The use of timber as trussed rafters for roof of buildings is a very popular alternative all over the world. The aim of this post is to show the design example of a timber roof truss (trussed rafter). As a direct product of nature, timber has so many variable properties that are more complex than that of concrete, steel, bricks, or aluminium. Some of the characteristics which influence the structural behaviour of timber are;

- moisture content
- direction of applied load (perpendicular or parallel to the grain)
- duration of loading
- strength grading of the timber

Quickly in this post, I am going to carry out a very simple design example of timber roof truss using BS 5268. A lot of information regarding timber as a structural material can be obtained from specialist textbooks. It is worth knowing that the most current design code for timber structures is Eurocode 5.

*Note:*BS 5268 is based on permissible stress design. When using permissible stress design, the margin of safety is introduced by considering structural behaviour under working/service load conditions and comparing the stresses thereby induced with permissible values. The permissible values are obtained by dividing the failure stresses by an appropriate factor of safety. The applied stresses are determined

using elastic analysis techniques, i.e.

Stress induced by working loads ≤ (failure stress / factor of safety)

Since BS 5268 is a permissible stress design code, mathematical modelling of

the behaviour of timber elements and structures is based on assumed elastic behaviour.

**Solved Example**

Let us design the roof truss of a building subjected to the following medium-term loads. The configuration of the roof truss is as shown above.

**Data**Span of roof truss = 4.8m

Spacing of the truss = 2.0m

Nodal spacing of the trusses = 1.2m

Service class of roof truss: Service class 2

**Load Analysis**

**(i) Dead Loads**

**On rafter (top chord)**Self weight of long span aluminium roofing sheet (0.55mm gauge thickness) = 0.019 kN/m

^{2}

Weight of purlin (assume 50mm x 50mm African Mahogany hardwood timber)

Density of African Mahogany = 530 kg/m

^{3}= 0.013 kN/m = (0.013 × 2m)/(2m × 1.2m) = 0.0108 kN/m

^{2}

Self weight of rafter (assume) = 0.05 kN/m

^{2}

**Total = 0.0885 kN/m**

^{2}Weight on plan = 0.0885 × cos 17.35 = 0.08 kN/m

^{2}

*On Ceiling Tie Member (bottom chord)*Weight of ceiling (10mm insulation fibre board) = 0.077 kN/m

^{2}

Weight of services = 0.1 kN/m

^{2}

Self weight of ceiling tie = 0.05 kN/m

^{2}

**Total = 0.227 kN/m**

^{2}**Therefore the nodal permanent load on rafter (Gk) = 0.08 kN/m**

^{2}× 2m × 1.2m = 0.192 kN

Therefore the nodal permanent load on ceiling tie (Gk) = 0.227 kN/m

^{2}× 2m × 1.2m = 0.5448 kN

**(ii) Live Load**

Imposed load on top and bottom chord (qk) = 0.75 KN/m

^{2 }(treat as medium-term load on plan)

Therefore the nodal permanent load on rafter (Gk) = 0.75 kN/m

^{2}× 2m × 1.2m = 1.8 kN

**Analysis of the rafter (top chord)**

Span Length = 1.257m

Load = (0.0885 + 0.75) × 2m = 1.667 kN/m

**Results**

Analysis of the structure for the loads gave the following results;

All load values are medium term loads;

Medium term load is defined in this case by:

*Dead load + temporary imposed load*

*Top Chord Result*Axial force = 10.1 kN (Compression)

Bending Moment = 0.2 kNm

Length of member = 1.26m

**Design of the Top Chord**

Let us try 38mm x 100mm timber

Strength class C18

Compression parallel to grain (Ïƒ

_{c,g,||}) = 7.1 N/mm

^{2}

Ïƒ

_{c,adm,||}= Ïƒ

_{c,g,||}× k

_{2 }× k

_{3}× k

_{8}× k

_{12}

Bending parallel to grain (Ïƒ

_{m,g,||}) = 5.8 N/mm

^{2}

Ïƒ

_{m,adm,||}= Ïƒ

_{m,g,||}× k

_{2 }× k

_{3}× k

_{6}× k

_{7 }× k

_{8}

k

_{2 }= wet exposure (does not apply in this case)

k

_{3 }= duration factor = 1.25 (medium-term loading)

k

_{6 }= shape factor = 1.0 (rectangular section)

k

_{7 }= Depth of section 72mm < h < 300mm

k

_{7 }= (300/h)

^{0.11}= (300/100)

^{0.11}= 1.128

k

_{8 }= Load sharing factor (does not apply since the spacing of the rafters exceed 610 mm).

*Section Properties*Area = 3.8 × 10

^{3 }mm

^{2}

Z

_{xx }= 63.3 × 10

^{3 }mm

^{3}

_{yy}= 24.1 × 10

^{3 }mm

^{3}

I

_{xx}= 3.17 × 10

^{6 }mm

^{4}

I

_{yy}= 0.457 × 10

^{6 }mm

^{4}

r

_{xx}= 28.9 mm

r

_{yy}= 11 mm

Applied bending stress

Ïƒ

_{m,a,|| }= M/Z = (0.2 × 10

^{6})/(63.6 × 10

^{3}) = 3.144 N/mm

^{2}

Axial compressive stress

Ïƒ

_{c,a,|| }= P/A = (10.1 × 10

^{3})/(3.8 × 10

^{3}) = 2.657 N/mm

^{2}

**Check for slenderness**

Effective length (Le) = 1260 mm (assuming pin end connection)

Î» = Le/r = 1260/28.9 = 43.598 < 52 Ok (clause 2.11.4)

**Medium-term load**

Compression parallel to grain (Ïƒ

_{c,g,||}) = 7.1 N/mm

^{2}

E

k

Ïƒ

E/Ïƒ

Slenderness Î» = 43.598

We can obtain the value of k

We are interpolating for E/Ïƒ

E/Ïƒ

600 0.774 0.692

700 0.784 0.711

On interpolating (bivariate interpolation);

k

Ïƒ

Ïƒ

Ïƒ

Ïƒ

Euler critical stress Ïƒ

Ïƒ

Ïƒ

Ïƒ

Ïƒ

Ïƒ

Ïƒ

[3.144/(6.699 × 0.9034)] + [2.657/6.699] = 0.919 < 1.0

Therefore, 38mm x 100mm GS C18 Timber is adequate for the rafter

Bending moment = 0.28 kN.m

Axial load (taking the average at that joint) = (10.81 + 8.43)/2 = 9.62 kN

Applied bending stress

Ïƒ

Axial compressive stress

Ïƒ

At node point, Î» < 5.0, and the rafter is designed as a short column at this point;

Ïƒ

Ïƒ

The interaction formula for this scenario is given below;

[Ïƒ

[4.40 / 8.178] + [2.531 / 8.875] = 0.8232 < 1.0

This shows that the section is satisfactory for rafter.

Span Length = 1.2m

Load = (0.227 + 0.75) × 2m = 1.954 kN/m

Axial force = 9.74 kN (tension)

Bending Moment = 0.22 kNm

Length of member = 1.2m

Let us still try 38mm x 100mm timber

Strength class C18

Tension parallel to grain (Ïƒ

Ïƒ

(width of section) k

Ïƒ

Bending parallel to grain (Ïƒ

Ïƒ

[Ïƒ

[3.459 / 8.178] + [2.563 / 4.935] = 0.9422 < 1.0

This is ok.

Bending moment = 0.3 kN.m

Axial load (taking the average at that joint) = (9.5 + 9.74)/2 = 9.62 kN

Applied bending stress

Ïƒ

Axial tensile stress

Ïƒ

[Ïƒ

[4.7169/ 8.178] + [2.531 / 4.935] = 1.089

In this case, the tie element may be increased to 38mm x 175mm or the grade of the timber could be changed to accommodate the combined flexural and axial stress in the member.

Deflection of trussed rafter under full load = 6.095mm (calculated on Staad)

Permissible deflection = 14mm

Deflection is ok.

_{min}= 6000 N/mm^{2}k

_{3}= 1.25 (Table 17)Ïƒ

_{c,||}= 7.1 × 1.25 = 8.88 N/mm^{2}E/Ïƒ

_{c,||}= 6000/8.88 = 675.67Slenderness Î» = 43.598

We can obtain the value of k

_{12}by interpolating from Table 22 of the codeWe are interpolating for E/Ïƒ

_{c,||}= 675.67 and Î» = 43.598E/Ïƒ

_{c,||}40 50600 0.774 0.692

700 0.784 0.711

On interpolating (bivariate interpolation);

k

_{12}= 0.7545Ïƒ

_{c,adm,||}= Ïƒ_{c,g,||}× k_{2 }× k_{3}× k_{8}× k_{12}Ïƒ

_{c,adm,||}= 7.1 × 1.0_{ }× 1.25 × 1.0 × 0.7545 = 6.699 N/mm^{2}Ïƒ

_{m,adm,||}= Ïƒ_{m,g,||}× k_{2 }× k_{3}× k_{6}× k_{7 }× k_{8}Ïƒ

_{m,adm,||}= 5.8 × 1.0_{ }× 1.25 × 1.0 × 1.128 × 1.0 = 8.178 N/mm^{2}Euler critical stress Ïƒ

_{e }= Ï€^{2}E_{min}/Î»^{2}Ïƒ

_{e }= Ï€^{2}(6000)/(43.598)^{2 }= 31.154 N/mm^{2}*For combined bending and compression*Ïƒ

_{m,a,|| }= 3.144 N/mm^{2}Ïƒ

_{c,a,|| }= 2.657 N/mm^{2}Ïƒ

_{c,adm,||}= 6.699 N/mm^{2}Ïƒ

_{m,adm,||}= 8.178 N/mm^{2}Ïƒ

_{e }= 31.154 N/mm^{2}[3.144/(6.699 × 0.9034)] + [2.657/6.699] = 0.919 < 1.0

Therefore, 38mm x 100mm GS C18 Timber is adequate for the rafter

*Consider portion over nodes (at supports)*Bending moment = 0.28 kN.m

Axial load (taking the average at that joint) = (10.81 + 8.43)/2 = 9.62 kN

Applied bending stress

Ïƒ

_{m,a,|| }= M/Z = (0.28 × 10^{6})/(63.6 × 10^{3}) = 4.40 N/mm^{2}Axial compressive stress

Ïƒ

_{c,a,|| }= P/A = (9.62 × 10^{3})/(3.8 × 10^{3}) = 2.531 N/mm^{2}At node point, Î» < 5.0, and the rafter is designed as a short column at this point;

Ïƒ

_{c,adm,||}= Ïƒ_{c,g,||}× k_{2 }× k_{3}× k_{8}Ïƒ

_{c,adm,||}= 7.1 × 1.0_{ }× 1.25 × 1.0 = 8.875 N/mm^{2}The interaction formula for this scenario is given below;

[Ïƒ

_{m,a,||}/ Ïƒ_{m,adm,||}] + [Ïƒ_{c,ma,||}/ Ïƒ_{c,adm,||}] ≤ 1.0[4.40 / 8.178] + [2.531 / 8.875] = 0.8232 < 1.0

This shows that the section is satisfactory for rafter.

**Analysis for Tie Element**Span Length = 1.2m

Load = (0.227 + 0.75) × 2m = 1.954 kN/m

**Results**Axial force = 9.74 kN (tension)

Bending Moment = 0.22 kNm

Length of member = 1.2m

**Design of the Bottom Chord (ceiling tie)**Let us still try 38mm x 100mm timber

Strength class C18

Tension parallel to grain (Ïƒ

_{t,g,||}) = 3.5 N/mm^{2}Ïƒ

_{t,adm,||}= Ïƒ_{t,g,||}× k_{2 }× k_{3}× k_{8}× k_{14}(width of section) k

_{14 }= (300/h)^{0.11}= (300/100)^{0.11}= 1.128Ïƒ

_{t,adm,||}= 3.5 × 1.0_{ }× 1.25× 1.0 × 1.128 = 4.935 N/mm^{2}Bending parallel to grain (Ïƒ

_{m,g,||}) = 5.8 N/mm^{2}Ïƒ

_{m,adm,||}= 5.8 × 1.0_{ }× 1.25 × 1.0 × 1.128 × 1.0 = 8.178 N/mm^{2}_{}

Applied bending stress

Ïƒ

Axial tensile stress

Ïƒ

Ïƒ

_{m,a,|| }= M/Z = (0.22 × 10^{6})/(63.6 × 10^{3}) = 3.459 N/mm^{2}Axial tensile stress

Ïƒ

_{c,a,|| }= P/Effective Area = (9.74 × 10^{3})/(3.8 × 10^{3}) = 2.563 N/mm^{2}**: When ceiling tie is connected to rafter by the means of a bolt, the projected area of the bolt hole must be subtracted from the gross area of the section.***Note**Combined tension and bending*[Ïƒ

_{m,a,||}/ Ïƒ_{m,adm,||}] + [Ïƒ_{t,ma,||}/ Ïƒ_{t,adm,||}] ≤ 1.0[3.459 / 8.178] + [2.563 / 4.935] = 0.9422 < 1.0

This is ok.

*Consider portion over nodes (at supports)*Bending moment = 0.3 kN.m

Axial load (taking the average at that joint) = (9.5 + 9.74)/2 = 9.62 kN

Applied bending stress

Ïƒ

_{m,a,|| }= M/Z = (0.3 × 10^{6})/(63.6 × 10^{3}) = 4.7169 N/mm^{2}Axial tensile stress

Ïƒ

_{c,a,|| }= P/Effective Area = (9.62 × 10^{3})/(3.8 × 10^{3}) = 2.531 N/mm^{2}*Combined tension and bending*[Ïƒ

_{m,a,||}/ Ïƒ_{m,adm,||}] + [Ïƒ_{t,ma,||}/ Ïƒ_{t,adm,||}] ≤ 1.0[4.7169/ 8.178] + [2.531 / 4.935] = 1.089

In this case, the tie element may be increased to 38mm x 175mm or the grade of the timber could be changed to accommodate the combined flexural and axial stress in the member.

**Check for deflection**Deflection of trussed rafter under full load = 6.095mm (calculated on Staad)

Permissible deflection = 14mm

Deflection is ok.

**That is it for now. Thank you so much for visiting Structville today and God bless you. Remember to share with your folks.**
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ReplyDeleteGreat knowledge acquired. Thanks.

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