Flexural Strengthening of Beams and Slabs - Strengthening Design of Reinforced Concrete with FRP - page 143

Compression steel is in tension (not yielded),

f s = 29,000 × 0.00168 = 48.72 ksi

M

( ) ( ) ( )

u

φ ==×= −β +Ψ

M 24 12 A fd c fd −β + cAfd − β c n s y f ffdf ss

( ) ( )

288 = 0.22 × ×− × 60 60.361.069 + ××− 0.85 A f 250 80.361.069 ×

( )

+× ×− × 0.22 48.72 20.361.069

196.57

0.121in.

2

.14in.

2

A f = ××− = < Unconservative

( )

0.85 250 80.361.069 ×

This is the second example in which the FRP debonding model shows uncon-
servative results.

Chapter Problems

Problem 5.1
A library building has a simply supported rectangular concrete beam reinforced
with four No. 6 bars in tension, two No. 3 bars in compression and No. 3 stirrups at
6 in. (152 mm) on center in shear. The details of the beam are shown in Figure 5.P.1.
As part of the library upgrade, the beam is subjected to a 40% increase in live load,
as shown in Figure 5.P.2. Determine the CFRP area that needs to be bonded in flex-
ure. Use VWRAP CFRP sheets with a modulus of 33,000 ksi (227,527 MPa) and a
tensile strength of 550 ksi (3,792 MPa) based on the net fiber area that has a net fiber
thickness of 0.0065 in. (0.165 mm).

2#3

#3 @ 6" c/c

20"

4#6

1.5"

12"

FIGURE 5.P.1

W DL + W LL

20 ft

f c = 4 ksi
f y = 60 ksi

FIGURE 5.P.2