Simplified Design for Structural Walls - Simplified Design of Reinforced Concrete Buildings - page 207

(b) For one-20 ft-8 in. wall segment at 3rd floor level:

P u = 162 kips
M u = 1367 ft-kips
Check No. 3 @ 10 in. (required shear reinforcement above 2nd floor):

A st = 0.13 × 20.67 = 2.96 in. 2

4.96
248

60
4 =

⎛ ⎞

ω=

⎜ ⎟

0.020

× 8

162

α=

4 =

0.020

248 × 8 ×

c
w =

0.020
20.020

0.020 +

) + 0.72 = 0.053 (

162
2.96

⎛ ⎞

( )

M n = 0.5 × 2.69 × 60 × 2.48 1

⎜

+

⎟

1 − 0.053 /12 = 3163 ft − kips

× 60

( ) = 2847 ft − kips > M u = 1367 ft − kips O.K.

φ M n = 0.9 3163

The required shear reinforcement for the 20 ft-8 in. wall segments is adequate for moment strength for full
height of building.

(4) Summary

Required shear reinforcement determined in Example 6.4.2 can be used for the flexural reinforcement
except for the 8 ft wall segments within the 1st floor where No. 5 @ 10 in. are required (see Fig. 6-5).

For comparison purposes, the shearwall was investigated using the program spColumn. 6.3 For the reinforcement
shown in Fig. 6-5 at the 1st story level, the shearwall was analyzed for the combined factored axial load (due
to the dead loads) and moments (due to the wind loads) about each principal axis. The results are shown for
the x and y axes in Figs. 6-6 and 6-7, respectively. As expected, the load combination point (represented by
point 1 in the figures) is in the lower region of the interaction diagram, with the applied axial load well below
the balanced point. Since spColumn uses the entire cross-section when computing the moment capacity (and
not only certain segments as was done in the steps above), the results based on the reinforcement from the
approximate analysis will be conservative.