Flexural members - Reinforced Concrete with FRP Bars: Mechanics and Design - page 95

Iteration 2: c = 3.3 in.

ε c = 0.00176
α 1 β 1 = 0.643
c = 3.7 in.

Iteration 3: c = 3.7 in.

ε c = 0.00202
α 1 β 1 = 0.688
c = 3.5 in. : Convergence is achieved
β 1 = 0.759
α 1 = 0.906
M n = 3,151 kip-in. = 262.6 ft-kip

2. Approximate method

From the previous example it is obvious that the exact value
of c is not an important factor in the overall accuracy of the
solution. Therefore, when FRP rupture governs the failure, it is
allowed to assume c = c b conservatively , which results in

′

− ω

fb

2

M =ω 1 f bd

n f c

1.7

Example 4.5

Calculate the nominal flexural strength of the beam in Example 4.4,
using the approximate method.

ω f = 0.1082
ω fb = 0.1667 > ω f : Failure by FRP rupture
M n = 3,023 kip-in. = 251.9 ft-kip

Solution:

Example 4.6

Repeat Example 4.5 with f ′ c = 5.0 ksi.

ω f = 0.0866
ω fb = 0.1569 > ω f : Failure by FRP rupture
M n = 3,044 kip-in. = 253.6 ft-kip

Solution:

This example shows that, in the FRP rupture mode, the strength of
concrete has little effect on the flexural strength of the member.

Example 4.7

In previous examples, the assumed rupture stress of the FRP rein-
forcement is equal to the yielding stress of ordinary reinforcing steel