Homework Help: Statically indeterminate or statically determinate

1. Sep 23, 2016

1. The problem statement, all variables and given/known data
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 ...

2. Relevant equations

3. 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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2. Sep 23, 2016

PhanthomJay

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.

3. Sep 23, 2016

Only 1 reaction at joint B which is internal to the system , right ? which is only HB ?

4. Sep 23, 2016

Can you list out what is the reaction force ,the reaction force that i gt are : HA , VA , VB and VC ?

5. Sep 23, 2016

PhanthomJay

What reactions are you talking about? External at the supports ? Or both internal and external. Also don't forget the moment reaction at A.

6. Sep 23, 2016

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 cant 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 ?

7. Sep 24, 2016

PhanthomJay

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.

Last edited: Sep 24, 2016
8. Sep 24, 2016

I'm getting more confused now , can you state which are the 4 reactions ?

9. Sep 24, 2016

It's stated in the beginning of the question ? there are 6 reactions exist in beam

10. Sep 24, 2016

PhanthomJay

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.

11. Sep 24, 2016

why when we look at the entire part , we dont have to consider thw 2 reactions at joint B ?

12. Sep 25, 2016

PhanthomJay

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.

13. Sep 25, 2016