reticular structure reactions

[stef_design]

Hi.
I thought that the hinges in the various points indicated by the brakes did not break the beam in various rods (such as you see in pdf).
to say it in simple words, the hinges are not applied in half-carry of the rods but external.

 

[volaff]

if it is a reticular structure the hinges connect more screw rods from classical definition!

 

[PierArg]

I press that I am also following the thread with particular interest.
Unfortunately I am not prepared to answer because I am a hosp in reticolar structures:biggrin

At least I can say that stef is right: those hinges put in that way do not break the beams

 

[volaff]

I press that I am also following the thread with particular interest.
Unfortunately I am not prepared to answer because I am a hosp in reticolar structures:biggrin

At least I can say that stef is right: those hinges put in that way do not break the beams

starting from the thread title: "reactions structure reticular", hi

 

[stef_design]

At least I can say that stef is right: those hinges put in that way do not break the beams

even if I

 

[crunch90]

even if the gdl counts the structure is hyperstatic, I think you can get the reactions with the overlap of the effects.

the two horizontal forces are discharged equally on the two supports, which will exercise a horizontal reaction equal to f and a vertical f*h of opposite sign between them so as to balance the moment around the cart.

the distributed load will be discarded for half on the cart and for a quarter on each external zipper

by adding contributions to the individual constraints you can find:

on cern sx: fx= fy=q*l-f*2l

on cern sx: fx= fy=q*l+f*2l

on cart: fx=0 fy=2*q*l

 

[PierArg]

starting from the thread title: "reactions structure reticular", hi

You're right.

But I can tell you that the hinges have a different meaning depending on whether they are drawn below/sopra/dx/sx the beam or in the middle.

 

[volaff]

You're right.

But I can tell you that the hinges have a different meaning depending on whether they are drawn below/sopra/dx/sx the beam or in the middle.

He's fine!
the only thing we have understood is that the starting design is ambiguous!

 

[stef_design]

even if the gdl counts the structure is hyperstatic, I think you can get the reactions with the overlap of the effects.

the two horizontal forces are discharged equally on the two supports, which will exercise a horizontal reaction equal to f and a vertical f*h of opposite sign between them so as to balance the moment around the cart.

the distributed load will be discarded for half on the cart and for a quarter on each external zipper

by adding contributions to the individual constraints you can find:

on cern sx: fx= fy=q*l-f*2l

on cern sx: fx= fy=q*l+f*2l

on cart: fx=0 fy=2*q*l

Hi.
I solved the structure with the fem and returns the values found in pdf.
the image at the top of the pdf are the various incognite reactions that are 5 (4 for the two hinges and 1 for the cart).

but the numerical values do not return with your reasoning. Am I wrong? ! ?
Hi.
View attachment REAZIONI VINCOLARI.pdf

 

[volaff]

with the only cardinal euqaizoni of statics for plane systems can not solve the problem.
I'll explain.

with the cardinal equations of the static x plane systems you can split two equations of balance to the translation (horizontal and vertical) + a equation of balance to the rotation.

in this case the unknowns are 5 (2 horizontal reactions and vert x 2 hinges and 1 reaz. vert for the cart).
You have a system of three equations in five unknowns that you can't solve.
what you could do if the struttutra was isostatic (for example with a less zipper)

 

[stef_design]

what you could do if the struttutra was isostatic (for example with a less zipper)

I had also thought of removing a hinge, example that of dx and analyzing only half structure so much then it is symmetric.
But I don't know if it works. .

 

[studio.avarino]

we give for granted that the structure

 

[studio.avarino]

Where did the message go? while I wrote it, he said, "Save it automatically" and then he doesn't send it! I hope you help me recover it, I had written for half an hour.

 

[volaff]

to solve it, it is advisable to use the method of forces using a "supporting" isostatic structure and writing congruence eqautions.
if you want I can attach some pdf with theory and exercises (if I find them).

 

[stef_design]

to solve it, it is advisable to use the method of forces using a "supporting" isostatic structure and writing congruence eqautions.
if you want I can attach some pdf with theory and exercises (if I find them).

Okay, thank you. You'd be so kind to tell me how you would do it. isostatic structure "support".
cation

 

[crunch90]

Hi.
I solved the structure with the fem and returns the values found in pdf.
the image at the top of the pdf are the various incognite reactions that are 5 (4 for the two hinges and 1 for the cart).

but the numerical values do not return with your reasoning. Am I wrong? ! ?
Hi.
View attachment 36337

Forgive me I was wrong in writing:

on cern sx: fx= fy=q*l-f

on cern sx: fx= fy=q*l+f

on cart: fx=0 fy=2*q*l

sw fem results are quite consistent with what I proposed. the only difference lies in the horizontal reaction caused by the vertical distributed load. for how the beam and the inner hinges are drawn, in my opinion, you can consider the beam with the stroke more often as a unique body within which the dots are inserted. if so were, the horizontal reaction is justified and interpretable as the reaction to the bending of the whole body, as from the theorist of the bending of the beam.. definitely the sw manages to evaluate this aspect. I can't help you with a mathematical theory. depends on your purposes

 

[stef_design]

Forgive me I was wrong in writing:

on cern sx: fx= fy=q*l-f

on cern sx: fx= fy=q*l+f

on the cart: ♪

I'm sorry because in the two trolleys like fy you also put the f force?
f does not act only horizontally?

 

[stef_design]

You think I could break the structure and consider it as an isostatic like this:http://design.rootiers.it/strutture/node/619

 

[Legs]

I'm sorry because in the two trolleys like fy you also put the f force?
f does not act only horizontally?

attention: the two forces f produce a global moment at the base worth 2f * 2l = 4 *f * l.
This means that it is necessary that a pair of the same value be born that is given precisely by the vertical components that are 4*l.
dividing. the previous term for this arm is precisely the +/- f vertical values that appear in the binding reactions.
If possible, if you cannot see it in this way consider the balance to the rotation around the trolley in the center.

 

[studio.avarino]

All right, you're upset about losing the message I've already written, I'm back here!
there is no doubt that the structure is hyperstatic so the solution is influenced by the mechanical properties of the rods. in the first line and to simplify things we imagine that all the rods have to, and, j equal or, at least, that the structure does not have geometric dissimmetries type a piece of the current with properties other than those of the other pieces.
the structure is symmetrical only geometrically: for horizontal loads is antisymmetric, only for vertical loads is symmetric! this means that any “measurements” must be different for the two load patterns!
If you want to manually solve the system exactly as it is, that is, considering the external hinges as well as drawings, the thing is really difficult because the unknowns are too many (there are all moments of continuity in knots) and there is a lot to write. to “nase” the method of deformations should give less unknown than that of forces.
if, instead, you want to find an approximate solution of comparison with the exact method the thing could be solved so:
First I see to solve external hyperstaticity, that is the fact that the beam has three supports, considering the similarity between reticular beams and full wall beams (valid for what I said on the sections of the rods). for vertical loads, the two external supports react with vertical forces up to 3/16 of the total vertical load (4*q*l) and the central one with 10/16 of the total vertical load (so 0.75*q*l and 2.5*q*l). what can be traced in any manual presenting the patterns of continuous beams (or if one has a good memory). the simplification that someone suggested (q*l and 2*q*l) and that we all use in doing the analysis of the loads (cariques brought by the beams on which they rest of the continuous floors), would be valid only if we neglect the most important thing in this case that is the continuity, or for a scheme with the zipper in the central support: in our case it could be true if the bridles were interrupted by hinges in half-works or, in the case of triangular knits, when the rod of the upper current is missing in correspondence to the central cart.
As for horizontal forces the explanation has already given it legs: the central support does not contribute while the hinges react with vertical forces equal to f (left down and right up) and with horizontal forces equal to f, directed to the left (equality comes from geometric symmetry).
solved the external hyperstatic can find the internal forces at the auctions to make a maximum sizing: I would use the method of the retter sections. As for the double diagonal in the rectangles, I would simply freghere: to the nose lumen the effort in the actual diagonal rods should be halved regarding what acts in the rods of the simplified scheme with only one diagonal (analogy of the cut between reticular beams and full wall beams).
if one the two diagonals are slender and then the compressed one goes out cause I consider the effort of the scheme with only one diagonal, if I want to make a more rigid structure dimensional the diagonals with the halved effort, but I have to consider the tip load.
remains to be dimensional the lower bridle: will be subject to both normal efforts determined by the method of rettering and even to the flender moments of the continuous beam of 4 equal spans (each of light l) evenly loaded. This time I challenge anyone to remember the results but any manual gives us the solution (e.g. alasia-pugno manual construction and. six).
Can I go get the coffee that went through yesterday for missing the message?