Civil Engineering - Soil Mechanics and Foundation Engineering - Discussion
Discussion Forum : Soil Mechanics and Foundation Engineering - Section 1 (Q.No. 40)
40.
A soil has bulk density 2.30 g/cm3 and water content 15 per cent, the dry density of the sample, is
Discussion:
9 comments Page 1 of 1.
Neha said:
4 years ago
Yd = Y/1+w.
= 2.4/1+.2.
= 2.
= 2.4/1+.2.
= 2.
(2)
Hemanta Kr Nath said:
7 years ago
Given,
The water content, W = 15%,
Bulk density, Ym = 2.30 gm/cc,
Dry density, Yd =?
We know that, Yd = 100/(100+W)*Ym.
= 100/(100+15)*2.30,
= 2.0 gm/cc.
The water content, W = 15%,
Bulk density, Ym = 2.30 gm/cc,
Dry density, Yd =?
We know that, Yd = 100/(100+W)*Ym.
= 100/(100+15)*2.30,
= 2.0 gm/cc.
(1)
CHANDAN said:
8 years ago
How is it possible? Explain in detail.
Rohit kumar said:
8 years ago
We know that Bulk unit weigh[Yb] =Total weight/Total volume.
So, Yb = W/V =Ww + Ws/V.
NOW, Take common Ws, Hence Ws[Ww/ Ws + 1]/V.
NOW Ws / V = Yd [dry unit weight] .
Hence Yb = Yd [w = 1] because Ww/Ws = w[water content].
Now Yd = Yb /[1 + w] = 2.30/[1 + 0.15] =2 g/centi meter cube.
So, Yb = W/V =Ww + Ws/V.
NOW, Take common Ws, Hence Ws[Ww/ Ws + 1]/V.
NOW Ws / V = Yd [dry unit weight] .
Hence Yb = Yd [w = 1] because Ww/Ws = w[water content].
Now Yd = Yb /[1 + w] = 2.30/[1 + 0.15] =2 g/centi meter cube.
(1)
Skkk said:
8 years ago
Dry density = y/1+w.
=> 2.30/1+0.15.
=> 2 g/cm2.
=> 2.30/1+0.15.
=> 2 g/cm2.
(1)
SAMBA said:
9 years ago
DRY DENSITY = 2.3/1+0.15.
Shyam sunder said:
1 decade ago
Dry density = 2.3/1.15 = 2.00 gm/cm3.
M.PAGAVAN DOSS said:
1 decade ago
Dry density = Grw/1+e.
Bulk density = Grw (1+w)/1+e.
rt = rd*(1+w).
rd = 2.3/1+.15 = 2gm/cubic centimeter.
Bulk density = Grw (1+w)/1+e.
rt = rd*(1+w).
rd = 2.3/1+.15 = 2gm/cubic centimeter.
E.Narasimhulu said:
1 decade ago
Dry density = ratio of bulk density to water content with adding one.
D.D = 2.3/1+ 0.15 = 2 g/cm3.
D.D = 2.3/1+ 0.15 = 2 g/cm3.
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