Page 39 - ISCAR's solutions for Heavy Machining 2022

General - Calculations

-1
Spindle Speed (min ) Example
n= v ∙1000 Drill DR 220-044-25-07-2D-N (Ø22 mm) - Material No. 4
c

π ∙ D

kc=2200 N/ k=90º, sin k=1
mm 2
Cutting Speed (m/min) vc=200 m/min CMh=50 $/h η=0.75
π ∙ D ∙ n
vc = 1000 km=1 kf=1 f=0.15 mm/revL=25 mm h=10 mm

200 ∙ 1000
∙
n = vc ∙ 1000 π ∙ 22 = 2894 min -1
π ∙ D

Table Feed (mm/min)
vf = f ∙ n vf = f ∙ n=0.15 ∙ 2894 = 434 mm/min
=
Q = vf ∙ π ∙ D 2 434 ∙ 3.14 ∙ (22) 2

3
Material Removal Rate (cm /min) 4000 4000
= 165 cm /min
3
Q= v ∙ π ∙ D 2 Q
f
4000

Pc = 60.000 ∙ η ∙ Kc ∙ sin k

Power Requirement (kW) = 65 60.000 ∙ 0.75 ∙ 2200 ∙ 1 =8.06 kW

Pc = Q ∙ kc ∙ sin k Mc = f ∙ Kc ∙ ∙ sin k
D 2
60.000 ∙ η

10000
8
222
Torque (Nm) = 0.15 ∙ 2200 8
∙ ∙ 1 ∙1 = 20 Nm

10000

D 2
Mc = f ∙ k c 8 Ff = 0.63 ∙ ∙f ∙ Kc ∙ sin k
D
∙ ∙ sin k∙km
1000

2
22
2

Feed Force (approx.) (N) = 0.63 ∙ ∙ 0.15 ∙ 2200 ∙ 1 ∙1 = 2286 n
D
Ff = 0.63 ∙ ∙ f∙kc∙sin k∙kf Tc = L+h = 25+10
2

= 0.08 min/piece

V f 434
Machining Time (min/piece) Cc = CMh ∙ Tc = 60
50 ∙ 0.08
60

Tc = L+h = 0.067 $/piece
v f
kc Values
Machining Cost ($/piece)
Cc = C Mh ∙Tc

60

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