Weight-Volume Relationships
and Plasticity
Weight-Volume Relationships
- relationships between
a.Unit weight
b.Void ratio
c.Porosity
d.Moisture content
e.Specific gravity
Volume and Weight Relationships
V=𝑉𝑠 + 𝑉𝑣 = 𝑉𝑠 + 𝑉w + 𝑉𝑎
Where :
V=total volume of the soil
sample
𝑉
𝑠=volume of soil soil solids
𝑉
𝑣=volume of voids
𝑉
w=volume of water
𝑉
𝑎=volume of air
𝑊 = 𝑊w + 𝑊𝑠
Where:
W =total weight of soil sample
𝑊
w=weight of water
𝑊𝑠=weight of soil
e=void ratio
𝑒 =
𝑉𝑣
𝑉𝑠
n=porosity
𝑛 =
𝑉𝑣
𝑉
S=degree of saturation
𝑉w
S =
𝑉𝑣
w=moisture content or water
content
𝑤 =
𝑊w
𝑊𝑠
𝛾 = 𝑢𝑛𝑖𝑡 𝑤𝑒𝑖𝑔ℎ𝑡 of soil per volume
𝛾 =
𝑊
𝑉
𝛾 =
𝖶
𝑉
𝖶 +𝖶
𝑉
= 𝑠 w
=
𝖶𝑠 1+ Ww
W𝑠
𝑉
= 𝑠
𝖶 1+w
𝑉
𝛾𝑑 = 𝑑𝑟𝑦 𝑢𝑛𝑖𝑡 𝑤𝑒𝑖𝑔ℎ𝑡
𝑑
𝛾 =
𝑊𝑠
=
𝛾
𝑉 1 + 𝑤
𝜌 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑠𝑜𝑖𝑙
𝜌 =
𝑚
𝑉
𝜌𝑑 = 𝑑𝑟𝑦 𝑑𝑒𝑛𝑠𝑖𝑡𝑦
𝑚
𝜌𝑑 = 𝑠
𝑉
𝑊𝑠 = 𝐺𝑠𝛾w
𝑊w = 𝑤𝑊𝑠 = 𝑤𝐺𝑠𝛾w
𝐺𝑠 = speciEic gravity of soil solids
𝛾 = 𝑉
= = =
𝖶 𝖶𝑠+𝖶w 𝐺𝑠𝛾w+w𝐺𝑠𝛾w 1+w 𝐺𝑠𝛾w
𝑉 1+𝑒 1+𝑒
𝛾𝑑 =
𝑊𝑠
=
𝐺𝑠𝛾w
𝑉 1 + 𝑒
Ws
Total
weight
= W
Total
volume
= V
W
Vs
Va
V
(a) (b)
Air Water Solid
W
V
V
𝑉w =
𝛾w
=
𝑊w 𝑤𝐺𝑠𝛾w
𝛾w
+ 𝑤𝐺𝑠
𝑉 𝑤𝐺
𝑆 = w
= 𝑠
𝑉
𝑣 𝑒
𝑆𝑒 = 𝑤𝐺𝑠
V = V = e
Ws = Gsg
W
Vs = 1
V = 1 + e
Weight Volume
W = eg
Water Solid
11 + w2Gsrw
Density = r =
1 + e
d
Dry density = r =
Gsrw
1 + e
Saturated density = r sat
1Gs + e2rw
=
1 + e
𝛾𝑠𝑎𝑡
𝑉 𝑉
=
𝑊
=
𝑊𝑠 + 𝑊w
=
𝐺𝑠𝛾w + 𝑒𝛾w
=
𝐺𝑠 + 𝑒 𝛾w
1 + 𝑒
1 + 𝑒
𝑒 = 𝑤𝐺𝑠
Ww = wWs = wGsgw11 —n 2
Ws = Gsgw11 —n 2
W = Gsg (1 —n)
Ws = Gsg (1 —n) Vs = 1 —n
V = 1
V = n
Weight Volume
Air Water Solid
Ws
gd =
V
=
Gsgw11 —n 2
1
= Gsgw11 —n 2
g =
Ws + Ww
V
= Gsgw11 —n 211 + w2
gsat =
Ws + Ww
V
=
11 — n 2Gsgw + ngw
1
= 311 — n 2Gs + n 4
gw
Ww
w = =
ngw
Ws 11 —n 2gwGs
n
=
11 —n 2Gs
Vv
n =
V
Example
A moist soil has these values: V=7.08x10-3 m3, m=13.95 kg, w=9.8%and Gs=2.66
Determine the following:
a. Density
b. Dry density
c. Void ratio
d. Porosity
e. Degree of saturation (%)
f. Volume occupied by water
In the natural state, a moist soil has a volume of 0.30 m3 and weighs 5500 N. The oven dry
weight of the soil is 4911 N. If𝐺𝑠= 2.74, calculate the following:
a. moisture content
b. moist unit weight
c. dry unit weight
d. void ratio
e. Porosity
f. degree of saturation
A representative soil specimen collected from the field weighs 1.8 kN and has a volume of
0.1 m3. The moisture content as determined in the laboratory is 1.2%. Given Gs=2.71,
determine the following:
a. Moist unit weight
b. Dry unit weight
c. Void ratio
d. Porosity
e. Degree of saturation
A saturated soil has a dry unit weight of 16.2 kN/m3. Its moisture content is 20%. Determine:
a. Saturated unit weight
b. Specific gravity
c. Void ratio
The following data are given for a soil: porosity =0.45, specific gravity of the soil solids=2.68,
and moisture content =10%. Determine the mass of water to be added to 10 m3 of soil for
full saturation.

Weight - Volume Relationships and Plasticity

  • 1.
  • 2.
    Weight-Volume Relationships - relationshipsbetween a.Unit weight b.Void ratio c.Porosity d.Moisture content e.Specific gravity
  • 3.
    Volume and WeightRelationships V=𝑉𝑠 + 𝑉𝑣 = 𝑉𝑠 + 𝑉w + 𝑉𝑎 Where : V=total volume of the soil sample 𝑉 𝑠=volume of soil soil solids 𝑉 𝑣=volume of voids 𝑉 w=volume of water 𝑉 𝑎=volume of air 𝑊 = 𝑊w + 𝑊𝑠 Where: W =total weight of soil sample 𝑊 w=weight of water 𝑊𝑠=weight of soil e=void ratio 𝑒 = 𝑉𝑣 𝑉𝑠 n=porosity 𝑛 = 𝑉𝑣 𝑉 S=degree of saturation 𝑉w S = 𝑉𝑣 w=moisture content or water content 𝑤 = 𝑊w 𝑊𝑠
  • 4.
    𝛾 = 𝑢𝑛𝑖𝑡𝑤𝑒𝑖𝑔ℎ𝑡 of soil per volume 𝛾 = 𝑊 𝑉 𝛾 = 𝖶 𝑉 𝖶 +𝖶 𝑉 = 𝑠 w = 𝖶𝑠 1+ Ww W𝑠 𝑉 = 𝑠 𝖶 1+w 𝑉 𝛾𝑑 = 𝑑𝑟𝑦 𝑢𝑛𝑖𝑡 𝑤𝑒𝑖𝑔ℎ𝑡 𝑑 𝛾 = 𝑊𝑠 = 𝛾 𝑉 1 + 𝑤 𝜌 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑠𝑜𝑖𝑙 𝜌 = 𝑚 𝑉 𝜌𝑑 = 𝑑𝑟𝑦 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑚 𝜌𝑑 = 𝑠 𝑉
  • 5.
    𝑊𝑠 = 𝐺𝑠𝛾w 𝑊w= 𝑤𝑊𝑠 = 𝑤𝐺𝑠𝛾w 𝐺𝑠 = speciEic gravity of soil solids 𝛾 = 𝑉 = = = 𝖶 𝖶𝑠+𝖶w 𝐺𝑠𝛾w+w𝐺𝑠𝛾w 1+w 𝐺𝑠𝛾w 𝑉 1+𝑒 1+𝑒 𝛾𝑑 = 𝑊𝑠 = 𝐺𝑠𝛾w 𝑉 1 + 𝑒 Ws Total weight = W Total volume = V W Vs Va V (a) (b) Air Water Solid W V V 𝑉w = 𝛾w = 𝑊w 𝑤𝐺𝑠𝛾w 𝛾w + 𝑤𝐺𝑠 𝑉 𝑤𝐺 𝑆 = w = 𝑠 𝑉 𝑣 𝑒 𝑆𝑒 = 𝑤𝐺𝑠
  • 6.
    V = V= e Ws = Gsg W Vs = 1 V = 1 + e Weight Volume W = eg Water Solid 11 + w2Gsrw Density = r = 1 + e d Dry density = r = Gsrw 1 + e Saturated density = r sat 1Gs + e2rw = 1 + e 𝛾𝑠𝑎𝑡 𝑉 𝑉 = 𝑊 = 𝑊𝑠 + 𝑊w = 𝐺𝑠𝛾w + 𝑒𝛾w = 𝐺𝑠 + 𝑒 𝛾w 1 + 𝑒 1 + 𝑒 𝑒 = 𝑤𝐺𝑠
  • 7.
    Ww = wWs= wGsgw11 —n 2 Ws = Gsgw11 —n 2 W = Gsg (1 —n) Ws = Gsg (1 —n) Vs = 1 —n V = 1 V = n Weight Volume Air Water Solid Ws gd = V = Gsgw11 —n 2 1 = Gsgw11 —n 2 g = Ws + Ww V = Gsgw11 —n 211 + w2 gsat = Ws + Ww V = 11 — n 2Gsgw + ngw 1 = 311 — n 2Gs + n 4 gw Ww w = = ngw Ws 11 —n 2gwGs n = 11 —n 2Gs Vv n = V
  • 8.
    Example A moist soilhas these values: V=7.08x10-3 m3, m=13.95 kg, w=9.8%and Gs=2.66 Determine the following: a. Density b. Dry density c. Void ratio d. Porosity e. Degree of saturation (%) f. Volume occupied by water
  • 9.
    In the naturalstate, a moist soil has a volume of 0.30 m3 and weighs 5500 N. The oven dry weight of the soil is 4911 N. If𝐺𝑠= 2.74, calculate the following: a. moisture content b. moist unit weight c. dry unit weight d. void ratio e. Porosity f. degree of saturation
  • 10.
    A representative soilspecimen collected from the field weighs 1.8 kN and has a volume of 0.1 m3. The moisture content as determined in the laboratory is 1.2%. Given Gs=2.71, determine the following: a. Moist unit weight b. Dry unit weight c. Void ratio d. Porosity e. Degree of saturation
  • 11.
    A saturated soilhas a dry unit weight of 16.2 kN/m3. Its moisture content is 20%. Determine: a. Saturated unit weight b. Specific gravity c. Void ratio
  • 12.
    The following dataare given for a soil: porosity =0.45, specific gravity of the soil solids=2.68, and moisture content =10%. Determine the mass of water to be added to 10 m3 of soil for full saturation.