Wall Anchorage Forces (ASCE 7-16 Section 12.11.2.1)

[Structure101]

I am working on a tilt up concrete panel building and am trying to calculate the wall anchorage forces on a single concrete panel for embed plate with headed stud anchors design.
- I calculated the seismic anchorage forces based on ASCE Section 12.11.2.1 and I got Fp = 4.3 kips.
- However, at this location, wind governs and my diaphragm and concrete panels are designed based on in-pane and out of plane wind forces.
- I calculated the out of plane wind force for each embed plate and I got Fp = 7.6 kips. I have attached a image below.

Q1) Is the Seismic Anchorage Forces based on ASCE Section 12.11.2.1 out of plane or in-plane. If it is out of plane does that mean the diaphragm shear along that grid line is in-plane?
Q2) ASCE Section 12.11.2.1 states Fp is the design force in individual anchors. If I have 4 or 2 headed stud anchors in a single embed plate, does that mean I should multiply the Fp (7.6 kips) x No. of Headed Stud Anchors?
Q3) Am I calculating the out of plane wind force correctly?

 

[Aesur]

Q1: 12.11.2.1 is Out of plane seismic forces, normal to the wall as stated in 12.11.1. This is the force you need to transfer into the building lateral system to prevent the wall from separating from the structure. Diaphragm shear would be parallel to the wall and typically transferred through a ledger to embed plates over the middle half of the panel.
Q2: I believe the wording in the code here may be bad, by anchor it is my opinion they mean anchor system, ie the plate + headed studs. It would not make sense to multiply the force by the number of anchors.
Q3: This is hard to say without more information, ie dimensions, wind speed, etc. Based on your profile location, I would assume you are in a location with Vult of 105 mph? I'm not sure if you need to design for higher forces for tornados, etc. as I don't practice in that area. If not, the wall pressure looks in the range I would expect, the parapet C&C loading however looks high. I would also sum the area per embed plate, because if your area is greater than 700 sq-ft (30.2.3) then you can use MWFRS instead of C&C. Note that trib area isn't always height x width, sometimes you can use 1/3H*H which often times gives a larger trib area, therefore lower loading. I don't recall the last time I had 100 psf on a parapet, but it was a while ago, and it was due to some unique dimensions/properties. Normally I expect in the 50 to 60 psf range. Your method/formula you use for wind is commonly used, but is technically incorrect and non-conservative. If you have access to EnerCalc or other software (or want to deal with the pain of doing it by hand) run the loading in beam analysis with the cantilever built in and you will get a more accurate reaction.