Chapter 7 - Timber Design

8/10/2019 Chapter 7 - Timber Design

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Timber Design


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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 (f y=275N/mm 2)

Tension and compression strength


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Moisture content The behaviour of timber is significantly influenced by the existence and

variation of its moisture content The moisture content

w=100 (m 1-m 2)/m 2Where:

m 1 is the mass of the test piece before drying (in g)m 2 is the mass of the test piece after drying (in g)

Moisture contained in green timberis 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


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The controlled drying timber is known as seasoning

Air seasoning stacked and layered

with air space open sided shed

Kiln drying dry out in a heated ,

ventilated and humidified oven


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Defects in timber Seasoning defect

Cause by uneven exposure to drying agent such as wind, sun etc. Defect: twisting, cupping , bowing and cracking


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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


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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 dampenvironment


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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 favourably with other structural material and itsoften better than most.

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)


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Classification of Malaysian timber

Classification of Malaysian timber

Heavy hardwood

Over 880 kg/m 3

Constructionaltimbers

i. Balau

ii. Cengal

iii. Merbau

Mediumhardwood

720-880 kg/m 3

Moderately heavyto heavy

construction

i. Keruing

ii. Mengkulang

iii. Tualang

Light hardwood

below 720 kg/m 3

General utilitiestimber

i. Nyatoh

ii. Meranti

ii. Rubberwood

Softwood

i. Damar minyak

ii. Podo

iii. sempilor

* For detail applicationand physical appearanceof timber, please refer tonotes


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Malaysian Timber Applications and PhysicalAppearance

http://www.hangtuahfurniture.com/timbers/


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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 strengthclass 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 ofreduction strength ratio after grading

Type of Forcec = compression

m = bendingt = tension

Significanceg = grade

a = appliedadm = permissible

e = effective

Mean arithmetic = mean

Geometry// = parallel

= perpendicular tograin

Example :

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

Solid Timber beam design

Less defect


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Solid Timber beam design

1. Permissible stress design Introducing the safety margin by considering structural behaviour

under working/service load condition and comparing the stressesthereby induced with permissible values

The applied stress are determine using elastic analysis ( assume thestructure 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 Hookes law

safetyof factor stress failure

loadsworking byinduced stress Permissiblestress value


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The value of tensile strength is greater than that of the compressionstrength 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 recovercompletely from any deformation after the removal of load

The modulus elasticity is the same in tension andcompression. The value is much lower when the load isapplied perpendicular to the grain than when it is appliedparallel 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


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In designing strength ( axial bending or shear strength ) must satisfiedbelow relationship

applied stress permissible stress

where :Applied stress is determined using elastic theoryPermissible stress is determined by:

permissible stress , adm = Grade stress( g) x modification factors(K)(excluding for permissible deflection)


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i. Duration of loading,K 1

( 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



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


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ii. Load sharing system, K 2 (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 adequateprovision for the lateral distribution load, K 2 = 1.1

Emean is used for K 2 = 1.1 and E minimum for K2 = 1.0 ( no load sharing)


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iii. Bearing stress, K 3 (Table 6 : MS 544 : Part 2) At any bearing on the side of timber, the permissible stress in

compression perpendicular to the grain is depended on the lengthand position of the bearing

For bearing length < 150 mm and located 75 mm or more from endof member, K 3 should be determined according to Table 6

Bearing stress of any length and bearing located at any place andhave length > 150 mm, K 3 = 1.0

Refer toTable 6 : MS 544Page 22


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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


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Principle considerations in design all beams:i. Bendingii. Sheariii. Bearingiv. 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

failurec) Second moment of area

To limit the deflection induced by bending and/or shear action toacceptable limits


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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


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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 shearAu= the area of the beam above the level at which is being calculatedy= the distance from the neutral axis of the beam to the centre of the area AuIx = the complete second moment of area of the beam at the cross-section being consideredb= the breath of the beam at the level at which is being calculated

If FvA

uy/I

xis evaluated, this gives the total shear force parallel to grain above the level

being considered per unit length of beam

x

uv

bI y A F


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Bearing The behaviour of timber under the action of concentrate loads , e.g at

positions of support , is complex and influenced by both the lengthand 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


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The actual bearing stress is determined from:c,a, = P/Ab

Where :P applied concentrated loadAb actual bearing area provided

The actual bearing area is the net area of the contact surface andallowance must be made for any reduction in the width of bearing dueto wane, as shown in Figure below

In timber engineering, pieces of wood with wane are frequently not

used and consequently this can be ignore


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iv. Deflection In the absence of any special requirements for deflection in building, it is

customary to adopt arbitrary limiting value based on experience andgood 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 11.7MS 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 thebeam ignoring the effects of shear deflection ( when designing ply-webbeams)


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The maximum shear deflection induced in single span simply supportedbeam of either rectangular or square cross-section may be determinedfrom:

where:A = the cross sectional area of the beamMmax = the maximum bending moment in the beam


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The critical value of bending moment which induces this type of failure isdependent on several parameters, such as: the relative cross-sectiondimensions (i.e. aspect ratio), shape, modulus of elasticity ( E), shearmodulus (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 asimplified approach based on practical experience, in which limiting ratiosof maximum depth to maximum breadth are given relating to differingrestraint conditions. In Table 7 of MS 544 Part2, values of limiting ratiosare 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 fullylaterally restrained


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Example 7.1: Timber beam Design

A main beam of 3 m length spans over an opening 2.8m wide andsupports a flooring system which exerts a long-duration loading of3.9kN/m, including its own self-weight over its span. The beam issupported by 50 mm wide walls on either side. Carry out design checks toshow that (75 x 200) mm deep sawn section of strength group 4 of timberat 19% of moisture content with standard grade.


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Example 7.2: Timber beam design

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

a) The ends are not notchedb) The bottoms at each end have a 25 mm deep and 150 mm long


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Example 7.3: Timber beam design

A timber floor spanning 3.8m c/c is to be designed using timber joist at400mm centres. The floor is subjected to a domestic imposed load of1.5kN/m 2 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.


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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

f


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Design of compression member

The main considerations for compression members are: Slenderness ratio

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

Axial compression and bending stress The effective length L e ( 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 any

restraint and or fixing moment (s). the effective length isconsidered as the distance between adjacent points of zerobending moment which the member is in single curvature


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Slenderness ratio () (Clause 12.4 of MS 544: Part 2) Slenderness ratio is calculated by using formula below:

WhereLe = effective lengthi = radius of gyration

= (I/A)Simplify of i

i = b/ 12Where:

b = the least lateral dimension


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The slenderness ratio should not exceed ( Clause 12.4 )a value of; 180

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

any compression member, however loaded, which by itsdeformation will adversely affect the stress in anothermember carrying dead and imposed loads other than wind

250 any member normally subjected to tension or combined and

bending arising from dead and imposed load, but subjected toa reversal of axial stress solely from the effect of wind any compression member carrying selfweight and wind loads

only


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Modification factor for compression members, (K 8 ) 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, E min = slenderness ratio

= 0.005

2/1

2

2

2

2

2

2

8 5.12

3)1(

2/1(3

)1(2/1

ccc

E E E K

M b bj d i l i l


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Member subjected to axial compression only

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

c,a,ll = P/AWhere:

P = axial compressive load

A = cross-sectional area

Permissible compressive stress, ( c, adm.ll ) (clause 12.5 MS 544)For 5:

c, adm.ll =

c, g.ll x K

1x K

2For 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


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j pbending

Clause 12.6 of MS 544 Members subject to eccentric

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


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

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 K6c,a,ll = applied compression stress = P/A

c,adm,ll = Permissible compression stress = c,g,ll x K1 x K2 x K8e = Euler critical stress = and E = E min

15.1

1 ,,

,,

8,,,,

,,

ll ad mc

ll ac

e

ll acll ad mm

ll am

K

2

2

)/( i Le

E

E l 7 4 l d i


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Example 7.4: column design

A Nyatoh timber column with standard grade at 19% of moisture contentis 4 m in height. The column is restrained at both ends in position but notin direction.

a) Determine an appropriate size of rectangular cross-section columnso 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 of12kN and a bending moment of 0.8kNm about its x-x axis.


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Method 1: using Table 10

E/ c, ll = Le/i

140 160

700 0.195 0.154

800 0.217 0.172

16.763,ll c

E

85.153 yy

e

i L

Table 10. MS 544


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Method 2: Equation (Appendix D)

177.0

)212.0473/0(688.0

5.12

3)1(

2/1(3

)1(2/1

2/1

2/1

2

2

2

2

2

2

8

ccc

E E E K


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Example 7 5


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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 moisturecontent. The loading is combination of DL+FLL+WL. Determine themaximum allowable concentric load, P that the column can adequatelysupport.


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Example 7 6


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Example 7.6

Check that a 100 mm x 250 mm rough sawn section as shown in Figurebelow is adequate as a column if the load is applied 90 mm eccentric to itsx-x axis. The column is 3.75m in height and has its ends restrained inposition 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


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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% maximummoisture content . The loading is a combination of DL+ FLL + WL.Determine the maximum allowable concentric load, P that the column canadequately support


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Example 7.9 : Other shape


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Example 7.9 : Other shape

Round columns design of a column with round cross section can bebased on the design calculations for a square column of the same crosssectional area.

ExampleA 150mm diameter log, 4m long is used as a compression member in afoundation system. The member is considered to be pinned at the top , butfixed 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 normalduration . Determine the maximum concentric load for the member


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Assignment 2

Redesign beam in Example 7.1 and must satisfied all the verification(Individual assignment)

Dateline of submission : 21 st December 2012 before 5.00pm

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Chapter 7 - Timber Design