Calculation of Unit Weight ૪, ૪sat, ૪d

Problem 1:  Find the unit weight of saturated soil ૪_sat. Unit weight of dry soil ૪_d = 110 lb/ft³ and specific gravity G_s = 2.7?
Exam Name: PGCL 2017

Solution:
We know that,
               ૪_d = (G_s ૪_w)/ 1+e

        Here, ૪_d = 110 lb/ft³
                 Gs = 2.7
                  ૪_w= 62.4 lb/ft³

 ∴ 110 = (2.7 * 62.4)/ (1+e)
 ⇒    e = 0.53

And,
   ૪_sat = (G_s +e)૪_w/ 1+e
          = {(2.7 + 0.53)* 62.4}/ (1+0.53)
          = 131.8 lb/ft³

Answer: ૪_sat =131.8 lb/ft3

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Problem 2. In a saturated soil w= 30%, G=2.65. Determine dry unit weight.
Exam Name: Easteni Refinery Ltd. 2017 

Solution:
We know that,
     ૪_d = (G_s ૪_w)/ 1+e
          = (G_s ૪_w) / (1+wGs)

Here w = 30%
         G =2.65
       ૪_w = 62.4 lb/ft³
  ∴   ૪_d = (2.65* 62.4) / (1 + 0.3*2.65)
             = 92.12 lb/ft³

Answer: 92.12 lb/ft³

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Problem 3. In a field a hole is made, it's cutoff volume 1.1 ft3 and wet mass of hole is 130 lb, dry mass of hole 119 lb. Determine degree of saturation if specific gravity is 2.7
Exam Name: BWDB 2016

Solution:
      ૪ = W /V
         = 130 / 1.1
         = 118.18 lb/ft³

    ૪_d = 119/ 1.1
         = 108.18 lb/ft³

  ∴  ૪_d =  ૪ / 1+w
108.18 = 118.18 / (1 + w)
        w = 0.0924
            = 9.24 %
We know, 
            ૪_d = (G_s ૪_w)/ 1+e
      108.18= (2.7* 62.4) / (1 + e)
              e = 1.3
Again , eS = wG_s
         ∴   S = (0.0924 * 2.7) / 1.3
                  = 0.19

Answer: Degree of saturation 0.19

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Problem 4. A soil sample is dry weight 40 lb and moist weight 50 lb. Volume of soil 0.21 ft³. Find the NMC, bulk unit weight and dry unit weight?
Exam Name: Gas Transmission Company Ltd. 2016, PGCB 2015

Solution:  
w = (50 - 40) / 40
    = 0.25
    = 25 % (Ans.)

  ૪ = 50/ 0.21
     =  238.1 lb/ft³ (Ans.)

૪_d = 40 / 0.21 lb/ft³ (Ans.)

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Problem 5. Percent of void is 50% . If specific gravity of soil solid is 2.70 then calculate submerged unit  weight?
Exam Name: BPDB 2015

Solution:
Specific gravity,

 G= ρ_s / ρ_w
∴ 2.70 = ρ_s / 1000

 ρ_s = 2700 kg/m³
  ૪ = (2700 * 9.81) / 1000
     = 26.49 kN / m³

૪_sat =   ૪(1 + w)
       = 26.49 (1 + 0.5)
       = 39.73 kN / m³
   ૪' = ૪_sat - ૪_w
       = 39.73 - 9.81
       = 29.92 kN / m³
Answer: 29.92 kN / m³

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Problem 6. Void ratio and volume of soil is given 0.876 and 40 cm3. Determine the amount of volume of void and volume of solid?
Exam Name: WASA 2014


Solution:
Void ratio e= V_v / V_s
                  = V_v / (V - V_v)
        0.866 = V_v / (40 - V_v)
       ∴ V_v = 18.67 cm³(Ans.)

V_s = 40 - 18.67
     = 21.33 cm³(Ans.)

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