Baltimore Truss Analysis Problem

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Discussion Overview

The discussion revolves around solving a homework problem related to the analysis of forces in a Baltimore truss, specifically focusing on determining the force in member JQ. The conversation includes the application of the method of sections and the method of joints within the context of static equilibrium in truss structures.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant describes the method of sections as cutting the truss into two portions to analyze forces in equilibrium.
  • Another participant points out that the chosen section intersected four unknown members, complicating the analysis and suggesting a different section could yield a solvable scenario.
  • A participant provides calculations for reactions at supports and moments, but expresses difficulty in progressing further.
  • There is a suggestion to analyze member HQ by inspection as an alternative approach.
  • One participant notes that the removal of a diagonal member changes the behavior of the panel and suggests that this could help in finding the force in member PQ.
  • Another participant encourages exploring different sections or techniques to resolve the unknown forces at joint J.

Areas of Agreement / Disagreement

Participants express differing views on the effectiveness of the chosen method of sections and the approach to solving for unknown forces. There is no consensus on a single method or solution, and the discussion remains unresolved.

Contextual Notes

Participants mention specific angles and dimensions, but the discussion does not clarify all assumptions or dependencies on definitions, leaving some mathematical steps unresolved.

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Homework Statement



Find the force in member JQ for the Baltimore truss where all angles are 30º, 60º, 90º or 120º.


Homework Equations



Length from A to N can be taken as 6 units.

The Attempt at a Solution



To use method of sections.
 

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What is your understanding of the "method of sections"? How do you think it might apply in this case?
 
Well.. method of sections is like, we have to cut the truss into two portions. the section we take should pass along the member whose force has to be determined. Both the portions will be in equilibrium and we can use either one of the two portions to determine the force on the member.

I have tried to solve this question.. but I am getting stuck.

Assuming the distance from A to N as 6units, then the height of the full truss will be √3 units.

RA (6)= 2.5 (100) + 100 (2)
RA (6)= 450
RA = 450 / 6
RA = 75kN

RN = 100 + 100 - RA
RN = 100 + 100 - 75
RN = 125kN

Figure : (See image attached)

Taking Moments at Y,

\SigmaMJ = 0
Therefore, 125(2) + FXY (√3) = 0
FXY = - 250 / √3 (tension)
FXY = 144.5 (compression)

\SigmaFY = 0
Therefore, 100 = 125 + FXJ + FQJ . (cos30)

\SigmaFX = 0
Therefore, 144.5 = FHJ + FQJ . (cos60)
 

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In general, in 2-dimensional problems of equilibrium, you have available 3 equations of equilibrium, and that is what you have applied, without success. The reason here is that your chosen section intersected 4 members whose forces you didn't know, and so you need an additional equation. For example, if the first section you chose passed through WX, GQ and GH, you would be able to solve for the force in WX in particular. Then, a second section through WX, QX, QJ and HJ would have 3 unknowns and is therefore solvable. In your case you have found the force in XY correctly, but that is not immediately helpful. You could also analyse member HQ by inspection. Incidentally, the method of joints is also a specific application of the method of sections. So the method of sections is valuable. The Baltimore truss is statically determinate, but is a special case that requires a stepwise process to unravel it. There are other examples.
 
thanks a loot :D i was able to solve it with your help :)
 
can you finish how to solve for F(qj)
 
If you look at the rectangular panel XYJL, and imagine the diagonal member PY removed, you will see that the panel becomes rhombic and behaves like a mechanism. The triangle JPL just stiffens the bottom chord and doesn't contribute to the shear resistance of the panel. So PY is the only member resisting the shear in that panel, namely the 125 reaction. So now you can find the force in PQ. If you understood that, you can now look at joint J, where there are 3 unknowns and not enough equations. So you have to find one of those unknowns, again indirectly. Can you try some other sections or the technique I have just suggested to unravel one of HJ, QJ or XJ?
 

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