Neat result for yield line analysis of circular anchor plate 2

[bugbus]

I've been tasked to check some circular plates welded to the ends of reinforcement to be used as anchorages. See below:

I wasn't quite sure how to check the bending failure of the circular plate, so I decided to start from first principles using yield line analysis.
Not knowing how many yield lines occurs for the critical case (though I suspected it would be n = infinite), I started with the below yield line model, where the collapse mechanism resembles an 'n'-sided pyramid:

The working follows below, but the main take-away is that the critical pressure 'p', which causes the plate to collapse, is simply equal to p_crit = 3*Mu/r^2, where Mu is the plastic section capacity per unit length of plate, and occurs for the case of n = infinity.

The working may be hard to follow, and I used WolframAlpha to help with some of the algebra, but hopefully it helps.

Note: to get within 10% of the critical failure pressure, you would need to use about 7 yield lines equally spaced.
Using 3 yield lines overestimates p_crit by close to 100%, whereas 4 yield lines overestimates it by about 40%.

[URL unfurl="true"]https://res.cloudinary.com/engineering-com/image/upload/v1677735073/tips/8a5e6518-db17-40cb-b611-599d1b5e93f8_fqkebt.pdf[/url]

 

[HTURKAK]

I did not look all the calculation but looked the last page ..You found qcr = 68.4 mPa..

I checked your calculation assuming the end plate is circular ftg ( with r=26.5 mm and the rod dia. = 0.0 ??? )

Wp=(53*16**2)/4=3392 mm3

Steel grade S 250 fy=250 mPa , half area A=1103 mm2 , moment arm ( centroid dist) x=4r/(3pi)=11.24 mm

Mp = qcr*1103*11.24= 250*3392 ⇒ qcr=68.40 mPa..

However, IMO , this approach is not reasonable and my points are,

- the stiffness of the plate is more important and the suggested thickness is t ≥ 1.5 D and diameter is 3-4 D of rod.

- You may look ( THEORY OF PLATES AND SHELLS by Timoshenko and Gere page 62 ) for elastic stresses ,

- Another book ( Roark’s Formulas for Stress and Strain )..

- You may prefer FEM for more refined calculation


In past , i preffered Roark’s Formulas for designing the bearing plates ..








Not to know is bad;
not to wish to know is worse.

NIGERIAN PROVERB

 

[bugbus]

HTURKAK, thanks a lot for your response.

I've been thinking about this the last few days and have done a bit more work on this, so just wanted to share some comments:

[li]I agree that 'bar diameter ~ 0' is a pretty gross simplification, but one that I expected would lead to a more conservative failure load of the anchor plate. Introducing another yield line around the base of the bar gives the following critical pressure on the plate: p_crit = [48*Mu*r]/[(2*r-db)^2*(4*r+db)], where db is the bar diameter. For db = 0, this brings us back to the original result of 3*Mu/r^2 as would be expected. For the example above, this leads to a critical pressure of 122 MPa compared to the previous estimate of 68 MPa, about 1.8x larger. This is more significant than I originally had thought and is more in line with what the FEA shows (see below).[/li]

[li]As you correctly pointed out, assuming the simplest case (n = 2), i.e. the plate folding in half along a diametrical yield line, the critical load is found to be the same as if you assume an infinite number of yield lines (in the example, 68.4 MPa). However, I don't think this is a realistic failure mode. I believe the 'dishing' failure mode predicted by n = infinity is most realistic and is reflected in the FEA (below). Still, it's interesting that it predicts the same collapse load. It's also worth pointing out that the case you checked (n = 2) is not 'pyramidal' so it can't be calculated by the method I proposed (which is not to say it isn't valid). I have gone through my working again, and indeed the case for small n (i.e. = 3, 4, 5, etc.) all fairly well overestimate the collapse load. It is not until you get to about 7+ yield lines that you would get something close to the critical case. By this point, the yield lines have turned more into a yield 'cone', which is what the FEA predicts.[/li]


[li]Could you provide a reference for the suggested thickness being t ≥ 1.5 D? The anchor diameter being 3.3x the bar diameter gives a net bearing area of 10x the bar area, which I believe a lot of the codes specify to achieve the short anchorage. Looking at some of the proprietary products, e.g. Ancon bar anchors, the anchor thickness would be 20 mm for a 16 mm bar, but possibly this is to achieve enough thread length rather than being a stiffness or strength requirement? As you will see from the FEA below, for a plate thickness of just 0.6xdb, the tip displacement is in the order of 0.1mm, which I would consider a reasonable target. It is comparable to the 0.1mm slip allowed for a threaded coupler.[/li]


[li]Another thickness concern is punching shear of the plate. By my quick calculations, the plate thickness should be at least t >= 0.4*(fy_b/fy_p)*db, where fy_b and fy_p are the yield strength of the bar and plate respectively. Assuming Grade 500 bar and Grade 250 plate, this requires a plate thickness of about 0.8*db.[/li]


[li]I agree in principle with your comment about the plate stiffness being important (I would not say more important). But I would say that for any anchor plate that has been designed to satisfy strength (i.e. satisfies the minimum punching shear and collapse load), the required proportions will basically guarantee that it will deflect a very small amount (<0.1 mm).[/li]


As per your suggestion, I have checked a few cases with FEA using solid elements and elastic-perfectly plastic material behaviour to best replicate the yield line analysis.
Bar is assumed to be fully butt welded to the plate (as per sketch below), with a small radius (2 mm) added to the model to give it a smooth transition.

Example: 16 mm bar, Grade 500; 53 dia x 6 thick anchor plate, Grade 250

I deliberately under-sized the plate thickness to cause the dishing-type failure, as it was proving hard to recreate this failure mode using the previous 16 mm plate example (without first yielding the rebar)
Critical pressure according to p_crit = 3Mu/r^2 is equal to 9.6 MPa.
Critical pressure according to p_crit = [48*Mu*r]/[(2*r-db)^2*(4*r+db)] is equal to 17.1 MPa.

You can see the graph of plate pressure vs tip deflection below.

The critical pressure is pretty close to the 17.1 MPa predicted by the more refined yield line model.
Tip displacement is in the order of 0.2 mm at the ultimate load, which may considered large at least compared to the 0.1 mm slip allowed for threaded couplers, for example.

Deformed shape and von Mises stress shown below:

Based on the more refined model, a ~10 mm thick plate (~0.6*db) would be required to satisfy strength. The tip deflection for this plate is close to 0.1 mm at the ultimate state, which in my view would be reasonable. However, the punching shear requirement is actually more onerous, and would required a plate thickness of 0.8*db > 12.8 mm.


Anyway, to sum all this up - I believe that the bare minimum plate thickness should be as follows (assuming Grade 500 bar and Grade 250 plate, and a net bearing area of 10*Ast):
[li]To avoid plastic collapse of the plate ~ t = 0.6*db[/li]
[li]To satisfy punching shear in the plate ~ t = 0.8*db[/li]
[li]To satisfy ~0.1 mm tip deflection ~ about 0.25*db^(4/3), assuming the tip deflection is inversely proportional to t^3[/li]

Of course, I don't suggest that a critical element like a reinforcement end anchor should be designed to its bare limit. I would always be more than happy to adopt similar anchor proportions to the commonly available proprietary products, but it's at least good to know what's up your sleeve.

 

[HTURKAK]

Q= Could you provide a reference for the suggested thickness being t ≥ 1.5 D?

A= Yes ... pls look ANCHORAGE DESIGN FOR PETROCHEMICAL FACILITIES ( ASCE TASK COM..)


- Apparently it is groove weld .. which i will not suggest .. ( You may prefer the end section threaded , the bearing plate has a hole and fixed to the rod with two nuts or , drill hole and insert the anchor rod and fillet weld from both sides


- If dia 53 mm, and the tip displacement 0.1mm , the concrete strain would be in the order of 4 E-3..

As far as i understand , your FEM model loading is based on uniform loading of the bearing plate which does not include the concrete bearing resistance variation acc. to displacement of bearing plate..


My opinion...











Not to know is bad;
not to wish to know is worse.

NIGERIAN PROVERB

 

[bugbus]

Thanks again, HTURKAK. The butt welded (groove welded as you called it) option is one of the prequalified weld to AS 1554.3. I’ve also seen these passed through a hole and welded from the other side, which might be easier to fabricate. I’ll look into this.

I agree that the concrete strain will be large, possibly in the order of 0.004 as you said. The concrete at the bearing face of the plate will have an average stress at the ultimate state of around 50 MPa, assuming Grade 500 reinforcement and a 10x bearing area. Naturally the strain will need to be somewhere in the order of 0.003 (maybe 0.004 as you said) to achieve this in the concrete for a strength grade of 40-50 MPa or so, which is common in my world. But I disagree that this is to do with deflection of the plate. Even for a rigid plate, the stress in the concrete will still be an average of 50 MPa and so the strain will still be in the order of 0.003 locally. Given the anchorage would be well confined with additional reinforcement in the transverse directions, I wouldn’t see an issue with such a large local stress.

Regarding the uniformity of the bearing stress, you’re correct that it might be slightly larger near the bar and slightly less near the edge. But with more practical sized plates, the deflection is closer to 0.01 mm (checked using Roark’s), so possibly it’s not so significant. But the uniform stress assumption should give a slightly conservative prediction in any case.

 

[structSU10]

just out of curiosity why are you not using an off the shelf solution? erico has many options proportioned to develop the strength of the bar.

 

[bugbus]

structSU10, it’s a good point. But it’s not my design, I’m just checking it.

I like the off the shelf products too. Not all of them provide the 10x bearing area, which is not necessarily a problem, but can be nice to have when you need a very short anchorage length.

I’m also wondering about the relative fatigue resistance of the forged and threaded bar of the proprietary system vs a butt welded bar. For rail bridges we have run into problems of congestion due to the quite stringent stress limits for threaded couplers (I imagine it’s the same for a threaded end anchor). Fully butt welded bars seem to be somewhat more resilient to fatigue but I don’t have all my sources at hand so I can’t be sure.