Civil Engineering - RCC Structures Design - Discussion


A reinforced concrete cantilever beam is 3.6 m long, 25 cm wide and has its lever arm 40 cm. It carries a load of 1200 kg at its free end and vertical stirrups can carry 1800 kg. Assuming concrete to carry one-third of the diagonal tension and ignoring the weight of the beam, the number of shear stirrups required, is

[A]. 30
[B]. 35
[C]. 40
[D]. 45
[E]. 50

Answer: Option C


No answer description available for this question.

Abhi said: (Dec 28, 2014)  
What is the formula?

Shamoa Sadhukhan said: (Aug 6, 2015)  
How it proof please explain?

Devi said: (Sep 9, 2015)  
Please explain me.

K.K said: (Oct 5, 2015)  
How to solve?

Rahul said: (Aug 23, 2016)  
Please explain.

Akash said: (Sep 28, 2016)  
Min spacing criteria are 1)7.5d, 2)300 we use a min of above.

Here the diameter of stirrups which is 6mm min so 6 x 7.5 =45 #so spacing we conveniently take 40.

Shailendra said: (Dec 5, 2016)  

Spacing is 0.75d where d is effective depth not Diameter.

Raju said: (Apr 16, 2017)  
Explain the answer.

Elumalai said: (May 20, 2017)  
Please explain the answer with formula.

Gubendhiran said: (Aug 22, 2017)  
When stress less than 5kg/cm2.

The Spacing of strippers is taken as the Lever's arm so 40cm is correct.

Ritss said: (Dec 7, 2017)  
Please explain in detail.

Sunil Kumar Meher said: (Apr 1, 2018)  
Please explain it.

Kinagi said: (Apr 5, 2020)  
Please explain in detail anyone?

Bilal Butt said: (Dec 16, 2020)  
Clear Spacing from either sides of the beam is L/4, i.e, 360/4 = 90 mm.
Now, No. Of Stirrups = L/ spacing = 3600/90 = 40.
The answer is C.

Engr Zaid said: (Apr 18, 2021)  
Please explain.

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