Simplified Design for Structural Walls - Simplified Design of Reinforced Concrete Buildings - page 200

6.4.2 Example 2: Design for Shear

For Alternate (2) of Building #2 (5-story flat plate), select shear reinforcement for the two shearwalls. Assume
that the total wind forces are resisted by the walls, with slab-column framing resisting gravity loads only.

(1) E-W direction
Total shear force at base of building (see Chapter 2, Section 2.2.2.1):

V = 6.9 + 13.4 + 12.9 + 12.2 + 12.6 = 58 kips

For each shearwall, V = 58/2 = 29 kips

8" typ.

V u = 46.4 kips

8'-0"

Factored shear force (use ACI Eq. (9-4) for wind load only):

V u = 1.6(29) = 46.4 kips

For the E-W direction, assume that the shear force is resisted by the two 8 ft flange segments only. For
each segment:

φ V c = 7.3 8 = 58.4 kips (see Table 6-5)

Since V u for each 8 ft segment = 46.4/2 = 23.2 kips which is less than V c /2 = 58.4/2 = 29.2 kips, provide

minimum wall reinforcement from Table 6-2 For 8 in. wall, use No. 4 @ 12 in. horizontal reinforcement
and No. 3 @ 11 in. vertical reinforcement.

φ

(2) N-S direction

Total shear force at base of building (see Chapter 2, Section 2.2.2.1):

V = 16.2 + 31.6 + 30.6 + 29.2 + 30.7 = 138.3 kips

For each shearwall, V = 138.3/2 = 69.2 kips