Simplified Design for Beams and One-Way Slabs - Simplified Design of Reinforced Concrete Buildings - page 96

A s = 584.9/4(17) = 8.6 in. 2

Use 11-No. 8 bars (A s = 8.6 in. 2 )

(d) Bottom bars in interior span:

A s = 324.9/4(17) = 4.78 in. 2

Use 7-No. 8 bars (A s = 5.53 in. 2 )

(6) Reinforcement details shown in Fig. 3-22 are determined directly from Fig. 8-3(a).* Provided 2-No. 5

top bars within the center portion of all spans to account for any variations in required bar lengths due
to wind effects.

Since the column line beams are part of the primary wind-force resisting system. ACI 12.11.2 requires
at least one-fourth the positive moment reinforcement to be extended into the supporting columns and
be anchored to develop full fy at face of support. For the end spans: A s /4 = 8/4 = 2 bars. Extend 2-No. 8
center bars anchorage distance into the supports:

￿ At the exterior columns, provide a 90° standard end-hook (general use). From Table 8-5, for
No. 8 bar:

˜ n = 14 in. = 16 - 2 = 14 in. O.K.

￿ At the interior columns, provide a Class B tension splice (ACI 13.2.4). Clear space between
No. 8 bars = 3.4 in. = 3.4d b . From Table 8-2, length of splice = 1.0 30 = 30 in. (ACI 12.15).

(7) Design of Shear Reinforcement

Design shear reinforcement for the end span at the interior column and use the same stirrup
requirements for all three spans.

Check governing load combination:

￿ gravity load only
V u = at interior column = 105.5 kips (governs)

￿ gravity + wind loads:
V u = 1.2(1.15)(3.9)(28.54)/2 + 0.8(6.64) = 82.1 kips

or

V u = 1.2(1.15)(3.9)(28.54)/2 + 0.5(1.15)(1.09)(28.54)/2 + 1.6(6.64) = 95.7 kips

* The bar cut-off points shown in Fig. 8-3(a) are recommended for beams without closed stirrups. The reader may consider
determining actual bar lengths using the provisions in ACI 12.10.