f3 floor joists

7/25/2019 F3 Floor Joists

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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


4/6

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

f3 floor joists

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