Flexural Strengthening of Beams and Slabs - Strengthening Design of Reinforced Concrete with FRP - page 133

Et
Cf

167,000 1.3
23.56

×

ff
ctm

l = = = 174.62 mm

b ,max

×

2

l = 550 mml > b b ,max

Since b f is unknown a priori, assume k b = 1.0 in Equation (5.91).

N =α C kkbEtf

fa ,max 1 cb ffctm

N fa =×××× 0.90.641.0 1.0 200 167,000 ×× = 1.33.56 101,276.2N = 101.28 kN

,max

N
AE

101.28

fa
ff

,max

ε= =

×× = →

0.00311 by estimating A fcd f

150 1.3 167

This is an obvious disadvantage of this model in solving design problems.

33

ε= 0.41 = 0.00505

fd

1 × 167,000 × 1.3

2, 906
167,000

0.9

C

ε= × 0.9 0.95 × = 0.0149

fu

ε controls

fcd

Unstrengthened beam capacity:

0.85 fb cAf β+ = Af

c 1 s s s y

+ × π ×× −=× π ××

600

( )

2 2

0.85 ×× × 33 200 0.826 c 2 14 c 37 2 14 540

4 c 4

2

4,633.86 c + 184,725.65 c − 6,834,848.98 − 166, 253.10 c =

c

2

+ 3.986 c − 1474.98 = 0

3.986 3.986

2

4 1474.98

c = −

+ +×

= 36.46 mm

2

c ⊕ Ignore the effect of compression steel.

− β

c 0.826 × 36.46

1

MAfd = = 307.9 × 540 × 163 −

n s y

2 2

= 24,597,727.9N-mm = 24.6kN-m

70
2

M n =×= 0.7 24.5kN-mO.K. ,exp

M
M

38.5
24.6

n

= = 1.565

n

M
M

f
f

d
d

ρ
ρ

n
n

ff
sy

f

= 0.7815 += 11.565

λ= ρ
ρ

f
f

d
d

0.565 × 0.009445 × 540 × 163

f fe
sy

f

ρ= = 0.00579

f

0.7815 × 167,000 × 0.00311 × 200