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

chip area = 6.908 mm^2

material removal rate 828.96 cm^3/min

power = 40644 kW

MACHINE TIME = 1.585 min

Step-by-step 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* 10^(-2) * 135 * 100 = 828.96 cm^3/min`

c) power = F* Vc

`Ks = (F)/((\pi)/(4) Do^2)`

`F = 2300 * (\pi)/(4) 100^2 = 18064 kN`

`p = 18064 * (135)/(60) = 40644 kW`

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

`N =(V)/(\pi D) = (135 * 1000)/(\pi * 100) = 429.963 rpm`

`tm = (150)/(0.22 * 430) = 1.5856 min`