Simplified Design for Beams and One-Way Slabs - Simplified Design of Reinforced Concrete Buildings - page 92

Use average web width for shear strength calculations

b w = 9 + (20.5/12) = 10.7 in.
( φ V c + φ V s ) max = 0.48 b w d = 0.48(10.7)(18.0) = 92.4 kips

φ V c = 0.095 b w d = 0.095(10.7)(18.0) = 18.3 kips

φ V c /2 = 0.048 b w d = 9.3 kips

Sym about L

10'-8" 10'-8"

#7 @ 10" (1st interior)
#6 @ 10" (interior)

7'-11"
4 @ 9"

#3 @ 13"

4 .5"

1-#8

1-#7

16"

6" Single leg

#3 stirrups

1-#7 3'-5" 3'-0 "

3'-5" 1-#7

0"

1' -8"

10'-6" 10'-6"

10'-6"

#3 @ 9" single leg
stirrups

#3 @ 9" single leg
stirrups

#3 @ 9" single leg stirrups

8" 30'-0" 30'-0"

Class B splice

End Span Interior Span (typ)

Figure 3-20 Reinforcement Details for 6 ft-3 in. Wide-Module Joist Floor System [Building #1—Alternate (2)]

Beam size is adequate for shear strength since 92.3 kips > 22.9 kips.

V c , more than minimum shear reinforcement is required. Due to the sloping face of the
joist rib and the narrow widths commonly used, shear reinforcement is generally a one-legged stirrup
rather than the usual two. The type commonly used is shown in Fig. 3-21. The stirrups are attached to
the joist bottom bars.

Since V u > φ

Stirrups

Alternate hook direction at
every other stirrup

Figure 3-21 Stirrup Detail

Single leg No. 3 @ d/2 is adequate for full length where stirrups are required. Length over which
stirrups are required: (25.3 - 9.3)/1.8 = 8.9 ft. Stirrup spacing s = d/2 = 18.0/2 = 9.0 in.

Check A v(min) = 50 b w s/f y = 50 x 10.7 9.0/60,000 = 0.08 in. 2

Single leg No.3 stirrup O.K.

Use 14-No. 3 single leg stirrups @ 9 in. Use the same stirrup detail at each end of all joists.