Members subjected to combined axial load and bending moment - Reinforced Concrete with FRP Bars: Mechanics and Design - page 181

5.9.6 Example—Shear strength for circular columns
Example 5.4

Repeat Example 5.2 for a circular column with the same concrete area
( D = 27 in.) and reinforcement with no P u effect:

Concrete: f ´ c = 4.0 ksi

E c (ksi) = 57√ f ´ c (psi) = 3600 ksi

Longitudinal reinf.: f ffu = f fd = 60 ksi

E f = 6000 ksi
A f = 12#8 = 9.42 in. 2
A f 1 = A f 2 = 3#8 = 2.36 in. 2

Shear
reinforcement:

f ffu = 60 ksi

E ffv = 6000 ksi
A ffv = #3@8 in.

Size: D = 27.0 in. Diameter of the column

c c = 2.5 in. Concrete cover to the center of #8

Solution:
Concrete contribution, V c :
d = 0.8 D = 21.6 in.
b = D = 27 in.
ρ f 1 = ρ f 2 = A f 1 /( bd ) = 0.00405
γ = ( D-c c )/ D = 0.907

Assuming k = 0.2

ρ f = 0.00766
n f = E f / E c = (6000 ksi)/(3600 ksi) = 1.67
n f ρ f = 0.01280
And k can be calculated according to Equation (5.56) as:
k = 0.148
Assuming k = 0.148 and using Equation (5.58):
ρ f = 0.00789
n f ρ f = 0.01317
k = 0.150
c = k c d = 3.24 in.
V c = 27.7 kip
Shear reinforcement contribution, V f :
A ffv = #3@8 in. = 0.22 in. 2
s = 8.0 in.
f ffv = 0.004 E ffv ≤ f ffb
f ffb = (0.05 r b / d b + 0.3) f ffu ≤ f ffu

Continued