CNC Machine Geek
Plastic
- Joined
- Dec 21, 2016
Hi PM,
Calculating thermal expansion in a 4140 HT&Q, 28-32 Rc ring: 6" OD x 4.75" ID by 1.00" Thick. Using a spreadsheet based calculator I found on line it says a 20 degree temperature differential I should compensate the ID of 4.775 by .0006" (.00057"). Is this correct?
Also I was wondering if a thin walled ring would change more, less or be the same as a solid disk of the otherwise same dimensions. Clearly the disk grows 'linearly' across the diameter. A ring is the same as a bar , right? So unroll it and you have a bar the length of the circumference. Run the calc for the bar length & rewrap it to find the delta diameter. But would one use the ring OD, ID, centerline to calc the bar length? TIA
William
Calculating thermal expansion in a 4140 HT&Q, 28-32 Rc ring: 6" OD x 4.75" ID by 1.00" Thick. Using a spreadsheet based calculator I found on line it says a 20 degree temperature differential I should compensate the ID of 4.775 by .0006" (.00057"). Is this correct?
Also I was wondering if a thin walled ring would change more, less or be the same as a solid disk of the otherwise same dimensions. Clearly the disk grows 'linearly' across the diameter. A ring is the same as a bar , right? So unroll it and you have a bar the length of the circumference. Run the calc for the bar length & rewrap it to find the delta diameter. But would one use the ring OD, ID, centerline to calc the bar length? TIA
William