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# stopping force/ cmu wall ftgs

• To: <seaint(--nospam--at)seaint.org>
• Subject: stopping force/ cmu wall ftgs
• From: "Andrew Kester" <akester(--nospam--at)bbma.com>
• Date: Wed, 8 Oct 2003 09:09:33 -0400

Title: Message
Stopping force on a log:

I think Scott pretty much had it. You need to know how much time it takes to stop the log (deacceleration). I believe this would be independent of the stiffness or type of stop. The term for this force is IMPULSE.

Basic equation:

m1*v1+F*dt = m1*v2

m= W/g,  g=32.3ft/s^2, W=lb
v1= velocity before
v2=velocity after

Since velocity after the impulse is zero, just solve for F. Assume dt= 1 second:

F= (1/s)*(6200lb/ 32.2ft/s^2)*(200ft/60s) = 642lb

Smaller the dt, larger the impulse. Same theory with traffic control barriers, extend the dt to lower the force.

Wall ftgs under CMU wall:

By most codes even interior walls need to be designed for a Plat=5psf. This can go up if you have a partially enclosed/ open building. This creates a moment, which means you have to reinforce the wall. By ACI 530 at least 1/4 of vertical wall reinforcement must be fully developed, which means you need a footing of some type. This reason is in addition to all the other great reasons others have mentioned. I have always put CMU walls of any tpe on some type of reinforced ftg, any additional expense is worth the cost and contractors are used to this. Metal stud non-brg may or may not require anything but a slab, I usually throw a small TE if I can. Make sure the top of your wall is braced off properly to the roof structure, you do not want a cantilever.

HTH,

Andrew Kester, EI
Longwood, FL