Hi all.
I have to cut sae 1070 Harden and tempered to 55 HRC, blue spring strip, 0.5mm thick 6 mm diameter hole.
It is done by hiting a 6 mm punch , made from HSS, to the such strip. I did it by a hammer about 1000 grams by my arm force.
It Works.
But I want to do it by a free fall weigth, or potential energy.
So my question is : how much energy is need to cut such hole.
For a 55 HRC the ultimate yield is about 212 Kg/mm2, so for a 6 mm diameter there is 6 mm * pi* 0.5 mm = 9.42 mm2 area
and for this ultimate yield 9.42 mm2 * 212 Kg/mm2 = 1997.04 Kg
But this is weigth , not energy.
As the work to do is seldom used I want to make the free fall hammer . The allowed heigth will be 2.00 meter , what will be the hammer weigth.
It seem to be a lazy student home work , but not .
Furthermore that all we are allways student,I'm 55 old mechanics and want to know the theory on this fact.
Pardal
I have to cut sae 1070 Harden and tempered to 55 HRC, blue spring strip, 0.5mm thick 6 mm diameter hole.
It is done by hiting a 6 mm punch , made from HSS, to the such strip. I did it by a hammer about 1000 grams by my arm force.
It Works.
But I want to do it by a free fall weigth, or potential energy.
So my question is : how much energy is need to cut such hole.
For a 55 HRC the ultimate yield is about 212 Kg/mm2, so for a 6 mm diameter there is 6 mm * pi* 0.5 mm = 9.42 mm2 area
and for this ultimate yield 9.42 mm2 * 212 Kg/mm2 = 1997.04 Kg
But this is weigth , not energy.
As the work to do is seldom used I want to make the free fall hammer . The allowed heigth will be 2.00 meter , what will be the hammer weigth.
It seem to be a lazy student home work , but not .
Furthermore that all we are allways student,I'm 55 old mechanics and want to know the theory on this fact.
Pardal
