Civil Engineering - Soil Mechanics and Foundation Engineering - Discussion
Discussion Forum : Soil Mechanics and Foundation Engineering - Section 2 (Q.No. 15)
15.
260 g of wet soil was taken in a pycnometer jar of weight 400 g in order to find the moisture content in the soil, with specific gravity of soil particles 2.75. The weight of soil and remaining water filled in pycnometer without air bubbles was 1415 g and the weight of pycnometer filled with water alone was 1275 g. The moisture content in the soil is
Discussion:
7 comments Page 1 of 1.
Abhay said:
9 years ago
WATER CONTENT = ([{(W2 - W1)/(W3 - W4)}{(G-1)/G]-1)*100.
Where,
W1 = wt. of empty Pycnometer = 400 g.
W2 = wt. of the Pycnometer with wet soil = 260 + 400 = 660.
W3 = wt. of the Pycnometer and soil, filled with water = 1415.
W4 = wt. of Pycnometer filled with water only = 1275.
G = Specific gravity of solids = 2.75.
Water content = ([(660 - 400)/(1415 - 1275)][(2.75 - 1)/(2.75)] - 1)*100.
= [(260/140)(1.75/2.75) - 1]*100 = 18.18% answer.
Where,
W1 = wt. of empty Pycnometer = 400 g.
W2 = wt. of the Pycnometer with wet soil = 260 + 400 = 660.
W3 = wt. of the Pycnometer and soil, filled with water = 1415.
W4 = wt. of Pycnometer filled with water only = 1275.
G = Specific gravity of solids = 2.75.
Water content = ([(660 - 400)/(1415 - 1275)][(2.75 - 1)/(2.75)] - 1)*100.
= [(260/140)(1.75/2.75) - 1]*100 = 18.18% answer.
(13)
Anji said:
1 decade ago
Can be calculated by below formula.
MOISTURE content = {[(w2-w1)/(w3-w4)][(Gs-1)/Gs]-1}*100.
MOISTURE content = {[(w2-w1)/(w3-w4)][(Gs-1)/Gs]-1}*100.
Hatem said:
1 decade ago
Gs = {(260/140)*((2.75-1)/2.75 )}-1.
Gs = 1.857*(.636)-1.
Gs = .181 = 18.1%.
Gs = 1.857*(.636)-1.
Gs = .181 = 18.1%.
(1)
Vishal singh said:
1 decade ago
How will 260 come (w2-w1) = 140? Explain please?
Ravi Kant said:
10 years ago
(Weight of pycnometer jar)-(Weight of wet soil)
Shashank Agnihotri said:
6 years ago
Good explanation @Abhay.
Guru said:
5 years ago
Thanks @Abhay.
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