Flexural Strengthening of Beams and Slabs - Strengthening Design of Reinforced Concrete with FRP - page 88

Assume ductile crushing failure mode:

Af + Af

fb

sy f fe

c = Equation (10.12) of ACI 440.2R-08

0.85 β

c 1

Using Equation (2.17):

5000
1000

β= − 1.05 0.05 × = 0.8

ε= bi 0 The beam was not loaded during strengthening (Rasheed et al. 2010)

dc
c

−

f

ε= 0.003 −ε Equation (10.3) of ACI 440.2R-08

fe bi

0.003 16.36

− c

=

c

0.003 16.36

− c

99 16.36

− c

f =ε= E 33,000 × =

fe f fe

c c

Force equilibrium:

16.39 − c

0.8550.8 ×× ×=× 10 c 1.76 83.50.286 + × 99

c

Multiply the equation by c :

2

34 c − 146.96 c − 463.22 + 28.31 c = 0

2

34 c − 118.65 c − 463.22 = 0

2

118.65 + 118.65 434(463.22)

234

− ××−
×

c = = 5.83 in

0.003 16.36 5.83
5.83

−

ε= = 0.00542 ≤ ε

fe fd

f
nE t

5000
23310 .0065

c
ff

ε= 0.083 = 0.083 = 0.00896 ≤ 0.9 C ε= 0.01197

fd e u

6

×× ×

dc
c

−

0.003 16.25 5.83
5.83

−

ε= 0.003 = = 0.00536 > 0.005 φ= 0.9

s

Ductile crushing failure confirmed

f = 33,000 × 0.00542 = 178.86 ksi

fe

− β

c

− β

c

1 1

MAfd = +Ψ Af d n sy f f fe f

2 2

0.85.83
2

×

M = 1.76 × 83.5 × 16.25 − n

0.85.83
2

×

+ 0.85 × 0.286 × 178.86 × 16.36 −

M = 2655.34k-in. = 221.30 k-ft

n