Last Updated on July 16, 2025 by Maged kamel
Solved problem 7-2 for k value for frame 2/3.
Side-sway Uninhibited frame calculation for the solved problem 7-2.
K value for frame-K value for column CD.
The author provides the Ix/L values for both columns and beams, placing them within brackets. For verification purposes, two elements are randomly selected for a check. The first column, DC, has a section of W12x40 and a height of 15′. The second element is beam DH, which has a section W18x50 and a length of 30′.
For column Dc with W12x40,Ix value=307.0 inch4,vL=15′, the Ix/h=307/15=20.466 inch4/ ft.
For the beam of W18x50, Ix=800 inch4, l=30, then Ix/l for the beam=800/30=26.666=26.67 inch4/ft.
First, for the G at Joint D, there is only one column at Joint D, so the G value is the sum of (EI/L) for one column/ sum(EI/L ) for one girder, G=20.47/26.67 =0.7675, at Joint C.
There are two columns and one girder for joint C, so the G for joint C is the v sum (20.47 + 23.20)/(70) =0.6238. Please refer to the following slide image.
Since column CB exists in the unbraced part of the frame, we use Monograph for the uninhibited frame using the value of GC and GD. We mark the points on the G scale and join them; we obtain the K value, the effective length factor for column CD, which is 1.23.
Using the French equation k=sqrt((1.6*GA*Gb+4*(GA+Gb)+7.5))/ sqrt(Ga+Gb+7.5), for the unbraced frame, substitute the values of GD as 0.7675 and GC=0.6238 in both the numerator and denominator. We obtain k = 1.2473, which is close to k = 1.23, and the error value is small.
Side-sway inhibited frame calculation for problem 7-2, part 2.
K value for frame- for column BC.
We check the other column BC, Gc=0.6238 for joint B, two columns and one girder Gb=sum(23.2+23.2)/70=0.6628; the web plane is in the frame direction if we have a look at the elevation of the section.
The two flanges are on the left and right. This section will intersect with the girder section’s two webs, and the column and girder will intersect in the frame elevation. I have written the values of Ix for the different sections.
If we divide the inertia value over the length or -Ix/L for the girder, we get the same figures between brackets. After estimating Gb and Gc, knowing that the frame is a braced frame at that level since the joints C and B are not moving, the next slide shows the calculation for Gc and Gb.
Here is the graph. The values of both Gc and Gb can be found as follows: Gc = 0.6238, Gb = 0.6628. The k value will be obtained from the Nomograph for a braced frame by marking the different values of Gc and Gb on the graph.
K value for frame- column BA.
Now, we will consider column BA, for which there is no sway Relative to each other. GA = 10, since the support is hinged.
For column Gb, as estimated earlier, Gb (23.2 + 23.2)/70=0.6628, GA=10. Now, we will consider column BA, for which, Relative to each other, there is no sway. The value of GA is 10, since the support is hinged.
In the next slide, we will estimate the k value by both methods, the nomograph and the French equation.
We use the Nomograph for the braced frame. Here, G=0.6628, Gb=10. The line joining represents column BA. It will intersect with the middle line, and then k = 0.83.
Using the French equation, substitute with 0.6628 and 10 for the braced frame, and we get k=0.8344.
We will check column HG and the rest of other columns and continue the next time in part 3 of the solved problem 5-2.
A handy external link is Chapter 7: Concentrically Loaded Compression Members from
The following post is Solved Problem 7-2, For K Factor part 3.