The following results were obtained by a mechanical analysis. Classify the soil using the AASHTO classification system and give the group index. Sieve Analysis, % Finer than No. 10: 99%, than No. 40: 85%, than No. 200: 71%; liquid limit = 55; plastic limit = 21.
What will be an ideal response?
Use Table 17.1 to classify this soil.
Since more than 35% of the material is passing No. 200 sieve, the liquid limit is
greater than 41, the plasticity index is greater than 11, the soil is Group A-7.
Since the plasticity index (34) is greater than the liquid limit minus 30, the
subgroup is A-7-6.
Determine the Group Index using Equation 17.18.
GI = (F – 35)[0.2 + 0.005(LL – 40)] + 0.01(F – 15)(PI – 10)
GI = (71 – 35)[0.2 + 0.005(55 – 40)] + 0.01(71 – 15)(34 – 10)
GI = 23.34
Use GI = 23
The soil can be classified as A-7-6(23).
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