Last Updated on June 18, 2024 by Maged kamel

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

In the second part of the solved problem 8-22, we will verify our nominal load values as we have estimated in post 35- part 1/2.

This is a reminder of the solved problem 8-22; we have a fixed end beam of W24x62 under nominal load. The yield stress is 50 ksi, Zx equals 153.0 inch3, so the nominal or the plastic moment will be 50×153=7650 inch. Kips, and in ft., kips will be equal to 637.50. But in MASTAN 2, the Zx equals 154.0inch3, which will lead to Mp= 7700 inch. kips and 641.66 ft.kips.

### The data for plastic hinges for solved problem 8-22 from MASTAN-2.

In the next slide image, we have a snapshot of the section of W24x62 and the relevant Zxof 154.0 inch3. The applied load is 100 kips; I have used 30 increments, each increment 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 the nominal load of 171.10 kips will create three plastic hinges.

### The location of the first plastic Hinge and relevant data.

With the help of MASTAN-2, we can find the location of 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 multiplier of 100 kips.

This is a reminder of our previous discussion about the first plastic hinge, in which we included that a nominal load of 136.0 kips will cause the first hinge to be created, and the values of shear at A and B are VA=43.032 kips and VB=92.968 kips. the relating MA=382.50 ft. kips and the Mb=Mp=637.50 ft.kips.

The next slide image contains the shear force diagram and the bending moment diagram for the first plastic hinge’s nominal load of 136.9 kips.

There is a slight difference between the VA and vb values created by MASTAN-2 and our calculation for shear because of the Zx value difference. The Mp value is 7700 inch. 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 160.72 kips that will cause the second hinge to be created.

The shear values at A and B are VA=54.47 kips and VB=106.25 kips. The corresponding MA=452.04 ft. kips and the Mb=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 multiplier of 100 kips. The shear value 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 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 VB=106.25 kips. The corresponding MA=452.04Mp=637.50 ft. kips and the Mb=Mp=637.50 ft. kips. The second plastic hinge is at C, with M= 637.50 Ft. kips.

Thanks a lot. I hope you find the post intersecting. Here’s a link to the first part of this post.

Have more information about the structural analysis –III.

For the next post, A Solved problems 8-32 for the plastic Nominal Uniform load.