Flexural Strengthening of Beams and Slabs - Strengthening Design of Reinforced Concrete with FRP - page 117

Thus, the failure mode is expected to be rupture of FRP.
Assuming interior exposure for CFRP,

*

0.9 .9 f = C f = 0.90.95 × × 3000 = 2565 MPa

fu E u

Finding ε bi based on the self-weight of the beam only (experimentally tested beam),

300 × 400

w = 23.5kN/m

3

× = 2.82 kN/m

DL

6

10

2 2

wL × 2.82 × 2.5 DL n

M = = = 2.203 kN-m < M

DL cr

8 8

9

fI
y

0.62 30 ××

−

1.710 rgt

M = = = 29024614 N-mm = 29 kN-m

cr

( )

400 201.1

bot

( ) ( )

= ××+−×

300 400 200 n 1 265.550 × +−× n 1 398.2 × 350

y = 201.1mm

top

( ) ( )

300 ×+−× +−× 400 n 1 265.5 n 1 398.2

200000
4700

n = = 7.77

30

3

300 400

12

×

( )

2

( ) ( )

2

I = +×× 300 400 200 − +−× 201.1 n 1 265.5 × 201.150 −

gt

( ) ( )

2

1 398.2 201.1 350 1.710mm

9 4

+−× n × − = ×

( ) ( )

ε= ×− =

Mhy
EI

6

2.203 ×× −

×× = →

10 400 201.1

DL top

0.00001 negligible bi

9

4700 30 1.710 cgt

Assuming the beam loaded with dead load only during strengthening,

3000
400000

max *
ε= ε+ε= × 0.9 C 0.9 0.95 × + 0.00001 = 0.00642 fu E u bi

f
E

30
4700

c
c

ε= =× = 1.71 1.71 0.00199 ≈ 0.002 c

30

First, use the exact fifth-degree polynomial ignoring compression steel:

f
f

d
d

y

Q 1 =ρ = 0.0376

s

cf

M
fbd

d
d

u
cf

Q = φ +Ψ−= Q 0.06397

2 1 f

2

f

AQ = 3( (1 −Ψ− ε− εε )9 ) QQ

3

3

2max

1 f 2 c 2 cfu

( )

3 2

= 30.0376 × −× 0.15 90.06397 × 0.002 − × 30.06397 × 0.002 × 0.00642

1.861 10

− 8

=− ×