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RE: P-Delta
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- Subject: RE: P-Delta
- From: "John Peterson" <jpeterson(--nospam--at)hcdaengineering.com>
- Date: Tue, 6 Apr 2004 08:21:13 -0600
Thanks for the detailed explanation. I believe you are saying exactly what my book is showing as far as the additional force of P*delta/length of column. Also, you are saying that these forces ARE additive at the base horizontal reactions and are not just there to increase the column and beam design due to an additional load. John Peterson -----Original Message----- From: Allen Adams [mailto:aadams(--nospam--at)ramint.com] Sent: Monday, April 05, 2004 4:25 PM To: seaint(--nospam--at)seaint.org Subject: P-Delta 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 ******* ****** ******* ******** ******* ******* ******* *** * 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. To * subscribe (no fee) or UnSubscribe, please go to: * * http://www.seaint.org/sealist1.asp * * Questions to seaint-ad(--nospam--at)seaint.org. Remember, any email you * send to the list is public domain and may be re-posted * without your permission. Make sure you visit our web * site at: http://www.seaint.org ******* ****** ****** ****** ******* ****** ****** ********
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- From: Allen Adams
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