bfanger
Mechanical
- Aug 23, 2012
- 3
I'm working at a company that uses jackhammers to drive our product into the ground. Not much is known about the forces involved except that this size works for this application. We are trying to design a new drive rod and would like to be a little more scientific than we have been in the past. To the best of my understanding jackhammers are rated by weight and impact energy. Neither of which directly leads to a force. The “hammer” is an electric hammer drill with a maximum impact energy of 11Nm. I tried an energy method of-
11 Nm= 97 in lb = ½ k x2
With k being the spring rate of a .625” steel rod 36 inches long – 255,000 lb/in
Then F= x k =7530 lb
This seems quite high for a relatively light piece of equipment
Is there a better way to do this that I’m missing?
11 Nm= 97 in lb = ½ k x2
With k being the spring rate of a .625” steel rod 36 inches long – 255,000 lb/in
Then F= x k =7530 lb
This seems quite high for a relatively light piece of equipment
Is there a better way to do this that I’m missing?