Solved Problem 5-10-available strength for short column-slender W section.
We are going to see, the solved problem- 5-10, which is From the unified design of steel structure, for local buckling, our steel section for the column is a W16x26 section. Kl at y direction =5.00 ft.
It is required to estimate the available strength of that column, the column is a slender column as we can see from the following picture after checking the Flange and web local buckling parameters.
This is the table of shapes and corresponding Fy, Fult values.
This is the λ2 expression used for earlier versions, for which λ2=Fy/Fe, where Fy= yield stress of the section, Fe is the Euler stress, and for the general buckling equation, the Fcr equations are adjusted accordingly.
A solved example 5-10 for the available strength.
This is a newly solved problem 5-10 from the unified design of steel structures by Louis F. Geshwendener, it is required to determine the available strength of a compression member with a slender web, the section is W16x26 as a column with lcy =5.0 ft.
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)=4.70SQRT(29000/50)=113.43.
Step-2-Get the Area of the section, ry radius of gyration about y, for the solved problem 5-10.
From the relevant table of W 16×26, we get the data for the area, try,d, h, t-web, the (K*l/r) at the y- direction=(1*5*12/1.12)=53.57, which is < 113.43, then the column is short.
The data can be obtained from Table 1-1 for the given section. Bf, tf, h, t web, section W 16×26 has a footnote c, which means that it does not meet the requirement of h/tw.
Is a column elastic or inelastic?
Step-3- Check the column, whether elastic or inelastic column, for both flange and web, from the previous image we get the values of Bf/2tf=7.97.
We will check against the value of 0.56*sqrt(E/fy), which gives 13.49>bf/2tf of the flange for the solved problem 5-10.
While for w 16×26(hw/tw)=56.80 from the table, the limiting value is 1.49*sqrt(E/fy), which gives 33.72, this value is <56.80, so the section is slender-web section.
Fcr value for the column.
Step-4: Make an estimate for λ2, which is Fy/FE, and get the relevant Fcr, consider Q=1, where Q is the reduction factor, Fe value=99.705 ksi, while Fy=50 ksi.
λ2=(50/99.705)=0.5014, we will evaluate Fcr by using the equation fcr=0.658^(λ2Q)(Q*Fy).
We will evaluate fcr by using the equation Fcr=0.658^(λ2Q)(QFy)=0.658^(0.50141)(150)=40.50 ksi.
The value of the effective area.
Step-5 we will estimate the value of he from the E-7-17. From the Aisc equation E7-17, we have Fcr=f=40.50 ksi .b/t=h/tw=56.80, we check that he is <h web, the value obtained=10.80″<14.206″.
Step-6 we will estimate the value of the A eff after deduction of the shortage in the area. A eff is obtained from Agross- Aw+(he*tw)=7.68-(14.20*0.25-10.80*0.25)=6.83 inch2.
Step-7 Use the value of A eff and the Fcr to get the LRFD and ASD values, we have A eff=6.83, fcr=40.50 ksi.
Estimate the Strength value for the column as LRFD value =Φc*Pn=Φc*For*A=0.90*40.50*6.83=250.0 kips. While for the ASD value=AFcr/Ω=7.6840.50/1.67=166.0.0 kips, that was the final result for the solved problem 5-10.
This is a link for the pdf file used in the illustration of this post without more iterations.
This is the link to the next post is Solved Problem 5-10 with more iterations.
This is a link to A very useful external link Chapter 7 – Concentrically Loaded Compression Members.