A150 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].

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

speed of cutting is v  135 m/min

feed rate fr = 0.22 min

Ks = 2300 N/mm^2

a) chip area  

b) material removal rate = chip area * Vc

c) power = F* Vc

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

when jerome paddle in flowing river towards east the net speed of jerome will increase

so the net speed of jerome will be given as

now the speed of paul will be same as it is given as he is running on the ground

so

so here we can say since the speed of jerome is more so he will win the race

now as we know the distance so the time difference is given as

so jerome will reach first and the time difference will be 203.55 s or 3.4 min