Last Updated on October 20, 2025 by Maged kamel
- CB value-bracing at the midpoint of a beam-uniform load.
- CB value-bracing at the midpoint of a beam-uniform load-simple beam case.
- CB value-bracing at the midpoint of a beam with a central load for a simple beam, two cases
- CB value-bracing for a fixed-end beam with a uniform load with two exterior braces.
- CB value-bracing for a fixed-end beam with a Central load and with three braces.
CB value-bracing at the midpoint of a beam-uniform load.
The next slide image shows the content of this post.
In this post, we will estimate the Cb value-bracing at the midpoint of a beam-uniform load—the case where we have three braces, one at the midpoint and two at the supports.
In the previous post, number 17, we estimated the CB value or the coefficient for a moment for a supported beam with two braces at the supports under uniform load.
The next slide image shows the different cases for a simply supported beam with different CB values based on the bracing locations in the span.
CB value-bracing at the midpoint of a beam-uniform load-simple beam case.
We can see that we have a brace in the mid-span of a simple beam under uniform load W. The midpoint is point C.
To estimate the CB value-bracing at the midpoint of a beam-uniform load, we divide part Ac into four quarters and calculate the value of moments at the different points.
What is the maximum value for a moment?
The moment value at point A’ is estimated as equal to ((7/(16*8))*wL^2, where L is the span and w is the uniform load per kip or meter based on the used units.
The moment value at point B’ is estimated as equal to (3/32)*wL^2, and the moment value at point C’ is estimated as equal to ((15/(16*8))*wL^2. The moment at point c is equal to (1/8)*wL^2. Exploring the moment values for segment AC, we find that the maximum value is WL^2/8.
Now, we will apply the equation to get the Cb value-bracing at the midpoint of a beam-uniform load. The maximum moment is w*l2/8, which will be multiplied by 12.50 for the numerator. For the denominator, we will sum 2.5*Max +3Ma’, 4Mb’, and 3Mc’.
The Cb value for a simply supported beam will be found to be equal to 1.3, which matches the Cb value for the second case in Table 3-1.
The Cb value for segment Bc will be equal to 1.3 because of the symmetry.
If you wish to compare the value of Cb based on the equation to the old expression for CB, we will find that the old value of Cb will be equal to 1.75 since M1/M2 will be equal to zero. Please refer to the following slide image for more illustration. The next slide details how to find the old Cb value.
CB value-bracing at the midpoint of a beam with a central load for a simple beam, two cases
For the cases that we will include in post 18b, I have highlighted the two cases of Cb for a simple beam under central load. The difference in the CB values depends on the number of braces for each beam.
The first case is the case of two bracings at the edge for the simple beam, in which the maximum positive value equals PL/4, where P is the Load and L is the span, and it is located at the midpoint of the beam. The span AB is divided into four quarters between the two supports, and we can see that MA’=PL/8 which is the same value as Mc’.
We will find the Cb value-bracing at the midpoint of a beam under the central load. The Cb value is estimated based on the equation, the Cb for part AB = 1.32, which matches Table 3.1.
The second case is the case of the same previous beam under a central load, but with two bracings at the edge of the simple beam, and an additional brace at the midpoint of the beam.
The maximum positive value equals PL/4, where P is the Load and L is the span, and it is located at the midpoint of the beam. The mid span AB is divided into four quarters between the two supports, and we can see that MA’=PL/8 which is the same value as Mc’.
The following slide image illustrates the detailed estimate of the Cb value for the beam with a central load and three braces. The final value equals 1.67.
This is a reminder of the CB equation and the different moment values for the various quarter points.
CB value-bracing for a fixed-end beam with a uniform load with two exterior braces.
The following case is a fixed-end beam with two braces at the exterior supports.
We will find the Cb value-bracing at the midpoint of a beam under a uniform load. As we know from our structural analysis study for a fixed-end beam under uniform load, the fixed-end moment of such a beam will be equal to W*L^2/12, where W is the uniform load, and L is the span of the beam.
As we know from our structural analysis study for a fixed-end beam under uniform load, the fixed-end moment of such a beam will be equal to W*L^2/12, where W is the uniform load, and L is the span of the beam.
The middle moment can be found by adding the positive WL2/8 and the fixed end moments of W*L2/12.
To estimate the CB value-bracing at the midpoint of a beam-uniform load, we divide part AC into four quarters and assess the value of moments at the different points. The next slide image includes the steps to find M’a and M’b moment values.
The next step is to estimate the moment for C’ in the third quarter for segment Ac. The value of the moment at C’ will be equal to (13/384)*w*L^2. The value of mc is w*L^2/24.
I have listed all the values for moments at points A, a’, b’, c’, and point c. Considering the maximum value of the estimated moment value, which will be found to be equal to Wl2/12, we consider the absolute value of moments.
The denominator values are prepared using (2.5M max+3*Ma’+4*Mb’+3Mc’) as a multiplier of w*L2. The final value is (7/16)*w*L2.
Apply the equation for Cb to get the coefficient of moment for the fixed-end beam with three supports, two at the supports and one at the middle point. The value is 2.38.
CB value-bracing for a fixed-end beam with a Central load and with three braces.
The following case is a fixed-end beam with two three braces, the acting load is a central load of value P.
We will divide the part AC, where we have braces at A and C, into four quarters. For each point, we will estimate the value of the moment, we have MA’=-P*L/16, while for Mb’=0 and for Mc’=+P*L/16. Please refer to the following slide image.
The following slide shows the details of the denominator of the Cb equation.
Apply the equation for Cb to get the coefficient of moment for the fixed-end beam with three supports, and acted upon by a central load P. The CB value is 2.27. Thanks a lot.
This is the PDF for the data used for the preparation of this post.
Download an easy illustration for the CB values for different loadings and End conditions [PDF]Please refer to Figure 8.2.1.7 Computation of Cb from A Beginner’s Guide to Structural Steel Manual, 16th 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.
3-18a-Old Cb Coefficient of bending for steel beams-post 18a-Previous post
4-Cb value-bracing at the midpoint of a beam-uniform load-Post 18b-This post.
5-Cb value bracing at third points of a beam-U load-Post-18C-Next post.