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Analysis of Orthogonal Machining

Common Data:I...

Question

Common Data:
In an orthogonal machining operation:
Uncut thickness $= 0.5 m m$
Cutting speed $= 20 m / m i n$
Rake angle $= 15^{o}$
Width of cut $= 5 m m$
Chip thickness $= 07 m m$
Thrust force $= 200 N$
Cutting force $= 1200 N$
Assume Merchant's theory.

The coefficient of friction at the tool-chip interface is

0.23

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0.46

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0.85

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0.95

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Solution

The correct option is B $0.46$
We know that

$μ = \frac{F}{N} = \frac{F_{C} s i n α + F_{T} c o s α}{F_{C} c o s α - F_{T} s i n α}$

$= \frac{1200 s i n 15^{o} + 200 c o s 15^{o}}{1200 c o s 15^{o} - 200 s i n 15^{o}}$

$μ = \frac{F}{N} = \frac{F_{C} s i n α + F_{T} c o s α}{F_{C} c o s α - F_{T} s i n α}$

$= \frac{310.6 + 193.2}{1159.1 - 51.76}$

$= 0.46$

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Common Data: In an orthogonal machining operation: Uncut thickness =0.5mm Cutting speed =20m/min Rake angle =15o Width of cut =5mm Chip thickness =07mm Thrust force =200N Cutting force =1200N Assume Merchant's theory. The coefficient of friction at the tool-chip interface is

The correct option is B 0.46We know that μ=FN=FCsinα+FTcosαFCcosα−FTsinα =1200sin15o+200cos15o1200cos15o−200sin15o μ=FN=FCsinα+FTcosαFCcosα−FTsinα =310.6+193.21159.1−51.76 =0.46