° for a prestressed beam

[sdz]

AS3600 has rules for calculating the capacity reduction factor (CRF) in Table 2.2.2 which prescribed linear interpolation between Nuot and Muo for axial tension, and between Muo and Nub for axial compression.

I have found for a prestressed beam that Nub can be tension, where I am taking Nu as the externally applied load. This means that applying the AS3600 rules that at Nub the CRF will be greater than 0.6, since Nu is tension at this point.

Is this correct? I think AS3600 needs more clarification on this point.

 

[IDS]

I'm surprised there have been no responses to this, so I'll have a go.

I think sdz is right, the rules as written give the phi factor increasing from a minimum of 0.6 at zero axial load to 0.8 at ultimate axial load, so a beam in tension will have a phi greater than 0.6, even if Nu is greater than Nub.

This doesn't seem to make a lot of sense, compared with the logic of increasing phi between zero and Nub when Nub is in compression. It would probably be a good idea to use the lesser of Nub and zero as the load at which phi starts to increase.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[apsix]

I haven't the time to investigate, but presumably it will be conservative to use phi = 0.6 anyway.

 

[rapt]

I have discussed this with sdz privately.

It only happens when the amount of prestress is enormous and, in reality, impractical for the concrete section in the member. In his case, P/A was 16MPa (no bending included) with 40MPa concrete. In that case, for P/A of less then 8MPa, the AS3600 phi calculation method was still logical.

This whole problem is a consequence of using capacity reduction factors rather than material factors. With capacity reduction factors, the code writers have to manipulate the factors to suit all of the different possible situations of concrete dominating and steel dominating capacity. With material factors as are used in Eurocodes, this all comes out automatically from the material factors.

 

[IDS]

This whole problem is a consequence of using capacity reduction factors rather than material factors. With capacity reduction factors, the code writers have to manipulate the factors to suit all of the different possible situations of concrete dominating and steel dominating capacity. With material factors as are used in Eurocodes, this all comes out automatically from the material factors.

Yes, I prefer the material factor approach myself, but maybe that's just how I was brought up .

I think it's also worth mentioning that the implementation of the reduction factors in the Australian Codes is much more conservative than either the Eurocode or the ACI code for loads below the balance load. That's a consequence of using a lumped factor (compared with the Eurocode), and starting to increase the reduction factor at zero axial load (compared with ACI).



Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[rowingengineer]

maybe someone on the committee should look at the implications of changing the standard.

ANY FOOL CAN DESIGN A STRUCTURE. IT TAKES AN ENGINEER TO DESIGN A CONNECTION.”

 

[rapt]

RE,

There are pros and cons to both methods. Though as I get older I tend to lean towards the materials factor logic. In terms of columns, the material factor logic is by far the easiest to use. My one problem with it is that you basically have a different stress/strain curve for concrete for service and ultimate conditions. There is no gradual transition in how things behave as load increases, so a moment curvature diagram is impossible to create logically!

It was discussed early in the last code development with Prof Warner. I am not sure why they never proceeded with the change. Maybe it was too "radical" for the overall committee at the time. It will be on the agenda for next time!

 

[sdz]

Thanks to all who replied. As rapt pointed out this seems to occur only in extreem cases but it does show that the rules for calculating ø do break down in some circumstances.

Perhaps the answer should be that we can calculate ø as
ø = 0.6 ? (1.19 ? 13ku/12) ? 0.8
with the interpolation rules of TABLE 2.2.2 (c) and (d)allowed as an alternative so that ku does not need to be calculated in each case.

 

[rapt]

TABLE 2.2.2 (c) and (d) are dependant on

ø = 0.6 ? (1.19 ? 13ku/12) ? 0.8

It is always calculated first.

The (c) and (d) conditions should still result in ø within the limits of .6 and .8!

I do not think that anyone on the committee ever considered to possibility that Nub would be a tension force! It can only happen with a very very highly prestressed and grossly over-reinforced PT section.

 

[IDS]

There is no gradual transition in how things behave as load increases, so a moment curvature diagram is impossible to create logically!

I don't see the problem there. If you want the M/C relationship for upper bound stiffness you wouldn't apply the materials factor anyway (or maybe apply an overstrength factor rather than a reduction), and if you want lower bound stiffness I don't see the problem with applying the reduction factor to a parabolic/rectangular stress block. Am I missing something?

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[sdz]

In table 2.2.2 (b) ø is calculated for "Bending without axial tension or compression" as
0.6 ? (1.19 ? 13kuo/12) ? 0.8 - note kuo

What I am suggesting is that at any combination of moment and axial load use
0.6 ? (1.19 ? 13ku/12) ? 0.8 - note ku without subscript o

This obviously gives the same value of ø for the pure moment case but avoids any problems with interpolation at other loads, even for grossly over-reinforced PT sections.

 

[rapt]

sdz,

I actually think the use of kuo is wrong anyway, even in the current formulae and in clause 8.1.5. Ductility should be dependant on the centroid of the tension force, not the depth to the extreme layer of reinforcement.

This is another item that is currently under discussion for review.
Hopefully we will go to material factors in the next code and all of this will disappear, and we will then have another set of problems. One interesting one is that developemnt length in Eurocode is calculated on the force in the steel reduced by the material factor, so, Fsy / 1.15! So developemnt lengths are nominally 13% less!

 

[IDS]

I have just completed a comparison of AS 3600 with ACI 318 and EC 2 and two UK codes (BS 5400 and 8110) which might be of interest. The main outcome is that AS 3600 is often very conservative and ACI is conservative compared with the European codes with concrete up to 50 MPa for loads at and above the balance load, but is unconservative for high strength concrete, especially if the factors for spiral reinforcement are used.

http://newtonexcelbach.wordpress.co...einforced-concrete-as-aci-and-european-codes/

Note that for the AS 3600 results I used the EC 2 parabolic-rectangular stress block with AS 3600 reduction factors. This is compared with the AS 3600 rectangular stress block here:

http://newtonexcelbach.wordpress.co...f-reinforced-concrete-to-as-3600-and-aci-318/

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/