# Civil Engineering - RCC Structures Design - Discussion

Discussion Forum : RCC Structures Design - Section 3 (Q.No. 2)
2.
If the maximum bending moment of a simply supported slab is M Kg.cm, the effective depth of the slab is (where Q is M.R. factor)
Explanation:
No answer description is available. Let's discuss.
Discussion:
10 comments Page 1 of 1.

Ujjawal said:   2 years ago
If we want M in N-mm then.

1kg=10N.
1cm=10mm.

Then if we consider B as a unit dimension and ignore it we will come to.
D=10√(M/Q).

B.Singh said:   2 years ago
Moment of resistance/maximum B.M.
(M) = QBd2.
Where,
Q = Moment resistacne factor.

B = width of slab = 1 m = 100 cm,
[since m was given in kg.cm],
D = depth of slab.
√ M = Q * 100 * d^2 (kg-cm).

Rajveer said:   4 years ago
M = 0.138Fckbd*2.
M = Qd*2(0.138fckb , M.R.factor),
So, option C is correct.
(1)

Indu said:   5 years ago
What is Q here?

Keshav Kaushik said:   5 years ago
None of the above is the correct answer.

According to me,
M= Qbd^2.

Here, ignoring b.
M= Qd^2.
So, d= √(M/Q).

Now M is given in kg.cm.
And 1kg = 9.81N or roughly 10N,
and 1cm = 10mm.

Therefore;
1kg.1cm = 10N.10mm =100Nmm.

So, d= √(100M/Q) = 10 √(M/Q).

Karan Panjabi said:   5 years ago
Moment Resistance
M = Qd^2.
M/(10x10) = Qd^2.
d = √M/100Q.

So option E is correct.

Monu said:   6 years ago
E is correct.

In this 100 is breadth in cm.

Garry said:   7 years ago
According to me, both C and D can be correct.

SUBHANKAR BAG said:   7 years ago
How it is possible? Explain.

Amrit Raj said:   7 years ago
d= √ M/1000Q is the correct answer.