Strengthening of Columns for Confinement - Strengthening Design of Reinforced Concrete with FRP - page 191

ε fu 0.029 0.029

f yt 89 ksi 614 MPa

Hoop spacing 3.5 in. 88.9 mm

Hoop diameter #3 9.5 mm

Longitudinal bars 14#4 14 ϕ 12.7 mm

f y 65 ksi 448 MPa

Draw the interaction diagram per ACI 440.2R-08 then use KDOT Column Expert.

Solution:
Using Equation (7.7):

2 nt E

D

ε

= ×× ×

2 60.03 2696 × 0.586 × 0.029 fffe

f = = 1.3745 ksi

fl

12

f
f

1.3745
5

fl
c

= = 0.275 > 0.08, stress-strain curve is ascending

Point A:
According to Equation (7.6),
=+ ××× f cc 50.953.3 11.3745 = 9.31ksi

0.45

0.586 0.029
0.002

×

ε= ×+×× × 0.002 1.51210.275 = 0.0203 > 0.01

ccu

ε = 0.01

ccu _ new

Recalculate f cc based on the original E 2 and the new ε ccu .

f − f 9.31 5

0.0203

− cc c

E = = = 212.315 ksi

2

ε

ccu

f =+ε f E = + 5 212.315 × 0.01)7.123 ksi =

cc _ newc 2 ccu _ new

Following Equation (7.1) with and without ϕ factors,

π
4

P n = 0.85 × ××−× +×× 7.123 12

2

14 0.20 65 14 0.20 = 849.8kips

φ= × P n 0.75 0.85 × 849.8 = 541.75 kips

Point B:

2 nt E

D

ε

= ×× ×

2 60.03 2696 × 0.004

fffe

f = = 0.3235 ksi

fl

12

This f fl will not be used to determine whether the stress-strain curve is
ascending or descending, since it is computed for eccentric points.

f cc =+ ××× 50.953.3 10.3235 = 6.014 ksi

E = 4030.51ksi, E = 198.042 ksi, ε= 0.002609 c 2 t

0.45

0.004
0.002

ε= ×+×× 0.002 1.51210.0647 × = 0.005121

ccu