IBC 2000 Seismic Forces - whatis "component operating weight"?

[imsengr]

I am trying to calculating seismic forces in IBC 2000. In Equation 16-17 there is a term "Wp" which is "component operating weight". The definition is not very clear. I was told that for brick wall, it is 50. Could someone please elaborate what "component operating weight" stands for, and how did you arrive at that conclusion? What are the units? And what are the Wp for brick, metal siding, EIFS, etc.? Thanks in advance.

GAEngr05

 

[SacreBleu]

I don't follow this - don't you mean Eq. 16-60? Wp is the component weight. I don't know where you got the term "operating", unless it was unrelated to your wall, and referred to an HVAC unit.
I would use simply the dead load, in pounds, as closely as you can calculate or ascertain from a table of building materials weights.

 

[TPNY]

I think you mean Wp (as in a capital W). That is equation 16-67 through 16-69. I am assuming that these are not the equations you want since the operating weight of the architectural, electrical, or mechanical component (all non-structural elements) is not meant for walls.
The wp (as in lower case w, mentioned by Sacrebleu) is the weight of the STRUCTURAL element, and by "50" they may have meant 50% of the weight of the wall acts on the floor/diaphragm above and the other 50% acts on the floor/diaphragm below. Make sure you read the Section Headers before jumping to the equations.

 

[imsengr]

Thanks, SacreBleu for your prompt response. Use the "dead load"? But how much of the dead load to use? Density of brick is about 150 PCF, I believe.

The equation is Equation 16-67 in IBC 2000. I am not really sure where it is in IBC 2003.

The equation is used to determine seismic forces, Fp, acting on a wall of a building - lateral loads. One of the terms of this equation is "Wp" and it was stated below the formula that "Wp" is the "component operating weight". Those are the exact words, "component operating weight".

I take it to mean the weight of the wall finish. If my take on this term is correct, I am still not sure how big of the wall should I consider. Should I consider the weight of one square foot of the wall? I was told to use "Wp" as "50". For a one square foot wall with a thickness of 4 in. (or 0.33ft.) and a brick density of 150 PCF, it would work out to be 50#. Am I correct? I am not sure, hence my post.

Any one out that who has worked on lateral seismic forces acting on walls before, appreciate your advice[/color red]. Thanks again.

 

[imsengr]

Thanks, Zo40. Yes, I did mean Wp (with the capital W[/color red].)

I am trying to calculate the seismic forces acting laterally on the brick wall of a building, and Equation 16-67 on Page 376 of the IBC 2000[/color red] provided that.

The term Wp was defined as "component operating weight, and I take it that the value of Wp has to do with what type of wall finish it is. For EIFS or metal siding exterior walls, presumably the value of Wp would be different.

I was just trying to get a "clearer definition"[/color blue] of what the term means? Can anyone out there in Structural Land help, please?

 

[JAE]

GAEngr05:

The Wp term is the weight (full weight) of whatever you are dealing with. So if you have a wall, it has a weight - perhaps a weight per square foot. That weight is entered into Equation 16-67, 68, or 69, and you get a lateral seismic force developed by that component.

For example, suppose you have a wall that weighs 50 lbs/sq. ft. You use the three equations to get something in the form of:

Fp = P x Wp

The P is the percent of Wp that is thrown sideways in the seismic event.

Suppose your P value is 0.25. That means that 25% of the wall weight is the lateral force that is applied horizontally in the seismic load case. So your lateral load would be 0.25 x 50 = 12.5 psf on the wall, either perpendicular to the wall or parallel to the wall. Every square foot of the wall produces 12.5 lbs. of lateral force in either direction.

 

[imsengr]

Thanks, JAE.

Can I take it that the Wp term refers to the weight of one square foot of the wall?

Say, the density of brick wall is 150 PCF and the thickness of the brick wall is 4 in. or 0.333 feet. Then, the weight per square feet of th e wall is:

150 PCF * 0.333' = 50 PSF

The math is not complicated, but the term was not defined clearly. For someone new at the game, the definition "component operating weight" doesn't help a lot.

If the definition was written as, "weight of one square foot of the wall with the units PSF", there wouldn't be too much doubt at all.

Did I get it right?

 

[JAE]

Well, keep in mind that the Wp just doesn't only apply to walls. Check out the applicable tables - 1621.2 and 1621.3. These are the "items" that apply to your three formulae for Fp. It all depends on how you are dealing with the load.

For a mechanical unit, you may have a single weight - say 500 lbs. Then the Fp would be in lbs, not psf. Its just how you keep track (bookkeeping) of the loads and their path through the structure.

The thing to remember is that all seismic loads are generated by mass and the the resultin seismic load is always placed at the center of mass of the "item" you are dealing with. For your wall, its a big flat uniform thing. So applying a uniform load in psf terms is the natural result. For a mechanical unit, the load would most likely be a point load.

 

[SacreBleu]

GA,
If you are designing the brick wall veneer tie to support 20 sq. ft of brick wall, then Wp is equal to the dead load of 20 sq. ft. of the brick.
Essentially, the Fp seismic load value is 20 times that generated by one sq. ft. of brick wall.

 

[01AudiS4]

I didnt read all of the posts, however, since you are in the components section, I am assuming you are designing for an architectural element of some type to be added to a building, I dont know. Wp "component operating weight" is the TOTAL weight of the component your are designing. For instance, if it were a 10" block wall, it would be something like 160 pcf. Its as easy as taking the volume and multiplying. Hope that helps!

 

[imsengr]

Thank you for all yours replies. I have read each and every one of them and talked to some guys here at the office and this is what I concluded. The term, "Wp[/color red]" - component operating weight[/color red] - actually should be the total weight of the component that I am designing for[/color blue]. 01AudiS4 hit the nail on the head on this one. However, I believe the units should be in POUNDS.

I am actually designing wall studs at 16" apart and the height of the wall stud is 17 feet tall. It is a brick wall and let's take the wall self weight (brick, studs, sheating, etc.) as 54 PSF.

Therefore, my "component operating weight[/color red]", Wp is:

54 PSF * 16"/12 * 17 = 1224#

With this value, I can work out Fp, the seismic forces, and then compare Fp with the forces caused by the wind load. In my case, the wind load controls.

Did I get it right this time?

 

[UcfSE]

Keep track of your units. That looks like pounds per 16", not pounds, even though the units in the equation you wrote work out to pounds. It's important to keep track of just how much wall you used to get that number. I think you should go for force per wall or force per unit length, such as 54psf*17ft=918 pounds per foot of length (plf).

When dealing with roof top units (rtu's) the manufacturer will provide the operating weight of the unit. This info you can request from the architect.

Also, try the ASCE 7 for some typical dead load estimates for materials and wall assemblies, brick, cmu, and others.

If you have seismic and wind loads to consider, you may hvae to design with the wind force but I believe you will still have to detail your structure for seismic force effects.

 

[TPNY]

Correct me if I'm wrong, but I think GAEngr05 was determining the Weight of the wall as if it acted as a point load on the individual wall stud. He (or she) then plans to compare the lateral seismic load caused by the brick wall with wind loads and design per the higher value as recommended in 1621.1.4. Sounds ok to me.
Remember to check your wall/structure for longitudinal Fp as well as lateral Fp.

 

[UcfSE]

I may have interpretted calculation intent wrong as per Zo40

 

[SacreBleu]

I don't think seismic will govern. It is just a time-consuming exercise the IBC forces us to do.
In my local area, a seismic event has never happened (in recorded hisory). However, the seismic coefficients are much higher than what would be expected. I understand that is based on geological studies.

 

[imsengr]

I hardly ever check for seismic forces myself unless I am working on a project that is in a known active seismic zone, e.g. South Carolina, North Carolina, Memphis, etc. Hence, my unfamiliarity with "Wp" and calculating the seismic forces.

The "component operating weight[/color red]" I got is for the "tributary area" covered by one wall stud, i.e. 16" wide and 17' high. And the Fp[/color green] I got (in POUNDS) is apparently a point load acting on the c.g. on one of my wall studs. I compared this to what equivalent force I would get from the wind, and the larger force controls.

Zo40, could you please elaborate what is longitudinal Fp[/color blue]? I did check the lateral Fp[/color green] for wall element, body of wall panel connections and fasteners of the connecting system, but I don't think I checked the longitudinal Fp[/color green].

 

[JAE]

GAEngr05,

Since the outbreak of the IBC Code, seismic load combinations do indeed control in many cases where wind load used to always control. Be careful to not only compare the basic seismic loads, but also seismic demands on connections which are many times required to resist overstrength seismic loads (see IBC 2000 section 1605.4).

Also, your use of Fp as a concentrated load at the c.g. of the stud is technically incorrect. All seismic loads act where the mass is, so technically, the seismic load is a uniform lateral load along the length of the stud. Lumping all the mass, and the resulting lateral seismic load, at the mid-height of the stud is a bit conservative. It probably won't affect your design result that much in this case but I point it out so that you can understand the concept for any future "bigger" designs that you attempt.

Also, I forgot to ask you: Is this a load-bearing stud wall? If so, then you should be in IBC Section 1620, not 1621. And also - what is your Seismic Design Category?

Don't forget to include with the seismic the applicable load combinations of Section 1605. This also requires a determination of E as defined in 1617.1 which includes a vertical seismic effect = .2 x S(DS) x D.

Lastly, you mentioned the "longitudinal" Fp. Keep in mind that the seismic forces work in any direction, both perpendicular (lateral) and parallel (longitudinal) to the wall. For a wall, this longitudinal condition simply forces the wall to act like a shearwall.

 

[imsengr]

JAE,

Thanks for the explanation, especially the bit that Fp "does not act as a concentrated load on the c.g. of the stud", but rather "is a uniform lateral load along the length of the stud[/color red].

In response to the questions you asked, the stud is not a load-bearing stud wall. My seismic design category is C[/color red].

Some other information:[/color purple]
Seismic Use Group II
Spectral Response Coefficients:
Sds = 0.24
Sd1 = 0.15
Site Class = C
Basic Struct. Syst. and Seismic Resisting Syst. = Ordinary Steel Concentrical Braced Frame
Design Base Shear, V = 329 Kips
Analysis Procedure = Equivalent Lateral Force Procedure