# 25-Part 1-4 Solved problem 9-9-6, how to find service LL?

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

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

This is the content of the lecture and the sequence of items for calculations of the solved problem 9-9-6.

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 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 a 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.

We want to check the compactness of both the flange and web and find out what type of F equation we will use. Please refer to the details of the different F types for doubly symmetric sections based of Chapter F of the specification.

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

We cannot take as λr=1*sqrt(E/Fy) as for case no.1  for flanges of Rolled I-shaped sections. We will use case No .11 to get the values for λp and λr for the flange of the Built-up section. For the web parameters please refer to case Number 15 of Table B4.1b.

This is a clear copy of Case 11 for the Built-up section, where, For the Flange 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.

This is a clear copy of Case 15 for the Built-up section, where, For the web λp=3.76*sqrt(E/Fy), while  λr=5.70*sqrt(Kc*E/Fy).

This is the calculation for the coefficient of the Kc value. This is the Built-up section,16″x5/8″, and 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.

### Find λp and λr for the Flange Local buckling parameters.

The Built-Up section is given and we have λp for flange equals 0.38*sqrt(E/Fy). Since E=29000 psi and Fy=65 ksi. the value of λrF will be equal to 8.026.
While for flange we have λr which is equal to 0.95*sqrt(Kc*E/Fy). Since E=29000 psi and Fy=65 ksi. t value of λrF will be equal to 0.950.95*sqrt(Kc*E/Fl)= 0.95 *sqrt(0.44*29000/0.7*65)=15.90.

Bf/2Tf =16/2*5/8=12.80. The Value of Bf/2tf is bigger than λp but less than λr. Thus the flange is Non-compact. The Mn will be less than Fy*Zx and bigger than 0.75Fy*Sx.

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

What is the value of Mp and also the value of 0.70*Fy*Sx? We need to estimate the Zx and Sx values for the solved problem 9-9-6.
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 in 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.

### Find λp and λr for the Flange Local buckling parameters.

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. this is the end of part 1 for the solved problem 9-9-6.

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

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