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Rigid diaphragm Spreadsheet[Subject Prev][Subject Next][Thread Prev][Thread Next]
- To: <seaint(--nospam--at)seaint.org>
- Subject: Rigid diaphragm Spreadsheet
- From: "Dennis S. Wish PE" <wish(--nospam--at)cwia.com>
- Date: Mon, 23 Nov 1998 13:08:24 -0800
I received a spreadsheet for rigid diaphragm analysis about a year or so ago from James Lord SE in Northern California. At the very least it can be used as a model to design a personal spreadsheet. The spreadsheet will also graph all walls and determine center of mass as well as distribution of load (if I remember it correctly. I intended to use it as an example an upcoming SEAint Online Tutorial but will place it on the SEAint Website sometime this week. It is in Excel format and should be of use to many of you. Regards Dennis Wish PE -----Original Message----- From: Mlcse(--nospam--at)aol.com [mailto:Mlcse(--nospam--at)aol.com] Sent: Sunday, November 22, 1998 5:47 PM To: seaint(--nospam--at)seaint.org Subject: Re: Rigid plywood diaphragms In a message dated 11/22/98 5:13:56 AM EST, itsekson(--nospam--at)jps.net writes: See inserted comments: << I will in fact use tributary area method as the base for my horizontal shear distribution (it does not take that much time and it seems to be still working). It should not be that difficult to put together the spreadsheet using UBC Standards formulas for shear wall deflection either. I am wondering though if it would give me the realistic shear distribution ( I don't think so) and would it be any closer to the actual performance of the system than the customary method above.>>> Putting together the spread sheet is not a problem for shear walls in the same line, except how you want to handle the the holdown assembly uplift. I found that to solve this you have to use an iterative solution until all shear walls along a given line have approximately the same deflection (we use +/- .005 inches). As you keep adjusting shear loads between shear walls, you keep changing holdown sizes (we added a data base of Simpson holdown sizes and deflections under maximum loads for the program to use in the iterative solution.) You may want to include in your program the ability to say what size holdowns you want to use on each wall instead of letting your program pick the size, the nail spacing to use and also plywood on two sides of the wall. This will help if you are trying to evaluate existing wall conditions. You may also want to consider tie-rods. I believe the UBC formula for shear wall deflection is fairly accurate up to first yield of the shear wall panel. I am not sure about it though from the first yield point to the wall's ultimate strength. John Rose would be able to answer this question more accurately. The distribution of shear to individual walls in the same line will vary depending upon wall length initially (Your program might give the following results: short walls 250 plf while longer walls are 500 plf). When the slack in the holdown assembly is taken up by overturning (you increased the loads on these wall while keeping the same size holdowns and nailing), then the shear in all the walls will approach the same (say 500 plf). <<<Let's not forget that we are talking about the RELATIVE RIGIDITY of shear walls along different lines of resistance. For the sake of the argument let's look at the direction when framing is parallel to shear walls. So the overturning moment and boundary member forces are the function of the unit shear times the height of the wall. I can drop out several components of the total deflection from my relative rigidity calcs that would cancel out. These elements may include deflection due to oversized holes in holdowns (which in fact are 2 to 3 times more than the holdown deformations itself), plywood shear deformations, slip in sill bolts. >> As you mentioned above, what is the relative rigidity of shear walls along different lines of resistance. I would still calculate the deflection of all shear wall lines using the UBC formula, I don't think you will be able to drop out certain components, in fact I think you would have to use the rigidity of the entire shear wall line since all walls along that line have the same deflection (tied together by the double top plate). There would be less work by using wall lines instead of individual walls in terms or calculating rigidity (similar to rigidity of walls with openings) and probably more accurate. Depending upon the deflection of each shear wall line, I would redistribute the lateral loads to adjacent shear walls (consider semi-rigid diaphragm). If the deflection in one wall line exceeds the deflection of an adjacent wall line by more than 20% (an arbutary value) I would redistribute the loads. Proposed additional force to add to wall line which deflects less: (D1 - D2) additional force = ------------- x Vwall(i) x 0.5 D1 D1 = Larger shear wall line deflection D2 = Smaller shear wall line deflection Vwall(i) = Total shear force from shear wall line with larger deflection (0.5 since taking some of the lateral force away from one wall line, it will now deflect less, hopefully less than 20% difference in deflection between this wall and the adjacent shear wall which is receiving the additional load ). I would then recalculate the deflection for the wall line which deflected less with the additional force added. ( I would not recalculate the wall line which I took the force away from, since it will now have more reserve capacity and should already meet code deflection requirements). The 20% may seem high, but if one shear wall line of resistance deflects 0.4 inches, the next adjacent shear wall line would have to deflect more than 0.48 inches before you would consider a redistribution. 0.8 inches of additional deflection for a shear wall line which say is offset by 15 feet or more from the wall which deflects 0.4 inches is relatively small. I would think that the 20% value, or some other value, is something which would have to be agreed upon by others. Just some additional thoughts, Michael Cochran
- Re: Rigid plywood diaphragms
- From: Mlcse
- Re: Rigid plywood diaphragms
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