15.3 Timber Design Principles, Working Stress, and Sawn Lumber Beams

15.3 Timber Design Principles, Working Stress, and Sawn Lumber Beams

Timber is one of the oldest and most versatile construction materials in the Philippines. As an organic cellular polymer formed by the biological growth of trees, wood possesses unique structural characteristics that distinguish it fundamentally from isotropic materials like structural steel and concrete. Wood is orthotropic (anisotropic)—its mechanical strength, stiffness, shrinkage, and thermal expansion differ across three mutually perpendicular axes: longitudinal (parallel to grain), radial (perpendicular to growth rings), and tangential (tangent to growth rings).

In Philippine civil engineering practice and licensure examinations, structural timber is designed strictly under the Working Stress Design (WSD) / Allowable Stress Design (ASD) methodology codified in NSCP 2015 Chapter 6 (derived from the National Design Specification for Wood Construction - NDS).

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1. Wood Anisotropy and Grain Orientation

Because wood fibers (tracheids in softwoods, vessels and fibers in hardwoods) are oriented along the trunk axis, wood exhibits maximum strength and stiffness parallel to the grain:

• Compression Parallel to Grain ($F_c$): Wood cells act as hollow micro-tubular columns that resist compressive forces efficiently until cell wall buckling occurs ($F_c \approx 20 - 45\text{ MPa}$ for Philippine hardwoods).
• Compression Perpendicular to Grain ($F_{c\perp}$): Compressive loads crush the hollow tubular cells transversely. Elastic deformation is small; failure occurs by progressive cell wall flattening at allowable stresses only $20\% - 30\%$ of parallel compression ($F_{c\perp} \approx 4 - 10\text{ MPa}$).
• Tension Parallel to Grain ($F_t$): Extremely high ultimate capacity, but governed in service by slope of grain, knots, and splits.
• Tension Perpendicular to Grain ($F_{t\perp}$): Tensile loads pull wood fibers apart laterally. Capacity is extremely low and unpredictable ($< 1.0\text{ MPa}$). Designs that induce tension perpendicular to the grain should be avoided wherever possible.
• Shear Parallel to Grain ($F_v$): Also known as horizontal shear, this limit state governs flexural beams. Shear stresses attempt to slide horizontal wood fibers past one another ($F_v \approx 1.5 - 3.5\text{ MPa}$).

Moisture Content and Fiber Saturation Point

Wood is hygroscopic, exchanging water vapor with the atmosphere. Water exists in two states: free water in cell cavities and bound water within cell walls. When all free water has evaporated while cell walls remain fully saturated, the wood is at the Fiber Saturation Point (FSP) (typically $25\% - 30\%$ moisture content). Above FSP, changes in moisture content do not affect mechanical strength. Below FSP, drying strengthens the wood cells and causes shrinkage. Lumber is classified as dry / seasoned when moisture content is $\le 19\%$, and green / unseasoned when $> 19\%$.

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2. Philippine Timber Classifications (NSCP / FPRDI)

NSCP Chapter 6 tabulates allowable stresses and moduli species by species for Philippine woods, based on research by the Forest Products Research and Development Institute (FPRDI). Values differ by species, moisture condition and stress grade, so board problems almost always state the values to use. For orientation, common structural species fall roughly into three groups:

The worked examples below use stated allowable stresses as given data. Use the NSCP table values for the species and grade specified in an actual design.

Note: Yakal and Guijo are widely specified in coastal bridges and wharf structures for high decay resistance, while Apitong and Lauan are standard for roof trusses, purlins, and floor framing.

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3. Working Stress Design Adjustment Factors

Reference design values ($F$) obtained from testing standardized defect-free specimens must be modified by environmental, geometric, and loading condition factors to determine the adjusted allowable design value ($F'$):

$F' = F \cdot \prod C_i = F \cdot C_D \cdot C_M \cdot C_t \cdot C_L \cdot C_F \cdot C_{fu} \cdot C_r$

1. Load Duration Factor ($C_D$)

Wood exhibits viscoelastic rheological behavior: it can sustain higher stresses for short time durations than for long durations. Reference design stresses are calibrated to a normal cumulative duration of 10 years ($C_D = 1.00$). The load duration factor applies to all strength properties ($F_b, F_t, F_c, F_v$), but never to modulus of elasticity ($E$) or compression perpendicular to grain ($F_{c\perp}$):

2. Wet Service Factor ($C_M$)

Applies when structural lumber is exposed to outdoor weathering or high humidity where moisture content exceeds $19\%$. For sawn lumber, $C_M = 0.85$ for flexure ($F_b$), $C_M = 0.80$ for compression parallel ($F_c$), and $C_M = 0.90$ for modulus of elasticity ($E$).

3. Size Factor ($C_F$)

Accounts for the volume effect: larger timber beams have a higher statistical probability of containing critical grain flaws. For visually graded sawn lumber dimension sizes ($d > 300\text{ mm}$ or $12\text{ in}$):

$C_F = \left( \frac{300}{d} \right)^{1/9}$

4. Repetitive Member Factor ($C_r = 1.15$)

When three or more parallel sawn lumber members (such as floor joists, roof rafters, or purlins) are spaced not more than $600\text{ mm}$ ($24\text{ in}$) on center and joined by sheathing that transfers load transverse to the members, load sharing prevents individual member overload. An allowable bending stress increase of $15\%$ ($C_r = 1.15$) is permitted.

5. Flat Use Factor ($C_{fu}$)

When rectangular sawn lumber is loaded on its wide face ($d < b$, flatwise bending), fiber stress is redistributed more favorably, yielding $C_{fu} > 1.0$ ($1.10 - 1.22$).

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4. Sawn Lumber Beam Design (Flexure and Horizontal Shear)

Flexural Strength Verification

For a rectangular sawn lumber beam of actual width $b$ and depth $d$, the section modulus is $S = \frac{b d^2}{6}$. The actual bending stress ($f_b$) must not exceed the adjusted allowable bending stress ($F'_b$):

$f_b = \frac{M}{S} = \frac{6 M}{b d^2} \le F'_b = F_b \cdot C_D \cdot C_M \cdot C_t \cdot C_L \cdot C_F \cdot C_r$

Horizontal Shear Stress in Rectangular Beams

From the mechanics of materials, the horizontal shear stress at distance $y$ from the neutral axis is $f_v = \frac{V Q}{I b}$. For a homogeneous solid rectangular cross-section, maximum shear occurs at the neutral axis ($Q_{\max} = \frac{b d^2}{8}$, $I = \frac{b d^3}{12}$):

$f_{v,\max} = \frac{V \left(\frac{b d^2}{8}\right)}{\left(\frac{b d^3}{12}\right) b} = \frac{3 V}{2 b d} = 1.5 \frac{V}{A} \le F'_v$

Critical Shear Reduction ($d$-Distance Rule)

Because loads near a support are carried directly into the bearing by compression, NSCP Chapter 6 (following the NDS) permits neglecting uniformly distributed loads within a distance equal to the beam depth ($d$) from the face of the support when calculating the design shear force $V$. For a uniformly distributed load $w$ on a simple span $L$:

$V_{\text{design}} = R - w \cdot d = \frac{w L}{2} - w d$

Bearing Stress Perpendicular to Grain

At beam supports, the reaction force $R$ induces compressive bearing perpendicular to wood fibers over bearing area $A_b = b \cdot l_b$:

$f_{c\perp} = \frac{R}{b \cdot l_b} \le F'_{c\perp} = F_{c\perp} \cdot C_M \cdot C_t \cdot C_b$

Where $C_b = \frac{l_b + 9.5}{l_b}$ (for metric units, with bearing length $l_b \ge 38\text{ mm}$ not closer than $75\text{ mm}$ to member ends).

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5. Solid Sawn Lumber Column Design

Solid wood columns are designed as axially loaded rectangular compression members with dimensions $b \times d$ (where $d$ is the smaller cross-sectional dimension). The effective length is $l_e = K L$.

Slenderness Limitations

For solid sawn lumber columns, NSCP Chapter 6 (following the NDS) limits the slenderness ratio to:

$\frac{l_e}{d} \le 50$

Column Stability Factor ($C_P$ via Ylinen Formula)

The allowable compressive stress parallel to grain ($F'_c$) is determined by applying the column stability factor ($C_P$):

$F'_c = F_c^* \cdot C_P$

Where $F_c^* = F_c \cdot C_D \cdot C_M \cdot C_t$ is the reference compressive stress multiplied by all applicable factors except $C_P$.

The elastic Euler buckling stress for wood is:

$F_{cE} = \frac{0.822 E_{\min}'}{(l_e / d)^2}$

Where $E_{\min}' = E_{\min} \cdot C_M \cdot C_t$ is the 5th percentile modulus of elasticity. The column stability factor $C_P$ is derived from the Ylinen equation:

$C_P = \frac{1 + (F_{cE} / F_c^*)}{2 c} - \sqrt{\left[ \frac{1 + (F_{cE} / F_c^*)}{2 c} \right]^2 - \frac{F_{cE} / F_c^*}{c}}$

• $c = 0.80$ for solid sawn lumber columns.
• $c = 0.90$ for glued laminated timber (glulam).
• $c = 0.85$ for round timber poles and piles.

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6. Comprehensive Worked Examples

Worked Example 1: Sawn Timber Floor Joist Design

Problem: A residential floor framing system uses Apitong sawn timber joists with given allowable values $F_b = 16.5\text{ MPa}$, $F_v = 1.75\text{ MPa}$, $E = 8,500\text{ MPa}$ on a simple span of $L = 4.0\text{ m}$. The joists are spaced $400\text{ mm}$ on center and covered with tongue-and-groove plywood sheathing. Total design service dead plus live load is $w = 3.6\text{ kN/m}$ per joist. Joist cross-section is $50\text{ mm} \times 200\text{ mm}$ rough sawn. Service condition is dry ($C_M = 1.0$), normal occupancy duration ($C_D = 1.0$). Check: (a) bending stress, (b) horizontal shear stress at distance $d$ from support, and (c) support bearing length for an allowable $F_{c\perp} = 3.5\text{ MPa}$.

Solution:

• Step 1: Section Properties: $b = 50\text{ mm}, \quad d = 200\text{ mm}$ $S = \frac{b d^2}{6} = \frac{50 \times (200)^2}{6} = 333,333\text{ mm}^3 = 3.333 \times 10^{-4}\text{ m}^3$ $A = b d = 50 \times 200 = 10,000\text{ mm}^2$

• Step 2: Flexural Stress Check: Maximum mid-span moment: $M = \frac{w L^2}{8} = \frac{3.6\text{ kN/m} \times (4.0\text{ m})^2}{8} = 7.20\text{ kN}\cdot\text{m} = 7.20 \times 10^6\text{ N}\cdot\text{mm}$ Actual bending stress: $f_b = \frac{M}{S} = \frac{7.20 \times 10^6}{333,333} = 21.60\text{ MPa}$ Adjusted allowable bending stress (with repetitive member factor $C_r = 1.15$ since joists are spaced $\le 600\text{ mm}$): $F'_b = F_b \cdot C_D \cdot C_r = 16.5\text{ MPa} \times 1.0 \times 1.15 = 18.98\text{ MPa}$ Check: $f_b = 21.60\text{ MPa} > F'_b = 18.98\text{ MPa}$. The $50\text{ mm} \times 200\text{ mm}$ joist is overstressed by $13.8\%$ in bending; a deeper joist (e.g., $50\text{ mm} \times 250\text{ mm}$) is required.

• Step 3: Horizontal Shear Check (at distance $d = 200\text{ mm}$): End reaction $R = \frac{w L}{2} = \frac{3.6 \times 4.0}{2} = 7.20\text{ kN}$. $V_{\text{design}} = R - w \cdot d = 7.20\text{ kN} - (3.6\text{ kN/m} \times 0.20\text{ m}) = 7.20 - 0.72 = 6.48\text{ kN}$ Maximum horizontal shear stress: $f_v = \frac{3 V_{\text{design}}}{2 b d} = \frac{3 \times 6,480\text{ N}}{2 \times 10,000\text{ mm}^2} = 0.972\text{ MPa}$ Check: $f_v = 0.972\text{ MPa} \le F'_v = 1.75\text{ MPa}$. (Adequate in shear).

• Step 4: Minimum Bearing Length: Reaction at support $R = 7.20\text{ kN} = 7,200\text{ N}$. $A_{\text{bearing,req}} = \frac{R}{F'_{c\perp}} = \frac{7,200\text{ N}}{3.5\text{ MPa}} = 2,057.1\text{ mm}^2$ $l_b = \frac{A_{\text{bearing,req}}}{b} = \frac{2,057.1\text{ mm}^2}{50\text{ mm}} = 41.14\text{ mm} \approx 42\text{ mm}$

Worked Example 2: Timber Column Capacity

Problem: A Yakal solid wood column ($150\text{ mm} \times 150\text{ mm}$) carries an axial compression load. The unbraced length is $L = 3.0\text{ m}$ with pinned ends ($K = 1.0$). Given reference values: $F_c = 15.8\text{ MPa}$, $E_{\min} = 7,200\text{ MPa}$. Normal duration ($C_D = 1.0$), dry service ($C_M = 1.0$). Compute the allowable axial load $P_{\text{allow}}$.

Solution:

• Step 1: Slenderness Check: $l_e / d = \frac{1.0 \times 3,000\text{ mm}}{150\text{ mm}} = 20.0 \le 50 \quad (\text{Satisfied})$

• Step 2: Euler Buckling Stress for Wood: $F_{cE} = \frac{0.822 E_{\min}}{(l_e / d)^2} = \frac{0.822 \times 7,200}{(20.0)^2} = \frac{5,918.4}{400} = 14.80\text{ MPa}$

• Step 3: Column Stability Factor ($C_P$ with $c = 0.80$): $F_c^* = 15.8\text{ MPa} \implies \frac{F_{cE}}{F_c^*} = \frac{14.80}{15.80} = 0.9367$ $\frac{1 + (F_{cE} / F_c^*)}{2 c} = \frac{1 + 0.9367}{2(0.80)} = \frac{1.9367}{1.60} = 1.2104$ $C_P = 1.2104 - \sqrt{(1.2104)^2 - \frac{0.9367}{0.80}} = 1.2104 - \sqrt{1.4651 - 1.1709} = 1.2104 - \sqrt{0.2942}$ $C_P = 1.2104 - 0.5424 = 0.6680$

• Step 4: Allowable Axial Compressive Load: $F'_c = F_c^* \cdot C_P = 15.8\text{ MPa} \times 0.6680 = 10.55\text{ MPa}$ $P_{\text{allow}} = F'_c \cdot A = 10.55\text{ MPa} \times (150 \times 150\text{ mm}^2) = 10.55 \times 22,500 = 237,375\text{ N} = 237.4\text{ kN}$

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7. Licensure Exam Pitfalls & Review Notes