Recent content by jmart157

Are you saying I have to do substeps with the rod from outside of the tube and push it throughout the entire tube?

I'm looking at 8600 sub steps then. I am aware it will be running for awhile - x6 1100t, fx4800, raid 0 across 6 hard drives, 16gb ddr3. Hopefully that will take this from a couple to one. JRM

I'm trying to flush out the top of the rod with the top of the tubing (faces are on the same plane). Would like to do a substep every 2". If the rod moves 12", then 6 substeps. JRM

Yes I have, you just posted in the thread that I revived. It is the exact one I was looking at as I feel this is a similar situation. I have had no success. The convergence line and criterion never meet. I have revised and tried probably 10 times with changing one thing at a time.

https://www.physicsforums.com/showthread.php?p=3445330#post3445330 I thought I did last night. Different forum, my mistake!

I am and have been trying to do a specific simulation for the last several weeks. It seems like I take one step forward two steps back. See below: See the attached files. I'm on Ansys Workbench 13. [PLAIN]www.plastics-consulting.com/rod.stp[/URL]...

Reviving an old thread because I need help with something similar! PM me!

Looking to figure out the allowable stress (psi) of a shaft which has two pin holes in it on the transverse axis of the shaft. tau = allowable stress, psi T = torque, in-lb c = distance from center of gravity to extreme fiber J = polar moment of inertia D = diameter of the shaft d =...

I'm back boys... Why doesn't this need to account for the amount of surface area in contact with the rotating shaft? I feel like if I had the same force applied on the shaft through more contact surface area, it would slow down more... As I just typed that... I realized the answer to my...

it looks like we figured this one out... thank jackaction. now the only problem is datalogging motor voltage and amperage with the DAQ so we can get this calculations done in real time :)

thanks jackaction. Hmmmm. Wouldn't the normal force (if we go back to the known loaded masses) be the 6x the 2.84 then? there is essentially 17.16 lbs of force into the spinning barrel/shaft... I followed everything else. I had made an error one of the formulas in my excel sheet. CoF =...

alright, this came too easy. please confirm... torque = Radius x force which means rotational force = torque/radius loaded rotational force = 41.83/2" = 20.92 lbsf no load rotational force = 38.25/2" = 19.123 lbsf resistive force from load = 20.92 lbsf - 19.123 lbs f = 1.7928 lbs...

alright, this came too easy. please confirm... torque = Radius x force which means rotational force = torque/radius loaded rotational force = 41.83/2" = 20.92 lbsf no load rotational force = 38.25/2" = 19.123 lbsf resistive force from load = 20.92 lbsf - 19.123 lbs f = 1.7928 lbs...

Here's my givens... i need to find the coefficient of friction on the material. No load torque on motor - 3.19 Nm Loaded torque on motor - 3.49 Nm Surface area in contact with motor shaft - .1885" Mass of loaded material on shaft - 17.16 lbs = 76.33 N Rads/sec - 31.42Looking to find the...

@jack action thank you for your input, its valued. I followed this thread (i thought) to get the CoF, but I may have misunderstood it. https://www.physicsforums.com/showthread.php?t=151516 I corrected my torque calculations, thank you. Please do give more input if you have time to...