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.

 

[cooperDBM]

Or a ratio of 0.56, using Steve's spreadsheet, for my example which may be more typical of a precast beam or end support. Just showing that there is a greater variation.

 

[rapt]

On the same run at a pinned end support at the critical section
- 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

 

[cooperDBM]

Sorry, in my example the ratio of 0.56 was at a point a bit away from the critical shear section to avoid beta 3 in AS-2009 to suit the spreadsheet. At the critical shear section the ratio for Vuc is about 0.8, due to the beta 3 factor in AS-2009. So the Vuc ratio between AS-2009 and CSA/MFCT can vary because MFCT is sensitive to the state of stress in the section.

 

[rapt]

cooperDBM,

I had set beta3 as 1!

 

[Toby43]

To add another example to highlight some differences in the design outcomes using AS3600:2009 vs AS3600:2018 (General Method)
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

 

[Settingsun]

Hi Toby,
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?)

 

[Agent666]

I'm not really familiar with newest version of AS3600, but have been following along to see if someone can explain the differences and if these differences are intended or an oversight in behaviour.

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 suspended beam 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.

 

[IDS]

Note that Cl. 8.2.6 (c) of the new code requires shear reinforcement in all members >= 750 mm deep.

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/

 

[rapt]

Toby43

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.

 

[cooperDBM]

In the Canadian code the V* < Vc requirement check is dependent on how Vc is calculated. If minimum stirrups are provided Vc is based on a crack spacing of 300 mm. If at least minimum stirrups are not provided the crack spacing is taken as dv (usually) which leads to a reduced value for Vc (for 300 mm < dv). So when checking for the requirement for minimum stirrups you would be using the reduced value of Vc (rapt, that may be what they meant). It's not as conservative as 0.5Vc. AS3600-2018 (based on draft) doesn't have that dependency on minimum stirrups for Vc, so a V* < Vc limit may be even less conservative.

 

[rapt]

cooperDBM,

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.

 

[cooperDBM]

Sounds like I'm not of help here, so I'll leave you to it.

 

[DOUGHBOY23]

Sorry I might have missed this, but there doesn’t seem to be any discussion about additional longitudinal reinforcement to resist shear forces when using MCFT vs 2009 code equation. If this is not being reviwed e.g when full bending capacity is used + shear without additional longitudinal reinforcement you don’t have a section that complies with MCFT... Make sure your comparing apples with apples..

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.

 

[Settingsun]

Are you referring to Delta-F.td from clause 8.2.7? Figure 8.2.8 makes it look as though offsetting the bending moment diagram by D ~ 1.5*D meets this requirement, ie just size reinforcement for the maximum bending moment alone, and therefore fully-utilised at continuous supports. Doesn't look like a great change from past practice at first glance.

 

[GB1000]

Apolgies for lateness into this conversation, but I'm I correct in the sense that in AS3600-2018 shear reinforcement is only required if V*>phiVuc and not AS3600-2009 0.5Phi.V?

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

 

[rapt]

Eurocode2 requires shear reinforcement at all locations in all beams (except beams of minor importance such as short span lintels) no matter what the value of Vc!

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.

 

[NTCONLINE]

Hi Steveh49,

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

 

[rapt]

The Simplified method assumes that the reinforcement required at the peak moment region is extended over the moment zone, so as you move away from the support, you have a lot more reinforcement than is required for flexural strength.

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!

 

[Settingsun]

I'm not in front of the code, but is this the same question as 13 March above? I never got around to looking into it properly as it turns out I haven't done much concrete design this year.