## AS3600-2018 Shear Capacity: General Method vs Simplified

## AS3600-2018 Shear Capacity: General Method vs Simplified

(OP)

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.

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.

## RE: AS3600-2018 Shear Capacity: General Method vs Simplified

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.

## RE: AS3600-2018 Shear Capacity: General Method vs Simplified

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.

## RE: AS3600-2018 Shear Capacity: General Method vs Simplified

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.

## RE: AS3600-2018 Shear Capacity: General Method vs Simplified

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?

## RE: AS3600-2018 Shear Capacity: General Method vs Simplified

There has been contact and questions asked.

## RE: AS3600-2018 Shear Capacity: General Method vs Simplified

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.

## RE: AS3600-2018 Shear Capacity: General Method vs Simplified

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:

## CODE

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:

- f'c = 50 MPa
- 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.)
- p = 0.5% * bv*dv and p = 1.0% * bv*dv (Longitudinal tension steel)

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):

This tallies fairly well with my spreadsheet, so that is attached warts and all.

https://files.engineering.com/getfile.aspx?folder=1339e1f4-20d1-4b11-988d-d1b7f0be2708&file=AS3600-2018_Shear_Comparison_(20190107).zip

## RE: AS3600-2018 Shear Capacity: General Method vs Simplified

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

## RE: AS3600-2018 Shear Capacity: General Method vs Simplified

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

## RE: AS3600-2018 Shear Capacity: General Method vs Simplified

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.

## RE: AS3600-2018 Shear Capacity: General Method vs Simplified

## RE: AS3600-2018 Shear Capacity: General Method vs Simplified

## RE: AS3600-2018 Shear Capacity: General Method vs Simplified

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).

## RE: AS3600-2018 Shear Capacity: General Method vs Simplified

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!

## RE: AS3600-2018 Shear Capacity: General Method vs Simplified

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).

## RE: AS3600-2018 Shear Capacity: General Method vs Simplified

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!

## RE: AS3600-2018 Shear Capacity: General Method vs Simplified

## RE: AS3600-2018 Shear Capacity: General Method vs Simplified

New version upcoming?

## RE: AS3600-2018 Shear Capacity: General Method vs Simplified

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!

## RE: AS3600-2018 Shear Capacity: General Method vs Simplified

## RE: AS3600-2018 Shear Capacity: General Method vs Simplified

- with reinforcement required at the critical section, the ratio is still 1.23.

- with maximum +ve moment reinforcement in the span developed as required at the end support, the ratio drops to .9

## RE: AS3600-2018 Shear Capacity: General Method vs Simplified

## RE: AS3600-2018 Shear Capacity: General Method vs Simplified

I had set beta3 as 1!

## RE: AS3600-2018 Shear Capacity: General Method vs Simplified

Consider the following

6m simply supported beam carrying ultimate design load of 30kN/m

25MPa Conc. 600mmX600mm cross-section with 4-N16 tensile reinforcement (500MPa) - Ast=800mm2 at d=530mm

Moment capacity approx=165kNm>135kNm Max demand - OK

Shear demand at d from face = 74kN

Shear Capacity to 2009 phiVuc= 104kN

2009 requires min shear reinforcement when V*>0.5phiVuc -- Thus provide

Shear Assessment to 2018 - Moment at d = 43kNm

Shear Capacity (2018 General method) phiVuc=206kN

2018 doesn't require shear reinforcement unless V*>phiVuc -- Thus not required (indeed V*<0.5phiVuc)

Thus the new MCFT provides a much less conservative design in this instance. Now I would never not provide shear reinforcement for a beam overhead, but this "beam" could indeed be a strip footing that supports a pre-cast panel on shims during erection (rigid footing assumption leading to 30kN/m UDL). Obviously depending on methodology employed the reinforcement details would be vastly different.

Regards

Toby

## RE: AS3600-2018 Shear Capacity: General Method vs Simplified

That's a good example showing a real difference in the design outcome. I wonder if our bridge colleagues are already well-aware of it. The first time I came across MCFT, it was a method that had potential for the state road agencies to load-rate (successfully) their older bridges for SM1600 loading. I understand there are quite a few pseudoslab bridges (contiguous pretensioned beams) with no stirrups as the prestress results in sufficient shear capacity. Since that was my first impression and it stuck, I was surprised by Rapt's feedback from Canada that MCFT should give lower capacity.

I think Rapt suggested in another topic that the 0.5*phi.Vuc limit for no stirrups may be re-introduced.

(Rapt, are you on the code committee, or one of the colourful concrete identities with contacts there?)

## RE: AS3600-2018 Shear Capacity: General Method vs Simplified

Thought I'd throw in a New Zealand code comparison point using Toby43's example numbers, NZS3101 results in a phiVc of:-

111.9kN when minimum shear reinforcement is provided,

or 104.9kN when minimum shear reinforcement is not provided.

In NZ the 4-N16 bars is less than minimum longitudinal steel requirements though.

We also have the V*<=0.5phiVuc requirement with no min shear reinforcement needing to be provided. This ratio applies except for a few specific cases relating to shallow beams and slabs, and then its revised to V*<=phiVc. Any member deeper than 250mm always requires at least minimum shear reinforcement if V*>0.5phiVc.

Having said that though I've never actually seen anyone design/detail a

suspendedbeam of any depth without actually providing minimum shear steel, irrespective of the actual V* vs phiVc ratio. Surprises me that it was dropped altogether in the 2018 AS code.## RE: AS3600-2018 Shear Capacity: General Method vs Simplified

In my opinion it is prudent to apply the old 0.5Phi.V for members < 750 mm deep, whether this makes its way back into the code or not.

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: AS3600-2018 Shear Capacity: General Method vs Simplified

The minimum reinforcement rule is one that is being fixed (hopefully).

Canadian code is the only one that does not use the .5. Their justification is that the Vc is much lower, but as discussed above, that is not always the case.

Using Vu for slabs is ok as slabs normally have alternate load paths if there is a shear crack. Because beams do not have an alternate load path, it has always been accepted that .5Vu is a logical point to add minimum shear reinforcement as we are basically relying on concrete in tension to provide strength in a member where redistribution is not possible.

AASHTO using MCFT uses .5Vu as the point at which minimum is required.

Eurocode uses 0 and requires it in all beams.

## RE: AS3600-2018 Shear Capacity: General Method vs Simplified

## RE: AS3600-2018 Shear Capacity: General Method vs Simplified

Yes I realise all of that.

And AASHTO still wants .5Vu based on the Vc without stirrups. Logically, if there are no stirrups, that is the Vc available so it is the one used to determine minimum. But you are still relying on concrete tension until minimum stirrups are provided. For something that is a brittle failure until stirrups are provided.

## RE: AS3600-2018 Shear Capacity: General Method vs Simplified

## RE: AS3600-2018 Shear Capacity: General Method vs Simplified

Also regarding the 2009 being unconservative, a large part of this is the beta 1 factor which applies size effect. When section size is increased 1.5m+ the 2009 code became more unconservative with increasing section thickness.

## RE: AS3600-2018 Shear Capacity: General Method vs Simplified

## RE: AS3600-2018 Shear Capacity: General Method vs Simplified

I believe this is correct and now gives an answer which is similar to Eurocodes (EC2)

## RE: AS3600-2018 Shear Capacity: General Method vs Simplified

That clause in AS3600-2018 is under review at the moment. Personally I think it is wrong and the .5 Vuc should be reinsatted. The only code with a similar rule is the Canadian.

## RE: AS3600-2018 Shear Capacity: General Method vs Simplified

Thank you for uploading your spreadsheet.

My personal experience is that for slab design, the simplified method is way too conservative. I haven't tried it for the beam yet. I will go through your spreadsheet and give a deep dive into it.

However, at first glance, I did realise that in your calculation of Muo,

you used all available Ast. Since your calculated phiVus (209kN for simplified method and 247kN for general method) are less than 2*Veq (=2* 239kN)

if you look at Clause 8.2.7 a (AS3600-2018) on page 121, deltaFtd > 0 therefore, definitely you would need some additional longitudinal reinforcement due to shear and/or torsion. Therefore, I would have thought that you need to exclude this additional amount of longiditunal reo when calculating Muo?

Could you please comment on this?

Regards

NTC

## RE: AS3600-2018 Shear Capacity: General Method vs Simplified

You should get similar results from both methods if you adopt the same logic for the General method which would mean basing the calculation on the final reinforcing pattern, not the reinforcement required for flexure.

That is the justification I have been given for it by the experts!

## RE: AS3600-2018 Shear Capacity: General Method vs Simplified