Material-removal rate and cutting force in turning

A 125 mm-long, 10 mm-diameter 304 stainless-steel rod is being reduced in diameter to 9 mm by turning on a lathe. The spindle rotates at N = 360 rpm, and the tool is traveling at an axial speed of 1.75 mm/min. Calculate the cutting speed, material-removal rate, cutting time, power dissipated, and cutting force.

Question

Material-removal rate and cutting force in turning

A 125 mm-long, 10 mm-diameter 304 stainless-steel rod is being reduced in diameter to 9 mm by turning on a lathe. The spindle rotates at N = 360 rpm, and the tool is traveling at an axial speed of 1.75 mm/min. Calculate the cutting speed, material-removal rate, cutting time, power dissipated, and cutting force.

Accepted Answer

The cutting speed is the tangential speed of the workpiece. The maximum
cutting speed is at the outer diameter, [latex] D_{o}[/latex], and is obtained from the expression

[latex]V = \pi D_{\omicron }N.[/latex]

Thus

[latex]=(\pi )(10)(1000)(360) = 11.3[/latex] m/min.

The cutting speed at the machined diameter is

[latex]=\pi \left\lgroup\frac{9}{1000} ight group (360)=10.2[/latex] m /min.

From the information given, note that the depth of cut is

[latex]d=\frac{(10-9)}{2}=0.5 [/latex] mm

and the feed is

[latex]f=\frac{175}{360} =0.49[/latex] mm/ rev.

Thus, according to Eq. (8.38),

[latex]MRR=\pi D_{avg}dfN.[/latex] (8.38)

the material-removal rate is

MRR =[latex]\pi (9.5)(0.5)(0.49)(360)= 2610.08[/latex] [latex]mm^{3}[/latex]/min.

The actual time to cut, according to Eq. (8.39)

[latex]t=\frac{l}{fN}. [/latex] (8.39)

is

[latex]t=\frac{125}{(0.5) (360)}=0.70 [/latex] min.

We can calculate the power required by referring to 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.

and taking an average value for stainless steel as 4 W-s/[latex]mm^3[/latex]. Therefore, the power dissipated is

Power = [latex](4)\left\lgroup\frac{2610}{60} ight group=178 [/latex] N-m s,

and since the power dissipated is 178 N-m/s. The cutting force, [latex]F_{c}[/latex], is the tangential force exerted by the tool. Since power is the product of torque, T, and rotational speed in radians per unit time, we have

[latex]T=\frac{(178)}{\left\lgroup\frac{360}{60} ight group (2\pi )}= 4.72 [/latex] N-m.

Since [latex]T=(F_{c})(D_{avg}/2)[/latex], we obtain

[latex]F_{c}=\frac{(4.72)(2) imes 1000}{9.5}=994 [/latex] N.

Question 8.4: Material-removal rate and cutting force in turning A 125 mm-...

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