Design Considerations for Earthquake Forces - Simplified Design of Reinforced Concrete Buildings - page 305

For No. 6 @ 15 in.:

A st = 0.0037 96(8) = 2.82 in. 2

2.82
96

60
4 =

⎛ ⎞

ω=

⎜ ⎟

0.055

× 8

109.8
96

α=

4 =

0.036

× 8 ×

c
w =

0.055 + 0.036

) + 0.72 = 0.109 (

2 0.055

109.8
2.82

⎛ ⎞

( )

M n = 0.5 × 2.82 × 60 × 96 1

⎜

+

⎟

1 − 0.109 /12 = 994 ft − kips

× 60

( ) = 894 ft − kips < M u = 3560 ft − kips.

φ M n = 0.9 994

In order to compensate for the big difference between M n and M u the thickness and reinforcement of this seg-

ment of the wall need to be increased. Increasing the thickness to 10 in. and use two layers off reinforcements.

φ

Use two layers No. 6 @ 4 in. spacing

ρ = 2 0.44/(10 4) = 0.022

A st = 0.0022 96(10) = 21.12in. 2

21.12
96

60
4 =

⎛ ⎞

ω=

⎜ ⎟

0.33

× 10

109.8
96

α=

4 =

0.33

× 8 ×

c
w =

0.029
20.33

0.33 +

( ) + 0.72 = 0.259

109.8
21.12

⎛ ⎞

( )

M n = 0.5 × 21.12 × 60 × 96 1

⎜

+

⎟

1− 0.259 /12 = 4079 ft − kips

× 60

( ) = 3671 ft − kips > M u = 3560 ft − kips. O.K.

φ M n = 0.9 4079