Structville

A hub for civil engineering related designs, analyses, discussion, information, and knowledge...

Trending Posts:

Search This Blog

Wednesday, April 11, 2018

Design of Timber Roof Truss to British Code: Solved Example


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/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/m(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 (σc,g,||) = 7.1 N/mm2
σc,adm,|| = σc,g,|| ×  k× k3 × k8 × k12

Bending parallel to grain (σm,g,||) = 5.8 N/mm2
σm,adm,|| = σm,g,|| ×  k× k3 × k6 × k× k8

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




Section Properties
Area = 3.8 × 10mm2
Zxx = 63.3 × 10mm3
Zyy = 24.1 × 10mm3
Ixx = 3.17 × 10mm4
Iyy = 0.457 × 10mm4
rxx = 28.9 mm
ryy = 11 mm

Applied bending stress
σm,a,|| = M/Z = (0.2 × 106)/(63.6 × 103) = 3.144 N/mm2

Axial compressive stress
σc,a,|| = P/A = (10.1 × 103)/(3.8 × 103) = 2.657 N/mm2


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

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

Medium-term load
Compression parallel to grain (σc,g,||) = 7.1 N/mm2
Emin = 6000 N/mm2

k3 = 1.25 (Table 17)

σc,|| = 7.1 × 1.25 = 8.88 N/mm2

E/σc,|| = 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/σc,|| = 675.67 and  λ = 43.598

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



On interpolating (bivariate interpolation);
k12 = 0.7545

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

σm,adm,|| = σm,g,|| ×  k× k3 × k6 × k× k8
σm,adm,|| = 5.8 ×  1.0 × 1.25 × 1.0 × 1.128 × 1.0 = 8.178 N/mm2

Euler critical stress σ=  π2Emin2

σ=  π2(6000)/(43.598)= 31.154 N/mm2

For combined bending and compression

σm,a,|| =  3.144 N/mm2
σc,a,|| = 2.657 N/mm2
σc,adm,|| =  6.699 N/mm2
σm,adm,|| =  8.178 N/mm2
σ=  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
σm,a,|| = M/Z = (0.28 × 106)/(63.6 × 103) = 4.40 N/mm2

Axial compressive stress
σc,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;

σc,adm,|| = σc,g,|| ×  k× k3 × k8
σc,adm,|| = 7.1 ×  1.0 × 1.25 × 1.0  = 8.875 N/mm2

The interaction formula for this scenario is given below;

m,a,|| / σm,adm,||] + [σc,ma,|| / σc,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 (σt,g,||) = 3.5 N/mm2
σt,adm,|| = σt,g,|| ×  k× k3 × k8 × k14
(width of section) k14 = (300/h)0.11 = (300/100)0.11 = 1.128
σt,adm,|| = 3.5 × 1.0 × 1.25× 1.0 × 1.128 = 4.935 N/mm2

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

Applied bending stress
σm,a,|| = M/Z = (0.22 × 106)/(63.6 × 103) = 3.459 N/mm2

Axial tensile stress
σc,a,|| = P/Effective Area = (9.74 × 103)/(3.8 × 103) = 2.563 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
m,a,|| / σm,adm,||] + [σt,ma,|| / σt,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
σm,a,|| = M/Z = (0.3 × 106)/(63.6 × 103) = 4.7169 N/mm2

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

Combined tension and bending
m,a,|| / σm,adm,||] + [σt,ma,|| / σt,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.

2 comments:

  1. Wow ranks, u r doing a tremendous work here I must confess. Your post has been colossal source of my bank of knowledge. God will reward u greatly. Thanks fr d post.

    ReplyDelete
  2. Great knowledge acquired. Thanks.

    ReplyDelete