 # 35-Solved problem 8-22-for Plastic nominal load.

## Solved problem 8-22-Plastic nominal load.

### Brief content of the video.

It is required to evaluate the Plastic nominal load for a beam that is fixed at both ends of 32 Feet. The section is given as W24x62, so we can use the table to get the necessary data of Zx, Fy=50ksi.

Two methods are used to get the Plastic nominal load, the first method is the statical method or the lower bound theory, in which the plastic hinges are assumed and the two-moment diagrams are laid together, the place of a maximum positive moment to be estimated at that point one plastic hinge to be placed, the other two hinges are located at the places of maximum negative moments, at the fixed supports. The video has a subtitle and 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.

We are to continue our subject by solving new solved problems. We will apply our information on structural steel by using the yield strength to estimate Zx and M-nominal and get the required plastic nominal load W nominal.

The videos that cover our subjects were Civil 120-8 & Civil 120-9. The last videos were Civil 120-32 & civil 120-33, in which I have discussed the upper bound and lower bound and the difference in between.

At the start, we will check the table that includes a list of shapes and the ASTM designation for ASTM A992, the yield stress Fy in ksi is between 50-65 KSI, and these are the shapes included in the table and their names.

In the first solved problem 8-22, we use the given section of ASTM A992 and plastic theory. Determine the value of P-nominal and W nominal as indicated. We have several examples for M nominal, including concentrated loads P loads, and the others, containing uniformly distributed loads, W nominal to be estimated. This example is from Prof. McCormick’s chapter 8.

We have a beam that is fixed from both sides at A and B, the concentrated load Pn acts on point c.
The section is given as W24x62, so we can use the table. The Pn acts at point c, which is located 20′ apart from the left support A. and at a distance=12′ from the right support B.

This is the table for w shapes. We have W24x62 as given in our example, we have obtained the flange and the overall height of the section, the width of the flange=7.04″, and its thickness=0.59″.

The thickness of the web=0.43″ and the overall height=23.70″, this will show the detailed section of the W- section.

The other part of the same table gives the Sx and Zx values for this section for axis X-X. The data includes Ix, Sx, rx, and Zx and the compact section criteria, to check the local buckling, by using the ratios of bf/tf and h/tw for flange and web, and lateral-torsional buckling check. We have Zx=153.0 inch3.

If we multiply by Fy, Fy*Zx= (KSI/inch^2)*(inch3), then 50*153, we get the result in Kip. inch, for  Ft. kips, divide by 12=50*153/12= 637.50 Ft.kips. The nominal moment=637.50 Ft.kips.

### Solved problem 8-22-Plastic nominal load by using the lower bound theorem.

For the next slide, in our system, we will solve by using the two methods.

The first method is the lower-bound theory. or the statical method, in which, we check the number of indeterminacy and then convert the system to an unstable system, we have three reactions at A+ 3 reactions at B the total number is 6, to get the degree of indeterminacy.

We have three reactions, horizontal, vertical, and moment and the same at the other support B. The number of indeterminacy=3+3-3=3 degrees of indeterminacy.

We will covert the system to Statically determinate, we need to add some hinges=6-3=3. For the three hinges, one hinge at A, the second one at B, and the third one are at C, since no horizontal load, the system needs 3 hinges only to be unstable.

We need to estimate M positive and the M negative and lay the two diagrams together then check the maximum positive M plastic value since the collapse will occur at the maximum value of M plastic.

For the beam shown, we estimate the reactions at A and B, RA=Pc*12/32, Rb=Pc*20/32, Mc=RA*20’=Pc*20*12/32=240*Pc/32, this is the moment as a function of Pn.

The next step is to include Mp negative at support A. Mp negative at the other support B, and drop the value of M-positive.
The hatched triangle below the datum represents the M=ve value, for which the maximum value is the Mp value.

We have Mp+Mp=M+ve= 240*Pn/32, Mp=120*Pn/32, substitute MP by the value of Mn, which =637.50 Ft.kips, already estimated from Zx and Fy. Pn=(32/120)*(637.50)=170.0 Klbs. Mp value is the maximum value of Moment=637.50 Ft.kips.

### Solved problem 8-22-Plastic nominal load by the upper bound theorem.

In the next slide, this is the Upper bound, with the mechanism with an assumed hinge location under the load, there is a slope of θ value at the left and θ1 at the right support.

If we consider the deflection at C=Δ, θ=tan θ=Δ/20, θ1=tan θ1=Δ/12.
We have Mp, which is a positive plastic moment with an angle=θ+θ1, and two plastic moments at the two supports, at support A, for which the angle is θ, while at support B, the angle is θ1.

The internal Work =Mp*θ+Mp*θ1+Mp*(θ+θ1). For θ+θ1=(Δ/20)+(Δ/12)=32*Δ/240.

The external work Pn*Δ=Mp*(2*θ+2*θ1)=2*Mp*(θ+θ1)=2*Mp*(32*Δ/240, Delta Δ goes with Δ, Pn=Mp*(64/240).
Pn=(8/30)*Mp, since Mp=Mn=637.50 Ft.kips.

The plastic nominal load  Pn=170.0 kips, which is the same result as the value obtained from the lower bound method.

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