15- Solved problem 6-19-4 for local Buckling of column-2010.

Last Updated on September 19, 2026 by Maged kamel

Solved problem 6-19-4 for local Buckling of column-2010.

Solved problem 6-19-4: estimate the Area and Inertia values of the given section.

The next slide shows a brief introduction to this Post.

 What is the main Chapter for designing compression members? The main Chapter is Chapter E, and the relevant sections for the built-up section are E6 and E7 for slender members.

 Determine the nominal axial Strength for the Non-standard shape in Figure 6.19.4 for an effective length kl = 8 ft, using fy = 100 ksi. The I-section is composed of three plates; the upper plate breadth is 10 inches. The thickness is 1/2 inch.

The same dimensions apply to the lower plate, t, as the middle plate has a height of 11 inches and a width of 1/4 inch; thus, the overall height is 12 inches.
Given k*L is 8 feet. Since this section is built up, we cannot obtain information from any tables. The following slide image shows the calculation of Ix, the Moment of Inertia about the x-x axis; the Ix Value is 358.56 in. 4.

For local Buckling, we have to estimate Ix and Iy. The Value of rx is much higher than ry; the minor direction will govern Buckling.

We will estimate Iy, since Buckling will be about the minor axis. the Area of section equals 12.75 inch2, the Moment of Inertia about Y axis equals 83.35 inch4. The radius of Gyration equals 2.56 inches. Please refer to the following slide image for more details.

Determine which direction controls the design.

We can find the Value of the radius of Gyration about the x-x, which is sqrt(358.56/12.75)=5.30 inches. Find the values of (Kl/rx) and Kl/ry. For Kl/rx = 96/5.3 = 11.62.

While Kl/ry=(96/2.56)=37.50. Since Kl/ry is bigger than Kl/rx, Buckling about the Y-axis will control the design.

Check b/t for the unstiffened Ff-part, f, f-parFlange, and the reduction factor Qs for the unstiffened part.

The b/t ratio is half the Flange length divided by the Flange upper plate thickness, and equals (10/2/0.50) = 10. Check this ratio against B/t as given by the E7-7 Equation, which gives this relation: b/t <= 0.64*sqrt(E*kc/Fy).

The kc value equals 4/sqrt(h/tw)=4/sqrt(11/4)=0.603, since hw=11″ and web thickness, middle plate=0.25 inches. The required b/t =0.64*sqrt(29000*0.603/100)=8.46.

Since the section b/t is greater than 8.46, Qs is not equal to 1. We need to proceed to the E7-8 equation to find the equation for Qs.

Estimate b/t based on the value of 1.17*sqrt(E*kc/Fy), which equals(1.17*sqrt(29000*0.603/100)=15.47.

The actual b/t,10, is bigger than 8.46 but less than 15.47. The Qs equation 1.145-0.65*(b/t)*sqrt(Fy/E*kc)=1.145-10*sqrt(100/29000*0.603)=0.923. Please refer to the slide image below.

For the stiffened part, the ratio (h/tw) = 11/0.25 = 44.

Find the effective b Value and the effective Area based on Q = 1.0.

Now, we will proceed to calculate Qa. For the slender stiffened element, h/tw=11/0.25=44 will be compared with (1.49*sqrt(E/Fy),  which gives the Value of  1.49*sqrt(29000/100)=25.40.

Our h/tw for the section is> 25.4; since h/tw > λr, Qa < 1. Then the modified h will be based on E7-17. But we need to find the Value of the stress f in the equation; consider Qa = 1.0.

Find the Euler stress and critical stress for the column.

The column is inelastic since Kl/r y value which equals 37.50, is less than 4.71*sqrt(E/Q*Fy) which equals 80.208, The facr=0.658^(Q*fy/Fe)*(Q*Fy). The Euler stress=Pi^2*E/(Kl/r)^=204 ksi,

First, consider Qa=1, our calculated Qs=0.923, then q=1*0.923=0.923,for the first trial, Qfy=0.923*100=92.3 ksi, Fcr=Qfy*(0.658^Qfy/fE)=92.3* 0.658^(92.3/204)=76.38 ksi.

The new graph for the short column is shown as a dotted line.
Take the fcr Value, 76.38 ksi, and plug it back into equation E 7-17 as follows.

Find the effective b-value and effective Area based on Q = 0.923.

Then be=1.92*(tw)*sqrt (E/f)*(1-(0.34/bt*Sqrt(E/f), tw=0.25 *sqrt(29000/f), the trial f=Qa*Fcr as estimated earlier, the trial f=0.923*76.28, Trial f=0.923*76.38=70.50 ksi, the term 1-(0.34/bt*Sqrt(E/f)=1-(0.34/11/0.25*sqrt(29000/70.5)=8.209 inches<11 inches which is the web height.

The effective area=Ag-hw*hw+be*tw=12.75-11*0.25+8.209*0.25=12.05 inch2. Next, we will estimate Qa.

Reestimate the Qa=Aeff /Agr=12.05/12.75=0.945. For the new Value of Qa. Qa=12.05/12.75=0.9452. So,,w new QQnewQ = Qs*Qa; Qa = 0.945 while Qs = 0.923.

Finally, Q new = 0.923*0.945 = 0.8722, back to Fcr = 72.94 ksi. Again, multiply by Q = 0.8722 to get another Value of stress: = 0.8722*72.94 = 63.603.60 ksi.

Multiply by Q to get the new Value: 0.8722*72.9 = 63.66 ksi.

Find the effective b Value and the effective Area based on Q=0.8722.

Evaluate the modified Value of (be) for the stiffened Web, be = 1.92*(0.25)*sqrt(29000/63.6), by the other bracket—finally, hw eff = 8.557 inches.

Recalculate the A eff=10*0.5*2+8.557*0.25, Aeff =12.14 inch2. The corresponding effective Area is 12.14 in^2.

Find the effective b-value and effective Area based on Q = 0.872.

Qa new =A eff/Ag=12.14/12.75=0.945, To get the  Q=Qs*Qa=0.923*0.945= 0.872,  fcr recalculated with the new Q Value, Fcr=72.94 ksi. Multiply by Q to get the new f: 0.872*72.94=63.6 ksi. Evaluate the modified Value of (be) for the stiffened Web, be = 1.92*(0.25)*sqrt(29000/63.6), by the other bracket—finally, hw eff = 8.557 inches, same as estimated earlier.

Solved problem 66-19-4: estimate the nominal Load.

The A eff=12.75-(11*0.25)+(8.557*0.25)=12.14 inch2. The Final Qa value=12.14/12.75=0.952. The final Q value=0.923*0.952=0.879. Estimate fcr=0.658^(0.879*100/204)*87.90=73.0 ksi. there Nominal Load =fcr *Ag=73*12.75=931 kips.

In this LRFD-only example, φc = 0.9, and φc*Pn = 0.9*931 = 838 kips. This concludes our solution to problem 6-19-4. Thanks a lot.

You can view and download the PDF for this Post from the following Link.

For a good A Beginner’s Guide to the Steel Construction Manual, 14^th ed. Chapter 7 – Concentrically Loaded Compression Members.

For a good A Beginner’s Guide to the Steel Construction Manual, 15^th ed. Chapter 7 – Concentrically Loaded Compression Members.

For a good A Beginner’s Guide to the Steel Construction Manual, 16^th ed. Chapter 7 – Concentrically Loaded Compression Members.

For the next Post, 16, here is the Link: A Solved Problem 5-10 for the Available Strength.