2nd Degree Inderteminacy for Structure Using Force Method

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    Force Structure
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Discussion Overview

The discussion revolves around calculating support reactions for a specific structural frame using the force method, also known as the unit load method or virtual work method. Participants are exploring the displacement calculations for both horizontal and vertical loads applied to the frame, with a focus on understanding the vertical deflection resulting from a horizontal unit load.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant describes the process of removing redundants at the pinned support to create a statically determinant frame and applies unit loads to analyze displacements.
  • The participant expresses confidence in calculating horizontal displacement but encounters difficulty in determining vertical deflection due to a horizontal unit load.
  • Another participant points out that the diagrams provided lack clarity and detail, making it difficult to follow the calculations.
  • A later reply critiques the accuracy of the deformation depicted in the diagrams, suggesting that certain joints cannot move as shown and that the structural behavior may not align with real-world expectations.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the clarity of the diagrams or the accuracy of the depicted structural behavior. There are competing views regarding the correctness of the deformation assumptions and the calculations presented.

Contextual Notes

The discussion highlights limitations in the clarity of visual aids and potential misunderstandings regarding the structural constraints and behavior under load. Specific assumptions about joint movement and deformation are not fully resolved.

Tygra
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Homework Statement
Calculating Deflections
Relevant Equations
In question
Dear all

I am trying to find the support reactions on the following structure:

Structure.png


The frame member in the horizontal is 8m long and in the vertical the member is 5m.

To do this I am using the force method (or unit load method or virtual work method).

Firstly, I removing the redundants at the pinned support to make a statically determinant frame as shown below:

Removing redundant.png


Next, I apply units loads in place of the pinned support that looks like this:

Unit loads.png



Firstly, lets consider the horizontal unit load. I need the displacement in the parallel and perpendicular directions as a result of this horizontal unit load.

Calculating the horizontal displacement as a result of this load is no problem. It is simply the sumof the integration of the bending moments.

Virtual structure Moment.png


So, the moment functions are: Mx = -1*x and Mx = 5. Hence, the integration to compute the delection in the horizontal direction is

1728644942535.png


The area where I am a little stuck is computing the vertical deflection as a result of the horizontal unit load.

If you see here from the software the vertical displacement is 4.074 mm.

Virtual structure displacement.png

So my question is: how do I calculate this displacement using the force method?


Many thanks in advance.
 

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I can follow your description, but it is impossible to see details of the posted diagrams.
 
Hi Lnewqban,

Sorry for this.

Is this any better?

Frame:

Frame.png


Statically Determinant Frame
Primary Structure.png


Application of Unit Loads:
Unit loads.png


Bending Moment diagram for horizontal unit load:

bending moment.png


Displacement for horizontal unit load:

displacement.png


Bending moment diagram for vertical unit load:

bending moment 2.png


Displacement for vertical unit load:

displacement2.png
 

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The numbers can't be clearly seen still.
I see that the diagrams shown the deformation do not match how the structure would really deform.
Hinge 3 can't move horizontally and joint 2 can't move upwards.
The beam located next to 1 can't remain horizontal due to the distributed load.
The angle of joint 2 should remain more or less 90°.
 

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