Design a round-tied column for axially loaded. PD=700kN and PL=500kN. Then check ACI-318 Code-2014 requirements for columns. fc'=28MPa and f_y=420MPa.

Steps to Solve

1. Calculate the factored axial load, $P_u$

Use the load combination formula from ACI code: $P_u = 1.2 P_D + 1.6 P_L$ where $P_D$ is the dead load and $P_L$ is the live load. $$P_u = 1.2 \times 700 , \text{kN} + 1.6 \times 500 , \text{kN} = 840 , \text{kN} + 800 , \text{kN} = 1640 , \text{kN}$$

2. Assume a reinforcement ratio, $\rho_g$

For columns, a typical reinforcement ratio ranges from 1% to 8%. Assume $\rho_g = 0.02$ (2%)

3. Calculate the required gross area, $A_g$

For tied columns, $\phi = 0.65$ and $\alpha = 0.80$ $$P_u = \phi \alpha [0.85 f_c' (A_g - A_{st}) + f_y A_{st}]$$ Where $A_{st} = \rho_g A_g$, substitute this into the equation: $$P_u = \phi \alpha [0.85 f_c' (A_g - \rho_g A_g) + f_y \rho_g A_g]$$ $$P_u = \phi \alpha A_g [0.85 f_c' (1 - \rho_g) + f_y \rho_g]$$ Rearrange and solve for $A_g$: $$A_g = \frac{P_u}{\phi \alpha [0.85 f_c' (1 - \rho_g) + f_y \rho_g]}$$ $$A_g = \frac{1640 \times 10^3 , \text{N}}{0.65 \times 0.8 [0.85 \times 28 , \text{MPa} \times (1 - 0.02) + 420 , \text{MPa} \times 0.02]}$$ $$A_g = \frac{1640 \times 10^3}{0.65 \times 0.8 [0.85 \times 28 \times 0.98 + 420 \times 0.02]} = \frac{1640 \times 10^3}{0.52 [23.8 + 8.4]} = \frac{1640 \times 10^3}{0.52 \times 32.2} = \frac{1640 \times 10^3}{16.744} \approx 97957 , \text{mm}^2$$

4. Determine the column diameter, $D$

Since $A_g = \pi (D/2)^2 = \frac{\pi}{4} D^2$, then $D = \sqrt{\frac{4 A_g}{\pi}}$ $$D = \sqrt{\frac{4 \times 97957}{\pi}} = \sqrt{\frac{391828}{\pi}} \approx \sqrt{124727} \approx 353 , \text{mm}$$ Round up to a practical value, say $D = 360 , \text{mm}$

5. Calculate the actual gross area, $A_g$

Using the rounded diameter, $A_g = \frac{\pi}{4} D^2 = \frac{\pi}{4} (360)^2 = \frac{\pi}{4} \times 129600 \approx 101788 , \text{mm}^2$

6. Calculate required steel area, $A_{st}$

$A_{st} = \rho_g A_g = 0.02 \times 101788 = 2035.76 , \text{mm}^2$

7. Select reinforcing bars

Use 8 bars of 20mm diameter ($8 \phi 20$) Area of one $\phi 20$ bar = $\pi (20/2)^2 = \pi (10)^2 = 314.16 , \text{mm}^2$ Total area of steel provided = $8 \times 314.16 = 2513.28 , \text{mm}^2$

8. Check the reinforcement ratio

Actual $\rho_g = \frac{A_{st}}{A_g} = \frac{2513.28}{101788} \approx 0.0247$ This is between 1% and 8%, so it is acceptable. ($0.01 < 0.0247 < 0.08$)

9. Design of lateral ties

Diameter of ties: Use $\phi 10$ ties (as the longitudinal bar size is $\phi 20$ which is smaller than $\phi 32$) Spacing of ties: The spacing should be the smallest of: a) 48 times the tie diameter: $48 \times 10 = 480 , \text{mm}$ b) 16 times the longitudinal bar diameter: $16 \times 20 = 320 , \text{mm}$ c) Least dimension of the column: $360 , \text{mm}$

Therefore, use $\phi 10$ ties @ $320 , \text{mm}$ spacing.