Background Knowledge - Strengthening Design of Reinforced Concrete with FRP - page 20

Assume the tensile steel has yielded at ultimate capacity

Af
fb

1.76 ×

×× =

60 sy

a 3.11in.

= =

0.85 0.85 410

c

a

β= 0.85 c = β = 3.65 in.

1

1

c
d

3.65
15.625

= = 0.234 < 0.375 tensioncontrolledfailure φ = 0.9

t

Or determine the actual strain in tension steel using strain compatibility:

ε
−

0.003

s

= ε = 0.00984 > 0.005 φ= 0.9 s

dc c

t

ε= 0.00984 > 0.004 O.K.

s

Computing the section moment capacity:

MAfd a u

3.11
2

=φ −=× 0.91.7660 × × 15.625 − = 1337.2k-in. = 111.43 k-ft s y

2

3 f 200

c

A = bd ≥ bd

s min w w

f f y y

3 f 200

c

= 0.00316 < = 0.00333 thelattercontrols

f f

y y

2 2

A = 0.00333 × 10 × 15.625 = 0.52 in. < .76in. O.K.

s min

Example 2.2: Design

For the reinforced concrete beam section shown in Figure 2.12, design the doubly
reinforced section to resist a moment capacity of M u = 220 k-ft, knowing that the
primary steel is composed of #7 bars and the compression steel is composed of #4
bars. Assume the shear stirrup size is #4 bar.

f
f

=
=

4ksi
60 ksi

c

y

A s

#4

18"

Stirrups

1.5"

A s

10"

FIGURE 2.12 Cross-section details for Example 2.2.