Question: An orthogonal cutting operation is being carried out under the following conditions:

• Cutting speed = 2 m/s
• Depth of cut = 0.5 mm
• Chip thickness = 0.6 mm

Calculate the chip velocity. [ESE 2003]

Solution: To solve this question we need to assume two things—(i) orthogonal rake angle (γ_O) of the cutting tool is 0° as it is the case of orthogonal cutting, and (ii) the given depth of cut is basically the uncut chip thickness. This problem can be solved with the help of velocity triangle in machining; however, there exists a simple way to solve this, as discussed below.

During machining, no material loss or gain takes place. Hence the rate of volume flow (i.e. MRR measured in mm³/s) of the workpiece material remains constant before and after cut. If the uncut chip thickness is a₁, the width of uncut chip is b₁, the cutting velocity is V_c, the chip thickness is a₂, the width of uncut chip is b₂, and the chip flow velocity is V_f, then applying the constant MRR approach, the following expression can be written for before and after cut.

\[{a_1}{b_1}{V_C} = {a_2}{b_2}{V_f}\]

During machining as the tool compresses a thin layer of workpiece material, the later undergoes shearing and thus thickness of chip becomes larger than the same before shearing (i.e. a₂ > a₁). Width of the chip can be considered as constant before and after shearing (i.e. b₁ = b₂) as no force is applied in this direction (so no deformation in width). Thus the above expression can be reduced to as follows.

\[{a_1}{V_C} = {a_2}{V_f}\]

\[\frac{{{a_1}}}{{{a_2}}} = \frac{{{V_f}}}{{{V_C}}}\]

\[{V_f} = \frac{{{V_C}{a_1}}}{{{a_2}}}\]

Calculate the chip velocity

Given, cutting speed (V_c) = 2 m/s

Uncut chip thickness (a₁) = 0.5 mm

Chip thickness (a₂) = 0.6 mm

Therefore, the chip velocity be:

\[{V_f} = \frac{{2 \times 0.5}}{{0.6}} = 1.67\]

Hence, the chip velocity for the given case is 1.67m/s.