A rectangular column of dimensions $(300 \times 450)$ mm subjected to an ultimate axial load of $1000 $KN. Design the isolated footing for the column assuming SBC to be $190 KN/m^2 .$

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written 8.8 years ago by

teamques10 ★ 69k

• modified 8.8 years ago

Column size $(300 \times 450)$ mm

$P = 1000KN \\ Pu = 1.5 \times 1000 = 1500 KN \\ SBC = 190 KN/m^2$

Area of footing $(A_f)=\dfrac {P+10\% \text {extra assel fwt,}}{SBC}\\ =\dfrac {1000+\frac{10}{100}\times 1000}{190}\\ = A_f=5.81 m^2$

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$C_x=\dfrac {B-0.3}2 ;C_y =\dfrac {L-0.45}2$

$Now, C_x=C_y \\ \dfrac {B-0.3}2=\dfrac {L-0.45}2\\ L=B+0.15\\ A_f = L\times B\\ 5.5 =(B+0.15)\times B\\ B=2.33 m\space \&\space L=2.48 m\\ Now, \\ \text { Upward soil pressure (w)} =\dfrac {Pu}{A_f\space provider} \\ W=\dfrac {1500}{2.33\times 2.48}=259.48 KN/m^2\\ C_x=\dfrac {2.33-0.3}2=1.015 \\ C_y=1.015\\ M_{ux}= L\times w\times \dfrac {C_x}2 \\ M_{ux}=259.48 \times 2.48\times \dfrac {1.015}2=393 KNm\\ M_{uy}= w\times B\times \dfrac {C_y^2}2 =259.48\times 2.33\times \dfrac {1.015}2=369 KNm \\ M_{u\space max}=0.318 f_ckbxd^2 \\ 504.07\times 10^6=0138\times 20\times 2480\times D_1^2\\ \therefore d_1=239.6mm ..... (1)$

Two way shear:

$Z_{cpermi}=Z_{uc}^1\times Ks$

$Z_{uc}^1=1.12 N/mm^2\\ Ks=0.5+\dfrac {0.3}{0.45}=1.16 \gt 1\\ Z_{uc}=Z_{uc}^1 \times Ks=1\times 1.12=1.200 N/mm^2=1200 KN/m^2$

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

$1200 =\dfrac {1500-\{259.48\times [(0.45+d)\times (0.3+d)]\}}{2\times [0.45+d+0.3+d]\times d}.... (2)\\d_2=0.392=392.3 mm \\ \text {Provide d}=400mm\\ Ast_x=\dfrac {0.5\times 20\times 2480\times 400}{415}\times [1-\sqrt{1-\dfrac{4.6\times 393\times 10^6}{20\times 2480\times 400^2}}]\\ Ast_x =2898 mm^2\\ \text {Provide 15 no. of 16 mm }\theta \\ Ast_p=3016 mm^2 \\Ast_y=\dfrac {0.5\times 25\times 2330\times 400}{415}\times [1-\sqrt{1-\dfrac{4.6\times 369\times 10^6}{25\times 2330\times 400^2}}]z\\ Ast_y =2721.18 mm^2\\ \text {Provide 14 no. of 16 mm }\theta \\ Ast_p=2814.86 mm^2\\ Ast_{min} =\dfrac {0.12}{100}bD =\dfrac {0.12}{100}\times 2480\times 475\\ Ast_{min}=1413.6mm^2 \lt Ast_x \space \& \space Ast_y \therefore$

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