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Sunday, 8 April 2018

Design a isolated footing for a column of 400mm*400mm subjected to an axial load of 1800KN. The safe bearing capacity of foundation soil is 150KN/MM.

ANS:-
Given Data
1.Size of column:- 400mm*400mm
2. Given load (p) :- 1800KN
3. SBC :- 150KN/mm
4. Assume m20 grade of concrete. And Fe 415 grade of
Steel.     
 STEP 1:-
Area of footing
Increase 10% of given load
Area of footing :- 10% of p+p /SBC
                             10*1800+1800 /100
                             1980 / 150
                              13.2mm
As we design square footing then,
                           A2 = 13.2
                           a = 3.63 =4m
Hence, provide 4*4mfooting size


STEP :-2
Calculate Soil Reaction
qu (soil reaction) = pu /a*a
 standard formula :-  pu = 1.5* p
                                          1.5* 1800
                                  pu = 2700KN
Therefor:- qu = 2700 / 4*4
                          168.75 KN/mm

                          0.168N/mm
STEP:-3
Calculate for depth of footing
The critical reaction is at the distanced ‘d’ from the face of the column as per IS code

Vu =qu B ( B-b/2-d)
Vu = 0.168*4000 (4000-400/2-d)
Vu = 672 (1800-d)
Assume 0.2% of steel for m20 grade of concrete.
Tc = 0.32 N/mm2…..(from IS code ..Design Strength Of
                                    Concrete)
 minimum depth
Tc Bd = Vu
0.32* 4000 * d = 672 (1800-d)
[d = 619.67 = 620]
STEP :-4
Check for Bending moment
Mulimit = 0.138 fck Bd2 …….                      (for 415)
                 0.138 * 20 *4000 *(620)2
                         4243.7*10^6 N.mm
Mu =qu B (B-b)2 /8
         0.168 *4000 ( 4000 – 400)2 /8
         1088.64 *106 N.mm
  Mulimit grater than Mu …(Hence provided depth is
                                               sufficient)
STEP :- 5
Critical Section
Parameter of critical section :-
                 4(400+620)
                   4080mm
Area of critical section :-
               Parameter * Depth
               4080 * 620 =2.52 N.mm
Upward Pressure
Area of Section
(0.168 * (40002 – 10202/ 2052
= 0.99 N/mm^2
According to IS Code…
Maximum shear stress = 0.25*fck
                                        0.25 * 20
                                        1.118N/mm^2
STEP :-6
Calculate Area Of Steel
Mu =0.87 fy Ast d [1-Ast/Bd * fy/Fck]
1080.64*106= 0.87*415*620 Ast [1- {Ast/4000*620}*{415/20}]
Ast = 1.114627 mm2
          2. 5078.70 mm2
Ast =5078.70 mm2
STEP :-7
 Calculate Spacing
Assume 16mm dia.
 Spacing = {Area of bar / Area of steel}*B
               = 3.14/4 *(16)2 / (5078.78) *{4000}
               = 158.27 ~160mm
Submitted by- Sneha Dhangar


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