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5-A Solved Problem 4-9 For available compressive strength.

A Solved Problem 4-9 For available compressive strength.

Brief content Content of the video.

Now with a new solved problem, Solved Problem 4-9 from prof Segui‘s book. We have a W12x 58, height is 24 feet long, is pinned at both ends and braced in the weak direction at the third points, in the x-direction, the column is braced at top and bottom, hence k=1  and in the y-direction is bracing at the third point, which means bracing for one-third of the column for each length is  8 feet. 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.

A Solved problem 4-9 for critical stress for a given W-section.

We are continuing With the strength solved problems for compression members and how to use tables to get the value for LRFD and ASD.

Now with a new solved problem that is solved problem 4-9 from Prof Segui’s handbook. We have a W12x 58, height is 24 feet long, is pinned at both ends and braced in the weak direction at the third point, in the x-direction, the column is braced at the top and bottom, hence k=1. 

A Solved problem 4-9 for critical stress for a given W-section.

The y-direction is bracing at the third point, which means bracing for one-third of the column for each length is  8 feet.

We have two points, the aim purpose of this solved problem is to make k*L/r in the x-direction the main criteria for evaluating the buckling load, and it is sometimes the x-direction slenderness ratio may control the strength by controlling the bracing in the y-direction of the column.

We log in the table of 1-1 to determine the properties of W12x58 to get the values of Ix,rx, Iy, and ry, as shown in the slide if we estimate  (Kl/r)  in the direction of x, we will find (Kl/r)x = (1x24x12/ 5.28)=54.54.

If we move to the other slide, we find the k value which is equal to 1, but the height getting shorter and becoming = 8  feet, then k*L/r in the y-direction will be estimated as 38.25 as shown.                                              
 
From the table of 1-1, from the properties of W-12×58 to get the values of  Lx,rx , Iy and ry, as shown in the slide, if we estimate  (Kl/r) for both the x and y direction and compare these values to find out the bigger value, which is major x-direction., then the major direction controls the buckling.

Buckling mode in the major axis.

Since kl/r in the x-direction is= 54.55 while the k l/r in the y-direction will be 38.25 since we are selecting the larger value, then the major axis governs since the slenderness value is higher and giving lower stress.

let us examine the fake value of k*L in the x-direction.h length is  8 feet. The value of K*L/r at the y-direction, the minor axis 

How to convert Covert kx*Lx/rx to Ky*Ly/ry?

This is the quote from prof Segui’s book, The available strength given in the column load tables is based on the effective length concerning the same as described.

The same result can be obtained after getting the (Ky L/ry)  fake for the major axis value=kx*Lx/rx, this value to be compared with the Ky*Ly/ry, then  KyLy/ry fake=1*24/(5.28/2.51=11.409′, For (Kl)fake/ry=11.409*12/2.51=54.55,  which is the original Kx*lx/rx

Small sketch for the KyLy/ry.

This is a mini-map, for how to covert the buckling (KL)x about the major axis to a fake value (Kl)y, used to compare with the original(KL)y about the minor axis to decide what is the controlling factor for buckling .kxlx/rx

Compare the maximum value of KL/r with the limiting value, we will find that KL/r at the major direction, which is 54.55, which is the same value of kx*lx/rx.

The maximum value which is equal to 54.55 will be compared to the limiting value 4.71*sqrt(E/fy). the limiting value for fy equals 50 ksi and e equals 29000ksi can be found to be equal to 113.43.Our column is not in the elastic region. hence the critical stress Fcr to be estimated from the relation Fcr=0.658^(Fy/Fe)*Fy.

Solved problem 4-9-Check whether the column is short or long.

To estimate the value of fcr we need to estimate the euler’s stress based on the buckling in the x- direction as shown in the next slide image.

LRFD and ASD values for Critical stress for solved problem 4-9-General equation.

Once we have estimated the Euler stress, we will proceed to estimate the critical stress value, plug in all the data , we will get the value of Fcr=40.23 KSi.

The LRFD value of nominal compressive stress can be estimated by multiplying phi*Fcr * area. The area is 17.00 inch2. The Phi value is 0.90. We can get the value of 616 kips for the LRFD design.

The ASD value of nominal compressive stress can be estimated by multiplying (1/omega)*Fcr * area. The area is 17.00 inch2. The (1/omega) value is (1/1.67). We can get the value of 410 kips for the ASD design.

LRFD and ASD compression strength by calculation

The factored LRFD & ASDF values for Fcr can be computed from the general equation, these values are shown in the next image.

LRFD and ASD values for Critical stress for solved problem 4-9-Table 4-22.

We can use table 4-22, from the construction manual -14, and specification AISC 360-10. The table will give the value of the critical stress for both LRFD and ASD designs. The controlling parameter is based on the (Kl/r) in the y-direction.

We have estimated the controlling value as equal to 54.44,ky*Ly/ry, we use this value from the left side and intersect it with the column for the value of yield stress Fy=50 ksi

LRFD and ASD compression strength from table 4-22.

The factored LRFD & ASDF values for Fcr can be computed from table 4-22, this time, these values are shown in the next image and are matching with the ones obtained from the general equation.

This is the pdf file used in the illustration of this post.
This is the next post, the Alignment chart for columns.
This is a link to an important resource that is Concentrically Loaded Compression Members.

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