Purlin Design
Design data
Span of the purlin L = 6 m
Spacing of purlins S = 1.2 m
Type of the purlin = C (C/Z)
Material yeild strength fy = 450 N/mm2 (280/350/450)
Tensile strength of sag rod ft = 400 N/mm2
Diameter of sag rod Φs = 12 mm
Purlin data Depth (D) = 150 mm
width (B) = 65 mm
Thickness (t) = 1.96 mm
Coating thickness (tc) = 0.04 mm
Lip height (d) = 14 mm
Internal radius .(r) = 3 mm
Elastic modulus (Zxx) = 28 cm3
(Zyy) = 7.2 cm3
Radius of gyration (rx) = 6 cm
(ry) = 2.2 cm
Inaertia Ixx = 190 cm4
Cross section area ( A) = 5.21 cm2
Modulus of Elasticity .(E) = 205 109 N/m2
Load data
Dead Load
Zn/Al roofing sheets (0.47mm coating thick) = 0.07 kN/m2
Area weight of steel purling = 0.02 kN/m2
Light weight ceiling in mineral = 0.1 kN/m2
Thermal insulations = 0.01 kN/m2
AC Ducting = 0.15 kN/m2
Total dead load = 0.35 kN/m2 gk = 0.35
kN/m2
Imposed load = 0.25 kN/m2 qk= 0.25
Point loads kN/m2
Resultant point load = 0.9 kN Min. 0.9 kN
BS 6399 pt 3 Ultimate resultant point load = 1.44 kN
Cl 4.3.1 Distance from left edge = 3 m
Wind load
CP3 Chap.V Calculated dynamic wind pressure q = 0.5 kN/m2
Table 8 h = 24.2 w = 14.4 h/w = 1.68
Roof angle = 10
Cpe = -0.7
Appendix E Cpi = 0.2
Pressure normal to roof F/A = (Cpe-Cpi)q
= -0.45 kN/m2 wk= -0.45
Page 14
Load case 1 - ( 1.4gk+1.6qk) = 0.89 kN/m2 0.89 kN/m2
Load case 2 - ( 1.4gk+1.4wk) = -0.1 kN/m2 0.14
Load case 3 - ( 1.2gk+1.2qk+1.2wk) = 0.18 kN/m2 0.18
Ultimate UDL (1.4 x0.35+ 1.6 x 0.25) = 0.89 kN/m2
UDL per purlin = 1.1 kN/m
Number of restraints (Sag rods) Lr = 1
Internal support width N = 100 mm
Constrction type = 2
(1 -Single span , 2- Double span, 3 -Not sure)
Analysis
Case 1 - Single span beam (Single span construction)
0kN/m 3m 0
6m 6m
Sagging Moment = 0 kNm + Sagging Moment = 0 kNm
Hogging Moment = 0 kNm Hogging Moment = 0 kNm
Shear force = 0 kN Shear force = 0 kN
Case 2 - Two continous span beam (Double span construction)
1.1kN/m 3m 1.44kN 3m 1.44kN
+
6m = 6m =
Reynolds HB Sagging Moment = 2.77 kNm + Sagging Moment = 2.16 kNm
Table 36 Hogging Moment = 4.95 kNm Hogging Moment = 2.6 kNm
Shear force = 4.13 kN Shear force = 1.58 kN
Max. design sagging moment = 4.93 kNm
Max. design hogging moment = 7.55 kNm
Design shear force at support = 5.7 kN
Design shear force at span = 1.6 kN
BS5950 pt5 Design stregth of purlin (fy/1.15) = 391 N/mm2
Cl 5.2.2.3
Page 15
Section analysis
BS5950 pt5 a.) B /t = 65 / 1.96 = 33.2 < 60 ; Hence Ok Section Ok
Cl 4.2 b.) min. height of lip (d) > B/5
65 / 5 = 13 < 14Hence;Ok
c.) 45t < Overall height < 100t
45x1.96 < 150 < 100x 1.96
88.2 < 150 < 196 Hence Ok
Limiting stress for stiffned web in bending
Limiting comptressive strength p0 = 370 N/mm2
py 1.13-0.0019D/t√(fy/280)
Max.allwable stress in the compression flange
fc = 370 N/mm2
BS5950 pt5 Efefctive width of compression flange
Table B.2 b2 ( = D - 2t - 2r) = 140 mm
b1 (= B - 2t - 2r) = 55.1
h = b2/b1 h = 140.08 / 55.08
= 2.54
3
K1 = 5.4 - 1.4h - 0.02h
0.6+h
= 3.94 or min 4
= 4
BS 5950 pt 5
Cl 4.3 p cr = 939 N/mm2
fc/pcr = 0.39 N/mm2 > 0.123
0.9841 b eff / b = 0.9841
beff = 54.20 mm
Equivalent length of corners (2 t) = 3.92 mm
effective width of the compression flange = 62.9 mm
Page 16
Centerline length of purlin
62.9
12
148
12
63.0
Calculation of effective section modulus
2
Element Ai yi Aiyi Ig +Aiyi
Top lip 23.5 68.0 1599.4 109038.7 1.96
Top flange 123.3 74.0 9125.9 675357.3
Web 290.1 0.0 0.0 529492.7
Bot. flange 123.5 -74.0 -9137.5 676216.0
Bot. lip 23.5 -68.0 -1599.4 109038.7
Total 583.9 -11.6 2099143.5
Vetical shift of Nutral axis = -11.6 = -0.02 mm
584
The second moment area of the effective section = 199 cm4 at po = 391
The effective section modulus Zxr = 199
(74 +0.02) /10
= 26.8 cm3
Design for Bending Bending Ok
The moment resistance if the retrained beam(Mc) = Zxr.py
= 26.8x1000 x 370 10-6
= 9.92 kNm
> 7.55kNm ; Ok
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Checking for lateral bracings L.T.B. Not Ok
BS5950 pt 5 Number of restraints (Sag rods) = 1
Cl 5.6.3 The effective length of the purlin LE = 0.8 Lr
= 0.8 x 6/2
= 2.4 m
LE = 109
ry
ƞ = My = Zxx . Py
= 10.4 kNm
Cl 5.6.2.1 Cb = 1 ME =
ƞ = 0.002 (2.4x100 - 40 x1)
2.2
= 0.1 6643.1716 1.05 ME = 7 kNm
Φb = My +(1+ƞ )ME Φb = 10.4+(1+0.1)x 7 = 9.1
2 2
Buckling resistance moment (Mb) =
= 7 x 10.4
9.1+ √82.81-7x10.4
= 5.94 kNm
> design moment of 4.93 ;Hence Ok
Design for Shear Shear Ok
BS5950 pt 5 Shear yeild strength pv = 0.6py
Cl 5.4.3. = 0.6x 391.3
= 235 N/mm2
2
Shear buckling strength qcr = 1000t
D
2
= 1000x 1.96
150
= 171 N/mm2
Max. shear force, Fvmax = 5.7 kN
Shear area = Dt
= 150x1.96
= 294 mm2
Average shear stress = 5.7x 1000
294
= 19.4 N/mm2
< qcr,pv = 171; Ok
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Web crushing at internal support Web crushing Ok
BS5950 pt 5 At the internal support, the bearig length ,N,is considered 100 mm.
Table 8 Rection at support Rs = 2 x shear force (=4.125)
Cl 5.3. = 2 x4.125 = 8.25 kN
for C> 1.5D , N/t = 100/1.96 = 51
Toal web resistance pw =
k = fy
228 γm
= 450 = 1.72
228 x1.15
C1 = 1.22 -0.22k
= 1.22-(0.22 x 1.72) = 0.80
C2 = 1.06 - 0.06 r/t
1.06- 0.06x 3/1.96= 0.97
2
C12 = 0.7+0.3(θ / 90)
2
= 0.7+0.3(90/90) 1
Pw = 4x1.72x0.8x0.97x1x {3350-4.6x(150/1.96)} x {1+0.007 x51}
Pw = 20.9 kN
> Rs (=8.25) ;Ok
Deflection check Deflection ok
Servicibility Loads
The unfactored imposed load UDL = 0.3 kN/m
The unfactored concentrated imposed load PL = 0.9 kN
Deflection due to UDL δ1 = wL4 /185EI
388800 72057500 = 5.4 mm
DEflection due to PL δ2 = WL3 /65EI
= 5.0 mm
Total Deflection δ1+δ2 = 10.4 mm
BS5950 pt 1
Table 8 Allowable span /depth (span /200) = 6000 /200
= 30 mm
> (= 10.4 ); Deflection ok
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Combined effects - Bending and Web Crushing Bend+crushing Ok
Concentrated web load or reaction Fw = 8.25 kN
Concentrated load resistance Pw = 20.9 kN
Applied bending moment M = 7.55 kNm
Moement capacity Mc = 9.92 kNm
Fw = 8.25 = 0.4 < 1;Ok
Pw 20.9
M = 7.55 = 0.76 < 1;Ok
Mc 9.92
1.2 Fw + M <= 1.5
Pw Mc
8.25 + 7.55 = 1.16 < 1.5;Ok
20.9 9.92
Combined effects - Bending and Shear Bend+Shear Ok
pvDt = 235x150x1.96 qcr Dt = 171x150x1.96
= 69.1 kN = 50.3 kN
Applied shear force Fv = 5.7 kN
Shear capacity Pv = 50.3 kN
2
Fv 1 + M
2
Pv Mc
2
5.7 1 + 7.55
2
= 0.59 < 1.0;Ok
50.3 9.92
Design of Sag rod Sag rod -OK
Design bending moment = 4.93 kNm
BS5950 pt 5 Balancing compression force (4.93/150) F = 32.9 kN
Cl 5.6.1 Maximum tension force on sag rod (2x 0.03F) = 1.97 kN
Tensile strength of sag rod = 400 N/mm2
No of sag rods = 1
Area of sga rod provided = 113 mm2
Required steel area = 5.67 mm2
<113 mm2
Sag rod -OK
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