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reticular structure reactions

  • Thread starter Thread starter stef_design
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I forgot to say that for the beam systems in which there are diagonal elements the static regime is influenced much more by the normal effort than at the moment flenching since the bending deformations are greater than those of traction/compression. I therefore believe that the moments of continuity, neglected by resolving the reticular with ritter as if the hinges cut all the rods, do not affect the normal efforts of the rods in an appreciable manner.
 
the structure is symmetrical only geometrically: for horizontal loads is antisymmetric, only for vertical loads is symmetric!

but horizontally I don't have 2 f forces at the 2 extremes. is it not symmetrical according to the axis of symmetry that passes through the cart?
 
It would be symmetrical if the forces were directed one to the left and the other to the right, you don't think.
 
Where do we drink it?
I'm a bit far away, but if you want to take a ride from the mountaineering parts (dry tip) I'll jump to offer him that I'm standing there
 
a consideration if we divide this beam into two. the first consisting of the left zipper to the cart the second from the cart until the dx zipper, the two beams are solved and then the results are added on the central cart? Can it be an idea or is it bullshit?
I'm a bit far away, but if you want to take a ride from the mountaineering parts (dry tip) I'll jump to offer him that I'm standing there
cabbage we are distant, even if I happen to come in Sicily 3-4 times in a year.

enigma
 
It's just that with the cell I couldn't erase the written msg
You know, I've proven I don't know how to use the forum well even from the fixed computer! (I still look for the message "Saved automatically", if someone told me where they are saved automatically would not be bad!).
 
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).
Are you sure it's 3/16 and 10/16?04-04-2014 08-06-23.webp
 
Are you sure it's 3/16 and 10/16?
Sorry stef, if I were one of my pupils I would have caxed you.. the scheme that I have considered takes into account only the three supports constituted by the external constraints, but in our case, between an external bond and the other the distance is 2*l and not simply as in the figure you post. I wrote 3/16 and 10/16 of the total vertical load. You make a little count and then you answer me: 2 times 0.75*q*l and 2.5*q*l does exactly 4*q*l which is the total load. stay tuned:cool:
Don't take it, when prof. takes over, I become a "bad beast." Bye!
 
regarding the photo of your profile I do not know if I agree to be so bad: You look great, big and muscular! :smile:
 
Okay!

if I removed the two concentrated loads f and kept only the distributed load q what would be the horizontal reactions of the two hinges?
 
I'm sorry how much you give me to come and do the classes you're attending? :Smile: apparently the teacher should not be so clear and available to deserve the assignment!
 
I'm sorry how much you give me to come and do the classes you're attending? :Smile: apparently the teacher should not be so clear and available to deserve the assignment!
No, no, no. but it is simply fem and therefore does not predict theory.
Was it just my question? Maybe he has a simple answer for you and now I miss him:
Hi.
 

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