Strengthening of Columns for Confinement - Strengthening Design of Reinforced Concrete with FRP - page 204

Point A:

( )

2

ρ= ×π

840.77
12

= 0.0259

s

× 12

b
h

h
b

( )

2

( )

2

hr 2 br 2

− + −

c c

2

− ×−×
×

[2 (1220.64)]

3

1 − −ρ

1 − 0.0259

s

A
A

3
1

A

144

g

e
c

= =

−ρ 10.0259 − s

= 0.454

2

A
A

b
h

e
c

κ= = 0.454

a

0.5

A
A

h
b

e

κ= = 0.454

b

c

According to Equation (7.6):

f cc = 6.19 + ×× × 0.95 3.30.454 0.902 = 7.47 ksi

0.45

0.586 0.0228
0.002

×

ε= ×+× 0.002 1.5120.454 × 0.146 × = 0.00674 ccu

Following Equation (7.2) with and without ϕ factors:

× ××−× π × + ×× π ×

2 2

P n = 0.85 7.47 (12128 4 0.77 ) 67.43 8 0.77 = 1141.87 kips

4

φ= ×× = P n 0.65 0.8 1141.87 593.77 kips

Point B:

2 nt E

D

2 40.05 2864.33 0.004

ε

= ×× ×

×

fffe

f = = 0.270 ksi

fl

12

2

12

2

+

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 = 6.19 + ×× × 0.95 3.30.454 0.270 = 6.574 ksi

E = 4484.56 ksi, E = 105.323 ksi, ε= 0.002827

c 2 t

0.45

0.27
6.19

0.004
0.002

ε= ×+× 0.002 1.5120.454 × × = 0.003649

ccu

P = 709.18 kips, M = 127.90 kip-ft

nB , nB ,

Point C:

P = 353.5kips, M = 169.8kip-ft

nC , nC ,

Point D:
=

P 0kips, M = 97.37 kip-ft

nD , nD ,