Chapter 6, Problem 6.2.1P

To determine

(a)

The satisfaction of AISC interaction equation using LRFD.

Answer to Problem 6.2.1P

The member satisfies the AISC interaction equation.

Explanation of Solution

Given:

The load is $250\text{kips}$.

The length of member is $14\text{feet}$.

The value of $k_{\text{x}}$ and $k_{\text{y}}$ are $1.0$ and $1.0$.

The flexural load is $240\text{ft}\cdot \text{kips}$.

Concept Used:

Write the LRFD interaction equation.

$\frac{P_{\text{u}}}{{\varphi }_{\text{c}}{P}_{\text{n}}}+\frac{8}{9}\left(\frac{M_{\text{ux}}}{{\varphi }_{\text{b}}{M}_{\text{nx}}}+\frac{M_{\text{uy}}}{{\varphi }_{\text{b}}{M}_{\text{ny}}}\right)\le 1.0$ ...... (I)

Here, the factored load is $P_{\text{u}}$, the axial compressive design strength is ${\varphi }_{\text{c}}P$, the flexural load about x and y-axis are $M_{\text{ux}}$ and $M_{\text{uy}}$ and the nominal flexural strength about x and y-axis are ${\varphi }_{\text{b}}{M}_{\text{nx}}$ and ${\varphi }_{\text{b}}{M}_{\text{ny}}$.

Calculation:

Calculate the factored load.

$P_{\text{u}}=1.2\left(DL\right)+1.6\left(LL\right)$ ...... (II)

Here, the dead load is $DL$ and the live load is $LL$.

Substitute $125\text{kips}$ for $DL$ and $125\text{kips}$ for $LL$ in Equation (II).

$\begin{array}{c}{P}_{\text{u}}=1.2\left(125\text{kips}\right)+1.6\left(125\text{kips}\right)\\ =150\text{kips}+200\text{kips}\\ =350\text{kips}\end{array}$

Calculate the effective length of the member.

$L_{\text{e}}=k_{\text{y}}\times L$ ...... (III)

Here, the unsupported length is $L$ and the effective length factor is $k_{\text{y}}$.

Substitute $14\text{ft}$ for $L$ and $1.0$ for $k_{\text{y}}$ in Equation (III).

$\begin{array}{c}{L}_{\text{e}}=1.0\times 14\text{ft}\\ =14\text{ft}\end{array}$

Calculate the axial compressive design strength.

From the manual table, the axial compressive design strength of a $W12\times 106$ with $f_{\text{y}}=50\text{ksi}$ and $L_{\text{e}}=14\text{ft}$ is $1130\text{kips}$.

Calculate the nominal flexural strength about x-axis.

From the design table, calculate the nominal flexural strength about x-axis by using $C_{\text{b}}=1.0$ and $L_{\text{b}}=14\text{ft}$.

${\varphi }_{\text{b}}{M}_{\text{nx}}=597\text{ft}\cdot \text{kips}$

Calculate the flexural load about x-axis.

$M_{\text{ux}}=1.2\left(M_{\text{D}}\right)+1.6\left(M_{\text{L}}\right)$ ...... (IV)

Here, the flexural dead load is $M_{\text{D}}$ and the flexural live load is $M_{\text{L}}$.

Substitute $120\text{ft}\cdot \text{kips}$ for $M_{\text{D}}$ and $120\text{ft}\cdot \text{kips}$ for $M_{\text{L}}$ in Equation (IV).

$\begin{array}{c}{M}_{\text{ux}}=1.2\left(120\text{ft}\cdot \text{kips}\right)+1.6\left(120\text{ft}\cdot \text{kips}\right)\\ =144\text{ft}\cdot \text{kips}+192\text{ft}\cdot \text{kips}\\ =336\text{ft}\cdot \text{kips}\end{array}$

There is no bending about y-axis, therefore ${\varphi }_{\text{b}}{M}_{\text{ny}}=0\text{ft}\cdot \text{kips}$ and $M_{\text{uy}}=0\text{ft}\cdot \text{kips}$.

Write the equation to calculate the controlling interaction formula.

$r=\frac{P_u}{{\varphi }_c{P}_n}$ ...... (V)

Substitute $44\text{kips}$ for $P_{\text{u}}$ and $1130\text{kips}$ for ${\varphi }_c{P}_n$ in Equation (V).

$\begin{array}{c}r=\frac{350\text{kips}}{1130\text{kips}}\\ =0.3097\end{array}$

The value is greater than $0.2$.

Calculate the LRFD interaction equation.

Substitute $350\text{kips}$ for $P_{\text{u}}$, $1130\text{kips}$ for ${\varphi }_{\text{c}}{P}_{\text{n}}$, $336\text{ft}\cdot \text{kips}$ for $M_{\text{ux}}$, $0\text{ft}\cdot \text{kips}$ for $M_{\text{uy}}$, $597\text{ft}\cdot \text{kips}$ for ${\varphi }_{\text{b}}{M}_{\text{nx}}$ and $0\text{ft}\cdot \text{kips}$ for ${\varphi }_{\text{b}}{M}_{\text{ny}}$ in Equation (I).

$\begin{array}{c}\frac{350\text{kips}}{1130\text{kips}}+\frac{8}{9}\left(\frac{336\text{ft}\cdot \text{kips}}{597\text{ft}\cdot \text{kips}}+\frac{0\text{ft}\cdot \text{kips}}{0\text{ft}\cdot \text{kips}}\right)\le 1.0\\ 0.3097+\frac{8}{9}\left(0.5628\right)\le 1.0\\ 0.3097+0.5002\le 1.0\\ 0.8099\le 1.0\end{array}$

The interaction equation is satisfied.

Conclusion:

Therefore, the interaction equation is satisfied with the AISD interaction equation.

To determine

(b)

The satisfaction of AISC interaction equation using ASD.

Answer to Problem 6.2.1P

The member satisfies the AISC interaction equation.

Explanation of Solution

Concept Used:

Write the ASD interaction equation.

$\frac{P_{\text{u}}}{\frac{P_{\text{n}}}{{\Omega }_{\text{c}}}}+\frac{8}{9}\left(\frac{M_{\text{ux}}}{\frac{M_{\text{nx}}}{{\Omega }_{\text{c}}}}+\frac{M_{\text{uy}}}{\frac{M_{\text{ny}}}{{\Omega }_{\text{c}}}}\right)\le 1.0$ ...... (VI)

Here, the factored load is $P_{\text{u}}$, the allowed compressive strength is $\frac{P_{\text{n}}}{{\Omega }_{\text{c}}}$, the flexural load about x and y-axis are $M_{\text{ux}}$ and $M_{\text{uy}}$ and the nominal flexural strength about x and y-axis are $\frac{M_{\text{nx}}}{{\Omega }_{\text{c}}}$ and $\frac{M_{\text{ny}}}{{\Omega }_{\text{c}}}$.

Calculation:

Calculate the factored load.

$P_{\text{u}}=DL+LL$ ...... (VII)

Here, the dead load is $DL$ and the live load is $LL$.

Substitute $125\text{kips}$ for $DL$ and $125\text{kips}$ for $LL$ in Equation (VII).

$\begin{array}{c}{P}_{\text{u}}=125\text{kips}+125\text{kips}\\ =250\text{kips}\end{array}$

Calculate the allowed compressive strength.

From the manual table, the allowed compressive strength of a $W12\times 106$ with $f_{\text{y}}=50\text{ksi}$ and $L_{\text{e}}=14\text{f}$ is $755\text{kips}$.

Calculate the nominal flexural strength about x-axis.

From the design table, calculate the nominal flexural strength about x-axis by using $C_{\text{b}}=1.0$ and $L_{\text{b}}=14\text{f}$.

$\frac{M_{\text{nx}}}{{\Omega }_{\text{c}}}=398\text{ft}\cdot \text{kips}$

Calculate the flexural load about x-axis.

$M_{\text{ux}}=M_{\text{D}}+M_{\text{L}}$ ...... (VIII)

Here, the flexural dead load is $M_{\text{D}}$ and the flexural live load is $M_{\text{L}}$.

Substitute $120\text{ft}\cdot \text{kips}$ for $M_{\text{D}}$ and $120\text{ft}\cdot \text{kips}$ for $M_{\text{L}}$ in Equation (VIII).

$\begin{array}{c}{M}_{\text{ux}}=120\text{ft}\cdot \text{kips}+120\text{ft}\cdot \text{kips}\\ =240\text{ft}\cdot \text{kips}\end{array}$

There is no bending about y-axis, therefore $\frac{M_{\text{ny}}}{{\Omega }_{\text{c}}}=0\text{ft}\cdot \text{kips}$ and $M_{\text{uy}}=0\text{ft}\cdot \text{kips}$.

Write the equation to calculate the controlling interaction formula.

$r=\frac{P_u}{\frac{P_{\text{n}}}{{\Omega }_{\text{c}}}}$ ...... (IX)

Substitute $250\text{kips}$ for $P_{\text{u}}$ and $755\text{kips}$ for $\frac{P_{\text{n}}}{{\Omega }_{\text{c}}}$ in Equation (IX).

$\begin{array}{c}r=\frac{250\text{kips}}{755\text{kips}}\\ =0.331\end{array}$

The value is greater than $0.2$.

Calculate the ASD interaction equation.

Substitute $250\text{kips}$ for $P_{\text{u}}$, $755\text{kips}$ for $\frac{P_{\text{n}}}{{\Omega }_{\text{c}}}$, $240\text{ft}\cdot \text{kips}$ for $M_{\text{ux}}$, $0\text{ft}\cdot \text{kips}$ for $M_{\text{uy}}$, $398\text{ft}\cdot \text{kips}$ for $\frac{M_{\text{nx}}}{{\Omega }_{\text{c}}}$ and $0\text{ft}\cdot \text{kips}$ for $\frac{M_{\text{ny}}}{{\Omega }_{\text{c}}}$ in Equation (VI).

$\begin{array}{c}\frac{250\text{kips}}{755\text{kips}}+\frac{8}{9}\left(\frac{240\text{ft}\cdot \text{kips}}{398\text{ft}\cdot \text{kips}}+\frac{0\text{ft}\cdot \text{kips}}{0\text{ft}\cdot \text{kips}}\right)\le 1.0\\ 0.3311+\frac{8}{9}\left(0.6030\right)\le 1.0\\ 0.3311+0.536\le 1.0\\ 0.8671\le 1.0\end{array}$

The interaction equation is satisfied.

Conclusion:

Therefore, the interaction equation is satisfied with the ASD interaction equation.