of area of the cross-section above (or below) the level of interest, taken about the neutral axis (mm³) • I = moment of inertia of the entire cross-section about the neutral axis (mm⁴) • b = width of the cross-section at the level where τ is computed (mm) • τ = shear stress at that level (MPa) Q is computed as Q = A'·ȳ', where A' is the area of the portion beyond the level of interest and ȳ' is the distance from the neutral axis to the centroid of that area.

Question Type

short_answer

Answer Structure

• Line 1: State the formula τ = VQ/(Ib) [1 mark]
• Line 2: Define Q correctly as first moment of area with formula Q = A'ȳ' [1 mark]

Scoring Breakdown

Common Mark Deductions

• Defining Q as 'moment of inertia' instead of 'first moment of area'
• Not specifying that b is the width at the specific level where stress is computed
• Omitting the formula Q = A'ȳ' — just saying 'first moment' without the computation method

Key Phrases To Include

• τ = VQ/Ib
• first moment of area
• Q = A'ȳ'
• neutral axis
• width b at the level of interest
• MPa

Marks

2

Topic

Stress Distribution in Beams

Difficulty

easy

Template Id

T4

Examiner Tip

The word 'differ' in the question is a signal — the examiner expects a direct comparison. State the contrast explicitly: 'bending stress is max where shear stress is zero, and vice versa.'

Model Answer

Bending Stress Distribution: Bending stress σ = My/I varies linearly with depth. It is zero at the neutral axis (centroid) and maximum (tensile or compressive) at the extreme fibers (top and bottom). For a sagging moment, the bottom fiber is in tension and the top fiber is in compression. Shear Stress Distribution: Shear stress τ = VQ/(Ib) varies parabolically with depth. It is zero at the top and bottom free surfaces (where Q = 0) and maximum at the neutral axis. For a rectangle: τ_max = 1.5V/A. Key Difference: Bending stress is maximum at the extreme fibers and zero at the NA; shear stress is maximum at the NA and zero at the extreme fibers — they are opposite in distribution.

Question Type

short_answer

Answer Structure

• Part 1: Describe bending stress — linear, zero at NA, max at extreme fibers [1 mark]
• Part 2: Describe shear stress — parabolic, zero at extreme fibers, max at NA; state the opposite distribution [1 mark]

Scoring Breakdown

Common Mark Deductions

• Saying both stresses are maximum at the same location
• Not mentioning the parabolic nature of shear distribution
• Failing to state that shear stress is zero at the free surfaces

Key Phrases To Include

• linear distribution
• parabolic distribution
• zero at the neutral axis
• maximum at extreme fibers
• maximum at neutral axis
• zero at top and bottom surfaces
• τ_max = 1.5V/A

Marks

3

Topic

Flexure Formula — Rectangular Section

Difficulty

easy

Template Id

T5

Examiner Tip

For 3-mark numerical questions, each logical step is worth 1 mark. Even if your final answer is wrong due to arithmetic, you still earn marks for correctly identifying M_max and S formulas. Always show the formula before substituting.

Model Answer

Given: • b = 100 mm, h = 200 mm • Span L = 3 m, Concentrated load P = 10 kN = 10,000 N at midspan Required: Maximum bending stress σ_max Step 1 — Maximum Bending Moment: For a simply supported beam with midpoint load: M_max = PL/4 = 10,000 × 3,000 / 4 = 7,500,000 N·mm = 7.5 × 10⁶ N·mm Step 2 — Section Modulus: S = bh²/6 = 100 × (200)² / 6 = 100 × 40,000 / 6 = 666,667 mm³ Step 3 — Maximum Bending Stress: σ_max = M/S = 7.5 × 10⁶ / 666,667 ∴ σ_max = 11.25 MPa

Question Type

numerical

Answer Structure

• Line 1: List Given data and Required [0.5 mark implicit — shows method]
• Step 1: Correct M_max formula and value [1 mark]
• Step 2: Correct S = bh²/6 with numerical value [1 mark]
• Step 3: σ_max = M/S with correct answer and unit [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using M_max = wL²/8 (UDL formula) instead of PL/4 for a point load — wrong moment loses 1 mark
• Forgetting to convert kN to N or m to mm, causing unit inconsistency
• Computing I instead of S and dividing M by I without multiplying by c

Key Phrases To Include

• M_max = PL/4
• S = bh²/6
• σ_max = M/S
• MPa
• N·mm

Marks

3

Topic

Horizontal Shear Stress — Rectangular Section

Difficulty

easy

Template Id

T6

Examiner Tip

For rectangular sections, the shortcut τ_max = 3V/(2A) = 1.5V/A is the fastest route. Always state 'occurs at the neutral axis' — this shows conceptual understanding and may earn partial marks.

Model Answer

Given: • b = 100 mm, h = 200 mm • P = 10 kN = 10,000 N, L = 3 m Required: Maximum horizontal shear stress τ_max Step 1 — Maximum Shear Force: V_max = P/2 = 10,000/2 = 5,000 N (Maximum shear occurs at the supports for a midpoint load) Step 2 — Cross-sectional Area: A = b × h = 100 × 200 = 20,000 mm² Step 3 — Maximum Shear Stress (rectangular section formula): τ_max = 3V/(2A) = 3 × 5,000 / (2 × 20,000) τ_max = 15,000 / 40,000 ∴ τ_max = 0.375 MPa (occurs at the neutral axis)

Question Type

numerical

Answer Structure

• Step 1: Correct V_max = P/2 [1 mark]
• Step 2: Correct area A = bh [0.5 mark]
• Step 3: Apply τ_max = 3V/(2A) with correct answer and unit [1.5 marks]

Scoring Breakdown

Common Mark Deductions

• Using τ = VQ/(Ib) long method but computing Q incorrectly — forgetting to use Q for half the rectangle
• Omitting the factor 3/2 (or 1.5) in the shortcut formula
• Using V_max = P (total load) instead of P/2 (reaction at support)

Key Phrases To Include

• V_max = P/2
• τ_max = 3V/(2A)
• 1.5V/A
• neutral axis
• 0.375 MPa

Marks

3

Topic

Section Design — Flexure

Difficulty

medium

Template Id

T7

Examiner Tip

Design problems always require rounding UP to the next safe dimension. Rounding down is unsafe and examiners penalize it. Show the cube-root step clearly.

Model Answer

Given: • b = 0.4h (width in terms of depth) • M_max = 25 kN·m = 25 × 10⁶ N·mm • σ_allow = 12 MPa = 12 N/mm² Required: Minimum depth h Step 1 — Required Section Modulus: S_req = M_max / σ_allow = 25 × 10⁶ / 12 = 2.0833 × 10⁶ mm³ Step 2 — Express S in terms of h: S = bh²/6 = (0.4h)(h²)/6 = 0.4h³/6 = h³/15 Step 3 — Solve for h: h³/15 = 2.0833 × 10⁶ h³ = 15 × 2.0833 × 10⁶ = 3.125 × 10⁷ h = (3.125 × 10⁷)^(1/3) = 314.8 mm ∴ Use h = 315 mm (rounded up to next whole mm), b = 0.4 × 315 = 126 mm

Question Type

numerical

Answer Structure

• Step 1: Compute S_req = M/σ_allow [1 mark]
• Step 2: Express S = bh²/6 with b = 0.4h substituted [1 mark]
• Step 3: Solve for h, cube-root calculation, round up [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not converting M from kN·m to N·mm (25 × 10⁶ N·mm) causing wrong numerical answer
• Rounding h down instead of up — this yields an undersized (unsafe) section
• Forgetting to state the final b value alongside h

Key Phrases To Include

• S_req = M/σ_allow
• S = bh²/6
• b = 0.4h substituted
• h³ = ...
• round up
• mm³

Marks

5

Topic

Complete Beam Design Check — Bending and Shear

Difficulty

medium

Template Id

T8

Examiner Tip

5-mark questions expect a complete problem: Given, Required, solution steps, AND a conclusion. The adequacy check and conclusion together are worth 1 mark — never skip them. Think of 5-mark answers as a mini-report.

Model Answer

Given: • b = 150 mm, h = 300 mm, L = 4 m • w = 6 kN/m, σ_allow = 10 MPa, τ_allow = 0.70 MPa Required: (a) σ_max; (b) τ_max; adequacy check — Part (a): Maximum Bending Stress — Step 1: M_max = wL²/8 = 6(4)²/8 = 12 kN·m = 12 × 10⁶ N·mm Step 2: S = bh²/6 = 150(300)²/6 = 2.25 × 10⁶ mm³ Step 3: σ_max = M/S = 12 × 10⁶ / 2.25 × 10⁶ = 5.33 MPa Check: σ_max = 5.33 MPa < σ_allow = 10 MPa ✓ SAFE in bending — Part (b): Maximum Shear Stress — Step 4: V_max = wL/2 = 6(4)/2 = 12 kN = 12,000 N Step 5: A = 150 × 300 = 45,000 mm² Step 6: τ_max = 3V/(2A) = 3(12,000)/[2(45,000)] = 36,000/90,000 = 0.40 MPa Check: τ_max = 0.40 MPa < τ_allow = 0.70 MPa ✓ SAFE in shear Conclusion: The 150 × 300 mm section is adequate for both bending and shear under the given loading.

Question Type

numerical

Answer Structure

• Part (a) Step 1: Correct M_max = wL²/8 [1 mark]
• Part (a) Step 2-3: Correct S and σ_max with check [1.5 marks]
• Part (b) Step 4: Correct V_max = wL/2 [0.5 mark]
• Part (b) Step 5-6: Correct τ_max = 3V/(2A) [1.5 marks]
• Conclusion: Both adequacy checks stated correctly [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using V_max = wL (total load) instead of wL/2 (reaction) — most common error
• Forgetting to explicitly compare computed stress with allowable — the check is worth marks
• Not writing a conclusion statement — examiners want to see SAFE or UNSAFE stated

Key Phrases To Include

• M_max = wL²/8
• V_max = wL/2
• S = bh²/6
• σ_max = M/S
• τ_max = 3V/(2A)
• σ_max < σ_allow
• τ_max < τ_allow
• SAFE

Marks

5

Topic

T-Section — Unsymmetric Bending Stress

Difficulty

hard

Template Id

T9

Examiner Tip

T-section problems are board exam favorites precisely because students forget the NA is NOT at mid-depth. Step 1 (finding ȳ) is always first and worth the most marks. Circle ȳ, c_top, c_bot before proceeding — they flow into every subsequent step.

Model Answer

Given: • Flange: 200 mm × 50 mm (top) • Web: 50 mm × 200 mm • Total depth = 250 mm • M = 30 kN·m = 30 × 10⁶ N·mm (sagging) Required: σ_top, σ_bot Step 1 — Locate the Neutral Axis (from top): Flange: A₁ = 200(50) = 10,000 mm², ȳ₁ = 25 mm Web: A₂ = 50(200) = 10,000 mm², ȳ₂ = 50 + 100 = 150 mm ȳ = (A₁ȳ₁ + A₂ȳ₂)/(A₁ + A₂) ȳ = [10,000(25) + 10,000(150)] / 20,000 ȳ = (250,000 + 1,500,000) / 20,000 = 1,750,000 / 20,000 = 87.5 mm from top So: c_top = 87.5 mm, c_bot = 250 – 87.5 = 162.5 mm Step 2 — Moment of Inertia about NA (parallel-axis theorem): Flange: I₁ = (200)(50)³/12 + 10,000(87.5 – 25)² = 2.083×10⁶ + 10,000(62.5)² = 2.083×10⁶ + 39.063×10⁶ = 41.146×10⁶ mm⁴ Web: I₂ = (50)(200)³/12 + 10,000(150 – 87.5)² = 33.333×10⁶ + 10,000(62.5)² = 33.333×10⁶ + 39.063×10⁶ = 72.396×10⁶ mm⁴ I_total = 41.146×10⁶ + 72.396×10⁶ = 113.54×10⁶ mm⁴ Step 3 — Fiber Stresses: σ_top = M·c_top / I = 30×10⁶ × 87.5 / 113.54×10⁶ σ_top = 2,625×10⁶ / 113.54×10⁶ = 23.12 MPa (Compression, sagging) σ_bot = M·c_bot / I = 30×10⁶ × 162.5 / 113.54×10⁶ σ_bot = 4,875×10⁶ / 113.54×10⁶ = 42.94 MPa (Tension, sagging) ∴ σ_top = 23.1 MPa (C), σ_bot = 42.9 MPa (T) Note: σ_bot > σ_top because the NA is closer to the top — a key feature of an asymmetric T-section.

Question Type

numerical

Answer Structure

• Step 1: Correct NA location from top = 87.5 mm using centroid formula [2 marks]
• Step 2: Correct I_total using parallel-axis theorem for each part [2 marks]
• Step 3: Correct σ_top and σ_bot with correct sign/type (C or T) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Assuming NA is at mid-depth (125 mm) — fatal error for asymmetric sections, loses 2 marks immediately
• Computing ȳ from the bottom instead of consistently from one reference — sign errors in d for parallel-axis theorem
• Not identifying which fiber is in compression (top, sagging) and which is in tension (bottom, sagging)

Key Phrases To Include

• locate neutral axis
• ȳ = ΣAȳ/ΣA
• c_top = 87.5 mm
• c_bot = 162.5 mm
• parallel-axis theorem
• I = Io + Ad²
• σ = Mc/I
• Compression
• Tension

Marks

5

Topic

Combined Axial and Bending Stress — Eccentric Loading

Difficulty

hard

Template Id

T10

Examiner Tip

Eccentric loading problems always need the kern check as the final step — it shows you understand the physical implication. Write 'kern = h/6' explicitly. If e > h/6, state 'load is outside kern, tension occurs.'

Model Answer

Given: • Section: 200 mm × 200 mm • P = 300 kN = 300,000 N (compressive) • e = 40 mm (eccentricity) Required: σ_max, σ_min; check for tension Step 1 — Cross-sectional Properties: A = 200 × 200 = 40,000 mm² I = bh³/12 = 200(200)³/12 = 1.333 × 10⁸ mm⁴ c = 200/2 = 100 mm (extreme fiber distance) Step 2 — Equivalent Loading: M = Pe = 300,000 × 40 = 12 × 10⁶ N·mm Step 3 — Axial Stress (uniform): σ_axial = P/A = 300,000 / 40,000 = 7.5 MPa (Compression) Step 4 — Bending Stress at Extreme Fibers: σ_bending = Mc/I = 12 × 10⁶ × 100 / 1.333 × 10⁸ = 9.0 MPa Step 5 — Combined Stresses: σ_max = P/A + Mc/I = 7.5 + 9.0 = 16.5 MPa (Compression) ← fiber on eccentricity side σ_min = P/A – Mc/I = 7.5 – 9.0 = –1.5 MPa (Tension) ← opposite fiber Step 6 — Kern Check: Kern limit = h/6 = 200/6 = 33.3 mm e = 40 mm > h/6 = 33.3 mm → load is OUTSIDE the kern ∴ σ_max = 16.5 MPa (C), σ_min = 1.5 MPa (T) Tension DOES occur — confirmed by kern check. The section design should be reviewed if tension is unacceptable (e.g., unreinforced masonry).

Question Type

numerical

Answer Structure

• Step 1: Correct A, I, c for 200×200 section [1 mark]
• Step 2-3: Correct M = Pe and axial stress P/A [1 mark]
• Step 4: Correct bending stress Mc/I [1 mark]
• Step 5: Correct superposition for σ_max and σ_min with signs [1 mark]
• Step 6: Kern check h/6 = 33.3 mm, conclusion on tension [1 mark]

Scoring Breakdown

Common Mark Deductions

• Adding bending stress instead of subtracting for the minimum fiber — confusing which fiber is 'near' vs 'far' from load
• Not performing the kern check — loses 1 mark
• Using wrong I (e.g., I = bh/12 instead of bh³/12)

Key Phrases To Include

• σ = P/A ± Mc/I
• M = Pe
• kern
• middle-third rule
• h/6 = 33.3 mm
• e > h/6
• Tension occurs
• superpose

Marks

2

Topic

Shear Flow — Built-Up Beams

Difficulty

medium

Template Id

T11

Examiner Tip

Shear flow q is in N/mm (force per length); shear stress τ is in N/mm² (MPa). Knowing this distinction is a key mark-earner. State units clearly.

Model Answer

Shear flow q is the shear force per unit length transmitted along the interface between two connected parts of a built-up beam: q = VQ/I Where V = shear force, Q = first moment of the connected area about the neutral axis, I = moment of inertia of the entire section. Nail Spacing: If each nail can carry a force F, the spacing s between nails is: s = F/q For nails on both sides (two rows), use s = 2F/q or treat each nail line separately. Example of use: A built-up box beam with V = 5 kN, Q = 1.5×10⁵ mm³, I = 4×10⁷ mm⁴: q = VQ/I = 5,000 × 1.5×10⁵ / 4×10⁷ = 18.75 N/mm For F = 400 N: s = 400/18.75 = 21.3 mm (round down for safety).

Question Type

short_answer

Answer Structure

• Line 1: Define shear flow and write q = VQ/I [1 mark]
• Line 2: State nail spacing formula s = F/q with brief explanation [1 mark]

Scoring Breakdown

Common Mark Deductions

• Confusing q = VQ/I (shear flow) with τ = VQ/(Ib) (shear stress) — they differ by the factor b
• Not mentioning to round DOWN the nail spacing (rounding up means nails are too far apart — unsafe)
• Failing to identify Q as the first moment of the connected (attached) area

Key Phrases To Include

• shear flow
• q = VQ/I
• shear force per unit length
• s = F/q
• nail capacity F
• round down for safety

Marks

2

Topic

Maximum Shear Stress — Standard Sections

Difficulty

easy

Template Id

T12

Examiner Tip

These two formulas are memory items on the board exam. Create a mnemonic: 'Rectangle = 3/2, Circle = 4/3' (both greater than 1, rectangle is larger factor). The question always specifies 'at the neutral axis.'

Model Answer

(a) Rectangular section: τ_max = 3V/(2A) = 1.5 × (V/A) The maximum shear stress is 1.5 times the average shear stress, occurring at the neutral axis. (b) Solid circular section: τ_max = 4V/(3A) ≈ 1.333 × (V/A) The maximum shear stress is 4/3 times the average shear stress, also at the neutral axis.

Question Type

very_short_answer

Answer Structure

• Line 1: Rectangle — τ_max = 3V/(2A) = 1.5(V/A) [1 mark]
• Line 2: Circle — τ_max = 4V/(3A) = (4/3)(V/A) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Reversing the factors (using 4/3 for rectangle and 3/2 for circle)
• Not expressing the answer as a multiple of V/A when specifically asked
• Stating the formula without noting that maximum occurs at the neutral axis

Key Phrases To Include

• τ_max = 3V/(2A)
• 1.5 times average
• τ_max = 4V/(3A)
• 4/3 times average
• neutral axis

Marks

3

Topic

Shear Flow — Nail Spacing in Built-Up Beams

Difficulty

medium

Template Id

T13

Examiner Tip

Nail spacing must always be rounded DOWN for safety (smaller spacing = more nails = stronger connection). This is a code principle consistent with timber design practice and is an easy 0.5 mark to earn if you state the reason.

Model Answer

Given: • V = 8 kN = 8,000 N • I = 1.2 × 10⁸ mm⁴ • Q = Q_flange = 8.0 × 10⁵ mm³ (first moment of flange about NA) • Nail capacity: F = 600 N per nail Required: Nail spacing s Step 1 — Shear Flow at Flange-Web Interface: q = VQ/I = 8,000 × 8.0 × 10⁵ / 1.2 × 10⁸ q = 6.4 × 10⁹ / 1.2 × 10⁸ q = 53.33 N/mm Step 2 — Nail Spacing: One nail per row: s = F/q = 600 / 53.33 = 11.25 mm Note: Since there is one row of nails (single line), use s = F/q directly. Round down for safety: s = 11 mm (or specify at most 11 mm spacing). ∴ Maximum nail spacing s = 11 mm

Question Type

numerical

Answer Structure

• Step 1: Correct shear flow q = VQ/I = 53.33 N/mm [2 marks]
• Step 2: Correct spacing s = F/q = 11 mm (round down) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using τ = VQ/(Ib) instead of q = VQ/I — wrong approach for spacing problems
• Rounding up the spacing (e.g., writing 12 mm instead of 11 mm) — this is structurally unsafe
• Not including units for shear flow (must be N/mm, not MPa)

Key Phrases To Include

• q = VQ/I
• shear flow
• N/mm
• s = F/q
• round down
• maximum spacing

Marks

2

Topic

Kern — Combined Axial and Bending

Difficulty

medium

Template Id

T14

Examiner Tip

The kern for a rectangle extends h/6 from center in each direction (total h/3 width). State 'e ≤ h/6' clearly. Examiners frequently set trap answers with h/3 — do not confuse total kern width with the eccentricity limit.

Model Answer

The middle-third rule (kern rule) states that for a rectangular section to remain entirely in compression (no tension anywhere on the cross-section), the resultant axial load P must act within the kern — the central region of the cross-section. For a rectangular section of width b and depth h: • Kern extends h/6 from the centroidal axis in the direction of eccentricity • No tension condition: e ≤ h/6 Physical basis: Combined stress is σ = P/A ± Mc/I. For no tension on the far fiber: P/A ≥ Mc/I → P/A ≥ (Pe)(h/2)/I → e ≤ I/(Ac) = (bh³/12)/[(bh)(h/2)] = h/6 Application: Important in unreinforced masonry walls, concrete footings, and short columns where tensile capacity is negligible.

Question Type

short_answer

Answer Structure

• Line 1: State the kern/middle-third rule and the no-tension condition e ≤ h/6 [1 mark]
• Line 2: Physical basis — superposition σ = P/A ± Mc/I, derive or cite e ≤ h/6 [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating kern = h/3 instead of h/6 (a very common error — the total kern width is h/3, but e must be ≤ h/6)
• Not connecting the kern rule to the no-tension condition
• Failing to mention the practical context (masonry, footings) where this matters

Key Phrases To Include

• kern
• middle-third rule
• e ≤ h/6
• no tension
• σ = P/A ± Mc/I
• masonry
• footing
• h/6 from centroid

Marks

3

Topic

Steel Beam — W-Section Bending Stress

Difficulty

medium

Template Id

T15

Examiner Tip

For W-shapes, always state 'c = d/2 because the section is doubly symmetric about the centroidal x-axis.' This shows code awareness (AISC 360, NSCP Chapter 5) and earns confidence marks from the examiner.

Model Answer

Given: • I_x = 178 × 10⁶ mm⁴ • d = 400 mm, so c = d/2 = 200 mm (W-shape is doubly symmetric) • M = 250 kN·m = 250 × 10⁶ N·mm Required: (a) σ_max; (b) S_x Part (a) — Maximum Bending Stress: σ_max = Mc/I = (250 × 10⁶ × 200) / (178 × 10⁶) σ_max = 50,000 × 10⁶ / 178 × 10⁶ ∴ σ_max = 280.9 MPa For A36 steel (F_y = 248 MPa, AISC 360/NSCP), allowable flexural stress = 0.66F_y = 163.7 MPa. The computed stress 280.9 MPa exceeds the allowable — the section is INADEQUATE for A36 steel. Part (b) — Section Modulus: S_x = I_x/c = 178 × 10⁶ / 200 ∴ S_x = 890,000 mm³ = 890 × 10³ mm³ Alternate check: σ_max = M/S = 250 × 10⁶ / 890,000 = 280.9 MPa ✓ (consistent)

Question Type

numerical

Answer Structure

• Step 1: Identify c = d/2 = 200 mm for W-shape [0.5 mark]
• Step 2: σ_max = Mc/I = 280.9 MPa [1.5 marks]
• Step 3: S_x = I/c = 890 × 10³ mm³ and verify σ = M/S [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using c = d (full depth) instead of d/2 for the W-section — critical error, doubles the stress
• Not converting M to N·mm (250 kN·m = 250 × 10⁶ N·mm)
• Mixing formulas: σ_max = M/I (missing c) or σ_max = M/S (correctly using S but not computing S first when asked)

Key Phrases To Include

• c = d/2
• doubly symmetric
• σ_max = Mc/I
• S_x = I_x/c
• MPa
• 280.9 MPa
• 890 × 10³ mm³