Solved problem 7-2-frame K value 3/3.
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Solved problem 7-2-frame K value 3/3, the un-braced portion of the frame-column GH.
We will continue discussing our solved problem 7-2 for frames, by God’s will; the last time we stopped by the hatched yellow color column GH, the column is an unbraced column, starting with joint H and continuing with the other joints.
At joint H no upper column, G=sum of (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, and 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 the far joint, the denominator =(1*70+1.5*21.25), Gg=0.5118. The French equation for the unbraced frame was used, with the values of Gh=0.675 and Gg=0.5118.
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)Â Gf=0.3471.
The Joint G has a girder ending 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.
The K-value from the French equation for the braced frame is 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-frame K value 3/3-column FE.
Let us continue with a new column. The last column, FE, 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. Joint E is fixed at the support with G value=1.
This frame is braced, using the 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.
From the French equation, the K=0.7112 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-frame K value 3/3-Author solution.
I have included the author’s solution for 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.
This is a full detail of the value estimated by the author.
Chapter 7 – Concentrically Loaded Compression, Members is a useful external link.
Please refer to the previous post link to check part 1 of the same example.
This is the next post, Stress Reduction Factor For Inelastic Columns
