LVL Handbook EUROPE

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9. CALCULATION EXAMPLES OF LVL STRUCTURES
w_(inst,q)=(5〖∙q〗_(d,SLS)∙s∙L^4)/(〖384∙E〗_mean∙I)+〖〖6/5 ∙q〗_(d,SLS)∙s∙L〗^2/(〖8∙G〗_mean A)=5,97
mm+0,38 mm=6,35 mm
w_inst = 2,86 mm+6,34 mm = 9,2 mm
Requirement: w_inst≤L/400=4500/400=11,3 mm→OK
Final deflection
w_(net,fin) = (1+k_def)∙w_(inst,g) + (1+ψ_2∙k_def)∙w_(inst,q) (4.73)
For the load category A: ψ2 = 0,3
w_(net,fin) = (1+0,6)∙2,86 mm + (1+0,3∙0,6)∙6,35 mm = 12,1 mm
Requirement: w_(net,fin)≤L/300=4500/300=15 mm→OK
Vibration design
Lowest natural frequency f 1
f_1=π/(2l^2 ) √((EI)_l/m) (4.79)
m = g_1+g_2+30 kg/m^2 = 60 kg/m^2+30 kg/m^2+30 kg/m^2 = 120 kg/m^2
Note: In Finnish NA the share of live load qk in the frequency calculation is 30kg/m2
〖(EI)〗_l = EI ∙(1000/s) = 7,15∙1011 Nmm^2 ∙(1000/400 mm)
〖(EI)〗_l = 1,79∙〖10〗^6 Nm^2/m
f_1=π/(2∙(4,5m)^2 ) √((1,79∙〖10〗^6 Nm^2/m)/(120 kg/m^2 ))=9,5Hz>8Hz→OK
→ The floor can be analyzed as a high frequency floor.
Floor stiffness perpendicular to the span direction based on 22 mm chipboard decking:
(EI)/m = 3500 N/mm^2∙1000 mm∙〖(22 mm)〗^3/12 = 3,11∙〖10〗^3 Nm^2/m
For a rectangular floor with overall dimensions b x l, simply supported along all four edges, the impulse
velocity response v [m/Ns2] value may, as an approximation, be taken as:
v=4(0,4+0,6n_40 )/(mbl+200) (4.80)
n_40={((40/f_1 )^2-1) (b/l)^4 (EI)_l/(EI)_b }^0,25
(4.81)
n_40={((40/9,5Hz)^2-1) 〖∙(5m/4,5m)〗^4∙(1,79∙〖10〗^6 Nm^2/m
v=(4 (0,4+0,6 n_40 ))/(mbl+200)=(4 (0,4+0,6∙11))/(120∙5∙4,5+200
When a high value b = 150 is chosen from the Figure 4.28 and a conservative damping value ξ = 0,01 is
used, the requirement for v is
v≤〖150〗^((f_1 ξ-1) )=0,011→OK (4.78)
inst = inst,g + inst,q
inst,g = 5∙ d,SLS∙ ∙ 4
384∙ mean∙ + 6 5∙ d,SLS∙ ∙ 2
8∙ mean (4.74)
inst,g = 2,69 mm + 0,17 mm = 2,86 mm
inst,q 5 ∙ d,SLS ∙ ∙ 4
384 ∙ mean ∙ + 6/5 ∙ d,SLS ∙ ∙ 2
8 ∙ mean = 5,97 mm + 0,38 mm = 6,35 mm
inst = 2,86 mm + 6,34 mm = 9,2 mm
Requirement: inst ≤ 400 = 4500 400= 11,3 mm → OK
net,fin = (1 + def) ∙ inst,g + (1 + 2 ∙ def ) ∙ inst,q (4.73)
For the load category A: ψ2 = 0,3
net,fin = (1 + 0,6) ∙ 2,86 mm + (1 + 0,3 ∙ 0,6) ∙ 6,35 mm = 12,1 mm
Requirement: net,fin ≤ 300 = 4500 300= 15 mm → OK
inst = inst,g + inst,q
inst,g = 5∙ d,SLS∙ ∙ 4
384∙ mean ∙ + 6 5∙ d,SLS∙ ∙ 2
8∙ mean (4.74)
inst,g = 2,69 mm + 0,17 mm = 2,86 mm
inst,q = 5 ∙ d,SLS ∙ ∙ 4
384 ∙ mean ∙ + 6/5 ∙ d,SLS ∙ ∙ 2
8 ∙ mean = 5,97 mm + 0,38 mm = 6,35 mm
inst = 2,86 mm + 6,34 mm = 9,2 mm
Requirement: inst ≤ 400 = 4500 400= 11,3 mm → OK
net,fin = (1 + def) ∙ inst,g + (1 + 2 ∙ def ) ∙ inst,q (4.73)
For the load category A: ψ2 = 0,3
net,fin = (1 + 0,6) ∙ 2,86 mm + (1 + 0,3 ∙ 0,6) ∙ 6,35 mm = 12,1 mm
Requirement: net,fin ≤ 300 = 4500 300= 15 mm → OK
i t i t, i t,
i t,g 5∙ ,SL ∙ ∙ 4
38 ∙ m an∙ + ∙ ,S ∙ ∙ 2
8∙ ean .
inst,g , m + , 7 m = ,
i t,q = ∙ , ∙ ∙
∙ mea ∙ ∙ , ∙ ∙
∙ , , ,
i t = , , 4 m = ,2
ir 0 45 0 11, m → O
ne ,fin = ( d ) ∙ in ,g + ( ∙ def ∙ i t,q .7
or t l ad t g r : = ,3
, , , 0,3 , , = ,
i t: n t,fin 4500 1 m →
1 = 2 2 �( )l (4.79)
= 1 + 2 + 30 kg/m2 = 60 kg/m2 + 30 kg/m2 + 30 kg/m2 = 120 kg/m2
( )l = ∙ (1000/ ) = 7,15 ∙ 1011 Nmm2 ∙ (1000/400 mm)
( )l = 1,79 ∙ 106 Nm2/m
1 = π 2 ∙ (4,5m)2 � 1,79 ∙ 106 Nm2/m
120 kg/m2 = 9,5Hz > 8Hz → OK
( )/ = 3500 N/mm2 ∙ 1000 mm ∙ (22 mm)3/12 = 3,11 ∙ 103 Nm2/m
= 4(0,4+0,6 40)
+200 (4.80)
40 = ���401�2 − 1� � �4 ( )l ( )b�0,25 (4.81)
40 = ��� 40 9,5Hz �2 − 1� ∙ � 5m 4,5m �4 ∙ 1,79 ∙ 106 Nm2/m
3,11 ∙ 103 Nm2/m �0,25 = 11
= 4 (0,4 + 0,6 40 )
+ 200 = 4 (0,4 + 0,6 ∙ 11)120 ∙ 5 ∙ 4,5 + 200 = 0,010
≤ 150( 1 −1) = 0,011 → (4.78)
∙ 2
42∙ ∙( )l (4.82)
1 = 2 2 �( )l (4.79)
= 1 + 2 + 30 kg/m2 = 60 kg/m2 + 30 kg/m2 + 30 kg/m2 = 120 kg/m2
( )l = ∙ (1000/ ) = 7,15 ∙ 1011 Nmm2 ∙ (1000/400 mm)
( )l = 1,79 ∙ 106 Nm2/m
1 = π 2 ∙ (4,5m)2 � 1,79 ∙ 106Nm2/m
120 kg/m2 = 9,5Hz > 8Hz → OK
( )/ = 3500 N/mm2 ∙ 1000 mm ∙ (22 mm)3/12 = 3,11 ∙ 103Nm2/m
= 4(0,4+0,6 40)
+200 (4.80)
40 = ���401�2 − 1� � �4 ( )l
( )b�0,25 (4.81)
40 = ��� 40 9,5Hz �2 − 1� ∙ � 5m 4,5m �4 ∙ 1,79 ∙ 106Nm2/m
3,11 ∙ 103Nm2/m �0,25 = 11
= 4 (0,4 + 0,6 40)
+ 200 = 4 (0,4 + 0,6 ∙ 11)120 ∙ 5 ∙ 4,5 + 200 = 0,010
≤ 150( 1 −1) = 0,011 → (4.78)
∙ 2
42∙ ∙( )l 3 (4.82)
1 = 2 2 �( )l (4.79)
= 1 + 2 + 30 kg/m2 = 60 kg/m2 + 30 kg/m2 + 30 kg/m2 = 120 kg/m2
( )l = ∙ (1000/ ) = 7,15 ∙ 1011 Nmm2 ∙ (1000/400 mm)
( )l = 1,79 ∙ 106 Nm2/m
1 = π 2 ∙ (4,5m)2 � 1,79 ∙ 106 N 2/m
120 kg/m2 = 9,5Hz > 8Hz → OK
( )/ = 3500 N/mm2 ∙ 1000 mm ∙ (22 mm)3/12 = 3,11 ∙ 103 Nm2/m
= 4(0,4+0,6 40)
+200 (4.80)
40 = ���401�2 − 1� � �4 l ( )b�0,25 (4.81)
40 = ��� 40 9,5Hz �2 − 1� ∙ � 5m 4,5m �4 ∙ 1,79 ∙ 106 Nm2/m
3,11 ∙ 103 Nm2/m �0,25 = 11
= 4 (0,4 + 0,6 40 )
+ 200 = 4 (0,4 + 0,6 ∙ 11)120 ∙ 5 ∙ 4,5 + 200 = 0,010
≤ 150( 1 −1) = 0,011 → (4.78)
∙ 2 l
(4.82)
1 = 2 2 �( )l (4.79)
= 1 + 2 + 30 kg/m2 = 60 kg/m2 + 30 kg/m2 + 30 kg/m2 = 120 kg/m2
( )l = ∙ (1000/ ) = 7,15 ∙ 1011 Nmm2 ∙ (1000/400 mm)
( )l = 1,79 ∙ 106 Nm2/m
1 = π 2 ∙ (4,5m)2 � 1,79 ∙ 106Nm2/m
120 kg/m2 = 9,5Hz > 8Hz → OK
( )/ = 3500 N/mm2 ∙ 1000 mm ∙ (22 mm)3/12 = 3,11 ∙ 103Nm2/m
= 4(0,4+0,6 40)
+200 (4.80)
�� 401�2 − 1� � �4 ( )l ( )b�0,25 (4.81)
40 = ��� 40 9,5Hz �2 − 1� ∙ � 5m 4,5m �4 ∙ 1,79 ∙ 106Nm2/m
3,11 ∙ 103Nm2/m �0,25 = 11
= 4 (0,4 + 0,6 40)
+ 200 = 4 (0,4 + 0,6 ∙ 11)120 ∙ 5 ∙ 4,5 + 200 = 0,010
≤ 150( 1 −1) = 0,011 → (4.78)
1 = 2 2 �( )l (4.79)
= 1 + 2 + 30 kg/m2 = 60 kg/m2 + 30 kg/m2 + 30 kg/m2 = 120 kg/m2
( )l = ∙ ( 000/ ) = 7,15 ∙ 1011 Nmm2 ∙ (1000/400 mm)
( )l = 1,79 ∙ 106 Nm2/m
1 = π 2 ∙ (4,5m)2 � 1,79 ∙ 106Nm2/m
120 kg/m2 = 9,5Hz > 8Hz → OK
( )/ = 3500 N/ m2 ∙ 1000 mm ∙ (22 mm)3/12 = 3,11 ∙ 103Nm2/m
= 4(0,4+0,6 40)
+200 (4.80)
40 = ��� 401�2 − 1� � �4 ( )l ( )b�0,25 (4.81)
40 = ��� 40 9,5Hz �2 − 1� ∙ � 5m 4,5m �4 ∙ 1,79 ∙ 106Nm2/m
3, 1 ∙ 103Nm2/m �0,25 = 11
= 4 (0,4 + 0,6 40)
+ 200 = 4 (0,4 + 0,6 ∙ 11)120 ∙ 5 ∙ 4,5 + 200 = 0,010
≤ 150( 1 −1) = 0,011 → (4.78)
2
1 = 2 2 �( )l (4.79)
= 1 + 2 + 30 kg/m2 = 60 kg/m2 + 30 kg/m2 + 30 kg/m2 = 120 kg/m2
( ) l = ∙ (10 / ) = 7,15 ∙ 101 Nm 2 ∙ (10 /40 m )
( ) l = 1,79 ∙ 106 Nm2/m
1 = π 2 ∙ (4,5m)2 � 1,79 ∙ 1 6 Nm2/m
120 kg/m2 = 9,5Hz > 8Hz → OK
( )/ = 350 N/mm2 ∙ 10 m ∙ (22 m ) 3 /12 = 3,11 ∙ 103 Nm2/m
= 4(0,4+0,6 40)
+2 0 (4.80)
40 = ���401�2 − 1� � �4 ( )l ( )b�0,25 (4.81)
40 = � � 40 9,5Hz �2 − 1� ∙ � 5m 4,5m �4 ∙ 1,79 ∙ 106 Nm2/m
3,11 ∙ 103 Nm2/m �0,25 = 1
= 4 (0,4 + 0,6 40)
+ 200 = 4 (0,4 + 0,6 ∙ 11)120 ∙ 5 ∙ 4,5 + 200 = 0, 10
≤ 150( 1 −1) = 0, 11 → (4.78)
2
11
217 (255)
i , 450 4 0 = 11,3 m → OK
Final deflection
e , i = ) ∙ in , 2 ∙ ∙ i , (4.
For the load category A: ψ = 0,3
, , , , , , mm = 12,1 mm
i net,fi , 450 mm 300 = 15 m → OK
Vibration design
owest natural frequency f1
1 = 2 2 �( )l (4.79)
= 1 + 2 + 30 kg/m2 = 60 kg/m2 + 30 kg/m2 + 30 kg/m2 = 120 kg/m2
Note: In Finnish NA the share of live load qk in the frequency calculation is
30kg/m2 ( )l = ∙ (1000/ ) = 7,15 ∙ 1011 Nmm2 ∙ (1000/400 mm)
( )l = 1,79 ∙ 106 Nm2/m
1 = π 2 ∙ (4,5m)2 � 1,79 ∙ 106 Nm2/m
120 kg/m2 = 9,5Hz > 8Hz → OK
→ The floor can be analyzed as a high frequency floor.
Floor stiffness perpendicular to the span direction based on 22 mm chipboard
decking:( )/ = 3500 N/ m2 ∙ 1000 mm ∙ (22 mm)3/12 = 3,11 ∙ 103 Nm2/m
For a rectangular floor with overall dimensions b x l, simply supported along all
four edges, the impulse velocity response v [m/Ns2] value may, as an
approximation, be taken as:= 4(0,4+0,6 40)
+200 (4.80)
40 � �401�2 − 1� � �4 ( )l ( )b�0,25 (4.81)
40 = ��� 4 9,5Hz �2 − 1� ∙ � 5m 4,5m �4 ∙ 1,79 ∙ 106 Nm2/m
3,11 ∙ 103 Nm2/m �0,25 = 11
= 4 (0,4 + 0,6 40 )
+ 200 = 4 (0,4 + 0,6 ∙ 11)120 ∙ 5 ∙ 4,5 + 200 = 0,010
When a high value b = 150 is chosen from the F igure 4.28 and a conservative
damping value ξ = ,01 is used, the requirement for is
217 (255)
i , ,
i l l i
, , ,
r l r :
, , , , , ,
,
i ti i
l e cy 1
� l 9
2
N i i li l i l l i is
= ,
= , m
, ) , ,
l c n be an lyze i ncy floor.
l i i l i i i
i m 00 , m
l l i ll i i l i l l ll
i l l i p 2 l
i i be taken as:
, , 0
4 � − 1� � l , (4.81)
, , , 2
, m m � ,
, , , , , = 0,010
i l i i i
i l i i i
LVL Handbook Europe 181