Find forces acting on a combined beam with load.

  • Thread starter Thread starter dreamliner
  • Start date Start date
  • Tags Tags
    Beam Forces Load
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
4 replies · 2K views
dreamliner
Messages
16
Reaction score
0
I was wondering if someone could have a look at my calculations. I believe I'm on the right track but I keep getting wrong answers, but I might be mixing + and - somewhere. I'm also a little unsure about the right way to incorporate the force working on top of the beam.

The task reads: A combined beam ABCD is resting on a hinged support in A, and on rollers in B and D. In C there is a frictionless joint. There is an evenly distributed load q on top of the beam. Calculate Ay, B and D.







The Attempt at a Solution


Due to the number of unknowns I split the beam in C and looked at the two halves independently.

I've attached my calculations as a file to this post. I'd appreciate it if someone could take a look.
 

Attachments

  • beam.jpg
    beam.jpg
    18.8 KB · Views: 468
Physics news on Phys.org
Yes I got a little impatient there. I tried to delete my first post and use this one instead since it was basically the same question, but no go there.

Anyway. Looking at Beam ABC from my drawing. If you are supposed to use the length from cg of the load to A, would this be correct:

q*4,3(the entire length of beam q is working on)*3,35(the length from cg of the load to A)?
 
dreamliner said:
Yes I got a little impatient there. I tried to delete my first post and use this one instead since it was basically the same question, but no go there.

Anyway. Looking at Beam ABC from my drawing. If you are supposed to use the length from cg of the load to A, would this be correct:

q*4,3(the entire length of beam q is working on)*3,35(the length from cg of the load to A)?
Yes, that is correct. The resultant of the distributed load, q*4,3, acts at the cg of the distributed load, 2,15 m from C, which is thus located 3,35 m from A.