## Part 1/4 of the solved Problem 9-9-6, How To Find service LL?

### Brief content of the video.

In this solved problem 9-9-6, we have the advantage of dealing with a built-up section. We will be able to review all the information about the Sx that is the elastic section modulus and also will be able to estimate the value of Zx, the plastic section modulus?

How to estimate the lateral -Torsional buckling as well as the local buckling of the flange and web?

How to estimate the Cb, the coefficient of bending?

We can not use tables. That is why we have to evaluate all the necessary data manually by using The LRFD design factor. There is an important point to consider that we have a built-up section then the lambda factor λr for non-compact slender.

We cannot take as λr=1*sqrt(E/Fy) as for case no.1. Flanges of Rolled I- shaped sections. This is a part of the video that 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.

### Solved problem 9.9.6 From Prof. Salmon’s Book.

This is the content of the lecture and the sequence of items calculations.

A solved problem 9.9.6 from Prof. Charles Salmon’s book. A solved problem 9.9.6 from Prof. Charles Salmon’s book.*Given the welded I beam section*, Used as a 45 ft simply supported beam laterally supported at the one-third point, please have a look at the small sketch.

The I beam section consists of three plates welded together, the Upper plate that represents the flange is a plate with the dimension of (16″x5/8″). The web part is a plate with the dimension of (26″x5/16″).

The lower flange plate is similar to the upper plate (16″x5/8″), the yield stress Fy is given as 65 ksi, E the modulus of elasticity is 29×10^6 psi or sometimes written as 29000 ksi.

Loadings are shown in the next image, bracings are located at the third point.

These are the equations relating to the value of Mn, The nominal moment.

### What is the local buckling λ parameter for flange?

**We cannot take as λr=1*sqrt(E/Fy) as for case no.1 for f**langes of Rolled I-shaped sections.

λp has the same value, as the first case, λp=0.38*sqrt(E/Fy), while λr=0.95*sqrt(Kc*E/Fy) which has a new factor that is called Kc. We start by evaluating the value of λr.

This is a quote from table 7.4.1, this is the Built-up section, 16″x5/8″, the inner height is 26″ with thickness=5/16″.

### The provision of AISC for M nominal

This is the Built-up section,16″x5/8″, the inner height is 26″ with thickness =5/16″. kc factor=4/sqrt(h/tw), h/tw=26/(5/16)=83.20,then kc=4/sqrt(83.20)=0.4385. kc should be >0.35 but must be < 0.763, In our case, we have 0.4385.

F-L=or **FL=0.7*Fy.λr. For Bf/2Tf =0.95*sqrt(Kc*E/Fl)= 0.95 *sqrt(0.44*29000/0.7*65).

λr=15.90, as we will see in the next slide.

If it was taken as 1.0*sqrt(E/Fy) as in the case of rolled I beam, it will be λr=21.10 for non-compact.

This is the AISC table B4.1B for the width to thickness ratio for stiffened element for the web of the built-up section it is item no.11.This case is the case included in the solved Problem 9-9-6

This is the value for λr in terms of kc &E and Fl.

This is the calculation for the coefficient of Kc value.

The next steps are shown as follows Check the local buckling Whether, for the flange or the web, I put them in one graph, on the left side, while the right-side graph is for the lateral-torsional buckling.

The same two diagrams as shown are identical except that the x-axis for the left diagram is for the λ values for Flange and Web sketch at for the relation between Lb and Mn at the right, the graph begins with a horizontal portion till Lb =LP, then an inclined line joining between Mn=Mp and 0.70*Fy Sx at Lb =Lr.

This is for the Compact and Non-compact slender sections.

This graph is for the Lateral torsional buckling. This part is for Lb=Lp and the inclined part from Lb >Lp and Lb=Lr.

The other graph is for the Local buckling for flange and web.

The x-axis is for lambda value, I put the values for both flange and web above each other.

The values are for the Welded I beam section The λr is adjusted.

For the flange width=16″ with thickness=5/8″ for local buckling consider half of the breadth of the flange, then bf/2tf=(16/2)*(1/5/8)=12.80 compare with λp=0.38*sqrt(E/Fy), Fy=65 ksi, E=29000 ksi.

λp= 0.38*sqrt(29000/65)=8.026.

bf/2tf is >λp, but<λr, the section is non-compact for the flange, Mn value will be in between Mp and 0.70*Fy*Sx.

The value of Mn is<Mp.

For the web the h is=26′, tw=5/16″, λw=26/(5/16)=83.20. λwp=3.76*sqrt(E/fy)= 3.76*sqrt(29000/65)=79.46.

λw is >λwp.

Let us check the value of λw-r= 5.7*sqrt(E/Fy)=5.7*sqrt(29000/65)=120.40, λw is >λwp. but <λwr, the section is also non-compact for the web for the local buckling.

The section is non-compact for both the flange and the web due to λ values >λp. but <λr Mn will be < Mp.

Mn is not the product of Fy*ZX.

### Estimation of Zx for the section for the solved problem 9-9-6.

What is the value Mp and also the value of 0.70*Fy*Sx? We need to estimate the Zx and Sx values.

what is the formula for Zx?

This is the total section due to bending, there will be a compression force acting on the CG of the one half of the section, the neutral axis N -A is at the middle, but the force, there are two forces C1 and C2, C2 acting on the web side till the N-A, C1, and C2 have equivalent forces as T1 and T2, these forces develop the plastic moment Mp.

Zx=At*(2*y bar)/2, 2*y-bar is the distance between the Cg of Compression force and Tension force. We call it Yct or the distance between compression and tension forces.

The plastic section modulus Zx=At*Y bar. The formula is widely used for irregular shapes.

AT/2 is the area of the top flange and half area of the web AT/2=A1+A2.

AT/2=(16*5/8)+(13*5/16))=10+4.06=14.06 inch2. Y1 is the distance from CG of the first area A1 to the N A.

Y1=13+0.50*(5/8)=13+(5/16)=13.3125″.

For A2, the height=13″ and thickness=5/16, then Y2=13/2″. AT/2*Y bar=A1*y1+A2*y2, then ybar=10*13.3125+4.06*6.50)/14.062. ybar=11.344″.

Zx=At*y bar=2*(14.06)*11.344=319.06 inch3. For the upper point of the graph Mp=Fy*Zx.

This is the end of part 1 of the solved problem 9-9-6. In the next post, we will continue in part 2.

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

For the next post, part 2/4 of the Solved problem 9-9-6, how to find LL for a given slender section?

As an external resource –**A Beginner’s Guide to Structural Engineering**