35-Solved problem 8-22-for Plastic nominal Load( 1/2).

Last Updated on September 15, 2026 by Maged kamel

Solved problem 8-22-Plastic nominal Load (1/2).

In the first solved problem 8-22, we use the given section of ASTM A992 and plastic theory. Determine the Value of the P-nominal and W-nominal as indicated. We have several examples for M nominal, including concentrated loads, P loads, and others with uniformly distributed loads, and W nominal to be estimated.

This example is from Prof. McCormick’s Chapter 8. We have Fy = 50 ksi, the Beam section is W24x62, and the Zx Value is 153 in³, according to CM#15, pages 1-19. I aim to evaluate the Nominal Load values for each plastic hinge.

How many plastic hinges are needed for collapse?

In the next slide image, we will examine the degree of indeterminacy and add one to find the number of possible plastic hinges after we ignore the horizontal reactions. As we showed, there are three possible plastic hinges: two at A and B, and one at point C.

Solve for the Value of the nominal Load by the lower-bound theory.

We will estimate the reactions at A and B and find the maximum Moment at point C, which will equal 60*P/8. For the three plastic hinges created, we have a 2Mp Value equal to 60*P/8, which, in terms of Zx and Fy, equals 2*Zx*Fy. From the first slide, Zx = 153 in^3, and Fy = 50 ksi.

Evaluate the nominal Moment Value, which is equal to Fy*Zx = 50*153 = 7650 in. Kips. Dividing by 12 gives Mp = 637.50 ft·kips. We can find that 60*Pn/8=2*637.50. The final Value of Pn is 170.0 kips. This Load creates three plastic hinges: two at the ends and one under the Load.

Find Pn using the upper-bound theorem.

In the next slide, this is the Upper bound, with the mechanism’s assumed hinge location under the Load. There is a slope of θ Value at the left and θ1 at the right support. Equate the external work to the internal work. The external work equals Pn*δ, while the internal work equals Mp*Δ—the Value of θ1=Δ/20, θ2=Δ/12.
We have Mp, a positive plastic Moment at angle θ + θ1, and two plastic moments at the supports: at support A, the angle is θ, and at support B, the angle is θ1.

The internal Work =Mp*θ1+Mp*θ2+Mp*(θ1+θ2). For 2*(θ1+θ2)=2*(Δ/20)+(Δ/12)=64*Δ/240. The Mp Value is 637.50 ft. We find that Pn = 170.0 kips.

Fixed-end momentvalues.

On the next slide, we can find the fixed-end moments at both ends A and B; in our case, a = 20 feet and b = 12 feet.

The fixed-end Beam with maximum Moment values is important. These points will have plastic hinges at the nominal Load so we will treat them as fixed ends for the working Load.

Consider a fixed-end Beam with a working concentrated Load P at a distance of 20 feet from the left support A.

The fixed Moment at A equals P*12^2*20/32^2=45*P/16. The fixed Moment at C equals P*20^2*12/32^2=75*P/16. The positive Moment Value at point B equals 60*P/8.

Finally, we sketch the moments at A, B, and C. The Moment is largest at Point C, followed by the Moment at B, and smallest at Point A. The first plastic hinge will occur at point C, the second plastic hinge at point B, and the third at point A.

The final positive Value is 3.51155P. Please refer to the image on the next slide for more details.

What Load creates the first plastic hinge?

We found that Pn = 170 kips results in three hinges, whereas a lower Load results in only the first and second hinges. We will Return to the working Load condition. The maximum Moment Value is at point C. We will equate the end Moment at C to Mp, which is 637.50 Ft-Kips, and we find that Pn = 136.0 kips.

We substitute Pn = 136.0 kips to obtain MA and MB.

When we consider P = 136 Kips, we find that the fixed Moment at A is 382.50 ft-kips, while the positive Moment is 45/16 * P = 382.50 ft-kips. These moments are less than Mp, so there is no second plastic hinge yet at Pn = 136.0 kips. Please refer to the slide image below below.

Shear and Moment values for P = 136 kips.

For a Load of P = 136 kips, there will be two end moments: 382.50 ft-kips at the left support and 637.50 ft-kips at the right support. The positive Moment will be 478.125 ft · kips. We can find the Shear values VA = 43.031 kips and Vb = 92.969 kips.

What Load creates the second plastic hinge?

The second hinge will be at point B. However, we will treat the beameam as a propped cantilever, since C has a plastic hinge and cannot be considered fixed. When the Moment at B equals 637.50 ft-kips, we will increase the Load by more than 136 kips and estimate the corresponding positive Moment. At P = 16060 kips, the fixed Moment at A is 382.5 ft-kips. Kips. At 136 kips, the Moment at C remains unchanged, and the positive Moment at C will be 478.125 ft—kips, which is less than 637.50 ft-kips.

Little time remains for the second plastic hinge to form. The next slide image shows that we need an additional reaction Load of 13.281 Kips to have a second plastic Hinge at B.

To have a Plastic Moment at B, we need a reaction at C equal to 13.281 kips, since we are dealing with a propped cantilever. We refer to the table of propped-cantilever moments and reactions to obtain the additional P Value for the second plastic hinge at B.

The additional Load P to be added to the 160.0 kips Load can be estimated, since the hinged support reaction is 13.28 kips.

Value of the Load required for the second plastic Hinge.

Once we have the Value of 28.269 Kips, we will add it to 136 kips. The Load on the second hinge is 164.629 kips. The additional Moment to be added to end support A will be 147.618 Ft.kips.

We will summarise the loads on the first and second hinges and the Moment Value at A. Since we have a second plastic hinge, there is no propped cantilever, but the system is considered a cantilever over the length of AB. The Moment at A is 530.118 Ft-Kips. To have a third plastic hinge, we need delta M = 107.381 Ft-Kips.

The Third plastic hinge nominal Load

We estimate the Load on delta P2 to be 107.381/20 feet, the distance between A and B. We are dealing with a cantilever; the Value equals 5.369 Kips. Add it to 164.629 Kips, and we get exactly 170 kips, the required amount for three plastic hinges.

Consider the following slide, which shows a graph of Shear and Moment.

Thanks a lot. I hope to see you in the next Post, in which we will verify our solution using MASTAN-2. Here is a Link to part-2, Post 35A.

You can view or download the PDF used for this Post from the following document.

Have more information about the structural analysis -Link to III.
For other solved problems, see solved problems 8-32 for the plastic Nominal ULoadrm load.

Here is the link to Chapter 8 – Bending Members, section, A Beginner’s Guide to the Steel Construction Manual, 14^th ed.

Here is the link to Chapter 8 – Bending Members, section, A Beginner’s Guide to the Steel Construction Manual, 15^th ed.

Here is a link to Chapter 8 – Bending Members, section, A Beginner’s Guide to the Steel Construction Manual, 16^th ed.