Flexural Strengthening of Beams and Slabs - Strengthening Design of Reinforced Concrete with FRP - page 119

Applying the approximate solution equation,

For

c ε<

0.0015

A

= 0.3447 × 0.0376 × 0.15 − 0.06397 − 0.0417 × 0.85 =− 0.0975

B = 0.0561 × 0.85 − 0.3447 × 0.0376 × 0.15 − 366.67 × 0.85 × 0.00642 = − 1.9552

3

D = 126.3911 × 0.85 × 0.00642 − 0.0144 × 0.85 = 0.67748

3

2 3

c
d

c
d

c
d

0.06397 − 0.0975 − 1.9552 + 0.67748 = 0

f f f

c
d

= 0.1607

f

c = 64.29 mm

f c

d

f

y

c

ρ=α − ρ f s

0.9 f 0.9 f

fu fu

c
dc

64.29
400.09

max

ε= − ε= × 0.00642 = 0.00123 (exact = 0.00116) cf fu

− 64.29

f

2

α= ε
ε

ε
ε

cf cf

− = 0.4889 (exact 0.4679)

3

2

c c

α= 500 ×ε − 83333 ×ε=

2

0.4889

approx cf cf

30
2565

64.29
350

340
2565

0.4889 0.003792 5.477 10

− 4

(exact 4.557 10

-4

)

ρ= × × − × = × = × f

A d =ρ = 57.51 mm

2

( exact = 47.85mm)

2

f f

A
b

57.51
300

f
f

t == = 0.192 mm > t = 0.17 mm O.K. f f actual

(exact = 0.16 mm) forsinglyreinforced section (with doubly .17is expected) t >

f

The third approach is by using the statistically correlated linear equation:
Unstrengthened beam:

0.003

( )

0.85 fb cAE β+ cd −= Af c 1 s s s y

c

2

6502.5 c + 265.5 × 600 c − 265.5 × ×− × 600 50 398.2 340 c = 0

6502.2 c

2

+ 23912 c − 7965000 = 0

2

= −

23912 + 23912 +× × 4 6502.5 7965000

c = 33.21 mm

2 × 6502.5

0.003
33.21

( )

f =× − − E 33.21 50 303.34 MPa

s s