SME
Manufacturing Processes
Machining
Use this for “A 50 mm steel bar is turned at 150 m/min, feed 0.25 mm/rev and depth 2 mm, over 200 mm. Find the spindle speed, removal rate and cutting time.”
Diagram
v = πDN = π × 0.05 m × 955 rpm
v = 150 m/min
MRR = vfd = 75 cm³/min.
Tₘ
=
L
fN
=
0.838
min.
Depth and feed are enlarged in the detail.
This is an example. Type your own values to start a new problem.
v (m/min)
=
π
×
D (mm)
×
N (rpm)
MRR (cm³/min)
=
v (m/min)
×
f (mm/rev)
×
d (mm)
Tₘ (min)
=
L (mm)
f (mm/rev)
×
N (rpm)
Formulas
v
=
π
×
D
×
N
1,000
150
=
π
×
50
×
954.9297
1,000
MRR
=
v
×
f
×
d
75 = 150 × 0.25 × 2
Tₘ
=
L
f
×
N
0.8378
=
200
0.25 × 954.9297
Assumptions
•
v is the surface speed at the uncut diameter D, so N = v ÷ (πD) with D in meters.
•
One pass along L at constant feed f and depth d; 1 cm³ = 1000 mm³.
•
The cutting time leaves out the tool’s approach and overtravel.
Step-by-step
WE KNOW
D = 50 mm, v = 150 m/min, f = 0.25 mm/rev, d = 2 mm, L = 200 mm
FIND
spindle speed (N), removal rate (MRR), cutting time (Tₘ)
Step 1 · Find spindle speed (N)
Use v = π × D × N ÷ 1,000
The cutting speed in mm each minute, shared over one circumference πD.
N
=
1,000
×
v
π ×
D
N
=
1,000
×
150
π × 50
N
=
150,000
50π
N
=
954.9297
rpm
Step 2 · Find removal rate (MRR)
Use MRR = v × f × d
Speed × feed × depth: m/min × mm × mm is 1000 mm³/min, which is 1 cm³/min.
MRR = 150 × 0.25 × 2
MRR = 37.5 × 2
MRR = 75 cm³/min
Step 3 · Find cutting time (Tₘ)
Use Tₘ = L ÷ (f × N)
The
tool
moves
f
along
the
bar
each
turn
and
N
turns
each
minute.
Tₘ
=
200
0.25 × 954.9297
Tₘ
=
200
238.7
Tₘ = 0.8378 min
Check
Put the numbers back in. Both sides should match.
150
=
π
×
50
×
954.9297
1,000
✓
75 = 150 × 0.25 × 2 ✓
0.8378
=
200
0.25 × 954.9297
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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