38- Solved Problem 10-1- Design of Steel Continuous Beam-1-4.

Last Updated on September 15, 2026 by Maged kamel

Solved Problem 10-1: Design of a steel continuous Beam 1-4.

 In solved problem 10-1, design of a steel continuous Beam 1-4. It requires designing the steel section part A using plastic analysis, for which the compression Flange has full lateral support.

A-Estimate ultimate concentrated and Uniform loads.

We will estimate the ultimate concentrated and Uniform loads, then 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 30′ with a concentrated Load of Pd = 15 kips and a live Load of 20 kips, and it is 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 is 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 by checking the end span.

By multiplying the dead Load by 1.20 and the live Load by 1.60 and summing these loads, for the first span, PnUl = 1.2*0.15 + 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-, we convert the ultimate loads to Nominal loads by dividing by 0.90 for solved problem 10-1, part 1 of 4, or by multiplying 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 (supported) and M1, with end moments MA and Mb. To form a mechanism, we need three plastic hinges for collapse.

C-Estimate the plastic Moment using the upper bound.

We will use the upper-bound method to evaluate the Mp Value. We have three hinges, two at the edges and one at the mid-span. The Deflection Δ occurs at mid-span. The external work is due to the nominal Uniform Load and the nominal concentrated Load, which can be estimated by multiplying Pn by delta and the nominal Uniform Load by span*delta/2.

So we can express the external work as 6.66*(0.50*δ)*(30) + 55.55*δ = 155.45δ. The internal work is due to two end moments with Mp Value for each and Mp at the midspan.

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 work 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 the lower bound to get the Mp Value.

Using a static method to verify our estimate of the MP Value. Maximum positive bending Moment = Pn*L/4 + Wn*L^2/8, M0 = 1165.88 ft-kips. The end span has two Mp values at both ends and a positive Mp value.

These values are taken equally, with the maximum value being a plastic moment. The estimated maximum positive Moment 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 of the Upper and lower methods.

This is the same Value obtained from the upper-bound method.

You can review or download the PDF file for this Post from the following document.

For a useful Link on Structural Analysis III, please find it here.

For the next Post, 38a: solved problem 10-1-design of a steel continuous Beam-2-4.

Here is the Link to Chapter 8 – Bending Members, section, A Beginner’s Guide to the Steel Construction Manual, 14^th ed.

Here is the Link to Chapter 8 – Bending Members, section, A Beginner’s Guide to the Steel Construction Manual, 15^th ed.

Here is the Link to Chapter 8 – Bending Members, section, A Beginner’s Guide to the Steel Construction Manual, 16^th ed.