# COLD FORMED STEEL SECTIONS - II

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COLD FORMED STEEL SECTIONS - II

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

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

Teaching Resources for Steel Structures
Axially Compressed Columns 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 (Aeff) is the first step in analysing column behavior the ultimate load (or squash load) of a short strut is Pc s = Aeff . fyd = Q. A. fyd Q = the ratio of the effective area to the total area of cross section at yield stress ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG © IIT Madras, SERC Madras, Anna Univ., INSDAG Calcutta

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

Teaching Resources for Steel Structures
Axially Compressed Columns - 3 Column Strength (non- dimensional) for different Q factors 1.0 e / ry pc / fy Q = 0 .9 Q = 0 .8 Q = 0 .7 Q = 0 .6 Q = 0 .5 Q = 0 .4 Q = 0 .3 Q = 1.0 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG © IIT Madras, SERC Madras, Anna Univ., INSDAG Calcutta

Axially Compressed Columns - 4
Teaching Resources for Steel Structures 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 es and a consequent moment P. es . This would cause an additional compression on flange AB Neutral axis A B Load point es Effective section neutral axis ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG © IIT Madras, SERC Madras, Anna Univ., INSDAG Calcutta

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

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

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

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 , PTF where PEX = ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG

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

Axially Compressed Columns - 10
x0 is the distance from shear centre to the centroid measured along the x axis (mm)  J St Venants' Torsion constant (mm4) =  warping constant for all section. ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG

Axially Compressed Columns - 11
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. ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG

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

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

Combined Bending and Compression- 2
Local Capacity Check Fc = applied axial load Pcs = short strut capacity defined by Aeff.Pyd Mx, My = applied bending moments about x and y axis Mcx = Moment resistance of the beam about x axis in the absence of Fc and My Mcy = Moment resistance of the beam about y axis in the absence of Fc and Mx. ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG

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

Combined Bending and Compression- 4
where   Pc = axial buckling resistance in the absence of moments PEX, PEY = flexural buckling load in compression for bending about the x- axis and for bending about the y-axis respectively. Cbx, Cby = Cb factors for moment variation about x and y axis respectively. Mb = lateral buckling resistance moment about the x axis. ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG

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 (Pt) is evaluated from Pt = Ae . Py When a member is subjected to both combined bending and axial tension, the capacity of the member should be ascertained as ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG

where Ft = applied load Pt = tensile capacity . Tension Members - 2

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

Empirical Methods Z Purlins
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 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG

ratio should be less than 35
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 -        ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG

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

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

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:  80  25  4.0 mm Using mid-line dimensions for simplicity. Internal radius of the corners is 1.5 t. ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG

Effective breadth of web ( flat element ) h = B2 / B1 = 60 / 180 = 0
196 76 23 Mid-line dimension 200 180 800 60 25 15 Exact dimension 4 6 mm ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG

= 5.71 pcr = K1 ( t / b )2 = N / mm2 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG

Effective width of flanges ( flat element ) K2 = K1 h2 ( t1 / t2 )2
or beff =  = mm Effective width of flanges ( flat element ) K = K1 h2 ( t1 / t2 )2 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG

K2 = K1 h2 ( t1 / t2 )2 = K1 h (  t1 = t2 ) =  = or 4 (minimum) = 4 pcr =  4  ( 4 / 60 ) = N /mm2 beff = mm ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG

= 1 beff = 15 mm Effective width of lips ( flat element)
K = ( conservative for unstiffened elements) pcr =   ( 4 / 15 )2 = N /mm2 = 1 beff = 15 mm ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG

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. 196 76 23 Gross section Reduced section 96.5 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG

The calculation for the area of gross section is tabulated below:
The area of the gross section, A = mm2 The calculation of the area of the reduced section is tabulated below: A i (mm2) Lips 2 * 23 * 4 184 Flanges 608 Web 784 Total 1576 2 * 76 * 4 196 * 4 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG

The area of the effective section , Aeff = 1490.2 mm2
Lips 2 * 15 * 4 = 120 Flanges Web Total 176 * 94 * 4 = 707.8 4 * 45* 6 = 182.4 2 * 60 * 4 = 480 Corners 1490.2 The area of the effective section , Aeff = mm2 Therefore, the factor defining the effectiveness of the section under compression ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG

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

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 :  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) ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG

Only the compression flange is subject to local buckling.
Using mid-line dimensions for simplicity. Internal radius of the corners is 1.5t. 218 63 14 Mid-line dimension 220 210.08 65 55.08 15 10.04 Exact dimension ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG

Thickness of steel(ignoring galvanizing), t = 2 - 0.04 = 1.96 mm
Internal radius of the corners =  = 3 mm Limiting stress for stiffened web in bending and py = / = N / mm2 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG

which is equal to the maximum stress in the compression
= N / mm2 which is equal to the maximum stress in the compression flange, i.e., fc = N / mm2 Effective width of compression flange h = B2 / B1 = 3.8 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG

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

Effective section in mid-line dimension:
beff =  = Effective section in mid-line dimension: The equivalent length of the corners is 2.0  = 4 mm The effective width of the compression flange =  = Neutral axis of Gross section reduced section 31.2 218 14 63 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG

The calculation of the effective section modulus is tabulated as below:
Elements Ai yi(mm) Ai yi(mm3) Ig + Ai yi2 (mm2) (mm4) Top lip Compression flange Web Tension Bottom lip Total ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG

The vertical shift of the neutral axis is
The second moment of area of the effective section is Ixr = (  )  10-4 = cm4 at p0 = N / mm2 or = at py = 280 / 1.15 N / mm2 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG

The effective section modulus is,

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 RA RC RB ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG

Coefficients for Reactions
Bending Moment 0.375 0.438 Two spans loaded One span loaded Coefficients for Reactions 1.25 0.625 -0.063 4 kN/m 4.5 m ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG

Maximum hogging moment = 0.125  29.3  4.5 = 16.5 kNm
Maximum sagging moment =  29.3  = kNm Shear Force Two spans loaded : RA =  = 11 kN RB =  = kN One span loaded : RA =  = kN ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG

 Maximum reaction at end support, Fw,max = 12.8 kN
Maximum shear force , Fv,ma = = kN Try 180  50  25  4 mm Double section ( placed back to back) 180 50 255 6 mm ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG

Material Properties : E = 205 kN/mm2 py = 240 / 1.15 = 208.7 N/mm2
Section Properties : t = mm D = mm ryy = mm Ixx =  518  104 mm4 Zxx =  103 mm3 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG

Only the compression flange is subject to local buckling
Limiting stress for stiffened web in bending and py = / = N / mm2 = N / mm2 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG

which is equal to the maximum stress in the compression
flange, i.e., fc = N / mm2 Effective width of compression flange h = B2 / B1 = 160 / = 5.3 = or 4 (minimum) = 4 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG

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

The compression flange is fully restrained over the
Ixr =  518  104 mm4 Zxr =  103 mm3 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. ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG

The moment resistance of the restrained beam is: Mcx= Zxr py
=  103 ( 240 / 1.15) 10-6 = 24 kNm > 16.5 kNm O.K Shear Resistance Shear yield strength, pv = 0.6 py =  240 / = N/mm2 ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG

Shear buckling strength, qcr = = = 493.8 N/mm2
Maximum shear force, Fv,max = kN Shear area =  = mm2 Average shear stress fv = < qcr O.K . ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG

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 = and restrained section . ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG

C is the distance from the end of the beam to the load or reaction.

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 = / = and restrained section. ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG

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

A coefficient of is used to take in account of
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 = / = kN ©Teaching Resource in Design of Steel Structures – IIT Madras, SERC Madras, Anna Univ., INSDAG

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

THANK YOU ©Teaching Resource in Design of Steel Structures –

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