Brief description -post 38 -steel beam

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

Spread the love

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

 In the A Solved problem-10-1-design of steel continuous beam-1-4. It required the design of a steel section part -A using the plastic analysis for which the compression flange has a full lateral support.

A-Estimate ultimate concentrated and uniform loads.

We will estimate the ultimate concentrated and uniform loads, and then we will convert them to nominal loads to proceed with the plastic analysis.

In the solved problem 10-1, part 1 of 4. We start with the first span AB. The end span is a 30′ span with a concentrated load of 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 continuous on the other; the second span has a distributed Wd and L. The second span length is 40′, while the third span has a distributed Wd and WL. The third span length=30′, and the third span is continuous at one end and hinged at the other.

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

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

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=1.2*015+1.60*20=50.0 kips. For the uniformly distributed loads, Wult=1.20Wd+1.60Wl=1.20*1+1.60*3=6 kips/ft, are the ultimate loads for the first.

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.

The nominal loads Pn=Pult/0.90=50/09=55.55 Kips. The first span can be represented by the superposition of the M0 diagram, which is simply supported plus M1, where we have end moments MA and Mb. to create a mechanism we need to have three plastic hinges for colapse.

C-Estimate the plastic moment using the upper bound.

We will use upper bound method to evaluate Mpvalue. We have three hinges, two at the edges and one at the mid-span. The deflection Δ will occur at the mid-span. The external work is due to nominal uniform load and nominal concentrated load, can be estimated by multiplying Pn by delta and Nominal uniform load by span*delta/2.

So we can express the external work as equal to 6.66*(0.50*Δ)*(30)+55.55*Δ equals 155.45 Δ . The internal work is due to two end moments with Mp value for each and Mp 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=Δ(155.54). The external We equals the internal work Wi. 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.

D-Use lower bound to get Mp value.

Using static method to verify our estimate Mp value. 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 and a positive value of Mp.
These values are taken equally, with the maximum value being a plastic moment. The maximum value of the positive moment, which has been estimated, is 1165.88 ft. kips.

Equate 2*Mp to 1165.88 Ft.kips. the plastic moment value Mp=1165.88/2=582.94=583.0 Ft.kips. please refer to the next slide image for the expansion for the Upper and lower methods

Plastic moment for the first span

This is the same value obtained from the upper bound method.

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

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

Scroll to Top