Purlin and Roof Design

PURLIN AND ROOF DESIGN Terminology of Truss

Rafter (Top chord)

Purlins

Nodes

Span

Bottom chord

Design of roof truss system The loading subjected by a truss is transferred through the purlins, either directed onto the nodes or on the top members span. It is ideal if the loads can be transferred to the truss at the node position. General Procedures: 1. Start 2. Loading 3. Analyse the truss members  Tension member  Compression member 4. Analyse load bearing member such as rafter as continuous beam supported at the nodes and loaded by a purlins. If the load position are uncertain, the rafter moment may be taken as wL2/6 where L is the node to node length of rafter and w is the total load per unit length. 5. End Loading The loading subjected to a truss system could be dead loads, live loads and wind load. For roof truss system, the dead load may be consisting of cladding, insulation, self-weight of trusses and purlins, services, etc.... For live load, according to BS 6399-2 or CP3: Ch V: Part 2, 0.75 kN/m2 may used where entranceto the roof is available for services purposes. Otherwise, use 1.5 kN/m2 may be used if more than that.

Purlin Design Refer to Cl. 4.12.4.3 Purlins are those members in truss system which carrying the roof sheets and transferring load to rafters. It is normally placed perpendicular to the rafters and sag rods may be added (to reduce purlin size of purlins) see Table 27 The purlins are not necessary to be analyzed as complicated as the other structural members. The satisfaction of purlins is approached by the empirical rules suggested in cl 4.12.4.3 as: (a) The slope of roof should less than 30o from horizontal (b) The loading on the purlin should substantially uniformly distributed. Not (c) The limitations of section modulus Z about its axis parallel to the plane of the cladding, member dimensions D perpendicular to the plane of caldding, B parallel to the plane of cladding are given in Table 27. Example:

Purlins 6.324 m

2m

12 m Figure Q3/ [Rajah Q3] A plane truss as Figure Q3 is arranged all purlins on its nodes. With the following data - Spacing between trusses =5m - Weight of roof sheet, insulation and purlins (on slope) = 0.35 kN/m2 - Self-weight of truss (on slope) = 0.20 kN/m2 - Imposed load (on plan) = 0.75 kN/m2 - Section Modulus of Purlins, Zp = WPL/1800 (Refer Table 27) (a) Design the purlins using single angle sections Dead load Imposed load Spacing of purlins Wpurlin Roof slope

= 0.35 kN/m2 (on slope) = 0.75 x 6/6.324 = 0.71 kN/m2 (on slope) = 6.324/3 = 2.11 m = (0.35 + 0.71) x 2.11 x 5 = 11.18 kN = tan-1 (2/6) = 18.4 < 30o

Refer Table 27: Zp = WpL/1800 D = L/45 = 5000/45 B = (L/2)/60 = (5000/2)/60

= 11.18 x 5000/1800 = 31.06 cm3 = 111.11 mm = 41.67 mm

From table of properties of single angle: use: 125 x 75 x 10L (Z = 36.5 cm3) (b) Analysis the truss. Loading transferred to the truss Dead Load = 0.35 +0.2 = 0.55 kN/m2 (on slope) Imposed Load = 0.71 kN/m2 (on slope) Total dead load = 0.55 x 2.11 x 5 = 5.80 kN Total imposed load = 0.71 x 2.11 x 5 = 7.49 kN Design load = 1.4 Gk + 1.6 Qk = 1.4 (5.8) + 1.6 (7.49) = 20.10 kN 20.10 20.10 20.10 10.05

Please analysis the truss by yourself. Tq

20.10 20.10 10.05

Designing a truss member 1. Tension Member Refer Cl. 4.6.1 & 4.6.2 Pt = pyAe a2 = Ag-Dt Pt = py(Ag-0.3a2). If higher than Ft. OK If less, choose new size (larger size) Example of sizing the truss member Force of tension members: Ft = 260 kN py = 275 N/mm2 Area needed = 260 x 103/275 = 945.45 mm2 Refer Table of properties of angle L types: Try 100 x 65 x 7L where Ag = 11.2 cm2 Dt= 100x7 = 700 mm2 a2 = 1120 – 700 = 420 mm2 Pt = py(Ag-0.3a2) = 275 (1120-0.3(420) = 275.4 kN > 260 kN  OK 2. Compression members Refer Cl. 4.7.10 Example of sizing the truss member Force of compression members, Fc = 60 kN Assume D = 24 mm, t = 8 mm Ah = Dt = 24 x 8 = 192 mm2 Assume pc = 0.4 py Ag = Fc/0.4py + Ah = 60 x 103/0.4(275) + 192 = 737.5 mm2 Try 60 x 60 x 8L (Ag = 9.03 cm2) Check like u check the column. Use Pc = Agpc If Pc > Fc Ok Otherwise redesign the section