lifting attachment

[Fulvio Romano]

but at the end of the fair, what load can the ring bear? can keep on the machine, with what degree of safety?

a saperlo...

@fulvio Roman was the real lifting test made? What happened?

Not yet.. You'll be the first to know. :

 

[Fulvio Romano]

Moreover the characteristics of the s235 should be understood because if it is cut from a sheet is one thing....if it was a diameter round 150 turned is like butter.

That's what I say. the calculations are made on the tables of the materials, the tables are precautionary and refer to the material less structurally valued at equal chemistry. probably s355 from round ha y=355, from sheet is more on. If you do the calculations you have to report to the tables, but if you do the tests with the real material (from which the normalized tables) you get higher numbers.

 

[Fulvio Romano]

other hypothesis, but this easily verifiable to the fem, is that straight traits help the distribution of tensions.

I would say the hypothesis is wrong.
the two fems report practically the same tensions both for the round, and for the oval, indeed for the oval are slightly higher. the maximum tensions are high for the way to apply constraints and efforts, but looking at the istograms I would say that the voltage is 625mpa for the round and 760mpa for the oval with an applied force of 1600kg, that is that reported in the table of the #29.

these rings declare that 1600kg is the wll, or 2.5 times less than the yield y=930mpa

deduces that alternatively:

• the fem returns a voltage 2* times the real, or
• the real material is 2* times more resistant than the y=930mpa model, or
• linear elastic simulation at the limits of yielding is an unenforceable model.

In cases 1 and 3 I feel quiet with that ring. if instead it is the case 2 and then everything depends on the real material of the ring.

(*) if the wll is 2.5 the yielding, it means that the real link has a voltage of 930/2.5 = 372 mpa which is the middle of the 760 mpa that returns the fem.

 

[meccanicamg]

I can tell you that a linear fem continues its straight s-e infinity and this is not real when we go to work in the elasto-plastic field.

As for the comparison between elastic and plastic field, I can tell you that for rectangular steels the elastic force is 2/3 compared to the total plastic.Moreover you must also consider that the engineering curve and the real one differs not by little.so it is a moment to have double values.

 

[Mattymecc]

If you and @meccanicamg don't happen I have to be explained grossly and franticly. I try to explain myself better. we have a ring in carpentry and a normalized link. the two objects have mainly three differences: different material, different section, express load as wll = 1/2.5 times the yield. As geometries are analogous I tried to calculate three scale factors for the three conversions (the ratio between the yields of the two materials, the ratio between the inertia moments of the two sections and... 2.5) so that the product of the three factors gives me the factor of scale between the ring model and the model-link, thus putting them in the same conditions of material, section and test conditions.
the result is that the ring, under the same conditions as the link, should yield to 2270kg instead of 1200kg of the fem (two different models). the explanation is:

1. the fem is wrong for some inscrutable reason
2. the ring and the link differ for some additional aspect besides the three shaped by the scale factor.

assuming that it is true the 2 I launched in a couple of considerations knowing that the wll of the link is not calculated, but it is the result of a test on a real material. and the real materials, especially when laminated, have characteristics even much higher than the precautions of the tables (which typically refer to the fork between casting in source and continuous casting).
other hypothesis, but this easily verifiable to the fem, is that straight traits help the distribution of tensions.

Maybe I know what you mean: you have taken the maximum load of the commercial ring (or, however, the nearest commercial piece) and have estimated the maximum load of your ring by making proportions based on the yielding loads and the inertia of the two sections. I think it cannot work very much because changing geometry also change the values of the stressful moments, not directly proportional to the characteristic size of the structure:

to better explain what I mean, in a uniform load beam, resting at the extremes the maximum moment is mmax = pl^2/8.

double l and quadruple mmax.

 

[Mattymecc]

If you give me a reference geometry for the ring with straight strokes I can try to see what happens with the fem...