Example :- A Short RCC column of size 400*400 mm as to cary axial factor load and 1800KN design the column factor .

Example :- A Short RCC column of size 400*400 mm as to cary axial factor load and 1800KN design the column factor .

Solution :- 

Ag  = 400*400

Ag =160000=16*10^4

Pu= .4*Fck*Ag + (.67*Fy + .4Fck)Asc

1800*10^3= .4*20*16*10^4 + (.67*500+.4*20)Asc

Asc=1590.21mm^2

By Assuming 16 mm Diameter bar 

The area  of 1 bar single bar is = (3.14*D^2)/4

                                                 = (3.14*16^2)/4

                                                 = 201mm^2

No. of bar requires   =    (Asc)/(Area of 16mm dia. bar)

                                  =(1590.21)/(201)

                                   = 7.91  = 8 no's

Required 

Pc    min=.8%  , max=4%

N    min= 4 no's

TR  min dia. = 12mm

here ,  Pc = (as*100)/bD

           Pc = (1590.21*100)/400*400

           Pc = .99%

Calculate transverse reinforcement 

1.  pitch 

   1. 16d = 16*16 = 256mm

   2. minimum lateral dimension =400mm 

   3. 300mm  

so here 256mm  then 250mm put 

2. Diameter of the transverse reinforcement 

    1. .25d = .25*16 =4mm

    2. 6mm 

        8mm put 

So the final reinforcement use of 8 no’s of 16mm diameter main reinforcement and 6mm diameter bars of 250 mm center to center distance is used in the 400mmX400mm size with 1800kN point load.