Question 7.3: Compute the compactive effort (CE) for a Standard Proctor Co...
Compute the compactive effort (CE) for a Standard Proctor Compaction Test (ASTM D698).
From Equation (7.3) and based on data from Table 7.1, the compactive effort (CE) for a standard Proctor compaction test can be obtained as:
\gamma _{o} =\frac{\gamma _{m} }{1+\omega } (7.3)
CE=\frac{\left(24.5 N\right)\left(0.305 m\right) \left(3 layers\right) \left(25 blow/layer\right) }{\left(0.000942 m^{3} \right) \left(1000\right) } =594.8 kJ/m^{3}
In SI units, the compactive energy equals to kilo-joules per cubic meter (kJ/m^{3}). In fps unit, 1 ft lb/ft^{3} = 0.04796 kJ/m^{3}, then
CE = 594.8 / 0.04796 = 12,400 ft lb/ft^{3}The compactive effort of compaction in the field, is often related to the number of passes of rolling equipment (e.g. a steel drum or sheepsfoot roller).Figure 7.3 presents the interrelationship of dry unit weight, optimum moisture content, and compactive effort.
| Table 7.1 Standard and modified laboratory compaction test procedure and their corresponding compactive efforts | ||
| Variables | Standard compaction (ASTM D698; AASHTO T99) |
Modified compaction (ASTM D1557; AASHTO T188) |
| Mold size | 4 in. (10.16 cm) | 4 in. (10.16 cm) or 6 in. (15.24 cm) |
| Volume of mold | 1/30 ft^{3} (0.000942 m^{3}) |
1/30 or 1/13.33 ft^{3} (0.000942 or 0.00212 m^{2}) |
| Hammer weight | 5 lb (24.5 N) | 10 lb (44.5 N) |
| Height of drop | 12 in. (0.305 m) | 18 in. (45.7 cm) |
| Layers | 3 | 5 |
| No. of blows | 25 | 55 |
| Soil | > #4 Sieve | > #4 Sieve |
| Compactive effort | 595 kJ/ m^{3} (12,400 lb ft/f t^{3}}) |
2698 kJ/ m^{3} (55,250 lb ft/f t^{3}) |
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