Brief description of post 16-compression.

16-A Solved Problem 5-10-Available Strength for short column.

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.

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.

Steel grades and their yield stress and ultimate stress values.

This is the table of shapes and corresponding Fy, Fult values.

Critical stress versus slenderness ratio for columns.

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 new 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.

Solved problem 5-10 to estimate the available strength for a W section with slender web


Step-1- Check whether the column is long or short, by using the limiting ( k*L/r) value for the solved problem 5-10, estimated from the formula (K*L/r)=4.71*sqrt(e/FY)=4.70*SQRT(29000/50)=113.43.

Check whether W section is slender or not.

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.

Estimate the slenderness ratio for w section in the solved problem 5-10  

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 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.

Estimate the value Euler stress for the column and Lambda^2.

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.

Estimate the value Euler stress for the column and Lambda^2.

λ2=(50/99.705)=0.5014, we will evaluate Fcr by using the equation fcr=0.658^(λ2*Q)*(Q*Fy).

Estimate The preliminary critical stress for Q=1 for the given W section.

We will evaluate fcr by using the equation Fcr=0.658^(λ2*Q)*(Q*Fy)=0.658^(0.5014*1)*(1*50)=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 .

Estimate the effective area of the web.

b/t=h/tw=56.80, we check that he is <h web, the value obtained=10.80″<14.206″.

Estimate the effective area of the web.

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.

Estimate The available strength of the section for the Solved problem 5-10

While for the ASD value=A*Fcr/Ω=7.68*40.50/1.67=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.

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