Flexural Strengthening of Beams and Slabs - Strengthening Design of Reinforced Concrete with FRP - page 141

Solution:

Assume ductile crushing failure mode.

f
f

d
d

0.22
86

60
7.95

6
8

y

Q =ρ = × × = 0.0259 s

×

cf

M
fbd

+Ψ−= ×

d
d

24 12 6

8

u

Q = φ Q

×× +

0.0259 × −= 0.85 0.0733

2 1 f

2

7.95 88

2

cf f

2

a
d

a
d

( )

−− × 20.208 0.0259 × + 2.77 × 0.0733 = 0

f f

2

a
d

1.995 − 1.995 − 11.08 × 0.0733

= = 0.1078 a = 0.862 in . β

1

2

f

= 0.6525 c = 1.321 in.

0.003 0.003

1.321

( ) ( )

ε= −−ε= 8 c 81.321 − − = 00.01516

fe bi

c

f
nE t

7,950
120,000

c
ff

ε= 0.083 = 0.083 = 0.0125 < 0.9 ε

fd fu

× × 0.0175

f
E

fu

0.9 .9 ε= ××= C 0.01368 debonding controls

fu E

f

Using the statistically correlated linear equation for initial estimate of ρ f :
Unstrengthened beam capacity:

0.85 fb cAf 1 β+ = Af c s s s y

87

( )

0.85 × 7.95 × × 80.6525 c + 0.22 × c − 2 .2260 = ×

c

35.274 c

2

+ 19.14 c − 38.28 = 13.2 c

35.274 c

2

+ 5.94 c − 38.28 = 0

2
c = −+ +× ×
×

5.94 5.94 4 35.274 38.28

= 0.961 in.

2 35.274

0.003
0.961

( )

ε= 0.961 −=− 2 .00324 (tension) > ε

s y

f = f s y

0.85 fb cAf β= + Af

c 1 s y s y

0.44 60

×

c = = 0.748 in.

0.85 × 7.95 × × 80.6525

− β

c β c

1 1

MAfd = − Afy − d n s y s

2 2