Contribute
EN / USD
Log in / Join
32
of 76
TEP , The Engineering Projects , Image

syedzainnasir

TEP , The Engineering Projects , Rating 7.5 7.5 / 10
TEP , The Engineering Projects , Icon Level: Moderator
TEP , The Engineering Projects , Icon Joined: 20 Mar 2022
TEP , The Engineering Projects , Icon Last Active: 2:21 PM
TEP , The Engineering Projects , Icon Location: TEP , The Engineering Projects , Flag
TEP , The ENgineering Projects , Icon TEP , The ENgineering Projects , Icon TEP , The ENgineering Projects , Icon TEP , The ENgineering Projects , Icon
Horizontal Rod Size Required
TEP , The Engineering Projects , Calender Question: 09-Mar-2017
TEP , The Engineering Projects , Category In: Mechanical Projects
Many, many years ago I graduated college as a Mechanical Engineering Technologist but my career path took me elsewhere. I therefore can use and comprehend most of the "tools of the trade" but I'm very rusty in that usage. I therefore don't feel fully confident in my own calculations and would like some input.

I want to use a 12" long Stainless Steel solid rod to support approximately 200 lbs of weight equally distributed between two mounted bearings on each end with the rod itself only supported in the center.

In addition to the rod supporting the weight, it needs to remain straight enough that the bearings do not bind due to a bend in the rod.

For unrelated reasons the rod cannot exceed 2.375" diameter But I'm hoping the actual required size is much less than that.

I'd appreciate either an answer with some supporting calculations or a link to somewhere the info can be found.
TEP , The Engineering Projects , Icon Answer: 0 TEP , The Engineering Projects , Icon Views: 150 TEP , The Engineering Projects , Icon Followers: 85
Small Bio
TEP , The Engineering Projects , Tags
PLC
Robot
STM32
Arduino
AI
ESP32
Ladder Logic
PLC Projects
Programming
Communicates STM32
PLC Projects
Communicates PLC
Font Style
Alignment
Indenting and Lists
Insert Media
Insert Items

Want to leave an answer!

Word Count :0 Draft Saved at 12:42 am.