Machine Tool Guideways A milling machine is used in factories and metalworking shops to cut slots and grooves in metal workpieces (Figure 6.8). The material to be machined is held and moved beneath a rapidly rotating cutting tool. The workpiece and its holder slide over smooth guideways that are

Question

Machine Tool Guideways

A milling machine is used in factories and metalworking shops to cut slots and grooves in metal workpieces (Figure 6.8). The material to be machined is held and moved beneath a rapidly rotating cutting tool. The workpiece and its holder slide over smooth guideways that are lubricated with oil having a viscosity of 240 cP. The two guideways are each of length 40 cm and width 8 cm (Figure 6.9). While setting up for a particular cut, a machinist disengages the drive mechanism, applies 90 N of force to the table holding the workpiece, and is able to push it 15 cm in 1 s. Calculate the thickness of the oil film between the table and the guideways.

Approach
When the machinist pushes the table, the oil film is sheared in a manner similar to that shown in Figure 6.7. The table is pushed at the speed ν = (0.15 m)/(1 s) = 0.15 m/s. The area of contact between the table and the
guideways is A = 2(0.08 m)(0.4 m) = 0.064 [latex]m^2[/latex]. We will calculate the film thickness by applying Equations (6.1) and (6.2)

[latex] au =\frac{F}{A} [/latex] (6.1)

[latex] au =\mu \frac{\upsilon }{h} [/latex] (6.2)

to relate the force, speed, contact area, and film thickness.

Accepted Answer

We first convert the oil’s viscosity to dimensionally consistent units using the definition of the unit centipoise:

[latex]\mu =(240cP)\left\lgroup0.001\frac{kg/(m.s)}{cP} ight group [/latex]

[latex]=0.24(\cancel{cP})\left\lgroup\frac{kg/(m.s)}{\cancel{cP}} ight group[/latex]

[latex]=0.24\frac{kg}{m.s} [/latex]

By applying Equation (6.1), the shear stress in the layer of oil is

[latex] au =\frac{90N}{0.064m^2} \longleftarrow \left[ au =\frac{F}{A} ight] [/latex]

[latex]=1406\frac{N}{m^2} [/latex]

The thickness of the oil film then follows from Equation (6.2):

[latex]h=\frac{(0.24kg/(m.s))(0.15 m/s)}{1406N/m^2} \longleftarrow [ au =\mu \frac{\upsilon }{h}][/latex]

[latex]=2.56 imes 10^{-5}\left(\frac{kg}{m.s} ight)(\frac{m}{s} )(\frac{m^2}{N} ) [/latex]

[latex]=2.56 imes 10^{-5}\left(\frac{\cancel{kg}}{\cancel{m}.\cancel{s}} ight)(\frac{m}{\cancel{s}} )(\frac{\cancel{m^2}.\cancel{s^2}}{\cancel{kg}.\cancel{m}} ) [/latex]

[latex]=2.56 imes 10^{-5}m[/latex]

Since this is a small numerical value, we apply an SI prefix from Table 3.3

Table 3.3 Order-of-Magnitude Prefixes in the SI

Name	Symbol	Multiplicative Factor
tera	T	[latex]1,000,000,000,000 = 10^{12}[/latex]
giga	G	[latex]1,000,000,000 = 10^9[/latex]
mega	M	[latex]1,000,000 = 10^6[/latex]
kilo	k	[latex]1000 = 10^3[/latex]
hecto	h	[latex]100 = 10^2[/latex]
deca	da	[latex]10 = 10^1[/latex]
deci	d	[latex]0.1 = 10^{-1}[/latex]
centi	c	[latex]0.01 = 10^{-2}[/latex]
milli	m	[latex]0.001 = 10^{-3}[/latex]
micro	[latex]μ[/latex]	[latex]0.000,001 = 10^{-6}[/latex]
nano	n	[latex]0.000,000,001 = 10^{-9}[/latex]
pico	p	[latex]0.000,000,000,001 = 10^{-12}[/latex]

to represent a factor of one-millionth. The thickness becomes

[latex]h=(2.56 imes 10^{-5}m)(10^6\frac{\mu m}{m} ) [/latex]

[latex]=25.6(\cancel{m}) (\frac{\mu m}{\cancel{m}} )[/latex]

[latex]=25.6\mu m[/latex]

Discussion
When compared to the thickness of a human hair (approximately 70–100 μm in diameter), the oil film is thin indeed, but not atypical of the amount of lubrication present between moving parts in machinery. By inspecting Equation (6.2), we see that the shear stress is inversely proportional to the oi film’s thickness. With only half as much oil, the table would be twice as hard to push.

[latex]h [/latex]= [latex]25.6μm[/latex]

Question 6.1: Machine Tool Guideways A milling machine is used in factorie...

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