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Following data correspond to an orthogonal turning of a 100 mm diameter rod on a lathe. Rake angle: +15o; Uncut chip thickness: 0.5 mm; nominal chip thickness after the cut: 1.25 mm. The shear angle (in degrees) for this process is _________ (correct to two decimal places).
    Correct answer is '23.26'. Can you explain this answer?
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    Given data:


    • Diameter of the rod (D) = 100 mm

    • Rake angle (α) = 15°

    • Uncut chip thickness (t₀) = 0.5 mm

    • Nominal chip thickness after the cut (t₁) = 1.25 mm



    Calculation:


    1. Shear plane angle (φ):

    The shear plane angle (φ) can be calculated using the following formula:

    φ = tan⁻¹((t₀ - t₁)/(πD))

    Substituting the given values:

    φ = tan⁻¹((0.5 - 1.25)/(π * 100))

    φ = tan⁻¹((-0.75)/(3.14 * 100))

    φ = tan⁻¹(-0.00239)

    φ ≈ -0.14° (rounded to two decimal places)


    2. Shear angle (β):

    The shear angle (β) can be calculated using the following formula:

    β = α - φ

    Substituting the given values:

    β = 15° - (-0.14°)

    β = 15.14° (rounded to two decimal places)


    Therefore, the shear angle for this process is approximately 15.14° (rounded to two decimal places).
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    Following data correspond to an orthogonal turning of a 100 mm diameter rod on a lathe. Rake angle: +15o; Uncut chip thickness: 0.5 mm; nominal chip thickness after the cut: 1.25 mm. The shear angle (in degrees) for this process is _________ (correct to two decimal places).Correct answer is '23.26'. Can you explain this answer?
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