## Cb the coefficient of bending part-3 for steel beams.

### What are the cases where the old CB value equals 1?

Based on the old equation of CB the value is given by the equation Cb=(1.75+1.05*(M1/M2)+0.30*(M1/M2)^2)<=2.30.

There are two cases for the CB old value to be equal to 1, the first case is the case of a cantilever beam without a brace. The second case for Cb equals 1, is the case of a simply supported beam of length L and acted upon by a uniform load w, that is braced at the two ends but there is a moment value at the middle equals wl^2/8 that is greater than M1 and M2. Please refer to the next slide image for more information.

If we consider a simply supported beam and bracing is placed at the midpoint of the beam, to evaluate the value of Cb based on the old equation, for the part from left support to the brace at the middle, CB value equals =1.75+1.05*(0/wl^8)+0.30*(0/wl^2/8)^2=1.75.

The same value of 1.75 will be given for the right part from the brace to the right support of the beam.

If we consider the case of a continuous beam of span L, where the braces are located at the supports, there are two equal and opposite moments. The two moments are opposite in direction so the ratio of M1/M2 is equal to -1.

If we check the value of Cb we find that it equals to +1.

I have included a solved problem from Schaum’s book, it is required to determine Cb for the continuous beam shown in figure 5-12.

The first case is the case where there are two braces provided only at the supports.

There are three values of moments, the first-moment value is at the left support, point a which equals 500 ft. kips, the second-moment value is at the midpoint c and equals 200 ft. kips. The third-moment value equals 500 ft. kips at the right support point B.

For case a , we can find that the old CB value equals 1.

In the Cb The Coefficient Of Bending Part-3, for the second case there are three braces two at the supports and one brace at the midpoint of the span. consider each part separately for part AC M2 equals 500 kips which is bigger than M of 200 ft. kips, the ratio of M1/M2 equals +0.40. We will apply the equation to get the old Cb value, we will find the CB value will be equal to 2.22. the detailed solution is shown in the next slide image.

In the Cb The Coefficient Of Bending Part-3, we will check a summary of the part of the NCEES handbook reference 9.50 page 163, He started with LRFD and E=29000 ksi.

Beams for doubly symmetric compact I -shaped members, Φb=0.90, at the yielding Mn=Mp=Fy*Zx.

Where Zx is the plastic section modulus and explored the lateral-torsional buckling. when Lb<=Lp, the Mn=Mp, the limit state of lateral-torsional buckling does not apply. When lb>Lp and less than Lr, then we will use the straight-line equation, Mn=cb(Mp-0.70*Fy*Sx*(lb-Lp)/(Lr-Lp)), where cb=12.50*Mmax/(2.5*M-max+3*MA+4*Mb+3*Mc).

### Table for the CB-coefficient of bending values for different loading conditions.

In the Cb The Coefficient Of Bending Part-3, we will check the different values of the bending coefficient based on the different loading conditions. This is part of the values of the CB, for which, I have included a snapshot.

Next is a brief discussion of the nominal flexural strength, I quote, the nominal flexural of the w shape is illustrated as a function of the unbraced length Lb.

The available strength is determined as of the as Φb*Mn or Mn/Ωb.

The ultimate load to be evaluated as 1.2 W_{d}+1.6 L, for the LRFD or the total load as estimated D+L in the case of ASD.

Then, the Φb*Mn >M-ult in the LRFD, or in the case of ASD, where Mn/Ωb> Mt.

Chapter F is dealing with flexural strength due to the moment in the AISC. Note F1-1 section outlines the sections of chapter F and the corresponding limit states applicable to each member type.

#### braced steel beam section.

For the braced and compact flexural Members, when flexural members are braced where Lb< Lp, where lb is the bracing distance, Lp is the plastic value for braces and the section is compact, λ<λp, whether in the case of flange or the case of the web.

Yielding must be considered in the nominal moment strength of the member, which Fy*Zx.

#### Ubraced steel beam section.

In the case of the unbraced member Lb>Lp, have flange- width to thickness ratios such that λ>λp or have a web -to – width ratios such that λ>λp lateral-torsional and elastic buckling effects must be considered in the calculation of the nominal moment strength.

This is the intermediate zone or the straight-line relation that starts from Mp and ends with 0.70*Fy*Sx, where λ>λp and lateral-torsional are to be considered after checking Lb value and evaluating the lr and then comparing. if lb>Lp Lb<Lr.

#### Noncompact or slender steel beam section.

If the section is a noncompact or slender section for members that width-to thickness such that λ>λp, local buckling must be considered Available flexural strength for weak-axis bending.

A flexural member subject to weak axis bending is similar to that for strong axis bending except that lateral-torsional buckling and web, local buckling does not apply this is bending about the y-axis will be dealt with with the same procedure as with the strong axis.

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

For more detailed illustrations in the CB, please follow this link. Flexural Limit State Behavior.

This is the complete list of all posts related to Cb:

1-Introduction to Cb-Bending coefficient part-1 for steel-post 17.

2- Cb-The coefficient of bending part 2 for steel beams-post 18-previous post.

4-Cb value-bracing at the midpoint of a beam-uniform load-Post 18b-Next post.

5-Cb value bracing at third points of a beam-U load-Post18C.