Flexural members - Reinforced Concrete with FRP Bars: Mechanics and Design - page 136

Shear reinforcement
contribution, V f :

A fv = #3@10 in. = 0.22 in. 2
s = 10.0 in.
f fv = 0.004 E fv ≤ f fb
f fb = (0.05 r b / d b + 0.3) f fu ≤ f fu
f fb = 0.45 f fu = 27.0 ksi
f fv = 24.0 ksi
V f = A fv f fv d / s = 11.6 kip

The nominal shear
strength:

V n = V c + V s = 33.2 kip

The design shear strength: ϕ V n = 0.75 V n = 24.9 kip

Example 4.14

Calculate the spacing of the stirrups for the beam in Example 4.13 so
that a design shear strength of 30 kip can be provided.

Solution:
ϕ V n = 0.75 V n = 30.0 kip
V n = V c + V f = 40.0 kip
V c = 21.6 kip
V f = 18.4 kip
V f = A fv f fv d / s = 18.4 kip
s = 6.3 in.
Use: #3@6 in.
V n = 41.0 kip
ϕ V n = 0.75 V n = 30.8 kip

Example 4.15
Compare the shear strength of the beams in Examples 4.13 and 4.14
with those of two similar beams with grade 60 steel flexural and shear
reinforcement.

Solution: Example FRP RC Steel RC

4.13 V c = 21.6 kip V c = 44.5 kip

V f = 11.6 kip V s = 29.0 kip

V n = 33.2 kip V n = 73.5 kip

ϕ V n = 24.9 kip ϕ V n = 55.1 kip

4.14 V c = 21.6 kip V c = 44.5 kip

V f = 19.4 kip V s = 48.4 kip

V n = 41.0 kip V n = 92.9 kip

ϕ V n = 30.8 kip ϕ V n = 69.7 kip