Chapter 4, Problem 4.8.4P

To determine

(a)

To select:

American standard channel for the given compression member using LRFD.

Answer to Problem 4.8.4P

$\text{Use C15}\times \text{33}\text{.9}$

Explanation of Solution

Given information:

Given compression member is :

Calculation:

Calculate the factored load by LRFD by using the equation.

$P_u=1.2D_L+1.6L_L$

Here $D_L$ is the dead load, $L_L$ is the live load.

Substitute

$D_L=30,L_L=70$

$\begin{array}{l}{P}_u=1.2D_L+1.6L_L\\ \text{ =1}\text{.2}\left(30\right)+1.6\left(70\right)\\ \text{ =148kips}\end{array}$

Try a C section $C15\times 33.9$

AISC must be used, as this shape is non slender and is neither a double angle nor a tee shape

Check the effective slenderness ratio about y-axis using the formula.

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

Here K is the effective length factor

L is the length of the member between the supports.

r is the radius of gyration

Take the properties steel from the AISC steel table. K value depends on the end conditions

$K=0.65,\left(12\times 12\right){\text{in,r}}_y=0.901$

$\begin{array}{l}\text{slenderness ratio=}\frac{KL}{r}\\ \text{ =}\frac{0.65\left(12\times 12\right)}{0.901}\\ \text{ =103}\text{.9<200}\left(\text{OK}\right)\end{array}$

Calculate the elastic buckling stress using the formula.

$\begin{array}{l}{\text{F}}_e=\frac{{\pi }^{2}E}{{\left(KL/r_x\right)}^2}\\ \text{Substitute E=29000,}KL/r=103.9\\ {\text{F}}_e=\frac{{\pi }^{2}E}{{\left(KL/r\right)}^2}\\ \text{ =}\frac{{\pi }^{2}29000}{{103.9}^2}\\ \text{ =26}\text{.51ksi}\end{array}$

Check for slenderness ratio by using the formula.

$\text{Slenderness ratio=4}\text{.71}\sqrt{\frac{E}{F_y}}$

Here $F_y$ is the yield strength

Substitute $F_y=50\text{ksi,E=29000}$

$\begin{array}{l}\text{Slenderness ratio=4}\text{.71}\sqrt{\frac{29000}{50}}\\ \text{ =113}\text{.4}\end{array}$

Since $103.9$< $113.4$, so calculate critical buckling stress using the formula

$\begin{array}{l}{\text{ F}}_{cr}\text{=}{0.658}^{\left(\frac{F_y}{F_e}\right)}{F}_y\\ \text{ =}{0.658}^{\left(\frac{36}{26.51}\right)}\left(36\right)\\ \text{ =20}\text{.39ksi}\end{array}$

Calculate the nominal compressive strength of column using the formula.

$P_n=F_{cr}{A}_g$

Substitute, $F_{cr}$ = $20.39$, $A_g$ = $10.0$

$P_n=F_{cr}{A}_g$

= $20.39(10.0)$

= $203.9\text{kips}$

Calculate design strength of the column using by LRFD method

$P_u\text{=}{\varphi }_c{P}_n$

Here we have $P_n=203.9$, $\varphi =0.9$

$\begin{array}{l}{P}_u\text{=}{\varphi }_c{P}_n\\ \text{ =0}\text{.9}\left(203.9\right)\\ \text{ =184kips>148kips(OK)}\end{array}$

Check the effective slenderness ratio about x-axis using the formula.

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

Substitute $K=0.65,\left(12\times 12\right)\text{in},r_x=5.62$

$\begin{array}{l}\text{slenderness ratio=}\frac{KL}{r}\\ \text{ =}\frac{0.65\left(12\times 12\right)}{5.62}\\ \text{ =16}\text{.65}\end{array}$

From the manual companion CD:

$\begin{array}{l}{\overline{r}}_o=5.94\text{in}\\ \text{H=0}\text{.920}\end{array}$

Calculate the elastic buckling stress using the formula.

$\begin{array}{l}{\text{F}}_e=\frac{{\pi }^{2}E}{{\left(KL/r_x\right)}^2}\\ \text{Substitute E=29000,}KL/r=16.65\end{array}$

$\begin{array}{l}{\text{F}}_e=\frac{{\pi }^{2}E}{{\left(KL/r\right)}^2}\\ \text{ =}\frac{{\pi }^{2}29000}{{16.65}^2}\\ \text{ =1032ksi}\end{array}$

Calculate the value of $F_{ez}$

$\begin{array}{l}{F}_{ez}=\left(\frac{{\pi }^{2}E{C}_w}{{\left(K_{z}L\right)}^2}+GJ\right)\frac{1}{A_g{\overline{r}}_o{}^2}\\ \text{ =}\left[\frac{{\pi }^2\left(29000\right)\left(358\right)}{{\left(0.65\times 12\times 12\right)}^2}+\left(11200\left(1.01\right)\right)\right]\frac{1}{10.0{\left(5.94\right)}^2}\\ \text{ =65}\text{.21ksi}\end{array}$

Calculate the total stress by equation

$\begin{array}{l}\text{Total stress}=F_{ey}+F_{ez}\\ \text{ =1032+65}\text{.21}\\ \text{ =1097ksi}\end{array}$

Calculate the value of $F_{cr}$ by using the AISC equation

$\begin{array}{l}{F}_{cr}=\left(\frac{F_{cry}+F_{crz}}{2H}\right)\left[1-\sqrt{1-\frac{4F_{cry}{F}_{crz}H}{{\left(F_{cry}+F_{crz}\right)}^2}}\right]\\ \text{ =}\left(\frac{1097}{2(0.920)}\right)\left[1-\sqrt{1-\frac{4\left(1032\right)\left(65.21\right)\left(0.920\right)}{{1097}^2}}\right]\\ \text{ =64}\text{.88ksi}\end{array}$

In order to determine which compressive strength equation to be use, compare the value of $F_e$

$\begin{array}{l}0.44F_y=0.44\left(36\right)\\ \text{ =15}\text{.8ksi}\end{array}$

Since $\text{64}\text{.88>15}\text{.8}$, so calculate critical buckling stress using the formula.

$\begin{array}{l}{F}_{cry}={0.658}^{\left(\frac{F_y}{F_e}\right)}{F}_y\\ \text{ =}{0.658}^{\left(\frac{36}{64.88}\right)}\left(36\right)\\ \text{ =28}\text{.54ksi}\end{array}$

Calculate the maximum strength by using the formula.

. $\begin{array}{l}{P}_n=F_{cr}+A_g\\ \text{ =28}\text{.54}\left(10.0\right)\\ \text{ =285}\text{.4kips}\\ \text{calculate the design strength by using the formula }\\ {\varphi }_c{P}_n=0.90\left(285.4\right)\\ \text{ =257kips>148kips(OK)}\end{array}$

Conclusion:

$\begin{array}{l}\text{Therefore, the American channel is calculated using this formula :}{P}_n=F_{cr}+A_g\\ \end{array}$.

To determine

(b)

To select:

American standard channel for the given compression member using ASD.

Answer to Problem 4.8.4P

$\text{ C15}\times \text{33}\text{.9}$

Explanation of Solution

Given information:

Given compression member is

Calculation:

Calculate the factored load by LRFD by using the equation.

$P_u=1.2D_L+1.6L_L$

He re $D_L$ is the dead load, $L_L$ is the live load.

Substitute

$D_L=30,L_L=70$

$\begin{array}{l}{P}_u=D_L+L_L\\ \text{ =30}+70\\ \text{ =100kips}\end{array}$

Try a C section $C15\times 33.9$

AISC must be used, as this shape is non slender and is neither a double angle nor a tee shape

Check the effective slenderness ratio about y-axis using the formula

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

Here K is the effective length factor

L is the length of the member between the supports

r is the radius of gyration

Take the properties steel from the AISC steel table. K value depends on the end conditions

$K=0.65,\left(12\times 12\right){\text{in,r}}_y=0.901$

$\begin{array}{l}\text{slenderness ratio=}\frac{KL}{r}\\ \text{ =}\frac{0.65\left(12\times 12\right)}{0.901}\\ \text{ =103}\text{.9<200}\left(\text{OK}\right)\end{array}$

Calculate the elastic buckling stress using the formula.

$\begin{array}{l}{\text{F}}_e=\frac{{\pi }^{2}E}{{\left(KL/r_x\right)}^2}\\ \text{Substitute E=29000,}KL/r=103.9\\ {\text{F}}_e=\frac{{\pi }^{2}E}{{\left(KL/r\right)}^2}\\ \text{ =}\frac{{\pi }^{2}29000}{{103.9}^2}\\ \text{ =26}\text{.51ksi}\end{array}$

Check for slenderness ratio by using the formula.

$\text{Slenderness ratio=4}\text{.71}\sqrt{\frac{E}{F_y}}$

Here $F_y$ is the yield strength

Substitute $F_y=50\text{ksi,E=29000}$

$\begin{array}{l}\text{Slenderness ratio=4}\text{.71}\sqrt{\frac{29000}{50}}\\ \text{ =113}\text{.4}\end{array}$

Since $103.9$< $113.4$, so calculate critical buckling stress using the formula.

$\begin{array}{l}{\text{ F}}_{cr}\text{=}{0.658}^{\left(\frac{F_y}{F_e}\right)}{F}_y\\ \text{ =}{0.658}^{\left(\frac{36}{26.51}\right)}\left(36\right)\\ \text{ =20}\text{.39ksi}\end{array}$

Calculate the nominal compressive strength of column using the formula.

$P_n=F_{cr}{A}_g$

Substitute, $F_{cr}$ = $20.39$, $A_g$ = $10.0$

$P_n=F_{cr}{A}_g$

= $20.39(10.0)$

= $203.9\text{kips}$

Calculate design strength of the column using by ASD method.

$P_u\text{=}\frac{P_n}{\Omega }$

Here we have $P_n=203.9$, $\Omega =1.67$

$\begin{array}{l}{P}_u\text{=}\frac{P_n}{\Omega }\\ \text{ =}\frac{203.9}{1.67}\\ \text{ =122kips>100kips(OK)}\end{array}$

Check the effective slenderness ratio about x-axis using the formula.

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

Substitute $K=0.65,\left(12\times 12\right)\text{in},r_x=5.62$

$\begin{array}{l}\text{slenderness ratio=}\frac{KL}{r}\\ \text{ =}\frac{0.65\left(12\times 12\right)}{5.62}\\ \text{ =16}\text{.65}\end{array}$

From the manual companion CD:

$\begin{array}{l}{\overline{r}}_o=5.94\text{in}\\ \text{H=0}\text{.920}\end{array}$

Calculate the elastic buckling stress using the formula.

$\begin{array}{l}{\text{F}}_e=\frac{{\pi }^{2}E}{{\left(KL/r_x\right)}^2}\\ \text{Substitute E=29000,}KL/r=16.65\end{array}$

$\begin{array}{l}{\text{F}}_e=\frac{{\pi }^{2}E}{{\left(KL/r\right)}^2}\\ \text{ =}\frac{{\pi }^{2}29000}{{16.65}^2}\\ \text{ =1032ksi}\end{array}$

Calculate the value of $F_{ez}$

$\begin{array}{l}{F}_{ez}=\left(\frac{{\pi }^{2}E{C}_w}{{\left(K_{z}L\right)}^2}+GJ\right)\frac{1}{A_g{\overline{r}}_o{}^2}\\ \text{ =}\left[\frac{{\pi }^2\left(29000\right)\left(358\right)}{{\left(0.65\times 12\times 12\right)}^2}+\left(11200\left(1.01\right)\right)\right]\frac{1}{10.0{\left(5.94\right)}^2}\\ \text{ =65}\text{.21ksi}\end{array}$

Calculate the total stress by equation

$\begin{array}{l}\text{Total stress}=F_{ey}+F_{ez}\\ \text{ =1032+65}\text{.21}\\ \text{ =1097ksi}\end{array}$

Calculate the value of $F_{cr}$ by using the AISC equation :

$\begin{array}{l}{F}_{cr}=\left(\frac{F_{cry}+F_{crz}}{2H}\right)\left[1-\sqrt{1-\frac{4F_{cry}{F}_{crz}H}{{\left(F_{cry}+F_{crz}\right)}^2}}\right]\\ \text{ =}\left(\frac{1097}{2(0.920)}\right)\left[1-\sqrt{1-\frac{4\left(1032\right)\left(65.21\right)\left(0.920\right)}{{1097}^2}}\right]\\ \text{ =64}\text{.88ksi}\end{array}$

In order to determine which compressive strength equation to be use, compare the value of $F_e$

$\begin{array}{l}0.44F_y=0.44\left(36\right)\\ \text{ =15}\text{.8ksi}\end{array}$

Since $\text{64}\text{.88>15}\text{.8}$, so calculate critical buckling stress using the formula.

$\begin{array}{l}{F}_{cry}={0.658}^{\left(\frac{F_y}{F_e}\right)}{F}_y\\ \text{ =}{0.658}^{\left(\frac{36}{64.88}\right)}\left(36\right)\\ \text{ =28}\text{.54ksi}\end{array}$

Calculate the maximum strength by using the formula.

$\begin{array}{l}{P}_n=F_{cr}+A_g\\ \text{ =28}\text{.54}\left(10.0\right)\\ \text{ =285}\text{.4kips}\\ \text{calculate the design strength of the column by using the formula }\\ P_n=\frac{P_n}{\Omega }\\ P_n=285.4,\Omega =1.67\\ P_n=\frac{P_n}{\Omega }\\ \text{ =}\frac{285.4}{1.67}\\ \text{ =171kips>100kips(OK)}\end{array}$

Conclusion: