Simplified Design for Structural Walls - Simplified Design of Reinforced Concrete Buildings - page 203

c
w =

ω+α

β 1 ,whereβ 1 = 0.85 for

f c = 4000psi ʹ

2 ω+ 0.85

f y
ʹ

⎛
⎜

⎞
⎟

A st
w h

ω=

f c

P u
w h

α=

f c ʹ

h = thickness of wall, in.

φ = 0.90 (strength primarily controlled by flexure with low axial load)

Note that this equation should apply in a majority of cases since the wall axial loads are usually small.

6.5.1 Example: Design for Flexure

For Alternate (2) of Building #2 (5-story flat plate), determine the required amount of moment reinforcement
for the two shearwalls. Assume that the 8 ft wall segments resist the wind moments in the E-W direction and
the 20 ft-8 in. wall segments resist the wind moments in the N-S direction.

Roof: DL = 122 psf

Floors: DL = 142 psf

(1) Factored loads and load combinations

When evaluating moment strength, the load combination given in ACI Eq. (9-6) will govern.

U = 0.9D + 1.6W

(a) Dead load at first floor level:

Tributary floor area = 12 40 = 480 sq ft/story

Wall dead load = (0.150 3392)/144 = 3.53 kips/ft of wall height (see Sect. 6.3.1)

P u = 0.9[(0.122 480) + (0.142 480 4) + (3.53 63)] = 498 kips

Proportion total P u between wall segments:

2-8 ft segments: 2 96 = 192 in. 192/440 = 0.44

1-20 ft-8 in. segment: 248 in. 248/440 = 0.56

For 2-8 ft segments: P u = 0.44(498) = 219 kips

1-20 ft-8 in. segment P u = 0.56(498) = 279 kips

96"

(b) Wind moments at first floor level: