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Soil Mechanics Quiz

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SOIL MECHANICS QUIZ NO.1
SOLUTION

1. Determine the void ratio of the soil in percent.

γm=G+Se1+eγw
Alternative 1
Use 105.73 and 50
105.73lbft3=G+0.5e1+e(62.4lbft3)

G=105.73+74.53e62.4
Substitute to
112.67lbft3=G+0.75e1+e(62.4lbft3)
Then
112.67lbft3=105.73+74.53e62.4+0.75e1+e(62.4lbft3)
SHIFT + SOLVE e=0.8013856813 in percent: 80.14%

Alternative 2
Use 112.67 and 75
112.67lbft3=G+0.75e1+e(62.4lbft3)

G=112.67+65.87e62.4
Substitute to
105.73lbft3=G+0.5e1+e(62.4lbft3)
Then
105.73lbft3=112.67+65.87e62.4+0.5e1+e62.4lbft3
SHIFT + SOLVE e=0.8013856813 in percent: 80.14%

NOTE: Rounding of values to nearest hundredths is permissible only in your final answer. It must be avoided in the computation. (e.g. 1.81) 2. Determine the specific gravity of the soil solids.

Alternative 1
G=105.73+74.53e62.4
G=2.651558891 or 2.65

Alternative 2
G=112.67+65.87e62.4
G=2.651558891 or 2.65

3. Determine the porosity of the soil in percent.

n=e1+ex100=0.80141.8014x100=44.49% n=44.49% 4-7. Noted that the container is filled with water. Therefore Vv=Vw

Total weight of soil mass (Wt): 0.0174g or 1.74x10^-5 kg
Weight of sample (Ws): 0.014g or 1.4x10^-5 kg
Weight of water (Ww) = Wt-Ws= 0.0034g or 3.4x10^-6kg x 9.81m/s^2
= 3.3354x10^-5 N or 3.3354x10^-8kN

γ=WV
VV=3.3354x10-8kN9.81kNm3
If Vv=Vw
Vw=3.4x10-9m3
Total Volume (Vt): 0.089cm3 or 8.9x10^-9m3
Volume of solids (Vs): Vt-Vw= 5.5 x 10^-9 m3 e=VvVs=0.62 n=e1+e=0.621.62=0.38 w=WwWs=3.3354x10-6kg1.4x10-5kgx100=24.29% G=eSw=0.6210.2429=2.55

8-10. γm=G+Se1+eγw γsat=2.55+0.621.629.81=19.2kNm3 or 1.957 g/cm3 γeff=19.2-9.81=9.39kNm3 or 0.957 g/cm3 γdry=2.551.629.81=15.44kNm3 or 1.57 g/cm3

11.
Water content of sample A: 41.79%
Water content of sample B: 43.5%

MODE 3-4 X | Y | 28 | 41.79% | 20 | 43.5% |

AC – input <25> -shift 1 5 5
25ŷ
LL=42.37%

INTERPOLATION log20-log25log20-log28=43.5-X43.5-41.79 SHIFT + SOLVE
LL=42.37%

12.
PL=23.13%+23.54%2=23.335%
13.
NWC=38.43%+38.05%2=38.24%
14.
PI=LL-PL
PI=43.37-23.335=19.035%
15.
LI=NWC-PLLL-PL=38.24-23.33519.035=0.783

16.
Dr=emax-eemax-eminx100
0.78=0.75-e0.75-0.46 e=0.5238 17. γm=G+Se1+eγw Gw=eS γm=2.68+2.680.091+0.52389.81 γm=18.806kN/m3

18. γdry=2.681+0.52389.81 γdry=17.25kN/m3

19.
PI=LL-PL
PI=61-30=31%
20.
Solve for water content first:
WC=96.2-71.971.9-20.8x100=47.55%
LI=WC-PLLL-PL=47.55-3031=0.566

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