Last Updated on September 15, 2026 by Maged kamel
Solved problem 10-1: Design -1: continuous steel Beam (2-4).
An estimated elastic Moment for the second span.
This is the second part of the solved problem 10-1. In the previous part, we considered the first span of the continuous Beam and estimated the plastic Moment. For more details, refer to part 1.
Estimate the nominal maximum moments for the second span using the mechanism method.
The next step in designing a steel continuous Beam 2-4 is to estimate Mp for the next span by superimposing a simple Beam and a Beam with two moments at each end.
We have three plastic hinges to create a collapse mechanism.
For the static method, we equate 2Mp to wn*L^2/8; Wn = 6.66 kips/ft and L = 30 ft. The Mp Value is 666 ft-kips.
For the upper bound, for the design of steel continuous Beam-2-4, the external work We=Wn*0.50*Δ*span, the span = 40 ‘.We=6.66*0.50*Δ*40=133.20*Δ.
We have Mp at the edges and Mp at the mid-span—the internal work Wi=Mp*α+Mp*α+Mp*2*α, α= tan α=Δ/20.
Wi=Mp*(Δ/20)+Mp*(Δ/20)+2*Mp*(Δ/20). Wi=Mp*4*(Δ/20)=Mp*Δ/5.
We=133.20*Δ=We=Mp*Δ/5.Mp=5*133.20=666 Ft.kips. This is the same plastic moment for the second span as obtained from static or lower-bound methods.
Estimate the nominal maximum moments for the third span using the statical method.
We will proceed to the third span to determine Mp.
In designing steel continuous Beam-2-4, to create a mechanism, we consider two hinges to cause collapse: one at the edge and one at the point of maximum Moment or zero Shear.
How do we derive the Expression for the point of the maximum positive Moment for the third span?
Check the Beam CD for a span of 30 Ft. It has Mp at support C, and a 6.66 kip/Ft Load acts on the span. The reaction at D=99.9-Mp/30. Suppose the point of zero Shear is at x distance from the right support d.
Consider the sum of vertical forces =0; we have 99.9-mp/30-6.66x=0. There is another sketch for the CD part; at the plastic hinge, we can create a relation between Mp and x by taking a Moment at D, 6.66*x^2/2=Mp; substitute into equation I; after simplifying, we get another equation for x^2+60x-900=0.
Solve for x; we get x = 12.426 ft and Mp = 3.33*x^2 = 514.204 ft-kips. We get the same Mp value from the BM sketch, where 1.414 Mp equals 749.25 Ft-Kips.
We will continue on the next slide. For the upper bound, the point of maximum Deflection is at x=12.42′ from the right support.
The remaining distance=30-12.42=17.58′. We have θ1 and θ2, the values are for θ1=tan θ1=Δ/17.58, θ2=tan θ2=Δ/12.42.
The external work We=Wn*0.5*Δ*span= 6.66*0.5*Δ*30=99.90*Δ. The internal work Wi = Mp*θ1 + Mp*(θ1 + θ2). Wi= Mp*(2*θ1+θ2)=Mp*(42.42*Δ/218.34).
Mp = 514.203 ft-kips, the same Value as the static method.
The final step for solving problem 10-1 is to design a continuous steel Beam, 2-4.
The next slide contains a sketch showing the different plastic Moment values for each span, Mp. The design of the steel continuous Beam (2-4) is the second of four parts.
For the first span, Mp = 583.0 ft-kips, while Mp for the second span = 666 ft-kips, and for the last span, Mp = 514.28 ft-kips.
We will select the largest design Value, Mp = 666 ft-kips.
The corresponding Mult = 600.0 ft-kips, which is 0.9*666 = 600 ft-kips. The required Zx is Mp/Fy = 666*12/50 = 159.84 in³ (inch-kips/kip/inch^2).
From Table 3-2, sorted by Zx, the lightest W section is W21x68, with Zx = 160.0 in³. the lRfd Value φbMp=0.9050*160/12=600.0 ft.kips=Mult. Our design section is W21x68, based on Lr = 0.0.
You can review or download the PDF file for this Post from the following document.
Provide more information about the structural analysis – Link to III.
The next Post will be “Moment values for continuous Beam by three-moment equations” 3/4.
Here is the Link to Chapter 8 – Bending Members, section, A Beginner’s Guide to the Steel Construction Manual, 14th ed.
Here is the Link to Chapter 8 – Bending Members, section, A Beginner’s Guide to the Steel Construction Manual, 15th ed.
Here is the Link to Chapter 8 – Bending Members, section, A Beginner’s Guide to the Steel Construction Manual, 16th ed.