`Limit State of Collapse in Compression Design of an Axially Loaded Short Column

Transcription

1 `Limit State of Collapse in Compression Design of an Axially Loaded Short Column Cl.39, p-70, IS: Members in compression are called columns and struts. The term column is reserved for members which transfer loads to the ground and the term strut is applied to a compression member in any direction, as those on a truss. Column members whose height is not more than three times its lateral dimension are called pedestal, while the term wall is used to compression members whose breadth is more than four times the thickness of the wall. as, According to Cl , p-41, IS: , a compression member may be considered i) short when slenderness ratio lex / D & ley / b (both) <= 12 ii) slender when slenderness ratio lex / D & ley / b (both) > 12 where, lex = effective length in respect of the major axis ley = effective length in respect of thor axis D = depth in respect of the major axis b = breadth in respect of thor axis According to Cl , p-42, IS: , a unsupported length L, of a compression member is defined as clear distance between the end restrains According to Cl.25.2, IS: , the unsupported length between end restrains shall not exceed 60 times the least lateral dimension of a column, which is known as slenderness limits for column. Minimum Eccentricity emin = L/500 + h/30 but < 20mm Where, L = unsupported length h = lateral dimension of the column perpendicular to the axis of bending The major axis of bending x-x, is taken as an axis bisecting the depth of the column and minor axis of bending y-y is one which bisects the width of the column.

2 According to Annex E, p-92, IS: , a effective length le, of a column is a length between points of zero bending moments or between the points of contra flexure of a buckled column. The main Assumptions made for limit state design of columns failing in pure compression, as given in IS: , Cl.39.1 are: i) plane sections remain plain in compression ii) the maximum compressive strain in concrete in axial compression is iii) the design stress strain curve of steel in compression is taken to be the same as in tension iv) the stress strain curve of concrete as given in IS: is such that the maximum strength attained in concrete is 0.45fck. Derivation of Design Formula for Short Axially Loaded Column The ultimate failure is assumed to be reached when the section reaches a uniform compression strain of Ultimate load is given by the expression, Ultimate Load = Load carried by Concrete + Load carried by Steel So, P = AC * C + AS * S Where, C & S are the stresses in the materials at a uniform strain of AC = area of concrete & AS = area of longitudinal steel The compression in concrete C at failure is given by C = 0.45 fck The compression in steel S at failure will be, S = 0.87 fy for Fe-250 S = 0.75 fy for Fe-415

3 Hence, the ultimate load carrying capacity of the column PU is given by the expression, For Fe-415 steel, PU = 0.45 fck AC fy AS For Fe-250 steel, PU = 0.45 fck AC fy AS However, it is never possible to apply the load centrally on a column. Accidental eccentricities are bound to happen. Indian and British codes allow an accidental eccentricity of 5% of the lateral dimension of the column in the plane of bending (0.05D) in the strength of column itself. For this purpose, the ultimate load PU derived above is reduced by 11%. For Fe-415 steel, PU = 0.40 fck AC fy AS For Fe-250 steel, PU = 0.40 fck AC fy AS Above equations can be slightly modified by, Total Gross cross section area = AG = Area of Concrete AC + Area of Steel AS So, For Fe-415 steel, PU = 0.40 fck AG + (0.67 fy 0.40 fck) AS Checking Accidental Eccentricity: Accidental eccentricities are caused by imperfections in construction, inaccuracy in loading etc. IS: Cl.25.4 gives an expression for the possiblimum eccentricity as, emin = L/500 + h/30 but < 20mm Where, L = unsupported length h = lateral dimension of the column perpendicular to the axis of bending e.g. For a column 600 x 450 of unsupported height 3.0m, considering the long direction according to IS formula, emin = L/500 + D/30 = 3000 / / 30 = 26mm, which is greater than 20mm so, use emin = 26mm [if emin is coming less than 20mm then take 20mm] 26 Then, = = 0.043, which is less than D 600 considering the short direction according to IS formula, emin = L/500 + b/30 = 3000 / / 30 = 21mm, which is greater than 20mm so, use emin = 21mm [if emin is coming less than 20mm then take 20mm] Then, = = 0.047, which is less than b Both, these values are less than the specified ratio of 0.050, and hence the simple column formula is applicable to the above column. Formula are, For Fe-415 steel, PU = 0.45 fck AC fy AS For Fe-250 steel, PU = 0.45 fck AC fy AS If, the ratio of emin / D & emin / b, both or any of one is coming greater than 0.050, then 11% strength reduction formula is applicable. Formula are, For Fe-415 steel, PU = 0.40 fck AC fy AS For Fe-250 steel, PU = 0.40 fck AC fy AS

4 Design of Longitudinal Steel: The area of the longitudinal steel is to be calculated by the strength formula given above. However, there are a number of other factors that should be taken into account in choosing the longitudinal steel and they are given in Cl., IS: The important provisions for design of longitudinal steel are, i) thimum diameter of longitudinal steel should be 12mm ii) there should be atleast four longitudinal bars in a rectangular column and at least six of them in a circular column iii) the % of longitudinal steel should not be less than 0.8, nor should it be greater than 6% of gross cross sectional area of the section. But, normally the % of steel in columns should not exceed 4 iv) Minimum cover is 40mm v) Spacing of the longitudinal bars measured along the periphery of the column should not be more than 300mm [IS: , Cl ] Design of Lateral Ties: All longitudinal bars (as these are in compression) should be properly restrained by ties, tied at proper intervals, so that the steel bars do not in their turn act as long columns. Detailed rules are given in IS: , Cl regarding provision of transverse steel. Briefly stated, these rules are, 1) links should be so arranged that every corner and alternate longitudinal bar, if spaced not more than 75mm, should have lateral support provided by the corner of a link having an internal angle of not more than 135 degrees. For circular columns the support is calculated adequate if they are provided by circular ties touching the longitudinal steel. 2) the diameter of the links should be at least ¼ of the largest diameter of the longitudinal steel. In any case, the links should not be less than 6mm in diameter. 3) the spacing of the links should not exceed the least of the following: i) the least lateral dimension of column ii) sixteen times the diameter of the smallest longitudinal bar iii) 300mm 4) proper cover should be provided for the links. Nominal cover of 40mm can be reduced to 25mm if column size is 200mm or less and main steel 12mm in diameter is used.

5 Type of Problem1: Calculate the area of reinforcement required Data Given: Grade of concrete & steel, size of column, ultimate or characteristic axial load, Effective length or Unsupported length of the column. Steps: i) Determinimum eccentricity emin considering the long direction according to IS formula, emin = L/500 + D/30 or 20mm [whichever is greater] considering the short direction according to IS formula, emin = L/500 + b/30 or 20mm [whichever is greater] Find the ratio of D & b If both ratios are coming less than then use the equations, For Fe-415 steel, PU = 0.45 fck AG + (0.75 fy 0.45 fck) AS For Fe-250 steel, PU = 0.45 fck AG + (0.87 fy 0.45 fck) AS If both ratios or any of the one are coming greater than then use the equations, For Fe-415 steel, PU = 0.40 fck AG + (0.67 fy 0.40 fck) AS ii) Find the value of AS (area of longitudinal reinforcement). Assume the diameter of bar (min. 12mm) & put the nos. of reinforcement (min. 4 nos.) & draw the longitudinal section of a column. iii) Check th. % of reinforcement, that is 0.8% of Gross cross section area of column. If this is coming greater than the value of AS, which we have found from the above step, then put this as AS. iv) Also check the max. % of reinforcement, that is 4% of Gross cross section area of column. If AS, which we have found from step no. ii) is coming greater than this value then revise the cross section of column or increase the grade of concrete. Type of Problem2: Calculate the size & area of reinforcement required Data Given: Grade of concrete & steel, ultimate or characteristic axial load, Effective length or Unsupported length of the column. Steps: i) Assume the % of longitudinal steel Pt in column (from 1% to 4%) ii) Pt = area of longitudinal steel AS / Gross cross section area of column AG iii) Consider the column is subjected to accidental eccentricity iv) use the equations, For Fe-415 steel, PU = [0.40 fck + (0.67 fy 0.40 fck) Pt] AG For Fe-250 steel, PU = [0.40 fck + (0.77 fy 0.40 fck) Pt] AG v) find AG. Define the size of column either square or rectangle vi) With the use of same equation mention in step iv) determine the PU, with consideration of the provided AG. vii) Compare the value of PU, which we have found from above step & value which is given in data. If the found value of PU is coming greater than the value of PU given in data i.e. the % of steel which we have assumed that is right, otherwise increase the % of steel and repeat the complete procedure. viii) and for finding out the area of longitudinal reinforcement AS, use the equations For Fe-415 steel, PU = 0.40 fck AG + (0.67 fy 0.40 fck) AS ix) draw the longitudinal section of a column.