Machining: speed, feed and the life of a tool edge
引入| English | 释义 | 拼音 |
|---|---|---|
| casting/ˈkæstɪŋ/ | 铸造 | zhù zào |
| machining | 切削加工 | qiē xiāo jiā gōng |
| cutting speed | 切削速度 | qiē xiāo sù dù |
| feed | 进给量 | jìn jǐ liàng |
| depth of cut | 切削深度 | qiē xiāo shēn dù |
| material removal rate | 材料去除率 | cái liào qù chú lǜ |
| Taylor's tool-life equation | 泰勒刀具寿命公式 | tài lēi dāo jù shòu mìng gōng shì |
Microns no casting can give
- A bearing seat needs $\pm10$ µm and a fine finish. Machining removes metal to get there, and every second and every tool edge costs money.
- In turning, three settings control the cut: cutting speed $v = \pi DN$, feed $f$ per revolution, and depth of cut $d$.
Rate, power and time
- Material removal rate $\mathrm{MRR} = vfd$; power $P = u\cdot\mathrm{MRR}$ with specific energy $u$; time per pass $t = L/(fN)$.
- The tool's nose radius leaves scallops, so roughness is $R_a \approx f^2/(32r)$: halve the feed and the roughness falls to a quarter, but the time doubles.
Which setting moves it?
Link each outcome to the cutting setting that controls it.
Roughness with f = 0.25 mm and nose radius 0.8 mm (two decimals, micrometres)?
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f squared over 32 r.
Halving the feed makes the roughness:
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R_a grows with f squared.
One pass on the shaft
- A 50 mm bar at $v = 150$ m/min: $N = 1000v/\pi D = \mathbf{955}$ rpm.
- With $f = 0.25$ mm and $d = 2$ mm: MRR $= \mathbf{1250}\ \mathrm{mm^3/s}$, and with $u = 2.8\ \mathrm{J/mm^3}$ the power is $\mathbf{3.5}$ kW.
- Over 200 mm: $t = 200/(0.25 \times 955) = \mathbf{0.84}$ min; with $r = 0.8$ mm, $R_a \approx \mathbf{2.44}$ µm.
Spindle speed for a 50 mm bar at 150 m/min (rpm, whole number)?
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1000 v over pi D.
Power with MRR = 1250 mm3/s and u = 2.8 J/mm3 (one decimal, kW)?
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u times MRR = 3500 W.
Taylor's trade
- Taylor's tool-life equation $vT^n = C$: faster cutting shortens tool life sharply. With $n = 0.25$ and $C = 400$, $T = 50.6$ min at 150 m/min but only $\mathbf{16.0}$ min at 200 m/min.
- Each pass is 25% faster, but an edge now lasts 25 shafts instead of about 60. Whether it pays depends on tool-change cost against machine time.
Tool life at 200 m/min with n = 0.25 and C = 400 (one decimal, min)?
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(400/200) to the fourth.
A higher cutting speed always lowers the cost per part.
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It shortens tool life sharply; the balance depends on tool-change cost.
$n = 0.25$, tool change 2 min, edge €4, machine €1/min: cost-optimal tool life (min)?
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$T_{opt} = (1/0.25 - 1)(2 + 4/1) = 3 \times 6 = 18$ min.
With $C = 400$: the matching cutting speed (m/min, nearest whole)?
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$v = 400/18^{0.25} = 400/2.06 = 194$ m/min.
The cheapest tool life
- Machine time costs $C_m$ per minute; each edge costs $C_t$ and $t_c$ minutes to change. Minimising cost per part gives $T_{opt} = (1/n - 1)(t_c + C_t/C_m)$.
- With $n = 0.25$, $t_c = 2$ min, $C_t = €4$ and $C_m = €1$/min: $T_{opt} = 3 \times 6 = \mathbf{18}$ min, so $v_{opt} = 400/18^{0.25} = \mathbf{194}$ m/min.
At 200 m/min: 0.63 min machining at €1/min plus €6 of tooling shared over 25 shafts. Cost per shaft (two decimals, €)?
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$0.63 + 6/25 = 0.63 + 0.24 = 0.87$.
Worked example: 150 or 200 m/min?
- At 150 m/min: a pass is 0.84 min and an edge lasts 60 shafts, so each shaft carries $(2 + 4)/60 = €0.10$ of tooling. Total $€0.94$.
- At 200 m/min: 0.63 min and 25 shafts per edge, $6/25 = €0.24$. Total $€0.87$, cheaper, and close to the 194 m/min optimum. Faster pays here because machine time is dearer than edges.
Three machining slips
- The finished diameter for $N$. Use the diameter being cut.
- m/min mixed with mm/s in MRR. Convert before multiplying.
- Faster is always cheaper. Tool life falls with the fourth power here.
Recap
- $v = \pi DN$, MRR $= vfd$, $P = u\cdot$MRR, $t = L/(fN)$: 955 rpm, 3.5 kW, 0.84 min.
- $R_a \approx f^2/32r = 2.44$ µm.
- $vT^n = C$: 50.6 min at 150 m/min, 16.0 min at 200.