CELE Steel & Timber Design — Timber DesignRevision Notes

Formulas

Exam Tips

• Memorize which C factors apply to Fb, Fv, Fc — draw a quick C-factor table on scratch paper at the start of the exam.
• In multi-part problems, compute F' once and use it for all sub-questions to save time.
• When the problem says 'all other factors = 1.0', simply multiply the stated factors — no need to hunt for missing ones.
• Always confirm units: M in N·mm, S in mm³ → fb in MPa; V in N, A in mm² → fv in MPa.

Key Points

• ASD principle for timber: F' = F × (product of all applicable adjustment factors C). Actual stress f must satisfy f ≤ F'.
• Reference design values (Fb, Fv, Fc, Fc⊥, Ft, E, Emin) are species- and grade-specific; NSCP 2015 Table 6 lists Philippine species groups.
• Adjustment factors are applied differently for each stress type — not every C applies to every F.
• The adjusted modulus of elasticity for stability calculations is E'min = Emin × CM × Ct (plus other factors as applicable).
• ASD timber is fundamentally different from LRFD used in AISC 360 for steel — do not mix load combinations or resistance factors.
• NSCP 2015 Chapter 6 is the governing code for timber in Philippine practice (alongside RA 544 for professional responsibility).

Definitions

Section Title

1. Allowable Stress Design (ASD) Framework for Wood

Common Mistakes

• Using the reference value F directly as the allowable without applying adjustment factors — this is the single most common exam error.
• Applying adjustment factors for one stress type (e.g., bending) to a different stress type (e.g., compression) where they do not apply.
• Confusing ASD timber with LRFD steel design — do not use load factors (1.2D + 1.6L) for wood ASD checks.
• Forgetting that E'min (not E') is used in stability calculations (CL, CP).

Formulas

Exam Tips

• In CELE problems, the CD value is usually stated directly or implied by the load type (e.g., 'floor dead load only' → CD = 0.9).
• If the problem mentions 'normal loading' or '10-year,' use CD = 1.0.
• For combined loads: use CD for the shortest-duration load in the combination (most critical governs).
• Cr = 1.15 is an easy free bonus — always check if the problem mentions 'repetitive' or '3 or more joists'.

Key Points

• CD (Load Duration): Most important timber-specific factor. Short-duration loads allow higher stresses because wood fibres recover. Values: permanent (CD = 0.9), 10-year (1.0), 2-month (1.15), 7-day (1.25), wind/seismic (1.6), impact (2.0).
• When multiple load types act simultaneously, use the CD for the shortest-duration load in the combination.
• CM (Wet Service): Reduces reference values when in-service moisture content > 19% for sawn lumber. Tabulated in NSCP 2015.
• Ct (Temperature): Reduces values for sustained temperatures above 38°C.
• CF (Size Factor): Adjusts for actual sawn lumber dimensions; larger cross-sections often have lower Fb per unit area.
• CL (Beam Stability Factor): Reduces Fb for beams susceptible to lateral-torsional buckling; analogous to Cb in steel design.
• CP (Column Stability Factor): Reduces Fc for slender columns; analogous to Fcr/Fy in steel design.
• Cr (Repetitive Member Factor): Cr = 1.15 for bending only, applied when three or more members spaced ≤ 600 mm share load (e.g., floor joists).
• Ci (Incising Factor), CH (Horizontal Shear Adjustment), Cb (Bearing Area Factor) — secondary factors occasionally tested.

Definitions

Section Title

2. Adjustment Factors — The C-Factor System

Common Mistakes

• Applying CD to E or Emin — CD does NOT affect modulus of elasticity.
• Applying CD to Fc⊥ (bearing perpendicular to grain) — CD does NOT apply to Fc⊥.
• Using CD = 1.0 for wind/seismic instead of the correct 1.6.
• Forgetting Cr = 1.15 for repetitive members — this 15% bonus can change a section selection.
• Applying CL and CF simultaneously at full value — for sawn lumber, only the lesser of CL or CF applies to Fb (not both multiplied together, unless both < 1.0 per code logic).

Formulas

Exam Tips

• Always convert moments: 1 kN·m = 10⁶ N·mm = 10⁶ N·mm. Write this conversion step explicitly.
• For a simply supported beam with uniform load w (kN/m) and span L (m): M = wL²/8; V = wL/2 — these are the most common CELE loading patterns.
• When asked to 'select a section,' find Sreq first, then pick from standard lumber sizes (e.g., 50×, 75×, 100×, 150×, 200× mm series in NSCP).
• If CL and CF are both given and both < 1.0, check which governs (use the smaller product as guided by NSCP 2015 Section 606).

Key Points

• Actual bending stress: fb = M/S where S = bh²/6 for a rectangular section (b = width, h = depth, both in mm).
• Adjusted allowable: F'b = Fb × CD × CM × Ct × CF × CL × Cr (apply only applicable factors).
• Check: fb ≤ F'b — if satisfied, beam is adequate in bending.
• Required section modulus: Sreq = M / F'b → select standard lumber with S ≥ Sreq.
• CL (beam stability) is required when the compression face of the beam is not laterally braced — long spans with no floor sheathing on top.
• For fully laterally braced beams (common in floor systems), CL = 1.0.
• Deeper sections are more efficient in bending (S ∝ bh²) but may require CF < 1.0.
• Glulam beams have different reference values and volume factor CV (not CF) for bending.

Definitions

Section Title

3. Bending Design of Wood Beams

Common Mistakes

• Using M in kN·m and S in mm³ without converting — always convert M to N·mm (multiply kN·m by 10⁶).
• Confusing b (width) and h (depth) in the S formula — h is the full depth, b is the breadth across the grain.
• Omitting CF for deep members (h > 305 mm = 12 in) — this can make the section appear adequate when it is not.
• Forgetting that S = bh²/6 uses the full depth h, not the net depth at notches.

Formulas

Exam Tips

• The shear formula 3V/2A is the same 1.5-parabolic-distribution factor used in mechanics of materials for rectangular sections.
• Always check: is the beam notched? If yes, use the notch formula. If the problem is silent, assume no notch.
• F'v for Philippine wood species is very low (often ~0.7–1.0 MPa) — shear frequently controls in short, heavily loaded beams.
• If both bending and shear are asked, check bending first (often it governs) then shear. If shear controls, increase depth or add a bearing block.

Key Points

• Horizontal shear (shear parallel to grain) is the critical shear failure mode in wood — not vertical shear.
• For a rectangular section: fv = 3V / (2A) = 1.5 V/A, where A = bh (gross cross-sectional area).
• Adjusted allowable: F'v = Fv × CD × CM × Ct (Cr and CF do not apply to shear).
• Check: fv ≤ F'v.
• CRITICAL: At notched supports, use the net depth dn and amplify: fv = (3V/2bdn)(d/dn) — notching dramatically reduces shear capacity.
• Moving loads: NSCP 2015 allows neglecting loads within a distance d (depth) from supports for shear calculation.
• Short, deep beams (small span-to-depth ratio) are most likely to fail in shear — this is a classic CELE exam setup.
• F'v is typically 0.7–1.2 MPa for most Philippine species — very low compared to bending allowables.

Definitions

Section Title

4. Shear Design of Wood Beams (Horizontal/Parallel-to-Grain Shear)

Common Mistakes

• Using A = gross area but the beam has a notch — at notches, use net area and apply the (d/dn) amplification factor.
• Applying Cr = 1.15 to shear — Cr applies to Fb (bending) only, never to Fv.
• Forgetting to convert V from kN to N before dividing by A in mm².
• Using the formula V/A (without the 1.5 factor) — this is the average shear, not the maximum horizontal shear in a rectangle.

Formulas

Exam Tips

• Step-by-step column drill: (1) Compute ℓe/d; (2) Compute F*c; (3) Compute FcE; (4) Compute β; (5) Compute CP; (6) F'c = F*c × CP; (7) P_allow = F'c × A.
• If the problem gives 'F'c directly' (already adjusted), skip to P = F'c × A — read carefully.
• For quick checks: if ℓe/d < 10, CP ≈ 1.0 (stocky); if ℓe/d > 40, CP will be small (< 0.5).
• The CP formula has a ± sign in the general form — for design, always take the MINUS sign (smaller root gives CP ≤ 1.0).
• Bring a scientific calculator — the CP computation involves squaring, subtraction inside a square root, and division. Practice this 5 times before the exam.

Key Points

• Actual compressive stress: fc = P/A ≤ F'c = F*c × CP.
• F*c = Fc × CD × CM × Ct × CF — all factors EXCEPT CP applied first.
• CP (Column Stability Factor) accounts for buckling, analogous to steel column curves.
• Euler critical stress for wood: FcE = 0.822 E'min / (ℓe/d)².
• Slenderness ratio ℓe/d ≤ 50 (maximum allowed for timber columns per NSCP 2015).
• c = 0.8 for sawn lumber; c = 0.9 for glulam; c = 0.85 for round timber.
• For a rectangular column: use the larger ℓe/d ratio (controlling axis — smaller d controls).
• When ℓe/d is large (slender column), CP → 0 and Fc' → FcE (Euler governs).
• When ℓe/d is small (stocky column), CP → 1.0 and F'c → F*c (material governs).
• Short columns (small ℓe/d): CP ≈ 1.0. Long columns: CP << 1.0.

Definitions

Section Title

5. Compression Parallel to Grain — Timber Column Design

Common Mistakes

• Including CP in the computation of F*c — F*c deliberately excludes CP to avoid circular computation.
• Using E instead of E'min in the FcE formula — always use the minimum modulus (Emin) adjusted for service conditions.
• Using the wrong value of c — 0.8 for sawn lumber, 0.9 for glulam, 0.85 for round timber.
• Using the larger dimension d instead of the smaller dimension in ℓe/d — always use the controlling (smaller) dimension.
• Forgetting to check ℓe/d ≤ 50 — exceeding this limit violates NSCP 2015 and the column cannot be used.
• Arithmetic error in the CP formula — expand the square root term carefully; a common error is computing (1+β)/(2c) incorrectly.

Formulas

Exam Tips

• When a problem mentions 'beam seat,' 'bearing plate,' or 'sill plate,' check Fc⊥.
• CD does not apply to Fc⊥ — memorize this exception; it is a classic distracter in CELE options.
• Bearing problems often pair with bending/shear problems in multi-part CELE questions — do all three checks.

Key Points

• Compression perpendicular to grain (Fc⊥) applies at bearing areas: beam seats on a plate, joist on a ledger, column on a sill.
• Actual stress: fc⊥ = P / Ab where Ab = bearing area.
• CD does NOT apply to Fc⊥ — this is one of the few cases where load duration does not help.
• Adjusted allowable: F'c⊥ = Fc⊥ × CM × Ct × Cb (bearing area factor).
• Cb (Bearing Area Factor): Cb = (ℓb + 9.5) / ℓb for ℓb ≤ 150 mm; where ℓb = bearing length (mm).
• Cb > 1.0 for short bearing areas (stress is distributed into a larger volume of wood, increasing capacity).
• Fc⊥ is typically 1.5–3.0 MPa — much lower than Fc parallel. Always check bearing at supports of heavily loaded beams.

Definitions

Section Title

6. Compression Perpendicular to Grain and Bearing

Common Mistakes

• Applying CD to Fc⊥ — duration factor does NOT apply to bearing perpendicular to grain.
• Using the full beam cross-section area instead of the actual bearing area Ab.
• Forgetting Cb for short bearing lengths — this factor can meaningfully increase the allowable.