Calculating Support Reactions for a Beam with Inclined Resting Position

  • Thread starter Thread starter dbag123
  • Start date Start date
  • Tags Tags
    Beam Incline
AI Thread Summary
The discussion focuses on calculating support reactions for a beam with a pin connection on one side and a roller on the other. The user has determined that the vertical reaction at the pin support (Fy) is 9kN but is uncertain about the horizontal reaction (Fx) and how to distribute the loads. Two methods are proposed: one involves using tangents of point loads to find reactions, while the other involves reorienting the beam to horizontal and applying moment equations. The latter method yields close results to the expected solution, confirming the validity of the approach. Ultimately, the moment balance calculations help clarify the distribution of forces at the supports.
dbag123
Messages
76
Reaction score
3
Homework Statement
Calculate the support reactions
Relevant Equations
ΣFx =0
ΣFy=0
ΣMa=0
ΣMb=0
Hello
244096
I have a problem with calculating the support reactions for a beam. Lefts side of beam has a pin connection so it takes both Fx,Fy. Right side of the beam has a horizontal roller and it takes only Fy in the direction of the wall. Therefore at the pin support Fy=9kN, but how do i figure out the reactions for X-direction?

I thought about the X-component as a tangent to the 2 point loads so 5kN/tan(21.8);4kN/tan(21.8)5kN/tan(21.8);4kN/tan(21.8) and they give me 12,5 and 10kN respectively. I don't know if they are correct and if they are how do they distribute to each support?

I can also think of another way of doing this and that is by making the beam horizontal and turning the roller support 68.2 degrees, then calculating the reactions ΣMa=0: -4kN*1.5m-5kN*4m+By*5m=0 5,2kN By and 13kN for Bx and ΣMb=0: 4kN*3.5m+5kN*1m-Ay*5m=0 it comes out to 3,8kN for Ay 9,5kN to Ax. These components are parallel to the beam itself. Is either of these methods correct?
Thanks
 
Physics news on Phys.org
Your second idea of taking moments makes sense, but why do you need to reorient the beam to the horizontal position? What does the moment balance about the lower support look like?
 
reorienting the beam was just a test to see if i got the same results as the solution suggests and its quite close and as far as the moment about pin support A goes and By then is the reaction perpendicular to the beam at the roller support:
ΣMa=0: -4kN*1.5m-5kN*4m+By*5m=0
 
dbag123 said:
reorienting the beam was just a test to see if i got the same results as the solution suggests and its quite close and as far as the moment about pin support A goes and By then is the reaction perpendicular to the beam at the roller support:
ΣMa=0: -4kN*1.5m-5kN*4m+By*5m=0

The moment arm for B is 2 meters.
 
  • Like
Likes CivilSigma and dbag123
Chestermiller said:
The moment arm for B is 2 meters.
and that is how the roller get its 13kN. Thank you very much.
 
Thread 'Have I solved this structural engineering equation correctly?'
Hi all, I have a structural engineering book from 1979. I am trying to follow it as best as I can. I have come to a formula that calculates the rotations in radians at the rigid joint that requires an iterative procedure. This equation comes in the form of: $$ x_i = \frac {Q_ih_i + Q_{i+1}h_{i+1}}{4K} + \frac {C}{K}x_{i-1} + \frac {C}{K}x_{i+1} $$ Where: ## Q ## is the horizontal storey shear ## h ## is the storey height ## K = (6G_i + C_i + C_{i+1}) ## ## G = \frac {I_g}{h} ## ## C...
Back
Top