Flexural members - Reinforced Concrete with FRP Bars: Mechanics and Design - page 96

( f fu = f y = 60 ksi). Compare the nominal strength of the beams in previ-
ous examples with their steel reinforced counterparts.

Solution: Example FRP RC Steel RC *

1: Concrete
crushing

M n = 428.8 ft-kip M n = 487.6 ft-kip

2: Concrete
crushing

M n = 477.6 ft-kip M n = 501.9 ft-kip

3: Balanced
failure

M nb = 388.2 ft-kip M nb = 826.1

ft-kip **

4: FRP rupture M n = 262.6 ft-kip M n = 261.6 ft-kip

5: FRP rupture M n = 251.9 ft-kip M n = 261.6 ft-kip

6: FRP rupture M n = 253.6 ft-kip M n = 265.2 ft-kip

* Note that for steel RC beams in all cases M n ≤ M nb and therefore all
beams fail in the same mode of steel yielding followed by concrete
crushing.
** M nb for steel RC has a reinforcement ratio 2.6 times greater than
that of GFRP RC.
From this comparison, it also appears that FRP reinforcement is at its
most effective when the failure is governed by FRP rupture. However,
for reasons of safety and serviceability, which are elaborated in their
respective chapters, this is not always the most desirable mode of failure.
Case b—calculation of design parameters of a rectangular flexural member
(b1) Applied moment and dimensions are known, reinforcement area is
unknown:
This is probably the most common case, as in order to perform the
structural analysis and obtain the internal moments, it is normally
required to assume the dimensions of the members. Since the reinforce-
ment area and subsequently the failure mode are not known before-
hand, the formulae for Case a need to be altered as

M
fbd

n
c

m =

0.85 ′

2

( )

ω= 0.85 112 −− m

f

0.85 β

1

− 1

ω

f

f =

e

ω

f

: if f ≤ 1

f

ω=

f

0.85 m

: if f > 1

− ω

fb

1

1.7