Load Combinations When Using 25% Storage Load For Seismic

[P1ENG]

ASCE 7-10, section 12.7 says to use 25% of the live load for live loads that are considered storage. My question is what happens to the load combinations (ASD for the sake of this conversation)? I have to have live load in order for 25% of that load to be considered seismic mass, so D+0.7E no longer applies unless it were (D+0.25L)+0.7E -OR- (D+L)+0.7E. In order to know which of the latter (2), I have to know if the requirement of 12.7 considers only 25% of the live load to be present and 100% of that load is seismic [(D+0.25L)+0.7E] -OR- if 100% of the live load is present but only 25% of that load is seismic [(D+L)+0.7E].

Then what do I do with D+0.75(L+0.7E)? This load combination already has a live load component. I would tend to think in either condition (100% of 25% -OR- 25% of %100) I wouldn't modify this one.

This matters to me because I have relatively light construction (mezzanine) that is a cantilever column system. I have large moments at the base but not a lot of weight if I only use actual dead loads in the load combinations. The extra gravity load by including all/some of the live load would greatly help with footing sizes due to overturning safety factor.

Juston Fluckey, SE, PE, AWS CWI
Engineering Consultant

 

[P1ENG]

Celt:
I have no problems with E. I've got the 25% baked into E.

Juston Fluckey, SE, PE, AWS CWI
Engineering Consultant

 

[MotorCity]

Let me muddy the waters with this.....0.6D+E.....which reflects an overturning condition. In this case it would be unconservative to assume that some portion of the live load was present

 

[P1ENG]

Correct, but I'm not assuming. Some portion of the live [highlight #F57900]IS[/highlight] present. It has to be or you couldn't have 25% of it acting laterally!

Juston Fluckey, SE, PE, AWS CWI
Engineering Consultant

 

[Celt83]

I am saying that:
E=seismic forces inclusive of effects of 25% of storage live loads
--> Ed=seismic forces only of dead load (25% of storage live loads ignored) <-- no E is E and must always include the 25% of storage live load

Open Source Structural Applications:

https://github.com/buddyd16/Structural-Engineering

 

[P1ENG]

Again, I am not arguing E. I have the 25% in it. The Ed I proposed was to maintain the load combination D+0.7Ed. It won't ever control and I didn't need to include it. If there is no live load D+0.7Ed would be present. If there is live load D+0.25L+0.7E would be present.

Juston Fluckey, SE, PE, AWS CWI
Engineering Consultant

 

[Celt83]

I don't think you need to add your .25L, the .25Lstorage for the weight in E is to capture the effective of the static nature of storage live load.
I'm sure the DL+0.75(L+0.7E) takes into account the statistically loading probabilities such that the storage live load will not in all likely hood be the full 125 psf on the whole floor area.

I guess if you want to do your combo it should probably be:
D + 0.25L,storage + E not total Live load





Open Source Structural Applications:

https://github.com/buddyd16/Structural-Engineering

 

[JAE]

Here's some thoughts:
1. D+0.7Ed [blue]code does not mandate this. It requires you to use E here, not Ed.[/blue]
2. (D+0.25L)+0.7E <--- 25% live load present to produce 25% effective seismic mass [blue]code does not mandate this[/blue]
3. (D+L)+0.7E <--- 100% live load present to produce 25% effective seismic mass [blue]code does not mandate this[/blue]
4. D+0.75(L+0.7E) [blue]correct - one of the mandated load combinations[/blue]
5. 0.6(D+0.25L)+0.7E <--- 25% live load present to produce 25% effective seismic mass [blue]code does not mandate this[/blue]
0.6(D+L)+0.7E <--- 100% live load present to produce 25% effective seismic mass -[blue]code does not mandate this[/blue]

You are missing a code required combination:
7. 0.6D + 0.7E. [blue](with E based on D + 0.25L for derivation of W)[/blue]

Here's the combinations strictly required by ASCE 7-10 (negating Lr, S, W, etc.)
1. D
2. D + L [blue](100% of Live Load)[/blue]
3. D + 0.7E - [blue](with E based on D + 0.25L for derivation of W)[/blue]
4. D + 0.75(L + 0.7E) - [blue](your combination 4 - with E based on D + 0.25L for derivation of W)[/blue]
5. 0.6D + 0.7E - [blue](with E based on D + 0.25L for derivation of W)[/blue]

In all of the above, per a strict following of the text of ASCE 7, E is always based on W = D+.25L (where the 25% LL requirement applies)
Section 12.7 only states that (for storage conditions) 25% of LL must be included in W - there are no exceptions or alterations to W such as what you are proposing.


HOWEVER,
I see where your logic is - that if you include 0.25L in deriviing E, then there MUST be some Live Load present in the combination.
So for a D+E combo - there's no LL present and therefore W = D only.
For a D+L+E combo - you'd use W = D + 0.25L.

But the code doesn't ever suggest that.

You can, of course, include this in your load combinations, but I'm not sure what it gets you.

For strength calculations, your Load Combo 3 would generally control over your Load Combo 2.
For overturning checks, your missing Load Combo 7 (my LC 5) would generally control over your Load Combo 5.







Check out Eng-Tips Forum's Policies here:
faq731-376

 

[Celt83]

in the ASD combos:
E is actually Eh + Ev or Eh - Ev

Ev is 0.2 Sds D, exclusive of the live load effect
Eh is per Qe , inclusive of the live load effect

so the vertical component of D+.7E only includes D
The vertical component of D+0.75(L+E) includes D and L

then there are the other basic combos:
(1+0.14Sds)D+0.7Qe
(1+0.105Sds)D + 0.525Qe + 0.75L
(0.6-0.14Sds)D+0.7Qe

Open Source Structural Applications:

https://github.com/buddyd16/Structural-Engineering

 

[P1ENG]

JAE,
Thank you. I understand what I am doing has no code guidance. I am using judgement because I believe this is an oversight. You used the key word: "LOGIC". I will not consider 0.6D+0.7E. I am replacing that with 0.6(D+0.25L)+0.7E (i.e. 0.6W+0.7E). This all started because of overturning: 0.6D+0.7E was causing such a problem and I thought, "Hey, I'm using live load in my E. That live load is a mass with gravity". I understand D+0.7E would provide the worst-case (very much so!), but you can't have the increased E without using some L (25% L actually). Basically, any load combination that does not already contain an "L" component will have D replaced with W (effective seismic weight):
W+0.7E
D+0.75(L+0.7E)
0.6W+0.7E

For me, there is no longer:
D+0.7E
0.6D+0.7E
(unless E ignores the 25% storage live load and then there is no point in checking these load combinations).

Juston Fluckey, SE, PE, AWS CWI
Engineering Consultant

 

[Celt83]

P1ENG:

I think the main issue with using W in place of D in those equations is both D and L are conservative guesses at the actual in service loads
the load combos are built with this in mind so to go outside of those combos in an unconservative manner as you are proposing seems risky, especially on the 0.6D+0.7E case.

Open Source Structural Applications:

https://github.com/buddyd16/Structural-Engineering

 

[P1ENG]

Celt:
Good point! I am using the effective seismic weight in the vertical component of seismic. Again, staying true to my "if the mass is present, it has to be present" argument, I should be using the effective weight in my modified load combinations. I overlooked that D vs W on the first go-around. Now I am curious if I exclude 25% live load from the vertical seismic component, if that puts actual LC's back near the results that I am trying to achieve with my modified load combinations (answer: better, but no {see results below}).

My Sds=1.31, DL=15psf, LL=125psf
My modified LC (net vertical): 0.6(D+0.25L)-0.14Sds(D+0.25L)= 0.6(46.25 psf) - 0.14(1.31)(46.25 psf) = 19.27 psf [down]
Code LC (net vertical): 0.6D-0.14Sds(D) = 0.6(15 psf)-0.14(1.31)(15 psf) = 6.25 psf [down]
My incorrect application of Code LC's (net vertical): 0.6D-0.14Sds(D+0.25L)= 0.6(15 psf)-0.14(1.31)(46.25 psf) = 0.52 psf [down]


Perhaps if I had the correct Ev in the Code Load Combinations, I wouldn't have started this endeavor. I almost had a net downward load of 0 psf which means all of my base plate loads were moment and no gravity load causing overturning nightmares. However, if maximum gravity load is my goal, my modified load combination is still beneficial to me. I could use it per engineering judgment not only because it is beneficial, but logical as well.

Juston Fluckey, SE, PE, AWS CWI
Engineering Consultant

 

[sandman21]

sandman:
I don't know if you understood my question. If I had 1 kip dead and 4 kip live in a column and laterally had 0.1 kip from dead and 0.1 kip from 25% live then per a D+0.7E load combination my column would be 1 kip axial and 0.14 kip lateral. How can I get the extra 0.07 kip lateral from live load in a load combination that only contains dead load? I think, and Motorcity confirmed, the column should have 2 kip (1+0.25*4) axial load and 0.14 kip lateral (0.07[D]+0.07[0.25L]). I simplify, but I would also be considering the vertical components of seismic.

I do not understand why is has created such an issue. The code is extremely clear on what is included in the effective seismic weight, dead load as defined in section 3.1 and the listed loading. So when you are determining the W in V. You are including 25% of the floor live in a storage area. Lets say that you have a 5000sf floor, dead load is 50psf, and 2500psf has 100psf storage load. The W at the floor is 5000*50 + (.25)*2500*100psf=312.5kips. This is the load that load be used in base shear V. This does not change the load combinations, it does not add to load combinations, it will be included in the base shear V acting on an element.

 

[P1ENG]

I think the main issue with using W in place of D in those equations is both D and L are conservative guesses at the actual in service loads the load combos are built with this in mind so to go outside of those combos in an unconservative manner as you are proposing seems risky, especially on the 0.6D+0.7E case.

But I disagree with the assumption it is unconservative. The factor of 0.6 on the D (or in my case W) component is to account for those uncertainties. You take 90% of the dead load as an assumed correction to an overestimate of the dead load, and then a safety factor of 1.5 is applied to that: 0.9/1.5 = 0.6.

Juston Fluckey, SE, PE, AWS CWI
Engineering Consultant

 

[P1ENG]

I do not understand why is has created such an issue. The code is extremely clear on what is included in the effective seismic weight, dead load as defined in section 3.1 and the listed loading. So when you are determining the W in V. You are including 25% of the floor live in a storage area. Lets say that you have a 5000sf floor, dead load is 50psf, and 2500psf has 100psf storage load. The W at the floor is 5000*50 + (.25)*2500*100psf=312.5kips. This is the load that load be used in base shear V. This does not change the load combinations, it does not add to load combinations, it will be included in the base shear V acting on an element.

It was an issue to me because my storage area is not confined to one portion of the floor area. It is over the entire floor area. Compounded with light framed construction, I have (or had {see my last post}) very small gravity loads at my foundation. No gravity load with large moments = no bueno for overturning.

Juston Fluckey, SE, PE, AWS CWI
Engineering Consultant

 

[sandman21]

As I mentioned in my post 12.7 does not create new load combinations, your modified LC is incorrect. Using your created LC would be unconservative and without justification in the code. Per 12.4.2.2 vertical seismic is based on D not the effective weight.

 

[P1ENG]

I know I have no justification from the code. My modified LC is incorrect in the sense that it doesn't conform to the code, not because it doesn't make sense. My logic has provided it is correct. My judgement (which I will clearly state in my calculations is in opposition to or unaddressed by the code) is that my modified LC's are not unconservative. The commentary of 12.7.2 states the reason for the 25% portion of the storage live load. Inertial forces *could* be developed by closely packing stored items. I would argue you would have to be close to the full rated live load before you would get the closely packed condition that would produce inertial forces; so if anything, I am being conservative that I am only using 25% of the live load in my modified combinations. The code does not state it or explicitly state otherwise, but I think the effective seismic mass should be classified a dead load and therefore my modified LC's would conform to code: D = W = Dead + 0.25*LL_storage. Mass is mass and will provide force if accelerated from seismic shaking or constant gravity. I will not let the seismic force magically increase laterally due to a live load if I am not allowed to also consider that same live load gravitationally. Regardless of the opposition, I think this has been a good discussion and I have decided to proceed with my modified LC's. This isn't a high risk structure, it is a single-story mezzanine. Beams-to-columns are assumed pinned (LFRS = ordinary cantilever column), but in reality there is a partial fixity between them reducing the overturning moment I am trying to reduce anyway. I have no fear placing my seal on this analysis as I am proposing.

I'm not some recent graduate. I have passed the SE 16-hour exam and have 12 years experience. I have given this much thought and am allowing my judgement to have some weight. Mass cannot magically act in one dimension only. I know I am shaking my fist at the code. The code isn't always correct or is open to interpretation. An engineer is allowed to have a brain and question the code. Don't check your thoughts at the door just because the code doesn't say you can do it.

Juston Fluckey, SE, PE, AWS CWI
Engineering Consultant

 

[Celt83]

P1ENG:
My argument to contrary and this is going to sound ridiculous is that that 25% of live load storage weight is static and motionless at the onset of the seismic event thus contributes to the mass and Qe. Once the building hits is peak movement bouncing back all those once static storage items are now moving and moving independent of the building motion. so how are you capturing where the new center of live load mass is relative to the dead load mass in your new combo for overturning.

assuming your D and LL are uniform over the whole floor area
Floor plate at
t=0, D and Lstorage load applied at center of mass - top image
t=period/2, D still at center of mass but now mobile storage items shift center of LL mass
t=period, again D still at center of mass but now mobile storage items shift center of LL mass

with this same line of thinking you might argue that the moving storage items would provide some dampening, just picture a floor full of three shelf high storage draws opening all at once at peak acceleration and applying a restoring force of F=MA of all that paper wanting to continue to move as the building wants to accelerate in the opposite direction.

Open Source Structural Applications:

https://github.com/buddyd16/Structural-Engineering

 

[sandman21]

Your modified LC is incorrect and would be unconservative, as you are assuming that the LL is permanently attached and unmovable. You have no idea what part of the LL will be contributing to the effective seismic forces, as far as you know a 50% loaded mezzanine could exceed the 25% of a fully loaded mezzanine. Which is why we have load cases with and without LL to check components and determine the worst case. It's not checking "thoughts at the door", it's more like having an understanding of when, why, where to be conservative with seismic loading.

 

[P1ENG]

with this same line of thinking you might argue that the moving storage items would provide some dampening, just picture a floor full of three shelf high storage draws opening all at once at peak acceleration and applying a restoring force of F=MA of all that paper wanting to continue to move as the building wants to accelerate in the opposite direction.

Moving (i.e. sliding on the floor) storage items would not be producing the inertial forces and would be like any other live load. The 25% (by definition of inertial forces) cannot move relative to its support. If it moves, it doesn't contribute to the 25%. "Tightly packed" in the commentary that explains the inertial forces means the live load will not fall over. I've not run into a seismic case where 0.14Sds exceeds the 0.6 factor on gravity loads, which would provide a net uplift on the mass, so the live load should not bounce off the floor because the vertical component never exceeds the 0.6 factored gravity. Yes, if storage items were not "tightly packed", then tipping and falling over would occur.

I also want to say that I am taking your posts very seriously because you have taken quite a bit of time with your points. Drawings and lengthy posts. I appreciate that!

... you are assuming that the LL is permanently attached and unmovable.

Of course I am. What is the definition of inertia? Read the commentary on 12.7.2. The 25% live load comes from inertial forces. Inertia = stays were it's at.

Juston Fluckey, SE, PE, AWS CWI
Engineering Consultant

 

[MrHershey]

But I disagree with the assumption it is unconservative. The factor of 0.6 on the D (or in my case W) component is to account for those uncertainties. You take 90% of the dead load as an assumed correction to an overestimate of the dead load, and then a safety factor of 1.5 is applied to that: 0.9/1.5 = 0.6.

If the goal is complete consistency between load combination and weight calculation, let's reverse this and go the other way for a minute. Why are we using the full 1.0D in our weight calculation in conjunction with the 0.6D+0.7E combination? If your load combination and weight calculation are supposed to be consistent, shouldn't we only be using 0.6D in our weight calculation when using this load combination? Or maybe 0.9D so you maintain the 1.5 safety factor as you've noted? Don't think I'd ever argue this, but if we're trying to be consistent then why not?

I get what you're saying and it makes a bit of sense to me. But regardless of whether you're theoretically correct, you also have to convince other people (code reviewers, building departments, etc.) of the same. And most of those types are inclined to point to the code and say that if it's not in there you can't do it. I honestly have a hard time seeing you win that argument.

Personally, this would be the kind of code gymnastics I'd consider when I'm trying to prove an existing structure is adequate. I don't think I'd employ it on a new structure.