dimensioning a screw

[maxopus]

So...
the company has bought millions and millions of screws m6 class 5.8 , for various reasons (office purchases ) of course and I repeat obviously if you could change , I didn't even write.. Right.
here we have a condition "must" that is so point!! not because I say it but because the company imposes it... now as there are safety coefficients that allow to verify the mechanical organs.... I was wondering if there was a soul pia that explained to me please how to do, so to tell the customer the vine holds only if you respect these conditions... but not the buzzard, and the office head that mounts the secretary etc...etc.. .
...
. .
cordially

Sebastiano

I suspected it was like this
now they have to use m6 also to spread the snow if necessary.

 

[maxopus]

I wanted to add... as I have to renovate my house, how much do they do to me if I have a milionata?

 

[Exatem]

So...
... the office head that mounts the secretary etc...etc.. .

depends very much on the secretary.
In my case, I'd prefer the buzzard. .

 

[MBT]

of theories and methods to dimensional the bolted joints are to iosa, more or less scientific, more or less empirical.
and, incredibly, all come to similar results
then, we start from the weight to support.
130 kg, to be supported with 6 screws
are 21,6 kg each screw
as security, we say multiply it by 1.5 to consider the dynamic factor and 1.5 to consider the fact that they are hand-held
are therefore 21.6*1.5*1.5=48.6 kg each screw must support
it supports it by friction. the friction coefficient you defined it in 0.15. therefore, the axial load that must provide the screw will be equal to
48.6/0.15=324 kg
Now, you can make two real.
on the tables around the screws, find the tightening and precarious values for the screws
in the table that I use, a m6 in class 5.8 is tightened up to 6.8 nm and get a preload of 6.1 kn (622 kg). This leads you to say that the screw holds with an additional security factor of 1.9
the other verification is to say that the screw in class 5.8 has yield load of 500*0.8=400 n/mm2 (40.7 kg/mm2)
if I hold it to 80%, the tension with 40,7*0.8=32.56 kg/mm2
Since for a m6 large pitch the resistant section is 20 mm2, the axial load that generates the screw is equal to 651 kg (note that is gross similar to those 622 kg previously). as you have calculated that you need 324, even here you have an additional security factor of about 2

where is the "truck" for which you suggest to change vines?
simple... In the meantime a screw like this is a moment to ruin it while you're smiling, so your calculations of loads and precarious go to ladies
second consideration. . 130 kg are "little" and the influence of external factors is felt. between the serious and the facto, if one leans on the picture immediately adds another 50 kg. and if you think that from that painting we have to get out of the electrical cables (which weigh... and so much!) you will understand that the true safety is down!

However, I repeat the concept, you are in a beautiful dilemma.
What do you mean "you can't change, the situation is like this." ?
If the calculation said "don't hold", what would you have done? kept so why can't you change?
the calculation now tells you "rules, but it is not as safe as the experience suggests"
So once again... .

or deny yourself (of the series you tell the customer "better to use the screws m8 class 8.8) but you are sure to sleep at night
or confirm the vines and experience that on that painting never make the nest a buzzard, otherwise it risks coming down with all the nest and the eggs

 

[MBT]

depends very much on the secretary.
In my case, I'd prefer the buzzard. .

Here... depends:
we say that with someone it would be worth checking the electric picture!:biggrin:

 

[Exatem]

Here... depends:
we say that with someone it would be worth checking the electric picture!:biggrin:

you have not seen our secretaria pantegana leopardata.. . .

 

[MBT]

and you should see our letter or the lawyer;-)

 

[Sebastianora]

Thanks really, now it's all clear!!!!

 

[Enigma]

and you should see our letter or the lawyer;-)

che invidia :frown:

enigma

 

[PierArg]

of theories and methods to dimensional the bolted joints are to iosa, more or less scientific, more or less empirical.
and, incredibly, all come to similar results
then, we start from the weight to support.
130 kg, to be supported with 6 screws
are 21,6 kg each screw
as security, we say multiply it by 1.5 to consider the dynamic factor and 1.5 to consider the fact that they are hand-held
are therefore 21.6*1.5*1.5=48.6 kg each screw must support
it supports it by friction. the friction coefficient you defined it in 0.15. therefore, the axial load that must provide the screw will be equal to
48.6/0.15=324 kg
Now, you can make two real.
on the tables around the screws, find the tightening and precarious values for the screws
in the table that I use, a m6 in class 5.8 is tightened up to 6.8 nm and get a preload of 6.1 kn (622 kg). This leads you to say that the screw holds with an additional security factor of 1.9
the other verification is to say that the screw in class 5.8 has yield load of 500*0.8=400 n/mm2 (40.7 kg/mm2)
if I hold it to 80%, the tension with 40,7*0.8=32.56 kg/mm2
Since for a m6 large pitch the resistant section is 20 mm2, the axial load that generates the screw is equal to 651 kg (note that is gross similar to those 622 kg previously). as you have calculated that you need 324, even here you have an additional security factor of about 2

where is the "truck" for which you suggest to change vines?
simple... In the meantime a screw like this is a moment to ruin it while you're smiling, so your calculations of loads and precarious go to ladies
second consideration. . 130 kg are "little" and the influence of external factors is felt. between the serious and the facto, if one leans on the picture immediately adds another 50 kg. and if you think that from that painting we have to get out of the electrical cables (which weigh... and so much!) you will understand that the true safety is down!

However, I repeat the concept, you are in a beautiful dilemma.
What do you mean "you can't change, the situation is like this." ?
If the calculation said "don't hold", what would you have done? kept so why can't you change?
the calculation now tells you "rules, but it is not as safe as the experience suggests"
So once again... .

I'm talking about the big one!

 

[gtemp]

and where, of grace, is such a thing affirmed? What is machine regulations?


first of all a question. Have you ever seen a life break? and instead have you ever seen a dowel parade from the wall, or the wall break?
the weak link of the chain must be checked, and if you have bolts to a wall, it is the wall the first to break. then you have to dimensional the dowel according to the masonry, not the screw. I think using m6 is like hanging clothes with the ball pins. I would not go under m12 rupture of the dowel during the screwing.

if you want to dimensional a bolting as if it were structural, then it proceeds like this:
- the weight to support is 'p'
- the friction force to support the weight 'p' will be fa = mu*p*k, where mu is the friction coefficient and k the safety factor (attrite)
- divide makes for the number of screws
- I find the screw whose section resistant equivalent 'a' (to look at catalog!!!) is such that: sigmay = does * k' / to where sigmay is the yielding sigma and k' is the safety factor (for yielding)

the screws that can be cut are only those with rectified stem, if the cutting work takes place on the rectified part of the stem. Of course, we're not talking about self-taxing. . .

Clear?

I'm sorry if they intervene "by argument" but I noticed that in these formulas there is something wrong, especially in the formula of the friction force, mu should be to the denominator.

 

[Er Presidente]

Yes, it happened but of course you can't be quiet.

p.s: Have you ever thought about why the customer asks you this question by questioning the choice of m6?
He obviously doubts her. :tongue:

Why? He can't see them! think they are the locks to keep the door closed during transport.

 

[Er Presidente]

I think we missed a moment. The question is this:
6 m6 screws are good to support a load of 130 kg static-fixing on steel friction coefficient 0.15 ??? ?

I don't think there are any other answers right? ?

Thank you.

)

No, because of the six vines, it's gonna be a miracle if three of them work together.
the holes made "hand" are never aligned and in the end they always work 3 screws (the first three that "pen" and cut because they're badly tight).
Invent an excuse (the warehouse error or better than "integrated standard management system") and tell him to put 12 of screws, hoping that in addition to having (maybe) 6 screws at work, you will also be happy with your boss, which consumes all the m6 in circulation.

 

[Fulvio Romano]

( warehouse error or better than "integrated standard management system")

I'll be useful. .

 

[Fulvio Romano]

if you want to dimensional a bolting as if it were structural, then it proceeds like this:
- the weight to support is 'p'
- the friction force to support the weight 'p' will be fa = mu*p*k, where mu is the friction coefficient and k the safety factor (attrite)
- divide makes for the number of screws
- I find the screw whose section resistant equivalent 'a' (to look at catalog!!!) is such that: sigmay = does * k' / to where sigmay is the yielding sigma and k' is the safety factor (for yielding)

I'm sorry if they intervene "by argument" but I noticed that in these formulas there is something wrong, especially in the formula of the friction force, mu should be to the denominator.

Yes, it is all wrong. . .
I meant that the friction force does have to be:

♪

Obviously equal to weight, more something for safety.

the force f of traction of the vines, which generates it will be:

f = fa / mu

and it is f that should be divided by the number of screws.

I have to take it easy with alcohol... it makes me write strange things...

 

[Enigma]

I have to take it easy with alcohol... it makes me write strange things...

do not say so... do not pass with them (the asses) remain faithful to the dark side of alcoholism http://www.cad3d.it/forum1/attachment.php?attachmentid=28568&stc=1&d=1344165484enigma

 

[New Rider]

excuse me if I interfere (everyone I am so ashamed even in doing so).. Anyway. .
will I need to understand the tension that arises when the screw reacts by opposing the ''force weight' of the electric picture? If the friction force that arises between the screw and the wall will be higher than the one that will be present between the screw and the painting (binding reaction) can I do without checking the cut?
I don't know if I was clear in expressing my concept. .
thank you all.. .

 

[PierArg]

excuse me if I interfere (everyone I am so ashamed even in doing so).. Anyway. .
will I need to understand the tension that arises when the screw reacts by opposing the ''force weight' of the electric picture? If the friction force that arises between the screw and the wall will be higher than the one that will be present between the screw and the painting (binding reaction) can I do without checking the cut?
I don't know if I was clear in expressing my concept. .
thank you all.. .

Hi.
screws usually never have to work to cut.

In case this is inevitable, we should use calibrated screws (they cost an eye of the head) or insert the screw into a bag that "protects" by acting like a huge hollow plug.

When you're thirsting for a vine, you imprison an axial force that tends to hold the parts together. this force, multiplied by the friction coefficient, is the resulting in radial direction that opposes the sliding (cut).
you have to do so to put a number "n" of screws so that the resulting radial force balances the flow (of course taking into account the safety factor).

 

[meccanicamg]

Hi.
screws usually never have to work to cut.

In case this is inevitable, we should use calibrated screws (they cost an eye of the head) or insert the screw into a bag that "protects" by acting like a huge hollow plug.

When you're thirsting for a vine, you imprison an axial force that tends to hold the parts together. this force, multiplied by the friction coefficient, is the resulting in radial direction that opposes the sliding (cut).
you have to do so to put a number "n" of screws so that the resulting radial force balances the flow (of course taking into account the safety factor).

But you want to put screws m6 to keep on an electrical picture... which hurt if they work to cut? You don't need those calibrated in this sector, you don't need to go a little bit.

it seems strange but in the world of electrics it is all overwhelmed and it is always on everything. as when they mount together 2 or 3 walk-in closet with screws m6... maybe only 4 and two brackets on the roof with 4 screws m6. but nothing has ever broken... it deforms before the sheet of the picture.

the application coefficient must be assessed. . .using the logic of the real application.
attached the good formulas of the now withdrawn one cnr 10011 which is always good to understand how to make the accounts.