Design of timber members using Eurocode 5 (EN1995):

Clause 6.1.2 Tension parallel to the grain

Criterion to be used for tension (Ales course project)

All references to tables etc. are to EN 1995-1-1 unless otherwise stated.

kmod

kmod from Table 3.1 is a strength modification factor to take account of the service class and the load duration class.

The service classes, defined in Clause 2.3.1.3, are dependent on the moisture content of the timber. The three services classes are basically:

· Service class 1: Temperature of 20o C and relative humidity only exceeding 65% for a few weeks per year.

· Service class 2 - as class 1 but with the relative humidity only exceeding 85% for a few weeks per year

· Service class 3 - for all moisture contents greater than service class 2

The load duration classes, defined in Table 2.1 For a roof truss self weight is the dominant load and therefore the permanent load duration class should be used.

Use solid timber for which gM (Table 2.3) is 1.3

How to choose the strength class

In a design context in practice, one would normally make a decision based on information (from a timber supplier) on the availability and cost of timber with respect to the strength classes. In the context of Assignment 2 make a choice of a softwood class.

Value of E in the analysis model

Clause 2.4.1 (2) states that Ed = Emeam/gM where Ed is the design value for the E value. Emean is obtained from Table 1 of EN 338

Example for strength of a tie

Solid timber tie 50 x 50 mm cross section has a factored axial load of 23 kN. Check adequacy in relation to Equation (6.1).

st,0,d = 23000/(50*50) = 9.2 N/mm2

Use strength class C20 ft,0,k = 12 N/mm2 (Table 1 EN 338)

Assume Service Class 1 (internal trusses with well ventilated roof space )

Assume load duration class Permanent

From Table 3.1 kmod = 0.6

gM = 1.3 (table 2.3)

ft,0,d = kmod ft,0,k/gM = 0.6*12/1.3 = 5.5 N/mm2

Section is not adequate.

Criterion to be used for compression (Ales course project)

The rule for compression parallel to the grain for a stocky strut (i.e. one for which lateral instability can be ignored) is as for tension but with the subscript ‘t’ replaced by ‘c’ for compression. i.e.

The process for using (6.2) is the same as for (6.1) except that fc,0,k is used from Table 1 of EN 338. Use fc0,k = 20 N/mm2


Clause 6.1.6 Bending

Criterion to be used for bending (Ales course project)

For uniaxial bending

Where the subscript ‘m’ denotes bending and y refers to the (major) axis of bending.

In this case the design stress s is calculated using s = M/Z where M is the bending moment and Z is the elastic modulus of the section

(Z = I/ymax = bd2/6 for a rectangular section).

For example if the bending moment due to factored loading on a 100 x 50 section is 0.6 kN m the design bending stress is:

M/Z = 0.6E6/(50*100^2/6) = 7.2 kN/mm2

The process for using (6.11) is the same as for (6.1) except that fm,0,k is used from Table 1 of EN 338.

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