 # 38- Solved problem-10-1-design of steel continuous beam-1-4.

## Solved problem-10-1-design of steel continuous beam-1-4.

### Brief content of the video.

For the solved problem 10-1. It is required to design a steel section for a continuous beam and select the lightest w section by using the plastic analysis, the solved problem is 10-1 from Prof. Mccormack’s textbook.

The procedure of solution for the first part(1/3), is to use the upper bound theorem to derive the Plastic moment value and get the appropriate Zx, the plastic section modulus.

For the design of the steel section for the continuous beam, we will estimate the plastic moment for each of the three spans and get the largest value. The video has a subtitle and closed caption in English.

### Solving for the lightest W section for the continuous beam by the upper bound theorem.

First, this is the list of topics that will be discussed.

This is the table for the various shapes of the ASTM designation and the corresponding yield stress and ultimate stress.

### An estimate of the ultimate concentrated loads and ultimate uniform load for the first span.

In the solved problem 10-1, part 1 of 4. We will discuss the different parts, we start with the first span AB. The end span is a 30′ span with a concentrated load Pd=15 kips and P live load of 20 kips and acted upon by uniformly distributed loads over the three spans where Wd=1.0 kips/ft and Wl=3.0 kips/ft.

The first span is fixed at one end and is continuous on the other end, the second span has a distributed Wd and L. The second span length=40′, while the third span has a distributed Wd and WL. The third span length=30′, third span is continuous at one end and is hinged at the other end.

First, we write the given information as Fy=50 KSi.We will start to check the end span.

The approach of Prof. Salmon by converting the ultimate loads to plastic loads, we will start calculating the ultimate loads as follows:

#### A-Convert D and L to Ultimate loads.

By multiplying the dead load by 1.20 and the live load by 1.60 and summing these loads, so for the first span PnUl=1.2015+1.6020=50.0 kips. For the uniformly distributed loads, Wult=1.20Wd+1.60Wl=1.201+1.603=6 kips/ft, are the ultimate loads for the first.

in Solved problem-10-1, we will convert the ultimate loads to Nominal loads by dividing by 0.90, for the solved problem 10-1 part 1 of 4. or as if we multiply by 1.111. So we can proceed with the mechanism and estimate the plastic moment. For those who will review Prof. Mccormacl’s solution, they will find out that he has used the ultimate loads and has estimated the collapse load as M ultimate, but later divide by 0.90 to get the nominal loads.

I prefer the method of Prof. Salmon’s approach since it is much clearer to estimate the plastic moment and then equate it to Fy*Zx, then it is easy to estimate the Zx value. Wn=Wult/0.90=6/0.90=6.66 kips/ft for the uniformly distributed load.

The nominal loads Pn=Pult/0.90=50/09=55.55 Kips. For the first span, it can be represented by the superposition of the M0 diagram, which is simply supported plus M1 where we have end moments MA and Mb.

### Solved problem-10-1. An estimate of the Plastic moment value for the first span.

#### C-Estimate the Nominal maximum moments for the first span by the statical method.

Maximum positive bending moment =Pn*L/4+Wn*L^2/8, M0=1165.88 Ft.kips. The end span has two moments of Mp value at both ends.
These values are taken equally with the maximum value as a plastic moment. The maximum value of the positive moment, which has been estimated= 1165.88 ft. kips.

We have Mp due to the plastic hinge at the mid-span, so 2*Mp=1165.88 Ft.kips. Mp=1165.88/2=582.94=583.0 Ft.kips.

#### D-Estimate the Nominal maximum moments for the first span by the mechanism method.

We have thus obtained the Mp value from the statical method or the Lower bound method for Solved problem-10-1. We will proceed to use the other method, which is the mechanism method, to check the Mp value. We have three hinges, two at the edges, and one at the mid-span. The deflection Δ will occur at the mid-span.

The internal work=2*Mp*θ+Mp(2*θ), the slope at both side=θ=tanθ=Δ/15, 2*θ=2*Δ/15.The external work is due to nominal uniform load and nominal concentrated load, we have 6.66*(0.50*Δ)*(30)+55.55*Δ.

The external work=Δ(155.54). The external work We=Wi=Mp(Δ/15)+Mp(Δ/15)+2Mp(Δ/15))=Mp(4Δ)/15, 155.54Δ=Mp(4Δ)/15.
Mp= 15*155.54/4, Mp=583 Ft.kips.

This is the same value obtained from the statical method or the lower bound method.

This is the pdf file used in the illustration of this post. 