Custom Search




(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


Integrated Publishing, Inc.
6230 Stone Rd, Unit Q
Port Richey, FL 34668
Phone For Parts Inquiries: (727) 4930744 Google + 