Flexural Strengthening of Beams and Slabs - Strengthening Design of Reinforced Concrete with FRP - page 114

The failure mode seems to be FRP rupture:

α=+ fbcAf A f (0.9 )

c s y f fu

max

ε= ε

c
dc

c fu

= 0.00665 5

cf

− − c

f

f
E

c
c

− 5

ε= =× 1.71 310 6480 = 0.00241

c

2

2

α= ε
ε

1
3

ε
ε

c 1

3

c

cf cf

− = 2.76 5 − 2.76 5

− c − c

c c

2

c c

α= 2.76 5 − 2.54

( )

2

− c

5 − c

α× ×= ×+ × 6.48 3 .051 c 75 2.38 0.0079 × 179.55 = 7.2

2

c c

2

2.76 − 2.54 × 19.44 c = 7.2 multiply by (5 − c )

( )

( )

2

5 − c

5 − c

( ) ( )

2

53.65 c

2

5 −− = c 49.38 c

3

7.25 − c

103.03 c

3

− 268.25 c

2

+ −+ = 180 72 c 7.20 c

2

3 2

103.03 c − 261.05 c − += 72 c 180 0

c = 0.828" (Goal-Seek in Excel)

0.828
50.828

ε= 0.00665 = 0.00132 < 0.003

cf

−

0.00665 4.37 0.828
50.828

−

ε= = 0.00565 > 0.005 φ= 0.9

s

−

6480

ε= 0.083 = 0.0145

fd

1 × 26980000 × 0.0079

max

ε= <ε 0.00665

fu fd

Thus,FRP ruptureisconfirmed

− β

c

− β

c

1 1

max

MAfd =φ +φΨ Af d u s y f f fu f

2 2

f

max

=×ε= × E

max

26980 0.00665 = 179.42 ≈ 0.9 f

fu f fu fu

0.726 × 0.828

M =× ×× − 0.90.051 75 4.37

u

2

0.726 × 0.828

+× × 0.90.852.380.0079 × × 179.42 × 5 −

2

14.01 12.13 26.14 k-in 2.178 k-ft

= + = =

from thepaper, M = 34.94 k-in = 2.912 k-ft

n