LVL Handbook PDF
LVL Handbook PDF
LVL Handbook PDF
EUROPE
Layout Faktor Oy
Printed by Punamusta, Finland 2020
Paper LumiPress Silk by Stora Enso 115 g/m2
Cover LumiPress Silk by Stora Enso 300 g/m2
Authors
1.1 INTRODUCTION
Laminated veneer lumber (LVL) is an engineered wood product • Easy to drill, cut, fasten and fit, only standard wood working
used in a diverse range of building and bridge construction tools needed.
applications. LVL beams, columns and panels have become • Precision-engineered and easily tailored.
established as essential components in modern timber • Can be produced to exact dimensions, minimizing cross cut-
construction due to their numerous advantages, versatility ting and sawing waste.
and proven structural performance. This handbook describes • Wide range of sizes: product dimensions not limited by raw
the state of the art of LVL from properties and applications to material size.
production and design methods. • Light and highly portable.
LVL is made of 3 mm thick veneers bonded together with • Dry from factory, moisture content 8-10% ensured minimal
weather-resistant phenolic adhesive. This means that the di- shrinkage in situ.
mensions of the final LVL product are not limited by the di- • Easily combined with other construction products.
mensions of the raw material, and even small-diameter logs can • Off-site production reduces construction time.
be used to produce large LVL beams and panels. When veneer • Entirely traceable, renewable, recyclable wood from certi-
logs are peeled to produce veneer, any natural defects in the fied sources.
wood, such as knots, are dispersed as small fragments across • Environmentally friendly carbon store: 1 m3 LVL contains
the veneer. This, together with the lamination effect, eliminates stored carbon equivalent to 789 kg CO2.
the impact of defects and results in exceptionally homogeneous • Production cost is higher than sawn timber, but less mate-
material properties. Although the production costs of LVL, like rial is needed to meet design specifications when building
all engineered wood products, are higher compared to sawn with LVL 1, 2.
timber, with LVL the same constructions can be designed with
smaller dimensions and LVL can also be used for applications
where suitable sawn timber sizes are not available.
The low deviation of LVL’s high strength and stiffness
Metsä Wood
means that these properties can be fully utilized as character-
istic values in structural design. In addition, due to the lack of
sizeable defects, the strength to weight ratio of LVL is extremely
high – LVL is twice as strong as steel in proportion to weight.
Due to its laminated structure, LVL is also dimensionally sta-
ble and free of warps, splinters and splits. LVL also comes dry
from the factory, eliminating the risk of shrinkage on site or in
the ready building, as long as the LVL members are protected
against weather exposure.
LVL is manufactured to exact dimensions, minimizing
cross cutting and sawing waste. The resulting low material
waste and uniform quality of LVL improve overall material
and time efficiency, especially in industrial applications and
off-site production of construction elements. LVL is easy to
drill, cut, fasten and fit – only standard wood working tools are
needed. LVL components are also highly portable due to their
light weight. LVL can also be easily combined with other wood
products and construction materials.
LVL is produced entirely from traceable, renewable, recy-
clable wood. It is a natural material from certified sources. LVL
also serves as a carbon store in buildings: 1 m3 of LVL contains
stored carbon equivalent to 789 kg of CO2, making it an envi-
ronmentally friendly choice 1, 2, 3.
Industrial applications
• Support structures and moulds for concrete formwork
• Scaffolding
• Door- and window frames
• Furniture components
• Packaging industry
1. LVL-P load-bearing wall stud: straight and 4. LVL-P floor joist: strong and rigid. 8. LVL-P/C sole plate: fits stud dimensions,
precise dimensions. 5. LVL-P roof rafter (or LVL-C when height- thin members reduce settling.
2. LVL-C rim board: dimensionally stable, thickness ratio is high): space for thermal 9. LVL-P lintel over garage door opening:
minimal settling. insulation of low-energy buildings. large openings possible.
3. LVL-C bracing panel: narrow, but robust 6. Double LVL-P ridge beam: strong and 10. LVL-P lintel in wall to carry roof loads:
panel next to large openings in walls rigid. straight and rigid.
when there is no space for large panel 7. LVL-C roof panel: fast installation, fewer 11. LVL-P ledger beam for canopy: straight
fields. joints, roof overhangs without additional and rigid, easy details.
supports.
1. LVL-C panel structure of volumetric 5. LVL-P roof beams/elements: space 10. LVL-C panel structures for balcony walls
bathroom module: light structure, short for thermal insulation of low energy (separate cladding needed): simple and
construction time on site. buildings, long lengths available. rigid structure.
2. LVL ribbed slab intermediate floor 6. LVL-C balcony floor slab: simple structure. 11. LVL-C panel structures for lift shafts: panel
elements: long spans with suitable floor 7. LVL-C corridor floor slab for limited spans: sizes available for full building height or
thickness. simple structure. storey-high elements. Acts as a part of
3. LVL-P wall studs (small dimensions the bracing system of the building.
8. LVL-C rim beams / Lintels: straight and
for non-load-bearing walls, larger rigid structure, simple geometry of 12. LVL-C roofing panels: large sizes, fast
dimensions for load-bearing walls): element joints. installation, fewer joints
straight and precise dimensions.
9. LVL-C mezzanine floor slab for loft spaces:
4. LVL-C bracing panel: stable and rigid minimal structural depth, better use of
building. room height.
Figure 1.8. LVL applications in hall constructions. LVL elements can be used together with any type of main frame, e.g. precast concrete beams,
steel trusses or glulam beams.
1. LVL king post or queen post roof 6. LVL-P purlins, multiple span: long
trusses: impressive appearance. lengths available.
2. LVL columns: fit together with LVL roof 7. LVL-C bracing panels for roofs: simple
trusses. and robust.
3. LVL-P and LVL-C portal frames: large 8. LVL-P studs for high walls: straight and
clear height. precise.
4. LVL lintels for door and window 9. LVL-P horizontal beams for walls: large
openings: strong and rigid. spacing between main frames.
5. LVL-P purlins, single-span: strong and 10. LVL-C bracing panels for walls: simple
rigid. and robust structure.
Figure 1.10. Renovation applications, left: LVL floor beam reinforcement, right: attic frames for room in roof space with LVL reinforced roof trusses.
for each case. For example, we have replaced concrete eleva- get a lot of feedback on the technical feasibility, assembly and
tor shafts with LVL-C panel structures and built intermediate cost efficiency of the structures in practice.
floors from offsite-produced LVL elements with 7-metre spans. We use LVL in all kinds of applications, such as the renova-
Floor elements have been developed so that the required mass tion and reinforcement of existing wooden floors, where slim
for acoustics is realized with dry screed panels pre-installed on cross-section LVL reinforcement brings structural enhance-
the elements already in the factory. This has further increased ment to existing joist structures. We also often use ribbed panel
the prefabrication rate and minimized the amount of site work and box panel structures to achieve light and long-spanning
needed. floors. In cross-laminated timber buildings, we use LVL lintels
where small cross-sections with high loadbearing capacity are
Structural engineer’s view of LVL, required. Occasionally, we use LVL-C panels for walls or floors
Wenzel von Fragstein, Germany without separate stud or joist structures. Long overhangs from
thin structural panels are commonly desired for front roofs,
As structural designers focusing on wooden buildings, we de- and LVL-C panels can be used to achieve them as they have
sign entire wood and wood-hybrid constructions. From sin- better strength properties than most other structural panels.
gle-family houses to wooden multi-storey buildings and in- Their high strength is a result of their glued veneer structure,
dustrial hall constructions, our activities cover the whole range which eliminates the influence of individual defects, such as
of wooden buildings. Renovation of churches and historical knots. LVL-C with cross veneers has significantly higher re-
building as well as church towers of new buildings have al- sistance to splitting than solid wood, so by choosing LVL-C for
so been part of our project portfolio. We typically prepare the members that have high tension perpendicular to grain stress-
work specifications and production drawings for manufacture es, separate reinforcements can be avoided.
all the way down to the steering files for CNC machining. Be- LVL enriches the product range of wood-based materials
ing involved in the whole construction process in this way, we significantly.
Puuinfo
Table 1.3. Global LVL production. Active manufacturers of structural LVL produce about 3.9 million cubic metres per year 4, 5.
Carter Holt
Steico 1 160 Weyerhaeuser 4 530 1 100
Harvey
Lousiana
Stora Enso 1 100 2 260 Nelson Pine 1 100
Pacific
Pacific
MLT 1 100 1 220 First plywood 1 100
Woodtech
Forex Amos
LVL Ugra 1 40 1 140 Keyteck 1 60
Inc.
West Fraser 1 90
LVL 01, Figure 1.11
RedBuilt 1 70
Global LVL 1 20
Raute Oyj
GLOBAL LVL PRODUCTION
4,5
4,0
3,5
3,0
Million m3 / year
2,5
2,0
1,5
1,0
0,5
0,0
1965 1970 1975 1980 1985 1990 1995 2000 2005 2010 2015 F2020
Figure 1.14. Development of structural LVL production volume globally 1965 – 2020 5 .
2,5
Australia
2,0
Million m3 / year
Japan
1,5
New Zealand
1,0
Europe
0,0
2011 2012 2013 2014 2015 2016 2017
Glue 6 %
LVL
Moisture
10 %
By-products
and losses
Dry chips & saw dust Drying shrinkage loss Compression loss
Figure 1.16. Left: Distribution of wood material in LVL production. 2,5 m3 of logs (over bark) is needed to produce 1m3 of LVL. Right: Weight
distribution of wood fibres, glue and moisture in ready LVL product 5 .
1.4.3 Wood species used in LVL 1.5 SUSTAINABLE BUILDING WITH LVL
LVL is commonly produced from softwood and, in Europe,
spruce and pine are typically used. Spruce has the best strength 1.5.1 Traceable raw material
to weight ratio, and its low resin content is also advantageous
with respect to the production process. Pine veneer, on the oth- and sustainable sources
er hand, has higher density, which gives the product slight- Sustainable forest management and use of forest products play
ly higher mechanical properties. Some LVL manufacturers in a key role in the mitigation of global warming and contribute
Europe also use the hardwood species beech and birch. The to achieving climate policy objectives. Forest climate mitiga-
mechanical properties of hardwood LVL are higher due to the tion options include reducing greenhouse gas emissions from
higher density. However, higher density introduces additional deforestation and forest degradation, enhancing the carbon se-
machining requirements, such as possible predrilling for screw questration rate in existing and new forests, providing wood
connections. The surface of hardwood LVL is also more sensi- residues as a substitute for fossil fuels, and replacing non-re-
tive to mould growth in humid conditions. newable and energy-intensive materials with wood products,
North American LVL manufacturers use different pine particularly in the building sector.
species, Douglas fir, western hemlock, yellow poplar and red
maple. In Australia LVL is produced from different pine species Due diligence on the origin of wood
and karri (eucalyptus). In Japan, LVL is produced from larch Due diligence systems differentiate wood and wood products
and Japanese cedar (sugi). Other species that have the required from many other building materials by verifying the origin of
mechanical and gluability properties may also be used. the wood raw material. The European Union Timber Regula-
tion (No 995/2010) was enacted to ensure that all wood placed
1.4.4 Durable and inert gluing of LVL on the European market from internal or external sources is
covered by a due diligence system for verifying the legal origin
In structural LVL the veneers are bonded together with weath- of wood 6. The objective is to prohibit any product that con-
er- and boil-resistant phenol formaldehyde (PF) adhesive, tains wood raw material harvested in violation of national or
which is cured in the hot pressing process. Curing converts international laws from entering the EU market. The regulation
the adhesive to a high temperature resistant inert polymer that covers all wood and wood products, such as round wood, solid
does not dissolve or react with other materials in the surround- wood, engineered wood, pulp, paper and board.
ing environment. LVL also fulfils the most stringent formal- The EU Timber Regulation covers the entire wood val-
dehyde emission requirements, with emissions 3 times lower ue chain. An ‘operator’ who first places a wood product on
than the limit value of the E1 classification tested according to the European market is obliged to prove the legality of the
standard EN 717-1. The dry solids content of adhesive in LVL wood to a national EU Timber Regulation authority. A Due
is approximately 30 kg/m3, i.e. about 6% by weight. Diligence system is applied for this purpose, including access
to information on wood sources, risk assessment, and mit-
igation of risks. Chain of Custody systems (e.g. PEFC™ and
FSC®) can be third-party certified according to Due Diligence
requirements.
Promoting sustainable,
certified forest management
European LVL producers are well positioned in forest certifi-
cation as European forests owners have been active in apply-
ing forest certification schemes (most commonly PEFC™ and
FSC®). Forest certification schemes provide third-party ver-
ification of sustainable forest management practices and the
chain of custody from forest to product. They include require-
ments for sustained harvesting and forest regeneration prac-
tices, biodiversity protection, multiple and recreational uses of
forests, social sustainability, training of employees, and occu-
pational safety.
In order for a wood product to PEFC™ or FSC® labelled,
it must contain at least 70% wood raw material from certified
forests. When non-certified wood is used in the production of
certified goods, the wood must originate from forests covered
by an appropriate due diligence system.
1.5.2 Sustainable over the life cycle Environmental performance of buildings and
Wood products offer renewable and sustainable solutions for building products
construction. Wood as a renewable material has a lower glob- The European standard series ‘Sustainability of Construction
al warming impact compared to alternative, non-renewable Works’ (CEN/TC 350) guides the assessment of the sustaina-
building materials. The infinite carbon cycle between the at- bility of buildings and building products. The standard series
mosphere, growing trees and wood products distinguishes re- aims to enhance the supply and demand of products and build-
newable wood from non-renewable materials. ings that have as low environmental impact as possible. Envi-
Life cycle assessment (LCA) is a holistic approach for as- ronmental assessment of a building is based on the life cycle
sessing environmental impacts throughout a product’s or sys- approach, in which each of the different stages of the building’s
tem’s life cycle, from extraction of raw materials to disposal of life cycle are included and assessed (Figure 1.17).
the product. The principles of LCA have been internationally At the product and service level, the environmental
agreed and standardized with the ISO 14040 and ISO 14044 product declaration (EPD) applies the life cycle assessment
standards, which enables third-party verification of life cycle approach and presents quantified environmental information
LVL 01, Table
calculations. 1.4.A • muokattu
LCA compiles and evaluates the inputs, outputs over a product’s life cycle. EPDs enable comparison between
and potential environmental impacts of a product or system different products with the same functional purpose at the
throughout its life cycle. LCA helps manufacturers to identi- building level. In the case of LVL, the comparison is most ap-
fy opportunities to improve the environmental and climate propriately done at the structure type level, e.g. structures with
performance of a product and to inform customers and stake- the same load carrying capacity and stiffness. The EN 15804
holders. LCA includes four steps; defining of the goal and standard provides product category rules (PCR) for an envi-
scope, inventory of material and energy flows, assessment of ronmental product declaration for any construction product
impacts, and interpretation of results. or construction service (Table 1.4). Biogenic carbon content
Table 1.4. Life cycle stages of building environmental assessment based on EN 15978.
A4 Transport
Construction process stage
A5 Construction installation process
B1 Use
B2 Maintenance
Building B3 Repair
life cycle
information Use stage B4 Replacement
B5 Refurbishment
B6 Operational energy use
B7 Operational water use
C1 De-construction, demolition
C2 Transport
End of life stage
C3 Waste processing
C4 Disposal
Additional
information
outside the
Potential benefits and loads D Reuse, recovery, recycling potential
system boundary
calculation rules are provided in the EN 16485 standard. tal performance of new and existing buildings. The environ-
The most used environmental indicator of EPDs is global mental performance of wooden buildings derives from their
warming potential (GWP), also known as carbon footprint. light weight (compared to other building materials), energy
The GWP reflects the amount of greenhouse gas emissions in efficiency of materials and buildings, life-time carbon storage
each stage of a product’s life cycle and is mainly the outcome of in wood, and their renewable and sustainable origin. Wood-
fossil fuel use in the raw material supply stage and energy use en buildings typically achieve the same service time as other
in the production stage (A1–3). buildings, typically 50–100 years. Up to 100 years or longer
At the building level, the EN 15978 standard provides sys- service time is achievable with proper design and optimized
tematic calculation rules for the assessment of the environmen- maintenance.
1000
800
600
400
200
-200
-400
-600
-800
-1000
A1-A3 A4-A5 B1-B7 C1-C4 Total D
Product stage Construction Use stage End of life (module D Loads and benefits
stage not included) beyond system
boundary
Figure 1.19. Example global warming potential of different life cycle stages 3, 7, 8.
1.5.4 Global
LVL 01, Tablewarming
1.4. B impact of buildings
The life cycle of a building covers all life cycle stages from ‘cra- building type (a 4-storey residential building) 9, 10. The assess-
dle to grave’ as well as, optionally, also the benefits and loads ment was done according to the EN 15978 standard, with the
beyond the building’s life cycle. These life cycles stages are specific assumptions presented in Table 1.5.
presented as modules A1–C4 + D (Figure 1.17). The environ- The global warming potential of the whole life cycle and
mental performance of a wood-frame building, and a con- beyond is 798 kg CO2e/m2 for a wood-frame building and
crete-frame building were assessed based on a generic Finnish 1022 kg CO2e/m2 for a concrete-frame building (Figure 1.20).
Table 1.5. Scenarios of a wooden frame building and a concrete frame building 9, 10.
Concrete-frame Wood-frame
Concrete: crushing and recycling for ground Concrete: crushing and recycling for ground
End of life construction construction
Wood: chipping and energy recovery Wood: chipping and energy recovery
Figure 1.20. Global warming potential (GWP) of a wood-frame building vs. a concrete-frame building 10.
LVL
Green chips
Bark
Drying shrinkage
Veneer handling
Sawing dust
Core
Compression
Figure1.23. Log use distribution by volume: LVL and by-products Figure1.24. Harvesting of logs.
(incl. shrinkage and compression losses).
Figure 1.28. Peeling line, from block centring (left) to veneer mat clipping and stacking (right).
A colour camera technique is used to recognize and ana- vary widely, from 30% to over 150%. Sorting of veneers accord-
lyse even the smallest details on the veneer mat, such as knots, ing to moisture content enhances the efficiency of drying and
holes, splits, bark and rot. Based on this camera analysis, the attainment of the desired final moisture content.
system optimizes the points at which the mat is to be cut. The
mat is then clipped into sheets. Defects such as splits, fishtails 1.6.3 Drying and grading the veneer
and large holes are cut out according to predefined parameters.
The clipped veneers are stacked in different bins according The aim of the drying process is to dry the green veneers to a
to their size and moisture content. Softwood is typically sorted moisture content suitable for gluing. Too high veneer moisture
into two or three moisture grades as the moisture content can hampers gluing and generates steam during hot pressing.
Figure 1.30. VDA camera image for visual defect analysis. Figure 1.31. Dryer infeed.
The target moisture content is below 5%. Automatic regulation veneer are also carried out based on radio frequency analysis
maintains the speed, temperature and humidity inside the dry- and ultrasonic propagation time.
er at an optimal level. The energy required to dry the veneer Veneer pieces of different sizes – i.e. ‘randoms’ – can be
can be produced from the mill’s own by-products. composed into veneer sheets of required width. Composed ve-
A fully automatic machine vision system analyses veneer neers are used as core veneers. The aim is to increase veneer
defects, such as knots, splits, breakages, micro-splits, decay, yield by maximizing the use of peeled material. Randoms can
resin pockets and discolouration, at process speed. The image be collected from peeling line and broken veneers from the
data is analysed in milliseconds and used to stack the veneers drying line. Veneer can be composed green or dry. At the com-
into different bins at the stacker. The dried veneer is measured poser, the randoms are glued or taped together to form full-size
for moisture content to ensure the target moisture has been veneer sheets for LVL production.
achieved. Density measurement and strength grading of the
Figure1.39. Liquid extruder gluing (LEG): glue is applied to the top Figure1.40. Staggered layup ahead of the pre-press.
of the veneer as it passes on the belt conveyor.
Hot pressing is typically continuous to allow for variation resistant. With phenolic resins the pressing time ranges from
in product lengths. The maximum LVL length (18-25 m) is 15 to 90 minutes depending on the thickness of the product.
limited either by the mill building or by delivery by road to the Long pressing times and high pressures cause the wood fibres
customer. Hot pressing must ensure that all glue lines reach to compress, resulting in higher product density.
the proper curing temperature. Once cured, the glue becomes
resistant to melting and water insoluble, i.e. highly weather 1.6.5 Finishing
The LVL panels are normally cut to size according to the width
and length specifications of the customer. Additionally, the
LVL can be sanded or otherwise treated. Some manufacturers
use portal-type sawing units that enable inclined, diagonal or
chamfered cuts.
Finally, appropriate packaging protects the finished panels
against soiling, moisture and handling damage, keeping them
straight and easy to store and handle throughout the delivery
chain from mill to customer.
LVL manufacturers offer further processing of their products 1.7.2 Special cutting
according to customer specifications as value-added services.
This is done either directly at the mill or by subcontractors LVL panels or beams can be sawn to special shapes or sawn di-
equipped with special machinery for LVL processing. The val- agonally to produce tapered beams or columns. Manufacturers
ue-added services save time and minimize waste for the cus- may also have special cutting tolerances for tailored products,
tomer and on the building site. e.g. of industrial customers.
1 2 3
Metsä Wood
Figure 1.49. LVL-P roof beam cut to special shape.
Stora Enso
lifting devices. PU, MUF and PRF adhesives that do not need
hot pressing and that are approved for load-bearing structures
can be used for multiple gluing of GLVL.
Metsä Wood
Figure 1.54. Moisture protection treatment of LVL-P beams by Stora Figure 1.55. Multiple-glued GLVL beams and panels.
Enso.
Figure 1.56. Stressed-skin panels, large I-beams and box beams from LVL.
Metsä Wood
Table 1.6. Nominal product thicknesses and layups of LVL-P and LVL-C.
Number of cross
Thickness [mm] Number of veneers Layup of LVL-P Layup of LVL-C
veneers in LVL-C
24 8 IIIIIIII II-II-II 2
27 9 IIIIIIIII II-III-II 2
30 10 IIIIIIIIII II-IIII-II 2
33 11 IIIIIIIIIII II-IIIII-II 2
39 13 IIIIIIIIIIIII II-III-III-II 3
42 14 IIIIIIIIIIIIII -
LVL 01, Table
45 1.6 15 IIIIIIIIIIIIIII II-IIII-IIII-II 3
48 16 IIIIIIIIIIIIIIII -
51 17 IIIIIIIIIIIIIIIII II-IIIII-IIIII-II 3
57 19 IIIIIIIIIIIIIIIIIII II-III-IIIII-III-II 4
63 21 IIIIIIIIIIIIIIIIIIIII II-III-III-III-III-II 5
69 23 IIIIIIIIIIIIIIIIIIIIIII II-IIII-III-III-IIII-II 5
75 25 IIIIIIIIIIIIIIIIIIIIIIIII II-IIII-IIII-IIII-IIII-II 5
For structures that are sensitive to dimensional changes walls are 39x66 and 39x92 in lengths of 2550 mm, 2700 mm,
due to changing humidity conditions, LVL-C is the most suit- 3000 mm, 3600 mm and 6000 mm.
able choice. No pre-cambering: Because LVL members are cut from
LVL-C beams and panels are usually produced according straight billets, beams cannot be pre-cambered in normal pro-
LVL
to 01, Table
customer 1.7 and their sizes are thus not stand-
specifications duction and pre-cambering should therefore not be included
ardized. The beam heights given in Table 1.7, however, ensure in the design.
efficient use of material. Other common widths are 900 mm, Larger thicknesses are available as multiple-glued GLVL
1200 mm, 1800 mm and 2500 mm, although the maximum beams and panels. While standard sizes ensure the most effi-
panel width depends on the production line. cient material use, custom sizes of GLVL are also available from
Standard sizes for LVL-P studs are smaller than beams. GLVL suppliers for project-specific needs, but the availability
Thickness is typically 39 and 45 mm and the width is normally shall to be checked case by case.
limited to 200 mm. Standard stud sizes for non-load-bearing
Table 1.8. Standard sizes of multiple-glued GLVL beams and panels. GLVL can be manufactured from LVL-P or LVL-C laminas.
Metsä Wood
1.9 TOLERANCES
The tolerances of LVL members are defined in FprEN
14374:2018 and depend on the member sizes. The tolerance
values are shown in Table 1.9 and the dimension definitions
in Figure 1.61.
Table 1.9. Maximum deviations from nominal sizes and nominal angles for LVL, unsanded and not pressure treatment (FprEN 14374:2018).
Figure 1.61. Dimensions of LVL. b = width (H=height), l = length, t = thickness. Arrow shows the grain direction of the surface veneer.
Bottom: Example of the angle α deviation from the right angle of a cross section of LVL (FprEN 14374:2018).
Figure 1.62. Left: Example of CE-mark label in LVL package, right: Example of CE-mark label in LVL product.
Moisture
LVL products are delivered from the factory at a moisture con-
LVL(MC)
tent 01, of
Table 1.11
8-10%, which is close to the MC of service class 1
end uses. This significantly reduces initial dimensional chang-
es due to moisture in structures if the members are protected
against weather exposure. LVL swells when its moisture con-
tent increases and shrinks when its moisture content decreas-
es 18.
Table 1.10. Example dimensional changes due to a 3% increase in moisture content (MC) %.
Density, kg/m3
Mean value ρmean 510 440 510 440
Characteristic value ρk 480 410 480 410
Density, kg/m3
Mean value ρmean 380 420 460
Characteristic value ρk 320 385 400
1.12.3 Emissions and product safety are needed to keep the impact sound level at low frequencies
low enough.
Phenolic adhesives that are cured at high temperature and Despite the challenges of low frequencies, real-life feed-
wood raw material both contain small quantities of free for- back from people living in well-designed wooden multi-storey
maldehyde. In Europe, the formaldehyde emissions of LVL buildings has been positive. The buildings are considered silent
products are tested according to the standards EN 717-1 and the room acoustics of wooden buildings is usually regard-
(chamber method) or EN ISO 12460-3 (gas analysis method). ed to be comfortable. One reason for this is that the surfaces of
The Class E1 requirement according to EN 14374 correspond- timber elements are less dense compared to steel or concrete,
ing to ≤ 0,1 ppm (EN 717-1) can be easily achieved with LVL which is beneficial for sound absorption.
products. LVL manufacturers commonly report significantly Robust perforated LVL panels can be used together with
lower formaldehyde emission levels of ≤ 0,03 ppm, and this mineral wool insulation installed in cavities behind the panels
limit stated in the voluntary certification of some construction for sound absorption, e.g., in sport halls and schools. The good
product associations to demonstrate the low emissions of their impact resistance of these panels is also advantageous, e.g., in
products. For example, the German Qualitätsgemeinschaft walls of halls for ball games.
Deutscher Fertigbau (QDF) für Holzwerkstoffe QDF-Positiv-
liste criteria states a formaldehyde limit of ≤ 0,03 ppm. 1.12.5 Fire safety
For volatile organic compounds (VOC) a European clas-
sification is currently under preparation, but for the present When wood burns, a layer of char forms on the wood surface.
different classification systems are used in different countries This char layer serves as protective thermal insulation, inhibit-
either voluntarily or based on legislation. For example, in Fin- ing further burning of the remaining wood cross section. This
land, LVL products are certified to show that they fulfil the M1 makes the behaviour of wooden structures in fire predictable
emission classification requirements of the Finnish Building and their resistance to fire can be calculated based on the char-
Information Foundation RTS for building materials. The M1 ring rates defined in EN 1995-1-2 (Eurocode 5). The one-di-
classification criteria set limit values for total volatile organic mensional charring rate β0 of LVL is 0,65 mm/min, and the
compounds (TVOC), formaldehyde, ammonia, carcinogens notional charring rate βn for beams and columns is 0,70 mm/
and sensory evaluation 19. min when the characteristic density is ≥ 480 kg/m3.
After curing at high temperature, the adhesive bond be- As LVL cross sections are typically thin, with a product
tween the LVL veneers becomes an inert polymer that does not thickness of max. 75 mm, they usually require additional pro-
dissolve or react with other materials in the surrounding envi- tection to achieve the required fire resistance. This is usually
ronment. It is safe and non-hazardous to humans and animals. achieved with gypsum plasterboard panelling directly onto the
Standard LVL products do not contain more than 0,1% LVL members, or onto LVL frame structures with cavities that
of any of the Substances of Very High Concern (SVHC) listed are left empty or filled with mineral wool insulation. Eurocode
in the Candidate List of the European Chemicals Agency, as 5 provides instructions for calculating resistance to fire.
these substances are not intentionally added to the products 20. The risk of flame spread is controlled by reaction to fire
Manufacturers continuously monitor the Candidate List for class classifications of construction products. The class for un-
updates. treated LVL is D-s2,d0, which is the same as solid wood, where
LVL does not contain anything classified as hazardous D is the combustibility class, s is smoke production and d is
waste, and has the following waste code in the consolidated burning droplets. The classification may be improved with fire
European Waste Catalogue: retardant treatments up to class B-s1,d0 for some structures,
- 17 02 01 Wood (Construction and Demolition Wastes) mainly in indoor applications.
After use at the end of its life cycle, LVL can be utilized e.g. For more information about fire safety, please see Chap-
for bioenergy production 3. ter 6.
resist seismic loads. LVL-C members, in particular, are not sen- some stains of adhesive from the production process. Some
sitive to cracking or brittle failures of connections, so they can manufacturers provide higher grade face veneers on special
yield and absorb more energy. Full utilization of these proper- request.
ties and use of LVL in higher dissipative classes for structures On the front side of the product a light-coloured mela-
in seismic design requires testing in cyclic loading according mine adhesive is used for scarf-jointing the surface veneers.
to EN 12512 for combinations of LVL panels and dowel type On the reverse side the scarf-joints are glued with the same
connections. dark brown phenol resin adhesive used in the glue lines be-
tween veneers. The distance between scarf-joints is usually 1.9
1.12.7 Visual properties of LVL surface m or 2.5 m depending on the production process. When LVL
members are used in visible applications, the designer specifies
Softwood LVL is produced from peeled conifer softwood which side of the member is the visible surface. This must also
veneers. An inherent characteristic of conifer wood is that be taken into consideration especially when the LVL is to be
branches are located in star-like clusters along the stem (knot cut into special shapes.
ringlets) and therefore there may be frequent knots in a peeled Standard LVL is delivered unsanded. It can, however, be
veneer. sanded to improve the visual appearance of the LVL surface by
LVL is mainly used as a structural load-bearing product so-called optical sanding which cleans and smoothens the sur-
for non-visible applications. Therefore, the sorting of veneers face by removing any dark glue stains and equalizing any local
in production is based mainly on the strength properties of the colour differences of the veneers. Another alternative sanding
veneers, not on their visual properties. During peeling, small treatment is calibration sanding, which can be carried out to
peeling cracks are formed in the veneer. These cracks may be- achieve more precise thickness tolerances for, e.g., LVL door
come visible due to swelling or shrinkage caused by moisture components. Unless some non-transparent coating is used on
variations or sanding. In addition, the scarf-joints of face ve- the surfaces, the calibrated sanding is not recommended for
neers can sometimes overlap, leaving the joint slightly open. visible applications, because it can sand through the surface
Due to moisture variations this may also occur later on the sur- veneers revealing the dark glue line especially in the thicker
face of sanded products. Other possible surface defects include range of the products 11. For more information on sanding
resin pockets, bark, and splits. Unsanded surfaces may have specifications, see subsection 1.7.1.
Metsä Wood
1 2 3 4
5 6 7
1.12.8 Surface coating of LVL exposure to sunlight; the colour and darkness of the coating;
and the level of exposure to moisture. General guidelines for
LVL can be painted with fully pigmented non-transparent coating LVL:
paints or stained with translucent stain or varnish. Due to • Apply the coating as soon as possible to prevent the effects
peeling cracks, the surface of LVL is more demanding to paint of UV radiation on the wood surface. If needed, sand the
than, e.g., solid wood. The cracks occur in the face veneers as surface prior to coating.
a result of contraction and swelling of the product. In indoor • Fill any voids, such as fallen knots, with substitute wood
conditions these cracks are not usually problematic, but in ap- compound.
plications exposed to weather, a pigmented covering paint is • Choose a primer with blue stain and mould protection.
required to protect the LVL. If the moisture nevertheless pen- • Round the edges of the member slightly to ensure adher-
etrates the LVL member, for example at the penetration points ence of an adequate thickness of coating also on the edg-
of connectors, at the edges or other similar areas, a thick paint es. Edge coating is especially important for all veneer-based
coating will tend to flake. This is especially evident on large wood products.
continuous surfaces. If surface cracks are not problematic for • Apply the coating agent to an adequate thickness. Apply the
the application in question, a light non-film-forming surface coating agent in at least two layers to ensure proper drying
treatment can be a suitable alternative. This may require more of the layers.
frequent maintenance, but the maintenance treatment is eas-
ier to perform. In all cases of coating LVL the application should be discussed
The durability of the coating depends on various factors: in detail with the coating agent manufacturer to ensure the
the base material and how it has been prepared; the level of suitability of the product for the application 21.
Figure 1.67. Structurally supported joint and self-supporting joints of LVL-C panels.
Eurofins
Figure 2.2. LVL-P joist floor structures. Figure 2.3. Joist hanger connections of floor joists.
LVL joist floors can be built on a wooden, steel or concrete blockings to a transverse tension flange board installed under
frame or walls. The 1:8 width to depth ratio of rigid LVL-P the floor joists, see Figure 2.6.
joists is ideally suited to the structural depths of floor con- Multi-span structures reduce deflection and enable longer
structions even with long spans. The recommended minimum maximum span lengths compared to single-span structures.
width of supports and LVL floor joists is 45 mm in order to However, if the floor joists are continuous between separate
provide proper support for decking panels. rooms more conservative span lengths are recommended as
Floor elements can be prefabricated off-site. This saves people are more sensitive to vibrations originating from spaces
time on the construction site, reduces waste, and improves external to the space that they are occupying.
quality as the elements can be made in dry factory conditions. Floor vibration is controlled in the design based on: 1)
An element floor usually has a slightly higher material con- deflection under point 1kN load when the lowest natural fre-
sumption as the edge joists that seal the elements at the sides quency f1 is higher than 8-9 Hz depending on country-specific
form double joists in the completed floor structure. design rules and 2) acceleration when f1 is 4,5-8 Hz. It is rec-
Floor deflections and vibrations are not only dependent on ommended to design joist floors to f1 > 8 Hz, because achiev-
the floor joists themselves, but on the decking on the joists and ing the requirements of the acceleration criterion requires the
on the blockings installed for transverse bracing perpendicular addition of considerable extra weight to the structure. For more
to the span of the floor. Due to the rigidity of typical decking information on floor vibration design, see subsection 4.3.7.
panels, a joist spacing of c/c 400 mm is recommended. The Residential floors within apartments can be light, with-
spacing should not exceed 600 mm to avoid excessive deflec- out acoustic or fire resistance requirements, in which case de-
tion of the panels in the perpendicular direction to the joists. flection under 1kN point load is the governing requirement.
It is also recommended to glue the decking panels to the joists Intermediate floors between apartments require protective
with polyurethane adhesive for best results. Depending on the cladding underneath the floor and extra mass on the top of
gluing conditions (on-site or off-site) and country-specific re- the floor, which usually make the fundamental natural fre-
quirements, at least half of the advantage of the composite ef- quency f1 > 8-9 Hz the most critical requirement. Figures 2.7
fect of gluing can be utilized in structural calculations. and 2.8 gives the maximum span lengths for different joist
Blockings between the joists reduce deflection under point sizes. Span lengths up to ~6 m can be achieved with normal
loads, but they need to be fixed well to the joists to give the floor thicknesses, but with the biggest joists over 8 m spans
desired transverse stiffness and to avoid creaking in the long are possible.
term. The best improvement can be achieved by fixing the
1. Floor joist
2. Rim board
3. Blockings at the centre of a span
4. Blockings at support
5. Trimmer beam
6. Staircase opening, trimmer connection
7. Joist hanger connection
Figure 2.5. LVL joist floor with decking from offsite produced elements.
Figure 2.6. Transverse blocking and tension board below the floor joists for reducing deflection under point load and improved performance
against floor vibration. Two blocking lines at 1m spacing at the centre of the span are recommended for L>4m span length. 1. Decking panel,
2. Floor joists, 3. Transverse tension board (C18, min 22x100) under the joists fixed with 2,8x75 nails to the joists and blockings. 4. Blocking.
63x500 9,0
7,1
51x400 7,5
5,4
45x360 6,9
4,8
75x300 6,8
Floor joist size
4,6
45x300 6,0
4,2
45x260 5,1
3,5
4,8 Advanced
45x240 3,3
75x200 4,7 Basic
LVL 02, Figure 2.7 3,2
45x200 4,0
2,9
0 1 2 3 4 5 6 7 8 9
Maximum span length [m]
Figure 2.7. Maximum span lengths of LVL 48 P floor joists for predesign of residential floors. 2,0 kN/m2 live load, 0,3 kN/m2 partition load and
0,6 kN/m2 self-weight. The basic option has c/c400 mm joist spacing, 22 mm chipboard decking without gluing and no transverse bracings. The
advanced option has transverse bracing, glued deck panel and 45x45 c/c 400mm cross batten underneath the joists. Lowest natural frequency
f1 > 8 Hz and maximum deflection under 1kN point load is 0,5 - 0,8 mm depending on the span length (FI NA requirement for EN1995-1-1).
63x500 7,9
7,5
51x400 6,6
6,2
45x360 6,1
5,7
75x300 5,9
Floor joist size
5,6
45x300 5,6
5,0
45x260 4,6
4,4
4,2 Advanced
45x240 4,2
75x200 4,2 Basic
4,1
45x200 3,7
3,5
0 1 2 3 4 5 6 7 8 9
Maximum span length [m]
Figure 2.8. Maximum span lengths of LVL 48 P floor joists for predesign of compartment floors. 2,0 kN/m2 live load, 0,3 kN/m2 partition load
and 1,6 kN/m2 self-weight including a 50 mm screed in addition to the LVL members. The basic option has c/c400 mm joist spacing, 22 mm
chipboard decking without gluing and no transverse bracings. In advanced option has transverse bracing, glued decking and 45x45 mm c/c
400 mm cross batten underneath the joists. Lowest natural frequency f1 > 8 Hz and maximum deflection under 1kN point load is 0,5 - 0,8 mm
depending on the span length (FI NA requirement for EN1995-1-1).
INTERMEDIATE FLOOR
COMPARTMENT FLOOR
Figure 2.9. LVL joist floor structures for ground floors, intermediate floors and compartment floors.
Figure 2.10. Instructions for nailing and screwing patterns for multiple-member LVL beams.
35 50 75x600
45x360
2x45x360
30 45
51x400
2x51x400
25 40 63x500
Characterastic load [kN/m]
20 35 75x600
15 30 2x45x360
2x51x400
10 25
5 20
0 15
2,0 2,5 3,0 3,5 4,0 4,5 5,0 5,5 6,0 6,5 7,0
10 Maximum span length [m]
5
0
2,0 2,5 3,0 3,5 4,0 4,5 5,0 5,5 6,0 6,5 7,0
Maximum span length [m]
30 45x300
40
75x200
45x200
25
35 75x300
45x240
20 2x45x200
45x260
Characterastic load [kN/m]
30 2x45x240
45x300
15 2x45x260
75x200
25
2x45x300
75x300
10
20 2x45x200
5 2x45x240
15 2x45x260
0 2x45x300
10
1,5 2,0 2,5 3,0 3,5 4,0 4,5
Maximum span length [m]
5
Figure 2.11. Span and capacity table of LVL 48 P main beams for predesign of floor structures. Calculations according to EN 1995-1-
1:2004+A1:2008 0 and its Finnish National Annex. The permanent load is 20% of the characteristic load kN/m2. The service class is 1 or 2 and
the consequences class is CC2. The rafter has lateral torsional buckling supports on the top surface with spacing ≤ 600 mm and the loads are
1,5 2,0 2,5 3,0 3,5 4,0
located at the lateral torsional buckling supports. The support length shall be calculated separately. The initial deflection winst ≤ L/400 and net
4,5
Maximum span length [m]
final deflection wnet,fin ≤ L/300. γM = 1.2. The table does not replace project-specific structural design. Double beams are calculated as separate
beams in lateral torsional buckling.
57
Floor main beams
LVL Handbook Europe
8,0
2020_LVL_02.indd 57 7,5 12.3.2020 10:34:00
2. LVL STRUCTURES IN FLOORS, WALLS, ROOFS AND IN SPECIAL APPLICATIONS
A B
Figure 2.12. Rim boards and rim beams in different structural solutions of timer frame structures A) Rim board as egde binder and transferring
loads from upper walls; B) Rim beams acting at the same time as a lintel over a window opening; C) Rim beam in compartment wall connection;
D) Rim beam integrated in exterior wall element; E) Rim beam integrated to compartment wall element.
Figure 2.14. Robust LVL-C floor panels can be used for bracing and
enable larger spacing between floor joists (A). Thicker LVL-C panels
can be used as floor slabs (B).
Stora Enso
Figure 2.13. Installation of a multiple-glued GLVL-C floor element. Figure 2.15. Installation of LVL floor elements.
69
63 2,1 2,4
[mm]
thickness
63
57 1,9 2,1
thickness
57
51 1,7 1,9
Panel
51
45 1,5 1,7
Panel
45
39 1,3 1,5
39 1,3
0,0 0,5 1,0 1,5 2,0 2,5 3,0 3,5 4,0 4,5
0,0 0,5 1,0 1,5 2,0
Maximum 2,5 [m]3,0
span length 3,5 4,0 4,5
Maximum span length [m]
133
[mm]
133
120 3,9 4,2
thickness
120 3,9
thickness
108 3,5
108
96 3,2 3,5
Panel
96
84 2,9 3,2
Panel
84 2,9
0,0 0,5 1,0 1,5 2,0 2,5 3,0 3,5 4,0 4,5
0,0
1,5 0,5
Maximum 2,0 1,0 2,5 [m]3,0
span length 3,5 4,0 4,5
Maximum
Figure 2.17. Span table of LVL 36 C and multiple-glued GLVL span
36 C panel [m]
length of floors supported from the ends. Calculations are
for predesign
according to EN 1995-1-1:2004+A1:2008 and its Finnish National Annex including floor vibrations. Permanent load is 0.4 kN/m² + panel’s
own weight, imposed load is 2.0 kN/m² (category A). Service class is 1 or 2 and consequences class CC2. The support length is ≥ 45 mm.
Instantaneous deflection winst ≤ L/400 and net final deflection wnet,fin ≤ L/300. γM = 1,2. Double span structures may have 0,1 – 0,3 m longer
maximum span lengths. Lowest natural frequency f1 > 8 Hz and maximum deflection under 1kN point load is 0,5 - 0,8 mm depending on the
span length.
Rib slab
(T- section)
Box slab
Figure 2.19. LVL rib slab, box slab and open box slab for floor constructions.
Figure 2.20. LVL open box slabs of intermediate floor constructions Figure 2.21. Installation of an LVL open box slab element.
in multi-storey buildings.
475 7,2
T-section with 25 mm LVL 36 C top slab and
Height of rib slab [mm]
0,0 1,0 2,0 3,0 4,0 5,0 6,0 7,0 8,0 9,0
Maximum span length [m]
0,0 1,0 2,0 3,0 4,0 5,0 6,0 7,0 8,0 9,0
Maximum span length [m]
Figure 2.22. Span table of LVL rib slab and open box slabs for predesign of compartment floors. Loads: 2,0 kN/m2 imposed load, 0,3 kN/m2
partition load and 1,6 kN/m2 self-weight including 50mm screed in addition to the LVL members. Transverse bracings are according to Figure
2.5. Lowest natural frequency f1 > 8 Hz and maximum deflection under 1kN point load is 0,5 - 0,8 mm depending on the span length (Finnish
National Annex requirements for EN1995-1-1).
Joist floor Rib slab floor Open box or box slab floor
Figure 2.23. Benefits of rib slab, open box and box slab structures compared to joist floors. Savings in floor height or longer spans.
Stressed-skin panels can be supported from the bottom as however, that top slab supported elements may have slightly
with normal joist floors. Another alternative is top slab sup- smaller maximum spans or may need thicker top panels and
port where the LVL-C panels are longer than the ribs which higher ribs to ensure adequate capacity of the support con-
are anchored to the top panel with screws, and only the canti- nection.
lever parts of the top panels are on the supports. This detailing Stressed-skin panels can be CE marked based on the Euro-
facilitates site installation of the floor elements and simplifies pean Technical Assessment (ETA) of the element supplier 22, 23.
the geometry of the supporting wall elements. It must be noted,
Figure 2.24. A) Top slab support detail of LVL rib slab or open box slab floor structure on exterior wall and intermediate wall. B) Rib slab or
open box slab floor structure supported from the bottom on exterior wall and intermediate wall. Alternative A) has simpler geometry of the
wall elements and less wood loaded perpendicular to grain.
Figure 2.25. LVL-P joist fixed on one side or both sides of old
timber beams as reinforcements.
Figure 2.26. A) LVL-P reinforcement members connected with inclined screws to the sides of an existing joists, B) LVL-C panel reinforcement
connected with inclined screws on the top of an existing joists.
Metsä Wood
duced offsite. Large and robust LVL-C roof panels are fast to
install, they work effectively in diaphragms, and they make it
possible to create roof overhang eaves that have a light appear-
ance without supporting beams.
Figure 2.29. A) Ridge roof with LVL-P rafters, B) Pitched roof with multiple span LVL-P rafters, C) End section of LVL-P rafter structure, D) Side
section of LVL-P rafter structure.
75x600
63x500
51x400
45x360
45x300
c/c 600
45x260
c/c 900
45x240
c/c 1200
45x200
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Maximum span length [m]
75x600
63x500
51x400
45x360
45x300
c/c 600
45x260
c/c 900
45x240
c/c 1200
45x200
0 1 2 3 4 5 6 7 8 910 11 12 13 14 15 16
Maximum span length [m]
Figure 2.30. Span table of single-span roof rafters at c/c 600, 900 and 1200 mm spacing for predesign. The spans lengths are given as
horizontal projections between the centres of supports, not the inclined length of a rafter.
Metsä Wood
Figure 2.31. Notched rafter end at the front roof overhang.
Figure 2.32. Connections of roof rafters at supports. Notch details to be verified in structural design.
63x500
45
75x600
40 LVL-P beams can be used as load-bearing main beams of
2x45x360
35 ridge roofs. As LVL beams are slim, LVL ridge beams are nor-
2x51x400
30
mally multiple-member
2x63x500 beams. Rafters can be supported on
25
the ridge beam or they can be connected to the sides of the
2x75x600
20
15 ridge beam with, e.g., joist hangers. Section 2.2.2 and Figure
10 2.10. have instructions for the nailing and screwing patterns
5 of multiple-member LVL beams. In the case of side connec-
0 tions, nailing or screwing between the LVL plies must be ver-
2,0 2,5 3,0 3,5 4,0 4,5 5,0 5,5 6,0 6,5 ified
7,0 case-specifically.
7,5 8,0
Maximum span length [m]
Figure 2.33. LVL-P ridge beam. Due to the height/thickness ratio
recommendations, LVL ridge beams are usually multiple-member
beams.
30 Calculations according to EN
45x300
1995-1-1:2004+A1:2008 and
25 75x200 its Finnish National Annex.
75x300 Permanent load is 20% of the
20 2x45x200 total characteristic load, the
15 2x45x240 remainder is snow load. Wind
2x45x260 load not included. Service class
10 2x45x300 1 or 2; consequences class
CC2. Beam has lateral torsional
5 buckling supports on the top
0 surface at ≤ 1200 mm spacing
2,0 2,5 3,0 3,5 4,0 4,5 and the loads are located at
the lateral torsional buckling
Maximum span length [m]
(LTB) supports. Double beams
are analysed as separate beams
in LTB. Support length shall be
calculated separately. Net final
LVL 48 P Ridge beams h = 360 - 600mm for roofs deflection wnet,fin ≤ L/300. γM = 1,2
60 Does not replace project-specific
45x360
55 structural design.
51x400
50
Floor main beams
load [kN/m]
63x500
45
8,0 75x600
40
7,5 2x45x360
35
7,0 2x51x400
30
6,5
Characterastic
2x63x500
25
6,0
2x75x600
length [m]
5,5
20
5,0
15
4,5
10
Span
4,0
3,5 5
3,0 0
2,5 2,0 2,5 3,0 3,5 4,0 4,5 5,0 5,5 6,0 6,5 7,0 7,5 8,0
2,0 Maximum span length [m]
1,5
0 10 20 30 40 50 60 70 80 90
Characteristic load [kN/m]
Figure 2.34. Capacity
45x200 table45x260
45x240 [kN/m] 45x300
of single-span
45x360 LVL 48 P ridge
51x400 63x500beams for predesign.
75x200 75x300 75x600
2.3.3 Purlins
In hall constructions LVL-P purlins are suitable secondary
structures for wooden or other main frames. Since LVL is avail-
able in long lengths, multiple span purlins are structurally effi-
cient and fast to install. However, purlins often also act as sup-
ports against lateral torsional buckling of the main frames or
beams setting capacity requirements for connection detailing.
Therefore single-span purlins supported with hangers from the
sides of rafter beams are recommended for 3-pin frames.
With the exception of flat roofs, purlins are biaxially load-
ed. As LVL purlins are thin, they need to be side supported in
the weaker direction.
Figure 2.35. Single-span and multiple span purlins for hall construction.
Figure 2.36. Support details of purlins installed vertically or perpendicular to the roof surface.
Figure 2.37. LVL purlins installed perpendicular to roof surface. Bracing beams are needed in purlin width directions. A) Steel sheeting transfers
lateral torsional buckling support forces to the bracing beam B) Wooden batten side supports at the mid span transfer loads perpendicular to
the span to bracing beams.
Metsä Wood
Figure 2.38. Roof elements with LVL purlins for hall construction can be used on any type of main frame.
Figure 2.39. Cross section and detailing of roof elements with LVL purlins.
Metsä Wood
Figure 2.40. Installation of roof elements with LVL purlins. Metsä Fibre pulp terminal, Vuosaari, Helsinki, Finland; Karisma, Lahti, Finland;
Skanssi, Turku, Finland.
Stressed skin rib panels and box panels The ceiling panel thickness depends on the fire resistance
For long-span roof elements, the composite action of glued requirements. Depending on the reaction to fire requirements
stressed-skin panels can be utilized between LVL-P ribs and of the project, LVL-C bottom flange panels can be left visible or
LVL-C panels. Rib panels can be used, e.g., for non-heated covered with gypsum board. A great advantage of box elements
shelters, and insulated box elements for hall constructions. De- is that suspension installations, such as ventilation channels,
pending on the snow loads and height of the elements, span can be freely fixed with screws anywhere on the ceiling sur-
lengths can be 10-20 metres. face because the LVL-C panels provide a solid base for axially
The structure is simpler than elements with purlins as no loaded connections.
secondary beams or battens are needed. However, if the rib An LVL box element roof can be realized as a cold roof or
spacing is large, up to 1250 mm, the top panel must be thick- warm roof solution, but the building physics design needs to
er because the main direction of the LVL-C is along the ribs, be done separately based on the project-specific indoor and
and the panels need to transfer the snow load to the ribs in the outdoor conditions. Stressed-skin panels can be CE marked
secondary direction. based on the ETA assessments of the element suppliers 22, 23.
LVL 36 C ROOFING PANEL, SINGLE SPAN PARALLEL LVL 36 C ROOFING PANEL, SINGLE SPAN PARALLEL
TO GRAIN, SURFACE STRUCTURE 0,20 kN/m2 TO GRAIN, SURFACE STRUCTURE 0,60 kN/m2
75 75
69 69
63 63
Panel thickness [mm]
57 57
51 51
45 45
39 39
33 sk = 0,65 kN/m2 33 sk = 0,65 kN/m2
0,0 0,5 1,0 1,5 2,0 2,5 3,0 3,5 4,0 4,5 5,0 0,0 0,5 1,0 1,5 2,0 2,5 3,0 3,5 4,0 4,5 5,0
Maximum span length [m] Maximum span length [m]
LVL 36 C ROOFING PANEL, MULTIPLE SPAN PARALLEL LVL 36 C ROOFING PANEL, MULTIPLE SPAN PARALLEL
TO GRAIN, SURFACE STRUCTURE 0,20 kN/m2 TO GRAIN, SURFACE STRUCTURE 0,60 kN/m2
75 75
69 69
63 63
Panel thickness [mm]
57 57
51 51
45 45
39 39
33 sk = 0,65 kN/m2 33 sk = 0,65 kN/m2
0,0 0,5 1,0 1,5 2,0 2,5 3,0 3,5 4,0 4,5 5,0 0,0 0,5 1,0 1,5 2,0 2,5 3,0 3,5 4,0 4,5 5,0
Maximum span length [m] Maximum span length [m]
Figure 2.42. Span tables of LVL-C roof panels for predesign. Span length parallel to grain of surface veneers. Calculations according to EN 1995-
1-1:2004+A1:2008 and its Finnish National Annex. Service class 2, Loads: g2 = own weight of surface structure, sk = snow load at ground level;
0,8∙x sk = snow load at roof level. Wind load wk = 0,4kN/m2, maintenance load qH = 0,4 kN/m2. Point loads are not taken into account. Deflection
limit wnet,fin ≤ L/100.
Figure 2.43. LVL-C roof panel orientations. Left: single span and multiple span parallel to grain of surface veneers. Right: single span and
multiple span perpendicular to grain of surface veneers.
LVL 36 C ROOFING PANEL, SINGLE SPAN PERPENDICULAR LVL 36 C ROOFING PANEL, SINGLE SPAN PERPENDICULAR
TO GRAIN, SURFACE STRUCTURE 0,20 kN/m2 TO GRAIN, SURFACE STRUCTURE 0,60 kN/m2
75 75
69 69
63 63
Panel thickness [mm]
57 57
51 51
45 45
39 39
33 sk = 0,65 kN/m2 33 sk = 0,65 kN/m2
0,0 0,2 0,4 0,6 0,8 1,0 1,2 1,4 1,6 1,8 2,0 2,2 0,0 0,2 0,4 0,6 0,8 1,0 1,2 1,4 1,6 1,8 2,0 2,2
Maximum span length [m] Maximum span length [m]
LVL 36 C ROOFING PANEL, MULTIPLE SPAN PERPENDICULAR LVL 36 C ROOFING PANEL, MULTIPLE SPAN PERPENDICULAR
TO GRAIN, SURFACE STRUCTURE 0,20 kN/m2 TO GRAIN, SURFACE STRUCTURE 0,60 kN/m2
45 45
Panel thickness [mm]
Panel thickness [mm]
39 39
33 33
sk = 0,65 kN/m2 sk = 0,65 kN/m2
27 sk = 1,5 kN/m2 27 sk = 1,5 kN/m2
0,0 0,2 0,4 0,6 0,8 1,0 1,2 1,4 1,6 1,8 2,0 2,2 0,0 0,2 0,4 0,6 0,8 1,0 1,2 1,4 1,6 1,8 2,0 2,2
Maximum span length [m] Maximum span length [m]
Figure 2.44. Span tables of multiple span LVL-C roof panels for predesign. Span length perpendicular to grain of surface veneers. Calculations
according to EN 1995-1-1:2004+A1:2008 and its Finnish National Annex. Service class 2, Loads: g2 = own weight of surface structure, sk = snow
load at ground level; 0,8∙x sk = snow load at roof level. Wind load wk = 0,4kN/m2, maintenance load qH = 0,4 kN/m2. Point loads are not taken
into account. Deflection limit wnet,fin ≤ L/100.
Metsä Wood
Figure 2.45. Installation of LVL-C roof panels. Figure 2.46. LVL-C roof panels.
Metsä Wood
Figure 2.47. Eaves and roof overhangs from LVL-C panels. Figure 2.48. Surface treated eave panel.
LVL 36 C ROOF OVERHANG PANEL, SPAN PARALLEL LVL 36 C ROOF OVERHANG PANEL, SPAN PERPENDICULAR
TO GRAIN OF SURFACE VENEERS, TO GRAIN OF SURFACE VENEERS,
SURFACE STRUCTURE 0,20 kN/m2 SURFACE STRUCTURE 0,20 kN/m2
75 75
69 69
63
57
57
51
51
45
45
39
39 sk = 0,65 kN/m2
33 sk = 0,65 kN/m2
27 sk = 1,5 kN/m2 33 sk = 1,5 kN/m2
sk = 2,75 kN/m2 sk = 2,75 kN/m2
24 27
0,0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1,0 1,1 1,2 1,3 1,4 1,5 1,6 1,7 1,8 0,0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1,0 1,1 1,2 1,3 1,4 1,5 1,6 1,7 1,8
Maximum span length L c [m] Maximum span length L c [m]
LVL 36 C ROOF OVERHANG PANEL, SPAN PARALLEL LVL 36 C ROOF OVERHANG PANEL, SPAN PERPENDICULAR
TO GRAIN OF SURFACE VENEERS, TO GRAIN OF SURFACE VENEERS,
SURFACE STRUCTURE 0,60 kN/m2 SURFACE STRUCTURE 0,60 kN/m2
75 75
69 69
63
Panel thickness [mm]
63
Panel thickness [mm]
57
57
51
51
45
45
39
39
33 sk = 0,65 kN/m2 sk = 0,65 kN/m2
27 sk = 1,5 kN/m2 33 sk = 1,5 kN/m2
sk = 2,75 kN/m2 sk = 2,75 kN/m2
24 27
0,0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1,0 1,1 1,2 1,3 1,4 1,5 1,6 1,7 1,8 0,0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1,0 1,1 1,2 1,3 1,4 1,5 1,6 1,7 1,8
Maximum span length L c [m] Maximum span length L c [m]
Figure 2.50. Span tables of LVL-C roof overhang panels for predesign. Left: Span length parallel to grain of surface veneers. Right: Span length
perpendicular to grain of surface veneer. Calculations according to EN 1995-1-1:2004+A1:2008 and its Finnish National Annex. Service class 2,
Loads: own weight of surface structure 0,20 or 0,60 kN/m2, sk = snow load on ground level 0,65, 1,5 or 2,75 kN/m2 and snow load on roof level
= 0,8∙x sk. Wind load wk = 0,4kN/m2; maintenance load qH = 0,4 kN/m2. Point loads are not taken into account. Deflection limit wnet,fin ≤ LC/100.
75
69
Panel thickness [mm]
63
57
Calculations according to
51 EN 1995-1-1:2004+A1:2008 and
its Finnish National Annex. Service
45 class 2, Loads: own weight of
surface structure 0,20 or
39 sk = 0,65 kN/m2
0,60 kN/m2, sk = snow load on
33 sk = 1,5 kN/m2 ground level 0,65, 1,5 or 2,75 kN/m2
sk = 2,75 kN/m2 and snow load on roof level
27 = 0,8∙x sk. Wind load
wk = 0,4kN/m2; maintenance load
0,0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9
qH = 0,4 kN/m2. Point loads are not
Maximum span length L c [m] taken into account. Deflection limit
at the corner overhang
wnet,fin ≤ (√2∙LC)/100.
LVL 36 C ROOF PANEL, CORNER OVERHANG, SURFACE STRUCTURE 0,60 kN/m2
75
69
Panel thickness [mm]
63
57
51
45
39 sk = 0,65 kN/m2
33 sk = 1,5 kN/m2
27
sk = 2,75 kN/m2
0,0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9
Maximum span length L c [m]
Figure 2.51. Span tables of LVL-C roof corner overhang panels for predesign. Span length LC is the same parallel and perpendicular to grain of
the surface veneer.
Metsä Wood
2.3.8 Roof renovation applications
In the renovation of residential buildings, nail plate truss roof
structures can be converted to room-in-the-roof attics by re-
inforcing or replacing the top and bottom cords with LVL-P
beams and removing the truss diagonals. LVL-C or plywood
connection plates are nailed or screwed to the corners to build
a new rigid frame structure. The economical span range for the
solution is 8-10 m depending on snow loads and thermal insu-
lation requirements. It is recommended to leave at least a 100
mm ventilation gap between the thermal insulation and roof
covering. The building physics design must, however, be done
project-specifically based on national requirements.
Metsä Wood
Figure 2.55. Conversion of nail plate truss structure to room-in-roof attic. Top and bottom cords have been reinforced with LVL-P beams and
truss diagonals removed. LVL-C or plywood is used for the frame corner connections.
Metsä Wood
Figure 2.59. Multiple-glued GLVL beam and post structures. Pro Nemus, Äänekoski, Finland.
2.4.3 Lintels
Lintels are one of the most common applications of LVL due to
its ideal dimensions and good strength and stiffness properties.
LVL’s good shear strength also means slim cross sections can
be used as lintels for window openings. This improves the en-
ergy efficiency of timber frame wall structures as there is more
space for thermal insulation and the cold bridging effect can
be reduced. For large openings, e.g. of garage doors and car-
ports, rigid LVL-P lintels reduce deflection and enable longer
span lengths.
Figure 2.61. Lintels for openings. Left: Straight and stiff LVL lintel for long-span garage door opening in timber frame wall; the straight, slim
member has minimal deflection. Right: Lintel for window opening inside the timber-frame wall. Key benefit is larger adaptability of opening
positioning with fewer load-bearing columns and more space for thermal insulation to improve the energy efficiency of the building envelope.
Characterastic load
35
30 2x63x500
2x75x600
25
20
15
10
5
2. LVL STRUCTURES
0 IN FLOORS, WALLS, ROOFS AND IN SPECIAL APPLICATIONS
2,0 2,5 3,0 3,5 4,0 4,5 5,0 5,5 6,0 6,5 7,0
Maximum span length [m]
8,0
45 75x600
7,5
40
7,0 2x45x360
6,5
35 2x51x400
6,0
30 2x63x500
Characterastic
5,5 2x75x600
25
5,0
20
4,5
15
4,0
3,5
10
3,0
5
2,5
2,0 0
1,5 2,0 2,5 3,0 3,5 4,0 4,5 5,0 5,5 6,0 6,5 7,0
0 10 20 30 Maximum
40 span
50 length60
[m] 70 80 90
Characteristic load [kN/m]
45x200 45x240 45x260
Figure 2.62. Capacities [kN/m] of45x300 45x360
single-span 51x400
LVL 63x500
48 P lintel 75x200
beams 75x300 75x600
for predesign.
LVL 48 P lintel beams h = 200 - 300mm
40 45x200
45x240
35 2.4.4 Integrated rim beams
Metsä Wood
45x260
Characterastic load [kN/m]
30
Integrated45x300
rim beams of timber frame walls are an essential and
75x200
25 common 75x300
application for LVL-P in Europe. They are installed
20 on top of 2x45x200
notched wall studs and carry the loads from the roof
15
or other structures
2x45x240 above to the studs. Since the integrated rim
beams transfer the loads, the floor joist, roof rafter or roof truss
2x45x260
10 spacing can be independent of the wall stud locations. Inte-
2x45x300
Figure 2.64. LVL-P integrated rim beam for timber frame wall structures. The bracing and load-bearing beam fits the thickness of timber-frame
walls. Key benefit: More possibilities for opening positioning with fewer load-bearing columns.
Figure 2.65. Capacities [kN/m] of single-span LVL 48 P integrated rim beams for predesign.
1. Integrated rim
beam notched
to the wall studs
transferring loads
from roof structures
to the load-bearing
studs.
3. Lintel carrying
loads above a
window opening.
Figure 2.66. Different alternatives of rim beams, lintels and ledger beams in wall structures.
Figure 2.67. Ledger beam supporting a canopy. Can be fixed to the timber frame wall studs with nail or screw connections or notched in to the
studs.
Metsä Wood
Figure 2.69. LVL-C boards for installing window and cladding elements in concrete buildings. The boards may be a visual element of the
architectural detailing.
of robust diaphragms can be critical, especially if the sections The greatest advantage of LVL-C bracing can be achieved
are narrow. Small fastener spacings increase the capacity of the in buildings which have large openings, e.g. for windows and
bracing. 75-150 mm is a suitable spacing for nailing and 100- doors, and thus a limited space for bracing. As Figure 2.74. il-
200 mm for screwing at the panel edges. At the centre studs of lustrates, to achieve the bracing capacity of a 27 mm LVL 36C
the panels the fastener spacing can be doubled without reduc- panel diaphragm, a 15 mm plywood panel diaphragm needs
ing the bracing capacity. The capacities are based on the panel, a 1,5 times wider space and a 9 mm plasterboard panel dia-
nail and stud combinations specified in Section 5.7. phragm needs a 2,5 times wider space.
Metsä Wood
Figure 2.71. Multiple-glued GLVL-C bracing column. Figure 2.72. LVL-C bracing panels.
Figure 2.73. Robust LVL-C bracing panel integrated into the wall structure.
Table 2.1. Characteristic racking load capacity Fi,V,Rk [kN] of LVL 36C panels nailed to LVL 48 P frame for predesign. The height of the LVL 36 C
panel is 3,0 m. The distance between the frame studs should not be more than the panel width.
Panel thickness [mm] Nail size d x Lmin [mm] Racking load capacity Fi,V,Rk [kN] of wall panel
24 2,1x50 3,6 5,4 7,2 8,9 13 17
27 2,5x60 4,8 7,2 9,5 11,5 17 23
33 2,8x70 5,7 8,6 11,5 14 21 28
45 3,1x90 6,8 10 13 17 25 34
Table 2.2. Characteristic racking load capacity Fi,V,Rk [kN] of LVL 36C panels nailed to C24 frame for predesign. The height of the LVL 36 C panel is
3,0 m. The distance between the frame studs should not be more than the panel width.
Panel thickness [mm] Nail size d x Lmin [mm] Racking load capacity Fi,V,Rk [kN] of wall panel
24 2,1x50 3,3 4,9 6,6 8,2 12 16
27 2,5x60 4,4 6,6 8,8 10,5 16 21
33 2,8x70 5,3 7,9 10,5 13 19 26
45 3,1x90 6,2 9,4 12 15 23 31
Metsä Wood
Figure 2.74. A wall diaphragm made with 27 mm LVL 36C panels, 15 mm plywood panels and 9 mm plasterboard panels have equal capacity.
LVL-C panels are the best solution where space for bracing is limited; however, the increased anchoring force needs to be taken into account in
connection detailing 24.
Figure 2.76. LVL-C panels as load-bearing structure together with LVL-P wall studs.
Figure 2.77. LVL-C panels as load-bearing structure of sandwich wall element structure.
Keminmaan puurakenne
Figure 2.78. LVL-P bottom chord for nail plate trusses for attic floors and R30 fire resistant roof structures.
Figure 2.79. LVL-P bottom chord provides stiffness as a floor member of an attic truss.
Metsä Wood
Figure 2.80. Left: LVL queen post trusses, Fupicsa production hall, Spain. Right: King post trusses, Manese Wassström, Tammisaari, Finland.
Stora Enso
In roof structures that have higher resistance to fire re-
quirements than the nail plate trusses can achieve, an LVL-P
bottom chord can be designed to carry the loads in the case
of fire as a beam, allowing the rest of the frame to be designed
to normal temperature requirements. Sides are protected with
stone wool insulation and the top edge is supported against
lateral torsional buckling.
When roof trusses are left visible, LVL-P king post or
queen post trusses with dowel type connections are an aes-
thetically pleasing solution. They are the most competitive op-
tion for 15-22 metre spans in cases where double-tapered solid
beams are uneconomical due to the roof slope and where the Figure 2.81. LVL roof trusses. Ydalir, France.
fire resistance requirement is R15. With larger trusses, e.g. for
sports halls, LVL can be used for all truss components, as di-
agonals combined with glulam chords, or in hybrid structures
with steel members. wards the pin points: the ridge and foundations. 3-pin frames
A 3-pin frame structure is a good solution when the interi- are the most competitive when the capacity of a single LVL-P
or height of a hall needs to be maximized. The built-up column rafter is adequate for the loads. In areas with high wind loads a
part of the frame has two LVL-C panels which are screwed and small roof angle is preferable and in areas with high snow load
glued together with wooden battens to form a box structure. a larger roof angle is preferable. The economical span range is
The intermediate battens are shorter than the LVL-C panels 10-30 m 19.
which creates a space for the rafter part of the 3-pin frame to
sit in. The battens should therefore have the same width as the
rafter. The rafter part of the 3-pin frame is a single LVL-P beam,
box beam or a glulam beam. The moment-rigid corner is a tim-
ber-to-timber connection made with bolt, screw or dowel cir-
cle without separate steel parts. The size and number of fasten-
ers in the corner connection depend on the size of the hall and
the loads. The crosswise veneers in LVL-C panels provide good
connection strength and prevent the risk of corner cracking 24.
To optimize material use, it is recommended to choose
Figure 2.82. Principle of diagonal sawing of LVL panels to produce
column panel and rafter sizes that fit the LVL panel width when single-tapered column and rafter members for 3-pin frames. The
they are diagonally cut as a single taper. The wider end is at dimensions should be chosen so that the full panel width can be
the moment-rigid corner connection and the narrower end to- used.
Figure 2.83. 3-pin portal frame structures for hall constructions with moment-resistant bolt circle corner connection.
Figure 2.84. A) LVL-C core panel of fire and safety doors, B) LVL-P battens to ensure the straightness of door and window frames.
Figure 2.87. Light wooden extension floor to an existing multi- Figure 2.89. LVL-C panel structures for lift shafts. Panel sizes are
storey building from volumetric elements. Poissy, France. suited for full building height or storey-high elements.
Figure 2.88. Tammelan Kruunu, Lisa Voigtländer & Sung Bok Song, 1st prize in ’City Above the City’ architectural design competition, Metsä
Wood 26.
Figure 2.90. LVL panels bent flatwise in parallel to grain direction (left) and perpendicular to grain direction (right).
Metsä Wood
Figure 2.91. LVL-C arches for the roof structure of a windmill, UK. The structure consists of several curved members.
2.5.8 Bridges
Robust LVL-C panels can be used as pedestrian bridge decks
supported on, e.g., glulam main beams. A 75 mm thick panel
has the required resistance against maintenance vehicle wheel
point loads and the bracing design of the bridge is straightfor-
ward using the solid panels. LVL-C panels allow up to 1.8 m
spacing of main beams and secondary beams or, if the main
beam spacing is small, secondary beams are not needed. On-
ly two panels are needed for a typical 3.6 m bridge width. The
deck needs a protective layer against water ingress on the top
side, but impregnation against decay is not needed.
Figure 2.92. 75mm LVL-C deck on a pedestrian bridge. Matinpuro Espoo, Finland.
Metsä Wood
Figure 2.94. Wooden I-joist with LVL flanges.
Metsä Wood
Metsä Wood
Figure 3.2. Loading LVL packages with an overhead crane for truck
transport.
Stora Enso
• Machining drawing reference, if relevant
• Packaging:
• Package size: max. weight or package height
• Wrapping: LVL supplier will propose a suitable wrapping
type, but customer specific requirements are taken into
consideration
• Required certification (PEFC or FSC on request)
• Terms of delivery:
• Road, rail or ship
• For rail transport: if there are several delivery addresses
for the same load, a loading order is required
• Availability of special surface qualities to be verified sepa-
rately. As a default, all products have the same visual grade.
Stora Enso
Figure 3.4. Storage of LVL packages
Stora Enso
Storage
LVL products must be stored under cover. When storing the
products temporarily on site, a solid, straight and dry platform
should be used. The height of ground bearers (skids) must be
at least 30 cm. To avoid twisting or cracking of the product,
the bearers between packs must be aligned vertically with the
ground bears. Bearers must be of a suitable size and number
and spread evenly.
The plastic wrapping of each pack must be cut open from
underneath to enable air circulation and moisture to evaporate
from the bundles. If the products are stored on site for longer
than one week, the bundles must be covered with an additional
protective covering. The condition of the product and the pro-
tective cover should be monitored regularly during storage 2, 27.
Figure 3.5. Handling of LVL beams
Handling and processing
LVL product packs can be unloaded on site with either a fork-
lift or a crane. When unloading with a forklift, follow the in-
structions given in the ‘Transport’ section. When unloading by
crane, approved webbing slings in proper condition and of the 3.3 PROTECTING THE STRUCTURE
appropriate strength class must be used. Do not use chains or
wires. If unloading is done manually, open the pack and carry DURING CONSTRUCTION
the products one-by-one. When cutting the banding, beware LVL structures must be protected on construction site to avoid
of the band ends. LVL products should not be dragged, pushed dimensional changes and surface mould due to exposure to
or dropped. moisture. Exposure to rain, splashing, and wetting from water
LVL is a light-weight material and is easy to shape, which convection from other structures must be avoided. In addition,
means notable time and cost savings in construction. LVL the designer must ensure in the detailed design that no water
products can be processed with conventional wood working pockets can form on the product in situ.
and power tools. There is no need for separate specialist tools. The product may be exposed to weather for a short period
Surface-treated products should be unloaded individually. If during installation. During the erection of the building, struc-
needed, a cellular plastic padding that does not stain, should tural LVL products and elements, which are structurally glued
be used under the webbing slings 2, 27. from LVL components, have good resistance to temporary ex-
Stora Enso
Figure 3.6. LVL structures can be processed with conventional wood working and power tools.
posure to water without damage or decay, provided that it is is at least 850°C; the combustion air and gases are well mixed;
ensured that the products can afterwards dry to the desired the retention time of the combustion gases in the furnace is
moisture content before the structures are closed. The integrity over two seconds; and the residual oxygen content of the flue
of the adhesive bonds is maintained according to the designat- gases is over 6%. In these combustion conditions the flue gases
ed service class throughout the expected life of the structure 28. are identical to the gases produced in burning untreated wood.
Further information on durability is presented in Chapter 7. LVL has a gross heating value of 19.4 MJ/kg
Composting requires the panels to be chipped and the long
composting process should also be taken into consideration.
3.4 HANDLING AFTER USE OF LVL The products can also be taken to landfill, although LVL de-
After use, LVL products are to be disposed of according to na- grades very slowly 3.
tional regulations and directives. In general, LVL products can LVL does not contain anything classified as hazardous waste
be reused or recycled as the preferred handling options; alter- and has the following waste code under the consolidated Eu-
natively, they can be composted or burned for energy recovery. ropean Waste Catalogue:
LVL can be safely burned when the combustion temperature - 17 02 01 Wood (Construction and Demolition Wastes)
The structural design of LVL structures is similar in purpose design in practice, but they can be used for code calibration
and principle to any load-bearing wood-based structure: to when the requirements are defined for building regulations and
verify that the structure fulfils its strength, serviceability and in comparison between different construction materials. They
other structural requirements. The basis of limit state design also have an important role e.g. in verification of the safety level
is to verify, based on the partial safety coefficient method, that of simplified design methods, such as the partial safety factor
the design value of an action Ed is smaller than the design val- method used in Eurocodes 29, 30.
ue of the resistance Rd of a section, structural member or con- The primary purpose of the design rules for load-bearing
nection: structures is to prevent failures that can lead to the risk of hu-
man injury and to ensure buildings function properly accord-
Ed ≤ Rd ing to their intended use.
Building codes such as the Eurocodes provide approved
where verification methods for verifying that the requirements are
Ed is the design value of the effect of actions such as internal fulfilled. They also present the loads and loading combinations
force, moment or a vector representing several internal that the structures must resist. The Eurocodes have common
forces or moments parts that are applied in all countries in connection with na-
Rd is the design value of the corresponding resistance. tional annexes. The annexes specify country-specific safety fac-
tors and parameters that take into consideration climate and
geology specific conditions. In addition, national choices are
4.1 BASIS OF STRUCTURAL DESIGN made in some alternative verification methods or adjustment
The Eurocode design system has been used for this purpose factors within them.
in Europe since 2010 with country-specific adjustment factors The general design requirements according to the Euroc-
that are defined in national annexes. For timber construction odes are 29, 30:
the important parts of the Eurocode system are: • Choice of structural system, structural design and construc-
tion work must be done by sufficiently qualified and experi-
EN 1990 Eurocode 0 Basis of structural design enced persons.
EN 1991 Eurocode 1 Actions on structures • Work must be supervised adequately and quality assurance
EN 1993 Eurocode 3 Design of steel structures implemented throughout the construction process from de-
EN 1995 Eurocode 5 Design of timber structures sign offices and factories to workshops and building sites.
• Structures must be made of construction materials and prod-
Eurocodes are limit state design codes that have two main limit ucts defined in the Eurocodes or their harmonized standards
states: Ultimate limit state (ULS) and serviceability limit state or in other harmonized technical specifications.
(SLS). In ultimate limit state (ULS) design the requirement is • Buildings must be adequately and regularly maintained
to verify that a structure has adequate safety against failure dur- throughout their design working life.
ing its whole designed service life. What is regarded adequate • Buildings must be used for the purpose for which they have
is defined in the relevant building regulations. Serviceability been designed
limit state (SLS) design evaluates whether the structure is fit
for purpose. In most cases, the building regulations do not de- The next generation of Eurocodes is currently under develop-
fine exact limit values for this evaluation. They provide recom- ment and scheduled to be ready for use after some years.
mendations for the maximum level of deformation and, e.g.,
human-induced floor vibration, but it is ultimately up to the 4.1.1 Actions (Loads)
contractor and the client to agree what is acceptable.
The risk of structural failure is depending on the proba- Actions are loads that cause, e.g., bending moment, shear and
bility that the expected actions are exceeded and the probabil- axial stresses and deformation of the structures. The determin-
ity that resistance of the structure is lower than calculated in ing actions and their combinations for each structure are evalu-
ULS design. Normally it can be presumed that the actions and ated based on the type, magnitude and duration of the action at
the resistance of a structure are random variables. When their the most unfavourable locations of the structure. Different load
distribution functions are known, it is possible to calculate the combinations are determined as load cases. The load combina-
risk of failure by the methods of probability theories. These tions consist of the main loads combined with reduced values
methods, however are usually too complicated for structural of other loads that may be acting at the same time. Reduction
Order of accumulated
Load-duration class Examples of loading Notes
duration of characteristic load
Permanent More than 10 years • Self-weight
• Permanently installed machinery
• Compartment walls in some
countries
in load is defined by multiplying the characteristic value of a The values of γG and γQ are set in national annexes, but in com-
load Qk by factor ψ0, ψ1 or ψ2 depending on the case: mon ULS design cases for unfavourable actions γG = 1,15 - 1,35
• Characteristic combination (ψ0Qk) is used for verification of and γQ = 1,5 - 1,6. In SLS design γG and γQ are 1,0.
ultimate limit states and for irreversible (permanent) defor-
mations of a structure in serviceability limit states 4.1.2 Consequence class, reliability class
• Frequent combination (ψ1Qk) is used for verification of ulti-
mate limit states involving accidental actions and for verifi- and factor KFI
cation of reversible deformations of a structure in servicea- For the purpose of reliability differentiation, consequences
bility limit states classes (CC1-CC3) may be established by considering the con-
• Quasi-permanent combination (ψ2Qk) is used for verifica- sequences of a failure or malfunction of the structure. Class
tion of ultimate limit states involving accidental actions and CC1 is used for low consequence for loss of human life or when
for verification of reversible serviceability limit states. Qua- economic, social or environmental consequences are small or
si-permanent values are also used for the calculation of long- negligible. For example, agricultural buildings and storage
term effects. houses may belong to class CC1. CC2 is a normal class with a
medium consequence level and is used as a default class for res-
Factor ψ2 can be concluded as a factor that converts short-term idential and office buildings. CC3 is used for buildings where
loads to permanent loads which have a similar long-term influ- the consequences of failure are high such as concert halls or
ence in the calculation of creep deformations. similar monumental structures.
Loads are defined in EN 1991 and the load combinations The requirements for different consequence classes are set
in EN 1990. The rules in these standards define how permanent in the associated reliability classes RC1-RC3. They include re-
actions and variable actions shall be taken into consideration in quirements for the level of reliability index β, the supervision
load combinations. The general equation of load combination of design and execution of the structures and resistance prop-
in the ultimate limit state is: erties of materials and products. In partial safety factor design
Ed=∑j≥1γG,j ∙ Gk,j + γQ,1Q k,1 +∑i≥1γQ,i ∙ ψ0,i ∙ Qk,i the reliability classes are taken into account by the KFI factor
𝐸𝐸𝐸𝐸d = ∑j≥1 𝛾𝛾𝛾𝛾G,j ∙ 𝐺𝐺𝐺𝐺k,j + 𝛾𝛾𝛾𝛾Q,1 𝑄𝑄𝑄𝑄k,1 + ∑i≥1 𝛾𝛾𝛾𝛾Q,i ∙ 𝜓𝜓𝜓𝜓0,i ∙ 𝑄𝑄𝑄𝑄k,i
(4.1) (4.1) of KFI are given in national annexes, but
for actions. The values
where according to the default values of EN1990, in RC1 the actions
γG,j = partial safety factor for permanent actions j; in equation (4.1) of the ULS are multiplied by KFI = 0,9, in RC2
Gk,j = characteristic value of permanent load j; by KFI = 1,0 and in RC3 by KFI = 1,1. KFI is not used in SLS.
γQ,1 = partial safety factor for decisive variable action 1;
Qk,1 = characteristic value of decisive variable load 1;
γQ,i = partial safety factor for variable action i;
Qk,I = characteristic value of variable load i; and
ψ0,i = reduction factor in load combination for variable action i.
Table 4.3. Values of kdef for different LVL types in different service classes.
where
ufin is the final deflection including creep deformation;
uinst,G is the instantaneous deflection due to permanent Mean values for stiffness properties are used in SLS design
actions; and are determined for short-term loading in service class 1
uinst,Q,1 is the instantaneous deflection due to the leading conditions. The influence of creep deformation is taken into
variable action; and account by the kdef factor. Characteristic 5% stiffness values are
uinst,Q,i is the instantaneous deflection due to accompanying used for stability calculation in ULS design.
variable actions.
4.2 STRUCTURAL PROPERTIES OF LVL
LVL-C has a higher kdef value when the loading causes defor-
mation in the flatwise direction due to rolling shear deforma- AND STRENGTH CLASSES
tion of the cross veneers similar to plywood. When LVL-C pan- The mechanical properties of structural LVL are determined
el is used as a component of stressed-skin panels, the kdef value according to the harmonized product standard EN 14374. The
in the longitudinal direction of the element is the same as for properties are assessed and their constancy of performance
LVL-P, because the loading causes mainly axial stresses on the verified according to the AVCP system 1 of the EU construc-
panel. tion product regulations. LVL suppliers declare their individ-
ual product properties in their Declaration of Performance
4.1.6 Design resistance and stiffness (DoPs).
In the future EN 14374 LVL product categories will be
Design resistance in ULS design is determined from the design introduced but in the meanwhile LVL industry has decided
values of the strength properties by modifying the character- to launch them as LVL strength classes. Information about
istic properties with partial safety factor γM and modification the strength classes can be found from the Laminated Veneer
factor kmod. Lumber (LVL) bulletin: New European strength classes 15,
𝑘𝑘𝑘𝑘 ∙𝑓𝑓𝑓𝑓 FprEN 14374 Annex B, and these are described in the following
𝑓𝑓𝑓𝑓d = 𝑘𝑘𝑘𝑘mod
f_d=(k_mod∙f_k)/γ_M
k
∙𝑓𝑓𝑓𝑓 (4.3) (EC5 2.17) (4.3) (EC5
subsections 2.17)
4.2.1 and 4.2.2.
𝑓𝑓𝑓𝑓d = mod 𝛾𝛾𝛾𝛾M k (4.3) (EC5 2.17)
𝛾𝛾𝛾𝛾
M
The bending and axial strength and stiffness properties of
LVL-C with crossband veneers having different layups (i.e. ve-
where neer orientation) may be calculated from one set of tests ac-
fk is the characteristic 5% value of a strength property; cording to FprEN 14374 Annex A by applying layup factors
kmod is the modification factor that takes into account the according to Annex C for thickness ranges specified in Annex
duration of load and service class, see Table 4.2; and A. LVL-C properties are defined with the assumption that the
γM is the partial safety factor of the material. cross veneers are zero layers.
Table 4.4. Symbols for strengths, moduli of elasticity and shear moduli in different directions of LVL 15
fm,0,edge, s and Em,0,edge fm,90,edge and Em,90, edge fm,0,flat, sflat,m and Em,0,flat
fc,90,flat and Ec,90,flat fv,edge and Gedge fv,0,flat, sflat,v and G0,flat
4.2.1 Strength classes for LVL-P without 4.2.2 Strength classes for LVL-C with
LVL 04, Table
crossband 4.5
veneers crossband veneers
For structural LVL made of spruce or pine the most relevant For structural LVL made of spruce or pine the most relevant
class is LVL 48 P for beam applications. LVL 32 P is suitable for class is LVL 36 C for load-bearing panels. LVL 25 C is suita-
stud applications where mechanical property requirements are ble for panel applications where mechanical property require-
lower. LVL 80 P is made from beech hardwood. The symbols of ments are lighter. LVL 70 C and LVL 75 C are made from beech
properties related to different directions of LVL are described hardwood. The symbols of properties related to different direc-
in Table 4.4. tions of LVL are described in Table 4.4.
Table 4.5. Strength classes for structural LVL-P without crossband veneers 15.
Strenght class
Shear Edgewise parallel to grain fv,0,edge,k N/mm² 3,2 3,2 4,2 4,8 8
strength Flatwise, parallel to grain fv,0,flat,k N/mm² 2,0 2,3 2,3 3,2 8
Parallel to grain E0,mean e
N/mm² 9 600 12 000 13 800 15 200 16 800
Modulus of Parallel to grain E0,k f N/mm² 8 000 10 000 11 600 12 600 14 900
elasticity Perpendicular to grain, edgewise Ec,90,edge,mean g
N/mm² MDV c
MDV c
430 430 470
Perpendicular to grain, edgewise Ec,90,edge,k h N/mm² MDV c MDV c 350 350 400
Edgewise, parallel to grain G0,edge,mean N/mm² 500 i
500 i
600 650 760
Strenght class
Parallel to grain
ft,0,k N/mm² 14 15 18 22 45 51
(length 3 000 mm)
Tension
strength
Perpendicular to grain,
ft,90,edge,k N/mm² 4 4 5 5 16 8
edgewise
Perpendicular to grain,
fc,90,flat,k N/mm² 1,0 1,0 2,2 2,2 16 16
flatwise (except pine)
Perpendicular to grain,
fc,90,flat,k,pine N/mm² MDV c MDV c 3,5 3,5 –
d
–
d
flatwise, pine
Edgewise parallel to grain fv,0,edge,k N/mm² 3,6 3,6 4,5 4,5 7,8 7,8
Shear
Flatwise, parallel to grain fv,0,flat,k N/mm² 1,1 1,1 1,3 1,3 3,8 3,8
strength
Flatwise, perpendicular to grain fv,90,flat,k N/mm² MDV c MDV c 0,6 0,6 MDV c MDV c
Parallel to grain, edgewise E0,edge,mean e N/mm² 6 700 7 200 10 000 10 500 11 800 13200
Parallel to grain, edgewise E0,edge,k f N/mm² 5 500 6 000 8 300 8 800 10 900 12200
𝜎𝜎𝜎𝜎 𝜎𝜎𝜎𝜎
k_m
𝑘𝑘𝑘𝑘m σ_(m,y,d)/f_(m,y,d)
m,y,d
+ 𝑓𝑓𝑓𝑓m,z,d ≤ 1 +σ_( (4.5) (EC5 6.12) (4.5) (EC5 6.12)
𝑓𝑓𝑓𝑓m,y,d m,z,d
𝜎𝜎𝜎𝜎m,y,d 𝜎𝜎𝜎𝜎
𝑘𝑘𝑘𝑘where
m 𝑓𝑓𝑓𝑓 + 𝑓𝑓𝑓𝑓m,z,d ≤ 1 (4.5) (EC5 6.12)
m,y,d m,z,d
σm,y,d and σm,z,d are the design bending stresses about the
principal axes as shown in Figure 4.3;
fm,y,d and fm,z,d are the corresponding design bending
strengths. In addition to kmod and γM, for
LVL in edgewise bending the design value
is dependent on the member height h. That
is taken into consideration by factor kh
𝑘𝑘𝑘𝑘 = �300�𝑠𝑠𝑠𝑠 ≤ 1,1 which is defined as (4.6) (EC5 3.3)
h ℎ
300 𝑠𝑠𝑠𝑠
𝑘𝑘𝑘𝑘k_h=(300/h)^s≤1,1
h = � ℎ � ≤ 1,1
(4.6) (EC5 3.3) (4.6) (EC5 3.3)
where
h is the member height in bending
s is the size effect parameter. For LVL-P and LVL-C strength
classes it is 0,15, but manufacturer-defined values are
also possible
𝜏𝜏𝜏𝜏d ≤120𝑓𝑓𝑓𝑓v, d
2020_LVL_04.indd (4.7) (EC5 6.13) 12.3.2020 10:49:23
4. STRUCTURAL DESIGN OF LVL STRUCTURES
4.3.2 Shear
For shear with a stress component parallel to the grain, see
Figure 4.5(a, b, d and e), and for shear with both stress compo-
nents perpendicular to the grain, see Figure 4.5 (c and f), the
following expression shall be satisfied:
A D
τ𝜏𝜏𝜏𝜏dd ≤≤fv,d
𝑓𝑓𝑓𝑓
v, d (4.7) (EC5 6.13) (4.7) (EC5 6.13)
d ≤ 𝑓𝑓𝑓𝑓v, d
𝜏𝜏𝜏𝜏where (4.7) (EC5 6.13)
τd is the design shear stress;
fv,d is the design shear strength for the actual condition.
𝑏𝑏𝑏𝑏
befef ==kcr
𝑘𝑘𝑘𝑘cr∙b∙ 𝑏𝑏𝑏𝑏 (4.8) (EC5 6.13a) (4.8) (EC5 6.13a)
2h l A
Vred V 1
l
Figure 4.6. Conditions at a support, for which the concentrated force F may be disregarded in the calculation of the shear force. In the case of
uniformly distributed loads, the shear force maybe reduced to the value which it has at a distance of the member height h from a support 31.
𝜎𝜎𝜎𝜎t,90,d ≤≤ft,90,d
σt,90,d (4.11)
𝑓𝑓𝑓𝑓t,90,d where (4.11)
σc,0,d is the design compression stress along the grain;
where Fc,0,d is the design compressive force;
σt,90,d is the design tensile stress perpendicular to the grain; A is cross-sectional area of the member; and
ft,90,d is the design tensile strength perpendicular to the grain. Fc,0,d is the design compression strength along the grain.
Although in LVL-P the veneers are parallel to the main For LVL strength classes a separate, 20% higher fc,0,k value
direction of the product, there is a small difference between has been defined for service class 1 in Table 4.5 and 4.6. The
the grain directions of the veneers. This makes the product less kmod factor of Eurocode 5 is similar in SC1 and SC2, but ma-
sensitive to cracking and the tension strength perpendicular to terial testing has shown compression strength values to be dif-
the grain edgewise of LVL-P ft,90,k = 0,5-0,8 N/mm2 is slightly ferent in SC1 and SC2. The same phenomenon is also known
higher than solid wood or glulam ft,90,k = 0,4-0,5 N/mm2. for other load-bearing wood products, but the strength values
𝐹𝐹𝐹𝐹
𝜎𝜎𝜎𝜎𝑐𝑐𝑐𝑐,0,𝑑𝑑𝑑𝑑 = c,0,d ≤ 𝑓𝑓𝑓𝑓c,0,d (4.12)
𝐹𝐹𝐹𝐹 𝐴𝐴𝐴𝐴
𝜎𝜎𝜎𝜎𝑐𝑐𝑐𝑐,0,𝑑𝑑𝑑𝑑 = c,0,d
𝐴𝐴𝐴𝐴
≤ 𝑓𝑓𝑓𝑓c,0,d (4.12)
Figure 4.9. Connection between a cord and diagonal strut or secondary beams connected to the lower part of a main beam create tension
stresses perpendicular to the grain. When the cord member is LVL-C and the connection is on the face side, equation 8.4 of EN1995-1-1 is not
applied, since LVL-C is not sensitive to splitting (Modified from EC5 figure 8.1).
for their standards classes are not defined directly from testing
according to EN 408.
In addition to the compression resistance itself, usually the
stability design (buckling) of the members is more critical, see
subsection 4.3.9.
4.3.6 Compression perpendicular Figure 4.10. Compression parallel to grain of surface veneers.
to the grain
Compression perpendicular to the grain design is defined in
EN1995-1-1, Clause 6.1.5. The following expression shall be
satisfied:
Where
σc,90,d is the design compressive stress in the contact area
perpendicular to the grain;
Fc,90,d is the design compressive load perpendicular to
the grain;
Figure 4.11. Compression perpendicular to the grain of surface
Aef is the effective contact area in compression veneers.
perpendicular to the grain;
fc,90,d is the design compressive strength perpendicular to
the grain; and
kc,90 is a factor taking into account the load configuration,
the possibility of splitting, and the degree of
compressive deformation.
Table 4.7. kc,90-values and increase of the actual contact length for the design of compression strength perpendicular to the grain of LVL 15.
Loading direction kc,90 -values Increase of the actual contact length a [mm]
35
30
LVL-C edgewise
25
Load F [kN]
LVL-C flatwise
20
15
10
0
0 0,2 0,4 0,6 0,8 1 1,2 1,4 1,6
Deformation [mm]
Figure 4.13. Example of LVL-C specimens in compression perpendicular to grain testing. In the flatwise direction LVL-C exhibits a ductile
behaviour. In the edgewise direction orientation LVL-C is much stronger and stiffer, but undergoes more brittle failure due to buckling of the
veneers.
LVL beam
Product Contact Contact Increase in actual Effective contact Compression kc,90 Characteristic bearing
type width b length l contact length parallel l1 area strength [-] capacity
[mm] [mm] and perpendicular l2 Aef = b∙(l+l1) + (b∙l2) fc,90,edge,k Fc,k = Aef ∙ kc,90 ∙ fc,90,edge,k
[mm] [mm2] [N/mm2] [kN]
Figure 4.14. Example of an LVL beam supported on an LVL sole plate. Effective contact area is increased by 15 mm in the edgewise direction
on the beam and separately by 60 mm (2x30 mm) in the length direction and 15mm in width direction on the sole plate.
m,0,d 𝑓𝑓𝑓𝑓
𝜎𝜎𝜎𝜎m,α,d ≤ 𝑓𝑓𝑓𝑓m,0,d (4.15)
σ_(m,α,d)≤f_(m,0,d)/(f_(m,0,d)/f_(m (4.15)
𝑓𝑓𝑓𝑓
∙ sin2 𝛼𝛼𝛼𝛼 + m,0,d ∙ sin𝛼𝛼𝛼𝛼 ∙ cos𝛼𝛼𝛼𝛼 + cos2 𝛼𝛼𝛼𝛼
𝑓𝑓𝑓𝑓m,90,d 𝑓𝑓𝑓𝑓v,d
where
σm,α,d is the bending stress at an angle α to the grain
When the characteristic bending strength fm,90,k is not de-
clared for the product, it should be assumed that fm,90,k = ft,90,k.
In Figure 4.16, the edgewise bending strength values of LVL 48
P and LVL 36 C at different angles are shown as graphics with
this assumption.
𝑓𝑓𝑓𝑓
t,0,d
𝜎𝜎𝜎𝜎t,α,d ≤ 𝑓𝑓𝑓𝑓t,0,d
σ_(t,α,d)≤f_(t,0,d)/(f_(t,0,d)/f_(t,90,d) (4.16)
∙ sin2 𝛼𝛼𝛼𝛼 + t,0,d ∙ sin𝛼𝛼𝛼𝛼 ∙ cos𝛼𝛼𝛼𝛼 + cos〖2 𝛼𝛼𝛼𝛼∙ (4.16)
𝑓𝑓𝑓𝑓
𝑓𝑓𝑓𝑓t,90,d 𝑓𝑓𝑓𝑓v,d
where
σt,α,d is the tensile stress at an angle α to the grain
In Figure 4.17, graphically illustrates tensile strength val-
ues of LVL 48 P and LVL 36 C at different angles.
More advanced instructions on determining the effect of
an angle to the grain on LVL strength properties may be found
in the manufacturers’ technical documentation.
Characteristic edgewise bending strength fm,edge,α,k Characteristic edgewise tension strength ft,edge,α,k
at an angle α to grain at an angle α to grain
45 35
40 LVL 48 P
30 LVL 48 P
35 LVL 36 C
LVL 36 C
25
30
[N/mm2]
[N/mm2]
25 20
20 15
15
10
10
5
5
0 0
0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90
Angle α between stress and grain [°] Angle α between stress and grain [°]
Figure 4.16. Characteristic edgewise bending strength at an angle Figure 4.17. Characteristic edgewise tension strength at an angle α
α to grain. to grain.
Table 4.8. Buckling coefficient kc of different LVL classes for different slenderness ratios λ.
kc [-]
λ [-]
LVL 32 P LVL 48 P LVL 25 C LVL 36 C
0,6 LVL 32 P
𝜎𝜎𝜎𝜎m,d ≤ 𝑘𝑘𝑘𝑘crit ∙ 𝑓𝑓𝑓𝑓m,d (4.38) (E
0,5 𝜎𝜎𝜎𝜎m,d ≤ 𝑘𝑘𝑘𝑘crit ∙ 𝑓𝑓𝑓𝑓m,d LVL 48 P (4.38) (E
LVL 36 C
0,4
LVL 25 C
0,3
fm,d is the design bending strength;
0,2 kcrit is a factor that takes into account the reduced bending strength
buckling.
0,1
In the case where a combination of moment My about the strong axis y and
0,0 force Nc exists, the stresses should satisfy the following expression
0 25 50 75 100 125 150 175 200 225 250
2
𝜎𝜎 𝜎𝜎
(𝑘𝑘 𝜎𝜎𝜎𝜎m,d
Slendernes ratio λ𝑓𝑓m,d )� +
2 c,0,d
𝜎𝜎𝜎𝜎 ≤1 (4.39) (E
�𝑘𝑘𝑘𝑘critm,d𝑓𝑓𝑓𝑓 2 + 𝑘𝑘c,zc,0,d
𝑘𝑘𝑘𝑘
𝑓𝑓c,0,d ≤ 1
𝑓𝑓𝑓𝑓
(4.39) (E
𝜎𝜎𝜎𝜎 m,d
crit 𝜎𝜎𝜎𝜎 c,0,d
c,z
�𝑘𝑘𝑘𝑘 m,d � + c,0,d ≤ 1 (4.39) (E
For critbeams
Figure 4.18. Buckling coefficient kc of different LVL classes for different slenderness ratioswith
𝑓𝑓𝑓𝑓m,d
λ. an initial lateral deviation from straightness within the limits
𝑘𝑘𝑘𝑘c,z 𝑓𝑓𝑓𝑓c,0,d
Section 10 of Eurocode 5, kcrit may be determined from expression:
1 when 𝜆𝜆 ≤ 0,75
4.3.9.3 Beams subjected to either bending or 1, when 𝜆𝜆𝜆𝜆rel,mrel,m ≤ 0,75
1,56 − 0,75𝜆𝜆rel,m when 0,75 < 𝜆𝜆rel,m ≤ 1,4
𝑘𝑘 = { 1,56 − 0,75𝜆𝜆𝜆𝜆
k_crit={█(1,when 1, rel,m
when, when ≤ 0,75
𝜆𝜆𝜆𝜆rel,m0,75 < 𝜆𝜆𝜆𝜆rel,m ≤ 1,4 (4.40) (E
combined bending and compression – Lateral 𝑘𝑘𝑘𝑘crit
crit = � 1 (4.40) (E
1,56 −𝜆𝜆2rel,m
0,75𝜆𝜆𝜆𝜆1rel,m , when
when 1,4
0,75 << 𝜆𝜆𝜆𝜆𝜆𝜆rel,m
rel,m ≤ 1,4
torsional buckling (LTB) 𝑘𝑘𝑘𝑘crit = � 2 , when 1,4 < 𝜆𝜆𝜆𝜆rel,m (4.40) (E
𝜆𝜆𝜆𝜆rel,m
1
2 , when 1,4 < 𝜆𝜆𝜆𝜆rel,m (4.40) (EC5 6.34)
Lateral torsional stability shall be verified both in the case The factor kcrit may𝜆𝜆𝜆𝜆rel,m be taken as 1,0 for a beam where lateral displacement o
edge is prevented throughout its length and where torsional rotation is preve
where only a moment My exists about the strong axis y and The factor k may be taken as 1,0 for a beam where lateral dis-
supports crit
where a combination of moment My and compressive force placement of its compressive edge is prevented throughout its
Nc exists. length
The and where
relative torsional rotation
slenderness for bending is prevented
should be at its supports
taken as
In the case where a bending moment M exists only on one The relative slenderness for bending should be taken as
axis, the stresses should satisfy the following expression: 𝑓𝑓
𝜆𝜆rel,m = √𝜎𝜎𝑓𝑓𝑓𝑓m,k (4.41) (E
m,k
𝜆𝜆𝜆𝜆rel,m = �𝜎𝜎𝜎𝜎m,crit
λ_(rel,m)=√(f_(m,k)/σ_(m,crit) ) (4.41) (EC5 6.30) (4.41) (E
𝑓𝑓𝑓𝑓m,crit
𝜎𝜎𝜎𝜎m,d ≤ 𝑘𝑘𝑘𝑘≤
σ_(m,d)
𝜎𝜎𝜎𝜎 ∙ 𝑓𝑓𝑓𝑓m,d∙f_(m,d)
k_crit (4.38) (EC5 6.33) (4.38)
𝜆𝜆𝜆𝜆rel,m =
(4.38) �(EC5
(EC5
m,k
6.33)
6.33) (4.41) (E
m,d ≤ 𝑘𝑘𝑘𝑘crit ∙ 𝑓𝑓𝑓𝑓m,d
crit
where is the
𝜎𝜎𝜎𝜎m,crit
σm,crit critical bending stress calculated according to the classica
where σ is the critical bending stress calculated according
stability,m,crit
using 5-percentile stiffness values. The critical bending stress shou
where to the classical theory of stability, using 5-percentile stiffness
σm,d is the design bending stress; values. The critical𝜋𝜋bending
𝑀𝑀y,crit √𝐸𝐸0,05 𝐼𝐼𝑧𝑧 𝐺𝐺0,05 stress
𝐼𝐼tor should be taken as:
𝜎𝜎m,crit = 𝑀𝑀𝑀𝑀𝑊𝑊 = 𝜋𝜋𝜋𝜋�𝐸𝐸𝐸𝐸0,05𝑙𝑙 𝐼𝐼𝐼𝐼𝑧𝑧𝑧𝑧𝑊𝑊𝐺𝐺𝐺𝐺0,05 𝐼𝐼𝐼𝐼tor (4.42) (E
fm,d is the design bending strength; and 𝜎𝜎𝜎𝜎m,crit = y,crit y = ef y (4.42) (E
𝑊𝑊𝑊𝑊y
𝑀𝑀𝑀𝑀y,crit 𝜋𝜋𝜋𝜋�𝐸𝐸𝐸𝐸0,05𝑙𝑙𝑙𝑙ef
𝐼𝐼𝐼𝐼𝑧𝑧𝑧𝑧𝑊𝑊𝑊𝑊
𝐺𝐺𝐺𝐺0,05 𝐼𝐼𝐼𝐼tor
kcrit is a factor that takes into account the reduced bending σ_(m,crit)=M_(y,crit)/W_y
𝜎𝜎𝜎𝜎m,crit = 𝑊𝑊𝑊𝑊 =
y
=(π√(E_ (4.42) (EC5 6.31) (4.42) (E
strength due to lateral buckling. where y 𝑙𝑙𝑙𝑙ef 𝑊𝑊𝑊𝑊y
Table 4.9. Effective length as a ratio of the span (Modified from EC5 Table 6.1.).
where
c is 0,58 for LVL 48 P and 0,67 for LVL 36 C;
b is the beam thickness [mm]; and
h is the beam height [mm].
4.3.10 Notches
The effects of stress concentrations at the notch shall be tak-
en into account in the strength verification of members. The
effect of stress concentrations may be disregarded in the fol- Figure 4.19. Installation of notched rafter beam.
lowing cases:
• Tension or compression parallel to the grain;
• Bending with tensile stresses at the notch, if the taper is not
steeper than 1:i = 1:10, that is i≥10, see Figure 4.20 a);
• Bending with compressive stresses at the notch, see Figure
4.20 b)
A B
1,5 𝑉𝑉𝑉𝑉
𝜏𝜏𝜏𝜏d =
𝑏𝑏𝑏𝑏 ∙ ℎ
d
≤ 𝑘𝑘𝑘𝑘v ∙ 𝑓𝑓𝑓𝑓v,d (4.46) (E
ef
Figure 4.20. Bending at a notch A) with tensile stresses at the notch B) with compressive stresses at the notch (EC5 Figure 6.10 and 6.11).
1,1 𝑖𝑖𝑖𝑖1,5
𝑘𝑘𝑘𝑘n �1 + �
𝑘𝑘𝑘𝑘v = min �1 ; √ℎ
� (4.47) (E
𝑥𝑥𝑥𝑥 1
√ℎ ��𝛼𝛼𝛼𝛼 (1 − 𝛼𝛼𝛼𝛼) + 0,8 ℎ �𝛼𝛼𝛼𝛼 − 𝛼𝛼𝛼𝛼2 �
When the taper at the notch at tensile side is steeper than 1:10, where
it can be located only at the support. i is the notch inclination, see Figure 4.20 a);
For beams with rectangular cross sections and where the h is the beam depth in mm;
grain runs essentially parallel to the length of the member, the x is the distance from line of action of the support reaction to
shear stresses at the notched support should be calculated us- the corner of the notch, in mm; and
ing the effective (reduced) depth hef , see Figure 4.20 b).
ℎ
It should be verified that α=h_ef/h (4.48)
𝛼𝛼𝛼𝛼 = ℎef
1,5 𝑉𝑉𝑉𝑉
𝜏𝜏𝜏𝜏dd =
1,5 𝑉𝑉𝑉𝑉d ≤ 𝑘𝑘𝑘𝑘 ∙ 𝑓𝑓𝑓𝑓
𝜏𝜏𝜏𝜏τ_d=(1,5
= 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 ∙∙ ℎℎefdV_d)/(b
≤ 𝑘𝑘𝑘𝑘vv ∙ 𝑓𝑓𝑓𝑓∙v,d
h_ef )≤k_v∙f_(v,d) (4.46) (EC5 6.60) (4.46)
kn is (EC5
(4.46) 4,5
(EC5for6.60)
LVL in general. Note: manufacturers provide
6.60)
v,d
ef
product-specific information on the kn values of their products,
where kv is a reduction factor defined as follows: especially where the advantages LVL-C are evident.
− For beams notched at the opposite side to the support, see
Figure 4.20 b), kv = 1,0.
− For beams notched on the same side as the support, see Fig-
ure 4.20 a)
1,1 𝑖𝑖𝑖𝑖1,5
𝑘𝑘𝑘𝑘 1,1 𝑖𝑖𝑖𝑖1,5
𝑘𝑘𝑘𝑘nn�1
�1 +
+ √ℎ ��
k_v=min(1
𝑘𝑘𝑘𝑘
𝑘𝑘𝑘𝑘vv =
= min �1;;;(k_n (1 + (1,1 i^1,5)/√h)
min �1 √ℎ
� (4.47) (EC5 6.62)
� (4.47)
(4.47) (EC5
(EC5 6.62)
6.62)
𝑥𝑥𝑥𝑥 1 2
��𝛼𝛼𝛼𝛼 (1
(1 −
− 𝛼𝛼𝛼𝛼)
𝛼𝛼𝛼𝛼) + 0,8 ℎ �
+ 0,8
𝑥𝑥𝑥𝑥 1 − 𝛼𝛼𝛼𝛼 2�
√ℎ ��𝛼𝛼𝛼𝛼
√ℎ �𝛼𝛼𝛼𝛼 − 𝛼𝛼𝛼𝛼 �
ℎ 𝛼𝛼𝛼𝛼
ℎ
𝛼𝛼𝛼𝛼
𝛼𝛼𝛼𝛼 =
ℎef
= ℎℎef (4.48)
(4.48) LVL Handbook Europe 131
1
𝑘𝑘m,α =
2 2
𝑓𝑓m,d 𝑓𝑓
√1 + ( tan𝛼𝛼) + ( m,d tan2 𝛼𝛼)
𝑏𝑏 ∙ 𝑓𝑓v,d 𝑓𝑓c,90,d
(4.53)32
Where
Figure 4.21. Single-tapered beam. α is the angle between the tapered edge and the grain direction of the beam (EC5 Figure 6.8).
1,5 for LVL − P Figure 4.21. Strength reduction factor km,α for tensile or compression stress para
𝑏𝑏 = { (4.54)32
1,0 for LVL − C tapered edge. Left LVL 48 P, right LVL 36 C.
It is not necessary to take km,α into consideration in the resistanceFor
LVL 48 P
against lateral torsional
high pitched roof beams
LVL 36 C(α ≥ ~10°) the maximum shear stress 𝜏𝜏v,max,d and te
buckling 1,0
of the beam equation (4.38). The effects of combined axial perpendicularbending
force
1,0
and to the grain 𝜎𝜎90,max,d shall be calculated at the point of the maximu
moment 0,9 shall be taken into account. When the tapered edge is under 0,9tension stress,∙ ktan𝛼𝛼𝛼𝛼
m,α is
moment
𝜏𝜏𝜏𝜏v,max,d =stress with the
𝜎𝜎𝜎𝜎m,0,max,d equations:
used to reduce
0,8 the bending strength in the equations
km,α,tension for combined stresses
0,8 = 𝜎𝜎𝜎𝜎 equation∙ tan𝛼𝛼𝛼𝛼
𝜏𝜏𝜏𝜏v,max,d (4.17) km,α,tension
Reduction factor km,α
m,0,max,d
and (4.18). When the tapered edge is underkm,α,compression
0,7 compression stress, 𝜎𝜎𝜎𝜎km,α is used
0,7 to reduce∙ tan
= 𝜎𝜎𝜎𝜎m,0,max,d the2 𝛼𝛼𝛼𝛼 km,α,compression
0,6 90,max,d
0,6 2
bending 0,5strength in the equations for combined stresses equations (4.19)
𝜎𝜎𝜎𝜎90,max,d
0,5
and
= (4.20). ∙ tan 𝛼𝛼𝛼𝛼
𝜎𝜎𝜎𝜎m,0,max,d
0,4 0,4
It is recommended
0,3 to have the tapered edge on the compressive side, 0,3 especially for LVL-P,
since the0,2tension perpendicular to grain strength ft,90,edge,k is low, which 0,2 can lead to cracks and
brittle failure.
0,1 LVL-C may be used for special shapes, also when the tapered 0,1 edge is on the
0,0
tensile side, 0
as its
5
f t,90,edge,k is higher due to the cross veneers and it behaves more ductile.
10 15 20 25 30 35 40 45
0,0
0 5 10 15 20 25 30 35 40 45
Figure 4.21 shows the km,α factors α as a function of the angle α. α
Figure 4.22. Strength reduction factor km,α for tensile or compression stress parallel to the tapered edge. Left LVL 48 P, right LVL 36 C.
It is recommended to have the tapered edge on the com- For double-tapered, curved and pitched camber beams de-
pressive side, especially for LVL-P, since the tension perpendic- sign instruction are given in Eurocode 5 clause 6.4.3. Addition-
ular to grain strength ft,90,edge,k is low, which can lead to cracks al information to the clause:
and brittle failure. LVL-C may be used for special shapes, also • Factor kr is 1,0 for LVL in the edgewise direction, as the shape
when the tapered edge is on the tensile side, as its ft,90,edge,k is of the beam is cut directly from a panel and no reduction due
higher due to the cross veneers and it behaves more ductile. to bending of the laminates during production is needed.
Figure 4.22 shows the km,α factors as a function of the angle α. • km,α is not used together with the equations for checking the
For high pitched roof beams (α ≥ ~10°) the maximum stresses at the apex point.
shear stress τv,max,d and tension perpendicular to the grain • It is not necessary to take kl into consideration in the resist-
σ90,max,d shall be calculated at the point of the maximum bend- ance against lateral torsional buckling of the beam (4.38).
ing moment stress with the equations:
τ_(v,max,d)=σ_(m,0,max,d)∙tanα
𝜏𝜏𝜏𝜏
v,max,d = 𝜎𝜎𝜎𝜎m,0,max,d ∙ tan𝛼𝛼𝛼𝛼 (4.55)30 (4.55)30
𝜏𝜏𝜏𝜏v,max,d = 𝜎𝜎𝜎𝜎m,0,max,d ∙ tan𝛼𝛼𝛼𝛼2 (4.55)30
𝜎𝜎𝜎𝜎90,max,d = 𝜎𝜎𝜎𝜎m,0,max,d ∙ tan 𝛼𝛼𝛼𝛼
σ_(90,max,d)=σ_(m,0,max,d)∙tan^2 α (4.56)30
2
𝜎𝜎𝜎𝜎90,max,d = 𝜎𝜎𝜎𝜎m,0,max,d ∙ tan 𝛼𝛼𝛼𝛼 (4.56)30 (4.56)30
Figure 4.23. Stress distributions in single and double-tapered beams. When the angle between loading and the grain is large (α ≥ 10°), shear
stress at the point of maximum bending moment stress may become more critical than the shear stress at the support 30.
Figure 4.24. Stresses at the tapered edge of a beam: bending stress σm,α at the direction of the edge, bending stress at the grain direction σ0,
shear stress τ = σ0∙tanα and stress perpendicular to the grain σ90 = σ0 ∙tan2α 30.
4.3.12 Holes Figure 4.25. Geometrical boundary conditions of holes in beams 33 (Kuva_97_1 h
beams)
Eurocode 5 does not provide instructions for designing holes Tension stress perpendicular to the grain in verified by the
in beams, but such instructions are presented in the non-con- Tension
equation stress perpendicular to the grain in verified by the equation
flicting complementary instructions (NCCI) for Eurocode 5. 𝐹𝐹𝐹𝐹
𝐹𝐹t,90,d ≤ 𝑓𝑓𝑓𝑓t,90,d
t,90,d
𝜎𝜎𝜎𝜎t,90,d =
The design method presented in this subsection is based on the 𝜎𝜎 t,90,d = 0,5 ∙ 𝑙𝑙𝑙𝑙t,90 ∙ 𝑏𝑏𝑏𝑏 ∙ 𝑘𝑘𝑘𝑘t,90
σ_(t,90,d)=F_(t,90,d)/(0,5 ≤ 𝑓𝑓t,90,d (4.57)
0,5 ∙ 𝑙𝑙t,90 ∙ 𝑏𝑏 ∙ 𝑘𝑘t,90
Austrian NCCI document ÖNORM B 1995-1-1:2015, annex (4.57)
F 33 and it can it be applied to holes in LVL beams in service where
class 1 and 2 conditions. LVL suppliers have in their technical where 1
𝑘𝑘𝑘𝑘 0,5
documentation also their own tailored instructions for design- t,90 = 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑡𝑡𝑡𝑡 � 450
k_(t,90)=min{█(1@(450/h)^0,5 )┤ (4.58)
� ℎ1�
ing holes in LVL beams with different boundary conditions. 0,5
𝑘𝑘t,90 = 𝑚𝑚𝑚𝑚𝑚𝑚 { 450
For all beams with holes the bending, shear and tension/ σt,90,d is the design ( ℎ ) value of tension stress perpendicular to the
compression resistance shall be verified at the location of the grain [N/mm2];
hole. When the diameter d of the hole is ≥ 50 mm or ≥ h/10, the 𝜎𝜎
Ft,90,d is the design value of tension stress perpendicular to the grain [N/m
t,90,d is the design value of tension force perpendicular to the
resistance against tension perpendicular to the grain shall be
𝐹𝐹 grain [N]; is the design value of tension force perpendicular to the grain [N];
t,90,d
verified by equation (4.57), shear stress concentration shall be lt,90 is the load distribution length [mm], see Figure 4.26;
verified by equation (4.62). The bending stress at the location of 𝑙𝑙b t,90 is the beam is the load distribution
thickness [mm]; length [mm], see Figure 4.26;
the hole shall be verified by equations (4.64) and (4.65) for rec- bft,90,d is the design is thevalue
beam ofthickness
tension strength
[mm]; perpendicular to
tangular holes or (4.71) for round holes. The corners of rectan- 𝑓𝑓 the grain [N/mm 2]; and
is the design value of tension strength perpendicular to the grain [N
t,90,d
gular holes shall be a rounding radius r ≥ 15 mm. The bound- h is the beam height [mm].
ary conditions of the geometry are specified in Figure 4.25.
The verification of the resistance against tension perpen- The tension perpendicular to the grain force Ft,90,d depends on
dicular to the grain stresses can be the most critical condition the shear force Vd and bending moment Md at the edge of the
to fulfil in the design of holes in LVL-P beams. LVL-C beams, hole: 𝑉𝑉𝑉𝑉 ∙ℎ ℎ 2 𝑀𝑀𝑀𝑀
𝐹𝐹𝐹𝐹 = d d ∙ �3 − � d � � + 0,008 ∙ d
t,90,d 4∙ℎ ℎ ℎr
on the other hand, offer a significant advantage for beams with 𝑉𝑉𝑉𝑉 ∙ℎ
d d d ℎ 2
d 𝑀𝑀𝑀𝑀
holes, as the cross veneers act as reinforcement around the t,90,d = 4∙ℎ ∙ �3 − � ℎ � � + 0,008 ∙ ℎ
𝐹𝐹𝐹𝐹
F_(t,90,d)=(V_d∙h_d)/(4∙h)∙[3(4.59) r
holes preventing cracking due to tension stresses perpendic- where
ular to the grain. Their resistance is therefore superior and the h_r={█(min(h_ro;h_ru
min(ℎro ; )ℎrufor) for
rectang
rectangular holes (4.60)
larger hole size limit for reinforced holes specified in the Aus- ℎr = � min(ℎro ; ℎru ) for rectangular holes
trian NCCI document may be applied to LVL-C beams. min(ℎro + 0,15 ∙ 𝑑𝑑𝑑𝑑; ℎro + 0,15 ∙ 𝑑𝑑𝑑𝑑) for round holes
ℎr = �
min(ℎro + 0,15 ∙ 𝑑𝑑𝑑𝑑; ℎro + 0,15 ∙ 𝑑𝑑𝑑𝑑) for round holes
1,5 ∙ 𝑉𝑉𝑉𝑉
𝜏𝜏𝜏𝜏d = 𝑘𝑘𝑘𝑘τ ∙ (ℎ−ℎd ) ≤ 𝑓𝑓𝑓𝑓v,d
𝑏𝑏𝑏𝑏 ∙ d
1,5 ∙ 𝑉𝑉𝑉𝑉
𝜏𝜏𝜏𝜏d = 𝑘𝑘𝑘𝑘τ ∙ (ℎ−ℎd ) ≤ 𝑓𝑓𝑓𝑓v,d
𝑏𝑏𝑏𝑏 ∙ d
𝑎𝑎𝑎𝑎 ℎ 0,2
𝑘𝑘𝑘𝑘τ = 1,85 ∙ �1 + ℎ� ∙ � ℎd �
𝑎𝑎𝑎𝑎 ℎ 0,2
𝑘𝑘𝑘𝑘τ = 1,85 ∙ �1 + ℎ� ∙ � ℎd �
Figure 4.26. Tension perpendicular to the grain stresses at the hole edges. (1) Risk of cracks
due to the tension in perpendicular to grain 33. (Kuva_97_2 f t90 in rectangular hole,
Kuva_97_3 f t 90 in round hole)
The tension perpendicular to the grain force Ft,90,d depends on the shear force Vd and
bending moment Md at the edge of the hole:
𝑉𝑉 ∙ℎ ℎ 2 𝑀𝑀
𝐹𝐹t,90,d = 4∙ℎ d d
∙ [3 − ( ℎd ) ] + 0,008 ∙ ℎ d (4.59)
𝑉𝑉𝑉𝑉
𝑉𝑉𝑉𝑉d ∙ℎ
d ∙ℎd
ℎ 2 2 𝑀𝑀𝑀𝑀r
t,90,d = 4∙ℎ
𝐹𝐹𝐹𝐹t,90,d d ∙
�3 − �ℎℎdd � � + 0,008 ∙ 𝑀𝑀𝑀𝑀ℎ dd (4.59)
4∙ℎ ℎ ℎrr
where 𝑀𝑀𝑀𝑀d 𝑀𝑀𝑀𝑀o,d
𝑀𝑀𝑀𝑀d + 𝑀𝑀𝑀𝑀o,d
𝑊𝑊𝑊𝑊 n + 𝑀𝑀𝑀𝑀 𝑊𝑊𝑊𝑊o,d
min(ℎro ; ℎru ) for rectangular holes
𝑀𝑀𝑀𝑀d
𝑀𝑀𝑀𝑀
𝑊𝑊𝑊𝑊dn + 𝑀𝑀𝑀𝑀
𝑀𝑀𝑀𝑀d𝑓𝑓𝑓𝑓+
𝑊𝑊𝑊𝑊
𝑀𝑀𝑀𝑀
o
o,do ≤≤ 11
𝑊𝑊𝑊𝑊 𝑊𝑊𝑊𝑊o,d
𝑊𝑊𝑊𝑊n m,d
n𝑓𝑓𝑓𝑓+m,d
𝑊𝑊𝑊𝑊oo ≤≤ 1
1
ℎ r = {4.26. Tension )
𝑊𝑊𝑊𝑊n𝑓𝑓𝑓𝑓 𝑊𝑊𝑊𝑊o
≤ 1 due (4.60)
Figure ro ; ℎruru for rectangular
perpendicular holes at the hole edges.
to the grain stresses (1) Risk
min(ℎ ro 𝑓𝑓𝑓𝑓m,d of cracks
m,d to the tension in perpendicular to grain 33.
𝑀𝑀𝑀𝑀d𝑓𝑓𝑓𝑓m,d𝑀𝑀𝑀𝑀u,d
ℎrr = � min(ℎ ro + 0,15 ∙ 𝑑𝑑; ℎ ro + 0,15 ∙ 𝑑𝑑) for round holes 𝑀𝑀𝑀𝑀d + 𝑀𝑀𝑀𝑀u,d
𝑊𝑊𝑊𝑊 n + 𝑀𝑀𝑀𝑀 𝑊𝑊𝑊𝑊u,d (4.60)
𝑀𝑀𝑀𝑀d
𝑀𝑀𝑀𝑀
𝑊𝑊𝑊𝑊dn + 𝑀𝑀𝑀𝑀 𝑊𝑊𝑊𝑊
u
u,du ≤
≤ 11
min(ℎro ro + 0,15 ∙ 𝑑𝑑𝑑𝑑; ℎro ro + 0,15 ∙ 𝑑𝑑𝑑𝑑) for round holes
𝑀𝑀𝑀𝑀d𝑓𝑓𝑓𝑓+
𝑊𝑊𝑊𝑊 𝑀𝑀𝑀𝑀
𝑊𝑊𝑊𝑊u,d
holes h ≤=1 ≤
𝑊𝑊𝑊𝑊n m,d
n𝑓𝑓𝑓𝑓+ 𝑊𝑊𝑊𝑊u
hdd is the is height of the of holethefor rectangular holes. For For round (M_d/W_n
round ≤
1 + M_(u,d)/W_u
10,7d )/f_(m,d) ≤1 (4.65)
u
the height hole for rectangular holes. 𝑊𝑊𝑊𝑊n𝑓𝑓𝑓𝑓 m,d
𝑓𝑓𝑓𝑓m,d
m,d
𝑊𝑊𝑊𝑊u d may be
used in the equation
holes hd = 0,7d may be used in the equation (4.59). (4.59). where 𝑓𝑓𝑓𝑓m,d
where
where
where
where
Load distribution length l is where𝑏𝑏𝑏𝑏 ∙ �ℎ22−ℎd22�
Load distribution length lt,90 t,90
is 𝑊𝑊𝑊𝑊
𝑊𝑊𝑊𝑊nn =
𝑏𝑏𝑏𝑏 ∙ �ℎ2 −ℎ2d �
= 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 ∙∙ �ℎ
�ℎ66 2−ℎ
2−ℎ
2�
2�
d
𝑊𝑊𝑊𝑊
W_n=(b
𝑊𝑊𝑊𝑊 =
n = 𝑏𝑏𝑏𝑏 ∙ �ℎ ∙ 6(h^2-h_d^2
−ℎdd � ))/6 (4.66)
0,5 ∙ (ℎd + ℎ) for rectangular holes 𝑊𝑊𝑊𝑊nn = 6
𝐴𝐴𝐴𝐴6 𝑎𝑎𝑎𝑎
𝑙𝑙l_(t,90)=
= { {█(0,5∙(h_d+h
0,5 ∙ (ℎdd + ℎ) for rectangular holes
(4.61) 𝑀𝑀𝑀𝑀
𝑀𝑀𝑀𝑀o,d = 𝐴𝐴𝐴𝐴o
o ∙ 𝑉𝑉𝑉𝑉 (4.61)
d 2 ∙ 𝑎𝑎𝑎𝑎
o,d = 𝐴𝐴𝐴𝐴 o o ∙ 𝑉𝑉𝑉𝑉d ∙ 𝑎𝑎𝑎𝑎
t,90
u𝐴𝐴𝐴𝐴
𝐴𝐴𝐴𝐴+𝐴𝐴𝐴𝐴
0,35 ∙ 𝑑𝑑 + 0,5 ∙ ℎ for round holes M_(o,d)=A_o/(A_u+A_o
𝑀𝑀𝑀𝑀
𝑀𝑀𝑀𝑀 o,d = o o ∙ 𝑉𝑉𝑉𝑉 ∙ 𝑎𝑎𝑎𝑎
= 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴uu𝐴𝐴𝐴𝐴+𝐴𝐴𝐴𝐴
+𝐴𝐴𝐴𝐴
o o ∙ 𝑉𝑉𝑉𝑉d ∙ 2𝑎𝑎𝑎𝑎2
d(4.61)
)∙V_d∙a/2 (4.67)
𝑙𝑙𝑙𝑙t,90
t,90 = � 𝑀𝑀𝑀𝑀o,d o,d = +𝐴𝐴𝐴𝐴
u𝐴𝐴𝐴𝐴 o ∙ 𝑉𝑉𝑉𝑉d ∙ 𝑎𝑎𝑎𝑎2
𝐴𝐴𝐴𝐴u +𝐴𝐴𝐴𝐴 2
0,35 ∙ 𝑑𝑑𝑑𝑑 + 0,5 ∙ ℎ for round holes 𝑀𝑀𝑀𝑀 = u o
𝐴𝐴𝐴𝐴u ∙ 𝑉𝑉𝑉𝑉d ∙ 𝑎𝑎𝑎𝑎
Verification
Verification of of shear
shearstress stress concentration
concentration at the
at the holehole
edgeedge
shall shall fulfil
𝑀𝑀𝑀𝑀u,d u,dthe= 𝐴𝐴𝐴𝐴condition:
M_(u,d)=A_u/(A_u+A_o
𝑀𝑀𝑀𝑀 = u𝐴𝐴𝐴𝐴 u o ∙ 𝑉𝑉𝑉𝑉d ∙ 𝑎𝑎𝑎𝑎
𝐴𝐴𝐴𝐴+𝐴𝐴𝐴𝐴
u o ∙ 𝑉𝑉𝑉𝑉 ∙ 𝑎𝑎𝑎𝑎
2 )∙V_d∙a/2 (4.68)
𝑀𝑀𝑀𝑀u,d u,d = 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴uu𝐴𝐴𝐴𝐴+𝐴𝐴𝐴𝐴
+𝐴𝐴𝐴𝐴
u o ∙ 𝑉𝑉𝑉𝑉d d ∙∙ 2
2
𝑎𝑎𝑎𝑎
fulfil the condition: 𝑀𝑀𝑀𝑀 u,d = u +𝐴𝐴𝐴𝐴
𝐴𝐴𝐴𝐴u +𝐴𝐴𝐴𝐴o o ∙ 𝑉𝑉𝑉𝑉
d 2 2
1,5 ∙ 𝑉𝑉d 2 𝑏𝑏𝑏𝑏 ∙ ℎro
𝜏𝜏d = 𝑘𝑘τ ∙ 𝑏𝑏 ∙ (ℎ−ℎ ) ≤ 𝑓𝑓v,d 𝐴𝐴𝐴𝐴
𝐴𝐴𝐴𝐴oo = = 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 ∙∙ ℎℎro and(4.62)
and 𝑊𝑊𝑊𝑊 o = 𝑏𝑏𝑏𝑏 ∙ ℎ2ro 2
A_o=b∙h_ro ro andand W_o=(b
𝑊𝑊𝑊𝑊 o= 2∙ h_ro^2)/6
= 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 ∙∙66ℎℎro (4.69)
1,5 ∙∙ 𝑉𝑉𝑉𝑉
1,5 𝑉𝑉𝑉𝑉dd
𝐴𝐴𝐴𝐴 = 𝑏𝑏𝑏𝑏
𝐴𝐴𝐴𝐴oo = 𝑏𝑏𝑏𝑏 ∙ ℎro ∙ ℎ 𝑊𝑊𝑊𝑊 ro
ro and 𝑊𝑊𝑊𝑊oo = 𝑏𝑏𝑏𝑏 ∙66ℎro 2
𝜏𝜏𝜏𝜏dd = 𝑘𝑘𝑘𝑘ττ ∙ (ℎ−ℎ
τ_d=k_τ∙(1,5 ∙ d
V_d)/(b
)
≤ 𝑓𝑓𝑓𝑓v,d
v,d∙ (h-h_d ) )≤f_(v,d) (4.62) 𝐴𝐴𝐴𝐴 = 𝑏𝑏𝑏𝑏 ∙ ℎ and (4.62)
𝑊𝑊𝑊𝑊 =
𝑏𝑏𝑏𝑏 ∙
𝑏𝑏𝑏𝑏 ∙ (ℎ−ℎd d) o ro o 2
Where A_u=b∙h_ru
𝐴𝐴𝐴𝐴 and𝑊𝑊𝑊𝑊 W_u=(b
𝑏𝑏𝑏𝑏 ∙ 6ℎru
∙ h_ru^2)/6 (4.70)
𝐴𝐴𝐴𝐴uu = = 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 ∙∙ ℎℎru and u = 𝑏𝑏𝑏𝑏 ∙ ℎ2ru 2
ru and 𝑊𝑊𝑊𝑊 u = 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 ∙∙66ℎℎru
2
Where 𝐴𝐴𝐴𝐴
𝐴𝐴𝐴𝐴 u == 𝑏𝑏𝑏𝑏
𝑏𝑏𝑏𝑏 ∙
∙ ℎℎ ru and
and 𝑊𝑊𝑊𝑊
𝑊𝑊𝑊𝑊 u =
= 𝑏𝑏𝑏𝑏 ∙ 6 ℎru
2
𝑎𝑎 ℎ 0,2 u
𝐴𝐴𝐴𝐴u = 𝑏𝑏𝑏𝑏 ∙ ℎru and 𝑊𝑊𝑊𝑊u = 66 ru u ru
𝑘𝑘τ = 1,85 ∙ (1 + ℎ) ∙ ( ℎd ) Wo and Wu is the (4.63)
effective section modulus of the beam at
𝑎𝑎𝑎𝑎 ℎ 0,2
0,2
𝑎𝑎𝑎𝑎 ℎd
k_τ=1,85∙(1+a/h)∙(h_d/h)^0,2(4.63)
𝑘𝑘𝑘𝑘 τ
τ = 1,85 ∙ �1 +
ℎ
� ∙ �
ℎ
d�
the location
(4.63) of a hole [mm3]
ℎ ℎ
𝜏𝜏d is the design value of shear stress; fm,d is the edgewise bending strength [N/mm2]
τdτ is the design
𝑘𝑘 is the value
factorof shear stress; maximum shear stress due to stress concentration;
to determine
kτ is the factor to determine maximum shear stress due to Bending stress at the location of a round hole is verified by the
stress concentration; equations:
𝑀𝑀𝑀𝑀d
𝑀𝑀𝑀𝑀d
𝑊𝑊𝑊𝑊
n ≤ 1
𝑀𝑀𝑀𝑀d n
a is the length of a hole [mm], for round holes a = hd; and 𝑀𝑀𝑀𝑀
𝑊𝑊𝑊𝑊d
𝑓𝑓𝑓𝑓𝑊𝑊𝑊𝑊
𝑀𝑀𝑀𝑀 d ≤ 1
m,d
𝑓𝑓𝑓𝑓𝑊𝑊𝑊𝑊
n
≤1
n ≤ 1
n
f(v,d) is the design value of edgewise shear strength (M_d/W_n
𝑓𝑓𝑓𝑓 m,d
𝑊𝑊𝑊𝑊
m,d ≤ 1
𝑓𝑓𝑓𝑓m,d )/f_(m,d) ≤1 (4.71)
𝑓𝑓𝑓𝑓m,d
Bending stress at the location of a rectangular hole is verified The resistance of LVL-P beams at the location of holes maybe
by the equations: improved and larger hole sizes are allowed when they are rein-
𝑀𝑀𝑀𝑀d
+
𝑀𝑀𝑀𝑀o,d forced by gluing wood-based panels such as plywood to both
𝑊𝑊𝑊𝑊n 𝑊𝑊𝑊𝑊o
(M_d/W_n
𝑓𝑓𝑓𝑓
≤ 1 + M_(o,d)/W_o )/f_(m,d) ≤1 (4.64) sides of the beam(4.64)
around the holes. Detailed design instruc-
m,d
tions are given e.g. in chapter F3.2 of the Austrian ÖNORM B
𝑀𝑀𝑀𝑀d
+
𝑀𝑀𝑀𝑀u,d
1995-1-1:2015 33. As LVL beams are thin, internal reinforce-
𝑊𝑊𝑊𝑊n 𝑊𝑊𝑊𝑊u
𝑓𝑓𝑓𝑓m,d
≤1 ment with screws(4.65)
or glued-in-rods is not recommended.
where
LVL Handbook Europe 135
2
𝑏𝑏𝑏𝑏 ∙ �ℎ2 −ℎd �
𝑊𝑊𝑊𝑊n = 6
(4.66)
𝐴𝐴𝐴𝐴o 𝑎𝑎𝑎𝑎
𝑀𝑀𝑀𝑀o,d
2020_LVL_04.indd 135 = ∙ 𝑉𝑉𝑉𝑉d ∙ (4.67) 12.3.2020 10:49:36
𝐴𝐴𝐴𝐴u +𝐴𝐴𝐴𝐴o 2
4. STRUCTURAL DESIGN OF LVL STRUCTURES
4.3.13 Serviceability limit state design: assumption of a linear relationship between the actions and
the corresponding deformations, as a simplification of EN
Deflections 1990:2002, 2.2.3(3), the final deformation, ufin, may be taken as:
Instantaneous deflection of a member is calculated using the
actions in equation (4.1). In serviceability limit state (SLS) the 𝑢𝑢𝑢𝑢fin = 𝑢𝑢𝑢𝑢fin,G + 𝑢𝑢𝑢𝑢fin,Q1 + ∑ 𝑢𝑢𝑢𝑢fin,Q1
u_fin=u_(fin,G)+u_(fi (4.72) (EC5 2.2) (4.72) (EC5 2
partial safety factors γG and γQ of loads are 1,0. The following
serviceability limit state principles are defined in Eurocode 5, where
clause 2.2.3: ufin,G=uinst,G (1+kdef ) for a permanent action G;
The deformation of a structure resulting from the effects of ufin,Q1=uinst,Q1 (1+ψ2,1 kdef ) for the leading variable action, Q1;
actions (such as axial and shear forces, bending moments and ufin,Qi=uinst,Qi (ψ0,i+ψ2,i kdef ) for accompanying variable
joint slip) and from moisture shall remain within appropriate actions, Qi (i>1);
limits, having regard to the possibility of damage to surfacing uinst,G , uinst,Q1 , uinst,Q,i are the instantaneous
materials, ceilings, floors, partitions and finishes, and to the deformations for actions G,Q1,Qi
functional needs as well as any appearance requirements. respectively;
Instantaneous deformation, uinst, see Figure 4.27, should ψ2,1 , ψ2,i are the factors for the quasi-
be calculated for the characteristic combination of actions, see permanent value of variable
EN 1990:2002, clause 6.5.3(2) a), using mean values of the ap- actions;
propriate moduli of elasticity, shear moduli and slip moduli. ψ0,i are the factors for the combination
Final deformation ufin, see e.g. wfin in Figure 4.27, should value of variable actions; and
be calculated by superimposing the creep deformation ucreep kdef is the creep deformation factor for
calculated using the quasi-permanent combination of actions, timber and wood-based materials.
see EN 1990:2002, 6.5.3(2)(c), onto the instantaneous defor-
mation uinst calculated from 2.2.3(2). The creep deformation The components of deflection resulting from a combination
should be calculated using mean values of the appropriate of actions are shown in Figure 4.27, where the symbols are de-
moduli of elasticity, shear moduli and slip moduli and the rel- fined as follows:
evant values of kdef given in Table 4.3. − winst is the instantaneous deflection;
− wcreep is the creep deflection; and
Note: When calculating the creep deflection of LVL-C in the edge- − wfin is the final deflection.
wise direction, the kdef value is similar to LVL-P. In flatwise di-
rection, the kdef is larger due to the rolling shear deformation of Note: LVL is not pre-cambered. Only in some very special cases
the cross veneers, similar to plywood, see Table 4.3 of kdef values. LVL beams may be cut to a camber by special sawing from an
LVL billet.
If the structure consists of members or components having dif-
ferent creep behaviour, the long-term deformation due to the The net deflection below a straight line between the supports,
quasi-permanent combination of actions should be calculated wnet,fin, should be taken as:
using the final mean values of the appropriate moduli of elas-
ticity, shear moduli and slip moduli according to Eurocode 5, w
𝑤𝑤𝑤𝑤net,fin =w𝑤𝑤𝑤𝑤
net, fin = ++
instinst wcreep
𝑤𝑤𝑤𝑤creep (4.73)
clause 2.3.2.2 (1). The final deformation ufin is then calculated
by superimposing the instantaneous deformation, due to the Note: The recommended range of limiting values of deflections
difference between the characteristic and the quasi-permanent for beams with span l is given in Table 4.10 depending upon the
combination of actions on the long-term deformation. level of deformation deemed to be acceptable. Information on
For structures consisting of members, components and national limit values can be found in the National Annex for
connections with the same creep behaviour and under the Eurocode 5.
5 ∙ 𝑞𝑞𝑞𝑞
𝑤𝑤𝑤𝑤 = 384 ∙d,i,SLS
∙ 𝐿𝐿𝐿𝐿4
+
𝜁𝜁𝜁𝜁 ∙ 𝑞𝑞𝑞𝑞d,i,SLS ∙ 𝐿𝐿𝐿𝐿2
𝐸𝐸𝐸𝐸 mean ∙ 𝐼𝐼𝐼𝐼 8 ∙ 𝐺𝐺𝐺𝐺mean ∙ 𝐴𝐴𝐴𝐴
Table
The 4.10.
net
𝑤𝑤net, fin
Example limiting values
=deflection
𝑤𝑤inst + 𝑤𝑤below
creep
for beam
a straight line deflection. 4.3.14
between the supports, wnet,fin Serviceability
, should (4.73)as: limit state design:
be taken
where
b
vn v is the unitis is impulse
floor
the width
unit velocity
[m]; response
impulse velocity [m/Ns
response 2 ]; [m/Ns ]; 2 effective stiffness EIeff, normal stresses from bending moment,
in
in the theof number
number of
of first-order
first-order modes
modes with with natural
natural frequencies
frequencies up to to 40 40atHz;
n
m n4040 in the number
40 is the massfirst-order
[kg/m 2modes with natural
]; and and shearupstress Hz;
the glued joints – can be defined according
n
v
b 40 in
is
is the
floor number
unit
width impulseof
[m]; first-order
velocity modes
response with
[m/Nsnatural
2
]; frequencies up to 40 Hz;
b frequencies is floor up towidth 40 Hz;[m]; to equations (4.84) – (4.88).
lb b 40 is floor width is
is
in the
the floor
floor width
number span of [m].
[m]; first-order modes with natural frequencies The up to 40 stiffness Hz; EIeff of a glued composite cross sec-
the[m]; effective
n is
is the mass
mass [kg/m [kg/m2]; ]; and
2
m
m and
m
b m is the mass is [kg/m
the
floor mass2 ];
width and
[kg/m
[m];
2
]; and tion is calculated according to equation:
lThe
ll value of nis is40spanthe
the may be span
floor
floor calculated [m]. from:
is the floor [m].span [m].
lm is the floor massspan [kg/m[m]. 2
]; and
The 0,25 〖EI〗_eff=
eff = ∑i∑_i▒〖
𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼 𝐸𝐸𝐸𝐸i 𝐼𝐼𝐼𝐼i +E_(i 𝑒𝑒𝑒𝑒i2
𝐸𝐸𝐸𝐸i 𝐴𝐴𝐴𝐴)i I_i+E_i A_i e_i^2 〗 (4.84)
The value of
of n
2 may be4 calculated from:
value40 n40 40 may 𝑏𝑏 be calculated
(𝐸𝐸𝐸𝐸)𝑙𝑙 from:
lThe
𝑛𝑛 The40 value
= valueof
{(( of n40
) is
n −the may1)
may floor
be
( be ) span
calculated
calculated [m].
} from:
from:
𝑓𝑓1 2 40 𝑙𝑙 4 (𝐸𝐸𝐸𝐸)𝑏𝑏 0,25
40
40 2 𝑏𝑏
𝑏𝑏 4 (𝐸𝐸𝐸𝐸) 0,25
(𝐸𝐸𝐸𝐸)𝑙𝑙𝑙𝑙 0,25 from:(4.80) (EC5 7.7) where
The
𝑛𝑛
𝑛𝑛 =
n_40={((4
40 = value
{((
{(( of )
) n2 40−
− may
1)
1) (
( be )
) 4 calculated }} (4.80) (EC5 7.7) EIeff is the effective stiffness of the composite cross
40 40
𝑓𝑓
𝑓𝑓11 ) − 1) ( 𝑙𝑙 ) (𝐸𝐸𝐸𝐸)𝑏𝑏𝑏𝑏 } 𝑏𝑏
𝑙𝑙 (𝐸𝐸𝐸𝐸) 𝑙𝑙
𝑛𝑛
where 40 = {(( (EI)𝑓𝑓b1is section
an axis[Nmm 2]; to the
2the equivalent 𝑙𝑙 4 (𝐸𝐸𝐸𝐸)plate 0,25bending stiffness
(4.80) (EC5
(4.80) (EC5 7.7) 2 of the
7.7) floor about parallel
40 𝑏𝑏 (𝐸𝐸𝐸𝐸)𝑏𝑏𝑙𝑙
beam
𝑛𝑛40 = {(( direction ) − calculated
1) ( ) for a}1 metre wide (4.80)section
(EC5 7.7) [Nm /m] andEi (EI)is the modulus
b<(EI) l. of elasticity of a part i [N/mm2];
where
where (EI)b (EI) 𝑓𝑓 is the 𝑙𝑙
equivalent (𝐸𝐸𝐸𝐸)plate bending stiffness of the floor about
Ii is an axis parallel of to the
where (EI)b isis the
b 1 theequivalent equivalent plate plate bending
𝑏𝑏 bending stiffness
stiffness of the floor about anthe axismomentparallel inertia
to the of a part i [mm4], for
The
beam
where
floor about
beam deflection
direction
(EI)
direction is
an axis under
the calculated F
equivalent
parallel tofor
calculated =1 kN
for apoint
plate
theabeam 1 load
metre
bending
1 metre can
(4.80)
wide
direction be calculated
(EC5
section
stiffness
calculated
wide section 7.7)
[Nm
of the
[Nm from
2/m]
2
floor equations:
and
about(EI) an
for /m] and (EI)rectangular<(EI)
axis . parallel cross section to the Ii = bi∙hi3/12, where bi is the
b b<(EI)l.
b l
beam
a 1 metre direction
wide calculated
section [Nm /m] for a 1 metre
and bending wide
(EI)b<(EI) section
l.
[Nm 2
/m] and (EI) <(EI) .
I i width [mm] of moment
the topart the and hi is the
ofheight
a part[mm]
i [mmof 4 the
2 b l
where
The (EI)
deflection b is the 𝐹𝐹under ∙ equivalent
𝑙𝑙 2 F =1 kN plate
point load can stiffness
be of the floor
calculated from about
equations: an axis is the
parallel of inertia ], for rectangular cross secti
The The deflection
deflection under F
under F =1 =1 kN kN pointpoint load
loadcan be
cansectioncalculated
be calculated from equations:
part; 3
beam
The deflection direction calculated for a 1 metre wide [Nm 2
/m] and (EI) <(EI) . b ∙h /12, where b is the width [mm] of the part and hi is the height [m
42 ∙ 𝑘𝑘under F =1 kN point load can be calculated from equations: b l i i i
from equations: 𝛿𝛿 ∙ 2(𝐸𝐸𝐸𝐸)l A is the cross-sectional
part; area of a part i [mm 2]; and
𝑤𝑤 = min 𝐹𝐹 ∙
𝐹𝐹 ∙∙ 𝑙𝑙𝑙𝑙 3 𝑙𝑙 2 i
The deflection 𝐹𝐹 under
𝐹𝐹 2 F =1 kN point load can be calculated from equations: Aei i is the eccentricity is the cross-sectional of the part i = distancearea of abetween
part i [mm2]; and
42
42
48 ∙
∙ ∙𝑘𝑘𝑠𝑠𝛿𝛿𝛿𝛿 ∙ ∙∙∙𝑙𝑙 (𝐸𝐸𝐸𝐸)
𝑘𝑘 (𝐸𝐸𝐸𝐸)l
(𝐸𝐸𝐸𝐸)
𝑤𝑤 = min
𝑤𝑤 = min 42 ∙ 𝐹𝐹 { l
l e i the centreisofthe gravity
eccentricity of part iofand theneutral axis
part i = of the between the centre of grav
distance
𝑘𝑘𝛿𝛿 ∙∙ ∙𝑙𝑙𝑙𝑙 332(𝐸𝐸𝐸𝐸)l
𝐹𝐹 (4.81)
31 (EC5 NA,entire Finland) 31
𝑤𝑤 w=min{█((F
= min ∙ l^2)/( 3 (4.81) (EC5 NA, Finland) composite
and neutral crossaxis section
of [mm].
the entire composite cross section [mm].
or {
{4248 ∙ ∙∙𝐹𝐹
48 ∙ 𝑙𝑙 (𝐸𝐸𝐸𝐸)l
𝑘𝑘𝑠𝑠𝑠𝑠𝛿𝛿 ∙∙ (𝐸𝐸𝐸𝐸) l
𝑤𝑤 = min { 48 ∙ 𝑠𝑠 ∙ 3(𝐸𝐸𝐸𝐸)l (4.81)
(4.81) (EC5(EC5The NA, Finland)
NA,location
Finland)
31
31
𝐹𝐹 ∙ 𝑙𝑙 The location of the
of the neutral
neutral axis axisof ofaacomposite
compositecross
cross section related to the bottom
section
𝐹𝐹 ∙ 𝑙𝑙 2 (4.81) (EC5 NA,
section Finland)
is:
31
or
𝑤𝑤
or = related to the bottom of the section is:
{ 48 ∙ 𝑠𝑠 ∙ (𝐸𝐸𝐸𝐸)l
or 43,6 ∙𝐹𝐹 𝑘𝑘∙𝛿𝛿 𝑙𝑙∙22 (𝐸𝐸𝐸𝐸)l (4.81) (EC5 NA,∑Finland)33 31
165 (255)
ei is the eccentricity of the part i = distance between the centre of gravity of part i
and neutral axis of the entire composite cross section [mm].
The location of the neutral axis of a composite cross section related to the bottom of the
section is: Composite cross section. In thin-flanged beams axial stresses are checked at points 1, 3 and 5. Shear stresses are checked at points
Figure 4.29.
2, 3 and 4.
∑i 𝐸𝐸i ∙ 𝐴𝐴i ∙ 𝑎𝑎i
𝑒𝑒0 =
∑i 𝐸𝐸i ∙ 𝐴𝐴i
(4.85)
where
ai is the distance between the centre of gravity of part i and the bottom of the
entire composite cross section [mm].
𝑆𝑆𝑆𝑆
(z) ∙ 𝑉𝑉𝑉𝑉
Normal stressstressfrom frombending
bendingmomentmoment is is
calculated
calculated where
for for composite cross
𝜏𝜏𝜏𝜏(z)d =sections
𝐸𝐸𝐸𝐸i ∙ d
Normal 𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼eff ∙according
𝑏𝑏𝑏𝑏(z) to
composite
the equation:cross sections according to the equation: τ(z)d is the design value of the shear stress at coordinate z in
𝐸𝐸i ∙ 𝑒𝑒(z)i ∙ 𝑀𝑀d the section [N/mm2];
σ_(i,d(z)
𝜎𝜎 i,d(z) = )=(E_i 〖 ∙ e〗_(z)i ∙ M_d)/〖EI〗_eff (4.86) Ei is the modulus of elasticity of a part i [N/mm2];
𝐸𝐸𝐸𝐸eff
S(z)(4.86)
is the static moment at coordinate z [mm³];
where Vd is the design value of shear force at the evaluated
where
σi,d is the design value of normal stress at coordinate z in the location of the member [Nmm];
σ
i,d section [N/mm is the 2design
]; value of normal stress at coordinate z EI ineffthe
is section
the effective [N/mm 2
];
stiffness of the composite cross
Ei i is the modulus
E is the of elasticityofofelasticity
modulus a part i [N/mm2];
of a part i [N/mm ]; 2
section [Nmm ]; 2
e(z)i
e (z)i is the coordinate z of the point
is the coordinate z ofi the
wherepoint whereisthe stress is banalysed
thei stress (z) is the=width distance of thetosection
the at coordinate z [mm];
analysed neutral
= distanceaxistoofthethe
neutral
entireaxis of the entire
composite cross section [mm];
d composite
M is cross sectionvalue
the design [mm];of the bending moment at the evaluated S𝑆𝑆𝑆𝑆(z)
(z)=∑= i∑Aiilocation
∙e𝐴𝐴𝐴𝐴(z)i
i ∙ (4.88)
𝑒𝑒𝑒𝑒(z)i of the
Md is the design member value [Nmm];
of the bending
and moment at the
eff evaluatedislocation
EI of the member
the effective stiffness[Nmm]; and
of the composite Ai [Nmm
cross section is the 2cross-sectional
]. area of a part i [mm2]; and
EIeff is the effective stiffness of the composite cross e(z)i is the coordinate z of the point i where the stress is
Shear
stresses
section [Nmm at the
2]. glued joints of composite cross sections are calculated
analysedaccording = distanceto to the neutral axis of the entire
equation:
Shear stresses at the glued joints of composite cross sections are composite cross section [mm].
calculated according to equation:
𝑆𝑆(z) ∙ 𝑉𝑉d
𝜏𝜏τ_(z)d=E_i∙(S_((z)
(z)d = 𝐸𝐸i ∙ ) ∙ V_d)/(〖EI〗_eff ∙〖 b〗_((z) ) ) (4.87)
𝐸𝐸𝐸𝐸eff ∙ 𝑏𝑏(z)
(4.87)
where LVL Handbook Europe 139
τ(z)d is the design value of the shear stress at coordinate z in the section [N/mm ]; [N/mm2];
Ei is the modulus of elasticity of a part i [N/mm2];
2020_LVL_04.indd 139 12.3.2020 10:49:39
S(z) is the static moment at coordinate z [mm³];
4. STRUCTURAL DESIGN OF LVL STRUCTURES
Figure 4.31. GLVL I-beams, box beams and stress distribution in the cross section. Section 1-1 is glued joint between web and flange.
4.4.4 Rib slabs and box slabs Figure 4.32. Multiple-glued GLVL panels.
LVL rib slab and box slab elements have LVL-P ribs and LVL-C Table 4.11. Maximum effective width bef of flange LVL-C panel
panels that distribute the loads to the ribs and work as thin parallel to the ribs (Modified from EC5 Table 9.1).
flanges of the composite cross sections. Structural design is
made separately for each rib section. Their specific design in-
Shear lag Plate buckling in ULS design
structions are given in Eurocode 5, clause 9.1.2. The effective
width bef of the flange panel that can be taken into account I-section bef = bw + 0,1 ∙ l bef = bw + 20 ∙ hf,c
in the calculations is limited due to shear lag and due to plate
buckling in ULS design on the compression side according to C-section bef = bw + 0,05 ∙ l bef = bw + 10 ∙ hf,c
Table 4.11. Where bw is rib thickness, l is span length and hf,c is the
In ULS design the following verifications are required: thickness of the flange panel on the compression side.
• The mean axial compression stress σf,c,d and tension stress
σf,c,d shall be smaller than ff,c,d and ff,c,d strengths respective-
ly. Since the flange panels are thin, it is not required to verify
the resistance against tension and compression stresses at the
extreme fibres of the flanges. In SLS design, bending and shear deformations are tak-
• Resistance against shear stress for the ribs at the neutral ax- en into account. For simplicity, the shear rigidity GA may be
is and with the equation (9.14) of Eurocode 5 for the glued evaluated based on the ribs only. In that case, the kdef factor of
joints between the ribs and flange panels. The critical mate- LVL-P can also be used for the LVL-C flange panels in the flat-
rial property is typically the flatwise shear strength fv,flat,0,d wise direction in the calculation of final deformation, as the
of the flanges due to the rolling shear of the cross veneers thin flanges are mainly axially loaded.
in LVL-C. Rib slab suppliers also have more advanced design instruc-
• Resistance against bending and shear stress of the flange tions tailored and approved for their elements as part of their
panels in the perpendicular direction of the slab technical documentation.
Figure 4.33. Definitions of the parts of rib slab part (EC5, Figure 9.2). I-section is used for the intermediate rib and C-section for the edge rib of
the element.
Figure 5.2. Edgewise (edge face) and flatwise (wide face) Figure 5.3. A) Laterally loaded connection 34 B) Axially loaded
orientations and loading types of connections. Fax,d are forces of connection.
axially loaded and Fv,d are forces of laterally loaded connections 34.
Figure 5.4. Spacings (a1 and a2), end distances (a3,t and a3,c) and edge distances (a4,t and a4,c) for fasteners 34.
Figure 5.5. Spacings and end and edge distances: (A) Spacing parallel to grain in a row and perpendicular to grain between rows; (B) Edge and
end distances; (1) Loaded end, (2) Unloaded end, (3) Loaded edge, (4) Unloaded edge; 1 Fastener, 2 Grain direction (EC5 figure 8.7).
Figure 5.6. Moment-resisting multi-shear LVL-to-LVL flatwise connections with circular patterns of fasteners.
Figure 5.7. Definitions of angles α, β and ε for screws (A) in the wide face and (B) in the edge face of LVL. 1) wide face, 2) edge face and 3) end
grain. α is the angle between the load and grain direction of laterally loaded connections. β is the angle between the screw axis and wide face.
ε is the angle between screw axis and grain direction 32.
Figure 5.8. Spacings and end and edge distances and definitions of angles α, β and ε for axially loaded screws in LVL. α is the angle between
the shear plane and screw axis. β is the angle between the screw axis and wide face. ε is the angle between screw axis and grain direction.
(modified from EC5 Figure 8.11a).
Table 5.1. Minimum spacings, end distances and edge distances for nails and screws with outer thread diameter <12 mm 32.
Note: EN 1995-1-1:2004 (Eurocode 5) has a limit of def<6 mm which corresponds with 9 mm outer thread diameter.
Spacing a1 (parallel to grain) 0° ≤ α ≤ 360° d < 5 mm: (7 + 8│cos α│) d 5 + 2│cos α│) d (4 +│cos α│) d
(5 + 5│cos α│) d
d ≥ 5 mm:
(5 + 7│cos α│) d
Distance a3,t (loaded end) -90° ≤ α ≤ 90° (10 + 5 cos α) d (15 + 5 cos α) d (4 + 3 cos α) d (7 + 5 cos α) d b)
Distance a4,t (loaded edge) 0° ≤ α ≤ 180° d < 5 mm: d < 5 mm: (3 + 4 sin α) d d < 5 mm:
(5 + 2 sin α) d (7 + 2 sin α) d (3 + 2 sin α) d
d ≥ 5 mm: d ≥ 5 mm: d ≥ 5 mm:
(5 + 5 sin α) d (7 + 5 sin α) d (3 + 4 sin α) d
a)
When pointside penetration length is less than 10d, the rules in the column LVL or GLVL wide face apply
b)
for LVL-C or GLVL-C wide face and pointside penetration length of at least 10d: (4 + 3 cos α) d
c)
for LVL-C or GLVL-C wide face and pointside penetration length of at least 10d: 5d
Table 5.2. Maximum nail and screw size d [mm] for edge face (edgewise) connections
Axially loaded
Laterally loaded connections
connections
Without pre- With predrilling a2,CG ≥ 4d
LVL drilling a4,c ≥ 7d a4,c ≥ 3d
thickness Nails and screws Nails Screws Screws
LVL4505,
mm
Table 5.33,2 7,5
• MUOKATTU 29.8.7,5 5,6
Table 5.3. Minimum spacings, end distances and edge distances for bolts and screws with max outer thread diameter >12 mm with predrilled
holes 32. Note: EN 1995-1-1:2004 (Eurocode has a limit of def < 6 mm which corresponds with 9 mm outer thread diameter.
Distance a3,t (loaded end) -90° ≤ α ≤ 90° max (7d; 105 mm) b)
max (4d; 60 mm) c)
Distance a4,t (loaded edge) 0° ≤ α ≤ 180° max [(2 + 2 sin α) d; 3d] max [(2 + 2 sin α) d; 3d]
a1 (spacing on circle) 6d 4d 5d
Table 5.4. Minimum spacings, end distances and edge distances for 6-30 mm dowels 32.
Distance a3,t (loaded end) -90° ≤ α ≤ 90° max (7d; 105 mm) b) max (4d; 60 mm) c)
Distance a3,c (unloaded end) 90° ≤ α < 150 ° a3,t │cos α│ (3 + │cos α│) d
150° ≤ α < 210° 3d
210° ≤ α ≤ 270° a3,t │cos α│
Distance a4,t (loaded edge) 0° ≤ α ≤ 180° max [(2 + 2 sin α) d; 3d] max [(2 + 2 sin α) d; 3d]
ℎe
𝐹𝐹90,k = 14 ∙ 𝑏𝑏√ ℎ [𝑁𝑁] (5.3) (EC
(1 − e )
ℎ
Figure 5.9. Inclined force transmitted by a connection (modified from EC5 Figure
where8.1)
F90,Rk is the characteristic splitting capacity [Ν];
LVL Handbook Europe 149
he is the loaded edge distance to the centre of the most distant fa
h is the timber member height, [mm]; and
5.3.3 Block shear and plug shear failure Wood failure should be checked for at tension-loaded
member ends for connection force components that are par-
modes at multiple dowel-type steel-to- allel to the grain. There are two types of timber failure mode:
timber connections block shear and plug shear.
• Block shear needs to be checked for bolt and dowel connec-
Block shear and plug shear failure modes shall be checked for tions and for screw connections when the centre member
steel-to-timber connections and tension-loaded member ends is screwed from both sides and the screws are overlapping.
of double or multiple shear plane timber-to-timber connec- • Plug shear failure mode must be checked for steel-to-timber
tions. The wood failure capacity of the joint area can be calcu- connections with surface fasteners (nails, screws, nail plates
lated according to the method presented in the Finnish Hand- and shear plates). Plug shear must be checked also for ex-
book for EC5: RIL 205-1-2009, Section 8.2.4S 31. In addition, ternal lamellas in cases where the dowels are shorter than
the effective number of fasteners, nef, according to Section 5.3.2 the overall thickness of the members in the connection. For
are taken into account to prevent splitting and row shear fail- LVL-C wide face bolt and dowel connections, both block and
ure mode. This method cannot be used for edgewise LVL con- plug shear failure modes must be checked.
nections. • Block and plug shear capacities are not checked for connec-
tions where all fasteners are in a single row parallel to the
grain (n2 = 1).
• If the timber member t1 has fasteners from opposite sides
and the effective thickness tef ≥ 0.5t1, the block shear capac-
A ity of steel-to-timber connections should also be checked.
• Block shear need not to be checked for bolt and dowel con-
nections when:
• the member thicknesses are t1 ≥ 4d, ts ≥ 5d (inner
member)
• there are max. 4 fasteners in a row parallel to the grain, and
• the distance perpendicular to grain between bolts a2 ≥ 5d
Figure
Figure or5.10.
between
5.10. a)
a) Block
Block a2 ≥ 4d.,
dowelsshear
shear see Figure
failure
failure mode
mode b)5.11.
b) Plug
Plug shear
shear failure
failure mode
31
mode 31 (Kuva
(Kuva
block shear failure 190320, Kuva_118_2 plug shear failure
block shear failure 190320, Kuva_118_2 plug shear failure 190314 190314
The characteristic plug shear capacity is calculated using the
The
The characteristic
characteristic plug
equation: plug shear
shear capacity
capacity is
is calculated
calculated using
using thethe equation:
equation:
k_C
𝑘𝑘 1 𝑓𝑓or LVL
𝑑𝑑 − P
{█(1 for LVL-P and GLVL-P@min{█and GLVL − P smooth nails (5.15)shall not be used to resist permanent or long-
C{
𝑘𝑘C { for LVL-C min𝑑𝑑 {(𝑑𝑑−2)and for LVL − )┤
GLVL-C┤ C and GLVL − C (5.15) term (5.15)
axial loading.
min {(𝑑𝑑−2) for 3 LVL − C and GLVL − C
3 In the edge face of LVL-C and GLVL-C the minimum
• To prevent splitting failure mode, for one row of n nails parallel to the nailgrain,
diameter unless d is 3mm.the
• To prevent nails of splitting
that rowfailure are staggered mode, forperpendicular one row of n to nails
grainparallel
by at to the1d,
least grain,theunless load-carrying the
nailscapacity
of that row are staggered
parallel to the grain perpendicular
(see EC5, clause to grain by at least
8.1.2(4)) should 1d,be thecalculated
load-carrying using the
capacity effectiveparallelnumber to theofgrain fasteners (see EC5, nef, = clausenkef . kef8.1.2(4))
in Table should be calculated
8.1 of Eurocode 5 applies usingtothe LVL-P
effectivewide numberface. ForofLVL-C fasteners widenefface = nkef kef. =kef1inand Table for LVL8.1 of orEurocode
GLVL edge 5 applies
face: to LVL-P
wideLVLface.
152 Handbook For LVL-C wide
Europe 1 face kef = 1 and for LVL or GLVL edge face:
𝑘𝑘ef = min { 1 (5.16)
1 − 0,03(20 − 𝑎𝑎1 /𝑑𝑑)
𝑘𝑘ef = min { (5.16)
1 − 0,03(20 − 𝑎𝑎1 /𝑑𝑑)
• For smooth nails in predrilled holes in the edge face of LVL or GLVL the pointside
• For smooth penetration
2020_LVL_05.indd 152
nailslength in predrilled
should holes be at in leastthe12d edge face of LVL or GLVL the pointside 12.3.2020 12:44:53
5. STRUCTURAL DESIGN OF CONNECTIONS
The following information should be taken from the nail sup- made, the yield moment capacity of the smooth shank may be
plier’s DoP: used, if the smooth shank penetrates into the member contain-
• Characteristic yield moment My,k [Nmm] ing the point of the screw by not less than 4d.
• Characteristic withdrawal parameter fax,k [N/mm2] Unless otherwise stated in this section, the rules for bolts
• Characteristic head out-through parameter fhead,k [N/mm2] apply to screws with an outer thread diameter d > 12 mm in
• Characteristic tensile capacity ftens,k [kN] predrilled LVL / GLVL members. The embedding strength fh,k
• Nail diameter [mm] made,
should the yield as:
be taken moment capacity of the smooth shank may be used, if the s
• Nail head area [mm2] penetrates into the member containing the point of the screw by not less tha
0,082 ∙ (1 − 0,01𝑑𝑑𝑑𝑑) 𝜌𝜌𝜌𝜌k
• Nail length [mm] f_(h,k)=(0,082
𝑓𝑓𝑓𝑓 h,k = (𝑘𝑘𝑘𝑘otherwise
∙ (1 - 0,01d) ρ_k)/((k_(90 )∙ sin^2 α2 + cos^2 α)
2 𝜀𝜀𝜀𝜀) N/mm
2
Unless 2
90 ∙ sin 𝛼𝛼𝛼𝛼 + cos stated
2 𝛼𝛼𝛼𝛼) ∙ (𝑘𝑘𝑘𝑘in∙ this
C cos 𝛽𝛽𝛽𝛽section,
2 + sin2 𝛽𝛽𝛽𝛽) ∙the
(2,5 ∙rules for
cos 𝜀𝜀𝜀𝜀 + sinbolts apply to screws w
• For threaded nails also length of threaded part (lg) and length ∙ (k_(C diameter
thread ) 〖∙ cos〗^2d β>〖+ 12sin〗^2 mm in β) ∙ (2,5 ∙ cos^2
predrilled LVL /ε GLVL
+ sin^2members. ε) ) The embedd
of point (lp)
should be taken as: (5.22)
where
0,082 ∙ (1 − 0,01𝑑𝑑) 𝜌𝜌k
𝑓𝑓d h,k is=the outer thread diameter of the screw [mm]; N/mm2
5.5 SCREWED CONNECTIONS (𝑘𝑘90 ∙ sin2 𝛼𝛼 + cos2 𝛼𝛼) ∙ (𝑘𝑘C ∙ cos2 𝛽𝛽 + sin2 𝛽𝛽) ∙ (2,5 ∙ cos2 𝜀𝜀 + sin2 𝜀𝜀)
ρk is the characteristic density [kg/m³];
Design instructions for screwed connections are given in Eu- α is the angle between load and grain direction, see
where
rocode 5, Section 8.7. The definitions provided in the current Figure 5.7; for softwood LVL-C / GLVL-C and α > 45°,
section introduce some differences to the Eurocode 5 instruc- d α may be assumed is the outer as 45°; thread diameter of the screw [mm];
tions in order to improve the connection design of LVL. Screw ρ is the
β k is the angle between screw axis and characteristic density wide[kg/m³];
face, see
suppliers also have their own design instructions for their fas- α Figure 5.7;is the angle between load and grain direction, see Figure 5.7; f
teners documented in their ETAs and DoPs. These must be ε is the angleCbetween / GLVL-C screw andaxis α >and45°,grain α may be assumed as 45°;
direction,
treated as separate supplier-specific instructions unless they β see Figure is 5.7;
the angle between screw axis and wide face, see Figure 5.7
make direct reference to Eurocode design. εk90 is 1,15 + 0,015 is the forangle
softwood between LVL /screw
GLVL;axis andand grain (5.23) direction, see Figur
The following information should be taken from the screw k90 𝑑𝑑𝑑𝑑is 1,15 + 0,015 for softwood LVL / GLVL; and
supplier’s DoP: k_C=max{█(d/((d-2)
𝑘𝑘𝑘𝑘 C = max � )@1,15) for
(𝑑𝑑𝑑𝑑−2) for softwood LVL softwood LVL┤ (5.24)
• Characteristic yield moment My,k [Nmm] 1,15𝑑𝑑
• Characteristic withdrawal parameter fax,k [N/mm2] 𝑘𝑘C = max {(𝑑𝑑−2) for softwood LVL
1,15
• Characteristic head out-through parameter fhead,k [N/mm2] Unless otherwise 𝑑𝑑𝑑𝑑 stated in this section, the rules for nails apply
• Characteristic tensile capacity ftens,k [kN] toC screws
𝑘𝑘𝑘𝑘
Unless = max with a for
� (𝑑𝑑𝑑𝑑−2)
otherwise diameter
softwood
stated ind this
≤ LVL 12section,
mm or the screwsrulesin fornon-pre-
nails apply to screws w
• Screw outer thread diameter d [mm] 12 mm or 1,15
screws
drilled timber or LVL / GLVL members. in non-predrilled timber or TheLVL / GLVL members. The emb
embedding
• Screw inner thread diameter d1 [mm] should
strengthbe takenbeas:
should taken as:
0,082 ∙ 𝜌𝜌𝜌𝜌k ∙ ∙ 𝑑𝑑𝑑𝑑−0,3
• Screw head diameter dh [mm] -𝑓𝑓𝑓𝑓• h,k wwithout
ithout
= (𝑘𝑘𝑘𝑘 ∙ predrilled
predrilled
2
holes in softwood2 LVL
2 holes in 2softwood N/mm
LVLor GLVL
or2 GLVL
C cos 𝛽𝛽𝛽𝛽 + sin 𝛽𝛽𝛽𝛽) ∙ (2,5 ∙ cos 𝜀𝜀𝜀𝜀 + sin 𝜀𝜀𝜀𝜀)
• Screw length L [mm] f_(h,k)=(0,082 ∙ ρ_(k ∙)∙ d^(-0,3))/((k_C 〖∙
0,082 ∙ 𝜌𝜌k ∙ ∙ 𝑑𝑑−0,3
• Thread length LG [mm] 𝑓𝑓β) h,k = N/mm2
(𝑘𝑘C ∙ cos 2 𝛽𝛽 + sin2 𝛽𝛽) ∙ (2,5 ∙ cos2 𝜀𝜀 + sin2 𝜀𝜀)
0,082 ∙ (1 − 0,01𝑑𝑑𝑑𝑑) 𝜌𝜌𝜌𝜌
h,k =
𝑓𝑓𝑓𝑓 0,082 ∙ 𝜌𝜌𝜌𝜌 ∙ ∙ 𝑑𝑑𝑑𝑑−0,3 k N/mm22 (5.25)
5.5.1 Laterally loaded screws in LVL 𝑓𝑓𝑓𝑓
h,k
C
= (𝑘𝑘𝑘𝑘 ∙ cos2 𝛽𝛽𝛽𝛽 + sin2 𝛽𝛽𝛽𝛽) ∙k(2,5 ∙ cos2 𝜀𝜀𝜀𝜀 + sin2 𝜀𝜀𝜀𝜀) N/mm
(𝑘𝑘𝑘𝑘C ∙ cos2 𝛽𝛽𝛽𝛽 + sin2 𝛽𝛽𝛽𝛽) ∙ (2,5 ∙ cos2 𝜀𝜀𝜀𝜀 + sin2 𝜀𝜀𝜀𝜀)
-• wwith
ith predrilled
predrilledholes
holesin in
softwood LVL
softwood or GLVL
LVL or GLVL
The instructions specific to LVL presented in this subsection
0,082 ∙ (1 − 0,01𝑑𝑑) 𝜌𝜌k
are based on the document Design rules for LVL to Eurocode 5, 𝑓𝑓f_(h,k)=(0,082
h,k = 1 ∙ 2(1 -∙ 0,01d)
0,082 (1for LVL ρ_k)/((k_(C
− 0,01𝑑𝑑𝑑𝑑) − P and GLVL
2 𝛽𝛽) ∙𝜌𝜌𝜌𝜌k
) 2〖 − P 2 N/mm2
Proposal for discussion in CEN/TC250/SC5, Prof. Dr.-Ing. H.J. (𝑘𝑘
𝑓𝑓𝑓𝑓h,k = (𝑘𝑘𝑘𝑘 C∙ cos2 𝛽𝛽𝛽𝛽 + sin+2 𝛽𝛽𝛽𝛽)
∙ cos 𝑑𝑑𝑑𝑑 𝛽𝛽 sin (2,5 ∙ 2cosN/mm 2
𝜀𝜀 + sin 𝜀𝜀)
𝑘𝑘𝑘𝑘C = � C 2
∙ (2,5 ∙ cos 𝜀𝜀𝜀𝜀 + sin 𝜀𝜀𝜀𝜀)
min �(𝑑𝑑𝑑𝑑−2) 𝑓𝑓𝑓𝑓or LVL − C and GLVL − C
Blaβ and Dr. –Ing.M.Flaig, Blaβ & Eberhart GmbH, 30.6.2017, (5.26) 3
CEN/TC250/SC5/N0764 32 and differ partially from EN1995- where
where
1-1:2004. The effect of the threaded part of the screw shall be k_C={█(1 1 forforLVL LVL-P
− P and and GLVL
GLVL-− P
1 𝑑𝑑𝑑𝑑 for LVL − P and GLVL − P
taken into account in determining the load carrying capacity 2) )@3)
𝑘𝑘𝑘𝑘C = � for LVL-C 𝑑𝑑 and GLVL-C┤ )┤ (5.27)
𝑘𝑘C = { min �(𝑑𝑑−2) (𝑑𝑑𝑑𝑑−2) 𝑓𝑓𝑓𝑓or LVL − C and GLVL − C
by using the yield moment capacity of the screw determined in min { 3 𝑓𝑓or LVL − C and GLVL − C
accordance with EN 14592. The outer thread diameter d shall 3
be used to determine the embedment strength, the spacing,
edge and end distances and the effective number of screws. 5.5.2 Axially loaded screws
For smooth shank screws, the yield moment capacity of the where
smooth shank may be used for plastic hinges occurring within 0,8
𝐹𝐹𝐹𝐹ax,ε,Rk = ef isaxthe𝑘𝑘𝑘𝑘ax,90,k
d 𝑛𝑛𝑛𝑛 ∙ 𝑘𝑘𝑘𝑘 ∙ 𝑓𝑓𝑓𝑓 outer
∙ 𝑑𝑑𝑑𝑑 thread
∙ 𝑙𝑙𝑙𝑙ef 𝜌𝜌𝜌𝜌k diameter; and
�𝜌𝜌𝜌𝜌 �
the length of the smooth shank. Unless a detailed analysis is β a
d1 is the inner thread diameter
the characteristic withdrawal capacity should be taken as 32:
𝑛𝑛𝑛𝑛 𝑘𝑘𝑘𝑘ax ∙ 𝑓𝑓𝑓𝑓ax,90,k ∙ 𝑑𝑑𝑑𝑑 ∙ 𝑙𝑙𝑙𝑙ef 𝜌𝜌𝜌𝜌k 0,8
∙0,5∙𝜀𝜀𝜀𝜀
LVL Handbook Europe 153
= ef
𝐹𝐹𝐹𝐹ax,ε,Rk 0,5 + for 15° ≤ 𝜀𝜀𝜀𝜀 �<𝜌𝜌𝜌𝜌 45° �
𝑘𝑘𝑘𝑘ax = � 45° 𝑘𝑘𝑘𝑘β a
1 for 45° ≤ 𝜀𝜀𝜀𝜀 ≤ 90°
2020_LVL_05.indd 153 𝑘𝑘𝑘𝑘𝛽𝛽𝛽𝛽 = 1,5 ∙ cos 𝛽𝛽𝛽𝛽 + sin2 𝛽𝛽𝛽𝛽
2 12.3.2020 12:44:54
0,5∙𝜀𝜀𝜀𝜀
0,082 ∙ (1 − 0,01𝑑𝑑𝑑𝑑) 𝜌𝜌𝜌𝜌k
𝑓𝑓𝑓𝑓h,k = (𝑘𝑘𝑘𝑘 2 2 2 N/mm2 (5.22)
C ∙ cos 𝛽𝛽𝛽𝛽 + sin 𝛽𝛽𝛽𝛽) ∙ (2,5 ∙ cos 𝜀𝜀𝜀𝜀 + sin2 𝜀𝜀𝜀𝜀)
5.5.2 Axially loaded screws fhead,k is the characteristic pull-through parameter of the
screw determined in accordance with EN 14592 for
For connections in softwood timber or LVL/GLVL with ε ≥ 15° the associated density ρa
of screws in accordance with EN 14592 with: dh is the diameter of the screw head [mm]
• 6 mm ≤ d ≤ 12 mm
• 0,6 ≤ d1 /d ≤ 0,75 5.5.3 Inclined screw connections
where
d is the outer thread diameter; and Inclined screwing is an efficient way to connect LVL members
d1 is the inner thread diameter 185 (255)
together or to other types of timber members. Although the
connections transfer shear forces, the fasteners are axially load-
The characteristic withdrawal capacity should be taken as 32: ed. The instructions in this subsection are based on the Finnish
F_(ax,ε,Rk)=(n_ef
𝑛𝑛𝑛𝑛 ∙ 𝑘𝑘𝑘𝑘 ∙ 𝑓𝑓𝑓𝑓 ∙ k_(ax∙ 𝑑𝑑𝑑𝑑 )∙ 𝜌𝜌𝜌𝜌 0,8
f_(ax,90,k)
�𝑑𝑑k�∙ 𝑙𝑙 𝜌𝜌∙ d 0,8 ∙ l Handbook RIL205-1:2017 for Eurocode 5, Chapter 8.7.4S 31.
∙ 𝑙𝑙𝑙𝑙ef
𝐹𝐹𝐹𝐹ax,ε,Rk = 𝑛𝑛ef ∙ ax𝑘𝑘 𝑘𝑘𝑘𝑘ax,90,k
∙ 𝑓𝑓 ∙ 𝜌𝜌𝜌𝜌a
(5.24)
(5.28)
𝐹𝐹 ax,ε,Rk =
ef ax β ax,90,k ef
(
k
)
𝑘𝑘β 𝜌𝜌a These rules concern the design of single shear connections
where (5.28) according to Figure 5.11, where the screw inclination angle α
k_ax={█(0,5+(0,5∙ε)/(45°)
where
0,5∙𝜀𝜀𝜀𝜀
0,5 + 45° for 15° ≤ for𝜀𝜀𝜀𝜀 15°≤ε45°≤ε≤90°)┤
< 45° should be between 30°…60° in regard to the shear plane. The
𝑘𝑘𝑘𝑘
ax = � (5.29) screws are axially(5.25) loaded. The head side timber member (t1)
1 for 45° ≤ 𝜀𝜀𝜀𝜀 ≤ 90°
0,5∙𝜀𝜀
0,5 + 45° for 15° ≤ 𝜀𝜀 < 45° may be replaced with a steel plate if the screw head has a full
𝑘𝑘
𝑘𝑘𝑘𝑘𝛽𝛽𝛽𝛽ax==1,5
k_β=1,5∙cos^2 { ∙ cos2 𝛽𝛽𝛽𝛽 + sin2 𝛽𝛽𝛽𝛽β
β+sin^2 (5.30) bearing area on (5.29) the steel plate for a Figure 5.12 (b) tension
(5.26)
1 for 45° ≤ 𝜀𝜀 ≤ 90°
screw connection. The screws should be self-drilling and fully
F𝛽𝛽ax,ε,Rk
𝑘𝑘 is the
= 1,5 ∙ coscharacteristic
2
𝛽𝛽 + sin2 𝛽𝛽 withdrawal capacity of the threaded or partly (5.30)threaded with a smooth part diameter of
connection at an angle ε to the grain [N]; ds ≤ 0.8d, where d is the outer thread diameter.
Ffax,90,k
ax,ε,Rk is the characteristic
is the characteristic withdrawal withdrawal capacity of the connection
strength perpendicular at an angle
Different ε to the
or supplementary connection types and screw
grain [N];
to the grain determined in accordance with EN 14592 specifications differing from eurocode 5 may be used accord-
f ax,90,k for the is the characteristic
associated withdrawal strength perpendicular
density ρa [N/mm²]; ingto to the
theirgrain
ETA.determined in
accordance with EN
nef is the effective number of screws, nef = n where n is 14592 for the
0,9 associated density ρ a [N/mm²];
n ef the numberis theofeffective
screws acting number of screws,
together nef = n0,9 where n Cross
in a connection; screw connection
is the number of screws
acting together in
kax is a factor to consider the influence of the angle ε a connection; The cross screw connection is built up from symmetrical screw
k ax between is screw
a factor axistoandconsider the influence
grain direction and the Crosssee
of the angle ε between
pairs, screw
screw
Figure connection
axis
5.12and (a), in grain which one screw is under compres-
direction
long-term behavior; and the long-term behavior; sion
Cross and the
screw other
connectionunder tension. The up characteristic load-car-screw pairs, see Figure
The cross screw connection is built from symmetrical
llefef is the penetration
is the penetration length oflength of the threaded
the threaded part [mm];part [mm]; Cross
rying
whichcapacity
screw
one screw of the
connection cross screw
is under compressionconnection andisthe calculated
other underby tension. The charact
ρkk is the characteristic
is the characteristic density [kg/m³];density [kg/m³]; The
the crosscapacity
equation:
carrying screw connection
of the crossisscrew built up from symmetrical
connection is calculatedscrew bypairs, see Figure
the equation:
ρaa is the associated density for
is the associated fax,k [kg/m³];
density for fax,k [kg/m³]; which
The one
cross screw
screw is under
connection compression
is built up and
from the other
symmetrical under tension.
screw pairs, The
seecharact
Figure
carrying
which 0,9capacity
one screw of under
is the cross screw connection
compression and the is calculated
other under by the(5.28)
tension. equation:
The charact
kkββ is a factor
is aconsidering the influence
factor considering of the angle
the influence of β R_k=n_p^0,9
𝑅𝑅
the angle β between
k = 𝑛𝑛 (𝑅𝑅
p the C,k (R_(C,k)+R_(T,k)
screw+ 𝑅𝑅 )cos
axis and
T,k 𝛼𝛼 )cos ε (5.32)
between theLVL’sscrewwideaxis and carrying capacity of the cross screw connection is calculated by the equation:
the face;the LVL’s wide face; 𝑅𝑅 0,9
(5.28)
k = 𝑛𝑛p (𝑅𝑅C,k + 𝑅𝑅T,k )cos 𝛼𝛼
Where
ε is the angle between the screw axis and the grain where
ε is the angle between the screw axis and the grain direction, 𝑅𝑅k = 𝑛𝑛p0,9with (𝑅𝑅C,kε+≥ 𝑅𝑅15°, see𝛼𝛼
T,k )cos (5.28)
direction,Figurewith5.7;ε ≥ 15°,
andsee Figure 5.7; and np is the number
Where is the of screw
number pairs in the joint;
of screw pairsand in the joint; and
β is the angle between the screw axis and the LVL’s wide α is the angle between screw axis and the shear plane
Where
is the nαp face, withis 0°≤ the βangle
number between
of screw screw
pairsaxis in theand the and
joint; shear plane (30° ≤ α ≤ 60°
β face, with 0°≤angle
β ≤ 90°, between
see figure the5.7.
screw axis and the LVL’s wide (30° ≤ α ≤ 5.11 60°), (a) see≤Figure90°, 5.12 (a)
see figure 5.7. n
α p is the number of screw pairs in the
is the angle between screw axis and the shear plane (30° ≤ α ≤ 60° joint; and
Note: Failure modes in the steel or in
Note: Failure modes in the steel or in the timber around the screw the timber around the screwαTheare characteristic
brittle,
The characteristic 5.11
i.e.
is with
the compression
(a) minimal
angle
compression between capacity
capacity screw of theof the
axis screw
and
screw the
is is calculated
shear
calcu- by the
plane (30° ≤ αequatio
≤ 60°
ultimate deformation and therefore
are brittle, i.e. with minimal ultimate deformation and thereforehave a limited possibility for stress redistribution.
latedcharacteristic
The 5.11
by the equation: (a)
𝑓𝑓ax,ε,1,kcompression
𝑑𝑑 𝑙𝑙g,1 capacity of the screw is calculated by the equatio
havescrews
For a limited inpossibility
LVL, the for stress redistribution.
characteristic withdrawal parameter may be The assumed
characteristic as
𝑓𝑓 f 𝑑𝑑
ax,90,k
𝑅𝑅C,k = min {𝑓𝑓ax,ε,2,k 𝑑𝑑 𝑙𝑙 g,2 𝑙𝑙=15
compression capacity of the screw is calculated by the equatio
N/mm², when ρa = 500 kg/m³ and screws 6 mm ≤ d ≤ 12 mm in softwood LVL/GLVL.
R_(C,k)=min{█(f_(ax,ε,1,k)
ax,ε,1,k g,1 d l_(g,1)@f_(ax
0,8 𝑓𝑓tens,k
For screws in LVL, the characteristic withdrawal parameter 𝑅𝑅
(g,2)@0,8〖 {𝑓𝑓ax,ε,1,k
C,k = min f〗_(tens,k)ax,ε,2,k 𝑑𝑑 𝑙𝑙)┤ g,1
g,2 (5.33)
The characteristic pull-through
may be assumed as fax,90,k =15 N/mm², when ρa = 500 kg/m³ resistance of connections with axially
𝑅𝑅 loaded
= min { screws
𝑓𝑓 0,8 𝑓𝑓 should
𝑑𝑑 𝑙𝑙
The
C,k characteristic withdrawal
ax,ε,2,k tens,k g,2 capacity of the screw is calculated by the equation
be taken as: 0,8 𝑓𝑓tens,k
and screws 6 mm ≤ d ≤ 12 mm in softwood LVL/GLVL.
0,8 The characteristic withdrawal capacity 𝜌𝜌of the screw is calculated by the equation
0,8
𝐹𝐹ax,ε,Rk The=characteristic
𝑛𝑛ef ∙ 𝑓𝑓head,k ∙ pull-through
2 𝜌𝜌
𝑑𝑑h (𝜌𝜌 )k resistance of connections The characteristic withdrawal
(5.31)
𝑓𝑓ax,ε,1,k 𝑑𝑑 𝑙𝑙g,1 + 𝑓𝑓capacity 𝑑𝑑h2of
(𝜌𝜌the
𝑘𝑘 screw is calculat-
) the
with axially loaded screws should be taken as: The
ed characteristic withdrawal capacity
head,k of
a 0,8 screw is calculated by the equation
𝑅𝑅T,kby=the equation:
a
min { 𝜌𝜌
𝑑𝑑 𝑙𝑙g,2𝑑𝑑h2 ( )
𝑓𝑓ax,ε,1,k 𝑑𝑑 𝑙𝑙𝑓𝑓g,1 + 𝑓𝑓head,k
ax,ε,2,k
𝑘𝑘
𝜌𝜌 0,8
where 2 𝜌𝜌𝜌𝜌k
0,8
𝑅𝑅 = min {𝑓𝑓ax,ε,1,k 𝑓𝑓 2 𝜌𝜌a𝑘𝑘
𝐹𝐹𝐹𝐹ax,ε,Rk = 𝑡𝑡𝑡𝑡ef ∙ 𝑓𝑓𝑓𝑓head,k ∙ 𝑑𝑑𝑑𝑑h �𝜌𝜌𝜌𝜌 �
F_(ax,ε,Rk)=n_ef∙f_(head,k)∙d_h^2 (ρ_k/ρ_a )^0,8 (5.31) R_(T,k)=min{█(f_(ax,ε,1,k)
T,k (5.27) 𝑑𝑑 𝑙𝑙𝑓𝑓g,1 +tens,k
d𝑓𝑓head,k
𝑑𝑑 𝑙𝑙g,2 𝑑𝑑h (𝜌𝜌a )
l_(g,1)+f_(head,k)
ax,ε,2,k
Fax,ε,Rk is the characteristic a
pull-through capacity of the connection 𝑅𝑅
ρ_a = min at an
{ angle ε to
T,k )^0,8@f_(ax,ε,2,k) the
where 𝑓𝑓d〖 𝑓𝑓l〗_(g,2)@f_(tens,k)
ax,ε,2,k
tens,k𝑑𝑑 𝑙𝑙g,2 )┤ (5.34)
grain [N], with ε ≥ 30°
where 𝑓𝑓tens,k
fFhead,k is the characteristic pull-through parameter of the screw fax,ε,1,kdetermined
where is the in characteristic withdrawal strength parameter for a screw at th
ax,ε,Rk is the characteristic pull-through capacity of the
accordance with EN 14592 for the associated density
where ρ member of the connection at an angle ε to the grain direction [N/mm
connection at an angle ε to the grain [N], with ε ≥ 30° a
fax,ε,1,k is the characteristic withdrawal strength parameter for a screw at th
dh is the diameter of the screw head [mm] ffax,ε,1,k
ax,ε,2,k is the
the characteristic
member
is characteristic
of the connection withdrawal
withdrawal strength
at anstrength parameter
angle ε parameter
to for
the grainfor aa screw
screw[N/mm
direction at th
at th
member
member of the connection
of the connection at an angle
at anstrength ε to the
angle ε parameter grain
to the grainfor direction
direction [N/mm
154 LVL5.5.3 Handbook Europe Inclined screw connections
fax,ε,2,k is the characteristic withdrawal a screw[N/mmat th
d
fax,ε,2,k is the
member outer of threaded
the connection diameter at an [mm];
angle
is the characteristic withdrawal strength parameter for a screw at th ε to the grain direction [N/mm
Inclined screwing is an efficient way to connect LVL members together dlg,1 or to othermember
is thetypes outer ofofthe
penetration connection
threaded length ofatthe
diameter an[mm];
angle ε topart
threaded theingrain direction
the head side[N/mm
mem
timber members. Although the connections transfer shear forces,dg,2 the fastenersisare the axially
outer threaded diameter [mm]; pointside memb
lg,1 is the penetration length of the threaded part 12.3.2020 in the head
12:44:55 side mem
loaded. The instructions in this subsection are based on the Finnish Handbook RIL205-
2020_LVL_05.indd 154
31 lg,1 is the penetration length of the threaded part in the head side mem
𝑓𝑓𝑓𝑓ax,ε,1,k 𝑑𝑑𝑑𝑑 𝑙𝑙𝑙𝑙g,1
𝑅𝑅𝑅𝑅C,k = min �𝑓𝑓𝑓𝑓ax,ε,2,k 𝑑𝑑𝑑𝑑 𝑙𝑙𝑙𝑙g,2 (5.29)
0,8 𝑓𝑓𝑓𝑓tens,k
5. STRUCTURAL DESIGN OF CONNECTIONS
𝜌𝜌𝜌𝜌 0,8
𝑓𝑓𝑓𝑓ax,ε,1,k 𝑑𝑑𝑑𝑑 𝑙𝑙𝑙𝑙g,1 + 𝑓𝑓𝑓𝑓head,k 𝑑𝑑𝑑𝑑h2 �𝜌𝜌𝜌𝜌𝑘𝑘𝑘𝑘 �
where a
𝑅𝑅𝑅𝑅T,k = min � 𝑓𝑓𝑓𝑓ax,ε,2,k 𝑑𝑑𝑑𝑑 𝑙𝑙𝑙𝑙g,2 strength parameter A (5.30) B
fax,ε,1,k is the characteristic withdrawal
𝑓𝑓𝑓𝑓
for a screw at the head side member of the connection
tens,k
at an angle ε to the grain direction [N/mm2];
fax,ε,2,k is the characteristic withdrawal strength parameter
for a screw at the pointside member of the connection
at an angle ε to the grain direction [N/mm2];
d is the outer threaded diameter [mm];
lg,1 is the penetration length of the threaded part in the
head side member [mm];
lg,2 is the penetration length of the threaded part in the
pointside member [mm];
ftens,k is the characteristic tensile capacity of the screw
determined in accordance with EN 14592 [N];
fhead is the characteristic pull-through parameter of the
screw for the associated density ρa [N/mm2];
dh is the head diameter [mm];
ρk is the characteristic density of LVL [kg/m3]; and
ρa is the associated density for fhead,k [kg/m3].
The characteristic load-carrying capacity of the tension screw connection, see Figure 5.11
(b), is calculated by the equation:
2020_LVL_05.indd 155 12.3.2020 12:44:56
𝑅𝑅k = 𝑛𝑛0,9 𝑅𝑅T,k (cos 𝛼𝛼 + 𝜇𝜇 ∙ sin 𝛼𝛼) (5.37)
5. STRUCTURAL DESIGN OF CONNECTIONS
0,082 ∙ (1 − 0,01𝑑𝑑𝑑𝑑)𝜌𝜌𝜌𝜌
Connection detailing 𝑓𝑓𝑓𝑓h,k = (𝑘𝑘𝑘𝑘
where 2 2
k
2 2
90 ∙ sin 𝛼𝛼𝛼𝛼 + cos 𝛼𝛼𝛼𝛼) ∙ (𝑘𝑘𝑘𝑘𝐶𝐶𝐶𝐶 ∙ cos 𝛽𝛽𝛽𝛽 + sin 𝛽𝛽𝛽𝛽)
Members should be pre-drilled when the diameter d of the ρk is the characteristic density [kg/m³];
screw exceeds 8 mm or the diameter of the smooth shank ds α is the angle between load and grain direction; for
exceeds 6 mm. The diameter of pre-drilled holes for non-self- softwood LVL-C / GLVL-C and α > 45°, α may be
drilling screws should be D = 0.5d - 0.7d, but not more than assumed as 45°;
the inner diameter of the threaded part di. β is the angle between bolt axis and wide face;
d is the bolt diameter 189 (255) [mm];
The thickness of the member should be at least: k90 is 1,15 + 0,015 for softwood LVL / GLVL; and
5𝑑𝑑𝑑𝑑
t=max{█(5d@(10d-30)
𝑡𝑡𝑡𝑡 = max �(10𝑑𝑑𝑑𝑑 − 30) 𝜌𝜌𝜌𝜌ρ_k/400)┤
k (5.38) (5.34)
k_C=max{█(d/( 𝑑𝑑𝑑𝑑
400 for softwood LVL / GLVL (5.40)
Members should be pre-drilled when the diameter d of the screw 𝑘𝑘𝑘𝑘 =
exceeds
C max 8 mm or for
� (𝑑𝑑𝑑𝑑−2) thesoftwood LVL / GLVL
diameter
where of the smooth shank ds exceeds 6 mm. The diameter of pre-drilled 1,15
holes for non-
ρself-drilling screws should be D = 0.5d - 0.7d, but not more than the inner diameter of the
k is the characteristic density [kg/m ]; and
3
threaded part d .
d is the screw diameter [mm].
i
5.7 FASTENING LVL PANELS TO FRAMES
The thickness of the member should be at least:
General spacings and end and edge distances are present- Connections of LVL-C panels can be designed according to Eu-
ed in Table 5.1. 5𝑑𝑑 values are valid for cross screw connec-
These rocode
𝑡𝑡 = max {(10𝑑𝑑 − 30) k 𝜌𝜌 (5.38)5 instructions for timber-to-timber connections with
tions when the compressed
400 and tensioned fasteners of the commonly available fasteners. The connection capacities and
screw pairs are placed in separate longitudinal rows parallel to minimum thickness of the frame depend on the panel, fastener
where
the grain, so that the spacing between the rows a2 is 4d and the and frame material combinations. Capacities and dimension
staggered
ρk distance
is thebetween the screw
characteristic heads[kg/m
density of a screw
3
]; andpair is for connections of LVL-C with LVL-P, glulam or solid timber
not more than 3t1 parallel to the grain. frames are presented in Table 5.5.
d Different types is theorscrew diameter [mm].
sizes of screws must not be combined
General
in the samespacings and end
connection. and edge
All screws mustdistances are presented
be positioned at the in Table 5.1. These values are
valid for cross screw connections when the
same inclination angles ε and β in a member. The screws must compressed and 5.8 SPECIAL
tensioned fasteners ofCASES the
screw pairs are placed in separate longitudinal
be positioned centrally to the connection force and screwed rows parallel to the grain, so that the spacing
Many nail plate manufacturers have tested the anchoring
between
deep enoughthe so
rows 2 is 4d and the staggered distance between the screw heads of a screw
thatathe screw head is in full contact with the strength of their products for LVL-P members and the values
pair is not more than 3 t1 parallel to the grain.
member surface. The minimum pointside penetration depth of are of similar levels to C30 structural timber. The values are in-
the threaded
Different partorshould
types sizes be 6d. The members
of screws must not should be com-in the same
be combined cluded in their design
connection. software for trusses. LVL bottom cords
All screws
pressed
must betogether
positioned so that
at theno gaps
sameareinclination
present 31.angles ε and β in a member. are usedThe in nail plate must
screws trussesbeto give additional stiffness for the
positioned centrally
LVL suppliers to thetheir
provide connection
own specificforce instructions
and screwed fordeep enough so that
floor part thetrusses
of attic screwand fire resistance in cases where the
head is in full contact with the member
inclined screw connections with their products. surface. The minimum pointside penetration
rest of the trusses may depth of
be exposed to fire.
the threaded part should be 6d. The members should be compressedUsually together the so that
anchoring no devices are fastened to bracing pan-
gaps are present 31. els with laterally loaded connections to the wide face of the LVL
5.6 BOLTED AND DOWELED
panels.
LVL suppliers provide their own specific instructions for inclined screw connections with their However, glued-in rod or glued-in screw connections
CONNECTIONS on the edge face of LVL panels can be an efficient solution for
products.
For bolted and dowelled connections the design rules are de- anchoring large bracing panels. They however require sepa-
5.6 for connections
fined Bolted and doweled
perpendicular to connections
the grain of the face rate type testing and production quality assurance and, in some
veneer. The design of LVL connections follows Eurocode 5 Sec- countries, also separate product certification. Glued-in rod or
For bolted
tions 8.5 and and
8.6dowelled connections
with the exception of thetheembedment
design rules are definedglued-in
calcu- for connections
screw connections at the edge face require a panel
perpendicular to the grain of the face veneer.
lation. The rules for connection geometry are given in Section The design of LVL thickness of atfollows
connections least 66 mm to fulfil the edge distance require-
Eurocode
5.2 and the5rules
Sections 8.5 failure
for wood and 8.6 with the
modes exception
in Section ofthis
5.3 of the embedmentment ofcalculation.
the connection.The rules
for connection
handbook. geometry are given in Section 5.2 and the rules for wood failure modes in
Section 5.3 of this handbook.
In laterally loaded connections of bolts and dowels up to
30 mm diameter
In laterally loaded arranged perpendicular
connections of bolts to andthedowels
grain, the fol-30 mm diameter arranged
up to
lowing characteristic
perpendicular embedment
to the grain, strengthcharacteristic
the following values shouldembedment
be strength values should be
used
usedfor forLVL
LVL3232: :
where
B: LVL 36 C panels nailed with smooth round nails to GL30c or C24 frame
h 1) d 2) Lmin 3) ≥ a 4) ≥ b 5) ≥ c 6) Rk 7)
[mm] [mm] [mm] [mm] [mm] [mm] [N]
24 2,1 50 11 11 45 440
27 2,5 60 13 13 63 580
33 2,8 70 14 14 63 700
45 3,1 90 16 16 63 840
57 3,4 100 17 17 75 950
69 4,2 125 21 21 90 1350
Figure 6.3. Influence of temperature on the mechanical properties of softwood. Left: Reduction of modulus of elasticity parallel to grain, Right:
Reduction of strength parallel to grain (EN1995-1-2:2004, Figure B.4 and B.5).
The requirements of reaction to fire classes for different ap- protect underlying products against damage. The classes may
plications are defined in national building regulations. In gen- be achieved by LVL panels, when the minimum thicknesses are
eral, LVL panels and structures may be left exposed in a build- according to the Table 6.1. However, in some applications only
ing under the same conditions as other solid wood products. non-combustible materials are approved for the encapsulation.
In the classification, the first letter A-E denotes combusti-
bility, with D being typical for wood products. The second sym- Note: the same layer in a structure can give fire protection in ac-
bol, s1-s3, denotes smoke production and the third symbol, cordance with the K-classes and at the same time be taken into
d0-d3, denotes the risk of flaming droplets. The class for LVL account for fire resistance time of the structure.
is similar to most untreated wood products. Some manufactur-
ers specify for some applications a smoke production class of 6.4 FIRE RESISTANCE OF LVL
s1 based on separate testing, which is slightly better than the
typical class s2. The flaming droplets class d0 assigned to wood- STRUCTURES
based products means that no flaming droplets or particles are The fire resistance of LVL structures can be calculated ac-
produced. Although the European system has been in use since cording to the structural fire design specification of Eurocode
the beginning of 2000, national classification systems are still 5 (EN1995-1-2) and its National annexes. The model of the
in parallel use in some countries. structural system adopted for design shall reflect the perfor-
The reaction to fire classification can be improved by fire mance of the structure in the fire situation.
retardant treatments or with inorganic surface laminates up to
a class B-s1,d0, which is the highest class for combustible ma- 6.4.1 Fire resistance design process
terials. The fire retardants are mainly salt-based chemicals that
are usually hygroscopic, meaning that they absorb humidity The fire resistance design process has the following steps:
from the surrounding air. It is therefore essential that the du- 1. Determination of the charring depth
rability of the treatment is verified for the intended use class for The charring depth is the distance between the outer sur-
the whole service life of the product. face of the original member and the position of the char line.
LVL manufacturers provide fire retardant treatment as an It is calculated using the time of fire exposure and the relevant
additional service for their products, details of which can be charring rate which depends on the material of the structure
LVL 06, Table 6.1
found from their product information material. and possible additional protection layers on the structure. The
original size reduced by the charring depth on exposed sides is
6.3 FIRE PROTECTION ABILITY, defined as the residual cross section.
Table 6.1. Classes of fire protection ability performance of LVL – laminated veneer lumber according to EN14374. The table has been published
as a draft annex of (EC Ref. Ares(2017)2463446 - 15/05/2017) pending formal issuing as a European Commission Delegated Regulation 38.
Figure 6.4. Left: One-dimensional charring of panel or wide cross section when fire exposure is below on one side, Right: Charring depth dchar,0
for one-dimensional charring and notional charring depth dchar,n which takes into account the rounding of corners.
where
d0 is 7 mm;
dchar,n is the notional design charring depth, see equation
(6.4); and
k0 is in the case of unprotected surfaces t/20, when
t< 20 min and 1,0 when t>20 min. In the case of
protected surfaces the value of k0 is given in Figure 6.6.
Figure 6.6. (a) Variation of k0 for unprotected members and protected members where tch ≤ 20 minutes and (b) for protected members where
tch > 20 minutes.
For timber surfaces facing a void cavity in a floor or wall 4. Determination of the design values of actions
assembly (normally the wide sides of a stud or a joist),the fol- The design effect of actions Ed,fi for the fire situation is
lowing applies: determined in accordance with EN 1991-1-2:2002, including
• Where the fire protective cladding consists of one or two the effects of thermal expansions and deformations. In typical
layers of gypsum plasterboard type A, wood panelling or cases of timber structures where the own weight is relatively
wood-based panels, at the time of failure tf of the cladding, low, the design values of actions Ed,fi are 0,2-0,4 times the ac-
k0 should be taken as 0,3. Thereafter k0 should be assumed tions Ed in normal temperature design.
to increase linearly to 1,0 during the following 15 minutes;
• Where the fire protective cladding consists of one or two lay- 5. Verification that design resistance is larger than design
ers of gypsum plasterboard type F, at the time of start of char- action
ring tch, k0 is 1. For times t < tch, linear interpolation should It shall be verified for the required duration of fire expo-
be applied, see Figure 6.6 (b). sure t that: Ed,fi ≤ Rd,t,fi
The effective cross section should be used for the calculation of According to Eurocode 5, clause 4.3 Simplified rules for analysis
the stiffness and fire resistance of an LVL member. of structural members and components, compression perpen-
Note: The effective cross section method is recommended. dicular to the grain and shear resistance may be disregarded.
However, depending on the National Annex, the reduced prop-
erties method of Eurocode 5 may also be used. 6.4.2 Charring rates of LVL
3. Determination of design values of strength and stiffness There are two different types of charring rates β0 and βn. For
For the calculation of the design values of mechanical re- panels and wide cross sections one-dimensional charring rate
sistance Rd,t,fi in a fire situation, the design values of strength β0 is used in the calculations. This is also used as the basis value
properties shall be determined by the equation: in some more advanced calculation methods. When the char-
acteristic density of LVL is ρk ≥ 480 kg/m3, the one-dimension-
f_(d,fi)=k_(mod,fi) 𝑓𝑓𝑓𝑓20
d,fi = 𝑘𝑘𝑘𝑘mod,fi 𝛾𝛾𝛾𝛾
𝑓𝑓𝑓𝑓 f_20/γ_(M,fi) (6.2) (EC5 2.1) al charring
(6.2) (EC5 β0 is 0,65 mm/min.
rate2.1)
M,fi
The design charring depth for one-dimensional charring
where dchar,0 [mm] should be calculated as follows when the surface is
fd,fi is the design strength in fire; unprotected throughout the time of fire exposure:
f20 is the 20 % fractile of a strength property at normal
temperature. It can be calculated as f20 = kfi ∙ fk. For dchar,0=β0 t (6.3) (EC5 3.1)
LVL kfi is 1,1, so f20 is 1,1 times the characteristic
strength fk; where t [min] is the time of fire exposure and β0 [mm/min] is
kmod,fi is the modification factor for fire. It replaces the the one-dimensional charring rate.
modification factor for normal temperature design For all other structures that are exposed from multiple
kmod given in EN 1995-1-1. kmod,fi is 1,0 in most sides, generally columns and beams, the notional charring rate
cases, except when the method of annex C of β0 is used in the calculations of the notional depth dchar,n. When
EN 1995-1-2 is used; and the characteristic density of LVL is ρk ≥480 kg/m3, the notional
γM,fi is the partial safety factor for timber in fire. The charring rate βn is 0,70mm/min.
recommended factor for material properties in fire is The design charring depth for notional charring dchar,n
γM,fi = 1,0. Information on national choice may be should be calculated as follows when the surface is unprotect-
found in the national annex. ed throughout the time of fire exposure:
For example, the design value of bending strength for LVL- dchar,n=βn t (6.4) (EC5 3.2)
36 C:
N where t [min] is the time of fire exposure and βn [mm/min] is
𝑘𝑘𝑘𝑘fi ∙ 𝑓𝑓𝑓𝑓m,k
f_(m,d,fi)=k_(mod,fi)∙(k_fi∙f_(m,k))/γ_(M,fi) 1,1 ∙ 36 N N/
2=1,0∙(1,1∙36 the notional charring rate
𝑓𝑓𝑓𝑓m,d,fi = 𝑘𝑘𝑘𝑘mod,fi ∙ = 1,0 ∙ mm = 39,6
mm^2 )/1,0=39,6 N/mm^2 𝛾𝛾𝛾𝛾M,fi 1,0 mm 2 In the test report VTT-S-04746-16 the one dimension-
al charring rate of different wood products was evaluated in
For stability calculations, the characteristic values of stiffness 120min fire exposure according to a standardized time-tem-
properties at normal temperature are used. perature exposure curve (EN 1363-1:2012) 39. According to the
report, wood products behaved predictably and, for LVL, the
one-dimensional charring rate β0 = 0,65mm/min can be used
for an extended fire exposure. The results were similar in both
face side and edge side exposure specimens. This gives the nec-
essary information and confidence for fire designers in assessing
80
Note: For special cases where more advanced design methods are
used, the report VTT-S-04746-16 also has information on the
70
charring rateOne-dimensional
Figure 6.7. β0 in a test based on to a more
charring of stringent
LVL-C ishydrocar-
linear in a 120 minute fire exposure test
according
bon to the standardized
(HC) time-temperature time-temperature
exposure curve. Blue and red curves: exposure
curve (EN 1363-2:1999). on the
panels
Note: For special cases where more advanced design methods are used, the report
40 VTT-S-
04746-16 also has information on the charring rate β0 in a test based on to a more stringent
Since
hydrocarbon LVL beams (HC)are typically slenderexposure
time-temperature structurescurve
with (ENlarg-1363-2:1999). 30
est available beam thicknesses up to 75 mm without multi-
ple gluing, unprotected
6.4.3 Design of unprotected LVL beams cannot beamsbe anddesigned
panels for 20
higher than 15 min fire resistance time requirements. The ze-
Since
ro LVL beams
strength layer (k0are ∙ d0typically
) reducesslender structures
the thickness of an with
effectivelargest available beam thicknesses
10 up
to 75 mm without multiple gluing,
cross section significantly, making the beam even more slender unprotected LVL beams cannot be designed for higher than
15 min fire resistance time requirements. The zero strength layer (k0 ∙ d0) reduces the 0
in lateral torsional buckling analysis.
thickness of an effective cross section significantly, making the beam even more slender 0 in 40
20 60 80 100 120
lateral torsional buckling analysis. TIME [min]
Example: 63x300 mm LVL-P beam and 33 mm LVL-C panel Wide face 2 Wide face 1
in 15 minute
Example: fire exposure:
63x300 mm LVL-P beam and 33 mm LVL-C panel in 15 minute fire exposure: Edge face 1 Edge face 1
Beam: Beam:
d_(ef,beam)=β_n∙t+k_0∙d_0=0,70 mm mm/ 15min Figure 6.7. One-dimensional charring
𝑑𝑑ef,beam = 𝛽𝛽n ∙ 𝑡𝑡 + 𝑘𝑘0 ∙ 𝑑𝑑0 = 0,70
min∙15min+15min/20min∙7mm=15,75mm ∙ 15min + ∙ 7mm = 15,75mm
min 20min of LVL-C is linear in a 120 minutes
Size of effective cross section of the beam in 3-side fire exposure: fire exposure test according to the
Size
Width b: of effective cross section
63 mm - 2∙15,75 mm = 31,5 mm of the beam in 3-side fire expo- standardized time-temperature curve.
sure: Blue and red curves: exposure on the wide
Height h: 300 mm - 15,75 mm = 284 mm face of the specimens. Green and grey
Width b: 63 mm - 2∙15,75 mm =31,5 mm curves: exposure on the edge face of the
Design Height
valueh: of bending 300 mmstrength - 15,75 mm = 28448
for LVL mm P: specimens 39.
0,15
300mm 0,15 N
Design value of bending 𝑘𝑘fi ∙ 𝑘𝑘h ∙strength 1,1 ∙ (300mm
for LVL )0,15 ∙ 44 N 2
𝑓𝑓m,k
𝑓𝑓m,d,fi = 𝑘𝑘mod,fi ∙ 𝑘𝑘𝑘𝑘fi ∙ 𝑘𝑘𝑘𝑘h ∙ 𝑓𝑓𝑓𝑓m,k = 1,0 ∙1,1 1,1 ∙48 P:
284mm
�300mm � ∙ 44 mm
N 2 = 44,3 N N
f_(m,d,fi)=k_(mod,fi)∙(k_fi∙〖k_h∙f〗_(m,k))/γ_(M,fi) 284mm
∙ �284mm � ∙ 44 mm2 = 44,3 mm
𝑓𝑓𝑓𝑓m,d,fi = 𝑘𝑘𝑘𝑘mod,fi ∙𝑘𝑘𝑘𝑘fi ∙ 𝑘𝑘𝑘𝑘𝛾𝛾hM,fi ∙ 𝑓𝑓𝑓𝑓m,k = 1,0 ∙ 1,00,15 mm N 2
m,d,fi = 𝑘𝑘𝑘𝑘mod,fi ∙
𝑓𝑓𝑓𝑓=1,0∙(1,1∙(300mm/284mm)^0,15∙44 𝛾𝛾𝛾𝛾M,fi = 1,0 ∙N/mm^2 300mm 1,0 N = 44,3 mm22
𝑘𝑘𝑘𝑘fi ∙𝛾𝛾𝛾𝛾𝑘𝑘𝑘𝑘M,fi
h ∙ 𝑓𝑓𝑓𝑓m,k
1,1 ∙ �284mm)/1, 1,0
� ∙ 44 2 mmN
𝑓𝑓𝑓𝑓Panel: mm
m,d,fi = 𝑘𝑘𝑘𝑘mod,fi ∙ = 1,0 ∙ = 44,3
𝛾𝛾𝛾𝛾M,fi 1,0 mm2
Panel:
mm 15min
d _ ( e f ,=
𝑑𝑑𝑑𝑑ef,panel p a𝛽𝛽𝛽𝛽n ∙e𝑡𝑡𝑡𝑡l +
) =𝑘𝑘𝑘𝑘β ∙_𝑑𝑑𝑑𝑑0 ∙ = k _ mm
t +0,65 0 = 0 , 6+515min ∙ m
0 ∙ d _∙ 15min 7mmm / = 15 mm
𝑑𝑑𝑑𝑑ef,panel = 𝛽𝛽𝛽𝛽00 ∙ 𝑡𝑡𝑡𝑡 + 𝑘𝑘𝑘𝑘00 ∙ 𝑑𝑑𝑑𝑑00 = 0,65 min∙ 15min + 20min∙ 7mm = 15 mm
min∙15min+15min/20min∙7mm=15 mmmm
min 20min
15min
𝑑𝑑𝑑𝑑ef,panel = 𝛽𝛽𝛽𝛽0 ∙ 𝑡𝑡𝑡𝑡 + 𝑘𝑘𝑘𝑘0 ∙ 𝑑𝑑𝑑𝑑0 = 0,65 ∙ 15min + ∙ 7mm = 15 mm
min 20min
mm
𝑑𝑑𝑑𝑑ef = 𝛽𝛽𝛽𝛽n ∙ 𝑡𝑡𝑡𝑡 + 𝑘𝑘𝑘𝑘0 ∙ 𝑑𝑑𝑑𝑑0 = 0,70mm ∙ 30min + 1,0 ∙ 7mm = 28mm
2020_LVL_06.indd 𝑑𝑑𝑑𝑑
163 ef = 𝛽𝛽𝛽𝛽n ∙ 𝑡𝑡𝑡𝑡 + 𝑘𝑘𝑘𝑘 0 ∙ 𝑑𝑑𝑑𝑑0 = 0,70 min∙ 30min + 1,0 ∙ 7mm = 28mm 12.3.2020 10:54:20
min
mm
6. PERFORMANCE OF LVL IN FIRE
Table 6.2. Effective thickness of LVL 36 C panels after 15 - 60 min fire exposure on one side.
Table 6.3. Minimum thickness of LVL 36 C panel to give a fire exposure protection time 30 - 90 minutes for an underlying wooden structure.
6.4.4 LVL-C panel as a protection against 6.4.5 Summary of LVL-C panels for fire
fire exposure protection
When the LVL-C panel thickness is according to Table 6.3, the Table 6.10 specifies the minimum thickness hp of LVL-C panel
panel protects a wooden structure behind it for a certain fire when used as a covering with fire protection ability for the un-
resistance time t [min]. Other fire design calculations of the derlying materials (column A), fire protection of the structures
remaining wooden structure are not needed unless the panel (column B) or when used as a ceiling structure which has a fire
is a part of the load-bearing system also in the structural fire resistance requirement EI (column C). All of the different fire
design. In the case of timber frame assemblies, the LVL-C pro- protection specifications also have requirements for the detail-
tection panel thickness hp1 [mm] is calculated for the required ing of, e.g., joints between the panels.
fire protection time t based on EN1995-1-2, equation (4.1) and
(C.7) or (D.3):
LVL 06, Figure 6.10
Minimum panel thickness hp1 = β0 ∙ (t+4min)+7mm(6.5)
A B C
Underlying structure
Increase Increase
250 °C 140°C
(max 270°C) (max 180°C)
hp
hp
hp
Panel as K2 classified covering Panel as fire protection Panel as ceiling with El requirement
Figure 6.11. LVL joist floor structure with rock fibre filled cavities. Wooden battens fixed to the joists create a shelf that keeps the rock fibre
insulation in place protecting the sides of the joists from charring.
resistant to strong acids, but its resistance to bases is good. and balcony structures, balustrades, stairs and piers where the
Lignin, on the other hand, is easily dissolved in bases, whereas products are not in direct ground contact (= Use class 3.1 con-
it is resistant to most strong acids. For these reasons, wood is ditions).
quite resistant to moderate chemical effects. AB class impregnation agents are aggressively corrosive and
LVL has good resistance to mild acids and acid salt solutions. therefore only stainless steel connectors should be used in
Alkalis, however, cause softening of the wood. Direct contact structures made of impregnated LVL.
with oxidizing agents such as chlorine, hypochlorites and ni- The surface of pressure-impregnated LVL is rougher than
trates should be avoided. regular LVL. The impregnation and drying processes cause
Wood is generally quite resistant to organic substances. swelling and shrinkage, which open some of the peeling cracks
However, organic solvents such as acetone, benzene, alcohol in the surface veneers. Local cracking around knots and minor
etc. dissolve resins, fats and waxes, causing similar effects to separation of scarf joints in small areas of the surface veneer
water, i.e., producing swelling and a slight reduction in strength might occur.
properties. Petroleum oils have no effect on strength properties Impregnated LVL products swell in the process and their
but cause discolouration. equilibrium moisture content and specific weight are a cou-
Chemical resistance can be improved with various types of ple of per cent higher than untreated products. The increased
coatings 42. weight shall be taken into account, but otherwise structural
design shall be made with the nominal dimensions of the prod-
uct with the strength and stiffness reduction factors according
7.4 CHEMICAL WOOD PROTECTION to service class 3 of Eurocode 5. In some countries additional
7.4.1 Surface treatment reduction factors given by their national requirements shall be
taken into account 44.
LVL can be protected against temporary weather exposure by
surface treatments that repel rainwater, and therefore reduce
the amount of moisture absorbed by the product, but allow wa-
ter vapour movement to and from the product. This improves
dimensional stability and reduces swelling of the treated LVL
product during the construction time.
Use in high relative humidity conditions may result in mould
growth on the surface of structural LVL products. If the prod-
ucts are exposed to outdoor air humidity conditions (e.g. struc-
tures in unheated spaces) or wetting during the logistic chain
and construction time, a brushable or sprayable treatment
should be applied on the surface of the LVL components to
reduce the risk of mould growth. In some cases LVL compo-
nents are treated at the factory, but the default practice is for
treatment to be done during offsite production of timber ele-
ments or on the construction site. If there is mould growth on
the surface of LVL products, the mould must be removed, e.g.
by sanding, before closing the structure.
The surface treatments do not affect the product’s strength
properties, but their compatibility with end finishes, such as
paints, shall be verified separately 28, 43.
Figure 8.1. Roof overhang protects the building against weather exposure and solar radiation, Kindergarten Vekara, Pukkila, Finland.
When leaving the factory, the moisture content ω of the LVL product is appr
10%. Due to changes in ambient temperature and relative humidity, the mo
the product will continuously change. In service class 1 the moisture conten
between 6 and 10%, while in service class 2 it usually varies between 10 an
LVL products are delivered from the factory at a moisture content that is clo
conditions.
LVL beams, studs and panels can be used in structures simi- Product
Product moisture
moisture content
content ω defined
ω is is definedasasfollows:
follows:
lar to all other timber and engineered wood products. Build-
𝑚𝑚 −𝑚𝑚
ing physics analyses of walls and roofs do not need any special ω=(m_ω-m_0
𝜔𝜔 = 𝜔𝜔 0 )/m_0
𝑚𝑚
(8.1) (8.1)
0
methods and normal design tools are suitable for their ther-
mal resistance calculation and dew point analyses. As LVL has where
where mω is the product mass at moisture content ω; and
fairly low thermal conductivity, the cold bridging effect of LVL m0 ismass
mω is the product the product dry content
at moisture mass. ω; and
beams or studs is minimal. In special cases LVL-C panels may m0 is the product dry mass.
be used alone as water vapour barriers with no separate plastic The average equilibrium moisture content of LVL products in different relativ
conditions (RH%) can be estimated using a sorption isotherm Figure 8.1. W
membranes required. The average equilibrium moisture content of LVL products in
drying (desorption) its equilibrium moisture content is higher in the same rel
different
than when relative humidity
the wood conditions
is wetting (RH%) canThis
(absorption). be estimated
phenomenon is called hys
using a sorption isotherm Figure 8.1. When wood is drying
8.1 LVL AND MOISTURE (desorption) its equilibrium moisture content is higher inMoisture
the absorption and desorption of s
30
LVL is a hygroscopic material similar to other wood-based same relative humidity than when the wood is wetting (absorp-
products. Therefore, the moisture content of LVL products is tion). This phenomenon is called hysteresis. 25 Absorptio
Desorptio
dependent on the relative humidity (RH%) and, more specif-
ically, on the direction of moisture content change (drying / 8.1.2 Measuring the moisture content 15
wetting). LVL products swell when their moisture content in-
10
creases and shrink when their moisture content decreases. A Moisture meters based on electrical resistance give somewhat
part of the swelling is permanent and the extent of these di- too high results for LVL due to the glue lines of LVL products.
5
mensional changes depends on the grain direction. Wetting For exact determination of moisture content in a LVL0sample,
0 10 20 30 40 50 6
can cause permanent deformations, and impair the visual ap- an oven drying test can be conducted according to EN 322. Relative humidity (R
pearance of surface veneers, such as colour changes due to wa- Surface moisture meters (non-invasive) are recommend-
ter staining, surface cracks and falling of knots due to drying ed for measuring
Figure the moisture
8.1. Left: Average content moisture
equilibrium of LVL products.
content Theof softwood LVL in di
shrinkage after wetting, see also subsection 7.3.1. measurements
humidity at 20should be taken
°C. Right: perpendicular
Absorption to the grain
and desorption di-
isotherms of softwood L
The hygroscopic surface of LVL has also advantages. If the based on
rection weather
from the facecycling
veneerstests,
at anRH 65% → 92%
undamaged → 40%
location. E.g.,45.
surface is untreated or the surface treatment does not form a through sanded areas of the face veneers cannot be measured
membrane on the surface, LVL can have a moisture buffering 8.1.2 To obtain
reliably. Measuring
the mostthe moisture
reliable content
results, the moisture meter
function absorbing humidity from the air when the RH% is should be calibrated using samples of known moisture content
Moisture meters based on electrical resistance give somewhat too high resu
high and releasing it when the RH% is low. This action levels measured, for example,
the glue lines by oven drying.
of LVL products. For exact determination of moisture content
out the peaks and can help to create pleasant indoor air condi- an oven drying test can be conducted according to EN 322.
tions. In unheated storage spaces, the hygroscopicity prevents Note: Examples of suitable moisture meters for measuring the
water condensation on cold surfaces and thus the risk of wa- Surface moisture
moisture content of meters (non-invasive)
spruce LVL are the Deltaare recommended
2000H (setup: H3 for measuring th
ter dropping from, e.g., roof structures is smaller than in, e.g., content and
Spruce) of LVL
the products. The measurements
Doser Messgerät should group
HD5 (setup: material be taken perpendicular
steel structures. direction
3) 18. from the face veneers at an undamaged location. E.g., through san
face veneers cannot be measured reliably. To obtain the most reliable resul
meter should be calibrated using samples of known moisture content measu
8.1.1 Moisture content of LVL by oven drying.
When leaving the factory, the moisture content ω of the LVL Note: Examples of suitable moisture meters for measuring the moisture con
product is approximately 8 to 10%. Due to changes in ambient LVL are the Delta 2000H (setup: H3 Spruce) and the Doser Messgerät HD5
temperature and relative humidity, the moisture content of the group 3) 18.
product will continuously change. In service class 1 the mois-
ture content usually varies between 6 and 10%, while in service
class 2 it usually varies between 10 and 16%. Thus, the LVL
products are delivered from the factory at a moisture content
that is close to the end use conditions.
25
20
contentcontent
20
15
Moisture
15
10
Moisture
10
5
5
0
0 10 20 30 40 50 60 70 80 90 100
0 Relative humidity (RH%)
0 10 20 30 40 50 60 70 80 90 100
Relative humidity (RH%)
Moisture absorption and desorption of softwood LVL, t = 20°C
30
Moisture absorption and desorption of softwood LVL, t = 20°C
30 Absorptio
25
Desorptio
MC (%)MC (%)
Absorptio
25
20
Desorptio
contentcontent
20
15
Moisture
15
10
Moisture
10
5
5
0
0 10 20 30 40 50 60 70 80 90 100
0 Relative humidity (RH%)
0 10 20 30 40 50 60 70 80 90 100
Relative humidity (RH%)
Figure 8.2. Left: Average equilibrium moisture content of softwood LVL in different relative humidity at 20 °C. Right: Absorption and
desorption isotherms of softwood LVL at 20 °C based on weather cycling tests, RH 65% 92% 40% 45.
where
Dimension LVL-P LVL-C Δω is the change in product moisture content [%];
αH is the product’s dimensional variation coefficient, see
Thickness t 0,32 0,32
Table 8.1 for values and Figure 8.3 for directions; and
Height h (or width of a panel) 0,32 0,03 L is the product dimension in the corresponding direction.
Length l 0,01 0,01
Notably, due to its cross band veneers LVL-C has a very
low αH factor in the member width direction: only 10% of the
value specified for LVL-P products. This advantage can be uti-
lized in structures that are sensitive to dimensional changes
due to moisture.
LVL products can warp if the moisture content of oppo-
site surfaces is not equal, for example if one surface is exposed
to a higher relative humidity than the other. LVL-P products
are more sensitive to such warping than LVL-C, especially if
the height of the product is more than 8 times the thickness
(h > 8t). Therefore, it is normally recommended to limit the
slenderness of the LVL-P beams to this ratio. If careful mois-
ture management of the components and structures can be as-
sured throughout the logistic chain and construction process,
e.g. in off-site element production, a h/t ratio of max ~12 may
be considered.
LVL 08, Table 8.2
Table 8.2. Example of dimensional changes due to moisture : If the relative humidity changes from 50% to 85%, the moisture content of a LVL
beam increases by approximately 7%. The resulting effect on beam dimensions is as follows:
Table 8.3. Water vapour resistance factor μ and water vapour diffusion coefficient in air δp of softwood LVL.
Water vapour resistance factor μ [-] Water vapour diffusion coefficient in air δp [kg/(Pa·s·m)]
Density ρmean Dry cup Wet cup Dry cup Wet cup
440 kg/m3 180 65 0,73 · 10-12
2,3 · 10-12
510 kg/m3 200 70 0,96 · 10-12 2,7 · 10-12
The dry cup values are tested in 23°C - 0/50 RH % and apply when the mean relative humidity across the material is less
than 70 %. The wet cup values are tested in 23°C - 50/93 RH % and apply when the mean relative humidity across the
material is greater than or equal to 70 %.
8.3 AIRTIGHTNESS
LVL-C panels are airtight beyond what can be measured. In
the building physics design of structures, it is essential pay at-
tention to the joints and seals between the panels and other
structures to ensure the airtightness of the entire building en-
velope. This can be achieved, e.g., with careful installation of
sealant tapes that are durable enough for the design service life
of the building.
LVL-C panels have been utilized, e.g., in 3-layer CLT wall
panels as the middle layer of the product to make them airtight.
These structural calculation examples for LVL are based on Eurocodes (EN1990, EN1991 and EN1995) and
the additional instructions given in Chapters 4-5. Where information from National annexes is required, the
Finnish annex or the default values of Eurocodes have been used. The calculations make references to the
equation numbers of the Chapters 4 - 6. The examples are chosen to demonstrate the calculations methods.
Therefore some of the component sizes may not necessarily ideal for practice and those cases have comments
of possibly better suitable sizes at the end of each example.
Joist properties:
Bending strength edgewise fm,0,edge,k = 48 N/mm2
Shear strength edgewise fv,0,edge,k = 4,2 N/mm2
Compression perpendicular to the grain edgewise fc,90,edge,k = 6 N/mm2
Modulus of elasticity E0,mean = 13800 N/mm2
Modulus of rigidity G0,edge,mean = 600 N/mm2
Area of cross section A = b∙h = 10800 mm2
Section modulus W = b∙h /6 2 = 4,32∙105 mm3
Moment of inertia I = b∙h /12
3 = 5,18∙107 mm4
Moment stiffness of the joist EI = 13800 N/mm ∙ 5,18∙10 mm = 7,15∙1011 Nmm2
2 7 4
Loading combinations
The most
𝐸𝐸𝐸𝐸 critical ultimate limit state (ULS) load combination: (4.1)
d,ULS = 𝛾𝛾𝛾𝛾G ∙ (𝑔𝑔𝑔𝑔1,k + 𝑔𝑔𝑔𝑔2,k ) + 𝛾𝛾𝛾𝛾Q ∙ 𝑞𝑞𝑞𝑞k
E_(d,ULS)
𝐸𝐸𝐸𝐸d,ULS = 𝛾𝛾𝛾𝛾G= ∙γ_G∙(g_(1,k)+g_(2,k))+
(𝑔𝑔𝑔𝑔1,k + 𝑔𝑔𝑔𝑔2,k )2+ 𝛾𝛾𝛾𝛾Q ∙ 𝑞𝑞𝑞𝑞k γ_Q∙q_k
2 2 (4.1) (4.1)
𝐸𝐸𝐸𝐸d,ULS = 1,15 ∙ (0,6 kN/m + 0,3 kN/m ) + 1,5 ∙ 2,0 kN/m
E_(d,ULS)
𝐸𝐸𝐸𝐸d,ULS = 1,15 =1,15∙(0,6
∙ (0,6 kN/m kN/m^2+0,3
2
+ 0,3 kN/m^2
kN/m 2
) +)+1,5∙2,0
1,5 ∙ 2,0 kN/m^2
kN/m2
2
𝐸𝐸𝐸𝐸d,ULS = 4,03
E_(d,ULS)= kN/m
4,03 kN/m^2
𝐸𝐸𝐸𝐸d,ULS = 4,03 kN/m2
Note: Safety factors γG and γQ are according to Finnish National annex of Eurocode 0.
The most critical serviceability limit state (SLS) load combination:
𝐸𝐸𝐸𝐸d,SLS = 𝛾𝛾𝛾𝛾=
E_(d,SLS) ∙ (𝑔𝑔𝑔𝑔1,k + 𝑔𝑔𝑔𝑔2,k ) + 𝛾𝛾𝛾𝛾G ∙ 𝑞𝑞𝑞𝑞k γ_G∙q_k (4.1)
G γ_G∙(g_(1,k)+g_(2,k))+ (4.1)
𝐸𝐸𝐸𝐸d,SLS = 𝛾𝛾𝛾𝛾G ∙ (𝑔𝑔𝑔𝑔1,k + 𝑔𝑔𝑔𝑔2,k2) + 𝛾𝛾𝛾𝛾G ∙ 𝑞𝑞𝑞𝑞k 2 (4.1)
E_(d,SLS)
𝐸𝐸𝐸𝐸d,SLS = 1,0 =1,0∙(0,6
∙ (0,6 kN/m kN/m^2+0,3
+ 0,3 kN/mkN/m^2 ) +)+1,0∙2,0 kN/m^2
1,0 ∙ 2,0 kN/m 2
2 2 2
𝐸𝐸𝐸𝐸d,SLS = 1,02,9
E_(d,SLS)= ∙ (0,6
kN/m^2kN/m
2 + 0,3 kN/m ) + 1,0 ∙ 2,0 kN/m
𝐸𝐸𝐸𝐸d,SLS = 2,9 kN/m
𝐸𝐸𝐸𝐸d,SLS = 2,9 kN/m2
ULS design
Bending moment resistance
𝑀𝑀𝑀𝑀d = 𝐸𝐸𝐸𝐸d,ULS ∙ 𝑠𝑠𝑠𝑠 ∙ 𝐿𝐿𝐿𝐿22/8 = 4,03kN/m22 ∙ 0,4m ∙ (4,5m)22/8 = 4,1 kNm
𝑀𝑀𝑀𝑀d == 𝐸𝐸𝐸𝐸E_(d,ULS)∙s∙L^2/8
M_d d,ULS ∙ 𝑠𝑠𝑠𝑠 ∙ 𝐿𝐿𝐿𝐿 /8 = = 4,03kN/m ∙ 0,4m ∙ (4,5m) /8 = 4,1
4,03kN/m^2∙0,4m∙〖(4,5m)〗^2/8 = 4,kNm
σ_(m,d)=M_d/W=(4,1
𝑀𝑀𝑀𝑀d 4,1 kNm kNm)/(4,32〖∙10〗^5 mm^3 )=9,5 N/mm^2
𝜎𝜎𝜎𝜎m,d = 𝑀𝑀𝑀𝑀d = 4,1 kNm = 9,5 N/mm2
𝜎𝜎𝜎𝜎m,d = 𝑊𝑊𝑊𝑊 = 4,32 ∙ 1055mm33 = 9,5 N/mm2
𝑊𝑊𝑊𝑊 4,32 ∙ 10 mm ∙k_h∙f_(m,0,edge,k)
f_(m,0,edge,d)=k_mod/γ_M
𝑘𝑘𝑘𝑘
𝑓𝑓𝑓𝑓(4.3) = 𝑘𝑘𝑘𝑘𝛾𝛾𝛾𝛾mod
mod ∙ 𝑘𝑘𝑘𝑘h ∙ 𝑓𝑓𝑓𝑓m,0,edge,k (4.3) 216 (255)
𝑓𝑓𝑓𝑓m,0,edge,d
m,0,edge,d = M ∙ 𝑘𝑘𝑘𝑘h ∙ 𝑓𝑓𝑓𝑓m,0,edge,k (4.3)
f_(m,0,edge,d)=0,8/1,2∙1,034∙44
𝛾𝛾𝛾𝛾M N/mm^2 =30,3 N/mm^2
σ_(m,d)≤f_(m,0,edge,d) 0,8 →OKN
𝑓𝑓𝑓𝑓m,0,edge,d = 0,8 ∙ 1,034 ∙ 44 N 2 = 30,3 N/mm22
𝑓𝑓𝑓𝑓m,0,edge,d = 1,2 ∙ 1,034 ∙ 44mm 2 = 30,3 N/mm
1,2 mm
0,8 N
𝜎𝜎𝜎𝜎𝑓𝑓𝑓𝑓m,d ≤ 𝑓𝑓𝑓𝑓m,0,edge,d
𝜎𝜎𝜎𝜎m,0,edge,d
≤ 𝑓𝑓𝑓𝑓 = ∙→ OK ∙ 44
1,034
→ OK 2
= 30,3 N/mm2
m,d 1,2
m,0,edge,d mm
𝜎𝜎𝜎𝜎m,d ≤ torsional
Lateral 𝑓𝑓𝑓𝑓m,0,edge,dbuckling
→ OK is prevented by the fixing of the decking.
𝑤𝑤𝑤𝑤 inst,g =
𝑤𝑤𝑤𝑤inst,g = 2,692,69 mm mm + + 0,17 0,17 mm
mm =
5∙𝑔𝑔𝑔𝑔d,SLS ∙𝑠𝑠𝑠𝑠∙𝐿𝐿𝐿𝐿4
= 2,86 2,866∙𝑔𝑔𝑔𝑔 mm
mm d,SLS ∙𝑠𝑠𝑠𝑠∙𝐿𝐿𝐿𝐿
2
𝑤𝑤𝑤𝑤inst,g = 384∙𝐸𝐸𝐸𝐸 +5 (4.74)
mean ∙𝐼𝐼𝐼𝐼 8∙𝐺𝐺𝐺𝐺 mean 2 𝐴𝐴𝐴𝐴
44 2
5 5 ∙ ∙ 𝑞𝑞𝑞𝑞
𝑞𝑞𝑞𝑞
w_(inst,q)=(5〖∙q〗_(d,SLS)∙s∙L^4)/(〖384∙E〗_mean∙I)+〖〖6/5
d,SLS
d,SLS ∙∙ 𝑠𝑠𝑠𝑠
𝑠𝑠𝑠𝑠 ∙∙ 𝐿𝐿𝐿𝐿
𝐿𝐿𝐿𝐿 6/5
6/5 ∙∙ 𝑞𝑞𝑞𝑞
𝑞𝑞𝑞𝑞
d,SLS
d,SLS ∙∙ 𝑠𝑠𝑠𝑠
𝑠𝑠𝑠𝑠 ∙∙ 𝐿𝐿𝐿𝐿
𝐿𝐿𝐿𝐿 ∙q〗_(d,SLS)∙s∙L〗^2/(〖8∙G〗_mean A)=5,97
𝑤𝑤𝑤𝑤 inst,q =
𝑤𝑤𝑤𝑤inst,q = ++ == 5,97
5,97 mm mm + + 0,38
0,38 mm
mm = 217mm
= 6,35
6,35 (255)
mm
mm+0,38 mm=6,35
384
384 ∙∙𝑤𝑤𝑤𝑤𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸inst,g
mean
mean mm =∙∙ 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼2,69 mm 88 ∙+ 0,17
∙ 𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺mean
mean𝐴𝐴𝐴𝐴
mm
𝐴𝐴𝐴𝐴 = 2,86 mm
w_inst = 2,86 mm+6,34 mm = 9,2 mm4
𝑤𝑤𝑤𝑤 inst =
𝑤𝑤𝑤𝑤inst
Requirement: = 2,86 2,86 mm mm + + 6,34 6,34
w_inst≤L/400=4500/400=11,3 5 mm ∙mm =
𝑞𝑞𝑞𝑞d,SLS= ∙9,2 𝑠𝑠𝑠𝑠 ∙ 𝐿𝐿𝐿𝐿mm
9,2 mm 6/5 ∙ 𝑞𝑞𝑞𝑞d,SLS ∙ 𝑠𝑠𝑠𝑠 ∙ 𝐿𝐿𝐿𝐿2
𝑤𝑤𝑤𝑤inst,q = + mm→OK = 5,97 mm + 0,38 mm = 6,35 mm
3844500
𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿
∙ 𝐸𝐸𝐸𝐸mean
4500
4500 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 ∙ 𝐼𝐼𝐼𝐼 8 ∙ 𝐺𝐺𝐺𝐺mean 𝐴𝐴𝐴𝐴
Requirement:
Requirement: 𝑤𝑤𝑤𝑤 𝑤𝑤𝑤𝑤inst inst ≤ ≤ 400 ,= = 400 400
= = 11,3 =
11,3 11,3 mm
mm mm → → OK→OKOK
400 400
𝑤𝑤𝑤𝑤inst = 2,86 mm + 6,34 mm = 9,2 mm
Final deflection
Final deflection Requirement: 𝑤𝑤𝑤𝑤 𝐿𝐿𝐿𝐿 4500
inst ≤ 400 = 400 = 11,3 mm → OK
w_(net,fin)
𝑤𝑤𝑤𝑤 net,fin =
𝑤𝑤𝑤𝑤net,fin = (1 (1 =+ (1+k_def)∙w_(inst,g)
+ 𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘def def)) ∙∙ 𝑤𝑤𝑤𝑤 𝑤𝑤𝑤𝑤inst,ginst,g + + (1 (1 + +(1+ψ_2∙k_def)∙w_(inst,q)
+ 𝜓𝜓𝜓𝜓
𝜓𝜓𝜓𝜓22 ∙∙ 𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘def
def)) ∙∙ 𝑤𝑤𝑤𝑤 𝑤𝑤𝑤𝑤inst,q
inst,q
(4.73)
(4.73) (4.73)
For the load category A: ψ2 = 0,3
For
For the
w_(net,fin) the load load=category category
(1+0,6)∙2,86 A:
A: ψ ψ mm 22 == 0,3
+0,3 (1+0,3∙0,6)∙6,35 mm = 12,1 mm
Requirement: w_(net,fin)≤L/300=4500/300=15 𝑤𝑤𝑤𝑤net,fin = (1 + 𝑘𝑘𝑘𝑘def ) ∙ 𝑤𝑤𝑤𝑤inst,g + mm→OK (1 + 𝜓𝜓𝜓𝜓2 ∙ 𝑘𝑘𝑘𝑘def ) ∙ 𝑤𝑤𝑤𝑤inst,q (4.73)
𝑤𝑤𝑤𝑤 net,fin =
𝑤𝑤𝑤𝑤net,fin = (1 (1 + + 0,6) 0,6) ∙∙ 2,86 2,86 mm mm + + (1 (1 + + 0,3 0,3 ∙∙ 0,6) 0,6) ∙∙ 6,35 6,35 mm mm = = 12,1
12,1 mm
mm
For the load 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿
category
4500
4500
A: ψ2 = 0,3
Requirement:
Requirement: 𝑤𝑤𝑤𝑤 net,fin ≤
𝑤𝑤𝑤𝑤net,fin ≤ 300 ,= =4500 300
300
mm
= = 15 =
1515 mmmm mm →→
→ OKOK
OK
300 300
𝑤𝑤𝑤𝑤net,fin = (1 + 0,6) ∙ 2,86 mm + (1 + 0,3 ∙ 0,6) ∙ 6,35 mm = 12,1 mm
Vibration design
Vibration design 𝜋𝜋𝜋𝜋 (𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)l
𝑓𝑓𝑓𝑓 = 𝜋𝜋𝜋𝜋 �(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼) 𝐿𝐿𝐿𝐿 4500 (4.79)
𝑓𝑓𝑓𝑓11 = 2𝑙𝑙𝑙𝑙𝜋𝜋𝜋𝜋22 � l
𝑚𝑚𝑚𝑚 l Requirement: 𝑤𝑤𝑤𝑤net,fin ≤
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼) = = 15 mm → OK (4.79)
𝑓𝑓𝑓𝑓
Lowest
Lowest = � 𝑚𝑚𝑚𝑚
2𝑙𝑙𝑙𝑙 2 natural frequency f 300 300 (4.79)
1 2𝑙𝑙𝑙𝑙 natural 𝑚𝑚𝑚𝑚 frequency1f1
𝜋𝜋𝜋𝜋 𝜋𝜋𝜋𝜋 (𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
𝑓𝑓𝑓𝑓1𝑚𝑚𝑚𝑚
𝑓𝑓𝑓𝑓1== = �
𝑔𝑔𝑔𝑔 1� + l l
𝑔𝑔𝑔𝑔 + 30 kg/m2 = 60 kg/m2 + 30 kg/m2 + 30 kg/m2 =(4.79)
𝑚𝑚𝑚𝑚2l + 30 kg/m22 = 60 kg/m22 + 30 kg/m22 + 30 kg/m22 = 120 kg/m22
2 (4.79)
120 kg/m2
𝑚𝑚𝑚𝑚
𝑓𝑓𝑓𝑓 =
2 𝜋𝜋𝜋𝜋1 +
=2𝑙𝑙𝑙𝑙2𝑙𝑙𝑙𝑙𝑔𝑔𝑔𝑔
2
� 𝑚𝑚𝑚𝑚 𝑔𝑔𝑔𝑔
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
(4.79)
𝑚𝑚𝑚𝑚 =
f_1=π/(2l^2
1 𝑔𝑔𝑔𝑔
𝜋𝜋𝜋𝜋 + 𝑔𝑔𝑔𝑔 + 30
𝑚𝑚𝑚𝑚)2l √((EI)_l/m)
(𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸) kg/m = 60 kg/m + 30 kg/m + 30 kg/m = 120 kg/m (4.79)
𝑓𝑓𝑓𝑓1 = 2𝑙𝑙𝑙𝑙221� 𝑚𝑚𝑚𝑚 (4.79)
𝑚𝑚𝑚𝑚m 𝑚𝑚𝑚𝑚===g_1+g_2+30
2𝑙𝑙𝑙𝑙
𝑔𝑔𝑔𝑔1𝑔𝑔𝑔𝑔1 ++𝑔𝑔𝑔𝑔2𝑔𝑔𝑔𝑔2++3030 kg/m^2
kg/m kg/m 2 2= 60 kg/m^2+30
==6060kg/m kg/m 22
++30kg/m^2+30
30kg/m kg/m 2 2 kg/m^22 =
++3030kg/m kg/m2=120 =120kg/m^2
120kg/m
kg/m22
𝑚𝑚𝑚𝑚 = 𝑔𝑔𝑔𝑔1 + 𝑔𝑔𝑔𝑔2 + 30 kg/m2 = 60 kg/m2 + 30 kg/m2 + 30 kg/m2 = 120 kg/m2
𝑚𝑚𝑚𝑚 = 𝑔𝑔𝑔𝑔1 + 𝑔𝑔𝑔𝑔2 + 30 kg/m2 = 60 kg/m2 + 302 kg/m2 + 30 kg/m2 = 120 kg/m2
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼) = 𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼 ∙ (1000/𝑠𝑠𝑠𝑠) = 7,15 ∙ 1011 Nmm ∙ (1000/400 mm)
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)ll = 𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼 ∙ (1000/𝑠𝑠𝑠𝑠) = 7,15 ∙ 1011 Nmm22 ∙ (1000/400 mm)
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
Note:l In
Note: = Finnish
In 𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼 ∙ (1000/𝑠𝑠𝑠𝑠)
Finnish NA NA the share = 7,15 of live ∙ 1011 loadload Nmm
qk inq the ∙ (1000/400
frequency calculation mm) is 30kg/m2
6 the2 share of live k in the frequency calculation is
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
30kg/m
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼) = 2 1,79 ∙ 10 6 Nm 2 /m 11 22
l ll== =𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼1,79
𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼∙ (1000/𝑠𝑠𝑠𝑠)
∙ (1000/𝑠𝑠𝑠𝑠)
∙ 10 6 Nm= =7,15 7,15∙ 1011 ∙ 1011Nmm Nmm ∙ (1000/400 ∙ (1000/400mm) mm)
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼) l
2/m
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)ll =
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
〖(EI)〗_l = = 1,79 ∙ 10 Nm=
EI∙ (1000/𝑠𝑠𝑠𝑠)
𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼 ∙(1000/s) /m
=7,15∙1011 7,15 ∙ 1011 Nmm^2 Nmm2∙(1000/400 2
∙ (1000/400 mm)mm)
11
(𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸)
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼) l = 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 ∙∙(1000/𝑠𝑠𝑠𝑠) 66 2 2= 67,152∙ 10 Nmm ∙ (1000/400 mm)
〖(EI)〗_l l l== 1,79 =π1,79 10 ∙ 10
1,79∙〖10〗^6 Nm
1,79 Nm ∙/m 10 /m
Nm^2/m 6 Nm Nm2 /m
𝑓𝑓𝑓𝑓(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
= = π 1,79 ∙ �
10 1,79
6
Nm ∙ 210 /m 6 Nm2/m = 9,5Hz > 8Hz → OK
1= l
𝑓𝑓𝑓𝑓(𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸) π � 1,79 ∙ 10
120 kg/m22 /m = 9,5Hz > 8Hz → OK ))=9,5Hz>8Hz→OK
𝑓𝑓𝑓𝑓f_1=π/(2∙(4,5m)^2 � )Nm √((1,79∙〖10〗^6 Nm^2/m)/(120
= 9,5Hz > 8Hz kg/m^2
1 = 2= ∙ (4,5m) 2 6 2
1 l2 ∙ (4,5m) 1,79 2∙210 120/m kg/m 2
→ OK
2 ∙ (4,5m) ππ 1,79
1,79 120
∙ 10 ∙ 10 6kg/m Nm 6 Nm 2 /m 2 /m
𝑓𝑓𝑓𝑓1𝑓𝑓𝑓𝑓1== π 2 2 1,79 � � ∙ 10 6 Nm 2
2 22 /m
= = 9,5Hz 9,5Hz > >8Hz8Hz → →OK OK
𝑓𝑓𝑓𝑓1 =2 2∙ (4,5m) ∙ (4,5m)
π 2
�1,79 120 120 ∙ 10 kg/m kg/m 6 Nm
2
/m = 9,5Hz > 8Hz → OK
𝑓𝑓𝑓𝑓1 = 2 ∙ (4,5m)2 � 120 kg/m2 = 9,5Hz > 8Hz → OK
2 ∙ (4,5m) 120 kg/m
→ The floor can be analyzed as a high frequency floor.
Floor stiffness perpendicular to the span direction based on 22 mm3 chipboard decking:
→ The floor
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)/𝑚𝑚𝑚𝑚 = 3500 can be N/mm analyzed 2
∙ 1000 asmm a high ∙ (22frequency mm)3 /12floor. = 3,11 ∙ 103 Nm22 /m
(EI)/m ==
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)/𝑚𝑚𝑚𝑚 3500 3500 N/mm^2∙1000 N/mm22 ∙ 1000 mm∙〖(22mm ∙ (22 mm)33/12==3,11∙〖10〗^3
mm)〗^3/12 3,11 ∙ 10 3Nm Nm^2/m
2/m
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)/𝑚𝑚𝑚𝑚
Floor stiffness = 3500 N/mm ∙ 1000
perpendicular to mm the span ∙ (22 mm) direction /12 based = 3,11 on∙ 10 Nm chipboard
22 mm /m
22 33 33 22
decking:
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)/𝑚𝑚𝑚𝑚
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)/𝑚𝑚𝑚𝑚 = = 3500 3500 N/mm N/mm
For a rectangular floor with overall dimensions b x l, simply supported along ∙ 1000
∙ 1000 mm mm ∙ (22
∙ (22 mm) mm) /12/12 = = 3,11
3,11 ∙ 10
∙ 10NmNm /m/m all four edges, the impulse
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)/𝑚𝑚𝑚𝑚 = 3500 N/mm22 2∙ 1000 mm ∙ (22 mm)33 /12 = 3,11 ∙ 1033 Nm22 /m
velocity response
(𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸)/𝑚𝑚𝑚𝑚 = 3500 N/mm v [m/Ns ∙ ]1000 valuemm may,∙ (22 as anmm) approximation,
/12 = 3,11be∙ 10 taken
Nmas:/m
For a4(0,4+0,6𝑛𝑛𝑛𝑛
rectangular
𝑣𝑣𝑣𝑣v=4(0,4+0,6n_40
= 4(0,4+0,6𝑛𝑛𝑛𝑛 40 )
40 )
floor with overall dimensions b x l, simply supported
)/(mbl+200) along all
(4.80) (4.80)
𝑣𝑣𝑣𝑣four = 4(0,4+0,6𝑛𝑛𝑛𝑛
edges,
𝑚𝑚𝑚𝑚𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏+200 the
40 ) impulse velocity response v [m/Ns2] value may, as an (4.80)
𝑣𝑣𝑣𝑣n_40={((40/f_1
= 𝑚𝑚𝑚𝑚𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏+200 )^2-1) (b/l)^4 (EI)_l/(EI)_b }^0,25
𝑚𝑚𝑚𝑚𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏+200
(4.80)
approximation,
4(0,4+0,6𝑛𝑛𝑛𝑛
4(0,4+0,6𝑛𝑛𝑛𝑛 ) ) be taken as: 0,25
𝑣𝑣𝑣𝑣 𝑣𝑣𝑣𝑣== 4(0,4+0,6𝑛𝑛𝑛𝑛 40 22
4040
𝑏𝑏𝑏𝑏 4 (𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼) 0,25
) 1� �𝑏𝑏𝑏𝑏 �4 (𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)ll �0,25
(4.80)
(4.80)
𝑡𝑡𝑡𝑡𝑣𝑣𝑣𝑣40==𝑚𝑚𝑚𝑚𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏+200
���40
𝑚𝑚𝑚𝑚𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏+200 � 40 − (4.81)
(4.80)
𝑡𝑡𝑡𝑡(4.81)
40 = ���
4(0,4+0,6𝑛𝑛𝑛𝑛 40𝑓𝑓𝑓𝑓1� 240− 4
) 1� � 𝑏𝑏𝑏𝑏𝑙𝑙𝑙𝑙� (𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)bl � (4.81)
𝑡𝑡𝑡𝑡
𝑣𝑣𝑣𝑣40 = = ���𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓1 � − 1� �𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 � (𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)b0,25
𝑚𝑚𝑚𝑚𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏+200 (𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼) � 0,25 (4.81)
(4.80)
n_40={((40/9,5Hz)^2-1)
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑙𝑙𝑙𝑙+200
4040 1 2 2 4 4
𝑏𝑏𝑏𝑏 𝑏𝑏𝑏𝑏 (𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)〖∙(5m/4,5m)〗^4∙(1,79∙〖10〗^6
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)b Nm^2/m
0,25
𝑡𝑡𝑡𝑡40𝑡𝑡𝑡𝑡 ==��� ��� � �− 2− 1�1� 2� �� �4
l l
�5m �0,254 (0,4+0,6∙11))/(120∙5∙4,5+200
∙ 1066 Nm22 /m 0,25 (4.81)
(4.81)
v=(4 40 (0,4+0,6 𝑓𝑓𝑓𝑓140 𝑓𝑓𝑓𝑓1 40n_40 2 ))/(mbl+200)=(4 (𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
𝑙𝑙𝑙𝑙 𝑏𝑏𝑏𝑏𝑙𝑙𝑙𝑙 (𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼) l 0,254 1,79
𝑡𝑡𝑡𝑡40 = ��� ���40 40
�2 − � 1�
2− � 1� �4 ∙ � b5m
(𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸)
b
� � 4∙ 1,79 ∙ 10 6 Nm 2 /m � 0,25
= 11 (4.81)
𝑡𝑡𝑡𝑡40 = ���
𝑛𝑛𝑛𝑛 ���𝑓𝑓𝑓𝑓9,5Hz 1� 40−�1�−� 1� 𝑚𝑚𝑚𝑚
𝑙𝑙𝑙𝑙� ∙(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
� 4,5m 5m
b� � ∙ 1,79
l
3,11 ∙∙ 10 103Nm Nm2/m /m�� = 11 (4.81)
𝑡𝑡𝑡𝑡40 = ���𝑓𝑓𝑓𝑓9,5Hz 1 2
� 2 − 𝑙𝑙𝑙𝑙1� ∙(𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸)
�4,5m b 4
� 4 ∙ 3,11 0,25 = 11
∙ 106 336Nm2 222/m 0,25
9,5Hz
4040 4,5m
5m 5m 1,79 3,11
1,79∙ 10∙∙ 10
10Nm Nm
Nm/m /m
/m
𝑡𝑡𝑡𝑡40𝑡𝑡𝑡𝑡40== 4��� ��� +40
(0,4 0,6 � �𝑡𝑡𝑡𝑡− 2− 1�) 1�∙ �∙4�(0,4 �+ �∙40,6
∙ 1,79
∙ 11) 6 2 � �0,25==1111
(0,4 9,5Hz 9,5Hz
0,6 �𝑡𝑡𝑡𝑡240−) 1� 40 44,5m 5m
4,5m
(0,4 3,11
3,11 ∙
∙ 10 10
3
10=Nm
∙ 10 3 Nm
6 Nm 2 2
/m /m
2 /m 0,25
= 4=
𝑣𝑣𝑣𝑣𝑡𝑡𝑡𝑡40 4 ���
(0,4
+40
+ 0,6 𝑡𝑡𝑡𝑡 ) = ∙4 � 5m
(0,4
+� 0,6
+
4
∙
0,6
∙ 11)
1,79
∙ 11) ∙ 0,010
Nm /m � = 11
𝑣𝑣𝑣𝑣𝑛𝑛𝑛𝑛40
= 𝑚𝑚𝑚𝑚𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏9,5Hz + 200 � 40 = ∙120 � 4,5m ∙ 5 ∙ 4,5 + 200
∙ 3,11 = 0,010
∙ 10 3 Nm 2 /m�
𝑣𝑣𝑣𝑣 = = ��� 𝑚𝑚𝑚𝑚𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏9,5Hz + 200 − 1� = 120 4,5m ∙ 5 ∙�4,5 + 200
3,11 ∙ 10=3 Nm0,010
2 /m
= 11
4 4(0,4 𝑚𝑚𝑚𝑚𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏
(0,4 ++0,6 +0,6 200 𝑡𝑡𝑡𝑡40 ) ) 4120
𝑡𝑡𝑡𝑡40 4(0,4 (0,4 ∙ 5++ ∙ 0,6
4,5
0,6∙+11)
∙ 200
11)
𝑣𝑣𝑣𝑣 𝑣𝑣𝑣𝑣== 4 (0,4 + 0,6 𝑡𝑡𝑡𝑡40 )== 4 (0,4 + 0,6 ∙ 11)==0,010 0,010
When
𝑣𝑣𝑣𝑣 = 4 (0,4𝑚𝑚𝑚𝑚𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏
𝑚𝑚𝑚𝑚𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏
a high + + 200
+ 0,6 value 200 𝑛𝑛𝑛𝑛40b)==150 120 120
4 (0,4 ∙ 5∙ 5∙
is chosen 4,5
∙ 4,5 +
+ 0,6 from+ 200
200the
∙ 11) Figure 4.28 and a conservative damping value ξ = 0,01 is
= 0,010
𝑣𝑣𝑣𝑣used,
= the 𝑚𝑚𝑚𝑚𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 + 200 =
requirement 120
for120 v is∙∙ 55 ∙∙ 4,5 4,5 + 200 = 0,010
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 + 200 + 200
(𝑓𝑓𝑓𝑓1 𝜉𝜉𝜉𝜉−1)
𝑣𝑣𝑣𝑣 ≤ 150(𝑓𝑓𝑓𝑓 = 0,011 → 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂
1 𝜉𝜉𝜉𝜉−1) = 0,011 → 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂
(4.78)
𝑣𝑣𝑣𝑣When ≤ 150 a (𝑓𝑓𝑓𝑓 high 1 𝜉𝜉𝜉𝜉−1) value b=→ 150 is chosen from the Figure 4.28 and a conservative (4.78)
𝑣𝑣𝑣𝑣v≤〖150〗^((f_1
≤ 150 =ξ-1) 0,011 )=0,011→OK 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 (4.78) (4.78)
damping (𝑓𝑓𝑓𝑓1(𝑓𝑓𝑓𝑓 value
𝜉𝜉𝜉𝜉−1)
1 𝜉𝜉𝜉𝜉−1) ξ = 0,01 is used, the requirement for 𝑣𝑣𝑣𝑣 is
𝑣𝑣𝑣𝑣 𝑣𝑣𝑣𝑣≤≤150 150 =
𝐹𝐹𝐹𝐹∙𝑙𝑙𝑙𝑙 2
= 0,011 0,011 → → 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂
𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 (4.78)
(4.78)
𝑣𝑣𝑣𝑣 ≤ 150(𝑓𝑓𝑓𝑓1 𝜉𝜉𝜉𝜉−1) 𝐹𝐹𝐹𝐹∙𝑙𝑙𝑙𝑙 22= 0,011 → 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 (4.78)
42∙𝑘𝑘𝑘𝑘𝐹𝐹𝐹𝐹∙𝑙𝑙𝑙𝑙
∙(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)l LVL Handbook Europe 181
𝑤𝑤𝑤𝑤 = min �42∙𝑘𝑘𝑘𝑘𝛿𝛿𝛿𝛿𝛿𝛿𝛿𝛿∙(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼) 3 l (4.82)
𝑤𝑤𝑤𝑤 = min � 42∙𝑘𝑘𝑘𝑘 𝐹𝐹𝐹𝐹∙𝑙𝑙𝑙𝑙∙(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
𝛿𝛿𝛿𝛿2 32 l (4.82)
𝑤𝑤𝑤𝑤 = min � 𝐹𝐹𝐹𝐹∙𝑙𝑙𝑙𝑙𝐹𝐹𝐹𝐹∙𝑙𝑙𝑙𝑙 𝐹𝐹𝐹𝐹∙𝑙𝑙𝑙𝑙 3 (4.82)
48∙𝑠𝑠𝑠𝑠∙(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
2 l 𝐹𝐹𝐹𝐹∙𝑙𝑙𝑙𝑙
42∙𝑘𝑘𝑘𝑘 𝐹𝐹𝐹𝐹∙𝑙𝑙𝑙𝑙
42∙𝑘𝑘𝑘𝑘
∙(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
∙(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
48∙𝑠𝑠𝑠𝑠∙(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
𝛿𝛿𝛿𝛿 𝛿𝛿𝛿𝛿 ll l
𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤==minmin � �42∙𝑘𝑘𝑘𝑘 48∙𝑠𝑠𝑠𝑠∙(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼) 3 l
3 ∙(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼) (4.82)
(4.82)
𝐹𝐹𝐹𝐹∙𝑙𝑙𝑙𝑙𝐹𝐹𝐹𝐹∙𝑙𝑙𝑙𝑙
𝛿𝛿𝛿𝛿 l
2020_LVL_09.indd 181
𝑤𝑤𝑤𝑤 = min �b 𝐹𝐹𝐹𝐹∙𝑙𝑙𝑙𝑙43 3,11∙10
4 (𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼) 3 (4.82) 12.3.2020 13:12:07
48∙𝑠𝑠𝑠𝑠∙(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
48∙𝑠𝑠𝑠𝑠∙(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼) (4.84)
𝑘𝑘𝑘𝑘𝛿𝛿𝛿𝛿 = 44�(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼) b = 4�3,11∙10 l l 3 = 0,2
𝑘𝑘𝑘𝑘𝛿𝛿𝛿𝛿 = � (𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼) =� 1,79∙1036 = 0,2
4 3,11∙10
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)bl48∙𝑠𝑠𝑠𝑠∙(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼) (4.84)
40 5m 1,79 ∙ 10 Nm /m
𝑡𝑡𝑡𝑡40 = ��� � − 1� ∙ � � ∙ � = 11
9,5Hz 4,5m 3,11 ∙ 103 Nm2 /m
4 (𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼) 4 3,11∙103
𝑘𝑘𝑘𝑘𝛿𝛿𝛿𝛿 = � (𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)b = �1,79∙10
k_δ=∜((EI)_b/(EI)_l = 0,2
=)6∜((3,11∙〖10〗^3)/(1,79∙〖10〗^6 ))=0,2 (4.84) (4.84)
l
1kN ∙ (4,5m)2
⎧ 6 2
⎪42 ∙ 0,2 ∙ 1,79 ∙ 10 Nm
⎪ 1,3 mm
m
𝑤𝑤𝑤𝑤 = min = min � = 1,3 mm
w=min{█((1kN∙(4,5m)^2)/█(42∙0,2∙1,79∙(〖10〗^6
⎨ 1kN ∙ (4,5m)3 Nm^2)/m@ )@(1kN∙(4,5m)^3)/(48∙0,4∙1,79∙(〖10〗^6
2,7 mm
Nm^2)/m))┤=min{█(1,3
⎪
⎪ mm@ 6 @2,7
2 mm)┤=1,3 mm
10 Nm
⎩48 ∙ 0,4 ∙ 1,79 ∙ m
According to the Finnish national Annex of EC5, for the span length L= 4500mm the requirement is w ≤ 0,6
mm and according to the Austrian national annex w ≤ 0,5 mm which are not fulfilled by the structure.
The performance can be improved by transverse bracing by 2 blockings lines with tension boards close to
the centre of the span and cross battens 45x45 c/c 400 mm underneath the floor.
The stiffness (EI)bracing of the bracing lines can be estimated according to EN1995-1-1, Annex B Mechanical-
ly jointed beams. For simplification, also the top flange of the transverse bracing is modelled as a 22x100 C14
board and the stiffness of the 2 bracing lines are divided for the whole floor span length.
ℎ0 3
�𝑏𝑏𝑏𝑏0 ∙ 12 + 𝛾𝛾𝛾𝛾 ∙ 𝑏𝑏𝑏𝑏0 ∙ ℎ0 ∙ (ℎ/2 + ℎ0 /2)2 �
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)bracing = 𝑡𝑡𝑡𝑡 ∙ 2 ∙ 𝐸𝐸𝐸𝐸0,mean ∙
𝑙𝑙𝑙𝑙0
(EI)_bracing=n∙2∙E_(0,mean)∙((b_0∙〖h_0〗^3/12+γ∙b_0∙h_0∙(h/2+h_0/2)^2 ))/l_0
where
n = number of bracing lines in each span = 2
E0,mean = mean modulus of elasticity of C14 flanges 7000 N/mm2
b0 = width of tension/compression chord = 100 mm
h0 = thickness of tension/compression chord = 22 mm
h = depth of joist = 240 mm
s0 = Spacing of the nailing between blocking and flanges = 200 mm
l0 = span length of the bracing line = width of the floor= 5 m
ρk = characteristic density (C14: 350 kg/m3)
d = diameter of the nail 2.5 mm for nails 2,5x60 mm
3
0 ℎ
𝑑𝑑𝑑𝑑0,8 �𝑏𝑏𝑏𝑏0 ∙ 12 + 𝛾𝛾𝛾𝛾 ∙ 𝑏𝑏𝑏𝑏0 ∙ ℎ0 ∙ (ℎ/2 + ℎ0 /2)2 �
1,5
𝑂𝑂𝑂𝑂
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼) = 𝜌𝜌𝜌𝜌
serbracing ∙ = 454 N/mm
k = 𝑡𝑡𝑡𝑡 ∙ 2 ∙ 𝐸𝐸𝐸𝐸0,mean ∙
30 𝑙𝑙𝑙𝑙0
1 1
𝛾𝛾𝛾𝛾 = =
𝜋𝜋𝜋𝜋 2 ∙ 𝐸𝐸𝐸𝐸0 ∙ ℎ0 ∙ 𝑏𝑏𝑏𝑏0 ∙ 𝑠𝑠𝑠𝑠0 2 2
π ∙ 7000 N/mm ∙ 22 mm ∙ 100 mm ∙ 200 mm
1+ 1+
𝑂𝑂𝑂𝑂ser ∙ 𝐿𝐿𝐿𝐿0 2 454N/mm ∙ (5000 mm)2
1,5
𝑑𝑑𝑑𝑑0,8
𝑂𝑂𝑂𝑂 er = 𝜌𝜌𝜌𝜌kpanel,
1.sDecking ∙ 2. Floor
= 454 N/mm
joists, 3. Transverse tension board (C18, min 22x100) under the joists fixed with 2,8x75 nails to the joists and
30
blockings, 4. Blocking
1 1
𝛾𝛾𝛾𝛾 = =
𝜋𝜋𝜋𝜋 2 ∙ 𝐸𝐸𝐸𝐸0 ∙ ℎ0 ∙ 𝑏𝑏𝑏𝑏0 ∙ 𝑠𝑠𝑠𝑠0 π2 ∙ 7000 N/mm2 ∙ 22 mm ∙ 100 mm ∙ 200 mm
182 LVL Handbook
1 + Europe 1+
𝑂𝑂𝑂𝑂ser ∙ 𝐿𝐿𝐿𝐿0 2 454N/mm ∙ (5000 mm)2
The stiffness of the 45x45 mm cross battens C14 solid wood underneath the floor:
22
NN 1000 1000N mm
mm 1000 4545mmNmm mm ∙(45
(45
∙100045 mm)mm)
mm
mm 2(45 (45mm
∙245 mm) 2
(45
2 2∙∙∙10
25,98 33 mm) 22
2
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)cross crossbattens = = 7000
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)cross (𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
battens 7000 N N= 1000
2 2
∙ ∙
N
7000 1000
N mm N 1000
= mm
100045
70001000
∙ ∙
mm∙mm
45
mm mm
mm45
∙ (45
45∙ mm∙
45 (45
mm
∙ mm)mm
∙ (45
mm)
∙ ∙ (45
mm)
∙ ==mm)
mm) 5,98 3
10=3
Nm
Nm
5,98
2 32
/m
/m
∙ 10
3 = 3
32 5,98
2Nm
2
103 Nm2 /m
2 ∙/m
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
cross (𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
cross
battens (𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
cross
battens
cross =
cross 7000
battens=battens
battens
battens
7000==7000 mm
mm
cross
=7000
∙ 7000 ∙ 400
battens 400 ∙2mm mm
∙mm∙∙ ∙400
2 mm ∙ mm∙ ∙ 12
2 12 mm= 12
400 5,98
= 5,98∙=10
=5,98
∙=
5,98
1210
Nm5,98
∙ Nm
10∙ 10
/m ∙ Nm
10
/mNm Nm
/m /m/m
(EI)_(cross battens)=7000 N/mm^2 ∙(1000 mm)/(400 mm)∙(45 mm ∙(45 mm)^2)/12=5,98∙〖10〗^3 Nm^2/m
mm 2
mm 2
400
mm mm 2
400
mm 2
400
mm400 400
mm mm mm 12 12 12 12 12
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)bb== (𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
〖(EI)〗_b= (𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
〖(EI)〗_decking
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)decking = (𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
decking
b(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
++ b(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
= +bracings
decking 〖(EI)〗_bracings
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
+decking
bracings ++bracings
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼) (𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
+ (𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
cross ++bracings
cross 〖(EI)〗_(cross
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)cross
battens
battens battens)
+ (𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
battenscross battens
b =(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼) b(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
= (𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
b(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼) ==
decking b(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
= (𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
+ (𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
+ (𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
++(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)+(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
+bracings
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
+ (𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
++(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
+
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
〖(EI)〗_b =b 3,11∙〖10〗^3
decking decking
decking
decking bracings
Nm^2/m bracings bracings bracings
+ 6,44∙〖10〗^4 cross cross
battens
cross
battens
cross
Nm^2/m cross
battens
battens
battens
+ 5,98∙〖10〗^3 Nm^2/m
33 22 3 44Nm 2 22/m 33 2 22
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
〖(EI)〗_b bb = = 3,11 3,11
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
== ∙b3∙10
10
7,35∙〖10〗^4 =3Nm Nm
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
3,11 /m
/m
32b3∙Nm^2/m
=10 +3+
32 3,11 2Nm
6,44
6,44
2 ∙ ∙10
2 ∙/m 10 10
+Nm Nm
4 6,44
2 /m
/m
+4 2+
4 24∙ 10
4+ 5,98
6,44Nm 5,98
2 ∙ ∙10
103410
2 ∙ /m +
Nm Nm
Nm
325,98
/m /m
/m+32 35,98
32 3∙ 10 2Nm
2
2 ∙/m 103 Nm2 /m
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
b =(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)b (𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
=(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
3,11 b 3,11 b∙ 10 =
b 3,11
∙=3,11
10
Nm ∙ 2Nm
3,11 10
/m
∙ 10 ∙ Nm
10
/m
+Nm 6,44
Nm/m
+ /m 6,44
/m
∙+10 +6,44∙4+6,44
10
Nm 6,44
∙ 10
Nm
∙/m10 ∙ 10
Nm
/m
+Nm +2 /m
5,98
Nm
/m 5,98
/m
∙+10+5,98
∙+
5,98
10
Nm 5,98
∙ Nm
10
/m
∙ 10∙ Nm
10
/m
Nm Nm
/m/m /m
k_(δ,2)=∜((EI)_b/(EI)_l
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)bb == 7,35
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼) 7,35 44
Nm4222b/m
=) ∜((6,44∙〖10〗^4)/(1,79∙〖10〗^6 ))=0,45
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)∙b4∙10 10=4Nm 27,35
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼) /m
4∙ =10 4
42 7,35 2Nm
2
2 ∙/m 104 Nm2 /m
b =(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼) 7,35
=(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
b (𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼) b 7,35 b=∙ 10 =
b 7,35
∙=7,35
10
Nm 7,35
∙ Nm
10
/m
∙ 10 ∙ Nm
10
/m Nm Nm/m /m/m
w_2=min{█((1kN∙(4,5m)^2)/█(42∙0,45∙1,79∙(〖10〗^6 4 4
Nm^2)/m@ )@(1kN∙(4,5m)^3)/(48∙0,4∙1,79∙(〖10〗^6
44 (𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)bb 444 (𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼) 6,44
6,44b∙4∙10 10
4 4(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼) 4 4 4
Nm^2)/m))┤=min{█(0,6 � mm@ 6,44@2,7 ∙ 10 6,44
mm)┤=0,6 ∙ 10 mm→OK
� 4�(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
b4�b6,44
𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝛿𝛿𝛿𝛿,2 =(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼) (𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼) 4 b
4 4 4
𝛿𝛿𝛿𝛿,24= b𝑘𝑘𝑘𝑘4𝛿𝛿𝛿𝛿,2b4= = =6,44
� ∙b410= 6,44
∙= 6,44
10
4 6,44
�6� ==
6∙ 10
0,45
∙ 10 0,45
∙ 10
= �6 = 0,45 6 = 0,45
� �𝑘𝑘𝑘𝑘1,79
4 4 4
𝑘𝑘𝑘𝑘𝛿𝛿𝛿𝛿,2𝑘𝑘𝑘𝑘= 𝛿𝛿𝛿𝛿,2� 𝑘𝑘𝑘𝑘�
𝑘𝑘𝑘𝑘=𝛿𝛿𝛿𝛿,2 (𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
𝑘𝑘𝑘𝑘= =� == l l�=
� =1,79
=�=
𝛿𝛿𝛿𝛿,2
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼) �∙6∙10
10
�
=(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
0,45
= 60,45 = = 0,45
=
0,45 0,45
𝛿𝛿𝛿𝛿,2
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼) 𝛿𝛿𝛿𝛿,2
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼) (𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
1,79(𝐸𝐸𝐸𝐸𝐼𝐼𝐼𝐼)
1,79
∙ 10 l1,79
∙ 1,79
10 6∙ 1,79
1,79 10∙ 10 6∙ 10
l∙ 10 6 1,79 ∙ 10
l l l l l
22
1kN∙ ∙(4,5m)
1kN (4,5m) 1kN ∙ (4,5m)
2 21kN
2
2 ∙ (4,5m)
2
⎧⎧ 1kN1kN ∙ (4,5m)
⎧1kN 1kN ∙2(4,5m)
∙ (4,5m)
1kN ⎧∙266(4,5m)
∙ (4,5m) 22
⎧ ⎪⎧ ⎪ ⎧ ⎧ ⎧ 10
10 Nm
Nm 10 6
2Nm
2 10mm6 Nm2
⎪42
42⎪∙ ∙0,45
0,45 ⎪
∙ ∙1,79
1,79 ∙6∙ 10⎪
⎪42⎪⎪ 42 10
∙ 0,45 Nm 6
422
∙m 60,45
10∙10
Nm
1,79 2Nm
6∙ Nm
10 62∙Nm1,7920,6
∙ mm
0,6
⎪∙42
0,45 ⎪
∙ 42 ⎪
0,45
∙42 ⎪
1,79
∙ 42
0,45
∙ 0,45
1,79
∙ 0,45
∙ 1,79
∙m ⎪
1,79 m
∙ 1,79
∙m∙ ∙mm m0,6 m mm m 0,6 mm 0,6 mm
𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤22==min ⎪ ⎪ ⎪ ⎪
min𝑤𝑤𝑤𝑤2 = min𝑤𝑤𝑤𝑤2 = min ==min min��0,6=mm 0,60,6mm
0,6
==�mmmm
0,6
0,6mm
mm�=→ OK
→0,6
OKmm=→
min = min 0,6
OKmm → OK
𝑤𝑤𝑤𝑤2 =𝑤𝑤𝑤𝑤2min =𝑤𝑤𝑤𝑤2min
𝑤𝑤𝑤𝑤=
2𝑤𝑤𝑤𝑤=
min
⎨
2⎨=minmin 1kN
1kN ⎨∙ ∙(4,5m)
(4,5m) 1kN ⎨33 = min
∙ (4,5m)
1kN= min
3�=
∙ =min
� =
min
2,7
2,7
(4,5m) min
�
mm3=
mm� 0,6
� = mm
0,6
==mm
0,6
→=0,6
OK
mm
0,6
→mm OK
mm
→→OK
→
OKOK
⎨ ⎪⎨ ⎨ ⎨ ⎨
∙ (4,5m)∙ (4,5m)3
∙ (4,5m) 3
∙ (4,5m) 3
(4,5m) 3 3 2,7 mm 2,7 mm
⎪ ⎪ 1kN
⎪⎪ 1kN ⎪1kN
⎪
1kN 1kN
10
10⎪⎪6∙6NmNm62226 10 2,7 mm
2,7 mm
2,7
6 Nm2 106 Nm2
2,7mm
2,7mmmm
⎪ ⎩⎪ ⎪
⎩ 48⎪
48 ⎪∙ ∙⎪
0,4
⎪ ⎪
0,4
⎪∙ ∙1,79
1,79
4810 ∙∙6∙0,4
10
Nm 6∙ 2
m 10
Nm
4810
1,79
m∙ ∙ ∙ 10
Nm
0,4
∙ 62∙Nm
Nm 2 2 ∙
1,79
⎩0,4 ∙ ⎩1,79 m m
⎩ 48 ⎩ ∙48
0,4
⎩∙⎩48
∙0,4
1,79
48
⎩∙ ∙48∙1,79
0,4
∙ ∙∙0,4
1,79
∙m 1,79 ∙m
mm m
The transfer bracing lines and cross battens underneath the joists improve the floor stiffness under a 1kN
point load significantly and the vibration design requirements are fulfilled. The deflection could be decreased
~0,2 mm more, by gluing the decking onto the joists.
Beam properties:
Bending strength edgewise fm,0,edge,k = 44 N/mm2
Shear strength edgewise fv,0,edge,k = 4,2 N/mm2
Compression perpendicular to grain edgewise fc,90,edge,k = 6 N/mm2
Modulus of elasticity E0,k = 11 600 N/mm2
Modulus of elasticity E0,mean = 13 800 N/mm2
Modulus of rigidity G0,edge,k = 600 N/mm2
Modulus of rigidity G0,edge,mean = 400 N/mm2
Area of cross section A = b∙h = 13500 mm2
Section modulus Wy = b∙h /6 2 = 6,75∙105mm3
Moment of inertia Iy = b∙h /12
3 = 1,01∙108 mm4
Moment of inertia Iz = h∙b /12
3 = 2,28∙106 mm4
Torsion moment of inertia Itor = 0,3∙h∙b
3 = 8,20∙106 mm4
Moment stiffness of the joist EIy = 13800 N/mm ∙ 1,01∙10 mm = 1,40∙1012 Nmm2
2 8 4
Loading combinations
Snow load at roof level qk = μ1 ∙ Ce ∙ sk. Form factor μ1=0,8, when roof angle is less than 30° and in normal
conditions Ce = 1,0 → qk = 0,8 ∙ 1,0 ∙ 2,75 N/m2 = 2,2 kN/m2.
The most critical ultimate limit state (ULS) load combination:
E_(d,ULS
𝐸𝐸𝐸𝐸 )=
d,ULS = 𝛾𝛾𝛾𝛾 G ∙γ_G∙g_k+
𝑔𝑔𝑔𝑔k + 𝛾𝛾𝛾𝛾Q ∙γ_Q∙q_k 𝑞𝑞𝑞𝑞k (4.1) (4.1)
E_(d,ULS
𝐸𝐸𝐸𝐸 )=
d,ULS = 𝛾𝛾𝛾𝛾 G ∙1,15∙(5m∙1,0
𝑔𝑔𝑔𝑔k + 𝛾𝛾𝛾𝛾Q ∙ 𝑞𝑞𝑞𝑞kkN kN/m^2 )+1,5∙5m∙2,2 kN/m^2=22,3 kN/m (4.1)
𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸d,ULS = 𝛾𝛾𝛾𝛾𝛾𝛾𝛾𝛾GG ∙∙ 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔∙kk�5m
d,ULS = 1,15
d,ULS
+ ∙∙ 𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞kk � + 1,5 ∙ 5m ∙ 2,2 kN/m2 = 22,3 kN/m (4.1)
+ 𝛾𝛾𝛾𝛾𝛾𝛾𝛾𝛾∙QQ1,0 (4.1)
kN2
m 2
𝐸𝐸𝐸𝐸d,ULS = 1,15 ∙ �5m ∙ 1,0 kN kN � + 1,5 ∙ 5m ∙ 2,2 kN/m = 22,3 kN/m
𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸d,ULS = 1,15 ∙ �5m ∙ 1,0 m2 � + 1,5 ∙ 5m ∙ 2,2 kN/m22 = 22,3 kN/m
d,ULS = 1,15 ∙ �5m ∙ 1,0 m22 � + 1,5 ∙ 5m ∙ 2,2 kN/m = 22,3 kN/m
Note: Safety factors γG and m γQ are according to Finnish National annex of Eurocode 0.
ULS design
𝑀𝑀𝑀𝑀d =moment
Bending 𝐸𝐸𝐸𝐸d,ULS ∙resistance 𝑠𝑠𝑠𝑠 ∙ 𝐿𝐿𝐿𝐿2 /8 = 22,3kN/m ∙ (2,3m)2 /8 = 14,7 kNm
𝑀𝑀𝑀𝑀d = 𝐸𝐸𝐸𝐸d,ULS ∙ 𝑠𝑠𝑠𝑠 ∙ 𝐿𝐿𝐿𝐿22 /8 = 22,3kN/m ∙ (2,3m)22 /8 = 14,7 kNm
𝑀𝑀𝑀𝑀dd =
𝑀𝑀𝑀𝑀 = 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝑀𝑀𝑀𝑀
d,ULS
𝑑𝑑𝑑𝑑
d,ULS
∙∙ 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 ∙14,7
∙ 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿2/8/8kNm == 22,3kN/m
22,3kN/m ∙∙ (2,3m)
(2,3m)2/8 /8 = = 14,7
14,7 kNm
kNm
𝜎𝜎𝜎𝜎M_d=m,d = E_(d,ULS)∙s∙L^2/8 = = = 21,8 N/mm2
22,3kN/m∙〖(2,3m)〗^2/8 = 14,7 kNm
𝑀𝑀𝑀𝑀
𝑊𝑊𝑊𝑊𝑑𝑑𝑑𝑑 6,75 14,7 ∙ 10 kNm
6 mm 3
𝜎𝜎𝜎𝜎σ_(m,d)=M_d/W=(14,7
m,d = 𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 = 14,7 14,7 kNm kNm = 21,8 N/mm22 mm^3 )=21,8 N/mm^2
kNm)/(6,75〖∙10〗^6
𝜎𝜎𝜎𝜎𝜎𝜎𝜎𝜎m,d =
= 𝑊𝑊𝑊𝑊 = = 6,75 ∙ 1066 mm33 = = 21,8 N/mm
N/mm2
21,80,8
f_(m,0,edge,d)=k_mod/γ_M
m,d
𝑊𝑊𝑊𝑊 𝑘𝑘𝑘𝑘6,75
𝑊𝑊𝑊𝑊 mod ∙∙ 10
6,75 10 mm6 mm ∙k_h∙f_(m,0,edge,k)=0,8/1,2∙1,00∙44
3 N N/mm^2 =29,3 N/mm^2
𝑓𝑓𝑓𝑓σ_(m,d)≤f_(m,0,edge,d)
m,0,edge,d = ∙ 𝑘𝑘𝑘𝑘 h ∙ 𝑓𝑓𝑓𝑓
→OK
m,0,edge,k = ∙ 1,00 ∙ 44 = 29,3 N/mm2
𝑘𝑘𝑘𝑘𝛾𝛾𝛾𝛾modM 0,8
1,2 N2
mm
𝑓𝑓𝑓𝑓m,0,edge,d = 𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘mod mod ∙ 𝑘𝑘𝑘𝑘h ∙ 𝑓𝑓𝑓𝑓m,0,edge,k = 1,2
0,8
0,8 ∙ 1,00 ∙ 44 N N 2 = 29,3 N/mm222
𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓m,0,edge,d
m,0,edge,d
== 𝛾𝛾𝛾𝛾 M ∙∙ 𝑘𝑘𝑘𝑘
𝑘𝑘𝑘𝑘 ∙∙ 𝑓𝑓𝑓𝑓
𝑓𝑓𝑓𝑓
hh m,0,edge,k
m,0,edge,k
=
= ∙∙ 1,00
1,00 ∙∙ 44
44 mm =
= 29,3
29,3 N/mm
N/mm
𝜎𝜎𝜎𝜎m,d ≤ 𝑓𝑓𝑓𝑓m,0,edge,d 𝛾𝛾𝛾𝛾𝛾𝛾𝛾𝛾M
M → OK
1,2
1,2 mm22
mm
𝜎𝜎𝜎𝜎m,d ≤ 𝑓𝑓𝑓𝑓m,0,edge,d → OK
𝜎𝜎𝜎𝜎𝜎𝜎𝜎𝜎m,d ≤ 𝑓𝑓𝑓𝑓m,0,edge,d →
m,d ≤ 𝑓𝑓𝑓𝑓m,0,edge,d → OK OK
The lintel beam is laterally supported to wall studs in 600mm spacing and the load is applied via them.
𝑀𝑀𝑀𝑀y,crit 𝜋𝜋𝜋𝜋�𝐸𝐸𝐸𝐸0,05 𝐼𝐼𝐼𝐼z 𝐺𝐺𝐺𝐺0,05 𝐼𝐼𝐼𝐼tor
𝜎𝜎𝜎𝜎Therefore
m,crit = 𝑊𝑊𝑊𝑊
the effective = length is Lef = 600mm (See table 4.9). (4.42)
𝑀𝑀𝑀𝑀y,crity 𝜋𝜋𝜋𝜋�𝐸𝐸𝐸𝐸0,05 𝑙𝑙𝑙𝑙ef𝐼𝐼𝐼𝐼z𝑊𝑊𝑊𝑊 𝐺𝐺𝐺𝐺y0,05 𝐼𝐼𝐼𝐼tor
𝜎𝜎𝜎𝜎m,crit = 𝑀𝑀𝑀𝑀 = 𝜋𝜋𝜋𝜋 � 𝐸𝐸𝐸𝐸
𝜋𝜋𝜋𝜋�𝐸𝐸𝐸𝐸0,05 𝐼𝐼𝐼𝐼 𝐺𝐺𝐺𝐺 𝐼𝐼𝐼𝐼 (4.42)
𝜎𝜎𝜎𝜎𝜎𝜎𝜎𝜎 = 𝑀𝑀𝑀𝑀𝑊𝑊𝑊𝑊 y,crit
y,crit
= 𝑙𝑙𝑙𝑙ef𝐼𝐼𝐼𝐼z𝑊𝑊𝑊𝑊
0,05 z 𝐺𝐺𝐺𝐺y0,05
0,05 𝐼𝐼𝐼𝐼tor tor
(4.42)
m,crit = 𝑊𝑊𝑊𝑊
σ_(m,crit)=M_(y,crit)/W_y
m,crit y = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙ef 𝑊𝑊𝑊𝑊 =(π√(E_0,05 I_z G_0,05 I_tor ))/(l_ef W_y (4.42)
) (4.42)
𝑊𝑊𝑊𝑊yy ef y 𝑊𝑊𝑊𝑊y 400N
π�10600 N/mm2 ∙ 2,28 ∙ 106 mm4 ∙ 2 ∙ 8,20 ∙ 10 6 ∙ mm4
mm
2 ∙ 2,28 ∙ 106 mm4 ∙ 400N ∙ 8,20 ∙ 106 ∙ mm4
𝜎𝜎𝜎𝜎σ_(m,crit)=
m,crit = π�(π√(10600 10600 N/mm N/mm^2∙2,28∙〖10〗^6 mm^4∙400N/mm^2 ∙8,20∙〖10〗^6∙
π � 10600 N/mm 22600mm∙ 2,28 ∙ ∙ 6,75
10 66mm 44 ∙ 5400N
∙ 10 400N
mm223 ∙∙ 8,20
mm ∙ 1066 ∙ mm44
𝜎𝜎𝜎𝜎〗^5 = ) π � 10600 N/mm ∙ 2,28 ∙ 10 mm ∙ mm 23 8,20 ∙ 10 ∙ mm
m,crit
𝜎𝜎𝜎𝜎𝜎𝜎𝜎𝜎m,crit mm^3= 600mm ∙ 6,75 ∙ 105 mm3 5mm
m,crit = 2 600mm
𝜎𝜎𝜎𝜎σ_(m,crit)=72,2
m,crit = 72,2 𝑁𝑁𝑁𝑁/𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 N/〖mm〗^2 600mm ∙∙ 6,75 6,75 ∙∙ 10
105mm mm3
2
𝜎𝜎𝜎𝜎λ_rel=√(f_(m,k)/σ_(m,crit)
m,crit = 72,2 𝑁𝑁𝑁𝑁/𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 22 )=√((44 N/mm^2)/(72,2N/mm^2 ))= 0,78
𝜎𝜎𝜎𝜎𝜎𝜎𝜎𝜎m,crit
m,crit = 72,2
= 72,2 𝑓𝑓𝑓𝑓 𝑁𝑁𝑁𝑁/𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
𝑁𝑁𝑁𝑁/𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
44 N/mm2
𝜆𝜆𝜆𝜆𝑟𝑟𝑟𝑟𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 = �𝜎𝜎𝜎𝜎 m,k = �72,2N/mm22 = 0,78 (4.41)
𝑓𝑓𝑓𝑓m,crit 44 N/mm
𝜆𝜆𝜆𝜆𝑟𝑟𝑟𝑟𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 = �𝜎𝜎𝜎𝜎𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓m,k m,k = � 44
44N/mm N/mm22
2 = 0,78 (4.41)
𝜆𝜆𝜆𝜆(4.41) = �𝜎𝜎𝜎𝜎m,crit
𝑟𝑟𝑟𝑟𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 = �
𝜆𝜆𝜆𝜆𝑟𝑟𝑟𝑟𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 m,k
== �
� 72,2N/mm
22 = = 0,78
0,78 (4.41)
(4.41)
when 0,75 < 𝜆𝜆𝜆𝜆rel,m 72,2N/mm
𝜎𝜎𝜎𝜎m,crit
m,crit
72,2N/mm
≤ 1,4 , 𝑘𝑘𝑘𝑘crit = 1,56 − 0,75 ∙ 𝜆𝜆𝜆𝜆rel,m = 1,56 − 0,75 ∙ 0,78 = 0,97
when 0,75 < 𝜆𝜆𝜆𝜆rel,m ≤ 1,4 , 𝑘𝑘𝑘𝑘crit = 1,56 − 0,75 ∙ 𝜆𝜆𝜆𝜆rel,m = 1,56 − 0,75 ∙ 0,78 = 0,97
when
𝑘𝑘𝑘𝑘when
crit ∙ 𝑓𝑓𝑓𝑓
0,75
0,75 m,d =
<< 0,97𝜆𝜆𝜆𝜆𝜆𝜆𝜆𝜆rel,m
rel,m∙ 29,3
≤≤ 1,4 1,4N/mm ,,𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘crit =
2 1,56 −
crit ==1,56
0,75
0,75 ∙∙ 𝜆𝜆𝜆𝜆2𝜆𝜆𝜆𝜆rel,m
28,6−N/mm = 1,56 − 0,75 ∙ 0,78 = 0,97
rel,m = 1,56 − 0,75 ∙ 0,78 = 0,97
𝑘𝑘𝑘𝑘crit ∙ 𝑓𝑓𝑓𝑓m,d = 0,97 ∙ 29,3 N/mm2 = 28,6 N/mm22 2
𝑘𝑘𝑘𝑘𝜎𝜎𝜎𝜎
𝑘𝑘𝑘𝑘when
crit
m,d
∙ 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓m,d
crit ∙≤0,75<λ_(rel,m)≤1,4
𝑘𝑘𝑘𝑘
m,d
= 0,97
crit=∙ 0,97 𝑓𝑓𝑓𝑓m,d → ∙∙ 29,3
29,3𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 N/mm N/mm 2 = 28,6 N/mm 2
= 28,6 N/mm
,k_crit=1,56-0,75∙λ_(rel,m)=1,56-0,75∙0,78=0,97
𝜎𝜎𝜎𝜎m,d ≤ 𝑘𝑘𝑘𝑘crit ∙ 𝑓𝑓𝑓𝑓m,d → 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂
𝜎𝜎𝜎𝜎𝜎𝜎𝜎𝜎m,d
m,d ≤ 𝑘𝑘𝑘𝑘crit
≤ 𝑘𝑘𝑘𝑘 crit ∙∙ 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓m,d → 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂
m,d → 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂
𝑓𝑓𝑓𝑓m,k 44 N/mm2
𝜆𝜆𝜆𝜆𝑟𝑟𝑟𝑟𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 = �𝜎𝜎𝜎𝜎 = �72,2N/mm2 = 0,78 (4.41)
m,crit
when 0,75 < 𝜆𝜆𝜆𝜆rel,m ≤ 1,4 , 𝑘𝑘𝑘𝑘crit = 1,56 − 0,75 ∙ 𝜆𝜆𝜆𝜆rel,m = 1,56 − 0,75 ∙ 0,78 = 0,97
2 2
𝑘𝑘𝑘𝑘crit ∙ 𝑓𝑓𝑓𝑓m,d = 0,97
9. CALCULATION ∙ 29,3 N/mm
EXAMPLES = 28,6
OF LVL N/mm
STRUCTURES
𝜎𝜎𝜎𝜎m,d ≤ 𝑘𝑘𝑘𝑘crit ∙ 𝑓𝑓𝑓𝑓m,d → 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂
𝑉𝑉𝑉𝑉red,d = 17,2kN
3 ∙ 𝑉𝑉𝑉𝑉d 3 ∙ 17,2 kN
𝜏𝜏𝜏𝜏v,d = = = 1,92 N/mm2
2 ∙ 𝐴𝐴𝐴𝐴 2 ∙ 13500mm2
2ℎ + 𝑙𝑙𝑙𝑙support 2 ∙ 300mm + 150mm
𝜏𝜏𝜏𝜏V_(red,d)=V_d∙(1-(2h+l_support)/l)=25,6kN∙(1-(2∙300mm+150mm)/2300mm)
𝑉𝑉𝑉𝑉 red,d<=𝑓𝑓𝑓𝑓v,0,edge,d
m,d 𝑉𝑉𝑉𝑉d ∙ �1 −→2ℎOK + 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙support � = 25,6kN ∙ �1 − 2 ∙ 300mm + 150mm�
𝑉𝑉𝑉𝑉 2ℎ + 𝑙𝑙𝑙𝑙support � = 25,6kN ∙ �1 − 2 ∙ 300mm 2300mm
+ 150mm�
red,d = 𝑉𝑉𝑉𝑉d ∙ �1 −
V_(red,d)=17,2kN
𝑉𝑉𝑉𝑉red,d = 𝑉𝑉𝑉𝑉d ∙ �1 − 𝑙𝑙𝑙𝑙 � = 25,6kN ∙ �1 − 2300mm �
τ_(v,d)=〖3∙V〗_d/(2∙A)=(3∙17,2 𝑙𝑙𝑙𝑙 kN)/(2 ∙13500mm^2 )=1,92 N/mm^2 2300mm
𝑉𝑉𝑉𝑉 red,d = 17,2kN
τ_(m,d)<f_(v,0,edge,d)
𝑉𝑉𝑉𝑉 ==17,2kN →OK
𝐹𝐹𝐹𝐹red,d
𝑉𝑉𝑉𝑉 c,90,d = 𝑉𝑉𝑉𝑉 = 25,6 kN
red,d 3 ∙17,2kN
𝑉𝑉𝑉𝑉dd 3 ∙ 17,2 kN
𝜏𝜏𝜏𝜏v,d = 3 ∙ 𝑉𝑉𝑉𝑉 = 3 ∙ 17,2 kN 2 = 1,92 N/mm2
𝜏𝜏𝜏𝜏𝜎𝜎𝜎𝜎v,d ==
c,90,d
32 ∙∙𝐹𝐹𝐹𝐹𝑉𝑉𝑉𝑉𝐴𝐴𝐴𝐴
d
c,90,d 2 ∙ 13500mm 𝐹𝐹𝐹𝐹c,90,d
d = = 3 ∙ 17,2 kN 2 = 1,92 N/mm2
2
(4.14)
𝜏𝜏𝜏𝜏v,d = 2 ∙ 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴ef= 2 𝑏𝑏𝑏𝑏∙�𝑙𝑙𝑙𝑙
∙ 13500mm support +152mm� = 1,92 N/mm
𝜏𝜏𝜏𝜏m,d < 2𝑓𝑓𝑓𝑓v,0,edge,d
∙ 𝐴𝐴𝐴𝐴 2 → ∙ 13500mm
OK
𝜏𝜏𝜏𝜏m,d < 𝑓𝑓𝑓𝑓v,0,edge,d →25,6kN OK
𝜏𝜏𝜏𝜏𝜎𝜎𝜎𝜎m,d < 𝑓𝑓𝑓𝑓
c,90,d =v,0,edge,d
→ OK = 3,4 N/mm2
45mm
Compression perpendicular to grain ∙ (150mm + 15mm)
𝐹𝐹𝐹𝐹c,90,d = 𝑉𝑉𝑉𝑉d = 25,6 kN 𝑘𝑘𝑘𝑘mod 0,8
𝑘𝑘𝑘𝑘F_(c,90,d)
𝐹𝐹𝐹𝐹 c,90 ∙ 𝑓𝑓𝑓𝑓=
c,90,d 𝑉𝑉𝑉𝑉d= V_d
c,90,edge,d = 25,6 = =𝑘𝑘𝑘𝑘c,90 kN∙ kN ∙ 𝑓𝑓𝑓𝑓c,90,edge,k = 1,0 ∙
25,6 ∙ 6 N/mm2 = 4 N/mm2
𝐹𝐹𝐹𝐹c,90,d = 𝐹𝐹𝐹𝐹𝑉𝑉𝑉𝑉 d = 25,6 kN
𝐹𝐹𝐹𝐹 𝛾𝛾𝛾𝛾M 1,2
σ_(c,90,d)=F_(c,90,d)/A_ef
𝜎𝜎𝜎𝜎 c,90,d = 𝐹𝐹𝐹𝐹 𝐴𝐴𝐴𝐴
c,90,d
= 𝑏𝑏𝑏𝑏∙�𝑙𝑙𝑙𝑙 𝐹𝐹𝐹𝐹c,90,d =F_(c,90,d)/(b∙(l_support+15 mm) ) (4.14)
c,90,d c,90,d+15 mm�
𝜎𝜎𝜎𝜎
(4.14)
𝜎𝜎𝜎𝜎𝜎𝜎𝜎𝜎 =
c,90,d ≤ 𝑘𝑘𝑘𝑘c,90,d
𝐹𝐹𝐹𝐹
ef
∙ =
𝑓𝑓𝑓𝑓
support
𝐹𝐹𝐹𝐹c,90,d→ OK (4.14)
(4.13)
c,90,d = 𝐴𝐴𝐴𝐴ef = 𝑏𝑏𝑏𝑏∙�𝑙𝑙𝑙𝑙support +15 mm�
c,90,d c,90
𝐴𝐴𝐴𝐴 m,0,edge,d
𝑏𝑏𝑏𝑏∙�𝑙𝑙𝑙𝑙 +15 mm� (4.14)
σ_(c,90,d)=25,6kN/(45mm∙(150mm+15mm))=3,4
ef 25,6kNsupport N/mm^2
2
𝜎𝜎𝜎𝜎 c,90,d =
k_(c,90)∙f_(c,90,edge,d)=k_(c,90)∙k_mod/γ_M 25,6kN + 15mm) = 3,4 N/mm ∙f_(c,90,edge,k)=1,0∙0,8/1,2∙6 N/mm^2=4 N/m
𝜎𝜎𝜎𝜎c,90,d = 45mm ∙ (150mm 25,6kN = 3,4 N/mm22
σ_(c,90,d)≤k_(c,90)∙f_(m,0,edge,d)
𝜎𝜎𝜎𝜎 c,90,d = 45mm ∙ (150mm + 15mm) →OK
= 3,4 N/mm
45mm ∙ (150mm𝑘𝑘𝑘𝑘+ 15mm) 0,8
mod
𝑘𝑘𝑘𝑘c,90 ∙ 𝑓𝑓𝑓𝑓c,90,edge,d = 𝑘𝑘𝑘𝑘c,90 ∙ 𝑘𝑘𝑘𝑘 ∙ 𝑓𝑓𝑓𝑓c,90,edge,k = 1,0 ∙ 0,8 ∙ 6 N/mm2 = 4 N/mm2
𝑘𝑘𝑘𝑘c,90 ∙ 𝑓𝑓𝑓𝑓c,90,edge,d = 𝑘𝑘𝑘𝑘c,90 ∙ 𝑘𝑘𝑘𝑘mod 𝛾𝛾𝛾𝛾
modM ∙ 𝑓𝑓𝑓𝑓 = 1,0 ∙ 1,2
0,8 ∙ 6 N/mm22 = 4 N/mm22
𝑘𝑘𝑘𝑘c,90 ∙ 𝑓𝑓𝑓𝑓c,90,edge,d = 𝑘𝑘𝑘𝑘c,90 ∙ 𝛾𝛾𝛾𝛾M ∙ 𝑓𝑓𝑓𝑓c,90,edge,k c,90,edge,k = 1,0 ∙ 1,2 ∙ 6 N/mm = 4 N/mm
𝜎𝜎𝜎𝜎c,90,d ≤ 𝑘𝑘𝑘𝑘c,90 ∙ 𝑓𝑓𝑓𝑓m,0,edge,d →𝛾𝛾𝛾𝛾M OK 1,2 (4.13)
𝜎𝜎𝜎𝜎c,90,d ≤ 𝑘𝑘𝑘𝑘c,90 ∙ 𝑓𝑓𝑓𝑓m,0,edge,d → OK (4.13)
𝜎𝜎𝜎𝜎(4.13)
c,90,d ≤ 𝑘𝑘𝑘𝑘c,90 ∙ 𝑓𝑓𝑓𝑓m,0,edge,d → OK (4.13)
25,6kN
𝜎𝜎𝜎𝜎c,90,d = = 3,4 N/mm2
45mm ∙ (150mm + 15mm)
𝑘𝑘𝑘𝑘mod 9. CALCULATION
0,8 EXAMPLES OF LVL STRUCTURES
𝑘𝑘𝑘𝑘c,90 ∙ 𝑓𝑓𝑓𝑓c,90,edge,d = 𝑘𝑘𝑘𝑘c,90 ∙ ∙ 𝑓𝑓𝑓𝑓c,90,edge,k = 1,0 ∙ ∙ 6 N/mm2 = 4 N/mm2
𝛾𝛾𝛾𝛾M 1,2
SLS𝜎𝜎𝜎𝜎design
c,90,d ≤ 𝑘𝑘𝑘𝑘c,90 ∙ 𝑓𝑓𝑓𝑓m,0,edge,d → OK (4.13)
222 (253)
223 (255)
Instantaneous deflection
6
5∙𝑔𝑔 ∙𝑠𝑠∙𝐿𝐿4 ∙𝑔𝑔d,SLS ∙𝑠𝑠∙𝐿𝐿2
𝑤𝑤
𝑤𝑤𝑤𝑤inst,g
inst = = 𝑤𝑤𝑤𝑤 d,SLS
inst,g + 𝑤𝑤𝑤𝑤+inst,q
5
(4.74)
w_inst =5∙𝑔𝑔
w_(inst,g)
384∙𝐸𝐸 mean4∙𝐼𝐼 +
∙𝑠𝑠∙𝐿𝐿
6 w_(inst,q)
∙𝑔𝑔
8∙𝐺𝐺mean 𝐴𝐴2
∙𝑠𝑠∙𝐿𝐿
5 d,SLS
d,SLS
𝑤𝑤w_(inst,g)=(5〖∙g〗_(d,SLS)∙s∙L^4)/(〖384∙E〗_mean∙I)+〖〖6/5∙g〗_(d,
inst,g = 384∙𝐸𝐸 + 68∙𝐺𝐺 (4.74)
∙𝐼𝐼 𝐴𝐴 2
5
5∙𝑔𝑔𝑔𝑔∙ d,SLS
𝑔𝑔mean
d,SLS∙𝑠𝑠𝑠𝑠∙𝐿𝐿𝐿𝐿4∙ 𝑠𝑠 ∙ 5𝐿𝐿∙𝑔𝑔𝑔𝑔4d,SLS 6/5
mean∙𝑠𝑠𝑠𝑠∙𝐿𝐿𝐿𝐿
∙ 𝑔𝑔d,SLS ∙ 𝑠𝑠 ∙ 𝐿𝐿2
𝑤𝑤𝑤𝑤
𝑤𝑤 inst,g = 384∙𝐸𝐸𝐸𝐸
inst,g
= + + = 1,30 mm + 0,49 mm = (4.74)
1,79 mm (4.74)
∙𝐼𝐼𝐼𝐼 48∙𝐺𝐺𝐺𝐺 mean 𝐴𝐴𝐴𝐴 8 ∙ 𝐺𝐺
w_(inst,g)=1,30 5 384
∙ 𝑔𝑔 d,SLS
∙
mean𝐸𝐸 ∙
mm+0,49
mean 𝑠𝑠 ∙ ∙
𝐿𝐿 𝐼𝐼 6/5
mm=1,79 ∙ 𝑔𝑔 d,SLS mm
mean ∙ 𝑠𝑠𝐴𝐴 ∙ 𝐿𝐿 2
𝑤𝑤inst,g = + = 1,30 mm + 0,49 mm = 1,79 mm
384 ∙ 𝐸𝐸mean ∙ 𝐼𝐼
w_(inst,q)=(5〖∙q〗_(d,SLS)∙s∙L^4)/(〖384∙E〗_mean∙I)+(6/5∙〖q_(d,SLS)∙s∙L〗
𝑤𝑤 8 ∙ 𝐺𝐺mean 𝐴𝐴
inst,g = 1,30 mm + 0,49 mm = 1,79 mm
𝑤𝑤𝑤𝑤 inst,g = 1,30
mm+1,08 mm=3,95 mm +mm 0,49 mm = 1,79 mm
𝑤𝑤𝑤𝑤𝑤𝑤inst,g
inst,g= =1,30
1,30mm+3,95 mm
mm++0,49 0,49 4 mm
mm==mm 1,79
1,79mm mm
w_inst=1,79 5 ∙ 𝑞𝑞d,SLS ∙ 𝑠𝑠 ∙ 𝐿𝐿mm=5,5 4 6/5 ∙ 𝑞𝑞d,SLS ∙ 𝑠𝑠 ∙ 𝐿𝐿22
𝑤𝑤inst,q = 5 ∙ 𝑞𝑞𝑞𝑞d,SLS ∙ 𝑠𝑠𝑠𝑠 ∙ 𝐿𝐿𝐿𝐿 + 6/5 ∙ 𝑞𝑞𝑞𝑞 d,SLS ∙ 𝑠𝑠𝑠𝑠 ∙ 𝐿𝐿𝐿𝐿 = 2,87 mm + 1,08 mm = 3,95 mm
𝑤𝑤𝑤𝑤inst,q =55384
∙384 ∙ 𝐸𝐸mean
∙𝑞𝑞𝑞𝑞𝑞𝑞d,SLS ∙ ∙𝑠𝑠 ∙
𝑠𝑠𝑠𝑠 ∙𝐿𝐿∙𝐿𝐿𝐿𝐿4𝐼𝐼4 +6/5 6/5 ∙8∙𝑞𝑞𝑞𝑞𝑞𝑞∙d,SLS
𝐺𝐺mean 𝑠𝑠𝑠𝑠𝐴𝐴∙ ∙𝐿𝐿𝐿𝐿𝐿𝐿22 = 2,87 mm + 1,08 mm = 3,95 mm
∙ ∙𝑠𝑠𝐴𝐴𝐴𝐴
𝑤𝑤𝑤𝑤𝑤𝑤inst,q ∙ 𝐸𝐸𝐸𝐸mean ∙ 𝐼𝐼𝐼𝐼 ++
d,SLS 8 ∙ 𝐺𝐺𝐺𝐺mean
d,SLS
inst,q= = ==2,872,87mm mm++1,081,08mm mm==3,95
3,95mm
mm
𝑤𝑤𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 = 1,79 384
384mm ∙ ∙𝐸𝐸𝐸𝐸𝐸𝐸mean
+ 3,95
mean ∙ ∙𝐼𝐼𝐼𝐼𝐼𝐼 mm =885,5 ∙ ∙𝐺𝐺𝐺𝐺𝐺𝐺mean
mm𝐴𝐴𝐴𝐴𝐴𝐴
mean
𝑤𝑤𝑤𝑤𝑖𝑖𝑖𝑖𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 = 1,79 mm + 3,95 mm = 5,5 mm
𝑤𝑤𝑤𝑤𝑤𝑤𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖
𝑖𝑖𝑖𝑖𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 ==1,79
1,79mm mm++3,95 3,95mm mm==5,5 5,5mm mm
Final deflection
Final deflection
Final deflection
𝑤𝑤net,fin = (1 + 𝑘𝑘def ) ∙ 𝑤𝑤inst,g + (1 + 𝜓𝜓2 ∙ 𝑘𝑘def ) ∙ 𝑤𝑤inst,q (4.73)
𝑤𝑤𝑤𝑤 net,fin = (1
w_(net,fin) =+ 𝑘𝑘𝑘𝑘def ) ∙ 𝑤𝑤𝑤𝑤inst,g + (1 +
(1+k_def)∙w_(inst,g) +(1+ψ_2∙k_def)∙w_(inst,q)
𝜓𝜓𝜓𝜓2 ∙ 𝑘𝑘𝑘𝑘def ) ∙ 𝑤𝑤𝑤𝑤inst,q (4.73) (4.73)
𝑤𝑤𝑤𝑤𝑤𝑤net,fin
net,fin = = (1 (1++𝑘𝑘𝑘𝑘𝑘𝑘def def))∙ ∙𝑤𝑤 inst,g +
𝑤𝑤𝑤𝑤inst,g + (1 (1++𝜓𝜓𝜓𝜓𝜓𝜓22∙ ∙𝑘𝑘𝑘𝑘𝑘𝑘def def))∙ ∙𝑤𝑤
𝑤𝑤𝑤𝑤inst,q
inst,q (4.73)
(4.73)
Note: For the snow load in Finnish National annex: ψ2 = 0,2
Note: For
Note: For the
thesnow snowload loadininFinnish Finnish National
National annex: annex: ψ2 =ψ0,2. = 0,2
𝑤𝑤net,fin = (1 + 0,6) ∙ 1,79 mm + (1 + 0,2 ∙ 0,6) ∙ 3,952mm = 7,3 mm
𝑤𝑤𝑤𝑤net,fin = (1 + 0,6) ∙ 1,79 mm + (1 + 0,2 ∙ 0,6) ∙ 3,95 mm = 7,3 mm
𝑤𝑤𝑤𝑤𝑤𝑤net,fin
net,fin = = (1 (1++0,6) 0,6)∙ ∙1,79 1,79mm mm ++ (1 (1+L+0,2 0,22300 ∙ ∙0,6)0,6)∙ ∙3,953,95mm mm == 7,3 7,3mmmm
When
w_(net,fin) the requirement
= (1+0,6)∙1,79 is 𝑤𝑤mm net,fin ≤ 300
L , 2300 = mm
+ (1+0,2∙0,6)∙3,95 7,7 mm = 7,3→ OK
mm
When the requirement is 𝑤𝑤𝑤𝑤 net,fin ≤ LL
= 300 = 7,7 mm
2300
2300mm
→ OK
300
WhenWhenthe
When requirement
requirementisisw_(net,fin)≤L/300=2300/300=7,7
therequirement 𝑤𝑤𝑤𝑤𝑤𝑤net,fin
net,fin≤ ≤ ,= 300 == 7,77,7 mm mm →mm→OK
→
OKOK
The lintel beam fulfils the design requirements. 300 However,
300 300in practice the required support
300
lengths
The lintelare quite
beam fulfils longthe and for windows
design requirements. a more However, strict deflection in practice limit can be required.
the required supportsupport lengths are quite
The lintel beam fulfils the design requirements. However, in practice the required
Therefore a double lintel 2x45x260 mm or a 69x300 mm lintel from LVL 36 C could be a
lengths arelong
quite long and for windows a more strict deflection limit can be required.
more suitable and choice. for windows a more strict deflection limit can be required. Therefore a double lintel 2x45x260 mm
Therefore ora double
a 69x300 lintel
mm2x45x260lintel frommm LVLor36aC69x300 could be mm a more lintelsuitable
from LVL 36 C could be a
choice.
more suitable choice.
9.3 Double LVL 48 P ridge beam for roof
9.3 Double LVL 48 P ridge beam for roof
Single-span ridge beam of the roof in a one family house is LVL 48 P double beam 2x51x400
mm. Span length is L = 4000 mm, width of the loading area 6000 mm and roof rafters
Single-span ridge beam of the roof in a one family house is LVL 48 P double beam 2x51x400
connected
mm. to the issides
Span length of themm,
L = 4000 beam at spacing
width s = 1200
of the loading mm.
area Support
6000 length
mm and roofisrafters
120 mm.
Snow load sk is 2,5 kN/m2, own weight of the roof structure is 1,0 kN/m2 and own weigh of
connected to the sides of the beam at spacing s = 1200 mm. Support length is 120 mm.
the beam
Snow load is
s 0,2 kN/m.
is 2,5 kN/m2, own weight of the roof structure is 1,0 kN/m2 and own weigh of
k
the beam is 0,2 kN/m. Service class SC1.
Beam properties:
Beam properties:
Bending strength edgewise f = 44 N/mm2
m,0,edge,k
Shear strength
Bending strengthedgewise
edgewisefv,0,edge,k
fm,0,edge,k ==44
4,2N/mm
N/mm2 2
Compression perpendicular
Shear strength edgewise fv,0,edge,k to grain edgewise fc,90,edge,k ==4,2
6 N/mm
N/mm
22
Modulus of elasticity
Compression perpendicularE 0,k to grain edgewise fc,90,edge,k = 611
= 600 2N/mm2
N/mm
Modulusofofelasticity
Modulus elasticityEE0,k0,mean ==11
13600
800N/mm
N/mm2 2 LVL Handbook Europe 187
Modulusofofelasticity
Modulus rigidity GE0,edge,k
0,mean ==13
600 N/mm
800 N/mm2
2
Modulus of rigidityGG0,edge,k
Modulus of rigidity 0,edge,mean ==600
400N/mm
N/mm2 2
2020_LVL_09.indd 187 Modulus of rigidity G0,edge,mean = 400 N/mm2 12.3.2020 13:12:11
9. CALCULATION EXAMPLES OF LVL STRUCTURES
Beam properties:
Bending strength edgewise fm,0,edge,k = 44 N/mm2
Shear strength edgewise fv,0,edge,k = 4,2 N/mm2
Compression perpendicular to grain edgewise fc,90,edge,k = 6 N/mm2
Modulus of elasticity E0,k = 11 600 N/mm2
Modulus of elasticity E0,mean = 13 800 N/mm2
Modulus of rigidity G0,edge,k = 600 N/mm2
Modulus of rigidity G0,edge,mean = 400 N/mm2
Area of cross section A =2 x b∙h = 40800 mm2
Section modulus Wy = 2 x b∙h /6
2 = 2,72∙106 mm3
Moment of inertia Iy = 2 x b∙h /12
3 = 5,44∙108 mm4
Moment of inertia Iz = 2 x h∙b /12
3 = 8,84∙106 mm4
Torsion moment of inertia Itor = 2 x 0,3∙h∙b
3 = 3,18∙107 mm4
Moment stiffness of the joist EIy = 13800 N/mm ∙ 5,44∙10 mm = 7,51∙1012 Nmm2
2 8 4
Loading combinations
Snow load at roof level qk= μ1∙Ce∙sk. Form factor μ1=0,8, when roof angle is less than 30° and in normal
conditions Ce = 1,0 → qk= 0,8∙1,0∙2,5 N/m2 = 2,0 kN/m2.
Most critical ultimate limit state (ULS) load combination:
E_(d,ULS)
𝐸𝐸𝐸𝐸 = 𝛾𝛾𝛾𝛾G=∙γ_G∙(g_(1,k)+g_(2,k))+
(𝑔𝑔𝑔𝑔1,k + 𝑔𝑔𝑔𝑔2,k ) + 𝛾𝛾𝛾𝛾Q ∙ 𝑞𝑞𝑞𝑞k γ_Q∙q_k (4.1) (4.1)
𝐸𝐸𝐸𝐸d,ULS
d,ULS = 𝛾𝛾𝛾𝛾G ∙ (𝑔𝑔𝑔𝑔1,k + 𝑔𝑔𝑔𝑔2,k ) + 𝛾𝛾𝛾𝛾Q ∙ 𝑞𝑞𝑞𝑞k (4.1)
E_(d,ULS)
𝐸𝐸𝐸𝐸 = 1,15 =1,15∙(6m
∙ (6m ∙ ∙1,0kN/m
1,0 kN/m^2+0,2
2
+ 0,2 kN/m ))+1,5∙6m∙2,0
kN/m + 1,5 ∙ 6m ∙ kN/m^2
2,0 kN/m2
d,ULS
𝐸𝐸𝐸𝐸d,ULS = 1,15
E_(d,ULS)= ∙ (6m
25,1 kN/m∙ 1,0 kN/m2 + 0,2 kN/m ) + 1,5 ∙ 6m ∙ 2,0 kN/m2
𝐸𝐸𝐸𝐸d,ULS = 25,1 kN/m
𝐸𝐸𝐸𝐸d,ULS = 25,1 kN/m
Note: Safety factors γG and γQ are according to Finnish National annex of Eurocode 0.
The most critical serviceability limit state (SLS) load combination:
𝐸𝐸𝐸𝐸d,SLS = 𝛾𝛾𝛾𝛾G ∙ (𝑔𝑔𝑔𝑔1,k + 𝑔𝑔𝑔𝑔2,k ) + 𝛾𝛾𝛾𝛾Q ∙ 𝑞𝑞𝑞𝑞k (4.1)
E_(d,SLS)
𝐸𝐸𝐸𝐸d,SLS = 𝛾𝛾𝛾𝛾=
G ∙γ_G∙(g_(1,k)+g_(2,k))+
(𝑔𝑔𝑔𝑔1,k + 𝑔𝑔𝑔𝑔2,k ) + 𝛾𝛾𝛾𝛾Q ∙ 𝑞𝑞𝑞𝑞k γ_Q∙q_k (4.1) (4.1)
2 2
𝐸𝐸𝐸𝐸d,SLS = 1,0
E_(d,SLS) ∙ (6m ∙ ∙1,0
=1,0∙(6m 1,0 kN/m 2 + 0,2 kN/m
kN/m^2+0,2 kN/m ))+6,0∙1,0∙2,0
+ 6,0 ∙ 1,0 ∙kN/m^2
2,0 kN/m2
𝐸𝐸𝐸𝐸d,SLS = 1,0 ∙ (6m ∙ 1,0 kN/m + 0,2 kN/m ) + 6,0 ∙ 1,0 ∙ 2,0 kN/m
E_(d,SLS)=
𝐸𝐸𝐸𝐸 = 18,2 18,2kN/m kN/m
𝐸𝐸𝐸𝐸d,SLS
d,SLS = 18,2 kN/m
ULS design
Bending moment resistance
𝑀𝑀𝑀𝑀d = 𝐸𝐸𝐸𝐸d,ULS ∙ 𝑠𝑠𝑠𝑠 ∙ 𝐿𝐿𝐿𝐿2/8 = 25,1 kN/m ∙ (4m)22 /8 = 50,3 kNm
𝑀𝑀𝑀𝑀d = 𝐸𝐸𝐸𝐸d,ULS ∙ 𝑠𝑠𝑠𝑠 ∙ 𝐿𝐿𝐿𝐿2/8 = 25,1 kN/m ∙ (4m) /8 = 50,3 kNm
M_d = 𝑀𝑀𝑀𝑀 E_(d,ULS)∙s∙L2/8
d 50,3 kNm= 25,1 kN/m∙〖(4m)〗^2/8 = 50,3 kNm
𝜎𝜎𝜎𝜎m,d = 𝑀𝑀𝑀𝑀d =
σ_(m,d)=M_d/W=(50,3 50,3 kNm = 18,5 N/mm22〖 mm〗^3 )=18,5 N/mm^2
kNm)/(2,72〖∙10〗^6
6 mm 3
𝜎𝜎𝜎𝜎m,d = 𝑊𝑊𝑊𝑊 = 2,72 ∙ 10 = 18,5 N/mm
𝑊𝑊𝑊𝑊 2,72 ∙ 106 mm3∙k_h∙f_(m,0,edge,k)=0,8/1,2∙0,96∙44 N/mm^2 =28,1 N/mm^2
f_(m,0,edge,d)=k_mod/γ_M
𝑘𝑘𝑘𝑘mod
σ_(m,d)≤f_(m,0,edge,d) →OK 0,8 N
𝑓𝑓𝑓𝑓m,0,edge,d = 𝑘𝑘𝑘𝑘mod ∙ 𝑘𝑘𝑘𝑘h ∙ 𝑓𝑓𝑓𝑓m,0,edge,k = 0,8 ∙ 0,96 ∙ 44 N 2 = 28,1 N/mm22
𝛾𝛾𝛾𝛾
𝑓𝑓𝑓𝑓m,0,edge,d = M ∙ 𝑘𝑘𝑘𝑘h ∙ 𝑓𝑓𝑓𝑓m,0,edge,k = 1,2 ∙ 0,96 ∙ 44 mm = 28,1 N/mm
𝛾𝛾𝛾𝛾M 1,2 mm2
𝜎𝜎𝜎𝜎m,d ≤ 𝑓𝑓𝑓𝑓m,0,edge,d → OK
𝜎𝜎𝜎𝜎m,d ≤ 𝑓𝑓𝑓𝑓m,0,edge,d → OK
The ridge beam is loaded by the roof rafters connected at the sides of the beam at 1200 mm spacing and
they act as supports against lateral torsional buckling, so the effective length is Lef = 1200 mm.
𝑀𝑀𝑀𝑀y,crit 𝜋𝜋𝜋𝜋�𝐸𝐸𝐸𝐸0,05 𝐼𝐼𝐼𝐼𝑧𝑧𝑧𝑧 𝐺𝐺𝐺𝐺0,05 𝐼𝐼𝐼𝐼tor
𝜎𝜎𝜎𝜎σ_(m,crit)=M_(y,crit)/W_y
m,crit = 𝑀𝑀𝑀𝑀y,crit = 𝜋𝜋𝜋𝜋�𝐸𝐸𝐸𝐸0,05 𝐼𝐼𝐼𝐼 𝐺𝐺𝐺𝐺 =(π√(E_0,05
𝐼𝐼𝐼𝐼 I_z G_0,05 I_tor ))/(l_ef W_y (4.42)
) (4.42)
𝜎𝜎𝜎𝜎m,crit = 𝑊𝑊𝑊𝑊 y = 𝑙𝑙𝑙𝑙ef𝑧𝑧𝑧𝑧𝑊𝑊𝑊𝑊y0,05 tor (4.42)
σ_(m,crit)= y(π√(10600 efN/mm^2∙8,84∙〖10〗^6
𝑊𝑊𝑊𝑊 𝑙𝑙𝑙𝑙 𝑊𝑊𝑊𝑊 y 〖 mm〗^4∙400N/m
〖2,72∙10〗^6 π�10600〖 mm〗^3 N/mm ) 2 ∙ 8,84 ∙ 106 mm4 ∙ 400N/mm2 ∙ 3,18 ∙ 107 ∙ mm4
𝜎𝜎𝜎𝜎m,crit = π�10600
σ_(m,crit)=34,8 N/mm^2 N/mm2 ∙ 8,84 ∙ 106 mm4 ∙ 400N/mm 2 ∙ 3,18 ∙ 107 ∙ mm4
𝜎𝜎𝜎𝜎m,crit = 1200 mm ∙ 2,72 ∙ 1066 mm33
1200 mm ∙ 2,72 ∙ 10 mm
2
𝜎𝜎𝜎𝜎m,crit = 34,8 N/mm2
𝜎𝜎𝜎𝜎m,crit = 34,8 N/mm
𝑓𝑓𝑓𝑓 44 N/mm2
λ_rel=√(f_(m,k)/σ_(m,crit)
m,k
𝜆𝜆𝜆𝜆rel = � 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓m,k
m,k = � 44 N/mm22)=√((44
44 N/mm = 1,12N/mm^2)/(34,8 N/mm^2 ))= 1,12 (4.41) (4.41)
𝜆𝜆𝜆𝜆rel
rel = � 𝜎𝜎𝜎𝜎𝜎𝜎𝜎𝜎m,crit = � 34,8 34,8 N/mm222 = 1,12 (4.41)
when 0,75<λ_(rel,m)≤1,4
𝜎𝜎𝜎𝜎 m,crit
m,crit 34,8 N/mm
N/mm,k_crit=1,56-0,75∙λ_(rel,m)=1,56-0,75∙1,12=0,72
k_crit∙
when 0,75 f_(m,d)=0,72
0,75 < 𝜆𝜆𝜆𝜆𝜆𝜆𝜆𝜆rel,m ≤∙28,1
≤ 1,4 N/mm^2=20,1 N/mm^2 = 1,56 − 0,75 ∙ 1,12 = 0,72
when
σ_(m,d)≤k_crit∙ < 1,4
f_(m,d)→OK
rel,m
rel,m ,,𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘crit
crit = 1,56 − 0,75 ∙ 𝜆𝜆𝜆𝜆rel,m =
crit = 1,56 − 0,75 ∙ 𝜆𝜆𝜆𝜆rel,m
rel,m 1,56 − 0,75 ∙ 1,12 = 0,72
𝑘𝑘𝑘𝑘crit ∙ 𝑓𝑓𝑓𝑓m,d = 0,72 ∙ 28,1 N/mm222 = 20,1 N/mm222
crit ∙ 𝑓𝑓𝑓𝑓m,d
𝑘𝑘𝑘𝑘crit m,d = 0,72 ∙ 28,1 N/mm = 20,1 N/mm
Shear resistance
V_d==𝐸𝐸𝐸𝐸E_(d,ULS)∙s∙L/2 = 25,1kN/m∙4,0m/2 = 50,3 kN kN
𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉ddd = d,ULS ∙ 𝑠𝑠𝑠𝑠 ∙ 𝐿𝐿𝐿𝐿/2 = 25,1kN/m ∙∙ 4,0m/2
𝐸𝐸𝐸𝐸d,ULS
d,ULS ∙ 𝑠𝑠𝑠𝑠 ∙ 𝐿𝐿𝐿𝐿/2 = 25,1kN/m
4,0m/2 == 50,3
50,3 kN
τ_(v,d)=〖3∙V〗_d/(2∙A)=(3∙50,3 kN)/(2 ∙40 800 mm^2 )=1,9 N/mm^2
3 ∙ 𝑉𝑉𝑉𝑉d
f_(v,0,edge,d)=k_mod/γ_M 3 ∙ 50,3 kN∙f_(v,0,edge,k)=0,8/1,2∙4,2 N/mm^2 =2,8 N/mm^2
𝜏𝜏𝜏𝜏v,d = 3 ∙ 𝑉𝑉𝑉𝑉dd = 3 ∙ 50,3 kN 2 = 1,9 N/mm222
𝜏𝜏𝜏𝜏τ_(m,d)≤f_(v,0,edge,d)
v,d
v,d = =
2 ∙ 𝐴𝐴𝐴𝐴 2 ∙ 40 800 →OK mm = 1,9 N/mm
2 ∙ 𝐴𝐴𝐴𝐴 2 ∙ 40 800 mm22
𝑘𝑘𝑘𝑘mod 0,8 N
𝑓𝑓𝑓𝑓v,0,edge,d = 𝑘𝑘𝑘𝑘modmod ∙ 𝑓𝑓𝑓𝑓v,0,edge,k = 0,8 ∙ 4,2 N = 2,8 N/mm222
v,0,edge,d = 𝛾𝛾𝛾𝛾M ∙ 𝑓𝑓𝑓𝑓v,0,edge,k
𝑓𝑓𝑓𝑓v,0,edge,d v,0,edge,k = 1,2∙ 4,2 mm222 = 2,8 N/mm
𝛾𝛾𝛾𝛾M
M 1,2 mm
𝜏𝜏𝜏𝜏𝜏𝜏𝜏𝜏m,d
m,d ≤ 𝑓𝑓𝑓𝑓v,0,edge,d →
m,d ≤ 𝑓𝑓𝑓𝑓v,0,edge,d
v,0,edge,d
→ OK
OK
𝜎𝜎𝜎𝜎𝜎𝜎𝜎𝜎c,90,d
c,90,d ≤ 𝑘𝑘𝑘𝑘c,90 ∙∙ 𝑓𝑓𝑓𝑓
𝑓𝑓𝑓𝑓m,0,edge,d →
→ OK
OK (4.13)
(4.13)
(4.13)c,90,d ≤ 𝑘𝑘𝑘𝑘c,90
c,90 m,0,edge,d
m,0,edge,d
50,3kN
𝜎𝜎𝜎𝜎c,90,d = = 3,7 N/mm2
2 ∙ 51mm ∙ (120mm + 15mm)
9. CALCULATION EXAMPLES OF LVL STRUCTURES
𝑘𝑘𝑘𝑘mod 0,8
𝑘𝑘𝑘𝑘c,90 ∙ 𝑓𝑓𝑓𝑓c,90,edge,d = 𝑘𝑘𝑘𝑘c,90 ∙ ∙ 𝑓𝑓𝑓𝑓c,90,edge,k = 1,0 ∙ ∙ 6 N/mm2 = 4 N/mm2
𝛾𝛾𝛾𝛾M 1,2
SLS design
𝜎𝜎𝜎𝜎c,90,d ≤ 𝑘𝑘𝑘𝑘c,90 ∙ 𝑓𝑓𝑓𝑓m,0,edge,d → OK (4.13)
Instantaneous deflection
w_inst
𝑤𝑤𝑤𝑤inst ==𝑤𝑤𝑤𝑤
w_(inst,g)
inst,g + 𝑤𝑤𝑤𝑤+ w_(inst,q)
inst,q
w_(inst,g)=(5〖∙g〗_(d,SLS)∙s∙L^4)/(〖384∙E〗_mean∙I)+〖〖6/5∙g〗_(d,SLS)∙s∙L〗^2/(〖8∙G〗_mean A)=2,76m-
5 ∙ 𝑔𝑔𝑔𝑔d,SLS ∙ 𝑠𝑠𝑠𝑠 ∙ 𝐿𝐿𝐿𝐿4 6/5 ∙ 𝑔𝑔𝑔𝑔d,SLS ∙ 𝑠𝑠𝑠𝑠 ∙ 𝐿𝐿𝐿𝐿2
m+0,61mm=3,36mm
𝑤𝑤𝑤𝑤inst,g = + = 2,76mm + 0,61mm = 3,36mm
384 ∙ 𝐸𝐸𝐸𝐸mean ∙ 𝐼𝐼𝐼𝐼 8 ∙ 𝐺𝐺𝐺𝐺mean 𝐴𝐴𝐴𝐴
w_(inst,q)=(5〖∙q〗_(d,SLS)∙s∙L^4)/(〖384∙E〗_mean∙I)+〖〖6/5∙q〗_(d,SLS)∙s∙L〗^2/(〖8∙G〗_mean A)=5,33m-
m+1,18mm=6,51mm 4 2 226 (255)
5 ∙ 𝑞𝑞𝑞𝑞d,SLS ∙ 𝑠𝑠𝑠𝑠 ∙ 𝐿𝐿𝐿𝐿 6/5 ∙ 𝑞𝑞𝑞𝑞d,SLS ∙ 𝑠𝑠𝑠𝑠 ∙ 𝐿𝐿𝐿𝐿
w_inst=3,36
𝑤𝑤𝑤𝑤inst,q =
mm+6,51 mm=9,9 + mm = 5,33mm + 1,18mm = 6,51mm
384 ∙ 𝐸𝐸𝐸𝐸mean ∙ 𝐼𝐼𝐼𝐼 8 ∙ 𝐺𝐺𝐺𝐺mean 𝐴𝐴𝐴𝐴
Final deflection
Final deflection
w_(net,fin) = (1+k_def)∙w_(inst,g) + (1+ψ_2∙k_def)∙w_(inst,q)
𝑤𝑤𝑤𝑤𝑤𝑤net,fin = (1 + 𝑘𝑘def ))∙ ∙𝑤𝑤𝑤𝑤𝑤𝑤inst,g + (1 + 𝜓𝜓2 ∙ ∙𝑘𝑘𝑘𝑘𝑘𝑘def ) ∙ 𝑤𝑤inst,q (4.73)
(4.73)
net,fin = (1 + 𝑘𝑘𝑘𝑘def
(4.73)
net,fin def inst,g + (1 + 𝜓𝜓𝜓𝜓2
inst,g 2 def ) ∙ 𝑤𝑤𝑤𝑤inst,q
def inst,q
Note:
For the Forsnow
the snow load
load in in Finnish
Finnish National
national annex:annex: ψ2 = 0,2
ψ = 0,2
Note: For the snow load in Finnish National annex:22ψ2 = 0,2.
w_(net,fin)
𝑤𝑤𝑤𝑤𝑤𝑤 = (1=+(1+0,6)∙3,36 mm +
0,6) ∙ 3,36 mm + (1+0,2∙0,6)∙6,51
(1 + 0,2 ∙ 0,6) ∙mm
6,51=mm
12,7=mm12,7 mm
net,fin = (1 + 0,6) ∙ 3,36 mm + (1 + 0,2 ∙ 0,6) ∙ 6,51 mm = 12,7 mm
net,fin
net,fin
𝐿𝐿 4000 mm
Whenthe
When therequirement
requirement ≤ 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 ,, 4000
requirementisisw_(net,fin)≤L/300=4000/300=13,3mm→OK
𝑤𝑤𝑤𝑤𝑤𝑤net,fin 4000 = 13,3mm → OK
When the net,fin ≤300 = 300 = 13,3mm → OK
net,fin 300
300 300
300
Beam properties:
Bending strength edgewise fm,0,edge,k = 44 N/mm2
Bending strength flatwise fm,0,flat,k = 48 N/mm2
Shear strength edgewise fv,0,edge,k = 4,2 N/mm2
Compression parallel to grain fc,0,SC2,k = 29 N/mm2
Compression perpendicular to grain edgewise fc,90,edge,k = 6 N/mm2
Modulus of elasticity E0,k = 11 600 N/mm2
Modulus of elasticity E0,mean = 13 800 N/mm2
Modulus of rigidity G0,edge,k = 600 N/mm2
Modulus of rigidity G0,edge,mean = 400 N/mm2
Area of cross section A = b∙h = 10800 mm2
Section modulus Wy = b∙h /6 2 = 4,32∙105 mm3
Section modulus Wz = h∙b /6 2 = 8,10∙104 mm3
Moment of inertia Iy = b∙h /12
3 = 5,18∙107 mm4
Moment of inertia Iz = h∙b /12
3 = 1,82∙106 mm4
Torsion moment of inertia Itor = 0,3∙h∙b 3 = 6,56∙106 mm4
Moment stiffness of the joist EIy = 13800 N/mm ∙ 6,56∙10 mm = 7,15∙1011 Nmm2
2 6 4
Loading combinations
Own weight in z-direction gk,z: cos15° ∙ 0,9m ∙ 0,3 kN/m2 = 0,26 kN/m
Own weight in y-direction gk,y: sin15° ∙ 0,9m ∙ 0,4 kN/m2 = 0,07 kN/m.
Snow load at roof level qk = μ1 ∙ Ce ∙ sk Form factor μ1 = 0,8, when roof angle is less than 30° and in normal
conditions Ce = 1,0.
qk = 0,8 ∙ 1,0 ∙ 2,5 N/m2 = 2 kN/m2 (horizontal projection).
qk,z = cos15° ∙ cos15° ∙ 2kN/m = 1,68 kN/m
qk,y = cos15 ∙ sin15° ∙ 2kN/m = 0,45 kN/m
The most critical ultimate limit state (ULS) load combination:
Ed,z,ULS = γG ∙ gk,z + γQ∙qk,z
Ed,z,ULS = 1,15 ∙ 0,26 kN/m2 + 1,5 ∙ 1,68 kN/m2 = 2,82 kN/m
Ed,y,ULS = γG ∙ gk,y + γQ ∙ qk,y
Ed,y,ULS = 1,15 ∙ 0,07 kN/m2 + 1,5 ∙ 0,45 kN/m2 = 0,76 kN/m
Axial compression Nc,d = γQ ∙ Nc,k = 1,5 ∙ 3 kN/m2 = 4,5 kN
Note: Safety factors γG and γQ are according to Finnish national annex of Eurocode 0.
ULS design
Bending moment resistance in y-direction
𝑘𝑘𝑘𝑘mod 0,8 N
𝑓𝑓𝑓𝑓m,0,edge,d = ∙ 𝑘𝑘𝑘𝑘h ∙ 𝑓𝑓𝑓𝑓m,0,edge,k = ∙ 1,034 ∙ 44 = 30,3 N/mm2
𝛾𝛾𝛾𝛾M 1,2 mm2
Bending moment resistance in z-direction at centre support of a 2-span beam
𝑀𝑀𝑀𝑀d,y = 𝐸𝐸𝐸𝐸d,y,ULS ∙ (L/2)222 /8 = 0,76 kN/m ∙ (4 m/2)222 /8 = 0,38 kNm
𝑀𝑀𝑀𝑀 d,y=
𝑀𝑀𝑀𝑀d,y =𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸d,y,ULS
d,y,ULS∙ ∙(L/2) (L/2) /8 /8 == 0,760,76kN/m
kN/m∙ ∙(4
(4m/2)
m/2) /8
/8 == 0,38
0,38kNm
kNm
𝑀𝑀𝑀𝑀y,d 0,38 kNm
𝜎𝜎𝜎𝜎m,z,d =𝑀𝑀𝑀𝑀 y,d = 0,38
𝑀𝑀𝑀𝑀y,d 0,38kNm kNm = 4,7 N/mm222
M_(d,y)=E_(d,y,ULS)∙〖(L/2)〗^2/8
𝜎𝜎𝜎𝜎𝜎𝜎𝜎𝜎 m,z,d=
m,z,d = 𝑊𝑊𝑊𝑊z == 8,10 ∙ 10444 mm333==4,7 = 0,76
4,7 N/mm
N/mmkN/m∙〖(4 m/2)〗^2/8 = 0,38 kNm
𝑊𝑊𝑊𝑊
𝑊𝑊𝑊𝑊
σ_(m,z,d)=M_(y,d)/W_z zz 8,10
8,10∙ ∙10 mm
mm kNm)/(8,10〖∙10〗^4 mm^3 )=4,7 N/mm^2
10 =(0,38
𝑘𝑘𝑘𝑘mod 0,8 N
𝑓𝑓𝑓𝑓f_(m,0,flat,z,d)=k_mod/γ_M
m,0,flat,z,d =𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘mod mod ∙ 𝑓𝑓𝑓𝑓m,0,flat,z,k =0,8
∙f_(m,0,flat,z,k)=0,8/1,2∙48
0,8 ∙ 48 NN 2 = 32,0 N/mmN/mm^2
2 =32,0 N/mm^2
𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓m,0,flat,z,d
m,0,flat,z,d = = 𝛾𝛾𝛾𝛾 M ∙ ∙𝑓𝑓𝑓𝑓
𝑓𝑓𝑓𝑓
m,0,flat,z,k
m,0,flat,z,k== 1,2∙ ∙48
48 mm =
= 32,0
32,0N/mm
N/mm 22
𝛾𝛾𝛾𝛾𝛾𝛾𝛾𝛾MM 1,2
1,2 mm
mm22
The purlin is loaded from the compression side and supported against torsion at the main supports and in
the middle of the span. According to Table 6.1 of EN1995-1-1, for uniformly distributed load, the effective
length is Lef = 2000 mm+2∙240 mm = 2480 mm.
𝑀𝑀𝑀𝑀 z,crit 𝜋𝜋𝜋𝜋�𝐸𝐸𝐸𝐸0,05 𝐼𝐼𝐼𝐼𝑧𝑧𝑧𝑧 𝐺𝐺𝐺𝐺0,05 𝐼𝐼𝐼𝐼tor
𝜎𝜎𝜎𝜎σ_(m,y,crit)=M_(z,crit)/W_y
m,y,crit =𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀z,crit =𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋��𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸0,05 =(π√(E_0,05 I_z G_0,05 I_tor ))/(l_ef W_y (4.42)
) (4.42)
𝜎𝜎𝜎𝜎𝜎𝜎𝜎𝜎m,y,crit =
= 𝑊𝑊𝑊𝑊
z,crit
y = =
0,05 𝑙𝑙𝑙𝑙𝐼𝐼𝐼𝐼ef
𝑧𝑧𝑧𝑧𝐼𝐼𝐼𝐼𝑧𝑧𝑧𝑧𝐺𝐺𝐺𝐺
𝑊𝑊𝑊𝑊 𝐺𝐺𝐺𝐺0,05
0,05
y
𝐼𝐼𝐼𝐼tor
𝐼𝐼𝐼𝐼tor
(4.42)
(4.42)
m,y,crit
σ_(m,y,crit)=yy (π√(10600efefN/mm^2∙1,82∙〖10〗^6
𝑊𝑊𝑊𝑊
𝑊𝑊𝑊𝑊 𝑙𝑙𝑙𝑙 𝑙𝑙𝑙𝑙 𝑊𝑊𝑊𝑊 𝑊𝑊𝑊𝑊 yy 〖 mm〗^4∙400 N/mm^2∙6,5
mm∙〖4,32∙10〗^5 π�10600 mm^3 N/mm ) 2 ∙ 1,82 ∙ 106 mm4 ∙ 400 N/mm2 ∙ 6,56 ∙ 106 ∙ mm4
𝜎𝜎𝜎𝜎m,y,crit = π�10600
π�10600
σ_(m,y,crit)=21,6N/mm^2 ∙ ∙1,82 N/mmN/mm 22 1,82∙ ∙10 1066mm
mm44∙ ∙400 N/mm22∙ ∙6,56
400N/mm 1066∙ ∙mm
6,56∙ ∙10 mm44
m,y,crit=
𝜎𝜎𝜎𝜎𝜎𝜎𝜎𝜎m,y,crit = 2480 mm ∙ 4,32 ∙ 105 mm3
λ_rel=√((k_h∙f_(m,k))/σ_(m,y,crit) 2480 )=√((1,03∙44
2480mm
mm∙ ∙4,32 1055mm
10 mm33
4,32∙ ∙N/mm^2)/(21,6 N/mm^2 ))=1,45
𝜎𝜎𝜎𝜎m,y,crit = 21,6N/mm222
m,y,crit=
𝜎𝜎𝜎𝜎𝜎𝜎𝜎𝜎m,y,crit =21,6N/mm
21,6N/mm
𝜆𝜆𝜆𝜆(4.41)
𝑘𝑘𝑘𝑘h ∙𝑓𝑓𝑓𝑓m,k 1,03∙44 N/mm2
rel = �𝑘𝑘𝑘𝑘𝜎𝜎𝜎𝜎
𝑘𝑘𝑘𝑘hh∙𝑓𝑓𝑓𝑓∙𝑓𝑓𝑓𝑓m,k = �1,03∙44
1,03∙44N/mm
N/mm222 = 1,45 (4.41)
𝜆𝜆𝜆𝜆𝜆𝜆𝜆𝜆rel m,k 21,6 N/mm (4.41)
(4.41)
rel==��𝜎𝜎𝜎𝜎𝜎𝜎𝜎𝜎m,y,crit==�� 21,6 N/mm22
21,6N/mm
==1,45
1,45
m,y,crit
m,y,crit
1 1
192 LVL Handbook
whenEurope
1,4 < 𝜆𝜆𝜆𝜆rel,m , 𝑘𝑘𝑘𝑘crit = 11 2 ∙= 11 2 = 0,48 (4.40)
when
when1,41,4<<𝜆𝜆𝜆𝜆𝜆𝜆𝜆𝜆rel,m
rel,m, ,𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘crit
crit= ∙= 1,4522==0,48
= 𝜆𝜆𝜆𝜆rel,m22∙= 0,48 (4.40)
(4.40)
𝜆𝜆𝜆𝜆𝜆𝜆𝜆𝜆rel,m
rel,m 1,45
1,45
𝑘𝑘𝑘𝑘crit ∙ 𝑓𝑓𝑓𝑓m,y,d = 0,48 ∙ 30,3 N/mm222 = 14,4 N/mm222
𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘crit
crit∙ ∙ 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓
m,y,d=
m,y,d =0,48
0,48 ∙ ∙30,3
30,3N/mm
N/mm ==14,4
14,4N/mm
N/mm
2020_LVL_09.indd 192 12.3.2020 13:12:14
𝜎𝜎𝜎𝜎m,y,d ≤ 𝑘𝑘𝑘𝑘crit ∙ 𝑓𝑓𝑓𝑓m,y,d → OK (4.38)
𝑀𝑀𝑀𝑀𝑊𝑊𝑊𝑊
𝑊𝑊𝑊𝑊
z,crit 𝜋𝜋𝜋𝜋�𝐸𝐸𝐸𝐸0,05 𝑙𝑙𝑙𝑙ef𝐼𝐼𝐼𝐼𝑊𝑊𝑊𝑊
𝑧𝑧𝑧𝑧 𝐺𝐺𝐺𝐺y0,05
𝑧𝑧𝑧𝑧𝑊𝑊𝑊𝑊 𝐼𝐼𝐼𝐼tor
𝜎𝜎𝜎𝜎m,y,crit
m,y,crit = 𝑊𝑊𝑊𝑊
z,crit
yy = 𝑙𝑙𝑙𝑙ef
0,05 y0,05 tor (4.42)
𝑊𝑊𝑊𝑊 y
y 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙ef 𝑊𝑊𝑊𝑊y
ef 𝑊𝑊𝑊𝑊y
N/mm22∙∙1,82
π�10600N/mm
π�10600 1066mm
1,82∙∙10 mm44∙∙400 N/mm22∙∙6,56
400N/mm 1066∙∙mm
6,56∙∙10 mm44
𝜎𝜎𝜎𝜎𝜎𝜎𝜎𝜎m,y,crit = π�10600 N/mm22 ∙ 1,82 ∙ 1066 mm44 ∙ 40055N/mm
m,y,crit = 33
2
2 ∙ 6,56 ∙ 1066 ∙ mm44
𝜎𝜎𝜎𝜎m,y,crit 2480mm
2480 mm∙∙4,32
4,32∙∙10
10 mmmm
m,y,crit = 5
2480 mm ∙ 4,32 ∙ 105 mm3 3
𝜎𝜎𝜎𝜎𝜎𝜎𝜎𝜎m,y,crit 21,6N/mm222
=21,6N/mm
m,y,crit =
m,y,crit = 21,6N/mm
𝜎𝜎𝜎𝜎m,y,crit 2
Axial compression
𝑁𝑁𝑁𝑁c,d
𝑁𝑁𝑁𝑁
N_(c,d)/A=
c,d 4,5kN
4,5kN
4,5kN/(10800 mm^2 2)=0,42 N/mm^2
𝑁𝑁𝑁𝑁c,d == 4,5kN ==0,42 N/mm22
0,42N/mm
f_(c,0,d)=k_mod/γ_M
𝐴𝐴𝐴𝐴
c,d 10800
𝐴𝐴𝐴𝐴 = 10800 mm mm22 = ∙f_(c,0,SC2,k)=0,8/1,2∙29
0,42 N/mm 2 N/mm^2 =19,3 N/mm^2
𝐴𝐴𝐴𝐴 10800 mm22
𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘mod
mod 0,8
0,8 NN
𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓c,0,d = 𝑘𝑘𝑘𝑘mod
c,0,d = ∙∙𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓c,0,SC2,k = 0,8∙∙29
c,0,SC2,k = 29 N22 ==19,3 N/mm22
19,3N/mm
𝑓𝑓𝑓𝑓c,0,d = 𝛾𝛾𝛾𝛾mod
𝛾𝛾𝛾𝛾M M ∙ 𝑓𝑓𝑓𝑓 c,0,SC2,k = 1,2
1,2 ∙ 29 mm
mm = 19,3 N/mm22
c,0,d 𝛾𝛾𝛾𝛾M c,0,SC2,k 1,2 mm2 2
M
𝑘𝑘𝑘𝑘k_y=0,5(1+β_c
y = 0,5 1 +
(λ_(rel,y)-0,3)+λ_(rel,y)^2
e ,y 0,3 + 2e ,y )
= 0,5 ∙ (1 +
𝑘𝑘𝑘𝑘k_y=0,5∙(1+0,1∙(0,92-0,3)+(0,92)^2=0,95 0,1 ∙ (2,45 0,3) +2 (2,45)2 = 3,61
z 1 + 0,3 + 2
𝑘𝑘𝑘𝑘yz = 0,5 ∙ (1 + 0,1 ∙e(2,45 ,y 0,3) +e (2,45)
,y = 3,61
𝑘𝑘𝑘𝑘k_(c,z)=1/(k_z+√(k_z^2-λ_(rel,z)^2
y = 0,5 ∙ (1 + 0,1 ∙ (2,45
z (0,92 0,3) + 2
))=1/(3,61+√(〖3,61〗^2-〖2,4
(2,45)
(0,92) 2
= 3,61
0,95
𝑘𝑘𝑘𝑘yy = 0,5 ∙1(1++ 0,1 ∙e(0,92 0,3 + 2e (0,92)
2
𝑘𝑘𝑘𝑘y = 0,5 1 + e ,y
,y
0,3 0,3) + + 2 e
,y
,y
= 0,95
𝑘𝑘𝑘𝑘y,z==0,5 1 +
𝑘𝑘𝑘𝑘 =,y 0,3 + e ,y = 0,16
e 2 (4.32)
𝑘𝑘𝑘𝑘y,z==0,5 ∙ (1 + 0,1, ∙=(0,92 0,3) + (0,92)
𝑘𝑘𝑘𝑘 y = 0,5 ∙ (1 + 0,1 ∙ (0,923, 0,3)
3, +
2, 5 = 0,16 2 = 0,95 (4.32)
𝑘𝑘𝑘𝑘 3, 3, 2, (0,92)
5 = 0,95
𝑘𝑘𝑘𝑘 y = 0,5 ∙ (1 + 0,1, ∙ (0,92 0,3) + (0,92)2 = 0,95
𝑘𝑘𝑘𝑘 ,z = 1 = 1 0,16
= (4.32) (4.32)
𝑘𝑘𝑘𝑘k_(c,y)=1/(k_y+√(k_y^2-λ_(rel,y)^2
,z =
,y 1 , = 3, = 3, =
1 0,16
2
= 0,83
2, 5 ))=1/(0,95+√(〖0,95〗^2-〖0,
2 = 0,83
(4.32)
𝑘𝑘𝑘𝑘 ,z = 2 = 2 3, = 0,95 3, + 2, 50,95 = 0,16 0,92 (4.32)
,y 𝑘𝑘𝑘𝑘y + 𝑘𝑘𝑘𝑘y , e3,,y 2
𝑘𝑘𝑘𝑘y + 𝑘𝑘𝑘𝑘1y2 , 2e ,y 0,95 + 0,95
3, 2, 5
1 0,922
𝑘𝑘𝑘𝑘 ,y = 1 = 1 = 0,83
𝑘𝑘𝑘𝑘 ,y = 12 2 = 0,95 + 0,95 1 2 0,922 = 0,83
𝑘𝑘𝑘𝑘
𝑘𝑘𝑘𝑘 ,y = 𝑘𝑘𝑘𝑘y + 𝑘𝑘𝑘𝑘y2 + 𝑘𝑘𝑘𝑘 2e ,y = 0,95 + 0,95 2 0,92 = 0,83
2
y y 2 0,922
𝑘𝑘𝑘𝑘y2 , , 2e ,y , , 0,95 + 0,95
e ,y
, , 𝑘𝑘𝑘𝑘y +
+ 𝑘𝑘𝑘𝑘 m ∙ + 1 (4.29)
, ∙ , , , , + 𝑘𝑘𝑘𝑘 ,, ,, + ,, ,, (4.29)
The following ∙ m∙
expressions shall be
1
satisfied with km = 0,7 for rectangular cross sections:
, , , , , , ,
, , N , , , , N N
0,42
, , +N𝑘𝑘𝑘𝑘m 2 ∙∙ , , + 13,1 , , N1 2 4,7 N 2 (4.29)
0,42 + 𝑘𝑘𝑘𝑘m
, ∙ , , mm
σ_(c,0,d)/〖k_(c,z)∙f〗_(c,0,d) + , , +
, ,0,7 ∙ 13,1 mm 1 + 4,7 mm = 0,14 +
, , +k_m∙σ_(m,y,d)/f_(m,y,d)
0,30 + 0,15 =(4.29)
+σ_(m,z,d)/f_(m,z,d)
0,59 ≤1OK (4.29)
, ∙ , , , , mm , , ,,
2
+ 𝑘𝑘𝑘𝑘mN∙ + 0,7 , +∙ , mmN1 + 2 mm N = 0,14 + 0,30 + 0,15 =(4.29)
2
0,59 OK
(0,42
0,16 , ∙ N/mm^2
∙, 19,3
,
N N2 , , )/(0,16∙19,3 30,3
, , N/mm^2
N 2 32,0 mm
N )+0,7∙(13,1
NN 2 N/mm^2 )/(30,3 N/mm^2 )+(4,7 N/
0,16 0,42 ∙ 19,3Nmm
)=0,14+0,30+0,15=0,59→OK 2 2 30,3 mm
13,1 N 22 32,0 4,7 N 22
0,42 mm mm
N 2 , , + 0,7 ∙ , , mm 13,1 mm
mmN2+ 4,7 mm
mm N 2 = 0,14 + 0,30 + 0,15 = 0,59 OK
0,42 mm
σ_(c,0,d)/〖k_(c,y)∙f〗_(c,0,d) N1 2 + 4,7 mm
13,1 +σ_(m,y,d)/f_(m,y,d) +k_m∙σ_(m,z,d)/f_(m,z,d)
+ 0,30 + 0,15 = 0,59 ≤1 OK
0,16
, ,
∙ 19,3 mm+ 2N N
+ 𝑘𝑘𝑘𝑘0,7
+ m ∙∙ 30,3 mmN 32,0 mm
N2 = 0,14
N 2 = 0,14 + 0,30 + 0,15 =(4.30)
∙ , , , ,
+ 𝑘𝑘𝑘𝑘0,7 ∙ , , 2 + 0,59 OK
0,16 ,
∙, , +
19,3 mm N
, , 2+
m ∙ 30,3, ,
mm 1
N2 32,0 mm N2 (4.30)
, ,2
0,16 , ∙ , ,
∙ 19,3 mm 2 30,3, , mm
2 32,0 mm2
, , Nmm , , , N, mm mm N
,0,42 + N 2, , + 𝑘𝑘𝑘𝑘13,1 m∙ , , 2 1 4,7 N 2 (4.30)
, N
,
, ∙ , , +
0,42 mm , , + 𝑘𝑘𝑘𝑘m ∙
+ mm
13,1 ,, ,, 2 + 0,7 ∙ 4,7
, 1 mm = 0,03 + 0,43 + 0,10 =(4.30)
0,56 OK
, ∙, , , , + 2, ,,
mm N ++𝑘𝑘𝑘𝑘m ∙ mm 2 (4.30)
(4.30) ,
N +1 0,7 ∙ mm N = 0,03 + 0,43 + 0,10 = 0,56 OK
0,83 , ∙ ,∙ ,19,3 , ,
N N 2 30,3 N
, ,
N 2 32,0 N
N 2
(0,42
0,83 0,42 19,3Nmm
∙N/mm^2 2 30,3 mm
)/(0,83∙19,3
13,1 N N/mm^2
2 4,7 mm
)+(13,1
32,0 NN/mm^2
2 )/(30,3 N/mm^2 )+0,7∙(4,7 N
Nmm + 13,1 mm mm
2 2 2
0,42 mm mm N 2 + 0,7 ∙ 4,7 mm N 2 = 0,03 + 0,43 + 0,10 = 0,56 OK
)=0,03+0,43+0,10=0,56→OK
0,42, ,mm2N2 + 13,1
2 mm N 2 + 0,7 ∙ 4,7 mm N2 = 0,03 + 0,43 + 0,10 = 0,56 OK
0,83 ∙ 19,3
(σ_(m,y,d)/(k_crit mm N2 2++〖∙f〗_(m,o,edge) 30,3 mm
, ,
N 1
2 + 0,7 32,0 mm N 2 = 0,03 + 0,43 + 0,10
))^2+σ_(c,0,d)/(k_(c,z)∙f_(c,0,d)
∙ )≤1=(4.39)
0,56 OK
0,83∙ ∙ 19,3 , ,, , mm N 2+ , , ,mm 30,3∙ , , mm N 21 32,0 mmN2 (4.39)
0,83∙ ∙ 19,3 mm 30,3∙ 32,0 mm
, ,
mm 2 , , ,
mm 2 mm 2
2
, , N 2+ 2 , , 1 N (4.39)
13,1
∙ , , N 22 + 2, ∙ , , , , 0,42
, ,
1 N2 (4.39)
∙ , , , , mm2
13,1 +∙ , , , , 0,42 mm2 = 0,82 + 0,14 = 0,96 OK
+ ∙ , 1 mm N (4.39) LVL Handbook Europe 193
∙ , , mm N + = 0,82 + 0,14 = 0,96 OK
0,48 ∙ 30,3N N 2 2 0,16 ∙ 19,3N N 2 , , ,
N N
N 2 LVL STRUCTURES
N N N
0,42
9. CALCULATION 13,1
0,42 mm22 EXAMPLES
13,1 mmOF 4,7
4,7 mm22
+ 2 mm +
mm
N + + 0,7 ∙∙ mm = = 0,03 +
+ 0,43
0,43 +
+ 0,10
0,10 =
= 0,56
0,56 →
→ OK
OK
0,83 ∙ 19,3 N 2 30,3 N
N 2 0,7 32,0 N N 2 0,03
0,83 ∙ 19,3 mm2 30,3 mm2 32,0 mm2
mm mm mm
2
𝜎𝜎𝜎𝜎m,y,d 2 𝜎𝜎𝜎𝜎
��𝑘𝑘𝑘𝑘 𝜎𝜎𝜎𝜎∙𝑓𝑓𝑓𝑓m,y,d �2 + 𝜎𝜎𝜎𝜎c,0,d
c,0,d ≤1 (4.39)
(4.39)
(4.39)
crit m,o,edge � + 𝑘𝑘𝑘𝑘c,z ∙𝑓𝑓𝑓𝑓c,0,d ≤ 1
𝑘𝑘𝑘𝑘crit
m,y,d c,0,d
crit ∙𝑓𝑓𝑓𝑓m,o,edge
m,o,edge 𝑘𝑘𝑘𝑘c,z
c,z ∙𝑓𝑓𝑓𝑓c,0,d
c,0,d
((13,1 N/mm^2 )/(0,48∙30,3 N/mm^2 ))^2+(0,42 N/mm^2 )/(0,16∙19,3 N/mm^2 )
2
N 2 N
13,1 N 2
13,1 0,42 N2
0,42 mm
� mm
mm 2 � + 2 = 0,82 + 0,14 = 0,96 → OK
� N mm
0,48 ∙ 30,3 N 2 � + 0,16 ∙ 19,3 N N = 0,82 + 0,14 = 0,96 → OK
0,48 ∙ 30,3 mm2 0,16 ∙ 19,3 mm22
mm mm
Shear resistance
𝑉𝑉𝑉𝑉d,y =
𝑉𝑉𝑉𝑉 𝐸𝐸𝐸𝐸
=E_(d,ULS)∙s∙L/2
𝐸𝐸𝐸𝐸d,ULS ∙ 𝑠𝑠𝑠𝑠 ∙ 𝐿𝐿𝐿𝐿/2 = 2,92 kN/m ∙∙ 44 m/2 = 6,2 6,2 kN
V_(d,y) d,y = d,ULS ∙ 𝑠𝑠𝑠𝑠 ∙ 𝐿𝐿𝐿𝐿/2 = =2,92 2,92kN/m∙4 kN/m m/2 m/2 = 6,2=kN kN
τ_(v,d)=〖3∙V〗_(d,y)/(2∙A)=(3∙6,2
3 ∙∙ 𝑉𝑉𝑉𝑉
𝑉𝑉𝑉𝑉d,y 3 kN)/(2 ∙10 800〖 mm〗^2 )=0,9 N/mm^2
3 3 ∙∙ 6,2 6,2 kN
kN = 0,9 N/mm2
𝜏𝜏𝜏𝜏𝜏𝜏𝜏𝜏v,d 2 = 0,9 N/mm2 N/mm^2 =2,8 N/mm^2 229 (253)
=
f_(v,0,edge,d)=k_mod/γ_M
=
d,y =
= ∙f_(v,0,edge,k)=0,8/1,2∙4,2
v,d 2 ∙
2 ∙ 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 2
2 ∙→OK ∙ 10
10 800 mm 800 mm 2
τ_(m,d)≤f_(v,0,edge,d) 230 (255)
𝑘𝑘𝑘𝑘mod 0,8 N
𝑓𝑓𝑓𝑓
𝑓𝑓𝑓𝑓v,0,edge,d = 𝑘𝑘𝑘𝑘mod ∙ 𝑓𝑓𝑓𝑓v,0,edge,k = 0,8 ∙ 4,2 N = = 2,8
2,8 N/mm N/mm2
2
v,0,edge,d = 𝛾𝛾𝛾𝛾M ∙ 𝑓𝑓𝑓𝑓v,0,edge,k = 1,2 ∙ 4,2 mm2
𝛾𝛾𝛾𝛾M 1,2 mm2 229 (253)
𝐹𝐹c,90,d = 𝑉𝑉d,y = 6,2 kN
𝜏𝜏𝜏𝜏
𝐹𝐹c,90,d𝜏𝜏𝜏𝜏m,d ≤ 𝑓𝑓𝑓𝑓
= 𝑉𝑉 𝑓𝑓𝑓𝑓v,0,edge,d
= 6,2 kN → OK
m,d ≤d,y v,0,edge,d → OK
𝐹𝐹c,90,d 6,2 kN
𝜎𝜎c,90,d =
𝐹𝐹c,90,d = 6,2 kN = 1,2 N/mm2
𝐴𝐴 45 mm ∙ (15 mm + 45 mm)
𝜎𝜎Compression
𝐹𝐹c,90,d = = perpendicular = = 6,2 kN to grain
ef = 1,2 N/mm2
c,90,d
𝐴𝐴ef 𝑉𝑉d,y45 mm ∙ (15 mm + 45 mm)
𝐹𝐹𝐹𝐹c,90,d =
𝐹𝐹𝐹𝐹 = 𝑉𝑉𝑉𝑉 𝑉𝑉𝑉𝑉d,y = 6,2 kN 𝑘𝑘 0,8
𝑘𝑘c,90 c,90,d∙ 𝑓𝑓c,90,edge,d d,y = 6,2 kN ∙ mod ∙ 𝑓𝑓 ∙ 6,0 N/mm2 = 4 N/mm2
F_(c,90,d) c,90,d = 𝑘𝑘c,90
= 𝐹𝐹V_(d,y)=6,2 𝑘𝑘 kNmod 𝛾𝛾M
6,2 kN c,90,edge,k = 0,8 1,0 ∙
1,2 22
𝜎𝜎 =
𝑘𝑘σ_(c,90,d)=F_(c,90,d)/A_ef = = 1,2 N/mm N/mm2
c,90 ∙ c,90,d 𝑓𝑓c,90,edge,d 𝐹𝐹𝐹𝐹
𝐹𝐹𝐹𝐹𝐴𝐴 = 𝑘𝑘c,9045∙ mm
c,90,d =(6,2∙ ∙(15
𝑓𝑓c,90,edge,k
6,2
kN)/(45
6,2 mm kN
kN + = 1,0
mm∙(15
45 mm) ∙ mm+45 ∙ 6,0 N/mm mm))=1,2 = 4N/mm^2
𝜎𝜎𝜎𝜎c,90,d = c,90,d = ef 𝛾𝛾M = 1,2 N/mm22
1,2
𝜎𝜎𝜎𝜎𝜎𝜎c,90,d
c,90,d ≤ = 𝑘𝑘c,90
σ_(c,90,d)≤k_(c,90)∙f_(m,0,edge,d) 𝐴𝐴𝐴𝐴
𝐴𝐴𝐴𝐴ef ∙ 𝑓𝑓= 45
ef m,0,edge,d 45 mm mm→ ∙∙ (15
(15 OKmm →OK+
mm + 45
45 mm) mm)
= 1,2 N/mm
𝑘𝑘 0,8
𝜎𝜎c,90,d𝑘𝑘 ≤ 𝑘𝑘∙c,90 ∙ 𝑓𝑓m,0,edge,d = 𝑘𝑘→ OK mod 2 2
c,90 𝑓𝑓c,90,edge,d c,90 ∙ mod ∙ 𝑓𝑓c,90,edge,k = 1,0 ∙ 0,8 0,8 ∙ 6,0 N/mm 2 = 4 N/mm 2
SLS design 𝑘𝑘𝑘𝑘 c,90 ∙ 𝑓𝑓𝑓𝑓c,90,edge,d = 𝑘𝑘𝑘𝑘 c,90 ∙ 𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝛾𝛾mod mod M ∙ 𝑓𝑓𝑓𝑓c,90,edge,k = 1,0 ∙ 1,2 ∙ 6,0 N/mm22 = 4 N/mm 22
c,90 c,90,edge,d c,90 c,90,edge,k
SLS design c,90 c,90,edge,d c,90
𝛾𝛾𝛾𝛾𝛾𝛾𝛾𝛾M
M
M
c,90,edge,k 1,2
1,2
Instantaneous𝜎𝜎c,90,d ≤ 𝑘𝑘c,90 ∙ 𝑓𝑓m,0,edge,d → OK
deflection
Instantaneous deflection𝜎𝜎𝜎𝜎c,90,d
𝜎𝜎𝜎𝜎 c,90,d ≤
c,90,d
≤ 𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘c,90
c,90 ∙∙ 𝑓𝑓𝑓𝑓
c,90
𝑓𝑓𝑓𝑓m,0,edge,d
m,0,edge,d →
m,0,edge,d
→ OK OK
SLS design 𝑤𝑤inst = 𝑤𝑤inst,g + 𝑤𝑤inst,q
𝑤𝑤inst = 𝑤𝑤inst,g + 𝑤𝑤inst,q
SLS design
Instantaneous deflection
Instantaneous5deflection ∙ 𝑔𝑔d,z,SLS ∙ 𝐿𝐿4 6/5 ∙ 𝑔𝑔d,z,SLS ∙ 𝐿𝐿2
𝑤𝑤inst,g inst5 = ∙=𝑔𝑔d,z,SLS
𝑤𝑤inst,g∙ + 𝐿𝐿4 𝑤𝑤inst,q + ∙ 𝐿𝐿2 𝐴𝐴 = 1,62 mm + 0,13 mm = 1,75 mm
𝑤𝑤𝑤𝑤inst = 384
𝑤𝑤𝑤𝑤 ∙ 𝐸𝐸mean + 𝑤𝑤𝑤𝑤 ∙6/5𝐼𝐼 ∙ 𝑔𝑔d,z,SLS 8 ∙ 𝐺𝐺mean
𝑤𝑤inst,g𝑤𝑤𝑤𝑤 =
inst = 𝑤𝑤𝑤𝑤 inst,g
inst,g + +
𝑤𝑤𝑤𝑤 inst,q
inst,q = 1,62 mm + 0,13 mm = 1,75 mm
w_instinst =384
w_(inst,g) mean ∙+𝐼𝐼 w_(inst,q)
∙ 𝐸𝐸inst,g inst,q
8 ∙ 𝐺𝐺mean 𝐴𝐴
(4.74) 66
w_(inst,g)=(5〖∙g〗_(d,z,SLS)∙L^4)/(〖384∙E〗_mean∙I)+mm+0,13
5∙𝑔𝑔𝑔𝑔
5∙𝑔𝑔𝑔𝑔 d,z,SLS∙𝐿𝐿𝐿𝐿
44
∙𝐿𝐿𝐿𝐿
∙𝐿𝐿𝐿𝐿4 ∙ 𝐿𝐿54
6∙𝑔𝑔𝑔𝑔
∙𝑔𝑔𝑔𝑔 d,z,SLS∙𝐿𝐿𝐿𝐿
∙𝑔𝑔𝑔𝑔d,z,SLS
d,z,SLS ∙𝐿𝐿𝐿𝐿22
∙𝐿𝐿𝐿𝐿2 mm=1,75 mm
(4.74) 𝑤𝑤𝑤𝑤inst,ginst,g
5∙𝑔𝑔𝑔𝑔
= 5384∙𝐸𝐸𝐸𝐸
5 ∙∙ d,z,SLS
𝑔𝑔
d,z,SLS
𝑞𝑞 d,z,SLS +∙ 𝐿𝐿
5
5
4 6/5
6/5 ∙∙ 𝑔𝑔 1,62 ∙∙mm
𝑞𝑞=d,z,SLS
d,z,SLS 𝐿𝐿22 + 0,13 mm = 1,75 mm (4.74)
𝐿𝐿
(4.74)
𝑤𝑤 inst,g
inst,g = 384∙𝐸𝐸𝐸𝐸
384∙𝐸𝐸𝐸𝐸mean d,z,SLS
mean ∙𝐼𝐼𝐼𝐼
mean ∙𝐼𝐼𝐼𝐼
∙𝐼𝐼𝐼𝐼 +
8∙𝐺𝐺𝐺𝐺
8∙𝐺𝐺𝐺𝐺
8∙𝐺𝐺𝐺𝐺mean 𝐴𝐴𝐴𝐴
mean
mean 𝐴𝐴𝐴𝐴
𝐴𝐴𝐴𝐴 = 1,62 mm + 0,13 mm = = 8,45
1,75 mm
mm
𝑤𝑤inst,q 5 ∙=𝑞𝑞d,z,SLS
384 ∙∙ 𝐸𝐸 𝐿𝐿4 ∙∙6/5 𝐼𝐼4𝐼𝐼4 + ∙ 𝑞𝑞d,z,SLS 8 = 7,83 mm + 0,62 mm
w_(inst,q)=(5〖∙q〗_(d,z,SLS)∙L^4)/(〖384∙E〗_m
384
55 ∙∙ 𝑞𝑞𝑞𝑞
𝑞𝑞𝑞𝑞 𝐸𝐸∙ mean
mean ∙∙ 𝐿𝐿𝐿𝐿
𝐿𝐿𝐿𝐿 4 6/5
6/5 8 ∙∙∙∙ 𝑞𝑞𝑞𝑞𝐺𝐺
𝐺𝐺
𝑞𝑞𝑞𝑞
∙ 𝐿𝐿2 𝐴𝐴
mean
mean 𝐴𝐴∙∙ 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿222
𝑤𝑤inst,q = d,z,SLS+
d,z,SLS
d,z,SLS d,z,SLS= 7,83 mm + 0,62 mm = 8,45 mm
d,z,SLS
d,z,SLS
𝑤𝑤𝑤𝑤inst,q
(4.74) 384 = 𝐸𝐸mm+8,45 + = 7,83 mm + 0,62 mm = 8,45 mm
w_inst
𝑤𝑤inst inst,q
inst,q = ∙1,75
=1,75 384
384
mean
∙∙mm
∙ 𝐼𝐼 + mm=10,2
𝐸𝐸𝐸𝐸mean
𝐸𝐸𝐸𝐸 mean
mean
8,45 8 ∙ 𝐺𝐺mean
∙∙ 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 mm88=∙∙ 𝐺𝐺𝐺𝐺
mm 𝐴𝐴
𝐺𝐺𝐺𝐺10,2
meanmm
mean
mean
𝐴𝐴𝐴𝐴
𝐴𝐴𝐴𝐴
𝑤𝑤inst = 1,75 5mm + 8,45∙ 𝐿𝐿mm = 4 10,2𝑞𝑞d,z,SLS mm ∙ 𝐿𝐿 2
𝑤𝑤𝑤𝑤inst,q = 1,75 1,75∙ 𝑞𝑞d,z,SLSmm + + 8,45 8,45 6/5
mm ∙= = 10,2 10,2 mm mm = 7,83 mm + 0,62 mm = 8,45 mm
Final deflection 𝑤𝑤𝑤𝑤𝑤𝑤 inst =
inst
inst = mm + mm
384 ∙ 𝐸𝐸mean ∙ 𝐼𝐼 8 ∙ 𝐺𝐺mean 𝐴𝐴
Final deflectionFinal deflection
𝑤𝑤net,fin
w_(net,fin) inst == =1,75 (1 + mm
(1+k_def)∙w_(inst,g)𝑘𝑘def +)8,45 ∙ 𝑤𝑤inst,g mm + =+(1 10,2 + 𝜓𝜓mm 2 ∙ 𝑘𝑘def ) ∙ 𝑤𝑤inst,q
(1+ψ_2∙k_def)∙w_(inst,q (4.73)
𝑤𝑤net,fin = (1 + 𝑘𝑘def ) ∙ 𝑤𝑤inst,g + (1 + 𝜓𝜓2 ∙ 𝑘𝑘def ) ∙ 𝑤𝑤inst,q (4.73)
(4.73)
Final deflection 𝑤𝑤𝑤𝑤net,fin
𝑤𝑤𝑤𝑤 net,fin = = (1 (1 + + 𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘def
def)) ∙∙ 𝑤𝑤𝑤𝑤 𝑤𝑤𝑤𝑤inst,g
inst,g + + (1 (1 + + 𝜓𝜓𝜓𝜓
𝜓𝜓𝜓𝜓222 ∙∙ 𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘def
def)) ∙∙ 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤inst,q
inst,q (4.73)
(4.73)
Note: For the
net,fin snow def loadinst,g in Finnish national annex:
def ψ2 = 0,2
inst,q
Note: For the snow load in Finnish national annex: ψ2 = 0,2
Note:𝑤𝑤For Forthethe= snow
snow (1 + load
load 𝑘𝑘defin
0,6) in Finnish
)∙ Finnish
∙1,75 𝑤𝑤inst,g mm national
national
+ +(1(1++ 𝜓𝜓annex:
annex:
0,2
2 ∙ 𝑘𝑘∙def ψ2)ψ=∙ 222∙𝑤𝑤0,2
0,6) =inst,q
8,450,2mm = 12,9 mm (4.73)
net,fin
𝑤𝑤net,fin = (1 + 0,6) ∙ 1,75 mm + (1 + 0,2 ∙ 0,6) ∙ 8,45 mm = 12,9 mm
𝑤𝑤𝑤𝑤net,fin
Note: For = the (1 + snow 0,6) ∙ 1,75Linmm + (1national
+ 0,2 ∙ 0,6) ∙ 8,45 ψ2mm = 12,9 mm
𝑤𝑤net,finload mm +Finnish annex: = 0,2
net,fin 4000
w_(net,fin)
Requirement:net,fin = (1+0,6)∙1,75 ≤ ,(1+0,2∙0,6)∙8,45 = 16 mm mm → OK = 12,9 mm
L 250
4000 mm 250
Requirement: 𝑤𝑤
Requirement: w_(net,fin)≤L/250=4000/250=16
net,fin ≤ 250 , LL L 250 4000 4000 = 16 mmmm→OK → OK
Requirement:
𝑤𝑤 net,fin = (1 + 𝑤𝑤𝑤𝑤0,6)
net,fin
net,fin
net,fin ∙ 1,75≤ 250 mm = 4000 + (1 =+160,2 mm → OK
∙ 0,6) ∙ 8,45 mm = 12,9 mm
9.5 Wall stud 250
250 250
250 250 250
250
Stud properties:
Bending strength edgewise fm,0,edge,k = 27 N/mm2
Shear strength edgewise fv,0,edge,k = 4 N/mm2
Compression parallel to grain fc,0,k = 26 N/mm2
Modulus of elasticity E0,k = 9 600 N/mm2
Modulus of elasticity E0,mean = 8 000 N/mm2
Modulus of rigidity G0,edge,k = 600 N/mm2
Modulus of rigidity G0,edge,mean = 400 N/mm2
Area of cross section A = b∙h = 5400 mm2
Section modulus Wy = b∙h2/6 = 1,08∙105mm3
Moment of inertia Iy = b∙h3/12 = 6,48∙106 mm4
Moment stiffness of the joist EIy = 13800 N/mm2 ∙ 1,08∙105 mm4 = 8,94∙1010 Nmm2
Shear rigidity of the joist GA = 600 N/mm2 ∙ 5400 mm2 = 3,24∙106 N
Modification factor kmod for medium-term, SC1 = 0,8
Modification factor kdef for SC1 = 0,6
Material safety factor γM (default value in EC5) =1,2
Size effect factor kh = (300/120)0,15 = 1,15
Loading combinations
The most critical ultimate limit state (ULS) load combination:
Ed,ULS = γG∙gk+ γQ∙qk = 1,15∙5,0 kN+1,5∙11 kN = 22,3 kN
Note: Safety factors γG and γQ are according to Finnish National annex of eurocode 0.
ULS design
Axial compression
𝑁𝑁𝑁𝑁c,d 𝐸𝐸𝐸𝐸
N_(c,d)/A=E_(d,ULS)/A=
𝑁𝑁𝑁𝑁 𝐸𝐸𝐸𝐸 𝑁𝑁𝑁𝑁c,d = 𝐸𝐸𝐸𝐸d,ULS 22,3 (22,3
22,3 kN kN)/(5400
kN mm^2 )=4,1 N/mm^2
kN 222,3 kNN/mm
d,ULS 2
𝑁𝑁𝑁𝑁c,d
c,d == 𝐸𝐸𝐸𝐸d,ULS
d,ULS == 22,3= = 4,1
= 4,1 N/mm = 4,12
2 N/mm2
𝑁𝑁𝑁𝑁𝐴𝐴𝐴𝐴
𝐴𝐴𝐴𝐴
c,d =
f_(c,0,d)=k_mod/γ_M 𝐸𝐸𝐸𝐸 d,ULS𝐴𝐴𝐴𝐴
𝐴𝐴𝐴𝐴
𝐴𝐴𝐴𝐴 = 5400
5400 22,3
𝐴𝐴𝐴𝐴 mm
kN
∙f_(c,0,SC1,k)=0,8/1,2∙26
mm 2
5400
2
= 4,1
mm N/mm
2 N/mm^2 =17,3 N/mm^2
𝐴𝐴𝐴𝐴 = 𝐴𝐴𝐴𝐴 = 5400 mm = 4,1 N/mm 2
𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 5400 mm 2
𝑘𝑘𝑘𝑘 0,8 N
= 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘mod = 0,8 26 0,8 N N N/mm22
mod
𝑓𝑓𝑓𝑓c,0,d =
𝑓𝑓𝑓𝑓 𝑘𝑘𝑘𝑘𝑓𝑓𝑓𝑓mod
mod ∙∙= 𝑓𝑓𝑓𝑓c,0,SC1,k
𝑓𝑓𝑓𝑓 ∙ 𝑓𝑓𝑓𝑓
= 0,8 ∙∙ 26 N 2∙ 26 = 17,3
17,3 2 N/mm2
c,0,d = 𝑘𝑘𝑘𝑘𝛾𝛾𝛾𝛾
𝑓𝑓𝑓𝑓c,0,d 𝛾𝛾𝛾𝛾mod
M ∙ 𝑓𝑓𝑓𝑓c,0,SC1,k
c,0,d
𝛾𝛾𝛾𝛾
c,0,SC1,k 1,2 ∙ 26=mm
=c,0,SC1,k
0,8
1,2 mm N
1,2 2
2
=
= 17,3
mm
N/mm
= 17,3
2N/mm
𝛾𝛾𝛾𝛾 M
𝑓𝑓𝑓𝑓c,0,d = M ∙ 𝑓𝑓𝑓𝑓c,0,SC1,k = M 1,2 ∙ 26 mm = 17,3 N/mm2
𝛾𝛾𝛾𝛾M 1,2 mm2
Buckling, buckling length lc = 2700 mm in z-direction
λ_y=√12 (l_c/h)=3,46∙(2700 mm)/(120 mm)=78
𝑙𝑙𝑙𝑙 2700 mm
𝜆𝜆𝜆𝜆
(4.37)
𝜆𝜆𝜆𝜆 y = √12√12 �𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙ccc �� √12 = 3,46 3,46 � ℎc �∙∙∙ 2700
𝑙𝑙𝑙𝑙 2700 mm =270078mm = 78 (4.37) (4.37)
(4.37)
𝜆𝜆𝜆𝜆yy = = √12𝜆𝜆𝜆𝜆y��𝑙𝑙𝑙𝑙ℎℎℎ= = = 3,46
mm∙= 78
� = 3,46 120
120
2700 mm =120
mm
mm 78mm (4.37)
λ_(rel,y)=λ_y/π
𝜆𝜆𝜆𝜆y = √12 � ℎ√(f_(c,0,k)/E_0,05 c
� = 3,46 ∙ 120 mm )=78/3,14∙√((26
= 78 N/mm^2)/(8000 N/mm^2 ))=1,41
(4.37)
120 mm
(4.35)
𝜆𝜆𝜆𝜆rel,y =
𝜆𝜆𝜆𝜆y 𝑓𝑓𝑓𝑓c,0,k
𝜆𝜆𝜆𝜆
= 𝜆𝜆𝜆𝜆𝜋𝜋𝜋𝜋𝜆𝜆𝜆𝜆y � 𝜆𝜆𝜆𝜆
= �78c,0,k
78 26 N/mm22 2
(4.35) (4.35)
∙∙∙ �
�= 2 = 1,41
y 𝑓𝑓𝑓𝑓
�𝑓𝑓𝑓𝑓𝐸𝐸𝐸𝐸c,0,k y 78𝑓𝑓𝑓𝑓 26 78
N/mm 26 N/mm
𝜆𝜆𝜆𝜆
k_y=0,5(1+β_c
𝜆𝜆𝜆𝜆rel,y ==
c,0,k
�(λ_(rel,y)-0,3)+λ_(rel,y)^2
26 N/mm∙ �8000 2 = 1,41 = 1,41
)=0,5∙(1+0,1∙(1,412-0,3)+(1,41)^2=1,56(4.35)
(4.35)
rel,y = 𝜆𝜆𝜆𝜆𝜋𝜋𝜋𝜋 = 3,14 �8000 2 = 1,41 2
rel,y 0,05 8000 N/mm 22
𝜋𝜋𝜋𝜋y 𝑓𝑓𝑓𝑓 𝐸𝐸𝐸𝐸0,05 𝜋𝜋𝜋𝜋 3,14
𝐸𝐸𝐸𝐸c,0,k 78
3,14 𝐸𝐸𝐸𝐸0,05 3,14
26
8000 N/mm
N/mm
N/mm N/mm
𝜆𝜆𝜆𝜆 =
k_(c,y)=1/(k_y+√(k_y^2-λ_(rel,y)^2
rel,y � 0,05 = ∙ � = 1,41
))=1/(1,56+√(〖1,56〗^2-〖1,41〗^2
2 (4.35)
))=0,45
𝜋𝜋𝜋𝜋 𝐸𝐸𝐸𝐸0,05 3,14 8000 N/mm
2 2
𝑘𝑘𝑘𝑘
𝑘𝑘𝑘𝑘 = 0,5�1
y (4.33) + 𝛽𝛽𝛽𝛽 �𝜆𝜆𝜆𝜆rel,y − 0,3� + 𝜆𝜆𝜆𝜆2 =
0,3��� + 0,52 ∙ (1 + 0,1 ∙ (1,412
∙∙ (1,412 − 0,3)
∙ (1,412+ (1,41)
(1,41) 2 = 1,562 = 1,56
(1,41)
𝑘𝑘𝑘𝑘yy = 𝑘𝑘𝑘𝑘y +
= 0,5�1 = 𝛽𝛽𝛽𝛽
𝛽𝛽𝛽𝛽0,5�1 + 𝛽𝛽𝛽𝛽 − �𝜆𝜆𝜆𝜆
0,3� +
+−𝜆𝜆𝜆𝜆 = 𝜆𝜆𝜆𝜆0,5
rel,y∙∙ �
(1
(1=+
+0,50,1
0,1∙ (1 + 0,1 − 0,3)
0,3) +
+− 0,3) +
2 = 1,56
c �𝜆𝜆𝜆𝜆 rel,y
0,5�1 + c �𝜆𝜆𝜆𝜆rel,y
c rel,y −c 0,3� rel,y 𝜆𝜆𝜆𝜆22rel,y
rel,y � = 0,5 (1,412 − (1,41) 2
= 1,56
𝑘𝑘𝑘𝑘y = 0,5�1 + 𝛽𝛽𝛽𝛽c �𝜆𝜆𝜆𝜆rel,y1 − 0,3� + 𝜆𝜆𝜆𝜆rel,y �1= 0,5 ∙ (1 + 0,1 ∙ (1,412 − 0,3) + (1,41) = 1,56
𝑘𝑘𝑘𝑘c,y =
(4.31)
𝑘𝑘𝑘𝑘 = 𝑘𝑘𝑘𝑘 11= =1
=
1
=1 22 −1,4122 =
1= 0,45 (4.31) (4.31)
(4.31)
c,y = 𝑘𝑘𝑘𝑘
𝑘𝑘𝑘𝑘c,y 𝑘𝑘𝑘𝑘y +�𝑘𝑘𝑘𝑘 c,y 2 −𝜆𝜆𝜆𝜆2
1
y 2
2 −𝜆𝜆𝜆𝜆𝑘𝑘𝑘𝑘 =21,56+�1,56
1,56+�1,56
2 1 2 −1,41
1,56+�1,56 2
=20,45
0,45
−1,412
= 0,45 (4.31)
+�𝑘𝑘𝑘𝑘
𝑘𝑘𝑘𝑘c,y = 𝑘𝑘𝑘𝑘yy +�𝑘𝑘𝑘𝑘yy2 −𝜆𝜆𝜆𝜆rel,y
σ_(c,0,d)/〖k_(c,z)∙f〗_(c,0,d)
rel,y
2y
rel,y
+�𝑘𝑘𝑘𝑘 = −𝜆𝜆𝜆𝜆
y 1,56+�1,56
=(4,1
rel,y −1,41 = 0,45
N/mm^2
2 2 )/(0,45∙17,3 N/mm^2 (4.31)
)=0,52≤1,0→OK
𝑘𝑘𝑘𝑘y +�𝑘𝑘𝑘𝑘y2 −𝜆𝜆𝜆𝜆2rel,y 1,56+�1,56 −1,41
N
𝜎𝜎𝜎𝜎c,0,d 4,1 N
4,1 N 2 4,1 N
𝜎𝜎𝜎𝜎
𝜎𝜎𝜎𝜎c,0,d =𝜎𝜎𝜎𝜎c,0,d 4,1 mm
mm N 22 mm 2
= 0,52
0,52 ≤=1,0
1,0 →≤OK1,0 → OK
𝑓𝑓𝑓𝑓c,0,d𝑘𝑘𝑘𝑘 =
c,0,d
𝑘𝑘𝑘𝑘c,z𝜎𝜎𝜎𝜎c,0,d
∙ 𝑓𝑓𝑓𝑓 = ∙0,45 4,1=mm2 N N = = 0,52 ≤ 0,52
N ≤ 1,0 → OK
→ OK
𝑘𝑘𝑘𝑘
𝑘𝑘𝑘𝑘c,z ∙
∙ 𝑓𝑓𝑓𝑓 𝑓𝑓𝑓𝑓 ∙ mm
17,3 N
c,0,d = 0,45 ∙∙ 17,3
c,0,d c,z c,0,d 0,45 mm ∙ 17,3
2 = 0,522≤ 1,0 → OK
𝑘𝑘𝑘𝑘c,z ∙ 𝑓𝑓𝑓𝑓c,0,d 0,45 17,3
c,z
mm N 22 mm
0,45 ∙ 17,3 mm2
mm
Bending moment resistance
𝑀𝑀𝑀𝑀
𝑀𝑀𝑀𝑀dd = = 𝐸𝐸𝐸𝐸 𝐸𝐸𝐸𝐸d,ULS
𝑀𝑀𝑀𝑀d = ∙∙∙ 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝐸𝐸𝐸𝐸
z = = 22,3 22,3 ∙ 𝑒𝑒𝑒𝑒zkN kN= ∙∙∙22,3
0,03kN
0,03 m ∙= = 0,67 kNm
𝑀𝑀𝑀𝑀 d = 𝐸𝐸𝐸𝐸d,ULS d,ULS z = 22,3
z d,ULS kN 0,03 m m = 0,03
0,67
0,67m kNm
= 0,67 kNm
kNm
𝑀𝑀𝑀𝑀d = 𝐸𝐸𝐸𝐸d,ULS ∙ 𝑒𝑒𝑒𝑒z = 22,3 kN ∙ 0,03 m = 0,67 kNm
M_d=E_(d,ULS)∙e_z= 𝑀𝑀𝑀𝑀d 0,67
𝑀𝑀𝑀𝑀d22,3 kNm kN∙0,03 m
𝜎𝜎𝜎𝜎m,d =
𝜎𝜎𝜎𝜎 = 𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀𝜎𝜎𝜎𝜎dd == = 0,67 0,67 =kNm
kNm
kNm5 mm3 =
0,67=kNm 6,2=N/mm
6,2
0,67 kNm
N/mm
= 6,2
2
2
2 N/mm2
𝜎𝜎𝜎𝜎m,d
m,d =
σ_(m,d)=M_d/W=(0,67 𝑊𝑊𝑊𝑊
𝑀𝑀𝑀𝑀
𝑊𝑊𝑊𝑊 d =
m,d 1,08
1,08 0,67
𝑊𝑊𝑊𝑊 ∙
∙ 10
10kNm)/(1,08〖∙10〗^5
5
1,08
5 mm 3
∙
3 10= 56,2
mm N/mm
3 mm^3 )=6,2 N/mm^2
𝜎𝜎𝜎𝜎m,d = 𝑊𝑊𝑊𝑊 = 1,08 ∙ 105 mm3 = 6,2 N/mm2
f_(m,0,edge,d)=k_mod/γ_M 𝑊𝑊𝑊𝑊 𝑘𝑘𝑘𝑘1,08 ∙ 10 mm ∙k_h∙f_(m,0,edge,k)=0,8/1,2∙1,15∙27
0,8 N N/mm^2 =20,7 N/mm^2
σ_(m,d)≤f_(m,0,edge,d)
𝑓𝑓𝑓𝑓 = 𝑘𝑘𝑘𝑘
𝑘𝑘𝑘𝑘
mod
mod ∙ →OK
𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘∙mod
𝑓𝑓𝑓𝑓 = 0,8
0,8 ∙ 1,150,8
∙ 27 N
N = N N/mm22
20,7
𝑓𝑓𝑓𝑓
m,0,edge,d 𝑓𝑓𝑓𝑓 = mod ∙=
𝛾𝛾𝛾𝛾mod 𝑘𝑘𝑘𝑘h ∙∙ 𝑓𝑓𝑓𝑓 ∙ 𝑘𝑘𝑘𝑘h ∙ 𝑓𝑓𝑓𝑓m,0,edge,k
m,0,edge,k = 1,2 ∙∙ 1,15
= ∙∙ 27
∙ 1,15
mm 2∙ 27
= 20,7 N/mm 2 N/mm2
= 20,7
𝑓𝑓𝑓𝑓m,0,edge,d
m,0,edge,dm,0,edge,d = 𝑘𝑘𝑘𝑘𝛾𝛾𝛾𝛾 M ∙ 𝑘𝑘𝑘𝑘h h 𝑓𝑓𝑓𝑓
𝛾𝛾𝛾𝛾 m,0,edge,k
m,0,edge,k = 0,8
1,2 1,151,227 mmN 2
2
= 20,7
mm 2N/mm
𝑓𝑓𝑓𝑓m,0,edge,d = 𝛾𝛾𝛾𝛾M ∙ 𝑘𝑘𝑘𝑘h ∙ 𝑓𝑓𝑓𝑓m,0,edge,k = 1,2 ∙ 1,15 ∙ 27 mm2 = 20,7 N/mm2
M M
𝛾𝛾𝛾𝛾M 1,2 mm
𝜎𝜎𝜎𝜎m,d ≤
𝜎𝜎𝜎𝜎 ≤ 𝑓𝑓𝑓𝑓 𝑓𝑓𝑓𝑓m,0,edge,d
𝜎𝜎𝜎𝜎m,d ≤ 𝑓𝑓𝑓𝑓m,0,edge,d → OK → OK
→
𝜎𝜎𝜎𝜎m,d ≤ 𝑓𝑓𝑓𝑓m,0,edge,d
m,d m,0,edge,d → OK OK
𝜎𝜎𝜎𝜎m,d ≤ 𝑓𝑓𝑓𝑓m,0,edge,d → OK
The following
𝜎𝜎𝜎𝜎c,0,d
expressions 𝜎𝜎𝜎𝜎m,y,d shall be satisfied with km = 0,7 for rectangular cross-sections:
𝜎𝜎𝜎𝜎 +
σ_(c,0,d)/〖k_(c,z)∙f〗_(c,0,d)
𝜎𝜎𝜎𝜎c,0,d 𝑘𝑘𝑘𝑘
𝜎𝜎𝜎𝜎 c,0,d ∙ 𝜎𝜎𝜎𝜎 m,y,d ≤ 1
∙∙ 𝜎𝜎𝜎𝜎𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓m,y,d
𝜎𝜎𝜎𝜎m,y,d
+k_m∙σ_(m,y,d)/f_(m,y,d) ≤1 (4.30) (4.30)
(4.30)
𝑘𝑘𝑘𝑘c,z ∙𝑓𝑓𝑓𝑓c,0,d +
c,0,d
+ 𝑘𝑘𝑘𝑘
𝑘𝑘𝑘𝑘
m
m + 𝑘𝑘𝑘𝑘m
m,y,d ≤
≤ ∙11 ≤1 (4.30)
𝑘𝑘𝑘𝑘
𝑘𝑘𝑘𝑘c,z ∙𝑓𝑓𝑓𝑓c,0,d + 𝑘𝑘𝑘𝑘 ∙ 𝑓𝑓𝑓𝑓m,y,d ≤ 1𝑓𝑓𝑓𝑓m,y,d
𝜎𝜎𝜎𝜎
c,z ∙𝑓𝑓𝑓𝑓c,0,d 𝑘𝑘𝑘𝑘 c,z ∙𝑓𝑓𝑓𝑓
m c,0,d𝜎𝜎𝜎𝜎 m,y,d
(4.30) c,0,d
𝑘𝑘𝑘𝑘c,z ∙𝑓𝑓𝑓𝑓c,0,d m 𝑓𝑓𝑓𝑓 (4.30)
(4,1 N/mm^2 N)/(0,45∙17,3 m,y,d
N )+0,7∙(6,2
4,1 N
4,1 mmN 2 4,1 N N/mm^2 6,2 N
6,2 mmN 26,2 N N/mm^2 )/(20,7 N
4,1 mm N2 2 mm
+ 0,72 6,2
0,7 ∙ + 0,7 mmN∙ 22 = mm2+
4,1 mm2 N N + 6,2 mm
+ 0,7N∙∙ 20,7 N2 = = 0,52
0,52
0,52 +=0,7
0,7 ∙∙ 0,30
0,52 0,30 =
= 0,73
+ 0,7 ∙ 0,30→
0,73 →=OK
OK
0,73 → OK
0,45 ∙ mm
17,3 N mmN
N N + 0,7 ∙ 0,30 = 0,73 → OK
0,45
0,45 ∙ 17,3 mm ∙ 17,3
0,45 mm ∙ 17,3
2
2 + 0,7 ∙ 20,7
20,7
2 mm 20,7
2
2 = 0,52 +
2 0,7 ∙ 0,30 = 0,73 → OK
N2 mm mm
N2 mm
0,45 ∙ 17,3 mm2 20,7 mm2
mm mm
𝐹𝐹𝐹𝐹
𝐹𝐹𝐹𝐹 =𝐹𝐹𝐹𝐹𝑁𝑁𝑁𝑁
= 𝑁𝑁𝑁𝑁d == 22,3 kN
𝑁𝑁𝑁𝑁dd ==22,3
𝑁𝑁𝑁𝑁d kN
= 22,3 kN
c,90,d
𝐹𝐹𝐹𝐹c,90,d
c,90,d = c,90,d 22,3 kN
𝐹𝐹𝐹𝐹c,90,d = 𝑁𝑁𝑁𝑁d = 22,3 kN
𝜎𝜎𝜎𝜎c,0,d 𝜎𝜎𝜎𝜎m,y,d
+ 𝑘𝑘𝑘𝑘m ∙ ≤1 (4.30)
𝑘𝑘𝑘𝑘c,z ∙𝑓𝑓𝑓𝑓c,0,d 𝑓𝑓𝑓𝑓m,y,d
N N
4,1 6,2
9. CALCULATION EXAMPLES OF LVL STRUCTURES
mm2
+ 0,7 ∙ mm 2
= 0,52 + 0,7 ∙ 0,30 = 0,73 → OK
N N
0,45 ∙ 17,3 20,7
mm2 mm2
The wall studs is installed on a 120 mm wide LVL 48P sole plate, which has compression perpendicular to the
grain flatwise strength fc,90,flat,k = 2,2 N/mm2. The resistance is
𝐹𝐹𝐹𝐹c,90,d = 𝑁𝑁𝑁𝑁d = 22,3 kN
F_(c,90,d) = N_d 𝐹𝐹𝐹𝐹c,90,d
= 22,3 kN 22,3 kN 22,3 kN
σ_(c,90,d)=F_(c,90,d)/A_ef
𝜎𝜎𝜎𝜎c,90,d = = =(22,3 kN)/(b∙(l_support+2∙30 = m
𝐴𝐴𝐴𝐴ef 𝑏𝑏𝑏𝑏 ∙ (𝑙𝑙𝑙𝑙support + 2 ∙ 30 mm) 120 mm ∙ (45 mm + 2 ∙ 30 mm)
σ_(c,90,d)=1,77 N/mm^2
k_(c,90)∙f_(c,90,flat,d)=k_(c,90)∙k_mod/γ_M
𝜎𝜎𝜎𝜎c,90,d = 1,77 N/mm2 ∙f_(c,90,flat,k)
k_(c,90)∙f_(c,90,flat,d)=1,4∙0,8/1,2∙2,2 N/mm^2 =2,05 N/mm^2
𝑘𝑘𝑘𝑘mod
𝑘𝑘𝑘𝑘c,90 ∙ 𝑓𝑓𝑓𝑓c,90,flat,d = 𝑘𝑘𝑘𝑘c,90 ∙ ∙ 𝑓𝑓𝑓𝑓c,90,flat,k
𝛾𝛾𝛾𝛾M
0,8 N
(k 𝑘𝑘𝑘𝑘c,90 ∙ 𝑓𝑓𝑓𝑓c,90,flat,d = 1,4 ∙ ∙ 2,2 = 2,05 N/mm2 (kc,90 from Table 4.7) c,90 from Table 4.7)
1,2 mm2
σ_(c,90,d)≤k_(c,90)∙f_(m,0,flat,d) →OK
𝜎𝜎𝜎𝜎c,90,d ≤ 𝑘𝑘𝑘𝑘c,90 ∙ 𝑓𝑓𝑓𝑓m,0,flat,d → OK
Loading combination
The most critical ultimate limit state (ULS) load combination for the connection:
Note: Safety factor γG for the case of only permanent loads according to Finnish National annex of
Eurocode 0.
Screw properties
Size 5,0 x 60 mm screws, 2 screws / connection
Threaded length lg = 50 mm
Head diameter dh = 10 mm (Head pull-through capacity is not governing in steel-to-timber
connection)
Tensile strength ttens,k = 7 kN determined in accordance with EN 14592. ttens,k > Ed,ULS → OK
Withdrawal strength fax,k = 10 N/mm2 at the edge face of LVL 48 P, determined in accordance
with EN 14592
Geometry conditions:
Minimum distance to edge aCG,2 in LVL edge ≥ 4d = 4∙5,0 mm = 20 mm. Beam thickness 45 mm/2 =
22,5 mm → Screw size 5,0x50 mm is OK for the beam.
Min. screw spacing a1 ≥ 10d = 50 mm
Min. distance from the beam end a1,CG ≥ 12d = 60 mm
Min. pointside penetration length of the threaded part lg ≥ 6d = 30mm, OK
The axial loaded screw connection can be made with two 5,0 x 50 mm size screws/connection positioned at
the middle of the edge face of the beam with 50 mm screw spacing. The screws shall not be closer than 60 mm
from the beam end.
Loading combinations
The most critical ultimate limit state (ULS) load combination for each connection between the beam and
stud:
Ed,ULS = s ∙ (γG ∙ gk + γQ ∙ qk)
Ed,ULS = 0,6m ∙ (1,15 ∙ 0,3 kN/m + 1,5 ∙ 3,5 kN/m) = 3,4 kN
Note: Safety factors γG and γQ are according to Finnish National annex of Eurocode 0.
Screw properties
Size 6,0x140mm full threaded screw
Threaded length lg = 123 mm
Unthreaded length lu = 17 mm
Head diameter dh = 12 mm
Tensile strength ttens,k = 10 kN, determined in accordance with EN 14592.
Head pull-through strength fhead,k = 13 N/mm2, when ρa = 350 kg/m3
Modification factor kmod for medium-term load, SC2 = 0,8
Material safety factor γm for connections (default value in EC5) =1,3
Geometry conditions:
Minimum distance to edge a2,CG in stud ≥ 4d = 4∙6,0 mm = 24 mm. Stud thickness 51 mm/2 = 25,5 mm
→ Screw size 6,0x140 mm is OK for the stud
Min. screw spacing a1 in stud ≥ 10d = 60 mm
Min. screw spacing a2 in beam ≥ 5d = 30 mm
→ a1 in the stud is more critical
Distance to edge of the beam a2,CG ≥ 4d = 24 mm. When the screwing angle is 45°, the beam thickness
t1/2 = 25,5 mm gives the minimum distance.
→ 2 screws are chosen for the connection, so that the heads of the screws are 20 mm and 110 mm from
the bottom edge of the beam.
Minimum distance to the beam end a1,CG ≥ 10d = 60 mm. Therefore the end of the bean shall exceed the
stud edge.
Effective penetration length lef,1 in ledger beam is
t1t1 5151mmmm
𝑙𝑙𝑙𝑙ef,1==𝑙𝑙𝑙𝑙g,1
𝑙𝑙𝑙𝑙ef,1 𝑙𝑙𝑙𝑙g,1== −−l l == −−17 17mm mm==55 55mm
mm
sin 45° u u sin
sin45° sin45°45°
l_(ef,1)=l_(g,1)=t_1/sin〖45°〗 -l_u=(51 mm)/(sin 45°)-17 mm=55 mm
𝑡𝑡𝑡𝑡1𝑡𝑡𝑡𝑡1 5151
mmmm
Penetrationlength
Penetration
Penetration lengthin
length ininwall
wallstud
wall studl_(g,2)=l-t_1/sin〖45°〗
𝑙𝑙𝑙𝑙g,2==𝑙𝑙𝑙𝑙 −
𝑙𝑙𝑙𝑙g,2 𝑙𝑙𝑙𝑙 −sinsin ==140140mm
mm−−
=140 mm-(51 ==68
68mm
mm
45° 45° sin45° mm)/(sin45°)=68 mm
sin45°
For the beam the angles in the connections are: α = 45°, β = 45° and ε = 90° and for the stud they are:
α = 45°,0,9 β0,9= 0° and ε = 45°.
𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅
k k==𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅
T,k T,k(cos
(cos𝜀𝜀𝜀𝜀 𝜀𝜀𝜀𝜀++𝜇𝜇𝜇𝜇 𝜇𝜇𝜇𝜇sin
sin𝜀𝜀𝜀𝜀)𝜀𝜀𝜀𝜀) (5.33)
(5.33)
Connection capacity
The characteristic load-carrying capacity of the tension screw connection, see Figure 5.11(b), is c alculated
by the equation:
The characteristic density ρk is 480 kg/m3 for LVL 48 P and 410 kg/m3 for LVL 32 P.
ffax,90,k
ax,90,k is
is the the characteristic
characteristic withdrawal withdrawal strength strength parameterparameter for
for aa screw
screw
fax,90,k is theperpendicular characteristic withdrawal to the grain strength
direction parameter
[N/mm 2for
]. a screw
For perpendicular
screws in to the grain direction
in LVL,
LVL, the
2
perpendicular to the grain direction [N/mm ]. For screws the
[N/mm characteristic withdrawal parameter may be assumed as fax,90,k =15 assumed as
characteristic 2]. For screws in
withdrawal LVL, the characteristic
parameter may withdrawal
be assumed parameter
as f may
ax,90,k =15 be
fax,90,kN/mm²
N/mm²
= 15 N/mm² for ρ
for ρaa = =for 500
500 ρa = kg/m³
500 kg/m³
kg/m³ and screws
and screws
and screws 6 mm≤
6 mm≤ 6 mm≤d≤
d ≤ 12 d ≤mm
12 mm
12 mmin softwood
in softwood
in softwood LVL/GLVL.
LVL/GLVL.
LVL/GLVL. 0,8
3 0,8
2
480 kg/m23 480 kg/m 3 0,8
0,8
𝑓𝑓𝑓𝑓ax,90°,1,k = 15𝑓𝑓𝑓𝑓N/mm
ax,90°,1,k ∙
= � 15 N/mm 2 �∙ 480
�
480 = kg/m
14,5
kg/m 3N/mm� = 2
14,5 N/mm 2
2 ))^0,8=14,5 N/mm^2
𝑓𝑓 f_(ax,90°,1,k)=15
𝑓𝑓ax,90°,1,k = 15
= 15500 N/mm
N/mm N/mm^2∙((480
kg/m 2 ∙(
∙
3 ( 500 kg/m kg/m^3)/(500
3 )
)
3 = 14,5
= 14,5 N/mm kg/m^3
N/mm 2
ax,90°,1,k 500
500 kg/m kg/m3
When ε = 45°,When kax = ε1 =and 45°, when kax =β1=and 0°, when β = 0°,
When
Whenεε ε=
When ==45°, 45°,kkkax
45°, ax =
ax ==1 11and
andandwhen when β
when β= 0°,
= 0°,
0°,
2 2 3 0,8 0,8
f_(ax,45°,2,k)=(1∙15 N/mm^2)/(1,5〖∙cos〗^2 1 ∙ 15 N/mm 2 1 ∙ 15 410kg/m
N/mm 3 0°+sin^2
0,8410kg/m 0°) 3(( 410kg/m^3)/(500kg/m^3 ))^0,8=8,5 N/
𝑓𝑓𝑓𝑓ax,45°,2,k = 𝑓𝑓𝑓𝑓1 ∙ 15 N/mm = 2 � 410kg/m 3 �0,8� = 8,5 N/mm � 2
= 8,5 N/mm2
mm^2
𝑓𝑓ax,45°,2,k =
𝑓𝑓 = 1,5 1 ∙ ∙ 152 0°
ax,45°,2,k
cos 2
N/mm + 1,5
sin 2∙ 0°
2 cos (
(
410kg/m
2500kg/m
0° + sin 230°
3 )
) =
500kg/m
= 8,5
8,5 N/mm
N/mm
3 2
2
ax,45°,2,k 1,5
1,5 ∙∙ cos cos2 0° 0° + + sinsin2 0° 0° 500kg/m
500kg/m3
TheThe different
different
𝑓𝑓𝑓𝑓ax,α,1,k 𝑙𝑙𝑙𝑙conditions
conditions
∙ 𝑑𝑑𝑑𝑑 ∙ conditions 𝑓𝑓𝑓𝑓ax,α,1,k
= 14,5 ∙ 𝑑𝑑𝑑𝑑 of∙N
of the
the 𝑙𝑙𝑙𝑙g,1 ∙equation
equation
6,0
= 14,5 mm ∙(5.31)
N
(5.31)
552mm ∙ 6,0give
give
=mm aa
4,8 characteristic
characteristic capacity
capacityRT,kR : :
The different g,1 of
mm the 2 equation mm (5.31) give a∙ characteristic
kN
55 mm = 4,8 kN
capacity T,k :
RT,k
1〖∶ f〗_(ax,90°,1,k)∙ 0,8 d∙ l_(g,1)=14,5 𝜌𝜌𝜌𝜌k 0,8N
N 2 N/mm^2 ∙6,0 mm∙55 mm=4,8
3 0,8 kN
480 kg/m3
0,8
1
1
𝑓𝑓𝑓𝑓 ∶∶ 𝑓𝑓
𝑓𝑓 ∙ 𝑑𝑑𝑑𝑑
ax,90°,1,k 2
∙ � ∙∙𝜌𝜌𝜌𝜌𝑓𝑓𝑓𝑓k𝑑𝑑
𝑑𝑑
� ∙∙ 𝑙𝑙𝑙𝑙g,1
=∙ =
=
𝑑𝑑𝑑𝑑 2 14,5
14,5
13,0N/mm
∙ � � = ∙ (12
∙ 6,0
6,0
13,0N/mm mm
mm
mm) ∙∙ 255
55∙ � ∙
480=kg/m
2 mm
mm(12 = 4,8
4,8
mm) kN
2
kN�∙ � = 2,4 kN � = 2,4 kN
2∶ f_(head,k)
head,k ax,90°,1,k h
𝜌𝜌𝜌𝜌a〖∙d〗_h^2∙(ρ_k/ρ_a
head,k g,1 h mm2 )^0,8=13,0〖N/mm〗^2∙(12
𝜌𝜌𝜌𝜌amm
2
350 kg/m3 350 mm)^2∙((480
kg/m3 kg/m^3)/(350 kg/m^3
))^0,8=2,4 kN 𝜌𝜌k 0,8 0,8 480 3 0,8
0,8
3∶
2 𝑓𝑓 ∙ 𝑑𝑑h22 ∙∙ ((𝜌𝜌k )) N=
∶∶ f_(ax,45°,2,k)∙d∙l_(g,2)=8,5 13,0N/mm N/mm^2 N22 ∙∙∙6,0
(12 mm∙68
mm) 2
2
480 kg/m
∙∙mm=3,5
( kg/m
kN 3 )
3
= 2,4 kN
2 𝑓𝑓head,k
𝑓𝑓𝑓𝑓ax,α,2,khead,k ∙ 𝑑𝑑𝑑𝑑 ∙∙𝑙𝑙𝑙𝑙𝑑𝑑g,2h𝑓𝑓𝑓𝑓ax,α,2,k= 𝜌𝜌 8,5 ∙ 𝑑𝑑𝑑𝑑 =
∙ 𝑙𝑙𝑙𝑙 13,0N/mm
∙ 6,0 = mm
8,5 ∙ 68 (12
mm
∙ 6,0 mm)
= mm 3,5 ∙ kN (350
68 mm kg/m
= 3,53)kN = 2,4 kN
4∶ f_(tens,k)=10kN mm 𝜌𝜌 a
a 2 g,2
mm 2 350 kg/m
R_(T,k)=min{█(max(4,8 kN;2,4 N kN)@3,5 kN@10 kN)┤=3,5 kN
𝑓𝑓𝑓𝑓tens,k
3
3 ∶∶ 𝑓𝑓 = 10kN∙𝑓𝑓𝑓𝑓𝑑𝑑tens,k
𝑓𝑓ax,45°,2,k ∙ 𝑙𝑙 == 8,5 N 2 ∙∙ 6,0
10kN 6,0 mm mm ∙∙ 68 68 mm mm = = 3,5 kN
ax,45°,2,k ∙ 𝑑𝑑 ∙ 𝑙𝑙g,2 g,2 = 8,5 mm2 3,5 kN
mm
max(4,8 kN; 2,4 max(4,8 kN) kN; 2,4 kN)
4
4 ∶∶ 𝑓𝑓
𝑓𝑓= min=
tens,k =� 10kN 10kN min = 3,5
𝑅𝑅𝑅𝑅T,k tens,k 𝑅𝑅𝑅𝑅T,k = 3,5 kN� 3,5 kNkN = 3,5 kN
max(4,8 10 kN 10 kN
max(4,8 kN; kN; 2,4 2,4 kN) kN)
𝑅𝑅T,k =
𝑅𝑅 = min min {{ 3,5
3,5 kN kN = 3,5
= 3,5 kN kN
T,k
10 kN
10 kN
𝑘𝑘𝑘𝑘mod 0,9 𝑘𝑘𝑘𝑘mod 0,9
Design
𝑅𝑅𝑅𝑅d = resistance
Design
Design resistance
resistance ∙ 𝑡𝑡𝑡𝑡 𝑅𝑅𝑅𝑅d ∙= of
of the
𝑅𝑅𝑅𝑅T,k
the
the (cos connection: sin (cos
∙ 𝑡𝑡𝑡𝑡𝛼𝛼𝛼𝛼 +∙ 𝜇𝜇𝜇𝜇𝑅𝑅𝑅𝑅T,k
connection: 𝛼𝛼𝛼𝛼) 𝛼𝛼𝛼𝛼 + 𝜇𝜇𝜇𝜇 sin 𝛼𝛼𝛼𝛼)
𝛾𝛾𝛾𝛾M 𝛾𝛾𝛾𝛾M connection:
𝑘𝑘
0,8 0,8 0,9
= 𝑘𝑘mod
mod
𝑅𝑅
𝑅𝑅𝑅𝑅dd =
𝑅𝑅 𝑛𝑛0,9
∙ 2∙∙0,9
𝑛𝑛 𝑅𝑅𝑅𝑅∙ d3,5
0,9
∙∙=𝑅𝑅T,k
𝑅𝑅 kN (cos(cos
∙∙(cos45°
2 𝛼𝛼 𝛼𝛼∙+ +3,5 𝜇𝜇+sin
𝜇𝜇 sin
kN0,26𝛼𝛼)
∙ (cos45°
𝛼𝛼) ∙ sin45°) + 0,26
= 3,6∙ sin45°)
kN = 3,6 kN
1,3𝛾𝛾
𝛾𝛾M T,k
1,3
M
0,8 0,9
𝑅𝑅
𝑅𝑅 = 0,8
𝐸𝐸𝐸𝐸d,ULS ≤ 𝑅𝑅𝑅𝑅 →𝐸𝐸𝐸𝐸∙∙OK
∙ 2d0,9 3,5 kN
kN≤ ∙∙𝑅𝑅𝑅𝑅(cos45°
d → OK+
(cos45° + 0,26
0,26 ∙∙ sin45°)
sin45°) =
= 3,6
3,6 kN
kN
d = 1,3 ∙ 2 3,5
d d,ULS
R_d=k_mod/γ_M
1,3 ∙n^0,9 〖 ∙R〗_(T,k) (cos α+μ sinα )
R_d=0,8/1,3∙2^0,9∙3,5 kN∙(cos45°+0,26∙sin45°)=3,6 kN
𝐸𝐸 d,ULS ≤
𝐸𝐸d,ULS ≤ 𝑅𝑅
𝑅𝑅dd →
→ OK OK
E_(d,ULS)≤R_d→OK
The
The canopy
canopy can
can be
be supported
supported on on a
a 51x200
51x200 mm mm LVL
LVL 48P
48P ledger
ledger beam
beam which
which is
is connected
connected to to
51 The32P
canopy can be supported on a 51x200 mm LVL 48P ledger beam which At
is connected to 51 mm LVL 32P
51 mm LVL 32P wall studs with 2pcs 6,0x140 mm full threaded inclined screws. At the
mm LVL wall studs with 2pcs 6,0x140 mm full threaded inclined screws. the ends
ends
the wall
the ledger studs
ledger beam with
beam shall 2pcs
shall exceed 6,0x140
exceed the mm
the studs full
studs edges threaded
edges atat least inclined
least 6060 mm screws.
mm -- 25,5 At
25,5 mm the
mm = ends
= 34,5 the
34,5 mm.
mm. ledger beam shall exceed the
studs edges at least 60 mm - 25,5 mm = 34,5 mm.
9.8
9.8 Laterally loaded
Laterally loaded nail
nail connection
connection
A
A canopy
canopy over
over the
the entrance
entrance of of a
a one
one family
family house
house is
is supported
supported to
to the
the external
external wall
wall by
by a
a
51x300 mm
51x300 mm LVL LVL 4848 PP ledger
ledger beam.
beam. The
The beam
beam is
is nailed
nailed to
to the
the edges
edges of
of 45
45 mm
mm thick
thick LVL
LVL 32
32
P
P wall
202
wall studs
studs which have
have spacing
whichEurope
LVL Handbook spacing s s== 600
600 mm.
mm. Line
Line load
load from
from own
own weight gkk is
weight g is 0,3
0,3 kN/m
kN/m and
and
imposed
imposed load
load from
from snow
snow q qkk is
is 3
3 kN/m.
kN/m.
Loading combinations
The most critical ultimate limit state (ULS) load combination for each connection between the beam and
stud:
Ed,ULS = s ∙ (γG∙gk + γQ∙qk)
Ed,ULS = 0,6m ∙ (1,15∙0,3 kN/m + 1,5∙3 kN/m) = 2,91 kN
Note: Safety factors γG and γQ are according to Finnish national annex of eurocode 0.
Nail properties
Size: 3,1x90 mm round nails
Tensile strength fu = 600 N/mm2
Modification factor kmod for medium-term, SC2 = 0,8
Material safety factor γM for connections (default value in EC5) = 1,3
Minimum distance to unloaded edge a4,c in stud ≥ 7d = 7∙3,1 mm = 21,7 mm. Stud thickness 45 mm/2 =
22,5 mm → Nail size 3,1x90 mm is OK for the stud.
Min. nail spacing a1 in stud ≥ [(7 + 8(cos α)] ∙ d = 15d = 46,5 mm
Min. distance to unloaded edge of the beam a4,c ≥ 5d = 15,5 mm
Min. distance to loaded edge of the beam a4,t ≥ (5 + 2∙sinα) ∙ d = 7d = 21,7 mm
Maximum number of nails in the connection:
1+((h –min a_(4,c)–min a_(4,t)))/(min a_1 ) =1+((300mm-15,5mm-21,7mm))/46,5mm=6,65
(ℎ – min 𝑎𝑎𝑎𝑎4,c – min 𝑎𝑎𝑎𝑎4,t ) (300mm − 15,5mm − 21,7mm)
1+ =1+ = 6,65
min 𝑎𝑎𝑎𝑎1 46,5mm
→ 6 nails are chosen for the connection, so that the edge distance a4,c and a4,t at the beam are 25mm and
the nail spacing a1 in the stud (which is a2 in the beam) is 50 mm.
Connection capacity
The embedment strength of 3,1x90 mm round nail in LVL 48P and LVL 32P
fh,LVL 48P,k = 0,082 ∙ ρk ∙ d - 0,3 = 0,082 ∙ 480 ∙ 2,5 - 0,3 = 28,0 N/mm2
fh,LVL32P,k = 0,082 ∙ ρk ∙ d - 0,3 = 0,082 ∙ 410 ∙ 2,5 - 0,3 = 23,9 N/mm2
Note: the angle β = 0° for the beam and 90° for the stud, so the simpler equation of embedment strength can
be used.
When the nails are produced from wire with tensile strength fu = 600 N/mm2, the characteristic value of
the yield moment My,k for round nails is
My,k = 0,3 ∙ fu ∙ d2,6 = 0,3 ∙ 600 ∙ (2,5)2,6 = 3410 Nmm
Note: My,k value should be checked from the DoP of the nail supplier.
The influence of rope effect based on the axial withdrawal capacity Fax,k of round nails is negligible. With
these properties FV,nail,Rk is (𝑎𝑎𝑎𝑎)
4,43 as the minimum of EN 1995-1-1, equation 8.6 failure modes (a)-(f)
⎧ (𝑏𝑏𝑏𝑏)
2,89 (𝑎𝑎𝑎𝑎)
4,43
⎪
⎧4,43 (𝑎𝑎𝑎𝑎)
(𝑐𝑐𝑐𝑐)
1,55 (𝑏𝑏𝑏𝑏)
2,89
𝐹𝐹𝐹𝐹V,nail,Rk = min ⎧
⎪2,89 (𝑏𝑏𝑏𝑏) = 0,85 kN
1,55 (𝑑𝑑𝑑𝑑)
⎪1,58
⎨ (𝑐𝑐𝑐𝑐)
𝐹𝐹𝐹𝐹V,nail,Rk = min 1,55 (𝑐𝑐𝑐𝑐) = 0,85 kN
F_(V,nail,Rk)=min{█(4,43
𝐹𝐹𝐹𝐹V,nail,Rk = min ⎪
⎨ 1,13
1,58 (a)@2,89
(𝑒𝑒𝑒𝑒)
(𝑑𝑑𝑑𝑑) (b)@1,55
= 0,85 kN (c)@1,58 (d)@1,13 (e)@0,85 (f))┤=0,85 kN
⎩1,58
⎨ 0,85 (𝑑𝑑𝑑𝑑)
⎪ 1,13 (𝑓𝑓𝑓𝑓)
(𝑒𝑒𝑒𝑒)
⎪
⎩0,85 (𝑒𝑒𝑒𝑒)
1,13 (𝑓𝑓𝑓𝑓)
⎩0,85 (𝑓𝑓𝑓𝑓)
Design resistance
𝑘𝑘𝑘𝑘mod of the connection:
𝑅𝑅𝑅𝑅d = 𝑘𝑘𝑘𝑘 ∙ 𝑡𝑡𝑡𝑡ef ∙ 𝐹𝐹𝐹𝐹V,nail.Rk
𝛾𝛾𝛾𝛾mod M
𝑅𝑅𝑅𝑅d = 𝑘𝑘𝑘𝑘mod ∙ 𝑡𝑡𝑡𝑡ef ∙ 𝐹𝐹𝐹𝐹V,nail.Rk
𝑅𝑅𝑅𝑅 d = 𝑡𝑡𝑡𝑡 𝑘𝑘𝑘𝑘𝛾𝛾𝛾𝛾efM ∙ 𝑡𝑡𝑡𝑡ef ∙ 𝐹𝐹𝐹𝐹V,nail.Rk
𝑡𝑡𝑡𝑡R_d
ef = = 𝛾𝛾𝛾𝛾k_mod/γ_M
M ∙n_ef∙F_(V,nail.Rk)
𝑡𝑡𝑡𝑡ef = 𝑡𝑡𝑡𝑡𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘ef
𝑡𝑡𝑡𝑡n_ef=n^(k_ef
ef = 𝑡𝑡𝑡𝑡 ef )
kef = 1, when a nail row staggered perpendicular to grain by at least 1d. Without staggering in LVL edge
face
1 50mm
𝑘𝑘𝑘𝑘ef = min � = 1 − 0,03 �20 − 50mm � = 0,88
1 −
k_ef=min{█(1@1-0,03(20-a_1/d) 0,03(20 1 − 𝑎𝑎𝑎𝑎1 /𝑑𝑑𝑑𝑑) )┤=1-0,03(20-50mm/3,1mm)=0,88
3,1mm
𝑘𝑘𝑘𝑘ef = min � 1 − 𝑎𝑎𝑎𝑎 /𝑑𝑑𝑑𝑑) = 1 − 0,03 �20 − 50mm � = 0,88
𝑘𝑘𝑘𝑘Nailing
ef = min �1 − 0,03(20
without staggering: 1 = 1 − 0,03 �20 − 3,1mm� = 0,88
1 − 0,03(20 − 𝑎𝑎𝑎𝑎1 /𝑑𝑑𝑑𝑑) 3,1mm
𝑘𝑘𝑘𝑘 0,8
R_d
mod
𝑅𝑅𝑅𝑅d == 𝑘𝑘𝑘𝑘k_mod/γ_M
∙ 𝑡𝑡𝑡𝑡ef ∙ 𝐹𝐹𝐹𝐹V,nail,Rk =0,8 ∙ 60,88 ∙ 0,85 kN = 0,62 ∙ 4,84 ∙kN=0,62∙4,84∙0,85
∙n_ef∙F_(V,nail,Rk)=0,8/1,3∙6^0,88∙0,85
0,88
0,85 kN = 2,55 kNkN=2,55 kN
𝑅𝑅𝑅𝑅d = 𝛾𝛾𝛾𝛾𝛾𝛾𝛾𝛾mod ef ∙ 𝐹𝐹𝐹𝐹V,nail,Rk =1,3 ∙ 6
M ∙ 𝑡𝑡𝑡𝑡 1,3 ∙ 0,85 kN = 0,62 ∙ 4,84 ∙ 0,85 kN = 2,55 kN
E_(d,ULS)>R_d→Not OK
M
𝐸𝐸𝐸𝐸d,ULS > 𝑅𝑅𝑅𝑅d → Not OK
𝐸𝐸𝐸𝐸d,ULS > 𝑅𝑅𝑅𝑅d → Not OK
The canopy can be supported on a 51x300 mm LVL 48P ledger beam which is connected to 45 mm LVL 32P
wall studs with 6pcs 3,1x90 mm round nails when the nail row is staggered perpendicular to grain by 1d.
Discussion:
The inclined screws connection has 16% higher capacity than the laterally loaded nailed connection in the
example 9.6 and the leger beam depth is 100 mm smaller. However, the LVL 32P stud needs to be thicker
due to the edge distance requirement a2,CG ≥4d of the screws. A laterally loaded screws connection would
not be possible for the combination of screw size and LVL beam stud thickness, since the edge distance
a4,c ≥ 7d at the LVL edge would not be fulfilled.
Geometry requirements:
→LVL-C type joist is needed to fulfil the requirement of the maximum hole depth hd.
Tension
𝜎𝜎𝜎𝜎t,90,dstress
= 0,5∙𝑙𝑙𝑙𝑙perpendicular
𝐹𝐹𝐹𝐹t,90,d to the grain is verified by the equation
≤ 𝑓𝑓𝑓𝑓t,90,d (4.57)
∙𝑏𝑏𝑏𝑏∙𝑘𝑘𝑘𝑘
t,90 t,90
𝐹𝐹𝐹𝐹t,90,d
𝜎𝜎𝜎𝜎𝜎𝜎𝜎𝜎
σt,90,d =
t,90,d
t,90,d = 𝐹𝐹𝐹𝐹t,90,d
=F_(t,90,d)/(0,5∙l_(t,90)∙b∙k_(t,90)≤≤𝑓𝑓𝑓𝑓t,90,d
𝑓𝑓𝑓𝑓t,90,d )≤f_(t,90,d) (4.57)
(4.57)
0,5∙𝑙𝑙𝑙𝑙
0,5∙𝑙𝑙𝑙𝑙 t,90
t,90
∙𝑏𝑏𝑏𝑏∙𝑘𝑘𝑘𝑘
∙𝑏𝑏𝑏𝑏∙𝑘𝑘𝑘𝑘 t,90
t,90
(4.57)
where 1
𝑘𝑘𝑘𝑘t,90 = min � 450 0,5 = 1,0 (4.58)
� ℎ11�
kt,90
𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘
t,90 =min{█(1@(450/h)^0,5
t,90== min�� 450
min 0,5 = 1,0 )=1,0┤
450 0,5 = 1,0 (4.58)
(4.58)
��ℎℎ��
(4.58)
σt,90,d
𝑙𝑙𝑙𝑙t,90 = =0,5design
∙ (ℎd value
+ ℎ) = of tension
0,5 ∙ (60stress
mm perpendicular
+ 240 mm) = to
150themm
grain [N/mm2] (4.61)
lt,90 = 0,5 ∙ (hd + h) = 0,5 ∙ (60 mm + 240 mm) = 150 mm (4.61)
Ft,90,d 𝑙𝑙𝑙𝑙t,90
𝑙𝑙𝑙𝑙t,90 === 0,5
0,5design(ℎddvalue
∙ ∙(ℎ ++ℎ)ℎ)= =tension
of 0,5 (60
0,5∙ ∙(60 mmperpendicular
force
mm ++240
240mm)
mm)==to150 mm
themm
150 grain [N] (4.61)
(4.61)
𝑘𝑘𝑘𝑘mod 0,8 N
ft,90,d =
𝑓𝑓𝑓𝑓t,90,d =k_mod/γ_M ∙f_(t,90,edge,k)=0,8/1,2∙5
∙ 𝑓𝑓𝑓𝑓t,90,edge,k = ∙5 2
N/mm2 =3,3 N/mm^2
= 3,3 N/mm^2
𝛾𝛾𝛾𝛾
𝑘𝑘𝑘𝑘 M 1,2
0,8 mmN
𝑘𝑘𝑘𝑘𝐹𝐹𝐹𝐹t,90,d
mod 0,8 N
𝜎𝜎𝜎𝜎t,90,d = == mod
𝑓𝑓𝑓𝑓t,90,d
𝑓𝑓𝑓𝑓t,90,d ≤ 𝑓𝑓𝑓𝑓t,90,d== ∙ ∙55
∙ ∙𝑓𝑓𝑓𝑓t,90,edge,k
𝑓𝑓𝑓𝑓t,90,edge,k ==3,3 N/mm2 2
3,3N/mm (4.57)
0,5∙𝑙𝑙𝑙𝑙𝛾𝛾𝛾𝛾
t,90𝛾𝛾𝛾𝛾MM∙𝑏𝑏𝑏𝑏∙𝑘𝑘𝑘𝑘t,90 1,2 mm
1,2 mm2 2
1
𝑘𝑘𝑘𝑘t,90 = min � 450 0,5 = 1,0 (4.58)
�ℎ �
𝑘𝑘𝑘𝑘mod 0,8 N
𝑓𝑓𝑓𝑓t,90,d = ∙ 𝑓𝑓𝑓𝑓t,90,edge,k = ∙5 = 3,3 N/mm2
𝛾𝛾𝛾𝛾M 1,2 mm2
𝑉𝑉𝑉𝑉d ∙ℎd ℎd 2 𝑀𝑀𝑀𝑀
𝐹𝐹𝐹𝐹t,90,d = ∙ �3 − � � � + 0,008 ∙ ℎ d (4.59)
4∙ℎ ℎ r
𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉 d ∙ℎ ℎℎ 22 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀
d ∙ℎ dd ∙ d d� dd (4.59)
𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹
t,90,d
t,90,d ==
4∙ℎ 4∙ℎ ∙ �3
�3 −−��
ℎℎ
� � �
+ + 0,008
0,008 ∙ ∙
ℎℎ
(4.59)
rr
ℎr = 90 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 (4.60)
ℎℎr r==90
90𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 (4.60)
(4.60)
Tension in perpendicular to grain force Ft,90,d depends on the shear force Vd and bending moment Md at the
edge of the hole:
𝑉𝑉𝑉𝑉 ∙ℎ ℎ 2 𝑀𝑀𝑀𝑀
F_(t,90,d)=(V_d∙h_d)/(4∙h)∙[3-(h_d/h)^2
d d d
t,90,d = 4∙ℎ ∙ �3 − � ℎ � � + 0,008 ∙ ℎ
𝐹𝐹𝐹𝐹 d
]+0,008∙M_d/h_r (4.59)
r
(4.59)
where
ℎ = 90
hrr=90 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
mm (4.60) (4.60)
Since the hole is at the centre line of the cross section, only one of the equations (4.64 and 4.65) needs to be
verified. The 140x60 mm hole 300 mm from the support edge fulfils the requirements in LVL 36 C joist.
The design load-carrying capacity FV,Rd (design racking resistance) under a horizontal force FV,Ed from wind
load acting at the top of a cantilever panel secured against uplift is determined using the simplified method A
of Eurocode 5 in section 9.2.4.2.
The penetration length t1 in LVL-C panel thickness should be at least 4d = 4∙2,5 mm = 10 mm → OK.
The unloaded edge distance a4,C should be at least 3d = 3∙2,5 mm = 8 mm in the face side of LVL-C panel
and 7d = 7∙2,5 mm =18 mm in the LVL-P edge face. The minimum stud thickness is 2∙7d = 36 mm for a
perimeter stud and 2∙(7d+3d) + 1 mm = 51 mm when the panel joint is at a stud location, see Table 5.5
of panel joints.
The embedment strength of 2,5x60 mm round nail in LVL 36 C and LVL 48P
When the nails are produced from wire with tensile strength fu = 600 N/mm2, the characteristic value of the
yield moment My,k for round nails is
The influence of rope effect based on the axial withdrawal capacity Fax,k of round nails is negligible. With these
properties FV,nail,Rk is as the minimum of failure modes (a)-(f)
⎧ 2,02 (𝑎𝑎𝑎𝑎)
2,02
2,02 (𝑎𝑎𝑎𝑎)
(𝑎𝑎𝑎𝑎)
⎧
⎧ 2,02
2,47 (𝑎𝑎𝑎𝑎)
(𝑏𝑏𝑏𝑏)
(𝑏𝑏𝑏𝑏)
⎪
⎧ 2,47
2,47 (𝑏𝑏𝑏𝑏)
⎪
⎪2,47 0,94 (𝑏𝑏𝑏𝑏)
(𝑐𝑐𝑐𝑐)
𝐹𝐹𝐹𝐹 = min ⎪ 0,94
0,94 (𝑐𝑐𝑐𝑐)
(𝑐𝑐𝑐𝑐) = 0,62 kN
𝐹𝐹𝐹𝐹V,nail,Rk = min
𝐹𝐹𝐹𝐹V,nail,Rk = min ⎨0,78 (𝑑𝑑𝑑𝑑)
V,nail,Rk 0,94
0,78 (𝑐𝑐𝑐𝑐) =
(𝑑𝑑𝑑𝑑) = 0,62
0,62 kNkN
𝐹𝐹𝐹𝐹V,nail,Rk = min ⎨ ⎨ 0,78 (𝑑𝑑𝑑𝑑) = 0,62 kN
⎪
⎨ 0,78
0,92
0,92 (𝑑𝑑𝑑𝑑)
(𝑒𝑒𝑒𝑒)
(𝑒𝑒𝑒𝑒)
⎪
⎪0,620,92 (𝑒𝑒𝑒𝑒)
⎩
⎪ 0,92 (𝑓𝑓𝑓𝑓) (𝑒𝑒𝑒𝑒)
⎩0,62
⎩ 0,62 (𝑓𝑓𝑓𝑓) (𝑓𝑓𝑓𝑓)
𝑘𝑘𝑘𝑘 ⎩ 0,62 (𝑓𝑓𝑓𝑓) 1,1
𝐹𝐹𝐹𝐹 = 𝑘𝑘𝑘𝑘
𝑘𝑘𝑘𝑘mod
mod ∙∙ 1,2 ∙∙ 𝐹𝐹𝐹𝐹 = 1,1
1,1 1,2 ∙∙ 0,62
1,1 ∙∙∙ 1,2 kN = 0,63 kN
mod
𝐹𝐹𝐹𝐹
𝐹𝐹𝐹𝐹f,Rd =
= 𝑘𝑘𝑘𝑘
F_(V,nail,Rk)=min{█(2,02
f,Rd 𝛾𝛾𝛾𝛾
mod ∙ 1,2
1,2 ∙ 𝐹𝐹𝐹𝐹
V,nail,Rk
𝐹𝐹𝐹𝐹
V,nail,Rk =
=
(a)@2,47
1,3 1,2 ∙ 0,62 kN
kN =
= 0,63
0,62 (c)@0,78
(b)@0,94 0,63 kN
kN
(d)@0,92 (e)@0,62 (f))┤=0,62 kN
f,Rd
𝐹𝐹𝐹𝐹f,Rd = 𝛾𝛾𝛾𝛾M 𝛾𝛾𝛾𝛾M V,nail,Rk 1,3
M ∙ 1,2 ∙ 𝐹𝐹𝐹𝐹V,nail,Rk = 1,3 ∙ 1,2 ∙ 0,62 kN = 0,63 kN
F_(f,Rd)=k_mod/γ_M 𝛾𝛾𝛾𝛾M 1,3
∙1,2∙F_(V,nail,Rk)=1,1/1,3∙1,2∙0,62 kN=0,63 kN
kmod = 1,1 for instantaneous load (wind load) in service class 1
γM = 1,3 for connections (default value in EC5)
F_(V,Rd)=(F_(f,Rd)∙b∙c)/s=(0,63 kN∙1200mm ∙∙mm∙0,96)/(100 mm)=7,3 kN
𝐹𝐹𝐹𝐹 =
𝐹𝐹𝐹𝐹
𝐹𝐹𝐹𝐹
𝐹𝐹𝐹𝐹f,Rd
∙∙ 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 ∙∙ 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 0,63
f,Rd ∙ 𝑏𝑏𝑏𝑏 ∙ 𝑐𝑐𝑐𝑐 =
f,Rd 0,63
0,63 kN
kN ∙∙ 1200
kN ∙ 1200
1200 mm
0,96
mm ∙ 0,96
0,96 = 7,3 kN
𝐹𝐹𝐹𝐹
𝐹𝐹𝐹𝐹V,Rd
V,Rd =
= 𝐹𝐹𝐹𝐹f,Rd ∙
𝑠𝑠𝑠𝑠 𝑏𝑏𝑏𝑏 ∙ 𝑐𝑐𝑐𝑐 =
= 0,63 kN ∙ 1200
100 mm mm ∙ 0,96 =
= 7,3
7,3 kN
kN
𝐹𝐹𝐹𝐹V,Rd
V,Rd =
𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 100
100 mm mm = 7,3 kN
𝑠𝑠𝑠𝑠 100 mm
Shear buckling of the panel may be disregarded, when the stud spacing bnet / t ≤ 100.
b_net/t=(600 mm)/(27 mm)=22 ≤100 →OK
𝑏𝑏𝑏𝑏
𝑏𝑏𝑏𝑏 600 mm
net = 600
𝑏𝑏𝑏𝑏net 600 mm mm = = 22 22 ≤ 100 100 → OK OK
𝑏𝑏𝑏𝑏net
𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 =
net = 600
27 mm =
mm 22 ≤ ≤ 100 → → OK
𝑡𝑡𝑡𝑡 = 27
27 mm mm = 22 ≤ 100 → OK
𝑡𝑡𝑡𝑡 27 mm
In order to with stand a horizontal force FV,Ed = 7,3 kN, the diaphragm panel shall be anchored at the bottom
corners for the external forces
245 (255)
𝐹𝐹𝐹𝐹
𝐹𝐹𝐹𝐹
V,Rd ∙∙ ℎ
ℎ 7,3kN
7,3kN ∙∙∙ 2500
2500 mmmm = 15,2
F_(t,Ed)=F_(c,Ed)=
𝐹𝐹𝐹𝐹 = 𝐹𝐹𝐹𝐹 = 𝐹𝐹𝐹𝐹
V,Rd
V,Rd (F_(V,Rd)∙h)/b=(7,3kN∙2500
∙ ℎ = 7,3kN 2500 mm mm)/(1200
kN mm)=15,2 kN
𝐹𝐹𝐹𝐹t,Ed
𝐹𝐹𝐹𝐹t,Ed =
= 𝐹𝐹𝐹𝐹 c,Ed
𝐹𝐹𝐹𝐹c,Ed = 𝐹𝐹𝐹𝐹 𝑏𝑏𝑏𝑏 ∙ ℎ =
c,Ed = V,Rd
= 7,3kN ∙ 2500 mm =
1200 = 15,2
15,2 kN
kN
𝐹𝐹𝐹𝐹t,Ed
t,Ed = 𝐹𝐹𝐹𝐹c,Ed =
𝑏𝑏𝑏𝑏
𝑏𝑏𝑏𝑏 = 1200 mm
1200 mm
mm = 15,2 kN
𝑏𝑏𝑏𝑏 1200 mm
The contactThearea
contact between perimeter
area between stud and
perimeter studthe
andend
the of
endthe
of horizontal solesole
the horizontal plate shall
plate shallbebe verified for compres-
verified for compression perpendicular
sion perpendicular to the grain. to the grain.
𝜎𝜎𝜎𝜎c,90,d ≤
𝜎𝜎𝜎𝜎 ≤ 𝑘𝑘𝑘𝑘
𝑘𝑘𝑘𝑘c,90 ∙∙ 𝑓𝑓𝑓𝑓
𝑓𝑓𝑓𝑓c,90,d
𝜎𝜎𝜎𝜎c,90,d
𝜎𝜎 ≤
c,90,d ≤ 𝑘𝑘𝑘𝑘
𝑘𝑘 c,90 ∙∙ 𝑓𝑓
c,90 𝑓𝑓𝑓𝑓
c,90,d
c,90,d
𝜎𝜎𝜎𝜎c,90,d
c,90,d 𝑘𝑘𝑘𝑘c,90
c,90 𝑓𝑓𝑓𝑓c,90,d
c,90,d
σ_(c,90,d)≤k_(c,90)𝐹𝐹𝐹𝐹c,90,d
𝐹𝐹 〖∙f〗_(c,90,d)
𝐹𝐹𝐹𝐹c,90,d
𝐹𝐹 c,90,d 15,2kN
15,2kN 2
𝜎𝜎𝜎𝜎c,90,d
𝜎𝜎 c,90,d = = c,90,d = = =
= = 1,3N/mm
1,3N/mm 2
σ_(c,90,d)=F_(c,90,d)/A_ef
𝐴𝐴𝐴𝐴
𝐴𝐴efef 𝑙𝑙 ∙ (𝑏𝑏 + 30 mm) 150 mm ∙ mm))=15,2kN/(150
𝑙𝑙𝑙𝑙 ∙ (𝑏𝑏𝑏𝑏 + 30 =F_(c,90,d)/(l∙(b+30
mm) 150 mm ∙ (51
(51 mm +
mm + 30 mm) = mm
30 mm) ∙ (51 mm+30 mm) )=1,3N/
mm^2
kkc,90 is 1,4
c,90 is
c,90 is 1,4for
1,4 forLVL-P
for LVL-Pflatwise
LVL-P flatwise
flatwise and
and
and ffc,90,k
fc,90,k is 2,2
is 2,2
c,90,k is 2,2
N/mmN/mm
N/mm 2. 22..
γM =1,2 (default
Material safety factor value in MEC5)
γ1,1 (defaultNvalue in EC5) = 1,2
k_(c,90)∙f_(c,90,flat,d)=1,4∙1,1/1,2∙2,2
𝑘𝑘𝑘𝑘c,90 ∙ 𝑘𝑘
𝑓𝑓𝑓𝑓c,90,flat,d
mod
= 1,4 ∙ ∙ 2,2
1,1 2 = N/mm^2
2,8
N N/mm =2,8
2 N/mm^2>σ_(c,90,d)→OK
> 𝜎𝜎𝜎𝜎c,90,d → OK
𝑘𝑘c,90 ∙ ∙ 𝑓𝑓c,90,flat,k =1,2 1,4 ∙ mm ∙ 2,2 = 2,8 N/mm 2
> 𝜎𝜎c,90,d → OK
𝛾𝛾M 1,2 mm2
AnchoringAnchoring
can be done with
can be e.g.with
done Rothoblaas WHT340
e.g. Rothoblaas brackets
WHT340 for tension
brackets loadsloads
for tension and and
Titan
Titan TFC200 brackets
TFC200 brackets for shear
for shear loads. loads.
Effective cross section size after 30 minutes fire exposure on all sides:
d_ef=β_n∙t+k_0∙d_0=0,70 mm/min∙30min+1,0∙7mm=28mm
mm
𝑑𝑑𝑑𝑑ef = 𝛽𝛽𝛽𝛽n ∙ 𝑡𝑡𝑡𝑡 + 𝑘𝑘𝑘𝑘0 ∙ 𝑑𝑑𝑑𝑑0 = 0,70 ∙ 30min + 1,0 ∙ 7mm = 28mm
Size of the effective cross section: min
Width b: 133 mm - 2∙28 mm = 77 mm
Height h: 400 mm - 2∙28 mm = 344 mm
Beam properties of the effective cross section after 30min fire exposure:
Bending strength edgewise fm,0,edge,k = 44 N/mm2
Shear strength edgewise fv,0,edge,k = 4,2 N/mm2
Compression perpendicular to grain edgewise fc,90,edge,k = 6 N/mm2
Modulus of elasticity E0,k = 11 600 N/mm2
Modulus of rigidity G0,edge,mean = 400 N/mm2
Area of cross section A =b∙h = 26488 mm2
Section modulus Wy = b∙h /6 2 = 1,52∙106 mm3
Moment of inertia Iy = b∙h /12
3 = 2,61∙107 mm4
Moment of inertia Iz = h∙b /12
3 = 1,31∙107 mm4
Torsion moment of inertia Itor = 0,3∙h∙b 3 = 4,71∙107 mm4
Modification factor kmod,fi = 1,0
Modification factor kfi = 1,1
Material safety factor γM,fi (default value in EC5) = 1,0
Size effect factor kh = (300/344)0,15 = 0,98
Loading combinations
Snow load at roof level qk = μ1 ∙ Ce ∙ sk. Form factor μ1 = 0,8, when roof angle is less than 30° and in normal
conditions Ce = 1,0 → qk = 0,8 ∙ 1,0∙2,5 N/m2 = 2,0 kN/m2.
Accidental load combination of fire in the ultimate limit state (ULS):
Ed,ULS,fi = γG ∙ (g1,k + g2,k) + ψ1 ∙ γQ ∙ qk
Ed,ULS,fi =1,0 ∙ (8m ∙ 1,0 kN/m2 + 0,2 kN/m ) + 0,4 ∙ 1,0 ∙ 8m ∙ 2,0 kN/m2
Ed,ULS,fi = 14,6 kN/m
Note: Safety factors γG, ψ1 and γQ are according to Finnish National annex of Eurocode 0.
ULS design
Bending moment resistance
M_d
𝑀𝑀𝑀𝑀 = E_(d,ULS,fi)∙s∙L2/8 = 14,6 kN/m∙(4m)^2/8 2
d = 𝐸𝐸𝐸𝐸d,ULS,fi ∙ 𝑠𝑠𝑠𝑠 ∙ 𝐿𝐿𝐿𝐿2/8 = 14,6 kN/m ∙ (4m)2 /8 = 29,2 kNm
= 29,2 kNm
𝑀𝑀𝑀𝑀 = 𝐸𝐸𝐸𝐸 ∙ 𝑠𝑠𝑠𝑠 ∙ 𝐿𝐿𝐿𝐿2/8
σ_(m,d)=M_d/W=(29,2 kNm)/(1,52〖∙10〗^6 〖 mm〗^3
d d,ULS,fi = 14,6 kN/m ∙ (4m) /8 = 29,2 kNm
)=19,2 N/mm^2
𝑀𝑀𝑀𝑀
f_(m,d,fi)=(k_(mod,fi)
d 29,2 kNm
〖∙k〗_fi∙k_h)/γ_(M,fi)2∙f_(m,edge,k)
𝜎𝜎𝜎𝜎m,d = 𝑀𝑀𝑀𝑀d = 29,2 kNm = 19,2 N/mm2
f_(m,d,fi)=(1,0∙1,1∙(300mm/344mm)^0,15)/1,0∙44
𝜎𝜎𝜎𝜎m,d = 𝑊𝑊𝑊𝑊 = 1,52 ∙ 106 mm3 = 19,2 N/mm
6 3 N/mm^2 =47,4 N/mm^2
𝑊𝑊𝑊𝑊 1,52 ∙ 10 mm
σ_(m,d)≤f_(m,d,fi) →OK
𝑘𝑘𝑘𝑘mod,fi ∙ 𝑘𝑘𝑘𝑘fi ∙ 𝑘𝑘𝑘𝑘h
𝑓𝑓𝑓𝑓m,d,fi = 𝑘𝑘𝑘𝑘mod,fi ∙ 𝑘𝑘𝑘𝑘fi ∙ 𝑘𝑘𝑘𝑘h ∙ 𝑓𝑓𝑓𝑓m,edge,k
𝑓𝑓𝑓𝑓m,d,fi = 𝛾𝛾𝛾𝛾M,fi ∙ 𝑓𝑓𝑓𝑓m,edge,k
𝛾𝛾𝛾𝛾M,fi
300mm 0,15
1,0 ∙ 1,1 ∙ �300mm�0,15 N N
𝑓𝑓𝑓𝑓m,d,fi = 1,0 ∙ 1,1 ∙ �344mm
344mm
� ∙ 44 N 2 = 47,4 N 2
𝑓𝑓𝑓𝑓m,d,fi = 1,0 ∙ 44 mm2 = 47,4 mm2
1,0 mm mm
𝜎𝜎𝜎𝜎m,d ≤ 𝑓𝑓𝑓𝑓m,d,fi → OK
𝜎𝜎𝜎𝜎m,d ≤ 𝑓𝑓𝑓𝑓m,d,fi → OK
Lateral torsional buckling
The beam is loaded on the top side and the purlins won’t act as supports against lateral torsional buckling
for 30min fire exposure. Therefore according to Table 4.9 and EN1995-1-2, clause 4.3.2 (1) and the effec-
tive length Lef of the beam is
𝐿𝐿𝐿𝐿ef = 0,9 ∙ 𝐿𝐿𝐿𝐿 + 2 ∙ ℎ = 0,9 ∙ 4000mm + 2 ∙ 344mm = 4288mm.
ef = 0,9 ∙ 𝐿𝐿𝐿𝐿+ +
𝐿𝐿𝐿𝐿L_ef=0,9∙L 2∙h2=∙ ℎ 0,9=∙ 4000mm 0,9 ∙ 4000mm + 2 ∙=344mm
+ 2 ∙ 344mm 4288mm.= 4288mm.
σ_(m,crit)=M_(y,crit)/W_y
𝑀𝑀𝑀𝑀
𝜎𝜎𝜎𝜎m,crit = 𝑀𝑀𝑀𝑀𝑊𝑊𝑊𝑊
y,crit 𝜋𝜋𝜋𝜋 � 𝐸𝐸𝐸𝐸0,05 𝐼𝐼𝐼𝐼
𝑧𝑧𝑧𝑧 𝐺𝐺𝐺𝐺 =(π√(E_0,05 I_z G_0,05 I_tor ))/(l_ef W_y (4.42)
0,05 𝐼𝐼𝐼𝐼
tor )
y,crit = 𝜋𝜋𝜋𝜋�𝐸𝐸𝐸𝐸0,05 𝐼𝐼𝐼𝐼 𝐺𝐺𝐺𝐺 𝐼𝐼𝐼𝐼
𝑙𝑙𝑙𝑙ef𝑧𝑧𝑧𝑧𝑊𝑊𝑊𝑊y0,05 tor
𝜎𝜎𝜎𝜎(4.42)
m,crit = 𝑊𝑊𝑊𝑊y = (4.42)
y 𝑙𝑙𝑙𝑙ef 𝑊𝑊𝑊𝑊y
σ_(m,crit)= (π√(10600 N/mm^2∙1,31∙〖10〗^7 〖 mm〗^4∙400N/mm^
π�10600 N/mm 2 7 4 2 ∙ 107 ∙ mm4
〖1,52∙10〗^6 〖 mm〗^3 ) 2 ∙ 1,31 ∙ 107 mm4 ∙ 400N/mm2 ∙∙ 4,71 4,71 ∙ 107 ∙ mm4
𝜎𝜎𝜎𝜎 m,crit = π�10600 N/mm ∙ 1,31 ∙ 10 mm ∙ 400N/mm 6 3
𝜎𝜎𝜎𝜎 m,crit =
σ_(m,crit)= 25,8 N/mm^2 4288 mm ∙ 1,52 ∙ 10 mm
4288 mm ∙ 1,52 ∙ 106 mm3
λ_rel=√(f_(m,k)/σ_(m,crit) )=√((44 N/mm^2)/(25,8 N/mm^2 ))= 1,36
𝜎𝜎𝜎𝜎m,crit = 25,8 N/mm22
𝜎𝜎𝜎𝜎m,crit = 25,8 N/mm
𝑓𝑓𝑓𝑓 2
44 N/mm
(4.41)
𝜆𝜆𝜆𝜆rel = � m,k = � = 1,36 (4.41)
𝜎𝜎𝜎𝜎m,crit 25,8 N/mm2
when 0,75<λ_(rel,m)≤1,4 ,k_crit=1,56-0,75∙λ_(rel,m)=1,56-0,75∙1,36=0,58
k_crit∙
when 0,75 f_(m,d,fi)=0,58
< 𝜆𝜆𝜆𝜆rel,m ≤ ∙47,4
1,4 , 𝑘𝑘𝑘𝑘N/mm^2=27,5 N/mm^2
crit = 1,56 − 0,75 ∙ 𝜆𝜆𝜆𝜆rel,m = 1,56 − 0,75 ∙ 1,36 = 0,58
σ_(m,d)=19,2 N/mm^2≤k_crit∙ f_(m,d)→OK
𝑘𝑘𝑘𝑘crit ∙ 𝑓𝑓𝑓𝑓m,d,fi = 0,58 ∙ 47,4 N/mm2 = 27,5 N/mm2
3 ∙ 𝑉𝑉𝑉𝑉d 3 ∙ 29,2 kN
𝜏𝜏𝜏𝜏v,d = = = 1,7 N/mm2
2 ∙ 𝐴𝐴𝐴𝐴 2 ∙ 26488 mm2
𝑘𝑘𝑘𝑘mod,fi ∙ 𝑘𝑘𝑘𝑘fi 1,0 ∙ 1,1 N
𝑓𝑓𝑓𝑓v,d,fi = ∙ 𝑓𝑓𝑓𝑓v,0,edge,k = ∙ 4,2 = 4,6 N/mm2
𝛾𝛾𝛾𝛾M,fi 1,0 mm2
when 0,75 < 𝜆𝜆𝜆𝜆rel,m ≤ 1,4 , 𝑘𝑘𝑘𝑘crit = 1,56 − 0,75 ∙ 𝜆𝜆𝜆𝜆rel,m = 1,56 − 0,75 ∙ 1,36 = 0,58
𝑉𝑉𝑉𝑉d = 𝐸𝐸𝐸𝐸d,ULS,fi ∙ 𝐿𝐿𝐿𝐿/2 = 14,6 kN/m ∙ 4,0m/2 =9.29,2 kN
CALCULATION EXAMPLES OF LVL STRUCTURES
𝑘𝑘𝑘𝑘crit ∙ 𝑓𝑓𝑓𝑓m,d,fi = 0,58 ∙ 47,4 N/mm2 = 27,5 N/mm 2
𝜏𝜏𝜏𝜏m,d ≤3𝑓𝑓𝑓𝑓∙v,d,fi
𝑉𝑉𝑉𝑉𝑓𝑓𝑓𝑓d → OK
3 ∙ 44
29,2 kN2
N/mm
𝜏𝜏𝜏𝜏𝜆𝜆𝜆𝜆v,d
rel = 1,7 N/mm2
= �𝜎𝜎𝜎𝜎 m,k= = �25,8 N/mm2 2==1,36 (4.41)
2 ∙ 𝐴𝐴𝐴𝐴
m,crit 2 ∙ 26488 mm
V_d = E_(d,ULS,fi)∙L/2=
𝑘𝑘𝑘𝑘mod,fi ∙ 𝑘𝑘𝑘𝑘fi ≤ 1,4 14,6 kN/m∙4,0m/2
1,0 = 29,2
∙ 1,1− 0,75 kN
𝑓𝑓𝑓𝑓when 0,75 < 𝜆𝜆𝜆𝜆rel,m , 𝑘𝑘𝑘𝑘crit= = 1,56 ∙N𝜆𝜆𝜆𝜆rel,m = 1,56 − 0,75
2 ∙ 1,36 = 0,58
v,d,fi =
τ_(v,d)=〖3∙V〗_d/(2∙A)=(3∙29,2 ∙ 𝑓𝑓𝑓𝑓v,0,edge,k kN)/(2 ∙ 4,2 mm^2 = )=1,7
4,6 N/mm
𝛾𝛾𝛾𝛾M,fi 1,0 ∙26488 mm 2 N/mm^2
𝑘𝑘𝑘𝑘f_(v,d,fi)=(k_(mod,fi)∙k_fi)/γ_(M,fi)
crit ∙ 𝑓𝑓𝑓𝑓m,d,fi = 0,58 ∙ 47,4 N/mm = 27,5 N/mm
2 ∙f_(v,0,edge,k)=(1,0∙1,1)/1,0∙4,2
2 N/mm^2 =4,6 N/mm^2
𝜏𝜏𝜏𝜏τ_(m,d)≤f_(v,d,fi)
m,d ≤ 𝑓𝑓𝑓𝑓v,d,fi → OK →OK mm
𝜎𝜎𝜎𝜎𝑙𝑙𝑙𝑙support,fi = 100mm2 − 0,70 ∙ 𝑓𝑓𝑓𝑓 ∙ 30min
m,d = 19,2 N/mm ≤ 𝑘𝑘𝑘𝑘crit min m,d → OK
+ 1,0 ∙ 7mm = 72mm (4.38)
Compression perpendicular to grain
𝐹𝐹𝐹𝐹c,90,d = 𝑉𝑉𝑉𝑉d = 29,2 𝑘𝑘𝑘𝑘𝑁𝑁𝑁𝑁
When the𝐹𝐹𝐹𝐹c,90,d main beam is 𝐹𝐹𝐹𝐹supported on a wooden column which has the notional charring rate
𝑉𝑉𝑉𝑉 = 𝐸𝐸𝐸𝐸
=d,ULS,fi ∙=𝐿𝐿𝐿𝐿/2the = 14,6 c,90,dkN/m ∙ 4,0m/2 = 29,2 kN
(4.14)
β𝜎𝜎𝜎𝜎dnc,90,d
= 0,70mm/min, 𝐴𝐴𝐴𝐴ef support
𝑏𝑏𝑏𝑏∙�𝑙𝑙𝑙𝑙support,fi +15length
mm� becomes
3 ∙= 𝑉𝑉𝑉𝑉d100mm 3 ∙− 29,2 mm
kN ∙ 30min + 1,02 ∙ 7mm = 72mm
𝑙𝑙𝑙𝑙𝜏𝜏𝜏𝜏support,fi 0,70
v,d =
l_(support,fi)=100mm-0,70 = 29,2kN min = 1,7 N/mm
mm/min∙30min+1,0∙7mm=72mm
2
𝜎𝜎𝜎𝜎c,90,d = 2 ∙ 𝐴𝐴𝐴𝐴 2 ∙ 26488 mm = 4,4 N/mm2
77mm
F_(c,90,d) = V_d = 29,2 kN ∙ (72mm + 15mm)
𝐹𝐹𝐹𝐹c,90,d =𝑘𝑘𝑘𝑘 𝑉𝑉𝑉𝑉d = ∙ 29,2 𝑘𝑘𝑘𝑘𝑁𝑁𝑁𝑁
mod,fi 𝑘𝑘𝑘𝑘fi
σ_(c,90,d)=F_(c,90,d)/A_ef 1,0 ∙ 1,1 N
=F_(c,90,d)/(b∙(l_(support,fi)+15 mm)
2 )
𝑓𝑓𝑓𝑓 v,d,fi = 𝑘𝑘𝑘𝑘 ∙ 𝑓𝑓𝑓𝑓v,0,edge,k
∙ 𝑘𝑘𝑘𝑘mod,fi ∙=𝑘𝑘𝑘𝑘fi 1,0 ∙ 4,2 mm2 = 4,6 N/mm
(4.14)
𝑘𝑘𝑘𝑘 ∙ 𝑓𝑓𝑓𝑓 𝐹𝐹𝐹𝐹 𝛾𝛾𝛾𝛾
c,90,d = M,fi
c,90 𝐹𝐹𝐹𝐹 c,90,d ∙ 𝑓𝑓𝑓𝑓
𝜎𝜎𝜎𝜎c,90,d = 𝐴𝐴𝐴𝐴 = 𝑏𝑏𝑏𝑏∙�𝑙𝑙𝑙𝑙 𝛾𝛾𝛾𝛾M,fi+15 mm�
c,90 c,90,d,fi c,90,edge,k (4.14)
σ_(c,90,d)=29,2kN/(77mm∙(72mm+15mm))=4,4
ef support,fi N/mm^2
𝜏𝜏𝜏𝜏m,d ≤ 𝑓𝑓𝑓𝑓v,d,fi → OK
k_(c,90)∙f_(c,90,d,fi)=(k_(c,90)∙k_(mod,fi)∙k_fi)/γ_(M,fi)
1,0 ∙ 1,0 ∙ 1,1 ∙f_
29,2kN 2 2
𝑘𝑘𝑘𝑘c,90,d
c,90 ∙ =
𝑓𝑓𝑓𝑓c,90,d,fi =
𝜎𝜎𝜎𝜎k_(c,90)∙f_(c,90,d,fi)=(1,0∙1,0∙1,1)/1,0∙6 ∙ 6 N/mm = = 6,6 N/mm
N/mm^2=6,6
4,4 N/mm 2 N/mm^2
77mm ∙ (72mm 1,0 + 15mm)
σ_(c,90,d)≤k_(c,90)∙f_(m,0,d,fi) →OK
𝜎𝜎𝜎𝜎c,90,d ≤ 𝑘𝑘𝑘𝑘c,90 ∙ 𝑓𝑓𝑓𝑓𝑘𝑘𝑘𝑘 → OK
c,90 ∙ 𝑘𝑘𝑘𝑘mod,fi ∙ 𝑘𝑘𝑘𝑘fi
m,0,d,fi (4.13)
𝑘𝑘𝑘𝑘c,90 ∙ 𝑓𝑓𝑓𝑓c,90,d,fi = ∙ 𝑓𝑓𝑓𝑓c,90,edge,k
𝛾𝛾𝛾𝛾M,fi
mm
𝑙𝑙𝑙𝑙support,fi = 100mm 1,0 ∙ − 1,00,70 ∙ 1,1 ∙ 30min2 + 1,0 ∙ 7mm = 72mm
𝑘𝑘𝑘𝑘c,90 ∙ 𝑓𝑓𝑓𝑓c,90,d,fi = ∙ 6 N/mm = 6,6 N/mm2
min
1,0
𝐹𝐹𝐹𝐹c,90,d = 𝑉𝑉𝑉𝑉d = 29,2 𝑘𝑘𝑘𝑘𝑁𝑁𝑁𝑁
𝜎𝜎𝜎𝜎c,90,d ≤ 𝑘𝑘𝑘𝑘c,90 ∙ 𝑓𝑓𝑓𝑓m,0,d,fi → OK (4.13)
𝐹𝐹𝐹𝐹
c,90,d 𝐹𝐹𝐹𝐹
c,90,d
𝜎𝜎𝜎𝜎c,90,d = 𝐴𝐴𝐴𝐴
(4.13) = 𝑏𝑏𝑏𝑏∙�𝑙𝑙𝑙𝑙 (4.14)
ef +15 mm�
support,fi
29,2kN 2
𝜎𝜎𝜎𝜎c,90,d =
Discussion = 4,4 N/mm
77mm ∙ (72mm + 15mm)
According to EN1995-1-2:2004, clause 4.3.1 it is not necessary to verify compression perpendicular to the grain
and 𝑘𝑘𝑘𝑘
shear∙ resistance 𝑘𝑘𝑘𝑘c,90
of ∙ 𝑘𝑘𝑘𝑘mod,fi
a beam ∙ 𝑘𝑘𝑘𝑘fistructural fire design. In this example they didn’t become critical, but
in the
c,90 𝑓𝑓𝑓𝑓c,90,d,fi = ∙ 𝑓𝑓𝑓𝑓c,90,edge,k
𝛾𝛾𝛾𝛾
in the detailing it shall be verified that the beam is securely supported also when the support length becomes
M,fi
smaller due to charring of the supports.
1,0 ∙ 1,0 ∙ 1,1
𝑘𝑘𝑘𝑘c,90 ∙ 𝑓𝑓𝑓𝑓c,90,d,fi = ∙ 6 N/mm2 = 6,6 N/mm2
1,0
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14374 with regard to their reaction to fire
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Standards
• EN 301:2017, Adhesives, phenolic and aminoplastic, • EN 15425:2017, Adhesives – One component polyurethane
for load-bearing timber structures – Classification and (PUR) for load-bearing timber structures – Classification
performance requirements and performance requirements
• EN 322:1993, Wood-based panels – Determination of • EN 15804:2012 + A1:2013. Sustainability of construction
moisture content works - Environmental product declarations - Core rules
• EN 335:2013, Durability of wood and wood-based for the product category of construction products
products – Use classes: definitions, application to solid • EN 15978:2011. Sustainability of construction works -
wood and wood-based products Assessment of environmental performance of buildings -
• EN 338:2016, Structural timber. Strength classes Calculation method
• EN 350:2016, Durability of wood and wood-based • EN 16485 (2014) Round and sawn timber. Environmental
products – Testing and classification of the durability to Product Declarations. Product category rules for wood and
biological agents of wood and wood-based materials wood-based products for use in construction.
• EN 408:2012, Timber structures – Structural timber and • EN ISO 10456:2007/AC:2009, Building materials and
glued laminated timber – Determination of some physical products. Hygrothermal properties -Tabulated design
and mechanical properties values and procedures for determining declared and design
• EN 717-1: 2004, Wood-based panels. Determination of thermal values (ISO 10456:2007)
formaldehyde release. Part 1: Formaldehyde emission by • EN ISO 12460-3:2015, Wood-based panels. Determination
the chamber method of formaldehyde release. Part 3: Gas analysis method
• EN 789:2004, Timber structures. Test methods. • EN ISO 14044:2006. Environmental management – life
Determination of mechanical properties of wood based cycle assessment – Requirements and guidelines
panels • CEN/TS 1099:2007, Plywood - Biological durability -
• EN 1990:2002+A1:2005+AC:2008, Eurocode 0. Basis of Guidance for the assessment of plywood for use in different
structural design use classes
• EN 1991, Eurocode 1. Actions on structures • ISO 14040: 2006. Environmental management – life cycle
• EN 1993:2005, Eurocode 3. Design of steel structures assessment – principles and framework.
• EN 1995-1-1:2004-A1:2008+A2:2014+AC:2006, Design of
timber structures, part 1-1: General, Common rules and
ruler for buildings
• EN1995-1-2:2004+AC:2009, Design of timber structures,
part 1-2: General. Structural fire design
• EN1998-1:2004 Eurocode 8, Design of structures for
earthquake resistance. Part 1: General rules, seismic actions
and rules for buildings
• EN 12512:2001+A1:2005, Timber structures. Test
methods. Cyclic testing of joints made with mechanical
fasteners
• EN 13501-1:2019, Fire classification of construction
products and building elements. Part 1: Classification using
data from reaction to fire tests
• EN 14374:2004, Timber structures. Structural laminated
veneer lumber. Requirements
• FprEN 14374:2018, Timber structures. Laminated veneer
lumber. Requirements
References of Figures
Figures which are not listed below are prepared by Tero Lahtela, Engineering office Lahtela Oy and Jouni
Hakkarainen, Eurofins Expert Services Oy. Names of the designer and photographer are mentioned, when
the information has been available.
Cover page. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Puuviikki, Helsinki, Finland, Metsä Wood, architect: Jari Viherkoski
..................................................................photographer: Hans Koistinen
Figure 1.1. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Wood City, Helsinki, Finland, architect: Anttinen-Oiva Arkitehdit,
..................................................................photographer Tiina Nykänen
Figure 1.2. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . LVL roof rafters, Kerto LVL QP-beams, Metsä Wood,
..................................................................photographer: Hans Koistinen.
Figure 1.3. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . LVL S beams by Stora Enso; Kerto LVL Q-panel, Metsä Wood,
..................................................................photographer: Tomi Aho
Figure 1.4. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . LVL–C roof panels, Kerto LVL Q-panels, Metsä Wood, Rauma,
..................................................................Finland
Figure 1.5. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . LVL–C panels, Kerto LVL Q-panels, Metsä Wood
Figure 1.11. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Off-site produced wooden volumetric elements, BoKlok, Vantaa,
..................................................................Finland, Stora Enso, photographer: Kuvatoimisto Kuvio Oy/
..................................................................Anders Portman & Martin
Figure 1.12. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Multi-storey houses, Honkasuo, Helsinki, Finland, Puuinfo,
..................................................................photographer: Kimmo Räisänen
Figure 1.13. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . LVL-P beam and LVL–C panels, Kerto LVL S and Kerto LVL Q,
..................................................................Metsä Wood
Figure 1.17. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Life cycle approach of building product assessment,
..................................................................design: Sirpa Outinen, Mainostoimisto Queens Oy and
..................................................................Anu Huovinen, Metsä Wood
Figures 1.21.-1.23. and 1.25.-1.45. . . . . . . LVL manufacturing process, Raute Oy
Figure 1.24. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Timber harvesting, Stora Enso, photographer: Lotta Forssell
Figure 1.46. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . LVL veneer surfaces, Kerto LVL, Metsä Wood
Figure 1.48. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Holes in LVL panels, LVL by Stora Enso,
..................................................................photographer: Maria Mattelmäki
Figure 1.49. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Kerto LVL S-beams with special shape, Metsä Wood,
..................................................................photographer: Hans Koistinen
Figure 1.50. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Hole in an LVL beam, LVL by Stora Enso,
..................................................................photographer: Per Kristiansen
Figure 1.52. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Kerto LVL Q-panels with edge profiling, Metsä Wood
Figure 1.53. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Moisture protection treatment of LVL-P beams, Kerto LVL,
..................................................................WeatherGuard® Metsä Wood, photographer: Hans Koistinen
Figure 1.54. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Moisture protection treatment of LVL-P beams; LVL by Stora Enso,
..................................................................photographer: Jussi Levonen
Figure 1.57. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Manufacture of structurally glued LVL roof elements, Kerto Ripa®,
..................................................................Metsä Wood, photographer: Hans Koistinen
Figure 1.58. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . LVL-C panel, Kerto LVL Q-panel, Metsä Wood
Figure 1.59. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . LVL-P walls studs, Kerto LVL T-stud, Metsä Wood,
..................................................................photographer: Hans Koistinen.
Figures 4.25. and 4.26. . . . . . . . . . . . . . . . . . . . . . . . . . Nationale Festlegungen zur Umsetzung der ÖNORM EN 1995-1-1,
..................................................................nationale Erläuterungen und nationale Ergänzungen, Annex F,
..................................................................Austrian Standards Institute,15.6.2015
Figure 4.28. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . EN 1995-1-1:2004, Figure 7.2
Figure 4.31. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Right side: EN 1995-1-1:2004, Figure 9.1
Figure 4.33. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . EN 1995-1-1:2004, Figure 9.2
Figure 5.1. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . LVL connections, Aika Stage, Aalto University, Finland, Stora Enso,
..................................................................photographer: Vesa Loikas
Figures 5.2.-5.4. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Kerto Manual, Screwed connections, Metsä Wood, April 2013
Figure 5.5. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . EN 1995-1-1:2004, Figure 8.7
Figure 5.7. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . CEN/TC 250/SC 5/N 764, Design rules for LVL to Eurocode 5,
..................................................................Proposal for discussion in CEN/TC250/SC5, Prof. Dr.-Ing. H.J. Blaβ
..................................................................and Dr. –Ing.M.Flaig, Blaβ & Eberhart GmbH , 30.6.2017
Figure 5.8. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Modified from EN 1995-1-1:2004, Figure 8.11a
Figure 5.9. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Modified from EN 1995-1-1:2004, Figure 8.1
Figure 5.11. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . RIL 205-1-2017 Design guide line of timber structure according to
..................................................................Eurocode 1995-1-1, Finnish Association of Civil Engineers, 2017,
..................................................................Figure 8.3
Figure 6.1. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Charring of wood, Fire safety in timber buildings - Technical guideline
..................................................................for Europe, 2010, SP Trätek
Figure 6.2. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Schaffer, FPL, US 1967 and Brandsäkra trähus 3, SP Trä, Sweden 2012
..................................................................and Puun hiiltymä, Charring of wood, VTT Research reports 689,
..................................................................Esko Mikkola, 1990
Figure 6.3. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . EN1995-1-2:2004, Figures B.4 and B.5
Figure 6.4. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . EN1995-1-2:2004, Figures 3.1 and 3.2
Figure 6.5. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . EN1995-1-2:2004, Figure 4.1
Figure 6.6. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . EN1995-1-2:2004, Figure 4.2
Figure 6.7. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . One-dimensional charring of solid timber, glued laminated timber,
..................................................................LVL and CLT, test report VTT-04746-16, VTT Expert Services Ltd,
..................................................................Finland, 25.11.2016
Figure 7.1. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Wooden pedestrian bridge, Matinpuro, Espoo, Finland, designer: WPS
..................................................................Finland Oy, Atte Mikkonen, photographer: Jouni Hakkarainen
Figure 7.3. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . WeatherGuard® treatment on Kerto LVL, Metsä Wood,
..................................................................photographer: Hans Koistinen.
Figure 7.4. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Pressure treated Kerto LVL Q-beams, Metsä Wood
Figure 8.1. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Roof overhang protects the building against weather exposure and
..................................................................solar radiation, Kindergarten Vekara, Pukkila, Finland, Puuinfo,
..................................................................architect: Klemetti & Räty, photographer: Kimmo Räisänen.
Figure 8.2. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Prepared based on the report Equilibrium moisture content of
..................................................................wood-based panels (in Finnish Puulevyjen tasapainokosteus ),
..................................................................Helsinki University of Technology (Aalto University), Laboratory of
..................................................................structural engineering and building physics, Report 60, Finland, 1997
Figure 9.1. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Wooden multi-storey building Lighthouse, Joensuu, Finland,
..................................................................Stora Enso, architect: Arcadia Oy Arkkitehtitoimisto, Samuli Sallinen,
..................................................................photographer: Tiina Tuomainen
Authors
Content (excluding Sections 1.5 and 1.6)
Jouni Hakkarainen, Eurofins Expert Services Oy
Technical drawings
Tero Lahtela, Engineering office Lahtela Oy
Jouni Hakkarainen, Eurofins Expert Services Oy