Answer:
a) the metal removal rate is 14.4 in³/min
b) the cutting time is 0.98 min
Explanation:
Given the data from the question
first we find the rpm for the spindle of the drilling tool, using the equation
Ns = 12V/πD
V is the cutting speed(120 fpm) and D is the diameter of the hole( 2 in)
so we substitute
Ns = 12 × 120 / π2
Ns = 1440 / 6.2831
Ns = 229.18 rmp
Now we find the metal removal rate using the equation
MRR = (πD²/4) Fr × Ns
Fr is the feed rate( 0.02 ipr ),
so we substitute
MRR = ((π × 2²)/4) × 0.02 × 229.18
MRR = 14.3998 ≈ 14.4 in³/min
Therefore the metal removal rate is 14.4 in³/min
Next we find the allowance for approach of the tip of the drill
A = D/2
A = 2/2
= 1 in
now find the time required to drill the hole
Tm = (L + A) / (Fr × Ns)
Lis the the depth of the hole( 3.5 in)
so we substitute our values
Tm = (3.5 + 1) / (0.02 × 229.18 )
Tm = 4.5 / 4.5836
Tm = 0.98 min
Therefore the cutting time is 0.98 min
