Last Updated on September 17, 2026 by Maged kamel
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 problem 7-1 from Prof. McCormac’s textbook; it requires determining the effective length factor.
For each frame column shown in Figure 7-4, assume the frame is not braced against sideways movement. Use the alignment chart of Fig. 7- 2 b. First, open Table 1-1, which contains the properties of W sections, including Inertia and radius of Gyration.
The Table has six columns and four girders. First, open Table 1-1 for the properties of W sections, including Inertia and radius of Gyration.
The first column is the member name, the second is the column section, the third is Ix (inch4), the fourth is the member length in inches, and the last is Ix/L.
The columns are ABC, D E F, and GHI, all hinged at the three bases.
A List of Ix/L for columns AB, BC, DE, and EF. Solved problem 7-1 for the column alignment chart.
The second image covers the remaining two columns, GH and HI—the Ix values, lengths, and Ix/L values.
A List of Ix/L for columns GH and HI is presented.
Solved problem 7-1: How do we get Ix/L values for the frame girders?
The following 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, and 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, and 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 gives 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: 1.217/(1.86 + 2.106) = 0.306.
We will move to joint E if we start with E. There is an upper column and a lower column; the summation of Ix/L of the columns / the sum (two beams), then (1.217 + 1.01414)/.333 + 4.861 ) = 0.272. For joint D, GD = 10.
For joint I, one column, HI, at joint I. For Gi, we have (0.689) / (2.106) = 0.327.
For joint H. We have two columns, so the numerator is (0.689 + 0.574), and the denominator is (4.861 + 0), which equals 0.260. Finally, we use Gg = 1d as a hinged support.
We have now completed the relevant G values for each joint.
For each column, we will write the G Value for each joint, as shown in the figure. Then, we will consider each column, starting with AB GA for the support = 10, and the other part GB = 0.379.
Columns BC and GB are repeated as 0.379, and GC is 0.369 for column DE; GD is 10, while GE is 0.272.
For EF, the joint Ge Value is 0.272, the Gf Value is 0.306, the Gg Value is 10 due to the support, the GH Value is 0.26, and Gi is 0.327.
The image below shows G’s Value for all joints.
The K values are determined from the chart for the unbraced columns as follows: for member AB, starting from the hinged support, the G Value is 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 for HI: 0.26 and 0.327, and approximately 1.1.
For member DE, we have GD = 10 and Ge = 0.272.
For member DE, we have GD = 10 and Ge = 0.272. The k Value is 1.73, close to 1.74.
Repeat the same procedure for all columns.
Alignment chart for columns, unbraced columns.
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