## N I

1. Actions on the beam

Permanent loading on the beam, Gk

Medium duration variable loading acting downwards on the beam, Qk.1

Instantaneous duration variable loading acting upwards on the beam, Qk.2

2. Partial safety factors

Table 2.2, f values (UKNA to EN 1990:2002 TableNA.Al.1):

Factor for the combination value of medium duration f01 = 0.7

variable action Qk.1, feu

Factor for the combination value of wind action Qk 2, f0 2 f0 2 = 0.5

(a) Partial factors for equilibrium ULS:

Table 2.8, equation EQU (a) (UKNA to EN 1990:2002 TableNA.A1.2(A)) For permanent favourable actions, yG.e yG.e = 0.9

For variable unfavourable actions, yQ.e.u YQ.e.u = 1.5

For variable favourable actions, Yq.ef YQ.e.f = 0

(b) Partial factors for strength ULS:

Table 2.8, equation STR (c),(i) (UKNA to EN 1990:2002 TableNA.A1.2(B)) For permanent unfavourable actions, yG.u yG.u = 1.35

For permanent favourable actions, yG.f yG.f = 1.00

Unfavourable variable actions, yq.u Yq.u = 1.5

Favourable variable actions, yQf yQf = 0

(c) Partial factors for serviceability deflection limit states: 2.2.25

Partial factors for permanent and variable actions = 1

3. Modification factors

Load duration factor for permanent duration loading kmod.perm = 0.6

at service class 1, kmod.perm

Load duration factor for medium duration loading kmod.med = 0.8

at service class 1, kmod.med

Load duration factor for instantaneous duration loading £mod.mst = 1.1

at service class 1, kmod.inst

4. Critical loading conditions for design effects

(a) Equilibrium states

Ignoring all load cases where the permanent action is unfavourable, eight alternative load cases have to be considered and the critical load case will be:

Critical load case: Permanent (favourable) Gk + leading variable (unfavourable) Qk,2 + accompanying variable (favourable) Qk1:

Equi1d = YG.e • Gk - YQ.e.u • Qk.2 + YQ.e.f • fo,1 • Qk.1

Equi1d = -0.36 kN/m i.e. loading is acting upwards on the beam

Ignoring all load cases where the permanent action is unfavourable, eight alternative load cases have to be considered and the critical load case will be:

Critical load case: Permanent (favourable) Gk+ leading variable (unfavourable) Qk,2+ accompanying variable (favourable) Qk>1:

Equi2d = YG.f • Gk - yq.u • Qk.2 + YQ.f • ^0.1 • Qk.1 Equi2d = -0.3 kN/m loading is acting upwards on the beam The design load case for equilibrium will be Equi1d.

It must also be noted that the supporting structure to which the beam is anchored must have sufficient static stability to withstand the uplift force generated at each end of the beam.

(b) Strength states

There will be eight load cases where the permanent action is unfavourable and eight where the permanent action is favourable. The design load case will be the load case generating the largest value when divided by the associated kmod, and must include for any reversal of stress condition. The design load case/kmod will be as follows: (i) For bending compression on the top face, for shear, for bearing:

(Permanent Gk (unfavourable) + leading variable (unfavourable) Qk1 + accompanying variable (favourable) Qk.2)/kmod.med NB: content of the last paragraph in 2.2.24:

„™ 1 G • Gk + Kq.u • Qk.1 - KQ.f • f0.2 • Qk.2 „™ 1 , 0,, STR1d =--—---- STR1d = 3.26 kN/m kmod.med

The design loading will therefore be:

DL1d = Yg.u • Gk + yq.u • Qk.1 - YQ.f • ^0.2 • Qk.2 DL1d = 2.61 kN/m

(ii) For bending compression on the bottom face (e.g. to check lateral torsional instability under stress reversal):

(Permanent Gk (favourable) + leading variable Qk,2 + accompanying variable (favourable) Qu)/kmod ,inst

STR2d = KG • f • Gk - KQ u kQk 2 + KQf • f(U ■ Qk1 STR2d = -0.27 kN/m kmod • inst

### The design loading will therefore be

DL2d = kg.f ■ Gk - kq.u ■ Qk. 2 + Kq.f ■ fa. 1 ■ Qk. 1 DL2d = -0 . 3 kN/m i.e the loading is acting upwards on the beam (c) Deflection states - adopting the irreversible limit state condition

Load case for determining the instantaneous downward deflection:

Using the characteristic combination, i.e. equation (2.24) in 2.2.25.2 (equation (6.14b) in EC0, four load cases have to be considered and the design condition will be:

Permanent + variable Qk,1:

Example 4.8.2 A 63 mm by 225 mm deep sawn timber beam in a domestic residence supports the characteristic loading shown in Figure E4.8.2. The beam has a clear span of 3.10 m, the bearing length has been restricted to 85 mm at each end, is of strength class C24 to BS EN 338:2003, and functions in service class 2 conditions. The beam is laterally restrained against lateral buckling along its length and is notched at its ends by 15 mm, extending 150 mm into the beam from the centre of each support position. The notches are on the same side as the beam supports.

### Given that

Gk.udl = 1-0 kN/m (characteristic uniformly distributed permanent action) Qk udl = 2.5 kN/m (characteristic uniformly distributed medium-term action) Gk p = 1.00 kN characteristic point load at mid-span carry out a design check to confirm that

(a) the beam will meet the ULS design requirements of EC5;

(b) the deflection criteria will also be acceptable, assuming that the structure is exposed without applied finishes.

Gk = 1.00 kN point load at mid-span

Gk = 1.00 kN point load at mid-span

210 mm

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