Solving Beam Theory & Statics Problems with Difficulties

[CalebP]

I'm having quite a bit of difficulty with the question posted - even from part 1.

I understand I have 6 unknowns and 2 seperable beams so we're statically determinate here. I can split the pictured beam into two parts at the hinge and apply the static equations twice.

On the right part I find that all forces are equal and opposite (Rbx = Rcx, Rby = Rcy, if drawn in the correct directions), however I notice I never used the moment equation or the fact that the reactions at B occur 3m higher than C - I think this is my first pitfall. Also, taking moments about any point on the RHS FBD results in all reactions = 0. For example, ƩM(c) = 0 ⇔ Rby*3 = 0 Rby = 0.

I'm around 99% sure my methods are wrong above. I run into similar problems on the LHS of C. I've tried splitting that one further into parts AD and DB, and replacing the bend by axial reactions Nad in both. I get a little further here, but again don't even use the 6m height of the bar, so I'm wrong somewhere.

I can't do part 2 without the reactions first. I'm sure I'd run into difficulty though due to those turns... how am I supposed to orientate SF and BM diagrams like this. On beams they purely run in the x-direction to the right, do I draw SF/BM diagrams along the y-axis for these beams?

I can probably get the displacement of D due to δ = NL/EA.

The last part is simply (I think) discovering the centroid of that cross-section - which I can do.

I think my greatest misunderstanding here is applying the sum of moments equation multidirectionally.

Thanks guys.

 

[tiny-tim]

Welcome to PF!

Hi CalebP! Welcome to PF!

On the right part I find that all forces are equal and opposite (Rbx = Rcx, Rby = Rcy, if drawn in the correct directions), however I notice I never used the moment equation or the fact that the reactions at B occur 3m higher than C …

Sometimes it's helpful to ignore the x and y components, and to consider the whole vector.

Using moments, what can you say about the line of application of the whole reaction force R_B ?

 

[CalebP]

Grrr still can't solve this question.

If I separate this into two FBDs the RHS of the structure tells me no matter what I try that all of the reactions in both directions are equal and opposite, and the same value.

I'm doing something wrong, but I don't know what, but every time I try to reapply my found values to the entire structure I know I get the wrong answer.

I've drawn this so many times now :( Is it true to say that there are no reactions in the x-direction? (this HAS to be true yeah?! Because the RHS of the hinge at B shows the x-directional forces at supports B and C are equal, and then the x-direction sum across the whole structure forces the x-reaction at A to be zero)

If the above is wrong I don't get this at all.

 

[tiny-tim]

Is it true to say that there are no reactions in the x-direction? (this HAS to be true yeah?! Because the RHS of the hinge at B shows the x-directional forces at supports B and C are equal, and then the x-direction sum across the whole structure forces the x-reaction at A to be zero)

No, the x-reaction at A will also be equal to those at B and C, won't it?

Take moments about C for the RHS …

what can you say about the moment (about C) of the reaction force at B? ​

 

[CalebP]

If I take moments about C on the RHS only, I get:

3Rby - 3Rbx = 0 => Rby = Rbx

If I sum the y's I get Rby = Rcy and Rcx = Rbx in the x-direction. The boundary conditions don't only for any moments...

Is there a concentrated moment = 3Rby where the beam turns?

 

[tiny-tim]

If I take moments about C on the RHS only, I get:

3Rby - 3Rbx = 0 => Rby = Rbx

[STRIKE]correct[/STRIKE]

oops! not correct, should be Rby = -Rbx

but an easier way of seeing this is to see that the total reaction force at B must have zero moment about C, so it must go through C (ie at 45°)

If I sum the y's I get Rby = Rcy and Rcx = Rbx in the x-direction. The boundary conditions don't only for any moments...

Is there a concentrated moment = 3Rby where the beam turns?

i don't understand any of this

you now know the total reaction force at B is at 45°

so using the vertical 5 kN, and the same principle, what is the direction of the total reaction force at A ?​

 

[CalebP]

The way I have it set-up up, the LHS has 3 unknowns as Rby = Rbx (or Rb@45deg)..

Forces in x: Rax = Rbx (=Rby)

Forces in y: 5 + Rby = Ray

Moments about B: 5*1.5 = 6Rax + 3Ray

That gets me 4.166 in the y-dirn and -0.833 in the x..

So the final resultant at A would be 4.249kN 78.7 deg from the x-axis..

I now have a solution where the LHS works out and so does the RHS, but as a whole it doesn't. I was under the impression the equilibrium equations needed to be satisfied for the whole structure as well (maybe this was my mistake)

 

[tiny-tim]

I now have a solution where the LHS works out and so does the RHS, but as a whole it doesn't. I was under the impression the equilibrium equations needed to be satisfied for the whole structure as well (maybe this was my mistake)

how does it not work out (A and C and 5kN) ?

 

[CalebP]

For the structure overall the sum of the forces do not equal zero. In the x-direction my calculated reactions are 0.833, -0.833 and 0.8333 at A, B and C respectively.

I have 4.167, 0.833, -0.833 at A, B and C with a -5 force also.

So my equations are satisfied on each side, but overall they do not :S

 

[CalebP]

Ok I don't know what I was doing. My algebra must suck.

I finally just grabbed all 6 equations and solved it online:

1 0 -1 0 0 0 0
0 1 0 1 0 0 5
0 0 6 3 0 0 7.5
0 0 0 0 3 -3 0
0 0 1 0 -1 0 0
0 0 1 0 0 -1 0

This gets me all reactive forces = 0 besides two vertical reactions at 2.5kn in the y-direction. Is this correct?

 

[tiny-tim]

but you can't include B, that's internal!

(or if you do, you must include B twice, in opposite directions, making zero)

 

[tiny-tim]

This gets me all reactive forces = 0 besides two vertical reactions at 2.5kn in the y-direction. Is this correct?

no!

your original result was correct

a quicker way of seeing this is to say that the three LHS forces must all go through the same point, which is clearly at (1.5,7.5), so that R_Ay must be 5 times R_Ax ​