Question: If $W_{L}$=51%, $W_{p}$=32% $W_{s}$=19% Soil specimen shrinks from volume 42.5 cc at $w_{L}$ to 26.8 cc of shrinkage limit.Determine gravity

0

0

Reduction in volume R=42.5-26.8

=15.7cc

=15.7gm

% reduction in water content =51 to 1g=32% w=\frac{M_{w}}{M_{s}}=0.32

$M_{s}=\frac{M_{w}}{0.32}=\frac{15.7}{0.32}$=49.06gm

Initial $M_{w}$ in soil sample=0.51 Ms=$0.51 \times 49.06$

=25.02gm Volume of water initially

$V_{s}$=volume of solids=42.5-25.05

=17.478 cc

G$\frac{M_{s}}{V_{s}\times P_{w}} =\frac{49.06}{17.478 \times 1}$

=2.806

Please log in to add an answer.