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Sagot :
Answer:
a) 24.07 m
b) 4 m
c) 14 number of drops
d) p = number of passes
e) Dcd = 2.27
0.69 m
Explanation:
Given data:
Depth ( D )= 7.6 m below ground surface
dynamic compaction ( w ) = 15-ton , diameter of tamper = 2.0 m , thickness = 1.4 m
Determine :
A) drop height ( H )
D = n √wH
therefore H = 361 / 15 = 24.07 m
where : D = 7.6 m , n = 0.4 , w = 15
B) Drop spacing
drop spacing = average of ( 1.5 to 2.5 ) * diameter of tamper
= 2 * 2.0m = 4 m
C) number of drops
since the applied energy for fine grained soils and day fills range from 250 - 350 kj/m^2 the number of drops can be calculated using the relation below
AE = [tex]\frac{NWHP}{SPACING ^2}[/tex]
w = 15, H = 24.07 , Np = ? , AE = 300 kj/m^2
∴ Np = 4800 / 361.05 = 13.3
the number of drops at one pass = 14
D) number of passes
p = number of passes
E) estimated crater depth and settlement
crater depth ( Dcd ) = 0.028 [tex]N_{d} ^{0.55} \sqrt{wtIt}[/tex]
Nd = 14 , wt = 15, It = 24.07
therefore : Dcd = 2.27
estimate settlement is within 3 to 5% therefore the improved settlement
= 2.27 * 0.04 * 7.6 = 0.69 m
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