Last Updated on February 12, 2026 by Maged kamel
Solved problem 8-22-for Plastic nominal load( 2/2).
In the second part of problem 8-22, we will verify our nominal load values as estimated in post 35, part 1/2.
This is a reminder about problem 8-22: we have a fixed-end beam of W24x62 under nominal load. The yield stress is 50 ksi, and Zx equals 153.0 in^3, so the nominal or plastic moment will be 50 x 153 = 7,650 in^3. Kips and ft. will be equal to 637.50. But in MASTAN 2, the Zx equals 154.0 in^3, which leads to Mp = 7,700 in. kips and 641.66 ft. kips.
The data for plastic hinges for solved problem 8-22 from MASTAN-2.
On the next slide, we have a snapshot of the W24x62 section and the relevant Zxof of 154.0 in^3. The applied load is 100 kips. I used 30 increments of 0.10, and the maximum load ratio is 3.
Three plastic hinges were created, the first of which has a lambda value of 1.369. The second plastic hinge has a lambda value of 1.657, and the third has a lambda value of 1.712.
Using the first-order inelastic analysis, a nominal load of (1.369*100)=136.90 kips will create the first plastic hinge. At Pn = 165.70 kips, we have two plastic hinges. Applying a nominal load of 171.10 kips will result in three plastic hinges.
The location of the first plastic Hinge and relevant data.
Using MASTAN-2, we can locate the first hinge. It will be located at support B under a load of 1.369*100 = 136.90 kips. This is the 14th increment, from 1.3 to 1.40, at a multiplier of 100 kips.
This is a reminder of our previous discussion about the first plastic hinge, in which we noted that a nominal load of 136.0 kips will cause the first hinge to form, and the shear values at A and C are VA = 43.031 kips and VC = 92.969 kips. the relating MA=382.50 ft. kips and the MC=Mp=637.50 ft.kips.
The next slide image shows the shear force and bending moment diagrams for the first plastic hinge at a nominal load of 136.9 kips.
There is a slight difference between the VA and VC values produced by MASTAN-2 and our shear calculation, due to a difference in the Zx value. The Mp value is 7700 inches. kips, while our estimation is Mp=7650 inch. kips.
The second plastic Hinge and relevant data.
This is a reminder of our previous discussion about the second plastic hinge, in which we included a nominal load of 164.629 Kips to initiate its formation.
The corresponding MA=452.04 ft. kips and the MC=Mp=637.50 ft. kips. The second plastic hinge is at C, with a value of 637.50 ft-kips.
The second hinge is created under a load of 1.657* 100 = 165.70 kips. This is the 17th increment, from 1.6 to 1.70, at a multiplier of 100 kips. The shear values are vA = 58.76 kips and Vb = 107.0 kips.
This is a snapshot of the moment diagram for the 2nd plastic hinge moments. The values are in inches. Kips.
The third plastic Hinge and relevant data.
The second hinge is created under a load of 1.7118* 100 = 171.18 kips. This is the 18th increment, from 1.7 to 1.80, at a multiplier of 100 kips. The shear value vA=64.18 kips and Vb=107.0 kips.
This is a reminder of our previous discussion about the third plastic hinge, in which we included a nominal load of 170 kips that will cause the third hinge to be created.
The shear values at A and B are VA = 63.75 kips and VC = 106.25 kips. The corresponding MA=452.04 Mp=637.50 ft. kips and the MC=Mp=637.50 ft. kips. The second plastic hinge is at B, with a moment of 637.50 ft-kips.
Thanks a lot. I hope you find the post interesting. Here’s a link to the first part of this post.
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Have more information about the structural analysis –III.
For the next post, A Solved problems 8-32 for the plastic Nominal Uniform load.