1. COLD FORMED STEEL SECTIONS - II
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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
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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.
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4. 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
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5. 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 ,
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6. 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
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7. 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
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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
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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.
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10. Axially Compressed Columns - 7 v
Shear centre
Original
position
Position after
Flexural - Torsional
buckling
Axis of symmetry
Column displacements during Flexural - Torsional buckling
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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 , PTF
where
PEX =
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12. 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
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13. 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.
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14. 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.
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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
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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.
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17. 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.
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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
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19. 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.
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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 (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
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21. where Ft = applied load Pt = tensile capacity . Tension Members - 2 .
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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
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23. 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
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24. 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 -
      
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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.
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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
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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:  80  25  4.0 mm
Using mid-line dimensions for simplicity. Internal
radius of the corners is 1.5 t.
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28. 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
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29. = 5.71
pcr = K1 ( t / b )2
= N / mm2
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30. 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
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31. K2 = K1 h2 ( t1 / t2 )2
= K1 h (  t1 = t2 )
=  = or 4 (minimum)
= 4
pcr =  4  ( 4 / 60 ) = N /mm2
beff = mm
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32. = 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
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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.
196 76 23
Gross section
Reduced
section
96.5
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34. 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
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35. 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
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36. The compressive strength of the member = Q A fy / m
=  1576  240 / 1.15
= kN
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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 :  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)
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38. 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
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39. 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
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40. 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
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41. = or 4 ( minimum) = 4
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42. 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
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43. 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
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44. 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
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45. The effective section modulus is,
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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 RA RC RB
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47. 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
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48. 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
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49.  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
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50. 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
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51. Only the compression flange is subject to local buckling
Limiting stress for stiffened web in bending
and py = / = N / mm2
= N / mm2
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52. 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
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53. i.e. the full section is effective in bending
= 1
beff = mm
i.e. the full section is effective in bending
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54. 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.
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55. 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
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56. Shear buckling strength, qcr = = = 493.8 N/mm2
Maximum shear force, Fv,max = kN
Shear area =  = mm2
Average shear stress fv = < qcr
O.K .
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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 = and restrained section .
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58. C is the distance from the end of the beam to the load or reaction.
Use
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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 = / = and restrained
section.
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60. C = ( k ) =  0.92
= > 0.6
C = ( m )
m = t / 1.9 = 4 / 1.9 = 2.1
C =  = 0.63
= kN > RB ( = 36 kN)
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61. 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
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62. 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.
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63. THANK YOU ©Teaching Resource in Design of Steel Structures –
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