Civil Engineering - RCC Structures Design - Discussion
Discussion Forum : RCC Structures Design - Section 4 (Q.No. 25)
25.
If permissible compressive stress in concrete is 50 kg/cm2, tensile stress in steel is 1400 kg/cm2 and modular ratio is 18, the depth d of the beam, is
Discussion:
8 comments Page 1 of 1.
Danish said:
7 years ago
1, d= √[B.M/b*q].
2, q=.5Cjk.
3, K= mC/(T+mC)
= 18*50/(1400+(18*50))
= 900/23,00
= .39.
4, J=1-k/3.
=1-0.39/3.
=.87.
Now, q=.5Cjk.
.5 * 50 * .87 * .39 = 8.482.
& d=√[B.M/(b*q)].
√[B.M/(b*8.482)],
√0[.177*B.M/b].
Hence option A is correct.
2, q=.5Cjk.
3, K= mC/(T+mC)
= 18*50/(1400+(18*50))
= 900/23,00
= .39.
4, J=1-k/3.
=1-0.39/3.
=.87.
Now, q=.5Cjk.
.5 * 50 * .87 * .39 = 8.482.
& d=√[B.M/(b*q)].
√[B.M/(b*8.482)],
√0[.177*B.M/b].
Hence option A is correct.
(2)
Ganesh said:
10 years ago
M = qbd^2.
Here, q = 0.5*50*j*n = 8.48.
j = 1-(n/3) = 0.39.
n = 1/(1+1400/18*50) = 0.87.
Here, q = 0.5*50*j*n = 8.48.
j = 1-(n/3) = 0.39.
n = 1/(1+1400/18*50) = 0.87.
(1)
Ganesh said:
10 years ago
B.M = qbd^2.
q = ???
q = ???
Aamir said:
8 years ago
Use 3 eq, mc/t=n/(d_n).
a=d-n/3.
BM=bnca/2.
a=d-n/3.
BM=bnca/2.
Aria Hous said:
8 years ago
How, can anyone explain it clearly?
Priya said:
8 years ago
Anyone Explain it.
Apcivilian said:
7 years ago
Very Nice explanation, Thanks @Danish.
Ryan said:
5 years ago
Thanks @Danish.
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