Solidworks simulation - stress anchor bolt

Hi all

I have a cantilevered platform fixed to the ground to design, and for this purpose I have to provide a calculation note to my client so that he can see if his slab is sufficient or if a mass must be poured for the anchors.

I briefly represent the platform, and put my virtual wall under the feet, I put my anchor bolts and I launch the simu in full solidarity with a distributed load of 250kg/m² (about 500kg in total).

Where I get stuck is that I find that the axial forces in my anchors (Y-component) are strangely low. I attach the model and the screenshots: 483N in traction at maximum (not even 50kg!). Given the construction I expected much more.

I just wanted to have an external opinion, the principle is correct and I'm making a fuss about nothing or is my way of doing things bad?

In the zip in PJ there are the solid files and I also put a screenshot.

Edit: go for the form I add a small screenshot in the environment.

Thank you in advance


pf_test.zip
capture_483n.jpg
mise_en_situation.jpg

Hello

I will allow myself to answer you not scientifically but logically.

I can't open your ASM because you have a higher version than the 2019. So I can't see if you used a virtual wall for your bolts

Personally, I would not proceed quite as you do, but that is only a point of view. ;-) ;-) (see the conclusions to save time   ;-)  ;-)     )

1°) I would not use a distributed load but would replace it with a mass at a distance of 1.10 m from the ground of the platform sheet which corresponds more or less to the CG of a human being.

Note : it is not surprising that you have low axial forces in your anchors, this is mainly due to the fact that you use a distributed load and also that the CG of the unloaded frame is not very far offset. In any case, it is not at the centre of the platform that receives the operator.
A vista de naz your frame must be in the 200 kg.

2°) from experience, the base of this load at the operator's distance must be on a circle or square of 60 cm (the load represented by the operator is transmitted by the operator's footings).
3°) you will have to make at least two separate  simulations with different remote ground locations (one at the extreme limit of the cantilever , in the center and in an angle as if the operator were leaning over the railing). It will also be necessary to do a simulation with 3 operators of 95 kg to get closer to the unusual but surely not so rare conditions of use. For this type of structure you need a resistance of 500 kg per m² (but you know that)

Another important point: you can also do your test without anchor bolts (even if bolt connectors are very useful to know the axial forces), just make a sketch of the size of the washers and make these eight areas fixed. This allows you to better see the deformation of the gantry footings.

On the other hand, your client should not make the classic mistake of taking the results of the simulation at face value.
The anchoring in the ground depends on the quality of the concrete, the size of the reinforced mass and especially the standards given by the suppliers for the diameters and depth of the anchor studs?
It depends on whether the anchors are made on a reinforced concrete slab or with solid stones. For unit beds, it depends on the quality of the soil in which they are poured (see the anchoring of its candelabra posts). So only a concrete engineer after sounding the soil can give the appropriate information.
Your simulation will only be an indication of the tensile forces and the forces on slopes.

For your question, you have to take the problem backwards because morality anchoring is not your problem because it is enough to tell it that a single SPIT anchor bolt (or HILTI)  for example for a diameter of 10 TRIGA Z XTREM V8-12/20 gives per powl 5.71 kN in tension in seismic zone C1 and  for cracked concrete. This means that even if you only put two studs per plant, it is not your structure that will be the problem. These data are  to be related to the assumption that you would only put two studs on each foot to calculate the leverage effect on the most stressed external studs, this would give 3000 N / 4 = 750 N  per studs, i.e. 0.75 kN (at the very least, because I don't have the dimensions of your frame or their weight, nor the position of the GC)

On the other hand, you must check with your customer that the stress cones of each stud and this when he has indicated the nature of the concrete and the depth of anchoring and its compatibility with the other parameters. It is this distance defined by the stress cones that must be used to define the centre distances of your two  fixing plates to the ground. Dowel manufacturers provide this information in their documentation.

Kind regards

 

 

  

 

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Hello. Unable to read the file for me. 

But I would like to give me a little advice. Equation of equilibrium around C in this image. You will immediately have an idea of the resultants in the supports (bolts in this case). Side A is favourable. Only the B side acts to tear off the studs. And there are 8 in total. Those closer to point C will be the most solicited. The one indicated at the back is less solicited so 50kg yes possible. But the best is the static calculation to check. And it's very simple in this case. Be careful with the units too.  

 

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Hello

I was talking earlier about the leverage effect very well illustrated by @soring  (which I salute).

Indeed, there is a very simple calculation to know the orders of magnitude, which allows by knowing the pull-out strength of a single anchor bolt by specification to respond to the customer. Then if you want to quibble the cabbage for peanuts you can do a PEF simulation.

Kind regards

 


calcul_arrachement___2021-05-30_15_24_33-window.jpg

Hello gentlemen

Thank you for your interventions, I'll elaborate a little more tonight when I have time for your first Zozo post in particular. But quickly by doing a little static calculation (see PJ) taking into account the mass of my structure.

For the efforts I took 250kg/m² on the 0.8m wide of the platform and I simplified the masses distributed by forces (F2 and F3). For F1 at the G-point it's the mass, and the reaction at the level of my anchors we are on F4

Edit: Zozo beat me to it :-)

We are still on 215kg (17261/8) per anchor in this case. Why is there such a difference with the basic result of solid? The fact that the distributed masses have been simplified?

Thank you in any case it's not the first time I find a very useful help here ;-) I put a small STEP of the FP


img_20210531_095736.jpg
pf.step
1 Like

Hello @Jwunder 

From your model!

Here is just for fun and with a single anchor bolt.    the results to be compared to the diameter of 10 TRIGA Z XTREM V8-12/20 which gives 5.71 kN in tension in C1 seismic zone and  for cracked concrete per stud.

With a short video


calcul_arrachement__sur_un_seul_gougeon__2021-05-30-window.jpg
deformation_emplacement_deux_gougeons___2021-05-31_12h34_39.mp4
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Hello @Jwunder 

I take care of what doesn't concern me: but are you sure of the dimensions of your HEA.

[HS On]

In addition, they seem very oversized for the load

[HS /Off]

No, don't hit the head

1 Like

Re, well now that the kids are at bedtime:

You will need to do at least two separate  simulations with different remote ground locations (one at the extreme limit of the cantilever , in the center and at an angle as if the operator were leaning over the railing). It will also be necessary to do a simulation with 3 operators of 95 kg to get closer to the unusual but surely not so rare conditions of use. For this type of structure you need a resistance of 500 kg per m² (but you know that)

So with a distributed load of 250kg/m² we cover the case of a 90kg person on 0.36m² (600x600 square). As a general rule I define 250 for a technical gateway, more depending on the uses, and I have a nice "maximum load"  plate engraved that I put on my gangways. For the 500, do you have a standard that imposes this? I go up as needed if motorized vehicles pass through

Another important point: you can also do your test without anchor bolts (even if bolt connectors are very useful to know the axial forces), just make a sketch of the size of the washers and make these eight areas fixed. This allows you to better see the deformation of the gantry footings.

But my goal is to get out of the effort to check the slab or size a bed. Impossible in this case I think?

On the other hand, your client should not make the classic mistake of taking the results of the simulation at face value. The anchoring in the ground depends on the quality of the concrete, the size of the reinforced mass and especially the standards given by the suppliers for the diameters and depth of the anchor studs?

Right. In fact, most of the time we have the design and installation of the equipment, so I do my purchases (steel, anchoring material, screws, everything from A to Z), I design, we manufacture, we have it validated by inspection bodies if necessary (lifting equipment for example) and we install. In our case, we will drill the slab to see the thickness of the latter, and if that is not enough, we will open to pour a mass of sufficient depth for the anchors that I would have defined. On the other hand, concrete is not our job so we subcontract this part, it's up to me to give the resulting efforts to the concrete worker so that it does its job. To do this, I give him as much information as possible, sometimes the moment at the base of my feet out of a beam simu if I want to do it quickly, sometimes in axial effort. Here for once I find my efforts ridiculous considering what the studs take ^^

I take care of what doesn't concern me: but are you sure of the dimensions of your HEA.

[HS On]

In addition, they seem very oversized for the load

[HS /Off]

No, don't hit the head

No worries, we're here to talk about that :-)

From experience, I oversize cantilevered structures to avoid the "spring" effect when the part is used in a vacuum. I already have a client who started jumping on a structure that started to vibrate, and I ended up having to reinforce, even if with calculation notes and misuse on his part. This for the time being, the statics do not inform it, and it is the only module we have. And then if I go down in hea size, I gain 5kg/m (from hea140 to hea120), or about 20 kilos, which represents almost nothing on the quote we are going to make to the customer. We work on the unitary project so the optimization we draw a line under it to reduce the study time but while building in a correct way because in the end the goal is to keep the customer.

There you have it! Thank you for your little video. I'm going to relaunch a static study with your hypotheses just to see tomorrow morning.

On the other hand, I noticed that the axial force in the studs increases  with the decrease in the screw passage diameters. Let me explain: I have a Ø20 hole to put Ø16 anchor rods (as a general rule we drill at 18 but depending on the accessibility for the perfo I increase the diameter). So in my simu on my diameter 20 I enter a rod of 16, a nut of Ø30 for example, it gives me an axial force. This effort increases if I keep the same assumptions but lower the Ø of passage to 18. Have you ever had the case?

Moreover, concerning the preload of the stud, I always enter it at 0 (torque or axial tightening is equal). Is this the right method? Because if I enter the tightening torque in the end, the von stress scale put in the results explodes the ceiling on the support surface of the nut!

HS- we can't quote on this forum!-

 @Jwunder

Thank you for all these clarifications, we are totally in sync.

- Indeed the material cost is not decisive, you do well to make the thing very slightly stronger.
That said, since I did the calculations on a single post and as the bending without bracing in my test only reveals 4 to 6 mm of bending, I think it was very strong! ;-) You would have to do the calculation in frequency (spring effect) but I don't have the PRO version of simulation.

-  I have the figure that was imposed on me 500Kg for a theater floor in reference to the operating load of balconies which is defined by the standard (EUROCODE 1). It is worth 350 kg/m² (i.e. 3.5 kN/m²) without a security coeff which must be 2.5 to 3.  Also the load of the guardrails (60 kg per meter of guardrail applied at the top).

For the simulation, take the bolt connector but with a hole of 17 so that it is as close as possible to the diameter of the stud (you just have to uncheck the adjusted rod). Look at the incidence of the adjusted rod or not http://help.solidworks.com/2021/french/SolidWorks/cworks/c_Example_of_a_Tight_Fit_Bolt_Connector.htm?id=076ce79608484e4caa6747b16af398fb#Pg0 
What you see is normal but a bit complicated to explain
http://help.solidworks.com/2021/french/SolidWorks/cworks/c_distributed_coupling_bolts.htm?id=469ca1d42bfb46c3ac41817e51e968e8#Pg0
It's a rule in simulation to be as close as possible to reality on your site: if your final hole is such and such a diameter then make the simu with this value. This is similar to all the technology of screwed joints for the overhead part. For the part in the concrete, it is the dowel suppliers who give the rules to be respected. But now you have all the experience.

For the prestressing it makes sense if we refer to the standards of  the stud supplier (between us on the site I rarely saw Mr. GrosBourrin use a torque  wrench;-)  There, your application is far from being critical AMHA. For von mises it is also better to look in the results at the "design dissection" choice which allows you to see the tensions inside the matter because the Von mises images only show the forces of external surfaces (even if Von Mises calculates the totality of the internal nodes in all the DDLs).

Remember that in my simulation examples I only use two studs instead of four and a single post instead of two.

Hold to distract you after a hard day's work, I made you a little video where I replaced the bolt connector with two spring connectors, this allows you to see what happens if the anchor is a little soft. I like to look at all the possible deformations, it shows a lot of things and it trains the brain to pick up on what is not always apparent    ;-)  ;-)

Kind regards

 

 


avec_connection_ressort_2021-06-01_20h18_26.mp4

Hello @Jwunder,

Some thoughts in the attached document, based on the simulation of your platform.

Kind regards.


plateforme.docx
3 Likes

Hello @m.blt

Thank you for the time!

I will continue your study on my side to see if I fall back on the same values that you give me.

A question comes to me with your hypotheses, on the supplier docs where I order I see that an M16 mechanical anchor in cracked concrete holds 16.7kN in tension.

It's never specified by any supplier, but we agree, it's the value if we hide the prestress? It would seem crazy to me to have to calculate the axial preload due to the tightening torque given in the anchor documentation to subtract it from the one we calculate.

So if we estimate that the tightening of the anchorage to the torque (torque given by the anchor supplier) induces an axial force of 20kN, with a tension of 16.7kN, we mean that the stud supports 20+16.7=36.7kN. Well, that's how I've always considered it, I'm just taking this opportunity to ask the question:)

 

Edit: Zozo's remark:

For the simulation, take the bolt connector but with a hole of 17 so that it is as close as possible to the diameter of the stud (you just have to uncheck the adjusted  rod). Look at the incidence of the adjusted rod on http://help.solidworks.com/2021/french/SolidWorks/cworks/c_Example_of_a_...
What you see is normal but a bit complicated to explain
http://help.solidworks.com/2021/french/SolidWorks/cworks/c_distributed_c...
It's a rule in simulation to be as close as possible to reality on your site: if your final hole is such and such a diameter then make the simu with this value. This is similar to all the technology of screwed joints for the overhead part. For the part in the concrete, it is the dowel suppliers who give the rules to be respected. But now you have all the experience.

Thank you for the links, interesting indeed!

1 Like

I have not consulted a supplier doc for the particular case of an anchor in cracked concrete. The value of 20 kN that I have imposed as preload (I prefer this term to that of prestress) in the definition of bolts is perfectly arbitrary...

So if we estimate that the tightening of the anchorage to the torque (torque given by the anchor supplier) induces an axial force of 20kN, with a tension of 16.7kN, we mean that the stud supports 20+16.7=36.7kN. Well, that's how I've always considered it, I'm just taking this opportunity to ask the question:)

This interpretation seems incorrect to me: from my point of view, the preload represents the axial force that is applied in the bolt at the time of assembly, assuming that the platform is not subjected to any external force: no passengers and the self-weight of the structure compensated by the handling crane. The preload can be "checked" by means of a torque wrench, but there is a fairly high degree of uncertainty due to a poor knowledge of the contact conditions of the screw and nut in the thread and with the assembled parts. It's always better than a tightening of dead reckoning...

During the simulation, the calculation considers the elastic behavior of the bolt and the assembled parts, and under the effect of the "operating" external loads, gravity and passengers for the platform, the axial forces evolve. The results given by SW Simulation represent the axial forces actually supported by the threaded elements.
In the case of your platform, the forces vary very little compared to the preload because the assembled parts are very rigid with regard to external forces (15 mm thick steel plates, 16 mm diameter threaded elements, etc.).

The value of 16.7 kN given by your supplier in the case of a concrete installation is the limit that must not be exceeded, either during assembly or in use. It is one of the advantages of simulation to check that this condition is respected... The limit is related to the resistance of the concrete in the anchoring area, as the screw itself is capable of supporting a load at least 5 times higher (for an ordinary quality, class 8.8).

Bolt Set... In theoretical models of connections by threaded elements, it is common to use the notion of pressure cone to define an equivalent of the area of the bonded parts stressed in compression. Even if this model is not the one used by SW, it illustrates the effect on the assembly, which is more "rigid" if the play is reduced, due to a larger cone cross-section. And the consequence on the efforts in the assembly.


Good night...

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Thank you all for your interventions, all the answers were useful to see more clearly!