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Re: '97 UBC Lateral Design - Envelope Solutions???? - Part 2[Subject Prev][Subject Next][Thread Prev][Thread Next]
- To: "'seaint(--nospam--at)seaint.org'" <seaint(--nospam--at)seaint.org>
- Subject: Re: '97 UBC Lateral Design - Envelope Solutions???? - Part 2
- From: "Swingle, Mark" <Mark.Swingle(--nospam--at)dgs.ca.gov>
- Date: Wed, 29 Sep 1999 09:36:17 -0700
- Cc: "'mswingle(--nospam--at)earthlink.net'" <mswingle(--nospam--at)earthlink.net>
-------------- PART 2 OF 2 -------------- Re: '97 UBC Lateral Design - Envelope Solutions???? Mark Swingle (MS) continues his response to Dennis Wish's (DW) post: (MS) Let's take the first wall mentioned above: "....in one wall, the flexible analysis yielded 21,371 pounds of shear into this lline of resistance. The same grid line, under torson analysis 1,523 pounds of shear." Let's see what the tributary area is that this wall is resisting: OK, I will make the following assumptions: 1. 3 story building 2. No irregularities (continuous 3-story wall) 3. DL of roof and 2 floors = 15 psf each 4. DL of int and ext walls tributary to each level (3 levels above the base) is 10 psf as a horiz distr load. 5. Zone 4 6. Na = 1.5 (near-fault) 7. R = 5.5 8. rho = 1 Therefore Total trib weight = (45 + 30) = 75 psf E = 0.30W or E/1.4 = 0.21W (by ASD method) So, the tributary force to any wall on the first floor is: F = 0.21(75 psf) = 16 psf = 0.016 ksf In your tributary analysis, therefore, THIS wall is resisting 21k/0.016ksf = 1,333 SF of tributary area. Now, assuming that this area is a rectangle with an aspect ratio of 2:1, this would yield a rectangle that is 15 ft x 30 ft. Now, assuming that the adjacent walls on each side are equidistant from this wall, that means that the diaphragm span on EACH side is 15 feet. ****(Note that if I had assumed a one-story building, or Zone 4, or a lighter building, or not near-fault, or any combination, this tributary area would be much LARGER. My assumptions are very conservative with regard to the points I will make below.)**** Let's assume the wall is sheathed with 1/2" Struct I, nailed w/ 10d @3" oc. With an allowable capacity of 665 plf this requires a wall with a length (per your flexible diap (tributary) analysis) of 21.3k/0.665klf = 32 feet. So this wall is the entire length of the rectangle. (Remember that your design base shear is probably lower and/or this may not be a 3-story building, so your area may be much larger. And also remember that if I assumed lighter nailing, the wall would be LONGER.) No matter how you slice it, this is a STIFF wall. But here's the point (at last, you say!!!!). You say that for the rigid case, the force is only 1.5 kips. For the purposes of this example, this is essentially equal to zero. That means that the adjacent walls (or some other walls somewhere in the builing) ARE SO MUCH STIFFER THAN THIS WALL that they take practically all of it's tributary load!!!! How can there possibly be other walls that are stiffer than a 32 ft wall (or 3 ten-ft walls, whatever)???? This is just not possible. In addition, this example means that for the rigid case, your wood diaphragm is so stiff that it can span 30 feet (skipping "over" the wall that has no force in it now) with almost no deflection (otherwise it would put load on the wall). This is just not possible. ------------------------------ Dennis continues: (DW) "If we follow your strict interpretation of the the code method so that if the diaphragm is "not-flexible" we distribute shear by torsional analysis - how many of us would feel justified in ignoring the flexible resulsts when the discrepancy between the two is 18 times larger than what we chose to resist. Remember, it is not a clear cut case of one analysis yielding more restrictive answers - this depends on the wall stiffness and the location of the applied loads. In my example there is sufficient different on some walls to warrant compliance with the flexible analysis while other walls require compliance with rigid." Mark responds: (MS) I never said that if the diaphragm fails the flexible test that one needs to analyze it by rigid diaphragm analysis. Reread what I said. I never said to ignore the flexible case. I said that in the worst case, one may want to adjust the design shears TOWARD the infinitely stiff case BECAUSE THE CODE REQUIRES IT. Even if your example above is correct, that the other walls ARE that much stiffer, my point is that no wood diaphragm is stiff enough to redistribute the shears ALL THE WAY to the 100% rigid analysis results. I think my example above demonstrates this, since none of us can imagine a wood diaphragm with a span of 30 feet (OR MORE) not deflecting very much. EVEN IF YOUR WALL STIFFNESSES ARE SO RADICALLY VARYING, AND EVEN IF THE DIAPHRAGM FAILS THE 2:1 TEST, IT WILL STILL BE FLEXIBLE ENOUGH TO DEFLECT AND THEREFORE TO LOAD THE WALL THAT YOU SAY HAS NO LOAD. Please take this in the spirit in which it was intended. I am simply looking for rationality. This is not a personal attack. I will repeat what I said at the beginning: Now that you have volunteered to be on the panel this weekend, you have to be prepared to stand by your statements more than ever before. You are now in a similar position to that of the committee members of the past who have been criticized by you (and others) for not being rational. ------------------ END OF PART 2 OF 2 ------------------ Mark Swingle Oakland, CA Disclaimer: These are my own opinions and are subject to change.
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