cockeyedMFer
Gold Member
- Joined
- Jul 25, 2011
- Messages
- 270
- Tractor
- 1959 MF 35 deluxe, gas
That fractional explanation simplifies it, thanks! I was trying to break it up by how much each beam supports but I always overcomplicate it - that's why I switched majors after a year or mechanical engineering LOL.
So the 16' beams and the ledger boards at the rear wall, which hold the 12' beams, are supporting the entire load. That means the 16' beams are sharing half the load along their entire length, and each 16' section supports 1/4 of the total?
If so, the total load at 40psf would equal (12x32)x40=15360. Divided in fours that's 3840# per 16' beam. That's the same number I came up with using another method. And its within spec according to the tables you provided.
What do you think?
So the 16' beams and the ledger boards at the rear wall, which hold the 12' beams, are supporting the entire load. That means the 16' beams are sharing half the load along their entire length, and each 16' section supports 1/4 of the total?
If so, the total load at 40psf would equal (12x32)x40=15360. Divided in fours that's 3840# per 16' beam. That's the same number I came up with using another method. And its within spec according to the tables you provided.
What do you think?