1

Timber Elements

By

Dr. Mahdi Damghani

2

Introduction

• Guidance on the design of timber structures is given in EC5

• EC5 uses limit state design method and considers the following load combinations:

kk

kkkQkG

QG

QGQG

00.100.1(ULS) stateslimit lity serviceabi checkingFor

5.135.1(ULS) stateslimit ultimate checkingFor

3

Introduction

• The partial safety factors used for materials are:

00.1 SLS checkingFor

below tablefrom ULScheckingFor

m

m

Material

Solid timber, treated or untreated 1.3

Glued laminated timber (Glulam) 1.25

Laminated veneer lumber (LVL), plywood, oriented strand board (OSB) 1.2

Fibreboard 1.3

Connections, except for punched metal fasteners 1.3

Punched metal plate fasteners, anchorage strength 1.3

Punched metal plate fasteners, plate (steel) strength 1.15

m

4

Introduction

• Member axes are as follow:– x-x is along the length of the member in the

direction of the grain– y-y and z-z axes are the major and minor bending

axes

5

Timber strength classes (softwood)

Note: Designers should normally specify either C16 or C24 timber as these are the most readily available.

6

Timber strength classes (hardwood)

7

Note • The softwood most commonly used in the UK is from the tree

Pinus Sylvestus and is known as Redwood• Other names include;

– Red Deal– Yellow Deal– Archangel Fir– Swedish Pine and Scots Pine– Cedar– Western Red– Douglas Fir– Hemlock– Parana Pine– Pitch Pine– Quebec Yellow Pine– Western White Pine– Sitka and Spruce. 

8

Note • Hardwood Trees have broad and flat leaves;– English Oak– Beech– Ash– Elm– Sycamore– Birch –Walnut.

9

Variation of timber stiffness & strength with load duration and service class

• Creep phenomenon– Any material loaded less than its ultimate strength will strain– If stress is maintained for some time at a constant level the

strain will increase

• All materials such as steel, concrete and masonry will creep, however timber is more susceptible

• Assume a timber after 10 years;– If kept dry will have 60% more stain than immediate strain– If kept wet will have 200% more stain than immediate strain

10

What does that mean?

• The strength of timber for long-term loads may be 50% less than its strength for short-term loads

• Dry timber is stronger than wet timber

• The design stiffness and design strength vary with the duration of the load and with moisture content of the timber

11

Load duration classes

• To deal with this problem EC5 classifies the load in terms of its duration

Approximate duration Examples of loading

Permanent More than 10 years Self-weight

Long-term 6 months to 10 years Storage loading (including in lofts), water tanks

Medium-term 1 week to 6 months Imposed floor load

Short-term Less than 1 week Snow, maintenance or man loading on roofs, residual structure after accidental event

Instantaneous Wind, impact loading, explosion

12

Service classes• Timber rapidly adjust its moisture content

according to the environment conditions• EC5 takes this into account and introduces

classes for the state of serviceabilityService class

Definition Examples

1 Temperature of 20˚C, relative humidity of surrounding air only exceeding 65% for a few weeks per yearMoisture content of timber generally not more than 16%

Warm roofs, intermediate floors, internal and party walls

2 Temperature of 20˚C, relative humidity of surrounding air only exceeding 85% for a few weeks per yearMoisture content of timber generally not more than 20%

Cold roofs, ground floors, external walls and external uses where timber is protected from direct wetting

3 Conditions leading to higher moisture contents than in service class 2

External uses, fully exposed

13

Commonly available lengths of softwood structural timber

Length (m)

2.4 3.00 4.20 5.10

3.30 4.50 5.40

3.60 4.80

3.90

14

Reminder

• Strength of timber = F(load duration, moisture content)

• Charecteristice strength values from tables of slides 5 & 6 should be multiplied by the appropriate kmod factor in the table of next slide

15

Values of kmod for timber and wood-based products at ULS

16

Example

• Consider two timber beams each carrying dead load and imposed load over a span of 4.5 m. The dead load is the same on both beams but the imposed load on beam 2 is much smaller than the imposed load on beam 1.

mkNQmkNG

mkNQmkNG

kk

kk

/5.2 load Imposed ,/0.10 load Dead :2 beam

/0.12 load Imposed ,/0.10 load Dead :1 beam

17

Load cases for design

• Beam 1– Dead load only• Ultimate load• Moment

• kmod factor = 0.6 for permanent actions

• Design strength required

– Dead load plus imposed load• Ultimate load• Moment

• kmod factor = 0.8 for permanent actions

• Design strength required

mkNGk /5.130.1035.135.1 kNm2.348/5.45.13 2

kNm0.576.0/2.34

mkNQG kk /5.310.1250.10.1035.15.135.1

kNm7.798/5.45.31 2

kNm6.998.0/9.79

18

Load cases for design

• Beam 2– Dead load only• Ultimate load• Moment

• kmod factor = 0.6 for permanent actions

• Design strength required

– Dead load plus imposed load• Ultimate load• Moment

• kmod factor = 0.8 for permanent actions

• Design strength required

mkNGk /5.130.1035.135.1 kNm2.348/5.45.13 2

kNm0.576.0/2.34

mkNQG kk /3.175.250.10.1035.15.135.1

kNm8.438/5.43.17 2

kNm8.548.0/8.43

19

Note

• For solid timber elements carrying dead load and imposed load, the dead-load-only ULS check is not required if the imposed load is more than 0.3 times the dead load

20

Timber Beams and Joists

21

Definition

• Horizontal timber members that resist vertical load primarily in bending are referred to as either beams or joists

• A single isolated member is usually called a beam

• Several similar parallel timbers, supporting a floor or roof are called joists

22

Check points

• Design for bending– Laterally restrained timber beam– Lateral Torsional Buckling

• Consideration of Bearing at the supports• Design for shear• Check deflection

23

Bending of Timber beams

24

Bending

1 and 1,,

,,

,,

,,

,,

,,

,,

,, dzm

dzm

dym

dymm

dzm

dzmm

dym

dym

ffk

fk

f

Design bending stresses about y-y & z-z axes

Design bending strength about y-y & z-z axes

Allowance for re-distribution of stresses & the in-homogeneities

sections) crossother (for 0.1

sections) squareor r rectangula(for 7.0

m

m

k

k

25

Note

• Tables on slides 5&6 apply to solid timber members with cross section dimensions of 150mm or more. Smaller cross sections have higher tension and bending strengths calculated using the kh factor from table below

Depth factor

Timber depth h (mm)

40 or less

50 60 70 80 90 100 110 120 130 140 150 or more

kh1.30 1.25 1.2 1.16 1.13 1.11 1.08 1.06 1.05 1.03 1.01 1.00

26

Note

• The design bending strength is calculated as

m

kmhdm

fkkf

,mod

,

27

Example

• A planed softwood beam of C16 timber is 72mm wide × 220mm deep, has an effective span of L = 4.25m and carries the following un-factored loads– Dead load of 2.0kN– imposed load of 3.5kN

the beam is part of the ground floor of an office building and is laterally restrained by the flooring boards. Check whether its bending strength at ULS is adequate

28

Solution

1. Imposed load > 0.3 × Dead load

2. Service class 2 for a ground floor

3. Load distribution class medium

kNkNkN 6.023.05.3

A ULS check under dead load only is not required

29

Solution

4. Elastic section modulus for a joist of size 72mm×220mm

5. Bending strength for C16 timber

6. Depth factor

333

108.580)2/220(

72220)12/1(mm

y

IWyy

2, /16 mmNf km

0.1hk

30

Solution

7. Partial safety factor for solid timber

8. Modification factor for solid timber in Service Class 2 and Medium-Term loading

9. Design bending strength

3.1m

8.0mod k

2,mod, /85.9

3.1

160.18.0mmN

fkkf

m

kmhdm

31

Solution

10.Ultimate moment of resistance

11.Ultimate load

12.Design is OK for bending because

kNmWfM yydmult 72.5108.58085.9 3,

kNmFLM

kNQGQG kkkQkG

22.48/25.495.78/

95.75.350.10.235.15.135.1

MM ult OK

32

Laterally un-restrained timber beams

• If the compression edge is not restrained against lateral deflection then the design compressive strength of the timber must be reduced

33

Laterally un-restrained timber beams

• Follow the maximum depth-to-breadth ratios of solid timber to avoid LTB

Lateral support Maximum ratio

None 2:1

Ends restrained against rotation 3:1

Ends restrained against rotation & member held in line as by purlins or tie rods at centres not more than 30 times the breadth of the member

4:1

Ends restrained against rotation and compression edge held in line, as by direct connection of sheathing, deck or joists

5:1

Ends restrained against rotation and compression edge held in line, as by direct connection of sheathing, deck or joists, together with adequate bridging or blocking at intervals not exceeding six times the depth of the member

6:1

Ends restrained against rotation and both edges held firmly in line 7:1

34

An example of Laterally Restrained timber beam

35

Bearing ULS

dccdc fk ,90,90,,90, Compressive stress perpendicular to the grain

Design bearing strength

la

h

lEnd support bearing a<=h/3 Internal support bearing

412

1250

38.21 90,

l

hlkc 4

61

25038.21 90,

l

hlkc

m

kcdc

fkf

,90,mod

,90,

36

Example

• A sawn softwood beam in C16 timber, 63mm wide × 200mm deep, is supported on 100mm wide bearings. The load is medium term and Service Class 2 applies. Find the ultimate capacity of the bearing:– If the bearing is at the end of the joist, so that a=0– If the bearing is internal, so that a ≥ 200/3=67mm

37

Solution

1. Bearing strength for C16 (from table)

2. Modification factor for solid timber in Service Class 2

3. Partial safety factor for solid timber

2,90, /2.2 mmNf kc

8.0mod k

3.1m

38

Solution 4. Design bearing strength

At End Bearing

5. Bearing capacity

6. Ultimate capacity of bearing

2,90,mod,90, /35.1

3.1

2.28.0mmN

fkf

m

kcdc

2,90,

90,

/12.335.131.2

31.210012

2001

250

10038.2

mmN

k

dc

c

kNNdc 7.19196601006312.3area bearing ,90,

39

Solution

At Internal Bearing

1. Bearing capacity

2. Ultimate bearing capacity

2,90,

90,

/56.335.164.2

64.21006

2001

250

10038.2

mmN

k

dc

c

kNNdc 4.22224301006356.3area bearing ,90,

40

Shear considerations

41

Introduction

• Maximum shear normally occurs at the supports and under point loads

F

VVFFz 0

42

Useful formula

efd bh

V5.1

m

kvdv

fkf

,mod

,

Maximum shear stress in a rectangular section

Design shear strength

hef

0

notch edge Top

vk

hef

notch edge bottom

h

h

h

hx

h

k efv

;11

0.8)1(

5

2

x

43

Example

• The end of a sawn softwood beam in C16 timber, 63mm wide × 200mm deep, is supported on a 100mm wide bearing. The load is medium term and service class 2 applies. Find the ultimate shear capacity at the support for three configurations;– No notch– Top notch 125mm long × 80mm deep– Bottom notch 125mm long × 80mm deep

44

Example

45

Solution

1. Shear strength for C16 timber

2. Modification factor for solid timber in service class 2 and medium-term loading

3. Partial safety factor for solid timber

4. Design shear strength

2, /8.1 mmNf kv

8.0mod k

3.1m

2, /11.1

3.1

8.18.0mmNf dv

46

Solution Not Notched

Notched at the top

Notched at the bottom

kNNbhfbh

Vbh

V efdvefd

efd 3.99324

5.1

2006311.1

5.15.1

5.1 ,

kNNbhfkbh

Vbh

V efdvvefd

efd 5.55544

5.1

1206311.11

5.15.1

5.1 ,

6.0200/120/;75;120 hhmmxmmh efef

1425.0

6.06.0

120075

8.0)6.01(6.0200

5

2

vk

kNNbhfkbh

V efdvvefd 4.223805.1

1206311.1425.0

5.15.1,

47

Deflection SLS considerations

48

Introduction

• Deflection of flexural members must be limited to avoid damage to– Finishes– Ceilings– Partitions

• Limits on vibration of floors may also be required

49

Checkpoints

• Vibration control• Final vibration

50

Deflection F (kN/m) P (kN)

EA

M

EI

FLwww vm

2.19

384

5 3

EA

M

EI

PLwww vm

2.19

48

3

Note: The total deflection relates to the deflection of middle of a timber beam under bending (denoted by subscript m) plus deflection due to shear (denoted by subscript v). M, A, E and I are bending moment, area of the cross section, modulus of elasticity and second moment of area, respectively.

51

Vibration of residential floors

m

EI

lf

21 2

Fundamental frequency (Hz)

Stiffness of the beams in Nm2/m

Mass of the unloaded floor in kg/m2

Span in meters

slidenext on ble ta

from value theexceednot doesspan

on the anywhere placed loadpoint

1kN aunder deflection immediate

theifry satisfacto isfloor The 8

required ision investigat specialA 8

1

1

Hzf

Hzf

52

Limiting immediate deflections under a 1kN point load

Span of floor in meters

Limit (mm)

4 or less

4.2 4.4 4.6 4.8 5 5.2 5.4 5.6 5.8 6

1.8 1.71 1.62 1.54 1.47 1.41 1.35 1.29 1.24 1.2 1.15

EI

lw

48

399 3

Deflection (bending plus shear) of a solid timber joist under 1kN load of decking

53

Illustration 1kN

We conservatively assume that each joist carries 0.38 x 1kN=380N just for the sake of vibration check

54

Final deflection

• Recommendation by EC5

Type of member Limiting value wfin final deflections of individual beams

A member of span/between two supports

A cantilever member of span l

Roof or floor member with plastered or plasterboard ceiling

l/250 l/125

Roof or floor member without plastered or plasterboard ceiling

l/150 l/75

55

Final deflection

• Assuming that a member is carrying one permanent load and two variable loads

)(

)1(

)1(

2,22,02,2,

1,21,1,

,,

,2,1,

defQinstQfin

defQinstQfin

defGinstGfin

GfinQfinQfinfin

kww

kww

kww

wwww

56

Final deflection

• Kdef is obtained from table below

Material Service class

1 2 3

Solid timber 0.6 0.8 2

Glulam 0.6 0.8 2

LVL 0.6 0.8 2

OSB 2.25 - -

57

Final deflection

• Ψ factors are obtained from below table

58

References

• T. Draycott, P. Bullman, Structural Elements Design Manual, Published by Elsevier Ltd. ISBN 978-07506-8668-6

• Chanakya Arya, Design of Structural Elements: concrete, steelwork, masonary, and timber designs, 3rd edition, published by Taylor & Francis, ISBN 978-0-415-46720-9