Civil Engineering - RCC Structures Design - Discussion
Discussion Forum : RCC Structures Design - Section 5 (Q.No. 11)
11.
The stresses developed in concrete and steel in reinforced concrete beam 25 cm width and 70 cm effective depth, are 62.5 kg/cm2 and 250 kg/cm2 respectively. If m = 15, the depth of its neutral axis is
Discussion:
18 comments Page 1 of 2.
Krishan Kumar said:
5 years ago
Yes, 55.26 is the right answer.
(4)
Fero said:
5 years ago
(1/(1+(280/3*steel stress)))*d.
N.A depth only depends on steel.
Or
k*d.
For fe250 k=0.4.
So N.A depth =0.4*70= 28cm= 30cm.
N.A depth only depends on steel.
Or
k*d.
For fe250 k=0.4.
So N.A depth =0.4*70= 28cm= 30cm.
(2)
Sikander said:
7 years ago
Mc ÷(Mc+s)
15*62.5÷ (15*62.5+1400).
=28.07 approx 30.
Here, 2500kg/cm2 steel stress so as IS 1400kg/cm2 is permissible stress.
15*62.5÷ (15*62.5+1400).
=28.07 approx 30.
Here, 2500kg/cm2 steel stress so as IS 1400kg/cm2 is permissible stress.
(2)
Ajay said:
7 years ago
55.26 cm is correct I think.
(2)
Mohsan Chhajro said:
3 years ago
By using formula mc/ (mc+t) = n/ (d-n).
And putting respective values answer comes out to be equal to 30.9 cm.
And putting respective values answer comes out to be equal to 30.9 cm.
(1)
Anji said:
7 years ago
What type of stresses developed in concrete and steel?
(1)
Gaja said:
8 years ago
mc/t=n/(d-n).
n= 55.26.
n= 55.26.
(1)
Sridhar said:
1 year ago
The Stress developed in steel is wrong at 250 kg/cm^2 instead of 1250 kg/cm^2.
put stress of steel = 1250 kg/cm^2.
the answer became 29 cm which is equal to 30 cm.
Stress in steel = 1250/15 = 83.33 kg/cm^2.
Now
62.5/x = 83.33/70-x.
x = 29 cm
= 30 cm.
put stress of steel = 1250 kg/cm^2.
the answer became 29 cm which is equal to 30 cm.
Stress in steel = 1250/15 = 83.33 kg/cm^2.
Now
62.5/x = 83.33/70-x.
x = 29 cm
= 30 cm.
Gani said:
10 years ago
How it comes?
Shubham sharma said:
7 years ago
In question stress in steel is given 250 this is wrong.
Take it as 1250. Then it comes 30.
Take it as 1250. Then it comes 30.
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