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 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. 

You can click on any picture to enlarge then press the small arrow at the right to review all the other images as a slide show.

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

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′, the 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.

B-Convert the Ultimate loads to nominal Loads.

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 to Fy*Zx, then it is easy to estimate the Zx value.Wn=Wult/0.90=6/0.90=6.66 kips/ft for uniformly distributed load.

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.

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.

For a useful link about the structural analysis -III, please find this link.

For the next post, 38a-solved problem 10-1-design of steel continuous beam-2-4.