Civil Engineering - RCC Structures Design - Discussion
Discussion Forum : RCC Structures Design - Section 1 (Q.No. 14)
14.
If the average bending stress is 6 kg/cm2 for M 150 grade concrete, the length of embedment of a bar of diameter d according to I.S. 456 specifications, is
Discussion:
34 comments Page 1 of 4.
Suraj kumar said:
2 years ago
130/0.6x4 = 55.17.
(1)
Rajkumar said:
4 years ago
It's 230/4 * 1 = 57.5.
(2)
Rakesh said:
4 years ago
You are absolutely right, Thanks @Dev Gaharwar.
(3)
Dishant Darji said:
5 years ago
Tbd = 6kg/cm^2= .6 n/mm^2.
Ld= (.87x250xd)/(4x1.25x.6)= 58d.
Ld= (.87x250xd)/(4x1.25x.6)= 58d.
(10)
Jagadeeshwar said:
5 years ago
For M40 and above tbd = 1.9.
And take the grade of steel 500.
Then length = 0. 87 * 500 * d/(4 * 1.9) = 58d.
And take the grade of steel 500.
Then length = 0. 87 * 500 * d/(4 * 1.9) = 58d.
(3)
Dev Gaharwar said:
5 years ago
For Fe250 bar dia less than 20 mm:
Fst = 140 N/mm2.
And as per WSM for M15 tbd = 0.6.
Hence:
Ld = fst x d / 4 x tbd.
Ld = 140 x d / 4 x 0.6,
Ld = 58.33d.
This could be a possibility.
Fst = 140 N/mm2.
And as per WSM for M15 tbd = 0.6.
Hence:
Ld = fst x d / 4 x tbd.
Ld = 140 x d / 4 x 0.6,
Ld = 58.33d.
This could be a possibility.
(5)
Laxman said:
6 years ago
It requires a grade of steel.
(1)
Ram said:
6 years ago
Ld=dia* σ st/4 permissible bond stress.
Ld=130*dia/4*.6=58dia.
Ld=130*dia/4*.6=58dia.
(1)
Pranjal jain said:
6 years ago
According to me, it's (700/(0.87fy+1100))*d.
K.K MAHATA said:
7 years ago
D*140/4*0.6= 58d.
140N/mm2 = stress of HYSD,
And 0.6 = bond stress of M15 in wsm.
140N/mm2 = stress of HYSD,
And 0.6 = bond stress of M15 in wsm.
(1)
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers