Axially Loaded Column Design

 

Axially Loaded Column Design

Axially loaded meaning is that the line of the resultant force of the loads supported by the column is coincident with the center of gravity line of column .

Axially loaded column is also known as short column. A short column may be considered to be short when its effective height does not exceeds 15 times its least breadth.

Example

Suppose a column is subjected to carry a load of 150 Ton having concrete of (1:2:4) and we have to design it as short column and size of column is also required. The effective height of column is 15 ft.

First of all we will put formulas of short column designs

1) Longitudinal steel = The cross sectional area of steel must not be less than 0.8% of gross cross sectional area of column and also not more than 8 % of gross  cross sectional area of column.

2) Concrete cover = The cover to outside of longitudinal bars must be a minimum of 1.5 inches or the diameter of the bar whichever is greater. In case of column minimum dimension of 7.5" or under, whose bars do not exceed 1/2" in diameter ,1" cover may be used.

3) Splices = The length of lap of bars in compression should not be less than =

The bar diameter x The compressive stress in bar/ Five times the permissible average bond stress

OR 24 bar diameter whichever is greater

For example if the compressive stress in the bars is 18000 Lbs per sq.in and the permissible average bond stress is 120 Lb. per sq.in. (nominal 1:2:4 mix ) the length of lap will be calculated as follows

Bar Diameter x 18000/(5x120)= Bar diameter x 30

4) Lateral Ties= The diameter of transverse reinforcement should not be less than one quarter the diameter of the main rods and in no case less than 3/16 in.

5) Pitch.  The distance apart of the ties must not be more than the least of following

a. least lateral dimension of column

b.Twelve times diameter of the smallest longitudinal bars.

c. Twelve inches

6)  P =  fcc.Ac+fsc.Asc

      P= fcc (Ag-Asc)+ fsc.Asc

fcc = permissible stress of concrete in direct compression (760 Lbs./sq.in for 1:2:4 mix)

fsc = permissible compressive stress for column bars (18000 Lbs./sq.in for Grade 60 steel)

Ac = cross sectional area of concrete excluding any finishing material and reinforcement steel i.e

Ac= Ag- Asc where

Ag= Gross cross sectional area of column

Asc= Cross sectional area of longitudinal bars

Now Putting values

Considering given load we will put values in formula no 6 , and will assume 2% of steel (As we can use range of steel from 0.8 % to 8% of gross cross sectional area of column.

      P= fcc (Ag-Asc)+ fsc.Asc

(150X2240) = 760( Ag - 2/100.Ag)+18000x2/100Ag)

336000 = 760Ag-15.2Ag + 360.Ag

336000= 1104.8 Ag

Ag= 304.127 Sq.inch

Considering Rectangular column having 12 inches as Least dimension of column does not exceeds 15 times of effective height of column = 1x15 =15 so it will be short column

Longer side of column = 304.127/12 = 25.34 say 26 inches so size of column will be 24"x12"

Longitudinal bars of columns = 24x12x 2/100= 5.76 Sq.In

Using 3/4" dia bars

No of bars = Asc /Area of bars = 5.76/0.44= 13 Nos

The length of lap of bars in compression should not be less than =

The bar diameter x The compressive stress in bar/ Five times the permissible average bond stress

OR 24 bar diameter whichever is greater

Lets try above rule to find out lap length

The bar diameterx18000/5X120 =30 Times of Diameter of bars

or 24 bar diameter

so we will use lap length = 30 times of diameter of bars

=30x 3/4" = 22.5 Inches

Pitch.  The distance apart of the ties must not be more than the least of following

a. least lateral dimension of column

b.Twelve times diameter of the smallest longitudinal bars.

c. Twelve inches

Pitch will be used as Twelve times of diameter of longitudinal bars = 3/4"x12=9" c/c