- Solved problem 4.2-how to find design compressive strength?
- Brief content Content of the video.
- Solved problem 4-20-how to find design compressive strength?
- Check whether the column is long or short.
- Estimate the required data for column from table 1-1.
- Check buckling controlling direction whether x or y.
- Estimate the required compressive strength for the given section.

## Solved problem 4.2-how to find design compressive strength?

### Brief content Content of the video.

This is part-2, solved problem 4-20 from Prof. Sequi’s book, for a w section W14x 74 of A992, the length of the column is 20 ft and pinned from both sides, compute the design compressive strength for LRFD and ASD.

This is an analysis for a given column of A W14x74 of A992, L=20′, and k=1, pinned-pinned, it is required to get the design compressive strength, for both LRFD and ASD. The content of this post starts from time 4:46 till the end. This video 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.

### Solved problem 4-20-how to find design compressive strength?

Now for the solved problem 4-20 from Prof. Sequi’s book, for a w section W14x 74 of A992, the length of the column is 20 ft and pinned from both sides, compute the design compressive strength for LRFD and ASD.

This is an analysis for a given column of A W14x74 of A992, L=20′, and k=1, pinned-pinned, it is required to get the design compressive strength, for both LRFD and ASD for design parameters.

### Check whether the column is long or short.

The following steps are to be performed.

1-The first step is to determine whether the column is long or short, recall if Kl/r>4.71*sqrt(E/Fy), column is long, else the column is short.

The length of the column L is given as equal to 20 feet. and the K factor =1, since the column is pinned.

2- We have Fy=50 ksi, E=29*10^3 ksi, then sqrt(29.3/50)*4.71=113.43.

### Estimate the required data for column from table 1-1.

3-We want to get sufficient information from table 1-1 for the given section of W14*74 for instance Ix, rx, area, Iy, ry and also the area.

### Check buckling controlling direction whether x or y.

4– Estimate Fe Euler stress from Fe=π^2 E/(KL/r)^2, for x-direction and y-direction, we will select the lesser value.

5- KL/r at the y-direction is used, which is >KL/r at the x-direction, to get the slenderness ratio to be compared with the limiting value which is 113.49 since KL/r at y is <113.49, the column is a short column.

6-The critical stress Fcr is governed by the equation =0.658^(Fy/FE)*Fy, KL/r at the y-direction is used, which is >KL/r at the x-direction, to get the slenderness ratio to be compared with the limiting value which is 113.49 since KL/r at y is <113.49

The column is a short column. Use the formula to get the value of Fcr.

### Estimate the required compressive strength for the given section.

Use the following Equation.

[latexpage]\begin{equation}

F_{c r}=0.658^{\left(\frac{F y}{F_{c}}\right)} f_{y}

\end{equation}

[latexpage]\begin{equation}

\begin{array}{l}

F_{c r}=0.658^{\left(\frac{F y}{F e}\right)} f_{y}=0.658\left(\frac{50}{30.59}\right)(50)=25.23 \ {ksi.} \\

P_{N}=A F_{c r}=21.80(25.23)= 550 kips. \end{array}

\end{equation}

For the LRFD, use Φ =0.9, while for the ASD use Ω =1.66. The final values and calculations are shown in the slide image.

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

The next post is How to compute critical stress-table 4-22?

For a good reference from Prof. T. Bart Quimby, P.E, Ph.D., F.ASCE, refer to this link for Limit State of Flexural Buckling for Compact and Non-compact Sections.