Flexural Strengthening of Beams and Slabs - Strengthening Design of Reinforced Concrete with FRP - page 139

Multiply this equation by (400 − c ) 2 ,

( ) ( )( ) ( )

2

2 3

22,590 c 400 −− + c 18,900 c 263,410 400 − −= − c c 50 439,484 400 c

9,036,000 c

2

− 22,590 c

3

− 18,900 c

3

+ 118,534,500 c − 263,410 c

2

− 5, 268,200,000

10 2

− 7.031744 × 10 + 351,587,200 c − 439,484 c = 0

− 41,490 c

3

+ 8,333,106 c

2

+ 470,121,700 c − 7.558564 × 10

10

= 0

c = 85.912 mm

85.912
400

c ε= 0.00497 × = 0.00136 < 0.003 O.K.

− 85.912

350 −

−

85.912

ε= 0.00497 × = 0.00418 < 0.005 Tension steel is in the transition zone.

400 85.912

0.25 0.00418

− 0.0017

φ= 0.65 + = 0.838

0.005 − 0.0017

f
E

340
200,000

y
s

where ε= = = 0.0017

y

85.912 − 50

ε= 0.00497 × = 0.00057 < ε y

No yielding of compression steel s

400 − 85.912

f =ε= E 200,000 × 0.00057 = 114 MPa

s s s

1
3

0.00136
12

−

× 0.00199

β= = 0.358

0.00136
30.00199

1 −

×

( ) ( ) ( )

MAfd c =φ −β +φΨ A fd −β+φβ− cAf cd

u s y f ffdf ss

( )

M = 0.838 × 398 × 340 350 − 0.358 × 85.912 + 0.838 × 0.85 × 153 × 1988 u

( ) ( )

× 400 − 0.358 × 85.912 + ××× 0.838 265 114 0.358 × 85.912 − 50

= 115,713,284.7N-mm = 115.713 kN-m

M n = 126.69 kN-m

,exp

M
M

u

φ = = 0.913 > 0.838 Slightly unconservative, should be < 0.838.

implied

n ,exp

Using the statistically correlated Equation (5.86),

λ= ρ
ρ

f
f

d
d

0.001457 ×

×

1988 400

350

f d f

2.569

= × =

0.00379 340

sy

M
M

n

= 0.7815 × 2.56913 += M =× = 3 48.365 145.47 > M = 126.69

n n ,exp

n

Where M n = 48.365 kN − m from Example 5.11, the prediction of equation (5.96) is
slightly unconservative.