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Re: P-Delta
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- Subject: Re: P-Delta
- From: mc.upadhyay(--nospam--at)jil.co.in
- Date: Wed, 7 Apr 2004 10:22:54 +0530
Hello Mr Adams the additional moment (PxDelta) creates a couple; which has two equal forces in opposite directions. So there is no net force on the system. why it should increase the base shear. Mukesh Upadhyay Delhi India "Allen Adams" <aadams@ramint.c To: <seaint(--nospam--at)seaint.org> om> cc: Subject: P-Delta 04/06/04 03:55 AM Please respond to seaint To expand on what Charlie Carter said, let me try to illustrate the situation with a simplified example. Imagine a one-bay frame consisting of a gravity column (pinned at both ends), a frame column (flag pole, pinned at the top and fixed at the bottom) and a beam spanning between them. There is a vertical load, P, applied to the top of the gravity column. Now imagine that the top of the frame column is displaced some distance. The beam pulls the top of the gravity column a distance Delta to match the frame column's displacement. Assuming that the frame column is strong enough not to buckle, this system will be in static equilibrium with a deflection at the tops of the columns equal to some value, Delta. The gravity column is now "leaning" on the frame column; the gravity column is pushing on the frame column, and the frame column is pushing back on the gvavity column, keeping the gravity column from just tipping over. The leaning gravity column is inducing a lateral force onto the frame column equal to P times Delta divided by the column height. This can be determined by taking a free-body diagram of the gravity column and taking moments about the top joint; the vertical reaction at the base is equal to the applied load P, which creates a couple equal to P x Delta. In order to be in equilibrium there must be a force couple acting in the other direction; this is produced by a horizontal reaction at the base, times the story height. Now look at the entire system. In order to remain in equilibrium, since there is a horizontal force at the base of the gravity column there must be an equal and opposite reaction at the base of the frame column. So even though there is no externally applied horizontal force, there is a horizontal shear induced into the frame system due to the P-delta effects. The "extra" base shear is the shear created by the P-delta effects. This is an oversimplification of what RAM Frame is doing, but hopefully it illustrates the concept. Allen Adams, S.E. RAM International Original Message: >I am using the RAM software to model ... a simple >rectangular, 3-bay, 2-story building with a rigid >frame on each end in each direction. When looking >at individual frames in the small model, with only >the wind loads applied in one direction (also the >innate mass load, which would be the "P" of P-delta), >the horizontal reactions at the base are greater >than the applied story forces for each direction. >I modeled one of the end frames using Visual Analysis >4.0 and did not get the same results as I did with >RAM despite using the P-delta analysis. The reactions >added up to the applied loads in VA. >I am hoping someone can explain to me why (or if) the >reactions are greater than the applied loads when using >a second-order analysis such as P-delta because it would >seem based on pure statics/dynamics, that the building >is no longer standing still. In the frame you modeled in RAM (a 3-by-3 bay, 2-story building), the middle two lines of framing lean on the outer two. Essentially, all the reactions required to stabilize the two interior frames against stability effects (or instability effects) are adding to the exterior two. I think that is why you are getting larger shear on the outer two. In the singular frame you modeled in Visual Analysis, you probably are not getting the additional effects of the interior frames. Does this explain the reasons? One important side note: when doing P-Delta (second order) analysis, you have to have the gravity loads on the frames in addition to the lateral loads. The second-order effects are generated by the effects of the gravity loads on the deflected frame. If the gravity loads are omitted in the analysis, you're not going to get second-order effects. Hope this helps. Charlie ******* ****** ******* ******** ******* ******* ******* *** * Read list FAQ at: http://www.seaint.org/list_FAQ.asp * * This email was sent to you via Structural Engineers * Association of Southern California (SEAOSC) server. 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