Chapter 4, Problem 4.7.10P

To determine

(a)

Axial Compressive design strength of column AB.

Answer to Problem 4.7.10P

$\text{728kips}$

Explanation of Solution

Calculation:

calculate the ratio of column stiffness to girder at each and column AB by using the equation.

$G_A=\frac{{\displaystyle \sum \frac{I_C}{L_C}}}{{\displaystyle \sum \frac{I_g}{L_g}}}$

Here we have

G=ratio of column stiffness to girder stiffness

$I_C$ =Moment of inertia of column

$L_g$ =Moment of inertia of girder

$L_C$ =Length of column

$L_g$ =Length of column

$G_B=10$

Here ratio of column stiffness to girder stiffness at end A is $G_A$

For Joint A:

$G_A=\frac{{\displaystyle \sum \frac{I_C}{L_C}}}{{\displaystyle \sum \frac{I_g}{L_g}}}$

Substitute

${\displaystyle \sum \frac{I_g}{L_g}}=\left(\frac{800}{18}+\frac{800}{20}\right),I_C=597,L_C=15$

$\begin{array}{l}{G}_A=\frac{{\displaystyle \sum \frac{I_C}{L_C}}}{{\displaystyle \sum \frac{I_g}{L_g}}}\\ =\frac{\left(\frac{597}{15}\right)}{\left(\frac{800}{18}+\frac{800}{20}\right)}\\ \text{ =0}\text{.471}\end{array}$

Refer the alignment chart for the value of $K_x=1.78$

Calculate the effective slenderness ratio for column by using the equation

Here $K_x$ is the effective length factor in X direction, $r_x$ is the radius of gyration in X direction, L is the length of the member between the supports

$\begin{array}{l}\text{Effective slenderness ratio=}\frac{K_{x}L}{r_x}\\ \text{ =}\frac{1.78\left(15\times 12\right)}{5.31}\\ \text{ =60}\text{.34}\end{array}$

Calculate the upper limit elasticity using the equation

$\begin{array}{l}\text{Upper limit elasticity=4}\text{.71}\sqrt{\frac{E}{F_y}}\\ \text{ =4}\text{.71}\sqrt{\frac{29000}{50}}\\ \text{ =113}\text{.4}\end{array}$

Since $60.34$ is less than $113.34$, the column is elastic

Calculate the factored load by LRFD by using the equation

$P_u=1.2D+1.6L$

He re D is the dead load, L is the live load

Substitute

$D=50,L=150$

$\begin{array}{l}{P}_u=1.2D+1.6L\\ \text{ =1}\text{.2}\left(50\right)+1.6\left(150\right)\\ \text{ =300kips}\end{array}$

Calculate the stress coming on the column

$\begin{array}{l}\text{Stress=}\frac{P_u}{A_g}\\ \text{Substitute }{P}_u=300,A_g=21.1\end{array}$

$\begin{array}{l}\text{Stress=}\frac{P_u}{A_g}\\ \text{ =}\frac{300}{21.1}\\ \text{ =14}\text{.22ksi}\end{array}$

Refer table $4-21$ form AISC steel manual ${\tau }_b=1.00$

No modification is necessary

Calculate effective slenderness ratio in y direction

$\begin{array}{l}\text{Effective slenderness ratio=}\frac{K_{y}L}{r_y}\\ \text{Substitute k=1}\text{.0,L=15ft,}{r}_y=3.04\\ \text{slenderness ratio=}\frac{K_{y}L}{r_y}\\ \text{ =}\frac{1\left(15\times 12\right)88}{3.04}\\ \text{ =59}\text{.21<60}\text{.34}\end{array}$

Calculate the buckling stress using the formula.

$\begin{array}{l}{\text{F}}_e=\frac{{\pi }^{2}E}{{\left(KL/r\right)}^2}\\ \text{Substitute E=29000,}KL/r=60.34\\ {\text{F}}_e=\frac{{\pi }^{2}E}{{\left(KL/r\right)}^2}\\ \text{ =}\frac{{\pi }^{2}29000}{{60.34}^2}\\ \text{ =78}\text{.61ksi}\end{array}$

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

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

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

Since $\text{60}\text{.34 is less than 113}\text{.4, so calculate the buckling stress using the formula}$

$\begin{array}{l}{F}_{cr}={0.658}^{\left(\frac{F_y}{F_e}\right)}{F}_y\\ \text{ =}{0.658}^{\left(\frac{50}{78.61}\right)}\left(50\right)\\ \text{ =38}\text{.31ksi}\end{array}$

Calculate the nominal compressive strength of column.

$\begin{array}{l}{P}_n=F_{cr}{A}_g\\ \text{Here we have }{A}_g=21.1,F_{cr}=38.31,\\ P_n=F_{cr}{A}_g\\ \text{ =38}\text{.31}\left(21.1\right)\\ \text{ =808}\text{.3kips}\\ \text{Calculate the design strength of the column by LRFD method}\text{.}\end{array}$

$\begin{array}{l}{P}_u=\varphi P_n\\ \text{ =0}\text{.9}\left(808.3\right)\\ \text{ =728kips}\end{array}$

Conclusion:

Hence, here the design strength is estimated using the formula: $P_u=\varphi P$

ii.

To determine

Themaximum axial compressive strength of column AB.

Answer to Problem 4.7.10P

$\text{484kips}$

Explanation of Solution

Calculation:

Calculate the factored load by ASD by using the equation

$P_a=D+L$

He re D is the dead load L is the live load

Substitute

$D=50,L=150$

$\begin{array}{l}{P}_a=D+L\\ \text{ =50}+150\\ \text{ =200kips}\end{array}$

Calculate stress coming on column.

$\begin{array}{l}\text{Stress=}\frac{P_u}{A_g}\\ P_u=200,A_g=21.1\\ \text{Stress=}\frac{P_u}{A_g}\\ \text{ =}\frac{200}{21.1}\\ \text{ =9}\text{.47ksi}\end{array}$

Refer table $4-21$ form AISC steel manual ${\tau }_b=1$

No modification is necessary

Calculate effective slenderness ratio in y direction

Effective $\text{slenderness ratio=}\frac{K_{y}L}{r_y}$

$K=1.0,L=15{\text{ft,r}}_y=3.04$

$\begin{array}{l}\text{slenderness ratio=}\frac{K_{y}L}{r_y}\\ \text{ =}\frac{1\left(15\times 12\right)88}{3.04}\\ \text{ =59}\text{.21<60}\text{.34}\end{array}$

Calculate the buckling stress using the formula

$\begin{array}{l}{\text{F}}_e=\frac{{\pi }^{2}E}{{\left(KL/r\right)}^2}\\ \text{Substitute E=29000,}KL/r=60.34\\ {\text{F}}_e=\frac{{\pi }^{2}E}{{\left(KL/r\right)}^2}\\ \text{ =}\frac{{\pi }^{2}29000}{{60.34}^2}\\ \text{ =78}\text{.61ksi}\end{array}$

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.

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

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

Since $\text{60}\text{.34 is less than 113}\text{.4, so calculate the buckling stress using the formula}\text{.}$

$\begin{array}{l}{F}_{cr}={0.658}^{\left(\frac{F_y}{F_e}\right)}{F}_y\\ \text{ =}{0.658}^{\left(\frac{50}{78.61}\right)}\left(50\right)\\ \text{ =38}\text{.31ksi}\end{array}$

Calculate the compressive strength of column.

$\begin{array}{l}{P}_n=F_{cr}{A}_g\\ \text{Here we have ,}{A}_g=21.1,F_{cr}=38.31\\ P_n=F_{cr}{A}_g\\ \text{ =38}\text{.31}\left(21.1\right)\\ \text{ =808}\text{.3kips}\end{array}$

Calculate the maximum strength by using the formula.

$\begin{array}{l}\text{ Maximum strength=}\frac{P_n}{\Omega }\\ \text{ =}\frac{808.3}{1.67}\\ \text{ =484kips}\end{array}$

Conclusion:

Therefore, the maximum strength is calculated using the formula: $\begin{array}{l}\text{ Maximum strength=}\frac{P_n}{\Omega }\\ \end{array}$