m). Using^' = 3000 lb/in^2 (20,685 kPa) and an allowable stress^, in the stirrups of 20,000
lb/in^2 (137,900 kPa), design web reinforcement in the form of vertical U stirrups.Calculation Procedure:- Construct the shearing-stress diagram for half-span
The design of web reinforcement by the working-stress method parallels the design by the
ultimate-strength method, given earlier. Let v = nominal shearing stress, lb/in
2
(kPa);
vc = shearing stress resisted by concrete; v' = shearing stress resisted by web reinforce-
ment.
The ACI Code provides two alternative methods of computing the shearing stress that
may be resisted by the concrete. The simpler method is used here. This sets
f>c=l.l(/cT^5 (33)The equation for nominal shearing stress is'- id
(34)
The shearing-stress diagram for a half-span is shown in Fig. 15. Establish the region
AD within which web reinforcement is required. Thus, vc = 1.1(3000)°^5 = 60 lb/in^2 (413.7
kPa). At the face of the support, V= 6.5(3800) + 1000 = 25,700 Ib (114,313.6 N); v =
25,700/[14(18.5)] = 99 lb/in^2 (682.6 kPa).
At midspan, V= 1000 Ib (4448 N); v = 4 lb/in^2 (27.6 kPa); slope of diagram = -(99 -
4)/78 = -1.22 Ib/(in^2 -in) (-0.331 kPa/mm). At distance d from the face of the support, v =
99 - 18.5(1.22) = 76 lb/in^2 (524.02 kPa); v' = 76 - 60 = 16 lb/in^2 (110.3 kPa); AC = (99 -
60)71.22 = 32 in (812.8 mm); AD = AC + d= 32 + 18.5 = 50.5 in (1282.7 mm).
- Check the beam size for compliance with the Code
Thus, vmax = 5(fc)Q-^5 - 274 lb/in^2 (1889.23 kPa) > 76 lb/in^2 (524.02 kPa). This is accept-
able.
Note: All dimensions are to A.FIGURE 15. Shearing-stress diagram.Foce of
supportNominalshearingstress