Flexural Strengthening of Beams and Slabs - Strengthening Design of Reinforced Concrete with FRP - page 130

Et
Cf

155000 1.2
25.55

×

ff
ctm

l = = = 129.45 mm

b, max

×

2

l = 638 mm > l

b b, max

N =α C kkbEtf

fa ,max 1 cb ffctm

120
155

2 −

k = 1.06 = 1.03 b

120
400

1 +

N =××× × 0.90.641.0 1.03 155 155000 × 1.25.55 × = 93431.73 N = 93.43 kN

fa ,max

Calculate the strain at cover delamination:

N
AE

93.432

fa
ff

,max

ε= =

×× =

0.00419

fcd

120 1.2 155

80

ε= 0.41 = 0.0085 >>ε Unconservative

fd fcd

1 × 155,000 × 1.2

α+=+ fbcAf Af Af c s s s y f f

2 2 2

2

α= ε
ε

ε
ε

= ε×
−×ε

c
dc

− ε×
−ε

c
dc

cf cf fcd fcd

−

( )

( )

3

2 2

3

c c f c

f c

f
E

f c c

ε= 1.71 = 1.71 4700 = 3.64 × 10

− 4

f = 0.00325 > 0.003 Crushing strain

c c

f

c c

Even though the strain ε > 0.003, still use it, since f c is very high (80 MPa).
Substituting ε and fc ε into the α equation,

2

c 1.662

3

c

α= 1.289 240 −

( )

2

c

− 240 c

−

2

c 1.662

3

c 0.00419

240

) (

)

1.289 240 80 155 c 226 204,000 c

− ×× +× × − 37

(

( )

2

− c − c

240 − c

3

=×+ 339 532 93.432 × 10 = 273,780

( )

2

Multiply theequationaboveby 240 − c

( ) ( ) ( )

2 3

15,983.6 c 240 −− + c 6,869.6 c 193,175.76 × −×− 240 c c 37

( )

2

− 273,780 × −= 240 c 0

2 3 3 2

3,836,064 c − 15,983.6 ,869.6 c − c − 193,175.76 c + 53,509,685.52 c

10 2

− 1, 715,400,749 − ×+ − 1.5769728 10 131,414,400 c 273,780 c = 0

3 2 10

− 22,853.2 ,369,108.24 c + c + 184,924,085.5 .748512875 c − × 10 = 0