Chapter 10, Problem 10.1PFS

To determine

To Select: The lightest section for the continuous beam in the figure using the LFRD method, use elastic method, and factored loads and $0.9$ rule.

Answer to Problem 10.1PFS

  $\text{W}21\times 62$

Explanation of Solution

Given:

The continuous beam is shown below.

  

And steel is $50\text{ksi}$ .

Calculation:

Compute the factored acting on the beam using the equation.

  $\begin{array}{l}{w}_u=1.2D+1.6L\\ w_u=1.2\left(1\right)+1.6\left(2\right)\\ w_u=4.4\text{k/ft}\end{array}$

Draw the bending moment diagram with maximum and minimum values.

  

Using $0.9$ rule, calculate the maximum factored negative moment of the beam.

  $\begin{array}{l}{M}_u(-)=0.9\times \left(-599\right)\\ M_u(-)=-539\text{ft-k}\end{array}$

Calculate the maximum factored positive moment of the beam.

  $\begin{array}{l}{M}_u(+)=309+\left(\frac{1}{10}\right)\left(\frac{599+544}{2}\right)\\ M_u(+)=366\text{ft-k}\end{array}$

Select a W-section from table 3.2 of the AISC Manual such that.

  $\left({\varphi }_b{M}_u\right)>M_u(-)$

Since,

  $\begin{array}{l}\left({\varphi }_b{M}_u\right)=540\text{ft-k}\\ \text{which is greater than 539 ft-k}\end{array}$

Therefore, use $\text{W}21\times 62$