5.10 Examples 5.10.1 Analysis of effective section under compression To illustrate the evaluation of reduced section properties of a section under axial compression. Section: 00 x 80 x 5 x 4.0 mm Using mid-line dimensions for simplicity. Internal radius of the corners is 1.5t. Effective readth of we (flat element) h = B / B 1 = 60 / 180 = 0. = 5.71 or 4 (minimum) = 5.71 1.8h K1 = 7 1.4h 0.15 + h 1.8x 0. = 7 1.4x 0. 0.15+ 0. p cr = 185000 K 1 ( t / ) = 185000 x 5.71 x (4 / 180) = 51.7 N / mm fcr 40 = = 0.4 > 0.1 p x γ 51.7 x1.15 cr m
eff f = + 1 14 cr 0.5 ( pcr xγm ) { } 0. 4 0. 4 = 1+ 14 0.4 0.5 = 0.98 or eff = 0.98 x 180 = 176.94 mm Effective width of flanges (flat element) K = K 1 h ( t 1 / t ) = K 1 h (t 1 = t ) = 5.71 x 0. = 0.6 or 4 (minimum) = 4 p cr = 185000 x 4 x ( 4 / 60 ) = 89 N /mm fc 40 = = 0.06> 0.1 p x γ 89 x1.15 cr m eff = 1 eff = 60 mm Effective width of lips ( flat element) K = 0.45 (conservative for unstiffened elements) p cr = 185000 x 0.45 x ( 4 / 15 ) = 5591 N /mm fc 40 = = 0.04> 0.1 p x γ 5591x1.15 cr m eff = 1 eff = 15 mm Effective section in mid-line dimension As the corners are fully effective, they may e included into the effective width of the flat elements to estalish the effective section.
The calculation for the area of gross section is taulated elow: A i (mm ) Lips x x 4 = 184 Flanges x 76 x 4 = 608 We 196 x 4 = 784 Total 1576 The area of the gross section, A = 1576 mm The calculation of the area of the reduced section is taulated elow: A i (mm ) Lips x 15 x 4 = 10 Corners 4 x 45.6 = 18.4 Flanges x 60 x 4 = 480 We 176.94 x 4 = 707.8 Total 1490. compression, The area of the effective section, A eff = 1490. mm Therefore, the factor defining the effectiveness of the section under Aeff 1490 Q = = = 0.95 A 1576
The compressive strength of the memer = Q A f y / γ m = 0.95 x 1576 x 40 / 1.15 = 1 kn 5.10. Analysis of effective section under ending To illustrate the evaluation of the effective section modulus of a section in ending. We use section: 0 x 65 x.0 mm Z8 Generic lipped Channel (from "Building Design using Cold Formed Steel Sections", Worked Examples to BS 5950: Part 5, SCI PUBLICATION P15) Only the compression flange is suject to local uckling. Using mid-line dimensions for simplicity. Internal radius of the corners is 1.5t. Thickness of steel (ignoring galvanizing), t = - 0.04 = 1.96 mm Internal radius of the corners = 1.5 x = mm Limiting stress for stiffened we in ending D f p 1.1 0.0019 p t 80 y 0 = y
and p y = 80 / 1.15 = 4.5 N / mm 0 80 80 p0 = 1.1 0.0019 x 1.96 80 1.15 =. N / mm Which is equal to the maximum stress in the compression flange, i.e., f c =. N / mm Effective width of compression flange h = B / B 1 = 10.08 / 55.08 =.8 1.4h K1 = 5.4 0.0h 0.6 + h 1.4 x.8 = 5.4 0.0 x.8 0.6 +.8 =.08 or 4 (minimum) = 4 1.96 pcr = 185000 x 4 x = 97 N / mm 55.08 fc. = = 0.4> 0.1 p 97 cr eff f = + 1 14 c 0.5 pcr { } 4 0. 0. 4 = 1+ 14 0.4 0.5 = 0.998 eff = 0.99 x 55 = 54.5 Effective section in mid-line dimension: The equivalent length of the corners is.0 x.0 = 4 mm The effective width of the compression flange = 54.5 + x 4 = 6.5 The calculation of the effective section modulus is taulated as elow:
Elements A i ( mm ) y i (mm) A i (mm ) y i I g + A i y i (mm 4 ) Top lip 7.44 10 799 448 + 85498 Compression flange 1.5 109 15.5 9. + 14554.5 We 47. 0 0 9. + 14554.5 Tension flange 1.5-109 -1459. 9.5 + 1467064 Bottom lip 7.4-10 -799 448 + 85498 Total 78. -106.8 518668.4 The vertical shift of the neutral axis is 106.8 y = = 0.15mm 78. The second moment of area of the effective section is I xr = (518668.4 + 78. x 0.15 ) x 10-4 = 518.7 cm 4 at p 0 =. N / mm or =. x 1.15 4 518.7 x = 475.5cm at p y = 80 / 1.15 N / mm 80 The effective section modulus is, Z xr 475.5 = = 4.56 cm ( 109 + 0.15) 10 5.10. Two span design Design a two span continuous eam of span 4.5 m suject to a UDL of 4kN/m as shown in Fig.1. Factored load on each span = 6.5 x 4.5 = 9. kn
Bending Moment Maximum hogging moment = 0.15 x 9. x 4.5 = 16.5 knm Maximum sagging moment = 0.096 x 9. x 4.5 = 1.7 knm Shear Force Two spans loaded: R A = 0.75 x 9. = 11 kn R B = 1.5 x 9. = 6.6 kn One span loaded: R A = 0.48 x 9. = 1.8 kn Maximum reaction at end support, F w,max = 1.8 kn Maximum shear force, F v,max = 9. - 11 = 18. kn
Try 180 x 50 x 5 x 4 mm Doule section (placed ack to ack) Material Properties: E = 05 kn/mm p y = 40 / 1.15 = 08.7 N/mm Section Properties: t = 4.0 mm D = 180 mm r yy = 17.8 mm I xx = x 518 x 10 4 mm 4 Z xx = 115.1 x 10 mm Only the compression flange is suject to local uckling Limiting stress for stiffened we in ending D f p 1.1 0.0019 p t 80 y 0 = y and p y = 40 / 1.15 = 08.7 N / mm 180 40 p0 = 1.1 0.0019 x x08.7 4 80 = 19. N / mm Which is equal to the maximum stress in the compression flange, i.e., f c = 19. N / mm Effective width of compression flange h = B / B 1 = 160 / 0 = 5.
= 1.1 or 4 (minimum) = 4 1.4h K1 = 5.4 0.0h 0.6 + h 1.4 x.8 = 5.4 0.0 x 5. 0.6 +.8 4 pcr = 185000 x 4 x = 1155 N / mm 0 fc 19. = = 0.017 > 0.1 p 1155 cr eff = 1 eff = 0 mm i.e. the full section is effective in ending. I xr = x 518 x 10 4 mm 4 Z xr = 115.1 x 10 mm Moment Resistance The compression flange is fully restrained over the sagging moment region ut it is unrestrained over the hogging moment region, that is, over the internal support. not critical. However unrestrained length is very short and lateral torsional uckling is The moment resistance of the restrained eam is: M cx = Z xr p y
O.K = 115.1 x 10 x (40 / 1.15) 10-6 = 4 knm > 16.5 knm Shear Resistance Shear yield strength, p v = 0.6 p y = 0.6 x 40 /1.15 = 15. N/mm Shear uckling strength, q cr = 1000t 1000 x 4 = D 180 = 49.8 N/mm Maximum shear force, F v,max = 18. kn Shear area = 180 x 4 = 70 mm Average shear stress f v = O.K 18. x10 70 = 5.4 N / mm < qcr We crushing at end supports Check the limits of the formulae. D 180 = = 45 00 O.K t 4 r 6 = = 1.5 6 O.K t 4 At the end supports, the earing length, N is 50 mm (taking conservatively as the flange width of a single section) For c=0, N/ t = 50 / 4 = 1.5 and restrained section. C is the distance from the end of the eam to the load or reaction. Use
Cr 1 750 m N { t } f y Pw = x t Cr 8.8 + 1.11 γ D = + t 45 = 1+ = 1.06 750 40 Pw = x 4 x1.06 x { 8.8 + 1.11 1.5} 10 1.15 We Crushing at internal support t the internal support, the earing length, N, is 100mm (taken as the flange width of a doule section) For c > 1.5D, N / t = 100 / 4 = 5 and restrained section. m m N { t } f y Pw = t C5C6 1. + 1.6 γ fy 40 k = = = 0.9 8 x γ 1.15 x 8 C 5 = (1.49-0.5 k) = 1.49-0.5 x 0.9 = 1.0 > 0.6 C 6 = (0.88-0.1 m) m = t / 1.9 = 4 / 1.9 =.1 C 6 = 0.88-0.1 x.1 = 0.6 { } 40 Pw = x 4 x1x 0.6x 1. + 1.6 5 10 1.15 = 89.8 kn > R B (= 6 kn) Deflection Check A coefficient of 84 is used to take in account of unequal loading on a doule span. Total unfactored imposed load is used for deflection calculation.
WL δ max = 84 E I I av W = 9. / 1.5 = 19.5 kn av I + I 106+ 106 = = = 106x10 mm xx xy 4 4 19.5 x10 x 4500 δ max = = 6.5mm 84 4 05 x10 x106 x10 Deflection limit = L / 60 for imposed load = 4500 / 60 = 1.5 mm > 6.5 mm O.K In the doule span construction: Use doule section 180 x 50 x 5 x 4.0 mm lipped channel placed ack to ack. 5.10.4 Column design Design a column of length.7 m for an axial load of 550 kn. Axial load P = 550 kn Length of the column, L =.7 m Effective length, le = 0.85L = 0.85 x.7 =. m Try 00 x 80 x 5 x 4.0 mm Lipped Channel section Material Properties: E = 05 kn/mm f y = 40 N/ mm p y = 40 / 1.15 = 08.7 N/mm Section Properties: A = x 1576 = 15 mm I xx = x 90 x 10 4 mm 4 I yy = [14 x 10 4 + 1576 x 4.8 ] = 44 x 10 4 mm 4
r min = 4 44 x10 x1576 = 7.4 mm Load factor Q = 0.95 (from worked example 1) The short strut resistance, P cs = 0.95?? 1576? 40/ 1.15 = 65 kn P = 550 kn < 65 kn O. K Axial uckling resistance Check for maximum allowale slenderness y le. x10 = = 61.5 < 180 O.K r 7.4 In a doule section, torsional flexural uckling is not critical and thus α = 1 Modified slenderness ratio, l α r λ= λ cs y e y E.05 x10 λ y = π = π = 98.5 p 08.7 1x 61.5 λ = = 0.6 98.5 Pc = 0.91 P y 5 P c = 0.91 x 65 = 569 kn > P O. K