Flexural Strengthening of Beams and Slabs - Strengthening Design of Reinforced Concrete with FRP - page 95

Solution:

20.110.22in. 2 A s =× =

f
E

83.5
29000

y
s

ε= = = 0.00288

y

3
8

1
2

3
8

d = ++×= 1 1.563

Assume ductile crushing failure mode:

0.85 fb cAf 1 β+ =+ Af Af c s s s y f fe

Assume yielding of compression steel:

16.36 − c

0.8550.8 ×× ×+× 10 c 0.22 83.51.76 = × 83.50.286 + × 99

c

2

34 c + 18.37 c − 146.96 c − 463.22 + 28.31 c = 0

2

34 c − 100.28 c − 463.22 = 0

c = 5.45 in

0.003
5.45

( )

ε= × 5.45 − 1.563 = 0.00214 < 0.00288 Compression steel does

s

not yield

Redo the force equilibrium accordingly:

c − 1.563 16.36 − c

34 0.22 87 146.96 28.31

c + × = +

c c

2

34 c + 19.14 c − 29.92 = 146.96 c + 463.22 − 28.31 c

2

34 c − 99.51 c − 493.14 = 0

c = 5.54 in

0.003 16.36 5.54
5.54

−

ε= = 0.00586 <ε= 0.00896 < 0.9 C ε = 0.01197 fe fd e u

f = 33,000 × 0.00586 = 193.38 ksi

fe

0.003 5.54 1.563
5.54

−

ε= = 0.002154

s

f = 62.45 ksi s

− β

c

− β

c β c

1 1 1

MAfd = +Ψ Af d + Af − d n s y f f fe f ss

2 2 2

0.85.54
2

0.85.54
2

× ×

M n = 1.76 × 83.5 × 16.25 − + 0.85 × 0.286 × 193.38 × 16.36 −

0.85.54
2

×

+ 0.22 × 62.45 × − 1.563 = 2736.33 k-in = 228.03 k-ft