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.
Shiv said:
7 years ago
Pt=50k^2 ÷ m(1-k).
k=mc ÷ mc+t.
k=mc ÷ mc+t.
Khan said:
7 years ago
Pt = K(t/c).
K = mc/(mc+t).
K = mc/(mc+t).
Malla said:
7 years ago
%At=m/(2r*(m+r))*100 where r=t/c.
r=1400/50 = 28
%At = 18/(2 * 28 * (18+28)) * 100 = 0.698%.
r=1400/50 = 28
%At = 18/(2 * 28 * (18+28)) * 100 = 0.698%.
Navya said:
6 years ago
Please explain the answer in detail.
Aswathy said:
6 years ago
BXX/2 = mAst (d-X) ..... (1)
X= critical depth of neutral axis
X = kd
X= (mc/mc+t)d...... (2)
We have,
m = 18
C = 50kg/cm2
t = 1400kg/cm2
From this by substituting in formula (2) we get X = 0.39d
And then by substituting X in (1) we get
Ast/bd =pt= 0.696%
X= critical depth of neutral axis
X = kd
X= (mc/mc+t)d...... (2)
We have,
m = 18
C = 50kg/cm2
t = 1400kg/cm2
From this by substituting in formula (2) we get X = 0.39d
And then by substituting X in (1) we get
Ast/bd =pt= 0.696%
Solver said:
10 years ago
Use the formula: 50mc^2/t(mc+t).
Aniket gawnde said:
5 years ago
P=50 * n1^2/m(1-n1) %.
n1 = m/(m+r).
r = tensile stress/compressive stress.
n1 = m/(m+r).
r = tensile stress/compressive stress.
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.
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