AS3600-2018 Shear Capacity: General Method vs Simplified

[Settingsun]

Hi all,
I've finally gotten around to putting General Method, Simplified Method and 2009 Method in a single spreadsheet for side-by-side comparison. When the beam has shear reinforcement, I find that the simplified method tends to give a higher capacity than the general method unless the bending moment is small. This is because theta,v is smaller in the simplified method (fixed at 36 degrees regardless of bending moment and shear force), giving a greater contribution from the fitments.

Does anyone know if this is the intention? I thought Standards Australia policy was for simplified methods to be conservative.

 

[rapt]

Steveh49,

Agreed, they are supposed to be.

You will have to blame Collins and Mitchell and the Canadian Code committee as it was assumed they they had it right! I will pass on the comment.

 

[Settingsun]

Rapt,
I saw your comment in another forum that AS3600-2018 isn't the BCA version yet. Is that just because BCA hasn't been updated since AS3600 was published?

In any case, the Simplified Method seems similar to the 2009 method so 2009 can be equally (or more) unconservative if the greater accuracy promised by the General Method is real.

I know very little of the theory behind the new method but would have thought 45 degrees (which gives kv=0.09) would be the simplified case as it has decades of track record. Maybe round off to kv=0.10 for simplicity even if this isn't quite consistent with 45 degrees.

 

[rapt]

Steveh49,

I have mentioned it to the person on the committee who did the shear section. Can you give us some comparative numbers, to my RAPT email address if possible.

The numbers used in the code are basically the same as CSA23, AASHTO and the 2010 Model code version of the Modified Compression Field Theory developed by Collins and Mitchell. It was assumed in putting it into AS3600 that they would be correct!

BCA will be up[dated later this year. Until then, the referenced code is 2009. previously when this has happened, if the new code has been more conservative in an area (eg development lengths last time), I have added it into RAPT immediately as it was known that the old code methodology was unconservative. If the new code is less conservative, technically you need to keep using the old code.

 

[cooperDBM]

In my long experience with the Canadian code I don't recall seeing the simplified method (within it's limits) being less conservative than the general method. For prestressed concrete the simplified method can be very conservative. I haven't seen the final version of AS3600-2018 yet but did review the Aug 2017 draft. If still relevant to the final version, the simplified methods were similar between AS and CSA for non-prestressed beams but AS uses the shear cracking limit for prestressed. The OP seems to be referring to non-prestressed. CSA makes no distinction between prestressed or not. Is the method shown in the draft still relevant?

In CSA the General Method can be very sensitive, particularly in the absence of shear reinforcing. It tries to estimate the crack spacing based on the longitudinal reinforcing which, as worded, I believe can lead to unconservative results. The AS (draft) avoided that.

Rapt, do you know if the committee is consulting with Michael Collins? He still chairs the Canadian shear committee.

Steve, I'd be interested in trying your test case with the Canadian code. Could you share the, hopefully simple, design details or your spreadsheet?

 

[rapt]

cooperBDM,

There has been contact and questions asked.

 

[Settingsun]

Rapt,
I'll send some comparisons and post the spreadsheet in a few days - I want to double-check my spreadsheet especially after CooperDBM's comment, especially the strain calculation. That will also give a few business days for others to chime in.

I've just done non-prestressed, rectangular, singly-reinforced (ie ignore compression reinforcement) cross-sections without axial force for the moment. Starting simple to get a feel for the new method.

 

[Settingsun]

I didn't have the enthusiasm to check formula-by-formula today, so I tried to reason whether Simplified Method would be expected to give a larger capacity than General Method. Here's what I came up with.

Simplified Method uses kv=0.15 and theta=36 degrees. Using equations 8.2.4.2(4) and 8.2.4.2(5), these correspond to strains (epsilon,x) of 1.11E-3 and 1.00E-3 respectively. So the simplified method should give a greater shear capacity if the strain is larger than about 1.1E-3. The limits on epsilon,x are -0.2E-3 to 3.0E-3, so it is certainly possible for the simplified method to give a larger capacity.

For non-prestressed concrete without axial force or torsion, the formula for strain is fairly simple and boils down to the sum of moment-strain and shear-strain:

epsilon,x = (M*/dv)/(2*Es*Ast) + V*/(2*Es*Ast).

Assuming M* is approximately phi.Muo, and that phi.Muo is approximately phi * Ast * Fy * dv, the moment-strain component is 0.425*yield strain = 1.06E-3 for Fy=500 MPa. This alone is equal to the transition point where the simplified method gives a larger capacity, so sections that are fully used in bending should have a higher shear capacity if the simplified method is used.

To calculate the shear strain that corresponds to shear capacity, we have to set V* = phi.Vu, which can vary widely depending on the amount of shear reinforcement provided and the concrete strength. For an order of magnitude of the shear-strain component, I've used:
[ul]
[li]f'c = 50 MPa[/li]
[li]Asv = Asv,min; and Asv = 5*Asv,min. (Asv,min was calculated using the simplified method, otherwise it's an iterative calculation using the general method.)[/li]
[li]p = 0.5% * bv*dv and p = 1.0% * bv*dv (Longitudinal tension steel)[/li]
[/ul]

phi.Vu,min = phi * 0.26 * sqrt(f'c) * bv * dv = 1.38 MPa * bv * dv. This gives a shear-strain component of 0.69E-3 for p=0.5% or 0.34E-3 for p=1.0%.

For Asv = 5*Asv,min, phi.Vu = 3.71 MPa * bv * dv, and the shear-strain component is 1.86E-3 (p=0.5%) or 0.93E-3 (1.0%).

My conclusions are (for 50MPa concrete at least):
[ul]
[li]The simplified method will typically give a higher capacity than the general method when the section's bending capacity is fully used.[/li]
[li]The simplified method will often give a higher capacity than the general method when the section has heavy shear reinforcement, even for small bending moment.[/li]
[/ul]


This tallies fairly well with my spreadsheet, so that is attached warts and all.
[URL unfurl="true"]https://files.engineering.com/getfile.aspx?folder=1339e1f4-20d1-4b11-988d-d1b7f0be2708&file=AS3600-2018_Shear_Comparison_(20190107).zip[/url]

 

[cooperDBM]

Thanks Steve,
I'll have a closer look at your numbers when I have a chance. My initial comments are that the MCFT is dependent on an analysis of the actual (factored) shear and corresponding moment on the section for the best estimate of the axial strain at mid-height in order to calculate the crack width. The crack width, and concrete strength, determines the amount of shear carried along crack through aggregate interlock. A high moment, or high concrete strength, would reduce of eliminate this part of the Vc. Using the flexural capacity of the section wouldn't give a realistic state of stress/strain. It would be like checking for simultaneous flexural and shear failure. So I suspect your model may show the general method results as too low instead the the simplified results as too high.
Dave

 

[cooperDBM]

Steve,

I've had a look at your spreadsheet and played with it. Further to my previous comment, consider a 10 m simple span with 10 kN/m DL and 9 kN/m LL, to fully use the flexural strength. At 1.2 m from the support, to avoid the development zone and beta 3 factor for the 2009 code, you get M* = 134.6 kNm and V* = 96.9 kN. With those numbers your spreadsheet gives,

phiVu (AS3600-2009) = 402 kN
phiVu (AS3600-2018 - Simplified) = 412 kN
phiVu (AS3600-2018 - General) = 522 kN

I agree with the 2009 number but without the 2018 code I can't check the other two. The Canadian code gives,

Vr (A23.3-14 General) = 524 kN (factors different but similar result)

The simplified method in A23.3-14 can't be used with a long. steel strength greater than 400 MPa which is common in Canada.

Hope that helps.
Dave

 

[Settingsun]

Hi Dave,
I agree it may not be an issue with fairly modest distributed loads and concentrated loads such as in many building floors. But consider a beam with load dominated by a concentrated load. This would give maximum bending moment and (almost) maximum shear force at the same cross section.

I also think the numbers in your previous post provide only an approximate comparison of the general method with other methods. Since the shear capacity in the general method depends on the shear force, the general method calculation only gives a pass/fail (not the capacity) unless you iterate until V* = phi.Vu, at which point you have the shear capacity. Keeping the bending moment as 96.9kNm, the shear capacity is 387kN in the general method - still a pass but more in line with the other numbers. But this is perhaps not a good example since the shear demand is low in this case: we're at ~minimum shear reinforcement but still has four times the capacity needed.

Another case to consider might be a continuous support designed with moment redistribution so the full bending capacity is used near the critical shear section. The bending moment will however drop off fairly sharply though so may end up similar to your numbers.

 

[cooperDBM]

The bottom line is that the MFCT method requires an accurate assessment of the state of stress of the section to determine the axial strain at mid-height. If your critical shear section coincides with the critical flexural section then the resulting shear strength will be low since the beam is already failing in flexure (lots of wide cracks). The simplified method wasn't calibrated for that situation.

 

[rapt]

Interesting numbers. The Canadians maintain that the PhiVu determined from the MCFT is significantly less than that calculated by the old design method. You are showing over 25% higher!

 

[cooperDBM]

rapt,

Would depend on V/M (location in span), prestressing, etc. In my test (uniformed loads) I found the CSA General method (MFCT) to be quite similar to AS3600-2009 with CSA getting more capacity nearer the supports (though again similar at the critical section, d_o from support). MFCT gives a much higher and more consistent capacity than ACI away from the supports. My example above would be at a point where the difference is the most (CSA being higher).

 

[rapt]

cooperDBM,

I just did a check for a fairly normal RC beam at the critical shear section at an interior support for AS 2009 and CSA23 with reinforcement fully utilized and the AS3600 Vuc is about 50% higher than the CSA23 value, 196KN versus 130KN!

 

[cooperDBM]

rapt,

When you say "reinforcement fully utilized" I assume you mean the long. steel with the section at it's flexural capacity as in Steve's approach. As I discussed this means the section is heavily cracked with a large mid-height axial strain. MCFT (General) will give a low concrete shear strength in that situation. Not the scenario I was checking and I'm not sure the use of the simplified method would be appropriate. The relative difference between AS 2009 and MCFT would be very dependent on the section loading and level of cracking (axial strain). So with your scenario what the Canadians told you would be correct.

Do you only consider this extreme case? Difference in design approach between Canada and Australia? Interesting stuff (for me at least).

 

[rapt]

cooperDBM,

This was using the General approach. I see no need to use the simplified approach now that we no longer use slide rules!

Not sure what you mean by the last sentence. We check for shear at all sections in the beam, I was just giving the figures at the critical shear section!

 

[cooperDBM]

rapt, I agree about the Simplified Method. If I understood "reinforcement fully utilized" right your figures are for the critical shear section which is also loaded to the full flexural capacity. That would be a lower bound value for the shear strength under MCFT. For a comparison of the results from MFCT versus AS-2009 I'm suggesting also looking at other models where the M*/phiMu ratio is lower at the critical shear section as in my example.

 

[Settingsun]

I interpreted Rapt's comments as meaning he's modelled a continuous beam (probably in his Rapt software), adjusted the loading and flexural reinforcement so the flexural capacity is fully used at the support, then checked shear capacity at the critical shear section (~d away from the support). In this case, the moment at the critical section would still be substantial (depending on the exact geometry) but less than the maximum value. The shear capacity would then be based on <100% flexural utilisation and reflect a lean design per current practice. Awaiting clarification with interest.

previously when this has happened, if the new code has been more conservative in an area (eg development lengths last time), I have added it into RAPT immediately as it was known that the old code methodology was unconservative.

New version upcoming?

 

[rapt]

steveh49.

Yes to both, but don't hold your breath for the 2108 code release. Still working some other things before finalising it.

CooperDBM,

Yes I imaging things would be different. If fact, looking at it quickly, if I double the tension reinforcement in the same design compared to what was required for ultimate strength, then I assume I am doing what you are asking for. In that case, the AS3600-2009/CSA23 strength ratio for Vuc drops from 1.5 to 1.43!