aid solution exercise

[elio71]

Good morning to all, this is my first post on this forum, I was looking for help to check the results related to this exercise:

you have a flange joint with the following characteristics: average radius of 40cm disc contact circular crown, dn10mm bolt diameter, 0.3 friction coeff; 8 bolts, circumference radius to which there are 39cm bolts, yielding voltage of the sigma sn 500mpa bolt material. determine the maximum transmissible moment from the joint both in case the bolts work to cut and in case they work to traction, considering the calibrated diameter of the bolt pairs to 1.1 times that of the dn hazel.
According to what I interpreted for traction calculation means the maximum shooting of bolts up to the sigma sn considering then friction between the disks, while by cutting it is obtained from the sigma sn multiplied by 0.7. I made the calculations, I just wanted a numerical match.
the formulas I have used are found on the vol i of the young people.
Thank you very much to all.
Helium.

 

[Fulvio Romano]

According to what I interpreted for traction calculation means the maximum shooting of bolts up to the sigma sn considering then friction between the disks, while by cutting it is obtained from the sigma sn multiplied by 0.7.

In both cases I would put a security factor. Otherwise, if the bolt deforms plastically, do not unscrew it anymore.

 

[MBT]

I made the calculations, I just wanted a numerical match.

Let's turn the problem down.. .
you put the solution and we will comment on it....:tongue::biggrin:

 

[MBT]

... maximum bolt shot to the sn sigma then considering friction between the discs...

generally never exceeds 80% of the yielding voltage:wink:

 

[elio71]

thank you guys for the attention.
It should be considered that it is an exercise not a practical application, no security factor or bolts that do not unscrew etc.
However to me it turns out about 3800 kg*m considering friction between the two flanges and about 2,7 times greater the value of the moment with bolt cut seal.
I asked you because not having experience on joints of this type they seem exaggerated values.
I also tried a solidworks simulation, but I can't handle the bullion issue well. . .

 

[PierArg]

hi, as already recommended by lightning I still put security factors on the exercises...the bad habit of dealing things only from the point of view "didactic" then leads to madornal horrors in practice.

 

[PierArg]

hi, as already recommended by lightning I still put security factors on the exercises...the bad habit of dealing things only from the point of view "didactic" then leads to madornal horrors in practice.

I also speak for personal experience. .. during the university they made me do the exercises as if I was in the "country of wonders", then in practice the accounts never came back because I ate the yields.

 

[elio71]

Of course, I agree with you, in this case I just need a numerical match regarding the exercise data.

 

[PierArg]

1) Traction bolts:
mt = 37680 nm



2) Cutting bolts:
mt = 74088 nm
for this case I considered tau amm = 0.5 sigma amm

 

[PierArg]

I forgot two things:

1) Why do you consider tau amm = 0.7 sigma amm?

2) My results are obtained from ideal values. Normally, as mbt said, 80% of sigma amm is used

 

[elio71]

I forgot two things:

1) Why do you consider tau amm = 0.7 sigma amm?

2) My results are obtained from ideal values. Normally, as mbt said, 80% of sigma amm is used

Hi.
on the young man calculates it as I have indicated, I carry from the text: a tau=0,7*sigma sn (snervamento).
Thanks again.
Helium.

 

[elio71]

1) Traction bolts:
mt = 37680 nm



2) Cutting bolts:
mt = 74088 nm
for this case I considered tau amm = 0.5 sigma amm

Perfect!
I saw the results now,
I get a bigger cut mt as I used a 0.7, the traction moment is very similar to mine.
Thank you very much, you were very kind and precious.
Helium.
ps: Do you always use a 0.5 factor for tau?

 

[PierArg]

ps: Do you always use a 0.5 factor for tau?

if it is not indicated by the norms or various theories (in this case it is clearly indicated by young people) it is assumed that tau amm is half the sigma amm.

In fact if you made the mohr circle of the traction test of the material you would get that the tau amm is the radius of the circumference. the sigma amm instead is the diameter, or 2 times tau amm.

kindly correct me if I said a str...ata! ! ! ! !

 

[PierArg]

Thank you very much, you were very kind and precious.

Please, I did it with pleasure!

the questions that ask others are a great method to repeat what I have studied

 

[elio71]

In fact if you made the mohr circle of the traction test of the material you would get that the tau amm is the radius of the circumference. the sigma amm instead is the diameter, or 2 times tau amm.

kindly correct me if I said a str...ata! ! ! ! !

if you consider an effort of pure traction (circle with origin on x positive, tangent the ordered) actually tau worth the radius then 1/2 sigma.
Helium.