Strengthening of Columns for Confinement - Strengthening Design of Reinforced Concrete with FRP - page 210

f =+Ψ× × f 3.3 kf

cc c f al

2

A
A

b
h

( )

2

e

k 0.458 0.833 0.318

= = = a

c

ρ= ×

10 1.27

480

= 0.0265

s

( )( ) ( (

2

)

2

0.833 24 21.2 1.22021.2

−× + − ×

1 − − 0.0265

A
A

3 × 480

e

0.458

= =

10.0265 −

c

7.584 =+×× 6.00.953.3 0.318 f

l

f = 1.589 ksi

l

2 nt E

bh

ε fffe

f

=

l

2 2

+

2 .015 n × × 33,000 × 0.586 × 0.0159

1.589 =

2 2

24 + 20

n = 5.38 plies, use6plies

f
f

1.589
6.0

l

0.265 0.08

= = >

c

0.5

A
A

h
b

e
c

k = = 0.458 1.20.502 = b

0.45

f
f

ε
ε

l
c

fe
c

ε=ε 1.512 + k ≤ 0.01

ccu c b

f
E

c
c

− 5

ε= =× = 1.71 310 f 0.00232

c c

0.45

0.586 0.0159
0.00232

×

ε= 0.00232 1.5120.502 +× × 0.265 = 0.0104 > 0.01

ccu

f cc needs to be adjusted to correspond to 0.01.
Recalculate f cc based on the original E 2 and the new ε ccu .

f − f 7.584 6

0.0104

− cc c

E = = = 152.26 ksi

2

ε

ccu

f =+ε f E = + 6 152.26 × 0.01 = 7.523 ksi

cc _ newc 2 ccu _ new

Example 7.6: Analysis

Determine the compression-controlled interaction diagram for the column in
Example 7.5 without FRP reinforcement. Use the simplified calculation approach.
Scale the ultimate unconfined interaction diagram to the design diagram using the
appropriate ϕ factors.