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A rectangular column of dimension $(275 \times 450)$ mm is subjected to an ultimate axial load of $900KN.$

Design the footing for the column assuming safe bearing capacity of the soil to the $210KN/m^2$ . Use M25 & TOR steel.

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Column size $(275 \times 450) mm \\ P = 900KN \\ Pu = 1.5 \times 900 = 1350 KN \\ SBC = 210 KN/m^2$

Area of footing $(A_f)=\dfrac {P+10 \% \text {extra assel fwt.}}{SBC}\\ =\dfrac {900+\frac {10}{100}\times 900}{210}\\ =A_f=4.71m^2$

$$C_x=\dfrac {B-0.275}2 ; C_y=\dfrac {L-0.45}2$$

$Now, C_x=C_y\\ \dfrac {B-0.275}2 =\dfrac {L-0.45}2 \\ L= B+ 0.175\\ A_f= L\times B\\ 5.5=(B+0.175)\times B\\ B=2.08m\space \&\space L=2.25m\\ Now, \\ \text {Upward soil pressure (w)}=\dfrac {Pu}{A_f\space provider}\\ W=\dfrac {1350}{2.08\times 2.25}=288.45KN/m^2\\ C_x=\dfrac {2.08-0.275}2=0.9\\ C_y=0.9\\ M_{ux}= L\times w\times \dfrac {C_x}2\\ M_{ux}=288.45\times 2.25\times \dfrac {0.9}2 =262.85 KNm\\ M_{uf}=w\times B\times \dfrac {C_y^2}2=288.46\times 2.08\times \dfrac {0.9}2=243 KNm\\ M_{u\space max}=0.318 f_Ckbxd^2 \\ 504.47\times 10^6=0.138\times 20\times 2250\times D_1^2\\ \therefore d_1=205.73 mm ..... (1)$

Two way shear:

$$Z_{cpermi} =Z_{uc}^1\times Ks$$ $Z_{uc}^1=1.25 N/mm^2\\ Ks=0.5 +\dfrac {0.275}{0.45} =1.11 \gt 1\\ Z_{uc}=Z_{uc}^1 \times Ks=1\times 1.25=1.25N/mm^2=1250 KN/m^2$

$$\text {Shear stress} =\dfrac {Pu- (w\times \text {Area of critical section}} {\text {Area of resting shear}}$$

$1250=\dfrac {1350-\{288.46\times [(0.45+d)\times (0.275+d)]\}}{2\times [0.45+d+0.275+d]\times d}.... (2)\\ d_2=0.342=342 mm \\ Provide \space d=342 mm \\Ast_x=\dfrac {0.5\times 20\times 2250\times 350}{415}\times [1-\sqrt{1-\dfrac{4.6\times 262.85\times 10^6}{20\times 2250\times 350^2}}]\\ Ast_x =2181.38 mm^2\\ \text {Provide 12 no. of 16 mm }\theta \\ Ast_p=2412 mm^2 \\Ast_y=\dfrac {0.5\times 25\times 2080\times 350}{415}\times [1-\sqrt{1-\dfrac{4.6\times 243\times 10^6}{25\times 2080\times 350^2}}]z\\ Ast_y =2016 mm^2\\ \text {Provide 11 no. of 16 mm }\theta \\ Ast_p=2211.68 mm^2\\ Ast_{min} =\dfrac {0.12}{100}bD =\dfrac {0.12}{100}\times 2250\times 425\\ Ast_{min}=1147.5mm^2 \lt Ast_x \space \& \space Ast_y \therefore safe$