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

## Cb-Bending coefficient part-1 for steel beams.

### The video I used for illustration.

in the video, we will focus on the CB-bending coefficient and how to estimate it for different types of beams based on end conditions and different types of loading. The video has a  subtitle and a closed caption in English.

You can click on any picture to enlarge, then press the small arrow to review all the other images as a slide show.

### How do we estimate the CB-bending coefficient values for different loading types and bracing locations?

We are going to talk about a new subject, which is how to determine the Cb- bending coefficient. A discussion of the calculation for different loading and bracing is provided.

For all the previous examples we take the value of Cb=1 whether for the estimation of nominal design strength or for the design of a given section.

The different values of the CB-bending coefficient are shown for the different types of loading for a simple beam.
The first figure shows a simply supported beam with a uniformly distributed load W-ult k/ft.
The bracings are at the support left end and also at the right support, each bracing is shown as x mark.
For the unbraced length =l, for the first case, the value of the Cb=1.14.

The next figure is for a simply supported beam, similar to the first case, but this time a bracing is added in the middle of the span. So in this case we have three bracings, at the left side, middle side, and at the right edge. The middle bracing is located at a distance =L/2 from the left support.

The uniformly distributed Load =W as Ultimate load Kips/Ft, for that case the cb-bending coefficient value is =1.30.

The third case if for a simply supported beam with a concentrated load acting in the middle, the cb-bending coefficient value for this case is =1.32, for two bracings only.

The fourth case is for a Simply supported beam, similar to the third case, but with three bracings. The acting concentrated load is Pult. The cb-bending coefficient for this case is 1.67.

There are several cases for two concentrated loads and also bending moments acting at the supports, Cb-bending coefficient value for the last case is =2.27.

For statically indeterminate structures we have a Fixed end beam under W-ult as uniform load, with two bracings at the two edges, the CB-bending coefficient value is 2.38.

For the same beam but with three bracings we have the CB -bending coefficient value is =2.38.
The third case is for the Fixed end supported beam with Pu load acting at the middle, the CB-bending coefficient value is =1.92.

The last case is for two Fixed ends supported beams with Pu load acting at the middle, but with three bracings, the CB-bending coefficient value is=2.27.

### How to estimate cb for a simply supported beam under a uniform loading?

The equation which we use for the evaluation of the cb value is  cb=12.50*M-max/(2.50* Max+3*MA+4*Mb+3*Mc), the sum of the denominator value=2.50+3+4+3=12.50.
A is the first quarter-point. B is the second quarter-point. C is the third quarter-point.

In the case of a simply supported beam under uniformly distributed W-ult k/Ft, for which we have two bracings, the first bracing is located at the left edge, while the second bracing is at the right support.

M-max value, as we are familiar with simply supported beam under uniform loading Mmax= W-ult*L^2/8, where L is the span distance.

This is M-max. for Ma value at the first-quarter point. First, estimate the reaction is W*L/2, the moment caused by the reaction= Wult*L/2*L4 opposed by a moment for loading =Wul*(L/4)*1/2*(L/4)= the moment value Wul*l^2/8-Wul*L^2*1/32.

Finally MA=3*Wult*L^2/32, due to symmetry Mc value is the same as the Ma, what is left is the Mb value.
For Mb value, Wult*L^2/8, now we have all the information to estimate the Cb value, as 12.50 *(Wul*L^2/8)/(2.50*(Wul*L^2/8)+(3+3)* Wult*3*L^2/32.

For the numertaor 12.50*Wult*L^2/8/(2.5*Wult*L^2/8)+3* (3*Wult*L^2/32) +(4*3*Wult*L^2/8) +(3*Wul*L^2/32).
Take Wult*l^2/8 as a common factor, for the denominator, we have (2.50+ 9/4+9/4+4).

The denominator sum=2.50+4+18/4)=11, cb=12.50/11=1.14, the same result as for the first case.

### How to estimate cb for a fixed ends beam under a uniform loading?

Let us check another case. For a beam with two fixed supports, L is the span length, under a uniform loading, with two bracings at the two edges.
For the fixed beam, the Fixed end moment M-fe=Wul*L^2/12, if the loading is a uniform load= W-ult k/Ft.
We will divide into A, B, C, A at a distance=L/4, b at the middle, and C is at L/4 from the right support.

We draw the bending moment diagram, but we need to estimate the reaction at the supports. We will perform the same exercise as we have done.

For the first point A. The total load =Wul*L, then the R-a, the reaction at A=WUlt*L/2=R-b, while the two fixed end moments are W*ult*L^2/12. Ma=moment from reaction-Moment from the W ult for a span of L/4-W*L^2/12 as resisting moments.

All resisting moments =Wult*(L/4)*1/2*(L/4)=Wult*L^2/32+W*l^2/12, we sum all moments, as shown, we get W-ult as a common factor.
We have wult*(-L^2/12+ L^2/8-L^2/32), make a common denominator as (32*12), so we have MA=-Wu(-32*L^2)+ +4*12*L^2 -12*L^2). So as a sum=(-32+48-12)=Ma=Wult*L^2(+4/(12*32)=+Wult*L^2/96, with a positive sign.

Due to symmetry, the Mc = +W ult*L^2/96. At the middle Mb=Wul*L^2/24, estimated as (Wult*L^2/8-W*L^2/12), we give the same value of Wult*L^2/24.
For cb=12.5*Mmax/(2.50*Mmax+3*MA+4*Mb+3*Mc) . Max moment. in this case is W*L^2/12>Wult*L^2/24.

The  numerator=12.50*Wul*L^2/12/(2.50*W*L^2/12+3*(1/96)*Wul*L^2+4*Wult*L^2/24+3*(Wul*L^2/96).

We will sum all the values, as(2.50/12)+(6/96)+(4/24), delete the item Wult*L^2, we are left with(12.50/12)/(make a common factor as 96.(8*2.50+6+ 16)/96.

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

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

For the next post, Cb The coefficient of bending– part-2.

Scroll to Top