Question

Compaction test

Determine the degree of saturation and the air voids at maximum dry density. Discuss the reasons for the shape of the graph of dry density versus water content Determine the compaction energy of the standard Proctor test. Discuss what effect would have on the maximum dry density and optimum water content, and the degree of saturation and air voids at maximum dry density if modified Proctor compaction test is conducted. . mass of wet sample 106.49 98.45 105.37 111.2996.27 and container mass of dry sample 97.5589.42 96.88 101.54 88.02 and container Mass of container 22.97 23.3423.4423.32 25.84 mass of 8.94 9.03 9.75 8.49 water (g) mass of dry 4.5866.0873.44 78.22 62.18 sample water content 11.98713 13.66525 11.56046 12.46484 13.26793 dry density 0.079004 (g/cmA3 0.07 0.077797 0.08286 0.065869 container 944 cm3 volume

Answer 1

8236 3lem2 optimum ater Content (w) = 12.464 ey % Aiy Voids -5% at man .、 From-the relationship 6 Se=LJel Oy density may vavies withvespect to the watey ContentC%). D14 density may incveade4 incneakes and them deeneases. here the peac point 6 dvy density hoas oceudedl i man "Any density, and Water Content w t e mon ry hensiry i catrd as optimum watey Content

Water conten Vs Dry density 0.09 0.08 0.07 E 0.06 0.05 0.04 0.03 0.02 0.01 13.5 14 11.5 12 12.5 13 Water Content (%)

の Compaction Energy Standard PYOC toy te st ×Height dvop hammer п(53)(1273) 99930S mrn3 99930S ,'. 가 MOCA;lied Pocto Y test l's Conducted mum dvy density will increases cheivid at lower water Content i.e. and a Tr the mocified procto test ates Cont ent o be reqwired in less Buantity to acheive oM c Coptimum moistuve Cantent) Degyee satavation at manimum dry cdantis uoill be incyeates ard air voics percontage W1'll be decre des ik modibid Proctor text Concluctool