- #1
yonese
- 15
- 1
- TL;DR Summary
- I don't understand how the radius and offset of the Mohr's circle can be simplified down to the principal Iq and Ip equations - where does the load F go? This was taken from a worked example with inverted L section carrying a load. My teacher hasn't explained how he jumps to these steps and I don't understand how the load F gets removed and the 6 simplifies to a 3. I have tried to solve for simultaneous equations using the radius and offset equation, but still stuck. Thanks.