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While turning a metal rod of diameter 150 mm, the feed force and the cutting force was noted about 200 N and 600 N respectively. The rake angle, principal cutting edge angle, and the friction angle of the tool used were 10 deg 25 deg and 5 deg respectively. Calculate the following: i. Shear angle (B) ii. Shear force (P_{s}) iii. Co-efficient of friction at the chip-tool interface (u) iv. Compressive force at the shear plane (P_{n})

Question

While turning a metal rod of diameter 150 mm, the feed force and the cutting force was noted about 200 N and 600 N respectively. The rake angle, principal cutting edge angle, and the friction angle of the tool used were 10 deg 25 deg and 5 deg respectively. Calculate the following: i. Shear angle (B) ii. Shear force (P_{s}) iii. Co-efficient of friction at the chip-tool interface (u) iv. Compressive force at the shear plane (P_{n})

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Solution

Para resolver este problema, utilizaremos las fórmulas y conceptos de la teoría de corte de metales. Vamos a calcular cada uno de los parámetros solicitados paso a paso.

i. Ángulo de cizalladura (B)

El ángulo de cizalladura (B) se puede calcular utilizando la fórmula de Merchant:

ϕ=π4α2+β2 \phi = \frac{\pi}{4} - \frac{\alpha}{2} + \frac{\beta}{2}

Donde:

  • ϕ\phi es el ángulo de cizalladura.
  • α\alpha es el ángulo de desprendimiento (rake angle).
  • β\beta es el ángulo de fricción (friction angle).

Dado:

  • α=10\alpha = 10^\circ
  • β=5\beta = 5^\circ

Sustituyendo los valores:

ϕ=π4102+52 \phi = \frac{\pi}{4} - \frac{10^\circ}{2} + \frac{5^\circ}{2} ϕ=455+2.5 \phi = 45^\circ - 5^\circ + 2.5^\circ ϕ=42.5 \phi = 42.5^\circ

Por lo tanto, el ángulo de cizalladura (B) es 42.5 grados.

ii. Fuerza de cizalladura (P_{s})

La fuerza de cizalladura se puede calcular utilizando la siguiente fórmula:

Ps=Fccos(ϕ)Ffsin(ϕ) P_s = F_c \cos(\phi) - F_f \sin(\phi)

Donde:

  • FcF_c es la fuerza de corte (cutting force).
  • FfF_f es la fuerza de avance (feed force).
  • ϕ\phi es el ángulo de cizalladura.

Dado:

  • Fc=600NF_c = 600 \, \text{N}
  • Ff=200NF_f = 200 \, \text{N}
  • ϕ=42.5\phi = 42.5^\circ

Sustituyendo los valores:

Ps=600cos(42.5)200sin(42.5) P_s = 600 \cos(42.5^\circ) - 200 \sin(42.5^\circ)

Calculando los valores de las funciones trigonométricas:

cos(42.5)0.7431 \cos(42.5^\circ) \approx 0.7431 sin(42.5)0.6691 \sin(42.5^\circ) \approx 0.6691

[ P_s = 600 \times 0.7431

This problem has been solved

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