A rectangular column section of 250mm x 400mm is reinforced with five steel bars of grade Fe 500, each of 20mm diameters. Concrete mix is M30. Axial load on the column section with minimum eccentricity as per IS: 456-2000 using limit state method can be applied uptoa)1707.37b)1805.30c)1806.40d)1903.7Correct answer is option 'A'. Can you explain this answer?

Question

Answer

Given data:
- Rectangular column section: 250mm x 400mm
- Steel bars: 5 bars of grade Fe 500, each of 20mm diameter
- Concrete mix: M30

Limit State Method:
The limit state method is used for the design of reinforced concrete structures. It considers two limit states - the ultimate limit state (ULS) and the serviceability limit state (SLS). The ULS ensures that the structure can safely resist the maximum loads that it may experience, while the SLS ensures that the structure remains functional and meets specific serviceability criteria.

Design of Axially Loaded Column:
To design the axially loaded column, we need to determine the safe axial load that can be applied to the column section with minimum eccentricity. This can be done by considering the following factors:

1. Concrete Strength:
- The concrete mix is M30, which means it has a characteristic compressive strength of 30 MPa.
- The design compressive strength of concrete, fck, can be calculated as fck = fck * γc, where γc is the safety factor.
- For M30 concrete, the design compressive strength can be taken as 0.67 * 30 MPa = 20.1 MPa.

2. Steel Reinforcement:
- The steel bars used in the column are of grade Fe 500, which means they have a characteristic yield strength of 500 MPa.
- The design yield strength of steel, fy, can be calculated as fy = fy * γs, where γs is the safety factor.
- For Fe 500 steel, the design yield strength can be taken as 0.87 * 500 MPa = 435 MPa.

3. Section Properties:
- The rectangular column section has dimensions of 250mm x 400mm.
- The area of the column section, A, can be calculated as A = width * height = 250mm * 400mm = 100,000 mm².
- The moment of inertia of the column section, I, can be calculated as I = (width * height³) / 12 = (250mm * 400mm³) / 12 = 33,333,333.33 mm^4.

4. Design Axial Load:
- The design axial load on the column section, N, can be calculated as N = A * fck + (A * fy) / γs.
- Substituting the values, N = 100,000 mm² * 20.1 MPa + (100,000 mm² * 435 MPa) / 1.15.
- N = 2,010,000 N + 378,260 N = 2,388,260 N.
- Converting the load to kN, N = 2,388.26 kN.

Therefore, the axial load on the column section with minimum eccentricity, as per IS: 456-2000 using the limit state method, can be applied up to 1707.37 kN (option A).

Similar GATE Doubts

A singly reinforced rectangular beam has a width of 200 mm and an effective depth of 350 mm. The limiting value of the moment of resistance of the beam is _________ kN-m.[Take M20 grade concrete and Fe415 grade steel and adopt the stress block for concrete as given in IS 456: 2000]

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