Civil Engineering - RCC Structures Design - Discussion
Discussion Forum : RCC Structures Design - Section 2 (Q.No. 12)
12.
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
Discussion:
17 comments Page 1 of 2.
Akash said:
9 years ago
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.
Here the diameter of stirrups which is 6mm min so 6 x 7.5 =45 #so spacing we conveniently take 40.
Bilal Butt said:
5 years ago
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.
Now, No. Of Stirrups = L/ spacing = 3600/90 = 40.
The answer is C.
(5)
Gubendhiran said:
8 years ago
When stress less than 5kg/cm2.
The Spacing of strippers is taken as the Lever's arm so 40cm is correct.
The Spacing of strippers is taken as the Lever's arm so 40cm is correct.
(3)
Prashant said:
1 year ago
They were asked no of shear stirrups required but not the amount of spacing (min or max).
(1)
Shailendra said:
9 years ago
@Akash.
Spacing is 0.75d where d is effective depth not Diameter.
Spacing is 0.75d where d is effective depth not Diameter.
Elumalai said:
8 years ago
Please explain the answer with formula.
(1)
Vinay said:
3 years ago
Gubendhiran.
Can you explain in depth?
Can you explain in depth?
Kinagi said:
5 years ago
Please explain in detail anyone?
Shamoa sadhukhan said:
1 decade ago
How it proof please explain?
Ritss said:
8 years ago
Please explain in detail.
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