Dead Load Distribution to Frames 2

[EngineerofSteel]

I am designing a simple steel frame building. There are 6 symmetrical frames; each is identical to the other frames.

The applied dead load is 3.5 psf. With 30 foot bays, I have .105 kips. This, however, overdesigns the end frames. (It is more economical to limit cross-sections on a job, so the end frames are not down-sized.)

My question: Is it valid engineering to distribute the total roof dead load equally to all the frames. In this case, I have 6 frames and 5 bays. This gives me .105 kips * 5 bays/ 6 frames = .088 kips per frame.

Is this valid?

Thanks, DD

 

[jdonville]

DairyDesigner,

From your post I am assuming the following configuration:

A 2-dimensional structure consisting of 6 columns and 5 beams.
C-C distance between columns is 30 ft.

You have a uniformly distributed load on all the beams of 3.5 psf (you don't specify a depth of the structure, but that's not terribly important at this stage - we can just adjust the units to reflect a 1 ft depth)

For "interior" columns:

Tributary area is 30 sq ft/ft (15 sq ft/ft on either side of column), therefore roof load on interior column is 15 * 2 * 3.5 / 1000 = .105 kip/ft

For end columns:

Tributary area is 15 sq ft/ft (15 sq ft/ft on one side of column), therefore roof load on exterior column is 15 * 1 * 3.5 / 1000 = .0525 kip/ft

You will need to multiply each column load by the tributary length in the 3rd dimension.

Jeff


Jeffrey T. Donville, PE
TTL Associates, Inc.

http://www.ttlassoc.com

 

[SperlingPE]

No
Check your units.
The nature by which the vertical load is distributed to the frames prevents this.
You should check your end frames for the reduced dead load if there are any lateral deflection or vertical deflection issues.

 

[EngineerofSteel]

jdonville,

I have 6 frames (=12 columns & 12 beams). The 6 frames have 5 bay spaces between them. Each bay is 30'.

The math you showed is the math I use on all my projects. However, I have a current project with a bending ratio on the columns of 1.04 - just hair over being acceptable. If I redistribute the DL, the bending ratio is acceptable. I have seen people do this with wind loads.

My question is: can I do this with a DL? To reword the question: Is there an equation which will show the true distribution of roof loads to the frames? It is easy to imagine how more than 1/2 a bay space's load will transfer to the end frame.

For example, if I heavily loaded just one frame on a completed structure in the field, the deflecting beams would pull on purlins and sheathing anchored to adjacent beams. So, the load would be more evenly distributed.

Thanks, DD

 

[Lutfi]

DD,

It is common engineering practice to use tributary areas or widths to determine the load on any structural member. It is true that your initial concept yield an over designed end frame; but you can design the frame for half the load should the savings be justified. Or you can leave it as designed for the full load; this way you can make an argument that the design has built robustness for future expansion! I would not use your reduced load approach for this particular application.

I have used your second approach once on a particular project for a stage. The original design stated that the platform to be designed for a total of 5000 pounds. When I done my analysis, I distributed that load over the entire area to arrive at SF value.

ASCE 7-98 corroborates your approach on page 13-footnote number 3, follows the same procedure for library floor loads. It reads something like this:

“The weight of books and shelving shall be computed using an assumed density of 65 pcf and converted to uniformly distributed load; this load shall not exceed 150 psf.”

My 2 cents worth


Regards,
Lutfi

 

[UcfSE]

What kind of roof framing are you using?

Whether you can evenly distribute vertical loads depends on how stiff your roof is. If it is so stiff as to deflect as a rigid body then you can distribute the dead load evenly. If it is flexible then it will distribute loads to its supports according to tributary areas. Your roof in all likely hood is not rigid enough to behave as a rigid body.

 

[EngineerofSteel]

UcfSE,
The roof sheathing is 26 gauge (.018") R-Panel on HSS 9x5x3/16 purlins. The purlins span between the beams to form a flush mounting surface for the R-Panel.

-DD

 

[CTW]

I would have to say that it is not valid to distribute the load that way. What you end up with is a dead load less than the 3.5 psf you started with.

 

[WillisV]

Agree with CTW - with the roof system you have the tributary width approach is the correct one.

 

[whyun]

I agree with everyone who said NO.

Half of each side of the 30 foot bay transfers to the center frames. End frame only sees half.

In an idealized situation where you have a roof diaphragm that is infinitely rigid in the vertical axis, the original method of equal distribution is valid but even for a very thick concrete diaphragm, it is flexible.

 

[EngineerofSteel]

Looks like the nays have it. Thanks, all!
-DD

 

[rowe]

I agree with jdonville first reply.

Although the structure MAY not collapse if you do distribute the DL equally to all 6 frames, the interior frames will deflect more than the exterior frames in order to redistribute the actual loading. This difference in deflections may result in roof leaks which are extremely aggravating to everyone (owner, contractor, and eventually the engineer)and may lead to costly litigation.

Good Luck.