Chapter 6, Problem 6.8.7P

To determine

(a)

The suitable $\text{W}$ shape of $\text{A992}$ steel on the basis of load resistance and factor design.

Answer to Problem 6.8.7P

The best suitable $\text{W}$ shape of $\text{A992}$ steel is on the basis of load resistance and factor design is $\text{W}12\times 72$.

Explanation of Solution

Given:

The axial load is $80\text{kips}$.

The dead load is the $25\%$ of the axial load.

The live load is the $75\%$ of the axial load.

The length of the column is $16\text{ft}$.

The moment at top in x-direction is $133\text{ft-kips}$.

The moment at top in y-direction is $43\text{ft-kips}$.

The moment at bottom in x-direction is $27\text{ft-kips}$.

The moment at bottom in y-direction is $91\text{ft-kips}$.

Calculation:

Write the equation to obtain the load factor.

$P_{\text{u}}=1.2P_{\text{D}}+1.6P_{\text{L}}$ ..... (I)

Here, load factor is $P_{\text{u}}$, axial live load is $P_{\text{L}}$ and axial dead load is $P_{\text{D}}$.

Calculate the dead load:

$\begin{array}{c}{P}_{\text{D}}=\frac{25}{100}\times 80\text{kips}\\ =20\text{kips}\end{array}$

Calculate the live load:

$\begin{array}{c}{P}_{\text{L}}=\frac{75}{100}\times 80\text{kips}\\ =60\text{kips}\end{array}$

Substitute $60\text{kips}$ for $P_{\text{L}}$ and $20\text{kips}$ for $P_{\text{D}}$ in Equation (I).

$\begin{array}{c}{P}_{\text{u}}=1.2\left(20\text{kips}\right)+1.6\left(60\text{kips}\right)\\ =24\text{kips}+\text{96}\text{kips}\\ =\text{120}\text{kips}\end{array}$

$$

Write the equation to obtain the factored bending moment at the bottom of the member for braced condition along x-axis.

$M_{\text{nbx}}=1.2M_{\text{Dbx}}+1.6M_{\text{Lbx}}$ ..... (II)

Here, factored bending moment at bottom along x-axis is $M_{\text{nbx}}$, bending moment due to live load at bottom along x-axis is $M_{\text{Lbx}},$ and the bending moment due to dead load at bottom along x-axis is $M_{\text{Dbx}}$.

Calculate the bending moment along x-direction at bottom due to dead load.

$\begin{array}{c}{M}_{\text{Dbx}}=\frac{25}{100}\times 27\text{kips}\\ =6.75\text{kips}\end{array}$

Calculate the bending moment along x-direction at bottom due to live load.

$\begin{array}{c}{M}_{\text{Lbx}}=\frac{75}{100}\times 27\text{kips}\\ =20.25\text{kips}\end{array}$

Substitute $20.25\text{ft-kips}$ for $M_{\text{Lbx}}$ and $6.75\text{ft-kips}$ for $M_{\text{Dbx}}$ in Equation (II).

$\begin{array}{c}{M}_{\text{nbx}}=1.2\left(6.75\text{ft-kips}\right)+1.6\left(20.25\text{ft-kips}\right)\\ =8.1\text{ft-kips}+32.4\text{ft-kips}\\ =40.5\text{ft-kips}\end{array}$

Write the equation to obtain factored bending moment at the top of the member for braced condition along x-axis.

$M_{\text{ntx}}=1.2M_{\text{Dtx}}+1.6M_{\text{Ltx}}$ ..... (III)

Here, factored bending moment at top along x-axis is $M_{\text{ntx}}$, bending moment due to live load at top along x-axis is $M_{\text{Ltx}},$ and the bending moment due to dead load at top along x-axis is $M_{\text{Ltx}}$.

Calculate the bending moment along x-direction at top due to dead load.

$\begin{array}{c}{M}_{\text{Dtx}}=\frac{25}{100}\times 133\text{kips}\\ =33.25\text{kips}\end{array}$

Calculate the bending moment along x-direction at top due to live load.

$\begin{array}{c}{M}_{\text{Ltx}}=\frac{75}{100}\times 133\text{kips}\\ =99.75\text{kips}\end{array}$

Substitute $99.75\text{ft-kips}$ for $M_{\text{Ltx}}$ and $33.25\text{ft-kips}$ for $M_{\text{Dtx}}$ in Equation (III).

$\begin{array}{c}{M}_{\text{ntx}}=1.2\left(33.25\text{ft-kips}\right)+1.6\left(99.75\text{ft-kips}\right)\\ =39.9\text{ft-kips}+159.6\text{ft-kips}\\ =199.5\text{ft-kips}\end{array}$

Write the equation to obtain factored bending moment at the bottom of the member for braced condition along y-axis.

$M_{\text{nby}}=1.2M_{\text{Dby}}+1.6M_{\text{Lby}}$ ..... (IV)

Here, factored bending moment at bottom along y-axis is $M_{\text{nby}}$, bending moment due to live load at bottom along y-axis is $M_{\text{Lby}},$ and the bending moment due to dead load at bottom along y-axis is $M_{\text{Lby}}$.

Calculate the bending moment along y-direction at bottom due to dead load.

$\begin{array}{c}{M}_{\text{Dby}}=\frac{25}{100}\times 9\text{kips}\\ =2.25\text{kips}\end{array}$

Calculate the bending moment along y-direction at bottom due to live load.

$\begin{array}{c}{M}_{\text{Lby}}=\frac{75}{100}\times 9\text{kips}\\ =6.75\text{kips}\end{array}$

Substitute $6.75\text{ft-kips}$ for $M_{\text{Lby}}$ and $2.25\text{ft-kips}$ for $M_{\text{Dby}}$ in Equation (IV).

$\begin{array}{c}{M}_{\text{nby}}=1.2\left(2.25\text{ft-kips}\right)+1.6\left(6.75\text{ft-kips}\right)\\ =2.7\text{ft-kips}+10.8\text{ft-kips}\\ =13.5\text{ft-kips}\end{array}$

Write the equation to obtain the factored bending moment at the top of the member for braced condition along y-axis.

$M_{\text{nty}}=1.2M_{\text{Dty}}+1.6M_{\text{Lty}}$ ..... (V)

Here, factored bending moment at top along y-axis is $M_{\text{nty}}$, bending moment due to live load at top along y-axis is $M_{\text{Lty}},$ and the bending moment due to dead load at top along y-axis is $M_{\text{Lty}}$.

Calculate the bending moment along y-direction at top due to dead load.

$\begin{array}{c}{M}_{\text{Dty}}=\frac{25}{100}\times 43\text{kips}\\ =10.75\text{kips}\end{array}$

Calculate the bending moment along y-direction at top due to live load.

$\begin{array}{c}{M}_{\text{Lty}}=\frac{75}{100}\times 43\text{kips}\\ =32.25\text{kips}\end{array}$

Substitute $32.25\text{ft-kips}$ for $M_{\text{Lty}}$ and $10.75\text{ft-kips}$ for $M_{\text{Dty}}$ in Equation (V).

$\begin{array}{c}{M}_{\text{nty}}=1.2\left(10.75\text{ft-kips}\right)+1.6\left(32.25\text{ft-kips}\right)\\ =12.9\text{ft-kips}+51.6\text{ft-kips}\\ =64.5\text{ft-kips}\end{array}$

Write the equation to obtain the ultimate moment along x-axis.

$M_{\text{ux}}=B_1{M}_{\text{ntx}}+B_2{M}_{\text{nbx}}$ ..... (VI)

Here, factor for braced condition is $B_1$, factor for side sway condition is $B_2$.

Substitute, $1$ for $B_1$, $199.5\text{ft-kips}$ for $M_{\text{ntx}}$, $0$ for $B_2,$ and $40.5\text{ft-kips}$ for $M_{\text{nbx}}$ in equation (VI).

$\begin{array}{c}{M}_{\text{ux}}=\left(1\right)\left(199.5\text{ft-kips}\right)+\left(0\right)\left(40.5\text{ft-kips}\right)\\ =199.5\text{ft-kips}\end{array}$

Write the equation to obtain the ultimate moment along y-axis.

$M_{\text{uy}}=B_1{M}_{\text{nty}}+B_2{M}_{\text{nby}}$ ..... (VII)

Here, factor for braced condition is $B_1$, factor for side sway condition is $B_2$.

Substitute, $1$ for $B_1$, $64.5\text{ft-kips}$ for $M_{\text{nty}}$, $0$ for $B_2,$ and $13.5\text{ft-kips}$ for $M_{\text{nby}}$ in equation (VII).

$\begin{array}{c}{M}_{\text{ux}}=\left(1\right)\left(64.5\text{ft-kips}\right)+\left(0\right)\left(13.5\text{ft-kips}\right)\\ =64.5\text{ft-kips}\end{array}$

The unbraced length and effective length of the member are same.

$\begin{array}{c}{L}_{\text{b}}=kL\\ =16\text{ft}\end{array}$

Try $\text{W}10\times 77$, therefore from table $\text{6-2}$, ${\varphi }_c{P}_{\text{n}}=684.93\text{kips}$, ${\varphi }_{\text{b}}{M}_{\text{nx}}=339.27\text{ft-kips,}$ and ${\varphi }_{\text{b}}{M}_{\text{ny}}=172.26\text{ft-kips}$.

Write the expression to determine which interaction equation to use.

$\frac{P_{\text{u}}}{{\varphi }_c{P}_{\text{n}}}>0.2$ ..... (VIII)

Here, load factor is $P_{\text{u}}$ and the factor which deals with the axial strength of member is ${\varphi }_c{P}_{\text{n}}$.

Substitute $120\text{kips}$ for $P_{\text{u}}$ and $684.93\text{kips}$ for ${\varphi }_c{P}_{\text{n}}$ in Equation (VIII).

$\begin{array}{l}\frac{120\text{kips}}{684.93\text{kips}}\\ 0.1752<0.2\end{array}$

Therefore, $\text{AISC}$ equation $\text{H1-1a}$ controls.

Write the expression $\text{AISC}$ equation $\text{H1-1a}$.

$\frac{0.5P_{\text{u}}}{{\varphi }_{\text{c}}{P}_{\text{n}}}+\frac{8}{9}\left(\frac{M_{\text{ux}}}{{\varphi }_{\text{b}}{M}_{\text{nx}}}+\frac{M_{\text{uy}}}{{\varphi }_{\text{b}}{M}_{\text{ny}}}\right)$ ..... (IX)

Here, load factor is $P_{\text{u}}$ and the factor which deals with the axial strength of member is ${\varphi }_c{P}_{\text{n}}$, moment factor in x-direction is ${\varphi }_{\text{b}}{M}_{\text{nx}}$, moment factor in y-direction is ${\varphi }_{\text{b}}{M}_{\text{ny}}$, ultimate moment in x direction is $M_{\text{ux}},$ and ultimate moment in y-direction is $M_{\text{uy}}$.

Substitute, $120\text{kips}$ for $P_{\text{u}}$, $684.93\text{kips}$ for ${\varphi }_c{P}_{\text{n}}$, $199.5\text{ft-kips}$ for $M_{\text{ux}}$, $339.27\text{ft-kips}$ for ${\varphi }_{\text{b}}{M}_{\text{nx}}$, $64.5\text{ft}\text{.kips}$ for $M_{\text{uy}},$ and $172.26\text{ft-kips}$ for ${\varphi }_{\text{b}}{M}_{\text{ny}}$ in Equation (IX).

$\begin{array}{l}\frac{\text{0}\text{.5}\times \text{120}\text{kips}}{684.93\text{kips}}+\frac{8}{9}\left(\frac{199.5\text{ft-kips}}{339.27\text{ft-kips}}+\frac{64.5\text{ft-kips}}{172.26\text{ft-kips}}\right)\\ 0.0876+\frac{8}{9}\left(0.588+0.374\right)\\ 0.0876+\frac{8}{9}\left(0.962\right)\\ 0.0876+0.8551\end{array}$

Further solve the above equation.

$0.942<1$

Hence, it is safe to use $\text{W}10\times 77$.

Try $\text{W}12\times 72$, therefore from table $\text{6-2}$, ${\varphi }_c{P}_{\text{n}}=709.21\text{kips}$, ${\varphi }_{\text{b}}{M}_{\text{nx}}=373.48\text{ft-kips}$ and ${\varphi }_{\text{b}}{M}_{\text{ny}}=184.41\text{ft-kips}$.

Write the expression to determine the interaction equation to be used.

$\frac{P_{\text{u}}}{{\varphi }_c{P}_{\text{n}}}>0.2$ ..... (X)

Here, load factor is $P_{\text{u}}$ and the factor which deals with the axial strength of member is ${\varphi }_c{P}_{\text{n}}$.

Substitute $120\text{kips}$ for $P_{\text{u}}$ and $709.21\text{kips}$ for ${\varphi }_c{P}_{\text{n}}$ in Equation (X).

$\begin{array}{c}\frac{P_{\text{u}}}{{\varphi }_c{P}_{\text{n}}}=\frac{120\text{kips}}{708.21\text{kips}}\\ =0.169<0.2\end{array}$

Therefore, $\text{AISC}$ equation $\text{H1-1a}$. controls.

Write the expression $\text{AISC}$ equation.

$\frac{0.5P_{\text{u}}}{{\varphi }_{\text{c}}{P}_{\text{n}}}+\frac{8}{9}\left(\frac{M_{\text{ux}}}{{\varphi }_{\text{b}}{M}_{\text{nx}}}+\frac{M_{\text{uy}}}{{\varphi }_{\text{b}}{M}_{\text{ny}}}\right)$ ..... (XI)

Here, load factor is $P_{\text{u}}$ and the factor which deals with the axial strength of member is ${\varphi }_c{P}_{\text{n}}$, moment factor in x-direction is ${\varphi }_{\text{b}}{M}_{\text{nx}}$, moment factor in y-direction is ${\varphi }_{\text{b}}{M}_{\text{ny}}$, ultimate moment in x direction is $M_{\text{ux}},$ and ultimate moment in y-direction is $M_{\text{uy}}$.

Substitute, $120\text{kips}$ for $P_{\text{u}}$, $708.21\text{kips}$ for ${\varphi }_c{P}_{\text{n}}$, $199.5\text{ft-kips}$ for $M_{\text{ux}}$, $373.48\text{ft-kips}$ for ${\varphi }_{\text{b}}{M}_{\text{nx}}$, $64.5\text{ft}\text{.kips}$ for $M_{\text{uy}}$ and $184.41\text{ft-kips}$ for ${\varphi }_{\text{b}}{M}_{\text{ny}}$ in Equation (XI).

$\begin{array}{c}\frac{0.5P_{\text{u}}}{{\varphi }_{\text{c}}{P}_{\text{n}}}+\frac{8}{9}\left(\frac{M_{\text{ux}}}{{\varphi }_{\text{b}}{M}_{\text{nx}}}+\frac{M_{\text{uy}}}{{\varphi }_{\text{b}}{M}_{\text{ny}}}\right)=\frac{\text{0}\text{.5}\times \text{120}\text{kips}}{708.21\text{kips}}+\frac{8}{9}\left(\frac{199.5\text{ft-kips}}{373.48\text{ft-kips}}+\frac{64.5\text{ft-kips}}{184.41\text{ft-kips}}\right)\\ =0.0847+\frac{8}{9}\left(0.5341+0.3497\right)\\ =0.0847+\frac{8}{9}\left(0.8832\right)\\ =0.0847+0.7850\end{array}$

Further solve the above equation.

$0.869<1$

Hence, it is safe to use $\text{W}12\times 72$.

Try $\text{W}14\times 68$, therefore from table $\text{6-2}$, ${\varphi }_c{P}_{\text{n}}=578.03\text{kips}$, ${\varphi }_{\text{b}}{M}_{\text{nx}}=460.56\text{ft-kips}$ and ${\varphi }_{\text{b}}{M}_{\text{ny}}=228.5\text{ft-kips}$.

Write the expression to determine the interaction equation to be used.

$\frac{P_{\text{u}}}{{\varphi }_c{P}_{\text{n}}}>0.2$ ..... (XII)

Here, load factor is $P_{\text{u}}$ and the factor which deals with the axial strength of member is ${\varphi }_c{P}_{\text{n}}$.

Substitute $120\text{kips}$ for $P_{\text{u}}$ and $578.03\text{kips}$ for ${\varphi }_c{P}_{\text{n}}$ in Equation (XII).

$\begin{array}{c}\frac{P_{\text{u}}}{{\varphi }_c{P}_{\text{n}}}=\frac{120\text{kips}}{578.03\text{kips}}\\ =0.2076>0.2\end{array}$

Therefore, $\text{AISC}$ equation $\text{H1-1a}$ controls.

Write the expression $\text{AISC}$ equation $\text{H1-1a}$.

$\frac{0.5P_{\text{u}}}{{\varphi }_{\text{c}}{P}_{\text{n}}}+\frac{8}{9}\left(\frac{M_{\text{ux}}}{{\varphi }_{\text{b}}{M}_{\text{nx}}}+\frac{M_{\text{uy}}}{{\varphi }_{\text{b}}{M}_{\text{ny}}}\right)$ ..... (XIII)

Here, load factor is $P_{\text{u}}$ and the factor which deals with the axial strength of member is ${\varphi }_c{P}_{\text{n}}$, moment factor in x-direction is ${\varphi }_{\text{b}}{M}_{\text{nx}}$, moment factor in y-direction is ${\varphi }_{\text{b}}{M}_{\text{ny}}$, ultimate moment in x direction is $M_{\text{ux}},$ and ultimate moment in y-direction is $M_{\text{uy}}$.

Substitute $120\text{kips}$ for $P_{\text{u}}$, $578.03\text{kips}$ for ${\varphi }_c{P}_{\text{n}}$, $199.5\text{ft-kips}$ for $M_{\text{ux}}$, $373.48\text{ft-kips}$ for ${\varphi }_{\text{b}}{M}_{\text{nx}}$, $64.5\text{ft}\text{.kips}$ for $M_{\text{uy}}$ and $138.45\text{ft-kips}$ for ${\varphi }_{\text{b}}{M}_{\text{ny}}$ in Equation (XIII).

$\begin{array}{c}\frac{0.5P_{\text{u}}}{{\varphi }_{\text{c}}{P}_{\text{n}}}+\frac{8}{9}\left(\frac{M_{\text{ux}}}{{\varphi }_{\text{b}}{M}_{\text{nx}}}+\frac{M_{\text{uy}}}{{\varphi }_{\text{b}}{M}_{\text{ny}}}\right)=\frac{\text{0}\text{.5}\times \text{120}\text{kips}}{578.03\text{kips}}+\frac{8}{9}\left(\frac{199.5\text{ft-kips}}{373.48\text{ft-kips}}+\frac{64.5\text{ft-kips}}{138.45\text{ft-kips}}\right)\\ =0.1038+\frac{8}{9}\left(0.5341+0.4622\right)\\ =0.0847+\frac{8}{9}\left(0.9963\right)\\ =0.0847+0.8856\end{array}$

Further solve the above equation.

$0.9703<1$

Hence, it is safe to use $\text{W}14\times 68$.

Further check $W12\times 72$ as it is lightest among all, therefore from table $\text{6-2}$, ${\varphi }_c{P}_{\text{n}}=781.25\text{kips}$, ${\varphi }_{\text{b}}{M}_{\text{nx}}=416.13\text{ft-kips,}$ and ${\varphi }_{\text{b}}{M}_{\text{ny}}=203.4\text{ft-kips}$ for final check.

Write the expression $\text{AISC}$ equation $\text{H1-1a}$.

$\frac{0.5P_{\text{u}}}{{\varphi }_{\text{c}}{P}_{\text{n}}}+\frac{8}{9}\left(\frac{M_{\text{ux}}}{{\varphi }_{\text{b}}{M}_{\text{nx}}}+\frac{M_{\text{uy}}}{{\varphi }_{\text{b}}{M}_{\text{ny}}}\right)$ ..... (XIV)

Here, load factor is $P_{\text{u}}$ and the factor which deals with the axial strength of member is ${\varphi }_c{P}_{\text{n}}$, moment factor in x-direction is ${\varphi }_{\text{b}}{M}_{\text{nx}}$, moment factor in y-direction is ${\varphi }_{\text{b}}{M}_{\text{ny}}$, ultimate moment in x direction is $M_{\text{ux}},$ and ultimate moment in y-direction is $M_{\text{uy}}$.

Substitute $128.8\text{kips}$ for $P_{\text{u}}$, $781.25\text{kips}$ for ${\varphi }_c{P}_{\text{n}}$, $299.4\text{ft-kips}$ for $M_{\text{ux}}$, $446.67\text{ft-kips}$ for ${\varphi }_{\text{b}}{M}_{\text{nx}}$, $43.4\text{ft}\text{.kips}$ for $M_{\text{uy}},$ and $203.4\text{ft-kips}$ for ${\varphi }_{\text{b}}{M}_{\text{ny}}$ in Equation (XIV).

$\begin{array}{c}\frac{0.5P_{\text{u}}}{{\varphi }_{\text{c}}{P}_{\text{n}}}+\frac{8}{9}\left(\frac{M_{\text{ux}}}{{\varphi }_{\text{b}}{M}_{\text{nx}}}+\frac{M_{\text{uy}}}{{\varphi }_{\text{b}}{M}_{\text{ny}}}\right)=\frac{\text{0}\text{.5}\times \text{128}\text{.8}\text{kips}}{781.25\text{kips}}+\frac{8}{9}\left(\frac{299.4\text{ft-kips}}{403.4\text{ft-kips}}+\frac{43.4\text{ft-kips}}{203.4\text{ft-kips}}\right)\\ =0.0824+\frac{8}{9}\left(0.7421+0.213\right)\\ =0.0824+\frac{8}{9}\left(0.9551\right)\\ =0.0824+0.849\end{array}$

Further solve the above equation.

$0.931<1$

Hence, it is safe to use $\text{W}12\times 72$.

Conclusion:

Thus, use $\text{W}12\times 72$.

To determine

(b)

The best suitable $\text{W}$ shape of $\text{A992}$ steel on the basis of Allowed Stress Design.

Answer to Problem 6.8.7P

The best suitable $\text{W}$ shape of $\text{A992}$ steel on the basis of Allowed Stress Design is $W12\times 72$.

Explanation of Solution

Calculation:

Write the equation to obtain the axial service load.

$P_{\text{u}}=P_{\text{D}}+P_{\text{L}}$ ..... (XV)

Here, load factor is $P_{\text{u}}$, axial live load is $P_{\text{L}},$ and axial dead load is $P_{\text{D}}$.

Calculate the dead load:

$\begin{array}{c}{P}_{\text{D}}=\frac{25}{100}\times 80\text{kips}\\ =20\text{kips}\end{array}$

Calculate the live load:

$\begin{array}{c}{P}_{\text{L}}=\frac{75}{100}\times 80\text{kips}\\ =60\text{kips}\end{array}$

Substitute $60\text{kips}$ for $P_{\text{L}}$ and $20\text{kips}$ for $P_{\text{D}}$ in Equation (XV).

$\begin{array}{c}{P}_{\text{u}}=20\text{kips}+60\text{kips}\\ =\text{80}\text{kips}\end{array}$

Write the equation to obtain factored bending moment at the bottom of the member for braced condition along x-axis.

$M_{\text{nbx}}=M_{\text{Dbx}}+M_{\text{Lbx}}$ ..... (XVI)

Here, factored bending moment at bottom along x-axis is $M_{\text{nbx}}$, bending moment due to live load at bottom along x-axis is $M_{\text{Lbx}},$ and the bending moment due to dead load at bottom along x-axis is $M_{\text{Lbx}}$.

Calculate the bending moment along x-direction at bottom due to dead load.

$\begin{array}{c}{M}_{\text{Dbx}}=\frac{25}{100}\times 27\text{kips}\\ =6.75\text{kips}\end{array}$

Calculate the bending moment along x-direction at bottom due to live load.

$\begin{array}{c}{M}_{\text{Lbx}}=\frac{75}{100}\times 27\text{kips}\\ =20.25\text{kips}\end{array}$

Substitute $20.25\text{ft-kips}$ for $M_{\text{Lbx}}$ and $6.75\text{ft-kips}$ for $M_{\text{Dbx}}$ in Equation (XVI).

$\begin{array}{c}{M}_{\text{nbx}}{}_{}=6.75\text{ft-kips}+20.25\text{ft-kips}\\ =27\text{ft-kips}\end{array}$

Write the equation to obtain factored bending moment at the top of the member for braced condition along x-axis.

$M_{\text{ntx}}=M_{\text{Dtx}}+M_{\text{Ltx}}$ ..... (XVII)

Here, factored bending moment at top along x-axis is $M_{\text{ntx}}$, bending moment due to live load at top along x-axis is $M_{\text{Ltx}},$ and the bending moment due to dead load at top along x-axis is $M_{\text{Ltx}}$.

Calculate the bending moment along x-direction at top due to dead load.

$\begin{array}{c}{M}_{\text{Dtx}}=\frac{25}{100}\times 133\text{kips}\\ =33.25\text{kips}\end{array}$

Calculate the bending moment along x-direction at top due to live load.

$\begin{array}{c}{M}_{\text{Ltx}}=\frac{75}{100}\times 133\text{kips}\\ =99.75\text{kips}\end{array}$

Substitute $99.75\text{ft-kips}$ for $M_{\text{Ltx}}$ and $33.25\text{ft-kips}$ for $M_{\text{Dtx}}$ in Equation (XVII).

$\begin{array}{c}{M}_{\text{ntx}}=33.25\text{ft-kips}+99.75\text{ft-kips}\\ =133\text{ft-kips}\end{array}$

Write the equation to obtain factored bending moment at the bottom of the member for braced condition along y-axis.

$M_{\text{nby}}=M_{\text{Dby}}+M_{\text{Lby}}$ ..... (XVIII)

Here, factored bending moment at bottom along y-axis is $M_{\text{nby}}$, bending moment due to live load at bottom along y-axis is $M_{\text{Lby}},$ and the bending moment due to dead load at bottom along y-axis is $M_{\text{Lby}}$.

Calculate the bending moment along y-direction at bottom due to dead load.

$\begin{array}{c}{M}_{\text{Dby}}=\frac{25}{100}\times 9\text{kips}\\ =2.25\text{kips}\end{array}$

Calculate the bending moment along y-direction at bottom due to live load.

$\begin{array}{c}{M}_{\text{Lby}}=\frac{75}{100}\times 9\text{kips}\\ =6.75\text{kips}\end{array}$

Substitute $6.75\text{ft-kips}$ for $M_{\text{Lby}}$ and $2.25\text{ft-kips}$ for $M_{\text{Dby}}$ in Equation (XVIII).

$\begin{array}{c}{M}_{\text{nby}}=2.25\text{ft-kips}+6.75\text{ft-kips}\\ =9\text{ft-kips}\end{array}$

Write the equation to obtain factored bending moment at the top of the member for braced condition along y-axis.

$M_{\text{nty}}=M_{\text{Dty}}+M_{\text{Lty}}$ ..... (XIX)

Here, factored bending moment at top along y-axis is $M_{\text{nty}}$, bending moment due to live load at top along y-axis is $M_{\text{Lty}},$ and the bending moment due to dead load at top along y-axis is $M_{\text{Lty}}$.

Calculate the bending moment along y-direction at top due to dead load.

$\begin{array}{c}{M}_{\text{Dty}}=\frac{25}{100}\times 43\text{kips}\\ =10.75\text{kips}\end{array}$

Calculate the bending moment along y-direction at top due to live load.

$\begin{array}{c}{M}_{\text{Lty}}=\frac{75}{100}\times 43\text{kips}\\ =32.25\text{kips}\end{array}$

Substitute $32.25\text{ft-kips}$ for $M_{\text{Lty}}$ and $10.75\text{ft-kips}$ for $M_{\text{Dty}}$ in Equation (XIX).

$\begin{array}{c}{M}_{\text{nty}}=10.75\text{ft-kips}+32.25\text{ft-kips}\\ =43\text{ft-kips}\end{array}$

Write the equation to obtain the ultimate moment along x-axis.

$M_{\text{ux}}=B_1{M}_{\text{ntx}}+B_2{M}_{\text{nbx}}$ ..... (XX)

Here, factor for braced condition is $B_1$, factor for side sway condition is $B_2$.

Substitute $1$ for $B_1$, $133\text{ft-kips}$ for $M_{\text{ntx}}$, $0$ for $B_2$ and $27\text{ft-kips}$ for $M_{\text{nbx}}$ in Equation (XX).

$\begin{array}{c}{M}_{\text{ux}}=\left(1\right)\left(133\text{ft-kips}\right)+\left(0\right)\left(27\text{ft-kips}\right)\\ =133\text{ft-kips}\end{array}$

Write the equation to obtain the ultimate moment along y-axis.

$M_{\text{uy}}=B_1{M}_{\text{nty}}+B_2{M}_{\text{nby}}$ ..... (XXI)

Here, factor for braced condition is $B_1$, factor for side sway condition is $B_2$.

Substitute $1$ for $B_1$, $43\text{ft-kips}$ for $M_{\text{nt}}$, $0$ for $B_2,$ and $9\text{ft-kips}$ for $M_{\text{nb}}$ in Equation (XXI).

$\begin{array}{c}{M}_{\text{ux}}=\left(1\right)\left(43\text{ft-kips}\right)+\left(0\right)\left(9\text{ft-kips}\right)\\ =43\text{ft-kips}\end{array}$

The unbraced length and effective length of the member are same.

$\begin{array}{c}{L}_{\text{b}}=kL\\ =16\text{ft}\end{array}$

Try $\text{W12}\times 72$, from table $\text{6-2}$, $P_{\text{n}}/{\Omega }_{\text{c}}=471.69\text{kips}$, $M_{\text{nx}}/{\Omega }_{\text{b}}=306.51\text{ft-kips}$ and $M_{\text{nx}}/{\Omega }_{\text{b}}=152.2\text{ft-kips}$.

Write the expression to determine which interaction equation to use.

$\frac{P_{\text{u}}}{P_{\text{n}}/{\Omega }_{\text{c}}}>0.2$ ..... (XXII)

Here, load factor is $P_{\text{u}}$ and the factor which deals with the axial strength of member is $P_{\text{n}}/{\Omega }_{\text{c}}$.

Substitute $80\text{kips}$ for $P_{\text{u}}$ and $471.69\text{kips}$ for $P_{\text{n}}/{\Omega }_{\text{c}}$ in Equation (XXII).

$\begin{array}{c}\frac{P_{\text{u}}}{P_{\text{n}}/{\Omega }_{\text{c}}}=\frac{80\text{kips}}{471.69\text{kips}}\\ =0.169<0.2\end{array}$

Therefore, $\text{AISC}$ equation $\text{H1-1a}$. controls.

Write the expression $\text{AISC}$ equation $\text{H1-1a}$.

$\frac{0.5P_{\text{u}}}{P_{\text{n}}/{\Omega }_{\text{c}}}+\frac{9}{8}\left(\frac{M_{\text{ux}}}{M_{\text{nx}}/{\Omega }_{\text{b}}}+\frac{M_{\text{uy}}}{M_{\text{ny}}/{\Omega }_{\text{b}}}\right)$ ..... (XXIII)

Here, load factor is $P_{\text{u}}$ and the factor which deals with the axial strength of member is $P_{\text{n}}/{\Omega }_{\text{c}}$, moment factor in x-direction is $M_{\text{nx}}/{\Omega }_{\text{b}}$, moment factor in y-direction is $M_{\text{ny}}/{\Omega }_{\text{b}}$, ultimate moment in x direction is $M_{\text{ux}},$ and ultimate moment in y-direction is $M_{\text{uy}}$.

Substitute $80\text{kips}$ for $P_{\text{u}}$, $471.69\text{kips}$ for $P_{\text{n}}/{\Omega }_{\text{c}}$, $133\text{ft-kips}$ for $M_{\text{ux}}$, $280.89\text{ft-kips}$ for $M_{\text{nx}}/{\Omega }_{\text{b}}$, $43\text{ft}\text{.kips}$ for $M_{\text{uy}},$ and $138.12\text{ft-kips}$ for $M_{\text{ny}}/{\Omega }_{\text{b}}$ in Equation (XXIII).

$\begin{array}{c}\frac{0.5P_{\text{u}}}{P_{\text{n}}/{\Omega }_{\text{c}}}+\frac{9}{8}\left(\frac{M_{\text{ux}}}{M_{\text{nx}}/{\Omega }_{\text{b}}}+\frac{M_{\text{uy}}}{M_{\text{ny}}/{\Omega }_{\text{b}}}\right)=\frac{\text{0}\text{.5}\times 80\text{kips}}{471.69\text{kips}}+\frac{9}{8}\left(\frac{133\text{ft-kips}}{280.89\text{ft-kips}}+\frac{43\text{ft-kips}}{138.12\text{ft-kips}}\right)\\ =0.0848+\frac{9}{8}\left(0.4734+0.3113\right)\\ =0.0848+\frac{9}{8}\left(0.7847\right)\\ =0.0848+0.8827\end{array}$

Further solve the above equation.

$0.9675<1$

Hence, it is safe to use $\text{W}12\times 72$.

Conclusion:

Thus, use $\text{W}12\times 72$.