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C 3 TIMBER FRAME WALL STUD
DESIGN DATA
Subject
It is required to check the adequacy of 38 x 89 mm Strength class C24 studs spaced at 600 mm
centres in a ground floor external domestic wall panel.
The wall supports 4.0 m span first floor
joists, the first floor wall panel and a trussed rafter roof spanning 8.0 m.
The stud length is 2.4 m
and the panel is sheathed externally with plywood and lined internally with plasterboard.
2X.r is a relatively complicated design problem because the stwls are subject to fou r
separate variable actions.
I t is necessary to ident& the source of each acti on, because
the values depend on the types of action concerned.
A quick method is given for checking the defl ection of a wind-l oaded column.
Service class
Service class 2
Clause 3.1.5
Properties of Strength class C24
BSEN 8
Properties for softwood Strength class C24 may be obtainedfr om BS EN 338 orfr om Table
I in the I ntr oduction to the Design Examples.
Bending strength
f
m.k
= 24.0
N/nUI?
Compression strength parallel to grain
f
c.0.k
=
21.0 N/tllrt?
Compression strength perpendicular to grain
f
0J.k
= 5.3 N/mm2
E mean parallel to grain
E
O,nrpn
= 11 000 N/mm2
Minimum modulus of elasticity parallel to grain Eo,05
= 7 400 N/mm*
Dimensions and section properties
Breadth of section
b =38 mm
Depth of section
h
=89 mm
Length of column L = 2400 mm
i%e
section size is a North American size, given in Table 4 of the I ntr oduction to the
Design Examples.
Area
A =
38 x 89 =
3 382 mm*
Section modulus
w,
38 x 84 =
50 170 mm3
6
Second moment of area
I, =
38 x 8g3 =
12
2 232 000 OM IT
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lhe major axis of a rectangular section is y-y and the minor axis z-z.
Figure 51.6
I t is assumed that the plywood sheathing restrai ns the stud against buckl ing abou t the
weaker z-z axis.
Actions
Self weight of:
rafters on slope 0.68 kN/m2
ceiling
first floor wall and ground floor wall construction
first floor joists + partition allowance (0.35 + 0.35)
Imposed loads on:
rafters
ceiling of roof
first floor
wind load say (C, + C,Jq
=
(0.7 + 0.3) x 0.87
=
On plan:
0.830 kNlm2
0.250 kN/m2
0.310 kN/m2
0.700 kN/m2
on plan:
0.53 kN/m2
0.25 l&/m2
1.50 kN/m2
0.87 kN/m2
There is dif ferentiation in EC5 between the duration of load for self weight (permanent),
storage load at ceil in g level (long-term), and imposed raf ter load and wind load (short-
Permanent load per stud
roof (rafters, ceiling etc)
first floor wall
first floor
= (0.830 + 0.25) x 0.6 x 8.0
2
= 0.31 x 0.6 x 2.40
= 0.70 x 0.6 x 4.00
2
ground floor wall (mid height)
0.31 x 0.6 x 2.4
=
2
ground floor wall (base)
= 0.31 x 0.6 x 2.4
EG,, (mid-height)
CG4, (base)
Long-term load per stud
roof ceiling
Q
0.25 x 0.6 x 8.0
141
=
2
Medium-term load per stud
first floor
Q
42
1.50 x 0.6 x 4.0
=
2
= 2.592
kN
= 0.446
kN
= 0.840
kN
= 0.223
kN
= 0.446
kN
= 4.101 kN
= 4.324
kN
= 0.600 kN
= 1.8OOkN
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C 3 TIMBER FRAME WALL STUD
Short-tern load per stud
roof rafter
Q
3
= 0.530
x
0.6
x
8.0
2
wind
Q
4
=
0.870 x 0.6 x 2.4
= 1.253 kN
= 1.272 kN
Partial safety factors and variable action coefficients
YG =
YQ =
Yh4 =
1.35
1.50
1.3
Table 2.3.3.1
Table 2.3.3.2
Values for the variable action coeficients or combination factors are given in Table 2
of the NAD and i n Table 7 of the I ntr oduction to the Design Examples.
Action
0 *I
Roof ceiling
441
0.5
0.4
First floor
Q
142
0.5
0.4
Roof rafter (snow) Qk3 0.7 0.2
Wind
Qt4
0 7
0 2
ULTIMATE LIMIT STATE
For the strength veri fi cation of wall s& s, it is usually necessary to check the beari ng
strength of the bottom plate, and the buckl ing resistance of the stud (combined bending and
compression) under wind load.
BEARING TRENGTHF BOTTOM LATE
The
earin g strength must be adequate for each of the our possibl e load cases: permanent
load only; permanent + long-term load; permanent + long-term + medium-term; and
permanent + long-term + medium-term + short-term. I n each load case, the strength
propert ies (i .e. the bearin g strength) are modif ied by the value of k_, which corr esponds
to the shortest durati on of load inclu ded in the load case.
3.1.7(2)
The cri ti cal load case may therefore be determined by divi ding the design value of the load
on the bottom plate for each l oad case by the corr esponding value of k_,.
The load case
which produces the largest quotient is the cri tical one.
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(2.3.2.2a)
C 3 TIMBER FRAME WALL STUD
Design value of actions
Nd = G G, +
YQ Qu + C o.i 7Q.i Qbi
i>l
Permanent
N+, =
YG (A.2
=
1.35 x 4.324 = 5.837 kN
Long-term
N
d,h8 = YG%.Z + YQQW
=
5.837 + 1.5 x 0.6 = 6.737 kN
Medium-term
There
are two vari able actions, Q,, and Q,,,
Since the value of ,, is the same for both
of them, the greater value of Nd s obtained by taking the lar ger vari able action as the
dominant one, i.e Qk,*
N
d,mcditon =
=
Short-term
yGG,2 + TQ (Q,, + v%,,,Q~,)
5.837 + 1.5 (1.8 + 0.5 x 0.6)
=
8.987 kN
Thr ee vari able actions, Qk,,, Qk,2 and Q,,. contr ibu te to the axial load, but in this load
case Q,, has a d@erent tiO alue. The dominant action may quickly be determi ned as the
one for which the product Qk,l (1 -+,,J is greatest.
Qw
(1 - Ad
=
0.6 (1 - 0.5) = 0.3
442 (1 - h.3
=
1.8 (1 - 0.5) = 0.9
Qw (1 - &,3
=
1.272 (1 - 0.7) =
0.38
Thus, Q,__ is the dominant action. Therefore
N
d&m = YG 2 + YQ (Q,, + k,Qw + ~o,~Q~J
=
5.837 + 1.5 (1.8 + 0.5 x 0.6 + 0.7 x 1.272)
For solid timber in Service class 2
=
10.32 kN
0.6
0.7
0.8
0.9
5.837
0.6
9.73
Table 3.1.7
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6.737 = 9.62
0.7
8.987 = 11.23
0.8
10.32 = 11.47
0.9
The short-term load case gives the maximum value, hence
Nd
=
Strength modification factors
Bearing strength
10.32 kN (short-term)
Clause 5. .
5
l he wall studs at each end of a wall panel share the load with the djacent stud in the
adjoin ing panels, so bearing beneath these end studs is not cri tical .
For 38 X 89 mm intermediate studs at 600 mm centres, the dimensions in EC5 Figure 5.1.5a are:
Distance to end:
a
> 1oomm
Bearing length:
e
= 38mm
Distance between studs: P, = (600 - 38) = 562
> 150 mm
From Table 5.1.5, for 4, > 150, a 1 100, 150 > P 2 15
k
c.90 =
1 + 150 - e =
1.659
170
Load sharing
Since only one bottom plate takes the verti cal loads in a wall panel, load shari ng is not
applicable to the verl jkation of beari ng strength.
Strength verification
Bearing stress
u
c.%d
=
Nd
A
10.32 x 1 000
=
3 382
=
3.05 N/mm
Bearing strength
f
c, d
k
c.90
k%d Q0.k
(2.2.3.2a)
YM
1.659 x 0.9 x
5.3
=
=
6.09 N/mm
1.3
f
c.5Q.d O,d Ei rvired
Bearing
strength
adequate
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C 3
TIMBER FRAME WALL STUD
BUCKLINGRESISTANCEF STUD
Design values of effects of actions
Since column design in volves a combined stress formula, combinat ions of stresses rather
than combin ations of loads must be investigated.
Axial compression stresses at mid-height
Design stress due to EG,,
Hence G,d
Design stress due to Qki
=
1.5 x 1 000 Qki =
3 382
Hence
,Ql .d
=
Lateral bending stress on major axis
Design
stress due to Qk4
=
Myd
w
Y
c.QZd
u
c,Q d
YG
CG,l
A
1.35 x 4.101
x 1 000
= 1.637
N/mm*
3 382
YQ Q, i
A
0.444 Qti
0.444 x 0.6
= 0.266
N/mm2
0.444 x 1.8
= 0.798
N/mm2
0.444 x 1.272
= 0.564
N/mm2
where
My,, =
YQ Q,2 L
8
Hence
,Q4,d =
1.5 x
1.253 1
000 X 2 400
=
11.24 N/mm2
8 x 50
170
Strength modification factors
Size
l e
characteristic bending strength of solid timber may be increased for depths in bending
of less than 150 mm.
Clause 3.2.2(4)
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k,, = 1.3
OR
Therefore k,,
= 1.11
Compression strength
The compression strength factor , k,, is most easily obtained from Table I in TRADA s
Design A id: Beam and Column Modif ication Factors.
(The alternati ve calculation
procedure is il lustrated in Design Example Cl ). The Design Aid also reproduces the
recommendations given in BS 5268 : Part 2 or the selection of effective length. I n this
example, the eflective length is taken as 0.85L.
For bending about the major axis
L
.=cY
=
0.85 x 2 400
=
2 040 mm
L
4Y
2 040
h =
89
From Figure 1 in the Design Aid
k
GY
Load sharing
=
22.9
=
0.445
The wall studs, in their support of the verti cal and hori zontal loads, may be considered to
act as part of a load distr ibu tion system. Clau se 54.6
k
IS
=
1.1
I
Since the short-term load case was criti cal for beari ng, it s evident that, with the addition
of another short-term variable load in the wind load, the short-term load case wil l be
I
I
critical for buckling also.
Design values of material properties
Compression strength
f
1.1 x 0.9 x 21.0
c,J,d
=
k,S kmcdfc,o,,
= =
15.99 N/mm2
YM
1.3
I
key is in troduced later i n the combined bendin g and compression formula.
I
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C 3 TIMBER FRAME WALL STUD
Bending strength
f
1.11 x 1.1 x 0.9 x 24.0
lsvP =
k, k,S kcd f,k _
-
=
20.29 N/mm2
YM
1.3
Combined bending and compression
I f either kc,Y or k,, is less than 1.0, it must be shown that the combined bending and
r
ompression requi rements given in Clause 5.2.
I(4)
are satisfied.
(Thi s requirement is not
given in EC5, but it is equivalent to the one given in Clause 5.2.1(3)).
k
=
C.Y
0.445, so for bending stresses applied to the major bending axis, it must be shown that
u
c.0.d
u
m.y.d
_
I
1
k,, fc,0,d fw,d
(5.2.1J must be evaluated accordin g to the combinati on ru les given by (2.3.2.2a) to
determine its limiting value.
CYGG, + yQ.1Qtg + W0.iQ.iQ i
(2.3.2.2a)
I t is evident that either Q,,, (as before or Qk,4
will be the dominant variable action.
With Qk2 as the dominant variable action, (5.2. If) and (2.3.2.2a) give
u
cGd +
. c,Qfd
+ ~O,lc,Ql,d + 0.3 4.Q3.d + ~0,4%,Q4.d
kc,y fc.O,d
f
my.d
1.637 + 0.798 + 0.5 x 0.266 +
0.7 x 0.564
+ 0.7 x 11.24
0.445 x 15.99
20.29
= 0.4164 + 0.3878 = 0.804 < 1.0 as required
With Q44 as the dominant variable action, (5.2. If) and (2.3.2.2a) give
U
c,G,d + GO.1 uc,Ql,d + 0.2 c,QZ,d + 0.3 uc,Q3,d + um,Q4,d
kc,y c.0.d
f
nqy,d
1.637 + 0.5 x 0.266 + 0.5
x 0.798 + 0.7
x
0.564 + 11.24
0.445 x 15.99
20.29
= 0.3603 + 0.5540 = 0.914 < 1.0 as required Section is
adequate to
resist buckling
@TRADA 1994. C
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c3 TlMBERFRAME WALL STUD
SERVICEABILITY LIMIT STATE - Deflection
FULL DESIGN METHOD
i ke in stantaneous deflection at mid-span caused by loading a lateral ly loaded pinned
column may be calcul ated as
uinrr =
uqbl [ EI -7.1pL ] + uh [E* YgL2 ]
where u,,,,~,
= instantaneous defI ection at mid-span due to lateral load only
ut__ = ini tial defl ection due to bow
E
= Eo,,
= mean modulu s of elasticity
4. (3)
I = second moment of area in direction of defl ection
P = Nxe7 = axial service load
L
= column length
Since the permitted in iti al bow in solid timber column members is L /300 (EC5 Clause
7.2.P(1)), it is neither appropriate not possible to incl ude u,, in the value of rr ,,
However , the additi onal deflection due to u,, p
educed by the axial l oad N,, is in cluded
and is represented by the second term in the above formula. Hence, using a value of
U - LNOO, the ormula for uin rr iven above may be rewritten as
bmv-
uincr =
1
[
+ O.ldL
1 - 0.14 u~im~
300
3
(1)
where t =
xec2
E I
0,~
I t may be assumed that the criti cal defl ection condition for a timberf rame wall stud will
be the instantaneous defl ection under wind load, with wind as the dominant variable
load. (Thi s is demonstrated in Design Example C2). For this condition, the defl ection
limitation recommended in EC5 is
u2,iWr
S L /300 Clause 4.3.1(2)
I nstantaneous defl ections are calcul atedfrom servi ce loads, i.e. the characteri stic loaak
un factor ed by ~c or yp The deflections thus calcul ated separately are combined
accordin g to the ru les given by:
+ Ctit,i Qtz,i
i I
(4.1.a)
The defl ection of timber fr me wall studs may be calcul ated using the eflective length of
2040 mm.
(p =
E_ f =
N,, x (2 400)2
=
0.0002346
o- Y
11 000 x 2232000
ls,
u],~,
is the in crease in ini tial bow caused by the dead weight alone.
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C 3 TIMBER FRAME WALL STUD
N =
ecI
EG,, =
4.101 kN
d
=
0.0002346 x 4.101 x 1 000
From equation (1) above
Ul,iEd =
1 O.lr L
[ I
-0.14 300
0.1 x 0.9621 x 2 400
= (1 - 0.1 x 0.9621) x 300
=
0.9621
=
0.85 mm
u~,~,~* s the deflection at mid-span caused by the wind alone.
F
Q
1.253 x 1 000
scr,udl =
-? =
2 400
Using Formula 2 from the Bending Formulae Design Aid
=
0.5221 x 2 4002 5 x 2 400* 89*
11 000 x 2 232 000 384
+T I
=
0.5221 N/mm
=
9.38 mm
uZ,inrrs the total instantaneous deflection caused by applyin g the fou r vari able loads i n
combinati on. I t is
c lcul t ed from
Equations (1) and (4.la) above as uin rr u,,~~.
With wind as the dominant variable action
= %,I + h.,
Qw
1,~
w
63 Qu
= 4.101 + (0.4 x 0.6) + (0.4 x 1.8) + (0.2 x 1.272) = 5.315 kN
= 0.0002346 x 5.315 x 1 000
=
1.247
=
1
I
0.14L
m 2,4,~+~
=
1
g 38
0.1 x 1.247 x
+
2 400
11.86 mm
*- 0.1 x 1.247 300
= u.
lnI
- ul,im*
= 11.86 - 0.85
=
11 01 m
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C 3 TIMBER FRAME WALL STUD
Recommended deflection limitation
Clause
4.3.1(2)
recommends that, in general, u,.,, 17.5, the deJection of a win d-loaded column should be acceptable
if it can be shown that
E
L2 crF
sI + P NJ 5 1.0
o-
I[
where F,, =
lateral (wind) service load (N)
N =
scr
axial service load (N)
( I I
P
Solid timber
General win d-loaded columns 3.9 0.20
Sheathed timber fr ame wall stu& 2.6 0.17
I he above formula should sati& the recommendations:
P
Glulam
0.16
0.14
for general columns
UZ.ilw
I L/300 and uM + I L /200
for timber rame wall stu&
u2,ilw
S L/2a) and uMr fi
I L/167
limber fr ame wall studs should be securely fastened to structural sheathin g materi al
which may be assumed to reduce the calcu lated defection to with in normal acceptable
limits.
L
=
24 mm
F
=
SC51
Q =
4
1
253 N
(dominant variable load)
= 4 101 + 0.4 x 600 + 0.4
x 1 800 + 0.2 x 1 272 = 5 315 N
=
; =
2.6
0.17
E
o,-
= 11 000 N/mm2
I = 2 232 000 mm4
L2
- I[ CXF
+
E
sr
@L]
2 4002
=
[2.6 x 1 253 + 0.17 x 5 315
11 000 x 2 232 000
]
o-
= 0.98 < 1.0 as
required.
Deflection
e
satisfactory
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