المحاظرة الخامسة / الجيلوجيا هندسية / مرحلة اولى

5-1 Derivation of Phase Relations
Bulk unit weight of rock and/or soil:
?=((G_s+S_r×e)/(1+e)) ?_w
Let consider Phase Diagrams as shown in figure (5-1) below:
For purpose of derivation, assume Vs=1, thus and from phase diagram, we can conclude:

e=V_v/V_s
?for Vs=1 ? e=Vv
Vt = Vs+Vv = 1+e
S_r=V_w/V_s
e=V_v/V_s
?S_r×e=V_w/V_v ×V_v/V_s
?S_r×e=V_w
W_w=V_w×?_w=S_r.e.?_w
W_s=V_s×?_s=1×?_s
W_s=G_s.?_w
Since the bulk unit weight of rock is :
?=W/V_t =(W_s+W_w)/V_t
??=(G_s.?_w+S_r.e.?_w)/(1+e)
?=((G_s+S_r.e)/(1+e)).?_w

Example 5-1:
A sandstone rock specimen consist of quartiz and feldspar. When the rock is saturated with water, it s weight is 21.4 N. While the dry weight is 20.3 N, and the spesific weight of rock is 2.63. Determine:
1- porosity of rock.
2- dry unit weight.
3- moist unit weight.
Solution:
n=(?G_s)/(1+?G_s )
?=W_w/W_s =(W_(sat.)-W_dry)/W_s =(21.4-20.3)/20.3=0.0541=5.41 %
n=(0.0541×2.63)/(1+0.135)=0.1253=12.53 %

?_dry=G_s/(1+e) ?_w
Since [n=e/(1+e)]
?e=n/(1-n)
e=0.1253/(1-0.1253)=0.1432=14.32 %
?_dry=2.63/(1+0.1432)×9.81=22.56 kN/m^3
?_dry=?_wet/(1+?)
22.56=?_wet/(1+0.0541)
?_wet=22.56(1+0.0541)
?_wet=23.78 kN/m^3