c3 timber frame wall stud

7/25/2019 C3 Timber Frame Wall Stud

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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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C 3

TIMBER FRAME WALL STUD

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

b TRADA 1994. C

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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)


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

a l-RADA 1994. C TRADA, Hughenden Valley, High Wycombc, Bucks. HP1 4 4ND. UK

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C3 TIMBER

FRAMJI WALL STUD

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)

ip TRADA 1994. C

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Wycombe,

Bucks. HP14 4ND. UK

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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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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

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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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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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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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@l-RADA 1994. C

c3 timber frame wall stud

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