Brief description of post 9-compression.

9-Two Solved Problems for steel column analysis.

Two Solved Problems For Column Analysis.

Brief content of the video includes two Solved problems for column analysis.

We are going to continue with solved problems, to perform column analysis, but after making some modifications, for solved problem13-28, we will call this solved problem 13-28a, which is covered from the start of the video to time 4:53.

The column is chosen to have a new section which is W14x61, The Fy is 50 KSI. What is the available Strength in axial compression in kips? Which value of the 4 alternatives, is listed from the options from A to D?

The support case of the column is case C, where the support is fixed at the bottom, and guided support at the top.

Our kx=ky=1.2 as recommended. For the needed information for W14x61, the area is 17.90 inch2.
Ix =640 inch 4, rx= 5.98 inch 2, Iy= 107 inch 4, and ry =2.45 inch2.

The procedure is to evaluate the kx and ky from the previously solved problem 14-28 is as follows,  our KyLy=18 feet.
We will proceed directly to table 4-1 from FE Exam Ref book.
We have our KyL= 1.2x 15=18 feet, mark the value our fy is 50 ksi, one remark, that the table is assigned only when the Fy =50 ksi.

The second solved problem starts from time 4:53 till the end of the video. That was a part of the video, which has a subtitle and closed caption in English.

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.

Two Solved problems for steel column analysis.

The content of lecture.

This is a brief description of the post content.

Two solved problems for column analysis, the first problem is a problem this is similar to the solved problem 13.28 which we have solved earlier in the previous post, but this time I have changed the dimension from W 12×50 to W14x61.
The column is fixed at one end and has a guide roller at the other end which is at the top.

Effective length factor table by AISC.

This is the table for the different values of k based on the end condition for columns. This problem is quoted from the printed problems from M Iqbal Book for F E exam review.

First, we will review the k values for the six cases of columns with different end conditions.

Revision for E-1 clause by the AISC.

This is the general provision E-1 from the AISC code, for the graph between Kl/r and Fcr/Fy.

Solved example 13-28 it is required to get the available compressive strength for W14x61

We are going to continue with the previously solved problem 13-28, it will be considered as 13-28a. But after making some  Modifications, for the solved problem 13-28a, the column is chosen to be W14x61,Fy is 50 KSI.

What is the available Strength in axial compression in kips? Which value of the 4 alternatives, listed from option A to option d.

The support case of a column is case c, where the support is fixed at the bottom and guided support at the top Our kx=ky=1.2 as recommended.

The Area& rx& ry Ix and Iy values for W14x61 section.

For the needed information for W14x61, the area is 17.90 inch 2, the value of Ix =640 inch 4,the radius of gyration about the major axis rx= 5.98 inch 2. Iy= 107 inch 4 and ry =2.45 inch2.

The procedure is to evaluate the kx and ky. From the previous example our KyLy=18 feet.

We will proceed directly to table  4.1, FE Exam Ref book. We have our KyL= 1.2x 15=18 feet, mark the value.
Fy is 50 ksi. One remark, that the table is assigned only when the yield stress Fy =50 ksi for other values use table 4-22, with the intersection of the horizontal line of 18′ with the vertical line for W14x61.

The LRFD  value from table.

We have completed solving the solved problem after changing the section to become W14x61. To prove that The result from the table coincides with the calculation.





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 4.71sqrt(E/fy), the column is short.
Proceed for Fe estimation which is =36.856 ksi.

But fcr will be estimated by using the formula, 0.658 ^ (50/36.856)* 50=28.337 ksi.
ry = 2.45 inch 2, for that section, but still, the k equivalent for the x major axis *(L) is s< ky* ly, so the weak axis controls the load, explained as follows:

A-Ky equivalent=k/(rx/ry)=(1.2/5.98/2.45)=0.4961, then again to be multiplied by the column height which is 15 then =0.4961*15=7.374 ft.
b-Our ky ly= 18 ‘ which is > 7.374 ‘. ky*ly/ry= 88.163 and 4.71*sqrt(E/fy)=113.43 which is constant for Fy=50 ksi.
C-The Euler value is =36.856 ksi. Fcr= 0.658 ^(50/36.856)* 50=28.337 ksi

Which axes will control the deflection, the major or the minor?

, will be multiplied by φ *area it will be equal to (0.9×17.9×28.337)=457 kips so option b is correct. This answer matches with the answer obtained previously.

Another solved problem 13-29 for column analysis.

Another solved problem is 13-29 for column analysis from M Iqbal’s book. The column given is 20 ft tall pinned at the bottom and top, but for the y-direction has roller support at the mid-height but regarding the x-direction, we have only supported at the bottom and top.
The section is W10x45, fy=50 ksi, its controlling slenderness ratio which is K *Lvalue/r is mostly,5 or 65 or 72 or 120.

Solved example 13-29 it is required to get the controlling slenderness ratio for W10x45

The solution is based on y-direction k is taken as 1.2.
while for the x, the column height is 20 feet and k was also taken as 1.2.
Why the selected value is 1.2?

A reference for the k value-based on different end conditions.

The different boundary conditions and the engineering effective length.

I have searched for the reason by referring to the e funda – engineering Fundamentals -includes free- free, Kl= 1.2L, while hinged -free, which is our case the KL is also 1.2 .

In the y-direction, hinged free, then 1.2 for- k is selected, but the length is 10 ft, while for k value for x is 1.2 but the length is the total length.

The information for W10x45 are as follows, the area is 13.3 inch 2, Ix= 24.8 inch4 and rx= 4.32 inch 2, Iy= 53.4 inch2 and ry= 2.01 inch 2,this is a section sketch in the y-direction.

Reduction factor for stiffened elements Qa.

The buckling shape can be imagined as pressing perpendicular to the y-axis, the buckling shape for x-direction can be imagined as if pressing perpendicular to axis x.

Again we will convert the kx into a fake or equivalent by/ rx/ ry, kx which is 1.2, refer to the values of rx,ry.
kx =1.20 / (4.32/2.01) = 0.558 *20= 11.17 ft .
While the ky = 1.2, Ky*Ly=1.2*10=12 ft, since the equivalent, Kx*Lx is < Ky *Ly.

Check whether the major or minor axis governs the deflection.

The buckling is controlled by the minor axis, then the controlling slenderness ratio is( ky*ly/ry).

Referring to the options numbers ky*ly/ry, which will give the value of 71.64 nearly 72, which is option C.

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

The next post is Introduction to Local Buckling.

For an external resource, this is a link for this post, Concentrically Loaded Compression Members.

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