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
Discussion:
13 comments Page 1 of 2.
Pooja said:
8 years ago
Stress = P/A.
=40000/(3.14 x 900)/4,
= 56.
=40000/(3.14 x 900)/4,
= 56.
(2)
Abcd said:
6 years ago
How can we find a modular ratio?
Please explain the steps.
Please explain the steps.
(2)
Jibin said:
6 years ago
Stress = P/ (Ac+mAs) = 4000/ ( (699.72+ (13.33*6.78) ) = 50.
(2)
Hima Bindu said:
9 years ago
Please Explain neatly.
(1)
Rakesh kumar meena said:
8 years ago
We are not able to calculate stress without modular ratio.
(1)
Abidha said:
6 years ago
Actually they should mention stress in what material steel or in concrete?
(1)
RANJAN PATRA said:
9 years ago
STRESS = 40000/(3.14 * 900) * 4 = 56.
So, the answer is option A.
So, the answer is option A.
Sajja said:
9 years ago
It is calculated by Load/area.
Shiva said:
8 years ago
Use M20 concrete.
Cal equivalent area of concrete.
= (Ac) + (m-1) * Asc.
Load ÷ equivalent area of concrete.
Cal equivalent area of concrete.
= (Ac) + (m-1) * Asc.
Load ÷ equivalent area of concrete.
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.
Ac+(m-1)xAsc, 706.5 + (13.33-1) x 6.78 =790.
Then, 40,000/790 = 50kg/cm2.
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