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 3 of 4.
SOUMEN DAS said:
8 years ago
The question is for M15 not M150.
For,
M15 58d
M20 47d
M25 40d
M30 38d In tension
For,
M15 58d
M20 47d
M25 40d
M30 38d In tension
Revathi said:
7 years ago
How this will come? Please explain.
Sameer sopori said:
7 years ago
M15 means 15Mpa (Mega pascal).
15Mpa = 150 bar.
(because 100000 pascal equal to 1 bar).
15Mpa = 150 bar.
(because 100000 pascal equal to 1 bar).
Jitendra jain said:
7 years ago
M15 means 15mpa (N/mm^2) = 150( kg/cm^2).
(1)
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)
Pranjal jain said:
6 years ago
According to me, it's (700/(0.87fy+1100))*d.
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)
Laxman said:
6 years ago
It requires a grade of steel.
(1)
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)
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)
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