Last Updated on January 7, 2026 by Maged kamel
Find the Fcr value for a slender column-Solved Problem 5-10 with more iterations.
A solved problem 5-10 with more iterations to find available strength.
This is the same solved problem 5-10 from the Unified Design of Steel Structures by Prof. Louis F. Geshwendener. It is required to determine the available strength of a slender-web compression member. The steel section is W16x26 as a column with a length of 5.0 ft.
I have solved this problem in accordance with the CM#14-2010 specifications.
The modulus of elasticity value E is equal to 29000 ksi, while the yield stress Fy equals 50 Ksi.
Step-1- Check whether the column is long or short, by using the limiting ( kL/r) value for the solved problem 5-10, estimated from the formula (KL/r)=4.71sqrt(E/FY)=4.70*SQRT(29000/50)=113.43. From the previous post, the Kl/r at y direction equals 53.57. Please refer to post 16 for more details
Confirm that the column is inelastic.
Step-2-Get the Area of the section, and the value of ry radius of gyration about y, for the solved problem 5-10 with more iterations. This is a reminder of the value of Kl/r at Y direction.
From the relevant Table 1-1 of W 16×26, we get the data for the area, try, d, h, t-web, and (K*l/r) in the y-direction (1*5*12/1.12)=53.57, which is < 113.43; thus, the column is inelastic.
From the previous post, post 16, we have estimated the Euler stress. It is found to be equal to 99.73 ksi. The lambda^2 value equals 0.50. Please refer to the following slide image for more details.
The estimated critical stress is 40.54 ksi, based on Q = 1.0.
From the previous post, LRFD value =Φc*Pn=Φc*Fcr*Ag=0.90*40.50*7.68=280.0 kips. For the ASD value, Ag*Fcr/Ω = 7.68*40.50/1.67 = 187.0 kips.
Change the value of Q based on the previous estimate.
From the previous post, the effective area is 6.83 in², while the gross area is 7.68 in². The Q value is the ratio Aef/Ag = 6.83/7.68 = 0.889. Based on the new Q value, we will re-estimate the critical stress, which will be 36.92 ksi. The calculation is shown in the slide.
Based on the new value of fcr, we will estimate the LRFD value =Φc*Pn=Φc*Fcr*Ag=0.90*36.92*7.68=256 kips. For the ASD value, Ag*Fcr/Ω = 7.68*36.92/1.67 = 170 kips.
We will estimate the new web height based on the latest Fcr value using equation E7-17.
Estimating the new of reduced web height he value after introducing fcr at the formula=36.90 ksi, we get he=11.20″<14.21″.
Find the estimated effective area for solved problem 5-10 with more iterations.
We can determine the effective area of the web by subtracting the ineffective web area from the total section area. The gross area equals 7.68 inch2, the web area equals 3.55 inch2, and the web effective area is 2.80 inch2.
We can find that the column’s effective area equals 6.93 inch2. Please refer to the following slide image.
We will estimate the new Q value and proceed to find Fcr value. The latest Q will be equal to (6.93/7.68)=0.90. Get the new value for Fcr by using the equation FCr=0.658^(λ2*Q)(Q*Fy)=37.26 ksi.
The value of compression Strength for the given column W16x26 as an LRFD value will be equal to Φc*Pn=Φc*Fcr*Ag=0.90*37.27*7.68=258 kips.
While for the ASD value, Pn/Ω = 37.27*7.68/1.67 = 171.0 kips.
Check our estimated factored available strength using Table 6-1- CM#14-2010.
We will verify our available strength using Table 6-1, per CM#14. In that table, we will multiply the tabulated value by 1000/p to the LRFD and ASD values. The p is the ratio of (1/φc*Pn) for the LRFD design, while for the ASD it is equal to (1/Ωc*Pn). Please refer to the following slide image.
We have two values for the LRFD; the first, for Kl about the Y-axis, equals zero, which is 1000/3.37 = 297 kips. For Kly equals 6 feet, the LRFD value equals 1000/4.21 = 237.50 kips.
Our estimated LRFD value for lc = 5 feet equals 258 kips, which falls between the two values and is acceptable.
We have two values for the ASD; the first, Kl, about the Y-axis, equals zero, which is 1000/5.06 = 198 kips. For Kly equals 6 feet, the ASD value equals 1000/6.33 = 158 kips.
Our estimated ASD value for lc = 5 feet equals 171 kips, which falls between the two values and is acceptable. Thanks a lot.
The Pdf of this post can be viewed and downloaded from the next document.
The following post is post 16b, which covers a solved problem for a slender W section. However, based on Cm#15, the height of the column is 6 Feet.
This is the next post, Alignment chart part 2.
This is a link to a handy external link Chapter 7 – Concentrically Loaded Compression Members.