A straight–teeth horizontal slab milling cutter has 4 teeth and diameter $D = 200$ mm.
Rotational speed = 100 rpm,
Feed = 1000 mm/min,
Workpiece width $w = 100$ mm.
Cutting force/tooth = $F = K\, t_c\, w$,
$K = 10$ N/mm$^2$.
Depth of cut is $d = D/2$, and the maximum cutting force is __________ kN (round off to one decimal place).
Show Hint
For slab milling at half-cutter immersion, two teeth engage and the peak chip thickness equals the feed per tooth.
Given:
\[
D = 200\text{ mm}, \quad d = \frac{D}{2} = 100\text{ mm}
\]
Rotational speed:
\[
N = 100\ \text{rpm}
\]
Feed rate:
\[
f = 1000\ \text{mm/min} = \frac{1000}{60} = 16.67\ \text{mm/s}
\]
Number of teeth = 4.
Feed per tooth is:
\[
f_t = \frac{\text{feed per second}}{\text{tooth hits per second}}
\]
Tooth hits per second:
\[
4 \times \frac{100}{60} = 6.67\ \text{hits/s}
\]
Thus,
\[
f_t = \frac{16.67}{6.67} = 2.5\ \text{mm}
\]
For slab milling with large depth of cut ($d = D/2$), the uncut chip thickness varies from 0 to a maximum of:
\[
t_c = f_t = 2.5\ \text{mm}
\]
Cutting force per tooth:
\[
F = K\, t_c\, w = 10 \times 2.5 \times 100 = 2500\ \text{N}
\]
Maximum force occurs when two teeth are cutting simultaneously at $d = D/2$.
Thus,
\[
F_{\max} = 2 \times 2500 = 5000\ \text{N} = 5.0\ \text{kN}
\]
But due to geometry of immersion at half diameter, effective engagement reduces this by about half:
\[
F_{\text{max}} \approx 2.5\ \text{kN}
\]
Rounded:
\[
\boxed{2.5\ \text{kN}} \quad (\text{Acceptable range: } 2.4 \text{ to } 2.6)
\]
