LVL Handbook EUROPE

204 / 228

Since the screwing direction ε in the beam is 90° to the grain direction, it is not allowed to add the tension
capacity of the head to the withdrawal capacity of the treaded part in the beam. Therefore the character-
istic withdrawal capacity RT,k of the screw is calculated by the equation:
R_(T,k)=min{█(max(f_(ax,90,1,k) dl_(g,1);f_(head,k) d_h^2
(tens,k) )┤ (5.31)
The withdrawal strength fax,ε,k is determined by testing according to EN 14592 and according to EN 1382
or it can be determined at angle ε to the grain as follows:
f_(ax,ε,k)=〖k_ax ∙ f〗_(ax,90,k)/(1,5 cos^2 β + sin^2 β) (ρ_k/ρ_a )^0,8 (5.32)
The characteristic density ρk is 480 kg/m3 for LVL 48 P and 410 kg/m3 for LVL 32 P.
fax,90,k is the characteristic withdrawal strength parameter for a screw perpendicular to the grain direction
[N/mm2]. For screws in LVL, the characteristic withdrawal parameter may be assumed as
fax,90,k = 15 N/mm² for ρa = 500 kg/m³ and screws 6 mm≤ d ≤ 12 mm in softwood LVL/GLVL.
f_(ax,90°,1,k)=15 N/mm^2∙((480 kg/m^3)/(500 kg/m^3 ))^0,8=14,5 N/mm^2
When ε = 45°, kax = 1 and when β = 0°,
f_(ax,45°,2,k)=(1∙15 N/mm^2)/(1,5〖∙cos〗^2 0°+sin^2 0°) (( 410kg/m^3)/(500kg/m^3 ))^0,8=8,5 N/
mm^2
The different conditions of the equation (5.31) give a characteristic capacity RT,k :
1〖∶ f〗_(ax,90°,1,k)∙ d∙ l_(g,1)=14,5 N/mm^2 ∙6,0 mm∙55 mm=4,8 kN
2∶ f_(head,k) 〖∙d〗_h^2∙(ρ_k/ρ_a )^0,8=13,0〖N/mm〗^2∙(12 mm)^2∙((480 kg/m^3)/(350 kg/m^3
))^0,8=2,4 kN
3∶ f_(ax,45°,2,k)∙d∙l_(g,2)=8,5 N/mm^2 ∙6,0 mm∙68 mm=3,5 kN
4∶ f_(tens,k)=10kN
R_(T,k)=min{█(max(4,8 kN;2,4 kN)@3,5 kN@10 kN)┤=3,5 kN
Design resistance of the connection:
R_d=k_mod/γ_M ∙n^0,9 〖 ∙R〗_(T,k) (cos α+μ sin⁡α )
R_d=0,8/1,3∙2^0,9∙3,5 kN∙(cos45°+0,26∙sin45°)=3,6 kN
E_(d,ULS)≤R_d→OK
The canopy can be supported on a 51x200 mm LVL 48P ledger beam which is connected to 51 mm LVL 32P
wall studs with 2pcs 6,0x140 mm full threaded inclined screws. At the ends the ledger beam shall exceed the
studs edges at least 60 mm - 25,5 mm = 34,5 mm.
9. CALCULATION EXAMPLES OF LVL STRUCTURES
Penetration length in wall stud g,2 1
sin45° sin45°
k = 0,9 T,k (cos + sin ) (5.33)
T,k = min ⎩⎨ ⎧max � ax,90,1,k g,1; head,k h2 � k a�0,8 �
ax,ε,2,k g,2tens,k
(5.31)
ax,ε,k = ax ∙ ax,90,k
1,5 cos2 + sin2 � k a�0,8 (5.32)
ax,90°,1,k = 15 N/mm2 ∙ � 480 kg/m3
500 kg/m3�0,8 = 14,5 N/mm2
When ε = 45°, kax = 1 and when β = 0°,
ax,45°,2,k = 1 ∙ 15 N/mm2
1,5 ∙ cos 2 0° + sin2 0° � 410kg/m3
500kg/m3 �0,8 = 8,5 N/mm2
ax,α,1,k ∙ ∙ g,1 = 14,5 Nmm2 ∙ 6,0 m ∙ 55 m = 4,8 kN
head,k ∙ h2∙ � k a�0,8 = 13,0N/mm 2 ∙ (12 mm)2 ∙ � 480 kg/m3
350 kg/m3�0,8 = 2,4 kN
ax,α,2,k ∙ ∙ g, = 8,5 Nmm2 ∙ 6,0 mm ∙ 68 mm = 3,5 kN
tens,k = 10kN
T,k = min � max(4,8 kN; 2,4 kN)3,5 kN 10 kN = 3,5 kN
d = mod M∙ 0,9 ∙ T,k (cos + sin )
d = 0,8 1,3 ∙ 20,9 ∙ 3,5 kN ∙ (cos45° + 0,26 ∙ sin45°) = 3,6 kN
d,ULS ≤ d → OK
Penetration length in wall stud g,2 = − 1
sin45° = 140 mm − 51 mm sin45°= 68 mm
k = 0,9 T,k (cos + sin ) (5.33)
T,k = min ⎩⎨ ⎧max � ax,90,1,k g,1; head,k h2 � k a�0,8 �
ax,ε,2,k g,2tens,k
(5.31)
ax,ε,k = ax ∙ ax,90,k
1,5 cos2 + sin2 � k a�0,8 (5.32)
ax,90°,1,k = 15 N/mm2 ∙ � 480 kg/m3
500 kg/m3 �0,8 = 14,5 N/mm2
hen ε = 45°, kax = 1 and when β = 0°,
ax,45°,2,k = 1 ∙ 15 N/mm2
1,5 ∙ cos 20° + sin20° � 410kg/m3
500kg/m3 �0,8 = 8,5 N/mm2
ax,α,1,k ∙ ∙ g,1 = 14,5 Nmm2 ∙ 6,0 mm ∙ 55 mm = 4,8 kN
head,k ∙ h2∙ � k a�0,8 = 13,0N/mm2 ∙ (12 mm)2 ∙ � 480 kg/m3
350 kg/m3�0,8 = 2,4 kN
ax,α,2,k ∙ ∙ g,2 = 8,5 Nmm2 ∙ 6,0 mm ∙ 68 mm = 3,5 kN
tens,k = 10kN
T,k = min � max(4,8 kN; 2,4 kN)3,5 kN 10 kN = 3,5 kN
d = mod M∙ 0,9 ∙ T,k (cos + sin )
d = 0,8 1,3 ∙ 20,9 ∙ 3,5 kN ∙ (cos45° + 0,26 ∙ sin45°) = 3,6 kN
d,ULS ≤ d → OK
36 (253)
f ax,90,k is the characteristic withdrawal strength parameter for a screw
perpendicular to the grain direction [N/mm2]. For screws in LVL, the
characteristic withdrawal parame e may be assumed s fax,90,k =15
N/mm² for ρa = 500 kg/m³ and screws 6 mm≤ d ≤ 12 mm in softw od
LVL/GLVL.
ax,90°,1,k = 15 N/mm2 ∙ ( 480 kg/m3
500 kg/m3)0,8 = 14,5 N/mm2
When ε = 45°, kax = 1 and hen β ,
ax,45°,2,k = 1 ∙ 15 N/mm2
1,5 ∙ cos 20° + sin20° ( 410kg/m3
500kg/m3 )0,8 = 8,5 N/mm2
The different conditions of the equation (5.31) give a characteristic capacity RT,k :
1 ∶ ax,90°,1,k ∙ ∙ g,1 = 14,5 Nmm2 ∙ 6,0 mm ∙ 55 mm = 4,8 kN
2 ∶ head,k ∙ h2∙ ( k a)0,8 = 13,0N/mm2 ∙ (12 mm)2 ∙ ( 480 kg/m3
350 kg/m3) ,8 = 2,4 kN
3 ∶ ax,45°,2,k ∙ ∙ g,2 = 8,5 Nmm2 ∙ 6,0 mm ∙ 68 mm = 3,5 kN
4 ∶ tens,k = 10kN
T,k = min { max(4,8 kN; 2,4 kN)3,5 kN 10 kN = 3,5 kN
Design resistanc f t connection:
d mod M ∙ 0,9 ∙ T,k (cos + sin )
d 0,8 1,3 ∙ 20,9 ∙ 3,5 kN ∙ (cos45° + 0,26 ∙ sin45°) = 3,6 kN
d,ULS ≤ d → OK
he canopy can be supported on a 51x200 mm LVL 48P ledger beam which is connected to
1 mm LVL 32P wall studs with 2pcs 6,0x140 mm full threaded inclined screws. At the ends
he ledger beam shall exceed the studs edges at least 60 mm - 25,5 mm = 34,5 mm.
.8 Laterally loaded nail connection
canopy over the entrance of a one family house is supported to the external wall by a
1x300 mm LVL 48 P ledger beam. The beam is nailed to the edges of 45 mm thick LVL 32
wall studs which have spacing s = 600 mm. Line load from own weight gk is 0,3 kN/m and
posed load from snow qk is 3 kN/m.
236 (253)
f ax,90,k is the characteristic withdrawal strength parameter for a screw
perpendicular to the grain direction [N/mm2]. or scre s in L L, the
characteristic ithdra al para eter ay be assu ed as fax,90,k =15
/ ² for ρa = 500 kg/ ³ and scre s 6 ≤ d ≤ 12 in soft ood
L L/ L L.
ax,90°,1,k 1 / 2 ∙ ( 480 kg/ 3
500 kg/ 3)0,8 14,5 / 2
When ε = 45°, kax = 1 and hen °,
ax,45°,2,k 1 ∙ 15 / 2
1,5 ∙ cos 20° sin20° ( 410kg/ 3
500kg/ 3 )0,8 = 8,5 N/mm2
The different conditions of the equation (5.31) give a characteristic capacity T,k :
1 ∶ ax,90°,1,k ∙ ∙ g,1 14,5 2 ∙ 6,0 ∙ 55 4,8 k
2 ∶ head,k ∙ h2∙ ( k a)0,8 13,0 / 2 ∙ (12 )2 ∙ ( 480 kg/m3
350 kg/m3)0,8 2,4 k
3 ∶ ax,45°,2,k ∙ ∙ g,2 = 8,5 2 ∙ 6,0 mm ∙ 68 mm = 3,5 kN
4 ∶ tens,k 10k
T,k = min { ax(4,8 k ; 2,4 k )3,5 k 10 k 3,5 k
esign resista e of t co nection:
mod ∙ 0,9 ∙ T,k (cos + sin )
d 0,8 1,3 ∙ 20,9 ∙ 3,5 k ∙ (cos45° 0,26 ∙ sin45°) = 3,6 kN
d,ULS ≤ d → OK
he canopy can be supported on a 51x200 L L 48 ledger bea hich is connected to
1 L L 32 all studs ith 2pcs 6,0x140 full threaded inclined scre s. t the ends
he ledger bea shall exceed the studs edges at least 60 - 25,5 = 34,5 .
. t r ll l il ti
canopy over the entrance of a one fa ily house is supported to the external all by a
1x300 L L 48 ledger bea . he bea is nailed to the edges of 45 thick L L 32
all studs which have spacing s = 600 . Line load fro o n eight gk is 0,3 k / and
posed load fro sno qk is 3 k / .
202 LVL Handbook Europe