Ingenious hardwood BauBuche Beech laminated veneer lumber Manual for design and structural calculation in accordance with Eurocode 5 3rd revised edition

Hans Joachim Blass, Johannes Streib

Ingenious hardwood BauBuche Beech laminated veneer lumber Manual for design and structural calculation in accordance with Eurocode 5

Hans Joachim Blass, Johannes Streib

This manual provides clear design assistance to the user when handling the new material “Laminated veneer lumber made from beech”. Relevant principles and regulations from Eurocode 5 are presented and explained in more detail to ­facilitate the design of members made of Beech laminated veneer lumber. ­Prac­tical design examples should also make it easier for engineers to apply these regulations.­ Please note that this manual does not constitute a replacement for design codes and approvals. When using the design regulations specified in this work, com­ pliance with current calculation standards must be verified at all times. In addi­ tion, the strength and stiffness values specified here for Beech laminated veneer lumber (Board BauBuche S / Q) and glulam made from Beech laminated veneer lumber (Beam BauBuche GL75) must always be compared with the values from the current approval / ETA or or the declaration of performance of the products used. This design assistance is based on Eurocode 5 (DIN EN 1995-1:2010-12). German national regulations will subsequently be listed in the relevant sections and iden­tified by shading in grey. Moreover, they prevail in all cases over the regulations of the main part of ­Eurocode 5. The numbering of the formulas is based on the system used in Euro­ code 5 or that of the National Annex, additional formulas are not numbered. The usage examples and tabular design aids use strength and stiffness values for BauBuche in accordance with current performance declarations (PM-005-2018, PM-008-2018) or the European Technical Assessment ETA-14/0354 as of 11.07.2018. We would like to thank the office of merz kley partner ZT GmbH for making ­suggestions and reviewing the 1st edition of the manual.

Karlsruhe, September 2019 Hans Joachim Blass, Johannes Streib

CONTENTS page 6 1 Product line 1.1 Board BauBuche 1.2 Beam BauBuche GL75 7 2 Principles of calculation and construction 2.1 Load-duration classes 2.2 Service class 2.3 Modification of the material properties 2.4 Verification in accordance with the partial factor method 2.5 Summary 9 3 Material properties 3.1 Strength properties for the Board BauBuche S and Q 3.2 Strength properties for Beam BauBuche GL75 3.3 Shrinking and swelling 3.4 Specific weights to calculate the dead load 3.5 Corrosiveness 13 4 Ultimate limit state 4.1 Verifications 4.2 Stability of members 4.3 Beams with variable cross-sections 4.4 Notched members 4.5 Step joints 23 5 Serviceability limit state 5.1 General points 5.2 Deflections 5.3 Vibrations 28 6 Connections with dowel-type metallic fasteners 6.1 Load-carrying capacity of connections with laterally loaded fasteners 6.2 Nailed connections 6.3 Stapled connections 6.4 Bolted and dowelled connections 6.5 Screwed connections 43 7 Glued components 47 8 Shear walls and diaphragms 8.1 General 8.2 Shear walls 48 9 Reinforcements and rehabilitation 9.1 Reinforcements for tensile stresses perpendicular to grain 9.2 Types of reinforcement 9.3 Applications 9.4 Cross-sectional reinforcements 9.5 Reinforced connection 60 10 Structural fire design 10.1 General 10.2 Requirements 10.3 Strength values 10.4 Actions 10.5 Design method 10.6 Charring 10.7 Connections with timber side members 64 11 References

66 12 Application examples 5 1. PRODUCT LINE

This approach also allows damaged areas to be cut out. The reduced thickness of veneers makes it easier to use the beech wood in a cost-effective 1.1 Board BauBuche manner. This approach also minimises both the The process of manufacturing BauBuche as lami­ time and thus cost involved in drying the wood. nated veneer lumber panels involves beech wood In accordance with the current declaration of veneers being bonded together. In the process, ­performance, Boards BauBuche can be manufac­ the veneers can either be bonded all parallel to tured up to a length of 35 m, a width of 1.85 m the grain in the main load-bearing direction (Board and thicknesses ranging from 21 to 66 mm. The BauBuche S) or with up to 30 % cross veneers deliverable panel dimensions must be confirmed (Board BauBuche Q). The Board BauBuche Q has with the manufacturer before planning. superior dimensional stability when exposed to changing climatic conditions as well as being 1.2 Beam BauBuche GL75 better able to resist in-plane tensile stresses The manufacture of the glulam Beam BauBuche ­per­pendicular to the grain. However, the arrange­ GL75 involves gluing together at least two lamina­ ment of cross veneers means a reduction in tions made of Board BauBuche S with a thickness ­bending strength as well as compressive and of 40 mm or 50 mm. Beam BauBuche GL75 may ­t­ensile strength parallel to the grain. The Board be manufactured with widths between 50 and BauBuche S is thus suited for forming linear 300 mm, heights between 80 and 1360 mm and members, while Board BauBuche Q is primarily lengths of up to 18 m. A maximum precamber for plates (e. g. wall panels). of the Beam BuBuche GL75 of up to L/100, larger The term veneers refers to sheets of wood around dimensions (a height of up to 2500 mm, a width 3 mm thick, which are extracted in the form of up to 600 mm and a length of up to 36 m) as of ­rotary peeled sections from beech trunks. well as block glued glulam are regulated in the By ­gluing together the chamfered ends veneers ­assessment documents, but should be confirmed of virtually unlimited length can be manufactured. with the manufacturer prior to planning.

You will find the available sizes of BauBuche in the product overview: http://my.pollmeier.com/overview

Figure 1: Products made of beech LVL: 6 Board BauBuche S and Q; Beam BauBuche S; Beam BauBuche GL75 and BauBuche Panel 2 PRINCIPLES OF CALCULATION AND CONSTRUCTION

DIN EN 1995-1-1, Chap. 2

Service class 2 is defined as an ambient climate with a temperature of 20 °C and relative humidity of up to 85 %. This applies to members in build­ ings which cannot be air-conditioned (not ­enclosed on all sides) but which are protected 2.1 Load-duration classes against weathering. Under certain circumstances The strength of timber declines with increasing and based on the planned usage, it may also be duration of load. Accordingly, the loads exerted necessary to classify closed buildings in service are categorised into various load-duration classes class 2 (e. g. greenhouses). In service class 2 (KLED). The classification is based on the accu­ the average wood moisture content tends to be mulated load duration, referencing the service less than 20 %. life of the construction. A total of five load-dura­ The fact that any adjustment of the level of wood tion classes are distinguished. Loads from the moisture content in line with the ambient climate dead weight are defined as load-duration class is delayed due to the slow rate of moisture trans­ “permanent”. Variable loads are classified in stag­ fer means the relative humidity may exceed the es in the load-duration classes as “long-term” values specified above for a few weeks in a year. (e. g. stored goods), “medium-term” (e. g. live loads in living spaces), “short-term” (e. g. snow) 2.3 Modification of the material properties and “instant­a­neous” (e. g. earthquake). More 2.3.1 Strength ­examples are included in Table NA.1. In cases Depending on the load-duration class, the charac­ in which no clear allocation is possible, the clas­ teristic strengths should be adapted with the modifi­ sification should be made by consulting jointly cation factors kmod in accordance with Table 1. When with the architect and building owner. connecting members with differing time-­dependent

behaviour, for kmod the square-root of the product

2.2 Service class of the individual kmod ­values is to be used. The level of humidity has a key impact on the me­ chanical strength and creep behaviour of wood, kmod = kmod,1 · kmod,2 (2.6) which is why it has to be taken into consideration when designing wooden members. Based on the If the load exerted by the exposures of various ­expected climatic conditions to which the mem­ load-duration classes is collectively applied, ber will be exposed throughout its period of use, the impact of the shortest load-duration classes

­classification is made into one of three service should be used to determine kmod . However, there classes. The use of load-bearing members made of is always a need to check whether the load case BauBuche is only permissible in service classes 1 “permanent loads” is governing the design. and 2. There is no need to differentiate the me­ chanical strength properties of BauBuche within 2.3.2 Creep behaviour service classes 1 and 2. In the event of BauBuche In structures made of members with different being securely used in service class 1, the charac­ creep behaviour, the final values of the mean teristic compressive strength may be increased moduli of elasticity, shear or slip moduli must (cf. Chap. 3.1.2). Conversely, the higher level of be used to calculate the final deformations. For creep behaviour of BauBuche in service class 2 this purpose, the mean values are divided by the compared to service class 1 must be taken into factor (1 + kdef ). The values for the deformation account. factor kdef are taken from Table 1 depending on Service classes 1 is defined as an ambient climate the service class. with a temperature of 20 °C and relative humidity If the internal forces or moments also depend on below 65 %. Members in closed or air-conditioned the individual stiffness values (when calculating buildings are generally allocated to service class 1. in accordance with second order theory), the In service class 1 the average wood moisture con­ mean moduli of elasticity, shear or slip moduli tent tends to be less than 12 %. should be divided by the factor (1 + ψ2 · kdef ). 7 For connections between members with the same Table 2: Partial factors coefficients for the ultimate

time-dependent behaviour, kdef must be doubled. limit state, * Recommendation For connections between members with differing Permanent Variable time-dependent behaviour, k amounts to def actions actions

Unfavourable kdef = 2 · kdef,1 · kdef,2 (2.13) effect γG,sup = 1.35 γQ = 1.50 Favourable 2.4 Verification in accordance with the partial effect γ = 0.90* – factor method G,inf To verify the members and connections, the actions

FE are compared with the resistances FR. The 2.5 Summary goal is to minimise the probability of any failure, Table 3 and Table 4 provide an overview of the namely an incident where the effect of actions stiffness values at the serviceability limit state exceeds the resistances, without rendering the (SLS) and the ultimate limit state (ULS). When cost of construction unfeasible. For this purpose, performing verifications of entire systems, mean the partial factor method multiplies the actions stiffness values can be assumed, since members

FE with the partial factors γ in accordance with with lower stiffness properties within a system

Table 2 and divides the resistances FR by the are offset by their more rigid peers. th ­partial factor γM for a material property and The 5 -percentile for the stiffness of connections

­multiplies it by the modification factor kmod. may be determined by reducing the mean value

For BauBuche, the value to be used for ongoing Kmean via the ratio E0,05 / Emean .

and temporary design situation is γM = 1.2, but in the event of an accidental design situation Table 3: Stiffness values for systems

(e. g. fire) γM = 1.0 may be used. For the persistent and transient design situation, SLS ULS

γM = 1.3 can be used. t = 0 t = ∞

Note: For the following examples, γM = 1.3 is Emean Emean used in accordance with the national annex for Members Emean ――――― ―――――――― γ γ · (1 + k ) Germany. M M def

Gmean Gmean Table 1: Modification factor kmod and deformation Gmean ――――― ―――――――― γM γM · (1 + kdef) factor kdef for BauBuche

Con- 2 · Kser 2 · Kser service kmod kdef nections Kmean ――――― ――――――――――― 3 · γ 3 · γ · (1 + k ) class Class of the load duration M M def continuous long medium short very short 1 0.60 0.70 0.80 0.90 1.10 0.60 Table 4: Stiffness values for individual members 2 0.60 0.70 0.80 0.90 1.10 0.80 SLS ULS t = 0 t = ∞ FRk FRd = kmod · ――― ( 2 . 1 7 ) E E γM 0,05 0,05 Members Emean ――――― ―――――――― γM γM · (1 + kdef) Both the actions FE as well as the resistances FR

are generally distribution functions of random G0,05 G0,05 variables. The reliability can be further enhanced, Gmean ――――― ―――――――― γ γ · (1 + k ) whereby instead of the mean values of these M M def

­random variables, upper (E) and lower (R) quantile Con- 2 · Kser · E0,05 2 · Kser · E0,05 values are used for the purpose of design. For nections Kmean ―――――――― ――――――――――――――― 3 · γ · E 3 · γ · (1 + k ) · E ­resistances, the 5th-percentile is generally used. M mean M def mean

8 3 MATERIAL PROPERTIES

The strength values must be taken from the current approval / ETA or declarations of performance for BauBuche. The strength values depend on the angle between the load and fibre direction and the member geometry.

3.1 Strength properties for the Board BauBuche Table 5: Definition of the Edgewise Flatwise S and Q strength designations When designing the Board BauBuche, it is impor­ for BauBuche tant to take into account the direction of the loading and the orientation of the cross-section precisely. For this reason, e. g. the values for f ­compressive strength perpendicular to the grain m,k Bending strength parallel to the fc,90,k differ depending on whether loading of the grain direction of the top layer wide or narrow surface is involved. The following considerations are based on the “German general f construction technique permit (Allgemeine Bau­ m,90,k Bending strength perpen- artgenehmigung)” no. Z-9.1-838 as of 19.09.2018 dicular to the grain direction and the performance declaration PM-005-2018 as of the top layer of 27.07.2018.

ft,0,k 3.1.1 Bending strength Tensile strength parallel The characteristic value of the bending strength to the grain direction of the fm,k is to be reduced for members of height h top layer ­between 300 mm and 1,000 mm with the coeffi­ ft,90,edge,k cient kh. Members with h larger than 1,000 mm Tensile strength perpendicular currently must not be subject to ­bending stress. to the grain direction of the top layer in the panel plane 300 0.12 kh = ――― (3.3) h f c,0,k Compressive strength parallel 3.1.2 Compressive strength to the grain direction of the

The values for compressive strengths fc,0,k and top layer fc,90,k show a significant negative correlation with f the wood moisture content. Provided the classifi­ c,90,edge,k Compressive strength perpen­ cation of the member in service class 1 is ensured, dicular to the grain direction of the compressive strengths may, in accordance the top layer in the panel plane with the details of Tables 8 to 11, be increased with a factor of 1.2. fc,90,flat,k Compressive strength perpen­

Table 6: Coefficient kh dicular to the grain direction of Member height the top layer and perpendicular

in mm kh to the panel plane 300 1.000 400 0.966 f 500 0.941 v,k Shear strength 600 0.920 700 0.903 800 0.889

900 0.876 fvR,k 1,000 0.865 Rolling shear strength 9 3.1.3 Tensile strength Table 8: Characteristic values The characteristic value of the tensile strength for Board BauBuche S with ft,0,k parallel to the grain is based on a length a nominal thickness of 3,000 mm. For larger or smaller lengths, the of 21 to 66 mm in N/mm2 coefficientℓ k shall be used. Characteristic strength values in N/mm2 s/2 a) 3,000 Bending fm,k (64.9) – 75.0 80.0 kℓ =min ― ― ――― where s = 0.12 (3.4) ℓ Tension f (51.7) – 60.0b) – (66.0) t,0,k 1.1 ft,90,edge,k 1.5 c) Compression fc,0,k 57.5 – (69.0) c) Table 7: Member length fc,90,edge,k 11.7 – (14.0) c) Coefficientℓ k in mm kℓ fc,90,flat,k 10.0 – (12.0)

500 1.100 Shear fv,k 8.0 1,000 1.068 Stiffness values in N/mm2 2,000 1.025 Modulus of E 16,800 3,000 1.000 0,mean elasticity E 14,900 4,000 0.983 0,05 E 470 5,000 0.970 90,mean E 400 6,000 0.959 90,05 Shear G 760 850 7,000 0.950 mean modulus G 630 760 8,000 0.943 05 9,000 0.936 Density values in kg/m3

10,000 0.930 ρk 730

20,000 0.892 ρmean 800 35,000 0.863

Table 9: Characteristic values for Board BauBuche Q where nominal thickness B ≤ 24 mm *) in N/mm2

Characteristic strength values in N/mm2 ) Bending fm,0,k * 70.0 ) fm,90,k * 32.0 b) b) Tension ft,0,k (39.7) – 46.0 – (50.6) (39.7) – 46.0 – (50.6)

ft,90,edge,k 15.0 15.0 c) c) Compression fc,0,k 57.0 – (68.4) 57.0 – (68.4) c) c) c) fc,90,edge,k 40.0 – (48.0) 40.0 – (48.0) 40.0 – (48.0) c) c) fc,90,flat,k 16.0 – (19.2) 16.0 – (19.2)

Shear fv,k 7.8 7.8 3.8

Rolling shear fvR,k 3.8

Stiffness values in N/mm2

Modulus of E0,mean 11,800 11,800 elasticity E0,05 10,900 10,900

E90,edge,mean 3,500 3,500

E90,edge,05 3,200 3,200

E90,flat,mean 470 470

E90,flat,05 400 400

Shear Gmean 820 820 430 430 modulus G05 540 540 360 360

Density values in kg/m3

ρk 730

10 ρmean 800 Table 10: Characteristic values for Board BauBuche Q with nominal thickness 27 mm ≤ B ≤ 66 mm in N/mm2

Characteristic strength values in N/mm2 a) Bending fm,0,k (51.1) – 59.0 81.0 a) fm,90,k (7.8) – 9.0 20.0 b) b) Tension ft,0,k (42.3) – 49.0 – (53.9) (42.3) – 49.0 – (53.9)

ft,90,edge,k 8.0 8.0 8.0 c) c) Compression fc,0,k 62.0 – (74.4) 62.0 – (74.4) c) c) c) fc,90,edge,k 22.0 – (26.4) 22.0 – (26.4) 22.0 – (26.4) c) c) fc,90,flat,k 16.0 – (19.2) 16.0 – (19.2)

Shear fv,k 7.8 7.8 3.8

Rolling shear fvR,k 3.8

Stiffness values in N/mm2

Modulus of E0,mean 12,800 12,800 elasticity E0,05 11,800 11,800

E90,edge,mean 2,000 2,000 2,000

E90,edge,05 1,800 1,800 1,800

E90,flat,mean 470 470

E90,flat,05 400 400

Shear Gmean 820 820 430 430 modulus G05 540 540 360 360

Density values in kg/m3

ρk 730

ρmean 800

Table 11: Characteristic values for Beam *) Board BauBuche Q with a nominal thickness BauBuche GL75 in N/mm2 of B ≤ 24 mm must not be used in edgewise bending Laminations flatwise edgewise a) for 300 mm < h ≤ 1,000 mm fm,k is to be 0.12 2 reduced by kh = (300/h) Characteristic strength values in N/mm b) d) a) ft,0,k is to be multiplied by kℓ = min Bending fm,k (65.0) – 75.0 – (91.7) (64,9) – 75.0 s/2 e) {(3,000/ℓ) ; 1.1} where s = 0.12 Tension ft,0,k (52.0) – 60.0 – (73.0) c) fc,0,k, fc,90,edge,k and fc,90,flat,k in service class 1 ft,90,k 0.6 1.5 g) may be ­multiplied with the factor 1.2 Compression fc,0,k service class 1: 59.4 – (70.0) d) 0.10 f) g) fm,k is to be multiplied by kh,m = (600/h) service class 2: 49.5 – (58.4) e) 0.10 ft,0,k is to be multiplied by kh,t = (600/h) ; fc,90,k service class 1: 14.8 service class 1: 14.0 where h is the longer side length service class 2: 12.3f) service class 2: 11.7f) f) h) fc,0,k and fc,90,k may be multiplied in service Shear fv,k 4.5 – (5.8) 8.0 class 1 by the factor 1.2 Stiffness values in N/mm2 g) fc,0,k may be multiplied by kc.0 = min Modulus of E0,mean 16,800 (0.0009 · h + 0.892; 1.18), if at least 4 elasticity E0,05 15,300 laminations are glued together E 470 h) 0.13 90,mean fv,k is to be multiplied by kh,v = (600/h) E90,05 400

Shear Gmean 850 760

modulus G05 760 630

Density values in kg/m3

ρk 730

ρmean 800 3.2 Strength properties for Beam BauBuche GL75 3.2.4 Compressive strength The following considerations are based on the Subject to continuous use of the product in ser­ European Technical Assessment ETA-14/0354 as vice class 1, the characteristic value of the com­ of 11.07.2018 and the declaration of performance pressive strength may be increased by 20 %. In

PM-008-2018 as of 11.07.2018. For the case ­addition, the value fc,0,k may be multiplied by the

“edgewise loading”, the basic material properties system coefficientc,0 k if at least four laminations of the Board BauBuche S in accordance with the are glued together. “German general construction technique permit

(Allgemeine Bauartgenehmigung)” no. Z-9.1-838 as kc,0 = min (0.0009 · h + 0.892 ; 1.18) of 19.09.2018 and the declaration of performance PM-005-2018 as of 27.07.2018 were assumed. where h is the member height in mm.

3.2.1 Bending strength Table 12: Coefficients The characteristic strength values may be modi­ Com­ fied in the event of bending and shear stress as Bending Tension Shear pression

well as tensile and compressive stresses parallel h in mm kh,m kh,t kh,v kc,0 to the grain, if the member height deviates from 80 1.22 1.22 1.30 1.00 600 mm. The reason for this is that the strength 120 1.17 1.17 1.23 1.00 values specified were determined on specimens 160 1.14 1.14 1.19 1.04 that were 600 mm high. For members higher than 200 1.12 1.12 1.15 1.07 600 mm, the following coefficients must be con­ 240 1.10 1.10 1.13 1.11 sidered. 280 1.08 1.08 1.10 1.14 For flatwise bending, the characteristic ­value of 320 1.06 1.06 1.09 1.18

the bending strength fm,k is to be multiplied by 360 1.05 1.05 1.07 1.18

the coefficient kh,m. 400 1.04 1.04 1.05 1.18 440 1.03 1.03 1.04 1.18 600 0.1 kh,m = ――― h = member height in mm 480 1.02 1.02 1.03 1.18 h 520 1.01 1.01 1.02 1.18 560 1.01 1.01 1.01 1.18 3.2.2 Tensile strength 600 1.00 1.00 1.00 1.18

The characteristic tensile strength value ft,0,k … … … … … parallel to the grain has to be multiplied by the co­ 1360 0.92 0.92 0.90 1.18

efficient kh,t depending on the larger side length h.

600 0.1 kh,t = ――― h = larger side length in mm 3.3 Shrinking and swelling h

When adjusting the tensile strength depending on Table 13: Degree of shrinking and swelling for BauBuche­ member length, section 3.1.3 applies analogously. Degree of shrinking/swelling in % for each 1 % 3.2.3 Shear strength change in moisture content below the fibre

The characteristic shear strength value fv,k has to ­saturation point (around 35 %)

be multiplied by the coefficient kh,v. Board S, Parallel to the grain direction 600 0.13 kh,v = ――― h = member height in mm GL75 of the top layer 0.01 h Perpendicular to the grain direction of the top layer 0.40 In the direction of the board thickness / member height 0.45 Board Q Parallel to the grain direction of the top layer 0.01 Perpendicular to the grain direction of the top layer 0.03 In the direction of the board thickness 0.45 12 The values in Table 13 describe the deformation 4 ULTIMATE LIMIT STATE behaviour respectively in the board plane or in the DIN EN 1995-1-1, Chap. 6 direction of the board thickness / member height. DIN EN 1995-1-1 /NA, NCI NA 6 The values are recommendations of the manu­ The loads for the verification at the ultimate limit state must be facturer. determined for the persistent and transient design situation. Generally, stresses triggered by climatic fluctua­ The strength values must be reduced by the partial factor γ tions must be taken into account. BauBuche is M and the modification factor k . ­delivered with a moisture content of 6 % (± 2 %) mod and has high degrees of shrinkage and swelling. To take changes in moisture content and possible 4.1 Verifications resulting damages into account, particularly dur- 4.1.1 General ing erection, an adequate protection of members The timber strength properties differ significantly and joint areas is required. Further information when load is applied parallel or perpendicular can be found in the brochures 03 “Building phys­ to the grain. When designing with BauBuche, it is ics” and 09 “Wood preservation and surface therefore important to accurately determine the treatment”. direction of the loading and the orientation of the cross-section. For this reason, e. g. the values

3.4 Specific weights to calculate the dead load for fc,90,k differ depending on whether loading of DIN EN 1991-1-1 does not specify any value for the wide or narrow surface is involved. the specific weights of laminated veneer lumber The strength values must be taken from the cur­ made from beech. We thus advise using the value rent approval / ETA or declaration of performance. from DIN 1055-1. Accordingly, the dead weight of The strength values in this case depend on the BauBuche members should be calcula­ted based orientation of the member relative to the loading on a specific weight of 8.5 kN/m3. and the member geometry.

3.5 Corrosiveness 4.1.2 Tension parallel to the grain In addition to the climatic conditions, the wood The tensile stresses must be verified using the net species used influences the risk of corrosion of cross-section. This means that weaknesses, which metallic fasteners. The key variables are the tannin may be caused e. g. by fasteners, must be taken content and the pH value of the wood. Beech into consideration. Additional moments are gener­ wood can be considered “slightly corrosive”. ated if tensile forces are introduced with an eccen­ Beech wood is more likely to cause corrosion tricity, which must be taken into consideration. than spruce, but far less likely than oak.

If using metallic fasteners, it is best to ensure a σt,0,d ≤ ft,0,d (6.1) zinc layer of minimum thickness corresponding to Table 4.1 of Eurocode 5 or equivalent corrosion Example 1: Beam BauBuche GL75 tension protection. If the approval for the selected fasten­ member er allows using a thinner zinc layer or alternative Load: F = 350 kN, k = 0.8 corrosion protection for installing in beech wood, Ed mod Dimensions: 80/120 mm, ℓ = 5 m deviations from the above reco mmendations are possible. The tensile stress is

3 350 · 10 N 2 σt,0,d = ―――――――――――― = 3 6 . 5 N / m m 120 mm · 80 mm

The design tensile strength may be increased by

the coefficient kh,t, since the member height is

below 600 mm. In addition, coefficientℓ k has to be taken into consideration, since the member length exceeds 3.0 m.

600 0.10 600 0.10 kh,t = ――― = ――― = 1.17 h 120

3,000 s/2 3,000 0.12/2 kℓ = min ― ― ――― = ― ― ――― = 0.97 = 0.97 ℓ 5,000 13

1.1 4.1.4 Compression perpendicular to the grain 0.8 2 2 ft,0,d = 1.17 · 0.97 · ――― · 60 N/mm = 41.9 N/mm Compressive forces acting on the wide or narrow 1.3 surfaces of BauBuche may be verified using an

σt,0,d 36.5 ­effective contact area. The contact area on both η = ―――― = ―――― = 0.87 ≤ 1.0 f 41.9 sides can be increased by a maximum of 30 mm t,0,d parallel to the grain, to take into consideration­ Tension at an angle the portion of adjacent fibres on the load-bearing For the Board BauBuche Q with cross layers, capacity (see Figure 2). An increase in the load- ­separate verification must be performed for stress carrying capacity perpendicular to the grain by the

at an angle to the grain direction of the top layer. factor kc,90 in accordance with DIN EN 1995-1-1, Chap. 6.1.5 is not possible for members made of

σt, ,d ≤ k · ft,0,d (NA.58) BauBuche. For Board BauBuche, it is important to dis­ where tinguish between loading on wide and narrow 1 surfaces. For loading of narrow surfaces, the k = ―――――――――――――――――――――――― ( N A . 5 9 ) strength is somewhat higher. ft,0,d ft,0,d ――― s i n 2 + ――― sin · cos + cos2 The verification of compressive stresses per­ f f t,90,d v,d pendicular to the grain reads as

σ ≤ f (6.3) 4.1.3 Compression parallel to the grain c,90,d c,90,d

σc,0,d ≤ fc,0,d (6.2)

Instability of members must be taken into ­account in accordance with Chapter 4.2.1.

ℓ 30 30 ℓ 30 30 ℓ 10 Example 2: Beam BauBuche GL75 compres-

sion member bef = ℓ + 30 bef = ℓ + 2 · 30 bef = ℓ + 30 + 10

Load: FEd = 850 kN, kmod = 0.8, Figure 2: Effective contact area; dimensions in mm service class 1 Dimensions: 160 /160 mm 4.1.5 Compression at an angle to the grain The compressive stress is For compressive stresses at an angle to the grain, stresses in both parallel and perpendicular to 3 850 · 10 N 2 σc,0,d = ―――――――――――― = 3 3 . 2 N / m m the grain directions appear. According to (6.16), 160 mm · 160 mm the compressive strength for an angle ­between The design value of the compressive strength the force and grain direction reads may be increased in service class 1 by the factor fc,0,k 1.2. In addition, an increase by the coefficient f = ――――――――――――――――――― ( 6 . 1 6 ) c, ,k 2 2 (fc,0,k/fc,90,k) · sin + cos kc,0 is also possible.

kc,0 = min (0.0009 · h + 0.892 ; 1.18) The verification of compressive stresses at an = min (0.0009 · 160 + 0.892 ; 1.18) ­angle to the grain reads as = min (1.04 ; 1.18) = 1.04

σc, ,d ≤ fc, ,d 0.8 2 2 fc,0,d = 1.2 · 1.04 · ――― · 49.5 N/mm = 38.0 N/mm 1.3 Figure 3 shows the decline in compressive

33.2 strength with increasing angle . For service η = ―――― = 0.87 ≤ 1.0 38.0 class 1, the values from Figure 3 are slightly

conser­vative, since the compressive strength

fc,0,k may be increased by the factor 1.2.

Due to the higher strength fc,90,k for loading on

narrow surfaces, the strength fc, ,k declines

more slowly here. The coefficientc,0 k was not considered in Figure 3. 14 0.10 0.10 2 600 600 fc, ,k in N/mm Wide surface BauBuche S kh,m = ――― = ――― = 1.04 h 400 60 Narrow surface BauBuche S GL24h 0.8 2 2 50 Wide surface BauBuche Q fm,d = 1.04 · ――― · 75 N/mm = 48.1 N/mm 1.3 Narrow surface BauBuche Q 40 BauBuche GL75 39.8 η = ―――― = 0.83 ≤ 1.0 48.1 30 The design value of the bending strength of 20 Board BauBuche­ S edgewise must be reduced by

the coefficient kh, since the member height is 10 ­between 300 and 1,000 mm.

300 0.12 300 0.12 0 kh = ――― = ――― = 0.97 h 400 0 ° 10 ° 20 ° 30 ° 40 ° 50 ° 60 ° 70 ° 80 ° 90 ° angle 0.8 2 2 Figure 3: fm,d = kh · ――― · 75 N/mm = 44.8 N/mm 1.3 Compressive strength fc, ,k at an angle to the grain

39.8 4.1.6 Bending η = ―――― = 0.89 ≤ 1.0 44.8 In the event of biaxial bending, the verification can be made with linear interaction of the bending stresses in accordance with equations (6.11) and 4.1.7 Shear

(6.12). The factor km = 0.7 allows the small but The shear strength of cross-sections made of highly stressed cross-sectional area for rec­ ­solid wood and glulam is significantly influenced tangular cross-sections to be positively taken into by cracks. BauBuche can be considered crack- consideration. For other cross-sectional shapes, free, which eliminates the need to reduce shear

km shall be set at 1.0. strength and the factor kcr can be used as 1.0. Shear stresses generated by individual loads near

σm,y,d σm,z,d supports can be discounted, due to the positive ―――― + k m · ―――― ≤ 1 (6.11) f f impact of simultaneously acting compressive m,y,d m,z,d stresses in a transverse direction. Loads in this

σm,y,d σm,z,d category include those within a distance h km · ―――― + ―――― ≤ 1 ( 6 . 1 2 ) f f ­(support height over centre of support) from the m,y,d m,z,d support edge.

Example 3: Comparison of bending member τ ≤ f (6.13) made of Beam BauBuche GL75 d v,d with Board BauBuche S edgewise Example 4: Shear design for

Moment: MEd = 85 kNm, kmod = 0.8 Beam BauBuche GL70

Cross-section: 80/400 mm Load: VEd = 60 kN, kmod = 0.8 Cross-section: 140/240 mm The bending stress is The shear stress is 85 · 106 N mm · 6 σ = ――――――――――――― = 3 9 . 8 N / m m 2 m,d 2 80 mm · (400 mm) Vd τd = 1 . 5 · ―――――― h · b · k cr The design value of the bending strength of Beam ­BauBuche GL75 may be increased by the 60 · 103 N = 1 . 5 · ――――――――――――――― = 2 . 6 8 N / m m 2 coefficient k , since the member height is 140 mm · 240 mm · 1.0 h,m under 600 mm. The design value of the shear strength may be in­

creased by the coefficient kh,v , since the member height is under 600 mm.

15 4.2 Stability of members 600 0.13 600 0.13 kh,v = ――― = ――― = 1.13 4.2.1 Buckling of columns h 240 Geometric and material imperfections are inevi­ table within static systems, which is why pure 0.8 2 2 fv,d = 1.13 · ――― · 4.5 N/mm = 3.12 N/mm (central) compression loads are never exerted. Im­ 1.3 perfections lead to eccentricity of the compres­ sive forces exerted relative to the system line and 2.68 η = ―――― = 0.86 ≤ 1.0 thus generate additional bending stresses. For 3.12 verifications involving determination of forces and With biaxial bending, shear stresses must be moments in accordance with first order theory, ­verified by quadratic interaction this is taken into consideration by using buckling curves that reduce the compressive strength. 2 2 τy,d τz,d ――― + ――― ≤ 1 (NA.55) When determining forces and moments in accor­ f f v,d v,d dance with second order theory or for cross-­ sections not exposed to buckling risk (compact 4.1.8 Torsion cross-sections that are continuously supported),

Torsional stresses must be verified in accor- the value for kc,y and kc,z in (6.23) and (6.24) can dance with equation (6.14). When detailing the be taken as 1.0. The term compact applies to support, torsional stress must also be taken into cross-sections with a relative slenderness ratio

consideration. λrel,y and λrel,z less than or equal to 0.3.

λ f λ f τtor,d ≤ kshape · fv,d (6.14) y c,0,k z c,0,k λrel,y = ― · ――― ; λrel,z = ― · ――― (6.21); (6.22) π E π E 0,05 0,05 For rectangular cross-sections, the torsional stresses amount to where

b Mtor,d τtor,d = 3 · 1 + 0.6 · ― · ―――― λy/z = ℓef / iy/z h h · b2

h b for rectangular The coefficient kshape can be calculated in accor­ iy = ――― ; iz = ――― 12 12 cross-sections dance with (6.15) for rectangular cross-sections

or read from Figure 4. ℓef = β · ℓ h 1 + 0.05 · ― kshape = min b (6.15) Verification of buckling 1.3

σc,0,d σm,y,d σm,z,d kshape ――――――― + ―――― + k m · ―――― ≤ 1 (6.23) k · f f f c,y c,0,d m,y,d m,z,d 1.3 σc,0,d σm,y,d σm,z,d ――――――― + k m · ―――― + ―――― ≤ 1 ( 6 . 2 4 ) k · f f f 1.2 c,z c,0,d m,y,d m,z,d where 1.1 1 kc,y = ――――――――――― ( 6 . 2 5 ) 2 2 1.0 ky + ky - λrel,y 1 2 3 4 5 6 7 8 9

1 ratio h/b kc,z = ――――――――――― ( 6 . 2 6 ) Figure 4: 2 2 kz + kz - λrel,z Coefficient k for rectangular cross-sections shape where

2 When the loading involves a combination of shear ky = 0.5 · (1 + βc · (λrel,y - 0.3) + λrel,y) (6.27) and torsion, the following condition must be met: k = 0.5 · (1 + β · (λ - 0.3) + λ2 ) (6.28) 2 2 z c rel,z rel,z τtor,d τy,d τz,d ―――――――― + ――― + ――― ≤ 1 (NA.56) kshape · fv,d fv,d fv,d where βc = 0.1 for glulam and laminated veneer lumber in accordance with (6.29). 16 Case 1 Case 2 Case 3 Example 5: Columns made of BauBuche GL75

Load: FEd = 50 kN, kmod = 0.9, Service class 2 N N N Ki Ki Ki Dimensions: 100/120 mm, ℓ = 4 m

The compressive stress is

ℓ β = ℓ β = 0.699ℓ 3 50 · 10 N 2 σc,0,d = ―――――――――――― = 4 . 1 7 N / m m 100 mm · 120 mm

where

4.00 m λz = ―――――――――― = 1 3 9 β = 2ℓ Case 4 0.10 m / 12

which can be read from Table 14 for kc,z at N Ki round 0.152. The stability verification is covered by

2 Figure 5: Buckling σc,0,d 4.17 N/mm β = ℓ/2 η = ―――――――― = ――――――――――――― = 0 . 8 0 ≤ 1 length coefficient β k · f · k 0.152 · 34.3 N/mm2 · 1,0 c,z c,0,d c,0 (Euler)

Table 14: Coefficientc,y/z k depending on λy/z for Board BauBuche and Beam BauBuche GL75

kc,y/z kc,y/z

λy/z Service class 1 Service class 2 λy/z Service class 1 Service class 2 GL75 Board S Board Q* GL75 Board S Board Q* GL75 Board S Board Q* GL75 Board S Board Q* 15 1.000 0.997 0.991 1.000 1.000 0.995 125 0.156 0.131 0.097 0.186 0.157 0.116 20 0.989 0.984 0.973 0.993 0.989 0.980 130 0.145 0.122 0.090 0.173 0.145 0.108 25 0.975 0.967 0.949 0.981 0.975 0.961 135 0.134 0.113 0.083 0.161 0.135 0.100 30 0.957 0.945 0.912 0.967 0.958 0.934 140 0.125 0.105 0.078 0.150 0.126 0.093 35 0.933 0.912 0.851 0.949 0.934 0.892 145 0.117 0.098 0.072 0.140 0.117 0.087 40 0.898 0.862 0.759 0.924 0.899 0.826 150 0.109 0.092 0.068 0.131 0.110 0.081 45 0.846 0.788 0.652 0.889 0.848 0.736 155 0.102 0.086 0.064 0.123 0.103 0.076 50 0.775 0.698 0.552 0.838 0.777 0.638 160 0.096 0.081 0.060 0.115 0.097 0.071 55 0.691 0.607 0.468 0.772 0.694 0.548 165 0.091 0.076 0.056 0.108 0.091 0.067 60 0.608 0.526 0.400 0.695 0.611 0.471 170 0.085 0.072 0.053 0.102 0.086 0.063 65 0.534 0.457 0.344 0.619 0.536 0.408 175 0.081 0.068 0.050 0.097 0.081 0.060 70 0.469 0.400 0.299 0.549 0.471 0.356 180 0.076 0.064 0.047 0.091 0.077 0.057 75 0.414 0.352 0.262 0.488 0.416 0.312 185 0.072 0.061 0.045 0.087 0.073 0.054 80 0.368 0.311 0.232 0.435 0.370 0.276 190 0.069 0.058 0.042 0.082 0.069 0.051 85 0.328 0.278 0.206 0.389 0.330 0.246 195 0.065 0.055 0.040 0.078 0.066 0.048 90 0.294 0.249 0.185 0.350 0.296 0.220 200 0.062 0.052 0.038 0.074 0.062 0.046 95 0.266 0.224 0.166 0.316 0.267 0.198 205 0.059 0.050 0.037 0.071 0.059 0.044 100 0.241 0.203 0.150 0.287 0.242 0.180 210 0.056 0.047 0.035 0.067 0.057 0.042 105 0.219 0.185 0.137 0.261 0.220 0.163 215 0.054 0.045 0.033 0.064 0.054 0.040 110 0.200 0.169 0.125 0.239 0.201 0.149 220 0.051 0.043 0.032 0.062 0.052 0.038 115 0.184 0.155 0.114 0.219 0.185 0.137 225 0.049 0.041 0.030 0.059 0.049 0.036 120 0.169 0.142 0.105 0.202 0.170 0.126 230 0.047 0.040 0.029 0.056 0.047 0.035

* Values apply for Board BauBuche Q with a nominal thickness of 27 mm < B < 66 mm as well as for Board BauBuche Q with a nominal thickness B < 24 mm 17 4.2.2 Lateral torsional buckling plied at the tensile edge, ℓef may be reduced by Similar to the case of buckling of compression 0.5 h. The prerequisite in each case is ensuring members, slender beams under vertical loading sufficient torsional support (torsional restraints). are prone to a lateral buckling of the compressed edge and thus torsion of the cross-section. Table 15: Effective length for members prone to Using buckling curves, the bending stresses are lateral torsional buckling compared with a reduced bending strength

­depending on the material and geometry of the Type Load ℓef/ℓ beam. Single-span Constant bending moment 1.0 The relative slenderness ratio for bending is beams Uniformly distributed load 0.9 Single load in the centre ℓef · Wy fm,k λrel,m = ―――――― · ―――――――――― ( 6 . 3 0 ) ; ( 6 . 3 1 ) of the span 0.8 Iz · Itor π E0,05 · G0,05 Cantilever Uniformly distributed load 0.5 beams Single load at the free For beams made of BauBuche GL75 the product cantilever end 0.8 of the 5 %-quantile of the stiffness variables

E0,05 · G0,05 may be multiplied by the factor 1.2. The coefficient for reduction of the bending When calculating the relative slenderness ratio, strength to take into consideration additional subdividing into a geometric and material coeffi­ stresses due to lateral buckling is thus cient is possible. The material coefficient 1.0 λ ≤ 0.75 rel,m kcrit = 1.56 - 0.75 · λrel,m ; 0.75 < λrel,m ≤ 1.4 (6.34) 2 κm = fm,k / (π · E0,05 · G0,05) thus amounts to e. g. 1 / λrel,m 1.4 < λrel,m

75 / (π · 15,300 · 760 · 1.2) =0.08. For bending stress only, the following verification must be met The geometric coefficient may, depending on the

ratio h to b, be read from Figure 6. Accordingly, σm,d ≤ kcrit · fm,d (6.33) the expression can be simplified to With combined bending and compressive stress,

λrel,m = ℓef · κm · κg (ℓef in mm) the result is

Geometric 2 σm,d σc,0,d ――――――― + ――――――― ≤ 1 ( 6 . 3 5 ) coefficient κg kcrit · fm,d kc,z · fc,0,d 0.44 h/b 0.40 10 0.36 8 Example 6: Lateral torsional buckling design 0.32 6 of Beam BauBuche GL75 0.28 4 Moment: M = 156 kNm, k = 0.9, 0.24 3 Ed mod 0.20 2 service class 1 0.16 Dimensions: 140/560 mm, ℓ = 10 m 0.12 0.08 The bending stress is 50 100 150 200 250 300 6 156 · 10 Nmm · 6 2 Member width b in mm σm,d = ―――――――――――――― = 2 1 . 3 N / m m 140 mm · (560 mm)2

Figure 6: Geometric coefficient κg for varying ratios h/b

depending on member width b The geometric coefficientg κ can be read in accordance­ with Figure 6 at around 0.175. The effective lengths of beams with load intro­ ­Accordingly, the related degree of slenderness duction in the centre of gravity are calculated in 3 this case in accordance with Table 15 from the λrel,m = 0.9 · 10 · 10 · 0.175 · 0.08 = 1.33 beam length or the distance of the bracing ele­ ments. For loads applied at the compressed edge, and

18 ℓef has to be increased by 2 h, and for loads ap­ kcrit = 1.56 - 0.75 · λrel,m = 0.56 for 0.75 < λrel,m ≤ 1.4 1 km, = ――――――――――――――――――――――――― ( 6 . 4 0 )

fm,d 2 fm,d 2 The stability verification is covered by 1 + ― ―――――― t a n + ― ――― ― t a n 2 1.5 · f f v,d c,90,d

σm,d η = ――――――――――――――― and of the Board BauBuche Q

fm,k kcrit · kmod · kh,m · ― ―― 1 γM km, = ―――――――――――――――――――――――――

fm,d 2 fm,d 2 21.3 1 + ― ――――― t a n + ― ――― t a n 2

= ――――――――――――――― = 0 . 7 8 ≤ 1 . 0 fv,d fc,90,d

75 0.56 · 0.9 · 1.01 · ――― 1.3 The taper angle is to be limited to 24°. For biaxial bending and cross-sectional ratios of The location of the governing cross-section x h/b ≤ 4, verification is performed as follows σ,max for a beam with a uniformly distributed load is at 2 σc,0,d σm,y,d σm,z,d ――――――― + ――――――― + ―――― ≤ 1 (NA.60) ℓ k · f k · f f xσ,max = ――――――――― c,y c,0,d crit m,y,d m,z,d 1 + h / h ap s 2 σc,0,d σm,y,d σm,z,d ――――――― + ――――――― + ―――― ≤ 1 (NA.61) where hap indicates the maximum member height kc,z · fc,0,d kcrit · fm,y,d fm,z,d and hs the minimum member height.

4.3 Beams with variable cross-sections Tension BauBuche S 4.3.1 Single tapered beams Compression BauBuche S For single tapered beams made of Board Bau­ Tension BauBuche Q Buche (edgewise) with tapered edge, the bending Compression BauBuche Q km, stresses are verified at point σ,maxx where the 1.00 stress peaks. 0.90 Additional shear stresses and stresses perpendic­ 0.80 ular to the grain are generated along the tapered 0.70 Compression edge. This stress interaction is taken into consid­ 0.60 eration by reducing the bending strength fm,k by 0.50 the factor km, . In the process, a distinction is 0.40 made between tensile and compressive stresses Tension 0.30 at the tapered edge. 0.20 0.10

σm, ,d ≤ km, · fm,d (6.38) 0.00 0 5 10 15 20 25 where km, denotes tensile stresses at the tapered Inclinazione in (º) angle in [°] edge of the Board BauBuche S

Figure 7: Coefficient km, for tapered edge subject to 1 km, = ――――――――――――――――――――――――― ( 6 . 3 9 ) tensile and compressive stresses of beams Board

fm,d 2 fm,d 2 BauBuche (edgewise) S and Q (B > 27 mm). The bending 1 + ― ―――――― t a n + ― ――― ― t a n 2 0.75 · f f strength was reduced acc. to (3.3) for a beam with v,d t,90,d height h = 1000 mm. and of the Board BauBuche Q 4.3.2 Double-tapered beams 1 km, = ――――――――――――――――――――――――― The member halves of double-tapered beams can

fm,d 2 fm,d 2 be considered as single tapered beams and verified 1 + ― ――――― t a n + ― ――― t a n 2

f f with regard to stress interaction in accordance v,d t,90,d with Chapter 4.3.1. or compressive stresses at the tapered edge of The variable beam height results in a non-linear the Board BauBuche S distribution of bending stress. The bending stress for the verification in the apex area is thus determined

with the coefficientℓ k in accordance with (6.43). 19 6 · Map,d σ = k · ――――――― ≤ f (6.41); (6.42) m,d ℓ 2 m,d b · hap hef = h h where ε 1 i 2 kℓ = k1 = 1 + 1.4 · tan ap + 5.4 · tan ap (6.43); (6.44)

X ℓA where hap indicates the member apex height and

ap the taper angle of the beam in the apex area. Figure 8: Notch with tensile / shear crack The kink of the member axis in the apex generates deviation forces, which lead to tensile stresses For beams with a notch on the opposite side of

­perpendicular to the grain. The following condition the support, kv = 1.0 may be assumed. The follow­ must be met ing applies for a beam notched on the support side

σt,90,d ≤ kdis · kvol · ft,90,d (6.50) 1.0 1.1 · i1.5 where kn 1 + ――――― h kv = min (6.62) kdis = 1.4 (6.52) x 1 h (1 - ) + 0.8 ― ― - 2 h

0.01 m3 0.2 kvol = ―――――― where V

i Incline of the notch 0.01 m3 0.2 = ―――――――――――――――――― (6.51) 2 h · b · (1 - 0.25 · tan ) ℓA ap ap i = c o t ε = ――――― h - h ef The maximum tensile stress perpendicular to the (i = 0 for right angle notches) grain due to bending stress is

kn = 4.5 for laminated veneer lumber (6.63)

6 · Map,d σ = k · ―――――― ( 6 . 5 4 ) t,90,d p 2 b · h hef Height h of the beam, reduced ap = ―― h height hef at the notched support where x Distance between notch corner and

kp = 0.2 · tan ap (6.56) support reaction

4.4 Notched members For beams with notches on the opposite side of

The main purpose of notches at supports is to the support, kv = 1. If x < hef, kv may be deter­ ­reduce the height of members. High tensile mined in accordance with equation (NA.62)

stresses perpendicular to the grain and shear h (h - hef) · x stresses are generated, which can lead to cracks kv = ――― · 1 - ―――――――― (NA.62) h h · h originating from the notch corner. The accelerat­ ef ef ed change of humidity via the end grain areas also Example 7: Comparison of rectangular notch exacerbates the cracking risk. in Beam BauBuche GL75 and The use of Board BauBuche Q (edgewise) ge­nerally ­Board BauBuche Q avoids the risk of cracks, since the transverse

­layers act as reinforcement and help ­transfer Load: governing shear force Vd = 6.0 kN,

­tensile forces perpendicular to the grain. kmod = 0.8 Dimensions: beam 50 /200 mm

1.5 · Vd height at support hef = 120 mm τd = ――――― ≤ k v · fv,d (6.60) b · h distance to the notch x = 75 mm ef

20 Beam BauBuche GL75: Note: The entire beam width is applied as the

Shear stress at the notch thickness of the reinforcement panel tr. For this purpose, the perpendicular to grain tensile 1.5 · V 1.5 · 6.0 · 103 N d 2 τd = ――――― = ―――――――――――― = 1 . 5 0 N / m m strength is assumed as ft,90,d. b · h 50 mm · 120 mm ef The reinforcing effect of the transverse layers means the shear stresses can be verified without Reduction coefficient kv the need to consider kv

1.0 1.50 4.5 η = ―――― = 0.31 ≤ 1.0 kv = min ――――――――――――――――――――――――――― 4.80 75 1 200 0.6 · (1 - 0.6) + 0.8 - 0.62 200 0.6 with the design value of edgewise shear strength

0.8 = 0.382 2 2 fv,d = ――― · 7.8 N/mm = 4.8 N/mm 1.3 where The transverse layers allow the load-carrying capa­ kmod fv,d = kv · ―――― · k h,v · fv,k city of the supports to be significantly increased. γ M

0.8 4.5 Step joints = 0.382 · ――― · 1.15 · 4.5 N/mm2 = 1.22 N/mm2 1.3 Step joints are “carpentry” connections to join

­inclined members, e. g. struts to timber chords. verification: In this case, the compressive force is introduced

1.50 by compression in the contact area and trans­ η = ――――― = 1.23 > 1.0!!! 1.22 ferred by shear stress. Classic versions employing

this approach include the single step and heel ­Board BauBuche Q: joint as well as the double step joint as a combi­ The transverse layers are considered reinforcing nation of both types. In addition, a version involv­ elements. ing multiple con­catenated heel joints (“multiple The tensile force to be transferred is calculated step joint”) was developed. The advantages of in accordance with Chapter 9.3.2. this shape include the low cutting depths, a ­central compressive force in the strut, the short F = 1.3 · V · [ 3 · (1 - )2 - 2 · (1 - )3] t,90,d d end length as well as a high connection stiffness. Multiple step joints may be designed in accor­ = 1.3 · 6.0 · [ 3 · (1 - 0.6)2 - 2 · (1 - 0.6)3] = 2.75 kN dance with Enders-Comberg and Blass (2014). (NA.77) Further information can be found in brochure 05 “Fasteners”. According to (NA.84), the transverse layers for the Note: Step joints are not covered in Eurocode 5. verification may only be taken into account within This document therefore specifies the calculation

principles of the NA-Germany. Where required, its ℓ ≤ 0.5 · (h - h ) = 0.5 · 80 = 40 mm r ef applicability outside Germany should be checked.

Accordingly, the verification of perpendicular to grain tensile stresses reads F 2 σt,d 1.38 N/mm 2.0 · ――― = 2.0 · ―――――――― = 0.56 ≤ 1.0 (NA.82) f 4.9 N/mm2 t,d β/2 where β/2 γ

3 Ft,90,d 2.75 · 10 N 2 tv1 t σt,d = ―――― = ――――――――――― = 1 . 3 8 N / m m γ/2 γ v2 t · ℓ 50 mm · 40 mm F1 r r F2 (NA.83) h

8 N/mm2 2 ft,90,d = 0 . 8 · ―――――――― = 4 . 9 N / m m 1.3 ℓv1 Figure 9: Double step joint ℓv2

21 Verification of compressive stresses in the Example 8: Double step joint for contact area: Beam BauBuche GL75

σc, ,d Load: Strut force F = 140 kN ―――― ≤ 1 ( N A . 1 6 1 ) d f k = 0.9, service class 1 c, ,d mod where Dimensions: Strut (edgewise) 120/120 mm Chord 120/200 mm Fc, ,d σc, ,d = ―――― ( N A . 1 6 2 ) Connection angle γ = 35° A Cutting depths tv,1 = 20 mm f c,0,d tv,2 = 25 mm fc, ,d = ―――――――――――――――――――――――――――――――――

fc,0,d 2 fc,0,d 2 ― ―――――― s i n 2 + ― ―――― s i n · cos + cos4 Design values for Beam BauBuche GL75: 2 · f 2 · f c,90,d v,d 0.9 2 2 (NA.163) fc,0,d = 1.2 · ――― · 49.5 N/mm = 41.1 N/mm 1.3

0.9 Note: For the Beam BauBuche GL75 (laminations 2 2 fc,90,d = ――― · 14.0 N/mm = 9.69 N/mm ­edgewise) the shear strength of f = 8.0 N/mm2 1.3 v,k can be used as the value of the basic material 0.9 2 2 2 (Board BauBuche S). When using 8.0 N/mm as fv,d = ――― · 8.0 N/mm = 5.54 N/mm 1.3 the characteristic value of the shear strength,

the ­coefficient kh,v > 1 must not be applied. In accordance with (NA.163) Accordingly, the capacity per contact area can 2 be determined fc,17,5°,d = 29.1 N/mm

f ·b · t c,γ / 2,d v,1 2 FR1,d = ――――――――― ( s i n g l e s t e p j o i n t ) fc,35°,d = 20.6 N/mm cos2 (γ/2)

The capacity of the single step area is fc,γ,d · b · tv,2 FR2,d = ――――――――― ( h e e l j o i n t ) -3 cos γ 29.1 · 120 · 20 · 10 FR1,d = ――――――――――――― = 7 6 . 8 k N cos2 (17.5°) FR1,d + FR2,d ≥ Fd

and of the heel joint area Verification of shear forces in the loaded end

-3 τd Fd · cos γ 20.6 · 120 · 25 · 10 ――― ≤ 1 where τd = ――――――― FR2,d = ――――――――――――― = 7 5 . 5 k N f b · ℓ cos 35° v,d v

Fd 140 The required loaded end lengths are thus η = ―――――――― = ―――――――― = 0 . 9 2 ≤ 1 . 0 FR1,d + FR2,d 76.8 + 75.5

FR1,d · Fd · cos γ ℓv,1 = ――――――――――――――― ( s i n g l e s t e p j o i n t ) (F + F ) · b · f R1,d R2,d v,d The required loaded end lengths amount to

FR2,d · Fd · cos γ ℓv,2 = ――――――――――――――― ( h e e l j o i n t ) 3 (FR1,d + FR2,d ) · b · fv,d 76.8 · 140 · 10 · cos 35° ℓv,1 = ――――――――――――――――― (76.8 + 75.5) · 120 · 5.54

For required loaded end lengths ℓv exceeding = 87.0 mm ≤ 8 · tv,1 = 160 mm 8 · tv, ­verification is deemed as non-compliant. The cutting depths t should meet the following v 75.5 · 140 · 103 · cos 35° conditions ℓv,2 = ――――――――――――――――― (76.8 + 75.5) · 120 · 5.54 (NA.160) h/4 für γ ≤ 50° t ≤ = 85.5 mm ≤ 8 · t = 200 mm v h/6 für γ > 60° or cutting from both sides v,2

In this example, the load-carrying capacity can

For double step joints, the cutting depth tv of the be increased by 12 % if “multiple step joints” heel joint should be selected to be larger than that are used whilst simultaneously reducing cutting of the single step joint, to ensure the end result is depth and loaded end length. Precise (CNC) 22 two separate shear surfaces in the timber chord. machining of the multiple step joint is required. 5 SERVICEABILITY LIMIT STATE

DIN EN 1995-1-1, Chap. 2.2 DIN EN 1995-1-1, Chap. 7

5.1 General points and for a cantilever beam with

In serviceability limit states, Eurocode 5 tends to 4 qk · ℓ employ “should” instead of “shall”. To guarantee uinst = ―――――――――― 8 · E · I the long-term and undisturbed use of a construc­ 0,mean tion, individual members must comply with re­ The total final deformation unet,fin is: quirements governing deformation and vibration behaviour as well as requirements in terms of the unet,fin = unet,fin,G + ∑unet,fin,Q,i - uc (NA.1) load-carrying capacity. Accordingly, deformation and vibration analyses in static calculations are where required. The limit values to be complied with should be agreed with the building owner. unet,fin,G = uinst,G · (1 + kdef)

5.2 Deformations unet,fin,Q,i = uinst,Q,i · ψ2,i · (1 + kdef)

The initial deformation uinst can be calculated

­using applicable design tables depending on uc = Camber the system and the characteristic loading. The mean value should always be used for the Eurocode 5, NA - NDP at 7.2(2), Tab. NA.13 modulus of elasticity, the shear and the slip specifies limit values for the deformations to be

­modulus (E0(90),mean, Gmean, Kmean). complied with. The creep of the wood exacerbates the defor­ mation of the member over the duration of the Table 16: Recommended limit values for deformation of load. This is taken into consideration by the bending members coefficient­ k . def u u u The initial deformation u is: inst fin net,fin inst Single-span beams ℓ / 300 ℓ / 200 ℓ / 300 u = u + ∑u inst inst,G inst,Q,i Cantilever beams ℓ / 150 ℓ / 100 ℓ / 150

The final deformation ufin is: Tab. NA. 13, row 2 applies for pre-cambered or ufin = ufin,G + ufin,Q,1 + ∑ufin,Q,i (2.2) secondary members. where Example 9: Deformation verification for a ­single-span girder u = u · (1 + k ) (2.3) fin,G inst,G def Uniformly distributed load, Beam BauBuche GL75, 120 / 240 mm ufin,Q,1 = uinst,Q,1 · (1 + ψ2,1 · kdef) (2.4) 2 Loads: Dead weight gk = 1.40 kN/m 2 ufin,Q,i = uinst,Q,i · (ψ0,i + ψ2,i · kdef) ; i > 1 (2.5) Live load (Cat. A) pk = 2.80 kN/m

kdef = 0.6; ψ2 = 0.3 ; service class 1 The initial deformation for a single-span girder Dimensions: Span ℓ = 6 m 8 4 ­with uniformly distributed load is calculated with Second moment Iy = 1.38 · 10 mm

4 of area 5 qk · ℓ uinst = ――― · ――――――― Beam spacing ℮ = 0.625 m 384 E · I 0,mean

23 Verification of elastic initial deformation: 5.3 Vibrations 5.3.1 General points

uinst = uinst,G + uinst,Q = 6.37 + 12.7 ≤ l/300 Disruptive vibrations are actually likely to occur in = 19.1 mm ≤ 20 mm one of the most popular examples of timber con­ where struction, namely lightweight floor constructions. 4 5 gk · ℓ The following section introduces two methods to uinst,G = ――― · ――――――― 384 E · I ­calculate the vibration behaviour of apartment 0,mean floors according to Blaß et al. (2005). 5 0.875 · 6,0004 Various checks may be required to verify the ser­ = ――― · ――――――――――――― = 6 . 3 7 m m 384 16,800 · 1.38 · 108 viceability of floors depending on the frequency

range. 4 5 pk · ℓ The Eigenfrequency of the floor can be deter­ uinst,Q = ――― · ――――――― 384 E · I mined in simplified form via the bending stiffness 0,mean of floor beams (without sheathing). In general, 5 1.75 · 6,0004 the bending stiffness of screed may be calculated = ――― · ――――――――――――― = 1 2 . 7 m m 384 16,800 · 1.38 · 108 assuming no composite action.

Verification of final deformation: π E · I f1 = kf · ―――― · ―――― 2 · ℓ2 m · e ufin = ufin,G + ufin,Q,1 ≤ ℓ/200 = 10.2 + 15.0 = 25.2 mm ≤ 30 mm where where m Mass under quasi-permanent load ufin,G = uinst,G · (1 + kdef) = 6.37 · (1 + 0.6) = 10.2 mm (g + ψ2 · p) in kg/m² ℓ Floor span in m u = u · (1 + ψ · k ) fin,Q inst,Q 2,1 def E · I Bending stiffness of floor beams in 2Nm

= 12.7 · (1 + 0.3 · 0.6) = 15.0 mm ℮ Beam spacing in m k Coefficient in accordance with Table 19 Verification of total final deformation: f

By considering floor beams as mechanically joint- u = u + u ≤ ℓ/300 net,fin net,fin,G net,fin,Q ed beams with an effective flange width of the = 10.2 + 6.10 = 16.3 mm ≤ 20 mm sheathing (see Chapter 9.4.2 γ-Method), improved where vibration calculations can be obtained.

unet,fin,G = uinst,G · (1 + kdef) = 6.37 · (1 + 0.6) = 10.2 mm Table 18: Damping ratio in accordance with SIA 265 and

unet,fin,Q = uinst,Q · ψ2,1 · (1 + kdef) ÖNORM B 1995-1-1:2015 + NA

= 12.7 · 0.3 · (1 + 0.6) = 6.10 mm Floor construction ξ Table 17 shows that deformations of a glulam Floors without floating screed resp. cross-section made of softwood, while having lightweight floor construction 0.01 the same dimensions and loads, exceed the Floors with floating screed 0.02 above-calculated values by around 50 %. Timber floors and mechanically laminated timber floors Table 17: Comparison of the deformations of the with floating screed 0.03 Beam BauBuche GL75 and softwood glulam beam GL24h in mm If no exact values for the modal damping ratio

uinst ufin unet,fin ­exist, the value ξ of 0.01 is recommended. BauBuche GL75 19.1 25.2 16.3 GL24h 27.7 36.5 23.6

24 5.3.2 Method in accordance with Blass et al. Since the natural frequency is below 8 Hz, (2005) ­examinations (3) and (4) are required. For floors with natural frequencies exceeding 8 Hz, the following load situations must be (3) Load case “heel impact” ­examined more closely: (1) Deflection due to single load F Initial deflection due to a vertical static single load (2) Speed due to unit impulse F (1 kN):

3 3 3 F · ℓ 1 · 10 N · (6,000 mm) Floors with natural frequencies of under 8 Hz u = ―――― = ―――――――――――――――――――――――― 48 · EI 48 · 1.68 · 104 N/mm2 · 1.38 · 108 mm4 ­require specific examinations. Examinations (3) and (4) according to Blass et al. (2005) are = 1.94 mm ­re­commended. (3) Examination of the vibration velocity for the With Table 21 b results as 80 when applying a to: load case “heel impact” u 1.94 mm (4) Acceleration; examination of resonance a ≥ ― = ―――――― = 1 . 9 4 m m / k N F 1 kN

The following section presents examinations of Accordingly, the vibration velocity v can be deter­ the vibration behaviour using two examples. mined:

55 Example 10: Vibration verification for a timber v ≈ ―――――――――――――― m · e · ℓ / 2 · γ + 50 floor (f1 < 8 Hz)

Single-span girder made of Beam BauBuche GL75 55 ≈ ――――――――――――――――――――――― = 0 . 1 1 5 m / s 120/240 mm 228 kg/m2 · 0.625 m · 6 m / 2 · 1.0 + 50

Loads: where 2 Dead weight gk = 1.40 kN/m 2 Live load (Cat. A) pk = 2.80 kN/m γ = 1.0 since single-span girder

ψ2 = 0.3 quasi-permanent The following limit value should be complied with: combination qk = 1.40 + 0.3 · 2.80 2 (f1 · ξ - 1) (5.4 · 0.01 - 1) = 2.24 kN/m vlimit = 6 · b = 6 · 80 = 0.096 m/s Input parameters: Mass m = 2.24/9.81 · 1,000 The vibration velocity is thus slightly above the = 228 kg/m2 limit value. If we consider the floor as a set of Beam span ℓ = 6 m ­mechanically connected beams, verification might Floor panel width b = 8 m be met. Beam spacing ℮ = 0.625 m 8 4 Second moment of area Iy = 1.38 · 10 mm (4) Acceleration; examination of resonance Damping ratio ξ = 0.01 (Table 18) Calculation of the dominant vertical acceleration Eigen frequency 56 1 56 1 a ≈ ―――――― · ― = ――――――――――――――― · ――― 2 π EIBeam m · b · ℓ · γ ξ 228 kg/m · 8 m · 6 m · 1.0 0.01 f1 = kf · ―――― · ―――――― 2 · ℓ2 m · e = 0.51 m/s2

10 2 -4 4 π 1.68 · 10 N/m · 1.38 · 10 m = 1 . 0 · ――――― · ――――――――――――――――――― According to Blass et al. (2005) the following limit 2 · (6 m)2 228 · 0.625 m values apply:

= 5.56 Hz < 8 Hz a < 0.1 m/s2 Well-being where a < 0.35 – 0.7 m/s2 Noticeable, but not kf = 1.0 for single-span girder unpleasant

a > 0.7 m/s2 Disruptive 25 Table 19: Coefficient kf to take into account the effect of continuous beams according to Blass et al. (2005); where

ℓ1 is the smaller span length and ℓ the longer span length

ℓ1 / ℓ 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

kf 1.00 1.09 1.15 1.20 1.24 1.27 1.30 1.3 3 1.3 8 1.42 1.56

Table 20: Coefficient depending on the span ratio to the adjacent span, according to Blass et al. (2005); where ℓ1 is the smaller span length and ℓ the longer span length

ℓ1 / ℓ 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 γ 2.00 1.40 1.15 1.05 1.00 0.969 0.951 0.934 0.927 0.918 0.912

For single-span girders, coefficients kf and γ are to be set at 1.0.

Table 21: Coefficient b according to Figure 7.2 DIN EN 1995-1-1

a 0.5 0.6 0.7 0.8 0.9 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0 3.2 3.4 3.6 3.8 4.0 b 150 144 138 132 126 120 112 104 96 88 80 77 74 71 68 65 62 59 56 53 50

Lower values of a indicate better vibration behaviour. If stringent demands are imposed on the member behaviour (e. g. public buildings), a should not exceed the value of 1.0.

Example 11: Vibration verification for a where ­timber floor (> 8 Hz) kf = 1.224 from Table 19 (interpolated) for ℓ1 / ℓ = 4 / Two-span beam made of Beam BauBuche GL75 6.25 = 0.64 160/280 mm For frequencies above 8 Hz, the aforementioned Loads: 2 ­requirements (1) and (2) must be met: Dead load gk = 1.50 kN/m 2 (1) Deflection due to single load F = 1 kN Live load (Cat. A) pk = 2.80 kN/m 3 ψ2 = 0.3 F · ℓ u = ―――― Quasi-continuous q = 1.50 + 0.3 · 2.80 48 · EI k combination = 2.34 kN/m2 3 3 1 · 10 N · (6,250 mm) Characteristic values: = ―――――――――――――――――――――――― 48 · 1.68 · 104 N/mm2 · 2.93 · 108 mm4 Mass m = 2.34/9.81 · 1,000 = 239 kg/m2 Long span ℓ = 6.25 m = 1,03 mm

Short span ℓ1 = 4 m Floor panel width B = 9 m The deflection is within the range between 0.5 and Beam spacing ℮ = 0.625 m 4.0 mm. This means requirement (1) has been met. 8 4 Second moment of area Iy = 2.93 · 10 mm (2) Unit impulse velocity Damping ratio ξ = 0.01 (Table 18) 1 v ≈ ―――――――――――――― m · e · ℓ / 2 · γ + 50 Natural frequency taking into account the effect of continuous beam 1 ≈ ――――――――――――――――――――――――― = 0.0019m/s 2 π EIBeam 239 kg/m · 0.625 m · 6.25 m / 2 · 1.02 + 50 f1 = kf · ―――― · ―――――― 2 · ℓ2 m · e where

π 1.68 · 1010 N/m2 · 2.93 · 10-4 m4 γ = 1.02 from Table 20 (interpolated) = 1 . 2 2 4 · ―――――――― · ――――――――――――――――――― 2 · (6.25 m)2 239 · 0.625 m u 1.04 mm a ≥ ―― = ―――――― = 1 . 0 4 m m / k N F 1 kN

= 8.93 Hz > 8 Hz using Table 21 a value for b = 120 is calculated

(f1 · ξ - 1) (8.93 · 0,01 - 1) 26 vlimit = b = 120 = 0.013 m/s > v 5.3.3 Method in accordance with ÖNORM B 22. Separate investigations are required for light­ 1995-1-1:2015 + NA weight floors with masses below 50 kg/m2 or with The following regulations are applicable to floors, a particular usage. which can be classified in accordance with Table

Table 22: Typical usage and structural requirements of floor classes in b Grit fill with a mass of at least 60 kg/m3 is called accordance with ÖNORM B 1995-1-1:2015, Tab NA.7.2.-E1 heavy grit fill.

Floor class 1 Floor class 2 Floor class 3 Typical usage – Floors between different – Floors within a unit – Floors under unused units of use (also continu- of use, rooms or undevel- ous floors) – Floors in normal oped attic spaces, – Apartment partitioning floors single-family homes – No requirements in multiple family units, in terms of vibration – Office floors with PCs or behaviour meeting rooms – Short-span corridors Structural requirements Floor made With (heavy) screed Floating with heavy grit fillb Floating – of timber (also without grit fill) beams Dry floor (e.g. with Separate verification Floating with heavy – gypsum fibre boards) required grit fillb Floor made With (heavy) screed Floating with heavy or light- Floating – of massive weight grit fillb (also without grit fill) timber Dry floor (e.g. with Floating with heavy grit fillb Floating with heavy – panels gypsum fibre boards) grit fillb

Tabella 23: Limit values for the frequency and stiffness criteria and vibration acceleration in accordance with ÖNORM B 1995-1-1:2015, Tab NA.7.2.-E4 and Tab NA.7.2.-E6

Floor class 1 Floor class 2 Floor class 3

Limit value for frequency criterion f1 ≥ flimit = 8 Hz f1 ≥ flimit = 6 Hz –

Limit value for stiffness criterion wstat ≤ wlimit = 0.25 mm wstat ≤ wlimit = 0.5 mm –

For floor constructions with 4.5 Hz ≤ f1 ≤ 8 Hz 4.5 Hz ≤ f1 ≤ 6 Hz – 2 2 Limit value for vibration acceleration arms ≤ alimit = 0.05 m/s arms ≤ alimit = 0.10 m/s –

The Eigenfrequency f1 should be considered for Table 24: Coefficientse,1 k to consider different support the verification of the frequency criterion, see also conditions section 5.3.1. Further information concerning Support conditions floors with transverse load distribution can be (without transverse load distribution) k taken from ÖNORM B 1995-1-1:2015. e,1 pinned – pinned 1.000 If the considered floor is not simply supported, fixed – pinned 1.562 the Eigenfrequency f may be multiplied by the 1 fixed – fixed 2.268 coefficient k in accordance with Table 24. e,1 fixed – free (cantilever) 0.356

The Eigenfrequency f1 of a two-span floor may be

multiplied by the coefficient kf in accordance with Table 19.

27 Verification 6 CONNECTIONS WITH DOWEL-TYPE A minimum value of f ≥ 4.5 Hz of the Eigen­ 1,min METALLIC FASTENERS frequency is required for floor classes 1 and 2 in accordance with Table 23. DIN EN 1995-1-1, Chap. 8 Vibrations of floor classes 1 to 3 in accordance with Table 22 are considered verified if the limit values for the frequency and stiffness criteria in accordance with Table 23 are met. The limit value for vibration acceleration must additionally be

met for floor constructions with f1,min ≤ f1 ≤ fgr. 6.1 Load-carrying capacity of connections with

The largest vertical initial deformation wstat caused laterally loaded fasteners by a single load F = 1 kN acting at the most un­ Depending on the geometry and bending resis­ favourable position and taking the effective width tance of fasteners as well as the embedding

bf into account must be calculated to verify the strength (resistance against the pushing-in of the stiffness criterion. For continuous floor systems, fastener) of the wood, the load-carrying capacity the largest floor span may be considered. of connections with laterally loaded, dowel-type

3 fasteners can be determined, assuming ductile F · ℓ wstat = ――――――――― NA.7.2-E2 failure and based on the Johansen theory. Plastic 48 · (E · I) · b ℓ F failure of both the wood under the impact of where embedding stress as well as the fasteners under

Wstat Vertical initial deformation caused by bending moments is presumed. Focusing on the F = 1 kN, in m following points during the design can help avoid F Single load F = 1 kN, in N sudden brittle failures: 2 (E · I)ℓ Bending stiffness of floor, in Nm /m – Use of slender fasteners – Use of lower strength steel grades

ℓ 4 (E · I)b – Avoiding low fastener spacing or end and ―― · ――――― bF bF = min 1.1 (E · I)ℓ edge distances

Floor width b Chapters 8.2.2 and 8.2.3 of Eurocode 5 specify The effective value of the vibration acceleration equations used to determine the characteristic of single-span and simply supported floors may load-carrying capacities of timber-to-timber be calculated as follows: ­connections and steel-to-timber connections. The failure modes taken into consideration in 0.4 · · F0 arms = ――――――――― NA.7.2-E4 this case are shown in Figure 10 and Figure 11. 2 · ξ · M*

where

arms Effective value of the vibration t1 t2 acceleration, in m/s² Fourier coefficient in dependence of the Eigenfrequency, with = ℮ - 0.4 · f1

F0 Load of a person walking on the considered floor (i. d. R.: F0 = 700 N), in N a b c d e f ξ Modal damping ratio (damping ratio acc.

to Lehr) in accordance to Table 18 t1 t2 t1 M* Modal mass in kg taking into account the

effective width fb in m ℓ M* = m · ―― · bf 2

g h j k Floor vibrations may also be verified by measuring occurring vibrations directly in the buildings. Figure 10: Failure modes for timber-to-timber connections

28 The equations for determining the load-carrying 6.1.1 Application and reductions for dowel-­ capacity of steel-to-timber connections depend type fasteners in Board BauBuche S/Q and on the thickness of the steel plate t compared Beam BauBuche GL75 to the fastener diameter d. A distinction is made Figure 13 to Figure 15 show a schematic illustra­ between thin (t < 0.5 · d) and thick (t ≥ d) steel tion of member surfaces where nails / screws or plates. Interim values may be linearly interpolated. bolts / dowels may be loaded in shear. In addition, The basis for this differentiation is the different the embedding strength must be reduced de­ support of the fastener in the steel plate. Thick pending on the installation situation and direction plates are considered as clamping supports, thin of loading. plates as a hinge. Internal steel plates, regardless The definition of the member surfaces is shown of the material thickness, are considered thick in Figure 16. plates, since no rotation of the fasteners within the plate is possible.

t1

Figure 12: Rope effect a b c d e f

t2 100 % 100 % 70 %

80 % 100 % g h j/l k m

Figure 11: Failure modes for steel-to-timber connections Figure 13: Application area and reduction factors of ­embedding strength for Board BauBuche S Fasteners with a withdrawal resistance lead to higher load-carrying capacities of connections 100 % (rope effect). For connections made of BauBuche, 100 % 70 % the rope effect can be taken into consideration only for screws and bolts, since other fasteners cannot be loaded in an axial direction. The rope effect can be described as follows: Under loading, 80 % 60 % the fastener deforms, is inclined in the area of the shear plane and is thus subject to axial load. This causes the members to be pressed against Figure 14: Application area and reduction factors of each other, while friction in the shear plane may ­embedding strength for Board Bau-Buche Q result in additional­ forces being transmitted (see Figure 12). 100 % This effect can be taken into consideration when 100 % making calculations, by increasing the lateral load-carrying capacity by a quarter of the axial load-carrying capacity. For screws, the increase due to the rope effect is limited to the load-carry­ 80 % in both ing capacity determined by the Johansen theory. directions for d ≥ 8 mm For bolts, the rope effect is limited to 25 % of the 100 % load-carrying capacity according to Johansen. for d ≥ 8 mm For failure mechanisms without any inclination of the fastener in the area of the shear plane, no rope effect applies. Figura 15. Application area and reduction factors of em­ bedding strength for Beam Bau-Buche GL75 29 wide surface The minimum timber thicknesses for single-shear connections in this case amount to

β My,Rk t1,req = 1.15 · 2 · ――― + 2 · ――――― ( N A . 1 1 0 ) 1 + β f · d end grain narrow surface h,1,k surface

1 My,Rk t2,req = 1.15 · 2 · ――― + 2 · ――――― ( N A . 1 1 1 ) 1 + β f · d Figura 16. Definition of the member surfaces h,2,k For double-shear connections, the following Simplified method for determining the load-­ ­applies for timber middle members carrying capacity of laterally loaded dowel-type

4 My,Rk fasteners t2,req = 1.15 · ―――― · ―――――― ( N A . 1 1 2 ) 1 + β f · d As an alternative to the equations in Chapters h,2,k 8.2.2 and 8.2.3, the NA-D allows a simplified where method to calculate the load-carrying capacity fh,2,k Ratio value of of connections with laterally loaded dowel-type β = ―――― f embedding strengths fasteners. Here, the load-carrying capacity for the h,1,k failure modes with two plastic hinges per shear When going below the minimum timber thick­ plane (cases f, k for timber-to-timber connections nesses, and cases e, h, m for steel-to-timber connections) t1 t2 2 · β are calculated.* Fv,Rk = min ――― ; ――― · ――― · 2 · My,Rk · fh,1,k · d t t 1 + β Accordingly, to force failure modes with two 1,req 2,req ­plastic hinges, pure embedding failure modes need to be avoided. The ratio of the timber Steel-to-timber connections thickness and dowel diameter must be large. Internal and external thick steel plates: This is achieved by defining minimum timber

thicknesses treq. Failure to comply with these Fv,Rk = 2 · 2 · My,Rk · fh,k · d (NA.115) ­minimum timber thicknesses means the load-­ carrying capacity needs to be reduced with the where

smaller ratio of t1/t1,req and t2/t2,req. The load- My,Rk carrying capacities determined using this treq = 1.15 · 4 · ――――― (NA.116) f · d ­approach are lower than the values determined h,k with the precise method. As far as compliance of External thin steel plates:

treq is concerned, there is no difference between

the precise and simplified methods. Fv,Rk = 2 · My,Rk · fh,k · d (NA.117) Note that in accordance with (NA.113), the design value of the load-carrying capacity is determined Minimum timber thicknesses for middle members

with the partial factor γM = 1.1 with fasteners in double-shear:

My,Rk * For connections with external thin steel plates, the treq = 1.15 · 2 2 · ――――― ( N A . 1 1 8 ) f · d failure mechanism with a plastic hinge (b and k) is h,k ­examined, since the hinged support in the thin plate or for all other cases: means a maximum of one plastic hinge per shear plane My,Rk can be formed. treq = 1.15 · 2 + 2 · ――――― ( N A . 1 1 9 ) f · d h,k Timber-to-timber connections If going below the minimum timber thicknesses

The characteristic value of the load-carrying FV,Rk must be reduced using ­capacity Fv,Rk per shear plane and per fastener is t1 t2 calculated as min ―――― ; ―――― t t 1,req 2,req 2 · β Fv,Rk = ―――― · 2 · My,Rk · fh,1,k · d (NA.109) 1 + β

30 6.2 Nailed connections splitting risk, meaning nef = n may be assumed. The following sections are based on regulations k given in EN 1995-1-1 and the technical assessment nef = n ef (8.17) documents of BauBuche. Deviating regulations are possible if they are explicitly described in Table 25: Coefficientef k for predrilled nail holes technical assessment documents of fasteners.

For instance, in accordance with ETA-13/0523, Nail spacing* kef round nails with a profiled shank may be inserted a1 ≥ 14 · d 1.0 without predrilling in a steel-to-timber connection a1 = 10 · d 0.85 * for interim values, and an axial capacity may be taken into account a1 = 7 · d 0.7 linear interpolation if the maximum penetration depth in BauBuche is a1 = 4 · d 0.5 is possible not exceeding 34 mm. To avoid any reduction in accordance with (8.17), 6.2.1 Laterally loaded nails the nails must be staggered by at least 1 · d 6.2.1.1 Design ­perpendicular to the grain.

The thickness t2 of the member with the nail tip is limited by the penetration depth in member 2. 6.2.1.2 Installation 3 Due to the high density (ρk > 500 kg/m ), nail Smooth-shank nails must penetrate at least 8 · d holes must be predrilled in BauBuche. Here, the into member 2, profiled nails at least 6 · d. borehole diameter should be 0.8 · d. To guarantee the full load-carrying capacity of Normally, round nails with a smooth or profiled the individual fastener, the spacing and distances shank (special or anchor nails) are used. The yield in accordance with Figure 17 and Table 26 must moment in this case is calculated as follows be complied with. The details apply both when ­attaching fasteners in the wide as well as in the M = 0.3 · f · d2.6 (8.14) y,Rk u narrow surface. is the angle between the force and the grain direction. The nails must be manufactured from wire with a minimum tensile strength of 600 N/mm2.

For predrilled nail holes, the embedding strength a4,c a3,c a2 Figure 17: does not depend on the angle between the force a4,t Definition of and grain direction a1 a3,t fastener spacing fh,k = 0.082 · (1 - 0.01 · d) · ρk (8.16) and distances

Nailed joints in the end grain surfaces of ­BauBuche are not permissible. For connections in the narrow surfaces of Board BauBuche Q, based on the current declaration of performance, the embedding strength should be reduced Table 26: Minimum spacing and distances according to by 60 %. In accordance with ETA-14/0354 for Figure 17 for nails in predrilled holes ­connections in the narrow surfaces of Beam ­BauBuche GL75, the embedding strength for Spacing and distances Minimum values 1) diameters d ≥ 8 mm is to be reduced to 80 %. Spacing a1 (parallel to the grain) (4 + | cos | ) · d 1) Nails, which penetrate the narrow surfaces of Spacing a2 (perpendicular to the grain) (3 + | sin | ) · d

Board BauBuche­ S, must have a minimum Distance a3,t (loaded end) (7 + 5 · cos ) · d

­diameter of 3.1 mm. Distance a3,c (unloaded end) 7 · d

When using nails with diameters exceeding 8 mm, Distance a4,t (loaded edge) d < 5 mm: (3 + 2 · sin ) · d the embedding strength must be calculated as for d ≥ 5 mm: (3 + 4 · sin ) · d bolts / dowels. Distance a4,c (unloaded edge) 3 · d For the load-carrying capacity parallel to grain of connections with multiple nails arranged in line Note: Board BauBuche Q may be considered as panel. Reducing a1 parallel to grain, an effective number of fasteners and a2 by the factor 0.85 is thus permissible.­ nef must be taken into account in accordance 1) with (8.17). The reason for this is the increased The minimum spacings a1 and a2 may be reduced for risk of timber splitting. For connections in the ­panel-to-timber connections by the factor 0.85 and for wide surfaces of Board BauBuche Q there is no ­steel-to-timber connections by the factor 0.7. 31 6.2.2 Axially loaded nails Table 27: Yield moment My,Rk and embedding strength

Smooth-shank nails in predrilled holes may not fh,0,k for nailed joints (where fu = 600 N/mm²) in BauBuche 3 be subject to withdrawal. with ρk = 730 kg/m d in mm 2.7 3.0 3.4 3.8 4.0 4.2 4.6 5.0 5.1 5.5 6.0 7.0 8.0

My,Rk in N mm 2,380 3,130 4,340 5,790 6,620 7,510 9,520 11,800 12,400 15,100 19,000 28,400 40,100 2 fh,k in N/mm 58.2 58.1 57.8 57.6 57.5 57.3 57.1 56.9 56.8 56.6 56.3 55.7 55.1

Table 28: Load-carrying capacity Fv,Rk per shear plane in accordance with (NA.109) and minimum wood thicknesses treq of timber-to-timber 3 connections with nails (BauBuche; nails in wide surface; ρk = 730 kg/m ) d in mm 2.7 3.0 3.4 3.8 4.0 4.2 4.6 5.0 5.1 5.5 6.0 7.0 8.0

Fv,Rk in kN 0,87 1,04 1,31 1,59 1,74 1,90 2,24 2,59 2,68 3,07 3,58 4,70 5,94 treq (single-shear) in mm 15,3 16,6 18,4 20,2 21,1 21,9 23,6 25,3 25,7 27,4 29,4 33,5 37,5 treq (double-shear) in mm 12,7 13,8 15,3 16,7 17,5 18,2 19,6 21,0 21,3 22,7 24,4 27,7 31,0

Example 12: Steel-to-timber nailed connec- Load-carrying capacity of a connection with a tion in Board BauBuche thin steel plate:

0.4 · fh,k · t1 · d (a) t1 t Fv,Rk = min 1.15 · 2 · M · f · d (b) y,Rk h,k

0.4 · 56.3 · 55 · 6 = min (8.9) 1.15 · 2 · 19,000 · 56.3 · 6

7,400 N = min = 4.1 kN 4,100 N

Load-carrying capacity of a connection with a thick steel plate:

fh,k · t1 · d (c) 4 · My,Rk fh,k · t1 · d · 2 + ――――――― - 1 (d) F f · d · t2 h,k 1 Fv,Rk = min

What we consider here is a single-shear steel- 2.3 · My,Rk · fh,k · d (e) to-timber connection. The steel plate thickness is t = 5 mm. Two rows with three nails 6 x 60 mm 56.3 · 55 · 6 per row are selected. 4 · 19,000 2 + ―――――――― - 1 According to Chapter 6.1, thick and thin steel 56.3 · 55 · 6 · 2 56.3 · 6 · 55 plates must be distinguished: = min – Thin steel plate: t < 0.5 · d = 3 mm 2.3 · 19,000 · 56.3 · 6 – Thick steel plate: t ≥ d = 6 mm

18,500 N The selected steel plate is between both limit = min 8,100 N = 5.8 kN (8.10) ­values. Linear interpolation between the design 5,800 N values for thick and thin steel plates is used to ­determine the load-carrying capacity. According to Table 27, the embedding strength Interpolating between the governing characteris­ 2 is fh,k = 56.3 N/mm and the yield moment tic values of the load-carrying capacity reveals

My,Rk = 19,000 Nmm. The thickness of the timber

member is 80 mm, kmod = 0.8 is taken into consi­ Fv,Rk = 4.1 kN + 2 / 3 · (5.8 kN - 4.1 kN) = 5.2 kN deration. 32 and the design value of the overall load-carrying For connections in the narrow surfaces of Board capacity of the connection: BauBuche, the embedding strength in accordance with the current declaration of per­formance is 0.8 Fv,Rd = ――― · 5.2 kN · 6 = 19.2 kN to be reduced to 70 % for in-plane loads and to 1.3 80 % for loads perpendicular to the panel plane.

Note: nef = n, since a staggered nail configuration, Interim values may be linearly interpolated. the use of Board BauBuche Q or a sufficiently For connections in the narrow surfaces of Beam large spacing a1 is assumed. ­BauBuche GL75, the embedding strength is ­reduced to 80 % in ­accordance with ETA-14/0354 6.3 Stapled connections for diameters d ≥ 8 mm. Connections in BauBuche with staples are not For the load-carrying capacity parallel to grain permissible, in accordance with the declaration of of connections with multiple fasteners arranged performance. in line parallel to grain, an effective number of

fasteners­ nef must be taken into account in accor­ 6.4 Bolted and dowelled connections dance with (8.34). The reason for this is the in­ 6.4.1 Laterally loaded bolts or dowels creased risk of the timber splitting. Connections 6.4.1.1 Design in the wide surfaces of Board BauBuche­ Q are

For dowels and bolts the yield moment is not subject to any splitting risk, meaning nef = n may be assumed. M = 0.3 · f · d2.6 (8.30) y,Rk u,k a 0.9 4 1 nef = min n ; n · ―――― (8.34) 13 · d where tensile strength fu,k according to Table 29 and Table 30.

In Table 35, values for nef are specified depending The yield moments of different fastener types are on the fastener diameter and the spacing. specified inTable 33. For between load and fibre direction of the wide surface of 0° to 90°, linear interpolation ­between

Table 29: Tensile strength fu,k for bolts n and nef is possible. Inserting fully threaded screws as transverse rein­ 2 Strength class fu,k in N/mm forcement eliminates the need to reduce the num­ 4.6 400 ber of fasteners. The screws should be arranged 5.6 500 on the loaded side of the bolts / dowels and should 8.8 800 be designed for axial loading equating 30 % of the 10.9 1,000 lateral load of the bolts / dowel.

Table 30: Tensile strength fu,k for dowels Example 13: Steel-to-timber connection with internal steel plate 2 Steel grade fu,k in N/mm

S235 360 F F1 F1 1 S275 430 F1 S355 490

The embedding strength for diameters up to a2 30 mm is: F a4,c

0.082 · (1 - 0.01 · d) · ρk f = ―――――――――――――――― ( 8 . 3 1 ) ; ( 8 . 3 2 ) h, ,k 2 2 k90 · sin + cos a1 a3,t where k90 = 0.90 + 0.015 · d in accordance with (8.33) for members made of BauBuche. Input parameters Beam BauBuche GL75 The values for the embedding strength in BauBuche­ 160/200 mm with a characteristic value of the den­sity of service class 1, kmod = 0.9 730 kg/m3 can be taken from Table 34. Thickness of Fasteners inserted parallel to the grain in the end the steel plate t = 12 mm grain surface of BauBuche are not permissible. Spacing a1 = 60 mm, a2 = a4,c = 50 mm 33 Version 1: 12 dowels (S235, d = 12 mm) F1,k ≥ 0.3 · Fv,Rk = 0.3 · 13.6 kN = 4.08 kN

The dowels are inserted into the narrow surfaces The minimum penetration length ℓ of the screws of the Beam BauBuche GL75. The load direction is ef corresponds to the edge distance a = 50 mm. parallel to grain. The embedding strength accord­ 4,c According to Table 45, the withdrawal capacity is ing to Table 34 must therefore be reduced to 80 % F = 10.5 kN. The selected screws are thus suf­ according to ETA-14/0354. ax,Rk ficient. 2 2 fh,1,k = 0.8 · 52.7 N/mm = 42.2 N/mm With the simplified design method, the load- According to Table 33, the yield moment for carry­ing capacity per fastener and shear plane ­dowels (S235) with the diameter of 12 mm is can be determined as follows:

My,Rk = 69,100 Nmm F = 2 · 2 · M · f · d The load-carrying capacity per shear plane is in v,Rk y,Rk h,k accordance with (8.11) = 2 · 2 · 69,100 · 42.2 · 12 = 11.8 kN

fh,1,k · t1 · d where 4 · My,Rk fh,1,k · t1 · d · 2 + ―――――――― - 1 f · d · t2 h,1,k 1 My,Rk Fv,Rk = min treq = 1 . 1 5 · 4 · ―――――― fh,k · d 2.3 · My,Rk · fh,1,k · d

69,100 42.2 · 74 · 12 = 1.15 · 4 · ―――――― = 53.8 mm ≤ tvorh = 74 mm 42.2 · 12 4 · 69,100 2 + ―――――――― - 1 42.2 · 74 · 12 · 2 The overall load-carrying capacity for the con­ 42.2 · 12 · 74 = min nection, taking into consideration the effective

2.3 · 69,100 · 42.2 · 12 number of fasteners nef is

2.74 0.9 Fv,Rd = 2 · 12 · ――― · ――― · 11.8 kN = 159 kN 37,400 N (f) 4 1.1

= min 16,800 N (g) = 13.6 kN and for the connection reinforced against splitting 13,600 N (h)

0.9 Fv,Rd = 2 · 12 · ――― · 11.8 kN = 232 kN 1.1 The governing failure mode is two plastic hinges per shear plane (Johansen case h). Version 2: 12 bolts (M12 – 4.6) with washer 44/4 For multiple dowels parallel to grain, the overall load-carrying capacity taking into consideration According to Table 33, the yield moment for bolts

the effective number of fasteners nef must be (4.6) with the diameter 12 mm is:

­calculated. According to Table 35 nef = 2.74.

Accordingly, this leads to the overall load-carrying My,Rk = 76,700 Nmm capacity The load-carrying capacity per shear plane is in 2.74 0.9 Fv,Rd = 2 · 12 · ――― · ――― · 13.6 kN = 155 kN accordance with (8.11) 4 1.3

If a reinforcement to eliminate the splitting risk fh,1,k · t1 · d is used (fully threaded screws), nef = n can be 4 · My,Rk 2 + ―――――――― - 1 ­assumed. This means a significant increase in the · 2 fh,1,k · t1 · d f · d · t h,1,k 1 overall load-carrying capacity of the connection Fv,Rk = min 2.3 · My,Rk · fh,1,k · d 0.9 Fv,Rd = 2 · 12 · ――― · 13.6 kN = 226 kN 1.3 42.2 · 74 · 12 In both timber side members, for each dowel row, 4 · 76,700 2 + ―――――――― - 1 a fully threaded screw (d = 6 mm) is screwed in. 42.2 · 74 · 12 · 2 42.2 · 12 · 74 The screws must be designed for 30 % of the lat­ = min eral load acting on the dowels. 2.3 · 76,700 · 42.2 · 12 34 37,400 N (f) It is recommended to install at least two fasteners = min 17,000 N (g) = 14.3 kN or four shear planes per connection. Connections 14,300 N (h) with only one fastener should only be taken into account with 50 % of the load-carrying capacity. The lateral load-carrying capacity may be in­ creased due to the rope effect. The characteristic 6.4.2 Axially loaded bolts compressive capacity under the washer according For bolts, the resistance in the axial direction to Table 32 is 57.9 kN. The design value of (withdrawal) is the minimum of the compressive the axial load-carrying capacity Ft,Rd is 24.3 kN capacity under the washer and the tensile capaci­

(see Table 31). For ease of calculation, this value ty ft,Rd of the bolts. Unlike spruce, the high com­ is converted to a characteristic value. For this pressive strength perpendicular to the grain of ­purpose, the value is multiplied with the ratio BauBuche means steel failure may be governing.

γM / kmod : 24.3 · 1.3 / 0.9 = 35.1 kN. The increase The compressive capacity under the washer is from the rope effect may be applied at ax,kF /4, calculated from the effective contact area and but not exceeding 25 % of Fv,Rk (governing here). three times the value of the compressive strength perpendicular to the grain f , see Table 32. 5 c,90,k Fv,Rk = ―― · 14.3 = 17.9 kN Dowels cannot transfer forces in axial direction. 4

Taking into consideration the effective number Table 31: Design value of the steel tensile capacity Ft,Rd of fasteners nef, the total load-carrying capacity of bolts in kN F of the connection can be found: v,Rd Strength class d in mm 4.6 5.6 8.8 10.9 2.74 0.9 Fv,Rd = 2 · 12 · ――― · ――― · 17.9 kN = 204 kN 12 24.3 30.3 48.6 60.7 4 1.3 16 45.2 56.5 90.4 113 20 70.6 88.2 141 176 Using a suitable reinforcement (as in the earlier 24 102 127 203 254 example), the full number of fasteners is effective:

0.9 Fv,Rd = 2 · 12 · ――― · 17.9 kN = 297 kN Table 32: Characteristic compressive capacity in kN 1.3 ­under washers depending on fc,90,k Depending on the bolt grade, the load-carrying 2 capacity of the connection, primarily due to the Bolts Washer fc,90,k in N/mm rope effect, can be increased by around 30 % d in mm Type 10.0* 12.3* 14.0 16.0* ­relative to the version with dowels. 12 44/4 41.3* 50.8* 57.9 66.1* 58/6 74.6* 91.8* 104,5 119.4* 6.4.1.2 Installation 16 56/5 66.7* 82.0* 93.3 106.7* The minimum spacing and distances in accor­ 68/6 101.3* 124.6* 141.8 162.1* dance with Table 36 must be complied with. 20 72/6 110.7* 136.2* 155.0 177.2* For tight-fitting bolts and dowels, the boreholes 80/8 139.4* 171.5* 195.1 223.0* must be drilled equivalent to the fastener dia­ 24 85/6 154.3* 189.8* 216.0 246.9* meter. For bolts, the borehole may be drilled a 105/8 242.6* 298.4* 339.6 388.2* maximum of 1 mm larger. For holes in steel plates, a tolerance of 1 mm is also permissible. External * may be multiplied by 1.2 in service class 1 steel plates must not be used with dowels.

Steel grade Diameter d in mm 6 8 10 12 16 20 24 30 Table 33: Yield

4.6 12,700 26,700 47,800 76,700 162,000 290,000 465,000 831,000 moment My,Rk 5.6 15,800 33,400 59,700 95,900 203,000 362,000 582,000 1,040,000 for dowels and 8.8 25,300 53,500 95,500 153,000 324,000 579,000 931,000 1,660,000 bolts in Nmm 10.9 31,600 66,900 119,000 192,000 405,000 724,000 1,160,000 2,080,000 S235 11,400 24,100 43,000 69,100 146,000 261,000 419,000 748,000 S275 13,600 28,700 51,400 82,500 174,000 311,000 500,000 894,000 S355 15,500 32,800 58,500 94,000 199,000 355,000 570,000 1,020,000 35 Diameter d in mm Table 34: 6 8 10 12 16 20 24 30 ­Embedding 0° 56.3 55.1 53.9 52.7 50.3 47.9 45.5 41.9 strength fh, ,k 15° 56.3 55.1 53.7 52.4 49.8 47.3 44.7 40.9 for dowels and 30° 56.3 55.1 53.2 51.6 48.6 45.6 42.7 38.5 bolts in N/mm2 45° 56.3 55.1 52.6 50.7 47.0 43.5 40.3 35.7 in the wide 60° 56.3 55.1 51.9 49.7 45.5 41.6 38.1 33.2 ­surface of 75° 56.3 55.1 51.5 49.0 44.5 40.4 36.6 31.6 BauBuche with 90° 56.3 55.1 51.3 48.8 44.1 39.9 36.1 31.0 3 ρk = 730 kg/m

Number n Spacing a1 as a multiple of the diameter d Table 35: Effec­ 5 · d 6 · d 7 · d 8 · d 10 · d 12 · d 14 · d 16 · d 18 · d 20 · d 24 · d 28 · d tive number nef 2 1.47 1.54 1.60 1.65 1.75 1.83 1.90 1.97 2.00 2.00 2.00 2.00 for multiple do­ 3 2.12 2.22 2.30 2.38 2.52 2.63 2.74 2.83 2.92 2.99 3.00 3.00 wels and bolts 4 2.74 2.87 2.98 3.08 3.26 3.41 3.55 3.67 3.78 3.88 4.00 4.00 arranged paral­ 5 3.35 3.51 3.65 3.77 3.99 4.17 4.34 4.48 4.62 4.74 4.96 5.00 lel to the grain 6 3.95 4.13 4.30 4.44 4.70 4.92 5.11 5.28 5.44 5.59 5.85 6.00 7 4.54 4.75 4.94 5.10 5.40 5.65 5.87 6.07 6.25 6.42 6.72 6.98 8 5.12 5.36 5.57 5.76 6.09 6.37 6.62 6.84 7.05 7.24 7.57 7.87 10 6.26 6.55 6.80 7.04 7.44 7.79 8.09 8.37 8.62 8.85 9.26 9.62 12 7.37 7.71 8.02 8.29 8.77 9.17 9.53 9.86 10.2 10.4 10.9 11.3 14 8.47 8.86 9.21 9.52 10.1 10.5 11.0 11.3 11.7 12.0 12.5 13.0 16 9.55 9.99 10.4 10.7 11.4 11.9 12.4 12.8 13.2 13.5 14.1 14.7

Table 36: Mini­ Spacing and distances Minimum values Minimum values mum spacing Bolts Tight-fitting bolts / dowels and distances Spacing a1 according to (parallel to the grain) 0° ≤ ≤ 360° (4 + | cos |) · d (3 + 2 · | cos |) · d

­Figure 17 for Spacing a2 bolts, tight-fit­ (perpendicular to the grain) 0° ≤ ≤ 360° 4 · d 3 · d ting bolts and Distance a3,t dowels (loaded end) -90° ≤ ≤ 90° max (7 · d ; 80 mm) max (7 · d ; 80 mm)

Distance a3,c

(unloaded end) 90° ≤ < 150° (1 + 6 · sin ) · d max [(a3,t · | sin |) ; 3 · d] 150° ≤ < 210° 4 · d 3 · d

210° ≤ ≤ 270° (1 + 6 · sin ) · d max [(a3,t · | sin |) ; 3 · d]

Distance a4,t

(loaded edge) 0° ≤ ≤ 180° max [(2 + 2 · | sin |) · d ; 3 · d] max [(2 + 2 · sin ) · d ; 3 · d]

Distance a4,c (unloaded edge) 180° ≤ ≤ 360° 3 · d 3 · d

36 Table 37: Load-carrying capacity Fv,Rk per shear plane in accordance with (NA.109) in kN and minimum timber thicknesses treq in mm 3 of bolts and dowels in timber-to-timber connections (wide surfaces; ρk = 730 kg/m ) d in mm 12 16 20 24 0° 90° 0° 90° 0° 90° 0° 90° 0° 90° 0° 90° 0° 90° 0° 90° ­ 90° ­ 0° ­ 90° ­ 0° ­ 90° ­ 0° ­ 90° ­ 0° β 1.00 1.00 0.68 1.48 1.00 1.00 0.65 1.54 1.00 1.00 0.63 1.60 1.00 1.00 0.60 1.66 Strength class 4.6

Fv,Rk 9.85 9.48 9.66 9.66 16.2 15.1 15.6 15.6 23.6 21.5 22.5 22.5 31.9 28.4 30.0 30.0 Single-shear t 42.9 45.3 55.0 60.3 67.0 76.2 42.9 45.3 1,req 43.3 45.0 55.7 59.5 68.3 74.8 43.3 45.0 t2,req 45.3 42.9 60.3 55.0 76.2 67.0 45.3 42.9 Double-shear t2,req 35.8 37.2 38.0 35.1 46.2 49.3 50.9 44.6 56.6 62.0 64.7 53.9 35.8 37.2 38.0 35.1 Strength class 8.8

Fv,Rk 13.9 13.4 13.7 13.7 22.8 21.4 22.1 22.1 33.3 30.4 31.8 31.8 45.1 40.2 42.4 42.4 Single-shear t 60.7 64.1 77.7 85.3 94.7 107.7 60.7 64.1 1,req 61.2 63.6 78.8 84.2 96.6 105.8 61.2 63.6 t2,req 64.1 60.7 85.3 77.7 107.7 94.7 64.1 60.7 Double-shear t2,req 50.7 52.7 53.7 49.7 65.3 69.7 72.0 63.1 80.0 87.6 91.5 76.3 50.7 52.7 53.7 49.7 Steel grade S235

Fv,Rk 9.34 8.99 9.16 9.16 15.3 14.4 14.8 14.8 22.3 20.4 21.3 21.3 30.2 26.9 28.4 28.4 Single-shear t 40.7 43.0 52.2 57.2 63.5 72.3 40.7 43.0 1,req 41.0 42.7 52.9 56.5 64.8 71.0 41.0 42.7 t2,req 43.0 40.7 57.2 52.2 72.3 63.5 43.0 40.7 Double-shear t2,req 34.0 35.3 36.0 33.3 43.8 46.8 48.3 42.3 53.7 58.8 61.4 51.2 34.0 35.3 36.0 33.3

d 12 16 20 24 Table 38: Load-carrying capacity

0° 90° 0° 90° 0° 90° 0° 90° Fv,Rk per shear plane in accordance

Strength class 4.6 Fv,Rk 13.9 13.4 22.8 21.4 33.3 30.4 45.1 40.2 with (NA.115) in kN and minimum

treq 50.7 52.7 65.3 69.7 80.0 87.6 95.0 106.6 timber thicknesses treq in mm of

Strength class 8.8 Fv,Rk 19.7 19.0 32.3 30.3 47.1 43.0 63.8 56.8 bolts and dowels­ in steel-to-timber

treq 71.7 74.5 92.4 98.6 113.1 123.9 134.3 150.7 ­connections with internal and

Steel grade S235 Fv,Rk 13.2 12.7 21.7 20.3 31.6 28.8 42.8 38.1 ­external thick steel plates (wide 3 treq 48.1 50.0 62.0 66.1 75.9 83.1 90.1 101.1 ­surfaces; ρk = 730 kg/m )

d 12 16 20 24 0° 90° 0° 90° 0° 90° 0° 90°

Strength class 4.6 Fv,Rk 9.85 9.48 16.15 15.13 23.5 21.5 31.9 28.4

Double-shear treq 35.8 37.2 46.2 49.3 56.6 62.0 67.1 75.4 Table 39: Load-carrying capacity

Single-shear treq 43.3 45.0 55.7 59.5 68.3 74.8 81.1 91.0 Fv,Rk per shear plane in accordance

Strength class 8.8 Fv,Rk 13.9 13.4 22.8 21.4 33.3 30.4 45.1 40.2 with (NA.117) in kN and minimum

Double-shear treq 50.7 52.7 65.3 69.7 80.0 87.6 95.0 107 timber thicknesses treq in mm of

Single-shear treq 61.2 63.6 78.8 84.2 96.6 106 115 129 bolts and dowels­ in steel-to-timber

Steel grade S235 Fv,Rk 9.34 8.99 15.3 14.4 22.3 20.4 30.2 26.9 ­connections with external thin

Double-shear treq 34.0 35.3 43.8 46.8 53.7 58.8 63.7 71.5 steel plates (wide surfaces; 3 Single-shear treq 41.0 42.7 52.9 56.5 64.8 71.0 76.9 86.3 ρk = 730 kg/m )

37 6.5 Screwed connections The values of the yield moments can be taken 6.5.1 General points from the approvals of the screws or estimated as For this chapter, the values of screws in accor­ 2.6 dance with ETA-11/0190 as of 23.07.2018 (Würth My,Rk = 0.15 · fu,k · d Assy) and ETA-12/0373 as of 03.11.2017 (Schmid RAPID Hardwood) are used for examples. In Screws made of carbon steel generally have a 2 general, all screws may be used for connections ­nominal tensile strength fu,k of 600 N/mm , in BauBuche, if their approval/ETA includes the while the strength of stainless steel screws is use in laminated veneer lumber made of beech 400 N/mm2. in accordance with EN 14374. Generally, screws in predrilled holes are used (see Table 44). Screws If the effective number of multiple screws arranged with drilling tips are not a substitute for predrilling. parallel to the grain shall equal the actual number, From a legislative point of view, the requirements screws must be arranged in a staggered configu­ given in the assessment documents concerning ration, at least 1 · d perpendicular to the grain rela­ predrilling diameters are binding. Extensive inves­ tive to each other. tigations carried out by the Technical University of Laterally loaded screwed connections in the Graz have shown that a predrilling diameter of up end grain surfaces of BauBuche are generally not to 0.8∙d does not have a significant influence on allowed. the withdrawal capacity of a self-drillling screw ETA-11/0190 considers the reduction of the em­ with a current thread shape inserted in BauBuche. bedding strength of screws inserted in the end Resulting larger predrilling diameters are already grain surfaces through the angle . For connec­ included in ETA-12/0197 as of 28.02.2019. Please tions with nails or screws in the narrow surfaces contact the company Pollmeier for further infor­ of Board BauBuche Q, the embedding strength mation. shall be reduced to 60% in accordance with the Further information concerning screws, whose current declaration of performance, see Figure 14. assessment documents allow for insertion For connections in the narrow surfaces of Beam ­without predrilling, can be found in brochure 05 BauBuche GL75, the embedding strength is “Fasteners”. ­reduced by parameter β in the equation given above. Table 40 and Table 41 give current em­ 6.5.2 Laterally loaded screws bedding strength values for screws acc. to ETA- 6.5.2.1 Design acc. to ETA-11/0190 11/0190, which are valid for predrilled and The embedding strength of Würth screws in pre­ non-predrilled BauBuche. drilled and non-predrilled members is calculated The minimum diameter of laterally loaded screws as for nails: inserted in the narrow surfaces of Board Bau­

-0,15 Buche S is 6.0 mm. 0,082 · pk · d fh,k = ――――――――――――――――――― (2,5 · cos2 + sin2 ) · k · k Table 40: Yield moment M and embedding strength ξ β y,Rk f for carbon steel screws in accordance with d Nominal diameter of the screw h, ,k ETA-11/0190 in narrow surface of BauBuche with ρk = ρk = 730 kg/m³ Characteristic density of 730 kg/m3 (β = 0° und = 90°) BauBuche Angle between screw axis and fibre direction Diameter d in mm k =(0.5 + 0.024 · d) · sin2 ε + cos2 ε ε 5 6 8 10 12 ε Angle between load and fibre direction and M k = 1.2 · cos2 β + sin2 β y,Rk β in Nmm 5,900 10,000 23,000 36,000 58,000 β Angle between screw axis and wide surface 2 ε fh,ε,k in N/mm 0° 39.2 38.1 36.5 35.3 34.4 .It is important to note that screws, which are 15° 40.2 39.1 37.3 35.9 34.9 ­inclined in the direction of loading, are mainly 30° 43.3 41.9 39.6 37.8 36.3 ­subject to axial stresses and are thus designed 45° 48.4 46.4 43.2 40.6 38.4 in accordance with Chapter 6.5.2. 60° 54.8 52.0 47.5 43.9 40.9 75° 60.7 57.1 51.2 46.6 42.8 90° 63.2 59.2 52.8 47.7 43.6

38 Table 41. Yield moment My,Rk and embedding strength d in mm tmin in mm Table 43: fh, ,k for carbon steel screws in accordance with ETA- < 8 24 Minimum member thickness

11/0190 in wide surfaces of BauBuche with ρk = 730 kg/ 8 30 tmin for screws in accordance m3 (β = 0° und = 90°) 10 40 with ETA-11/0190 12 80 Diameter d in mm 5 6 8 10 12 Table 44. Predrilling diameter for screws in BauBuche

My,Rk in Nmm 5,900 10,000 23,000 36,000 58,000 Predrilling di­ Predrilling di­ 2 ε fh,ε,k in N/mm ameter in mm, ameter in mm 0° 47.0 45.8 43.8 42.4 41.2 e.g. in accor­ in accordance 15° 48.2 46.9 44.7 43.1 41.8 d dance with with 30° 52.0 50.2 47.5 45.3 43.5 in mm ETA-11/0190 ETA-12/0197 45° 58.1 55.7 51.8 48.7 46.1 6 4.0 4.5 60° 65.8 62.4 57.0 52.6 49.0 8 6.0 6.5 75° 72.8 68.5 61.5 56.0 51.4 10 7.0 8.0 90° 75.8 71.0 63.3 57.3 52.3 12 8.0 9.0 14 9.0 11.0 See also section 6.5.1 6.5.2.2 Installation For screws inserted in predrilled BauBuche, 6.5.2 Axially loaded screws regardless of diameter, the minimum spacing and 6.5.2.1 Design distances for nailed joints (Table 26) should be For the load-carrying capacity of connections complied with. with axially loaded screws, the following points For screws inserted in non-predrilled BauBuche, must be considered: regardless of the diameter, the minimum spacing – Withdrawal and distances for nailed joints in higher density – Head pull-through timber (Tabelle 42) should be complied with. – Tensile capacity of the screw According to ETA-11/0190, the minimum member thickness in accordance with Table 43 must be For screws subject to compressive loads, the complied with. buckling of the screw should be considered rather than the tensile load-carrying capacity. Table 42: Minimum spacing and distances according to Figure 17 for screws in non-predrilled BauBuche in ac­ The withdrawal resistance of a screw in accor­ cordance with ETA-11/0190 dance with ETA-11/0190 in BauBuche is

Spacing and distances Minimum values nef · kax · fax,k · d · ℓef Fax, ,Rk = ―――――――――――――――― Spacing a k 1 β (parallel to the grain) (7 + 8 · | cos |) · d with Spacing a2 (perpendicular to the grain) 7 · d kax = 1.0 for 45° ≤ ≤ 90° respectively Distance a3,t (loaded end) (15 + 5 · cos ) · d 0.5 · kax = 0.5 + ――――― for 0° ≤ ≤ 45°; Distance a 45° 3,c (unloaded end) 15 · d 2 fax,k = 35 N/mm and Distance a4,t d < 5 mm: (loaded edge) (7 + 2 · sin ) · d k = 1.5 · cos2 β + sin2 β d ≥ 5 mm: β

(7 + 5 · sin ) · d Depending on: Distance a 4,c – Effective thread length ℓ (unloaded end) 7 · d ef Length of the threaded part in the respective member subject to withdrawal. For fully threaded screws, minus the length of the smooth parts at the screw tip and head. 39 – Outer thread diameter d For screws with a diameter of 8 mm, countersunk

d corresponds to the nominal diameter of the washers 45°with dhead = 25 mm and a thickness screws of at least 40 mm: – Angle between screw axis and grain direction F = n · 22,500 N For screws in the wide and narrow surfaces, ax, ,Rk ef corresponds to the insertion angle. In the end The tensile load-carrying capacity is calculated as grain surfaces, = 90° minus screw-in angle. follows: The current declaration of performance for F = n · f (8.40c) BauBuche limits the minimum angle to 45°, t,Rk ef tens,k

in accordance with the screw approvals / ETA, The tensile capacity ftens,k is to be taken from the smaller angles would mostly be permissible. approvals/ETA of the screws (see Table 46).

According to ETA-11/0190, Fax, ,Rd for angles between 45° and 90° need not be reduced. The partially threaded self-drilling screw RAPID – Angle β between screw axis and wide surface Hardwood of the company Schmid Schrauben

kβ = 1.0 for screws inserted perpendicular Hainfeld in accordance with ETA-12/0373 as of to the wide surface (β = 90°) 03.11.2017 was developed especially for the use

kβ = 1.5 for screws inserted perpendicular in hardwood species and has high load-carrying to the narrow surface (β = 0°) capacities. For instance, the head pull-through

Fax, ,Rd must not be reduced for angles ­capacity of a screw with a diameter of 8 mm and between 45° and 90° in accordance with a thread length of 100 mm is 32.1 kN. The RAPID ETA-11/0190. Hardwood screw has a high tensile capacity

– Effective number of screws nef ftens,k = 32.8 kN and a high yield moment My,k = For connections involving interaction among 42.8 Nm thanks to the increased core diameter.

multiple screws, nef is assumed. In particular

for connections involving steel plates, nef Table 45: Withdrawal capacity Fax,Rk in kN per 10 mm should be used, since a very direct transfer penetration depth of screws for insertion angles of load to the individual screws takes place, between 45° and 90° in accordance with ETA-11/0190 which prevents uniform load distribution among all fasteners. Diameter d in mm

nef is in accordance with Eurocode 5 β 5 6 8 10 12 0° (in narrow n = n0.9 (8.41) ef surface) 1.17 1.40 1.87 2.33 2.80 15° 1.19 1.43 1.91 2.39 2.86 For tensile forces, which act at an angle of 30° 1.27 1.53 2.04 2.55 3.05 ­between 30° and 60° to the screw axis, in 45° 1.40 1.68 2.24 2.80 3.36 ­accordance with ETA-11/0190 60° 1.56 1.87 2.49 3.11 3.73 0.9 nef = max n ; 0.9 · n 75° 1.69 2.03 2.71 3.39 4.06 90° (in wide must be taken into account. surface) 1.75 2.10 2.80 3.50 4.20

The head pull-through resistance of a screw is Table 46: Yield moment My,Rk in N/mm and tensile 0.8 2 ρk capacity ftens,k in kN in accordance with ETA‑11/0190 Fax, ,Rk = nef · fhead,k · d h · ――― (8.40b) 350 Carbon steel Stainless steel

The head pull-through parameter fhead,k must be d in mm My,Rk ftens,k My,Rk ftens,k taken from the approvals/ETA of the screws. 6 9,500 11.0 5,500 7.10 For screws with washers, the washer diameter 8 20,000 20.0 11,000 12.0 may be used instead of the head diameter. 10 36,000 32.0 20,000 18.8 For screws according to ETA-11/0190 with head 12 58,000 45.0 ­ ­

diameters dh ≤ 25 mm and a thickness of at least 40 mm:

2 Fax, ,Rk = nef · (40 - 0.5 · dh) · d h

40 6.5.3.2 Installation 6.5.4 Combined laterally and axially loaded Axially loaded screws may only be used subject screws to compliance with certain minimum dimensions If screws are simultaneously loaded laterally for the member thickness and minimum screw and axially, the verification is performed using spacing and distances. Since the timber con­ quadratic interaction struction industry nowadays uses almost exclu­ 2 2 Fax,Ed Fv,Ed sively screws according to approvals / ETA, the ―――― + ―――― ≤ 1 (8.28) F F minimum dimensions to be complied with should ax,Rd v,Rd be taken from the respective technical assess­ ment documents. These values are often lower Table 47: Minimum spacing and distances for screws in than those of Eurocode 5. accordance with ETA-11/0190

Screw spacing a1 parallel to grain screw inside cone of the wide surface 5 · d (7 + 8 · | cos |) · d

Screw spacing a2 perpendicular 45° 45° to grain of the wide surface 2.5 · d 7 · d

End distance a1,CG 5 · d (7 + 5 · | cos |) · d

Edge distance a2,CG to the wide 45° Type Q: d ≥ 6 mm and narrow surfaces 3 · d 3 · d screw inside cone Screw spacing for crossed Type Q: d ≥ 6 mm screw pairs 1.5 · d 1,5 · d screw outside cone

Figure 18: Permissible insertion angle and diameter for Example 14: screws in BauBuche in accordance with Eurocode 5 – Tensile connection 45° for screws in accordance with e.g. ETA-11/0190, more with fully threaded insertion angles are possible. screws under 45°

S When installing screws in laminated veneer lum­ ber, own values are often specified in accordance with the approval / ETA. Example minimum dis­ tances which must be complied with in accor­ a1 dance with Figure 19 from ETA-11/0190 are listed in Table 47. The screw spacing and distances F in this case are defined from the centre of gravity of the thread in the respective members. a2,CG

A steel plate (t = 10 mm) with four fully threaded a a a a a a 1 1 1,CG 2,CG 2 2,CG screws 8.0 x 120 mm in accordance with ETA- 11/0190 under 45° is connected to a column made S S a2,CG a2,CG of BauBuche. a S S 2,CG The effective thread length is calculated from the a 2,CG screw length minus the length in the steel plate

a1,CG ℓef = 120 mm - 10 · 2 = 106 mm a1 a1 a1,CG a2

S S a2,CG a2,CG S S a2,CG

a2,CG

a1,CG a2,CG a2,CG

Figure 19: Definition of the minimum distances for ­axially loaded screws

41 Linear interpolation, in accordance with Table 45 The total load-carrying capacity of the connection and considering β = 90° and = 45°, reveals a should be calculated with an effective number of

withdrawal capacity Fax,Rk of 106/10*2.8 = 29.6 kN inclined screws per screw. According to Table 46 the tensile 0.9 ­capacity is ftens,k = 22.0 kN. nef = max { 4 ; 0.9 · 4 } = 3.6

With kmod = 0.9, the design value of the load-­

carrying capacity in the axial direction results as: Fax,Rd,ges = 3.6 · 16.9 kN = 60.8 kN

0.9 22 Fax,Rd = min ――― · 29.6 ; ――― Taking into consideration the angle between the 1.3 1.3 load direction and the screw axis and the rope

= min { 20.5 ; 16.9 } = 16.9 kN ­effect, the design load-carrying capacity is

60.8 kN Head pull-through of the head is prevented by the FRd = 1.25 · ―――――― = 53.7 kN 2 steel plate and is thus not relevant.

Table 48: Load-carrying capacity Fv,Rk in kN and minimum wood thicknesses treq in mm of screws in accordance­ with ETA-11/0190 in timber-to-timber and steel-to-timber connections with external steel plates; screws are inserted perpendicular to the wide surfaces and the rope effect is not considered. d 6 8 10 12 ε 0° 90° 0°-90° 90°-0° 0° 90° 0°-90° 90°-0° 0° 90° 0°-90° 90°-0° 0° 90° 0°-90° 90°-0° β 1.00 1.00 1.55 0.65 1.00 1.00 1.45 0.69 1.00 1.00 1.35 0.74 1.00 1.00 1.27 0.79 Timber-to-timber connection

FV,Rk 2.34 2.92 2.58 2.58 4.01 4.83 4.36 4.36 5.53 6.42 5.92 5.92 7.57 8.53 8.01 8.01 Single-shear t 24.7 18.1 33.0 25.4 37.3 30.1 43.5 36.8 1,req 23.7 19.0 31.8 26.5 36.2 31.1 42.5 37.7 t2,req 18.1 24.7 25.4 33.0 30.1 37.3 36.8 34.5 Double-shear

t2,req 19.6 15.8 14.0 21.6 26.3 21.9 19.8 28.7 30.0 25.8 23.8 32.1 35.2 31.3 29.4 37.3 Steel-to-timber connection (thin steel plates)

FV,Rk 2.34 2.92 - - 4.01 4.83 - - 5.53 6.42 - - 7.57 8.53 - - treq 23.7 19.0 - - 31.8 26.5 - - 36.2 31.1 - - 42.5 37.7 - - Steel-to-timber connection (thick steel plates)

FV,Rk 3.32 4.13 - - 5.68 6.83 - - 7.81 9.08 - - 10.7 12.1 - - treq 27.7 22.3 - - 37.3 31.0 - - 42.4 36.5 - - 49.8 44.2 - -

42 7 GLUED COMPONENTS ports, additional bracing of the bottom flanges is possible by crosspieces to the higher-lying brac­ DIN EN 1995-1-1, Chap. 9 ing construction. The webs primarily transfer the shear forces. Rein­ forcement by using additionally glued-on Board BauBuche in areas with high shear forces, e.g. at the supports, thus may be required. In addition to shear forces, webs are also subject to bending stresses. For cross-sections in particular, in which Note: In accordance with the “German general the web passes through up to the upper edge, construction technique permit (Allgemeine Bau­art­ significant bending stresses occur in the webs. genehmigung)” no. Z-9.1-838, Boards BauBuche S The glued joints used to connect individual cross-­ may be used as webs and Boards BauBuche Q as sections transfer the shear forces from the webs sheathing for glued thin-flanged beams (stressed into the flanges. For design purposes, the glued skin panels) acc. to DIN 1052-10. Boards BauBuche connections are considered rigid, eliminating any may be used for further glued applications acc. impact of the glued joints on overall stiffness. to DIN 1052-10, for instance reinforcements with The strength of the glued joints may be considered glued-on boards. as at least equivalent to that of the neighbouring Currently, the bonding of Board BauBuche and Board BauBuche. Accordingly, only the local shear Beam BauBuche GL75 to glued components in stresses in the BauBuche members need to be normative terms is only regulated for cases, in checked. If only veneers parallel to the beam axis which the glue line pressure is applied by screws are glued together (Board BauBuche S), no rolling (see Chapter 9.2.1). Here, the thickness of the shear stress is exerted, meaning a shear strength

­BauBuche cross-sections to be glued together is value of fv,0 can be expected. limited to 50 mm. If the assembled partial cross-sections consist of­ Gluing individual cross-sections of Board BauBuche different materials, possible impacts on the verifica­ to T-beams, double T-beams or box girders allows tions in the final state due to differingdeformation ­ manufacturing sophisticated members with high behaviour (kdef) must also be taken into account. bending capacities. Box girders provide a bracing The following section focuses on symmetrical cross-­ effect due to the high torsional stiffness, which sections made up of individual parts of BauBuche­ may eliminate the need for roof bracing. Glued of the same type glued together. composite members comprise flanges (horizontal panels), webs (vertical panels) and glued joints. b2 b1 The manufacture of glued load-bearing members Figure 20: is subject to a whole series of requirements. Com­ 0.5 h1 A1 h1 panies executing such work must be able to pro­ Box girder cure special approvals, while the range of permis­ beam made of a1 sible climatic conditions and the moisture content y BauBuche of the members to be glued are both subject hw h2 A2 to strict limits. Moreover, only surfaces directly z planed or sanded before the gluing are permissible. Accordingly, glued members should be already manufactured in the factory. Under bending moments, it is primarily the flang­ b es which absorb tensile or compressive bending 1 stresses. As a general rule, the flanges in com­ Figure 21: A1 0.5 h1 pression are governing the design of cross-sec­ h1 Double T-beam tions. Economical cross-sections are thus those a1 b2 made of where the cross-sections of the flanges in com­ y 0.5 h2 a2 BauBuche pression are larger than those in tension. When h2 = hw A 2 z a3 used as roof or floor beams, the connection with 0.5 h2 the roof or floor panels generally provides suffi­ A 3 h cient bracing against any lateral displacement of 3 0.5 h3 the flanges in compression. For bottom flanges under compression, e.g. above intermediate sup­ b3 43 The effective bending stiffness (EI)ef is calculated where

as a composite cross-section with infinitely stiff hw Web height between flanges joints: n Number of webs

3 2 Elef = ∑ (Ei · Ii + Ei · Ai · a i ) Verification of local shear stresses at the glue

i=1 lines The distance of the centres of gravity of the indi­ Vd · E1(3) · S1(3) Vd · E1(3) · A1(3) · a1(3) vidual cross-sections from the centre of gravity of τk,d = ―――――――――― = ――――――――――――― ≤ f v,d (EI) · n · b (EI) · n · b the ­overall cross-section is calculated as follows ef KF,1(3) ef KF,1(3)

where E1 · A1 · ( h1 + h2 ) - E3 · A3 · ( h2 + h3 ) a = ――――――――――――――――――――――――― b Width of glued joints 2 3 KF 2 · ∑ Ei · Ai n Number of glued joints i=1 S First moment of area based on the overall h1 + h2 h2 + h3 a1 = ――――― - a 2 a3 = ――――― + a 2 centre of gravity 2 2

For cross-sections with webs covering the total Example 15: Bonded box girder

height, h1 and h3 should be used with negative

signs. It is presumed that a2 is positive and 40 ­smaller than or equal to h2 / 2. The following verifications must be performed at the ultimate limit state: 40 120 Verification of flanges Centre of gravity stresses: 40

E1(3) · Md · a1(3) σc(t),1(3),d = ―――――――――― (El)ef 250 Stresses in the extreme fibres: Loads:

E1(3) · Md · (a1(3) + h1(3) /2) Moment Md = 30 kNm σm,1(3),d = ―――――――――――――――――― (El) Shear force V = 15 kN ef d k = 0.8; service class 1 Verification of web mod Stress in the extreme fibres: Dimensions: Individual cross-sections made of E2 · Md · (a2 + h2 /2) σm,2,d = ―――――――――――――― Board BauBuche Q, panel thickness 40 mm (El)ef h2 = 120 mm; b1 = 250 mm; ℓ = 6 m

Shear stress: Note: The following calculations can be per­ 2 (E3 · A3 · a3 + 0.5 · E2 · b2 · h ) · Vd formed without consideration of the MOE-values, τ2,max,d = ――――――――――――――――――――――― b · (El) since the individual cross-sections have the 2 ef same MOE. where I = ∑ (I + A · a2 ) h ef i i i 2 6 6 2 h = ――― + a 2 = 2 · 5.76 · 10 + 2 · 1.33 · 10 + 2 · 10,000 · 80 2 = 1.42 · 108 mm4

Simplified buckling verification: 1. Verification of centre of gravity stresses in the flanges h + h 1 3 6 n · b h 1 + 0.5 · ―――― · f ; h ≤ 35b Md · a1(3) 30 · 10 · ± 80 2 · w v,d w 2 σ = ――――― = ―――――――― = ± 1 6 . 9 N / m m 2 hw c(t),1(3),d 8 FV,Rd ≤ I 1.42 · 10 ef h1+ h3 n·35·b2 1 + 0 . 5 · ―――― · f ; 35b ≤ h ≤ 70b 2 v,d 2 2 2 hw 16.9 η = ――― = 0.58 ≤ 1.0 (9.9) 28.9

44

h1 + h3 where Fv,Rd = n · b2 · hw 1 + 0 . 5 · ―――――― · fv,d hw

kℓ · ft,0,k f = k · ―――――― t,0,d mod 40 + 40 γM -3 = 2 · 40 · 120 1 + 0 . 5 · ―――――― · 4.80 · 10 120

0.96 · 49 N/mm2 2 = 61.4 kN = 0 . 8 · ――――――――――― = 2 8 . 9 N / m m 1.3

0.5 · 15 η = ―――――― = 0.12 ≤ 1.0 1.2 · 62.0 N/mm2 2 61.4 fc,0,d = 0 . 8 · ――――――――――― = 4 5 . 8 N / m m 1.3

6. Verification of local shear stresses at the glue line In service class 1, the characteristic value of the com­

V · A · a 15 · 103 · 10,000 · 80 pressive strength may be modified by the factor 1.2. d 1(3) 1(3) 2 τk,d = ―――――――― = ――――――――――― = 1 . 0 5 N / m m I · n · b 1.42 · 108 · 2 · 40 ef KF,1(3) 2. Verification of maximum stresses in the flanges 1.05 η = ――― = 0.45 ≤ 1.0 2.34 Md · (a1(3) + h1(3) /2) σm,1(3),d = ――――――――――――― I where ef

3.8 N/mm2 6 2 30 · 10 · (80 + 40/2) fv,d = 0 . 8 · ―――――――― = 2 . 3 4 N / m m = ――――――――――――― = 2 1 . 1 N / m m 2 1.3 1.42 · 108

Note: The verification of the box girder cross-­section, 21.1 η = ――― = 0.42 ≤ 1.0 as shown in Example 15 corresponds to a box girder 49.8 floor, where the same effective flange widths are used. where The use as a floor element requires additional check of the vibration behaviour. 81 N/mm2 2 fm,d = 0 . 8 · ――――――― = 4 9 . 8 N / m m 1.3 80 Example 16: Bonded Double T-beam 3. Verification of maximum stress in the web 180

6 Md · h2/2 30 · 10 · 120/2 σ = ―――――― = ―――――――――― = 1 2 . 7 N / m m 2 m,2,d 8 Ief 1.42 · 10 40 1,000

12.7 η = ――― = 0.35 ≤ 1.0 36.3 where 130

59 N/mm2 2 fm,d = 0 . 8 · ――――――― = 3 6 . 3 N / m m 1.3 Loads:

Moment Md = 850 kNm

4. Verification of shear stresses in the web Shear force Vd = 170 kN

ℓ = 20 m; kmod = 0.9, service class 1 τ = (10,000 · 80 + 0.5 · 80 · (120 /2)2) 2,max,d Dimensions: · 15 · 103 / (80 · 1.42 · 108) = 1.24 N/mm2 Web made of Board BauBuche Q 1.24 Flanges made of Board BauBuche S η = ――― = 0.26 ≤ 1.0 4.80 EI = 16,800 · 7.78 · 107 + 13,200 ·3.33 ·109 where ef + 16,800 · 2.93 · 107 + 16,800 · 28,000 · 375.92

7.8 N/mm2 2 2 2 + 13,200 · 40,000 ·34.1 + 16,800 · 20,800 · 469.1 fv,d = 0 . 8 · ――――――― = 4 . 8 0 N / m m 1.3 = 1.92 · 1014 N/mm2

a2 = (16,800 · 2 · 80 · 180 · (-180 + 1,000) 5. Simplified buckling verification of the web - 16,800 · 2 · 80 · 130 · (1,000 - 130))/(2 · (16,800 · 28,800 + 13,200 · 40,000 + 16,800 · 20,800)) hw = 120 mm ≤ 35 · b2 = 35 · 40 = 1,400 mm = 34.1 mm 45

- 180 + 1,000 39.7 a1 = ―――――――― ­ 34.1 = 375.9 m m η = ――― = 0.76 ≤ 1.0 2 51.9

where 1,000 - 130 a = ―――――――― + 34.1 = 469.1 m m 3 75 N/mm2 2 2 fm,d = 0 . 9 · ―――――――― = 5 1 . 9 N / m m 1.3 Subsequently, only the governing checks are ­performed. A beam sufficiently braced against 5. Verification of bending stress in the web flexural and torsional buckling is presumed. 13,200 · 850 · 106 · (34.1 + 1,000/2) 2 σm,2,d = ――――――――――――――――――― = 3 1 . 2 N / m m 1.92 · 1014 1. Verification of compressive stress in the ­upper flange 31.2 η = ――― = 0.88 ≤ 1.0 16,800 · 850 · 106 · (- 375.9) 2 35.3 σc,1,d = ――――――――――――――――― = ­ 2 8 . 0 N / m m 1.92 · 1014 where

28.0 0.865 · 59 N/mm2 2 η = ――― = 0.59 ≤ 1.0 fm,d = 0 . 9 · ―――――――――――― = 3 5 . 3 N / m m 47.8 1.3

where 6. Verification of shear stress in the web

1.2 · 57.5 N/mm2 2 fc,0,d = 0 . 9 · ――――――――――― = 4 7 . 8 N / m m τ2,max,d = (16,800 · 20,800 · 469.1 1.3 + 0.5 · 13,200 · 40 · 534.12) 2. Verification of tensile stress in the bottom · 170 · 103 / (40 · 1.92 · 1014) = 5.30 N / mm2 flange where

16,800 · 850 · 106 · 469.1 1,000 2 σt,3,d = ―――――――――――――――― = 3 4 . 9 N / m m h = ―――― + 34.1 = 534.1 mm 1.92 · 1014 2

34.9 5.30 η = ――― = 0.94 ≤ 1.0 η = ―――― = 0.98 ≤ 1.0 37.1 5.40

where where

0.892 · 60 N/mm2 7.8 N/mm2 2 2 ft,0,d = 0 . 9 · ―――――――――――― = 3 7 . 1 N / m m fv,d = 0 . 9 · ―――――――― = 5 . 4 0 N / m m 1.3 1.3

3. Verification of maximum stress in the upper 7. Simplified buckling verification of the web flange h = 1,000 mm ≤ 35 · b = 35 · 40 = 1,400 mm 16,800 · 850 · 106 · - (375.9 + 180/2) w 2 σm,1,d = ―――――――――――――――――――――― 1.92 · 1014 180 + 130 FV,Rd = 1 · 40 · 1,000 1 + 0 . 5 · ――――――――――― (1,000 - 180 - 130) = - 34.7 N/mm2 · 5.40 · 10-3 = 265 kN 34.7 η = ――― = 0.67 ≤ 1.0 51.9 170 η = ―――― = 0.64 ≤ 1.0 265 where 8. Verification of local shear stresses at the 75 N/mm2 2 fm,d = 0 . 9 · ―――――――― = 5 1 . 9 N / m m glue line to the bottom flange 1.3

170 · 103 · 16,800 · 20,800 · 469.1 4. Verification of maximum stress in the τk,d = ―――――――――――――――――――― 1.92 · 1014 · 2 · 130 bottom flange = 0.56 N/mm2 16,800 · 850 · 106 · (469.1 + 130/2) σm,1,d = ―――――――――――――――――――――― 1.92 · 1014 0.56 η = ――― = 0.21 ≤ 1.0 2.63 = 39.7 N/mm2 where

3.8 N/mm2 2 46 fv,d = 0 . 9 · ―――――――― = 2 . 6 3 N / m m 1.3

qd 8 SHEAR WALLS AND DIAPHRAGMS Hd DIN EN 1995-1-1, Chap. 9.2.3/4

h

8.1 General Board BauBuche Q may be used for constructing roof, floor and wall panels that supplyin ­ -plane ℓ stiffness and capacity. ℓ* Board BauBuche is produced in thicknesses of up Figure 22: Shear wall to 60 mm. Together with high shear and compres­ sive strength it is suitable for use as solid shear Floor diaphragm walls. Solid floors made of Board BauBuche are also possible. However, significant deflections limit Edge member their use beyond a span of around 3.5 m. Accordingly, only the formation and design of ­solid shear walls is presented. a 8.2 Shear walls Floor beam Walls are designed to accommodate vertical 40 loads from the dead weight, live loads and snow as well as horizontal bracing loads caused by 80 wind and earthquakes. The buckling of the wall is Wall generally the key factor when it comes to vertical loads. When designing for bracing loads, above Figure 23: Diaphragm all, there is a need to carefully examine the load to shear wall connection introduction, the way the individual wall elements are connected to each other and the way the shear forces are transferred to the foundations. Figure 23 shows an example whereby the floor shear forces from the floor are transferred to a 60 solid wall panel. For this purpose, the wall is notched at the top for the edge member of the floor and fixed to the same by horizontal nails 40 or screws. The notch is required, since laterally loaded nails / screws in the end grain of BauBuche Figure 24: Element connection with a rebate are not allowed (with the exception of e.g. section (screw d = 6 mm) 6.5.2) and ­inclined screws under axial loading are not feasible­ due to the edge distances having The shear flow in the wall panels is to be complied with. Hd Board BauBuche is produced up to 1.82 m wide. In sv,0,d = ――― ℓ general, walls must therefore be assembled from multiple individual elements. One possibility is to The tensile force at the wall ends is design a rebate with laterally loaded nails / screws Hd · h ℓ as the connection. Further information can be Zd = ――――― - g k · ―― ℓ* 2 found in brochure 05 “Fasteners”. ℓ* is the distance from the centre of gravity of the hold-down to the wall end.

47 Example 17: Shear wall consisting of two 9 REINFORCEMENTS AND REHABILITATION elements DIN EN 1995-1-1/NA, NCI NA 6.8 Board BauBuche Q, t = 60 mm, DIN EN 1995-1-1/NA, NCI NA 11.2.3 System see Figure 22 DIN EN 1995-1-1, Annex B DIN 1052-10 Loads: Hd = 60 kN, service class 1, KLED short

Dimensions: h = 2.7 m, ℓ = 3.6 m

Horizontal load: The shear flow is 9.1 Reinforcements for tensile stresses perpendi­ 60 kN sv,0,d = ――――― = 1 6 . 7 k N / m cular to grain 3.6 m Reinforcements of timber members are mostly For the rebate connection, screws 6 x 60 mm ­required due to the low tensile strength perpen­ in accordance with ETA-11/0190 are selected. dicular to the grain. Perpendicular to grain tensile The individual load-carrying capacity of a screw, stresses occur e. g. in the connection areas of

in accordance with Chapter 6 is Fv,Rk = 2.34 kN. members, at notched supports, holes and in the The required screw spacing therefore is apex of double tapered beams. BauBuche has far higher perpendicular to grain tensile strength 1.62 kN e = ――――――― = 0 . 1 0 m than solid or glulam softwood. Board BauBuche Q 16.7 kN/m with cross layers possesses a perpendicular to The shear capacity of the wall itself is grain ­tensile strength of 16 N/mm2 (for nominal thickness B ≤ 24 mm) or 8 N/mm2 (for nominal 0.9 2 fv,0,d = ―― · 7.8 N/mm · 30 mm = 162 kN/ m thickness 27 mm ≤ B ≤ 60 mm). By using members 1.3 made of BauBuche, as an alternative to solid or and the actual load is far less. glulam softwood, it is often possible to eliminate the need of reinforcement perpendicular to grain. The following tensile force must be anchored at Board BauBuche is ideal for use as external rein­ the wall ends into the foundation forcement for solid or glulam members subject to significant perpendicular to grain tensile stresses. 60 kN · 2.7 m 3.6 m Zd = ―――――――― ­ 0 . 9 · 1 . 3 0 k N / m · ―――― = 4 4.2 k N Perpendicular to grain reinforcement is not 3.5 m 2 ­covered in Eurocode 5. This document therefore Conservatively, only the dead weight of the wall is specifies the calculation principles of the NA - Ger­ considered in this case. many. Its applicability outside Germany should be checked. Vertical stress: According to Chapter 4.2.1, the buckling coeffi­ 9.2 Types of reinforcement

cient is kc,z In terms of reinforcement, internal and external reinforcements are distinguished. Examples 1 1 kc,z = ―――――――――― = ――――――――――――― = 0 . 1 3 of internal­ reinforcements include fully threaded k + k2 - λ2 4.27 + 4.272 - 2.702 z z rel,z screws, glued-in rods or reinforcement bars. where ­External reinforcements include glued-on boards or wood-based panels. The design of fully thread­ 3 ℓef β · ℓ 1.0 · 2.7 · 10 mm λz = ―― = ――――― = ―――――――――――――= 1 5 6 ed screws is covered by the screw’s approv­ i b/ 12 60 mm / 12 z al / ETA. For reinforcements subsequently applied with glued-on boards (rehabilitation), the method

λz fc,0,k 156 74.4 λrel,z = ―― ―――― = ――― ―――――― = 3 . 9 4 of using screws to apply the necessary gluing π E π 11,800 0,05 pressure (“screw gluing”) can be applied.

2 kz = 0.5 · (1 + 0.1 (λrel,z - 0.3) + λ rel,z)

= 0.5 · (1 + 0.1 (3.94 - 0.3) + 3.942) = 8.43

The vertical load capacity is thus

0.9 2 qd = 0.063 · ――― · 74.4 N/mm · 60 mm = 195 kN/m 1.3 48 9.2.1 Screw gluing Table 49: Conditions for screw gluing in accordance­ When retrofitting tensile cracks perpendicular to with DIN 1052-10, Chap. 6.2 the grain or reinforcing subsequently installed connections or openings after those have already Thickness of reinforcement panel been installed, the use of hydraulic presses is tmax = 50 mm (wood-based panel) generally unfeasible, due to the confined spaces. Fastener In this case, it is possible to manufacture load-­ Self -drilling partially threaded screw according bearing glued connections using “screw gluing”. to ­approval / ETA with:

In the process, the pressure is applied by self-­ (1) ℓsmooth ≥ treinforcement panel drilling partially threaded screws. Eurocode 5 (2) Thread length in timber member including does not cover screw gluing. Accordingly, screw tip

­reference is made at this point to DIN 1052-10. ℓef ≥ max(treinforcement panel; 40 mm) To ensure uniform pressure and thus ensure the (3) Nominal diameter d ≥ 4 mm quality of the glued joints, the thickness of the Arrangement wood-based reinforcement panel must be limited (1) Screw interval distances a1, a2 ≤ 150 mm 2 to a maximum of 50 mm. The spacing between (2) Glued area per screw a1 · a2 ≤ 15,000 mm the fasteners must not exceed 150 mm and the (3) Uniform grid when a1 = a2 = 120 mm glued area per screw is limited to 15,000 mm². Member The only feasible fasteners in this case are (1) Moisture content u ≤ 15 % partially threaded screws with approval / ETA. (2) Moisture difference Δu ≤ 4 % The length of the smooth shank must at least (3) Surface planed or sanded ­correspond to the thickness of the reinforcement Adhesive panel. The adhesive used must, according to Permissible for screw gluing the relevant approval, be suitable for use with Company load-bearing screw gluing. Accredited for gluing in accordance with DIN 1052-10

Glued joint t 9.3 Applications ≥ 0 9.3.1 Connections loaded at an angle to the grain ≥ 40 mm For tensile loads acting perpendicular to the d ≥ 4 mm ≥ t member axis, there is a risk of tensile failure ­perpendicular to grain, when forces are intro­ duced close to the loaded edge. The governing fastener is the fastener furthest away from the loaded edge. Figure 25: Geometry of screw gluing In the design of reinforcement perpendicular to the grain a cracked cross-section is assumed. In To avoid additional stresses in the glue joint, the other words, the reinforcing elements must be moisture content content in the members to be ­capable of transferring the complete tensile loads connected must not differ by more than 4 %. For perpendicular to the grain. rehabilitations, it is thus advisable, e. g. to place the reinforcement boards in the building for some Example 18: Connection loaded perpendicu- time before the actual gluing. lar to the grain using Beam BauBuche GL75

Loads:

FEd = 45 kN, kmod = 0.7, service class 2

Dimensions: Beam BauBuche GL75 140/240 mm

49 Connection: 9.3.2 Notched supports max. distance between Perpendicular to grain tensile stresses decline

fastener and loaded edge he = 150 mm very rapidly with increasing distance from the

Width of the fastener group ar = 50 mm notch. Accordingly, external reinforcing elements Number of the fastener rows n = 3 must be installed up to the notch corner. Subject Angle between load and to maintaining the minimum edge distances, grain direction = 90° ­internal reinforcing elements should be arranged Fastener diameter d = 10 mm as close as possible to the notch corner. For the same reason, only the first fastener row in the ­direction of the member axis may be taken into ar account. To reduce the penetration depth and h1 hi hn h hence the insertion resistance, constructive solu­ tions as shown in Figure 27 are possible. he Lreq Predrilling tpen diameter acc. Fv,Ed tpen to screw b L req assessment FEd Fv,Ed/2 Fv,Ed /2 documents Figure 26: Connection loaded perpendicular to the grain (see Table 44) L req

The load-carrying capacity of the non-reinforced Predrilling connection is diameter ≥ screw 18 · h2 e 0.8 F90,Rd = ks · kr · 6 . 5 + ―――― · (tef · h) · ft,90,d diameter h2 Figure 27: Notch with reduced penetration depth 18 · 1502 0.8 F90,Rd = 1.0 · 1.83 · 6 . 5 + ――――― · (120 · 240) · 0.32 2402 Example 19: Reinforcement of a notch with

glued-on Board BauBuche = 29,300 N (NA. 104) Loads: V = 10.3 kN, k = 0.9 where d mod

1.4 · ar Dimensions: GL28h, 100/250 mm ks = max 1 ; 0 . 7 + ――――― h Height at the support h = 145 mm ef Distance to the notch x = 100 mm 1.4 · 50 = max 1 ; 0 . 7 + ――――― = 1.0 (NA. 105) 240 X

n 3 kr = ―――――― = ―――――――――――― = 1 . 8 3 2 2 2 hef n h1 90 90 ∑ ―― 1 + ― ―― + ― ―― i=1 hi 190 140 (NA. 106) h

tef = min {b; 2 · tpen; 24 · d} = min {140; 2 · 60; 24 · 10} = 120 mm

Since F90,Rd < FEd the connection must be rein­ ℓr forced. The reinforcement consists of two fully Figure 28: Notch with reinforcement threaded screws, which must be designed for the following tensile force: Verification of the load-carrying capacity of the

2 3 notch he he Ft,90,d = 1 - 3 · ―― + 2 · ―― · FEd 2 h h τd 1.07 N/mm η= ―――― = ―――――――――――――――― = 1 . 3 1 > 1 . 0 2 kv · fv,d 0.472 · 0.9 / 1.3 · 2.5 N/mm 2 3 150 150 = 1 - 3 · ――― + 2 · ――― · 45 kN (NA.69) with 240 240

50 = 14.2 kN shear stress in the remaining cross-section 3 where 1.5 · Vd 1.5 · 10.3 · 10 N N τ = ―――― = ―――――――――――― = 1 . 0 7 ――― ( 6 . 6 0 ) d 2 F 5.1 · 103 N b · hef 100 mm · 145 mm mm t,90,d 2 σt,d = ――――― = ――――――― = 2 . 4 3 N / m m (NA.83) 2 · t · ℓ 2 · 20 · 52.5 r r and the reduction coefficient k v 60 N/mm2 2 ft,d = 0 . 9 · ―――――――― = 4 1 . 5 N / m m 1.0 1.3

6.5 kv = min ―――――――――――――――――――――――――――― The shear stress in the glue line is governing the 100 1 250 0.58 · (1 - 0.58) + 0.8 - 0.582 reinforcement design; the tensile strength of the 250 0.58 Board BauBuche­ is not attained. kv = 0.472 (6.62) 9.3.3 Holes To reinforce the notch, Board BauBuche S Note: The design of holes in beams made of 60 / 250 mm, t = 20 mm is applied on both sides ­BauBuche GL75 is currently excluded by the through screw gluing. declaration of performance.

According to (NA.84), the following condition Openings in beams with clear dimensions hd ­applies for the width of the reinforcement panel ­exceeding 50 mm are to be considered as holes. Perpendicular to grain tensile stresses at such ℓr 0.25 ≤ ――――― ≤ 0.5 (NA.84) openings are caused by shear forces and bending h - h ef moments. When mainly shear forces are present, This means that in this case, only a width of cracks tend to appear at points 1 and 2 in Figure 29, while when bending moment prevails, the

ℓr ≤ 0.5 · (h - hef) = 0.5 · 105 = 52.5 mm cracks only appear at the upper edge (points 1 and 3). When designing the reinforcing elements, may be taken into account for the verification. all points at risk must be examined. Determining the tensile force to be transferred by the reinforcement Ft,90,d 3 2 1

2 3 Ft,90,d = 1.3 · Vd · 3 · (1 - ) - 2 · (1 - ) (NA.77) hro

hd h Md 2 3 = 1.3 · 10.3 · 3 · (1 - 0.58) - 2 · (1 - 0.58) = 5.1 kN hru Vd

Verification of shear stresses in the glue line

2 Vd τef,d 0.46 N/mm η= ――― = ――――――――― = 0 . 8 8 ≤ 1 . 0 ( N A . 8 0 ) 2 ℓ f 0.52 N/mm A a ℓz k2,d ℓV where 3 2 1 3 Ft,90,d 5.1 · 10 N τef,d = ―――――――― = ――――――――― ( N A . 8 1 ) 2 · (h - h ) · ℓ 2 · 105 · 52.5 h + 0.15 h hro ef r ro d h h M = 0.46 N/mm2 d d hru + 0.15 hd hru Vd 0.75 N/mm2 2 fk2,d = 0 . 9 · ―――――――― = 0.52 N/mm 1.3 Vd in accordance with Table NA.12 ℓ A a ℓz Verification of tensile stress in the reinforcement ℓV panels

σt,d Figure 29: Openings with transverse tensile cracks η = kk · ―――― f t,d When designing reinforcements for openings, the 2.43 N/mm2 geometric conditions in accordance with Table 50 = 2 . 0 · ――――――――― = 0.12 ≤ 1.0 (NA.82) 41.5 N/mm2 must be complied with.

51 Table 50: Requirements for reinforced openings (3.58 + 3.57) · 103 N η = ――――――――――――――――――― = 2 . 1 0 > 1 , 0 ℓ ≥ h, however at least 300 mm 0.5 · 139 · 140 · 1 · 0.35 N/mm2 z ℓ ≥ h v where ℓA ≥ h/2

hro(ru) ≥ 0.25 h ℓt,90 = 0.353 · hd + 0.5 · h = 139 mm (NA.65) a ≤ h and a / hd ≤ 2.5 h ≤ 0.3 h (for internal reinforcement) d for round openings hd ≤ 0.4 h (for external reinforcement)

k = min (1 ; (450/h)0,5) = min (1 ; (450/240) 0.5) = 1.0 Example 20: Reinforcement of a round t,90

opening with glued-on ­ To transfer the perpendicular to grain tensile Board BauBuche stresses, panels 240 / 210 mm, t = 20 mm made of

Loads: Md = 45 kNm, Vd = 30 kN Board BauBuche S are glued on both sides.

service class 1, kmod = 0.9 Verification of the shear stress in the glue line (governing!) Opening: Glulam beam GL 24h 2 Beam width b = 140 mm τef,d 0.50 N/mm η = ――― = ――――――――― = 0 . 9 6 ≤ 1 . 0 ( N A . 8 7 ) Beam height h = 240 mm f 0.52 N/mm2 k2,d Remaining height h = 92.5 mm ro where Remaining height hru = 92.5 mm 3 Diameter hd = 55 mm Ft,90,d 7.15 · 10 N τef,d = ―――――― = ―――――――――――――― ( N A . 8 8 ) Distance to end grain ℓ > 1,500 mm 2 · a · h 2 · 83.4 mm · 85.8 mm V r ad Distance to support ℓ = 1,500 mm A = 0.50 N/mm2 hr = min{hro + 0.15 hd ; hru + 0.15 hd} = 101 mm 0.75 N/mm2 2 fk2,d = 0 . 9 · ―――――――――― = 0.52 N/mm 1.3

h1 hro h = h + 0.15 · h = 77.5 + 0.15 · 55 = 85.8 mm h ad 1 d d hd h h1 a ≤ 0.6 · ℓ = 83.4 mm (NA.91) hru r t,90

9.3.4 Apex area of beams with variable cross- ar a ar tr b tr sections For economic and aesthetic reasons, long glulam Figure 30: Reinforcement of a round opening beams are normally designed with a variable The design value of the tensile force perpendi­ beam height and with or without curvature. The cular to grain at the governing point is: kink in the beam axis in the apex generates per­ pendicular to grain stresses. The risk of cracks due

Ft,90,d = Ft,V,d + Ft,M,d = 3.58 + 3.57 = 7.15 kN to tensile stresses perpendicular to the grain is increased under unfavourable climatic conditions. where (NA.66)

2 Example 21: Pitched cambered beam with Vd · 0.7* · hd (0.7* · hd) Ft,V,d = ――――――――― 3 ­ ――――――― = 3.58 kN tensile cracks 4 · h h2

Repair of a crack in the apex area of a pitched * for round openings (NA.67) cambered beam

Md Md Loads: Md = 340 kNm, kmod = 0.9 Ft,M,d = 0 . 0 0 8 · ―― = 0 . 0 0 8 · ――――――――― = 3 . 5 7 k N h h + 0.15 · h r ru/ro d Dimensions: (NA.68) Material GL 28c, b = 200 mm

For non-reinforced openings, the following con­ Height in the apex hap = 1,462 mm dition must be complied with: Roof angle δ = 15° Angle of lower edge β = 9° Ft,90,d η = ―――――――――――――――――― ≤ 1 . 0 ( N A . 6 3 ) Length of apex area c = 2,200 mm 0.5 · ℓ · b · k · f t,90 t,90 t,90,d 52 V 2 σt,90,d · b · a1 F = ―― · ―――――――― hap t,90,d hr δ 3 n hx r h r t a 2 in β = ―― · 35.0 kN = 23.3 kN (NA.102) 3 x c a L Verification of glue line in the governing internal quarters Figure 31: Pitched cambered beam 2 τef,d 0.09 N/mm η = ――― = ――――――――― = 0 . 0 9 ≤ 1 ( N A . 9 7 ) For the reinforcement of the apex area glued-on f 1.04 N/mm2 k3,d panels made of Board BauBuche Q are used on where both sides.

Verification of perpendicular to grain tensile 2 · Ft,90,d τef,d = ――――――― stresses in the apex cross-section ℓ · ℓ r ad σ t,90,d 2 · 35.0 · 103 N η = ――――――――――― (6 . 5 0 ) 2 k · k · f = ――――――――――――― = 0.09 N/mm (NA.98) dis vol t,90,d 1,153 mm · 679 mm

0.29 N/mm2 = ――――――――――――――――――― = 1 . 2 6 > 1 1.5 N/mm2 2 2 1.7 · 0.39 · 0.9 / 1.3 · 0.5 N/mm fk3,d = 0 . 9 · ――――――― = 1.04 N/mm (Tab. NA.12) 1.3 where Verification of tensile stress in the glued-on rein­

6 · Map,d forcement panels (Board BauBuche Q, t = 20 mm) σ = k · ――――――――― t,90,d p 2 b · h ap 2 σt,d 1.52 N/mm η = ――― = ――――――――― = 0 . 0 5 ≤ 1 ( N A . 9 9 ) f 31.8 N/mm2 6 · 340 · 106 t,d 2 = 0 . 0 6 · ―――――――― = 0.29 N/mm (6.54) 200 · 1,4622 where

Ft,90,d σt,d = ――――― k = 1.7 (6.52) t · ℓ dis r r

35.0 · 103 N 0.2 0.2 2 0.01 0.01 = ――――――――――――― = 1.52 N/mm (NA.100) kvol ≈ ――― ≈ ――― = 0.39 (6.51) 20 mm · 1,153 mm V 1.10

46 N/mm2 2 kp = 0.06 where ft,d = 0 . 9 · ――――――― = 31.8 N/mm 1.3

k5 = 0.054, k6 = 0.035, k7 = 0.276 (6.56)–(6.59) 9.4 Cross-sectional reinforcements Verification of reinforcement When additional loads are imposed, e. g. due to To take into consideration the decrease of perpen­ changes in use, adding extra storeys to existing dicular to grain tensile stresses in the longitudinal constructions, or due to damage, there may be a direction of the beam, the tensile force in both need to reinforce individual members. the external quarters of the affected area may be reduced by a third. 9.4.1 Member reinforcement without ­con­nection The tensile force per reinforcement element is The simplest type of reinforcement involves add­

ing additional members. The loading qi, to which σt,90,d · b · a1 Ft,90,d = ――――――――― the individual cross-sections are exposed, can be n determined for beams via the ratio η of their 0.29 · 200 · 1,205 · 10-3 bending stiffnesses. = ――――――――――――――― = 3 5 . 0 k N ( N A . 1 0 1 ) 2

q = η · q ; q = (1 - η) · q and in the external quarters 1 2 1 where η = ―――――― (El)2 ――― + 1 (EI)1 53 The prerequisite is uniform loading, both of the The shear forces acting in the joint lines cause original member as well as the reinforcement, a ­deformation in the longitudinal direction of e. g. by battens or planks. In addition, all load-­ the beam. The flexibility of the joint line can be bearing members must be supported. ­described by the slip modulus K and the fastener It is important to note that subsequently installed spacing. By using inclined fully threaded screws, members cannot accommodate pre-existing more rigid­ joints can be established. For simplifi­ loads. This underlines the need to relieve the load cation, the screws can be considered purely on existing members as far as possible. axially loaded. The axial loading of the screws is calculated by dividing the shear flow in the joint Example 22: Strengthening of timber beams line by the cosine of the insertion angle. Values for the slip modulus of axially loaded fully thread­ The intention is to strengthen existing timber ed screws can be taken from the respective beams cross-section by adding a Board BauBuche. ­approvals / ETA. Loads: q = 10.0 kN/m, M = 31.3 kNm d d Annex B of Eurocode 5 includes a design method service class 1, k = 0.9 mod for mechanically jointed cross-sections in in Existing members: C 24, 140/240 mm, the form of the γ method. The applicability of the Reinforcement: Board BauBuche S, γ-method is limited to the following: 40/240 mm – Individual components non-abutted over the 1 total beam length η = ―――――――――――――――― = 0 . 7 0 16,800 · 4.61 · 107 – Cross -sections with constant geometry ―――――――――――― + 1 11,000 · 1.61 · 108 – Uniformly distributed load in z direction (sine or parabolic bending moment) The loading on the existing beams is reduced to – Cross -sections made of two or three individual 0.7 · 10.0 = 7.0 kN/m. The utilisation factor for the components (a maximum of two flexible joint existing beams is thus 98 %. The reinforcement lines) with Board BauBuche S is used up to 47 %. Strictly speaking, the γ-method only applies for 9.4.2 Reinforcement through mechanically pin-ended single-span beams. Multi-span beams, jointed members however, can be calculated as single-span beams 9.4.2.1 Mechanically jointed members of length ℓ = 0.8 · ℓ, whereby the shorter span By connecting additional members to existing is used for ℓ. For cantilever beams, the doubled members by mechanical fasteners, it is possible length should be used. to create a composite cross-section, with a The principle of the calculation method lies in

load-carrying capacity far exceeding the sum of ­determining an effective bending stiffness (EI)ef, the load-carrying capacities of the individual taking into account the flexibility of the connec­ cross-sections. Unlike members glued together in tion in the joint line. The flexibilities are expressed accordance with Chapter 7, this does not, how­ by the γ-values. These can be assigned values ever, lead to an “ideal” composite cross-section, ­between 0 (no connection) and 1 (rigid connection; since the flexible nature of the fasteners in the glued). γ depends on the slip modulus K of the joint line has a significant impact on the overall mechanical fasteners and their spacing. Efficient load-carrying capacity. To connect individual composite cross-sections are achieved by a staged members nails, screws, bolts or dowels are used. arrangement of the fasteners along the course of

Figure 32: b1 0.5 b1 Geometry of A1,I1,E1 0.5 h mechanically 1 A1,I1,E1 0.5 h1 h1 h1 jointed beams a1 0.5 h2 a b2 b2 1 y 0.5 h2 a y a h 2 2 A2,I2,E2 2 0.5 h a A2,I2,E2 z 2 3 z a3 0.5 h2 A3,I3,E3 h3 0.5 h A3,I3,E3 h3 3 0.5 h3

b3 54 0.5 b3 the shear force. Here, the greatest spacing must 4. Load-carrying capacity of the fasteners not exceed a value four times that of the smallest. γi · Ei · Ai · ai · si · Vd To facilitate the design, normally only the fastener Fi,d = ―――――――――――― (B.10) (EI) spacing in the external beam quarters and those ef in the internal quarters differ from each other. Note: For fasteners with larger diameters in par­ The parameters for the γ-method are calculated ticular, the cross-sectional reductions caused as follows: by fasteners in the tensile areas are taken into

Effective bending stiffness (EI)ef ­account. For this purpose, the centre of gravity

3 stresses are multiplied by Ai / Ai,net , and the bend­ 2 Elef = ∑ (Ei · Ii + γi · Ei · Ai · a i ) (B.1) ing stresses in the extreme fibres by iI / Ii,net .

i=1 If the conditions for the γ-method cannot be met, where the coefficient γ can be determined as fol­ the “shear analogy method” in accordance with lows Kreuzinger can be used (not explained in more de­ tail at this point). 1 γ = ―――――――――――――― f o r i = 1 a n d i = 3 ( B . 5 ) 1 2 π · Ei · Ai · si 1 + ―――――――――― 9.4.2.2 Mechanically jointed lateral reinforce- 2 Ki · ℓ ments

γ2 = 1.0 (B.4) The version for member reinforcement presented in Chapter 9.4.1 presumes direct load introduction The distance from the centre of gravity of the in the reinforcement members. In cases in which cross-section i to the centre of gravity of the total the spatial constraints do not allow the new cross- cross-section is sections to reach the upper edge of the existing member, additional cross-sections can still be γ1 · E1 · A1 · (h1 + h2) - γ3 · E3 · A3 · (h2 + h3) a2 = ――――――――――――――――――――――――― (B.6) ­laterally attached with mechanical fasteners. The 3 2 · ∑ γi · Ei · Ai calculation uses the γ-method shown in Chapter i=1 9.4.2.1. Apart from the shear forces from the composite h1 + h2 h2 + h3 a1 = ―――――― - a2 a3 = ―――――― + a2 action, the fasteners must also transfer the load 2 2 portion of the superimposed load passed on For cross-sections with webs over the total beam from the existing member to the reinforcement. height, h1 and h3 should have negative signs, For stacked partial members, this portion can

­provided a2 is positive and smaller than or the be disregarded, since the load transferred by the same as h2 / 2. reinforcement member is accommodated by ­direct contact due to the deflection of the mem­ The following checks must be performed at the ber on top. ultimate limit state: For lateral reinforcements, the centre of gravity 1. Stress at the centre of gravity in the res­ of the reinforcement should be as close as possi­ pective cross-sections ble to the centre of gravity of the existing member. This keeps the loading of the fasteners resulting γi · ai · Ei · Md σt(c),i,d = ―――――――――― (B.7) from the composite effect on a low level. (EI) ef Provided only one-sided reinforcements are pos­ 2. Maximum stress of the respective sible, additional torsional stresses in the cross- cross-sections sections, particularly in timber cross-sections, must be taken into consideration. 0.5 · Ei · hi · Md σm,i,d = ――――――――――― + σ t(c),i,d (B.8) (EI) ef Example 23: Reinforcement of a beam with 3. Shear stress laterally nailed-on BauBuche

2 sections (γ3 · E3 · A3 · a3 + 0.5 · E2 · b2 · h ) · Vd τ2,max,d = ――――――――――――――――――――― ( B . 9 ) b · (EI) 2 ef Caused by an extension, an existing purlin may be where subject to greater snow loads due to snow drift. This additional load cannot be accommodated by h2 h = ―― + a2 the existing cross-section. Since the thickness 2 of the roof should not be expanded, panels made 55 1.5 of Board BauBuche S are laterally connected with Kmean = 2/3 · ρ m · d/23 nails as a reinforcement. The existing installations 1.5 do not allow reinforcement boards to be estab­ = 2/3 · ( 420 · 800) · 3.8/23 = 1,540 N/mm lished up to the upper edge of the purlins, which means verification in accordance with Chapter The calculation of the composite cross-section is 9.4.1 is not possible. performed in accordance with the γ-method for the cross-section shown type shown in Figure 32

Loads: qd = 10.0 kN/m on the right and reduced to two parts. The stiff­

Md = 31.3 kNm ness values are determined without taking into

Vd = 25 kN consideration the safety coefficientsM γ , since

service class 1, kmod = 0.9 only the ratio of the cross-sectional stiffness Existing members: C 24, 140 x 240 mm, ℓ = 5.0 m ­values applies. The effect of creep deformations Reinforcement: Board BauBuche S, is also disregarded, since solid timber and Bau­

2 x 30/200 mm Buche have the same deformation coefficientdef k .

Connection: Nails 3.8 x 70 mm in two The effective bending stiffness (EI)ef of the com­ rows, predrilled posite cross-section is calculated from

1 1 γ = ―――――――――― = ――――――――――――――――――― = 0 . 2 6 3 2 2 π · Ei · Ai · si π · 16,800 · 12,000 · 55 1 + ――――――― 1 + ―――――――――――――― 2 2 Ki · ℓ 1,540 · 5,000 (B.5) 200 240 γ2 = 1.0 (B.4)

3 2 3 and

γ1 · E1 · A1 · (h1 + h2) - γ3 · E3 · A3 · (h2 + h3) 30 140 30 a2 = ――――――――――――――――――――――――― 3 2 · ∑ γi · Ei · Ai Figure 33: Reinforcement measure for a beam i=1

To ensure economic use of the fasteners, the -0.26 · 16,800 · 12,000 · (-200 + 240) = ―――――――――――――――――――――――――――――― spacing in the longitudinal direction of the purlin 2 · (1.0 · 11,000 · 33,600 + 0.26 · 16,800 · 12,000)

is increased in both the internal quarters of the purlin length. = -2.5 mm (B.6) Selected fastener spacing:

sexternal = 160 mm, sinternal = 400 mm h2 + h3 -200 + 240 a3 = ―――――― + a2 = ―――――――― - 2.5 = 17.5 mm 2 2

For an arrangement with two rows on both sides, to the calculated nail spacing results as: 3 2 Elef = ∑ (Ei · Ii + γi · Ei · Ai · a i ) (B.1) s = 40 mm, s = 100 mm external internal i=1 = 11,000 · 1.61 · 108 + 1.0 · 11,000 · 33,600 · 2.52 To determine the effective stiffness, an effective + 16,800 · 4.0 · 107 + 0.26 · 16,800 · 12,000 · 17.52 fastener distance is used: = 2.46 · 1012 Nmm2

s = 0.75 · s + 0.25 · s = 55 mm eff external internal The small difference in height of the cross-sec­ tions cause only low tensile and compressive where s ≤ 4 · s internal external stresses, meaning verifications of the centre of gravity stresses can be disregarded. The load-carrying capacity of a laterally loaded Verification of bending stresses in the extreme nail is F = 1.1 kN, the average slip modulus v,Rd fibres Kmean is Md · E2 h2 σm,2,d = ――――― · γ2 · a2 + ―― EI 2 ef 56 6 31.3 · 10 · 11,000 240 γ3 · E3 · A3 · a3 · sexternal · Vd (0) = ―――――――――――― · 1.0 · (-2.5) + ――― Fd(x = 0 m) = ―――――――――――――――――― 2.46 · 1012 2 (EI) ef = 16.4 N/mm2 (B.8) 0.26 · 16,800 · 12,000 · 17.5 · 40 · 25 = ―――――――――――――――――――――――― 16.4 2.46 · 1012 η = ――― = 0.98 ≤ 1.0 16.6 = 0.37 kN (without reinforcement: η = 1.40)

γ3 · E3 · A3 · a3 · sinternal · Vd (1.25) where Fd(x = 1.25 m) = ――――――――――――――――――――― (EI) ef 24 N/mm2 2 fm,2,d = 0 . 9 · ――――――― = 16.6 N/mm 1.3 0.26 · 16,800 · 12,000 · 17.5 · 100 · 12.5 = ―――――――――――――――――――――――― 2.46 · 1012

Md · E3 h3 σm,3,d = ――――― · γ3 · a3 + ―― = 0.47 kN EI 2 ef In addition, the fastener must transfer the load 6 31.3 · 10 · 16,800 200 portion, which is accommodated by the reinforce­ = ―――――――――――― · 0.26 · 17.5 + ――― 2.46 · 1012 2 ment panels. This corresponds to the ratio of

bending stiffness of the reinforcements boards = 22.3 N/mm2 (B.8) (EI)3 to the overall stiffness (EI)ef. 22.3 η = ――― = 0.43 ≤ 1.0 7 51.9 (EI)3 16,800 · 4.0 · 10 ――― = ―――――――――――――― = 0 . 2 7 (EI) 2.46 · 1012 where ef

75 N/mm2 2 fm,3,d = 0 . 9 · ――――――― = 51.9 N/mm (Note: Ignoring the distance of the centres of ­gravity) 1.3 The loading of the reinforcement boards is thus Verification of maximum shear stress 0.27 x 10.0 kN/m = 2.70 kN/m.

2 The fastener in the internal quarters and perpen­ (γ3 · E3 · A3 · a3 + 0.5 · E2 · b2 · h ) · Vd τ2,max,d = ―――――――――――――――――――――― dicular to the member axis is subject to loading of b · (EI) 2 ef 2.70 kN/m x 0.10 m = 0.27 kN. The fastener subject to maximum loading with­ (0.26 · 16,800 · 12,000 · 17.5) · 25 · 103 = ―――――――――――――――――――――――― stands 140 · 2.46 · 1012

F = 0.472 + 0.272 = 0.54 kN ≤ F = 1.1 kN (0.5 · 11,000 · 140 · 117.52) · 25 · 103 d,res v,Rd + ―――――――――――――――――――――― 140 · 2.46 · 1012 0.54 η = ――― = 0.49 ≤ 1.0 1.1 = 0.84 N/mm2 (B.9) where F = 1.1 kN 0.84 v,Rd η = ――― = 0.61 ≤ 1.0 1.38

Example 24: Timber beams with screwed-on where panel strips made of BauBuche 4.0 N/mm2 2 Loads: qd = 3.2 kN/m, kmod = 0.8, fv,d = 0.9 · kcr · ―――――――― = 1.38 N/mm 1.3 service class 1

M = 14.4 kNm, V = 9.6 kN Fastener verification d d Dimensions: (1) Board BauBuche S, h = 60 mm The nail loads from the composite effect depend f (2) C24 as a beam, 100/200 mm on the shear force and the fastener spacing. Span ℓ = 6 m

Vd (x) · γ3 · E3 · A3 · a3 · e (x) Connection: fully threaded screws 6.0 x 200 mm, Fd = ――――――――――――――――― ( B . 1 0 ) (EI) Screw crosses inserted under 45° ef Spacing: external quarter 120 mm, The loads must be checked at the point of maxi­ internal quarter 300 mm mum shear force and at the beginning of the ex­ Fax,Rd = 13.6 kN per screw cross in panded fastener spacing. direction of the shear plane, acc. to ETA-11/0190 57 The distances of the centres of gravity of the 60 mm cross-sectional parts from the centre of gravity of 1 the overall cross-section

γ1 · E1 · A1 · (h1+h2) - γ3 · E3 · A3 · (h2+h3) a2 = ―――――――――――――――――――――――― 3

2 · ∑ γi · Ei · Ai 200 mm i=1

0.542 · 16,800 · 6,000 · (60 + 200) = ―――――――――――――――――――――――――――― 2 2 · (0.542 · 16,800 · 6,000 + 11,000 · 20,000)

Figure 34: Timber beam = 25.9 mm (B.6) 100 mm with Board BauBuche

60 + 200 a1 = ――――――― - 25.9 = 104.1 mm Slip modulus of the fastener: 2

in axial direction per screw Verification of stresses in the centre of gravity

Kax,ser,1 = 30 · 6· ( 2 · 60) = 15,300 N/mm γ1 · a1 · E1 · Md σc,1,d = ――――――――― (El) ef Kax,ser,2 = 25 · 6 · (200 - 2 · 60) = 17,300 N/mm 0.542 · 104.1 · 16,800 · 14.4 · 106 = ――――――――――――――――――― 1.36 · 1012 1 Kax,ser,ges = ―――――――――――― = 8 , 1 0 0 N / m m 1 1 = 10.0 N/mm2 (B.7) ――――― + ――――― 15,300 17,300 10.0 η = ――― = 0.24 ≤ 1.0 in joint line per screw cross 42.5

K = 2 · K · cos(45º)2 = 2 · 8,100 · 0.5 ser ax,ser,ges where = 8,100 N/mm 1.2 · 57.5 N/mm2 2 fc,1,d = 0 . 8 · ――――――――――― = 42.5 N/mm K = 2/3 · K = 2/3 · 8,100 = 5,400 N/mm 1.3 mean ser

The effective fastener spacing is γ2 · a2 · E2 · Md σt,2,d = ―――――――――― (El)ef seff = 0.75 · sexternal + 0.25 · sinternal = 0.75 · 120 + 0.25 · 300 = 165 mm 1.0 · 25.9 · 11,000 · 14.4 · 106 = ――――――――――――――――― 1.36 · 1012 Effective bending stiffness (EI) ef 2 3 = 3.01 N/mm (B.7) 2 Elef = ∑ (Ei · Ii + γi · Ei · Ai · a i )

i=1 3.01 η = ――― = 0.35 ≤ 1.0 = 16,800 · 1.80 · 106 + 0.542 · 16,800 · 6,000 · 104.12 8.62

+ 11,000 · 6.67 · 107 + 1.0 · 11,000 · 20,000 · 25.92 where = 1.36 · 1012 Nmm2 (B.1) 14.0 N/mm2 2 ft,2,d = 0 . 8 · ―――――――― = 8.62 N/mm with the reduction coefficients γ 1.3

1 γ1 = ―――――――――――――――2 Verification of stresses in the extreme fibres π · E1 · A1 · s1 1 + ――――――――――2 K1 · ℓ 0.5 · E1 · h1 · Md σm,1,d = ―――――――――― + σc,1,d 1 (El) = ――――――――――――――――――――― = 0 . 5 4 2 ( B . 5 ) ef π2 · 16,800 · 6,000 · 165 1 + ――――――――――――――― 6 5,400 · 6,0002 0.5 · 16,800 · 60 · 14.4 · 10 = ―――――――――――――――― + 10.0 1.36 · 1012 γ2 = 1,0

58 = 15.4 N/mm2 (B.8) 13.0 6.81 η = ――― = 0.26 ≤ 1.0 η = ――― = 0.50 ≤ 1.0 49.2 13.6

where 0.542 · 16,800 · 6,000 · 104.1 · 300 · 4.8 Fd (x = 1.5 m) = ―――――――――――――――――――――― 1.36 · 1012 · cos (45º) 80 N/mm2 2 fm,1,d = 0 . 8 · ――――――― = 49.2 N/mm 1.3 = 8.52 kN

0.5 · E2 · h2 · Md σm,2,d = ―――――――――― + σt,2,d 8.52 (El) η = ――― = 0.63 ≤ 1.0 ef 13.6

0.5 · 11,000 · 200 · 14.4 · 106 = ――――――――――――――――― + 3.01 1.36 · 1012 9.5 Reinforced connection

= 14.7 N/mm2 (B.8) Board BauBuche can be used to increase the load-carrying capacity of connections with laterally­ 14.7 loaded fasteners. For this purpose, the Board η = ――― = 0.99 ≤ 1.0 (without reinforcement: 14.8 ­BauBuche is glued-on to the timber members in η = 1.46) the areas of the shear planes (see Figure 35). where Due to the higher embedding strengths of the Board BauBuche in comparison to the timber­ 24 N/mm2 2 fm,2,d = 0 . 8 · ――――――― = 14.8 N/mm members to be connected, the load-carrying 1.3 ­capacity of the connection can be considerably increased. Another very positive effect comes Verification of maximum shear stress in the reduced splitting risk, since the glued-on

2 reinforcement boards represent a form of (γ 3 · E3 · A3 · a3 + 0.5 · E2 · b2 · h ) Vd τ2,max,d = ―――――――――――――――――――――― cross-reinforcement for the timber. b ·(El) 2 ef Werner (1995) sets out design equations for the same. These are based on equations from Chapter (0.5 · 11,000 · 100 · 125.92) · 9.6 · 103 = ―――――――――――――――――――――― 8 of Eurocode 5 (Johansen theory). 100 · 1.36 · 1012

= 0.62 N/mm2 (B.9) F where

h2 200 h = ―― + a2 = ――― + 25.9 = 125.9 mm 2 2

0.62 η = ――― = 0.50 ≤ 1.0 1.23 Figure 35: Reinforced ­connection where F 4.0 N/mm2 2 fv,d = 0.8 ·kcr · ――――――― = 1.23 N/mm 1.3

Verification of fastener

γ1 · E1 · A1 · a1 · s1 · Vd (0) Fd (x = 0 m) = ―――――――――――――― (EI) · cos (45º) ef

0.542 · 16,800 · 6,000 · 104.1 · 120 · 9.6 = ――――――――――――――――――――――――― 1.36 · 1012 · cos (45º)

= 6.81 kN

59 10 STRUCTURAL FIRE DESIGN 10.3 Strength values The strength and stiffness design values for verifi­ DIN EN 1991-1-2 cations in the load case fire are to be determined DIN EN 1995-1-2 in accordance with equations (2.1) and (2.2). DIN EN 13501-1, 2

f20 fd,fi = kmod,fi · ――― ( 2 . 1 ) γ M,fi

S20 Sd,fi = kmod,fi · ――― ( 2 . 2 ) γ 10.1 General M,fi With regard to timber constructions and fire, Here, when using laminated veneer lumber and ­unfortunately, given the fla mmable property of the reduced cross-section method, both the

wood, many people conclude that it is not suit­ ­partial factor γM,fi as well as the mo­dification

able for use in buildings subject to fire resistance ­factor kmod,fi can be taken as 1.0. The 20 %-quan­ requirements. tile values are obtained from the 5 %-quantile

Although wood is combustible, the combustion ­values multiplied by the correction factor kfi (for

behaviour is slow and uniform above all. The com­ laminated veneer lumber kfi = 1.1). bustion is delayed by evaporation of water as well Accordingly, equations (2.1) and (2.2) can be as the formation of a protective charcoal layer. simplified­ to This means the load-bearing behaviour of timber f = 1.1 · f members subject to fire actions can be effectively d,fi k

predicted and thus calculated. Sd,fi = 1.1 · S0,05 In most cases, the premature failure of the metal fastener is thus the key factor governing the over­ where all load-carrying capacity. Establishing suitable S = E bzw. G construction details can thus significantly boost 0,05 0,05 0,05 the resistance duration. The design value of the load-carrying capacity of a protected connection R may be determined 10.2 Requirements d,fi from the characteristic load-carrying capacity at Whether individual members have to meet fire normal temperature R as: protection requirements and the nature of such k

requirements is to be taken by referencing state- Rd,fi = kfi · Rk (2.3) specific legislation. where Board BauBuche may be classified in material

class E - normally combustible in accordance with Table 52: kfi -values DIN EN 13501-1 and thus designed with the design 1.15 Connection with side members made of values for the combustion rate under Eurocode 5. timber or wood-based panels In Table 51, details of the fire resistance classes 1.05 Connection with external steel plates in accordance with DIN EN 13501-2 are provided. 1.05 Axially loaded fastener The figure indicates the time in minutes, for which the load-carrying capacity of the member must Note: For unprotected connections (not covered be maintained in the event of a fire. here) the conversion factor η must also be used.

Table 51: R30 Fire retardant 10.4 Actions Fire resistance R60 Highly fire retardant 10.4.1 Design values of actions classes R90 Fire resistant The actions for the verifications in the load case fire R120 Highly fire resistant are to be determined in accordance with EN 1991-1-2.

The material class B2 in accordance with DIN E = ∑ γ · G + (ψ or ψ ) · Q + ∑ ψ · Q 4102 corresponds to material class E in accor­ d,fi GA,j k,j 1,1 2,1 k,1 2,i k,i (+ P + A ) dance with DIN EN 13501-1. d The designation FXX of DIN 4102 for the fire Prestressing (P) is not covered in more depth at ­resistance classes corresponds to the designation this point. Accidental actions (A ) need not be RXX of DIN EN 13501-2. d taken into consideration for the load case fire. 60 The combination coefficient used for the leading In addition, the following simplifications can be variable load Qk,1 can be either ψ1,1 or ψ2,1. made for verifications ­Contrary to the Eurocode recommendation, how­ – Compression perpendicular to grain can be ever, in those cases in which wind or snow is the disregarded leading variable load, ψ1,1 should be used. – Shear can be disregarded for rectangular and round cross-sections 10.4.2 Simplified determination of actions – For braced members, either the functional

For simplicity, the design value for actions Ed,fi ­integrity of the bracing is to be verified or a may also be derived from the action at normal stability verification must be performed with temperature at full buckling length.

Ed,fi = ηfi · Ed (2.8)

reduction coefficient ηfi k0 · d0 0.8 dchar,n dchar,0 0.7 d ψ = 0.9 3 ef fi 2 0.6 ψfi = 0.7 1 0.5 ψfi = 0.5 0.4 1 original cross-section Figure 37: Definition of 2 remaining cross-section remaining and effective ψfi = 0.2 0.3 3 effective cross-section cross-section in a fire

0.2 10.6 Charring 0.0 0.5 1.0 1.5 2.0 2.5 3.0 10.6.1 Charring of unprotected members load ratio Qk,1 / Gk To determine the remaining cross-section under

Figure 36: Reduction coefficient ηfi depending on the fire, a distinction is made between one-dimen­ load ratio Qk,1 to Gk (Figure 2.1) sional charring dchar,0 and charring taking into

consideration corner roundings and cracks dchar,n.

The reduction coefficient ηfi may either be deter­ The design of bar-shaped members must be mined from the diagram in Figure 36 or simplified ­performed with the combustion depth dchar,n, to 0.6 or for category E live loads to 0.7. while for plate-like members, dchar,0 may be used.

10.5 Design method dchar,0 = β0 · t (3.1) For design of timber members for the load case fire, Eurocode 5 presents two simplified methods dchar,n = βn · t (3.2) – Reduced cross-section method – Method with reduced strength and stiffness The following values can be applied in this case values for BauBuche. The method with reduced properties is only Table 53: Charring rates for BauBuche ­app­licable to softwood and is thus not covered in ­further detail at this point. Bar-shaped members: βn = 0.70 mm/min beams, tension members, columns 10.5.1 Reduced cross-section method

For this method, verifications are performed on Plate-like members: β0 = 0.65 mm/min an effective cross-section. This is obtained by panels ­deducting the charring depth as well as a transi­ tional layer between the combustion and the The effective charring depth def for fire protection ­unburnt wood, which is assumed to have no lon­ verifications for a fire duration of more than ger any strength or stiffness. 20 min is The strength and stiffness of the remaining (effec­ tive) cross-section are assumed to be unchanged def = dchar + k0 · d0 = dchar + 7 mm (4.1) (see Chapter 10.3). 61 Table 54: R30 R60 R90 R120 Here, the simplified determination with ηfi = 0.6

Charring depths dchar,0 19.5 39.0 58.5 78.0 leads to uneconomical design. Determining via in mm of panels dchar,n 21.0 42.0 63.0 84.0 Figure 36, conversely, allows a more accurate cal­ and beams def 28.0 49.0 70.0 91.0 culation of the loads. made of β0 = 0.65 mm/min BauBuche 10.6.2 Charring of protected members for panels with d ≥ 20 mm (1st line); Applying cladding panels can delay or even pre­ β = 0.7 mm/min n vent the start of the charring. The charring of for bar-shaped members (2nd/3rd line) only initially protected members is calculated in accordance with β0 may be reduced for panel thicknesses smaller than 20 mm and densities exceeding 450 kg/m3. 0 ; t ≤ tf For Pollmeier Board BauBuche, no adaptation is min {βn · ta ; 25} possible, since currently no boards below 20 mm (t - tch) · ――――――――――― ; t f < t ≤ ta (t - t ) a ch board thickness are produced. dchar =

βn · (t - ta) + min {βn · ta ; 25} ; t > ta Example 25: Bending stress verification for a roof beam in the event of a fire Here, for βn the values of the member to be pro­

Beam BauBuche GL75, 160/240 mm, ℓ = 6.0 m, tected can be used. tch describes the time up to e = 2.0 m, fire on three sides, R30 the point of failure of the fire protection cladding.

2 For fire protection claddings without gaps (smaller Loads: Dead weight gk = 1.50 kN/m 2 than 2 mm) tch corresponds to the time of the Snow qk = 3.00 kN/m start of charring tf for the member to be protected.

2 The fire resistance duration cht of the cladding is pd = 1.35 · 1.5 + 1.50 · 3.0 = 6.53 kN/m calculated for wood-based panels of thickness hp 2 pd,fi = 1.0 · 1.5 + 0.2 · 3.0 = 2.10 kN/m and with a charring rate of

hp The verification in the event of a fire is performed tch = ――― ( 3 . 1 0 ) β based on a reduced cross-section. For a fire 0

­duration of t = 30 min, the charring depth def is with β0 of the wood-based panel. The time limit ta is

2 · t d = 30 min · 0.7 mm / min + 7 mm = 28 mm f ef 25 ta = min ―――― + t f (3.8) 2 · β With the reduced section modulus, the verification is n with β of the member to be protected. 18.9 · 106 Nmm n ――――――――――――――――――― 2 σm,d, fi 104 mm · (212 mm) / 6 η = ―――― = ――――――――――――――― = 0 . 2 6 ≤ 1 , 0 Figure 38 shows the qualitative progression of the 2 fd,fi 94.1 N/mm charring of an initially protected member. where The rapid increase in charring (steep line) in f = k · k · f = 1.1 · 1.14 · 75 N/mm2 = 94.1 N/mm2 ­accordance with the failure of the fire protection d,fi fi h,m m,k cladding is attributable to accelerated burning 2 m · 2.10 kN/m2 · (6 m)2 Md,fi = ―――――――――――――――― = 1 8 . 9 k N m of the preheated wood behind the cladding. From 8 a charring depth of 25 mm, the combustion pro­

bef = 160 mm - 2 · 28 mm = 104 mm gression is likely to be more normal, since by that time, a sufficiently thick charcoal layer is present, h = 240 mm - 28 mm = 212 mm ef which delays the progression of the fire. Alternatively, p can also be determined in sim­ d,fi Figure 38 shows that while a thin protection plified form with the reduction coefficient η = 0.6 fi layer initially has a positive impact, this does not from the design value of the governing load com­ last, with no further beneficial effect on the pro­ bination at normal temperature gression of the charring. 2 2 pd,fi = 0.6 · 6.53 kN/m = 3.92 kN/m To prevent premature failure (collapsing) of fire protection cladding, the fasteners must have a or from Figure 36 depending on the load ratio q / g k k penetration length into the non-charred member 2 2 62 pd,fi = 0.33 · 6.53 kN/m = 2.15 kN/m of ℓA = min {10 mm ; 6 · d}. Charring depth d Charring depth d in mm 50 Figure 39: Unprotected member Progression d = 25 mm 40 Plasterboard of charring for 30 BauBuche various fire ­protection 20 ­claddings 10 tf ta tf ta Time t

thin protection layer unprotected 0 0 10 20 30 40 50 60 Time t in min effective protection layer

Figure 38: Qualita­tive progression of the charring of an 10.7 Connections with timber side members initially protected member 10.7.1 Unprotected connections Unprotected connections are not covered here. Example 26: Charring of a member with cladding 10.7.2 Protected connections The following example shows the charring depths If fasteners are protected from the effects of fire for fire protection cladding made of BauBuche and by cladding, it is important to ensure that the fire fire resistant plasterboards. To facilitate comparison, resistance duration of the cladding tch exceeds the combustion of the unprotected cross-­section the required fire resistance duration of the con­ is specified. nection treq minus half the fire resistance duration of the unprotected connection t . Board BauBuche d,fi

hp β0 = 0.65 mm/min tch = ―― ≥ t req - 0.5 · td,fi cf. (3.10); (6.2) β 0 h = 20 mm p From this results the required thickness h of the 25 p ta = min 2 · 30.8 min ; ――――――― + 30.8 min fire protective cladding made of BauBuche to mm 2 · 0.7 ― ――― min hp ≥ β0 · (treq - 0.5 · td,fi) = min {61.5 ; 48.6} = 48.6 min with a charring rate β = 0.65 mm/min for ­Board t = h / β = 20 mm / 0.65 mm/min = 30.8 min 0 ch p 0 BauBuche. For laterally loaded fasteners in double-shear and Plasterboard with timber side members, td,fi = 15 min, and for h = 12.5 mm connections with dowels, t = 20 min. p d,fi 25 ta = min 2 · 21 min ; ――――――― + 21 min mm Example 27: Fire protection cladding 2 · 0.7 ― ――― min for nail groups

= min {42 ; 38.9} = 38.9 min The nail groups of the tensile connections of a truss are designed for the fire resistance duration t = 2.8 · h - 14 = 2.8 · 12.5 - 14 = 21 min ch p R30. For this purpose, a fire protection cladding made of Board BauBuche is selected to cover the Protected member nail group.

βn = 0.7 mm/min The required panel thickness hp is revealed as follows:­ For a fire duration of 30 min, the following char­ mm ring depths dchar,n(30) result hp ≥ 0.65 ――― · (30 min - 0.5 · 15 min) = 14.6 mm min – BauBuche 0.0 mm – Fire resistant plasterboard 12.6 mm where the fire duration t = 30 min and the fire

– Unprotected 21.0 mm ­resistance duration of the fastener td,fi = 15 min To fasten the fire protection cladding in accordance For a fire of 30-minute duration, the cross-section with DIN EN 1995-1-2, Gl. (3.16) it is important to can be completely protected by the Board BauBuche. ensure that it is not at risk of falling off before the In Figure 39, the progression of charring is shown start of charring t of the member to be protected. 63 for the various design versions. ch 11 REFERENCES

Literature Approvals / ETA / declarations of performance Enders-Comberg, M., Blass, H.J. Treppenversatz – PM-005-2018 Declaration of performance – Laminated Leistungsfähiger Kontaktanschluss für Druckstäbe. veneer lumber made from beech. Laminated veneer Bauingenieur Band 89, 04/2014, Springer-VDI-Verlag, lumber according to EN 14374:2005-02 for non-load Düsseldorf ­bearing, load bearing and stiffening elements as of Blass, H.J., Ehlbeck, J., Kreuzinger, H., Steck, G. 27.07.2018. Pollmeier Furnierwerkstoffe GmbH, Creuzburg Erläuterungen zu DIN 1052: PM-008-2018 Declaration of performance – BauBuche Entwurf, Berechnung und Bemessung von Holz­ GL75 beam. Glued laminated timber made of hardwood – bauwerken. 2005, Bruderverlag, Munich Structural laminated veneer lumber made of beech Kreuzinger, H. Verbundkonstruktionen. ­according to ETA-14/0354 of 11.07.2018. Pollmeier ­Holzbau-­Kalender 2002, Bruderverlag, Karlsruhe ­Furnierwerkstoffe GmbH, Creuzburg Werner, H. Empfehlungen für die Bemessung von ETA-14/0354 European Technical Assessment ­Verbindungen mit verstärkten Anschlussbereichen. ETA-14/0354 as of 11.07.2018. Glued laminated timber Bauen mit Holz 12/1995, Bruderverlag, Karlsruhe made of hardwood – Structural laminated veneer lumber made of beech. Austrian Institute of Construc­ Standards tion ­Engineering, Vienna DIN 1052-10 Design of timber structures – Part 10: ETA-11/0190 European Technical Assessment Additional provisions, May 2012 ETA-11/0190 as of 23.07.2018. Self-tapping screws for DIN 4102 Fire behaviour of building materials and use in timber constructions. DIBt Deutsches Institut ­building components, May 1998 für Bautechnik, Berlin DIN EN 1990 Eurocode 0: Basis of structural design: ETA-12/0197 European Technical Assessment German version, December 2010 ETA-12/0197 as of 2019/02/28. Screws for use in DIN EN 1990/NA National Annex Germany – timber constructions.­ ETA-Danmark A/S, Nordhavn Eurocode 0: Basis of structural design, December 2010 Z-9.1-838 “German general construction technique DIN EN 1995-1-1/NA National Annex Germany – ­permit” as of 19.09.2018: Allgemeine Bauartgenehmi­ Eurocode 5: Design of timber structures – Part 1-1: gung. Furnierschichtholz aus Buche zur Ausbildung General – Common rules and rules for buildings, stabförmiger und flä-chiger Tragwerke – „Platte August 2013 BauBuche S“ und „Platte BauBuche Q“. Deutsches DIN EN 1995-1-2 Eurocode 5: Design of timber ­Institut für Bautechnik, Berlin ­structures – Part 1-2: General – Structural fire design, December 2010 Brochures DIN EN 1995-1-2/NA National Annex – Eurocode 5: 03 Building physics Brochure BauBuche – Building Design of timber structures – Part 1-2: physics as of October 2018. Pollmeier Furnierwerkstoffe General – Structural fire design, December 2010 GmbH, Creuzburg DIN EN 13501-1 Fire classification of construction 05 Fasteners Brochure BauBuche – Fasteners as of March ­products and building elements – Part 1: Classification 2019. Pollmeier Furnierwerkstoffe GmbH, Creuzburg using data from reaction to fire tests, January 2010 09 Wood preservation Brochure BauBuche – Wood DIN EN 14374 Timber structures – Structural laminated preservation and surface treatment as of February 2019. veneer lumber – Requirements, February 2005 Pollmeier Furnierwerkstoffe GmbH, Creuzburg ÖNORM B 1995-1-1 Eurocode 5: Design of timber struc­ tures – Part 1-1: General – Common rules and rules for buildings, June 2015 ÖNORM B 1995-1-1/NA National Annex Austria – Euro­ code 5: Design of timber structures – Part 1-1: General – Gym, Islisberg Common rules and rules for buildings Architecture: Langenegger Architekten AG, Muri SIA 2003 SIA 265 Timber structures. Swiss Society of Structural design: Makiol Wiederkehr AG, Engineers and Architects, Zurich Beinwil am See Completion: Max Vogelsang AG, Wohlen 64 Photos: Yves Siegrist

12 APPLICATION EXAMPLES

Joint Figure 40:

280/180 1000

Trusses made of 280/160 280/160 280/160 280/160 280/100 280/100 Beam BauBuche 280/160 1800 GL75 Joint Joint

Deformations displayed tenfold enlarged 36 mm after completion of building without camber of trusses 58 mm after 10 years 91 mm after 10 years considerung snow load 7400 280/300 +348

25.000

New construction of office and Trusses: Beam BauBuche GL75 (originally production ­facilities – Trusses planned and built in beam BauBuche GL70), The following application example is bottom chord 280/160, upper chord 280/180, based on the static calculation of the diagonals 280/160, posts 280/100 (edgewise office of merz kley partner ZT GmbH. arrangement of panels) Joint Truss connections: Fixing system WS-T-7 of SFS intec AG (in accordance with declaration of ­performance No. 100144897)

service class 1, kmod = 0.9

Dowel Continuous joint Dowel Ø10 30 60 60 30 2x3x6 WS-T-7 Ø7x133 Dowel Ø10 20 60 80 80 60 20 2x4 WS-T-7 Ø7x133 60/160 C24 Figure 41: TG Ø8x160 e=500 predrilled

Connection de­ 60 20 20 20 20 10 10 tails of trusses 10 180 5x20 5x20 90 39 90 39 10 30 60 30 40 60 10 30 40 30 10 10 10 30 30 2 x FLA t=5 S235 3 VG Ø8x300 predrilled 30 5x20 30 5x20 no slot at the 10 2x2x6 WS-T-7 Ø7x133 30 30 5x20 bottom side! 128 30 5x20 30 2x2x6 WS-T-7 Ø7x133 2x1x6 WS-T-7 Ø7x113 60 30 60 5x20 30 10 2x1x6 WS-T-7 Ø7x113 30

4 x FLA t=5 S235

4 x FLA t=5 S235 2x3x6 WS-T-7 Ø7x133 2 WS-T-7 Ø7x133 1640

2x3x6 WS-T-7 Ø7x133 2x2x6 WS-T-7 Ø7x133

30 220 30

10 80 10 100x200x16 S235 5 VG Ø8x200 30 3x5 CNa Ø6x80 30 5x20 staggered by 6 mm parallel to grain direction 30 10 10 60 60 30 30 30 5x20 72 60 2x28 WS-T-7 Ø7x133 30 60 80 60 30 72 30 60 80 60 30

10 per side 10 5 5 5 5

60 10

25 5 5 5 5 10 40 50 60 160 150 140 150 40 10 10

20 45 30 10 Dowel Ø10 100 7x60 60 45 20 30 25 4x40 35 30 30 420 30 2x4x6 WS-T-7 Ø7x133 no slot at the bottom side! 20 no slot at the bottom side! 2x2x2 WS-T-7 Ø7x133 4 x FLA t=5 S235 4 x FLA t=5 S235 66 480 540 540 Production hall, elobau sensor technology, Probstzella, Thuringen Architecture: F64 Architekten BDA Structural design: merz kley partner ZT GmbH Completion: Holzbau Amann GmbH Photos: Michael Christian Peters

Dowel Continuous joint Dowel Ø10 30 60 60 30 2x3x6 WS-T-7 Ø7x133 Dowel Ø10 20 60 80 80 60 20 2x4 WS-T-7 Ø7x133 60/160 C24 TG Ø8x160 e=500 predrilled 60 20 20 20 20 10 10 10 180 5x20 5x20 90 39 90 39 10 30 60 30 40 60 10 30 40 30 10 10 10 30 30 2 x FLA t=5 S235 3 VG Ø8x300 predrilled 30 5x20 30 5x20 no slot at the 10 2x2x6 WS-T-7 Ø7x133 30 30 5x20 bottom side! 128 30 5x20 30 2x2x6 WS-T-7 Ø7x133 2x1x6 WS-T-7 Ø7x113 60 30 60 5x20 30 10 2x1x6 WS-T-7 Ø7x113 30

4 x FLA t=5 S235

4 x FLA t=5 S235 2x3x6 WS-T-7 Ø7x133 2 WS-T-7 Ø7x133 1640

2x3x6 WS-T-7 Ø7x133 2x2x6 WS-T-7 Ø7x133

30 220 30

10 80 10 100x200x16 S235 5 VG Ø8x200 30 3x5 CNa Ø6x80 30 5x20 staggered by 6 mm parallel to grain direction 30 10 10 60 60 30 30 30 5x20 72 60 2x28 WS-T-7 Ø7x133 30 60 80 60 30 72 30 60 80 60 30

10 per side 10 5 5 5 5

60 10

25 5 5 5 5 10 40 50 60 160 150 140 150 40 10 10

20 45 30 10 Dowel Ø10 100 7x60 60 45 20 30 25 4x40 35 30 30 420 30 2x4x6 WS-T-7 Ø7x133 no slot at the bottom side! 20 no slot at the bottom side! 2x2x2 WS-T-7 Ø7x133 4 x FLA t=5 S235 4 x FLA t=5 S235 480 540 540 67 Verification of bottom chord: The design value of the compressive strength

Nd = 857 kN (Tension), Md = 3.82 kNm may be increased in service class 1 by the The tensile stress is factor 1.2. 0.9 857 · 103 N f = 1.2 · 1.0 · · 49.5 N/mm2 = 41.1 N/mm2 σ = = 25.4 N/mm2 c,0,d 1.3 t,0,d (160 - 4 · 7) · (280 - 4 · 6) mm2 The stability verification of the posts is covered by The design value of the tensile strength may be 2 increased by the coefficient kh,t, since the mem­ σc,0,d 2.54 N/mm η = = 2 = 0.10 ≤ 1 ber height is below 600 mm. In addition, the kc,z · fc,0,d 0.57 · 41.1 N/mm coefficient­ k has to be taken into consideration, ℓ Verification of diagonals: The governing parame­ since the member length exceeds 3.0 m. ters are Nd = 317 kN (compression), Md = 1.04 kNm 600 0.10 600 0.10 The compressive stress is kh,t = = = 1.08 h 280 3 317 · 10 N 2 σc,0,d = = 7.08 N/mm 3,000 s/2 3,000 0.12/2 280 mm · 160 mm k = min = = 0.92 = 0.92 ℓ ℓ 11,350 where 1.1 0.9 3.58 m f = 1.08 · 0.92 · · 60 N/mm2 = 41.3 N/mm2 λ = = 77.5 t,0,d 1.3 y 0.16 m / 12 The bending stress is From Table 14 kc,y can be derived at around 0.39. 3.82 · 106 N mm · 6 The design value of the compressive strength in σ ≈ = 5.14 N/mm2 m,d (280 - 4 · 6) mm · (160 - 4 · 7 mm)2 service class 1 may be increased by the factor 1.2. In addition, an increase may be achieved by the The design value of the bending strength is coefficientc,0 k . 0.9 f = · 75 N/mm2 = 51.9 N/mm2 k = min (0.0009 · h + 0.892 ; 1.18) m,d 1.3 c,0 = min (0.0009 · 160 + 0.892 ; 1.18) The verification of tension and bending in the = min (1.04 ; 1.18) = 1.04 ­bottom chord is covered by 0.9 f = 1.2 · 1.04 · · 49.5 N/mm2 = 42.8 N/mm2 σ σ 25.4 5.14 c,0,d 1.3 η = t,0,d + m,d = + = 0.71 ≤ 1.0 f f 41.3 51.9 t,0,d m,d The bending stress is Verification of posts: N = 68.7 kN (compression) d 1.04 · 106 N mm · 6 The compressive stress is σ = = 0.87 N/mm2 m,d 280 mm · (160 mm)2 68.7 · 103 N σ = = 2.45 N/mm2 The design value of the bending strength is c,0,d 280 mm · 100 mm 0.9 where f = · 70 N/mm2 = 48.5 N/mm2 m,d 1.3 1.81 m λz = = 62.7 0.10 m / 12

From Table 14 kc,z can be derived at around 0.57.

68 The stability verification of the diagonal is The shear stress is covered by V τ = 1.5 · d σ σ d h · b · k η = c,0,d + m,y,d cr k · f f c,z c,0,d m,y,d 33.1 · 103 N = 1.5 7.08 N/mm2 0.87 N/mm2 (180 - 6 · 7) mm · (280 - 4 · 6) mm · 1.0 = + = 0.44 ≤ 1 0.39 · 42.8 N/mm2 48.5 N/mm2 = 1.41 N/mm2 Verification of the upper chord: N = 825 kN (compression), M = 13.5 kNm, d d The design value of the shear strength is Vd = 33.1 kN The compressive stress is 0.9 f = · 8.0 N/mm2 = 5.54 N/mm2 v,d 1.3 825 · 103 N σ = = 16.4 N/mm2 c,0,d 280 mm · 180 mm The verification of shear in the upper chord is ­covered by where 1.41 3.09 m η = = 0.25 ≤ 1 λy = = 59.5 5.54 0.18 m / 12 Verification of truss connections (example given

From Table 14 kc,y can be derived at around 0.62. for a connection loaded in tension in the bottom

The upper chord is braced by the roof panel. chord): Nd = 652 kN (tension) The design value of the compressive strength For the connections of the truss, according to the may be increased in service class 1 by the static calculation of office merz kley partner ZT factor 1.2. In addition, an increase may be GmbH, the fixing system WS-T-7 of SFS intec AG achieved by the coefficientc,0 k . was used. The design was conducted in accor­ dance with DIN EN 1995-1-1 with NA, paragraph 8 k = min (0.0009 · h + 0.892 ; 1.18) c,0 as a dowelled connection, taking into consider­ = min (0.0009 · 180 + 0.892 ; 1.18) ation the details of the manufacturer as well as = min (1.05 ; 1.18) = 1.05 the declaration of performance No. 100144897 0.9 of SFS intec AG. The design value of the load- f = 1.2 · 1.05 · · 49.5 N/mm2 = 43.2 N/mm2 c,0,d 1.3 carrying capacity of a connection with multiple shear planes and fasteners WS-T-7x133 mm is The bending stress is ­provided as an example for the connection loaded 6 13.5 · 10 N mm · 6 2 in tension in the bottom chord according to the σm,d = 2 = 8.93 N/mm 280 mm · (180 mm) static calculation, Fv,Rd = 19.3 kN. The verification of the connection loaded in tension (2 x 28 WS-T- The design value of the bending strength is 7x133 mm, 4 slotted-in steel plates) is met, taking 0.9 into consideration the effective number f = · 75 N/mm2 = 51.9 N/mm2 m,d 1.3 of fasteners where

The stability verification in the upper chord is N η = d ­covered by Fv,Rd · nef

σ σ η = c,0,d + m,d 652 k · f f = = 0.81 ≤ 1.0 c,y c,0,d m,d 19.3 · (2 · 2 · 5.86 + 2 · 2 · 4.52) 16.4 8.93 = + = 0.78 ≤ 1.0 The verification for the transfer of forces in the 0.62 · 43.2 51.9 steel plates is met with η ≤ 1.0.

69 70 Wooden skyscraper Suurstoffi 22, Risch Posts and beams in BauBuche GL75, Timber-concrete composite floors Architecture: Burkard Meyer Architekten BSA Structural design: MWV Bauingenieure AG Completion: Erne AG Holzbau 71 Photos: Bernhard Strauss 72 Office building euregon AG, Augsburg Frame construction: columns, beams from BauBuche GL75, Structural floor construction from Board BauBuche Q, floor assembly from BauBuche Floor Architecture: lattkearchitekten BDA Structural design: bauart konstruktions GmbH Completion: Gumpp & Maier GmbH 73 Photos: Eckhart Matthäus Carpentry Anton Mohr, Andelsbuch visible structural construction from Beam BauBuche GL75 Architecture: Andreas Mohr, Structural design: merz kley partner ZT GmbH Completion: Kaufmann Zimmerei Photos: Christian Grass

74 Parking deck with timber-concrete composite slabs, columns and beams from BauBuche GL75, Research project of TUM.Wood in cooperation of Professor Hermann Kaufmann, Florian Nagler, Stefan Winter, Klaus Richter, Jan-Willem van de Kuilen

75 Imprint

Publisher: Pollmeier Massivholz GmbH & Co.KG Pferdsdorfer Weg 6 99831 Creuzburg, Germany Phone +49 (0)36926 945-0, F -100 [email protected] www.pollmeier.com Project management: Dipl.-Ing. Jan Hassan

Authors: Univ.-Prof. Dr.-Ing. Hans Joachim Blass Dipl.-Ing. Johannes Streib Ingenieurbüro für Baukonstruktionen Blaß & Eberhart GmbH Pforzheimer Straße 15b 76227 Karlsruhe, Germany

Photos: Yves Siegrist, Muri, Switzerland Michael Christian Peters, Amerang, Germany Markus Bertschi, Zürich, Switzerland Eckhart Matthäus, Wertingen, Germany Christian Grass, Dornbirn, Austria

Visualisations: Hof 437, Thomas Knapp, Alberschwende, Austria

Layout: Atelier Andrea Gassner, Feldkirch, Austria Reinhard Gassner, Marcel Bachmann

3rd revised edition 2019

Set in Univers Next

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