## 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. 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, and the last column is for the **Ix/L value.**

The columns as ABC, D E F, and 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.

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, and EH, from table 1-1, we get all the necessary data for Ix.

A list of Ix/L for beams CF, FI, BE&EH. The beams are arranged in a similar column, the first column is the designation, and 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.

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 start with 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 intermediate, then 1.217 /( 1.86 + 2.106 )= 0.306.

We will move to joint E if we start with 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. We have two columns, so (0.689+0.574)as numerator/ (4.861+0) = 0.260, and 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. 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.

The value will be between 1.7 and 1.8.

Let us select another member GH, we have two values 0.26 and 0.327 for HI 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 is 1.73 close to 1.74.

The same procedure will be repeated for all the columns.

Alignment chart for columns, unbraced columns.

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.