Forces in surface grinding Assume that you are performing a surface-grinding operation on a low-carbon steel workpiece using a wheel of diameter D = 200 mm that rotates at N = 3600 rpm. The width of cut is w = 20 mm, depth of cut is d = 0.04 mm, and the feed rate of the workpiece is v = 1200 mm/min

Question

Forces in surface grinding

Assume that you are performing a surface-grinding operation on a low-carbon steel workpiece using a wheel of diameter D = 200 mm that rotates at N = 3600 rpm. The width of cut is w = 20 mm, depth of cut is d = 0.04 mm, and the feed rate of the workpiece is v=1200 mm/min. Calculate the cutting force, Fc (the force tangential
to the wheel), and the thrust force, [latex]F_n [/latex](the force normal to the workpiece).

TABLE 9.3

Approximate Specific-Energy Requirements for Surface Grinding
Workpiece material	Hardness	Specific energy (W-s/[latex]mm^3[/latex])
Aluminum	150 HB	7–27
Cast iron (class 40)	215 HB	12–60
Low-carbon steel (1020)	110 HB	14–68
Titanium alloy	300 HB	16–55
Tool steel (T15)	67 HRC	18–82

Accepted Answer

We first determine the material removal rate as follows:

[latex]MRR = dwv = (0.04)(20)(1200) = 9600 \ mm^3/min.[/latex]

The power consumed is given by

[latex]Power = (u)(MRR),[/latex]

where [latex] u [/latex] is the specific energy, as obtained from Table 9.3. For low-carbon steel, let’s estimate [latex]u [/latex] to be 41 [latex]W-s/mm^3.[/latex] Hence,

Power = [latex](41)\left\lgroup\frac{9600}{60} ight group =6.56 [/latex] kW or [latex]6.56 [/latex] kJ/s.

Since power is defined as

Power = Tω,

where T is the torque and is equal to [latex](F_c) (D/2) [/latex] and ω is the rotational speed of the wheel in radians per minute, we also have ω = 2[latex]\pi [/latex] N. Thus,

[latex]6560=(F_c)\left[\frac{200 imes \frac{1}{1000} }{2} ight] (2\pi ) \left\lgroup\frac{3600}{60} ight group [/latex]

and therefore, [latex]F_c = 174 [/latex]N. The thrust force, [latex]F_n[/latex], can be calculated by noting from experimental data in the technical literature that it is about 30% higher than the cutting force, [latex]F_c[/latex]. Consequently,

[latex]F_n = (1.3)(174) = 226 [/latex]N

Question 9.2: Forces in surface grinding Assume that you are performing a ...

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