Handbook of Civil Engineering Calculations

FIGURE 7. Shearing stress diagram

sion. The less precise and more conservative method restricts the shearing stress to a stip-
ulated value that is independent of the flexural stress.
For simplicity, the latter method is adopted here. A section of the Code sets </> = 0.85
with respect to the design of web reinforcement. Let vu = nominal ultimate shearing
stress, lb/in^2 (kPa); vc = shearing stress resisted by concrete, lb/in^2 (kPa); #J = shearing
stress resisted by the web reinforcement, lb/in^2 (kPa); Av = total cross-sectional area of
stirrup, in^2 (cm^2 ); Vu = ultimate vertical shear at section, Ib (N); s = center-to-center spac-
ing of stirrups, in (mm).
The shearing-stress diagram for half-span is shown in Fig. 7. Establish the region AF
within which web reinforcement is required. The Code sets the allowable shearing stress
in the concrete at

vc = 2<Kfy-^5 (15)

The equation for nominal ultimate shearing stress is

«•-£

(16)

Then, vc = 2(0.85)(3000)°^5 = 93 lb/in^2 (641.2 kPa).
At the face of the support, Vu = 9(10,200) = 91,800 Ib (408,326.4 N); vu =
91,800/[ 15(22.5)] = 272 lb/in
2
(1875.44 kPa). The slope of the shearing-stress diagram =
-272/108 = -2.52 Ib/(in^2 -in) (-0.684 kPa/mm). At distance d from the face of the support,
vu = 272 - 22.5(2.52) = 215 lb/in^2 (1482.4 kPa); 0J = 215 - 93 = 122 lb/in^2 (841.2 kPa).
Let E denote the section at which vu = vc. Then, AE = (272 - 93)/2.52 = 71 in (1803.4
mm). A section of the Code requires that web reinforcement be continued for a distance d
beyond the section where vu = vc\ AF= 71 + 22.5 = 93.5 in (2374.9 mm).

2. Check the beam size for Code compliance
   Thus, ^>max =W(Kfc)
   Q
   '
   5
   = 466 > 215 lb/in
   2
   (1482.4 kPa). This is acceptable.

Face of

support

Note: All dimensions are to A.

Nominal

shearing

stress