Question: single point cutting tool geometry in ASA system.
0 Where, $\alpha$b = Back rake angle $\theta$e = End relief angle

$\alpha$s = Side rake angle $\theta$s = Side relief angle

Ce = End cutting edge angle r = Nose radius.

Cs = Side cutting edge angle

ASA System : $\alpha$b - $\alpha$s - $\theta$e - $\theta$s - Ce - Cs - r

• The angle specified in ASA system are measured with respect to 3 mutually perpendicular planes.

• As the angles are measured w.r.t. 3 mutually perpendicular planes, the measurement of angle and indication will be easier.

• During machining operation the formation of chip and how behaviour of chip is mainly depends upon major cutting edge of tool. therefore basic geometric properties of tool are measured w.r.t. major cutting edge.

• Hence if the tool is designated in ASA system the analysis of machining will be difficult.

while machining steel with a tool of [0-10-6-6-8-75-1] ORS shape following observation were made,

1. Spindle speed 300 rpm

2. Work diameter 40 mm

3. Depth of cut 3.5 mm

4. Tool feed rate 70 mm / min

5. Cut chip thickness 0.55 mm

Determine :

1. Chip thickness ratio

2. Shear plane angle

3. Dynamic shear

4. Theoretical continuous chip length per minute

Solution :

Data :

From tool designation $\alpha$ = 10 degree

N = 300 rpm, D = 40 mm, t=w=3.5 mm, $t_2$ = 0.55 mm

f = $t_1$ = 70 $\frac{mm}{min}$ = $\frac{70}{300}$ $\frac{mm/min}{rev/min} = 0.23 mm/rev$

To find (1) r = ? (2) p = ? (3) $\mathcal{V}$ (4) $L_2$

Solution :

1) Chip thickness ratio (r) :

r = $\frac{t1}{t2}$

= $\frac{0.23}{0.55}$ = 0.42

r = 0.42 ------- Ans.

2) Shear plane angle (p) :

tan (p) = $\frac{r.CDS\alpha}{1-r sin \alpha}$

= $\frac{0.42 x cos 10}{1- 0.42xs in 10}$

tan(p) = 0.45

$\therefore$ (p) = tany (0.45)

$\therefore$ p = 24.25 degree ------ Ans

3) Dynamic shear strain :

$\mathcal{V}$ = $\frac{cos\alpha}{cos (p - \alpha) . sin p}$

= $\frac{cos10} {cos(24.25-10).sin 24.25 degree}$

$\mathcal{V}$ = 2.47 ------- Ans

4) Theoretical continuous chip length per minute :

chip length before cutting $(L_1)$

$\therefore$ $L_1$ = $\pi$.D.N

= $\pi$ x 40 x 300

= ?7.699 x $10^3$ mm/min

Now,

r = $\frac{L2}{L1}$ Here L2= Theoretic chip length per min.

0.42 = $\frac{L2}{37.699x10^3}$

$\therefore$ $L_2$ = 15.83x$10^3$ mm/min ---- Ans

renu • 216 views
 modified 5 weeks ago  • written 7 months ago by RB ♦♦ 110