# RE: Masonry Shear Pier Design - Balanced

• To: <seaint(--nospam--at)seaint.org>
• Subject: RE: Masonry Shear Pier Design - Balanced
• From: "Bill Allen" <Bill(--nospam--at)AllenDesigns.com>
• Date: Sat, 26 Aug 2000 11:16:31 -0700
```Roger-

Yeah, that's (three point) the curve I'm developing.

My section is as follows:

b=15.625 in
h=56 in
d1=4 in As1=0.88 in^2
d2=12 in As2=0.88 in^2
d3=20 in As3=0.88 in^2
d4=28 in As4=0.88 in^2
d5=36 in As5=0.88 in^2
d6=44 in As6=0.88 in^2
d7=52 in As7=0.88 in^2

(yes, I probably could reduce As3, As4 and As5 to 2*0.31, but the savings
are not worth the effort. I only have 8 piers).

F`m=1,500 psi
Fy=60,000 psi
F`g=3,000 psi (which is making me redesign the beam with smaller bars and
higher F`m. Sigh...)

With the above, I got the following results:

Po=1477 k * phi=0.6 --> phi*Po = 886.4 k

Pn(max)=0.8* Po = 1,181 k --> phi * Pn(max) = 709.1 k

Pb=545.9k, Mb=802.4 ft-k * phi = 0.65 --> phi * Mb = 521.5 ft.-k, phi * Pb =
354.8 k

at P=0, Mn=677.8 ft.-k * phi = 0.85 --> phi * Mn = 576.1 ft.-k

I've got enough capacity, so maybe I should just let it go. It's just that
the curve looks funny due to the phi phactors :o).

I've been using the ICBO Seminar #113 notes "1994 U.B.C. Strength Design of
Masonry" and the publication by Joe Uzarski "Design of Masonry Wall Frames"
as well as the CMACN 1997 "Design of Reinforced Masonry Structures".

Thanks,

Bill Allen, S.E. (CA #2607)
ALLEN DESIGNS
Consulting Structural Engineers
Laguna Niguel, CA
http://www.AllenDesigns.com
V (949) 365-5696
F (949) 249-2297

||-----Original Message-----
||From: Roger Turk [mailto:73527.1356(--nospam--at)compuserve.com]
||Sent: Saturday, August 26, 2000 10:14 AM
||To: seaint(--nospam--at)seaint.org
||Subject: RE: Masonry Shear Pier Design - Balanced
||
||
||Bill,
||
||The P-M diagram as you describe it does occur in some
||instances.  Although I
||have not had an occasion to develop such a diagram (by hand,
||of course), I
||recall seeing such as you described in handbooks, but I can't
||remember which
||handbook at the present (Amrhein's or ACI ???).
||
||Typically, I will do a 3-point P-M diagram; M = 0, Balanced,
||and P = 0 and
||connect them with straight lines (conservative).
||
||I will try to find the diagrams similar to what you described
||and send the
||references.
||
||A. Roger Turk, P.E.(Structural)
||Tucson, Arizona
||
||Bill Allen wrote:
||
||>>Sooo...I'm assuming by the responses I've gotten that no
||one has done this
||by hand, right?
||
||FWIW, based on the section I am looking at, the phi*M-sub-n at P=0 is
||greater than phi*M-sub-n at the balanced condition. This
||could be due to the
||amount of steel?? I dunno. The graph just looks wierd, that's
||all. I also am
||using a phi=0.85 at P=0 and a phi =0.60 at P-sub-b.
||
||BTW - the Paulay and Priestley reference sounds pretty good.
||I'll check it
||out.<<

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