Civil Engineering - RCC Structures Design - Discussion
Discussion Forum : RCC Structures Design - Section 2 (Q.No. 8)
8.
If the maximum shear stress at the end of a simply supported R.C.C. beam of 6 m effective span is 10 kg/cm2, the share stirrups are provided for a distance x from either end where x is
Discussion:
14 comments Page 1 of 2.
Lingaraj said:
5 years ago
Thank you @Supriya.
Supriya said:
5 years ago
The Formula is = L/2(1 - 5/qmax).
(2)
Behara said:
5 years ago
It means we need to provide stirrups 1.5 m length from either side of the beam end. why because the max shear force is at end of the beam.
Dibakar said:
5 years ago
Why not L/7 where the crank starts?
King said:
5 years ago
150cm is not stirrup spacing. Stirrup provides after 150cm from support. @Sameer.
Sameer said:
6 years ago
But 150cm stirrup spacing too much. Practically not possible.
Shah faisal Khan said:
7 years ago
At the ends of supports mean discontinuous case. So it will be 600/4 (L/4).
Er N.Sureshkumar said:
7 years ago
It is a general/detailing approach by considering the location of the point of contraflexure is at L/4 up to which we have to provide shear stirrups.
The industrial standard is L/3.
The industrial standard is L/3.
(1)
Khan G said:
7 years ago
Shear produced always at a Span/4 so,
Here Span is 6 m=600 cm, then
Shear stirrups are provided for a distance x from either end is Span /4= 600 / 4 = 150 cm.
Here Span is 6 m=600 cm, then
Shear stirrups are provided for a distance x from either end is Span /4= 600 / 4 = 150 cm.
(7)
Navi said:
9 years ago
Here, L/2(1-5/q max).
(2)
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