Load Combinations When Using 25% Storage Load For Seismic
Load Combinations When Using 25% Storage Load For Seismic
(OP)
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
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






RE: Load Combinations When Using 25% Storage Load For Seismic
RE: Load Combinations When Using 25% Storage Load For Seismic
Anywhere "L" appears in a load combination w/ "E", use 100% of "L".
Anywhere "D" appears in a load combination w/ "E", use "D + 0.25L"
RE: Load Combinations When Using 25% Storage Load For Seismic
You calculate E based on all the dead load and 25% of the live load.
Then you apply your dead, live, wind, seismic loads as they are to the load combinations.
The 25% is NOT saying that: "when you have a code seismic event, only 25% of the live load is present".
It IS saying: "when you have a code seismic event, including 25% of the live load in the effective seismic mass is an means of estimating the contribution of some of the present live load to the seismic demand."
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RE: Load Combinations When Using 25% Storage Load For Seismic
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.
MotorCity:
That was the way I was leaning. Thanks for the confirmation. Have you got published guidance or is that your standard practice? Please rethink D+0.75L+0.525E though. Per what you said, that would become D+L+0.525E. I would think that the 0.75L of that load combination is inclusive of the 0.25L that acts as the seismic mass, so there is no need for that load combination to change.
JAE:
See my response to sandman. If you include 25% of the live load in the lateral/vertical seismic forces, then you have to include that same 25% in the load combinations you are using. D+0.7E becomes (D+0.25L)+0.7E. In my query, it does matter if some or all of the live load has to be present to get the 25% effective seismic weight.
Juston Fluckey, SE, PE, AWS CWI
Engineering Consultant
RE: Load Combinations When Using 25% Storage Load For Seismic
RE: Load Combinations When Using 25% Storage Load For Seismic
No you don't.
Section 12.7 says ONLY - that you have to include 25% of the live load in the DERIVATION of W for determining your seismic demand, E.
This does not affect your live load at all - nor does it affect your inclusion of the full live load, or 0.75L, etc. in your combinations.
It ONLY affects how you get your E to use in the combinations.
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RE: Load Combinations When Using 25% Storage Load For Seismic
I'm sorry, but I [somehow by conveying it tactfully on a forum] reject that. If you assume a portion of live load is present as a mass, then that same portion of mass is present for both lateral and gravity. It doesn't become weightless. The 25% live load mass has acceleration laterally and vertically from the seismic event; that mass will also have acceleration of gravity acting on it.
Motorcity:
I did not replace 0.75L with L. I did (D+0.25L) + 0.75L + 0.525E = D + L + 0.525E.
Juston Fluckey, SE, PE, AWS CWI
Engineering Consultant
RE: Load Combinations When Using 25% Storage Load For Seismic
RE: Load Combinations When Using 25% Storage Load For Seismic
You could have 100% of your live load present during a code seismic event, but only 25% of it contributing to our code-estimated-static-equivalent seismic demand.
The other 75% could be jostled around, not stiffly secured to the floor, etc. and not contribute much of anything to the lateral demand.
Example - a floor with shelving and lots of weight - the floor shaking would cause the shelving to sway within the building - possibly in opposite directions/rhythm to the floor movement and actually reduce the seismic demand a bit. So correlating what live load I put into my seismic W has nothing to do with the live load I include in my various combinations.
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RE: Load Combinations When Using 25% Storage Load For Seismic
Right. So I am assuming people load is not present when altering my load combinations. The 25% of live load is assumed connected to the structure by inertial affects or attachment. If you have to have 100% of the live load to obtain the 25% inertial loads assumed by the effective weight, then you should have to include that 100% of live load in your gravity components of the load combinations. If you only have to have 25% of the live load to obtain the 25% inertial loads assumed by effective weight, then you should have to include that 25% of live load in your gravity components of the load combinations.
Juston Fluckey, SE, PE, AWS CWI
Engineering Consultant
RE: Load Combinations When Using 25% Storage Load For Seismic
Except that by your admission, you have a minimum of 25% of the live load present (i.e. the condition where 100% of the live load is seismic mass) or a maximum of 100% of the live load present (i.e. the condition where 25% of the live load is seismic mass). You have an upper and lower bound of gravity influenced mass contributing to the seismic forces. I still maintain that mass cannot be ignored in the load combinations.
Juston Fluckey, SE, PE, AWS CWI
Engineering Consultant
RE: Load Combinations When Using 25% Storage Load For Seismic
I'd agree with that.
So what I think you are suggesting is that:
With load combinations that include seismic, E, and Live Load, L, my E would include 25% of the live load and the L would include 100% of the live load.
With load combinations without Live Load, L, my E would be based only on the DL in deriving W, and not include the 25% Live Load?
Is that what you are saying?
I might agree with that in logical terms but need to think about it.
I'm not sure the technical-logical flow of the ASCE 7 text suggests this, though.
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RE: Load Combinations When Using 25% Storage Load For Seismic
for derivation of E:
v = Cs W
W should be all structure dead load (D) + 25% of of storage live load x storage floor area + partition weight or 10 psf + weight of permanent equipment + 20% of snow load where Pf exceeds 30 psf.
when checking a member:
D+E combo should only include the applied dead load and seismic load (storage live load component as well as partitions and excess snow are baked in)
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RE: Load Combinations When Using 25% Storage Load For Seismic
I am suggesting the following load combinations to be used with storage live loads:
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)
- D+0.7Ed
- (D+0.25L)+0.7E <--- 25% live load present to produce 25% effective seismic mass
- (D+L)+0.7E <--- 100% live load present to produce 25% effective seismic mass
- D+0.75(L+0.7E)
- 0.6(D+0.25L)+0.7E <--- 25% live load present to produce 25% effective seismic mass
- 0.6(D+L)+0.7E <--- 100% live load present to produce 25% effective seismic mass
I don't think 100% of the seismic load must be present in order to get the 25% effective seismic weight [else we would have a standard load combination of D+L+0.525E instead of D+0.75L+0.525E]. I think more realistically 75% of the live load would have to be present to obtain 25% effective seismic weight which corresponds to the already defined live load of D+0.75(L+0.7E). Obviously the upper/lower bounds of the live load required to obtain the 25% contribution would affect the above load combinations.Juston Fluckey, SE, PE, AWS CWI
Engineering Consultant
RE: Load Combinations When Using 25% Storage Load For Seismic
Baked in to what? E? D? or both? I am trying to argue that W should at least replace D in the load combinations.
Juston Fluckey, SE, PE, AWS CWI
Engineering Consultant
RE: Load Combinations When Using 25% Storage Load For Seismic
RE: Load Combinations When Using 25% Storage Load For Seismic
I think that goes directly against what AISC instructs you to do for derivation of the seismic weight.
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RE: Load Combinations When Using 25% Storage Load For Seismic
I agree! I don't know that I want these mandated, but applicable via engineering judgment. I want the extra weight. How much extra weight I can legitimately use depends on how the 25% of live load effective mass is figured. That is what I am trying to find out here. My arguments I feel are valid. If you are going to say some mass is present in the seismic force calculation, whether it helps you or hurts you, that same mass that affects the seismic is also there having gravity act on it.
Juston Fluckey, SE, PE, AWS CWI
Engineering Consultant
RE: Load Combinations When Using 25% Storage Load For Seismic
It's baked into E
My understanding is its trying to capture static masses so partitions, permanetly attached equipment, a portion of the storage room live load.
I can see the rational on the portion of storage room live load being that likely the items being stored will shift as the building moves to and fro, or the probability that the 125 psf applies everywhere as there is likely access walkways, etc.
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RE: Load Combinations When Using 25% Storage Load For Seismic
I have no problems with E. I've got the 25% baked into E.
Juston Fluckey, SE, PE, AWS CWI
Engineering Consultant
RE: Load Combinations When Using 25% Storage Load For Seismic
RE: Load Combinations When Using 25% Storage Load For Seismic
Juston Fluckey, SE, PE, AWS CWI
Engineering Consultant
RE: Load Combinations When Using 25% Storage Load For Seismic
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
RE: Load Combinations When Using 25% Storage Load For Seismic
Juston Fluckey, SE, PE, AWS CWI
Engineering Consultant
RE: Load Combinations When Using 25% Storage Load For Seismic
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
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RE: Load Combinations When Using 25% Storage Load For Seismic
1. D+0.7Ed code does not mandate this. It requires you to use E here, not Ed.
2. (D+0.25L)+0.7E <--- 25% live load present to produce 25% effective seismic mass code does not mandate this
3. (D+L)+0.7E <--- 100% live load present to produce 25% effective seismic mass code does not mandate this
4. D+0.75(L+0.7E) correct - one of the mandated load combinations
5. 0.6(D+0.25L)+0.7E <--- 25% live load present to produce 25% effective seismic mass code does not mandate this
0.6(D+L)+0.7E <--- 100% live load present to produce 25% effective seismic mass -code does not mandate this
You are missing a code required combination:
7. 0.6D + 0.7E. (with E based on D + 0.25L for derivation of W)
Here's the combinations strictly required by ASCE 7-10 (negating Lr, S, W, etc.)
1. D
2. D + L (100% of Live Load)
3. D + 0.7E - (with E based on D + 0.25L for derivation of W)
4. D + 0.75(L + 0.7E) - (your combination 4 - with E based on D + 0.25L for derivation of W)
5. 0.6D + 0.7E - (with E based on D + 0.25L for derivation of W)
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.
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RE: Load Combinations When Using 25% Storage Load For Seismic
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
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RE: Load Combinations When Using 25% Storage Load For Seismic
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
RE: Load Combinations When Using 25% Storage Load For Seismic
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.
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RE: Load Combinations When Using 25% Storage Load For Seismic
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
RE: Load Combinations When Using 25% Storage Load For 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.
RE: Load Combinations When Using 25% Storage Load For Seismic
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
RE: Load Combinations When Using 25% Storage Load For Seismic
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
RE: Load Combinations When Using 25% Storage Load For Seismic
RE: Load Combinations When Using 25% Storage Load For Seismic
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
RE: Load Combinations When Using 25% Storage Load For Seismic
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.
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RE: Load Combinations When Using 25% Storage Load For Seismic
RE: Load Combinations When Using 25% Storage Load For Seismic
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!
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
RE: Load Combinations When Using 25% Storage Load For Seismic
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.
RE: Load Combinations When Using 25% Storage Load For Seismic
We are in agreement on this point. My thought process on the timetable of a design seismic event is that at t=0 that storage live load is static and "tightly" packed and at any time > period/2 all of that storage live load are now fluid particles that have individual accelerations in various directions that do not necessarily align with the acceleration vector of the building structure.
This is a good discussion and I can see your point and logic behind your proposal, I just have a feeling that your proposed load combination will fall somewhere in between the code load combos ie. the 0.6D+0.7L will produce a higher overturning ratio than your combo....but maybe your (D+.25L,storage) + .75(L,storage+0.7E) will produce a more conservative design condition for main seismic force resisting members.
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RE: Load Combinations When Using 25% Storage Load For Seismic
RE: Load Combinations When Using 25% Storage Load For Seismic
The ASCE commentary mentions very little about the 25% other than it is to provide inertial forces of "tightly packed" stored items.
Juston Fluckey, SE, PE, AWS CWI
Engineering Consultant
RE: Load Combinations When Using 25% Storage Load For Seismic