Last Updated on September 14, 2026 by Maged kamel
Old Cb Coefficient of bending for steel beams.
This is a brief illustration of this Post’s content, showing how to determine the Value of old Cb using the Expression for old Cb, the bending coefficient. A practice problem is used. What are the values for Cb using Table 3-1?
What are the cases where the old CB Value equals 1?
Based on the old equation for CB, the Value is given by the equation Cb (1.75+1.05*(M1/M2)+0.30*(M1/M2)^2)<=2.30.
The old CB value of Cb must equal 1 in two cases. The first is a cantilever Beam without a brace.
The second case for Cb equals 1 is a supported Beam of length L acted upon by a uniform load w. It is braced at the two ends, but the Moment Value in the middle equals wL^2/8, which is greater than M1 and M2.
Please refer to the image on the next slide for more information.
If we consider a supported Beam with bracing placed at the midpoint to evaluate the Value of Cb based on the old equation, for the part from the left support to the brace in the middle, the CB Value is equal to( 1.75+1.05 (0/wL^28)+0.30(0/wL^2/8) 2=1.75.
The exact 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 equals -1.
If we check the Value of Cb, we find that it equals +1.
A solved problem for the old Cb for a case of continuous beams.
I have included a solved problem from Schaum’s book. It requires determining Cb for the continuous Beam shown in Figure 5-12.
The first case is where two braces are provided only at the supports.
There are three Moment values: the first Moment is at the left support, point A, and equals 500 ft-kips; the second Moment is at the midpoint, point C, and equals 200 ft-kips. The third-moment Value equals 500 ft-kipsat the right support, point B.
For case a, we can find that the old CB Value equals 1.
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: A, C = 500 kips, which is more significant than M1 = M of ft-kips; the ratif M1/M=ls +0.40.
We will apply the equation to find the old Cb Value; we find that CB equals 2.22. The detailed solution is shown in the next slide image.
NCEES handbook FE handbook 10.4.
We will check a summary of the NCEES handbook reference 10.4, page 281; it starts with LRFD and E = 29000 ksi.
Beams for doubly symmetric compact I-shaped members, Φb=0.90, Yielding Mn=Mp=Fy*Zx.
Where Zx is the plastic section Modulus, and lateral-torsional buckling is explored.
When Lb < = Lptn = Mp;= 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-0.70FySx(lb-Lp)/(Lr-Lp)), where cb=12.50*Mmax/(2.5M-max+3*MA+4*Mb+3*Mc).
Table 3-1 shows the CB coefficient of bending values for different loading conditions.
We will check the bending coefficient values for different loading conditions. This is part of the CB values, for which I have included a snapshot. The following table is in the NCEES handbook, reference 10.4, page 282. Table 3-1 is included in the CM#15 construction manual of structural steel on page 3-18.
Next is a brief discussion of the nominal flexural strength, I quote, that the nominal flexural strength of the W shape is illustrated as a function of the unbraced length Lb.
The available strength is determined as Φb*Mn or Mn/Ωb.
Evaluate the ultimate load as 1.2 Wd + 1.6 L for LRFD, or as the total load, D + L, in ASD.
Then, Φb*Mn >M-ult in LRFD, or, in the case of ASD, where Mn/Ωb> Mt.
Chapter F deals with flexural strength due to the Moment in the AISC. Section F1-1 outlines the sections of Chapter F and the corresponding limit states for each member type.
Braced steel Beam section.
For 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 the flange or the case of the web.
Yielding must be considered in the member’s nominal Moment strength, which is Fy*Zx.
Unbraced steel Beam section.
In the case of the unbraced member Lb>Lp, have flange-width-to-thickness ratios such that λ>λp or have 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 intermediate Zone, or the straight-line relation, starts from Mp.
It ends with 0.70*Fy*Sx, where λ>λp and lateral-torsional effects are to be considered after checking the Lb Value, evaluating the lr, and then comparing. if lb>Lp Lb<Lr.
Noncompact or slender steel Beam section.
If the section is noncompact or slender for width-to-thickness members 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 do not apply. Bending about the y-axis is handled using the same procedure as for the strong axis.
This PDF includes the content of this file.
As an external resource –A Beginner’s Guide to Structural Engineering–Chapter 8 – Bending Members- 14th edition.
As an external resource –A Beginner’s Guide to Structural Engineering–Chapter 8 – Bending Members-15th edition.
As an external resource –A Beginner’s Guide to Structural Engineering–Chapter 8 – Bending Members- 16th edition.
This is the complete list of all posts related to Cb:
1-Introduction to Cb-Bending coefficient part 1 for the steel Post 17-previous Post.
2- Cb-The coefficient of bending part 2 for steel beams-post 18-previous Post.
3-18a-Old Cb Coefficient of bending for steel beams-Post 18a-this Post.
4-Cb value-bracing at the midpoint of a beam-uniform load-Post 18b-Next Post.
5-5-Cb value-bracing at the third point of a beam-uniform load-Post 18C.