Timber Engineeing - Example 5A 2014-02-18
A5
Roof diaphragm:
q
w.k
1.3
kN
m
2
:= b 4.8m := H 4.8m := L 10m :=
Wind pressure:
From Appendix D6 in Actions on structures we get pressure coefficients for zone D and E.
Height/Depth ratio of the house is assumed to be 1.
C
pe.pressure
0.8 := C
pe.suction
0.5 :=
C
pe
C
pe.pressure
C
pe.suction
:=
0,8 0,3
0.5
Vertical wall, 90 deg towards
wind direction (zone D and E)
q
w.d
1.5 C
pe
q
w.k
:= q
w.d
2.54
kN
m
2
=
Hinged columns result in that half the load goes directly into roof diaghragm
F
d
q
w.d
H
2
L := F
d
60.84 kN =
q
d
q
w.d
H
2
:=
Floor and roof diaphragms
Simplified design procedure for floor diaphragms
For the simplified design procedure three conditions needs to be fullfilled.
(a) The span L 10m = must lie between 2 b 9.6m = and 6 b 28.8 m = OK !
(b) Failure in ULS must occur in fastners and not in beams and panels
(c) The panels are fixed to their supports, as defined in EC5 9.2.3.2 (1) and (4)
Spacing between nails along panel edges max 150 mm along the supports
Load effect:
qd
R
R
D
L
Nc
Nt
Vmax
Vmax
Lateral load = F
d
Nailed memebers
No splices next to each other
Plywood diaphragm beam
b
1/7
Timber Engineeing - Example 5A 2014-02-18
Bending strength of edge beams
--> Normally tension is decicive, when checking for the
compressive side think of buckling. Here the reduction
factor for the buckling is just ~4%
F
d
L
8 w h ( ) b
f
t.0.d
Dimensions of edge beams:
w
c16
0.045m := h
c16
0.120m :=
Tensile capacity of timber C16:
f
t.0.k.c16
10MPa :=
k
mod.C16
0.9 := (Solid timber C16, Service klass 2, short term action from wind)
m.C16
1.3 :=
Size effect of timber in tension:
k
h
min
150mm
h
c16
|
\
|
|
0.2
1.3 ,
(
(
(
h
c16
150mm if
1 otherwise
:= k
h
1.05 =
f
t.0.d
k
mod.C16
k
h
f
t.0.k.c16
m.C16
7.24 MPa = :=
F
d
L
8 w
c16
h
c16
( )
b
2.93 MPa = < f
t.0.d
7.24 MPa = OK
Shear strength of the diaphragms along the edges A-A
All shear stress must be taken by the panel material. The stress is
assumed to be uniform across the width of the diaphragm.
(Panel shear)
F
d
2 b t ( )
f
v.d
Thickness of panel:
t
ply
0.024m :=
Shear capacity of plywood
f
v.k
2.9MPa :=
k
mod.ply
0.9 := (Solid timber C16, Service klass 2, short term action from wind
Tables 2.3 and 3.1)
m.ply
1.2 :=
f
v.d
k
mod.ply
f
v.k
m.ply
2.17 MPa = :=
F
d
2 b t
ply
( )
0.26 MPa = < f
v.d
2.17 MPa = OK
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Timber Engineeing - Example 5A 2014-02-18
Lateral strength of the panel to support roof structure fixes along edges A-A
F
d
2
R
d
b
s
Lateral capacity of one individual fastener (R
d
)
Nail dimensions for square smooth nails
d
nail
2mm := l
nail
50mm := f
u
600MPa :=
Thickness of members, penetrationlengths
t
C16
l
nail
t
ply
26 mm = :=
t
1
t
ply
24 mm = := t
2
t
C16
26 mm = :=
Laterally loaded nails, General
my mm
0.4
:= Unit correction factor for mathcad
M
y.Rk
0.45 f
u
d
nail
2.6
my 1.64 kN mm = := [EC 5 eq. 8.14]
Densities of the wooden materials:
Timber C16:
Plywood P30:
k.C16
310
kg
m
3
:=
k.ply
410
kg
m
3
:=
Embedment strength:
m
emb
m
3
kg
MPa mm
0.3
:= Unit correction factor for mathcad
Timber C16:
Plywood P30:
f
h.k.C16
0.082
k.C16
d
nail
0.3
m
emb
20.65 MPa = :=
f
h.2.k
f
h.k.C16
:= [EC 5 eq. 8.15]
f
h.k.ply
0.11
k.ply
d
nail
0.3
m
emb
36.63 MPa = :=
f
h.1.k
f
h.k.ply
:= [EC 5 eq. 8.20]
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Timber Engineeing - Example 5A 2014-02-18
Panel-to-timber connection: [EC 5 eq. 8.6]
f
h.2.k
f
h.1.k
0.56 = :=
F
vRk1
f
h.1.k
t
1
d
nail
1.76 kN = :=
F
vRk2
f
h.2.k
t
2
d
nail
1.07 kN = :=
F
vRk3
f
h.1.k
t
1
d
nail
1 +
2
2
1
t
2
t
1
+
t
2
t
1
|
\
|
|
2
+
(
(
(
+
3
t
2
t
1
|
\
|
|
2
+ 1
t
2
t
1
+
|
\
|
|
(
(
(
0.58 kN = :=
F
vRk4
1.05
f
h.1.k
t
1
d
nail
2 +
2 1 + ( )
4 2 + ( ) M
y.Rk
f
h.1.k
d
nail
t
1
2
(
(
(
0.61 kN = :=
F
vRk5
1.05
f
h.1.k
t
2
d
nail
1 2 +
2
2
1 + ( )
4 1 2 + ( ) M
y.Rk
f
h.1.k
d
nail
t
2
2
(
(
(
0.48 kN = :=
F
vRk6
1.15
2
1 +
2 M
y.Rk
f
h.1.k
d
nail
0.48 kN = :=
The lowest failure load is the design load
F
vRk
min F
vRk1
F
vRk2
, F
vRk3
, F
vRk4
, F
vRk5
, F
vRk6
,
( )
0.48 kN = :=
F
vRd
F
vRk
k
mod.C16
k
mod.ply
max
m.C16
m.ply
,
( )
0.33 kN = :=
R
d
1.2 F
vRd
0.4 kN = := [EC 5: 9.2.4.2 (5)]
Spacing of nails: s
nails
50mm :=
F
d
2
R
d
b
s
nails
1 = 1 is true --> OK!!!
Utilization:
F
d
s
nails
2 b R
d
79.76 % =
4/7
Timber Engineeing - Example 5A 2014-02-18
Wall diaphragms [EC5: Racking load
Simplfed analysis of wall diaphragms - Method A
The simplifed method can be used if following demands are fullfilled.
- The panel has a tie-down at the end preventing overturning.
- The spacing of fasteners are constant along the perimeter of every sheet
- The width of each sheet is at least h/4
Number of panels if width of individual panel is 1,2m (b
i
)
b
i
1.2m := n
panels
b
b
i
4 = :=
For a wall made up of several wall panels, the design racking load-carrying
capacity of a wall should be calculated from:
F
v.Rd
F
i.v.Rd
=
[EC5 eq.9.20]
For one individual panel:
[EC5 eq.9.21]
F
i.v.Rd
F
f.Rd
b
i
c
i
( )
s
=
Thickness of wall panel
t
ply.wall
12mm :=
If the thickness of the outer panel is changed, the penetration length will change as well.
The lateral design capasity of an individual fastener is re-calculated with "new" values of
t1 and t2.
Thickness of members, penetrationlengths
t
timber
l
nail
t
ply.wall
38 mm = :=
t
1.wall
t
ply.wall
12 mm = := t
2.wall
t
timber
38 mm = :=
Panel-to-timber connection: [EC 5 eq. 8.6]
F
v.Rk.1
f
h.1.k
t
1.wall
d
nail
0.88 kN = :=
F
v.Rk.2
f
h.2.k
t
2.wall
d
nail
1.57 kN = :=
F
v.Rk.3.1
f
h.1.k
t
1.wall
d
nail
1 +
:=
F
v.Rk.3.2
2
2
1
t
2.wall
t
1.wall
+
t
2.wall
t
1.wall
|
\
|
|
2
+
(
(
(
+
3
t
2.wall
t
1.wall
|
\
|
|
2
+ 1
t
2.wall
t
1.wall
+
|
\
|
|
(
(
(
:=
F
v.Rk.3
F
v.Rk.3.1
F
v.Rk.3.2
0.58 kN = :=
5/7
Timber Engineeing - Example 5A 2014-02-18
F
v.Rk.4
1.05
f
h.1.k
t
1.wall
d
nail
2 +
2 1 + ( )
4 2 + ( ) M
y.Rk
f
h.1.k
d
nail
t
1.wall
2
(
(
(
0.38 kN = :=
F
v.Rk.5
1.05
f
h.1.k
t
2.wall
d
nail
1 2 +
2
2
1 + ( )
4 1 2 + ( ) M
y.Rk
f
h.1.k
d
nail
t
2.wall
2
(
(
(
0.65 kN = :=
F
v.Rk.6
1.15
2
1 +
2 M
y.Rk
f
h.1.k
d
nail
0.48 kN = :=
The lowest failure load is the design load
F
v.Rk
min F
v.Rk.1
F
v.Rk.2
, F
v.Rk.3
, F
v.Rk.4
, F
v.Rk.5
, F
v.Rk.6
,
( )
0.38 kN = :=
F
v.Rd
F
v.Rk
k
mod.C16
k
mod.ply
max
m.C16
m.ply
,
( )
0.27 kN = :=
F
f.Rd
1.2 F
v.Rd
0.32 kN = :=
[EC 5: 9.2.4.2 (5)]
Shape factor, c
i
b
0
H
2 2
1.2m = :=
c
i
1 b
i
b
0
if
b
i
b
0
otherwise
1 = :=
Spacing of nails: s
nails.wall
50mm :=
Capacity of one sheet:
F
i.v.Rd
F
f.Rd
b
i
c
i
s
nails.wall
7.66 kN = :=
Capacity of entire wall:
F
v.Ed
F
d
2
30.42 kN = :=
F
v.Rd.tot
F
i.v.Rd
n
panels
30.65 kN = :=
F
v.Rd.tot
F
v.Ed
1 =
1 is true --> sufficient with 50mm spacing
Utilization:
F
v.Ed
F
v.Rd.tot
99.26 % = OK!!!
6/7
Timber Engineeing - Example 5A 2014-02-18
Shear buckling of the sheet may be disregarded, provided that b net / t <= 100. Where bnet is
clearance between studs and t is thickness of sheet.
b
net
b
i
1.2m = :=
b
net
t
ply.wall
100 = OK! [EC5: 9.2.4.2 (11)]
Calculate the need of resistance against overturning (lifting force)
Horizontal load on one panel
F
i.v.Ed
F
v.Ed
n
panels
7.61 kN = :=
Force that need to be "achoraged" to the foundation.
Disregarding the self-weight
F
i.t.Ed
F
i.v.Ed
H
b
i
30.42 kN = :=
[EC5: eq 9.23]
Check of penetration length of nail: (EC 5: 8.3.1.2 (1))
pointside penetration should be at least 8d
8 d
nail
0.02 m = t
2.wall
0.04 m = OK!
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