Calculation of material-removal rate, power required, and cutting time in face milling

Refer to Fig. 8.54, and assume that D = 160 mm, $w$ = 70 mm, $l$ = 450 mm, d = 3 mm, $v$ =0.5 m/min, and N = 120 rpm. The cutter has 10 inserts and the workpiece material is a high strength aluminum alloy. Calculate the material- removal rate, cutting time, and feed per tooth, and estimate the power required.

Question

Calculation of material-removal rate, power required, and cutting time in face milling

Refer to Fig. 8.54, and assume that D = 160 mm, [latex]w[/latex] = 70 mm, [latex]l[/latex]= 450 mm, d = 3 mm, [latex]v/latex] =0.5 m/min, and N = 120 rpm. The cutter has 10 inserts and the workpiece material is a high strength aluminum alloy. Calculate the material- removal rate, cutting time, and feed per tooth, and estimate the power required.

Accepted Answer

The cross section of the cut is [latex] wd[/latex] = [latex](70)(3) = 210 mm^2.[/latex] Since the workpiece speed [latex] v[/latex] is 0.5 m/min = 500 mm/min, the material-removal rate is

MRR=[latex] (210)(0.5 × 1000) = 105,000 \ mm^{3}[/latex]/min.

The cutting time is given by Eq. (8.44) as

[latex]t=\frac{l+2l_{c}}{v} . [/latex]

Note from Fig. 8.54 that for this problem,

[latex]l_{c}=\frac{D}{2} [/latex] or 80 mm

so that, thus, the cutting time is

[latex]t=\frac{(450+2 imes 80)}{0.5 imes 1000} =1.22 [/latex] min

We obtain the feed per tooth from Eq. (8.43).

[latex]f=\frac{v}{Nn}[/latex] (8.43)

Noting that N = 120 rpm,

[latex]f=\frac{0.5 imes 1000}{(360)(10)} =0.138 \ mm /tooth. [/latex]

For this material, the unit power can be taken from Table 8.3

Table 8.3,

Approximate Specific-Energy Requirements in Machining Operations
Material	Specific energy* (W-s/mm[latex]^{3}[/latex])
Aluminum alloys	0.4–1.1
Cast irons	1.6–5.5
Copper alloys	1.4–3.3
High-temperature alloys	3.3–8.5
Magnesium alloys	0.4–0.6
Nickel alloys	4.9–6.8
Refractory alloys	3.8–9.6
Stainless steels	3.0–5.2
Steels	2.7–9.3
Titanium alloys	3.0–4.1
*At drive motor, corrected for 80% efficiency; multiply the energy by 1.25 for dull tools.

to be 1.1 W-s/[latex] mm^{3} [/latex]; hence, the power can be estimated as

Power = [latex](1.1)\left\lgroup\frac{105,000}{60} ight group= 1925 [/latex] W = 1. 925 kW.

Question 8.6: Calculation of material-removal rate, power required, and cu...

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