Question

2. In lab we cut an aluminum rod from 7.42 mm to 6.50 mm diameter with 50 rake angle (a), 1700 rpm spindle speed, 3.21 mm/s tool feedrate (vi) (this is the linear feedrate of the tool),82° major cutting edge angle (k), 10° minor cutting edge angle (K"), 0.68 mm measure cut chip thickness, 50 N cutting force (Fc), and 28 N tangential force (Ft) in single point turning. (a) Find uncut chip thickness, to in [mm/rev] (5 points) (b) Find the shear angle, 0 (5 points) (c) Find the friction energy, ur (hint: need F, V, and V.) (15 points)

Answer 1

(a) To find the uncut chip thickness (to), we can use the formula:

to = vi / (rpm * feedrate)

Given:
Spindle speed (rpm) = 1700
Tool feedrate (vi) = 3.21 mm/s

Substituting these values into the formula:

to = 3.21 mm/s / (1700 rpm * 3.21 mm/s)
to = 3.21 / 1700
to ≈ 0.00189 mm/rev

Therefore, the uncut chip thickness is approximately 0.00189 mm/rev.

(b) The shear angle (φ) can be determined using the formula:

tan(φ) = to / (pi * (di + df) / 2)

Given:
Initial diameter (di) = 7.42 mm
Final diameter (df) = 6.50 mm

Substituting these values into the formula:

tan(φ) = 0.00189 mm/rev / (pi * (7.42 mm + 6.50 mm) / 2)
tan(φ) = 0.00189 mm/rev / (pi * 6.96 mm / 2)
tan(φ) = 0.00189 mm/rev / 10.92 mm
φ ≈ 0.000173 radians

Therefore, the shear angle is approximately 0.000173 radians.

(c) The friction energy (Ur) can be calculated using the formula:

Ur = Ft * V * to

Given:
Tangential force (Ft) = 28 N
Uncut chip thickness (to) = 0.00189 mm/rev
Cutting speed (V) = (pi * (di + df) / 2) * rpm

Substituting these values into the formula:

V = (pi * (7.42 mm + 6.50 mm) / 2) * 1700 rpm
V = 3.1416 * 6.96 mm * 1700 rpm
V ≈ 35,240 mm/min

Ur = 28 N * 35,240 mm/min * 0.00189 mm/rev
Ur ≈ 1.003 J

Therefore, the friction energy is approximately 1.003 Joules.