CELE Steel & Timber Design — Timber DesignDetailed Explanation
This is the "office hours" version of Timber Design for the CELE 2026. No shortcuts, no hand-waving — just a full unpacking of why Professional Regulation Commission (PRC) — Board of Civil Engineering cares about each concept and how the Steel & Timber Design section items tend to play out on exam day. Read this once, then hit the practice questions with real understanding.
Exam context
Professional Regulation Commission (PRC) — Board of Civil Engineering runs the Civil Engineer Licensure Examination on May and November 2026. Its Steel & Timber Design section sits under a "Core" weighting, and Timber Design is the 5th chapter in the 5-chapter CELE Steel & Timber Design rotation. The CELE passing mark is 70% weighted average, no sub-test below 50%, and the most recent 2026 paper drew about a meaningful share of questions from Steel & Timber Design.
Timber Design - Detailed Explanation
Timber design is a recurring topic in the PRC Civil Engineer Licensure Examination (CELE) and forms part of the Steel & Timber Design cluster of the board exam. Unlike reinforced concrete (ACI 318) and structural steel (AISC 360 / NSCP Chapter 5), wood is designed exclusively by Allowable Stress Design (ASD) under NSCP 2015 Chapter 6. The fundamental principle is simple: compute the actual stress on a member, compare it with an adjusted allowable stress that accounts for species grade, load duration, moisture, size, and stability, and verify that the actual stress does not exceed the allowable. What makes timber design distinctly challenging on the board exam is the system of adjustment (C) factors — getting the right factors, applying them correctly, and knowing which ones interact. This chapter covers the full ASD framework for wood: the adjustment-factor system, bending, horizontal shear, and compression parallel to grain (columns), complete with worked board-style problems and exam strategy.
Concepts
Allowable Stress Design (ASD) Framework for Wood
All timber design under NSCP 2015 Chapter 6 follows the Allowable Stress Design method. The governing equation is: F' = F × (product of all applicable adjustment factors C) where F is the reference design value for a specific species and grade (tabulated in NSCP or supplier grading certificates), and F' is the adjusted (allowable) design value. The actual computed stress f must satisfy f ≤ F'. Reference design values you will encounter on the exam: • F_b — bending stress • F_v — horizontal (parallel-to-grain) shear stress • F_c — compression parallel to grain • F_c⊥ — compression perpendicular to grain (bearing) • F_t — tension parallel to grain • E — modulus of elasticity • E_min — modulus for stability calculations Each of these has its own set of applicable adjustment factors. Not all factors apply to every reference value — knowing which C applies where is itself an exam competency. The most critical distinction: C_P (column stability) applies only to F_c, and C_L (beam stability) applies only to F_b.
Examples
Each applicable factor is multiplied together. Since all C_M, C_t, C_L = 1.0, they do not change the result but must be identified as applicable. The 2-month load duration (construction phase) increases the allowable by 15% through C_D = 1.15. The size and repetitive-member factors add further capacity.
Scenario
A wood beam has a reference bending stress F_b = 16.5 MPa. For a 2-month construction load (C_D = 1.15), dry service (C_M = 1.0), normal temperature (C_t = 1.0), size factor C_F = 1.1, repetitive member (C_r = 1.15), and beam stability factor C_L = 1.0. Find F'_b.
Solution
F'_b = F_b × C_D × C_M × C_t × C_F × C_r × C_L F'_b = 16.5 × 1.15 × 1.0 × 1.0 × 1.1 × 1.15 × 1.0 F'_b = 16.5 × 1.15 × 1.1 × 1.15 F'_b = 16.5 × 1.455 F'_b = 24.01 MPa
Applications
- Sizing floor joists and roof rafters in light wood framing
- Designing timber formwork and falsework for concrete pours
- Checking scaffolding plank adequacy for temporary works
- Evaluating existing wood members for code compliance during retrofits
Misconceptions
- Thinking the reference value F is already the allowable stress — it is NOT; you must apply all C factors first
- Applying all C factors to every reference value — each F has specific applicable factors only
- Using C_D = 1.0 for all loads — permanent loads use C_D = 0.9, which REDUCES allowable capacity
- Forgetting that when C_M = 1.0 (dry), wet-service reduction does not apply; never skip identifying it
Related Concepts
- Species and grade of lumber (determines reference F values)
- Load combinations under NSCP 2015 Section 203
- Comparison with LRFD/strength design used in steel and concrete
Common Exam Questions
Example
Given F_b = 14 MPa, C_D = 1.25, C_M = 0.85, C_F = 1.0, all others = 1.0. Find F'_b. Answer: F'_b = 14 × 1.25 × 0.85 = 14.875 MPa ≈ 14.88 MPa
Approach
Identify all applicable C factors for the specific reference value (F_b, F_v, or F_c). Multiply them together: F' = F × C_D × C_M × C_t × C_F × (others). Be careful: C_L and C_r are bending-only; C_P is compression-only.
Question Type
Compute adjusted allowable stress given reference values and C factors
Example
Dead load only on a permanent structure: C_D = 0.9. Wind + dead: C_D = 1.6 (wind governs).
Approach
Memorize the C_D table: 0.9 = permanent, 1.0 = 10-year (occupancy), 1.15 = 2-month (construction), 1.25 = 7-day, 1.6 = wind/seismic, 2.0 = impact. When multiple loads act together, use the C_D for the shortest-duration load in the combination.
Question Type
Identify the correct C_D value for a given load type
Key Points To Remember
- F' = F × ∏C — adjusted allowable = reference value × product of adjustment factors
- The check is always: actual stress f ≤ adjusted allowable F'
- Reference values F are species-and-grade-specific; they are given in problems or must be looked up
- Not all adjustment factors apply to all reference design values — know which C goes with which F
- C_P applies only to F_c (compression/columns); C_L applies only to F_b (bending)
- NSCP 2015 Chapter 6 is the governing Philippine code for structural timber
Load Duration Factor C_D and Other Key Adjustment Factors
The load duration factor C_D is the most uniquely timber-specific adjustment factor and a frequent exam focus. Wood is a viscoelastic material: it can carry higher loads for short durations than for sustained loads, because long-term loading causes creep and cumulative fiber damage. C_D VALUES (memorize these): • 0.9 — Permanent load (> 10 years, e.g., self-weight of structure) • 1.0 — Normal (10-year, e.g., floor live loads) • 1.15 — 2-month (construction/snow loads) • 1.25 — 7-day (roof construction) • 1.6 — Wind or seismic • 2.0 — Impact (dynamic) Rule: When multiple loads act simultaneously, use the C_D corresponding to the shortest-duration load in the combination. OTHER KEY FACTORS: C_M (Wet Service): Reduces allowable stress when equilibrium moisture content exceeds 19% for sawn lumber. For dry conditions (indoor, sheltered), C_M = 1.0. Wet service significantly reduces F_b, F_v, F_c, and E. C_t (Temperature): = 1.0 for T ≤ 37°C. Reduces values for sustained high temperatures. C_F (Size Factor): Applies to sawn lumber bending, tension, and compression. Larger members have lower size factors because larger pieces are more likely to contain defects. It is tabulated by nominal width × depth. C_r (Repetitive Member): = 1.15 for bending only, when three or more parallel members are spaced ≤ 600 mm o.c. and are joined by a load-distributing element (floor sheathing, roof decking). Rationale: if one member is overloaded, adjacent members share the load. C_L (Beam Stability Factor): Accounts for lateral-torsional buckling of deep, unbraced beams. It is the bending analog of the column stability factor C_P. When full lateral support is provided along the compression face, C_L = 1.0.
Examples
The 2-month load has the shorter duration, so C_D = 1.15 applies to the entire combination. This is the NSCP rule for load combinations.
Scenario
A roof rafter carries dead load (permanent) + roof live load (construction, 2-month duration). Reference F_b = 12 MPa, C_M = C_t = C_F = C_r = C_L = 1.0. Determine the controlling C_D and F'_b.
Solution
The two loads present are: permanent (C_D = 0.9) and 2-month construction load (C_D = 1.15). When combined, use the SHORTER duration: C_D = 1.15 (2-month governs over permanent). F'_b = 12 × 1.15 × 1.0 × 1.0 × 1.0 × 1.0 × 1.0 = 13.8 MPa
Wet service conditions common in Philippine coastal construction reduce F_b significantly. This is a critical check for bahay-kubo style structures or exterior decking.
Scenario
A deck joist is fully exposed to rain in a coastal province (C_M = 0.85 for F_b, per NSCP Table). F_b = 14 MPa, C_D = 1.0, C_F = 1.0, C_r = 1.0, C_L = 1.0. Find F'_b.
Solution
F'_b = 14 × 1.0 × 0.85 × 1.0 × 1.0 × 1.0 × 1.0 = 11.9 MPa
Applications
- Roof framing design using C_D = 1.15 for construction loads
- Bleacher and grandstand design using C_D = 2.0 for crowd impact
- Classifying permanent vs. live wood loading in industrial timber warehouses
- Coastal/tropical building checks using C_M for wet conditions
Misconceptions
- Using C_D = 1.0 for dead loads — permanent loads use C_D = 0.9, reducing allowable by 10%
- Thinking C_r applies to axial compression — C_r is bending ONLY
- Applying C_L to columns — C_L is for beams; columns use C_P
- Forgetting to use C_D = 1.6 for seismic design, a significant allowable increase
Related Concepts
- Viscoelastic behavior and creep in wood
- Moisture content and wood shrinkage
- Lateral bracing requirements for beams (C_L = 1.0 condition)
Common Exam Questions
Example
A gymnasium floor carries dead load + crowd impact load. C_D for this combination = 2.0 (impact governs as shortest duration).
Approach
Identify all load types acting. Determine each load's duration category. When loads combine, select C_D for the shortest-duration component. Apply C_D to ALL reference design values in that load combination.
Question Type
Select the correct C_D for a described loading scenario
Example
F_b = 15 MPa, C_D = 1.25, C_M = 0.85, C_F = 1.1, others = 1.0. F'_b = 15 × 1.25 × 0.85 × 1.1 = 17.53 MPa
Approach
List all applicable factors. Set any not mentioned = 1.0. Multiply sequentially: F'_b = F_b × C_D × C_M × C_t × C_F × C_L × C_r. Check units — F'_b is in MPa.
Question Type
Calculate F'_b with multiple C factors given
Key Points To Remember
- C_D = 0.9 for permanent loads — this REDUCES capacity; many reviewees forget this and use 1.0
- C_D = 1.6 for wind/seismic — this INCREASES capacity by 60%
- When loads combine, use C_D of the shortest-duration load in the set
- C_r = 1.15 applies to bending ONLY, for repetitive members (≥3 parallel, ≤600 mm spacing)
- C_L = 1.0 when compression edge is continuously braced
- C_M reduces stresses for wet/exposed conditions; for sheltered indoor use, C_M = 1.0
Bending Stress in Timber Beams
Bending is the most common timber design check and is virtually guaranteed on the board exam. The flexure formula applies: f_b = M / S where: • f_b = actual bending stress (MPa) • M = maximum bending moment (N·mm) • S = section modulus (mm³) For a rectangular cross-section (standard for sawn lumber): S = bh² / 6 where b = breadth (width) and h = depth (height in the direction of bending). The design check: f_b ≤ F'_b = F_b × C_D × C_M × C_t × C_F × C_L × C_r NOMINAL vs. DRESSED DIMENSIONS: Philippine practice and NSCP typically use nominal dimensions unless stated otherwise. For precision, dressed (actual) dimensions are smaller — e.g., a 50 × 100 nominal may be 38 × 89 mm dressed. Board exams usually specify actual dimensions or state 'nominal' explicitly. BEAM STABILITY FACTOR C_L: When the compression face of a beam is laterally unsupported over a span, the beam can buckle sideways (lateral-torsional buckling). C_L reduces the allowable bending stress to account for this. C_L depends on the slenderness ratio R_B = √(ℓ_e × d / b²). In most exam problems, C_L = 1.0 is given (full lateral support assumed) unless the problem specifically asks you to compute it.
Examples
The beam just barely fails. A 100 × 325 mm section would give S = 100 × 325²/6 = 1.76 × 10⁶ mm³, giving f_b = 9.1 MPa < 10.5 MPa — adequate. On the exam, 'just over' or 'just under' conditions test whether you compute exactly.
Scenario
A simply-supported timber beam spans 4.0 m and carries a total uniformly distributed load of 8 kN/m (including self-weight). The beam is 100 mm wide × 300 mm deep. Check bending if F'_b = 10.5 MPa.
Solution
Step 1: Maximum moment for UDL on simple span: M = wL²/8 = 8 × (4.0)² / 8 = 8 × 16 / 8 = 16 kN·m = 16 × 10⁶ N·mm Step 2: Section modulus: S = bh²/6 = 100 × 300² / 6 = 100 × 90,000 / 6 = 1,500,000 mm³ = 1.50 × 10⁶ mm³ Step 3: Actual bending stress: f_b = M/S = 16 × 10⁶ / 1.50 × 10⁶ = 10.67 MPa Step 4: Check: f_b = 10.67 MPa > F'_b = 10.5 MPa → INADEQUATE in bending (overstressed by ~1.6%)
In design problems, compute S_req then trial sections. Note that changing depth is more efficient than width since S ∝ h² but S ∝ b linearly.
Scenario
Find the required section modulus for a timber beam with M = 20 kN·m, F_b = 16.5 MPa, C_D = 1.15, C_F = 1.1, all other C = 1.0.
Solution
F'_b = 16.5 × 1.15 × 1.1 = 20.87 MPa S_req = M / F'_b = 20 × 10⁶ / 20.87 = 958,308 mm³ ≈ 958 × 10³ mm³ Try 150 × 300 mm: S = 150 × 300²/6 = 2,250,000 mm³ > 958,000 mm³ ✓ (adequate but oversized) Try 100 × 350 mm: S = 100 × 350²/6 = 2,041,667 mm³ ✓ Try 125 × 250 mm: S = 125 × 250²/6 = 1,302,083 mm³ ✓ (more economical)
Applications
- Floor joist design in residential and commercial wood-framed buildings
- Timber bridge stringer design for rural road crossings
- Scaffolding plank adequacy check for construction loads
- Purlin design for metal roofing on timber frames
Misconceptions
- Using S = bh²/6 with b as the larger dimension — b is ALWAYS the width (horizontal), h is depth (vertical in bending plane)
- Forgetting to convert kN·m to N·mm: 1 kN·m = 1 × 10⁶ N·mm
- Using gross area for notched beams — at notches, use the reduced depth d_n
- Applying C_r to non-repetitive single beams — C_r = 1.15 only for 3+ parallel members with sheathing
Related Concepts
- Shear stress (companion check to bending)
- Deflection limits for serviceability
- Lateral-torsional buckling and C_L computation
Common Exam Questions
Example
100 × 250 mm beam, M = 7.5 kN·m, F'_b = 9.5 MPa. S = 100×250²/6 = 1,041,667 mm³. f_b = 7.5×10⁶/1,041,667 = 7.2 MPa < 9.5 MPa. ADEQUATE.
Approach
1) Compute S = bh²/6. 2) Compute f_b = M/S (convert M to N·mm). 3) Compute F'_b = F_b × applicable C factors. 4) State pass/fail.
Question Type
Given M and beam dimensions, find f_b and compare with F'_b
Example
b = 100 mm, M = 10 kN·m, F'_b = 12 MPa. h = √(6 × 10×10⁶ / (100 × 12)) = √(60,000,000/1200) = √50,000 = 223.6 mm → use 225 mm or next standard size
Approach
From f_b = M/S and S = bh²/6: h = √(6M / (b × F'_b)). Solve directly.
Question Type
Find the minimum depth h for a given width b and moment M
Key Points To Remember
- f_b = M/S, with S = bh²/6 for rectangular sections
- M must be in N·mm and S in mm³ for f_b to come out in MPa
- b = breadth (horizontal), h = depth (vertical, in the direction of load)
- If depth and breadth are switched, the section modulus changes by the square of the ratio — a large error
- C_L = 1.0 only when the compression face is continuously laterally braced
- For required section modulus: S_req = M / F'_b; then select a commercial size with S ≥ S_req
Horizontal Shear in Timber Beams
Horizontal shear (shear parallel to grain) is a critical check for wood because wood is significantly weaker in shear along the grain than perpendicular to it. The grain runs longitudinally, and horizontal shear stresses try to slide wood fibers past each other along this direction. For a rectangular cross-section (standard for sawn lumber), the maximum horizontal shear stress occurs at the neutral axis: f_v = 3V / (2A) = 3V / (2bh) where: • f_v = actual horizontal shear stress (MPa) • V = maximum shear force (N) • A = cross-sectional area = b × h (mm²) The factor 3/2 = 1.5 appears because of the parabolic shear stress distribution in a rectangular section (same factor as in reinforced concrete beams and steel beams — a common formula across materials). The design check: f_v ≤ F'_v = F_v × C_D × C_M × C_t Note: C_F, C_L, C_r do NOT apply to shear. C_D, C_M, and C_t do. SHEAR GOVERNS SHORT, DEEP BEAMS: The bending moment varies with L², while shear varies with L. For short spans with heavy loads, shear can govern over bending. NOTCHED BEAMS — A CLASSIC EXAM TRAP: When a beam is notched at the support (to fit flush with a ledger), the effective shear depth is reduced to d_n (the remaining depth below the notch). The shear stress at the notch is: f_v = (3V/2bd_n) × (d/d_n) or equivalently: f_v = 3Vd / (2bd_n²) This is significantly higher than the unnotched value because the stress concentrates at the notch root. Notched beams on the exam require the amplified formula — a very common board-exam pitfall.
Examples
The beam is exactly at the allowable limit. In practice, you would choose a larger section for safety margin. Note V must be in Newtons (12 kN = 12,000 N).
Scenario
A 100 × 200 mm timber beam (actual dimensions) carries a maximum shear V = 12 kN. The adjusted allowable shear stress F'_v = 0.9 MPa. Check shear adequacy.
Solution
A = b × h = 100 × 200 = 20,000 mm² f_v = 3V/(2A) = 3 × 12,000 / (2 × 20,000) f_v = 36,000 / 40,000 = 0.90 MPa f_v = 0.90 MPa = F'_v = 0.90 MPa → EXACTLY AT LIMIT (acceptable, just passes)
This is the classic board-exam trap: a beam that passes ordinary shear check fails at a notch. The ratio d/d_n = 300/240 = 1.25 amplifies the stress by (d/d_n)² = 1.5625 × (3V/2A) relative to the net section.
Scenario
A 100 × 300 mm beam is notched at the support: the notch removes the bottom 60 mm, leaving d_n = 240 mm. Total beam depth d = 300 mm. V = 15 kN. F'_v = 1.0 MPa. Check shear at the notch.
Solution
Using the notched beam formula: f_v = 3Vd / (2bd_n²) f_v = 3 × 15,000 × 300 / (2 × 100 × 240²) f_v = 13,500,000 / (2 × 100 × 57,600) f_v = 13,500,000 / 11,520,000 f_v = 1.172 MPa Since f_v = 1.172 MPa > F'_v = 1.0 MPa → INADEQUATE at notch (17% overstress) Unnotched check for comparison: f_v = 3 × 15,000 / (2 × 100 × 300) = 45,000/60,000 = 0.75 MPa < 1.0 MPa ✓ The beam would pass if unnotched but FAILS at the notch — the notch is the critical detail!
Applications
- Short-span timber floor beams in bahay na bato construction
- Timber girder shear check in pedestrian bridges
- Notched beam connections in traditional Filipino roof framing
- Scaffolding plank checks under concentrated construction loads
Misconceptions
- Using f_v = VQ/It instead of 3V/(2A) — for a rectangular section, these are equivalent, but 3V/(2A) is the shortcut; use it
- Applying C_F to shear — C_F does not apply to F_v
- Not checking notched beams separately — always use the amplified formula at notches
- Forgetting shear check entirely and only doing bending — both checks are required
Related Concepts
- Bending stress check (companion calculation)
- Beam reactions and shear diagrams
- Notch geometry and effective depth
Common Exam Questions
Example
150 × 300 mm beam, V = 20 kN, F_v = 0.8 MPa, C_D = 1.0, all others = 1.0. A = 45,000 mm². f_v = 3×20,000/(2×45,000) = 0.667 MPa < 0.8 MPa. ADEQUATE.
Approach
f_v = 3V/(2bh). Convert V to Newtons. Compute F'_v = F_v × C_D × C_M × C_t. Compare. Note: C_F and C_r do NOT apply.
Question Type
Compute f_v for a rectangular beam and compare with F'_v
Example
100 × 250 mm section, F'_v = 1.0 MPa. V_allow = 2 × 25,000 × 1.0/3 = 16,667 N = 16.67 kN
Approach
Rearrange: V = 2A × F'_v / 3 = 2bh × F'_v / 3.
Question Type
Find the allowable shear V for a given section and F'_v
Example
See Example 2 above — the notch formula gives a significantly higher stress than the plain formula.
Approach
Use f_v = 3Vd/(2bd_n²). Identify d (full depth) and d_n (remaining depth after notch). Do NOT use the simple 3V/(2A) formula.
Question Type
Notched beam shear check
Key Points To Remember
- f_v = 3V/(2A) — the 1.5 factor always applies to rectangular sections
- Horizontal shear (parallel to grain) governs in short, heavily loaded beams
- For notched beams at supports: f_v = 3Vd/(2bd_n²) where d_n = remaining depth
- C_L, C_F, and C_r do NOT apply to F_v — only C_D, C_M, C_t
- Wood reference shear values F_v are typically 0.6–1.2 MPa (much lower than bending)
- Always check both bending AND shear; they are independent checks
Compression Parallel to Grain — Timber Columns
Timber columns are designed for compression parallel to grain. The governing check is: f_c = P/A ≤ F'_c = F_c* × C_P where: • f_c = actual compressive stress = P/A (P = axial load, A = gross area) • F_c* = F_c × C_D × C_M × C_t × C_F (all adjustment factors EXCEPT C_P) • C_P = column stability factor (accounts for buckling) The column stability factor C_P is the timber equivalent of the buckling reduction in steel columns. It reduces the allowable stress as the slenderness ratio ℓ_e/d increases. STEP-BY-STEP C_P COMPUTATION: 1. Compute slenderness ratio: ℓ_e/d (ℓ_e = effective column length, d = least dimension of cross-section) Maximum allowable: ℓ_e/d ≤ 50 (NSCP) 2. Compute critical buckling stress: F_cE = 0.822 E'_min / (ℓ_e/d)² 3. Compute ratio: β = F_cE / F_c* 4. Compute C_P: C_P = [(1 + β)/(2c)] - √{[(1 + β)/(2c)]² - β/c} where c = 0.8 for sawn lumber, c = 0.9 for glulam. 5. Compute F'_c = F_c* × C_P 6. Check: f_c = P/A ≤ F'_c PHYSICAL MEANING OF C_P: • Short, stocky column (low ℓ_e/d, large β): C_P → 1.0 (no buckling reduction) • Long, slender column (high ℓ_e/d, small β): C_P → 0 (heavily penalized by buckling) EFFECTIVE LENGTH ℓ_e: Same as in steel column design. Both ends pinned: K = 1.0, ℓ_e = L. Fixed-pinned: K = 0.7. Fixed-fixed: K = 0.5. Conservative practice uses K = 1.0 unless fixity is guaranteed.
Examples
The slenderness ratio of 20 results in C_P = 0.781, meaning buckling reduces the column capacity to about 78% of its crushing strength. This is a moderately slender column. The same column with ℓ_e/d = 40 would have a much lower C_P (closer to 0.3–0.4).
Scenario
A 150 × 150 mm sawn timber column has an effective length of 3.0 m. Given: E'_min = 6,500 MPa, F_c* = 10.0 MPa, c = 0.8. Find C_P, F'_c, and the allowable axial load P_allow.
Solution
Step 1: Slenderness ratio ℓ_e/d = 3000/150 = 20 ✓ (< 50, acceptable) Step 2: Critical buckling stress F_cE = 0.822 × E'_min / (ℓ_e/d)² F_cE = 0.822 × 6500 / 20² F_cE = 5343 / 400 = 13.36 MPa Step 3: Ratio β = F_cE/F_c* = 13.36/10.0 = 1.336 Step 4: C_P (1 + β)/(2c) = (1 + 1.336)/(2 × 0.8) = 2.336/1.6 = 1.4600 C_P = 1.4600 - √(1.4600² - 1.336/0.8) C_P = 1.4600 - √(2.1316 - 1.670) C_P = 1.4600 - √(0.4616) C_P = 1.4600 - 0.6794 = 0.7806 ≈ 0.781 Step 5: Allowable stress F'_c = F_c* × C_P = 10.0 × 0.781 = 7.81 MPa Step 6: Allowable load A = 150 × 150 = 22,500 mm² P_allow = F'_c × A = 7.81 × 22,500 = 175,725 N ≈ 175.7 kN
Using the least dimension (100 mm, not 150 mm) for the slenderness ratio is critical — it gives the worst-case buckling condition. The column passes, with 25% reserve capacity.
Scenario
A timber column (sawn) has F_c* = 12 MPa. Check if the column is adequate for P = 80 kN given: 100 × 150 mm cross-section, ℓ_e = 2.5 m, E'_min = 7,000 MPa.
Solution
Critical dimension = 100 mm (least dimension) ℓ_e/d = 2500/100 = 25 F_cE = 0.822 × 7000/25² = 5754/625 = 9.206 MPa β = 9.206/12 = 0.767 (1+β)/(2c) = 1.767/(1.6) = 1.1044 C_P = 1.1044 - √(1.1044² - 0.767/0.8) = 1.1044 - √(1.2197 - 0.9588) = 1.1044 - √(0.2609) = 1.1044 - 0.5108 = 0.5936 F'_c = 12 × 0.5936 = 7.12 MPa A = 100 × 150 = 15,000 mm² f_c = P/A = 80,000/15,000 = 5.33 MPa 5.33 MPa < 7.12 MPa → ADEQUATE
Applications
- Timber posts and columns in traditional and modern Filipino residential construction
- Scaffolding and formwork shoring design
- Roof truss compression chord members
- Column-and-beam framing in low-rise commercial structures
Misconceptions
- Including C_P in F_c* — C_P is NOT part of F_c*; F_c* is computed WITHOUT C_P
- Using the larger dimension d in ℓ_e/d — always use the LEAST dimension for the critical buckling axis
- Using c = 0.9 for sawn lumber — c = 0.8 for sawn, 0.9 for glulam only
- Forgetting the slenderness limit of 50 — columns with ℓ_e/d > 50 are code non-compliant
- Confusing F_cE (buckling stress) with F'_c (allowable) — F_cE is an intermediate value, not the answer
Related Concepts
- Euler buckling stress for steel columns (analogous concept)
- Effective length factors K for different end conditions
- Combined axial and bending in beam-columns
Common Exam Questions
Example
See full Example 1 above — it is a model exam solution.
Approach
Follow the 6-step procedure: (1) ℓ_e/d, (2) F_cE, (3) β = F_cE/F_c*, (4) C_P formula, (5) F'_c = F_c* × C_P, (6) compare f_c = P/A with F'_c. This is the most complex timber calculation and frequently appears as a full problem.
Question Type
Full C_P computation and column capacity check
Example
From Example 1: P_allow = 7.81 × 22,500 = 175.7 kN
Approach
Compute A = b × d. Compute F'_c through full C_P procedure. Then P_allow = F'_c × A.
Question Type
Given dimensions, find allowable load P
Example
200 mm column, L = 12 m, both ends pinned: ℓ_e/d = 12000/200 = 60 > 50 — NOT ALLOWED as a column.
Approach
Check ℓ_e/d ≤ 50. If exceeded, the member is not permitted as a column under NSCP 2015.
Question Type
Verify if ℓ_e/d is within the allowable slenderness limit
Key Points To Remember
- F_c* = F_c × C_D × C_M × C_t × C_F (all factors EXCEPT C_P)
- C_P is always computed last and multiplied to F_c* only
- c = 0.8 for sawn lumber, c = 0.9 for glulam
- ℓ_e/d must not exceed 50 (slenderness limit for timber columns)
- Use the LEAST dimension d of the cross-section for the critical ℓ_e/d ratio
- β = F_cE/F_c* — when β >> 1, column is short; when β << 1, column is very slender
Practice Problems
The 150 × 250 mm section is quite conservatively sized for this load. In a real design, a smaller section might be tried. Note that depth (h = 250) is the larger dimension placed vertically — this maximizes S. Reversing b and h (using 250 × 150) would give S = 250 × 150²/6 = 937,500 mm³, reducing capacity by 40% — a critical orientation error.
Problem
PROBLEM 1 — Bending Check A 150 × 250 mm (actual) simply-supported timber joist spans 3.5 m and carries a uniformly distributed load w = 6 kN/m (total, including self-weight). The adjusted allowable bending stress F'_b = 11 MPa. (a) Find the actual bending stress f_b. (b) Is the joist adequate in bending?
Solution
(a) Maximum moment for simple beam with UDL: M = wL²/8 = 6 × (3.5)² / 8 = 6 × 12.25 / 8 = 9.1875 kN·m = 9.1875 × 10⁶ N·mm Section modulus (b = 150 mm, h = 250 mm): S = bh²/6 = 150 × 250² / 6 = 150 × 62,500 / 6 = 1,562,500 mm³ Actual bending stress: f_b = M/S = 9.1875 × 10⁶ / 1,562,500 = 5.88 MPa (b) Check: f_b = 5.88 MPa < F'_b = 11 MPa ✓ The joist is ADEQUATE in bending with a stress ratio of 5.88/11 = 53.5% (substantial reserve).
Even with wet service reducing the value (C_M = 0.85), the combination of short-duration load (C_D = 1.25) and repetitive-member factor (C_r = 1.15) results in an adjusted allowable higher than the reference value F_b. This illustrates how C factors can both increase and decrease allowable stresses depending on conditions.
Problem
PROBLEM 2 — Adjusted Allowable Stress Computation A timber beam has a reference bending value F_b = 14 MPa. The following conditions apply: 7-day roof construction load (C_D = 1.25), wet service (C_M = 0.85), normal temperature (C_t = 1.0), size factor C_F = 1.0, repetitive member C_r = 1.15, beam stability C_L = 1.0. Find the adjusted allowable bending stress F'_b.
Solution
F'_b = F_b × C_D × C_M × C_t × C_F × C_L × C_r F'_b = 14 × 1.25 × 0.85 × 1.0 × 1.0 × 1.0 × 1.15 F'_b = 14 × 1.25 = 17.50 17.50 × 0.85 = 14.875 14.875 × 1.15 = 17.106 MPa F'_b ≈ 17.11 MPa
This problem is the classic board-exam notch trap. The unnotched beam would pass shear, but the notch causes failure. Solutions: (1) increase d_n by using a shallower notch, (2) use a larger beam, (3) add a metal bearing plate. The amplification factor is (d/d_n)² = (300/225)² = (1.333)² = 1.778, applied to the net section shear.
Problem
PROBLEM 3 — Horizontal Shear Check (Notched Beam) A 100 × 300 mm timber beam is notched at the support. The notch is cut at the bottom, leaving a net depth d_n = 225 mm (removed 75 mm from 300 mm total depth d). The beam reaction (maximum shear at support) is V = 18 kN. The adjusted allowable shear stress F'_v = 1.0 MPa. Check shear at the notch.
Solution
Using the notched beam shear formula: f_v = 3Vd / (2b × d_n²) Substituting values: V = 18 kN = 18,000 N d = 300 mm (total depth) d_n = 225 mm (net depth at notch) b = 100 mm f_v = 3 × 18,000 × 300 / (2 × 100 × 225²) f_v = 16,200,000 / (2 × 100 × 50,625) f_v = 16,200,000 / 10,125,000 f_v = 1.600 MPa Check: f_v = 1.600 MPa > F'_v = 1.0 MPa → INADEQUATE (60% overstress!) For comparison, if unnotched: f_v = 3V/(2bh) = 3×18,000/(2×100×300) = 54,000/60,000 = 0.90 MPa < 1.0 MPa → would PASS The notch causes a 78% increase in shear stress (1.60 vs 0.90 MPa) — the beam must be redesigned.
Column design by direct sizing requires iteration — C_P depends on d, which is what we're solving for. The exam typically asks you to CHECK a given size (Part b), not iterate from scratch. Part (a) illustrates why checking a proposed size is more practical on the exam.
Problem
PROBLEM 4 — Timber Column Design (Full C_P Solution) A square sawn timber column supports P = 120 kN. Effective length ℓ_e = 3.6 m. Given: F_c* = 9.5 MPa, E'_min = 5,500 MPa, c = 0.8. Determine: (a) the required size if a square section is used, (b) check a 175 × 175 mm section.
Solution
(b) Check 175 × 175 mm section: Step 1: Slenderness d = 175 mm (both sides equal for square) ℓ_e/d = 3600/175 = 20.57 Check: 20.57 < 50 ✓ Step 2: Critical buckling stress F_cE = 0.822 × 5500 / (20.57)² = 4521 / 423.1 = 10.69 MPa Step 3: β ratio β = F_cE/F_c* = 10.69/9.5 = 1.125 Step 4: C_P (1+β)/(2c) = (1 + 1.125)/(2 × 0.8) = 2.125/1.6 = 1.3281 C_P = 1.3281 - √(1.3281² - 1.125/0.8) = 1.3281 - √(1.7638 - 1.4063) = 1.3281 - √(0.3576) = 1.3281 - 0.5980 = 0.7301 Step 5: Allowable stress F'_c = F_c* × C_P = 9.5 × 0.7301 = 6.936 MPa Step 6: Capacity check A = 175 × 175 = 30,625 mm² P_allow = F'_c × A = 6.936 × 30,625 = 212,340 N = 212.3 kN f_c = P/A = 120,000/30,625 = 3.92 MPa Check: f_c = 3.92 MPa < F'_c = 6.94 MPa ✓ The 175 × 175 mm column is ADEQUATE for 120 kN. Stress ratio = 3.92/6.94 = 56.5%. (a) For minimum size, set f_c = F'_c. This requires iteration since C_P depends on d. Starting estimate without C_P: A_req ≈ P/F_c* = 120,000/9.5 = 12,632 mm² → d ≈ 112 mm. Try 125 × 125: ℓ_e/d = 28.8, iterate C_P → likely around 0.55, giving F'_c ≈ 5.2 MPa, A_req = 23,077 mm² → d = 152 mm. Try 150 × 150: ℓ_e/d = 24, proceed with full C_P calculation as shown above to verify.
Both checks must be performed independently. The beam fails bending despite passing shear — neither check can be skipped. A 3.0 m span with a 25 kN concentrated load is relatively high loading for a 100 × 300 section, and the bending governs. Note: for a simply supported beam with midspan point load, M = PL/4 and V = P/2 are the critical values.
Problem
PROBLEM 5 — Combined Bending and Shear Check A 100 × 300 mm timber beam spans 3.0 m (simply supported) and carries a concentrated load P = 25 kN at midspan. F'_b = 12 MPa and F'_v = 1.0 MPa. Check the beam for both bending and shear.
Solution
Reactions: R = P/2 = 25/2 = 12.5 kN (symmetric simple beam) Maximum moment (at midspan): M = PL/4 = 25 × 3.0/4 = 18.75 kN·m Maximum shear (at supports): V = 12.5 kN BENDING CHECK: S = bh²/6 = 100 × 300²/6 = 1,500,000 mm³ f_b = M/S = 18.75 × 10⁶/1,500,000 = 12.50 MPa f_b = 12.50 MPa > F'_b = 12.0 MPa → INADEQUATE in bending (4.2% overstress) SHEAR CHECK: A = 100 × 300 = 30,000 mm² f_v = 3V/(2A) = 3 × 12,500/(2 × 30,000) = 37,500/60,000 = 0.625 MPa f_v = 0.625 MPa < F'_v = 1.0 MPa → ADEQUATE in shear CONCLUSION: Beam fails in bending but passes in shear. A 100 × 325 mm (or 125 × 300 mm) section should be tried next. For 100 × 325: S = 100 × 325²/6 = 1,760,417 mm³, f_b = 18.75×10⁶/1,760,417 = 10.65 MPa < 12 MPa ✓
Exam Preparation Tips
- MEMORIZE the C_D table cold: 0.9 (permanent), 1.0 (normal 10-yr), 1.15 (2-month), 1.25 (7-day), 1.6 (wind/seismic), 2.0 (impact). At least one question on every CELE will test this.
- Know which C factors apply to which reference value: C_D, C_M, C_t apply to ALL; C_F, C_L, C_r apply to F_b only (with C_L and C_r having specific conditions); C_F also applies to F_c and F_t; C_P applies ONLY to F_c.
- Always convert moments to N·mm before dividing by S in mm³ — the result is in MPa. Do NOT mix kN·m with mm³.
- For timber columns, the formula sequence is fixed: (1) ℓ_e/d → (2) F_cE → (3) β → (4) C_P → (5) F'_c = F_c* × C_P → (6) f_c = P/A ≤ F'_c. Practice this until it is automatic.
- Use the least cross-sectional dimension d in the ℓ_e/d slenderness ratio for columns — this gives the worst-case axis.
- c = 0.8 for sawn lumber, c = 0.9 for glulam — the exam usually specifies 'sawn'; when in doubt, use 0.8.
- Notched beams: When a problem mentions a notch at the support, use f_v = 3Vd/(2bd_n²), NOT f_v = 3V/(2bh). This is the most common shear trap on the CELE.
- S = bh²/6 requires that h is the bending dimension (depth). Sketch the cross-section and label b and h before computing.
- F_c* ≠ F'_c. F_c* is computed WITHOUT C_P; F'_c = F_c* × C_P. Many examinees confuse these.
- For design problems (find required section), compute S_req = M/F'_b, then select a section with S ≥ S_req. The most economical choice maximizes h over b since S ∝ h².
- Always check both bending AND shear independently. A beam that passes one check may fail the other. The exam frequently sets up both checks with different outcomes.
- Slenderness limit: ℓ_e/d ≤ 50 for timber columns. If this is violated, the member is code-non-compliant — mark it as inadequate without computing C_P.
- When load combinations include wind or seismic, C_D = 1.6 applies. This is a significant 60% increase in allowable capacity, which makes the combined check less critical — understand this physically.
- Practice the C_P formula repeatedly with different β values. Note: when β is very large (stocky column), C_P approaches 1.0. When β is very small (slender column), C_P approaches β/c (Euler-like control).
In summary
Timber design under NSCP 2015 Chapter 6 is one of the more straightforward topics on the PRC CELE once the ASD framework is internalized. The entire discipline reduces to three words: adjust, compute, compare. Adjust the reference design value using the appropriate C factors (F' = F × ∏C); compute the actual stress (f_b = M/S, f_v = 3V/2A, or f_c = P/A); and compare (f ≤ F'). The complexity lies in the details: knowing which C factors apply to which reference value, correctly identifying C_D for the governing load duration, applying the notched-beam shear formula when a notch is present, and executing the full 6-step C_P column procedure accurately. These four areas — C_D identification, C factor assignment, notched shear, and C_P computation — are where most examination points are won or lost. Review the worked examples in this chapter until each calculation is automatic. Understand the physics behind each factor: C_D reflects wood's viscoelastic response to sustained loads; C_L and C_P account for buckling instability; C_M reflects the weakening effect of moisture on wood fibers. With this physical understanding backed by procedural fluency, timber design becomes one of the more reliably scoring topics in the Steel & Timber Design component of the CELE. Mahal naming lahat — study hard and pass the board!
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