Use a dead load factor of $1\mathrm{.}25$ and live load factor of $1\mathrm{.}5.$ Use the Canadian Metric rebar sizes Design a square tied column $5\mathrm{.}5m$ tall with loads $LL=1,881kN$ and $DL=3,678kN$ using $F_y=$

$414$MPa steel and $f_c^{\text{'}}=35$MPa concrete. We will assume that the column has no intermediate

bracing and has an effective length factor $K=1\mathrm{.}0.$

Design load:

$1\mathrm{.}6(1,881kN)+1\mathrm{.}2(3,678kN)=7,423\mathrm{.}2kN$

Tied column design formula:

$P_{\text{m}\text{a}\text{x}}=R(0\mathrm{.}65)(0\mathrm{.}8)A_g[0\mathrm{.}85f_c^{\text{'}}(1-p_g)+F_y{p}_g]$

Assume that $R=1,p_g=0\mathrm{.}03.$

$7,423,200N=1\mathrm{.}0(0\mathrm{.}65)(0\mathrm{.}8)A_g[0\mathrm{.}85(35$MPa$)(1-0\mathrm{.}03)+4,141$MPa$(0\mathrm{.}03)]$

$A_g=\frac{7,423,200(N)}{1\mathrm{.}0(0\mathrm{.}65)(0\mathrm{.}8)[0\mathrm{.}85(35\text{M}\text{P}\text{a})(1-0\mathrm{.}03)+414\text{M}\text{P}\text{a}(0\mathrm{.}03)]}$

$=\frac{7,423,200(N)}{0\mathrm{.}52[29\mathrm{.}75-0\mathrm{.}892+12\mathrm{.}42\text{M}\text{P}\text{a}]}=345,839m{m}^2$

We have specified a square column; therefore:

$d=\sqrt[\phantom{2}]{345,839m{m}^2}=588mm=600mm$

$A_g=360,000m{m}^2$

Radius of gyration $r=0\mathrm{.}3(600mm)=180mm.$

Slenderness ratio $\frac{KL}{r}=\frac{(1\mathrm{.}0)5,500(mm)}{180(mm)}=30\mathrm{.}528$

$R$ must be calculated.

$R=1\mathrm{.}07-0\mathrm{.}008\frac{L}{r}=1\mathrm{.}07-0\mathrm{.}008\frac{5,500(mm)}{180(mm)}=0\mathrm{.}83$

Since $R$ is not as assumed in step $1,$ the actual required $p_g$ must be found by solving the col$-$

umn equation.

$P_{\text{m}\text{a}\text{x}}=R(0\mathrm{.}65)(0\mathrm{.}8)A_g(0\mathrm{.}85f_c^{\text{'}}(1-p_g)+F_y{p}_g)$

$7,423,200N=0\mathrm{.}83(0\mathrm{.}65)0\mathrm{.}8(360,000m{m}^2)[0\mathrm{.}85(35$MPa$)(1-p_g)+(14$MPa$)p_g]$

$=155,376m{m}^2[29\mathrm{.}75$MPa$(1-p_g)+(414$MPa$)p_g]$

$=155,376m{m}^2(29\mathrm{.}75$MPa$-(29\mathrm{.}75$MPa$)p_g+(414$MPa$)p_g)$

$=155,376m{m}^2(29\mathrm{.}75$MPa$+(384\mathrm{.}25$MPa$)p_g)$

$=4,622,436N+(59,703,228N)p_g$

$\frac{7,423,200(N)-4,622,436(N)}{59,703,228(N)}=p_g=0\mathrm{.}04690\mathrm{.}010\mathrm{.}08(okay)$

Determine the required steel area

Any of these selections would be appropriate. The most economical choice based on labor

would be the $1636M$ bars, which require less time to erect the system.

Select the ties. With $36M$ bars, the minimum tie diameter is $13M.$

Maximum tie spacing $=16d_b=16(36mm)=576mm$

Maximum tie spacing $=48d_t=48(13mm)=624mm$

Maximum tie spacing $=$ minimum column dimension $=600mm$

The first tie spacing value must be used, which can be rounded to $570mm.$ For lateral brac$-$

ing, longitudinal bars must comply with minimum distances from tie corners, as shown in the

illustration.

$36M$ bar dia. $=36mm$

Minimum spacing $=38mm$

$1\mathrm{.}5d_b=1\mathrm{.}5(36mm)=54mm>38mm$

Check bar spacing:

Cover $2(38mm)=76mm$

Five bars $(36mm)=180mm$

Min. spacing $4(54mm)=216mm$

$472mm<mn>6</mn><mn>0</mn><mn>0</mn><mi>m</mi><mi>m</mi><mtext></mtext><mi>(</mi>$okay$)$

The tie requirements indicate that no bar may be more than $150mm$ from a tie corner, so

we will need to insert a second tie at each location as shown to meet this requirement. A

"corner" is defined as an angle of $135$ or less.