calculation of uncertainty and safety index

[Joshua662]

Hi guys I write this topic only to have a confirmation on an exercise found on a book that I am reading i.e. project and construction of machines of shigley the text recità so.

the load of a structure is known with an uncertainty of + or - 20% and the load that causes the break is known with an uncertainty of + or - 15 %.
if the load causing breakage is rated at 10000 n determine the safety coefficient and the maximum permissible load to exceed all uncertainties.

I performed so:

10000+20%=12000
12000/10000 = 1.2

10000-15%=8500
8500/10000=0,85

nd= 1/0,85 / 1/1,2 = 1,4 about 1/0,85

maximum permissible load = 10000/1,4 = 7140 n

How to proceed is it right or is it wrong? I know that for you it is trivial as what, but for me that I have resumed my studies at 32 years it is very complicated.

with this wording 1/1,2 what is indicated? that every n has a break rate 1.2%?

ps. Excuse my ignorance... .

 

[meccanicamg]

Hi guys I write this topic only to have a confirmation on an exercise found on a book that I am reading i.e. project and construction of machines of shigley the text recità so.

the load of a structure is known with an uncertainty of + or - 20% and the load that causes the break is known with an uncertainty of + or - 15 %.
if the load causing breakage is rated at 10000 n determine the safety coefficient and the maximum permissible load to exceed all uncertainties.

I performed so:

10000+20%=12000
12000/10000 = 1.2

10000-15%=8500
8500/10000=0,85

nd= 1/0,85 / 1/1,2 = 1,4 about 1/0,85

maximum permissible load = 10000/1,4 = 7140 n

How to proceed is it right or is it wrong? I know that for you it is trivial as what, but for me that I have resumed my studies at 32 years it is very complicated.

with this wording 1/1,2 what is indicated? that every n has a break rate 1.2%?

ps. Excuse my ignorance... .

but the solution that you posted you weighed it from the book or is it your resolution attempt?

apart from that if you copied all the text, it is not very clear. I have reasoned in the attached way finding the worst condition of applied load and the lower limit that causes breakage. maximum scissor condition, i.e. material that breaks at the minimum of its resistance and force as much as possible to apply. of course for the safety coefficient should be imposed as follows:

1 = to obtain the limit value of fr0
1.5 = for ductile materials if fr0 is tied to the yielding load rs
3 = for fragile materials if fr0 is tied to rm traction breakage load

basically it seems to me a little intelligent exercise and that it doesn't need anything, but it doesn't matter, you've asked question and I have given you my interpretation.

I am waiting for comparison :biggrin:

 

[Joshua662]

the exercise was carried out in a partial way gave only the results I with a lot of patience and with other formulas came to that solution of the book but I did not know whether the procedure was right or not.... thanks always very kind, pearl workstation all in place the drivers for the video card released from nvidia still have some compatibility problem with the nvidia video cards ewin 7.

 

[meccanicamg]

the exercise was carried out in a partial way gave only the results I with a lot of patience and with other formulas came to that solution of the book but I did not know whether the procedure was right or not.... thanks always very kind, pearl workstation all in place the drivers for the video card released from nvidia still have some compatibility problem with the nvidia video cards ewin 7.

I hope I've helped you :biggrin: for the workstation I'm glad you've arranged at least the main things. for drivers we are always waiting for some really tested drivers to be released, then some tricks are always found :finger:

 

[Exatem]

Hi guys I write this topic only to have a confirmation on an exercise found on a book that I am reading i.e. project and construction of machines of shigley the text recità so.

the load of a structure is known with an uncertainty of + or - 20% and the load that causes the break is known with an uncertainty of + or - 15 %.
if the load causing breakage is rated at 10000 n determine the safety coefficient and the maximum permissible load to exceed all uncertainties.

a short break.
a certain ing. silvia, which dealt with naval constructions, defined the safety coefficient as a "coefficient of ignorance" pointing out that, as much as we know about the calculation of structures and the resistance of materials, the lower can be such a factor.
Greetings to everyone.

 

[Joshua662]

exatem what do you mean by this? I don't understand. as ignorant as I am and as you wrote....

 

[Exatem]

exatem what do you mean by this? I don't understand. as ignorant as I am and as you wrote....

I didn't make it up, but it's a step taken from U. Costaguta's "water sinks", but I feel like sharing it.
I think the silvia meant that, knowing perfectly both the characteristics of the structure, and the qualities of the material, you could give up the safety coefficient.
In the end, if you think well, adopt a safety coefficient, we only need to be careful in front of the "incognite", that is what we suppose but we do not know for sure.
Of course, thinking of being able to do without it is only a utopia, for example any material, despite knowing its characteristics of yielding, breaking, etc.etc. always maintains a certain risk that such characteristics are not uniform throughout the structure, or that the structure itself, is in its life, having to endure an unforeseen load, and so on...
I mean, we're cauteli because we ignore.

 

[Joshua662]

ahhhhhh now I understand well seen from this point of view is a nice thought, and I give it in part right. Thanks for the explanation, you were very kind.... have patience with me, I like to study I like to learn, I like to learn new things even if very difficult to understand for my imitated bases, but I try to learn and enter this fascinating field of mechanics in general... .

 

[peloritano]

I can't find the result in your formula... according to the short track is the known term and is equal to 10000 while fo is the unknown then applying your equation fo=7083 from which the safety coefficient is 10000/7083=1.4