Question

A 150 mm long normalized 1045 steel shaft is turned down from a diameter of 100 mm to 90 mm in a single pass. The side cutting edge angle σ is 0°, the cutting speed v is 135 m/min, and the feed per revolution fr is 0.22 mm. Determine (a) the chip area A [mm2], (b) the material removal rate Q [cm3/min], (c) the tangential component of the cutting force Ft [kN], (d) the power needed to drive the turning operation P [kW], and (e) the machining time t [min].

Answer 1

Answer:

chip area = 6.908 mm^2

material removal rate 828.96 cm^3/min

power = 40644 kW

MACHINE TIME = 1.585 min

Explanation:

GIVEN DATA:

Length 150 mm

diameter vary from 100 to 90 mm

cutting edge angle $\sigma = 0 degree$

speed of cutting is v  135 m/min

feed rate fr = 0.22 min

Ks = 2300 N/mm^2

a) chip area  $A = \pi (D_O -D_i) fr$

                        $= \pi (100-90) 0.22 = 6.908 mm^2$

b) material removal rate = chip area * Vc

 $Q = 6.908\times 10^{-2} \times 135 \times 100 = 828.96 cm^3/min$

c) power = F* Vc

$Ks = \frac{F}{\frac{\pi}{4} Do^2}$

$F = 2300 \times \frac{\pi}{4} 100^2 = 18064 kN$

$p = 18064 \times \frac{135}{60} = 40644 kW$

C) Machine time[/tex] tm  = \frac{L}{fN}[/tex]

$N =\frac{V}{\pi D} = \frac{135 \times 1000}{\pi \times 100} = 429.963 rpm$

$tm = \frac{150}{0.22 \times 430} = 1.5856 min$