Civil Engineering - RCC Structures Design - Discussion

Discussion Forum : RCC Structures Design - Section 2 (Q.No. 21)
21.
An R.C.C. column of 30 cm diameter is reinforced with 6 bars 12 mm φ placed symmetrically along the circumference. If it carries a load of 40, 000 kg axially, the stress is
49.9 kg/cm2
100 kg/cm2
250 kg/cm2
175 kg/cm2
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
13 comments Page 1 of 2.

Suresh said:   4 years ago
@Pooja and@Ranjan.

How can we calculate when force is in SI units and diameter is in cm so first, we must balance then units then we will be able to calculate so it's wrong, diameter is not 900 it is 0.90m.

Jibin said:   6 years ago
Stress = P/ (Ac+mAs) = 4000/ ( (699.72+ (13.33*6.78) ) = 50.
(2)

Abidha said:   6 years ago
Actually they should mention stress in what material steel or in concrete?
(1)

Abcd said:   6 years ago
How can we find a modular ratio?

Please explain the steps.
(2)

Titam said:   6 years ago
Assume M20 and 415 steel for modular ratio.

The idea of assumption for m20 answer 415 is that since the question states the RCC column of 30cm, so any mix less than M20 is not a recommendation. Same goes for the use of HYSD bars i.e. 415.

Arnab said:   7 years ago
@Ankit Bahuguna.

How do you find out the value of modular ratio?

Ankit Bahuguna said:   7 years ago
First, calculate the equivalent area of concrete.

Ac+(m-1)xAsc, 706.5 + (13.33-1) x 6.78 =790.
Then, 40,000/790 = 50kg/cm2.

Rakesh kumar meena said:   8 years ago
We are not able to calculate stress without modular ratio.
(1)

Pooja said:   8 years ago
Stress = P/A.
=40000/(3.14 x 900)/4,
= 56.
(2)

Shiva said:   8 years ago
Use M20 concrete.

Cal equivalent area of concrete.
= (Ac) + (m-1) * Asc.
Load ÷ equivalent area of concrete.


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