check ala trave heb140 for paranco

[Mello Dessi]

Hi, I'm new, and I'd like to thank for those who will answer me.
My question is this:
I should check the scope of a heb140 beam material s235, the 2 supports are welded at a distance of 4mt, on 2 plant.
this hoist with wheels, it must be mounted and flows on the lower wing of the heb140.
the hoist has a capacity of 2 tons.
I wanted to know if you have to calculate the size of the wing alone (how?) or just calculate the beam in the classic way of "trave doublely stuck with concentrated load"? !
to better understand I attach the photo:
Thank you in advance
greetings

 

[biz]

If you want to be precise in your calculations you can use the fem method.
But first you do a manual calculation.
Consider that the bond is not exactly an ink, the better are two incastrious.
consider it as a disaster for the calculation of stress.

 

[meccanicamg]

Hi, I'm new, and I'd like to thank for those who will answer me.
My question is this:
I should check the scope of a heb140 beam material s235, the 2 supports are welded at a distance of 4mt, on 2 plant.
this hoist with wheels, it must be mounted and flows on the lower wing of the heb140.
the hoist has a capacity of 2 tons.
I wanted to know if you have to calculate the size of the wing alone (how?) or just calculate the beam in the classic way of "trave doublely stuck with concentrated load"? !
to better understand I attach the photo:
Thank you in advance
greetings

Have you read the rules?
Did you show up?
did you search the forum?
discussion already open qui....e quiIf you look at the hoist builders provide you with the beams on which to mount them.
at least the beam is a zipper and cart because the two planters are sure to push. It won't be two inks.

 

[meccanicamg]

Meanwhile I am testing copilot that alone gave me general information while adding gpt-4 solved my calculation....but it wronged the units of measurement and so I had to try to make the request in yes but also here it was wrong and I had to do in n and in mm.
give control over the units of measurement and possibly make the account alone.
for your problem, you have to calculate the bending moment and normal voltage at the distance of 2000 mm from one of the ends of an isostatic heb140 beam with a concentrated load of 20000 n in 4000 mm long halfway. you can use the following formulas:

- flender moment: [math]m = \frac{p \cdot x}{2}[/math] where p is the applied load and x is the distance from the loading point.
- normal voltage: [math]\sigma = \frac{m \cdot y}{i}[/math] where y is the distance from the center of the section and i is the moment of inertia of the section.

Furthermore, you can find the geometric features of the heb140 section in the following table:
replacing known values in formulas, you get:

- flender moment: [math]m = \frac{20000 \cdot 2000}{2} = 20000000 \text{ nmm}[/math]- normal voltage: [math]\sigma = \frac{20000000 \cdot y}{2670000} = \frac{200 \cdot y}{267} \text{ mpa}[/math]Note that normal voltage depends on the y position along the height of the section, then you have to specify the point where you want to calculate it. For example, if you want to calculate the normal voltage at the top edge of the section, you must use y = 70 mm, which is half the height of the section. in this case, the normal voltage would be:
[math]\sigma = \frac{200 \cdot 70}{267} = 52.06 \text{ mpa}[/math]from here compare with the admissible voltage of your material and see if you are in favor of oversized or not.
I can use the following formula:

- brake: [math]\delta = \frac{p \cdot l^3}{48 \cdot e \cdot i}[/math] where p is the applied load, l is the length of the beam, and it is the young module of the material and i is the moment of inertia of section1.

replacing the known values in the formula, you get:

- Arrow: [math]\delta = \frac{20000 \cdot 4000^3}{48 \cdot 210000 \cdot 2670000} = 0.87 \text{ mm}[/math]note that the module of young steel is assumed pairs to 210000 mpa3. the arrow is the same at any point of the beam, because the load is applied in half-carry.


origin: conversation with bing, 21/2/2024
(1) calculation arrow supported beam - nuncio grieco - calculations. calculation arrow beam supported - nuncio grieco - calculations.
(2) calculation of beams supported with concentrated load - welcome on oppo.it. calculation of beams supported with concentrated load.
(3) metal beam supported - ediltool. metal beam supported - ediltool.
(4) calculation of the arrow of a beam - kinematic nuncio - calculations. calculation of the arrow of a beam - nuncio grieco - calculations.you can compare the arrow with l/1000 i.e. 4000/1000=4mm permissible.

you can also use ftool which is free and you study the whole structure.

 

[Mello Dessi]

Have you read the rules?
Did you show up?
did you search the forum?
discussion already open qui....e quiIf you look at the hoist builders provide you with the beams on which to mount them.
at least the beam is a zipper and cart because the two planters are sure to push. It won't be two inks.

I read it now the regulation, pardon!
next post I will follow the rules.
I did the research, the first discussion I found and read, but I wanted something more precise, the second one you posted just didn't find it.
Thank you.

 

[Esselle]

s

Meanwhile I am testing copilot that alone gave me general information while adding gpt-4 solved my calculation....but it wronged the units of measurement and so I had to try to make the request in yes but also here it was wrong and I had to do in n and in mm.
give control over the units of measurement and possibly make the account alone.
for your problem, you have to calculate the bending moment and normal voltage at the distance of 2000 mm from one of the ends of an isostatic heb140 beam with a concentrated load of 20000 n in 4000 mm long halfway. you can use the following formulas:

- flender moment: [math]m = \frac{p \cdot x}{2}[/math] where p is the applied load and x is the distance from the loading point.
- normal voltage: [math]\sigma = \frac{m \cdot y}{i}[/math] where y is the distance from the center of the section and i is the moment of inertia of the section.

Furthermore, you can find the geometric features of the heb140 section in the following table:
View attachment 70497replacing known values in formulas, you get:

- flender moment: [math]m = \frac{20000 \cdot 2000}{2} = 20000000 \text{ nmm}[/math]- normal voltage: [math]\sigma = \frac{20000000 \cdot y}{2670000} = \frac{200 \cdot y}{267} \text{ mpa}[/math]Note that normal voltage depends on the y position along the height of the section, then you have to specify the point where you want to calculate it. For example, if you want to calculate the normal voltage at the top edge of the section, you must use y = 70 mm, which is half the height of the section. in this case, the normal voltage would be:
[math]\sigma = \frac{200 \cdot 70}{267} = 52.06 \text{ mpa}[/math]from here compare with the admissible voltage of your material and see if you are in favor of oversized or not.
I can use the following formula:

- brake: [math]\delta = \frac{p \cdot l^3}{48 \cdot e \cdot i}[/math] where p is the applied load, l is the length of the beam, and it is the young module of the material and i is the moment of inertia of section1.

replacing the known values in the formula, you get:

- Arrow: [math]\delta = \frac{20000 \cdot 4000^3}{48 \cdot 210000 \cdot 2670000} = 0.87 \text{ mm}[/math]note that the module of young steel is assumed pairs to 210000 mpa3. the arrow is the same at any point of the beam, because the load is applied in half-carry.


origin: conversation with bing, 21/2/2024
(1) calculation arrow supported beam - nuncio grieco - calculations. calculation arrow beam supported - nuncio grieco - calculations.
(2) calculation of beams supported with concentrated load - welcome on oppo.it. calculation of beams supported with concentrated load.
(3) metal beam supported - ediltool. metal beam supported - ediltool.
(4) calculation of the arrow of a beam - kinematic nuncio - calculations. calculation of the arrow of a beam - nuncio grieco - calculations.you can compare the arrow with l/1000 i.e. 4000/1000=4mm permissible.

you can also use ftool which is free and you study the whole structure.

It would be interesting if I told us what question you asked to copilot. I noticed that sometimes the same question answers in different ways.

mark this resource too static calculations sizing beams

 

[pollo]

I usually use the annexx f of the en 15011
but there are many normative references in this regard
Hi.

 

[biz]

Have you read the rules?
Did you show up?
did you search the forum?
discussion already open qui....e quiIf you look at the hoist builders provide you with the beams on which to mount them.
at least the beam is a zipper and cart because the two planters are sure to push. It won't be two inks.

attention, "the knots are always incastrious." the beams communicate with each other through incastri.
the known scheme is as follows:

 

[Legs]

Hi, I'm new, and I'd like to thank for those who will answer me.
My question is this:
I should check the scope of a heb140 beam material s235, the 2 supports are welded at a distance of 4mt, on 2 plant.
this hoist with wheels, it must be mounted and flows on the lower wing of the heb140.
the hoist has a capacity of 2 tons.
I wanted to know if you have to calculate the size of the wing alone (how?) or just calculate the beam in the classic way of "trave doublely stuck with concentrated load"? !
to better understand I attach the photo:
Thank you in advance
greetings

Forget the inconvenience if you don't know how knots are made. Nothing changes.
by making two accounts of the servant you will have a moment of calculation:
med = 1,5x20x4/4 = 30 knm
which is widely covered by a steel heb s235 which has mrd=54,92 knm
so you have very wide margin.
I do not know what size has the trolley that supports the beam (100-200mm?) but I think reasonable an effective width of about 200mm (large way a diffusion at 45°: 66+100+66 rounded for defect at 200). We release the r wholesale connection between wing and soul and therefore we can consider a light of about 66mm.
I consider the uncentered load by applying a behavior coefficient 1.2
the moment of the wing will be: 1,2x1,5x20/2x0,066 = 1,18 knmm
the resistant moment (calculated in complete plasticization hypotheses at the last limit state) is worth:
mrd = 200x12^2/4x235/1,25x10^-6 = 1,35 knm > m
then it is true there are other problems such as braking forces and any torque effects and dynamic amplifications. the calculations shown are used to get the idea of the validity of the putrella.

there would be to verify the arrow but I think here it is important to have a better knowledge of the structure. We still have a semi-category, making the average between perfect hinges and hinges. the coefficient to be used is therefore the average between: 1/48 and 1/192 = 1/76,8
(76,8x210000x1509e4) = 5,3mm
on 4m is a ratio of 4000/5,3 = 754.
Let's say we're a little close. Usually for this type of works you consider a light / arrow ratio of 800. as I said, we must know the structure well to know exactly what is the right value.
@meccanicamg be careful that you used a wrong young module

 

[biz]

med = 1,5x20x4/4

What are 1.5x20x4/4?
and we know them. Certainly the structure cannot be assimilated to that of the manual, as a and b are not incastrious, but first approximation I would say it is okay.

 

[Legs]

1.5 is the partial load amplification safety coefficient
20 is the load expressed in kn
4 is the light in meters
and the last 4 is the divider coefficient to obtain the maximum bending moment of a beam with concentrated load at the center: m=p*l/4

 

[biz]

1.5 is the partial load amplification safety coefficient

by law ntc? That's why I didn't understand that 1.5

 

[biz]

the moment of the wing will be: 1,2x1,5x20/2x0,066 = 1,18 knmm

explain to me the moment of the wing. I do not know

 

[meccanicamg]

Forget the inconvenience if you don't know how knots are made. Nothing changes.
by making two accounts of the servant you will have a moment of calculation:
med = 1,5x20x4/4 = 30 knm
which is widely covered by a steel heb s235 which has mrd=54,92 knm
so you have very wide margin.
I do not know what size has the trolley that supports the beam (100-200mm?) but I think reasonable an effective width of about 200mm (large way a diffusion at 45°: 66+100+66 rounded for defect at 200). We release the r wholesale connection between wing and soul and therefore we can consider a light of about 66mm.
I consider the uncentered load by applying a behavior coefficient 1.2
the moment of the wing will be: 1,2x1,5x20/2x0,066 = 1,18 knmm
the resistant moment (calculated in complete plasticization hypotheses at the last limit state) is worth:
mrd = 200x12^2/4x235/1,25x10^-6 = 1,35 knm > m
then it is true there are other problems such as braking forces and any torque effects and dynamic amplifications. the calculations shown are used to get the idea of the validity of the putrella.

there would be to verify the arrow but I think here it is important to have a better knowledge of the structure. We still have a semi-category, making the average between perfect hinges and hinges. the coefficient to be used is therefore the average between: 1/48 and 1/192 = 1/76,8
(76,8x210000x1509e4) = 5,3mm
on 4m is a ratio of 4000/5,3 = 754.
Let's say we're a little close. Usually for this type of works you consider a light / arrow ratio of 800. as I said, we must know the structure well to know exactly what is the right value.
@meccanicamg be careful that you used a wrong young module

elastic steel module 210'000mpa.
Where's wrong, you tell me, please?

 

[Legs]

By the way, I neglected the weight of the beam in the calculations. to do things well should be inserted.
coefficient 1.5 applies to variable loads (ntc18).
in practice apply a multiplier coefficient for loads and a divider for resistors.
Note that for the current elements in steel the divider is worth 1,05 while for the verifications of detail (such as the wing of the beam or the knots) is used 1,25.

the moment in the wing is calculated in the hypothesis to divide in two the concentrated load.
half goes on the left wing and half on the right wing.
But I have speculated that the distribution between the wings is not exactly accurate and maybe goes a little more load on one instead than on the other. I have typically seen using behavior coefficients between 1.2 and 1.3. It is not a standard thing but it is a scruple to consider in a lumptive way that the load may not be centered.
in practice I put to the extreme of the wing the force: 1,2x20/2 = 12 kn then increased from the coefficient 1.5 to use for the variables = 1,5x12 = 18 kn.
the distance between the soul of the putrella and its extreme is 66mm (look at the silhouette).
therefore the moment on the wing is 18x0,066 = 1,188 knm. I realize now that it was the case of approximating to 1,19 knm. changes little.

then I considered an effective width perhaps a little small to evaluate the resistance.
Maybe the resistance is a little greater.

 

[Legs]

elastic steel module 210'000mpa.
Where's wrong, you tell me, please?

I'm sorry, I meant moment of inertia. It was a lapsus.
indicated in 2670 cm4 the moment of inertia of heb140. Actually it is worth 1590cm4.

 

[meccanicamg]

I'm sorry, I meant moment of inertia. It was a lapsus.
indicated in 2670 cm4 the moment of inertia of heb140. Actually it is worth 1590cm4.

Okay. Thank you.
the sheer is not quite mature to make calculations, take right data and assist.
Unless he took it from a source where it was written wrong.

 

[Mello Dessi]

Forget the inconvenience if you don't know how knots are made. Nothing changes.
by making two accounts of the servant you will have a moment of calculation:
med = 1,5x20x4/4 = 30 knm
which is widely covered by a steel heb s235 which has mrd=54,92 knm
so you have very wide margin.
I do not know what size has the trolley that supports the beam (100-200mm?) but I think reasonable an effective width of about 200mm (large way a diffusion at 45°: 66+100+66 rounded for defect at 200). We release the r wholesale connection between wing and soul and therefore we can consider a light of about 66mm.
I consider the uncentered load by applying a behavior coefficient 1.2
the moment of the wing will be: 1,2x1,5x20/2x0,066 = 1,18 knmm
the resistant moment (calculated in complete plasticization hypotheses at the last limit state) is worth:
mrd = 200x12^2/4x235/1,25x10^-6 = 1,35 knm > m
then it is true there are other problems such as braking forces and any torque effects and dynamic amplifications. the calculations shown are used to get the idea of the validity of the putrella.

there would be to verify the arrow but I think here it is important to have a better knowledge of the structure. We still have a semi-category, making the average between perfect hinges and hinges. the coefficient to be used is therefore the average between: 1/48 and 1/192 = 1/76,8
(76,8x210000x1509e4) = 5,3mm
on 4m is a ratio of 4000/5,3 = 754.
Let's say we're a little close. Usually for this type of works you consider a light / arrow ratio of 800. as I said, we must know the structure well to know exactly what is the right value.
@meccanicamg be careful that you used a wrong young module

first of all I would like to thank you/you for the explanation and for the time spent.
as you wrote, the carello has a size of supports (seat area) between 200/250 mm.
to be more precise and to have a clearer idea I posted the 3d of the structure, to better understand the knots.
the heb140 beam is the central one at the top, welded on the wings to the structure made with hea120.
to me they asked, in order to put a tag, the max flow of the heb140 where the hoist should be mounted regardless of the scope of the hoist.
Thank you

 

[meccanicamg]

I'm sorry, I meant moment of inertia. It was a lapsus.
indicated in 2670 cm4 the moment of inertia of heb140. Actually it is worth 1590cm4.

I searched the internet and found a site with lots of data for beams.
moment inertia bending axle best iy=1509cm4 and not 1590cm4. even on opposite is so indicated.

 

[meccanicamg]

attention, "the knots are always incastrious." the beams communicate with each other through incastri.
the known scheme is as follows:View attachment 70501

and to think that of my very zealous colleagues, on a similar project but with the things up, they made a structure with the legs and then the horizontal beams connected them with two thorns to the sides, thus flexing the beam with two hinges to the sides. and who knows why they had to increase the section of the beam.
certainly that if you use the recesses you need to be robust and well sized, but surely also the legs collaborate to react to the bending.