# 10a-Regions for Lateral-torsional buckling for beams -2/2.

## Regions for Lateral-torsional buckling for beams-2/2

We have three regions for Lateral-torsional buckling for beams, with similar shapes, to the previous zones at the local buckling for steel beams but instead of using lambda λ factor in the horizontal axis as λp plastic and λr elastic, we use Lp, plastic length Lr, which is the unbraced length Which is a boundary between elastic and inelastic lateral- Torsional buckling in Feet.

The first region of the three Regions for Lateral-torsional starts from Lb=0, where Lb stands for bracing length to lb= Lp. Lp is the plastic length between bracings. The second point which Lr for the unbraced elastic moment, which is=0.7Fy*Sx, where Sx is the elastic section modulus, the inclined line has a zone of inelastic buckling,

While the last curved zone is for elastic buckling, every line has a reference equation.

Each formula starts from F2-1 to f2-3 for the three regions for Lateral-torsional buckling for beams shown in the next slide image with the relevant equation.

### Cb-coefficient of bending.

A new factor will be considered which is the Cb= bending coefficient, defined in equation 5.10, This factor appears in the three Regions for Lateral-torsional buckling for beams for the different equations F2-1&F2-2 and F2-3.

The recent equation for the Cb equals to 12.50 *M-max/(2.50 Mmax+3Mb+4Mb+3Mc)*Rm<=3.0.

We will take about Cb in a later video, we notice that 12.5 in the numerator is equal to the sum of 2.50+3+4+3 in the denominator, the item M max is the maximum moment in the beam to be divided into three points A, b, c.

We will discuss later the Cb with solved examples, while the old equation was Cb=(1.75+1.05*M1/M2+0.30*(M1/M2)^2)<=2.30.

The first part of the curve for Cb=1, but the Cb due to bracing, can be >1, the three Regions for Lateral-torsional buckling for beams can be seen Mp, the second inclined line joining Mp with  Mr=0.70*Fy*Sx, while when cb>1.

The horizontal line extends horizontally the inclined line will be displaced in case of Cb>1.

### How to evaluate Mn in case of lateral-torsional buckling?

The most important item for us is how to determine the relationship between the bracing length and Lp and Lr, and how we can estimate the Lp and Lr? How to get the value of Mn in the case of lateral- torsional buckling?

For the case of Cb=1, for the horizontal line, it has a value of Mp, while for the inclined line joining between two points, the first point is Mp at the unbraced length of Lp, and the second point is 0.70*Fy*Sx.

### How to get Lp value?

There is a relation between Lp and ry or the radius of gyration in the y direction and Fy which is the yield stress of steel based on the designated ASTM.

Mn for any point, if we have a bracing length =Lb, in between Lp and Lr, can be evaluated as Mn= MP- slope by the difference,  then Mn=(Mp- 0.70*Fy*Sx) /Lr- Lp all *(Lb -LP)*Cb, Lp value =ry *300/ sqrt FY, the values of Fy are listed with the corresponding Lp values. the next slide lists the different values of Fy starting from 36ksi& 42 ksi&45 ksi &50ksi and finally Fy=65 ksi.

For instance for Fy=36 KSI, Lp=50*ry, where ry is the radius of gyration at the y-axis.  For Fy=42 KSI, then Lp=46.3*ry.

For Fy=45 KSI, then Lp=44.7*ry.

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

For a useful external source, Chapter 8 – Bending Members.
For the next post, Solved problem 4-5-How to design a steel beam?

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