Wood Shear Wall Deflection - Hold Down Elongation (4th Term, IBC Eq. 23-2) - FTAO Diekmann Method

[bones206]

I'm learning the Force Transfer Around Openings (FTAO) method for wood shear wall design. I followed along the example in APA Technical Note 555 (attached), and everything made sense until I got to the deflection calculations.

The theory:

According to T555, the deflection of each wall pier is calculated then the overall wall deflection is taken as the average of all pier deflections. The end pier deflecting away from the wall interior is assumed to be full height, with the other pier heights starting at the window sill. They also state an assumption that "the FTAO wall is controlled by the deflection of the full-height piers". Each pier deflection is calculated using the 4 term IBC Equation 23-2.

The implementation:

From what I can tell in the T555 example calculation, the wall pier deflection due to hold down elongation is calculated as follows:

1) Calculate overturning moment in pier based on the panel shear adjacent to the opening and the assumed pier height (either full height or partial height, depending on the loading direction and pier location). For pier 1 v = 481 plf, h = 8 ft and b = 4 ft. So M[sub]OT[/sub] = (v*b)*h = (481)(4)(8) = 15,392 ft-lb.

2) Calculate a tension/hold down force based on the pier's OT moment. T = M[sub]OT[/sub]/b = 15,392/4 = 3,848 lb.

3) Prorate the actual elongation based on the calculated tension relative to the published max tension and elongation values. In the example, the hold down manufacturer's allowable force is 2,145 lb at an elongation of 0.128 inches. So d[sub]a[/sub] = (T/Tmax)*d[sub]max[/sub] = (3,848/2,145)*0.128 = 0.2296 inches.

4) Multiply the prorated elongation by the pier's height-to-width ratio to get the lateral pier deflection. So deflection = 0.2296*(8/4) = 0.459 inches.

The head-scratchers:

- The calculated tension and elongation in pier 1 both exceed the hold down manufacturer's allowable values by a fair margin.

- The hold down tension force was previously calculated as 1,538 lb (ASD) using the overall wall dimensions. The deflection calculation used an LRFD factor, so to compare apples to apples I believe the hold down force would be 1,538/0.7 = 2,197 lbs. Again, this is much smaller than the value of 3,848 considered in the elongation/deflection calc.

- Each wall pier has a deflection component based on hold down elongation, even though hold downs are only provided at the ends of the walls (i.e., not all piers have hold downs).

- How can the overall deflection be "controlled by the full height pier", yet is calculated as an average of both full-height and partial-height piers?

I was wondering if anyone has better insight into this method of calculating wall deflection using the FTAO method. The treatment of the elongation term is fuzzy to me and I haven't been able to find much material on the subject.

 

[txeng91]

I’m pretty familiar with FTAO walls but not enough to answer your questions. If anyone could it would be Terry Malone with Woodworks. He is the author of “The Analysis of Irregular Shaped Structures: Diaphragms and Shear Walls” which is state of the art in terms of calculating forces and deflections wood diaphragms and shear walls. I believe he puts his contact info in all his slideshows if you look him up, if not let me know and I can send it to you. I’ve been to a few of his presentations and he insists that he would be more than happy to help anyone that reaches out to him.

 

[bones206]

I'm actually about halfway through that book right now. Part of my crash course on wood engineering so I can hopefully branch out and do some residential side work. I have to say that book is worth every penny.

I was actually planing to fly down to Charlotte next week to attend his seminar on cantilever diaphragm design, but couldn't end up making it. If nobody else chimes in on the forum I might try reaching out to him or maybe even Mr. Diekmann. I posed the question to the APA a few days ago but haven't received a response yet.

I think a more rational way to do it would be to average all the pier deflections without the 4th term, then add the 4th term to that average value. In that case, the 4th term hold-down deflection would be based on the full wall length and the same tension force used to size the hold-down.

In the APA example, the deflection due to hold-down elongation is almost 3x the combined deflection due to bending, shear and fastener slip. That seems out of whack to me... is deflection usually controlled by hold-down deflection?