Simplified Design for Beams and One-Way Slabs - Simplified Design of Reinforced Concrete Buildings - page 83

Minimum area of stirrups:

50b w s/f y = 50(12)(6)/60,000 = 0.06 in. 2

(7) Required additional longitudinal reinforcement:

A t
s

A = p h = 0.023(50) = 1.15in. 2 govern

5 f c A cp

f y

ʹ

A t
s

5 4000(337.8)
60000

⎛ ⎞

( )

Minimum = −

⎜ ⎟

p h = − 0.023 50 = 0.63in. 2

,

Place the longitudinal bars around the perimeter of the closed stirrups, spaced not more than 12 in. apart. Locate
one longitudinal bar in each corner of the closed stirrups (ACI 11.5.5.2). For the 20 in. deep beam, one bar is
required at mid-depth on each side face, with 1/3 of the total A ˜ required at top, mid-depth, and bottom of the
closed stirrups. A ˜ /3 = 1.7/3 = 0.57 in 2 . Use 2-No.5 bars at mid-depth (one on each face). Longitudinal bars
required at top and bottom may be combined with the flexural reinforcement.

Details of the shear and torsion reinforcement (at support) are shown in Fig. (3-16).

0.38 in. 2 + A s (required for flexure)

2-#4

#4 closed stirrups @ 6 in. o.c.

0.38 in. 2 + A s (required for flexure)

12"

Figure 3-16 Required Shear and Torsion Reinforcement (Example 3.7.2.1)

3.8 EXAMPLES: SIMPLIFIED DESIGN FOR BEAMS AND ONE-WAY SLABS

The following three examples illustrate the use of the simplified design data presented in Chapter 3 for propor-
tioning beams and slabs. Typical floor members for the one-way joist floor system of Building #1 are designed.