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, AÂ Solved problems 8-32 for the plastic Nominal Uniform load.