Statically indeterminate or statically determinate

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Discussion Overview

The discussion revolves around the classification of a beam as statically indeterminate or statically determinate, focusing on the number of reactions and the application of equations for determinacy. Participants analyze a specific problem involving a beam with multiple reactions and joints, exploring the implications of separating the beam into parts for solving the forces.

Discussion Character

  • Homework-related
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants argue that the author of the problem incorrectly states that the beam is statically determinate, suggesting that it should be considered statically indeterminate due to the number of reactions.
  • Others clarify that the reactions at joint B are internal to the system, emphasizing that only external reactions should be counted in the determinacy equations.
  • There is a discussion about the correct application of the equation r = n + 3, with some asserting it applies to the entire beam while others suggest it should be applied to parts of the beam.
  • Participants express confusion regarding the number of reactions, with some stating there are 6 reactions when considering the entire beam, while others maintain that only 4 external reactions should be counted.
  • Some participants highlight the need to break the beam into sections to solve for forces, while others assert that only the equations of equilibrium are necessary for statically determinate beams.
  • There is a question about why internal reactions at joint B are not considered when analyzing the entire beam, with responses indicating that only external supports are relevant in that context.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether the beam is statically determinate or indeterminate. Multiple competing views remain regarding the classification of reactions and the application of determinacy equations.

Contextual Notes

There are limitations in the clarity of the problem statement and the definitions of reactions, leading to confusion about the classification of the beam. The discussion reveals unresolved mathematical steps and differing interpretations of the problem's conditions.

chetzread
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Homework Statement


in this question , it's stated in the beginning that there are 6 reactions ... But , the author gave the equation r = n +3 , r = 4 , n = 1 ...It's statically determinate ...

Homework Equations

The Attempt at a Solution


IMO , the author's is wrong . his original 's intention is to show that the question is statically indetreminate where r > n+3 ( r = 6 , n =1 ) ...so, he broke up beam into 2 parts to solve this question?
 

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You must be careful using these equations for determinancy. The 2 reactions at the joint B are internal to the system. R is the number of external reactions, which is 4, n is the number of internal joints or hinges.
 
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PhanthomJay said:
You must be careful using these equations for determinancy. The 2 reactions at the joint B are internal to the system. R is the number of external reactions, which is 4, n is the number of internal joints or hinges.
Only 1 reaction at joint B which is internal to the system , right ? which is only HB ?
 
PhanthomJay said:
You must be careful using these equations for determinancy. The 2 reactions at the joint B are internal to the system. R is the number of external reactions, which is 4, n is the number of internal joints or hinges.
Can you list out what is the reaction force ,the reaction force that i gt are : HA , VA , VB and VC ?
 
What reactions are you talking about? External at the supports ? Or both internal and external. Also don't forget the moment reaction at A.
 
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PhanthomJay said:
What reactions are you talking about? External at the supports ? Or both internal and external. Also don't forget the moment reaction at A.
The static determinate beam refer to the beam that has been separated (part AB ) , If so , the reaction (r) in r = n+3 refer to Moment at A , HA , VA and VB ?

If I didnt get the beam separated , i would have 6 reactions , which are Moment at A , HA , VA, VB , HB , and VC ? this is statically indeterminate beam , where r = 6 > (n+3) , where n=1 , and n+3 = 4 ?
Since this is this is statically indeterminate beam , so we can't solve the problem ? So , we need to break up the beam into 2 different sections , so that we can solve the forces ?

Is my idea correct ?
 
chetzread said:
The static determinate beam refer to the beam that has been separated (part AB ) , If so , the reaction (r) in r = n+3 refer to Moment at A , HA , VA and VB ?
the equation r = n+3 applies to the entire beam, not its parts. When you look at AB separately, you left out HB as a possible reaction.
If I didnt get the beam separated , i would have 6 reactions , which are Moment at A , HA , VA, VB , HB , and VC ?
no, you just have 4.
this is statically indeterminate beam , where r = 6 > (n+3) , where n=1 , and n+3 = 4 ?
The beam is statically determinate. You just have to look at each part.
Since this is this is statically indeterminate beam , so we can't solve the problem ? So , we need to break up the beam into 2 different sections , so that we can solve the forces ?

Is my idea correct ?
the beam is statically determinate. You do have to break it into parts to solve, but you only need the 3 equations of equilibrium to solve . Indeterminate beams require other equations involving deflections/compatability analyses.
 
Last edited:
PhanthomJay said:
the equation r = n+3 applies to the entire beam, not its parts. When you look at AB separately, you left out HB as a possible reaction.
no, you just have 4. The beam is statically determinate. You just have to look at each part. the beam is statically determinate. You do have to break it into parts to solve, but you only need the 3 equations of equilibrium to solve . Indeterminate beams require other equations involving deflections/compatability analyses.
I'm getting more confused now , can you state which are the 4 reactions ?
 
PhanthomJay said:
the equation r = n+3 applies to the entire beam, not its parts. When you look at AB separately, you left out HB as a possible reaction.
no, you just have 4. The beam is statically determinate. You just have to look at each part. the beam is statically determinate. You do have to break it into parts to solve, but you only need the 3 equations of equilibrium to solve . Indeterminate beams require other equations involving deflections/compatability analyses.
It's stated in the beginning of the question ? there are 6 reactions exist in beam
 
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When you look at the entire 2 parts as one whole beam, there are 4 reactions, 3 at A and 1 at C. Those are the 4 that are used for the determinancy equation. When you split the beam into 2 parts, then the internal reactions at B come into play. Total of 6. The wording is not clear, The determinancy equations are also confusing, and there are others.
 
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  • #11
PhanthomJay said:
When you look at the entire 2 parts as one whole beam, there are 4 reactions, 3 at A and 1 at C. Those are the 4 that are used for the determinancy equation. When you split the beam into 2 parts, then the internal reactions at B come into play. Total of 6. The wording is not clear, The determinancy equations are also confusing, and there are others.
why when we look at the entire part , we don't have to consider thw 2 reactions at joint B ?
 
  • #12
chetzread said:
why when we look at the entire part , we don't have to consider thw 2 reactions at joint B ?
Because there is no external support at B. The supports are at A and C.
When you draw the free body diagram of the entire part, only applied forces and externall support reactions at A and C show in the diagram. Forces at B are internal and do not show. When you draw a ffree body diagram of each part, only then will the B forces show in the diagram.
 
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  • #13
PhanthomJay said:
Because there is no external support at B. The supports are at A and C.
When you draw the free body diagram of the entire part, only applied forces and externall support reactions at A and C show in the diagram. Forces at B are internal and do not show. When you draw a ffree body diagram of each part, only then will the B forces show in the diagram.
The hinge is the external support that we could see , right ?
 
  • #14
chetzread said:
The hinge is the external support that we could see , right ?
You can see them but they are not outside the beam . Imagine that the support at A is a wall that the beam is anchored into, and that the support at C is a wall that the beam is resting on. There is no wall at B, so there is no outside support at B,
 

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