Flexural Strengthening of Beams and Slabs - Strengthening Design of Reinforced Concrete with FRP - page 86

f =ε= E 5360 × 0.01116 = 59.82 ksi

fe f e

f
f

f
f

a
d

5
59.82

3.9
21.5

60
59.82

c y

0.01163 0.001223

ρ= 0.85 −ρ = 0.85 × × − × =

f s

fe fe

2

A = 0.001223 × 12 × 21.50.316 in =

f

A
t

0.316
0.04

f
f

b == = 7.9in. Use8" with one layer of FRP f

5000 psi
15,360,000 psi

ε= 0.083 = 0.0127 >ε= 0.01116

( ) (

fd

)

fe

0.04

No Debonding O.K.

ACI 440.2R-08used2layersofFRP of full beam width ( A .96in)

2

= f

Example 5.2: Analysis

Use the r e sults of Example 5.1 to check the moment capacity of the strengthened
section M = 294.4 k − ft = 399 kN − m .

u

Solution:

Using force equilibrium,

0.003 d

' f

0.85 fb cAf β= +ε =+ − AE f AE 0.003 − ε

c 1 s y f f e s y f f bi

c

0.072

40.8360 c =× + 0.32 × 5360 × − 0.00361

c

40.8 c

2

− 173.808 c − 123.49 = 0

2

173.808 ± 173.808 + × 4 40.8 × 123.49

c = = 4.88 in (positiveroot)

2 × 40.8

a = 3.904 , ε = 0.011144 f = 59.73 ksi fe f

3.904
2

3.904
2

M =×× 0.9360 21.5 − + 0.85 × 0.32 × 59.73 × 24 −

u

= 3489.16 k-in = 290.76 k-ft ≈ 294.4k-ft

Whythe small difference? Becauseofrounding theconstantsofEquation(5.19)

Check thesteel strain:

0.003
4.88

( )

ε= 21.54.88 .0102 − = > 0.005 φ = 0.9. . OK s