Chapter 5: Design of Timber Structures

5.1 Introduction

Timber is used as one of the construction material in areas where it is easily available and cheap. Timber
can also be used for temporary structures, formwork and scaffolding.
Timber is an organic material generally used in its natural state but it should be properly seasoned before
use. Of the many factors that determine the strength of timber, slope of grains, specific gravity, moisture
content and the natural characteristics of timber (like defects due to knots, checks & shakes) may have an
effect on the strength of timber. The life of timber structure is long if it is maintained either dry or wet
continuously.
Advantages of Timber
- Good resistance against chemicals, such as acid and salts.
- Easy to handle, even with simple tools.
- Strong material compared to its own weight (it has a low unit weight).
- Nearly no change in length due to temperature.
- Nearly it is thermal and electrical insulating material.
- It has a good ability to dampen vibration.
- Being a fibrous material, it is less sensitive to repeated loads (fatigue).
Disadvantages of Timber
- Its strength decrease by moisture.
- Change its volume or/and shape depending on its water content.
- It is inflammable.
- It is easily attacked by fungi, insects etc.
- It takes a lot of time until a tree can be used as timber.

5.2 Permissible stresses of Timber

Strength of timbers is extremely variable. It depends on natural defects, slopes of grains, specific gravity
and moisture content. Large safety factors are normally used to determine the permissible stresses of
timber. The following tables may provide approximate values of permissible stresses of some of local and
commercially available imported timbers.

Table1:- Allowable Stresses of some of Local Timbers

Stress Zigba/Tid Eucalyptus, round
(Bahir-zaf)
-Bending ,
F b || 4 .0 MPa 12.0 MPa
-Tension ,
F t || 2 .5 MPa 10.0 MPa
-Compression ,
F c || 4.0 MPa 10.0 MPa
F c ⊥¿ ¿ 1.5 MPa 2 .0 MPa
-Shear ,
F s || 0 .6 MPa 1.5 MPa
E
-Elastic Modulus, timber 5,500 to 7,000 MPa 10,000 MPa
-Unit Mass 600 to 750 kg/m3 850 kg/m3
Table2:- Allowable Stresses of some of commercially available Timbers

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 Stress Cedar, Fir (pine tree), Teak (tall Oak
Similar to Zigba similar to Tid east
Indian tree)

-Bending ,
F b || 10 .59 MPa 6 .47 MPa 11. 38 MPa 12 .16 MPa
-Tension ,
F t || 10.59 MPa 6 .47 MPa 11. 38 MPa 12.16 MPa
-Compression ,
F c || 6 .87 MPa 5.1 MPa 7 .65 MPa 7.85 MPa
F c ⊥¿ ¿ 3 .34 MPa 1.19 MPa 3 .04 MPa 3 .43 MPa

-Shear ,
F s || 1 .37 MPa 0 .78 MPa 1 .37 MPa 1 .37 MPa
E
-Elastic Modulus, timber
10 , 690 MPa 9 , 220 MPa 9 , 420 MPa 12 , 260 MPa
-Unit Mass 700 kg/m3 440 kg/m3 630 kg/m3 850 kg/m3
The permissible stresses given in the above table are used for moist out door structures and under medium
term loading. For other moisture condition and duration of loading, corrections are to be made as given
below. Modulus of elasticity of timber is not affected by moisture conditions of timber.
a) Correction for moisture content

Condition of Allowable stress

Moisture increases by factor
-Dry condition (indoor structure) 1.20
-Moist condition (outdoor structure) 1.00
-Wet condition (Structure in water) 0.80

b) Correction for duration of loading

Duration of Design load

Loading increases by factor
-Short-term loading (wind load etc.) 0.8
-Medium-term loading (live load) 1.00
-Long-term loading (dead load) 1.2

5.3. Design of Structural Members

5.3.1 Tension Members

Tensile loads are allowed to act only parallel to the grains. Minimum area of tension member is limited to
24mm x 48mm. Tension members are simple to design. The required cross-sectional area may be
determined using:
T
An ≥
F t ||
where
An is net area = (Gross area – largest of imperfections or fastener holes)
A =
For nailed timber pieces, n
0.9 A g , and for bolted timber pieces, A n = 0.8 A g .

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5.3.2 Compression Members
The cross-sections of compression members are generally rectangular some times circular. As in the case
of steel column, timber column are classified in to short, intermediate and long column.
un sup ported length (l u )
≤ 11
a) For short column, least lateral dim ension (b )
l
11 < u ≤ k c
b) For intermediate column, b
lu
k c < ≤ 50
c) For long column, b

Where
kc =
√ 0 . 45 Etimber
F c||
lu
--limiting b ratio dividing intermediate & long column.
-for column with circular section, least lateral dimension, ‘b’ is taken as the side of equivalent
square section.
Unsupported length of compression member made of timber may be determined depending end condition
as given below:

lu = l lu =0 . 85 l lu =0 . 7 l lu =2 .5 l lu =1 .5 l

The permissible column loads used for design according to American Forest and Paper Association
(AFPA) shall be calculated as follows:
a) For short columns
P
fc = ≤ F c||
A
b) For intermediate columns

[ ( )]
4
p 1 l u /b
fc = ≤ F c|| . 1 −
A 3 kc
c) For long columns
P 0 . 3 Etimber
fc = ≤ 2
A ( lu /b )

( )
2
P 2 F c|| kc
fc = ≤ .
A 3 l u /b
or

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Note: No need of reducing cross sectional area of compression member if holes are filled by a material
which has at least the same strength as the timber column.

5.3.3 Design of Timber Beam

Beams and their supports are designed to resist the developed bending, shear and bearing stresses. The
minimum width of the beams shall not be kept less than 50mm or 1/50 th of the span, whichever is greater.
The depth of beam shall not be taken more than three time its width without lateral stiffening. Typical
span range of timber beams are:
Roof beams (with simple and continuous span) 3m to 7.5m
Floor beams  1.8m to 6m (with simple span)
 3m to 12m (with continuous span)
a) Design for simple bending
M .c M
fb = = ≤ F b||
I S
When depth of beam larger than 300mm, allowable bending stress is modified as
F b||| = ( 300/d )1/9 . F b||
When circular section is used by timber beam, allowable bending stress is increased by form factor of
1.18 as
|
F b|| = 1 .18 ∗ F b||
When notches are located at or near to the middle of the span, the net depth (d-d 2) should be used in
determining the bending strength of the beam where d2 is depth of notch.

b) Design for shear

V . Q V . ∑ ( A . y)
fs = = ≤ F s||
I .t I .t
If the section of beam is rectangular,
3V
fs = ≤ F s||
2b . d
If beam notched at lower (tension) face at supports,
3V . d
fs = ≤ F s||
2b . d 2
1
where d1 --depth of beam at notch

c) Design for bearing stresses of support of beam

In order to transmit the load from the beam to the support, the bearing stresses developed on the support
should satisfy the following equation:
End Re action
fp = ≤ F c ⊥¿ ¿
Bearing Area of the Beam
When the direction of bearing stress is at angle to the direction of the grain in any structural member, the
permissible bearing stress in that member shall be calculated by the formula:
F c|| . F c ⊥
Fc =
F c|| . sin2 θ + F c ⊥ . cos 2 θ
where  --angle between bearing stress & the direction of grain of timber.

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Flexural member excepting roof timbers which are supported directly on masonry or concrete wall shall
have a length of bearing of not less than 75mm. Members supported on corbels, offsets and roof timbers
on a wall shall bear immediately on and be fixed to wall-plate not less than 75mm x 400mm.

d) Design for deflection

Apart from strength requirement, the serviceability of the beam under service load should be maintained.
One of the serviceability requirements is deflection. The maximum permissible deflection of timber
beam:
l
Δ≤
-for beams supporting brittle material (like partition wall), 360
l l
Δ≤ Δ≤
-for other cases of beam, 240 for interior span and 180 for cantilever span
e) Design of Beam for bi-axes bending
Mx My
fb = ≤ Fb +
Sx Sy
√
Δ = Δ 2 + Δ 2 ≤ Δ permissible deflection
x y

5.3.4 Members Subjected to Combined Action of Axial-load & Bending

a) Axial Tension and Bending

Interaction equation used for design of member subjected to combined axial tension and bending is given
as:
ft f
+ b ≤ 1.0
F t || F b||
where
f t , f b are tensile and bending stresses.
F t , F b|| are allowable tensile and bending stresses.

b) Axial Compression and Bending

Interaction equation used for design of member subjected to combined axial compression and bending is
given as:
f fb
C
+ ≤ 1.0
F c || | F b||
where
f c , f b are compressive and bending stresses.
F ||| F ||
c , b are allowable compressive and bending stresses.
Allowable compressive stress considering the buckling effect is determined as follow:
lu
≤ 11 , F ||| = F c||
-For b c

[ ( )]
4
l 1 lu / b
11 < u ≤ k c , F ||| = F c|| . 1 −
b c 3 kc
-For

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 lu 0 .3 Etimber
kc < ≤ 50 , F ||| = 2
b c
( lu / b )
-For

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