Shear and Diagonal Tension - Design of Reinforced Concrete - page 245

Figure 8.18

For an example using this table, reference is made to the beam and V u diagram
of Example 8.3, which was shown in Figure 8.12 where 60-ksi #3 stirrups were se-

22 2 lected for a beam with a d of in. For our closest spacing, d /4, we can calculate

71.6 k. Similar calculations are made for d /3 and d /2
spacings, and we obtain, respectively, 61.7 k and 51.8 k. The shear diagram is repeated
in Figure 8.18, and the preceding values are located on the diagram by proportions or
by scaling.
From this information we can see that we can use d /4 for the first 2.72 ft, d /3 for the
next 0.68 ft, and d /2 for the remaining 2.49 ft. Then the spacings are smoothed (preferably
to multiples of 3 in.). Also, for this particular beam we would probably use the d /4 spac-
ing on through the 0.68-ft section and then use d /2 the rest of the required distance.

V c 39.6 32.018 39.6

8.12 SHEAR FRICTION AND CORBELS

If a crack occurs in a reinforced concrete member (whether caused by shear, flexure,
shrinkage, etc.) and if the concrete pieces on opposite sides of the crack are prevented
from moving apart, there will be a great deal of resistance to slipping along the crack due
to the rough and irregular concrete surfaces. If reinforcement is provided across the crack
to prevent relative displacement along the crack, shear will be resisted by friction between
the faces, by resistance to shearing off of protruding portions of the concrete, and by
dowel action of the reinforcing crossing the crack. The transfer of shear under these cir-
cumstances is called shear friction .
Shear friction failures are most likely to occur in short, deep members subject to
high shears and small bending moments. These are the situations where the most nearly