Calculating Support Reactions for a Beam with Inclined Resting Position

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Discussion Overview

The discussion revolves around calculating support reactions for a beam with a pin connection on one end and a roller support on the other. Participants explore different methods for determining the reactions in both the vertical and horizontal directions, considering the effects of inclined loads and moment balances.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant describes a method for calculating the X-component of reactions using tangents to point loads, yielding values of 12.5kN and 10kN, but expresses uncertainty about their correctness and distribution to supports.
  • Another participant questions the necessity of reorienting the beam to a horizontal position for moment calculations and asks for clarification on the moment balance about the lower support.
  • A participant explains that reorienting the beam was a test to verify results against a suggested solution and provides a moment equation for the pin support, indicating that the reaction at the roller support is perpendicular to the beam.
  • Further clarification is provided regarding the moment arm for the roller support, which is stated to be 2 meters, and confirms a calculated reaction of 13kN at the roller.

Areas of Agreement / Disagreement

Participants express differing views on the methods for calculating support reactions, with no consensus reached on the correctness of the proposed approaches or the necessity of reorienting the beam.

Contextual Notes

Participants do not fully resolve the assumptions regarding the distribution of forces or the implications of the inclined loads on the calculations. The discussion includes various approaches that may depend on specific interpretations of the problem setup.

dbag123
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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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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.
 
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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.
 

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