LVL Handbook EUROPE

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4. STRUCTURAL DESIGN OF LVL STRUCTURES
Figure 4.26. Tension perpendicular to the grain stresses at the hole edges. (1) Risk of cracks due to the tension in perpendicular to grain 33.
h d is the height of the hole for rectangular holes. For round
holes h d = 0,7d may be used in the equation (4.59).
Load distribution length lt,90 is
l_(t,90)= {█(0,5∙(h_d+h (4.61)
Verification of shear stress concentration at the hole edge shall
fulfil the condition:
τ_d=k_τ∙(1,5 ∙ V_d)/(b ∙ (h-h_d ) )≤f_(v,d) (4.62)
Where
k_τ=1,85∙(1+a/h)∙(h_d/h)^0,2 (4.63)
τd is the design value of shear stress;
kτ is the factor to determine maximum shear stress due to
stress concentration;
a is the length of a hole [mm], for round holes a = hd; and
f(v,d) is the design value of edgewise shear strength
Bending stress at the location of a rectangular hole is verified
by the equations:
(M_d/W_n + M_(o,d)/W_o )/f_(m,d) ≤1 (4.64)
(M_d/W_n + M_(u,d)/W_u )/f_(m,d) ≤1 (4.65)
where
W_n=(b ∙ (h^2-h_d^2 ))/6 (4.66)
M_(o,d)=A_o/(A_u+A_o )∙V_d∙a/2 (4.67)
M_(u,d)=A_u/(A_u+A_o )∙V_d∙a/2 (4.68)
A_o=b∙h_ro and W_o=(b ∙ h_ro^2)/6 (4.69)
A_u=b∙h_ru and W_u=(b ∙ h_ru^2)/6 (4.70)
Wo and Wu is the effective section modulus of the beam at
the location of a hole [mm3]
fm,d is the edgewise bending strength [N/mm2]
Bending stress at the location of a round hole is verified by the
equations:
(M_d/W_n )/f_(m,d) ≤1 (4.71)
The resistance of LVL-P beams at the location of holes maybe
improved and larger hole sizes are allowed when they are rein-
forced by gluing wood-based panels such as plywood to both
sides of the beam around the holes. Detailed design instruc-
tions are given e.g. in chapter F3.2 of the Austrian ÖNORM B
1995-1-1:2015 33. As LVL beams are thin, internal reinforce-
ment with screws or glued-in-rods is not recommended.
t,90,d = d∙ℎ d 4∙ℎ ∙ �3 − �ℎd ℎ �2� + 0,008 ∙ d ℎr (4.59)
ℎr = � min(ℎro; ℎru) for rectangular holes
min(ℎro + 0,15 ∙ ; ℎro + 0,15 ∙ ) for round holes (4.60)
t,90 = � 0,5 ∙ (ℎd + ℎ) for rectangular holes
0,35 ∙ + 0,5 ∙ ℎ for round holes (4.61)
d = τ ∙ 1,5 ∙ d
∙ (ℎ−ℎd) ≤ v,d (4.62)
τ = 1,85 ∙ �1 + ℎ� ∙ �ℎd
ℎ �0,2 (4.63)
t, , d∙ d
∙ ∙ , ∙ (4.59)
ℎ i ro; l l
i , ∙ ; , ∙ l .
t, 0,5 ∙ (ℎd l l
, ∙ , ∙ l (4.
∙ , ∙
∙ , .
τ , ∙ ∙ � ℎ , .63)
d n+ o,d o m,d≤ 1 (4.64)
d n+ u,d u m,d≤ 1 (4.65)
where
2 2 (4.66)
d n+ o,d o m,d≤ 1
d n+ u,d u m,d≤ 1
wh re
n = ∙ �ℎ2−ℎd2� 6
o,d = o u
+ o ∙ d ∙ 2
u,d = u u + o ∙ d ∙ 2
o = ∙ ℎ o and o = ∙ ℎro2
6
u = ∙ ℎru and u = ∙ ℎru2 6
d n m,d≤ 1
d n+ o,d o m,d
d n u,d u m,d≤ 1
w re
n = ∙ �ℎ 2 ℎd2� 6
o,d = o u o∙ d ∙ 2
u,d = u u
+ o ∙ d ∙ 2
o ∙ ℎro and o = ∙ ℎro2
6
u = ∙ ℎru and u = ∙ ℎru2
6
d n ,d≤ 1
+ ,
,
,
, ≤
, o
, d
∙ ℎ and =
ru d 2
,
d n+ o,d o m,d≤ 1
d n+ u,d u m,d≤ 1
where
n = ∙ �ℎ2−ℎd2� 6
o,d = o u + o ∙ d ∙ 2
u,d = u u
+ o ∙ d ∙ 2
o = ∙ ℎro and o = ∙ ℎro2
6
u = ∙ ℎru and u = ∙ ℎru2
6
d n m,d≤ 1
d n+ o,d o m,d≤ 1
d n+ u,d u m,d≤ 1
where
n = ∙ �ℎ 2−ℎd2� 6
o,d = o u
+ o ∙ d ∙ 2
u,d = u u + o ∙ d ∙ 2
o = ∙ ℎro and o = ∙ ℎro2
6
u = ∙ ℎru and u = ∙ ℎru2 6
d n m,d≤ 1
d n+ ,d o m,d≤ 1
d n+ u,d u m,d≤ 1
where
n = ∙ �ℎ2−ℎd2� 6
o,d = o u
+ o ∙ d ∙ 2
u,d = u u + o ∙ d ∙ 2
o = ∙ ℎro and o = ∙ ℎro2
6
u = ∙ ℎru and u = ∙ ℎru2 6
d n m,d≤ 1
o,d
,d
d n , u ,d
= 2 2
, o ∙
, =
= 2
r
6
,
Figure 4.26. Tension perpendicular to the grain stresses at the hole edges. (1) Risk of cracks
due to the tension in perpendicular to grain 33. (Kuva_97_2 f t90 in rectangular hole,
Kuva_97_3 f t 90 in round hole)
The tension perpendicular to the grain force Ft,90,d depends on the shear force Vd and
bending moment Md at the edge of the hole:
t,90,d = d ∙ℎd 4∙ℎ ∙ [3 − (ℎd ℎ )2] + 0,008 ∙ d ℎ r (4.59)
where
ℎ r = { min(ℎro; ℎru) for rectangular holes
min(ℎro + 0,15 ∙ ; ℎro + 0,15 ∙ ) for round holes (4.60)
is the heig t of the hol for rectangular h les. For round holes hd = 0,7d may be
use in the quatio (4.59).
Load distribution length lt,90 is
t,90 = { 0,5 ∙ (ℎd + ℎ) for rectangular holes
0,35 ∙ + 0,5 ∙ ℎ for round holes (4.61)
erificati of shear stress concentration at the hole edge shall fulfil the condition:
d = τ ∙ 1,5 ∙ d ∙ (ℎ−ℎd) ≤ v,d (4.62)
Where
τ = 1,85 ∙ (1 + ℎ) ∙ (ℎd
ℎ )0,2 (4.63)
is the design value of shear stress;
τ is the fact r to determine maximum shear stress due to stress concentration;
LVL Handbook Europe 135