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Revised internal pressure calculation in AS/NZS 1170.2:2021 Part 2

Tomfh

Structural
Feb 27, 2005
3,645
This is a follow up from the following thread:

In it, someone asked:

For the Kl part, is it still only referring to cladding, cladding fixings, elements that supports cladding, and this fixing?

to which @rowingengineer replied:
For KL internal pressure it applies to all parts of the building. This has been clarified by the code committee.

Can anyone confirm/clarify this? Are local pressures at the locations of openings now applied to all internal surfaces simultaneously, including loading of primary framing members?
 
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5.4.4 Local pressure factor (Kℓ) for cladding
The local pressure factor (Kℓ) shall be taken as 1.0 in all cases except when determining the wind
actions applied to cladding, their fixings, the members that directly support the cladding, and the
immediate fixings of these members.

1.4.6 cladding
material that forms the external surface over the framing of a building or structure

Kℓ is applied only to external pressure on cladding and is combined with the appropriate internal pressure.

Hence pnett = qz(Kℓ Cpe - Cpi)
 
Table 5.1B says Kl applies to internal pressures when there are external openings greater than 0.5% which are located in the corresponding external local pressure zone.

The reply I quoted above suggests this applies internally not just to cladding and its supports (which is already quite drastic), but to the whole building, presumably meaning the structure itself. So I'm querying if I've understood this correctly, does it also apply to the main structural members?
 
Table 5.1B says Kl applies to internal pressures when there are external openings greater than 0.5% which are located in the corresponding external local pressure zone.

The reply I quoted above suggests this applies internally not just to cladding and its supports (which is already quite drastic), but to the whole building, presumably meaning the structure itself. So I'm querying if I've understood this correctly, does it also apply to the main structural members?
You're right. However;
For nominally sealed buildings you'll be using Table 5.1A. Table 5.1B only applies for dominant openings.

When there is a dominant opening;
When Kl affects internal pressure it affects entire surfaces so it is no longer a local pressure on those internal surfaces.
When the area of the opening A>a^2, Kl =1
When A>10m^2, Ka<1.0

If say, you're looking at an industrial building with truck size opening, then you will most likely have Kl =1 and Ka<1.0.

For internal walls the extreme combination of pressures on either side would only need to be used for structural walls.
 
Table 5.1B only applies for dominant openings.
That's what this thread (and the thread it follows on from) refers to.
When Kl affects internal pressure it affects entire surfaces so it is no longer a local pressure on those internal surfaces.
That's not what the table appears to say. Say you have a small (0.5%) opening at a Kl=1.5, or 2, or 3 external local pressure zone, then by the table this local factor applies to all internal surfaces. If we're reading this wrongly, it would be good to know.
 
Do you have this book? I thought they had an example in it, maybe not but if my memory serves me it did.

No. I’ll check it out.

In your reference to “clarified by the committee”, what specifically were they clarifying, and where did they clarify it?
 
That's what this thread (and the thread it follows on from) refers to.

That's not what the table appears to say. Say you have a small (0.5%) opening at a Kl=1.5, or 2, or 3 external local pressure zone, then by the table this local factor applies to all internal surfaces. If we're reading this wrongly, it would be good to know.
I think you are reading it wrong. Let's see what others say.

btw, in the bottom row what does "t5-1(b)-1" mean? I suspect it was meant to be
1748483390641.png

 
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T5.1(B) directs the reader to Cl 5.4.4 which clearly states that KL is 1.0 unless applied to cladding, their fixings, the members that directly support the cladding, and the immediate fixings of these members. Surely it does not apply primary members.

I watched a bit of the youtube video referenced in the other thread. It may have been clarified by the code committee and stated in the video that it is to be applied to all parts of the building, but that is fundamentally different to how the code is written.
 
You're right. However;
For nominally sealed buildings you'll be using Table 5.1A. Table 5.1B only applies for dominant openings.

When there is a dominant opening;
When Kl affects internal pressure it affects entire surfaces so it is no longer a local pressure on those internal surfaces.
When the area of the opening A>a^2, Kl =1
When A>10m^2, Ka<1.0

If say, you're looking at an industrial building with truck size opening, then you will most likely have Kl =1 and Ka<1.0.

For internal walls the extreme combination of pressures on either side would only need to be used for structural walls.
I didn't initially agree with sdz's interpretation, however, after reading the reference provided by rowingengineer; I believe they have got it correct.

See below from WIND LOADING HANDBOOKFOR AUSTRALIA & NEW ZEALAND: Background to AS/NZS 1170.2 Wind Actions
From Paragraph 3.2.1
View attachment 10107
From example B.3
View attachment 10108
 
T5.1(B) directs the reader to Cl 5.4.4 which clearly states that KL is 1.0 unless applied to cladding, their fixings, the members that directly support the cladding, and the immediate fixings of these members. Surely it does not apply primary members.

I watched a bit of the youtube video referenced in the other thread. It may have been clarified by the code committee and stated in the video that it is to be applied to all parts of the building, but that is fundamentally different to how the code is written.
"T5.1(B) directs the reader to Cl 5.4.4", how did you get that?
 
It may have been clarified by the code committee and stated in the video that it is to be applied to all parts of the building, but that is fundamentally different to how the code is written.
Yes, Clause 5.4.4 says that, but based on how the code authors discuss it (e.g. in the video), Kl in the context of internal pressures is meant more as a parameter for defining internal pressure conditions, rather than something applied in the usual patch load sense to cladding or its supports.
 
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Apologies, take 2 (although not sure I've got it right this time either...)

Paragraph 3.2.1
1748498841692.png

Example B.3
1748498971448.png
 
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Yes, Clause 5.4.4 says that, but based on how the code authors discuss it (e.g. in the video), Kl in the context of internal pressures is meant more as a parameter for defining internal pressure conditions, rather than something applied in the usual patch load sense to cladding or its supports.
Then they should add another note that simply says "KL when referenced by T5.1(B) is applicable to the whole structure and not just the elements directly supporting cladding." Problem solved.

As the code is written, that definition in the video goes directly against it. The reference to Cl 5.4.4 is just cladding elements and the members that support it.

Are the code writers expecting that engineers need to read online forums and watch random youtube videos to gain a full understanding of how the code is to be interpreted?
 
I can see the examples now. Thanks. They don't really seem to address the issues as the examples only cover the Kl=1.0 case, not what to do when Kl is greater than one, which is the point of contention here. If Kl=1.0 then you are basically ignoring local pressure effects.

There appears to be two key questions here:

1. Does Kl apply uniformly to all internal cladding surfaces? For example, if Kl = 1.5 (e.g. due to a window in the Kl=1.5 zone), does this Kl=1.5 factor apply across all internal cladding surfaces simultaneously, or is it applied locally internally in the usual patch loading manner? I thought the intention was to apply to across all internal cladding surfaces, but several of you here are saying NO to that.

2. If Kl is applied to all internal surfaces simultaneously, does this also apply to primary framing members as well? (My original question)

Are the code writers expecting that engineers need to read online forums and watch random youtube videos to gain a full understanding of how the code is to be interpreted?
I'm not a fan of the level of complexity in codes either. Regardless, it would be good to understand precisely what it is that the code expects of us, given we are expected to certify that structures comply with the code.
 
here is an internal pressure calculator I wrote as a result of this discussion. I'm pleased that the results match the example give above..
 

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  • Internal Pressure Calc.xlsx
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