Last Updated on June 2, 2025 by Maged kamel
Two Solved Problems For Column Analysis-FE Exam.
This is a brief description of the post content.
Two Solved problems for steel column analysis.
Two solved problems for column analysis. The first problem is similar to the solved problem 13.28, which we addressed earlier in the previous post. However, this time, I have changed the dimensions from W 12×50 to W14x61.
Solved problem 13-28-Use Provision E-1 from AISC-360-16.
The column is fixed at one end and has a guide roller at the other end, located at the top. This is the table for the different values of k based on the end condition for columns.
This problem is quoted from the M Iqbal Book for the F E exam review. We will continue with the previously solved problem 13-28, which will be considered as 13-28a. However, after making some modifications to the solved problem 13-28a, the column is chosen to be W14x61, with Fy = 50 ksi.
What is the available Strength in axial compression in kips? Which value of the four alternatives is listed from option A to option D?
Table C-A-7-1 gives six cases of support conditions and the corresponding and recommended effective value K values. We will choose the appropriate k value based on the data from problem 13-28A.
What is the selected K value?
The support case of a column is case c, where the support is fixed at the bottom and guided at the top. Our kx = ky = 1.2, as recommended. The Lex=Ley=1.2*15=18 feet.
Use Table 1-1 to get the area and radii of inertia values for 13.28a.
We will use Table 1-1 to get the area and the values of rx and ry, the radii of inertia. This is a table quoted from the FE Reference handbook in which there is a list of W sections, for which the area, depth, flange width & thickness, and inertias about axis x and axis y are given; for E value of 29000 ksi and yield stress of 50 ksi.
The following slide image shows the first part of Table 1-1.
For the required information for W14x61, the area is 17.90 in², the value of the radius of gyration about the major axis is rx = 5.98 in, and the value of the radius of gyration about the minor axis is ry = 2.45 in.
We will proceed directly to Table 4.1, FE Exam Ref book 10.40. We have our KyL = 1.2 x 15 = 18 feet; mark the value. The yield stress Fy is 50 ksi.
What is the requirement of Table 4-1?
Table 4-1 requires that the tabulated effective length concerning ry is the maximum value of ly required from the X, or the ly modified which is Lcx*(ry/rx), direction, and the ly value from the y-direction, the lc y required equals 7.37 feet; please refer to the following slide image.
The ly value equals 18 feet; the selected effective length is the maximum value of 7.37, and 18 feet, will be 18 feet.
What is the available strength from Table 4-1?
We use Ly equals 18 feet, we move horizontally, and we drop a line from the top W14x61 section’ the intersection will give the available strength, which is equal to 270 kips; the solution matches with option b).
We will proceed to Table 4-1 to get the available strength or the LRFd value. One remark is that the table is assigned only when the yield stress Fy =50 ksi, with the intersection of the horizontal line of 18′ with the vertical line for W14x61.
The available strength value equals 457 kips, which is nearly close to option B. Please refer to the following slide image for more details.
We could get the same result by using The E-1 provision of AISC-360-16, checking whether the given column is short or long, and then choosing the appropriate Fcr equation to use.
This curve we have already discussed, at the value for KL/r = 4.71* sqrt of (E/Fy), is the parameter that distinguishes between short columns and long columns.
If the value of Le/r= KL/r > 4.71* sqrt of (E/Fy), then the column is considered as a Long column, and hence the Euler Elastic stress. The division of pi^2 EI/ (kl)^2 over the area, then the stress will be (pi^2 E/ (le/r)^2).For the traditional way of calculation.
We will estimate whether the column is long or short by using the formula for the value of (k*l/r), which will give ( 4.71sqrt(E/Fy)), will give=4.71sqrt(50000/50)= 113.43.
The same procedure for our Ky*ly/ry from the previous calculation is smaller than the 4.71*sqrt(E/fy); the column is short.
In the next slide, there is a hint: what is Lcye from the column in the x direction as compared to the Lcy in the Y direction?
We will apply the Fcr equation for the short column; the critical stress equals 0.658^lambda2 *Fy, where lambda ^2=Fy/Fe.
The Euler stress equals 28.33 ksi, the Fcr=28.33 ksi, multiplied by phi by the area to get the LRFD value of the available strength, which equals 456.40, very close to option B.
Another solved problem 13-30 for column analysis.
Another solved problem is 13-30 for column analysis from M Iqbal’s book. The column given is a W10x49 column that carries a dead service load of 100 kips; the slenderness of the column is 120, and the yield stress is 50 ksi. it is required to select the most nearly value for the service live load from the given four options.
The controlling slender ratio is 113.43, and the slenderness ratio for the given section is 120, which is bigger than 113.43, so the column is short.
Use Table 1-1 to get the area and radii of inertia values for 13.30.
We use Table 1-1 to get the area and radii of inertia for W10x49; please refer to the following slide image.
Use Table 4-14 to get the factored critical load.
We use Table 4-14 to get the factored critical stress for a slenderness value of 120; the factored critical stress equals 15.70 ksi.
What is the final service load value?
We equate The ultimate Load to the factored Nominal load based on a dead Load of 100 kips. The final service load is found to be equal to 66 kips, which is most nearly equivalent to option D; please refer to the next slide image for more details.
The next post is Introduction to Local Buckling.
For an external resource, this is a link for this post, Concentrically Loaded Compression Members.