Q) The following data relate to orthogonal cutting of mild steel part:

Cutting speed = 200 m/min

Tool rake angle = 12°

Width of cut = 1.8 mm

Uncut thickness = 0.2 mm

Average value of the coefficient of friction by the tool and the chip = 0.55

Shear stress of work material = 390 $N/mm^2$

Calculate:

1] Shear angle ($\phi$)

2] Cutting force

3] Shear force

4] Feed force or Thrust force

Solution:

Given:

Cutting speed = V = 200 m/min

Tool rake angle = $\gamma$ = 12 ?

Width of cut = b = 1.8mm

Uncut thickness = t = 0.2mm, $\mu$ = 0.55

Shear stress of work = $\tau_{avg}$ = 390 $N/mm^2$

$\mu=\frac{F_{q}+F_{p} \cdot \tan \gamma}{F_{p}-F_{q} \cdot \tan \gamma}=\tan \beta$

$\therefore 0.55=\tan \beta$

$\therefore \beta=28.81^{\circ}$

$\therefore$ We know that,

$\emptyset=45^{\circ}+\frac{12^{\circ}}{2}-\frac{28.81^{\circ}}{2}=36.59^{\circ}$

$\therefore 0.55=\frac{F_{q}+F_{p} \tan (12)}{F_{p}-F_{q} \cdot \tan (12)}=\frac{F_{q}+F_{q}(0.2126)}{F_{p}-F_{q}(0.2126)}-------1-A$

Also,

$\tau_{a v g}=\frac{\left(F_{p} \cdot \cos \emptyset-F_{q} \sin \emptyset\right) \sin \emptyset}{b t}$

$\therefore 390=\frac{\left[F_{p} \cdot \cos (36.59)-F_{q} \cdot \sin (36.59)\right] \sin (36.59)}{(1.8)(0.2)}----B$

On solving equation A & B

$F_{q} = 114.3 N$

$F_{p} = 378.33 N$

$\begin{aligned} \therefore F_{s} &=F_{p} \cos \emptyset-F_{q} \sin \emptyset \\ &=378.33 \cos (36.59)-114.3 \sin (36.59) \\ \therefore F_{s} &=235.58 N \end{aligned}$