Structural Elements of Timber Structures

A timber column is a vertical compression member. Its shape is chosen so that the least dimension (and hence the slenderness) gives the required capacity.

Types of timber columns

1. Solid (simple) column: a single sawn piece, square or rectangular. Used when the length is short and the section is available (houses, posts).
2. Spaced column: two or more shafts placed parallel with a gap and joined by spacer blocks at the ends and middle, glued or bolted. It gives a larger radius of gyration for the same timber. The spacer blocks are placed at the ends and at the mid-length or third points, and are not more than 1.5 m apart.
3. Built-up (laminated) column: planks nailed, bolted or glued together. They can be made to large sizes from small timber.
4. Box column: four planks forming a hollow section; high radius of gyration for little timber.
5. Composite column: timber with a steel core or steel angles, used for heavier loads.

 Solid        Spaced        Built-up       Box
 +----+      +--+  +--+    +-+-+-+-+     +------+
 |    |      |  |__|  |    | | | | |     | +--+ |
 |    |      |  |  |  |    | | | | |     | |  | |
 +----+      +--+  +--+    +-+-+-+-+     +------+
         (spacer blocks)    (laminated)

Column bases

The base transfers the load to the footing and keeps the timber off the ground so that it does not rot.

1. Timber post on concrete/stone pedestal with steel shoe: a steel plate with side flanges or angle cleats holds the post in position. The shoe is raised 100–150 mm above floor level.
2. Bolted steel base plate with anchor bolts: a plate fixed to the footing by holding-down bolts; the column is fixed to the plate with angle cleats and bolts. It can give a degree of fixity.
3. Dowelled base: a steel dowel or pin projecting from the footing enters a hole at the foot of the column. It gives a hinged support and resists lateral displacement.

   | post |       | post |        | post |
   |      |      /|      |\       |  ||  |
 --+------+--   / +------+ \      +--||--+
 [  shoe  ]     [ plate  ]        [ dowel ]
 ##########     ####anchor###     ########
 pedestal         bolts           footing

A damp-proof layer or gap should always separate timber from the masonry or concrete.

Formulas (IS 883)

Slenderness ratio $\lambda = l_e/d$, with $d$ the least dimension. Define

$$K_8 = 0.702\sqrt{\frac{E}{\sigma_{cc}}}$$

• Short column: $l_e/d \le 11$: $\sigma_c' = \sigma_{cc}$
• Intermediate: $11 < l_e/d < K_8$: $\sigma_c' = \sigma_{cc}\left[1 - \frac{1}{3}\left(\frac{l_e/d}{K_8}\right)^4\right]$
• Long: $l_e/d \ge K_8$: $\sigma_c' = \dfrac{0.329\,E}{(l_e/d)^2}$

Common value

$$K_8 = 0.702\sqrt{\frac{9500}{7.5}} = 0.702 \times 35.59 = 24.98$$

Area $A = 100 \times 150 = 15000$ mm²; least dimension $d = 100$ mm.

(a) Both ends hinged

$l_e = L = 3000$ mm, so $l_e/d = 30 > K_8 = 24.98$: long column.

$$\begin{aligned} \sigma_c' &= \frac{0.329 \times 9500}{30^2} = 3.47\ \text{N/mm}^2 \\ P &= \sigma_c' A = 3.47 \times 15000 = 52\,090\ \text{N} \approx 52.1\ \text{kN} \end{aligned}$$

(b) One end fixed, other hinged

$l_e = 0.8 \times 3000 = 2400$ mm, so $l_e/d = 24 < 24.98$: intermediate column.

$$\begin{aligned} \sigma_c' &= 7.5\left[1 - \frac{1}{3}\left(\frac{24}{24.98}\right)^4\right] = 7.5\,[1 - 0.2845] = 5.37\ \text{N/mm}^2 \\ P &= 5.37 \times 15000 = 80\,570\ \text{N} \approx 80.6\ \text{kN} \end{aligned}$$

Answer: (a) 52.1 kN (long column); (b) 80.6 kN (intermediate column).

Method

The permissible stress depends on slenderness, which depends on the section, so we try sections and compare the capacity with 150 kN.

Effective length $l_e = L = 3600$ mm (hinged both ends). For a square section $d \times d$, $l_e/d = 3600/d$.

$$K_8 = 0.702\sqrt{\frac{E}{\sigma_{cc}}} = 0.702\sqrt{\frac{10500}{8}} = 0.702 \times 36.23 = 25.43$$

For $11 < l_e/d < K_8$ (intermediate):

$$\sigma_c' = 8\left[1 - \frac{1}{3}\left(\frac{l_e/d}{25.43}\right)^4\right]$$

Trial sections

Check of 160 mm × 160 mm

$$\begin{aligned} \frac{l_e}{d} &= \frac{3600}{160} = 22.5 < 25.43 \\ \sigma_c' &= 8\left[1 - \frac{1}{3}(0.8848)^4\right] = 8\,[1 - 0.2043] = 6.37\ \text{N/mm}^2 \\ \sigma_{c,\text{actual}} &= \frac{150\,000}{160 \times 160} = 5.86\ \text{N/mm}^2 < 6.37\ \text{N/mm}^2 \end{aligned}$$

The column is safe. Self-weight of the column (about 0.2 kN) is negligible.

Answer: Provide a 160 mm × 160 mm timber column (capacity 163 kN > 150 kN, $l_e/d = 22.5$).

Interaction rule

For members under combined direct compression and bending (IS 883), the condition is

$$\frac{\sigma_{c,cal}}{\sigma_c'} + \frac{\sigma_{bc,cal}}{\sigma_{bc}} \le 1$$

where $\sigma_c'$ is the permissible axial compressive stress allowing for slenderness. (Magnification of moment due to deflection is ignored here for simplicity.)

Step 1: Actual stresses

$$\begin{aligned} A &= 150 \times 200 = 30\,000\ \text{mm}^2 \\ \sigma_{c,cal} &= \frac{80\,000}{30\,000} = 2.67\ \text{N/mm}^2 \\ Z &= \frac{150 \times 200^2}{6} = 1.0 \times 10^{6}\ \text{mm}^3 \\ \sigma_{bc,cal} &= \frac{5.0 \times 10^6}{1.0 \times 10^6} = 5.0\ \text{N/mm}^2 \end{aligned}$$

Step 2: Permissible axial stress

Least dimension $d = 150$ mm, so $l_e/d = 3000/150 = 20$.

$$K_8 = 0.702\sqrt{\frac{9000}{8}} = 0.702 \times 33.54 = 23.55$$

Since $11 < 20 < 23.55$, the column is intermediate:

$$\sigma_c' = 8\left[1 - \frac{1}{3}\left(\frac{20}{23.55}\right)^4\right] = 8\,[1 - 0.1735] = 6.61\ \text{N/mm}^2$$

Step 3: Interaction check

$$\frac{2.67}{6.61} + \frac{5.0}{10.0} = 0.403 + 0.500 = 0.903 < 1$$

The column is safe; about 10 percent reserve remains.

Answer: Interaction value = 0.90 < 1.0, so the 150 mm × 200 mm column is safe.

Principle

In a flitched beam the timber and steel strains are equal at each fibre, so stresses are proportional to modulus. Convert steel to an equivalent timber section using the modular ratio

$$m = \frac{E_s}{E_t} = \frac{2\times 10^5}{9000} = 22.2$$

Equivalent moment of inertia (in timber units)

$$\begin{aligned} I_t &= 2 \times \frac{75 \times 250^3}{12} = 1.953 \times 10^8\ \text{mm}^4 \\ I_s &= \frac{10 \times 250^3}{12} = 1.302 \times 10^7\ \text{mm}^4 \\ I_{eq} &= I_t + m I_s = 1.953\times10^8 + 22.2 \times 1.302\times10^7 = 4.847\times10^8\ \text{mm}^4 \end{aligned}$$

Moment of resistance

Extreme fibre distance $y = 125$ mm.

• From timber: $M_t = \dfrac{\sigma_t I_{eq}}{y} = \dfrac{10 \times 4.847\times10^8}{125} = 38.8$ kN·m
• From steel: the steel stress is $m$ times the equivalent timber stress, so $M_s = \dfrac{\sigma_s I_{eq}}{m\,y} = \dfrac{140 \times 4.847\times10^8}{22.2 \times 125} = 24.4$ kN·m

The lower value controls: $M_R = 24.4$ kN·m (steel reaches its limit first).

Safe load

$$M = \frac{wL^2}{8} \;\Rightarrow\; w = \frac{8 \times 24.43}{4^2} = 12.2\ \text{kN/m}$$

Deflection check (limit $L/360 = 11.1$ mm):

$$\delta = \frac{5wL^4}{384 E_t I_{eq}} = \frac{5 \times 12.2 \times 4000^4}{384 \times 9000 \times 4.847\times10^8} = 9.3\ \text{mm} < 11.1\ \text{mm}$$

Comparison with timber alone

Timber only: $M = 10 \times 1.953\times10^8 / 125 = 15.6$ kN·m, giving $w = 7.8$ kN/m.

The steel plate raises the capacity by about 56 percent.

Answer: $M_R = 24.4$ kN·m; safe UDL $= 12.2$ kN/m (deflection 9.3 mm, OK).

Design loads

$w = 7.5$ kN/m, $L = 3.6$ m.

$$\begin{aligned} M &= \frac{wL^2}{8} = \frac{7.5 \times 3.6^2}{8} = 12.15\ \text{kN·m} \\ V &= \frac{wL}{2} = \frac{7.5 \times 3.6}{2} = 13.5\ \text{kN} \end{aligned}$$

Step 1: Depth for bending

$$Z_{req} = \frac{M}{\sigma_{bc}} = \frac{12.15\times10^6}{11} = 1.105\times10^6\ \text{mm}^3$$ $$d = \sqrt{\frac{6 Z_{req}}{b}} = \sqrt{\frac{6 \times 1.105\times10^6}{100}} = 257\ \text{mm}$$

Step 2: Depth for deflection

$$\delta_{allow} = \frac{3600}{360} = 10\ \text{mm}$$ $$I_{req} = \frac{5wL^4}{384 E \delta} = \frac{5 \times 7.5 \times 3600^4}{384 \times 9800 \times 10} = 1.674\times10^8\ \text{mm}^4$$ $$d = \left(\frac{12 I_{req}}{b}\right)^{1/3} = \left(\frac{12 \times 1.674\times10^8}{100}\right)^{1/3} = 272\ \text{mm}$$

Deflection controls. Provide $100 \times 275$ mm.

Step 3: Checks for 100 mm × 275 mm

$$\begin{aligned} Z &= \frac{100 \times 275^2}{6} = 1.260\times10^6\ \text{mm}^3 \\ \sigma_b &= \frac{12.15\times10^6}{1.260\times10^6} = 9.64\ \text{N/mm}^2 < 11 \quad \text{OK} \\ \tau &= \frac{3V}{2bd} = \frac{3 \times 13\,500}{2 \times 100 \times 275} = 0.74\ \text{N/mm}^2 < 1.0 \quad \text{OK} \\ I &= \frac{100 \times 275^3}{12} = 1.733\times10^8\ \text{mm}^4 \\ \delta &= \frac{5 \times 7.5 \times 3600^4}{384 \times 9800 \times 1.733\times10^8} = 9.66\ \text{mm} < 10\ \text{mm} \quad \text{OK} \end{aligned}$$

Step 4: Bearing and lateral stability

Bearing length is chosen so that bearing stress $V/(b \times l_b)$ does not exceed the permissible compression perpendicular to grain; a bearing length of 100 mm gives about 1.35 N/mm². Since $d/b = 2.75$, the ends must be held in position and the compression edge held by floor boards.

Answer: Provide a 100 mm × 275 mm timber beam (bending stress 9.64 N/mm², shear 0.74 N/mm², deflection 9.66 mm).

(a) Types of timber beams

1. Solid (simple) beam: a single sawn rectangular section. Cheap and common for floors and short spans.
2. Built-up beam: planks nailed or bolted to form I or box sections for longer spans or large loads.
3. Laminated (glulam) beam: layers of timber glued with grain parallel, may be curved or tapered. It can reach large sizes with high strength since defects are scattered.
4. Flitched beam: timber planks bolted to one or two steel plates; the steel adds stiffness and strength without increasing the depth.
5. Compound / spliced beam: two or more members joined along the length using keys, dowels or bolts.

 Solid   Built-up (box)   Flitched
 +--+    +--+----+--+     +--+#+--+
 |  |    |  |    |  |     |  |#|  |
 +--+    +--+----+--+     +--+#+--+

(b) Design checks

1. Bending: $\sigma_b = M/Z \le \sigma_{bc}$ (permissible bending stress, modified by depth and load duration factors).
2. Shear: maximum horizontal shear stress $\tau = \dfrac{3V}{2bd} \le \tau_{perm}$. Check near supports; notches and holes reduce the shear area and cause splitting.
3. Deflection: under total load, $\delta \le$ span/360 (or as per the code); under live load a lower limit is often used. Timber shows creep so long-term deflection is checked.
4. Bearing: compression perpendicular to grain at supports and under loads, $\sigma_{cp} = R/(b\,l_b) \le \sigma_{cp,perm}$.
5. Lateral stability: deep narrow beams may buckle sideways. The permitted depth-to-breadth ratio rises with lateral restraint provided by ends, joists, flooring or bridging.
6. Self-weight and span: the effective span is the clear span plus half the bearing at each end (or centre-to-centre of bearings, whichever is smaller).