Last Updated on September 14, 2024 by Maged kamel

- A Solved problem 4.2-how do we find design compressive strength?
- A Solved problem 4-2-how do we find design compressive strength?
- Check whether the column is long or short.
- Estimate the required data for the column from Table 1-1.
- Check the buckling controlling direction, whether it is x or y.
- Estimate Euler stress.
- Estimate the required compressive strength for the given section.

## A Solved problem 4.2-how do we find design compressive strength?

### A Solved problem 4-2-how do we find design compressive strength?

Now, for the solved problem 4-2 from Prof. Sequi’s book, for a w section W14x 74 of A992, the column is 20 ft long and pinned from both sides. Compute the design compressive strength for LRFD and ASD. This solved problem is included in the fifth and sixth versions of the book.

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 design parameters. Analysis problems are different from design problems. In analysis problems, you are dealing with a given section for which you can’t change.

### 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 that if Kl/r>4.71*sqrt(E/Fy), the column is long; otherwise, it 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, taken equals 113. In the fifth version, kl is used, while in the 6th version, the Lc expression is used. The buckling length is equal for the x and y directions. The lcx or (Kl) in the x direction equals 20 feet, and the length lcy =(Kl) in the y direction equals 20 feet.

Estimate the required data for the column from Table 1-1.

3-Table 1-1 provides all the necessary information for the given section of W14*74, for instance, Ix, rx, area, Iy, dry, and the area we will use to get the slenderness ratio and calculations for Euler strength.

### Check the buckling controlling direction, whether it is x or y.

4—Estimate Fe Euler stress from the equation Fe=π^2 E/(KL/r)^ 2 for the Aisc*360- 2010 or the modified equation Fe=π^2 E/(lc/r)^ 2 for Aisc 360-2016 . For the x-direction and y-direction, we estimate Fe and select the lesser value.

5- the slenderness ratio KL/r at the y-direction is used, which is >KL/r at the x-direction

, The value of(lc/ry)=96.77, while the value of (Lc/rx)=39.74. The slenderness ratio will be compared with the limiting value of 113 since KL/r at y is <113; the column is short. Since the column is short, use the formula to get the value of Fcr.

### Estimate Euler stress.

The value of the critical stress Fcr is governed by the equation =0.658^(Fy/FE)*Fy, where Fy equals 50 ksi, and The Euler stress equals 30.56 ksi for the Y direction as estimated. The ratio of Fe/Fy equals 0.61, which is higher than 0.39, which is also another sign of a short column. Please refer to the next slide image for more data.

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

The critical load estimated from the equation is 25.21 ksi. The nominal load is the product of the Area by the Fcr, which is 550 kips.

For the LRFD, use Φ =0.9; for the ASD, use Ω =1.67. The column’s LRFD design strength is 495 kips; this value can be compared to the bigger value of (1.4d or 1.2d+1.6L), where d is the dead load and L is the Live load. The allowable design strength of ASD is 329 kips, which can be compared with D+L. The detailed calculations are shown in the slide image.

The following 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.