CELE Steel & Timber Design — Timber DesignMisconception Buster
Avoid the most common Timber Design mistakes made by CELE reviewers. Each misconception here has been pulled from real CELE Steel & Timber Design questions where Professional Regulation Commission (PRC) — Board of Civil Engineering used it to separate strong reviewers from weak ones. Learn these before your next mock.
Exam context
The Civil Engineer Licensure Examination is conducted by Professional Regulation Commission (PRC) — Board of Civil Engineering and is scheduled for May and November 2026. The Steel & Timber Design subtest is marked as "Core" in the official pattern, and Timber Design appears in position 5th of 5 in the CELE Steel & Timber Design review rotation. Passing mark: 70% weighted average, no sub-test below 50%. Recent CELE 2026 papers have drawn roughly a meaningful share of questions from this subject.
Timber Design - Misconception Buster
Timber Design consistently appears on the PRC Civil Engineer Licensure Examination, yet it is one of the most mishandled topics by reviewees. The pitfalls are not in complex mathematics — they are in systematic misapplication of adjustment factors, misidentification of the governing formula, and confusion between wood ASD and the LRFD methods used for steel and concrete. A single missed adjustment factor can flip a 'pass' section to a 'fail,' costing you points on what should be a straightforward computation. This guide isolates the 12 most dangerous misconceptions in Timber Design (NSCP 2015, Chapter 6), explains why smart students fall for them, and provides trap questions that mirror actual board-exam style. Study these corrections until the right approach is automatic — the difference between the reference design value F and the adjusted allowable F-prime is the single most important distinction in the entire chapter.
Summary
The 12 misconceptions in this guide converge on four master principles for CELE timber design success. First, F is never F-prime: every reference design value must be multiplied by its applicable adjustment factors (CD, CM, Ct, CF, CL, CP, Cr) — the adjusted allowable F' is always the correct comparator against actual stress. Second, CD applies only to strength properties — never to E or Fc-perp — and for combined loads, the shortest-duration load's CD governs, never an average. Third, column stability is always computed: CP must be derived from the full formula using F*c (not F'c) and the slenderness-based FcE; c = 0.8 for sawn, 0.90 for glulam, 0.85 for poles — lateral bracing reduces ℓe but does not set CP = 1.0 automatically. Fourth, timber is pure ASD: service-level (unfactored) loads, no φ factors, no 1.2D + 1.6L — this is the most conceptually dangerous confusion for students who have just studied LRFD steel and concrete. Additional traps to avoid: use 3V/(2A) with dn and the (d/dn) amplification at notched supports; take the lesser of CF or CL for sawn lumber bending (not their product); and apply Cr = 1.15 only to closely-spaced groups of dimension lumber with load-sharing sheathing. Master these distinctions and timber questions on the CELE become straightforward computation problems rather than sources of lost marks.
Misconceptions
The reference design value F (e.g., Fb, Fv, Fc) from the species/grade table is already the allowable stress that can be used directly in design checks.
Tags
- common_error
- formula_confusion
- exam_trap
Topic
Allowable Stress Design — Adjustment Factor System
Severity
critical
Exam Impact
Students who use F directly instead of F-prime will either declare adequate sections as failing (conservative error) or pass dangerously under-designed sections (unconservative error), depending on whether the adjustment factors increase or decrease F. Either way, the numerical answer is wrong.
The Reality
In NSCP 2015 Chapter 6 ASD for wood, the tabulated reference design value F is NEVER directly the allowable stress. You must compute F-prime = F × (product of all applicable C factors). Applying even a single adjustment factor — say CD = 1.15 for a 2-month construction load — changes the allowable by 15%. Ignoring CM (wet-service) can overestimate the allowable by 20–40%. The adjustment-factor product can range from roughly 0.5 to 2.0, meaning the true allowable may be half or double the reference value.
Trap Question
Question
A wood beam has a reference bending stress Fb = 12 MPa. The load is a normal occupancy floor load (10-year duration, CD = 1.0), moisture content is above 19% (CM = 0.85), and all other factors are 1.0. What is the allowable bending stress F'b?
Explanation
The wet-service factor CM = 0.85 reduces the allowable by 15% when the moisture content exceeds 19% (the fiber-saturation threshold for most species). Ignoring CM overestimates the allowable by 1.75 MPa, which could mask an overstressed section.
Wrong Answer
F'b = 12 MPa (student reads the table value and uses it directly).
Correct Answer
F'b = 12 × 1.0 × 0.85 = 10.2 MPa.
Misconception Id
M1
Correct Vs Incorrect
Correct Approach
F'b = Fb × CD × CM × Ct × CF × CL × Cr = 16.5 × 1.15 × 1.0 × 1.0 × 1.1 × 1.0 × 1.0 = 20.9 MPa. Then check fb ≤ 20.9 MPa.
Incorrect Approach
F'b = Fb = 16.5 MPa (table value used directly). Then check fb ≤ 16.5 MPa. WRONG.
Why Students Believe It
Students are used to structural steel tables where Fy is a material property directly usable in limit-state equations. They pattern-match: 'I look up a stress value in a table, and I use it.' The timber reference table looks identical to a steel property table, so they skip the adjustment-factor system entirely.
The load duration factor CD applies equally to all reference design values, including modulus of elasticity E.
Tags
- common_error
- formula_confusion
- code_misread
Topic
Load Duration Factor CD — Scope of Application
Severity
critical
Exam Impact
In column stability problems, students compute E'min = Emin × CD instead of E'min = Emin × CM × Ct. This changes FcE and therefore CP, shifting the allowable column load by a potentially large margin.
The Reality
NSCP 2015 explicitly exempts the modulus of elasticity E and the compression-perpendicular value Fc-perp from the CD factor. CD is applied to Fb, Fv, Fc (parallel), Ft, and connection design values — strength properties. E is used for deflection and stability (Euler-type buckling via FcE), not strength, so load duration does not modify it. Applying CD to E is a code violation and produces incorrect stability calculations.
Trap Question
Question
A timber column is designed for wind load (CD = 1.6). The reference minimum modulus is Emin = 4,000 MPa. What is E'min used in computing the critical buckling stress FcE?
Explanation
NSCP 2015 Chapter 6 excludes E and Fc-perp from the load duration factor. Applying CD = 1.6 to E inflates FcE by 60%, artificially boosting CP and overstating the column's axial capacity — an unconservative and code-non-compliant result.
Wrong Answer
E'min = 4,000 × 1.6 = 6,400 MPa (student applies CD to E).
Correct Answer
E'min = 4,000 MPa (assuming CM = 1.0, Ct = 1.0). CD does not apply to E.
Misconception Id
M2
Correct Vs Incorrect
Correct Approach
E'min = Emin × CM × Ct (CD does NOT apply to E or Fc-perp). For dry service and normal temperature, E'min = Emin × 1.0 × 1.0 = Emin.
Incorrect Approach
E'min = Emin × CD × CM × Ct. Student multiplies E by CD = 1.6 for a wind load combination. WRONG.
Why Students Believe It
CD is introduced as a universal multiplier for wood design. Students logically generalize: 'Every property of wood is affected by how long the load acts, so CD must multiply everything.' This feels scientifically sound because creep does affect long-term stiffness.
Horizontal shear in a wood beam is computed the same way as in a steel wide-flange: fv = VQ/(Ib), using the actual cross-sectional geometry at the point of interest.
Tags
- common_error
- formula_confusion
- exam_trap
- notch_effect
Topic
Horizontal Shear — Notched Beams
Severity
critical
Exam Impact
Students who apply VQ/(Ib) at the wrong location under-predict the maximum shear stress and wrongly declare an overstressed beam as adequate. Students who forget the notch amplification overestimate shear capacity at notched supports — a classic CELE trap worth several marks.
The Reality
For rectangular wood cross-sections, NSCP 2015 Chapter 6 prescribes the simplified formula fv = 3V/(2A), which is equivalent to VQ/(Ib) evaluated at the neutral axis of a rectangle (the worst-case location). This formula must always be evaluated with the full cross-sectional area A = b × d, and V is the shear force AT the critical section. At a notched support, the critical section uses the net depth dn, not full d, AND an amplification factor (d/dn) is applied — significantly reducing apparent capacity.
Trap Question
Question
A 100 × 200 mm wood beam (b = 100 mm, d = 200 mm) has a square notch at the support reducing depth to dn = 150 mm. Maximum shear V = 10 kN. The adjusted allowable shear is F'v = 0.9 MPa. Is the beam adequate in shear at the notch?
Explanation
The notch amplification factor (d/dn) = 200/150 = 1.333 increases the effective shear stress from what it would be in an un-notched section of depth dn. Ignoring this factor yields 1.0 MPa, which is itself already above the allowable, but the amplified value of 1.333 MPa makes the overstress even more severe. Notched supports are a known failure mode in wood beams — NSCP 2015 Chapter 6 requires this check explicitly.
Wrong Answer
fv = 3(10,000)/(2 × 100 × 200) = 0.75 MPa < 0.9 MPa. Adequate. (Student uses full depth d = 200 mm — WRONG.)
Correct Answer
fv = [3V/(2bdn)] × (d/dn) = [3 × 10,000/(2 × 100 × 150)] × (200/150) = 1.0 × 1.333 = 1.333 MPa > 0.9 MPa. NOT adequate.
Misconception Id
M3
Correct Vs Incorrect
Correct Approach
fv = (3V/2) × (d/(b × dn²)) = (3V × d)/(2 × b × dn²) at a notched support, OR equivalently: fv = [3V/(2 × b × dn)] × (d/dn). Use dn and the amplification factor.
Incorrect Approach
At a notched support with dn = 150 mm and d = 200 mm: fv = 3V/(2 × b × d) = 3V/(2 × b × 200). WRONG — uses full depth instead of net depth.
Why Students Believe It
Students have just finished steel design where the shear-flow formula VQ/(Ib) is standard. Timber beams look structurally similar, so they apply the same formula. For a rectangular section, this would give fv_max = 1.5V/A, which is actually correct — but students often use Q for a point that is not the neutral axis, obtaining a non-maximum value and under-checking shear.
The column stability factor CP can be taken as 1.0 for any timber column that has adequate lateral bracing at mid-height.
Tags
- common_error
- conceptual_gap
- exam_trap
- buckling
Topic
Compression — Column Stability Factor CP
Severity
critical
Exam Impact
Setting CP = 1.0 when ℓe/d = 20 could overestimate the allowable axial load by 20–30%, leading to an unsafe and incorrect design that the exam will flag.
The Reality
CP is a computed value that depends on the actual slenderness ratio ℓe/d in the critical buckling direction, the adjusted E'min, and F*c. Bracing at mid-height halves the effective length in that plane, reducing ℓe/d, but CP must still be calculated from the formula. CP = 1.0 only in the limiting case where FcE/F*c >> 1 (very stocky column). For most practical timber columns with ℓe/d > 10, CP is meaningfully less than 1.0.
Trap Question
Question
A 100 × 100 mm sawn timber column, 4 m tall, is braced at mid-height against weak-axis buckling. F*c = 9 MPa, E'min = 5,500 MPa, c = 0.8. Is the allowable axial load equal to F*c × A = 9 × 10,000 = 90 kN?
Explanation
Even with bracing reducing effective length to 2 m, the slenderness ratio of 20 is still large enough that CP = 0.763, not 1.0. CP = 1.0 would require an extremely stocky column (ℓe/d approaching zero). This is the most dangerous misconception in timber column design.
Wrong Answer
Yes, because the column is braced at mid-height, so CP = 1.0 and P_allow = 90 kN.
Correct Answer
No. ℓe = 2.0 m (bracing halves effective length), ℓe/d = 2000/100 = 20. FcE = 0.822(5,500)/400 = 11.30 MPa. β = 11.30/9 = 1.256. (1+β)/(2c) = 2.256/1.6 = 1.410. CP = 1.410 − √(1.410² − 1.256/0.8) = 1.410 − √(0.419) = 1.410 − 0.647 = 0.763. F'c = 9 × 0.763 = 6.87 MPa. P_allow = 6.87 × 10,000 = 68.7 kN — significantly less than 90 kN.
Misconception Id
M4
Correct Vs Incorrect
Correct Approach
With bracing at mid-height, ℓe = 0.5L in that direction. Compute ℓe/d, then FcE = 0.822 E'min/(ℓe/d)², then β = FcE/F*c, then CP = (1+β)/(2c) − √[(1+β)/(2c)]² − β/c. Use F'c = F*c × CP.
Incorrect Approach
Column has lateral bracing at mid-height; therefore CP = 1.0; F'c = F*c × 1.0 = F*c. Use P_allow = F*c × A. WRONG.
Why Students Believe It
Students know from steel design that lateral bracing reduces the effective length, and with sufficient bracing, buckling is prevented and a stability factor of 1.0 applies. They incorrectly generalize this to timber: 'If I brace it, CP = 1.0.' Bracing does reduce the effective slenderness, but CP = 1.0 only when the column is so stocky (small ℓe/d) that buckling is not a concern — not simply because bracing exists at one point.
In the CP formula for timber columns, c = 0.8 is used for all types of timber (glulam, sawn, round poles).
Tags
- formula_confusion
- material_type
- memorization_error
Topic
Compression — Column Stability Factor CP (c values)
Severity
major
Exam Impact
In a glulam column problem, c = 0.8 vs c = 0.9 changes CP by a few percent, which changes P_allow by a few percent — often the difference between two close answer choices on a multiple-choice exam.
The Reality
NSCP 2015 specifies: c = 0.8 for visually graded sawn lumber, c = 0.90 for structural glued-laminated timber (glulam) and structural composite lumber, c = 0.85 for round timber poles. Using c = 0.8 for glulam underestimates CP (conservative error), but the examiner may specifically test if you know the correct value — and using the wrong c in a multi-part problem guarantees a wrong numerical answer.
Trap Question
Question
A glued-laminated (glulam) timber column has β = FcE/F*c = 1.5. Using the correct value of c, compute CP.
Explanation
Using c = 0.9 for glulam gives CP = 0.879 vs 0.811 for sawn lumber. The difference of 0.068 in CP translates directly to ~8% more allowable load — meaningful in design and potentially decisive in an exam problem.
Wrong Answer
c = 0.8 (sawn lumber default). (1+1.5)/(2×0.8) = 1.5625. CP = 1.5625 − √(1.5625² − 1.5/0.8) = 1.5625 − √(0.565) = 1.5625 − 0.751 = 0.811.
Correct Answer
c = 0.90 (glulam). (1+1.5)/(2×0.9) = 1.389. CP = 1.389 − √(1.389² − 1.5/0.9) = 1.389 − √(0.260) = 1.389 − 0.510 = 0.879.
Misconception Id
M5
Correct Vs Incorrect
Correct Approach
Glulam: c = 0.90. Sawn lumber: c = 0.80. Round poles: c = 0.85. Identify material first, then select c.
Incorrect Approach
Glulam column: use c = 0.8 in CP formula. WRONG.
Why Students Believe It
Students memorize 'c = 0.8' from lecture notes or review books that focus on the most common case (sawn lumber) without emphasizing that glulam and round poles use different values. Under exam pressure, the memorized single value is applied universally.
CF (size factor) and CL (beam stability factor) are always multiplied together to get the adjusted bending allowable F'b.
Tags
- formula_confusion
- conceptual_gap
- code_misread
Topic
Bending — Interaction of CF and CL
Severity
major
Exam Impact
Multiplying CF and CL together artificially lowers F'b below what the code intends, leading to over-design. More importantly, if the exam asks for a specific value of F'b, the wrong formula gives a wrong answer.
The Reality
NSCP 2015 specifies that for sawn lumber bending members, when BOTH CF and CL are applicable (i.e., the beam is both deep/narrow and laterally unbraced), you do NOT simply multiply both. Instead, you use the LESSER of: (1) Fb × CF (size-adjusted, with CL = 1.0) or (2) Fb × CL (stability-adjusted, with CF = 1.0). These two effects compete rather than combine because the size factor addresses section geometry and the stability factor addresses lateral-torsional buckling — applying both at full value would be double-counting the size penalty. Note: for glulam, the volume factor CV interacts with CL similarly — use the lesser.
Trap Question
Question
A sawn lumber beam has Fb = 15 MPa, CD = 1.0, CM = 1.0, Ct = 1.0, CF = 1.15 (size), and CL = 0.85 (stability). What is F'b?
Explanation
The beam stability factor CL = 0.85 governs. Using the 'multiply all' approach gives 14.66 MPa — an intermediate value that is neither physically correct nor code-compliant. The correct value of 12.75 MPa reflects the lateral-torsional buckling limit as the governing condition.
Wrong Answer
F'b = 15 × 1.0 × 1.0 × 1.0 × 1.15 × 0.85 = 14.66 MPa (all factors multiplied).
Correct Answer
Option 1: F'b = 15 × 1.15 = 17.25 MPa. Option 2: F'b = 15 × 0.85 = 12.75 MPa. Governing F'b = min(17.25, 12.75) = 12.75 MPa.
Misconception Id
M6
Correct Vs Incorrect
Correct Approach
Compute F'b option 1 = Fb × CD × CM × Ct × CF (with CL = 1.0). Compute F'b option 2 = Fb × CD × CM × Ct × CL (with CF = 1.0). Use the LESSER value.
Incorrect Approach
F'b = Fb × CD × CM × Ct × CF × CL. Student multiplies all six factors. If CF = 1.2 and CL = 0.8, F'b = Fb × ... × 1.2 × 0.8. WRONG.
Why Students Believe It
The general equation F' = F × ∏C suggests that all applicable factors are multiplied together. Students list CF and CL as both applicable to bending, multiply them both in, and proceed. This looks correct because the general formula supports it.
Wood design in the Philippines uses LRFD (Load and Resistance Factor Design) with factored loads, just like concrete (ACI 318) and steel (AISC 360).
Tags
- conceptual_gap
- design_philosophy
- common_error
- exam_trap
Topic
ASD vs LRFD — Design Philosophy
Severity
critical
Exam Impact
Applying 1.2D + 1.6L to a timber beam problem then checking against F'b guarantees a wrong answer. The actual demand used in ASD is simply D + L (service load).
The Reality
NSCP 2015 Chapter 6 designs wood members exclusively by ASD (Allowable Stress Design). Service-level (unfactored) loads are used to compute actual stresses, and these are compared against adjusted allowable stresses F'. There is no phi factor (φ) for wood in NSCP Chapter 6 ASD. Applying LRFD load factors to a wood ASD problem inflates the demand by 20–60%, causing correctly-sized members to appear overstressed.
Trap Question
Question
A wood floor joist carries dead load D = 2 kN/m and live load L = 4 kN/m on a 4 m simple span. What moment M is used to check bending stress in the ASD timber design?
Explanation
Timber ASD uses unfactored service loads. The LRFD approach would overstate the moment by 17.6/12 = 1.47 times, causing a serviceable joist to be rejected. NSCP 2015 Chapter 6 is ASD; there is no φ factor and no load factoring for wood member design.
Wrong Answer
Mu = (1.2×2 + 1.6×4)(4²)/8 = (2.4 + 6.4)(2) = 17.6 kN·m (student applies LRFD load factors).
Correct Answer
M = (D + L)(L²/8) = (2 + 4)(16/8) = 6 × 2 = 12 kN·m (service-level ASD).
Misconception Id
M7
Correct Vs Incorrect
Correct Approach
W = D + L = 3 + 5 = 8 kN/m (service load). M = W × L²/8. fb = M/S. Compare to F'b. This is ASD — use unfactored loads.
Incorrect Approach
Wu = 1.2(3) + 1.6(5) = 11.6 kN/m. Mu = Wu × L²/8 = ... fb = Mu/S. Compare to F'b. WRONG for timber ASD.
Why Students Believe It
Most of the CELE structural subjects use LRFD: steel uses LRFD per NSCP/AISC 360, and concrete uses factored loads per ACI 318. Students who have just studied these subjects assume wood follows the same framework and apply load factors (1.2D + 1.6L) before checking against reference values.
The CD value for a combined loading condition is the average of the CD values of the individual load types.
Tags
- common_error
- conceptual_gap
- code_misread
Topic
Load Duration Factor CD — Combined Loading
Severity
major
Exam Impact
Under-averaging CD underestimates the allowable stress. Over-applying the wrong dominant CD (e.g., using 1.6 for a combination that has no wind) overestimates the allowable. Both cause wrong answers.
The Reality
NSCP 2015 states that when multiple loads of different durations act simultaneously, the CD factor for the SHORTEST-duration load in the combination governs the entire combination. The logic is conservative: the design must be safe for the worst-case loading event, which is when the short-duration load is at its peak. For D + L: CD = 1.0 (10-year live load governs over the permanent dead load). For D + L + W: CD = 1.6 (wind governs). You never average or pro-rate CD.
Trap Question
Question
A wood rafter carries dead load (CD = 0.9), roof live load (CD = 1.25), and wind (CD = 1.6) simultaneously. What single CD value is used to compute F'b for this combined loading check?
Explanation
NSCP 2015 Chapter 6 requires that the CD corresponding to the load of shortest duration in the combination be used. Wind (CD = 1.6) has the shortest duration among the three loads, so CD = 1.6 governs for the combined D + Lr + W check. A separate check with D only uses CD = 0.9.
Wrong Answer
CD = (0.9 + 1.25 + 1.6)/3 = 1.25 (student averages the three values).
Correct Answer
CD = 1.6 (wind load governs — use the CD of the shortest-duration load in the combination).
Misconception Id
M8
Correct Vs Incorrect
Correct Approach
Load combination D + W: shortest duration is wind, so CD = 1.6 governs. F'b = Fb × 1.6 × ... Also check D only with CD = 0.9 if that controls another limit.
Incorrect Approach
Load combination D + W: CD_avg = (0.9 + 1.6)/2 = 1.25. F'b = Fb × 1.25 × ... WRONG.
Why Students Believe It
When dead load (CD = 0.9) and live load (CD = 1.0) act together, students reason that the combined effect should be somewhere between the two, so they average: CD = (0.9 + 1.0)/2 = 0.95. This seems logical — a weighted combination.
The section modulus S = bh²/6 for a wood beam uses the full nominal (commercial) dimensions b and h.
Tags
- common_error
- practical_application
- conceptual_gap
Topic
Section Properties — Nominal vs Actual Dimensions
Severity
major
Exam Impact
Using nominal instead of actual dimensions overestimates S and A, causing an overstressed section to appear adequate — unconservative and wrong. This is a subtle trap in problems that mention nominal sizes.
The Reality
For design purposes, the actual (dressed) dimensions must be used, not the nominal. A nominal 50 × 150 mm piece of sawn lumber may have actual dimensions of 38 mm × 140 mm after surfacing. However, in Philippine board exams (CELE), the problem will typically state 'actual dimensions' or give the size explicitly. If nominal dimensions are given, you must apply the appropriate reduction — or the problem will specifically state to use nominal. Always note whether dimensions given are 'actual' or 'nominal.' When in doubt, use what is given explicitly.
Trap Question
Question
A problem states a '50 mm × 200 mm nominal sawn lumber beam.' The actual (dressed) dimensions are 38 mm × 184 mm. Which dimensions should be used to compute the section modulus S for bending design?
Explanation
Design stresses act on the actual cross-section, not the nominal size. Using nominal dimensions overstates S by about 56% in this case (333,333 vs 214,251 mm³), which would allow a much smaller actual beam than is safe. CELE problems will usually specify 'actual dimensions' to avoid ambiguity, but awareness of this distinction is essential.
Wrong Answer
S = (50)(200²)/6 = 333,333 mm³ using nominal dimensions.
Correct Answer
S = (38)(184²)/6 = 214,251 mm³ using actual dressed dimensions.
Misconception Id
M9
Correct Vs Incorrect
Correct Approach
Always confirm: 'actual' or 'nominal'? If the problem states actual 100 × 200 mm, use S = 666,667 mm³. If nominal, use the actual dressed dimensions specified by the applicable lumber standard.
Incorrect Approach
Nominal 100 × 200 mm beam: S = (100)(200²)/6 = 666,667 mm³. Use this for fb check. (May be wrong if actual dimensions are different.)
Why Students Believe It
Timber is often specified by nominal dimensions (e.g., '2×6 lumber'). Students use 50 mm × 150 mm as b and h without realizing that Philippine commercial lumber, like American lumber, has dressed (actual) dimensions smaller than the nominal. Exam problems sometimes specify nominal sizes expecting students to recognize this.
Wood is strong in shear across the grain (perpendicular to grain), so horizontal (parallel-to-grain) shear rarely governs.
Tags
- conceptual_gap
- common_error
- exam_trap
Topic
Horizontal Shear — Governs for Short Beams
Severity
major
Exam Impact
Students who believe horizontal shear 'rarely governs' skip the shear check entirely or treat it as a formality. For short-span beams with large loads, horizontal shear often governs over bending — missing this costs points on both the shear check and the member selection.
The Reality
Wood is VERY weak in shear parallel to the grain (also called horizontal shear or longitudinal shear). Typical Fv values parallel to grain for Philippine structural lumber range from 0.7 to 1.1 MPa — far below the compressive and bending allowables. Shear perpendicular to grain (cross-grain shear) is rarely the design limit because wood almost never fails by being cut cleanly across the grain; it splits along the grain first. The formula fv = 3V/(2A) computes horizontal (parallel-to-grain) shear stress, and this is what governs short, heavily-loaded wood beams.
Trap Question
Question
A 100 × 300 mm wood beam spans 1.5 m simply supported with a central point load P = 30 kN. F'b = 12 MPa, F'v = 1.0 MPa. Which stress governs the design?
Explanation
For a 1.5 m span with a 30 kN point load, the span-to-depth ratio is only 5 — a very short, deep beam. Short beams are shear-critical. Always check both bending and shear for all wood beams; never assume one is automatically non-governing.
Wrong Answer
Check bending: M = PL/4 = 30(1.5)/4 = 11.25 kN·m; fb = 11,250,000/(100×300²/6) = 7.5 MPa < 12 MPa. OK. Bending governs, shear is fine (student skips shear).
Correct Answer
Bending: fb = 7.5 MPa < 12 MPa ✓. Shear: V = P/2 = 15 kN; fv = 3(15,000)/(2×100×300) = 0.75 MPa < 1.0 MPa ✓. Both pass, but both must be checked. If P were 40 kN: fv = 3(20,000)/(2×30,000) = 1.0 MPa = F'v exactly (shear governs).
Misconception Id
M10
Correct Vs Incorrect
Correct Approach
Always perform BOTH checks: (1) fb = M/S ≤ F'b for bending, and (2) fv = 3V/(2A) ≤ F'v for horizontal shear. For short spans or concentrated loads near supports, shear often governs.
Incorrect Approach
Student checks bending only (fb ≤ F'b) and skips shear check because 'wood doesn't usually fail in shear.' WRONG — horizontal shear controls short deep beams.
Why Students Believe It
Students intuitively think of shear as the force that tends to cut a beam in two vertically — which IS the strong shear direction for wood (cross-grain shear). They forget that the critical shear failure mode for wood beams is the horizontal splitting along the grain (horizontal shear parallel to grain), which is the weakest direction.
The repetitive member factor Cr = 1.15 applies to all wood members in a floor or roof assembly.
Tags
- formula_confusion
- code_misread
- minor_error
Topic
Bending — Repetitive Member Factor Cr
Severity
minor
Exam Impact
Applying Cr = 1.15 where not applicable inflates F'b by 15%, causing an overstressed section to appear adequate — a 1-point error in a problem but signals a fundamental gap in code knowledge.
The Reality
Cr = 1.15 applies ONLY to bending members that are: (1) sawn lumber 50–100 mm thick (dimension lumber), (2) spaced not more than 600 mm apart, (3) at least 3 members in the group, and (4) connected by a load-distributing element (sheathing, decking). It does NOT apply to: timbers (>125 mm thick), glulam, columns, axially loaded members, or single isolated beams. Applying Cr to a glulam girder or a timber post is a code violation.
Trap Question
Question
A 150 × 300 mm sawn timber beam is installed alone (single beam, no adjacent parallel members) supporting a concrete slab. Can Cr = 1.15 be applied?
Explanation
The repetitive member factor rewards load-sharing between multiple parallel members. A single isolated beam, regardless of size or use, has no neighboring member to share load with, so the repetitive member benefit does not apply.
Wrong Answer
Yes, Cr = 1.15 because it is a bending member in a floor system.
Correct Answer
No. Cr = 1.15 requires at least 3 closely spaced parallel members with load-distributing sheathing. A single isolated beam does not qualify; Cr = 1.0.
Misconception Id
M11
Correct Vs Incorrect
Correct Approach
Cr = 1.15 applies only to sawn dimension lumber (50–100 mm thick) spaced ≤ 600 mm, in groups of ≥ 3, with load-sharing sheathing. For a glulam girder, Cr = 1.0 (or CV factor applies instead).
Incorrect Approach
A 200 × 400 mm glulam girder: F'b = Fb × CD × CM × Ct × CL × Cr = ... × 1.15. WRONG — Cr does not apply to glulam.
Why Students Believe It
Students learn 'Cr = 1.15 for repetitive members' as a simple rule. Since most floors use closely spaced joists, they apply Cr = 1.15 to every wood member in the structure — including beams, columns, and posts.
In timber column design, F*c (F-star-c) is the same as F'c (F-prime-c — the final allowable compressive stress).
Tags
- formula_confusion
- notation_error
- conceptual_gap
- exam_trap
Topic
Compression — F*c vs F'c Notation in Column Stability
Severity
critical
Exam Impact
Using F'c where F*c is needed in the β ratio produces an iterative (circular) calculation that has no straightforward closed-form solution — students get stuck or use an incorrect value, losing all marks on the column stability computation.
The Reality
F*c is an intermediate value: it is Fc multiplied by ALL adjustment factors EXCEPT CP. Specifically: F*c = Fc × CD × CM × Ct × CF. F'c is the final adjusted allowable: F'c = F*c × CP. The CP formula requires F*c as an input (β = FcE/F*c), so using F'c in place of F*c makes the calculation circular and incorrect. The sequence is: (1) compute F*c from Fc with all factors except CP; (2) compute β = FcE/F*c; (3) compute CP; (4) get F'c = F*c × CP.
Trap Question
Question
A sawn timber column has Fc = 12 MPa, CD = 1.0, CM = 1.0, Ct = 1.0, CF = 1.0, and FcE = 9.6 MPa. Identify the correct intermediate value used in computing CP.
Explanation
F*c is the reference compression value adjusted for everything except column stability. It is a known, computable quantity at the start of the CP calculation. F'c is the end product. Confusing them makes the problem unsolvable and shows a fundamental misunderstanding of the NSCP timber column stability procedure.
Wrong Answer
Use β = FcE/F'c = 9.6/F'c. (Student cannot compute this without knowing F'c first — circular.)
Correct Answer
F*c = Fc × CD × CM × Ct × CF = 12 × 1.0 × 1.0 × 1.0 × 1.0 = 12 MPa. β = FcE/F*c = 9.6/12 = 0.80. Then compute CP, then F'c = F*c × CP.
Misconception Id
M12
Correct Vs Incorrect
Correct Approach
Step 1: F*c = Fc × CD × CM × Ct × CF (exclude CP). Step 2: β = FcE/F*c. Step 3: CP = [(1+β)/(2c)] − √{[(1+β)/(2c)]² − β/c}. Step 4: F'c = F*c × CP.
Incorrect Approach
β = FcE/F'c (student uses F'c — the final answer — in the formula that is used to derive F'c). Circular and WRONG.
Why Students Believe It
The notation F*c and F'c look very similar. Students see both in the CP formula derivation and conflate them, either using F*c as the final design value (skipping CP) or substituting F'c where F*c is needed in the CP computation — creating a circular reference.
Quick Self Check
Fb is the reference value. The allowable stress is F'b = Fb × (CD × CM × Ct × CF × CL × Cr ...). You must apply all applicable adjustment factors before comparing to the actual stress.
Statement
The reference design value Fb from a wood species table is the allowable bending stress that can be directly compared to the actual bending stress fb.
CD does NOT apply to E or to Fc-perp. NSCP 2015 Chapter 6 explicitly excludes these from the CD adjustment. Only strength properties (Fb, Fv, Fc, Ft) are modified by CD.
Statement
The load duration factor CD applies to the modulus of elasticity E used in computing FcE for column stability checks.
The formula fv = 3V/(2A) is the standard NSCP Chapter 6 expression for horizontal (parallel-to-grain) shear in rectangular wood sections. It equals VQ/(Ib) evaluated at the neutral axis of a rectangle, where Q/Ib = 3/(2A).
Statement
For a rectangular wood cross-section, the maximum horizontal shear stress is computed as fv = 3V/(2A), where A is the full cross-sectional area b × d.
Timber in NSCP 2015 Chapter 6 is designed by ASD using unfactored (service-level) loads. The actual moment is computed from D + L, not 1.2D + 1.6L. Load factors are for LRFD (steel, concrete) — not wood ASD.
Statement
Timber design per NSCP 2015 Chapter 6 uses LRFD load combinations (1.2D + 1.6L) to compute the design moment for bending checks.
For sawn lumber, when both CF and CL apply, use the LESSER of Fb×CD×CM×Ct×CF (with CL=1.0) and Fb×CD×CM×Ct×CL (with CF=1.0). They are not multiplied together — they represent competing geometric and stability limits.
Statement
When a wood beam has both CF (size factor) and CL (beam stability factor) applicable, the adjusted allowable bending stress is F'b = Fb × CD × CM × Ct × CF × CL.
NSCP 2015 specifies c = 0.90 for structural glued-laminated timber (glulam) and structural composite lumber. c = 0.80 is for visually graded sawn lumber. c = 0.85 is for round timber poles.
Statement
For a glulam timber column, the c value in the CP formula is 0.90.
CP must always be computed from the formula using the effective slenderness ratio ℓe/d. Bracing reduces ℓe (and hence ℓe/d), which increases CP, but CP = 1.0 only in the limiting case of a very stocky column where FcE/F*c >> 1. Practical timber columns with ℓe/d > 10 have CP < 1.0 regardless of bracing.
Statement
The column stability factor CP equals 1.0 whenever a timber column is adequately braced against lateral movement.
F*c = Fc × CD × CM × Ct × CF — all factors EXCEPT CP. It is the intermediate adjusted compressive value used to compute β = FcE/F*c, from which CP is then derived. F'c = F*c × CP is the final allowable, which includes CP.
Statement
F*c (F-star-c) in timber column design already includes the column stability factor CP.
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