Chapter 4, Problem 4.6.9P

To determine

(a)

A $\text{A992}$ W shape with a nominal depth of $\text{21 inches}$ using LRFD.

Answer to Problem 4.6.9P

$\text{W21}\times \text{62}$

Explanation of Solution

Given information:

$\text{D=90}$

$\text{L=260}$

Calculation:

$\begin{array}{l}{P}_u=1.2D+1.6L\\ P_u=1.2\times 90+1.6\times 260\\ P_u=524\text{kips}\end{array}$

Assume $F_{cr}=25\text{ksi}$

$\begin{array}{l}{A}_g>\frac{P_u}{{\varphi }_c{F}_{cr}}=\frac{524}{0.90\times 25}\\ A_g>\frac{P_u}{{\varphi }_c{F}_{cr}}=23.29\text{i}{\text{n}}^2\end{array}$

Try the web size $\text{W21}\times \text{93}$ (non-slender shape)

$A_g=27.3\text{i}{\text{n}}^2,r_y=1.84\text{in}$

Effective slenderness ratio is given by

$\begin{array}{l}\frac{kL}{r_y}=\frac{0.65\times \left(15.33\times 12\right)}{1.84}\\ \frac{kL}{r_y}=64.99<200\end{array}$

$\begin{array}{l}{F}_e=\frac{{\pi }^{2}E}{\left(\raisebox{1ex}{$KL$}\!\left/ \!\raisebox{-1ex}{$r^2$}\right.\right)}\\ F_e=\frac{{\pi }^2\times 29,000}{{64.99}^2}\\ F_e=67.76\text{ksi}\end{array}$

Check for slenderness ratio

$\begin{array}{l}4.71\sqrt{\frac{E}{F_y}}=4.71\sqrt{\frac{29,000}{50}}\\ 4.71\sqrt{\frac{E}{F_y}}=113.4>64.99\end{array}$

$\begin{array}{l}{F}_{cr}={0.658}^{{}^{\frac{F_y}{F_e}}}\times 50\\ F_{cr}={0.658}^{{}^{\frac{50}{67.76}}}\times 50\\ F_{cr}=36.71\text{ksi}\end{array}$

Nominal compressive strength is given by

$\begin{array}{l}{\text{P}}_{\text{n}}\text{=}{\text{F}}_{\text{cr}}{\text{A}}_{\text{g}}\\ {\text{P}}_{\text{n}}\text{=36}\text{.21}\left(\text{27}\text{.3}\right)\\ {\text{P}}_{\text{n}}\text{=1002kips}\\ {\varphi }_{\text{c}}{\text{P}}_{\text{n}}\text{=0}\text{.90×724}\text{.2}\\ {\varphi }_{\text{c}}{\text{P}}_{\text{n}}\text{=652kips>524kips(ok)}\end{array}$

Check for web buckling from dimensions and properties table in the manual the width thickness ratio of the web is given by

$\frac{h}{t_w}\text{=43}\text{.6}$

From the AISC section we have the equation

$\begin{array}{l}\frac{b}{t}\ge 1.49\sqrt{\frac{E}{f}}=1.49\sqrt{\frac{29,000}{36.21}}\\ \frac{b}{t}\ge 1.49\sqrt{\frac{E}{f}}=42.17\end{array}$

Since $\frac{h}{t_w}\text{>}1.49\sqrt{\frac{E}{f}}$,

Local buckling should be checked

Width of the web is given by

$\begin{array}{l}\text{b=d-2k}\\ \text{b=d-2×1}\text{.19}\\ \text{b=18}\text{.72in}\end{array}$

Effective width is given by

$\begin{array}{l}{b}_e\text{=1}\text{.92t}\sqrt{\frac{E}{f}}\left[1-\frac{0.34}{43.6}\sqrt{\frac{E}{f}}\right]\le b\\ b_e\text{=1}\text{.92}\times \text{0}\text{.430}\sqrt{\frac{29,000}{36.21}}\left[1-\frac{0.34}{43.6}\sqrt{\frac{29,000}{36.21}}\right]\le 18.72\\ b_e\text{=}18.21\le 18.72\end{array}$

Calculate effective area

$\begin{array}{l}{\text{A}}_{\text{eff}}\text{=A-}{\text{t}}_{\text{w}}\left(\text{b-}{\text{b}}_{\text{e}}\right)\\ {\text{A}}_{\text{eff}}\text{=20-0}\text{.430}\left(\text{18}\text{.72-18}\text{.21}\right)\\ {\text{A}}_{\text{eff}}\text{=19}\text{.78i}{\text{n}}^{\text{2}}\end{array}$

Critical stress is given by

$\begin{array}{l}4.71\sqrt{\frac{E}{Q{F}_y}}=4.71\sqrt{\frac{29,000}{\left(\frac{19.78}{20}\right)50}}\\ 4.71\sqrt{\frac{E}{Q{F}_y}}=114.1>66.43\end{array}$

Calculate the critical buckling stress as below

$\begin{array}{l}{F}_{cr}=Q\left({0.658}^{{}^{\frac{Q{F}_y}{F_e}}}\times 50\right)\\ F_{cr}=0.989\left({0.658}^{{}^{\frac{0.989\times 50}{64.86}}}\times 50\right)\\ F_{cr}=35.94\text{ksi}\end{array}$

Calculate nominal compressive strength of column

$\begin{array}{l}{P}_n=F_{cr}{A}_g\\ P_n=35.94\times 20\\ P_n=718.8\text{kips(ok)}\end{array}$

Try the web size $\text{W21}\times \text{62}$,

${\text{A}}_{\text{g}}\text{=18}\text{.3i}{\text{n}}^{\text{2}},r_y\text{=1}\text{.77in}$

Effective slenderness ratio is given by

$\begin{array}{l}\frac{\text{kL}}{\text{r}}\text{=}\frac{\text{0}\text{.65×}\left(\text{15}\text{.33×12}\right)}{\text{1}\text{.77}}\\ \frac{\text{kL}}{\text{r}}\text{=67}\text{.56ksi}\end{array}$

Calculate buckling stress

$\begin{array}{l}{F}_e=\frac{{\pi }^{2}E}{\left(\raisebox{1ex}{$KL$}\!\left/ \!\raisebox{-1ex}{$r^2$}\right.\right)}\\ F_e=\frac{{\pi }^2\times 29,000}{{\left(67.56\right)}^2}\\ F_e=113.4\text{ksi>67}\text{.56}\end{array}$

$\begin{array}{l}{F}_{cr}={0.658}^{{}^{\frac{F_y}{F_e}}}\times 50\\ F_{cr}={0.658}^{{}^{\frac{50}{62.71}}}\times 42\\ F_{cr}=35.81\text{ksi}\end{array}$

Calculate nominal compressive strength of column

$\begin{array}{l}{P}_n=F_{cr}{A}_g\\ P_n=35.81\times 18.3\\ P_n=655.3\text{kips(ok)}\end{array}$

$\begin{array}{l}{\varphi }_{\text{c}}{\text{P}}_{\text{n}}\text{=0}\text{.90×655}\text{.5}\\ {\varphi }_{\text{c}}{\text{P}}_{\text{n}}\text{=590kips>524kips(ok)}\end{array}$

Check for web buckling from dimensions and properties table in the manual the width thickness ratio of the web is given by

$\frac{h}{t_w}\text{=46}\text{.9}$

From the AISC section we have the equation

$\begin{array}{l}\frac{b}{t}\ge 1.49\sqrt{\frac{E}{f}}=1.49\sqrt{\frac{29,000}{36.21}}\\ \frac{b}{t}\ge 1.49\sqrt{\frac{E}{f}}=42.4\end{array}$

Since $\frac{h}{t_w}\text{>}1.49\sqrt{\frac{E}{f}}$,

Local buckling should be checked

Width of the web is given by

$\begin{array}{l}\text{b=d-2k}\\ \text{b=21-2×1}\text{.12}\\ \text{b=18}\text{.76in}\end{array}$

Effective width is given by

$\begin{array}{l}{b}_e\text{=1}\text{.92t}\sqrt{\frac{E}{f}}\left[1-\frac{0.34}{43.6}\sqrt{\frac{E}{f}}\right]\le b\\ b_e\text{=1}\text{.92}\times \text{0}\text{.400}\sqrt{\frac{29,000}{35.81}}\left[1-\frac{0.34}{46.9}\sqrt{\frac{29,000}{35.81}}\right]\le 18.76\\ b_e\text{=}17.35\le 18.76\end{array}$

Calculate effective area

$\begin{array}{l}{\text{A}}_{\text{eff}}\text{=A-}{\text{t}}_{\text{w}}\left(\text{b-}{\text{b}}_{\text{e}}\right)\\ {\text{A}}_{\text{eff}}\text{=18}\text{.3-0}\text{.400}\left(\text{18}\text{.76-17}\text{.35}\right)\\ {\text{A}}_{\text{eff}}\text{=17}\text{.74i}{\text{n}}^{\text{2}}\end{array}$

Critical stress is given by,

$\begin{array}{l}4.71\sqrt{\frac{E}{Q{F}_y}}=4.71\sqrt{\frac{29,000}{\left(\frac{17.74}{18.3}\right)50}}\\ 4.71\sqrt{\frac{E}{Q{F}_y}}=115.2>67.56\end{array}$

Calculate the critical buckling stress as below

$\begin{array}{l}{F}_{cr}=Q\left({0.658}^{{}^{\frac{Q{F}_y}{F_e}}}\times 50\right)\\ F_{cr}=0.9694\left({0.658}^{{}^{\frac{0.9694\times 50}{62.71}}}\times 50\right)\\ F_{cr}=35.07\text{ksi}\end{array}$

Calculate nominal compressive strength of column

$\begin{array}{l}{P}_n=F_{cr}{A}_g\\ P_n=35.07\times 18.3\\ P_n=641.8\text{kips}\\ {\varphi }_n{P}_n=0.90\times 641.8\\ {\varphi }_n{P}_n=578\text{kips>524kips(ok)}\end{array}$

Try the web size $\text{W21}\times \text{62}$,

Conclusion:

Therefore, size $\text{W21}\times \text{62}$ of $\text{A992}$ W shape with a nominal depth of $\text{21 inches}$ used LRFD.

To determine

(b)

Select a $\text{A992}$ W shape with a nominal depth of $\text{21 inches}$ using ASD.

Answer to Problem 4.6.9P

$122.90kips$

Explanation of Solution

Given information:

$\text{D=90}$

$\text{L=260}$

Calculation:

$\begin{array}{l}{P}_a=D+L\\ P_a=90+260\\ P_a=350\text{kips}\end{array}$

Assume $F_{cr}=25\text{ksi}$

$\begin{array}{l}{A}_g>\frac{P_u}{{\varphi }_c{F}_{cr}}=\frac{524}{0.90\times 35}\\ A_g>\frac{P_u}{{\varphi }_c{F}_{cr}}=16.63\text{i}{\text{n}}^2\end{array}$

Try the web size $\text{W21}\times \text{62}$,

$A_g=18.3\text{i}{\text{n}}^2,r_y=1.77\text{in}$

Effective slenderness ratio is given by

$\begin{array}{l}\frac{kL}{r_y}=\frac{0.65\times \left(15.33\times 12\right)}{1.77}\\ \frac{kL}{r_y}=67.56<200\end{array}$

$\begin{array}{l}{F}_e=\frac{{\pi }^{2}E}{\left(\raisebox{1ex}{$KL$}\!\left/ \!\raisebox{-1ex}{$r^2$}\right.\right)}\\ F_e=\frac{{\pi }^2\times 29,000}{{67.56}^2}\\ F_e=67.76\text{ksi}\end{array}$

Check for slenderness ratio

$\begin{array}{l}4.71\sqrt{\frac{E}{F_y}}=4.71\sqrt{\frac{29,000}{50}}\\ 4.71\sqrt{\frac{E}{F_y}}=113.4>67.56\end{array}$

$\begin{array}{l}{F}_{cr}={0.658}^{{}^{\frac{F_y}{F_e}}}\times 50\\ F_{cr}={0.658}^{{}^{\frac{50}{62.71}}}\times 50\\ F_{cr}=35.81\text{ksi}\end{array}$

$\begin{array}{l}{\text{P}}_{\text{n}}\text{=}{\text{F}}_{\text{cr}}{\text{A}}_{\text{g}}\\ {\text{P}}_{\text{n}}\text{=35}\text{.81}\left(\text{18}\text{.3}\right)\\ {\text{P}}_{\text{n}}\text{=655}\text{.3kips}\\ \raisebox{1ex}{${\text{P}}_{\text{n}}$}\!\left/ \!\raisebox{-1ex}{${\Omega }_{\text{c}}$}\right.\text{=}\raisebox{1ex}{$\text{655}\text{.3}$}\!\left/ \!\raisebox{-1ex}{$1.67$}\right.\\ \raisebox{1ex}{${\text{P}}_{\text{n}}$}\!\left/ \!\raisebox{-1ex}{${\Omega }_{\text{c}}$}\right.\text{=392kips>350kips(ok)}\end{array}$

Check for web buckling from dimensions and properties table in the manual the width thickness ratio of the web is given by

$\frac{h}{t_w}\text{=46}\text{.9}$

From the AISC section we have the equation

$\begin{array}{l}\frac{b}{t}\ge 1.49\sqrt{\frac{E}{f}}=1.49\sqrt{\frac{29,000}{35.81}}\\ \frac{b}{t}\ge 1.49\sqrt{\frac{E}{f}}=42.4\end{array}$

Since, $\frac{h}{t_w}\text{>}1.49\sqrt{\frac{E}{f}}$,

Local buckling should be checked

Width of the web is given by

$\begin{array}{l}\text{b=d-2k}\\ \text{b=21-2×1}\text{.12}\\ \text{b=18}\text{.76in}\end{array}$

Effective width is given by

$\begin{array}{l}{b}_e\text{=1}\text{.92t}\sqrt{\frac{E}{f}}\left[1-\frac{0.34}{43.6}\sqrt{\frac{E}{f}}\right]\le b\\ b_e\text{=1}\text{.92}\times \text{0}\text{.400}\sqrt{\frac{29,000}{35.81}}\left[1-\frac{0.34}{46.9}\sqrt{\frac{29,000}{35.81}}\right]\le 18.72\\ b_e\text{=}17.35\le 18.76\end{array}$

Calculate effective area

$\begin{array}{l}{\text{A}}_{\text{eff}}\text{=A-}{\text{t}}_{\text{w}}\left(\text{b-}{\text{b}}_{\text{e}}\right)\\ {\text{A}}_{\text{eff}}\text{=18}\text{.3-0}\text{.400}\left(\text{18}\text{.76-17}\text{.35}\right)\\ {\text{A}}_{\text{eff}}\text{=17}\text{.74i}{\text{n}}^{\text{2}}\end{array}$

Critical stress is given by

$\begin{array}{l}4.71\sqrt{\frac{E}{Q{F}_y}}=4.71\sqrt{\frac{29,000}{\left(\frac{17.74}{18.3}\right)50}}\\ 4.71\sqrt{\frac{E}{Q{F}_y}}=115.2>67.56\end{array}$

Calculate the critical buckling stress as below

$\begin{array}{l}{F}_{cr}=Q\left({0.658}^{{}^{\frac{Q{F}_y}{F_e}}}\times 50\right)\\ F_{cr}=0.9694\left({0.658}^{{}^{\frac{0.9694\times 50}{62.71}}}\times 50\right)\\ F_{cr}=35.07\text{ksi}\end{array}$

Calculate nominal compressive strength of column

$\begin{array}{l}{P}_n=F_{cr}{A}_g\\ P_n=35.07\times 18.3\\ P_n=641.8\text{kips(ok)}\\ \frac{P_n}{{\Omega }_c}=\frac{641.8}{1.67}\\ \frac{P_n}{{\Omega }_c}=384\text{kips>350kips(ok)}\end{array}$

Conclusion:

Therefore, size $\text{W21}\times \text{62}$ of $\text{A992}$ W shape with a nominal depth of $\text{21 inches}$ used ASD.