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RE: Lateral Load Due to Foot-Traffic

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Nels,

Perhaps this is analogous to the force on a pier from a berthing ship.  The kinetic energy, E=(1/2)(WVsq/g)(C)(H).  Assume the coef. C and H are 1.  E=Fd/2. Then the force, F=WVsq./dg.  You have to estimate the persons velocity and also the distance, d, in the original direction that it takes the person to change direction 90 degrees after he puts his foot down to pivot around.  Maybe that is 6” to a foot.

You have to play with the numbers to come up with something reasonable but this is a valid approach so long as your assumptions of V and d are correct.

Richard

 

-----Original Message-----
From: Nels Roselund [mailto:njineer(--nospam--at)att.net]
Sent: Tuesday, November 08, 2005 3:12 PM
To: seaint(--nospam--at)seaint.org
Subject: Lateral Load Due to Foot-Traffic

 

I have a 6-ft wide wood-framed exterior one-story stairway/exiting-corridor, no roof.  At about mid-height is a landing supported on 4x4 posts where the stairway makes a 90 degree turn.  I have the feeling that lateral loads due to wind and seismic loading will be relatively insignificant compared during emergency exiting with people turning the corner.  Any ideas on how to estimate or calculate the appropriate lateral load?   

 

Nels Roselund, SE

South San Gabriel, CA

njineer(--nospam--at)att.net