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# Re: shear design

• To: seaint(--nospam--at)seaint.org
• Subject: Re: shear design
• From: "Bill Sherman" <SHERMANWC(--nospam--at)cdm.com>
• Date: 13 Feb 99 12:51:36 -0500

```Here's my "proof" of the design for shear-friction reinforcement at
non-keyed
construction joints:

1. I concur with three general statements made in ACI 318 R11.7.7. which
suggest the following: i) where there is direct tension across a joint,
reinforcement must be provided to resist the tension in addition to the
shear-friction reinforcement; ii) where there is direct compression across a
joint, the shear-friction reinforcement may be reduced accordingly; iii)
when
considering flexure only on a joint, the tension and compression are in
equilibrium, i.e. T=C.

2. For shear alone, Av = V/uFy (I will leave "phi" out of the formulas to
simplify typing).  Av equals the total reinforcement across the joint
(tension
and compression faces).

3. For tension alone, As = T/Fy.  For tension plus shear, Total As = T/Fy +
V/uFy.

4. For compression plus shear, V = u(C + AvFy) or Av = V/uFy - C/Fy.

5. Combining shear plus compression plus tension: As = T/Fy + V/uFy - C/Fy.
If T and C are due to flexure only, T = C, so As = V/uFy = Av.  Thus,
shear-friction reinforcement is not affected by flexure.

```