## Cb value bracing at third points of a beam-uniform load.

In this post, we will estimate the Cb value bracing at third points of a beam-uniform load in the case where we have four braces wo at the ends and two at each third point.

We have estimated the CB value or the coefficient for a moment for a simply supported beam for the case of two braces at the supports, under uniform load in the previous post number 17.

The next slide image shows the different cases for a simply supported beam with different CB values based on the locations of bracing in the sthe pan.

### Cb value bracing at third points of a beam-uniform load-simple beam case.

To estimate CB value bracing at the third points of a beam-uniform load we divide part AC into four quarters and estimate the value of moments at the different points.

### What is the moment value at point C?

The moment value at point C is estimated as equal to ((1/9)*wL^2, where L is the span and w is the uniform load per kip or per meter based on the used units. To find out how we get this value, please refer to the next slide image.

The moment value at point A’ is estimated as equal to (11*/(12×24))*wL^2.

The moment value at point B’ is estimated as equal to (5/72)*wL^2.

The moment at point C’ is equal to (3/32)*wL^2. Exploring the moment values for segment AC, we find that the maximum value is WL^2/9 at point C. A sketch of the absolute values for the different points is drawn and shown in the next slide image.

### Cb value for part AC.

Now we will apply the AISC equation to get the Cb value bracing at the third points of a beam-uniform load. The maximum moment is w*L^2/9 will be multiplied by 12.50 for the numerator. For the denominator, we will add ( 2.5*Max +3*Ma’+4Mb’+3Mc’.)

The Cb value for a simply supported beam with four braces for part Ac can be approximated to 1.46.

### Cb value for part CD.

To estimate the Cb value bracing at third points of a beam, we divide part CD into four quarters and estimate the value of moments at the different points. Due to symmetry the moment at ends C and D are equal and equals (1/9)*WL^2.

The moment value at point d” which is located at the mid-span will be equal to w/*L^2/8.

The moment value for both Mc” and M’d’ will be equal to W*L^2/(35/288).

We will draw the portion of the beam for part CD and list all the absolute values for moments.

I have listed all the values for moments at points C, C”, D”‘, D’ and d. Considering the maximum value of the estimated moment value it will be found as equal to Wl^2/12, we take into consideration the absolute value of moments. use the multipliers of 2.5,3, 4 and 3 to get the value of the element.

The denominator values are prepared as we have (2.5M max+3*Ma’+4*Mb’+3Mc’) as a multiplier of w*L^2. The final value is (148/96))*w*L^2.The numerator value is 12.50*WL^&2/8. The Cb final value for part Cd can be found to be equal to 1.01.

Please refer to table 8.2.1.1 Computation of Cb from A beginner guide to structural steel manual 15th edition.

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.

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

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

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

This is the next post, Review shear stresses. The post is an introduction to shear stress for beams.