Last Updated on June 12, 2024 by Maged kamel

## Cb-The coefficient of bending part 2 for steel beams.

### CB- coefficient of bending values for a simply supported beam under two concentrated loads.

We will talk about a new subject: how to determine the Cb and the calculation for different loading and bracing.

Another example concerns a simply supported beam acted upon by two concentrated loads of Pult value for each.

According to the table, the beam has four bracings: two at both ends and two at one-third of the span. The CB at the middle third is 1.00, while the CB value for the first 1/3 of the span and the last third is 1.67. We want to prove that value.

Determine the reactions at A and B, then estimate the bending moment for beam AB. The Moment at a’=Moment at b’=Pu*L/3. Please refer to the next slide image for more details.

Assume we have 4 bracings at A, A’, B’, and B as shown in the sketch. We will estimate the value of Bending moments and multiply these values by the coefficients. The numerator has a 12.50 factor.

#### Cb value for parts A-A’ and B-B’.

The moment at A’=P*L/3 is a positive value for the Bending moment diagram. And the exact value of moment=P*L/3 is at point B’.

The Bending moment values are equal, and between points, A’ and B’ are a straight line. For the part from A to A’, we will enlarge the bending moment values and divide that part into four quarters.

The span length is L/3; after dividing by 4, each distance equals L/12 since we have 4 quarters.

Since the Moment at A’=Pu*L/3, the bending moment is in the form of a line, the values of the moment at each quarter can be found.

The moment values are M1=Pu*L/12 at 1& M2=Pu*L/6, while at 3, M3=P*L/4. We refer back to the equation of the CB. M-max is still Pu*L/3.

Apply the Cb equations, the 12.50 max equals 50/12*Pu*L. Find the sum of 3*M1+2.5*Mmax+4*M2 and 3*M3. the sum value equals 30* Pu*L*30/12.

The Cb value equals 50 Pu*L/12/30*Pu*L*30/12= equals 1.67. The CB value is for parts A-A’ and B-b’.

### Cb value for parts A’ -B’.

Let us consider the middle part, from A’ to B’. The bending moment is a constant value=Pu*L/3.

Again divided into four quarters. All moment values are the same=Pu*L/3,

The final value of Cb for the middle part A’-B’ will be equal to 12.50/12.50=1.0.

### Old Equation to Estimate CB, Coefficient Of Bending Value.

The old equation includes that, the cb=(1.75+1.05*(M1/M2)+0.30*(M1/M2)^2)<=2.30, where M1<M2, M1 smaller moment at the end of the unbraced length of the member. M2 is the larger moment at the end of the unbraced length of the member. If M1 rotates in the same direction as M2, then M1/M2 is considered positive.

But if both M1 and M2 rotate against each other, then M1/M2 is considered negative, this consideration will make a different calculation of the value of the CB-coefficient of bending. Since the sign of M1/M2 is important.

I quote, if M1 is the smaller and M2 is the larger end moment for the unbraced segment of the beam under consideration if the rotations due to end moments M1 and M2 are in opposite directions, then M1/M2 is negative, clockwise against anticlockwise otherwise, M1/M2 is positive. The value of the CB- the coefficient of bending equals 1.

The next slide image shows the plot of cb versum the ratio of M1/M2 for different values of M1/M2.

Thee discussion about Cb will be continued in the next posts.

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 the steel-post 17-previous post.

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

3-Cb-The coefficient of bending part-3 for steel beams-post 18a -next post.

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

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