Design of a one-way slab - Reinforced Concrete with FRP Bars: Mechanics and Design - page 214

First guess: x 02 := 0.1 df2

cu 2 := root (f 2 (x 02 ), x 02 )

The neutral axis depth is

cu 2 = 0.86·in.

The nominal bending moment capacity can be computed as follows:

( )

⋅ ε

c,y

c2 u2

2" ⋅σ

c

ε

c u2

∫

c0

( ) ( )

M:b =⋅ y ⋅ psidyTc + ⋅ dc −=⋅⋅ 7ftkip

n2 u2 u2

f2 f2

2

( )

ε c,y

0

c2 u2

1 +

ε

c0

The strain distribution over the cross section is shown next.

ε fu = 0.014 ε c2 (c u2 , c u2 ) = 0.00197

d f2 = 7·in

A f

ε f2 (c u2 ) = 0.014 ε cu = 0.003

The ϕ-factor is calculated according to Jawahery and Nanni [1]:

( )

ε c

f2 u2

0.65if 1.15 − ≤ 0.65

2 ε

fu

( )

ε c

f2 u2

φ : = 0.75if 1.15 − ≥ 0.65

b2

2 ε

fu

( )

ε c

f2 u2

1.15 − otherwise

2 ε

fu

φ = 0.65

b2