how to solve 7

Question

3.1. A 250mm diameter shaft is revolving in a bearing and exerting a force of 75kN on the bearing. The coefficient of friction between the shaft and the bearing is 0,02 and the shaft rotates at 350r/min. Calculate the power lost due to friction 3.2. in a cutting tooth operation on a milling machine, the horizontal force exerted by the milling cutter is 1,5kN. The milling cutter has a diameter of 125mm and rotates at 70r/min. The feed of the milling cutter is 0,175mm per tooth and the cutter has 45 teeth. Calculate the following if the average feed is equal to the tangential cutting force: 3.2.1. the cutting speed of the milling cutter The diameter of the shaft is 250 mm . The radius of the shaft is [ r2 ] 125 mm . Amount of force exerted by shaft on bearing is (F) 75 kN . The coefficient of friction between the shaft and the bearing is ( μ) 0.02 . The shaft rotates at the speed of 350 r/min . The shaft rotates at the speed of 350 r/min = ( 350/60 ) = 5.8333 revolutions per second . The axial load W is the amount of force exerted by shaft on bearing . So the axial load W is 75 kN . The frictional torque T is given by T = (2/3)μW [ r2 3 / r2 2 ] T = (2/3)μW [ r2 ] T = (2/3)(0.02) (75 * 1000 ) [ 0.00125 ] T = (0.667)(0.02) (75) [ 125 ] T = (0.0133) (9375) T = 125 . So the frictional torque is 125 Nm . Thus the power loss is p = T2 . p = T2 p = T * 23 ( revolutions per second ) p = 125 * 23 ( 5.8333 ) p = 125 * (36.6519 ) p = 4581.489 W . So thus the power loss is 4581.489 W . 3.2.2. the power consumed during the cutting action of the milling cutter 3.2.3. the power consumed for the feed of the milling machine

Answer 1

3.1)

The the power loss  due to friction can be evaluated using formula p = Tω .

Where T is frictional torque .

The frictional torque T is given by 

T = (2/3)μW [ r₂³ / r₂² ]

T = (2/3)μW [ r₂ ] 

Where r₂ is radius of the shaft is 125 mm .

Amount of force exerted by shaft on bearing is (F) 75 kN .

The axial load W is the amount of force exerted by shaft on bearing  .

So the axial load W is 75 kN .

 The coefficient of friction between the shaft and the bearing is ( μ) 0.02 .

T = (2/3)(0.02) (75 * 1000 ) [ 0.00125 ]

 T = (0.667)(0.02) (75) [ 125 ]

T = (0.0133) (9375) = 125 .

So the frictional torque is 125 Nm .

The shaft rotates at the speed of 350 r/min = ( 350/60 ) = 5.8333 revolutions per second . 

p = Tω 

p = T * 2π ( revolutions per second ) 

p = 125 * 2π ( 5.8333 )

p = 125 * (36.6519 )

p = 4581.489 W .

Hence the power loss is 4581.489 W . 

Answer 2

3.2.1)

Horizontal force of the milling cutter = 1.5 kN

Milling cutter diameter d = 125 mm

Milling cutter speed RPM = 70 rpm

The cutting speed of the milling cutter N = ?

Cutting speed = (RPM)πd/1000

Cutting speed = 70×π×125/1000

Cutting speed = 27.49 rev/min

The cutting speed of the milling cutter is 27.49 rev/min.