13

EXAMPLE 1 LVL ROOF BEAMIt is required to check the adequacy of a 4.00 m clear span Kerto LVL beam 75 mm x 400 mm supporting atrussed rafter roof. For the applied loads, see Figure 19. It can be assumed that the compression edge is fullyrestrained and adequate bearing is provided. For comparison purposes the beam will be checked using bothBS 5268 Part 2 and EC5 Part 1.1.

Fig ure 19

Snow load 4.12 kN/m

Ceil ing im posed load 1.31 kN/m

4.1 m

BS 5268 PART 2SERVICE CLASS

Mois ture con tent ≤ 20% ∴ Serv ice class 2 (Cl 1.6.4)

TIMBER PROPERTIES

GRADE STRESSESFrom Table 6

Bending parallel to grain as a joist = 13.9 N/mm2

Shear parallel to grain as a joist = 1.5 N/mm2

Modulus of elasticity mean = 12750 N/mm2

minimum = 10400 N/mm2

Shear modulus = E mean

20

DIMENSIONS OF SECTION PROPERTIES

Breadth of beam section b = 75 mm

Depth of beam section h = 400 mm

Span between bearing centres L = 4100 mm

Bearing length lbearing

= 100 mm

Area of cross section A = 75 400 30000× = mm 2

Section modulus Zx

= 75 400

6

2×

= 2 10 6× mm 3

Second moment of area Ixx

= 75 400

12

3×

= 400 10 6× mm 4

EC5 PART 1.1

Mois ture con tent ≤ 20% ∴Serv ice class 2 Cl 6.3a NAD

CHARACTERISTIC VALUES From Table 7

fm,k

= 51 N/mm2

fv,r,0,k

= 5.1 N/mm2

E0,mean

= 14000 N/mm2

E0,05

= 12400 N/mm2

G0,mean

= 960 N/mm2

A = 30000 mm2

Wy

= 2 x 106 mm3

Iy

= 400 x 106 mm4

Self weight of to tal struc ture 5.45kN/m

Note: In EC5 the ma jor bend ingaxis of a rec tan gu lar sec tion isdes ig nated y-y as shown be low

z

z

y y

x

x

14

LOADS

BS 5268 treats snow load as a medium term durationload and the storage and total self-weight as long term duration loads, hence the medium term case will becritical ie

w = 5.45 + 1.38 + 4.12 = 10.95 kN/m

STRENGTH CHECKBENDING CHECK

Bending moment M = 10 95 41

8

2. .×

= 23.01 kN/m

∴ Bending stress σm,a,//

= 23 01 10

2 10

6

6

. ××

= 11.50 N/mm2

Bending strength σm,adm,//

= K3 x K

7 x grade stress

Depth factor K7

= 0 81400 92300

400 56800

2

2.

( )

)

××

= 0.94

σm,adm,//

= 1.25 x 0.94 x 13.9

= 16.33 N/mm2

Bending stress

Bending strength=

1150

16 33

.

.= 0.70 ∴ OK

SHEAR CHECK

Shear force V = 10 95 41

2

. .×= 22.45 kN

Shear stress = 15 22 45 10

30000

3. .× ×

= 1.12 N/mm2

Shear strength = K3 x grade stress

= 1.25 x 1.5 = 1.88 N/mm2

Shear stress

Shear strength=

112

188

.

. = 0.60 ∴ OK

ACTIONS

EC5 treats the snow load as short term duration, thestorage load as long term and the total self weight aspermanent loads. Critical case is short term with thesnow load dominant and adopting the characteristicload combination given in EC1 Part 1

Fd = (1.35 x 5.45) + (1.5 x 4.12) + (0.7 x 1.5 x 1.38)

Cl 9.10 EC1.1

= 14.99 kN/m

Md

= 14 99 41

8

2. .× = 31.50 kN/m

σm,d

= 3150 10

2 10

6

6

. ××

= 15.75 N/mm2

fm,d

= k k k fh crit mod m.k

Mγ

Depth factor kh: Step lecture A9 recommends no

modification is made for depth effects.

Instability factor kcrit

: Effective length from TRADADA3. L

ef = 1.0 x L = 4100 mm

∴ σm,crit = 0 75 2. E

L h0,05

ef

bNAD Cl 6.5a

= 0 75 12400 75

4100 400

2. × ××

= 31.9

λrel,m = fm k

m crit

,

,σ =

51

319.= 1.26 Eq 5.2.2a

kcrit = 1.56 - 0.75 λrel,m Eq 5.2.2d

= 1.56 - 0.75 x 1.26 = 0.62

fm,d

= 0 62 0 9 5100

13

. . .

.

× ×= 21.89 N/mm2

15 75

2189

.

.= 0.72 ∴ OK

Vd

= 14 99 41

2

. .×= 30.73 kN

τd

= 15 30 73 10

30000

3. .× ×= 1.54 N/mm2

fv,d

= k fmod v, k

γ M

= 0 9 51

13

. .

.

×= 3.53 N/mm2

= 154

353

.

.= 0.44 ∴ OK

BS 5268 EC5

15

SERVICEABILITY

=

FUDL

5 4100

384 14000 400 10

12 4100

8 960 30000

4

6

2×× × ×

+ ×× ×

.

= 0.745 FUDL

To provide a low risk of cracking in a plasterboardceiling, the characteristic load combination given inEC1 Part 1 will be used. By inspection, the short termcase will be critical with the snow load dominant.

Fd1

= 5.45 kN/m Fd2

= 4.12 + 0.7 x 1.38

= 5.09 kN/m

∴u1, inst

= 0.745 x 5.45 u2,inst

= 0.745 x 5.09

= 4.06 mm = 3.79 mm

INITIAL DEFLECTION

Creep will be calculated using the quasi-permanentload combination using a k

def,perm of 1.0 Table 4.1

A value of ψ2 = 0.3 is assumed, for this example only,

for the long term ceiling imposed load pending avalue being given in EC1.1 NAD. This recognises thatit is likely that areas of the ceiling will be unloaded.

Fd.creep = 5.45 + (0 x 4.12) + (0.3 x 1.38)EC1.1 Eq 9.18

= 5.86 kN/m

ucreep

= 0.745 kdef,perm

Fd,creep

= 0.745 x 1.0 x 5.86 = 4.37 mm

For a UDL the elastic + shear deflection = 5 12 UDL L

384 EI

UDL L

8 G

4 2

0, mean

+ .

A

ufin

= u1,inst

+ u2,inst

+ ucreep

= 4.06 + 3.79 + 4.37 = 12.22 mm

u2,fin

= u2,inst

+ ucreep

= 3.79 + 4.37 = 8.16 mm

= w5 4100

384 12750 400 10

12 4100 20

8 12750 3000

4

6

2×× × ×

+ × ×× ×.

0

= 0.853w

Design load: By inspection, the medium term case iscritical

ie w = 10.95 kN/m as previously calculated.

∴ Deflection= 0.853 x 10.95

= 9.34 mm

The deflection limits for ufin

and u2,fin

are taken fromTRADA WI Sheet 4 - 24 Serviceability limit states fortimber in buildings.

ufin

≤ span

250=

4100

250= 16.4 mm

u2,fin

≤ span

350=

4100

350= 13.76 mm

Repeat calculations will show that a 63 x 400 mmdeep section is 1% over stressed which the designermay consider acceptable.

BS 5268 EC5

CREEP DEFLECTIONNot considered in BS 5268

TOTAL DEFLECTION= Initial deflection

= 9.34 mm

DEFLECTION LIMITS

Deflection ≤ 0.003 x span = 0.003 x 4100

= 12.3 mm

If smaller depth used ie 360 mm, deflection limits exceeded

If thinner section used ie 63 mm, depth to breadth ratio exceeds 6.3 see BS 5268 Table 16.

∴ use 75 x 400 mm Kerto LVL.