Chapter 11, Problem 11.1PFS

To determine

Whether a $W14\times 30$ tension member is satisfactory for the given conditionss.

Answer to Problem 11.1PFS

Satisfied

Explanation of Solution

Given:

  $\begin{array}{l}{F}_y=50ksi\\ F_u=65ksi\\ P_D=80k\\ P_L=100k\\ M_{Dx}=10ft-k\\ M_{Lx}=12ft-k\\ L_b=12ft\\ C_b=1.0\end{array}$

Calculation:

The sectional properties are as follows:

Refer Table 3-2, “W shapes properties” in the AISC steel construction manual.

  $\begin{array}{l}{L}_p=5.26ft\\ L_r=14.9ft\\ {\varphi }_b{M}_{px}=177kip-ft\\ \frac{M_{px}}{{\Omega }_b}=118kip-ft\\ {\varphi }_{b}BF=6.95kips\\ \frac{BF}{{\Omega }_b}=4.65kips\\ {\varphi }_b=0.9\\ {\Omega }_b=1.67\end{array}$

The ultimate load by LRFD is

  $\begin{array}{l}{P}_u=1.2P_D+1.6P_L\\ P_u=1.2\left(80\right)+1.6\left(100\right)\\ P_u=256kips\end{array}$

The design moment is

  $\begin{array}{l}{M}_{ux}=1.2M_{Dx}+1.6M_{Lx}\\ M_{ux}=1.2\left(10\right)+1.6\left(12\right)\\ M_{ux}=31.2k-ft\end{array}$

Refer Table 1-1, “W shapes properties” in the AISC steel construction manual.

  $A_g=8.85in{.}^2$

The available tensile strength is

  $\begin{array}{l}{P}_n=F_y{A}_g\\ P_n=50\times 8.85\\ P_n=442.5kips\end{array}$

The design strength is

  $\begin{array}{l}{P}_c={\varphi }_c{P}_n\\ P_c=0.9\times 442.5\\ P_c=398.25kips\end{array}$

Since, $L_p\le L_b\le L_r$ zone 2 will prevails.

The nominal moment strength is

  $\begin{array}{l}{M}_{cx}={\varphi }_b{M}_{nx}\\ M_{cx}=C_b\left[{\varphi }_b{M}_{px}-{\varphi }_{b}BF\left(L_b-L_p\right)\right]\\ {\varphi }_b{M}_{nx}=1.0\left[177-6.95\left(12-5.26\right)\right]\\ {\varphi }_b{M}_{nx}=130.16k-ft\end{array}$

Check:

  $\begin{array}{l}\frac{P_r}{P_c}\ge 0.2\\ \frac{P_r}{P_c}=\frac{P_u}{P_c}=\frac{256}{398.25}\ge 0.2\\ 0.64\ge 0.2\end{array}$

The section is subjected to both the bending and axial tension:

  $\begin{array}{l}\frac{P_r}{P_c}+\frac{8}{9}\left(\frac{M_{rx}}{M_{cx}}+\frac{M_{ry}}{M_{cy}}\right)\le 1.0\\ \frac{256}{398.25}+\frac{8}{9}\left(\frac{312}{130.16}+\frac{0}{0}\right)\le 1.0\\ 0.643+0.89\left(0.2396\right)\le 1.0\\ 0.856\le 1.0\end{array}$

Hence, it’s OK.

The ultimate load by ASD is

  $\begin{array}{l}{P}_a=P_D+P_L\\ P_a=80+100\\ P_a=180kips\end{array}$

The design moment is

  $\begin{array}{l}{M}_{ux}=M_{Dx}+M_{Lx}\\ M_{ux}=10+12\\ M_{ux}=22k-ft\end{array}$

The design strength is

  

Since, $L_p\le L_b\le L_r$ zone 2 will prevails.

The nominal moment strength is

  $\begin{array}{l}{M}_{cx}=\frac{M_{nx}}{{\Omega }_b}\\ M_{cx}=C_b\left[\frac{M_{px}}{{\Omega }_b}-\frac{BF}{{\Omega }_b}\left(L_b-L_p\right)\right]\\ {\varphi }_b{M}_{nx}=1.0\left[118-4.65\left(12-5.26\right)\right]\\ {\varphi }_b{M}_{nx}=86.7k-ft\end{array}$

Check:

  $\begin{array}{l}\frac{P_r}{P_c}\ge 0.2\\ \frac{P_r}{P_c}=\frac{P_a}{P_c}=\frac{180}{264.97}\ge 0.2\\ 0.68\ge 0.2\end{array}$

The section is subjected to both the bending and axial tension:

  $\begin{array}{l}\frac{P_r}{P_c}+\frac{8}{9}\left(\frac{M_{rx}}{M_{cx}}+\frac{M_{ry}}{M_{cy}}\right)\le 1.0\\ \frac{180}{264.97}+\frac{8}{9}\left(\frac{22}{86.7}+\frac{0}{0}\right)\le 1.0\\ 0.679+0.2555\le 1.0\\ 0.905\le 1.0\end{array}$

Hence, it’s OK.

Conclusion:

Therefore, the tension member $W14\times 30$ is satisfied for both LRFD and ASD.