8.7 Calculate the axial compression capacity of the reinforced concrete columns for the following cases: transverse reinforcement: square ties 0 P = 0.80 Ø [0.85 f'Ag + (fy − 0.85 fé)Ast] -0.65 transverse reinforcement: round spirals OP=0.850 [0.85 fAg+ (fy - 0.85 fe) Ast] $=0.75 a) Square column with tied transverse reinforcement: 24° 8814 vert, bars 0.00 Vert. bars b) Round column with spiral reinforcement: 200 Column Section 14-39 Vert. P5 spral 24° c) Elongated round column with spiral reinforcement: Column Section 2¹-6°Ø Column Section fc' = 4 ksi, (concrete compressive strength) fy = 60 ksi (steel yield strength) Ag - gross concrete area Ast = area of vertical steel

Answer 

(a) Calculation of axial compression capacity of the square column

To calculate the axial compression capacity with square ties we are given

$\varphi .P_n = \varphi *0.80(0.85*f\text{'}c*A_g+(f_y-0.85*f\text{'}c)*A_{st}))$

$\varphi$= strength reduction factor

$\varphi$= 0.65

f'c = compressive strength of concrete

f'c = 4 ksi

A_g = gross section area

A_g = 24*24

A_g = 576 in² 

f_y = yield strength of steel

f_y = 60 ksi

A_st = Area of steel bars

A_st =18 in²     (from areas of group of standard bars for 8-#14)

Now substitute the values to get '$\varphi .P_n$'

$\varphi .P_n = \varphi *0.80(0.85*f\text{'}c*A_g+(f_y-0.85*f\text{'}c)*A_{st}))$

$\varphi .P_n = 0.65*0.80(0.85*4*576+(60-0.85*4)*18))$

$\varphi .P_n$ = 1548.144 kips

" axial compression capacity of the square column = 1548.144 kips "