1. ERT352 FARM STRUCTURES TRUSS DESIGN
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2. INTRODUCTION
A truss is essentially a triangulated system of straight interconnected structural elements.
The most common use of trusses is in buildings, where support to roofs, the floors and internal loading such as services and suspended ceilings, are readily provided.
The individual elements are connected at nodes; the connections are often assumed to be nominally pinned.

3. INTRODUCTION(cont.)
The external forces applied to the system and the reactions at the supports are generally applied at the nodes.
The principal force in each element in a truss is axial tension or compression.

4. TYPE OF TRUSSES
Trusses comprise assemblies of tension and compression elements.
Under gravity loads, the top and bottom chords of the truss provide the compression and tension resistance to overall bending, and the bracing resists the shear forces.
A wide range of truss forms can be created. Each can vary in overall geometry and in the choice of the individual elements.

5. Some of the commonly used types are shown below.
Pratt truss ('N' truss)
Pratt trusses are commonly used in long span buildings ranging from 20 to 100 m in span.
Warren truss
The Warren truss has equal length compression and tension web members, and fewer members than a Pratt truss.

6. North light truss Saw-tooth truss Fink truss
North light trusses are traditionally used for short spans in industrial workshop-type buildings.
Saw-tooth truss
The saw-tooth truss is used in multi-bay buildings
Fink truss
This type of truss is commonly used to construct roofs in houses.

7. ANATOMY OF TRUSS

8. ANALYSIS OF TRUSS POINT LOADS AT THE NODES OF THE TRUSS
This condition occurs when purlins are located at the nodes of the truss
Purlin spacing is equal to node spacing; Sp = Sn

9. ANALYSIS OF TRUSS (cont.)

10. POINT LOADS AT THE NODES OF THE TRUSS (cont.)
Given Imposed load on plan, Qk and Dead load on plan, Gk
To obtain the point load, P;
Area of load transferred to each node, A = Sp x St
Design load, q = 1.35Gk + 1.5Qk
Point load, P = q x A
Calculate the tension and compression forces in each truss members.

11. ANALYSIS OF TRUSS (cont.)
There are possible combinations of steel sections that can be used for tension or compression truss members.
Single angle connected through on leg
Double angle back-to-back, connected to both sides of gusset plate
Double angle back-to-back, connected to one side of gusset plate

12. Example 1: Analysis of truss – Determining point loads at nodes
The loads from the roof sheets, transferred to the purlins, are as followed:
Dead load on slope: Corrugated steel roofing = 0.1 kN/m2
Lighting and insulation = 0.15 kN/m2
Self weight of purlins = 0.05 kN/m2
Self weight of trusses = 0.1 kN/m2
Total dead load on slope, Gk = 0.4 kN/m2
Imposed load on slope, Qk = 0.81 kN/m2
Spacing between trusses, St = 5 m
Purlin Spacing , Sp = 2 m
The purlins are arranged and located at the nodes of the truss as shown in figure. Determine the point loads that are imposed by the purlins at all the nodes.

14. Solution: Design load or factored load, q q = 1.35Gk + 1.5 Qk
= 1.35 (___) (___)
= _______ kN/m2
Area of load transferredto any intermediate nodes, A
A = Sp x St = ____ x ____ = _____ m2
Point load at node, P
P = q x A = ____ x ____ = _____ kN

15. Example 2: Analysis of truss – Determining the axial loads in the truss member
The following truss carries a uniform factored load of 26 kN/m. Determine the member forces.

16. Solution: Point load at each node, P P = w * purlin spacing (Sp)
P = 26 kN/m x 3 m
= 78kN
~ for external nodes, point load is equal to P/2 = 78/2 = 39kN
By using method of joint or method of section, member forces can be determined.

17. DESIGN OF TENSION MEMBER
The design value of tension force, NEd at each cross section shall satisfy as specified in clause EN :2005 ;
Where, NEd is design force ; Nt,Rd is design values of resistance to tension forces

18. DESIGN OF TENSION MEMBER (cont.)
For sections with holes, the design tension resistance, Nt,Rd should be taken as the smaller of
The design plastic resistance of the gross cross-section, Npl,Rd
Where, A is area of section; fy is yield strength; γm0 is resistance of cross-sections (= 1.0);
The design ultimate resistance of the net cross-section at holes for fasteners, Nu,Rd
Where, Anet is net area of cross-section; fu is ultimate strength (Table 3.1 of En :2005); γm2 is resistance of cross-sections in tension fracture (= 1.25);

19. Example: Design of tension member with single angle section
The maximum design tension forces in a truss member is 230kN. The member consists of an unequal angle of x 75 x 8 L of steel grade S275. Check the capacity of the tension member if the ends of the member are:
Connected by two bolts of 20mm nominal diameter
Welded with grade 42 electrode.

20. Solution: End members connected with 2 bolts
Tension force, NEd = 230 kN
Unequal angle; 100 x 75 x 8 L
From table of properties; Ag = 1350mm, iy = 40mm, iz = 20.9mm, iu = 42.1mm, iv = 16.3mm
From Table 3.1, t = 8mm < 40mm, steel grade S275
 fy = 275N/mm2, fu = 430N/mm2
Diameter of bolt, d = 20mm

21. Solution: (cont.)
For sections with holes, the design tension resistance, Nt,Rd should be taken as the smaller of
The design plastic resistance of the gross cross-section, Npl,Rd

22. Solution: (cont.)
The design ultimate resistance of the net cross-section at holes for fasteners, Nu,Rd
 therefore, Nt,Rd = 363.5kN

23. Solution: (cont.)
Section 100 x 75 x 8 L is adequate to carry the force.

24. Solution: (cont.) End member connected with welded
Tension force, NEd = 230 kN
In this case, there is no hole for the bolts, hence no reduction in cross sectional of the angle section
The design tension resistance, Nt,Rd should be taken as the smaller of
The design plastic resistance of the gross cross-section, Npl,Rd

25. Solution: (cont.)
The design ultimate resistance of the net cross-section , Nu,Rd
Since there is no reduction in the cross section of the angle, the
 therefore, Nt,Rd = 371.3kN

26. Solution: (cont.)
Section 100 x 75 x 8 L is adequate to carry the force.

27. DESIGN OF COMPRESSION MEMBER
FULL SECTION RESISTANCE OF COMPRESSION MEMBER
The full design resistance of a member that fails by yielding and not susceptible to buckling is given as Nc,Rd as specified in clause EN :2005
Where NEd is the design compression force and Nc,Rd is the full design resistance for section in classes 1,2, and 3.

28. DESIGN OF COMPRESSION MEMBER
BUCKLING RESISTANCEOF COMPRESSION MEMBER
The design of compression member susceptible to buckling shall satisfy as specified in clause EN :2005 :
The buckling resistance, Nb,Rd of a member carrying axially loaded compression force is given by:

29. BUCKLING RESISTANCE ABOUT Y-Y AXIS, Ny,b,Rd
Where; the capacity of reduction factor is given as
And λ1 = 93.9ε = 93.9 x 0.92 = 86.39

30. BUCKLING RESISTANCE ABOUT Z-Z AXIS, Nz,b,Rd
Where; the capacity of reduction factor is given as
And λ1 = 93.9ε = 93.9 x 0.92 = 86.39

31. Buckling Length, Lcr
Lcr of the top chord equals to the distance between nodes

32. Example-Design of Compression Member with Single Angle Section
The following figure shows a truss with one of the member carries compression force of 50kN. The member consists of a single angle section with both ends are connected to long leg. Design the compression member using steel grade S275 with end members are connected with two bolts.

33. Solution: Axial compression load, NEd = 50 kN
Try unequal angle 80 x 60 x 7 L
Ag = 938 mm2; T = 7 mm;
iy = 25.1mm; iz = 17.4mm; iu = 27.7mm; iv = 12.8mm
From Table 3.1: Steel grade S275, t = 7mm < 40mm
fy = 275 N/mm2

34. From Table 5.2(sheet 3 of 3) (page 44)
ε = (235/fy)0.5 = (235/275)0.5 = 0.92
For angle; h/t = 80/7 = 11.4 ≤ 15 ε = 15(0.92) = 13.8
(b + h)2t = (60+80)/(2x7) = 10.0 ≤ 11.5ε = 11.5(0.92) = 11.5
 Section is class 3 and not susceptible to local buckling

35. Buckling resistance about weakest axis, v-v axis, Nv,b,Rd
The smallest radius of gyration, iv = 12.8mm will contribute to the largest slenderness = Lcr/iv, hence v-v is the weakest axis.
*Note: the larger the slenderness, the weaker is the axis.
Buckling resistance about weakest axis, v-v axis, Nv,b,Rd

36. And λ1 = 93.9ε = 93.9 x 0.92 = 86.39
Lcr = 1.0L = 1.0 (2154) = 2154 mm

37. From Table 6.2: L – sectionshave buckling curve type ’b’
From Table 6.1:for type ‘b’ , imperfection factor, α = 0.34

38. Check force equilibrium
NEd =50kN < Nv,b,Rd = 51.6kN  OK!
Angle section 80 x 60 x 7 L is adequate.

39. THANK YOU
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