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24-Solved problem 7-2 for frames-part 3.

Solved problem 7-2 for frames-part 3-How to get K value?

The video I used for illustration.

We will continue estimating the k value for the given frames, for the solved problem 7-2, by using both Nomographs and the french equation, based on the condition of frames, whether braced or unbraced, this part of the video which has a subtitle and closed caption in English.

You can click on any picture to enlarge then press the small arrow at the right to review all the other images as a slide show.

Full description of the content.

Solved problem 7-2 for frames-part 3, the un-braced portion of the frame-column GH.

We are going to continue discussing our solved problem 7-2 for frames, by God’s well, The last time we stopped by the hatched yellow color column GH,  the column is unbraced column,  starting by joint H, and joint H at joint H no upper column, G=sum(EI/L) which is=20.47/ sum(EI/L) for girder Gh =26.67, Gh=(20.47/26.67)=0.675.

For Gg at joint G, we have two columns GH, GF, Gc=(20.47+31.67)/sum of(m*EI/L) for girders CG and GJ, m value=1 at GC and m= 1.5, for girder GJ because of the hinge at far joint, the denominator =(1*70+1.5*21.25), Gg=0.5118.

Solved problem 7-2 -Determine k value for a given frame.

Using the french equation for the unbraced frame, with the values of Gh=0.675 and Gg=0.5118.
We use the equation for the side-sway uninhibited, we substitute the result of the french equation for column HG will be=1.1238.
The calculation details are shown in the next slide image.

Solved problem 7-2 -Using the Nomograph for the un- braced frame.

Using the Nomograph for the un-braced frame to get the K value for column HG, we have, Gleft=0.6753, G right=0.5118, The K value will be between 1.1 and 1.20 very close to 1.20.
The column Dc for the unbraced frame is also drawn in the nomograph for a side-sway un-inhibited frame.

Solved problem 7-2 for frames-part 3, column GF.

We continue estimating G for the new column GF hatched with yellow color For joint G, the joint G value as Gg=0.5118
  For joint F, the G value expressed as Gf= sum(EI/L) for columns/ sum(m*EI/L) for girders, Gf=(31.67+31.67)/denominator,  since m=1 for Girder FB, for girder FI is m=2, because the far end is fixed, the value of 2 is for the braced frame.
The denominator=(70+2*56.25)  Gf=0.3471.

Solved problem 7-2 for frames-part 3

The Joint G has a girder ended with pinned support for which m value=1.50, substitute to get GG=0.5118.

Using the alignment chart for the braced frame, Our column is GF, the point at the left Gf =0.5118, while Gc=0.3471.

Solved problem 7-2 -Using french equation compared with the Alignment chart.

The K-value from the French equation, for the braced frame, k=0.672, while by using the Nomograph method, the k value =0.66, the k value from the french equation is quite near.

Solved problem 7-2 for frames-part 3, column FE.

Solved problem 7-2 -G value at joints F&E.

Let us continue with a new column, The last column FE,  this column is braced Gf=0.3471. For point F, we have used m=2 since it is connected to a fixed support, for A side-sway inhibited frame. While joint E is fixed at the support with G value=1.

This frame is a braced frame, using Nomograph of the braced frame, with Gf and GE values, Ge=1 and Gf=0.3471, as we can see, the marked point, k will be above 0.70.

We consider k=0.71.
While using the french equation k  value for column GE, which is braced, K=  ((3GA*Gb+1.4(GA+Gb)+0.64))/(3GA*Gb+2(GA+Gb)+1.28), substitute with Ge=1 and Gf=0.3471, as Ga and Gb, k=3(0.3471*1+1.4(0.3471+1)+0.64)/(3(0.3471*1)+2(0.3471+1+1.28)=0.7112.

Solved problem 7-2 -Using the french equation for k for column EF& FG.

From the french equation, the K=0.7112 which is close to 0.71 from the Nomograph chart, thus we have completed the evaluation of the k values for all the columns for the solved problem 7-2 part-2.

Solved problem 7-2 for frames-part 3, Author solution.

Solved problem 7-2 -The author's solution.

I have included the solution of the author, for the solved problem 7-2 as included in the slides, as a reference for the solution, this is the table at the end, where figures of k values are closed.

Solved problem 7-2 -The author's solution.

This is a full detail of the value estimated by the author.

This is the pdf file used in the illustration of this post.

A very useful external link is Chapter 7 – Concentrically Loaded Compression, Members.
To check part 1 of the same example, please refer to the previous post link.
This is the next post, Stress Reduction Factor For Inelastic Columns


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