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A punch press similar to that of the example given in class is to punch holes of diameter up to 0.75 inch through steel plate up to 0.375 inch thick. The shear strength of the steel will range up to 60,000 psi. The rated speed of the motor (maximum) is 1500 rpm and a 10% drop in motor speed is allowable. If holes are to be punched at a maximum rate of 1 per second, find the required flywheel inertia and motor power

Respuesta :

Answer:

power required = 1123.07 W

flywheel inertia  = 0.479 kg/m²

Explanation:

given data

diameter = 0.75 inch = 19.05 mm

thickness  = 0.375 inch = 9.525 mm

shear strength = 60,000 psi = 413.68 N/mm²

speed = 1500 rpm  

drop motor speed = 10%

punched at a maximum rate = 1 per second

solution

we know here drop motor speed = 10%

so speed will be = 1500 × 10%  = 150

so speed remaining will be

N = (1500 + 1350) ÷ 2  

N = 1425 rpm

and ω will be

ω = [tex]\frac{2\pi N}{60}[/tex]     .........1

put here value

ω = [tex]\frac{2 \times \pi \times 1425}{60}[/tex]  

ω = 149.22

and

area will be

Area = π × diameter × thickness    ....................2

Area = π × 19.05 × 9.525

Area =  570.045 mm²

and

maximum shear forced is required is

maximum shear force = area × shear force    ..........3

maximum shear force = 570.045 × 413.68  

maximum shear force = 235816.2156 N

and

energy required for punching that is

energy required = average maximum shear force × displace   ..............4

energy required = ( 235816.2156 × 9.525 × [tex]10^{-3}[/tex] )  ÷ 2

energy required = 1123.074727 N-m

so here power required will be

power required = energy required × no of hole per second      .............5

power required =  1123.07 ×  1

power required = 1123.07 W

and

flywheel inertia is express as

Energy  = I × ω²  × Cs    ....................6

here Cs = [tex]\frac{150}{1425}[/tex]  

Cs = 0.1052

put value in equation 6 we will get

1123.07  = I × (149.22)²  × 0.1052

solve it we get

flywheel inertia  = 0.479 kg/m²

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