Chapter 9, Problem 9.1PFS

To determine

The most economical section using LRFD and ASD.

Answer to Problem 9.1PFS

The safe designed section is $W30\times 90$

Explanation of Solution

Given Information:

  

LRFD:

The design load is

  $\begin{array}{l}{w}_u=1.2w_D+1.5w_L\\ w_u=1.2\times 1.2+1.6\times 3\\ w_u=6.24\text{k/ft}\end{array}$

The design moment is

  $\begin{array}{l}{M}_u=\frac{w_u{l}^2}{8}\\ M_u=\frac{6.24\times {36}^2}{8}\\ M_u=1010.88\text{k-ft}\end{array}$

Assuming self weight as 0.1 k/ft.

The moment due to self weight is

  $\begin{array}{l}M=\frac{0.1\times {36}^2}{8}\\ M=16.2\text{k-ft}\end{array}$

The total design moment is

  $\begin{array}{l}M=1010.88+16.2\\ M=1027.08\text{k-ft}\end{array}$

The design strength of beam is

  ${\varphi }_b{M}_n={\varphi }_b{M}_p={\varphi }_b{Z}_x{F}_y$

Taking

  ${\varphi }_b{M}_n=M_u$

The required section modulus is

  $\begin{array}{l}{Z}_x=\frac{1027.08\times 12}{0.9\times 50}\\ Z_x=273.9{\text{in}}^3\end{array}$

Selecting the section such that Z>Z_required

  $W30\times 90$

The plastic section modulus is 283 in.3

ASD:

The design load is

  $\begin{array}{l}{w}_a=w_D+w_L\\ w_a=1.2+3\\ w_a=4.2k/ft\end{array}$

Assuming self weight as 0.1 k/ft.

The total design load is

  $\begin{array}{l}w=4.2+0.1\\ w=4.3\text{k/ft}\end{array}$

The design moment is

  $\begin{array}{l}{M}_u=\frac{w{l}^2}{8}\\ M_u=\frac{4.3\times {36}^2}{8}\\ M_u=696.6\text{k-ft}\end{array}$

The design strength of beam is

  $\frac{M_n}{\Omega }=\frac{Z_x{F}_y}{1.67}$

The required section modulus is

  $\begin{array}{l}{Z}_x=\frac{696.6\times 12\times 1.67}{50}\\ Z_x=279.2in{.}^3\end{array}$

Selecting the section from AISC manual such that Z>Z_required

  $\begin{array}{l}W30\times 90\\ Z_x=283{\text{in}}^3\end{array}$

Conclusion:

Select $W30\times 90$ by both LRFD and ASD.