Deflections due to Wind Loads, Reduced Wind? 2

[IJR]

Dear pals

Someone discussed with me in SEAOC forum some years ago, and mentioned that wind loads for deflections should be taken as 10yr wind, which comes roughly to 0.75 times the 50yr wind

I need a code item for that(citing from any code), can not find it in the codes I use.

respects

IJR

 

[UcfSE]

What codes do you use?

If you convert 50-year wind from the ASCE to 10-year wind, the velocity will be multipled by 0.74 (Table C6-3 p308 of the 7-02). Pressure depends on the square of the velocity, so the pressure going from 50-year to 10-year will be multiplied by (0.74)², or approximately 0.55. Even so, most sources I've seen do not account for this and use a factor of 0.75 or 0.7 instead.

In the FBC 2004, which is based on the IBC 2003, Table 1604.3 footnote f allows the use of 0.7 times the C&C wind for checking deflection. I would imagine a similar note is in the IBC since the FBC is based on that.

The use of 10-year wind for deflection is an option. You may use full load if you want.

 

[WillisV]

See my response below to another similiar question a while back which has a bit more information as to the origination of the 0.7 factor:

"Back on the wind load issue, I believe it is helpful to think of the wind loads provided by ASCE in terms of their recurrence intervals rather than stamping them with a "service" or "ultimate" level. The VELOCITY levels provided by the code maps are in terms of a 50 year recurrence interval. This can be verified using the velocity recurrence interval multiplier in the commentary: Fc = 0.36 + 0.1 ln(12T). With T=50yrs, Fc = 1.0. Keep in mind this is a velocity multiplier and force levels are equal to velocity levels squared. In this case, 1.0 squared = 1.0.

Therefore if you used a 1.0 load factor you would have a 50 year wind. As demonstrated in the commentary, you can back calculate the return period for a certain load factor by plugging in the square root of the load factor as Fc and solving for T. Keep in mind that the return intervals are referring to wind VELOCITY and that force levels are based on velocity squared. So by removing the directionality factor from the typical 1.3 load factor = 1.3/0.85 = 1.53. And taking the square root of this to make it in terms of force: Fc = 1.23. Back solving for T results in a return period of approximately 500 years. Therefore using a 1.3 load factor with the directionality factor results in a 500 year "ultimate" return period which is in line with seismic design.

As an end note, the code recommends using 0.7W for "service" level deflection checks. This is based on bringing the 50 year map speeds down to 10 year speeds. By plugging T=10 into the Fc equation you get Fc = 0.838. Fc is the wind speed multiplier so to get the force multiplier you must square 0.838 = 0.7. "

 

[Murali27]

Willis V

Nice answer for the reduced wind indicated for serivceability checks.

Can we check the reduced wind with dead load alone?
Code says D + 0.5L + 0.7W (including live load)

Thanks
Murali

 

[WillisV]

Murali,

Typically Dead and Live load would only affect the lateral deflection of the building when considering second order P-Delta effects. It is fairly common to check servicability concerns with a first order effects, therefore only the 0.7W portion would be of concern. If you feel like secondary affects are of a major concern, then I would go with Dead plus 50% live as you have shown.

 

[IJR]

That combination(D+0.5L+0.7W) is defined as follows in the commentary

"Service load with a probability of exceedence of 0.05"

hinting that for serviceability checks with wind involved, that probability is just sufficient, or is a reasonable criterion

respects
IJR

 

[Murali27]

WillisV/IJR

Thanks for your input. We used to check for wind with 50 yrs recurrence for deflection and 10 yr recurrence for accleration.

Can we conclude like this

For design Load Factor * W (with 50 yrs recurrence)

For Serviceability Check

W (with 10yrs recurrence)

Thanks again

Murali