Last Updated on July 27, 2025 by Maged kamel
How to compute Critical Stress-Table 4-22& Table 4-1?-CM#14.
In the next slide, we will find a brief content of post 4- compression. What are the different Design tables for compression members to find the available strength based on CM#14 and CM#15? We have solved problem for which we need to estimate the nominal strength using Table 4-1 and Table 4-22.
How do we use Table 4-1?
The first table that is used for the available strength is Table 4-1, based on CM#14. But Table 4-1 requires the use of the larger value of the Y (Kl) equivalent for the x-direction, and the ( KL)y about the Y axis.
The equivalent (Kl)y from the x-axis is equal to (Kl/rx)*ry, where rx is the radius of gyration about the X-axis and ry is the radius of Gyration about the Y-axis. The following slide image explains the difference between ( Kl)y eq and ( Kl)Y. This table determines available strength as factored in the Nominal load.
But there is a change in CM#15-Aisc-360-16 for the table in the expression of Lc, where Lc replaces the Kl expression. For instance, Lcy is used instead of (Kl)y and the table is for W sections, with a yield stress of Fy=50 ksi.
Our subject as of today will be how to estimate critical stress -Table 4-22. The controlling factor that differentiates between short columns and long columns is the criterion of Kl/r.
Table 4-22 is a table that gives the value of available critical stress for various values of yield stress, Fy, from 35 ksi to 50 ksi. The table assumes that the governing (KL/r) is in the y-direction as being the bigger value compared with the value of (KL/r)x in the x-direction.
We evaluate the maximum kL/r value that will give the minimum compressive strength value Fcr. Then, we multiply by the area for the factored values of LRFD or ASD.
There is no Table 4-22 in CM#15; its replacement is Table 4-14.
The same solved problem explained earlier in the previous post, which was problem 4-2, included W14x74 of A992 steel, the height was 20 feet, and the column was hinged from both ends, computing the design compressive strength by LRFD and ASD.
The radius of Gyration about the X-direction equals 6.04 inches, while the radius of gyration in the Y direction is 2.48 inches. The value of Kl/rx=39.74, while Kl/ry=96.77, buckling about the y direction controls the design.
How to compute critical stress-Table 4-22?-LRFD design.
We are going to use Table 4-22, the yield stress Fy =50 ksi, first for LRFD, for the k*L/r =96.77, it is between 96 and 97, for LRFD, for 96, the value is 22.9. While for 97 the value is 22.6, so our value will be < 22.9.
The next slide image shows how we use linear interpolation to find the value for φ*Fcr of 22.67 ksi, we will multiply by the area to get the load for the LRFD there and estimated as =22.9 minus the difference between (22.9-22.6) * 0.77/ 1, which will give the value for φ*Fcr of 22.90 ksi, we will multiply by the area to get the load for the LRFD =494.18 kips.
How to compute critical stress-Table 4-22?-ASD design.
This is Table 4-22, we are checking the critical stress based on the ASD design. For ASD, for K*L/r= 96, (1/Ω)*Fcr= 15.30 ksi, while for K*L/r= 97, (1/Ω)*Fcr= 15.0 ksi. So (1/Ω)*Fcr for 96.77 is <15.30 ksi and estimated as (15.30 –(15.30-15.00)* (0.30/1)=15.07 ksi, we will multiply by the gross area A, to get the load for the ASD: (1/Ω)*Pcr =328.50 kips.
How to compute critical stress-Table 4-1?-LRFD and ASD design.
To use table 4-1, we need to find the equivalent (Kl)y=Kl*ry/rx=20*12*2.48/6.04=8.21 Feet, we compare will (KL), the effective length in y-direction which is equal to 20 feet, since Kl at y is bigger than (KL)y equivalent, we use the bigger value for Table 4-1. Please refer to the following slide image for more information.
For Fy=50 ksi, for W14x74, theφ*Pn=495 kips, while for the ASD design (1/Ω)*Pcr =329 kips. These figures are very close to the values that we have obtained from Table 4-22.
Thanks a lot, I hope the information is useful.
If you wish to review the Column Compressive strength by the general provision
This is the next post, A Solved Problem 4-9 For available compressive strength.