For f_{c}′ grerater than
4000 psi the value of β_{1} is calculated as given
below;
β_{1} = 0.85  0.05 {( f_{c}′
4000)/1000} = 0.8
Assume that the steel has already yielded when the strength is
reached (strain in concrete is 0.003).
Given that tension steel consists of 4 bars of #9 (dia 1.128 in.) .
Area of one bar of #9 = 1
in^{2}.
A_{s} = 4#9
bars. = 4 (1.00) = 4
in^{2}.
The internal forces acting on the section shown in figure
91(c) are calculated as given below;
C_{c} = 0.85 f_{c}′ ba = 0.85
(5) (15) a = 63.75 a
T = A_{s} f_{y}
= (4) (50) = 200 kips
Applying static equilibrium, we get C_{c} = T;
therefore 63.75 a = 200
Depth of Whitney stress block, a = 3.14
in.
Depth of neutral axis x = a / β_{1}
= 3.14/0.8 = 3.93 in
C_{c} = 63.75 (3.14) = 200.18 kips
