Chapter 3 Weight–Volume Relationships

Chapter 3

(Gs + e)γ w Gsγ w e γ w eγw ⎛ 1 + wsat ⎞
3.1 γ sat = = + = + n γw = n⎜⎜ ⎟⎟γ w
1+ e 1 + e 1 + e (1 + e) wsat ⎝ w sat ⎠

(Gs + e)γ w Gs γ w eγ w ⎛ e ⎞
3.2 γ sat = = + =γd +⎜ ⎟γ w
1+ e 1+ e 1+ e ⎝1+ e ⎠

Rearranging, γ sat (1 + e) = γ d (1 + e) + e γ w

γ sat − γ d
Therefore, e =
γ d − γ sat + γ w



⎛ 1 + wsat ⎞ ⎛ 1 + wsat ⎞ e γ w (1 + wsat )n γ w
3.3 γ sat = ⎜ ⎟G s γ w = ⎜ ⎟ =
⎝ 1+ e ⎠ ⎝ 1+ e ⎠ wsat wsat


Rearranging, wsat (γ sat − n γ w ) = n γ w

nγw
Therefore, wsat =
γ sat − n γ w



W 12.5
3.4 a. γ = = = 125 lb/ft3
V 0.1

γ 125
b. γ d = = = 109.64 lb/ft3
1+ w 1 + 0.14

Gs γ w (2.71)(62.4)
c. e = −1 = − 1 = 0.54
γd 109.64

e 0.54
d. n = = = 0.35
1 + e 1 + 0.54



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, ( w)(Gs ) (0.14)(2.71)
e. S = = = 0.702 = 70.2%
e 0.54

(γ − γ d )V (125 − 109.64)(0.1)
f. Volume of water = = ≈ 0.024 ft3
γw 62.4



⎛1+ w ⎞ (1 + 0.098)( 2.69)(9.81)
3.5 a. γ = ⎜ ⎟Gs γ w ; 19.2 = ; e = 0.51
⎝ 1+ e ⎠ 1+ e

Gs γ w (2.69)(9.81)
b. γ d = = = 17.48 kN/m3
1+ e 1 + 0.51

( w)(Gs ) (0.098)(2.69)
c. S = = = 0.517 = 51.7%
e 0.51



(Gs + Se)γ w (2.69)(9.81) + (0.9)(0.51)(9.81)
3.6 a. γ = = = 20.45 kN/m3
1+ e 1 + 0.51

Water to be added = 20.45 – 19.2 = 1.25 kN/m3

(Gs + e)γ w (2.69 + 0.51)(9.81)
b. γ sat = = = 20.78 kN/m3
1+ e 1 + 0.51

Water to be added = 20.78 – 19.2 = 1.58 kN/m3



π ⎛ 1 ⎞ W 9.56
3.7 a. V = (2.8) 2 (22)⎜ 3 ⎟ = 0.078 ft3; γ = = = 122.56 lb/ft3
4 ⎝ 12 ⎠ V 0.078

W − Ws 9.56 − 8.51
b. w = = = 0.1233 = 12.33%
Ws 8.51

Ws 8.51
c. γ d = = = 109.1 lb/ft3
V 0.078




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© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.