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RE: LRFD, ASD, and USD

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I believe the provision is to limit the width of the crack. At a given
strain level in the reinforcing, the crack width at the surface of the
concrete will be wider if there is more cover.

Regards,
Bill Allen

-----Original Message-----
From: Roger Turk [mailto:73527.1356(--nospam--at)compuserve.com]
Sent: Tuesday, July 28, 1998 9:25 AM
To: seaint(--nospam--at)seaint.org
Subject: Fw: LRFD, ASD, and USD


I have a problem understanding the crack control requirements of the ACI
code.  I understand the stated theory about the crack control requirements,
but don't understand why additional cover will require more reinforcing.  If
the width of the crack at the surface is the limiting factor, then increased
cover should have less effect on the exposure of the reinforcing to
corrosive
environments.  After all, we are still dealing with "plane sections remain
plane" and "Hooke's Law applies," aren't we?

Since the crack control requirements do not apply if Grade 40 steel is used,
that is the maximum amount of reinforcing that I will use in a footing,
regardless of whether or not Grade 60 reinforcing is used.

A. Roger Turk, P.E.(Structural)
Tucson, Arizona

T. Eric Gillham wrote:

. > Bill:
. >
. > It sounds like the foundation sections you were reviewing were
. > overdesigned (for strength), perhaps as the result of minimum steel
. > requirements?
. >
. > USD ASSUMES an ultimate state when computing the strength of a member.
. > The steel is assumed to be yielding, the concrete is assumed to have a
. > strain of .003. When designing using USD, if one were to design a member
. > "exactly", it would have just the right amount of steel such that when
. > the steel is yielding, the moment manifested in the member matches the
. > imposed moment (factored). Therefore, it sounds strange to me that the
. > strain in the concrete was way below .002. It would seem that if the
. > ultimate stength of the example foundation section was calculated, it
. > should be greater than that required by the factored loads.
. >
. > For example, computing the strength of a 12x24 beam, 2#8 tension steel,
. > d=22", f'c=4ksi, fy=60ksi using USD gives an ultimate strength of
165k-ft.
. >
. > Using a moment curvature program, no strain hardening in the steel,
gives:
. >
. > concrete strain=.0008   steel starts yielding, moment is 154.5k-ft
. >
. > concrete strain=.003     steel strain is .0234, moment is 160k-ft
(about
. > 3% different from USD).
. >
. > Max moment is 160k-ft, with concrete strain maxing out at .004.
. >
. > The MC program (SEQMC) takes into account the cracking of the section,
and
. > of course is based on equilibrium, and uses the Mander model for
concrete.
. >
. >
. > T. Eric Gillham PE
. >