Question

A sample of saturated clay was placed in a container and weighed. The weight was 6N. The clay in its container was placed in an oven for 24 hours at 105° C. The weight reduced to a constant value of 5N. The weight of the container is 1N. If G-2.7, determine the:
(a) water content;
(b) void ratio;
(c) bulk unit weight;
(d) dry unit weight;
(e) effective unit weight.

Answer 1

Answer is given below

Explanation:

given data

weight = 6N

temp = 105° C

weight reduced = 5N

solution

weight of container is 1N

SO W = (6-1) = 5

And Wd = 5 - 1 = 4

so

moisture content is

moisture content = $\frac{W-Wd}{Wd} \times 100$       .......1

moisture content = $\frac{5-4}{4} \times 100$

moisture content = 25%

and

as we know density of soil soild = 2700 kg/m³

density of water = 1000 kg/m³

and sp gravity of soil =  $\frac{2700}{1000}$  = 2.7

so

now we get here bulk unit weight

bulk unit wt = $Yw \times [\frac{G+e}{1+e}]$     ..........2

bulk unit wt = $9.01 \times [\frac{2.7 + 0.675}{1+0.675}]$

bulk unit wt = 19.766 KN/m³

and

so dry unit wt  will be

dry unit wt = $\frac{Ysat}{1+w}$           ..............3

dry unit wt = $\frac{19.766}{1+0.25}$

dry unit wt = 15.813 kN/m³