Question

3. Calculate the load carrying capacity and percentage of reinforcement for a short rectangular column of
cross section dimension 280 mm x 500 mm is reinforced with 4 bars of 25 mm diameter, 2 bars of 20
mm diameter and 2 bars of 12 mm diameter. Use M30 grade concrete and Fe 500 grade steel. Also
design a 4 legged ties necessary for this section.

Answer 1

Given data:-

Column Diamension (BXD) = 280 X 500 mm

Concrete grade (fck) = M30 = 30 N/mm2 ,

Steel grade (fy) = Fe500 = 500 N/mm2 ,

Steel reinforcement = 4 bar of 25 mm dia , 2 bar of 20 mm dia , 2 bar of 12 mm dia. is given

Determine:- Load carrying capacity (Pu) and percentage of reinforcement

From IS Code 456:2000

Limit state method,

In the question given that column is short.

We know that ultimate load carrying capacity of short column is given by as,

------------------Eq.A

Where,

fck = Strength of concrete = M30 = 30 N/mm2 ,

Ac = Area of concrete = Total column area - Area of steel = A - Asc ,

Asc = Area of steel

fy = yield Strength of steel = Fe500 = 500N/mm2 ,

Calculating area of steel :-

One bar area is given by as

Total Area of Steel

Area of Column (A) = (Width) X (Depth) = 280 X 500 = 140000 mm2 ,

Area of Concrete (Ac) = Area of column - Area of steel = A - Asc = 140000 - 2816.58 = 137183.42 mm2 ,

From Eq.A, put all known data

Hence ,

Ultimate load carrying capacity of column is 2589.75 kN.

Service load on column = (Ultimate load / Load factor )

= (2589.75 / 1.5 ) (By IS Code load factor in lomit state is 1.5)

= 1726.5 kN

So that,

Ultimate load (Pu) = 2589.75 kN

Service Load (P) = 1726.5 kN   Ans.

Percentage of Reinforcement :-

  Percentage of reinforcement (Pt)in column is given by as,

Where,

A = Column area = BXD = 140000 mm2

Asc  = Total area of longitudnal reinforcement steel = 2816.58 mm2 ,

Pt = 2.01 %

So that Percentage of reinforcement is 2.01%.

According to IS Code 456:2000, the Maximum percentage of steel in the column is 6%.

Then,

2.01% < 6% (O.K) (Maximum % of steel)

According to IS Code 456:2000 , Minimum percentage of steel in column is 0.8%.

Then,

2.01% > 0.8% (O.K) (Minimum % of steel )

Design of legged ties:-

According to IS Code 456:2000 Limit state method ,

The diameter of the legged ties bar is given by as,

Diameter of Tie bar is maximum of,

So that Dia. of the tie bar is Maximum of

In Present time 6 mm dia. of steel bar not available in the market. so that takes 8 mm dia of the tie bar.

Tie bar dia () = 8 mm

The spacing of Tie bar:-

According to IS code 456:2000, Column specification

The spacing of Tie bar is Minimum of

(1) Least lateral dimension =  280 mm

(2) 16 X Max. dia of Longitudinal bar = 16X25 = 400 mm

(3) 300 mm

that Spacing of the tie bar is 280 mm

Provide 4 legged tie bar 8 mm dia @280 mm c/c.