140 / 228
4. STRUCTURAL DESIGN OF LVL STRUCTURES
The calculations are made assuming that the floor is unloaded,
i.e. only the mass of the floor and other permanent actions are
accounted for. Note: in some National Annexes a part of the
live load is also taken into consideration, e.g. in Finland 30 kg/
m
2 31
. For a rectangular floor with span l, the fundamental fre-
quency
f
1
may be approximately calculated as follows:
f_1=π/(2l^2 ) √((EI)_l/m)
(4.78) (EC5 7.5)
where
m
is mass per unit area [kg/m
2
];
l
is the floor span [m]; and
(EI)
l
is the equivalent plate bending stiffness of the floor
about an axis perpendicular to the beam direction
calculated for 1 metre wide section [Nm
2
/m].
For a rectangular floor with overall dimensions b x l, simply
supported along all four edges, the value may, as an approxi-
mation, be taken as:
v=(4 ∙ (0,4 + 0,6 n_40 ))/(m ∙ b ∙ l + 200)
(4.79) (EC5 7.6)
where
v
is the unit impulse velocity response [m/Ns
2
];
n
40
in the number of first-order modes with natural
frequencies up to 40 Hz;
b
is floor width [m];
m
is the mass [kg/m
2
]; and
l
is the floor span [m].
The value of n40 may be calculated from:
n_40={((4
(4.80) (EC5 7.7)
where (EI)b is the equivalent plate bending stiffness of the
floor about an axis parallel to the beam direction calculated for
a 1 metre wide section [Nm
2
/m] and (
EI
)
b
<(
EI
)
l
.
The deflection under
F
=1 kN point load can be calculated
from equations:
w=min{█((F ∙ l^2)/(
(4.81) (EC5 NA, Finland)
31
or
w=(F ∙ l^2)/(43,6 ∙〖 k_(δ )∙ (
(4.82) (EC5 NA, Austria)
33
where
s
is the spacing of the floor beams [m]
k_δ=∜((EI)_b/(EI)_l )
(4.83)
with the limitation
k
δ
≤ b/l
For multiple span floor additional instructions can be
found, e.g., from the National annex of Austria
33
.
For residential floors with a fundamental frequency less
than 8Hz (
f
1
≤ 8Hz) a special investigation should be made.
Instructions for 4,5 Hz ≤
f
1
≤ 8 Hz cases are defined, e.g., in
the National Annexes of Austria
33
or Germany. In practice,
their requirements can be fulfilled only when the own weight
of the floor is >250 kg/m
2
, which is quite heavy for an LVL
floor structure.
4.4 COMBINED CROSS SECTIONS
4.4.1 Basic principles
Glued composite cross sections utilize the joints between
members, significantly increasing the stiffness and resistance
of the whole cross section compared to the members acting
separately. This composite action can be calculated for a me-
chanically jointed cross section, but the influence of joint slip
must then be taken into consideration and, therefore, the over-
all stiffness is much lower. The specific properties of the com-
posite cross section that are essential to the structural analysis –
effective stiffness
EI
eff
, normal stresses from bending moment,
and shear stress at the glued joints – can be defined according
to equations (4.84) – (4.88).
The effective stiffness
EI
eff
of a glued composite cross sec-
tion is calculated according to equation:
〖EI〗_eff= ∑_i▒〖 E_(i ) I_i+E_i A_i e_i^2 〗
(4.84)
where
EI
eff
is the effective stiffness of the composite cross
section [Nmm
2
];
Ei
is the modulus of elasticity of a part i [N/mm
2
];
Ii
is the moment of inertia of a part i [mm4], for
rectangular cross section Ii = bi∙hi3/12, where bi is the
width [mm] of the part and hi is the height [mm] of the
part;
A
i
is the cross-sectional area of a part i [mm
2
]; and
e
i
is the eccentricity of the part i = distance between
the centre of gravity of part i and neutral axis of the
entire composite cross section [mm].
The location of the neutral axis of a composite cross section
related to the bottom of the section is:
e_0=(∑_i▒〖E_i ∙ A_i ∙ a_i 〗)/(∑_i▒〖E_i ∙ A_i 〗)
(4.85)
where
a
i
is the distance between the centre of gravity of part i and
the bottom of the entire composite cross section [mm].
eff
= ∑
i i
+
i
i
i 2
i
I
i
is the moment of inertia of a part
i
[ m
4
], for rectangular c
b
i
∙h
i
3
/12
, where
b
i
is the width [mm] of the part and
h
i
is th
part;
A
i
is the cr ss-sectional are of a part
i
[mm
2
]; and
e
i
is the eccentricity of the part
i
= distance between the cen
and neutral axis of the entire composite cross section [m
he location of t e neutral axis of a composite cross section related to
section is:
0
= ∑
i
∙
i
∙
i
i
∑
i
∙
i
i
(4.8
where
a
i
is the distance between the centre of gravity of part
i
and
entire composite cross section [mm].
Normal stress from bending moment is calculated for composite cross
the equation:
i,d(z)
i
∙
(z)i
∙
d
(4.8
163 (255)
(4.78) (EC5 7.5)
where
m
is mass per unit area [kg/m
2
];
l
is the floor span [m]; and
(EI)
l
is the equivalent plate bending stiffness of the floor about an axis perpendicular
to the beam direction calculated for 1 metre wide section [Nm
2
/m].
For a rectangular floor with overall dimensions
b x l
, simply supported along all four dges,
the value may, as an approximation, be taken as:
= 4 ∙ (0,4 + 0,6
40
)
∙ ∙ + 200
(4.79) (EC5 7.6)
where
v
is the unit impulse velocity response [m/Ns
2
];
n
40
in the number of first-order modes with natural frequencies up to 40 Hz;
b
is floor width [m];
m
is the mass [kg/m
2
]; and
l
is the floor span [m].
The value of n
40
may be calculated from:
40
= {(( 40
1
)
2
− 1) ( )
4
( ) ( ) }
0,25
(4.80) (EC5 7.7)
where
(EI)
b
is the equivalent plate bending stiffness of the floor about an axis parallel to the
beam direction calculated for a 1 metre wide section [Nm
2
/m] and
(EI)
b
<(EI)
l
.
The deflection under
F =
1 kN point load can be calculated from equations:
= min { ∙
2
42 ∙
∙ ( )
l
∙
3
48 ∙ ∙ ( )
l
(4.81) (EC5 NA, Finland)
31
or
= ∙
2
43,6 ∙
∙ ( )
l
(4.82) (EC5 NA, Austria)
33
where
s
is the spacing of the floor beams [m]
= √ ( )
b l
4
(4.83)
163 (255)
(4.78) (EC5 7.5)
where
m
is mass per unit area [kg/m
2
];
l
is the floor span [m]; and
( )
is th equivale t plate bending stiffness of the floor about an axis perpendicular
to beam direction calculated for 1 metre wide section [Nm
2
/m].
For a rectangular floor with overall dimensions
b x l
, simply supported along all four edges,
the value may, as an approximation, be taken as:
= 4 ∙ (0,4 + 0,6
40
)
∙ ∙ + 200
(4.79) (EC5 7.6)
where
v
is the unit impulse velocity response [m/Ns
2
];
n
40
in the number of first-order modes with natural frequencies up to 40 Hz;
b
s floor width [m];
m
is the mass [kg/m
2
]; and
l
is the floor span [m].
The value of n
40
may be calculated from:
40
= {(( 40
1
)
2
− 1) ( )
4
( ) ( ) }
0,25
(4.80) (EC5 7.7)
)
b
is the eq ivale t late
i stif ess t floor about an axis parallel to the
beam direction calculat d for a 1 metre wide section [Nm
2
/m] and
(EI)
b
<(EI)
l
.
The deflection under
F =
1 kN point load can be calculated from equations:
= min { ∙
2
42 ∙
∙ ( )
l
3
48 ∙ ∙ ( )
l
(4.81) (EC5 NA, Finla d)
31
= ∙
2
43,6 ∙
∙ ( )
l
(4.82) (EC5 NA, Austria)
33
where
s
is the spacing of the floor beams [m]
= √ ( )
b
( )
l
4
(4.83)
163 (255)
(4.78) (EC5 7.5)
where
m
is mass per unit area [kg/m
2
];
l
is the floor span [m]; and
(EI)
l
is t equivale t plate bending stiffness of the floor ab ut an axis perpendicular
to the beam direction calculated for 1 metre wide section [Nm
2
/m].
For a rectangul r floor with overall dim nsio s
b x l
, simply supported along all four edges,
the value may, as an approximation, be taken as:
= 4 ∙ (0,4 + 0,6
40
)
∙ ∙ + 200
(4.79) (EC5 7.6)
where
v
is the unit impulse velocity response [m/Ns
2
];
n
40
in the number of first-order modes with natural frequencies up to 40 Hz;
b
is floor width [m];
m
i the mass [kg/m
2
]; and
l
is t floor span [m].
h value of n
40
may be calculated from:
40
= {(( 40
1
)
2
− 1) ( )
4
( ) ( ) }
0,25
(4.80) (EC5 7.7)
where
(EI)
b
is the equivalent plate bending stiffness of the floor about an xis parallel to the
bea direction calculated for a 1 metre wide section [Nm
2
/m] and
(EI)
b
<(EI)
l
.
The deflection under
F =
1 kN point load can be calculated from equations:
= min { ∙
2
42 ∙
∙ ( )
l
∙
3
48 ∙ ∙ ( )
l
(4.81) (EC5 NA, Finland)
31
or
= ∙
2
43,6 ∙
∙ ( )
l
(4.82) (EC5 NA, Austria)
33
where
s
is the spacing of the floor beams [m]
= √ ( )
b
( )
l
4
(4.83)
( 5)
( . ) (
. )
r
is
ss r it r [k /
2
];
l
is the floor span [m]; and
( I)
l
is t
iv l t l t
i
stiff ss f t fl r
t xis r
ic l r
t t
m ir cti
c lc l t f r tr i s cti
[
2
/ ].
F r r ct
l r fl r it v r ll i
si s
x l
, si
ly s rt l
ll f r
s,
t v l
y, s
r xi
ti , t ke s:
∙ ( ,
,
40
)
∙ ∙ + 200
( . ) (
. )
where
v
is t it i
ls v l city r s s [ / s
2
];
n
40
in the number of first-ord r
s it t r l frequencies up to 40 Hz;
b
is flo r i t [ ];
m
is the mass [k /m
2
];
l
i the fl r s [ ].
v l f
40
y c lc l t fr :
40
= {(( 40
1
)
2
− 1) ( )
4
( ) ( ) }
0,25
( . ) (
. )
I
b
is t
iv l
t l t
i stiff s f t fl r
t xis r ll l t t
be ir cti c lc l te f r tr i s cti [
2
/ ]
( I)
b
( I)
l
.
fl cti
r
k i t l
c calculated from equations:
= min {
∙
2
∙
∙ ( )
l
∙
3
48 ∙ ∙ ( )
l
(4.81) (
, i l
)
31
r
= ∙
2
43,6 ∙
∙ ( )
l
(4.82) (EC5 NA, Austria)
33
r
s
is the s ci f t fl r
s [ ]
= √ ( )
b
( )
l
4
( . )
163 (255)
(4.78) (EC5 7.5)
where
m
is mass per unit area [kg/m
2
];
l
is the floor span [m]; and
(EI)
l
i the equivalent plate bending stiffness of the floor about an axis perpendicular
to t beam direction calculated for 1 metre wide section [Nm
2
/m].
For a rectangular floor with overall dimensions
b x l
, simply supported along all four edges,
the value may, as an approximation, be taken as:
= 4 ∙ (0,4 + 0,6
40
)
∙ ∙ + 200
(4.79) (EC5 7.6)
where
v
is the unit impulse velocity response [m/Ns
2
];
n
40
in the number of first-order modes with natural frequencies up to 40 Hz;
b
is floor width [m];
m
is t mass [kg/m
2
]; and
l
is the floor span [m].
The value of n
40
may be calculated from:
40
= {(( 40
1
)
2
− 1) ( )
4
( ) ( ) }
0,25
(4.80) (EC5 7.7)
where
(EI)
b
is the equivalent plate bending stiffness of the floor about an axis parallel to the
beam direction calculated for a 1 metre wide section [Nm
2
/m] and
(EI)
b
<(EI)
l
.
The deflection under
F =
1 kN point load can be alculated from equations:
= min {
∙
2
42 ∙
∙ ( )
l
∙
3
48 ∙ ∙ ( )
l
(4.81) (EC5 NA, Finland)
31
or
= ∙
2
43,6 ∙
∙ ( )
l
(4.82) (EC5 NA, Austria)
33
where
s
is the spacing of the floor beams [m]
= √ ( )
b
( )
l
4
(4.83)
Figure 4.28. Recommended range of and relationship between a and b. Performance
improves in the arrow 1 directio and decrease i the arrow 2 directi n (EC5 Figure 7.2).
The calculations are made assuming that the floor is unloaded, i.e. only the mass of the floor
and other permanent actions are accounted for. Note: in some National Annexes a part of
the live load is also taken into consideration, e.g. in Finland 30 kg/m
2
31
. For a rectangular
floor with span
l
, the fundamental frequency
f
1
may be approximately calculated as follows:
1
= 2
2
√ ( )
l
138
LVL Handbook Europe