Flexural Strengthening of Beams and Slabs - Strengthening Design of Reinforced Concrete with FRP - page 134

where

A
bd

307.9
200

s
ρ= = = 0.009445

s

× 163

188.75 mm 2

A =ρ×= bd

f f

A
t

f
f

b == ≈ 145.2mm 150 mm Actual f

Use 3 plates of 50 × 1.3 mm at the soffit of the beam, as done experimentally.
Confirm that the failure mode is cover delamination and not concrete crushing.

f
f

c
d

f
f

c
fe

y
fe

ρ=α− ρ Ignoring compression steel

f s

α= ε×
−×ε

c
dc

− ε×
−ε

c
dc

2 2

2

fcd fcd

( )

( )

2

3

f c

f c

22

0.00311 c 0.00311

3 200

c 33

519.4

c

0.00579 = − × ×

( )

( )

2

200 −× c 0.00209

2

163

−× c 0.00209

540
519.4

− 0.009445 ×

where

f c

ε= 1.71 4,700 = 0.00209

c

f

c

f 167,000 0.00311 519.4MPa

= × = fe

c c

2

0.0156 = 1.488 × − 0.738 × × 0.0003898 c

( )

2

200 − c

200 − c

Multiply the previous equation by (200 − c ) 2 :

2 2 3

0.0156(40000 − 400 cc + )0.00058 = c (200 − c )0.000288 − c

3 2

0.000868 c − 0.1004 c − += 6.24 c 624 0

c

3

− 115.67 c

2

− 7188.94 c + 718894 = 0

c = 69.07 mm
0.00311 × 69.07

ε= = 0.00164 < 0.003 OK .. cf

( )

200 − 69.07

( )

0.00311 × 163 − c

540
200000

ε= = 0.00223 < = 0.0027 No yielding of tensionsteel s

( ) 200 − c