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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 $= 1.5 m m$

Chip speed along the tool rake face will be

0.83 m / s

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0.53 m / s

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1.2 m / s

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1.88 m / s

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Solution

The correct option is B $0.53 m / s$
Taking $d = t_{1} = 0.8$
$r = \frac{t_{1}}{t_{2}} = \frac{0.8}{1.5} = 0.533$

Chip thickness ratio:
$r = \frac{V_{f}}{V_{c}}$

Velocity of the chip along the rake face of the tool:
$V_{f} = r V_{c} = r V_{c}$
$= 0.533 \times 1 = 0.533 m / s$

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Similar questions

Q. In an orthogonal cutting test, the following observation were made
Cutting force

= 1200 N

Thrust force

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Tool rake angle

= z e r o

Cutting speed

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Depth of cut

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Chip thickness

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Friction angle during machining will be

Q. In an orthogonal cutting test on mild steel, the following data were obtained
Cutting speed :

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Depth of cut :

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Total rake angle :

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Thrust force :

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Using Merchant's analysis, the friction angle during the machining will be

Q. In an orthogonal cutting test with a tool of rake angle 15 deg, the following observations were made: chip thickness ratio = 0.37, horizontal cutting force = 1000 N, vertical cutting force = 1500 N.

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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 $= 07 m m$
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Assume Merchant's theory.

The coefficient of friction at the tool-chip interface is

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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 Chip speed along the tool rake face will be

The correct option is B 0.53m/sTaking d=t1=0.8 r=t1t2=0.81.5=0.533 Chip thickness ratio: r=VfVc Velocity of the chip along the rake face of the tool: Vf=rVc=rVc =0.533×1=0.533m/s

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 $= 1.5 m m$

Chip speed along the tool rake face will be

The correct option is B $0.53 m / s$
Taking $d = t_{1} = 0.8$
$r = \frac{t_{1}}{t_{2}} = \frac{0.8}{1.5} = 0.533$

Chip thickness ratio:
$r = \frac{V_{f}}{V_{c}}$

Velocity of the chip along the rake face of the tool:
$V_{f} = r V_{c} = r V_{c}$
$= 0.533 \times 1 = 0.533 m / s$