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RE: P-Delta

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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)]
Sent: Monday, April 05, 2004 4:25 PM
To: seaint(--nospam--at)
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

This is an oversimplification of what RAM Frame is doing, but hopefully it
illustrates the concept.
Allen Adams, S.E.
RAM International

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