Calculating Support Reactions for a Beam with Inclined Resting Position

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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
 
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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.
 
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Chestermiller said:
The moment arm for B is 2 meters.
and that is how the roller get its 13kN. Thank you very much.