Problem-5-3 Local buckling for square hollow steel section column

13-A Solved problem 5-3 for local buckling of columns.

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A Solved problem 5-3 for local buckling of columns.

Brief content Content of the video.

We are going to see, solved problem 5-3 for local buckling, From Prof. Jack McCormac’s handbook.
  An HSS section 16x16x1/2 inch, with Fy =46 ksi, is used for an 18 feet column with simple end supports. Determine Φc*Pn and Pn/Ωc, with the appropriate AISC equations, repeat part- a) using 4-4 in the AISC Manual.


Check whether the column is short or long column criteria, by using equation 4.71*sqrt(E/Fy)= 4.71*sqrt(29000/46)=118.26, while Kl/r,rx=ry=6.31 inch2, KL=1*18 *12, for inch, then (18*12/6.31)=34.23, which is < 118.26 the column is short, and non-slender element. We will estimate fcr =0.658^(46/Fe)*fy, no Q in the equation since Q=1.

We need to estimate Fe, Fe=Pi^2E/(KL/r)^2,  That was a part of the video, which has a subtitle and closed caption in English.

Review of the equation used to differentiate between long and short columns.

As a review, first, we have to check that, the column is long or short, by using equation 4.71sqrt(E/Qfy), in the case of KL/r >4.71sqrt(E/Qfy).

A reminder of the formula for Qs value.

The column is long. if kl/r> 4.71*sqrt(E/Q*Fy),  else the column is short if the  K*l/r<=4.71*sqrt(E/Q*fy).

If the column is short, proceed to check the local buckling. For the flange and web, in our case, the hollow steel section is a  stiffened element, only one equation will be used. Refer to item d, See B-3, which includes, how to measure the b length of HSS, and the definition of both b and h.

Section B-3 from the Aisc code.

A Solved problem 5-3 for local buckling.

For the solved problem 5-3, An HSS section 16x16x1/2 inch, with Fy =46 ksi, is used for an 18 feet column with simple end supports. Determine Φc*Pn and Pn/Ωc, with the appropriate AISC equations, repeat part a) Using 4-4 in the AISC Manual.

Data for the solved problem 5-3 from Prof. MCcormac’s Handbook.

We refer to Our table 1-12 for the properties of the Hollow sections, select HSS16x16x ½, wall thickness is 0.465 inch and the area is 28.3 inch 2, fy=46 ksi, to check,  b/t=h-2t/t, we have 16 inches as b, t= 0.465, then b-2t=(16-2*0.465)/0.465=32.41.

Ix =  1130 inch 4, and rx=6.31 inch2, According to item 6 for HSS, λr=1.4*sqrt(E/fy)=1.4*sqrt(29000/46)=35.15,then b/t <32.41 , the section will have QA=1 , we do not have Qs, no unstiffened section.

In the next slide, Check for short or long column criteria, by using equation 4.71*sqrt(E/fy)= 4.71*sqrt(29000/46)=118.26, while Kl/r,rx=ry=6.31 inch2, KL=1*18 *12, for inch, then (18*12/6.31)=34.23, which is < 118.26 the column is short, and non-slender element.

We will estimate Fcr =0.658^(46/Fe)*fy, with no Q in the equation since Q=1. We need to estimate Fe, Fe=Pi^2E/(KL/r)^2, Fe= (22/7)^2(29000)/(34.23)^2 =244.4744 ksi, back to fcr=0.658^(46/244.4744)(46)=42.516 ksi.

Check whether the given column- in the solved example 5-3 is long or short.

Then for LRFD, ΦFcr for ASD, use Fcr/Ωc. For the Pn= =FcrA=42.516*28.23=1200.2 kips, for LRFD, our Φc=0.9.

Use the general provision formula for the Solved problem 5-3.

Then Φ*Pn=0.9*1200=1080 kips. for the ASD, since Ωc=1.67,Pn/Ωc=1200/1.67=720 kips.

Using table 4-22 as an additional option for solved problem 5-3.

For table 4-22 for available critical stress for compression members, page 4-322 Aisc 14 edition manual, we have kl/r=34.23, from fy=46 ksi, From the left side of the table, log in with 34.24, in between 34 and 35, for kl/r=35, we have  38.3 ksi, while for kl/r=34 38.1, by interpolation our final value ΦFcr=38.254 ksi, that is the stress value

To convert it as a load multiplied by the area which is 28.3 inch 2, or Φ*Pn=1083 kips, can be approximated to 1080 kips, similar to the one estimated using the hand calculation for the LRFD.

Using table 4-22 for the critical stress for compression members for the solved problem 5-3.

From Table 4-22, The stress, Fcr/Ω is between 25.5 and 25.4, by interpolation= 25.477 ksi, by interpolation= 25.477 ksi, then multiply * area which is 28.3 inch2, Pn/Ω=721 kips, same as the estimate earlier. The values were driven first by using equations, then by using Table 4-22.

Using table 4-4 as part b for the solved problem 5-3.

Using table 4-22 for the critical stress for compression members for the solved problem 5-3.

For part b, where it is required to use table 4-4,  table 4-4 is for the square HSS, we have Fy=46 ksi, HSS is 16x16x1/2 inch, we have K*l at y direction =18′. We can log to table 4-4 and select kl at y=18′ then from the column 16x16x1/2″, we get the LRFD and ASD values for the column, please refer to the next two slides.

Using table 4-22 for the critical stress for compression members for the solved problem 5-3.

This is the calculation for the load based on the ASD design.

Using table 4-4 for the available strength for the tubular section for solved problem 5-3.

This is the pdf file used in the illustration of this post.

If you wish to refer to the previous post: A Solved problem 5-2 for local buckling of columns.

For a good external reference, find this linkChapter 7 – Concentrically Loaded Compression Members.

 This is a link for the next post, solved problem 6-8

 

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