Timber (Wood) Design · Study · PE Civil: Structural · FE → PE Prep
Timber (Wood) Design
4% of exam
NDS allowable-stress design of sawn-lumber and glulam beams and columns, adjustment factors, bending and horizontal shear, bearing, stability, and combined bending and axial loads.
3 concepts
D. Timber
NDS Timber Design: Adjustment Factors
How the NDS reference design values become adjusted (allowable) values F' through the chain of adjustment factors — CD, CM, Ct, CF, CL, CP, Cr and the rest.
Timber design on the PE Civil exam is, at heart, one disciplined piece of bookkeeping: you start with a tabulated reference design value — Fb, Fv
,
Fc
,
Fc⊥
,
Ft
,
E
,
Emin
— and you walk it through a chain of multiplicative adjustment factors to reach the adjusted (allowable) value
F′
that you actually compare against the applied stress. Almost every point lost in this area is lost not in the trigonometry or the section properties but in the factor chain: an examinee forgets the load-duration factor
CD
, applies the wet-service factor
CM
to a dry member, or multiplies in a size factor that does not apply to the species and size at hand. Because NCEES has you work wood in Allowable Stress Design (ASD), this whole concept lives in the National Design Specification for Wood Construction (NDS 2018) and its Supplement of design values — there is no timber design content in the NCEES PE Civil Reference Handbook, so cite the NDS, not a handbook page.
Reference design values and the master equation
The NDS Supplement tabulates reference design values for each species, grade, and product (visually graded sawn lumber, MSR lumber, glued-laminated timber, etc.). These are baseline stresses valid for one reference condition: normal load duration, dry service, normal temperature, and a benchmark size. Every adjusted value is that reference value multiplied by the applicable factors. The general form is identical for each property; you simply pick the factors that apply to that property. The adjusted modulus of elasticity for stability, Emin′, is what later feeds the beam- and column-stability calculations.
Fb′=FbCDCMCtCLCFCfuCiCr
Load duration C_D — wood gets stronger for short loads
Wood carries a higher stress for a load applied briefly than for one applied for years; the load-duration factor CD captures this. The reference values are normalized to ten years (the 'normal' duration that floor live load is taken to represent), so CD=1.0 there. Common values are CD=0.9 for permanent (dead-only) loading, 1.0 for occupancy live load, 1.15 for snow, 1.25 for construction or a seven-day load, 1.6 for wind or seismic, and 2.0 for impact. The governing rule: in any one load combination, use the CD of the shortest-duration load in that combination, and check every combination separately. CD applies to all strength properties but never to E, Emin, or Fc⊥.
Three factors correct for the environment. The wet-service factor CM reduces values when the in-service moisture content exceeds the reference threshold (above 19% for sawn lumber, 16% for glulam); for fully dry interior members CM=1.0 and is simply omitted. The temperature factor Ct reduces values for sustained service above 100∘F; ordinary buildings sit below that, so Ct=1.0. The incising factor Ci applies only when sawn lumber has been incised to improve preservative penetration. The exam usually states the service conditions explicitly — read for 'wet', 'exposed', 'sustained high temperature', or 'incised'; absent those words, these factors are unity, and a frequent error is dragging a CM<1 into an obviously dry, heated building.
Geometry factors C_F, C_fu, C_r
The size factor CF corrects Fb, Ft, and Fc of visually graded sawn lumber for members deeper or wider than the reference size, because larger members have more flaws and a lower characteristic strength — deep bending members get CF<1. For glulam, the analogous bending correction is the volume factor CV, which competes with the beam-stability factor CL (you use the smaller of the two, never both). The flat-use factor Cfu raises Fb when a member is loaded on its wide face. The repetitive-member factor Cr=1.15 rewards Fb when three or more parallel members (joists, studs, rafters) spaced no more than 24 in. are tied by a load-sharing deck. These geometry factors apply only to the properties named for each — applying CF to shear or to E is a classic slip.
Glulam bending: use min(CL,CV),never both
Stability factors C_L and C_P
Two factors are not table look-ups but computed quantities tied to slenderness. The beam-stability factor CL reduces Fb for a laterally unsupported compression edge (lateral-torsional buckling); it is 1.0 when the compression edge is continuously braced or the depth-to-breadth ratio is small. The column-stability factor CP reduces Fc for a slender column (flexural buckling); it falls toward zero as ℓe/d grows. Both are evaluated from the same Euler-type term FcE or FbE built on Emin′, and both have their own dedicated concepts. The key discipline here is sequencing: you compute Fb∗ (every factor except CL) and Fc∗ (every factor except CP) first, because the stability factor itself depends on that starred value through the ratio FcE/Fc∗.
Fc∗=FcCDCMCtCFCi,Fc′=Fc∗CP
Reading the NDS adjustment-factor applicability table
The NDS gives a single applicability table (NDS 2018 Table 4.3.1 for sawn lumber, 5.3.1 for glulam) with a checkmark grid: rows are design properties, columns are factors. The professional move is to reproduce the relevant row mentally for the property you need. E and Emin take essentially none of the stress factors — no CD, no CF — because elastic behavior is duration-independent. Fc⊥ (bearing) takes CM, Ct, Ci, and the bearing-area factor Cb, but not CD and not CF. Compression parallel Fc takes the column factor CP; bending Fb takes the beam factor CL; the two are never swapped. Build the chain property-by-property, not as one universal list.
Exam strategy
Make the factor chain a checklist you write down before computing anything: list CD, CM, Ct, CF (or CV), CL/CP, Cr, Cfu, Ci, Cb and strike out the ones the problem makes unity. Decide CD per load combination and use the shortest-duration load's value — and never apply CD to E, Emin, or Fc⊥. Compute the starred values Fb∗ and Fc∗ first so the stability factors CL and CP have something to reference. Default everything environmental to 1.0 unless the words 'wet', 'exposed', 'high temperature', or 'incised' appear. Finally, cite NDS 2018 (ASD) — the editions NCEES supplies are fixed, so write 'NDS 2018', not a later year.
Permanent, 10-yr (normal), snow, 7-day/construction, wind/seismic, impact. Use the shortest-duration load in each combination.
Repetitive-member factorCr=1.15
Applies to Fb only when 3+ parallel members spaced ≤24
ASD demand-to-capacityf≤F′⟺F′f≤1.0
Worked examples
Adjusted bending value of a roof rafter
Problem. A No. 2 Douglas Fir-Larch 2x10 rafter (Fb=900psi) carries snow load and is part of a system of rafters spaced 16in. on center sheathed with plywood. The attic is dry and unheated. The size factor for a 2x10 in bending is CF=1.1. Find the adjusted bending value Fb′.
Solution. Build the chain for Fb:
- Load duration: snow governs, so CD=1.15.
- Wet service: dry interior, CM=1.0
Fb′=900(1.15)(1.1)(1.15)=1,309psi
Choosing C_D across two load combinations
Problem. A No. 1 DF-L 6x6 post (Fc=1,000psi, dry, heated interior, not slender so take CP=0.70
Common pitfalls
•Citing the NCEES PE Civil Reference Handbook for timber. The handbook has NO wood-design content; the governing source is NDS 2018 (ASD). Attributing the factor chain to a handbook section is a sourcing error.
•Applying CD to the wrong properties. Load duration adjusts strength values (Fb, Fv, Fc, Ft) but NEVER E, Emin, or bearing Fc⊥.
•Using a single CD for all combinations. CD is the shortest-duration load IN THAT combination; check dead-only (0.9), live (1.0
•Multiplying CL and CV together for a glulam beam. Use only the SMALLER of the two — using both double-counts the strength reduction.
•Forgetting that the stability factors are computed last. CP depends on Fc∗ (all factors except CP
•Dragging CM<1.0 into a dry, heated building. The wet-service factor applies only above the moisture threshold (19% sawn, 16% glulam); absent 'wet/exposed', it is 1.0.
•Citing a later edition (NDS 2015 or a future year). NCEES supplies NDS 2018 for the PE Civil structural exam — cite that edition, in ASD, exactly.
References
AWC National Design Specification (NDS) for Wood Construction, 2018 edition (ASD method) — Ch. 2 general adjustment factors + Ch. 4/5 applicability tables
NDS Supplement, 2018 — Design Values for Wood Construction (reference $F_b$, $F_v$, $F_c$, $F_{c\perp}$, $E$, $E_{min}$ and size factors $C_F$)
The four ASD checks for a wood beam — bending f_b = M/S ≤ F'_b, horizontal shear f_v = 1.5V/A ≤ F'_v, bearing perpendicular to grain, and deflection.
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Timber Column & Combined Bending + Axial
The column stability factor C_P from slenderness l_e/d and F_cE, the adjusted F'_c, and the NDS interaction equation for combined axial compression plus bending.
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the size factor (sawn) or
CV
the volume factor (glulam — use only the smaller of
CL
,
CV
).
Fc∗ is every factor except CP; the column-stability factor CP is computed last from FcE/Fc∗.
Horizontal shear; takes duration and service factors but NOT CF, CL, or Cr.
Bearing perpendicular to grain; Cb is the bearing-area factor. Note: NO load-duration factor CD and NO size factor.
Feeds FcE and FbE. Elasticity is duration-independent, so no CD and no CF ever apply to E or Emin.
All factors except CL; the ratio FbE/Fb∗ then yields CL. Same idea for Fc∗ and CP.
in. share load through a deck (joists, studs, rafters).
The ASD check: actual stress f (e.g., fb=M/S) must not exceed the adjusted allowable F′. Never mix in LRFD ϕ-factors.
.
- Temperature: unheated but below
100∘F
,
Ct=1.0
.
- Size:
CF=1.1
(given for 2x10).
- Repetitive: rafters spaced
≤24
in.,
≥3
members, sheathed, so
Cr=1.15
.
- Beam stability: rafters braced by sheathing on the compression edge,
CL=1.0
.
Fb′=900(1.15)(1.0)(1.0)(1.0)(1.1)(1.15)=1,309psi
.
**Answer:
Fb′≈1,310psi
.** Sanity check: the reference
900psi
is boosted by snow duration and load sharing to a little over
1,300psi
— a
45%
increase, reasonable given
1.15×1.1×1.15=1.45
and every other factor unity.
) must be checked for two combinations: (a) dead + floor live load, and (b) dead + wind. Which combination governs the allowable compression value, and what is
Fc′
for each?
Solution. Only CD differs between the combinations; CM=Ct=CF=Ci=1.0 for a dry, heated, square post, and CP=0.70 is given.
(a) Dead + floor live: shortest-duration load is the 10-yr live load, CD=1.0.
Fc′=1,000(1.0)(0.70)=700psi.
(b) Dead + wind: shortest-duration load is wind, CD=1.6.
Fc′=1,000(1.6)(0.70)=1,120psi.
**Answer: each combination is checked separately; the allowable is 700psi for the live-load case and 1,120psi for the wind case.** Sanity check: the wind case allows 60% more stress (exactly the CD ratio 1.6/1.0), but it also carries a different factored demand, so 'governs' is decided by demand-to-capacity, not by Fc′ alone — never apply CD=1.6 to the gravity-only check.