Chapter 6, Problem 6.8.6P

To determine

(a)

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

Answer to Problem 6.8.6P

The best suitable $\text{W}$ shape of $\text{A992}$ steel is on the basis of load resistance and factor design is $\text{W}12\times 79$.

Explanation of Solution

Given:

The axial load is $92\text{kips}$.

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

The moment at top in x-direction is $160\text{ft-kips}$.

The moment at top in y-direction is $24\text{ft-kips}$.

The moment at bottom in x-direction is $214\text{ft-kips}$.

The moment at bottom in y-direction is $31\text{ft-kips}$.

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 $P_{\text{D}}$.

Substitute $\left(92/2\right)\text{kips}$ for $P_{\text{L}}$ as $50\%$ is the live load given and $\left(92/2\right)\text{kips}$ for $P_{\text{D}}$ as $50\%$ is the dead load in Equation (I).

$\begin{array}{c}{P}_{\text{u}}=1.2\left(\frac{92}{2}\text{kips}\right)+1.6\left(\frac{92}{2}\text{kips}\right)\\ =55.2\text{kips}+73.6\text{kips}\\ \text{=128}\text{.8}\text{kips}\end{array}$

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

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

Substitute $\left(160/2\right)\text{ft-kips}$ for $M_{\text{Lb}}$ as $50\%$ of the load is live load, and $\left(160/2\right)\text{ft-kips}$ for $M_{\text{Db}}$ as $50\%$ of the load is dead load in Equation (II).

$\begin{array}{c}{M}_{\text{nbx}}=1.2\left(\frac{160}{2}\text{ft-kips}\right)+1.6\left(\frac{160}{2}\text{ft-kips}\right)\\ =96\text{ft-kips}+128\text{ft-kips}\\ =224\text{ft-kips}\end{array}$

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 $M_{\text{Ltx}}$.

Substitute $\left(214/2\right)\text{ft-kips}$ for $M_{\text{Ltx}}$ as $50\%$ of the load is live load, and $\left(214/2\right)\text{ft-kips}$ for $M_{\text{Dtx}}$ as $50\%$ of the load is dead load in Equation (III).

$\begin{array}{c}{M}_{\text{ntx}}=1.2\left(\frac{214}{2}\text{ft-kips}\right)+1.6\left(\frac{214}{2}\text{ft-kips}\right)\\ =128.4\text{ft-kips}+171.2\text{ft-kips}\\ =299.6\text{ft-kips}\end{array}$

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 $M_{\text{Lby}}$.

Substitute $\left(24/2\right)\text{ft-kips}$ for $M_{\text{Lb}}$ as $50\%$ of the load is live load and $\left(24/2\right)\text{ft-kips}$ for $M_{\text{Db}}$ as $50\%$ of the load is dead load in Equation (IV).

$\begin{array}{c}{M}_{\text{nby}}=1.2\left(\frac{24}{2}\text{ft-kips}\right)+1.6\left(\frac{24}{2}\text{ft-kips}\right)\\ =14.4\text{ft-kips}+19.2\text{ft-kips}\\ =33.6\text{ft-kips}\end{array}$

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

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

Substitute $\left(31/2\right)\text{ft-kips}$ for $M_{\text{Lty}}$ as $50\%$ of the load is live load and $\left(31/2\right)\text{ft-kips}$ for $M_{\text{Dty}}$ as $50\%$ of the load is dead load in Equation (V).

$\begin{array}{c}{M}_{\text{nty}}=1.2\left(\frac{31}{2}\text{ft-kips}\right)+1.6\left(\frac{31}{2}\text{ft-kips}\right)\\ =18.6\text{ft-kips}+24.8\text{ft-kips}\\ =43.4\text{ft-kips}\end{array}$

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 $B_1$, $299.6\text{ft-kips}$ for $M_{\text{ntx}}$, $0$ for $B_2$ and $224\text{ft-kips}$ for $M_{\text{nbx}}$ in Equation (VI).

$\begin{array}{c}{M}_{\text{ux}}=\left(1\right)\left(299.6\text{ft-kips}\right)+\left(0\right)\left(224\text{ft-kips}\right)\\ =299.6\text{ft-kips}\end{array}$

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 $B_1$, $43.4\text{ft-kips}$ for $M_{\text{nty}}$, $0$ for $B_2$ and $33.6\text{ft-kips}$ for $M_{\text{nby}}$ in Equation (VII).

$\begin{array}{c}{M}_{\text{ux}}=\left(1\right)\left(43.4\text{ft-kips}\right)+\left(0\right)\left(33.6\text{ft-kips}\right)\\ =43.4\text{ft-kips}\end{array}$

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

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

Try $\text{W}10\times 100$ ; from table $\text{6-2}$, ${\varphi }_c{P}_{\text{n}}=900.9\text{kips}$, ${\varphi }_{\text{b}}{M}_{\text{nx}}=460.56\text{ft-kips}$ and ${\varphi }_{\text{b}}{M}_{\text{ny}}=228.5\text{ft-kips}$.

Write the expression to determine 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 $128.8\text{kips}$ for $P_{\text{u}}$ and $900.9\text{kips}$ for ${\varphi }_c{P}_{\text{n}}$ in Equation (VIII).

$\begin{array}{l}\frac{128.8\text{kips}}{900.9\text{kips}}\\ 0.142<0.2\end{array}$

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 $128.8\text{kips}$ for $P_{\text{u}}$, $900.9\text{kips}$ for ${\varphi }_c{P}_{\text{n}}$, $299.4\text{ft-kips}$ for $M_{\text{ux}}$, $460.56\text{ft-kips}$ for ${\varphi }_{\text{b}}{M}_{\text{nx}}$, $43.4\text{ft}\text{.kips}$ for $M_{\text{uy}}$ and $228.5\text{ft-kips}$ for ${\varphi }_{\text{b}}{M}_{\text{ny}}$ in Equation (IX).

$\begin{array}{l}\frac{\text{0}\text{.5}\times \text{128}\text{.8}\text{kips}}{900.9\text{kips}}+\frac{8}{9}\left(\frac{299.4\text{ft-kips}}{460.56\text{ft-kips}}+\frac{43.4\text{ft-kips}}{228.5\text{ft-kips}}\right)\\ 0.0714+\frac{8}{9}\left(0.65+0.189\right)\\ 0.0714+\frac{8}{9}\left(0.839\right)\\ 0.0714+0.745\end{array}$

Further solve the above equation.

$0.816<1$

Hence it is safe to use $\text{W}10\times 100$.

Try $\text{W}12\times 79$, 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 ${\varphi }_{\text{b}}{M}_{\text{ny}}=203.4\text{ft-kips}$.

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 $128.8\text{kips}$ for $P_{\text{u}}$ and $781.25\text{kips}$ for ${\varphi }_c{P}_{\text{n}}$ in Equation (X).

$\begin{array}{l}\frac{128.8\text{kips}}{781.25\text{kips}}\\ 0.1648<0.2\end{array}$

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

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

$\frac{0.5P_{\text{u}}}{{\varphi }_{\text{c}}{P}_{\text{n}}}+\frac{8}{9}\left(\frac{M_{\text{ux}}}{{\varphi }_{\text{b}}{M}_{\text{nx}}}+\frac{M_{\text{uy}}}{{\varphi }_{\text{b}}{M}_{\text{ny}}}\right)$ ..... (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 $128.8\text{kips}$ for $P_{\text{u}}$, $781.25\text{kips}$ for ${\varphi }_c{P}_{\text{n}}$, $299.4\text{ft-kips}$ for $M_{\text{ux}}$, $416.13\text{ft-kips}$ 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 (XI).

$\begin{array}{l}\frac{\text{0}\text{.5}\times \text{128}\text{.8}\text{kips}}{781.25\text{kips}}+\frac{8}{9}\left(\frac{299.4\text{ft-kips}}{416.13\text{ft-kips}}+\frac{43.4\text{ft-kips}}{203.4\text{ft-kips}}\right)\\ 0.0824+\frac{8}{9}\left(0.719+0.213\right)\\ 0.0824+\frac{8}{9}\left(0.932\right)\\ 0.0824+0.828\end{array}$

Further solve the above equation.

$0.910<1$

Hence, it is safe to use $\text{W}12\times 79$.

Try $\text{W}14\times 82$, from table $\text{6-2}$, ${\varphi }_c{P}_{\text{n}}=694.44\text{kips}$, ${\varphi }_{\text{b}}{M}_{\text{nx}}=460.56\text{ft-kips}$ and ${\varphi }_{\text{b}}{M}_{\text{ny}}=228.5\text{ft-kips}$.

Write the expression to determine which interaction equation to use.

$\frac{P_{\text{u}}}{{\varphi }_c{P}_{\text{n}}}>0.2$ ..... (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 $128.8\text{kips}$ for $P_{\text{u}}$ and $694.44\text{kips}$ for ${\varphi }_c{P}_{\text{n}}$ in Equation (XII).

$\begin{array}{l}\frac{128.8\text{kips}}{694.44\text{kips}}\\ 0.1854<0.2\end{array}$

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 $128.8\text{kips}$ for $P_{\text{u}}$, $694.44\text{kips}$ for ${\varphi }_c{P}_{\text{n}}$, $299.4\text{ft-kips}$ for $M_{\text{ux}}$, $462.96\text{ft-kips}$ for ${\varphi }_{\text{b}}{M}_{\text{nx}}$, $43.4\text{ft}\text{.kips}$ for $M_{\text{uy}}$ and $168.03\text{ft-kips}$ for ${\varphi }_{\text{b}}{M}_{\text{ny}}$ in Equation (XIII).

$\begin{array}{l}\frac{\text{0}\text{.5}\times \text{128}\text{.8}\text{kips}}{694.44\text{kips}}+\frac{8}{9}\left(\frac{299.4\text{ft-kips}}{462.96\text{ft-kips}}+\frac{43.4\text{ft-kips}}{168.03\text{ft-kips}}\right)\\ 0.0927+\frac{8}{9}\left(0.6467+0.258\right)\\ 0.0927+\frac{8}{9}\left(0.9049\right)\\ 0.0927+0.8043\end{array}$

Further solve the above equation.

$0.897<1$

Hence, it is safe to use $\text{W}14\times 82$.

Further check $\text{W}12\times 79$ as it is lightest among all, 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).

$\begin{array}{l}\frac{\text{0}\text{.5}\times \text{128}\text{.8}\text{kips}}{781.25\text{kips}}+\frac{8}{9}\left(\frac{299.4\text{ft-kips}}{403.4\text{ft-kips}}+\frac{43.4\text{ft-kips}}{203.4\text{ft-kips}}\right)\\ 0.0824+\frac{8}{9}\left(0.7421+0.213\right)\\ 0.0824+\frac{8}{9}\left(0.9551\right)\\ 0.0824+0.849\end{array}$

Further solve the above equation.

$0.931<1$

Hence, it is safe to use $\text{W}12\times 79$.

Conclusion:

Thus, use $\text{W}12\times 79$.

To determine

(b)

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

Answer to Problem 6.8.6P

The best suitable $\text{W}$ shape of $\text{A992}$ steel on the basis of Allowed Stress Design is $\text{W}12\times 79$.

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 $P_{\text{D}}$.

Substitute $\left(92/2\right)\text{kips}$ for $P_{\text{L}}$ as $50\%$ is the live load given and $\left(92/2\right)\text{kips}$ for $P_{\text{D}}$ as $50\%$ is the dead load in Equation (XV).

$\begin{array}{c}{P}_{\text{u}}=\left(\frac{92}{2}\text{kips}\right)+\left(\frac{92}{2}\text{kips}\right)\\ =46\text{kips}+46\text{kips}\\ =9\text{2}\text{kips}\end{array}$

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 $M_{\text{Lbx}}$.

Substitute $\left(160/2\right)\text{ft-kips}$ for $M_{\text{Lb}}$ as $50\%$ of the load is live load and $\left(160/2\right)\text{ft-kips}$ for $M_{\text{Db}}$ as $50\%$ of the load is dead load in Equation (XVI).

$\begin{array}{c}{M}_{\text{nbx}}=\left(\frac{160}{2}\text{ft-kips}\right)+\left(\frac{160}{2}\text{ft-kips}\right)\\ =80\text{ft-kips}+80\text{ft-kips}\\ =160\text{ft-kips}\end{array}$

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 $M_{\text{Ltx}}$.

Substitute $\left(214/2\right)\text{ft-kips}$ for $M_{\text{Ltx}}$ as $50\%$ of the load is live load and $\left(214/2\right)\text{ft-kips}$ for $M_{\text{Dtx}}$ as $50\%$ of the load is dead load in Equation (XVII).

$$

$\begin{array}{c}{M}_{\text{ntx}}=\left(\frac{214}{2}\text{ft-kips}\right)+\left(\frac{214}{2}\text{ft-kips}\right)\\ =107\text{ft-kips}+107\text{ft-kips}\\ =214\text{ft-kips}\end{array}$

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 $M_{\text{Lby}}$.

Substitute $\left(24/2\right)\text{ft-kips}$ for $M_{\text{Lb}}$ as $50\%$ of the load is live load and $\left(24/2\right)\text{ft-kips}$ for $M_{\text{Db}}$ as $50\%$ of the load is dead load in Equation (XVIII).

$\begin{array}{c}{M}_{\text{nby}}=\left(\frac{24}{2}\text{ft-kips}\right)+\left(\frac{24}{2}\text{ft-kips}\right)\\ =12\text{ft-kips}+12\text{ft-kips}\\ =24\text{ft-kips}\end{array}$

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 $M_{\text{Lty}}$.

Substitute $\left(31/2\right)\text{ft-kips}$ for $M_{\text{Lty}}$ as $50\%$ of the load is live load and $\left(31/2\right)\text{ft-kips}$ for $M_{\text{Dty}}$ as $50\%$ of the load is dead load in Equation (XIX).

$\begin{array}{c}{M}_{\text{nty}}=\left(\frac{31}{2}\text{ft-kips}\right)+\left(\frac{31}{2}\text{ft-kips}\right)\\ =15.5\text{ft-kips}+15.5\text{ft-kips}\\ =31\text{ft-kips}\end{array}$

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 $B_1$, $214\text{ft-kips}$ for $M_{\text{nt}}$, $0$ for $B_2$ and $160\text{ft-kips}$ for $M_{\text{nb}}$ in Equation (XX).

$\begin{array}{c}{M}_{\text{ux}}=\left(1\right)\left(214\text{ft-kips}\right)+\left(0\right)\left(160\text{ft-kips}\right)\\ =214\text{ft-kips}\end{array}$

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 $B_1$, $31\text{ft-kips}$ for $M_{\text{nt}}$, $0$ for $B_2$ and $24\text{ft-kips}$ for $M_{\text{nb}}$ in Equation (XXI).

$\begin{array}{c}{M}_{\text{ux}}=\left(1\right)\left(31\text{ft-kips}\right)+\left(0\right)\left(24\text{ft-kips}\right)\\ =31\text{ft-kips}\end{array}$

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

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

Try $\text{W}10\times 100$, from table $\text{6-2}$, $P_{\text{n}}/{\Omega }_{\text{c}}=598.8\text{kips}$, $M_{\text{nx}}/{\Omega }_{\text{b}}=306.51\text{ft-kips}$ and $M_{\text{nx}}/{\Omega }_{\text{b}}=152.2\text{ft-kips}$.

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

$\frac{P_{\text{u}}}{P_{\text{n}}/{\Omega }_{\text{c}}}>0.2$ ..... (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 $92\text{kips}$ for $P_{\text{u}}$ and $598.8\text{kips}$ for $P_{\text{n}}/{\Omega }_{\text{c}}$ in Equation (XXII).

$\begin{array}{l}\frac{92\text{kips}}{598.8\text{kips}}\\ 0.153<0.2\end{array}$

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

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

$\frac{0.5P_{\text{u}}}{P_{\text{n}}/{\Omega }_{\text{c}}}+\frac{8}{9}\left(\frac{M_{\text{ux}}}{M_{\text{nx}}/{\Omega }_{\text{b}}}+\frac{M_{\text{uy}}}{M_{\text{ny}}/{\Omega }_{\text{b}}}\right)$ ..... (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 $92\text{kips}$ for $P_{\text{u}}$, $598.8\text{kips}$ for $P_{\text{n}}/{\Omega }_{\text{c}}$, $214\text{ft-kips}$ for $M_{\text{ux}}$, $306.51\text{ft-kips}$ for $M_{\text{nx}}/{\Omega }_{\text{b}}$, $31\text{ft}\text{.kips}$ for $M_{\text{uy}}$ and $152.2\text{ft-kips}$ for $M_{\text{ny}}/{\Omega }_{\text{b}}$ in Equation (XXIII).

$\begin{array}{l}\frac{\text{0}\text{.5}\times 92\text{kips}}{598.8\text{kips}}+\frac{8}{9}\left(\frac{214\text{ft-kips}}{306.51\text{ft-kips}}+\frac{31\text{ft-kips}}{152.2\text{ft-kips}}\right)\\ 0.0768+\frac{8}{9}\left(0.6981+0.2036\right)\\ 0.0768+\frac{8}{9}\left(0.9017\right)\\ 0.0768+0.8015\end{array}$

Further solve the above equation.

$0.878<1$

Hence, it is safe to use $\text{W}10\times 100$.

Try $\text{W}12\times 79$, from table $\text{6-2}$, $P_{\text{n}}/{\Omega }_{\text{c}}=520.83\text{kips}$, $M_{\text{nx}}/{\Omega }_{\text{b}}=306.51\text{ft-kips}$ and $M_{\text{nx}}/{\Omega }_{\text{b}}=135.5\text{ft-kips}$.

Write the expression to determine which interaction equation to use.

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

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 $92\text{kips}$ for $P_{\text{u}}$ and $520.83\text{kips}$ for $P_{\text{n}}/{\Omega }_{\text{c}}$ in Equation (XXIV).

$\begin{array}{l}\frac{92\text{kips}}{520.83\text{kips}}\\ 0.176<0.2\end{array}$

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

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

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

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 $92\text{kips}$ for $P_{\text{u}}$, $520.83\text{kips}$ for $P_{\text{n}}/{\Omega }_{\text{c}}$, $214\text{ft-kips}$ for $M_{\text{ux}}$, $306.51\text{ft-kips}$ for $M_{\text{nx}}/{\Omega }_{\text{b}}$, $31\text{ft}\text{.kips}$ for $M_{\text{uy}}$ and $135.50\text{ft-kips}$ for $M_{\text{ny}}/{\Omega }_{\text{b}}$ in Equation (XXV).

$\begin{array}{l}\frac{\text{0}\text{.5}\times 92\text{kips}}{520.83\text{kips}}+\frac{8}{9}\left(\frac{214\text{ft-kips}}{306.51\text{ft-kips}}+\frac{31\text{ft-kips}}{135.50\text{ft-kips}}\right)\\ 0.0883+\frac{8}{9}\left(0.6981+0.2287\right)\\ 0.0883+\frac{8}{9}\left(0.9017\right)\\ 0.0883+0.8015\end{array}$

Further solve the above equation.

$0.8898<1$

Hence, it is safe to use $\text{W}12\times 79$.

Try $\text{W}14\times 82$, from table $\text{6-2}$, $P_{\text{n}}/{\Omega }_{\text{c}}=462.96\text{kips}$, $M_{\text{nx}}/{\Omega }_{\text{b}}=307.57\text{ft-kips}$ and $M_{\text{nx}}/{\Omega }_{\text{b}}=111.80\text{ft-kips}$.

Write the expression to determine which interaction equation to use.

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

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 $92\text{kips}$ for $P_{\text{u}}$ and $462.96\text{kips}$ for $P_{\text{n}}/{\Omega }_{\text{c}}$ in Equation (XXVI).

$\begin{array}{l}\frac{92\text{kips}}{492.96\text{kips}}\\ 0.1866<0.2\end{array}$

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

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

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

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 $92\text{kips}$ for $P_{\text{u}}$, $492.96\text{kips}$ for $P_{\text{n}}/{\Omega }_{\text{c}}$, $214\text{ft-kips}$ for $M_{\text{ux}}$, $307.57\text{ft-kips}$ for $M_{\text{nx}}/{\Omega }_{\text{b}}$, $31\text{ft}\text{.kips}$ for $M_{\text{uy}},$ and $111.80\text{ft-kips}$ for $M_{\text{ny}}/{\Omega }_{\text{b}}$ in Equation (XXVII).

$\begin{array}{l}\frac{\text{0}\text{.5}\times 92\text{kips}}{492.96\text{kips}}+\frac{8}{9}\left(\frac{214\text{ft-kips}}{307.57\text{ft-kips}}+\frac{31\text{ft-kips}}{111.80\text{ft-kips}}\right)\\ 0.0933+\frac{8}{9}\left(0.695+0.2772\right)\\ 0.0933+\frac{8}{9}\left(0.9722\right)\\ 0.0933+0.8641\end{array}$

Further solve the above equation.

$0.9574<1$

Hence, it is safe to use $\text{W14}\times 82$.

Further check $\text{W}12\times 79$ as it lightest among all, from table $\text{6-2}$, $P_{\text{n}}/{\Omega }_{\text{c}}=520.83\text{kips}$, $M_{\text{nx}}/{\Omega }_{\text{b}}=347.221\text{ft-kips,}$ and $M_{\text{nx}}/{\Omega }_{\text{b}}=111.80\text{ft-kips}$.

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

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

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 $92\text{kips}$ for $P_{\text{u}}$, $520.83\text{kips}$ for $P_{\text{n}}/{\Omega }_{\text{c}}$, $214\text{ft-kips}$ for $M_{\text{ux}}$, $347.22\text{ft-kips}$ for $M_{\text{nx}}/{\Omega }_{\text{b}}$, $31\text{ft}\text{.kips}$ for $M_{\text{uy}},$ and $111.80\text{ft-kips}$ for $M_{\text{ny}}/{\Omega }_{\text{b}}$ in Equation (XXVIII).

$\begin{array}{l}\frac{\text{0}\text{.5}\times 92\text{kips}}{520.83\text{kips}}+\frac{8}{9}\left(\frac{214\text{ft-kips}}{347.221\text{ft-kips}}+\frac{31\text{ft-kips}}{111.80\text{ft-kips}}\right)\\ 0.0883+\frac{8}{9}\left(0.6163+0.2772\right)\\ 0.0883+\frac{8}{9}\left(0.8935\right)\\ 0.0883+0.7942\end{array}$

Further solve the above equation.

$0.8825<1$

Hence, it is safe to use $\text{W}12\times 79$.

Conclusion:

Thus, use $\text{W}12\times 79$.