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Tipping force of Wind 1

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silverback1

Mechanical
Nov 12, 2010
3
I have a pretty basic question but i have hit a road block.
I have a concrete wall that is 14'Lx1'Wx7'T that is sitting on another reinforced concrete slab.

Using IBC 2009 for my region the Basic Wind Speed for design is: 90 mph (normal to wall surface). With this the Wind Pressure is: 20.7psf.

I calculated the tipping force of the wall to be 1,050lbf at the top of the wall with no slipping.
Sum of moments = 0, p=Tipping Force; W=Weight of Wall W=(98Cuft*150lb/Cuft=14,700lbs)

0=-P(7)+14,700(.5) P=1,050lbf

Where i am having trouble is determining if the pressure force is enough to tip the wall over. I know the wall area is 96sqft but i think 20.7*96=2,020lbf cannot be considered a concentrated load that would cause tipping cause the force on the bottom half of the wall is going to cause the wall to slide before it causes it to tip over. Do i only consider the top half of the wall in the tipping?

It seems illogical that a 90mph wind could tip a 1'thick concrete wall thats only 7'x14'...

I need help please!!!
 
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I checked the forum for help first...

The thread you posted just takes me to main board that I searched...

Also the wall is freestanding... it can be assumed to not be tied in on any side or top and is resting on a flat concrete surface.

Thanks!
 
It was meant to be a joke..

They're ripping a guy who posted a very fundamental question a new one on this thread:


What I mean is that you're likely to get some responses in the same tone as on that thread.

Post a sketch and it's likely you'll get a better response.

I don't really know what you mean by "tipping force". There are several failure mechanisms for the problem you describe. A 1 ft. wall is very large; where are you getting these dimensions? How is the wall reinforced? how is it connected to the slab below? How are you getting that wind load?
 
Assuming the wind pressure is 20.7 psf, the total wind load is 7*14*20.7 = 2028'#. The weight of wall is 14*7*145 = 14210#.

Overturning moment = 2028*7/2 = 7098'#
Stabilizing moment = 14210*0.5 = 7105'#

The wall is in imminent danger of tipping over. The factor of safety is 7105/7098 = 1.001 which is inadequate. It should be at least 1.5 to satisfy most building codes.

BA
 
BA.. you're assuming the wall isn't attached to the slab below.. I don't know that that's what he's implying..
 
frv,

He is asking if the wind pressure is enough to tip the wall over. Without reinforcement, it is just about enough to tip it. If there is reinforcement between the wall and the slab below, it could be a different story, but the amount and position of the reinforcement would need to be known in order to come up with a sensible answer.

BA
 
Just put the Walnuts on the backside of the wall and wait for the wind to blow.

Mike McCann
MMC Engineering
Motto: KISS
Motivation: Don't ask
 
....cause the force on the bottom half of the wall is going to cause the wall to slide before it causes it to tip over. Do i only consider the top half of the wall in the tipping?
Sliding and tipping are two deferent failure modes and should be treated separately.
You can’t alter the basic statics; for each failure mode destabilizing force with appropriate ultimate factor should be below stabilizing force with code recommended reduction factor.
When checking overturning:-
Tipping moment caused by wind pressure on full area of the wall x ultimate load factor + other lateral force that your code may require should be less than the resisting moment due to dead weight with ultimate reduction factor.
 
As usual BAretired nailed it. In theory you have no factor of safety as required by IBC (1.5 as BA noted). In reality, you might have enough construction irregularities to reduce your safety factor even further.

I live and practice in a high wind area. A 90 mph wind can do a lot of damage and can easily tip over a freestanding wall that is not sufficiently tied to its foundation or slab.

I'm assuming this is a fence or a screen wall. It might come under a lower category of structure, but nevertheless, it needs adequate resistance to overturning, which it currently does not have.
 
Thanks for the replies. My biggest issue came when placing the concentrated wind load properly for determining tipping oment of wind. I applied to top of wall and I see it should have been at middle.

This wall will be tied into slab below with #5 rebar. The wall also will have a double cage of #5 with 1' centers H&V.

I think this should be enough but I'll find out next week when I have my civil/structural guy take a look...

Your thoughts/opinions are welcome and will be heeded...
 
Keep in mind that the rebar needs to be as far as possible from the center of the wall, while still maintain adequate cover.
 
#5's @ 12" o.c. ew ea. face should be OK to develop your required moment.. However, you need to make sure that the #5's are developed into the slab below and you need to check the slab below as well.. Is it a Slab on grade, or a structural slab? I suspect this may be your failure point.
 
Does the 2009 IBC have specific provisions for wind forces on a freestanding wall or does it direct you to ASCE 7-05? ASCE 7-05 has specific requirements for walls. For instance, the resultant wind force should be applied at 0.55 x the height for walls at ground level. The code also specifies three load cases. One case requires higher wind forces on one end of the wall to account for oblique wind conditions. Also, the code prescibes load combinations to account for uncertainties in the loads.
 
The IBC will direct you to ASCE 7 which has a section in the wind chapters for solid signs and similar structures...which in my opinion this would fall under.

The wind pressure is applied at the mid-height of the wall under ASCE 7's case A. There is also a case B and C that might apply - Case B has the load shifted upwards slightly from the middle of the wall.

If you are the responsible party designing this wall the I would suggest you get someone (like you mentioned - a civil/structural) friend to look at it.

 
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