Q) The following data was recorded in orthogonal operation machining with carbide tipped roof.

Cutting speed = 120 m/min

Tool rake angle = 10°

Feed = 0.2mm/rev

Depth of cut = 2mm

Chip thickness = 0.45mm

Cutting force = 185kgs

Feed force = 75kg

Calculate:

Shear angle ($\phi$)

1] Chip thickness ratio

2] Resultant force

3] Shear angle

4] Length of shear plane

5] Frictional force

6] Friction angle

7] Normal compressive force

8] Coefficient of friction

9] Shear force

10] Shear velocity

Solution:

Given:

Cutting speed = V = 120 m/min = 2m/sec

Tool rake angle = r = $10^{\circ}$

Width of cut = b = 2mm

Uncut thickness = t = 0.45mm,

Fp = 185kg, Fq = 75Kg

$r_{c}=\frac{t}{t_{c}}=\frac{0.2^{\prime}}{0.45}=0.444^{\prime}$

$\tan \emptyset=\frac{r_{c} \cos \gamma}{1-r_{c} \sin \gamma}=\frac{0.444 \cos 10}{1-0.444 \sin 10}=\frac{0.4373}{0.9229}=0.4738$

$\emptyset=25.35^{\circ}$

$R=\sqrt{F_{p}^{2}+F_{q}^{2}}$

$R=\sqrt{185^{2}+75^{2}}=199.62 K g$

$F_{s}=F_{p} \cos \emptyset-F_{q} \sin \emptyset$

$=185 \cos 25.35-75 \sin 25.35$

$=135.08 \mathrm{kg}$

$F = F_p sinγ+F_q cosγ$

$=185 \sin 10+75 \cos 10$

$=106 N$

$N=F_{p} \cos \gamma-F_{q} \sin \gamma$

$=105 \cos 10-75 \sin 10$

$=169 N$

$\mu=\tan \beta$

$\mu=\frac{F_{q}+F_{p} \text { tany }}{F_{p}-F_{q} \text { tany }}=\frac{75+185 \tan 10}{185-75 \tan 10}=\frac{107.62}{171.78}=0.6265$

$\beta=32.07^{\circ}$

$V_{s}=\frac{V \cos \gamma}{\cos (\emptyset-\gamma)}=\frac{2 \times \cos 10}{\cos 15.35}=2.04 \mathrm{m} / \mathrm{sec}$