Simplified Design for Structural Walls - Simplified Design of Reinforced Concrete Buildings - page 201

8" typ.

8'-0"

Factored shear force:

V u = 1.6(69.2) = 110.6 kips

For the N-S direction, assume that the shear force is resisted by the 20 ft-8 in. web segment only. From
Table 6-5

φ V c = 7.3 20.67 = 150.9 kips

Since V c = 150.9 kips, provide minimum shear reinforcement from

Table 6-3 For 8 in. wall, use No. 4 @ 10 in. horizontal as well as vertical reinforcement.

φ V c /2 = 75.5 kips < V u = 110.6 kips < φ

(3) Check shear strength in 2nd story in the N-S direction

V u = 1.6(16.2 + 31.6 + 30.6 + 29.2)/2 = 86.1 kips

The minimum shear reinforcement given in Table 6-3 s still required in the 2nd story since φ V c /2 = 75.5

V c = 150.9 kips. For the 3rd story and above, the minimum wall reinforcement
given in Table 6-2 can be used for all wall segments (V u @ 3rd story = 62.7 kips <

kips < V u = 86.1 kips < φ

V c /2 = 75.5 kips). For
horizontal reinforcement, use No. 4 @ 12 in., and for vertical reinforcement, use No. 3 @ 11 in.

φ

(4) Summary of Reinforcement

Vertical bars: Use No. 4 @ 10 in. for 1st and 2nd stories*

No. 3 @ 10 in. for 3rd through 5th stories**

Horizontal bars: Use No. 4 @ 10 in. for 1st and 2nd stories

No. 4 @ 12 in. for 3rd through 5th stories

* For moment strength, No. 6 @ 10 in. are required in the 8 ft. wall segments within the first story (see Example 6.5.1).
** Spacing of vertical bars reduced from 11 in. to 10 in. so that the bars in the 3rd story can be spliced with the bars in the 2nd story.