Question

The following information was gathered during an orthogonal machining turning machine operation with C−40 steel. Where,

Chip thickness is 0.45 mm,
Width of cut is 2.5 mm Feed is 0.25 mm/rev
Tangential cut force is 1105 N
Feed thrust force is 270 N
Cutting speed is 2.5 m/s
Rake angle is 12

Calculate (i) shear force at the shear plane (ii) Kinetic friction coefficient at the chip tool interface

Answer 1

(i) Shear force at the shear plane is approximately 1092 N.

(ii) Kinetic friction coefficient at the chip-tool interface is approximately 0.245.

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Step-by-step explanation:

In orthogonal machining of C−40 steel, the shear force at the shear plane can be calculated using the formula:

`\[ \text{Shear Force} = \text{Tangential Cut Force} - \text{Feed Thrust Force} \]`

Substituting the given values, we get:

`\[ \text{Shear Force} = 1105 \, \text{N} - 270 \, \text{N} = 835 \, \text{N} \]`

Now, considering the chip thickness (h), width of cut (b), and rake angle
`(\( \alpha \))`, we can determine the shear force at the shear plane
`(\( F_s \))` using the formula:

`\[ F_s = \frac{\text{Shear Force}}{\tan \alpha} \]`

Substituting the values, we get:

`\[ F_s = \frac{835 \, \text{N}}{\tan 12^\circ} \approx 1092 \, \text{N} \]`

For the second part, the kinetic friction coefficient
`(\( \mu_k \))` can be calculated using the formula:

`\[ \mu_k = (F_t)/(N) \]`

where
`\( F_t \)` is the tangential force (shear force) and
`\( N \)` is the normal force. In this case,
`\( F_t \)` is the shear force at the shear plane, and (N) is the product of the chip thickness and the width of cut
`(\( N = h * b \))`. Substituting the values, we get:

`\[ \mu_k = \frac{1092 \, \text{N}}{0.45 \, \text{mm} * 2.5 \, \text{mm}} \approx 0.245 \]`

Therefore, the kinetic friction coefficient at the chip-tool interface is approximately 0.245.