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

## Cb the coefficient of bending part-3 for steel beams.

### A comparison between the new and the old equation using an example.

In the Cb The Coefficient Of Bending Part-3, there is a comparison, from Prof. Salmon’s book, for the Cb- coefficient of bending values for a bending moment graph, the author would like to compare the Cb evaluated from the recent equation, with the 12.50 coefficients,  and the old equation of M1/M2. Cb=(1.75+1.05*(M1/M2)+0.30*(M1/M2)^2)<=2.30.

If we consider a simply supported beam is acted upon by a uniformly distributed load, the moment at the mid-span will be M=Wult*L^2/8. From the old equation, cb=1.00.

These are the values based on the existing equation. we will consider one case if we have a bracing, at the left hinge, and the one-quarter point as shown in the left sketch, where cb=1.52.

I quote laterally braced at the end and at the first quarter-point bracing, here and here there are two bracings M max at point 2, so he evaluated the moment at point 2.

Let us determine these values. We consider a part of a simply supported beam with a span of L/4, where we have bracing art point d, then, the total load is w*L, the reaction at the left support= W*L/2. The moment at d Md=w*L/2*(L/4)- W*L/4*(L/8), then Md=3*W*L^2/32.

Then we indicated the bending moment graph, for the portion from support to point d, again we divide it into quarters, for the values of Ma and Mb.

The span is L/16 between A and b and 2*L/16 between B and the left support, and 3*L16 between c and the left support. Estimate MA=w*L/2*(L/16)- w*(L/16)*L/32, Ma=15*w*L^2/(16*32). For the Mb value= ((W*L/2)*2*L/16 -W*(2*L/16)*L/16)). Mb=7*W*L^2/(8*16), similarly Mc=39*W*L^2/(16*32).

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

The cb from the 12.50 coefficients equation, cb=12.50*(3*w*L^2/32)/(2.50*(3*w*L^2/32)+3*15*W*L^2/(16*32)+4*(7*W*l^2/8*16)+3*(39*W*l^2/16*32).

From the previous slide, We will make a common factor (16*32). the coefficient of bending cb=12.50*w*l^2*3/32)/(2.50* 16*3/(16*32)+ 45/16*32+112/(16*32)+3*39/(16*32), then cb= 32 goes with 32, we have 37.50.

The sum of denomiator=(120+45+112+117)=394. Cb value =1.52 as indicated in the graph.

In the Cb The Coefficient Of Bending Part-3, we will estimate the CB- coefficient of bending for a simple beam for which there are two bracings one at the left support and the other one at  L/4, but by using the M1/M2 equation. We have opposite direction, so the M1/M2 is -ve, but cb=1.75+1.05*(0)+0.30*(0)=1.75 as shown also in the graph.

If we have a continuous beam, how to evaluate the CB value?

By using the old equation for the two opposite Moments and equal moments, M1/M2=-1, since these moments are equal. CB=1.75+1.05*(-1)+0.30*(-1)^2 , cb=1.75+0.30 -1.05. Finally cb =1.

In the Cb The Coefficient Of Bending Part-3, we will check a summary of the part of the NCEES handbook reference  9.50  page 163, He started with LRFD and E=29000 ksi.
Beams for doubly symmetric compact I -shaped members, Φb=0.90, at the yielding Mn=Mp=Fy*Zx.

Where Zx is the plastic section modulus and explored the lateral-torsional buckling. when Lb<=Lp, the Mn=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-0.70*Fy*Sx*(lb-Lp)/(Lr-Lp)), where cb=12.50*Mmax/(2.5*M-max+3*MA+4*Mb+3*Mc).

### Table for the CB-coefficient of bending values for different loading conditions.

In the Cb The Coefficient Of Bending Part-3, we will check the different values of bending coefficient based on the different loading conditions.

This is the part of the values of the CB, for which, I have included a snapshot.

The last item is a brief discussion of the nominal flexural strength, I quote, the nominal flexural of the w shape is illustrated as a function of the unbraced length Lb.
The available strength is determined as of the as Φb*Mn or Mn/Ωb.
The ultimate load to be evaluated as 1.2 Wd+1.6 L, for the LRFD or the total load as estimated D+L in the case of ASD.
Then, the Φb*Mn >M-ult in the LRFD, or in the case of ASD, where  Mn/Ωb> Mt.

Chapter F is dealing with flexural strength due to the moment in the AISC. Note F1-1 section outlines the sections of chapter F and the corresponding limit states applicable to each member type.

For the 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 flange or the case of the web.
Yielding must be considered in the nominal moment strength of the member, which Fy*Zx.
In the case of the unbraced member Lb>Lp, have flange- width to- thickness ratios such that λ>λp or have a 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 is the intermediate zone or the straight-line relation that starts from Mp and ends with 0.70*Fy*Sx, where λ>λp and lateral-torsional to be considered after checking Lb value and evaluating the lr and then compare. if lb>Lp Lb<Lr.

If the section is a noncompact or slender section for members that width-to thickness 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 does not apply this is bending about the y-axis will be dealt with with the same procedure as with the strong axis.

This is the pdf file used for the illustration of this post and the previous post. For more detailed illustrations in the CB, please follow this link. Flexural Limit State Behavior
This is the next post, Review shear stresses. The post is an introduction to shear stress for beams.

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