Geotechnical Engineering

DHARM

INDEX PROPERTIES AND CLASSIFICATION TESTS 75

Specific gravity of kerosene at 27°C = 0.773

Determine the specific gravity of the soil solids.

What will be the value if it has to be reported at 4°C?

Assume the specific gravity of water at 27°C as 0.99654.

Let the weight be designated as W 1 through W 4 in that order.

Wt of dry clay sample, Ws = (W 2 – W 1 ) = (0.816 – 0.6025) N = 0.2135 N

By Eq. 3.2,

G =

WG

WWW

s k

s

.

−−() 34

Gk here is given as 0.773.

∴ G =

0.2135 0.773

0.2135 (2.5734 2.4217)

×

−−

≈ 2.67

If the value has to be reported at 4°C, by Eq. 3.3,

GG

G

TTG

w

w

T

T

21

2

1

=.

∴ G4° = G27°.^1

0.99654

2.67 1

0.99654

= × = 2.68.

Example 3.4: In a specific gravity test, the following observation were made:
Weight of dry soil : 1.04 N
Weight of bottle + soil + water : 5.38 N
Weight of bottle + water : 4.756 N
What is the specific gravity of soil solids. If, while obtaining the weight 5.38 N, 3 ml of
air remained entrapped in the suspension, will the computed value of G be higher or lower
than the correct value? Determine also the percentage error.
Neglect temperature effects.

Neglecting temperature effects, by Eq. 3.1,

G =

W

WWW

s

s−−() 34

.

It this case, Ws = 1.04 N; W 3 = 5.38; W 4 = 6.756 N

∴ G =

1.04

1.04 (5.38 4.756)−−

= 2.50

If some air is entrapped while the weight W 3 is taken, the observed value of W 3 will be
lower than if water occupied this air space. Since W 3 occurs with a negative sign in Eq. 3.1 in
the denominator, the computed value of G would be lower than the correct value.

Since the air entrapped is given as 3 ml, this space, if occupied by water, would have

enhanced the weight W 3 by 0.03 N.

∴ Correct value of G =

1.04

1.04 (5.41 4.756)

1.040

−− 0.386

= = 2.694