ERT 348 Controlled Environment Design 1

TRUSS DESIGN

ERT352FARM STRUCTURESIt doesnt matter what the subject is; once youve learnt how to study, you can do anything you want.INTRODUCTIONA 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. 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.

TYPE OF TRUSSESTrusses 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.

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 trussThe Warren truss has equal length compression and tension web members, and fewer members than a Pratt truss.

North light trussNorth light trusses are traditionally used for short spans in industrial workshop-type buildings.

Saw-tooth trussThe saw-tooth truss is used in multi-bay buildings

Fink trussThis type of truss is commonly used to construct roofs in houses.

ANATOMY OF TRUSS

ANALYSIS OF TRUSSPOINT LOADS AT THE NODES OF THE TRUSSThis condition occurs when purlins are located at the nodes of the trussPurlin spacing is equal to node spacing; Sp = Sn

ANALYSIS OF TRUSS (cont.)

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

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 legDouble angle back-to-back, connected to both sides of gusset plateDouble angle back-to-back, connected to one side of gusset plate

Example 1: Analysis of truss Determining point loads at nodesThe loads from the roof sheets, transferred to the purlins, are as followed:Dead load on slope: Corrugated steel roofing = 0.1 kN/m2Lighting and insulation = 0.15 kN/m2Self weight of purlins= 0.05 kN/m2Self weight of trusses= 0.1 kN/m2Total dead load on slope, Gk= 0.4 kN/m2

Imposed load on slope, Qk = 0.81 kN/m2

Spacing between trusses, St= 5 mPurlin 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.

Solution:Design load or factored load, q q = 1.35Gk + 1.5 Qk = 1.35 (___) + 1.5 (___) = _______ kN/m2

Area of load transferredto any intermediate nodes, AA = Sp x St = ____ x ____ = _____ m2

Point load at node, PP = q x A = ____ x ____ = _____ kN

Example 2: Analysis of truss Determining the axial loads in the truss memberThe following truss carries a uniform factored load of 26 kN/m. Determine the member forces.Solution:Point load at each node, PP = 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.

DESIGN OF TENSION MEMBERThe design value of tension force, NEd at each cross section shall satisfy as specified in clause 6.2.3 EN 1993-1-1:2005 ;

Where, NEd is design force ; Nt,Rd is design values of resistance to tension forces

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; m0 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 1993-1-1:2005); m2 is resistance of cross-sections in tension fracture (= 1.25);

Example: Design of tension member with single angle sectionThe maximum design tension forces in a truss member is 230kN. The member consists of an unequal angle of 100 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 diameterWelded with grade 42 electrode.Solution:End members connected with 2 boltsTension force, NEd = 230 kNUnequal 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

20Solution: (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

Solution: (cont.)The design ultimate resistance of the net cross-section at holes for fasteners, Nu,Rd

therefore, Nt,Rd = 363.5kN

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

Solution: (cont.)End member connected with weldedTension force, NEd = 230 kNIn 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

Solution: (cont.)The design ultimate resistance of the net cross-section , Nu,RdSince there is no reduction in the cross section of the angle, the

therefore, Nt,Rd = 371.3kN

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

DESIGN OF COMPRESSION MEMBERFULL SECTION RESISTANCE OF COMPRESSION MEMBERThe full design resistance of a member that fails by yielding and not susceptible to buckling is given as Nc,Rd as specified in clause 6.2.4 EN 1993-1-1:2005

Where NEd is the design compression force and Nc,Rd is the full design resistance for section in classes 1,2, and 3.

DESIGN OF COMPRESSION MEMBERBUCKLING RESISTANCEOF COMPRESSION MEMBERThe design of compression member susceptible to buckling shall satisfy as specified in clause 6.3.1 EN 1993-1-1:2005 :

The buckling resistance, Nb,Rd of a member carrying axially loaded compression force is given by:

BUCKLING RESISTANCE ABOUT Y-Y AXIS, Ny,b,Rd

Where; the capacity of reduction factor is given as

And 1 = 93.9 = 93.9 x 0.92 = 86.39

BUCKLING RESISTANCE ABOUT Z-Z AXIS, Nz,b,Rd

Where; the capacity of reduction factor is given as

And 1 = 93.9 = 93.9 x 0.92 = 86.39

Buckling Length, LcrLcr of the top chord equals to the distance between nodes

Example-Design of Compression Member with Single Angle SectionThe 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.

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 < 40mmfy = 275 N/mm2

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

34The 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

And 1 = 93.9 = 93.9 x 0.92 = 86.39Lcr = 1.0L = 1.0 (2154) = 2154 mm

From Table 6.2: L sectionshave buckling curve type bFrom Table 6.1:for type b , imperfection factor, = 0.34

Check force equilibriumNEd =50kN < Nv,b,Rd = 51.6kN OK!

Angle section 80 x 60 x 7 L is adequate.

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