Your boss needs to recommend either the in-situ soil or the borrow material. Although you do not have much background on soil be
havior, what is your recommendation?
In providing your recommendation, you want to consider constructability issues (such as ease of compaction, moisture content). Without doing any additional tests, and based on your experience with soils, what compaction conditions would you suggest to achieve low permeability? What compaction conditions would you recommend to reduce soil expansion (swelling)?
Table 1 Compaction Characteristics of Selected Soils.
Borrow Soil In-Situ Soil
STANDARD MODIFIED STANDARD MODIFIED
PROCTOR PROCTOR PROCTOR PROCTOR
Dry Unit Water Dry Unit Water Dry Unit Water Dry Unit Water
Weight Content Weight Content | Weight Content Weight content
Yd (kN/m³) w(%) Yd (kN/m³) w(%) Yd (kN/m³) w(%) Yd (kN/m³) w(%)
16.68 8.4 17.58 5.7 15.63 12.7 15.72 10.4
18.79 9.7 20.02 7.4 17.72 14.4 18.76 12.2
S3? S3? 20.60 9.3 17.46 16.0 18.64 14.0
18.62 13.4 19.62 11.3 17.15 16.7 18.15 15.0
18.05 15.0 18.93 13.0 16.30 18.2 17.48 16.4
In providing your recommendation, you want to consider constructability issues (such as ease of compaction, moisture content). Without doing any additional tests, and based on your experience with soils, what compaction conditions would you suggest to achieve low permeability? What compaction conditions would you recommend to reduce soil expansion (swelling)?
Table 1 Compaction Characteristics of Selected Soils.
Borrow Soil In-Situ Soil
STANDARD MODIFIED STANDARD MODIFIED
PROCTOR PROCTOR PROCTOR PROCTOR
Dry Unit Water Dry Unit Water Dry Unit Water Dry Unit Water
Weight Content Weight Content | Weight Content Weight content
Yd (kN/m³) w(%) Yd (kN/m³) w(%) Yd (kN/m³) w(%) Yd (kN/m³) w(%)
16.68 8.4 17.58 5.7 15.63 12.7 15.72 10.4
18.79 9.7 20.02 7.4 17.72 14.4 18.76 12.2
S3? S3? 20.60 9.3 17.46 16.0 18.64 14.0
18.62 13.4 19.62 11.3 17.15 16.7 18.15 15.0
18.05 15.0 18.93 13.0 16.30 18.2 17.48 16.4
1 answer:
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Answer:
Maximum number that can be represented by 13 bits is 8192 Instructions
Explanation:
number of instructions = 1000
number of bits = log(1000) x number of register
= 6 bits
Since the complete instruction must have 32 bits, then
remaining number of bits = 32 - 6 = 236
number of registers in instruction = 2
number of bits per register = 26/2 = 13
Maximum number that can be represented by 13 bits =
= 2¹³ = 8192