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

Last Updated on February 4, 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. My aim is 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 number of inderminancyId and add one to find the number of possible plastic hinges after we ignore the horizontal reactions. As we can see, there are three possible plastic hinges, two at A and B, and one at point C.

Solve for the value of 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. We have Zx from the first slide equal to 153 in^3, and Fy = 50 ksi.

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

Find the value of Pn by 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*delta, while the internal work equals Mp*Δ.The value of θ1=Δ/20, θ2=Δ/12.
We have Mp, a positive plastic moment with an angle θ + θ1, and two plastic moments at the two 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 moments values.

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 the fixed-end beam for the working load at these points.

Consider the case of a fixed-end beam with a working concentrated load P at a distance equal to 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 have the sketch of the moments at A, B, and C. The largest value of the moment is at Point C, followed by the moment at B, and the least 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 will be equal to 3.51155P. Please refer to the next slide image for more details.

What is the load that creates the first plastic hinge?

We have found that Pn=170 kips creates three hinges, whereas a lower load creates 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 the value of P as equal to 136 Kips, we will find out that the fixed moment at A will be equal to 382.50 ft-kips, while the positive moment will be equal to 45/16 * P = 382.50 ft-kips. These moments are less than the Mp value, so there is no second plastic hinge yet at Pn = 136.0 kips. Please refer to the following slide image.

Shear and moment values for P=136 Kips.

Due to a load of P=136 Kips, there will be two end moments, one at the left support with 382.50 Ft. kips and a right moment of the value of 637.50 ft-kips. 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 is the load that creates the second plastic hinge?

The second hinge will be at point B. But the beam will be treated as a propped cantilever, since C has a plastic hinge and cannot be considered as 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=160 kips, the fixed moment at A is equal to 382.50 ft. Kips, the moment at C is unchanged, and the positive moment at C will be equal to 478.125 ft. Kips, which is less than 637.50 Ft.kips.

There is little time left 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.

If we wish 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 for propped cantilever moments and reactions to get 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. We find that 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 value of the moment at A. Since we have a second plastic hinge, there will be no propped cantilever, but the system is considered a cantilever for the length of AB. The moment at A is equal to 530.118 Ft.kips. To have a third plastic hinge, we need a value of delta M equal to 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 the 170 kips required 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. This is a link to part -2 post 35A.

The PDF used for this post can be viewed or downloaded from the following document.

Have more information about the structural analysis –III.
for another solved problems, Solved problems 8-32 for the plastic Nominal Uniform load.