Bending Moment and Force calculations two beams connected

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Discussion Overview

The discussion revolves around calculating the bending moments and forces at the midpoint cross-sections of two connected beams: a horizontal beam subjected to a uniformly distributed load (UDL) and a vertical beam fixed at the ground. Participants explore methods for analyzing the system, particularly in the context of their coursework, which has primarily focused on single, simply supported beams.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related
  • Debate/contested

Main Points Raised

  • One participant suggests treating the beams as separate entities to find the reaction bending moment at the joint using the UDL.
  • Another participant mentions the importance of considering the reaction force at the joint in addition to the bending moment.
  • There is a discussion about whether the moment acts at the tip of the vertical beam or at the midpoint of the horizontal beam, with a suggestion to use beam centerlines for analysis.
  • Calculations are presented for the top beam, yielding a reaction force of 10 kN and a bending moment of -5 kNm at the midpoint.
  • For the vertical beam, similar calculations are provided, resulting in a bending moment of -5 kNm and a force of -10 kN at the midpoint.
  • One participant expresses confusion over discrepancies between their manual calculations and results obtained from finite element analysis (FEA), noting differences in bending moments and forces.
  • A response indicates that variations in results may arise from differences in modeling the joint and inputting member properties in the FEA software.

Areas of Agreement / Disagreement

Participants generally agree on the methods for calculating forces and moments but express uncertainty regarding the modeling of the joint and the resulting discrepancies in analysis outcomes. There is no consensus on the exact cause of the differences in results from manual calculations versus FEA.

Contextual Notes

Participants note that the accuracy of the FEA results may depend on how the end connections are modeled, which could affect the outcomes. The discussion also highlights that the solutions may only be accurate to one significant figure.

olski1
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Hello,

I need to find the bending moment and forces at the mid point cross-sections of two separate beams connected together. One is a horizontal beam connected to the top of a vertical beam that sort of look like this:

H========
H
H
H
H

There is a UDL of 10kn/m on the top member, with the bottom of the vertical beam is fixed to the ground.

I need to find the bending moment and Forces at the midpoint cross-section of both beams separately. How would I go about doing this?

So far in my course we have only dealt with single beams that are simply supported.

My first attempt at this question was to treat them as two separate beams and firstly find the reaction bending moment at the joining of the two beams using the udl. Then I would take that moment as a acting on the end of the vertical beam and work my way from there. Does that moment act at the very tip of the upright beam or half the width down of the horizontal beam?

But honestly I am quite confused with this one... any help on the method (not answer) would be greatly appreciated :)
 
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olski1 said:
Hello,

I need to find the bending moment and forces at the mid point cross-sections of two separate beams connected together. One is a horizontal beam connected to the top of a vertical beam that sort of look like this:

H========
H
H
H
H

There is a UDL of 10kn/m on the top member, with the bottom of the vertical beam is fixed to the ground.

I need to find the bending moment and Forces at the midpoint cross-section of both beams separately. How would I go about doing this?

So far in my course we have only dealt with single beams that are simply supported.
I assume you have also dealt with beams fixed at one support?
My first attempt at this question was to treat them as two separate beams and firstly find the reaction bending moment at the joining of the two beams using the udl.
you can do it this way, but don't forget the reaction force at that joint as well.
Then I would take that moment
..and force...
as a acting on the end of the vertical beam and work my way from there. Does that moment act at the very tip of the upright beam or half the width down of the horizontal beam?
For the purposes of this problem, you can use the beam centerlines for the analysis.
But honestly I am quite confused with this one... any help on the method (not answer) would be greatly appreciated :)
You should note that it it is not necessary to separate the beams when solving for the forces in the vertical beam. You can take a free body diagram at the mid point of the horiz beam to get it's internal moment and force at that point, and you can take a FBD at the midpoint of the vertical beam to get its internal force and moment
 
Okay thanks for your help,

For the top member which is 1m long I found:

ƩFy=0=-10kN + RAY => RAY=10KN

ƩMo=5<x>2 + MA => MA= -5<x>2

when x = 1 => MA=-5kNm

At the mid point section,

F=-5kN and M=-1.25KN

For the Vertical Memeber which is also 1 m long.

ƩFx=0=-10kN + RBx => RBx=10KN

ƩMo=-MA + Mo => Mo= -5kNm

hence at x=0.5, M=-5knm and F=-10kN.

Just to check I ran this through F.E.A with a beam with cross section 80mm high and 40mm wide, which is the dimensions given in the problem. But I am getting slightly different results.

For example, the top member moment is 1.133kNm and the Force is 4.733kN compared to 1.25kNm and 5kN.

Why are they different? where did I go wrong?
 
You didn't go wrong, your calculation was excellent. When you do the computer analysis, the solution is dependent on the end connections...for example, whether the top beam frames into the vertical beam onto its top, or onto its side, will yield slightly different results, depending on how you input the member properties and lengths into the computer, and how you modeled the joint. Heck, the solution is only accurate to 1 significant figure anyway.
 

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