Timber Elements Design
Transcript of Timber Elements Design
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Timber Elements
By
Dr. Mahdi Damghani
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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
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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
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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
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Timber strength classes (softwood)
Note: Designers should normally specify either C16 or C24 timber as these are the most readily available.
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Timber strength classes (hardwood)
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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.
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Note • Hardwood Trees have broad and flat leaves;– English Oak– Beech– Ash– Elm– Sycamore– Birch –Walnut.
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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
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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
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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
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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
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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
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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
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Values of kmod for timber and wood-based products at ULS
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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
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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
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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
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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
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Timber Beams and Joists
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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
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Check points
• Design for bending– Laterally restrained timber beam– Lateral Torsional Buckling
• Consideration of Bearing at the supports• Design for shear• Check deflection
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Bending of Timber beams
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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
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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
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Note
• The design bending strength is calculated as
m
kmhdm
fkkf
,mod
,
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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
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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
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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
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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
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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
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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
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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
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An example of Laterally Restrained timber beam
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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,
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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
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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
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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,
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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,
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Shear considerations
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Introduction
• Maximum shear normally occurs at the supports and under point loads
F
VVFFz 0
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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
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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
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Example
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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
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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,
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Deflection SLS considerations
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Introduction
• Deflection of flexural members must be limited to avoid damage to– Finishes– Ceilings– Partitions
• Limits on vibration of floors may also be required
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Checkpoints
• Vibration control• Final vibration
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Deflection F (kN/m) P (kN)
EA
M
EI
FLwww vm
2.19
384
5 3
EA
M
EI
PLwww vm
2.19
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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.
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Vibration of residential floors
m
EI
lf
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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
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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
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399 3
Deflection (bending plus shear) of a solid timber joist under 1kN load of decking
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Illustration 1kN
We conservatively assume that each joist carries 0.38 x 1kN=380N just for the sake of vibration check
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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
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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
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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 - -
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Final deflection
• Ψ factors are obtained from below table
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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