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(8) Compute pure axial load capacity from Equation (56).
Po = 0.85 f'dc (Ag  Ast) + Ast fdy
where Ag = 18 x 18
= 324 in2 and
Ast = 12 x 0.6
= 7.2 in2
Therefore,
Po = 0.85 x 5,000 (324  7.2) + 7.2 x 66,000
= 1,821,600 lb
(9) Ultimate capacity of the column from Equation (78):
1
1
1
1
= + 
Pu
Px
Py
Po
1
1
1
1
= +  = 1/lbs
808,775
1,148',662
1,821,600
641,830
1
= 1/lb
641,830
Pu = 641,830 lbs > 636,000
O.K.
(10) Provide ties.
For No. 7 longitudinal bars use No. 3 ties.
s < /= 16 (longitudinal bars) = 16 x 0.875
= 14 in
and
s < /= 48 (ties) = 48 x 0.375
= 18 in
and
s < /= h/2 = 9 in
Use two No. 3 ties at 9 inches arranged as shown in
Figure 40.
7.
NOTATION.
a

Depth of equivalent rectangular stress block, in
Ag

Gross area of section, in2
A*l 
Area of longitudinal torsion reinforcement, in2
As

Total area of tension reinforcement, in2
A's 
Total area of compression reinforcement, in2
Ast 
Total area of uniformly distributed longitudinal reinforcement,
in2
2.08124


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