CELE Steel & Timber Design — Timber DesignExam Answer Templates

Marks

1

Topic

Allowable Stress Design — ASD Framework

Difficulty

easy

Template Id

T1

Examiner Tip

One mark means one key idea. Name the formula and at least two C-factor symbols. Do not over-explain.

Model Answer

The adjusted allowable design value is: F' = F × (C_D × C_M × C_t × C_F × C_L …) where F is the reference design value for the species/grade and the C-factors are applicable adjustment factors per NSCP 2015 Chapter 6.

Question Type

very_short_answer

Answer Structure

• Line 1: Write the formula F' = F × ΠC with at least three C-factors named [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing F_allow = F_b only (omitting the C-factor product entirely)
• Confusing F' with LRFD phi-factor design values

Key Phrases To Include

• F' = F × C
• reference design value
• adjustment factors
• NSCP 2015 Chapter 6

Marks

1

Topic

Load Duration Factor C_D

Difficulty

easy

Template Id

T2

Examiner Tip

The board exam frequently gives a load type and asks for C_D. Memorise the five standard values in ascending order: 0.9, 1.0, 1.15, 1.25, 1.6 (and 2.0 for impact).

Model Answer

(a) Wind/seismic load: C_D = 1.6 (b) Permanent (dead) load: C_D = 0.9 Note: Higher C_D reflects wood's ability to carry greater stress for shorter durations.

Question Type

very_short_answer

Answer Structure

• Part (a): C_D = 1.6 for wind/seismic [0.5 mark]
• Part (b): C_D = 0.9 for permanent load [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Reversing the values (writing 0.9 for wind and 1.6 for dead load)
• Using C_D = 1.0 for wind (this is the 10-year normal load value)

Key Phrases To Include

• C_D = 1.6
• C_D = 0.9
• load duration factor
• shorter duration higher allowable

Marks

1

Topic

Section Properties — Bending

Difficulty

easy

Template Id

T3

Examiner Tip

For a rectangular section, always write S = bh²/6 as the first line. If the dimensions are given in mm, keep everything in mm — do not convert to m mid-computation.

Model Answer

S = bh²/6 S = (100)(300)²/6 S = (100)(90,000)/6 S = 9,000,000/6 S = 1,500,000 mm³ = 1.50 × 10⁶ mm³

Question Type

numerical

Answer Structure

• Line 1: Write formula S = bh²/6 [formula mark implicit]
• Line 2: Substitute b = 100 mm, h = 300 mm and solve [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using S = bh²/4 (wrong formula — that is for plastic modulus Z)
• Omitting mm³ units from the answer
• Confusing b and h (does not affect answer here but becomes critical for non-square sections)

Key Phrases To Include

• S = bh²/6
• 1.5 × 10⁶ mm³

Marks

2

Topic

Shear Stress — Rectangular Timber Section

Difficulty

medium

Template Id

T4

Examiner Tip

The 3/2 factor in f_v = 3V/2A comes from parabolic shear distribution in a rectangular section. Examiners will check this coefficient explicitly.

Model Answer

Given: b = 100 mm, h = 250 mm, V = 18 kN = 18,000 N, F'_v = 1.0 MPa Step 1 — Cross-sectional area: A = b × h = 100 × 250 = 25,000 mm² Step 2 — Actual horizontal shear stress (rectangular section): f_v = 3V / (2A) f_v = 3(18,000) / [2(25,000)] f_v = 54,000 / 50,000 f_v = 1.08 MPa Step 3 — Check: f_v = 1.08 MPa > F'_v = 1.0 MPa ∴ The beam is INADEQUATE in shear — the section must be increased.

Question Type

numerical

Answer Structure

• Line 1: Compute A = bh [implicit]
• Line 2: Write and apply f_v = 3V/2A [1 mark]
• Line 3: Compare f_v with F'_v and state conclusion [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using f_v = V/A instead of 3V/2A (forgetting the 3/2 factor for rectangular sections)
• Not writing the conclusion sentence — a common one-mark loss
• Using V in kN without converting to N, leading to a wrong answer in kPa instead of MPa

Key Phrases To Include

• f_v = 3V/2A
• 1.08 MPa
• INADEQUATE
• f_v > F'_v

Marks

2

Topic

Adjustment Factors — Bending

Difficulty

medium

Template Id

T5

Examiner Tip

Always list each C-factor on a separate line or clearly inline before multiplying. The examiner awards a mark for recognising C_D = 1.15 even if there is a downstream arithmetic error.

Model Answer

Given: F_b = 14 MPa, C_D = 1.15 (2-month load), C_M = 0.85, C_F = 1.05 Adjusted allowable bending stress (NSCP 2015 Ch. 6): F'_b = F_b × C_D × C_M × C_F F'_b = 14 × 1.15 × 0.85 × 1.05 F'_b = 14 × 1.027 F'_b = 14.38 MPa ∴ F'_b ≈ 14.38 MPa

Question Type

numerical

Answer Structure

• Line 1: Identify C_D = 1.15 for 2-month load [0.5 mark]
• Line 2: Write and apply F'_b = F_b × C_D × C_M × C_F [1 mark]
• Line 3: Final answer with units [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using C_D = 1.0 (normal load) instead of 1.15 for 2-month duration
• Omitting one of the C-factors from the product
• Arithmetic error in the chain multiplication

Key Phrases To Include

• C_D = 1.15
• 2-month load
• F'_b = F_b × C_D × C_M × C_F
• 14.38 MPa

Marks

3

Topic

Bending Stress Check

Difficulty

medium

Template Id

T6

Examiner Tip

The unit conversion M from kN·m → N·mm is a board-exam trap. Always write M = 9 kN·m = 9 × 10⁶ N·mm before substituting.

Model Answer

Given: b = 150 mm, h = 250 mm, M = 9 kN·m = 9 × 10⁶ N·mm, F'_b = 11 MPa Step 1 — Section modulus: S = bh²/6 = (150)(250)²/6 = (150)(62,500)/6 = 9,375,000/6 = 1,562,500 mm³ S = 1.5625 × 10⁶ mm³ Step 2 — Actual bending stress: f_b = M/S = (9 × 10⁶) / (1.5625 × 10⁶) f_b = 5.76 MPa Step 3 — Adequacy check: f_b = 5.76 MPa < F'_b = 11 MPa ✓ ∴ The 150 mm × 250 mm beam is ADEQUATE in bending.

Question Type

numerical

Answer Structure

• Step 1: Compute S = bh²/6 with correct substitution [1 mark]
• Step 2: Compute f_b = M/S with correct unit conversion (kN·m to N·mm) [1 mark]
• Step 3: Compare f_b with F'_b and state 'ADEQUATE' conclusion [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not converting M from kN·m to N·mm (using 9 instead of 9 × 10⁶)
• Skipping the conclusion sentence — direct one-mark loss
• Computing S = bh³/6 (cubing h instead of squaring)

Key Phrases To Include

• S = bh²/6
• f_b = M/S
• 5.76 MPa
• ADEQUATE in bending
• f_b < F'_b

Marks

2

Topic

Stability Factors — C_L and C_P

Difficulty

medium

Template Id

T7

Examiner Tip

Connect each factor to its physical phenomenon: C_L → lateral-torsional buckling (beam bends sideways), C_P → column buckling (compression instability). The examiner rewards the correct phenomenon name.

Model Answer

Beam Stability Factor C_L: C_L is an adjustment factor applied to the reference bending design value F_b. It accounts for lateral-torsional buckling of the wood beam when the compression edge is insufficiently braced. For a fully braced beam, C_L = 1.0; it decreases as the unsupported length increases. Column Stability Factor C_P: C_P is an adjustment factor applied to the reference compression-parallel-to-grain design value F_c. It accounts for column buckling (flexural instability) in wood columns. It is computed from the Euler critical stress F_cE and the adjusted compression value F_c*, analogous to a buckling reduction factor in steel column design.

Question Type

short_answer

Answer Structure

• Part 1: Define C_L as a bending reduction for lateral-torsional buckling [1 mark]
• Part 2: Define C_P as a compression reduction for column buckling, tied to F_cE [1 mark]

Scoring Breakdown

Common Mark Deductions

• Confusing C_L with C_F (size factor)
• Describing C_P as a simple safety factor rather than linking it to slenderness/Euler buckling

Key Phrases To Include

• lateral-torsional buckling
• C_L
• column buckling
• C_P
• F_cE
• slenderness

Marks

3

Topic

Bending Design — Required Section Modulus

Difficulty

medium

Template Id

T8

Examiner Tip

For design problems (find S_req), the answer is incomplete without a statement 'Choose a section with S ≥ [value].' That final line often carries the last half-mark.

Model Answer

Given: F_b = 16.5 MPa, C_D = 1.25, C_F = 1.1, M = 20 kN·m = 20 × 10⁶ N·mm Part (a) — Adjusted allowable bending stress: F'_b = F_b × C_D × C_F F'_b = 16.5 × 1.25 × 1.1 F'_b = 22.69 MPa Part (b) — Required section modulus: S_req = M / F'_b = (20 × 10⁶) / 22.69 S_req = 881,400 mm³ ≈ 8.81 × 10⁵ mm³ ∴ Select a section with S ≥ 8.81 × 10⁵ mm³ (e.g., a 150 × 300 section gives S = 150(300)²/6 = 2.25 × 10⁶ mm³ — more than adequate.)

Question Type

numerical

Answer Structure

• Part (a): Apply F'_b = F_b × C_D × C_F correctly = 22.69 MPa [1.5 marks]
• Part (b): Apply S_req = M/F'_b with correct unit conversion = 8.81 × 10⁵ mm³ [1.5 marks]

Scoring Breakdown

Common Mark Deductions

• Using C_D = 1.15 (2-month) instead of 1.25 for a 7-day load
• Forgetting to convert M from kN·m to N·mm before dividing
• Not stating the selection criterion S ≥ S_req

Key Phrases To Include

• F'_b = F_b × C_D × C_F
• C_D = 1.25 (7-day)
• S_req = M/F'_b
• 22.69 MPa
• 8.81 × 10⁵ mm³

Marks

3

Topic

Horizontal Shear — Timber Beams

Difficulty

medium

Template Id

T9

Examiner Tip

This is a theory question worth 3 marks, so structure it in 3 distinct paragraphs. Examiners scan for the three key ideas: (1) why timber is weak in this direction, (2) the formula, and (3) the short/deep beam and notch condition.

Model Answer

In timber beams, the material is significantly weaker in shear parallel to the grain (along wood fibres) than in shear perpendicular to the grain. Horizontal (longitudinal) shear stress is highest at the neutral axis and is given by: f_v = VQ / (Ib) = 3V / (2A) [for a rectangular section] This horizontal shear is resisted only by the relatively weak wood fibres sliding along each other. Timber's allowable shear parallel to grain (F_v) is typically 1–2 MPa — far lower than its bending or compression values. For SHORT, DEEP beams: - Shear force V is large relative to moment M - The small depth d reduces the moment arm and increases M-to-V ratio - As the span-to-depth ratio (L/d) decreases, shear governs over bending Typical board-exam rule: when L/d < 10, always check shear. At a notched support, use reduced depth d_n (net depth) because the effective area resisting shear drops sharply, making the shear check even more critical.

Question Type

short_answer

Answer Structure

• Para 1: Explain timber weakness in shear parallel to grain [1 mark]
• Para 2: Write formula f_v = 3V/2A for rectangular section [1 mark]
• Para 3: Explain why short/deep beams and notched supports make shear critical [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing f_v = V/A (forgetting the 3/2 factor)
• Not mentioning notched beams — a common CELE trap
• Confusing horizontal shear with vertical shear without explaining the parallel-to-grain weakness

Key Phrases To Include

• shear parallel to grain
• f_v = 3V/2A
• short, deep beams
• notched support
• neutral axis
• low F_v

Marks

5

Topic

Column Stability Factor C_P — Timber Column Design

Difficulty

hard

Template Id

T10

Examiner Tip

The C_P formula is the most calculation-intensive in timber design. Break it into three numbered sub-steps: (1) β, (2) α = (1+β)/2c, (3) C_P = α − √(α² − β/c). This structure earns partial credit even with minor arithmetic errors.

Model Answer

Given: b = d = 150 mm, ℓ_e = 3,600 mm, E'_min = 7,200 MPa, F_c* = 11.0 MPa, c = 0.8 (sawn lumber) Part (a) — Slenderness ratio: ℓ_e/d = 3,600 / 150 = 24 (Check: ℓ_e/d = 24 < 50 maximum, OK per NSCP 2015) Part (b) — Euler critical stress: F_cE = 0.822 × E'_min / (ℓ_e/d)² F_cE = 0.822 × 7,200 / (24)² F_cE = 5,918.4 / 576 F_cE = 10.27 MPa Part (c) — Column stability factor C_P: β = F_cE / F_c* = 10.27 / 11.0 = 0.934 α = (1 + β) / (2c) = (1 + 0.934) / (2 × 0.8) = 1.934 / 1.6 = 1.209 C_P = α − √(α² − β/c) C_P = 1.209 − √(1.209² − 0.934/0.8) C_P = 1.209 − √(1.462 − 1.168) C_P = 1.209 − √(0.294) C_P = 1.209 − 0.542 C_P = 0.667 Part (d) — Adjusted allowable compression stress: F'_c = F_c* × C_P F'_c = 11.0 × 0.667 F'_c = 7.34 MPa ∴ The adjusted allowable compression stress is F'_c = 7.34 MPa, representing a 33% reduction due to column buckling.

Question Type

numerical

Answer Structure

• Part (a): ℓ_e/d = 3,600/150 = 24 [0.5 mark]
• Part (b): F_cE = 0.822 × E'_min / (ℓ_e/d)² = 10.27 MPa [1.5 marks]
• Part (c): Compute β, then α = (1+β)/2c, then full C_P formula = 0.667 [2 marks]
• Part (d): F'_c = F_c* × C_P = 7.34 MPa with conclusion [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using c = 0.9 (glulam) instead of c = 0.8 for sawn lumber
• Computing (ℓ_e/d)² incorrectly — squaring only ℓ_e or only d
• Not defining β and α separately before plugging into C_P — loses partial credit if arithmetic is wrong
• Forgetting to multiply F_c* by C_P for final F'_c (stopping at C_P)

Key Phrases To Include

• ℓ_e/d = 24
• F_cE = 0.822 × E'_min / (ℓ_e/d)²
• β = F_cE/F_c*
• C_P formula
• c = 0.8 (sawn lumber)
• F'_c = F_c* × C_P
• 7.34 MPa

Marks

2

Topic

Timber Column — Basic Sizing

Difficulty

easy

Template Id

T11

Examiner Tip

For a first-pass column sizing ignoring C_P, the formula is simply A_req = P/F'_c. The board exam often includes this as a quick sub-question within a larger column problem.

Model Answer

Given: P = 80 kN = 80,000 N, F'_c = 8 MPa, square section (A = b²) Step 1 — Required area: A_req = P / F'_c = 80,000 / 8 = 10,000 mm² Step 2 — Required side dimension: b = √A_req = √10,000 = 100 mm ∴ Use a minimum 100 mm × 100 mm timber column. (In practice, select the next standard size ≥ 100 mm to allow for size and stability factors.)

Question Type

numerical

Answer Structure

• Step 1: A_req = P/F'_c = 10,000 mm² [1 mark]
• Step 2: b = √A_req = 100 mm with conclusion [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using P = 80 kN without converting to N, giving a wrong area in m²
• Not stating that a practical/standard size should be selected next

Key Phrases To Include

• A_req = P/F'_c
• 10,000 mm²
• b = √A_req
• 100 mm × 100 mm
• select next standard size

Marks

2

Topic

Column Slenderness Limit — Timber

Difficulty

easy

Template Id

T12

Examiner Tip

Board exams often test limits as trap MCQs. Remember: timber columns → ℓ_e/d ≤ 50; steel columns → KL/r ≤ 200. Do not mix them.

Model Answer

Maximum slenderness ratio for solid timber columns: ℓ_e/d ≤ 50 (per NSCP 2015 Chapter 6 / NDS provisions adopted in NSCP) Reason for the limit: Beyond ℓ_e/d = 50, a wood column becomes extremely susceptible to sudden lateral buckling failure, which is difficult to predict reliably due to wood's natural variability (knots, grain angle, moisture content). The limit also prevents excessive lateral deflection during construction, which could cause unsafe instability before the full load is applied. The C_P formula is calibrated for slenderness ratios within this range; extrapolation beyond 50 is not valid.

Question Type

short_answer

Answer Structure

• Line 1: State ℓ_e/d ≤ 50 as the NSCP 2015 limit [1 mark]
• Lines 2–4: Explain the reason — buckling susceptibility and formula calibration [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating ℓ_e/d ≤ 200 (steel column limit — wrong material)
• Giving the limit without any engineering explanation

Key Phrases To Include

• ℓ_e/d ≤ 50
• solid timber column
• lateral buckling
• NSCP 2015
• C_P formula

Marks

3

Topic

Shear at Notched Timber Supports

Difficulty

hard

Template Id

T13

Examiner Tip

Notched beams are a favourite CELE trap. Always flag: 'At a notched support, use net depth d_n.' If you use full d, you get a passing stress — completely wrong conclusion.

Model Answer

Given: b = 100 mm, d = 250 mm, d_n = 180 mm (net depth at notch), V = 14 kN = 14,000 N, F'_v = 1.0 MPa Critical note: At a notched support, shear is checked using the net depth d_n only. The effective area is reduced: Step 1 — Effective area at notch: A_n = b × d_n = 100 × 180 = 18,000 mm² Step 2 — Actual shear stress at notch (using d_n): f_v = 3V / (2A_n) f_v = 3(14,000) / [2(18,000)] f_v = 42,000 / 36,000 f_v = 1.167 MPa Step 3 — Adequacy check: f_v = 1.167 MPa > F'_v = 1.0 MPa ✗ ∴ The notched section is INADEQUATE in shear. The notch must be redesigned or the beam size increased. [Compare: Without notch, f_v = 3(14,000)/[2(100×250)] = 0.84 MPa — adequate. The notch alone makes it fail.]

Question Type

numerical

Answer Structure

• Step 1: Identify use of net depth d_n = 180 mm at notch [1 mark]
• Step 2: Compute f_v = 3V/(2A_n) = 1.167 MPa [1 mark]
• Step 3: Compare with F'_v and state INADEQUATE conclusion [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using full depth d = 250 mm instead of net depth d_n = 180 mm — the classic notched-beam trap
• Not writing the conclusion even after correct calculation
• Omitting comparison with un-notched case (which demonstrates why the notch matters)

Key Phrases To Include

• net depth d_n
• A_n = b × d_n
• f_v = 3V/2A_n
• INADEQUATE
• notched support
• 1.167 MPa

Marks

2

Topic

ASD vs. LRFD — Conceptual Comparison

Difficulty

medium

Template Id

T14

Examiner Tip

This comparative theory question tests your understanding across subjects. The key phrase examiners want: 'C_D can exceed 1.0 because wood is stronger under short loads.' That insight earns the mark.

Model Answer

Load Duration Factor C_D (Timber ASD): C_D is an UPWARD adjustment on the allowable stress — it increases the permitted stress for short-duration loads (e.g., C_D = 1.6 for wind) and decreases it for permanent loads (C_D = 0.9). It reflects the physical property of wood: under sustained load, wood undergoes creep and strength reduction ('mechano-sorptive' effect), so brief loads can be carried at higher stress than long-duration loads. Strength Reduction Factor φ (Concrete LRFD/USD per ACI 318): φ is always ≤ 1.0 and represents a DOWNWARD adjustment on the nominal capacity to account for variability in material strength, workmanship, and failure mode consequences (e.g., φ = 0.90 bending, φ = 0.75 shear for concrete). Fundamental Difference: C_D modifies the allowable STRESS based on load duration (can go UP or DOWN); φ modifies the nominal CAPACITY based on failure mode uncertainty (always reduces capacity). They operate under different design philosophies — C_D is unique to wood ASD, while φ is a reliability-based LRFD concept.

Question Type

short_answer

Answer Structure

• Para 1: Explain C_D — upward/downward adjustment on allowable stress based on load duration [1 mark]
• Para 2: Explain φ — downward factor on capacity for uncertainty; state the fundamental difference [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating that C_D is also ≤ 1.0 always (forgetting it goes up to 2.0 for impact)
• Not explaining the physical reason for C_D (wood's time-dependent strength)

Key Phrases To Include

• C_D increases allowable stress for short-duration loads
• C_D = 0.9 permanent
• C_D = 1.6 wind
• φ ≤ 1.0
• ACI 318
• different design philosophies

Marks

5

Topic

Combined Load — Bending Stress Check with C_D Selection

Difficulty

hard

Template Id

T15

Examiner Tip

The rule for C_D in combined loading: the C_D of the SHORTEST DURATION load in the combination governs (highest C_D). This is often tested in 5-mark problems. State it explicitly: 'C_D = 1.25 governs because the 7-day load is the shortest-duration load present.'

Model Answer

Given: b = 200 mm, h = 400 mm, F_b = 15 MPa M_D = 8 kN·m, M_L = 10 kN·m (7-day load), all other C = 1.0 Step 1 — Governing load combination and C_D: Combined moment: M_total = M_D + M_L = 8 + 10 = 18 kN·m = 18 × 10⁶ N·mm The governing C_D is that of the SHORTEST-DURATION load in the combination: M_L is a 7-day load → C_D = 1.25 (governs over C_D = 0.9 for dead load alone) Step 2 — Adjusted allowable bending stress: F'_b = F_b × C_D = 15 × 1.25 = 18.75 MPa Step 3 — Section modulus: S = bh²/6 = (200)(400)²/6 = (200)(160,000)/6 = 32,000,000/6 S = 5.333 × 10⁶ mm³ Step 4 — Actual bending stress: f_b = M_total / S = (18 × 10⁶) / (5.333 × 10⁶) f_b = 3.375 MPa Step 5 — Adequacy check: f_b = 3.375 MPa < F'_b = 18.75 MPa ✓ ∴ The 200 mm × 400 mm beam is ADEQUATE in bending with a stress ratio of 3.375/18.75 = 0.18 (very conservative — the section is oversized for this loading).

Question Type

numerical

Answer Structure

• Step 1: Combine moments and correctly identify governing C_D = 1.25 for 7-day load [1 mark]
• Step 2: Compute F'_b = 15 × 1.25 = 18.75 MPa [1 mark]
• Step 3: Compute S = bh²/6 = 5.333 × 10⁶ mm³ [1 mark]
• Step 4: Compute f_b = M/S = 3.375 MPa with unit conversion [1 mark]
• Step 5: Compare and state ADEQUATE conclusion with stress ratio [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using C_D = 0.9 (dead load only) when a live load is also present — wrong governing load
• Using C_D = 1.0 (normal) instead of 1.25 for a 7-day load
• Checking only dead load moment without combining
• Not converting M from kN·m to N·mm before computing f_b

Key Phrases To Include

• governing C_D = 1.25
• shortest-duration load governs
• M_total = 18 kN·m
• F'_b = 18.75 MPa
• S = 5.333 × 10⁶ mm³
• f_b = 3.375 MPa
• ADEQUATE