- Solved problem-8-34 for a nominal uniform load.

## Solved problem-8-34 for a nominal uniform load.

Another solved problem-8-34, we have a continuous beam with three spans, the first span length=24′ and the second span length=36′, while the last span length=24′, the beam has a distributed uniform loading, the section of the beam is * W27x84, Fy=50 KSI.* The plastic modulus Zx= 244 inch3.

If we consider the first span, which we want to convert to a statically determinate system, then estimate the number of plastic hinges for an unstable system. We need to have 3 Hinges. The first hinge will be at the left support.

The second hinge is at the point where there is a maximum positive moment. The last hinge will be at the right support. In our solved problem-8-34, for the first span, a question will be asked, whether it is an end span or a continuous span. For the case of the intermediate span, the parabola value is W*L^2/8 for the maximum positive moment.

We can get the plastic moment value by multiplying 50*244 and the result is 12200 inch. kips, to convert it to foot kips divide by 12. The Mp value will be equal to 1016.66 Ft. kips. This value is the value in the third span due to symmetry.

### Nominal uniform load for the first span by the lower bound method.

We can estimate the plastic moment for the first span by equating the moment from reactions and uniform load with the plastic moment at the middle of the first span. The Wn value obtained by using the lower bound method equals 28.24 kips/ft. Please refer to the next slide image for more details.

### Nominal uniform load for the second span by the lower bound method.

For the second span, we can get the value of Wn that creates three plastic hinges, two at the support and one at the mid-span, again we equate the moment from the reaction at the left support together with the moment from the uniform load to Mp.

The value of Wn is estimated as equal to 12.551 ft/kips. We will use the Upper bound theory to get the values of Wn for both the first and second spans.

#### Solved problem-8-34-Plastic nominal uniform load Wn for the first span by the upper bound theorem.

So the first span, in this case, is a continuous beam from both sides, for which if we wish to solve by the statical method,2Mp=W*L^2/8, then Mp=W*L^2/16. The same result can be obtained by using the upper bound theorem.

The place of the plastic hinge is at 0.50*L=12′.

The slope of each side due to the deflection delta Δ, θ= tan θ=Δ/12. The external work= internal work, the external work=Wn*(0.50*Δ*24)=12*Wn*Δ. We have three MP due to the three hinges. The internal work=Mp*θ+Mp*θ+Mp*(θ+θ).

The internal work=Mp*θ*412*Wn*Δ=Mp*θ*4=Mp*4*Δ/12. Δ goes with Δ.Wn=4*Mp/144.

We know the plastic moment value, we can get the Wn value as Wn=4*Mp/144.=4*1016.66/144=28.24 kips/ft.

### Wn for the second span by the upper bound theorem-solved problem-8-34.

Due to symmetry, the deflection is Δ at the mid-span.The slope at each angle=Δ/18. The external work= = internal work. W_{n}*36*(0.50*Δ)=18*Wn*Δ.

4Mp*θ, but our theta θ =tan θ=Δ/18, the internal work=4*Mp*Δ/18.

Wn=(1/18)*(4/18)*(1016.66)=12.55 kips/ft, we will select the smaller value between(28.24,12.55), which is *Wn=12.55 kips/ft.*

### Nominal uniform load for the first span using MASTAN 2.

We can use MASTAN 2, the first inelastic analysis to get the value for Wn the nominal uniform load. I have prepared a model in inches and applied a uniform load of 1.00 k/inch as a uniform load for the three spans I have added extra nodes for the places of the expected plastic hinges.

According to the next slide, there are three plastic hinges developed, the first hinge is located at the two supports with a value of 0.911 kip/ inch.

The program MASTAN 2 has indicated that at a load of 0.91099 kips/ inch, there will be a plastic hinge at the right support of the first span. I have estimated the shear force at the left and right supports as indicated in the next slide image.

In the next slide, I have included the shear diagram snapshot from MASTAN 2. The Load 0.91009 Kips/ inch cannot create a plastic hinge at the mid-span since the positive moment is less than the MP value. Please refer to the calculations attached herewith.

To develop a plastic hinge at the mid-span, increase the load to 1.0458 for all spans, this will cause the positive value to increase to be equal to 12200 inch. kip. This value matches our calculations.

This is the final moment diagram for the Wn value that creates three hinges. I hope this post presents a useful value. Thank You all.

This is a pdf file used in the illustration of this post and the previous post without the illustration of MASTAN 2 data.

Have more information about the structural analysis –III.

For the next post, 38-solved problem 10-1-Design of steel section for continuous beam part-1/3.