Civil Engineering - RCC Structures Design - Discussion

Discussion Forum : RCC Structures Design - Section 2 (Q.No. 10)
10.
With usual notations the depth of the neutral axis of a balanced section, is given by
t/mc = n/(d-n)
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
10 comments Page 1 of 1.

Debraj Swain said:   4 years ago
n=critical neutral axis
m=modular ratio
c=permissible stress in concrete
t=permissible stress in steel.

We know that n=[mc/(mc+t)] *d.
= n/d=mc/(mc+t).

Now reciprocating both sides,
= d/n=(mc+t) /mc,
= d/n=1+t/mc,
= d/n-1=t/mc,
= (d-n)/n=t/mc.

Now reciprocating again to get the required expression i.e.
= n/(d-n) = mc/t.
(4)

Vikas Guleria said:   5 years ago
It will be mc/t =d/d-n.
(1)

Manish dev said:   6 years ago
On solving for the value of n.

ie. neutral axis, we will get n = (m * c *d)/(m*c+t).

Talib said:   7 years ago
Please explain option D.

Shailendra said:   7 years ago
Here,
t= permissible tensile stress in steel,
c= permissible compressive stress in concrete,
m=modular ratio.

Sanju said:   8 years ago
Please explain anyone clearly.

ASD said:   8 years ago
(t/mc) = [(d÷n)-1].

SILVER HEAVEN said:   8 years ago
WHAT IS M, C, T?

Rahul Rathod said:   9 years ago
Both are right @Bipin.

Bipin said:   1 decade ago
What is difference between B & D?

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