Question

During orthogonal machining of a job at a cutting speed of 90 m/min with a tool of 10° rake angle, the cutting force and thrust force are 750 N and 390 N, respectively. Assume a shear angle of 35°. The power (in W) expended for shearing along the shear plane is ________.

Answer 1

Concept:

Shear Power = Fs.V_s

\(V_s=\frac{{V_c~\times~cos~α }}{{cos(ϕ~-~α)}}\)

Vs = shear Veloity, Vc = cutting velocity, Fs = shear force, α = Rake Angle, ϕ = Shear Angle

The cutting force in terms of Thrust force and cutting force is given by 

Fs = Fc cos ϕ - Ft sin ϕ

Calculation:

Given:

ϕ = 35, α = 10°, Vc = 90 m/min, F_C = 750 N, F_t = 390 N

Now,

Fs = Fc cos ϕ - Ft sin ϕ

Fs = 750 (cos35°) – 390 (sin35°)

Fs = 390.66 N

Now,

\(V_s=\frac{{V_c~\times~cos~α }}{{cos(ϕ~-~α)}}=\frac{{90~\times~cos~10}}{{cos(35~-~10)}}=\;97.79~m/min\)

Vs = 97.79 m/min

Vs = 1.62 m/s

Now,

Shear power = Fs.Vs

​Shear power = 390.66 × 1.62

Shear power = 632.88 N-m/s

∴ Shear Power \(\approx\) 633 W