Civil Engineering - RCC Structures Design - Discussion
Discussion Forum : RCC Structures Design - Section 2 (Q.No. 6)
6.
If the permissible compressive and tensile stresses in a singly reinforced beam are 50 kg/cm2 and 1400 kg/cm2 respectively and the modular ratio is 18, the percentage area At of the steel required for an economic section, is
Answer: Option
Explanation:
Taking Moment of area of tension and compression zone for a singly reinforced beam:
Discussion:
19 comments Page 2 of 2.
Vasantha devi said:
8 years ago
@Shilendra Singh.
Where is this formula given? Explain the formula.
Where is this formula given? Explain the formula.
(1)
Shilendra singh said:
8 years ago
P=50k^2/m (1-k).
Because
K = mc/(mc+t).
= 18 *50/(18 *50+1400).
= 0.391.
The value of k put in above eq.
= (50 *0.391^2)/18 (1-0.391),
= 0.697.
So option C is correct.
Because
K = mc/(mc+t).
= 18 *50/(18 *50+1400).
= 0.391.
The value of k put in above eq.
= (50 *0.391^2)/18 (1-0.391),
= 0.697.
So option C is correct.
Anjali said:
9 years ago
Yes, it is 0.698%.
Arnab said:
9 years ago
Here, p = (k * c * 100)/(2 * t).
Then k = (18*50)/((18*50) + 1400) = 9/23.
Putting value of k in p we get p.
So, p = 0.698%.
Then k = (18*50)/((18*50) + 1400) = 9/23.
Putting value of k in p we get p.
So, p = 0.698%.
Akash said:
10 years ago
M = es/ec = 18x50 = 1400, 900/1400 = 0.6.
Solver said:
10 years ago
Use the formula: 50mc^2/t(mc+t).
Rohit said:
10 years ago
First find value of n in form of d i.e. effective depth.
Then from relation moment of area of compression zone = moment of area of equivalent tensile zone we can find value of AST.
Then from relation moment of area of compression zone = moment of area of equivalent tensile zone we can find value of AST.
Dunya khleel said:
1 decade ago
Please explain to me.
Harshit said:
1 decade ago
0.5(B*X*F) = Area of steel*permissible stress in steel.
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