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 1 of 2.
Er. Taba Tallum said:
3 years ago
Alternate solution:
c=50kg/cm2 , t=1400kg/cm2 and m=18
Critical N.A depth, n=(m*c)xd/(t+m*c)=0.39*d.
Equating total compression force to the total tension force.
=> bxnxc/2 = Ast*t.
=>b*0.39d*50/2 = Ast*1400
=>Ast/(bd) = (0.39*25)÷1400 = 0.00696.
Percentage steel, Pt = Ast*100/(bd) = 0.00696*100 = 0.696%.
c=50kg/cm2 , t=1400kg/cm2 and m=18
Critical N.A depth, n=(m*c)xd/(t+m*c)=0.39*d.
Equating total compression force to the total tension force.
=> bxnxc/2 = Ast*t.
=>b*0.39d*50/2 = Ast*1400
=>Ast/(bd) = (0.39*25)÷1400 = 0.00696.
Percentage steel, Pt = Ast*100/(bd) = 0.00696*100 = 0.696%.
(2)
Adilmansuri said:
4 years ago
(Ast/Bd) = ((k*ca)/(ta*2)).
(1)
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.
Rashik Koundal said:
5 years ago
If the area of steel given than % of steel
p=100.ast/b.d
ast =area of steel
b.d= eff. Breadth and depth of the beam
But in this case, ast(area of steel) not given.
Then,
p= 50 * k^2 / m(1-k).
{k = N.A factor}
k= mc/mc+t.
m= modular ratio.
{ symbol/ mens divde by}.
{ k^2 means to raise to power 2}.
So now put values in formula find( k) 1st.
k= 18 * 50 / 18 * 50 + 1400 = 900/900+1400.
=900/2300,
k=0.39.
Now p= 50 * 0.39^2/18 (1-0.39) = 0.696 answer.
p=100.ast/b.d
ast =area of steel
b.d= eff. Breadth and depth of the beam
But in this case, ast(area of steel) not given.
Then,
p= 50 * k^2 / m(1-k).
{k = N.A factor}
k= mc/mc+t.
m= modular ratio.
{ symbol/ mens divde by}.
{ k^2 means to raise to power 2}.
So now put values in formula find( k) 1st.
k= 18 * 50 / 18 * 50 + 1400 = 900/900+1400.
=900/2300,
k=0.39.
Now p= 50 * 0.39^2/18 (1-0.39) = 0.696 answer.
(2)
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%
Navya said:
6 years ago
Please explain the answer in detail.
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%.
Khan said:
7 years ago
Pt = K(t/c).
K = mc/(mc+t).
K = mc/(mc+t).
Shiv said:
7 years ago
Pt=50k^2 ÷ m(1-k).
k=mc ÷ mc+t.
k=mc ÷ mc+t.
Harry bhai said:
8 years ago
And is 0.704166666%
Using Xc= 0.40d
Then,
PS= 50*x^2÷(md(d-x)).
Using Xc= 0.40d
Then,
PS= 50*x^2÷(md(d-x)).
(1)
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