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RE: Shear

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I believe that Michael described the differences between wood and concrete
pretty well.

As more shear reinforcement is placed at closer spacings, the diagonal
tension crack becomes more vertical to cause failure between the stirrups. I
believe this is where the spacing of stirrups reduces to d/4 (ACI 11.5.4.3)
for larger shear forces and an upper limit of 8 times the square root of F'c
for stirrups (ACI 11.5.6.8) comes from. I believe Phil Ferguson in his
textbook "Reinforced Concrete Fundamentals" referred to this maximum limit
as as 2 times the square root... for concrete plus 8 times the square
root.... for stirrups or the total limit of 10 times the square root of F'c.
Even though he states that he is a bit uncertain, he suggests that this may
be a the upper limit for designing keyways for shear. Later on he goes on to
state that shear friction, for the example, would be adequate to resist the
shear force.

Jim K.

-----Original Message-----
From: Michael Bryson [mailto:MBryson(--nospam--at)mhpse.com]
Sent: Monday, July 29, 2002 4:47 PM
To: 'seaint(--nospam--at)seaint.org'
Subject: RE: Shear


The failure of the concrete beam would be in shear friction. This is how you
design corbels for example.

Anyway, the difference between wood and concrete would be because the shear
failuires are completely different. Concrete fails in diagonal tension while
wood fails in horizontal shear. I believe there is some sort of clamping
action that prevents the horizontal shear from occurring near the support of
a wood member. For a concrete beam the failure would most likely be from
arching action.


-----Original Message-----
From: Mark Pemberton [mailto:Mark.Pemberton(--nospam--at)lbdg.com]
Sent: Monday, July 29, 2002 3:14 PM
To: 'seaint(--nospam--at)seaint.org'
Subject: RE: Shear


That's a good question.  I have always wondered
what the concrete shear strength would be for a
member if it were forced to failed in shear in
a plane parallel to the load applied.  Anyone
have any input on this?  Perhaps the reduced
failure area exactly compensates an increased
strength due to the forced shear plane?

Mark Pemberton, P.E.
Sacramento, CA

-----Original Message-----
From: Jim Kestner [mailto:jkestner(--nospam--at)somervilleinc.com]
Sent: Monday, July 29, 2002 1:42 PM
To: 'SEAINT'
Subject: Shear


Does anyone know why for Concrete design per ACI 318-95  11.1.3.1 you are
not allowed to ignore concentrated loads within a distance "d" from the
support for shear yet for Wood design, you can ignore all loads within a
distance "d" from the support per NDS '97 3.4.3a (Figure 3D). Shouldn't
proceedures for wood and concrete be the same? Is NDS ignoring some effect
that they should not be?

Jim K.



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