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Fw: LRFD, ASD, and USD

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> From: Bill Allen, S.E. <Bill(--nospam--at)>
> To: seaint(--nospam--at)
> Subject: RE: LRFD, ASD, and USD
> Date: Tuesday, July 28, 1998 1:22 PM
> in certain conditions. Also, some engineers I have had experience with
> (admittedly, the younger ones) don't have a clue about the strain
> transformation as the load increases, first starting elastically,
> and still elastic then into plastic deformation. I have reviewed some
> calculations of foundations (sized for soil pressure) for example and,
> with factored loads, the strain on the section is way below 0.002.
> the engineer still used the USD formulae.


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