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©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 1 COLD FORMED STEEL SECTIONS - II.

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Presentation on theme: "©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 1 COLD FORMED STEEL SECTIONS - II."— Presentation transcript:

1 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 1 COLD FORMED STEEL SECTIONS - II

2 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 2 COLD FORMED STEEL SECTIONS  Introduction  Design of axially compressed columns  Flexural Torsional buckling  Design for Combined bending and compression  Design of Tension members  Design on the basis of testing  Empirical methods  Conclusion

3 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 3 INTRODUCTION Diversity of cold formed steel shapes and multiplicity of purposes makes it a complex product Design of columns for axial compression, compression combined with bending and flexural- torsional buckling are discussed. Design procedures using prototype tests or empirical rules are discussed in summary form.

4 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 4 local buckling under compressive loading is an extremely important feature. a compressed plate element with an edge free to deflect does not perform as satisfactorily as a similar element supported along the two opposite edges determination of effective area (A eff ) is the first step in analysing column behavior Axially Compressed Columns the ultimate load (or squash load) of a short strut is P c s = A eff. f yd = Q. A. f yd Q = the ratio of the effective area to the total area of cross section at yield stress

5 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 5 In long column with doubly - symmetric cross section, the failure load depends on Euler buckling resistance and the imperfections present. The failure load is, Axially Compressed Columns - 2

6 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 6 Column Strength (non- dimensional) for different Q factors e / r y p c / f y Q = 0.9 Q = 0.8 Q = 0.7 Q = 0.6 Q = 0.5 Q = 0.4 Q = 0.3 Q = 1.0 Axially Compressed Columns - 3

7 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 7 Axially Compressed Columns - 4 Effective Shift of Loading axis Effective shift in the loading axis in an axially compressed column a) Channel section loaded through its centroid (b) The move of the neutral axis (due to plate buckling) causes an eccentricity e s and a consequent moment P. e s. This would cause an additional compression on flange AB Neutral axis A B Load point B A eses Effective section neutral axis

8 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 8 Axially Compressed Columns - 5 If a section is not doubly symmetric and has a large reduction of effective widths of elements, then the effective section may be changed position of centroid. This would induce bending on an initially concentrically loaded section The ultimate load is evaluated by allowing for the interaction of bending and compression using the equation

9 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 9 Axially Compressed Columns - 6 Torsional - Flexural Buckling Singly symmetric columns may fail either a) by Euler buckling about an axis perpendicular to the line of symmetry b) by a combination of bending about the axis of symmetry and a twist Purely torsional and purely flexural failure does not occur in a general case.

10 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 10 Axially Compressed Columns - 7  v Shear centre Original position Position after Flexural - Torsional buckling Axis of symmetry Column displacements during Flexural - Torsional buckling

11 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 11 Axially Compressed Columns - 8 For sections with at least one axis of symmetry (say x - axis) and subjected to flexural torsional buckling, BS5950, Part 5 suggests that the torsional flexural buckling load (in Newtons) of a column, P TF where P EX =

12 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 12 Torsional buckling load of a column ( in Newtons), P T, given by  is a constant given by r o = polar radius of gyration about the shear centre Axially Compressed Columns - 9

13 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 13 x 0 is the distance from shear centre to the centroid measured along the x axis (mm) J St Venants' Torsion constant (mm 4 ) =  warping constant for all section. Axially Compressed Columns - 10

14 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 14 Torsion Behaviour Cold formed sections are mainly formed with "open" sections and do not have high resistance to torsion. The total torsion may be regarded as being made up of two effects - St. Venant's Torsion or Pure Torsion - Warping torsion. Axially Compressed Columns - 11

15 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 15 Axially Compressed Columns - 12 St.Venant's torsion produces shear stresses. Warping torsion produces in-plane bending of the elements of a cross section. Cold formed sections they have very little resistance to St. Venant's Torsion Short beams with ends restrained exhibit high resistance to warping torsion The resistance to warping torsion becomes low for long beams with ends restrained

16 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 16 Combined Bending and Compression Compression members subjected to bending will have to be designed considering the effects of interaction The following checks are suggested for members which have at least one axis of symmetry: (i) the local capacity at points of greatest bending moment and axial load and (ii) an overall buckling check.

17 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 17 Local Capacity Check F c = applied axial load P cs = short strut capacity defined by A eff.P yd M x, M y = applied bending moments about x and y axis M cx = Moment resistance of the beam about x axis in the absence of F c and M y M cy = Moment resistance of the beam about y axis in the absence of F c and M x. Combined Bending and Compression- 2

18 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 18 Combined Bending and Compression- 3 Overall buckling check For members not subject to lateral buckling, the relationship to be satisfied is For beams subject to lateral buckling, the following relationship should be satisfied

19 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 19 Combined Bending and Compression- 4 where P c = axial buckling resistance in the absence of moments P EX, P EY = flexural buckling load in compression for bending about the x- axis and for bending about the y-axis respectively. C bx, C by = C b factors for moment variation about x and y axis respectively. M b = lateral buckling resistance moment about the x axis.

20 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 20 Tension Members a member connected in such a way as to eliminate any moments due to connection eccentricity, may be designed as a simple tension member The tensile capacity of a member (P t ) is evaluated from P t = A e. P y When a member is subjected to both combined bending and axial tension, the capacity of the member should be ascertained as

21 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 21 Tension Members - 2 where F t = applied load P t = tensile capacity.

22 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 22 Design on the basis of Testing large variety of shapes and complex interactions make it uneconomical to design members and systems completely on theoretical basis ensure that the test set up reflects the in-service conditions as accurately as possible. testing is probably the only realistic method of assessing the strength and characteristics of connections. there is a possibility that the tests giving misleading information. testing by an independent agency is widely accepted the manufacturers provide load/span tables

23 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 23 Empirical Methods members such as Z purlins are sometimes designed by time-tested empirical rules empirical rules are employed when - theoretical analysis may be impractical or not justified - prototype test data are not available Z Purlins B  B/5 t  L /60   L / 45  Z Purlins

24 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 24 Empirical Methods - 2 Empirical design rules for Z sections (BS 5950, Part 5 ) - overall depth should be greater than 100 t and not less than L /45. - overall width of compression flange / thickness ratio should be less than 35 - Lip width should be greater than B /5 - Section Modulus ( W in kN and L in mm ) for simply supported purlins for continuous or semi rigidly jointed purlins - 

25 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 25 The net allowable wind uplift in a direction normal to roof when purlins are restrained is taken as 50% of the (dead + imposed) load.

26 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 26 CONCLUSION A detailed discussion of design of elements made from cold rolled steel The most striking benefits of all forms of light steel framing are  speed of construction  ease of handling  savings in site supervision  elimination of wastage in site  elimination of shrinkage and movement of cracks  greater environmental acceptability  less weather dependency  high acoustic performance  high degree of thermal insulation

27 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 27 Section Properties Calculation PROBLEM 1 Analysis of effective section under compression To illustrate the evaluation of reduced section properties of a section under axial compression. Section: 200  80  25  4.0 mm Using mid-line dimensions for simplicity. Internal radius of the corners is 1.5 t.

28 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 28 Effective breadth of web ( flat element ) h = B 2 / B 1 = 60 / 180 = Mid-line dimension Exact dimension 4 6 mm

29 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 29 = 5.71 p cr = K 1 ( t / b ) 2 = N / mm 2

30 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 30 or b eff =  180 = mm Effective width of flanges ( flat element ) K 2 = K 1 h 2 ( t 1 / t 2 ) 2

31 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 31 K 2 = K 1 h 2 ( t 1 / t 2 ) 2 = K 1 h 2 (  t 1 = t 2 ) = 5.71  = or 4 (minimum) = 4 p cr =  4  ( 4 / 60 ) 2 = 3289 N /mm 2 b eff = 60 mm

32 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 32 Effective width of lips ( flat element) K = ( conservative for unstiffened elements) p cr =   ( 4 / 15 ) 2 = 5591 N /mm 2 = 1 b eff = 15 mm

33 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 33 Effective section in mid-line dimension As the corners are fully effective, they may be included into the effective width of the flat elements to establish the effective section Gross section Reduced section

34 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 34 The calculation for the area of gross section is tabulated below: The area of the gross section, A = 1576 mm 2 The calculation of the area of the reduced section is tabulated below:

35 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 35 The area of the effective section, A eff = mm 2 Therefore, the factor defining the effectiveness of the section under compression

36 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 36 The compressive strength of the member = Q A f y /  m = 0.95  1576  240 / 1.15 = 313 kN

37 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 37 PROBLEM 2 ANALYSIS OF EFFECTIVE SECTION UNDER BENDING To illustrate the evaluation of the effective section modulus of a section in bending. We use section : 220  65  2.0 mm Z28 Generic lipped Channel (from "Building Design using Cold Formed Steel Sections", Worked Examples to BS 5950: Part 5, SCI Publication P125)

38 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 38 Only the compression flange is subject to local buckling. Using mid-line dimensions for simplicity. Internal radius of the corners is 1.5t Mid-line dimension Exact dimension

39 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 39 Thickness of steel(ignoring galvanizing), t = = 1.96 mm Internal radius of the corners = 1.5  2 = 3 mm Limiting stress for stiffened web in bending and p y = 280 / 1.15 = N / mm 2

40 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 40 = N / mm 2 which is equal to the maximum stress in the compression flange, i.e., f c = N / mm 2 Effective width of compression flange h = B 2 / B 1 = 3.8

41 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 41 = 3.08 or 4 ( minimum) = 4

42 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 42 b eff = 0.99  55 = 54.5 Effective section in mid-line dimension: The equivalent length of the corners is 2.0  2.0 = 4 mm The effective width of the compression flange =  4 = 62.5 Neutral axis of Gross section reduced section

43 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 43 The calculation of the effective section modulus is tabulated as below: Elements A i y i (mm) A i y i (mm 3 )I g + A i y i 2 (mm 2 ) (mm 4 ) Top lip Compression flange Web Tension flange Bottom lip Total

44 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 44 The vertical shift of the neutral axis is The second moment of area of the effective section is I xr = (  )  = cm 4 at p 0 = N / mm 2 or = at p y = 280 / 1.15 N / mm 2

45 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 45 The effective section modulus is,

46 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 46 PROBLEM 3 Design a two span continuous beam of span 4.5 m subject to a UDL of 4kN/m as shown in Fig.1. Factored load on each span = 6.5  4.5 = 29.3 kN 6.5 kN/m 4.5 m RARA RCRC RBRB

47 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 47 Bending Moment Two spans loaded One span loaded Coefficients for Reactions kN/m 4.5 m

48 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 48 Maximum hogging moment =  29.3  4.5 = 16.5 kNm Maximum sagging moment =  29.3  4.5 = 12.7 kNm Shear Force Two spans loaded : R A =  29.3 = 11 kN R B = 1.25  29.3 = 36.6 kN One span loaded : R A =  29.3 = 12.8 kN

49 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 49  Maximum reaction at end support, F w,max = 12.8 kN Maximum shear force, F v,ma = = 18.3 kN Try 180  50  25  4 mm Double section ( placed back to back) mm

50 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 50 Material Properties : E = 205 kN/mm 2 p y = 240 / 1.15 = N/mm 2 Section Properties : t = 4.0 mm D = 180 mm r yy = 17.8 mm I xx = 2  518  10 4 mm 4 Z xx =  10 3 mm 3

51 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 51 Only the compression flange is subject to local buckling Limiting stress for stiffened web in bending and p y = 240 / 1.15 = N / mm 2 = N / mm 2

52 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 52 which is equal to the maximum stress in the compression flange, i.e., f c = N / mm 2 Effective width of compression flange h = B 2 / B 1 = 160 / 30 = 5.3 = 1.1 or 4 (minimum) = 4

53 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 53 b eff = 30 mm i.e. the full section is effective in bending = 1

54 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 54 I xr = 2  518  10 4 mm 4 Z xr =  10 3 mm 3 Moment Resistance The compression flange is fully restrained over the sagging moment region but it is unrestrained over the hogging moment region, that is, over the internal support. However unrestrained length is very short and lateral torsional buckling is not critical.

55 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 55 The moment resistance of the restrained beam is: M cx = Z xr p y =  10 3  ( 240 / 1.15) = 24 kNm > 16.5 kNm  O.K Shear Resistance Shear yield strength, p v = 0.6 p y = 0.6  240 /1.15 = N/mm 2

56 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 56 Shear buckling strength, q cr = = = N/mm 2 Maximum shear force, F v,max = 18.3 kN Shear area = 180  4 = 720 mm 2 Average shear stress f v = < q cr  O.K.

57 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 57 Web crushing at end supports Check the limits of the formulae At the end supports, the bearing length, N is 50 mm (taking conservatively as the flange width of a single section) For c=0, N/ t = 50 / 4 = 12.5 and restrained section.

58 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 58 C is the distance from the end of the beam to the load or reaction. Use

59 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 59 Web Crushing at internal support At the internal support, the bearing length, N, is 100mm (taken as the flange width of a double section) For c > 1.5D, N / t = 100 / 4 = 25 and restrained section.

60 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 60 C 5 = ( k ) =  0.92 = 1.0 > 0.6 C 6 = ( m ) m = t / 1.9 = 4 / 1.9 = 2.1 C 6 =  2.1 = 0.63 = 89.8 kN > R B ( = 36 kN)

61 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 61 Deflection Check A coefficient of is used to take in account of unequal loading on a double span. Total unfactored imposed load is used for deflection calculation. W = 29.3 / 1.5 = 19.5 kN

62 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 62 Deflection limit = L / 360 for imposed load = 4500 / 360 = 12.5 mm > 6.53 mm  O.K  In the double span construction : Use double section 180  50  25  4.0 mm lipped channel placed back to back.

63 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG 63 THANK YOU


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