Question: The following observations were made in a standard Practor's Test. Draw the dry density unit weight versus water content graph
0

and obtained MDD and OMC when volume of mould is 945 c.c and G = 2.7 . Also plot zero air void line.

Subject : Geo-technical Engineering 1

Topic : Compaction Of Soils & Soil Exploration

Difficulty : Moderate

mumbai university ge1(75) • 85 views
 modified 5 months ago  • written 5 months ago by Abhishek Tiwari ♦♦ 50
0

Volume of practor mould = 945 c.c

∴ γbulk = (Mass of soil) + (Volume of soil) = 945 cc | gm/c.c

∴ γdry = {( γbulk)/(1+ω)}

w = % with content

= (G .γω)/(1+ ωG) (For saturate sample or for zero air void line)

Trail Number 1 2 3 4 5
Mass of wet soil 1673 1814 1928 1985 1947
γbulk X k (gm/c.c) 1.77 1.97 2.04 2.1 2.06
ω = water content 9.7 11.6 13.3 16.2 18.7
γdry = {( γbulk)/(1+ω)} 1.613 1.72 1.8005 1.8076 1.735
Co–ordinate of zero air void line 2.099 2.017 1.949 1.843 1.76

γ Co–ordinate of zero void line

γdry = (G .γω)/(1+ ωG)

= (2.7 X 9.81)/(1+(0.097)X (2.7)) = 2.099 { γ = 0.981, (G = 2.7)} (ω = 9.7%)

= (2.7 X 9.81)/(1+(0.116)X (2.7)) = 2.017 ( ω= 11.6%)

= (2.7 X 9.81)/(1+(0.133)X (2.7)) = 1.949 ( ω= 13.3%)

= (2.7 X 9.81)/(1+(0.162)X (2.7)) = 1.843 ( ω= 16.2%)

= (2.7 X 9.81)/(1+(0.187)X (2.7)) = 1.76 ( ω= 18.7%)

Calculated :

OMC = 16.2 %

MDD = 1.8076