Question

design a square reinforced column of size 400mm to carry an ultimate load of 1000 KN at an eccentricity of 160mm.Use M20 grade of concrete and Fe415 grade of steel​

Answer 1

Answer:

please help

Explanation:

please help

Answer 2

Answer:

The answer is Pu= 794 kN

Explanation:

Pu = 0.40 fck Ac+ 0.67 x Fyx Asc for column with lateral or circular ties.

Where, Pu= ultimate load

fck = Characteristic strength of concrete

Ac, Asc = Area of concrete & steel respectively

Fy = yield strength of steel

When helical reinforcement is used, ultimate load capacity is increased by

5%

Pu= 1.05 (0.40 fck Ac +0.67 Fy x Asc)

This formula is valid only when emin ≤ 0.05 x D

Lun supported 500 20 mm emin = minimum eccentricity: = maximum + D 30

Calculation:

Given, D = 300 mm, Asc = 0.01 Ac

fck = 20, Fy = 415 Ac = × (300)² = 70685.83 mm²

Pu= 1.05 (0.40 x 20 x Ac +0.67 x 415 x 0.01 Ac)

Pu= 794 kN

But it does not satisfy the minimum eccentricity condition.

Other important points:

When minimum eccentricity condition is not satisfied,

Puz = 0.45 fck Ac + 0.75 x Fy x Asc

For helical reinforcement, Puz = 1.05 [0.45 fck Ac + 0.75 × Fy * Asc]

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