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RE: moment due to differential settlement

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I think this is supposed to be a classical beam on an elastic foundation.

My guess is:

15" is probably the width. (A typical back-hoe bucket is 15-16"
18" is probably the depth. (This the universal minimum depth of any
foundations isn't it?)
20' is about a short span bay.

Oh yeah, what are the BOUNDARY CONDITIONS??

Arvel


-----Original Message-----
From: Michael L. Hemstad [mailto:hemstad.ml(--nospam--at)tkda.com]
Sent: Monday, March 28, 2005 12:56 PM
To: seaint(--nospam--at)seaint.org
Subject: moment due to differential settlement


Chris Wright responded to Dave Adie, P.E.:


"On Mar 27, 2005, at 10:09 AM, DA wrote:

> there an equation to calculate the moment due to differential
> settlement for footing 15"x18"  . 20'  long  settled 1" can someone
> have it  handy remind me
This one is just as easy to figure out as the cantilever with the end
moment. Especially for a P.E. Get yourself a copy of Blodgett's 'Design
of Welded Structures' and check the beam bending load cases in the back.
Faster that repeated e-mail postings.

Christopher Wright P.E. |"They couldn't hit an elephant at
chrisw(--nospam--at)skypoint.com   | this distance" (last words of Gen.
.......................................| John Sedgwick, Spotsylvania
1864)"


Yeah, it would be a lot faster than the 6 or so emails over this
question, especially when you multiply it by the 15,000 or so engineers
scratching their heads trying to figure the question out.

Dave, I know you said "simple span," but what does the footing really
look like?  It sounds like a strip footing, but is it really 20 foot
long or is that the distance between points of no settlement?  If it is
20 feet, is the settlement at the end or in the middle?

What is the wall supporting?  If it's a block wall, the wall may help it
span, albeit while cracking badly.

You said the footing is 15"x18".  Not knowing otherwise, I assume it's
18" wide.  15" is an unusual depth for a strip footing assumed to be
uniformly supported; and a really unusual depth for a member trying to
span 20 feet.  If you use the gross moment of inertia in Steve Gordin's
equation (against his prudent recommendations), I would guess you'll
need some awfully big bars in that little footing.

Bear in mind that the reinforcing design methodology you'll use to
design the reinforcing assumes the member cracked, while Steve's
equation (the portion of it asking for I) assumes whatever you assumed
when you calculated a value for I.  What in your judgement should you
use for I?  If you divide 1" deflection by 10 feet of beam each side of
the deflected point, times 15" of depth, neglecting bending in the rest
of the beam, the crack is 1/4 inch wide.  Doesn't sound promising.

I apologize for always being the jerk on your case, Dave, but if we
can't understand your questions, we can't give you good answers.  It
doesn't seem like that much to ask.

Mike Hemstad, P.E.
St. Paul, Minnesota



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