The volume relationships commonly uses for the three phases in soil elements:
The weight relations are moisture content (w) and unit weight (gamma w). Moisture content is also reffered to as water content
w, mc
(unit weight/ moisture content)
w=\frac{W_w}{W_s}\\ \space \\mc=\frac{W_w}{W_s}Ww = weight of water
Ws = weight of soil solids
unit weight
(ɣ)
weight of soil per unit volume
\gamma=\frac{W_t}{V_t}Wt =
Vt =
ɣd
(dry unit weight)
\gamma_d=\frac{W_s}{V_t}Vw = volume of water in the voids
Vv = volume of voids
dry unit weight (ɣd)
dry unit weight & soil unit weight and moisture content (w) relationsh
\gamma=\frac{W}{V}=\frac{W_s+W_w}{V}=\frac{W_s(1+w)}{V}\gamma_d=\frac{W_s}{V}\begin{aligned} \gamma &=\frac{W_s(1+w)}{V} \\ \space \\ \frac{\gamma}{(1+w)}&=\frac{W_s\xcancel{(1+w)}}{V} \\ \space \\ \frac{\gamma}{(1+w)}&=\frac{W_s}{V} \\ \space \\ \frac{\gamma}{(1+w)}&=\gamma_d
\end{aligned}Ww = weight of water
Ws = weight of soil solids
n
(porosity)
porosity commonly expressed relationships
n=\frac{e}{1+e}=\frac{\frac{V_v}{V_s}}{1+\frac{V_v}{V_s}}e = void ratio
Vv = volume of voids
Vs = volume of soil solids
density
(ρ)
weight of soil per unit volume
\rho=\frac{M}{V}Wt =
Vt =
ρd
(dry density of soil)
\rho_d=\frac{M_s}{V_t}ρ = dry density of soil (lb/ft3, slug/ft3)
Ms= mass of soil solids in sample (lb, slug)
V = volume of soil sample (ft3, gal)
unit weight of water (ɣw)
= 62.43 lb/ft3
= 8.35 lb/gal
= .0361 lb/in3
= 1.94 slug/ft3
= .259 slug/gal
= .00112 slug/in3
relationships between: unit weight (ɣ), void ratio (e), moisture content (w), & specific gravity (Gs)
The tip for figuring out problems in this conceptual way, is to make V(soil solids) = 1; and if Vs = 1, Vv = e
Ws
(weight of soil solids)
W_s=G_s\gamma_w
Gs = specific gravity
ɣw = unit weight of water, (62.43 lb/ft3)
Ww
(weight of water)
\begin{aligned} W_w&=wG_s\gamma_w \\ &=wW_s\end{aligned}Gs = specific gravity
ɣw = unit weight of water, (62.43 lb/ft3)
Ws = weight of soil solids
Gs
(specific gravity)
specific gravity of solid soils
G_s=\frac{W_s}{V_s\gamma_w}
ɣw = unit weight of water, (62.43 lb/ft3)
volume relationships
(with Vv = e and Vs = 1)
V_v=V_w=V_a=e
V_s=1
V_t=1+e
V…
ɣd
(dry unit weight)
if the soil sample is saturated – that is, the void spaces are completely filled with water – the relationship for saturation unit weight (gamma sat) can be derived in a similar manner:
\gamma_{sat}=\frac{W_t}{V_t}=\frac{W_s+W_w}{V}=\frac{G_s\gamma_w+e\gamma_w}{1+e}=\frac{\gamma_w(G_s+e)}{1+e}Vw = volume of water in the voids
Vv = volume of voids
°S,
(degree of saturation)
\begin{aligned} \text{\textdegree}\text{S}&=\frac{V_w}{V_v} \\ \space \\ \text{\textdegree}\text{S}&=\frac{wG_s}{e} \\ \space \\ \text{\textdegree}\text{S}e&=wG_s
\end{aligned}and with °S = 1… (fully saturated)
e=wG_s
unit weight defintion, with Vt = 1+e
void ratio commonly expressed relationships
\gamma=\frac{W}{V}=\frac{W_s+W_w}{V}=\frac{G_s\gamma_w+wG_s\gamma_w}{1+e}=\frac{(1+w)G_s\gamma_w}{1+e}Vv = volume of voids
Vs = volume of soil solids
dry unight weight, with Vt = 1+e
\begin{aligned}\gamma_d&=\frac{W_s}{V} \\ \space \\\gamma_d&=\frac{G_s\gamma_w}{1+e} \end{aligned}e = void ratio
Vv = volume of voids
Vs = volume of soil solids
solving for void ratio (e), using unit weight and dry unit weight definitions of Vt = 1+e
void ratio commonly expressed relationships
\begin{aligned} e &=\frac{G_s\gamma_w}{\gamma_d} -1 \\ \space \\e &= \frac{G_s\gamma_w (1+w)}{\gamma}-1\end{aligned}Vv = volume of voids
Vs = volume of soil solids
dry unight weight, with V = 1 + e
\begin{aligned}\gamma_d&=\frac{W_s}{V} \\ \space \\\gamma_d&=\frac{G_s\gamma_w}{1+e} \end{aligned}e = void ratio
Vv = volume of voids
Vs = volume of soil solids
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