5.10

The maximum and minimum dry unit weights of a sand were determined in the laboratory to be 104 lb/ft³ and 93 lb/ft³, respectively. What would be the relative compaction in the field if the relative density is 78%? Given : maximum dry unit weight = 104 lb/ft3 minimum dry unit weight = 93 lb/ft3 relative density = 78% we know that, γd = Gγw/(1+e) where, G = specific gravity of soil γw = unit weight of water = 62.4 lb/ft3 e = void ratio γd = dry unit weight. we can observed that, γd is inversly propotional to void ratio,e. means, maximum γd at minimum 'e' and minimum γd at maximum 'e'. therefore, γd,max = Gγw/(1+emin) 104 = G*62.4/(1+emin) by solving this, we get ; emin= 0.6*G – 1 similarly, γd,min = Gγw/(1+emax) emax = 0.671*G – 1 formula for Relative Density, Dr = (emax - e)/(emax - emin) Dr = 78% = 0.78 put the all values in above equation,we get ; 0.78 = ((0.671*G - 1) - e)/((0.671*G - 1) - (0.6*G - 1)) e = 0.616*G – 1 this 'e' is void ratio of soil in the field. then, dry unit weight of soil in the field, γd = Gγw/(1+e) γd =G*62.4/(1+0.616*G-1) = G*62.4/(0.616*G) = 101.3 lb/ft3 Relative compaction, (R%) = (γd in the field/γd,max from the lab)*100. R% = (101.3/104)*100 = 97.4%