Civil Engineering - RCC Structures Design - Discussion

Discussion Forum : RCC Structures Design - Section 1 (Q.No. 27)
27.
A short column 20 cm x 20 cm in section is reinforced with 4 bars whose area of cross section is 20 sq. cm. If permissible compressive stresses in concrete and steel are 40 kg/cm2 and 300 kg/cm2, the Safe load on the column, should not exceed
4120 kg
41, 200 kg
412, 000 kg
none of these.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
39 comments Page 1 of 4.

A Shrivastava said:   3 years ago
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.
(9)

Numan said:   3 years ago
@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.
(5)

B.Singh said:   3 years ago
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.
(3)

Rakesh Singh Bisht said:   5 years ago
Safe load of column = Fc * Ac +fs * Ast.
(2)

Kitkupar said:   5 years ago
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.
(1)

Shivam Sundriyal said:   8 years ago
0.4 Fck Ac + 0.67 Fy Asc.

This formula is used when given parameters are:-
Fck i.e Characterstic strength of concrete
Fy i.e Yield strength of Bar

But
Given papameters are:-
Permissible compressive stress in concrete i.e sigma cc
Permissible compressive stress in steel i.e sigma sc

So formula used will be:-
Sigma cc * Ac + sigma sc * Asc.
Where Ac = cross sectional area of concrete.
Asc = cross sectional area of steel in compression.
I hope it's clear.
(1)

Hriday narayan said:   7 years ago
Here, Pc = .4Fck * Ac+.67Fy * mAst.

Naveen said:   7 years ago
@Ramesh.

0.67x300=201 not 1300.

Shubham said:   7 years ago
@Sankha.

Why 1300 taken in place of 300?

Sameer sopori said:   7 years ago
First we have to find out the permissible compressive stresses on steel and concrete area.
for concrete= 380 * 40 = 15200.

For steel = 20 * 300 = 6000.
So , the permissible stress on section = steel + concrete = 15200+6000 = 21200 kg/cm2
With respect to buckling only, the Allowable Load on the column, P for a Factor of Safety is F.S. = 1.95.

So the safe load on column is = 21200 * 1.95 = 41340 kg.


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