In a turning operation, spindle speed is set to provide a cutting speed of 1.8 m/s. The feed and depth of cut are 0.30 mm and 2.6 mm, respectively. The tool rake angle is 8°. After the cut, the deformed chip thickness is measured to be 0.49 mm.
Determine:
(a) shear plane angle, (b) shear strain, and (c) material removal rate.
Use the orthogonal cutting model as an approximation of the turning process.

(a) shear plane angle (∅) = 33.53°

(b) shear strain (y) = 1.989

(c) material removal rate (MRR) = 1404mm³/s

Explanation:

Cutting speed = 1.8m/s

Feed = 0.3mm

Depth = 2.6mm

Angle = 8°

Thickness = 0.49

(a) calculating the chip thickness ratio using the formula;

r = t₀/tc

 = 0.3/0.49

 = 0.6122

Calculating the shear angle using the formula;

∅ = tan⁻¹[rcosα/1-rsinα]

  = tan⁻¹[(0.6122*cos8)/(1-0.6122sin8)]

  = 33.53°

(b) Calculating the shear strain using the formula;

y = cot∅ + tan(∅-α)

   = cot 33.53 + tan(33.53-8)

   = 1.509 + 0.477

  = 1.989

(c) Calculating the material removal rate using the formula;

MRR =f*d*V

        = 0.3 * 2.6 *1.8 *1000mm

        = 1404mm³/s

answer:

my guess it 8 i'm probably wrong

Torque stick

Hope Helps

:)