Brief content of post- 7-compression members posts.

7- Solved problem 7-1 for alignment chart for columns.

Solved Problem For Alignment Chart for columns-7-1.

Solved problem 7-1 How to get Ix/L values for columns of the frame?

Our solved problem is solved problem 7-1 from Prof. Mccormac’s textbook, it is required to determine the effective length factor.

For each of the frame columns , shown in figure 7-4, if the frame is not braced against sideways. Use the alignment chart of fig 7-2 b.

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.

Solved problem 7-1- determine the effective length factor for a given frame.

First, we need to open table 1-1, for the properties of W sections for the inertia and radius of gyrations.

We have in this solved problem 6 columns and 4 girders, First, we need to open table 1-1, for the properties of W sections for the inertia and radius of gyrations.

The first column is for the member name, the second column for the column section, the third column is for it’s Ix value – inch4, the fourth column is for the length of the member in inches, the last column is for the Ix/L value.

The columns as ABC, D E F, GHI, all hinged at the three bases. We have in this example 6 columns and 4 girders.
A list of Ix/L for columns AB, BC, DE, and EF. A solved problem 7-1 for the alignment chart for columns.

Ix/L for two columns GH and HI.

The second image is for the remaining two columns GH and Hi.,the Ix values, lengths, and the value of  Ix/L.

A list of Ix/L for columns, GH&HI is presented.

Solved problem 7-1 How to get Ix/L values for Beams of the frame?

The next image shows the data for the four beams CF, FI, BE, EH, from table1-1, we get all the necessary data for Ix.
A list of Ix/L for beams CF, FI, BE&EH.

Ix/L for the four beams CH&FL&BE and EH.

The beams are arranged in a similar column, the first column is the designation, the second column is for the section of each beam.

The third column is for the inertia value of Ix , the last column is for the division of Ix over L which is  EI/L.

if we have finished writing all the values for beams, we will start joint by joint, first for joint A, since it is a hinge then Ga is taken as 10, and for point B it is a part of column AB and column Bc, standing at joint B, then the sum( Ix/L) for columns/ sum(  Ix/L of only one beam).

Adding (0.689 + 0.574 )/3.333 will give the value of 0.379.

G value at joint A&B and C.

We obtained it earlier please refer to the first image for an illustration of the numerator values.
We have one column and one beam, so (0.689/ 1.867 ) =0.369 for joint C. 

We will move to joint F  if we started by F, this joint. There is no upper column, we have only one column the summation of Ix/L of the column / (the summation of two beams) since this joint is an intermediate joint, then 1.217 /( 1.86 + 2.106 )= 0.306.

G value at joint F&D and E.

We will move to joint E  if we started by E, this joint There is an upper column, and lower column, the summation of Ix/L of the columns / the sum( two beams) then (1.217+1.014) /( 3.333 + 4.861)= 0.272. for joint D GD=10.

For joint I, one column HI at the joint I, For Gi we have  (0+0.689) /(0+2.106) = 0.327, For joint H.

G value at joint I&H and G.

We have two columns, so (0.689+0.574)as numerator/ (4.861+0) = 0.260,  finally Gg=10 as hinged support.

Now we have completed the related part of G values for each joint.

For Each column, we will write the G value for each joint as shown in the figure, then consider each column as, starting with AB GA for the supprt=10 and the other part GB 0.379.

The columns  BC, Gb are repeated as =0.379 and Gc is 0.369 for the column DE, the Gd is 10, While for GE is 0.272.

For EF the joint Ge value is 0.272, Gf value is 0.306,  Gg value is 10 due to the support, the GH value is =0.26, and Gi=0.327.
G’s value for all the joints is shown in the next slide image.

 k values for members Ab, BC, DF, EF, GH, and HI.

The K values are determined via the chart, for the unbraced columns as follows for member AB starting from hinged support with Gvalue = 10 and the other Gb value is 0.379, by interpolation, based on the graph, each division adds 0.10.

Alignment chart for side sway uninhibited.

The value will be between 1.7 and 1.8.

Let us select another member GH, we have two values as 0.26 and 0.327 for HI k=approximately 1.1.

We have for member DE, Gd=10 and Ge=0.272, for the member, for DE, 10 and 0.27 The k value 1.73 close to 1.74.
The same procedure will be repeated for all the columns.
Alignment chart for columns, unbraced columns.

Brief content of the video.

First we need to open table 1-1, for the properties of W-sections for the inertia and radius of gyrations, All The relative sections for each column are listed in columns, and from tables,  we write down the corresponding Ix for each column as chosen in the column of Ix, while for the length of the column, it will be included as well.

Since our equation is the sum of EI/L for columns divided by the sum of EI/L for beams, and girders.
The first column height is 12 feet. Then converted to inches by multiplying by 12, it will be 144 inches.
For column Bc, the length will be 120 inches. and so on for the other columns for lengths.

For each line, we will evaluate the corresponding Ix/L only since, the E value for both columns and beams are the same, the value for column Ab will be 82.7/144 the result will be 0.574 inch3. That was a part of the video, which has a subtitle and closed caption in English.

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

The next post is – Solved Problem For Compression Members-13-28.

For a valuable resource, please visit the following link: Concentrically Loaded Compression Members.

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