Design of Timber Roof Truss to British Code: Solved Example

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

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/m²
Weight of purlin (assume 50mm x 50mm African Mahogany hardwood timber)
Density of African Mahogany = 530 kg/m³ = 0.013  kN/m = (0.013 × 2m)/(2m × 1.2m) = 0.0108 kN/m²
Self weight of rafter (assume) = 0.05 kN/m²
Total = 0.0885 kN/m²
Weight on plan = 0.0885 × cos 17.35 = 0.08 kN/m²

On Ceiling Tie Member (bottom chord)
Weight of ceiling (10mm insulation fibre board) = 0.077 kN/m²
Weight of services = 0.1 kN/m²
Self weight of ceiling tie = 0.05 kN/m²
Total = 0.227 kN/m²

Therefore the nodal permanent load on rafter (Gk) = 0.08 kN/m² × 2m × 1.2m = 0.192 kN
Therefore the nodal permanent load on ceiling tie (Gk) = 0.227 kN/m² × 2m × 1.2m = 0.5448 kN

(ii) Live Load
Imposed load on top and bottom chord (qk) = 0.75 KN/m² (treat as medium-term load on plan)
Therefore the nodal permanent load on rafter (Gk) = 0.75 kN/m² × 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 (σ_c,g,||) = 7.1 N/mm²
σ_c,adm,|| = σ_c,g,|| ×  k₂ × k₃ × k₈ × k₁₂

Bending parallel to grain (σ_m,g,||) = 5.8 N/mm²
σ_m,adm,|| = σ_m,g,|| ×  k₂ × k₃ × k₆ × k₇ × k₈

k₂ = wet exposure (does not apply in this case)
k₃ = duration factor = 1.25 (medium-term loading)
k₆ = shape factor = 1.0 (rectangular section)
k₇ = Depth of section 72mm < h < 300mm
k₇ = (300/h)^0.11 = (300/100)^0.11 = 1.128
k₈ = Load sharing factor (does not apply since the spacing of the rafters exceed 610 mm).

Section Properties
Area = 3.8 × 10³ mm²
Z_xx = 63.3 × 10³ mm³

E_min = 6000 N/mm²

k₃ = 1.25 (Table 17)

σ_c,|| = 7.1 × 1.25 = 8.88 N/mm²

E/σ_c,|| = 6000/8.88 = 675.67
Slenderness λ = 43.598

We can obtain the value of k₁₂ by interpolating from Table 22 of the code
We are interpolating for E/σ_c,|| = 675.67 and  λ = 43.598

E/σ_c,||           40         50
600           0.774       0.692
700           0.784       0.711

On interpolating (bivariate interpolation);
k₁₂ = 0.7545

σ_c,adm,|| = σ_c,g,|| ×  k₂ × k₃ × k₈ × k₁₂
σ_c,adm,|| = 7.1 ×  1.0 × 1.25 × 1.0 × 0.7545 = 6.699 N/mm²

σ_m,adm,|| = σ_m,g,|| ×  k₂ × k₃ × k₆ × k₇ × k₈
σ_m,adm,|| = 5.8 ×  1.0 × 1.25 × 1.0 × 1.128 × 1.0 = 8.178 N/mm²

Euler critical stress σ_e =  π²E_min/λ²

σ_e =  π²(6000)/(43.598)² = 31.154 N/mm²

For combined bending and compression

σ_m,a,|| =  3.144 N/mm²
σ_c,a,|| = 2.657 N/mm²
σ_c,adm,|| =  6.699 N/mm²
σ_m,adm,|| =  8.178 N/mm²
σ_e =  31.154 N/mm²

[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
σ_m,a,|| = M/Z = (0.28 × 10⁶)/(63.6 × 10³) = 4.40 N/mm²

Axial compressive stress
σ_c,a,|| = P/A = (9.62 × 10³)/(3.8 × 10³) = 2.531 N/mm²

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

σ_c,adm,|| = σ_c,g,|| ×  k₂ × k₃ × k₈
σ_c,adm,|| = 7.1 ×  1.0 × 1.25 × 1.0  = 8.875 N/mm²

The interaction formula for this scenario is given below;

[σ_m,a,|| / σ_m,adm,||] + [σ_c,ma,|| / σ_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 (σ_t,g,||) = 3.5 N/mm²
σ_t,adm,|| = σ_t,g,|| ×  k₂ × k₃ × k₈ × k₁₄
(width of section) k₁₄ = (300/h)^0.11 = (300/100)^0.11 = 1.128
σ_t,adm,|| = 3.5 × 1.0 × 1.25× 1.0 × 1.128 = 4.935 N/mm²

Bending parallel to grain (σ_m,g,||) = 5.8 N/mm²
σ_m,adm,|| = 5.8 ×  1.0 × 1.25 × 1.0 × 1.128 × 1.0 = 8.178 N/mm²

Note: 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.

Combined tension and bending
[σ_m,a,|| / σ_m,adm,||] + [σ_t,ma,|| / σ_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
σ_m,a,|| = M/Z = (0.3 × 10⁶)/(63.6 × 10³) = 4.7169 N/mm²

Axial tensile stress
σ_c,a,|| = P/Effective Area = (9.62 × 10³)/(3.8 × 10³) = 2.531 N/mm²

Combined tension and bending
[σ_m,a,|| / σ_m,adm,||] + [σ_t,ma,|| / σ_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.