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Column design as Per BS 8110-1:1997 PHK/JSN

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Contents :- General Recommendations of the code Classification of columns Effective Length of columns & Minimum eccentricity Design Moments in Columns Design

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General Reco’s of the code m for concrete 1.5, for steel 1.05 Concrete strength – CUBE STRENGTH Grades of steel Fe250 & Fe460 Primary Load combination 1.4DL+1.6LL E of concrete E c = 5.5√f cu / m 10% less than IS Ultimate stress in concrete 0.67f cu / m Steel Stress-strain curve – Bilinear E of steel 200 kN/mm 2

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Classification of columns SHORT – both l ex /h and l ey /b < 15 for braced columns < 10 for unbraced columns BRACED - If lateral stability to structure as a whole is provided by walls or bracing designed to resist all lateral forces in that plane. else – SLENDER Cl.3.8.1.5else – UNBRACED

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Effective length &minimum eccentricity Effective length l e = ßl o ß – depends on end condition at top and bottom of column. e min = 0.05 x dimension of column in the plane of bending ≤ 20 mm

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Deflection induced moments in Slender columns M add = N a u where a u = ß a Kh ß a = (1/2000)(l e /b’) 2 K = (N uz – N)/(N uz – N bal ) ≤ 1 N uz = 0.45f cu Ac+0.95fyA sc N bal = 0.25f cu bd Value of K found iteratively Contd..

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Design Moments in Braced columns :- Maximum Design Column Moment Greatest of a) M 2 b) M i +M add M i = 0.4M 1 +0.6M 2 c)M 1 +M add /2 d) e min N Columns where l e /h exceeds 20 and only Uniaxially bent Shall be designed as biaxially bent with zero initial moment along other axis.

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Braced and unbraced columns

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Design Moments in UnBraced columns :- The additional Moment may be assumed to occur at whichever end of column has stiffer joint. This stiffer joint may be the critical section for that column. Deflection of all UnBraced columns in a storey a uav for all stories = Σ a u /n

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Design Moments in Columns Axial Strength of column N = 0.4f cu A c + 0.8 A sc f y Biaxial Bending Increased uniaxial moment about one axis M x /h’≥ M y /b’ M x ’ = M x + ß 1 h’/b’M y M x /h’≤ M y /b’ M y ’ = M y + ß 1 b’/h’M x Where ß 1 = 1- N/6bhf cu (Check explanatory hand book) Minimum P t =0.4% Max P t = 6%

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Shear in Columns Shear strength v c ’ = v c +0.6NVh/A c M To avoid shear cracks, v c ’ = v c √(1+N/(A c v c ) If v > vc’, Provide shear reinforcement If v ≤ 0.8√f cu or 5 N/mm²

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Design – Construction of Interaction Curve A 1 A 2 Section Stress Strain Distribution of stress and strain on a Column-Section d1d1 d h 0.5h f1 f2 M N x 0.9x 11 22 0.67f cu / m 0.0035

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Equilibrium equation from above stress block N = 0.402f cu bx + f 1 A 1 +f 2 A 2 M =0.402f cu bx(0.5h-0.45x)+f 1 A 1 (0.5h-d1)+f 2 A 2 (0.5h-d) f 1 and f 2 in terms of E andf 1 = 700(x-d+h)/x f 2 = 700(x-d)/x The solution of above equation requires trial and error method

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THANK YOU

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