Chapter 7, Problem 7.1PFS

To determine

The effective length factors for columns IJ, FG and GH of the given frame.

Answer to Problem 7.1PFS

The effective length factor for column IJ, FG and GH is 1.23, 1.11 and 1.10.

Explanation of Solution

Given:

The given frame is,

  

Formula used:

The rotational stiffness is given by,

  $G=\frac{{\displaystyle \sum \left(\frac{I}{L}\right)}}{\left(\frac{I}{L}\right)}$

Here, $I$ is the moment of inertia and $L$ is the length of the member.

Calculation:

From table 1-1, W-shapes properties in the AISC steel construction manual,

The moment of inertia for the section W8x31 is 110 in⁴.

The moment of inertia for the section W8x24 is 82.7 in⁴.

The moment of inertia for the section W18x35 is 510 in⁴.

The moment of inertia for the section W21x50 is 984 in⁴.

The moment of inertia for the section W16x26 is 301 in⁴.

The moment of inertia for the section W8x40 is 146 in⁴.

Calculate the ratio of moment of inertia to the length of member for frame.

For member IJ,

  $\begin{array}{c}\left(\frac{I}{L}\right)=\frac{110{\text{in}}^{\text{4}}}{12\text{ft}\times \frac{\text{12}\text{ft}}{1\text{in}}}\\ =0.763\end{array}$

Similarly calculate the ratio of moment of inertia to the length of member for frame and shown in table below.

  

Consider the column IJ .

Since column IJ is a pinned column so the rotational stiffness for A is 10.

Calculate the rotational stiffness for B .

  $\begin{array}{c} G_B=\frac{{\left(\frac{I}{L}\right)}_{IJ}+{\left(\frac{I}{L}\right)}_{JK}}{{\left(\frac{I}{L}\right)}_{FJ}}\\ =\frac{0.763+0.689}{3.416}\\ =0.43\end{array}$

Refer the figure 7.2, "Side sway uninhibited” (Moment frame) of the textbook to find the effective length factor for $G_A=10\text{and}{G}_B=0.43$ .

The effective length factor for column IJ is 1.23.

Consider the column FG .

Calculate the rotational stiffness for A .

  $\begin{array}{c} G_A=\frac{{\left(\frac{I}{L}\right)}_{FE}+{\left(\frac{I}{L}\right)}_{FG}}{{\left(\frac{I}{L}\right)}_{BF}+{\left(\frac{I}{L}\right)}_{FG}}\\ =\frac{1.013+0.916}{2.125+3.416}\\ =0.35\end{array}$

Calculate the rotational stiffness for B .

  $\begin{array}{c} G_B=\frac{{\left(\frac{I}{L}\right)}_{FG}+{\left(\frac{I}{L}\right)}_{GH}}{{\left(\frac{I}{L}\right)}_{CG}+{\left(\frac{I}{L}\right)}_{GK}}\\ =\frac{0.916+0.916}{3.125+3.416}\\ =0.33\end{array}$

Refer the figure 7.2, "Side sway uninhibited” (Moment frame) of the textbook to find the effective length factor for $G_A=0.35\text{and}{G}_B=0.33$ .

The effective length factor for column FG is 1.11.

Consider the column GH .

Calculate the rotational stiffness for A .

  $\begin{array}{c} G_A=\frac{{\left(\frac{I}{L}\right)}_{FG}+{\left(\frac{I}{L}\right)}_{GH}}{{\left(\frac{I}{L}\right)}_{CG}+{\left(\frac{I}{L}\right)}_{GK}}\\ =\frac{0.916+0.916}{3.125+3.416}\\ =0.33\end{array}$

Calculate the rotational stiffness for B .

  $\begin{array}{c} G_B=\frac{{\left(\frac{I}{L}\right)}_{GH}}{{\left(\frac{I}{L}\right)}_{HL}+{\left(\frac{I}{L}\right)}_{DH}}\\ =\frac{0.916}{1.770+1.254}\\ =0.30\end{array}$

Refer the figure 7.2, "Side sway uninhibited” (Moment frame) of the textbook to find the effective length factor for $G_A=0.33\text{and}{G}_B=0.30$ .

Conclusion:

Hence, the effective length factor for column IJ, FG and GH is 1.23, 1.11 and 1.10.