Civil Engineering - RCC Structures Design - Discussion

@All.

Super elevation, also known as cross slope or cant, refers to the tilting of a road or railway track to one side, used to counteract the effects of centrifugal force and ensure the stability of vehicles while negotiating curves. The calculation of superelevation involves several factors, including the design speed of the road or track, the radius of the curve, and the height of the vehicle.

The formula used to calculate super elevation is:

e = V^2 /(g * r).

where:
e = super elevation (in %)
V = design speed of the road or track (in km/h)
g = acceleration due to gravity (9.8 m/s^2)
r = radius of the curve (in meters)

The safe load on the column can be determined by calculating the maximum compressive stress that the column can resist, taking into account both the concrete and the steel reinforcement.

The concrete contributes to the overall strength of the column through its compressive strength, while the steel reinforcement increases the load-carrying capacity of the column. The maximum compressive stress in the concrete is limited by its permissible compressive stress of 40 kg/cm^2. The steel reinforcement also has a permissible compressive stress of 300 kg/cm^2.

The total compressive stress in the column is given by:

P = fc * Ac + fs * As.

where:
P = total compressive stress (kg/cm^2).
fc = compressive stress in the concrete (kg/cm^2).
fs = compressive stress in the steel (kg/cm^2).
A_c = cross-sectional area of the concrete (cm^2).
A_s = total cross-sectional area of the steel reinforcement (cm^2).

For a column with a cross-sectional area of 20 cm x 20 cm = 400 cm^2, and with 4 bars of cross-sectional area 20 cm^2 each, the total cross-sectional area of the steel reinforcement is 4 * 20 = 80 cm^2.

Substituting the values into the equation, we get:
P = 40 * 400 + 300 * 80.
P = 16000 + 24000,
P = 40000 kg/cm^2.

The safe load on the column, therefore, should not exceed 40,000 kg/cm^2.

Safe Load on Column = (Permissible Stress in Con. x Area of Con.) + (Permissible Stress in Steel x Area of Steel) -----> (i).


Given:
Area of Steel = 20 cm^2,
Per. Stress of Concrete = 40 kg/cm^2,
Per. Stress of Steel = 1300 kg/cm^2,
C/A of whole column 20 cm x 20 cm,

Then we find out the Area of concrete:

Area of Concrete = (Whole Area of the column - Area of steel).
= (20 x 20) - (20).
= 380 cm^2.

From Eq. (i).

Safe Load on Column = (Per. Stress in Con. x Area of Con.) + (Per. Stress in Steel x Area of Steel).
= (40 kg/cm^2 x 380 cm^2 ) + (1300 kg/cm^2 x 20 cm^2),
= (15200 kg) + (26000 kg),
= 41200 kg.

Safe Load on Column = (Permissible Stress in Con. x Area of Con.) + (Permissible Stress in Steel x Area of Steel) ------------ (i).


Given:
Area of Steel = 20 cm^2,
Per. Stress of Concrete = 40 kg/cm^2,
Per. Stress of Steel = 1300 kg/cm^2,
C/A of whole column 20 cm x 20 cm,

Then we find out Area of concrete:

Area of Concrete = (Whole Area of column - Area of steel).
= (20 x 20) - (20).
= 380 cm^2.
From Eq. (i).

Safe Load on Column = (Per. Stress in Con. x Area of Con.) + (Per. Stress in Steel x Area of Steel)
= (40 kg/cm^2 x 380 cm^2 ) + (1300 kg/cm^2 x 20 cm^2)
= (15200 kg) + (26000 kg)
= 41200 kg.

Pu = (0.45fck x Ac) + (0.67fy Asc) -----> (1 eqn.)

0.45 fck = 40Kg/cm^2.
0.67fy = 300Kg/cm^2.

Put values in equation 1.

= 40 x (400-20) + (300 x 20),
= 15200 + 6000,
Pu = 21200 Kg.

Compressive stress in concrete = 40kg/cm^2.
Compressive stress in steel = 300kg/cm^2,
Area of concrete = 400cm^2,
Area of steel=4*20 = 80cm^2 , Pu=40*400+80*300=40,000kg so answer (B) is correct.

Safe load of column = Fc * Ac +fs * Ast.

NOTE: Annexure B-3.1(IS 456-2000) Permissible axial Load P= σcc Ac + σscAsc.

Where, σcc = Permissible stress in concrete.
And σsc = Permissible compressive stress for column bars.

Therefore,
P = σcc Ac + σscAsc.
= 40 * 20*20 + 300 *(4* 20) = 40000 kg.

P=0.4FckXAc+0.67XFyXAst [ Hear-Fck=40 kg/sm2,Fy=300Kg/sm2, Ac= 20x20 sqcm,Ast= 4x20=80 sqcm]

P= 0.4x40x(20x20-4x20)+0.67x300x4x20=21200 Kg This factored load.

So the safe load which column can carry-= 21200/1.5=14133.33 kg. == 14133 kg.

You are correct, Thanks @Suraj.

P=sigma cc *(total area-area of steel)+sigma sc*area of steel.

Total area=20*20=400sq cm ..hence p=36800.

Hence nearly option B.