AsMIN = 0. 005Ag = (0.005)(244) = 1.22 in2

As = 6.00 in2 > AsMIN= 1.22 in2

...O.K.

AsMAX = 0. 04Ag = (0.04)(244) = 9.76 in2

As = 6.00 in2 < AsMAX = 9.76 in2

...O.K.

(3) Check the assumed pilaster for the given loadings at the top.

MeccT = Pe = 40(2) 80 in-kips

Fa = [0.18f'm + 0.65 pgFsc]R

Where:

pg = As/Ag

= (6.00 in2)/(244 in2) = 0.0246

R = The stress reduction factor. *(Note. *R is one at top of pilaster since stability is not a

consideration.)

Fa = [0.18(1350) + 0.65(0.0246)(24,000)]1.0 = 627 psi

fa =109 psi < Fa = 627 psi

...O.K.

Where:

d = 15.62 in - 3.5 in = 12.12 in; use d = 12 in.

AsT = The area of the reinforcement that is in tension, which is 3-#9 bars.

AsT = 3(1.00 in2 = 3.00 in2

p = AsT/bd = 3.00 in2/(15.62 in 12 in) = 0.016

np = 21.5(0.016) = 0.344

...O.K.

checking for flexural compression, the unity equation (equation 9-8) becomes:

...O.K.

(4) Check the assumed pilaster for the given loadings at mid-height.

MeccM = Pe/2 = 40(2)/2 = 40 in-kips

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