×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Are you an
Engineering professional?
Join Eng-Tips Forums!
• Talk With Other Members
• Be Notified Of Responses
• Keyword Search
Favorite Forums
• Automated Signatures
• Best Of All, It's Free!

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

#### Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

# Design Wind Pressure on Wall components and cladding

## Design Wind Pressure on Wall components and cladding

(OP)
I would appreciate your input on the following problem. Consider wind loads on components and cladding as defined by ASCE7-98. Let me start first with the no-brainer scenario, which would be for example the computation of NET uplift loads on a roof truss or joist. The design pressure includes a factor Kd, wind directionality factor, which is to be applied with load combinations only. In the case of a roof truss, you would do a free body diagram, Load combination=(Dead Load) – (Wind uplift pressure) = (Net uplift pressure). As a result you obtained the net uplift load and therefore the required uplift strap can be specified. Simple enough.

Consider now a window located on one of the walls of the building, the positive or negative pressure that the window has to resist is a direct result of the components and cladding wall pressure. In this case a free body diagram would tell you that there’s only two forces acting, the positive or negative pressure, and the rated capacity of the window itself. But this is not really a load combination, (i.e. D+L, or D+W, or D+L+W). As a result, I do not apply the Kd factor, resulting in required window pressures that are 15% higher.  (since Kd is 0.85 for components and cladding) My question is would you still apply Kd=0.85 in this case? I asked several colleagues and heard responses like: use the more conservative value, or Do use Kd since load combinations do exist but Dead=0, and Live=0, for some reason neither answer convinced me, appreciate your input.
Replies continue below

### RE: Design Wind Pressure on Wall components and cladding

Even though wind on a window is the only force, you are still ALWAYS designing elements of your structure under a combination....not an individual case.  Nowhere in any code (that I recall) have I ever seen where some element is to be designed for a single "case" but the intent of all codes is that the design is based on load combinations.

For your window case, you technically have to calculate ALL the combinations listed in the code (Chapter 16 of the UBC or IBC).  This is a bit anal, but in reality you have to do this...even though DL, LL, etc are all zero.

This gives you the required load to use.

Your question is based on the idea that, since WL is the only non-zero value applicable to the window, then I only use a single case:  WL to design.  This is incorrect.

You are really designing for DL+WL with DL=0.  Thus, the Kd factor may be applied.

### RE: Design Wind Pressure on Wall components and cladding

so now the question is under what conditions do you use kd=1.0?

V2

### RE: Design Wind Pressure on Wall components and cladding

(OP)
Thanks JAE and V2. That is a good question V2, the ASCE code states that the Kd factor is to be applied in conjunction with load combinations, but according to JAE this requirement is meet for the window case, and for that matter for any other component, i.e.  there wouldn't be a scenario under which there is no load combination. If that is the case, why was that note about Kd placed there, at this point I am trying to think of a specific case where there's unequivocably no load combinations, any ideas?

I was trying to idealize the framing srtructure of the window as a translation support, with one load, wind, as a result, reaction=(wind pressure*area), still see no load combinations, but i am the stubborn one. Thanks for your input

### RE: Design Wind Pressure on Wall components and cladding

OK...after doing a bit of reading, I found the answer on this Kd issue.

Check out the commentary in the back of ASCE 7 for section 6.5.4.4.

Basically, what they did was change the OLD load combinations which had 1.3W to the NEW combinations that have 1.6W x Kd where Kd generally = 0.85.

0.85 x 1.6 = 1.36 kinda equal to 1.3

So what they've done is taken OUT the Kd factor from the OLD combo (1.3W) and separated it so that "as new research becomes available, this factor can be directly modified without changing the wind load factor".

So the footnote about using ONLY with the comos of 2.3 and 2.4 is to guard against you taking the wind from ASCE 7 and using it in another code that might have different combination factors.

How this adds up with ASD (section 2.4) isn't clear to me but they specifically include it as well.

Hope this helps.

#### Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

#### Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Close Box

# Join Eng-Tips® Today!

Join your peers on the Internet's largest technical engineering professional community.
It's easy to join and it's free.

Here's Why Members Love Eng-Tips Forums:

• Talk To Other Members
• Notification Of Responses To Questions
• Favorite Forums One Click Access
• Keyword Search Of All Posts, And More...

Register now while it's still free!