Civil Engineering - RCC Structures Design - Discussion

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)

[A].
[B].
[C].
[D].
[E].

Answer: Option E

Explanation:

No answer description available for this question.

Amrit Raj said: (Apr 5, 2017)  
d= √ M/1000Q is the correct answer.

SUBHANKAR BAG said: (Apr 27, 2017)  
How it is possible? Explain.

Garry said: (Oct 12, 2017)  
According to me, both C and D can be correct.

Monu said: (Jan 12, 2018)  
E is correct.

In this 100 is breadth in cm.

Karan Panjabi said: (Aug 13, 2019)  
Moment Resistance
M = Qd^2.
M/(10x10) = Qd^2.
d = √M/100Q.

So option E is correct.

Keshav Kaushik said: (Sep 16, 2019)  
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).

Indu said: (Dec 3, 2019)  
What is Q here?

Rajveer said: (Jan 4, 2020)  
M = 0.138Fckbd*2.
M = Qd*2(0.138fckb , M.R.factor),
So, option C is correct.

B.Singh said: (Apr 3, 2022)  
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).

Ujjawal said: (Jun 15, 2022)  
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).

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