Chapter 4, Problem 4.7.3P

To determine

(a)

If the compression member is adequate to support the loads using LRFD.

Answer to Problem 4.7.3P

$\text{W}12\times 79\text{ is adequate}$

Explanation of Solution

Given information:

$\text{Dead load},DL=180\text{kips}$

$\text{Live load},LL=320\text{kips}$

$\text{Length of the column},L=28\text{ft}$

$\begin{array}{l}\text{W}12\times 79\\ K_x=1.0\\ K_y=1.0\\ b_f=12.1\text{in}\\ t_f=0.735\text{in}\\ h/t_w=20.7\end{array}$

$\text{Steel type}=\text{A572}$

$\text{Yield stress}=60\text{ksi}$

Concept used:

$\text{Slenderness ratio}=\frac{KL}{r}$

$F_e=\frac{{\pi }^{2}E}{{\left(KL/r\right)}^2}$

${\varphi }_c{P}_n={\varphi }_c\times F_{cr}\times A_g$

$P_u=1.2\left(DL\right)+1.6\left(LL\right)$

$\begin{array}{c}K\to \text{Effective length factor}\\ r\to \text{Radius of gyration}\\ F_e\to \text{Buckling stress}\\ F_{cr}\to \text{Critical stress}\end{array}$

$\begin{array}{c}E\to \text{Modulus of elasticity}\\ P_n\to \text{Nominal compressive strength}\\ A_g\to \text{Gross area}\\ {\varphi }_c\to \text{Resistance factor}=0.90\\ {\Omega }_c\to \text{Safety factor}=1.67\end{array}$

Calculation:

From AISC manual, the radius of gyration for the given material is 5.34 in. and 3.05 in.

$\begin{array}{c}\frac{K_{x}L}{r_x}=\frac{1.0\times 28\text{ft}\times \frac{\text{12}\text{in}}{\text{1}\text{ft}}}{5.34}\\ =62.92\end{array}$

$\begin{array}{c}\frac{K_{y}L}{r_y}=\frac{1.0\times 16\text{ft}\times \frac{\text{12}\text{in}}{\text{1}\text{ft}}}{3.05}\\ =62.95\end{array}$

Taking the higher value of slenderness ratio is 62.95.

$\begin{array}{c}{F}_e=\frac{{\pi }^{2}E}{{\left(KL/r\right)}^2}\\ =\frac{{\pi }^2\times 29,000}{{\left(62.95\right)}^2}\\ =72.23\text{ksi}\end{array}$

Check:

$\begin{array}{c}4.71\sqrt{\frac{E}{F_y}}=4.71\sqrt{\frac{29,000}{60}}\\ =103.55\end{array}$

$\begin{array}{c}\frac{K_{y}L}{r_y}<4.71\sqrt{\frac{E}{F_y}}\\ 62.95<103.55\end{array}$

Use AISC equation 3-2.

$\begin{array}{c}{F}_{cr}={0.658}^{\left(F_y/F_e\right)}{F}_y\\ ={0.658}^{\left(60/72.23\right)}\times 60\\ =42.38\text{ksi}\end{array}$

From AISC manual, the gross area for the given material is $23.2{\text{in}}^{\text{2}}$.

$\begin{array}{c}{\varphi }_c{P}_n={\varphi }_c\times F_{cr}\times A_g\\ =0.9\times 42.38\times 23.2\\ =884.89\text{kips}\end{array}$

Check for slender:

$\begin{array}{c}\lambda =\frac{b_f}{2t_f}\\ =\frac{12.1}{2\times 0.735}\\ =8.23\end{array}$

$\begin{array}{c}{\lambda }_r=0.56\sqrt{\frac{E}{F_y}}\\ =0.56\sqrt{\frac{29,000}{60}}\\ =12.31\end{array}$

$\begin{array}{c}\lambda <{\lambda }_r\\ 8.23<12.31\end{array}$

The flange is non-slender.

$\begin{array}{c}\lambda =\frac{h}{t_w}\\ =20.7\end{array}$

$\begin{array}{c}{\lambda }_r=1.49\sqrt{\frac{E}{F_y}}\\ =1.49\sqrt{\frac{29,000}{60}}\\ =32.76\end{array}$

$\begin{array}{c}\lambda <{\lambda }_r\\ 20.7<32.76\end{array}$

The web is non-slender.

$\begin{array}{c}{P}_u=1.2\left(DL\right)+1.6\left(LL\right)\\ =1.2\left(180\right)+1.6\left(320\right)\\ =728\text{kips}\end{array}$

$\begin{array}{c}{\varphi }_c{P}_n>P_u\\ 884.89\text{kips}>728\text{kips}\end{array}$

Hence, the section is safe.

Conclusion:

The section $\text{W}12\times 79$ is adequate.

To determine

(b)

The compression member is adequate to support the loads using ASD.

Answer to Problem 4.7.3P

$\text{W}12\times 79\text{ is adequate}$

Explanation of Solution

Given information:

$\text{Dead load},DL=180\text{kips}$

$\text{Live load},LL=320\text{kips}$

$\text{Length of the column},L=28\text{ft}$

$\begin{array}{l}\text{W}12\times 79\\ K_x=1.0\\ K_y=1.0\\ b_f=12.1\text{in}\\ t_f=0.735\text{in}\\ h/t_w=20.7\end{array}$

$\text{Steel type}=\text{A572}$

$\text{Yield stress}=60\text{ksi}$

Concept used:

$\text{Slenderness ratio}=\frac{KL}{r}$

$F_e=\frac{{\pi }^{2}E}{{\left(KL/r\right)}^2}$

$\frac{P_n}{{\Omega }_c}=\frac{F_{cr}\times A_g}{{\Omega }_c}$

$P_a=DL+LL$

$\begin{array}{c}K\to \text{Effective length factor}\\ r\to \text{Radius of gyration}\\ F_e\to \text{Buckling stress}\\ F_{cr}\to \text{Critical stress}\end{array}$

$\begin{array}{c}E\to \text{Modulus of elasticity}\\ P_n\to \text{Nominal compressive strength}\\ A_g\to \text{Gross area}\\ {\varphi }_c\to \text{Resistance factor}=0.90\\ {\Omega }_c\to \text{Safety factor}=1.67\end{array}$

Calculation:

From AISC manual, the radius of gyration for the given material is 5.34 in. and 3.05 in.

$\begin{array}{c}\frac{K_{x}L}{r_x}=\frac{1.0\times 28\text{ft}\times \frac{\text{12}\text{in}}{\text{1}\text{ft}}}{5.34}\\ =62.92\end{array}$

$\begin{array}{c}\frac{K_{y}L}{r_y}=\frac{1.0\times 16\text{ft}\times \frac{\text{12}\text{in}}{\text{1}\text{ft}}}{3.05}\\ =62.95\end{array}$

Taking the higher value of slenderness ratio is 62.95.

$\begin{array}{c}{F}_e=\frac{{\pi }^{2}E}{{\left(KL/r\right)}^2}\\ =\frac{{\pi }^2\times 29,000}{{\left(62.95\right)}^2}\\ =72.23\text{ksi}\end{array}$

Check:

$\begin{array}{c}4.71\sqrt{\frac{E}{F_y}}=4.71\sqrt{\frac{29,000}{60}}\\ =103.55\end{array}$

$\begin{array}{c}\frac{K_{y}L}{r_y}<4.71\sqrt{\frac{E}{F_y}}\\ 62.95<103.55\end{array}$

Use AISC equation 3-2.

$\begin{array}{c}{F}_{cr}={0.658}^{\left(F_y/F_e\right)}{F}_y\\ ={0.658}^{\left(60/72.23\right)}\times 60\\ =42.38\text{ksi}\end{array}$

From AISC manual, the gross area for the given material is $23.2{\text{in}}^{\text{2}}$.

$\begin{array}{c}\frac{P_n}{{\Omega }_c}=\frac{F_{cr}\times A_g}{{\Omega }_c}\\ =\frac{42.38\times 23.2}{1.67}\\ =588.75\text{kips}\end{array}$

Check for slender

$\begin{array}{c}\lambda =\frac{b_f}{2t_f}\\ =\frac{12.1}{2\times 0.735}\\ =8.23\end{array}$

$\begin{array}{c}{\lambda }_r=0.56\sqrt{\frac{E}{F_y}}\\ =0.56\sqrt{\frac{29,000}{60}}\\ =12.31\end{array}$

$\begin{array}{c}\lambda <{\lambda }_r\\ 8.23<12.31\end{array}$

The flange is non-slender.

$\begin{array}{c}\lambda =\frac{h}{t_w}\\ =20.7\end{array}$

$\begin{array}{c}{\lambda }_r=1.49\sqrt{\frac{E}{F_y}}\\ =1.49\sqrt{\frac{29,000}{60}}\\ =32.76\end{array}$

$\begin{array}{c}\lambda <{\lambda }_r\\ 20.7<32.76\end{array}$

The web is non-slender.

$\begin{array}{c}{P}_a=DL+LL\\ =180+320\\ =500\text{kips}\end{array}$

$\begin{array}{c}{\varphi }_c{P}_n>P_u\\ 500\text{kips}>588.75\text{kips}\end{array}$

Hence, the section is safe.

Conclusion:

The section $\text{W}12\times 79$ is adequate.