Timber Design

Introduction

• Structural timber design in Malaysia is using MS 544.

– Standard include beam, column, truss and connection

• Only cover design of beam and column.

• Comparison between timber and concrete/steel

Timber Concrete/steel

•Natural material

•Characteristic and properties are distinct and more

complex

•Strength depend on axis

- flexural strength

- Tension is parallel to grain

- Compression is parallel to grain

- Shear is parallel to grain

- Compression is perpendicular to grain

•Material produced from factory

•Strength can be determine

e.g. Gred S275 (fy=275N/mm2)

•Tension and compression strength

Moisture content

• The behaviour of timber is significantly influenced by the existence and variation of its moisture content

• The moisture content

w=100 (m1-m2)/m2

Where:

m1 is the mass of the test piece before drying (in g)

m2 is the mass of the test piece after drying (in g)

• Moisture contained in ‘green” timber is held both within cells (free water) and within the cell wall (bound water).

• Fibre saturated point( FSP)

– All free water has been removed but the cell walls are still saturated

– Moisture below FSP – properties considerable changes

– Moisture above FSP – properties remain constant

• Drying timber is control by seasoning

Air seasoning – stacked and layered with air space opensided shed

Kiln drying – dry out in a heated , ventilated and humidified oven

Defects in timber

– Seasoning defect

• Cause by uneven exposure to drying agent such as wind, sun etc.

• Defect: twisting, cupping , bowing and cracking

– Natural defect

• Most common: knot

– Will decreased the physical properties of timber such as tensile and compressive strength

• Others : wane defect and shake defect

Material properties

• Density

– Is expressed as mass per unit volume

– Principle properties effecting strength

– Wood with thick cell walls and small cell cavity (heavy species) have higher densities and strongest species.

– higher density give higher shrinkage, stiffness and hardness.

• Shrinkage

– Occur during drying process as absorbed moisture begin to leave the cell walls

– Width and thickness change but length remains the same

– Depends on initial moisture content value and the value at which it stabilises in service

– Result : Defect such as cupping, bowing etc

• Hygroscopic

– Can absorb moisture and can reached an equilibrium moisture content

• Anisotropy

– Characteristic of timber because of the long fibrous of the cells and their common orientation

• Direction of grain

– The elastic modulus of a fibre in a direction along its axis is considerably greater than the across it

– The slope of the grain can have an important effect on the strength of a timber member.

• Stress and strain

– The strain for a given load increase with moisture content

• Strain in a beam under constant will increased in damp environment

• Creep

– Demonstrate these behaviour as high stress levels induce increasing strains with increasing time

– The magnitude of long term strains increase with higher moisture content

– In structure where the deflection is important, the duration of the loading must be considered

– Reflect in MS 544-2-2001 by applying modification factor to admissible stresses depend on type of loading

• Fire resistance

– Generally compares favourable with other structural material and its often better

• Durability

– In resisting the effects of weathering, chemical or fungal attack

– E.g. Heartwood is more durable to fungal decay than the sapwood

• Presence organic compound ( toxic to fungi and insect)

Classification of Malaysian timber

Classification of Malaysian timber

Heavy hardwood

Over 880 kg/m3

Constructional timbers

i. Balau

ii. Cengal

iii. Merbau

Medium hardwood

720-880 kg/m3

Moderately heavy to heavy

construction

i. Keruing

ii. Mengkulang

iii. Tualang

Light hardwood

below 720 kg/m3

General utilities timber

i. Nyatoh

ii. Meranti

ii. Rubberwood

Softwood

i. Damar minyak

ii. Podo

iii. sempilor

* For detail application and physical appearance of timber, please refer to notes

Notation use in timber design

• Grade stress (g) is define as “ the stress which can safely be permanently sustained by material of a specific section size and of a particular strength class or species ( Table 1,2 or 4 in MS544:Part 2)

– Four grade depends on the defect (Basic, select, standard, common)

• Strength is graded by taking into account of defect by the process of reduction strength ratio after grading

Type of Forcec = compression

m = bendingt = tension

Significanceg = grade

a = appliedadm = permissible

e = effective

Mean arithmetic = mean

Geometry// = parallel

= perpendicular to grain

Example : m,a, = applied bending stress perpendicular to the grain

Solid Timber beam design

Less defect

Solid Timber beam design

1. Permissible stress design

– Introducing the safety margin by considering structural behaviour under working/service load condition and comparing the stresses thereby induced with permissible values

– The applied stress are determine using elastic analysis ( assume the structure in elastic behaviour) and refer to elastic theory

• The material is homogeneous (have same physical properties)

• The material is isotropic ( elastic properties same in all direction)

• The material obey Hooke’s law

safetyoffactor

stressfailureloadsworkingbyinducedstress

Permissible stress value

– The value of tensile strength is greater than that of the compression strength

• Both compression and tension have linear behaviour

– Design to resist combined bending and axial stresses

• Compression ductility is present before failure occurs, whilst in tension brittle, sudden failure occurs.

• With the assumption:

– The material is elastic, which implies that it will recover completely from any deformation after the removal of load

– The modulus elasticity is the same in tension and compression. The value is much lower when the load is applied perpendicular to the grain than when it is applied parallel to the grain

– Plane section remain plane during deformation.

» During bending this assumption is violated and reflected in non-linear bending stress diagram throughout cross-sections subject to a moment

• In designing strength ( axial bending or shear strength ) must satisfied below relationship

applied stress permissible stress

where :

Applied stress is determined using elastic theory

Permissible stress is determined by:

permissible stress, adm = Grade stress, g x modification factors, K

(excluding for permissible deflection)

2. Moisture content

– Properties of timber in MS 544 : part 2 categorized all the grade into dry and wet

– 19% of moisture is proposed to achieved higher strength

3. Modification factor

• Modification factor for stress grade in beam design

i. Duration of loading,K1

ii. Load sharing system, K2

iii. Bearing stress, K3

iv. Shear at Notched End, K4

v. Form factor, K5

vi. Depth factor, K6

Moisture content (MC) > 19 % ( wet) – use wet stress in calculationMoisture content (MC) < 19% ( Dry ) – use dry stress in calculation

i. Duration of loading,K1 ( Table 5, MS 544 : Part 2)

– Is used depend on the duration of loading being considered

Stress increased up to 25 %

No increase in stress

Stress increased up to 50 %

Stress increased up to 75 %

From building CP 3: Building factorA = element cladding, roof , etcB = all buildings with dimension > 50C = all building with dimension > 50

ii. Load sharing system, K2 (Clause 10 : Ms 544 : Part 2)

– If the number of element is 4 or more and the distance between element (spacing) is less than 610 mm, and which has adequate provision for the lateral distribution load, K2 = 1.1

– Emean is used for K2 = 1.1 and Eminimum for K2 = 1.0 ( no load sharing)

iii. Bearing stress, K3 (Table 6 : MS 544 : Part 2)

– At any bearing on the side of timber, the permissible sytress in compression perpendicular to the grain is depended on the length and position of the bearing

– For bearing length < 150 mm and located 75 mm or more from end of member, K3 should be determined according to Table 6

– Bearing stress of any length and bearing located at any place and have length > 150 mm, K3 = 1.0

Refer to Table 6 : MS 544Page 22

iv. Shear at Notched End, K4 ( Clause 11.4, MS 544 : Part 2)

– Square corned notches at the ends of a flexural member cause a stress concentration which should be allowed as follow;

Bottom notch

Top notch

v. Form factor, K5 ( Clause 11.4, MS 544 : Part 2)

– Grade bending stresses apply to solid timber members of rectangular cross section, K1 = 1.0

– For other shapes of cross section, the grade bending stresses should be multiplied by the modification factor K5:

Where ;

K5 = 1.18 for solid circular section

K5 = 1.41 for solid square sections loaded on a diagonal

vi. Depth factor, K6

– The grade bending stress is applied to timber member having a depth of h >3 00 mm.

– The grade bending stress should be multiplied by the depth modification factor

)56800(

)92300(81.0

2

2

6

h

hK For solid and glued laminated beams

• Lateral stability for solid and laminated beams

– The depth to breadth ratio of solid and laminated beams of rectangular section should be checked that there is no risk of buckling under design load.

– Alternatively the recommendation of Table 7: MS 544 : Part 2 should be followed:

Solid beam design : Flexural member

• Beam are the most commonly used structural element

– E.g : floor joist , trimmer joist around opening, rafter, etc

• The cross section of timber beam may be one of a number of frequently used sections such as those indicated in Figure

• Principle considerations in design all beams:

i. Bending

ii. Shear

iii. Bearing

iv. Deflection

v. Lateral stability

• Size of timber may be govern by the requirements of:

a) Elastic section modulus (z)

• To limit the bending stress and ensure that neither torsional buckling of the compression flange nor fracture flange induces failure

b) Cross-section

• To ensure the vertical and/or horizontal shear stress do not induce failure

c) Second moment of area

• To limit the deflection induced by bending and/or shear action to acceptable limits

• Bearing area

– Provided at the ends of beam is much larger than is necessary to satisfy the permissible bearing stress requirement

• Lateral stability

– Should be check

– Frequently provided to the compression flange of a beam by nailing of floor boards, roof decking etc

• Most timber beams are designed as simply supported and the effective span

i. Bending

– Applied bending stress is determined using simple elastic bending theory

m,a ll =

Where : m,a,ll maximum applied bending stress parallel to grain

Ma maximum applied bending moment

Z elastic section modulus about the axis of bending (usually the x-x axis)

– The permissible bending stress is given by

m,adm ll = m,g,ll x K1 x K2 x K5 x K6

Where : m,a,ll is grade bending stress parallel to the grain

– Must satisfy

m,a, ll m,adm, ll

Z

M a

ii. Shear

– The grade and hence permissible stresses given in the MS relate to the maximum shear stress parallel to the grain for a particular species or strength class

– Rectangular cross-section

• the maximum horizontal shear stress occurs at the level of the neutral axis and is equal to 1.5 x average value

Where:

a,ll maximum applied horizontal shear force

V maximum applied vertical shear force

A cross-sectional area

A

Vlla

5.1,

– The magnitude of a,ll must not exceed adm,ll given by

adm,ll = g,ll x K1 x K2 x K4

where g,ll grade stress parallel to the grain

And a,ll adm,ll

– For other type of section

Where = the shear parallel to grain stress at the level being considered

Fv = the vertical external shear

Au= the area of the beam above the level at which is being calculated

y= the distance from the neutral axis of the beam to the centre of the area Au

Ix = the complete second moment of area of the beam at the cross-section being considered

b= the breath of the beam at the level at which is being calculated

– If FvAuy/Ix is evaluated, this gives the total shear force parallel to grain above the level being considered per unit length of beam

x

uv

bI

yAF

• Bearing

– The behaviour of timber under the action of concentrate loads, e.g at positions of support , is complex and influenced by both the length and location of the bearing, as shown in Figure (s) and (b)

– The grade stress for compression perpendicular to the grain is used to determine the permissible bearing stress

– The actual bearing stress is determined from:

c,a,= P/Ab

Where :

P applied concentrated load

Ab actual bearing area provided

– The actual bearing area is the net area of the contact surface and allowance must be made for any reduction in the width of bearing due to wane, as shown in Figure below

– In timber engineering, pieces of wood with wane are frequently not used and consequently this can be ignore

iv. Deflection

– In the absence of any special requirements for deflection in building, it is customary to adopt arbitrary limiting value based on experience and good practice

– The combination deflection due to m (bending) and s(shear) should not exceed (0.003 x span) or 14 mm whichever is the lesser (clause 11.7 MS 544)

total (m+s) 0.003 x span or 14 mm

– Limitation to minimize the risk of cracking/ damage to brittle finished, unsightly sagging or undesirable vibration

– Deflection for solid beam is usually based on the bending action of the beam ignoring the effects of shear deflection ( when designing ply-web beams)

• The maximum shear deflection induced in single span simply supported beam of either rectangular or square cross-section may be determined from:

where:

A = the cross sectional area of the beam

Mmax = the maximum bending moment in the beam

v. Lateral stability

– A beam in which the depth and length are large in comparison to the width (i.e. a slender cross-section) may fail at a lower bending stress value due to lateral torsional buckling, as shown in figure below:

• The critical value of bending moment which induces this type of failure is dependent on several parameters, such as: the relative cross-section dimensions (i.e. aspect ratio), shape, modulus of elasticity (E), shear modulus (G), span, degree of lateral restraint to the compression flange, and the type of loading.

• This problem is accommodated in BS 5628-Part 2:2001 by using a simplified approach based on practical experience, in which limiting ratios of maximum depth to maximum breadth are given relating to differing restraint conditions. In ‘Table 7’ of MS 544 Part2, values of limiting ratios are given varying from ‘2’, when no restraint is provided to a beam, to a maximum of ‘7’, for beams in which the top and bottom edges are fully laterally restrained

Terms of Sawn Timbers

• Table B3A/3b: For Rough Sawn/ Full Sawn

– This timber is not dressed or surface, but sawed, edged and trimmed. The dimension is oversize to allow for shrinkage and before fully seasoned.

• Table B4: Dressed or Surface Timber

– This timber has been dressed by planning for smooth surface and uniform dimension. The dimension is measured at dry

Example 7.1: Timber beam Design

A main beam of 3 m length spans over an opening 2.8m wide and supports a flooring system which exerts a long-duration loading of 3.9kN/m, including its own self-weight over its span. The beam is supported by 50 mm wide walls on either side. Carry out design checks to show that (75 x 200) mm deep sawn section of strength group 4 of timber at 19% of moisture content with standard grade.

Example 7.2: Timber beam design

A floor beam carries a critical uniform load of DL + FLL of 2kN/m with moisture content more than 19%. The beam has a 9 clear span of 4.5m and is to be sawn and the width of both support are 150 mm. Using shearing stress criteria only, determine the required size for a select grade, yellow meranti if

a) The ends are not notched

b) The bottoms at each end have a 25 mm deep and 150 mm long

Example 7.3: Timber beam design

A timber floor spanning 3.8m c/c is to be designed using timber joist at 400mm centres. The floor is subjected to a domestic imposed load of 1.5kN/m2 and carries a dead loading including selfweight of 0.35kN/m2. Carry out design checks to show that a series of 50 mm x 200 mm strengthgroup 4 (SG4) sawn standard grade timber under dry condition is suitable or not.

Timber column design

• Introduction

– Example of axially members or members in members in combined axial force and bending are

• Post or columns

• Vertical wall studs

• Truss

• Bracing element

• Etc

Design of compression member

• The main considerations for compression members are:

– Slenderness ratio

• Relates to positional restrain of ends, lateral restraint along the length and cross-sectional dimensions of the member

• Axial compression and bending stress

• The effective length Le ( Clause 12.3)

– should be derived from either:

• Table 9 in MS 544: Part 2: 2001 for the particular end conditions

• The deflected form of compression member affected by ant restraint and or fixing moment (s). the effective length is considered as the distance between adjacent points of zero bending moment which the member is in single curvature

•

• From Table 9:MS 544: Part 2 : 2001 – Clause 12.3:

a) Restraint at both ends in position and in direction

b) Restrained at both ends in position and one end in direction

c) Restraint at both ends in position but not in direction

d) Restraint at one end in position and in direction and at the other end in direction but not in position

e) Restrained at one end in position and in direction and free at the other end

Le = coefficient x L

• Slenderness ratio (λ) (Clause 12.4 of MS 544: Part 2)

– Slenderness ratio is calculated by using formula below:

Where

Le = effective length

i = radius of gyration

= (I/A)

Simplify of i

i = b/12

Where:

b = the least lateral dimension

– The slenderness ratio should not exceed (Clause 12.4)a value of;

• λ 180

– any compression members carrying dead and imposed loads other than loads resulting from wind

– any compression member, however loaded, which by its deformation will adversely affect the stress in another member carrying dead and imposed loads other than wind

• λ 250

– any member normally subjected to tension or combined and bending arising from dead and imposed load, but subjected to a reversal of axial stress solely from the effect of wind

– any compression member carrying selfweight and wind loads only

• Modification factor for compression members, (K8)

– Can be determined using either Table 10 or calculated from the equation (Appendix D)

Where:

c = permissible stress (c,ll x K1)

E = minimum modulus of elasticity, Emin

λ = slenderness ratio

= 0.005λ

2/1

2

2

2

2

2

2

85.1

2

3

)1(2/1(

3

)1(2/1

ccc

EEEK

Member subjected to axial compression only

• The axial compressive strength (c,a.ll):

c,a,ll = P/A

Where:

P = axial compressive load

A = cross-sectional area

• Permissible compressive stress, (c, adm.ll) (clause 12.5 MS 544)

For λ 5:

c, adm.ll = c, g.ll x K1 x K2

For λ 5:

c, adm.ll = c, g.ll x K1 x K2 x K8

Verification

c,a.ll c, adm.ll

Member subject to axial compression and bending

• Clause 12.6 of MS 544

• Members subject to eccentric

– Load act through a point at a certain distance from the centroidal axis,

• Members which are restrained at both ends in position but not in direction and subject to bending and axial compression, should be proportioned that (must satisfy):

Where

m,a,ll = applied bending stress = M/Z

m,adm,ll = permissible bending stress = m,g,ll x K1 x K2 x K5 x K6

c,a,ll = applied compression stress = P/A

c,adm,ll = Permissible compression stress = c,g,ll x K1 x K2 x K8

e = Euler critical stress = and E = Emin

15.1

1,,

,,

8,,

,,

,,

lladmc

llac

e

llac

lladmm

llam

K

2

2

)/( iLe

E

Example 7.4: column design

A Nyatoh timber column with standard grade at 19% of moisture content is 4 m in height. The column is restrained at both ends in position but not in direction.

a) Determine an appropriate size of rectangular cross-section column so that the axial long term load of the column is more than 25kN

b) Check that the column is adequate to resist long term axial load of 12kN and a bending moment of 0.8kNm about its x-x axis.

Method 1: using Table 10

E/c, ll

λ = Le/i

140 160

700 0.195 0.154

800 0.217 0.172

16.763,

llc

E

85.153

yy

e

i

L

Table 10. MS 544

Method 2: Equation (Appendix D)

177.0

)212.0473/0(688.0

5.1

2

3

)1(2/1(

3

)1(2/1

2/1

2/1

2

2

2

2

2

2

8

ccc

EEEK

Example 7.5

A column is loaded and supported as indicated. The column is a surfaced (150 x 250)mm, common grade Penaga used at 16% maximum moisture content. The loading is combination of DL+FLL+WL. Determine the maximum allowable concentric load, P that the column can adequately support.

Example 7.6

Check that a 100 mm x 250 mm rough sawn section as shown in Figure below is adequate as a column if the load is applied 90 mm eccentric to its x-x axis. The column is 3.75m in height and has its ends restrained in position but not in direction.

Given :Timber (Dry) = SG5Strength grade = commonDesign load (long term) = 25 kNDesign load ( medium term) = 30 kN

Example 7.7: column with different length

• A column is loaded and supported as shown in figure below. The column has a surfaced 150 x 250, common grade Penaga used at 16% maximum moisture content. The loading is a combination of DL+ FLL + WL. Determine the maximum allowable concentric load, P that the column can adequately support

Example 7.9 : Other shape

Round columns – design of a column with round cross section can be based on the design calculations for a square column of the same cross sectional area.

Example

A 150mm diameter log, 4m long is used as a compression member in a foundation system. The member is considered to be pinned at the top, but fixed at the bottom. It is made of select grade Kempas and is not surfaced. It is used in an environment with high moisture content and normal duration. Determine the maximum concentric load for the member