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Doss [256]
3 years ago
11

A saturated 1.5 ft3 clay sample has a natural water content of 25%, shrinkage limit (SL) of 12% and a specific gravity (GS) of 2

.70. Determine the volume of the sample when the water content is 10%.

Engineering
1 answer:
Svetllana [295]3 years ago
7 0

79 f t^{3} is the volume of the sample when the water content is 10%.

<u>Explanation:</u>

Given Data:

V_{1}=100\ \mathrm{ft}^{3}

First has a natural water content of 25% = \frac{25}{100} = 0.25

Shrinkage limit, w_{1}=12 \%=\frac{12}{100}=0.12

G_{s}=2.70

We need to determine the volume of the sample when the water content is 10% (0.10). As we know,

V \propto[1+e]

\frac{V_{2}}{V_{1}}=\frac{1+e_{2}}{1+e_{1}}  ------> eq 1

e_{1}=\frac{w_{1} \times G_{s}}{S_{r}}

The above equation is at S_{r}=1,

e_{1}=w_{1} \times G_{s}

Applying the given values, we get

e_{1}=0.25 \times 2.70=0.675

Shrinkage limit is lowest water content

e_{2}=w_{2} \times G_{s}

Applying the given values, we get

e_{2}=0.12 \times 2.70=0.324

Applying the found values in eq 1, we get

\frac{V_{2}}{100}=\frac{1+0.324}{1+0.675}=\frac{1.324}{1.675}=0.7904

V_{2}=0.7904 \times 100=79\ \mathrm{ft}^{3}

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