## Local Buckling for stiffened and un-stiffened elements.

### Brief content of the video.

**Our subject** is the local buckling classification of sections for local buckling, for compression, sections are classified as a non-slender element or slender element.

For non-slender elements, We have two factors, one factor for flange and the other for the web, if the b flange / t flange < certain ratio λr, then the section is a non-slender. the ratio, which we have already talked about via tables.

** ****If** the h web/ t web < λr for a certain factor then the section is non-slender, if > λr, shown in the table, then the section is slender, then a reduction of the coefficient is needed.

There are two coefficients, the first one is Qs and the second coefficient is Qa, the Qs is for the flange, if the lambda λ, is > the ratio is given. While the Qa is for the (effective area/ A gross), Non -slender not exceeding λr, from table B4.1a. This a part of the Video that has a subtitle and a closed caption in English.

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.

## Local Buckling for stiffened and unstiffened elements.

Our subject is the local buckling classification of sections for local buckling, for compression, sections are classified as a non-slender element or slender element.

For non-slender elements, We have two factors, one factor for flange and the other for the web. if the b flange / t flange < certain ratio λr, then the section is the non-slender.the ratio, which we have already talked about via tables.

This is a brief description of the content of this post.

h web/ t web < λr for a certain factor then the section is non-slender, if > λr AS showeNin the table, then the section is slender, then a reduction of the coefficient.

There are two coefficients, the first one is Qs and the second coefficient is Qa, the Qs is for the flange, if the lambda λ, is bigger than the ratio given.

While the Qa is for the effective area/ A gross. Non-slender not exceeding λr, from table B4.1a.

Pn, the nominal compressive strength shall be the lowest value based on the applicable limit states of flexural buckling, pn=fcr*Ag.

First, the column to be checked whether a Long or short column, but the formula used, will be KL/r<4.71*sqrt(E/Qfy).

Where Q=Qs*Qa or Qfy/Fe< or=2.25,Then the Fcr can be Fcr=Q(0.658^Qfy/fy)*fy =Q(0.658^Qfy/fy)*fy,but if Kl/r>4.71*sqrt(E/Qfy), the Fcr=0.877 Fe.

Let us look at the Kl/r=4.71*sqrt(E/Qfy), How the equation was derived?

Consider QFy/Fy when Q=1, and when Fy/Fe=2.25,again when shifting( Kl/r)^2=2.25*(pi^2*E/fy).

Take the SQRT (Kl/r)^2, gives ( Kl/r)= SQRT(2.25*pi^2*E/fy)=SQRt of (50), this 50 = pi^2*2.25, which is = 7.41.

So we come to our equation of 4.71*sqrt(E/fy), which defines whether the column is short or long. K*l/r is shown as a vertical line in the graph.

If the column is long the fcr=0.877 *Fe, the table lists the different values of fy

for instance, for fy =36, then the limiting Kl/r=134, while for fy=50 ksi.

### Limiting Kl/r based on the different Fy values.

If the column is long then fcr=(0.877 *Fe), The table lists the different values of fy, for instance, for fy =36, then the limiting Kl/r=134, while for fy=50 ksi, the limiting Kl/r=113, for fy=60 ksi, then the limiting case is 104, the controlling factor =4.71*sqrt(E/fy), and the corresponding values of the values of 0.44 fy are all listed.

If the column is long, then the graph is the black shifted curve from the dotted Euler graph.0.877Fe, let us have a look at the Euler graph the ordinate is fcr when is =fy/2.25.

Draw a horizontal line from that point that intersects with the Euler curve at 7.41 *sqrt(E/fy), the ordinate with the intersection with the shifted curve, will be fy/2.25*(0.877)=0.39 Fy.

If the Kl/r< 7.41 *sqrt(E/Fy), then the equation E 7.2 will be used as fcr=0.658^(Fy/Fe)*Fy, For Q when the equation is modified, the Q will appear with fy.

If the column is long, then the graph is the black shifted curve from the dotted Euler graph.0.877Fe, let us have a look of the Euler graph the ordinate is fcr when is =fy/2.25.

Draw a horizontal line from that point that intersects with the Euler curve at 7.41 *sqrt(E/fy), the ordinate with the intersection with the shifted curve, will be fy/2.25*(0.877)=0.39 Fy.

If the Kl/r< 7.41 *sqrt(E/fy), then the equation E 7.2 will be used as fcr=0.658^(Fy/Fe)*fy. For Q when the equation is modified, the q will appear with fy.

### Parameters for local buckling for Unstiffened elements.

For the unstiffened elements, the elements, for instance, are free on one side, and fixed on the other side, same as flanges, the ratio of bf/2tf, the C channel, and the outer portion.

A- For flanges of I-shaped members and tees, the width b is (1/2)of the full-flange width, bf/2tf, as for legs of angles and flanges of channels and zees, the width b, is the full nominal dimension, b/tf.

B- C-channel has only one part, so no division by 2 as compared with I-shaped sections.

The factor for Qs for b/t is an indication for slender if exceed certain criteria, or if less then can be considered as non-slender.

For hot rolled when b/t<= 0.56 *sqrt (E/fy),then the qs is =1,but if b/t > = 0.56*sqrt (E/fy) and < 1.03 *sqrt (E/fy) and < 1.03 *sqrt (E/Fy.

The qs can be given as per the equation of E 7.5, but if b/t > = 1.03 *sqrt (E/fy) then Qs can be evaluated from E7.6.That was for the Qs for flanges.

These are the relative equations from the AISC code.

### Table B4.1a for un-Stiffened elements part.

This is a part of the un-stiffened elements is the web in an I-shaped section, the web is stiffened by the upper and lower flanges as shown.

The height for estimating the ratio, full height of the section -2 ks, where ks is the distance of the fillet or corner radius at each flange. From table λr = 1.49*sqrt (Fy/E).

### Table B4.1a for Stiffened elements part.

Refer to the table from case no 5 for I beam sections for stiffened elements.

The Qa is given by For E 7-16 &E7-17 is used for the QA, The Qa is the factor relating to the web. A stiffened part, Qa=Ae/Ag, effective area / gross area after checking b/t value.

For the web of I section > 1.49 sqrt(E/fy) or equal, be was b-2ks, will be modified, the be will be determined from equation E 7-17, this equation based on Qa =1.

First, we start with the flange to evaluate the relevant Qs, assuming that Q=1, from the graph we will estimate the f value as Fcr, as we will see later, f is taken as fcr with fcr calculated based on Q=1.

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

This is a link to Solved problem 5.2, from Prof. McCormac’s handbook, which will be our next post.

for a useful reference, Limit State of Flexural Buckling for Slender Sections