# 5.0 DESIGN CALCULATIONS

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3 Moment calcluation due to axial laod f<b f<b f<b f>b f>b f>b Case B : Full Raft with Moment laod condition Y2 =3/β^ β^ Y3 = Y Y6 = xβxβ Y Y7 =3XβXβ 0.64 Refer : Tall chimney by S.N.Manohar TABLE 6.10 Moment calcluation due to Moment laod. When θ is 0 cos 0 = 1 Sin 0 = 0.00 r f Mr Mt Mtor Qr Qt Qtor=1.6 xmtor/b m - Kn-m Kn-m Kn-m Kn Kn Kn f<b f<b f<b f>b f>b f>b q = M / Z Kn/m^2 Total (max at f=β ) Radial Moment Mr Total (max at f=β ) Tengantial Moment Mt Kn-m Kn-m When θ is 45 cos 45 = sin 45 = 0.71 Qtor=1.6 r f Mr Mt Mtor Qr Qt xmtor/b m - Kn-m Kn-m Kn-m Kn Kn Kn f<b f<b f<b f>b f>b f>b 7

4 When θ is 90 cos 90 = 0 sin 90 = 1.00 r f Mr Mt Mtor Qr Qt Qtor=1.6 xmtor/b m - Kn-m Kn-m Kn-m Kn Kn Kn f<b f<b f<b f>b f>b f>b Summery at different angle When θ is When θ is When θ is Sample calculation Grade of concrete = M25 Grade Of steel = Fe415 Design Moment and Reinforcment Calculation Dia of bar to be used = 25 r Mr Mu/bd^2 pt min pt ast spacing Provided m Kn-m SP-16 (TABLE 1) mm^2 required

5 Reinforcment Due to Radial Moment (max at f= 0.5) Mr Kn-m Factored Moment Kn-m Depth of Raft D mm Effective Depth d mm Mu/bd For the Mu/bd2 and fck pt 0.23 Using SP-16 (table-1) Pt Min 0.12 Ast = ptxbxd mm^2 Dia of bar to be used Spacing required mm Provide 25 mm dia 100 mm C/C Reinforcment Due to Radial Moment (at f = 0.67) Mr Kn-m Factored Moment Kn-m Depth of Raft D mm Effective Depth d mm Mu/bd For the Mu/bd2 and fck pt 0.09 Using SP-16 Pt Min 0.12 Ast = ptxbxd mm^2 Dia of bar to be used Spacing required mm Provide 25 mm dia 200 mm C/C Design Moment and Reinforcment Calculation Dia of bar to be used = 25 r Mt Mu/bd^2 pt min pt ast spacing Provided m Kn-m SP-16 (TABLE 1) mm^2 required Reinforcment Due to (Max) Tangential Moment Mt Kn-m Factored Moment Kn-m Depth of Raft D mm Effective Depth d mm Mu/bd For the Mu/bd2 and fck pt 0.09 Using SP-16 Pt Min 0.12 Ast = ptxbxd mm^2 Dia of bar to be used Spacing required mm Provide 25 mm dia 200 mm C/C 9

6 Check for One Way Shear :: Shear Force (max at d at any angle ), KN per m length Factored shear force, Vu 1500 KN per m length Calculated Shear Stress N/mm2 Min % Of Steel provide Shear strength of conc N/mm2 Enhancement of shear strength,2xdxtc/av As per clause Of IS-456 Hence Safe Check for Two-Way Shear :: Max. vertical load on Raft = W2 + W3 Shear Sress Permissible stress KN N/mm2.t c 0.25 X fck N/mm N/mm 2 ok Design of Pedestal Mg Kn-m Wg kn e = M / W m r 7 m e / r β 0 µ 0 value of φ 150 ο from chart 7.1 value of C Assume % of reinforcment 3 % Fc= (CW)/( r t (1-p)) 0.84 N/mm^2 <0.38xfck 9.5 ok Fs = (SxmxW)/(r t (1-p) S = C ((cos F) + cos m))/((cos b) - cos f)) Maximum steel stress, Fs Provide 20 mm dia 150 mm C/C 0.64 N/mm^2 <0.57fy ok 10

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