f3 floor joists
TRANSCRIPT
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F 3 FLOOR JOISTS
DESIGN DATA
Subject
It is required to check the adequacy of 47 x 200 mm regularised domestic floor joists of Strength
class C24 timber on an effective span of 3.600 m spaced at 600 mm centres.
Service class
EC5 defi nes hree serv i ce classes w hi ch el at e o he moi stur e ont ent of imber.
Serv i ce
class 1 corresponds o a moi sture ont ent n most of nvoods f 12 or ess, w hich s i kely
t o occur in cent ral l y heat ed buil dings. For members made of soli d or glued t km i nat ed
t im ber, designs whi ch are v ali d or Serv i ce class 1 w il l also be val id or Serv i ce class 2.
Service class 1
Timber properties
For C24 timber BS EN 338 gives
Shear strength
f
v.k
= 2.5 N/Il l I I?
Bending strength f
W.
=
24.0 N / I l l &
Compression strength perpendicular to grain
fc,$,k
= 5.3
N/ l I l I l12
Mean modulus of elasticity parallel to grain
E,,,, = 11 000 N/mm*
Mean shear modulus
G,.
= 690
N/rlUll2
Average density
Pm,
= 420
kg/m3
sectionroperties
For secti on propem es, t arget sizes are used.
Cl ause 3.2.3 P(I )
The UK preferr ed t arget si zes or sawn sofrw oods r e show n n Tables 3 and 4 i n the
I nt roducti on o he Design Lk unples.
For regular i sed im ber, a t ot al of 5 mm s planed
off he op and bott om u@aces f he ori ginal dept h exceeds 150 mm.
Breadth
b
=47 mm
Depth
h =195 mm
Effective span
L = 3600 mm
Bearing length
e
=50 mm
Area
Section modulus
Second moment of area
A = 47 x 195
w, =
47 x 195*
6
I, =
47 x 1953
12
=
9 165 mm2
=
297 900 mm3
=
29040ooomnl*
16of44
TRADA, HughendenVaUey, High Wycombe, Bucks. HP14 4ND. UK
OTR@DA19 M.F
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F 3 FLOOR JOISTS
Self-weight
=
9.81 p_A
109
9.81 X 420 X 9 165
=
= i o 038 IrN,m
109
Actions ,
Weight of floor decking, ceiling, etc.
= 0.40 kN/&; imposed load (domestic) = 1.50 kN/m*
Permanent load: floor decking etc =
0.40 x 0.600 = 0.240 kN/m
self weight
= 0.038 kN/m
Therefore
Gk
* 0.278
kN/m
Medium-term imposed load:
Q c =
1.50 x 0.60 L 0.900 kN/m
There s a dif ferent kt ion n EC5 betw een he durat i on f oad or sel f w e& ht permanent
durat i on) and for imposed occupancy loadi ng, e.g. dw el l i ngs, off i ces, school s, et c.
(medium-t erm urati on).
Partial safetyactors
Parti al afety actors or acti ons and material smay be
obt ined
rom T& & s 6 to 8 n he
I nt roducti on o he Desi gn Examples, or
from
Tabl es and 2 of t he Nm.
Permanent actions
Variable actions
Mater&l factor for timber
YG
= 1.35
YQ
= 1.5
Ytvl
= 1.3
ULTIMATE LIMlT STATE
Design values of actions
I Clause 2.3.2.2P l)
I
By nspecti on, t s evi dent hat he medi um-t erm oad case w il l be crit i cal.
I
Design load
Fd
= YG Gk YQ QI;
I
= 1.35 x 0.278 + 1.50 x 0.900
= 1.725 kN/m
Shear force
v, =
1.725 x 3.600
2
Bending moment M, =
1.725 x 3.600*
8
= 3.105lcN
= 2.795 kN/m
=TRADA1994.F
TRADA, Hughenden Valley, High Wycombe, Bucks. HP14 4ND. UK
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F 3 FLOOR JOISTS
Strength modification factors
k& for medium -t erm oad dura t i on n Serv i ce class 1 = 0.80 (see Tabl e 8)
Table 3.1.7
I t s assumed hat
he
fl oor deck const ructi on il l estrai n he compression dge of he oi st
against at eral displacement hrough he ength of t he beam - i n thi s condi ti on kht =
1.0.
Cl ause 5.2.2(5)
The loor const ructi on s presumed o be able o rov i de at eral di stri buti on f oad so he
load sharing actor k, = 1.1
Cl ause 5.4.6(2)
The dept h of t he secti on > 150 mm so k,, = 1.0
Cl ause 3.2.2a
The beari ng actor , k c.,, i ncreases he eff ecnv e ear i ng st rengt h, so t s unnecessar y o
calculate c,,
unl ess he unfactor ed beari ng str ength rov es to be inadequate.
Strength verifications
Shear stress
7d
Shear strength
fv.d
Bending stress
Bending strength
f
mY d
Compression perpendicular to grain
Bearing stress
%Q,d
lSV,
A
1.5 x 3.105 x ld
9 165
1.1 x 0.8 x 2.5
1.3
Md
w,
1.1 x 0.8 x 24.0
1.3
=
d
b0,
= 0.51 N/mm
= 1.69NtmmZ
Shear
strength
satisfactory
= 9.38 N/mm2
= 16.25 N/mm2 i BC?Ildillg
strength
satisfactory
=
3.105 x 103
47 x 50
= 1.32 N/mm2
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F 3 FLOOR JOISTS
Ekaring
strength
f
c.9Q.d
=
kodfc9ok
.
YM
=
0.8 x 5.3
1.3
= 3.2CN/mm2
The oad dist ri buti on actor , k,=, s not appl icable o he beari ng str ength of he bott om
rai l , because onl y one member s nvo l ved. I f necessary , t s calcul at ed beari ng str engt h
could have been i ncreased by he actor kc,,.
Cl ause 5.1.5
SWtiOll
adequate in
shear,
bending
nd
bearing
for
medium-term
loads
SERVICEABILITY LIMIT STATE - Deflection
Service loads
Wi th only one var i able oad, serv i ce lo& are equal to the unfact ored characterist ic
loads. (4.la)
Permanent service load
=
Gk
0.278 kN/m
Medium-term service load
Fdlti
=
Qk =
0.900
klWm
Deflection formulae
I t s necessary o calculat e eparat el y he defl ecti ons due t o permanent and medi um-t erm
loadi ngs, as he kdcractor s or t ime-r el at ed eform at i on creep) are di fferent or each oad
duration.
Flexural deflection caused by u.d.1.
=
5Fff, L4
3$%&y
Shear deflection due to u.d.1.
=
4 FudlL2
8
G-A
bTRADA1994.F
TRADA,
Hughenden
alley, High Wycombe, Bucks. HP14 4ND. UKI
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F 3 FLOOR JOISTS
t$ = form actor = 1.2 or rectangular section.
Separate values or G_ are given in BS EN 338 and are equi valent to E0,_J6.
lhe instantaneous d@lection is calcul ated using E,,_ and G_. Clause 4.1(3)
Al ternatively, the o efl ection may be calculated using Formula 1 n he ntr oduction o the
Design Examples, Bending and De&& on Formul ae.
uo
=
precamber (zero in thi s case)
4
=
defZction due to permanent loads
U2
=
def lection due to vari able (imposed) loads
U
=
lb?
UJ + u2 - ug
The ecommended de$Lxtion limit s are:
Clause 4.3.1(l)
instantaneous imposed load u2*
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F 3 FLOOR JOISTS
Finaldeflection
Final deflect i ons re obt ai ned by nstan taneous efect i ons x (1 + kW h as descri bed in
Cl auses 4.1(4) and 4.1(7)
I
Val ues of kW may be obtained from Tabl e 9, or rom Tabl e 4.1 in EC.4.
For acti ons of
permanent durat ion on sol id t imber n Serv i ce class 1, k, = 0.80. For acti ons of
medi um-t erm urat ion, kdcr 0.25.
Permanent load
Ul,fi
=
ul,imdl + kef =
1.99 x 1 + 0.80) = 3.58 mm
Medium-term load
UZ,fin
=
UZ.dl k efl =
6.44 x 1 + 0.25) =
8.05 mm
kt.fi
=
Ul.fi"
+
%in =
3.58 + 8.05 = 11.63 mm
Maximum recommended deflections
I
UZ.imt 5
3600
=
12.00 mm
Actual = 6.44 mm Instantaneous
300
deflection
satisfactory
hai 5
3600
= lS.OOmm
Actual=
200
11,63 mm Final deflection
I
satisfactory
ul,inst + U2,im