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- amilapsn

A punching machine has to carry out piercing operations at the rate of 12 holes per minute. The amount of work to be done in cutting 30mm diameter holes in 15mm thick plate is 8J/mm2 of the sheared area. The actual piercing operation takes 2s. The total frictional losses over a cycle are equivalent to one sixth of the work done during a single piercing operation. The machine shaft is driven by a constant power electric motor. The maximum fluctuation of the speed of the machine shaft is between 190 rev/min and 210 rev/min.
(a) Find the minimum required motor power to drive the piercing machine

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- amilapsn

- katieb

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- amilapsn

(b) Determine the required sum of the mass moments of inertia of all revolving parts on the machine shaft in order to keep the speed fluctuation within the given limits. State any assumption made.

- amilapsn

- amilapsn

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- dan815

:)

- amilapsn

I've got some problems to sort out regarding this problem @dan815

- dan815

okay what seems to be the problem

- amilapsn

During a piercing operation the shear area would be \(\pi r^2\) or \(s2\pi r h\) where r is the diameter of the hole while h is the depth...

- amilapsn

*\(2\pi rh\)

- dan815

why 2*pirh

- amilapsn

Then why the depth is given?

- dan815

not sure why, i dont think we need it

- dan815

it gives work per surface area though in mm^2 so

- amilapsn

Work done per hole = 8 x pi x r^2?

- amilapsn

Then another q:

- dan815

yep

- amilapsn

Am I right:
minimum power to drive the piercing machine=maximum power in a cycle

- amilapsn

|dw:1433649727575:dw|

- dan815

yeah

- amilapsn

But in a text book they've taken the average as the minimum required power...

- IrishBoy123

you need the thickness in shear area
Ex 5 on page 10: http://www.freestudy.co.uk/mech%20prin%20h2/stress.pdf

- amilapsn

Thank you so much! @IrishBoy123

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