TM 5-809-3/NAVFAC DM-2.9/AFM 88-3, Chap. 3
(a) The minimum reinforcement, AsMIN, is determined as follows:
AsMIN = 0. 005Ag = (0.005)(244) = 1.22 in2
As = 6.00 in2 > AsMIN= 1.22 in2
...O.K.
(b) The maximum reinforcement, AsMAX, is determined as follows:
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.
(a) The eccentric moment at the top, MeccT, is determined as follows:
MeccT = Pe = 40(2) 80 in-kips
(b) The axial compressive stress, fa, is determined as follows:
(c) The allowable compressive stress, Fs, is determined as follows:
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.
(d) The flexural compressive stress, fb, is determined as follows (Note: Assume a cracked section):
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.
(e) Check combined axial and bending compressive stresses using the unity equation. When
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.
(a) The eccentric moment at mid-height, MeccM, is determined as follows:
MeccM = Pe/2 = 40(2)/2 = 40 in-kips
(b) The wind loading moment at mid-height, Mwind, is determined as follows:
9-9
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