1. Axial Force ($P_n$)
$$ P_n = 0.85 f’_c (A_g – A_{st}) + \sum (A_{si} f_{si}) $$

2. Moment ($M_n$)
$$ M_n = \sum A_{si} f_{si} (d_i – \frac{h}{2}) + C_c (\frac{h}{2} – \frac{a}{2}) $$

Where:
$f_{si}$ = Stress in steel layer $i$ (Elastic-Perfectly Plastic based on strain).
$C_c = 0.85 f’_c a b$ (Equivalent Rectangular Stress Block)

ACI 318-19 Ref: 6.6.4.4.4 & 6.6.4.4.2 $$ (EI)_{eff} = \frac{0.2 E_c I_g + E_s I_{se}}{1 + \beta_{dns}} \quad \text{or} \quad \frac{0.4 E_c I_g}{1 + \beta_{dns}} $$ $$ P_{crit} = \frac{\pi^2 (EI)_{eff}}{(k L_u)^2} $$ $$ \text{Check: } \frac{k L_u}{r} \le 34 – 12(M_1/M_2) \le 40 \text{ (Neglect Slenderness)} $$

ACI 318-19 Ref: 6.6.4.5.2 & 6.6.4.5.3 $$ C_m = 0.6 + 0.4 \frac{M_1}{M_2} \ge 0.4 \quad (\text{Sing. Curv. } +, \text{Dbl. Curv. } -) $$ $$ \delta_{ns} = \frac{C_m}{1 – \frac{P_u}{0.75 P_{crit}}} \ge 1.0 $$ $$ M_c = \delta_{ns} M_u $$

Capacity & Biaxial Interaction $$ P_n = 0.85 f’_c (A_g – A_{st}) + \sum (A_{si} f_{si}) \quad \text{(Strain Compatibility)} $$ $$ \text{If } P_u \ge 0.1 f’_c A_g: \quad \frac{1}{\phi P_n} \approx \frac{1}{\phi P_{nx}} + \frac{1}{\phi P_{ny}} – \frac{1}{\phi P_o} \quad \text{(Bresler)} $$ $$ \text{If } P_u < 0.1 f'_c A_g: \quad \left(\frac{M_{ux}}{\phi M_{nx}}\right)^{1.15} + \left(\frac{M_{uy}}{\phi M_{ny}}\right)^{1.15} \le 1.0 \quad \text{(PCA Load Contour)} $$

ACI 318-19 Ref: 22.5.5.1 & 22.5.8.5.3 $$ V_c = 0.17 \lambda \sqrt{f’_c} b_w d $$ $$ V_s = \frac{A_v f_{yt} d}{s} \quad (\text{where } A_v = 2 \times A_{tie}) $$ $$ \phi V_n = \phi (V_c + V_s) \quad (\phi = 0.75) $$