Respuesta :
Answer:
159.1 ton
Explanation:
The solution is shown in the attached file
Answer:
The safe load is 159 ton
Explanation:
The safe load is equal to:
[tex]L=\frac{WHBV^{2/3} }{K}[/tex]
Where:
W = weight of hammer = 2.5 ton
H = drop height = 22 ft
B = number of blows used to drive the last batch = 36/4 = 9 ft³
K = dimensionless constant = 28
V = uncompacted volume = 27 ft³
Replacing values:
[tex]L=\frac{2.5*22*9*(27^{2/3}) }{28} =159ton[/tex]
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