Flexural Strengthening of Beams and Slabs - Strengthening Design of Reinforced Concrete with FRP - page 118

Q

( )

( ) ( )

3 1 2max

B = 93 QQ − 1 −Ψ ε+ − 3 2 Q 1 −Ψ−Ψεε 3 2 1 f c 2 f f cfu

4

( )

3

=× − 93 0.06397 0.0376 × 0.15 × 0.002

0.0376
4

2 − 7

+× − 32 0.06397 × −× × 0.1530.85 .002 × 0.00642 =− 1.733 × 10

Q

− Ψ

( )

( )

1 f

3 2max max2

D9 = Q 1 −Ψ− ε+ Q 3 1 +Ψ−ε ε+Ψεε 6 Q 6

1 f 2 c f 2 c fu fcfu

2 2

( )

= 90.0376 × 0.15 − 0.06397 × 0.002

3

0.0376
2

2

+ 3 × +× − 0.575 60.850.06397 × 0.002 × 0.00642

2

+× × 60.850.002 × 0.00642

− 7

= 8.05 × 10

Q 31

4

( ) ( )

3 1 2max max2 max3

E =− 31 Q −Ψ ε− −Ψ +Ψεε −Ψεε −Ψ ε 3 1 5

1 f c f f cfu f c u f fu

4

0.0376
4

3 2

=− × 30.0376 × 0.15 × 0.002 − 3 × 0.1550.85 .002 + × × × 0.00642

31
4

2 3 − 6

−× × 0.85 0.002 × 0.00642 − 0.85 × 0.00642 =− 1.0956 × 10

7
4

1
4

2max max2 max3

F =Ψεε +Ψεε +Ψε 3

fc fu fcfu f u

7
4

1
4

30.850.002

2

0.00642 0.85 0.002 0.00642

2 3

=× × × +× × × +× × 0.85 0.00642

− 7

= 2.4433 × 10

3 3 − 9

9 Q ε=× 9 0.06397 × 0.002 = 4.60584 × 10

2 c

The fifth-degree polynomial equation becomes

2 3

c
d

c
d

c
d

4.60584 × 10

− 9

− 1.861 × 10

− 8

− 1.733 × 10

− 7

+ 8.05 × 10

− 7

f f f

4 5

c
d

c
d

− 1.0956 × 10

− 6

+ 2.4433 × 10

− 7

= 0

f f

Multiplying the previous equation by 1 × 10 9 ,

2 3 4 5

c
d

c
d

c
d

c
d

c
d

4.60584 − 18.61 − 173.3 + 805 − 1095.6 + 244.33 = 0

f f f f f

c
d

c
dc

max

= 0.1532 c = 61.29 mm ε= − ε= 0.001161 cf fu

f f