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Merchant's Force Analysis - II

In an orthogo...

Question

In an orthogonal cutting test, the following observation were made
Cutting force $= 1200 N$
Thrust force $= 500 N$
Tool rake angle $= z e r o$
Cutting speed $= 1 m / s$
Depth of cut $= 0.8 m m$
Chip thickness $= 1.5 m m$

Friction angle during machining will be

22.6

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32.6

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Solution

The correct option is A $22.6$
$F_{C} = 12000 N, F_{T} = 500 N; α = 0$
$V_{C} = 1 m / s, d = b = 0.8, t_{2} = 1.5 m m$
$t a n (β - α) = \frac{F_{T}}{F_{C}}$
$β = α + t a n^{- 1} (\frac{F_{T}}{F_{C}}) = 0 + t a n^{- 1} (\frac{500}{1200}) = {22.62}^{o}$

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In an orthogonal cutting test, the following observation were made Cutting force =1200N Thrust force =500N Tool rake angle =zero Cutting speed =1m/s Depth of cut =0.8mm Chip thickness =1.5mm Friction angle during machining will be

The correct option is A 22.6FC=12000N,FT=500N;α=0 VC=1m/s,d=b=0.8,t2=1.5mm tan(β−α)=FTFC β=α+tan−1(FTFC)=0+tan−1(5001200)=22.62o

In an orthogonal cutting test, the following observation were made
Cutting force $= 1200 N$
Thrust force $= 500 N$
Tool rake angle $= z e r o$
Cutting speed $= 1 m / s$
Depth of cut $= 0.8 m m$
Chip thickness $= 1.5 m m$

Friction angle during machining will be

The correct option is A $22.6$
$F_{C} = 12000 N, F_{T} = 500 N; α = 0$
$V_{C} = 1 m / s, d = b = 0.8, t_{2} = 1.5 m m$
$t a n (β - α) = \frac{F_{T}}{F_{C}}$
$β = α + t a n^{- 1} (\frac{F_{T}}{F_{C}}) = 0 + t a n^{- 1} (\frac{500}{1200}) = {22.62}^{o}$