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## Solved problem 10-1 design of steel continuous beam-2-4.

### An estimate of the Plastic moment value for the second span.

This is the second part of the solved problem 10-1, in the previous part we have considered the first span of the continuous beam, for which we have estimated the plastic moment value. For more details refer to part-1.

#### E-Estimate the Nominal maximum moments for the second span by the mechanism method.

The next step for the design of steel continuous beam-2-4 is to estimate the Mp for the next span, which is a superposition of a simple beam plus a beam with two moments at each end.

We have three plastic hinges to create a mechanism of collapse.The maximum positive moment= Wn*L^2/8=6.66*(40)^2/8=1332.0 ft.kips.

We have 2Mp at the edges, and Mp at the mid-span, the sum of Mp+Mp=1332.0 ft. kips.

Mp=1332/2=666.0 ft. kips, this is the Mp value from the lower bound method.

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*θ1+Mp*θ1+Mp*2*θ1, θ1= tan θ1=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 value for the plastic moment for the second span as obtained from the statical method or the lower bound method.

### An estimate of the Plastic moment value for the third span.

#### F-Estimate the Nominal maximum moments for the third span by the statical method.

We will proceed to the third span, to get the value of Mp.

The third span can be considered as the superposition of a simple support beam plus one beam acted upon by M negative at one edge.

In the design of steel continuous beam-2-4 To create a mechanism, we consider the presence of three hinges to create collapse, two hinges are in the edge and, one hinge in the mid-span. We have proved earlier that for an end span, the point of maximum positive moment is located at a point, whose distance from the non-continuous joint=0.414*L, since the span length=30′, X=0.414*30=12.42′.

This third span has one Mp at one edge, that is why we used the 0.414*L, unlike, the first span, where 2 Mp exists at both ends.

For the maximum positive moment, Mmax=Wn*L^2/8=6.66*30^2/8=749.25 ft. kips.

The bending moments are drawn as shown in the sketch.

The place of the plastic joint should fulfill **that the Mp value is the maximum value.**If somebody considers the plastic joint is located at the mid-span, the Mp value can be estimated as Mp=749.25-0.50*Mp=Mp, Mp=(1/1.50)*(749.25)=4995.50 Ft.kips.Now consider that the plastic joint is at 0.414*L=0.414*30=12.42′.

First, we estimate the reaction as =6.66*15=99.90 kips.The moment at 0.414L =moment from the reaction- moment from the nominal uniform load M=99.9*(12.42)-6.66*(12.42)^2/2=727.08 Ft.kips.

For the Mp+(0.414*Mp)=727.08, Mp*(1.414)=727.08. Mp=727.08/1.414=Mp=514.20 ft.kips, which is >499.50 ft.kips.

As estimated as if the plastic joint is considered at mid-span.

### How to derive the expression for the point of the maximum positive moment for the third span?

Here is how to get the exact distance for the point of the plastic joint where maximum moment by assuming that the joint is apart by x distance from the free hinge.

We get the value of Mp as a function of x, then we can differentiate with respect to dx, and get the x value. **X=0.414*L is obtained.**

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 θ2 and θ3, the values are for θ2=tan θ2=Δ/17.58, θ3=tan θ3=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*θ2+Mp*(θ2+θ3).Wi= Mp*(2*θ2+θ3)=Mp*(2*(Δ/17.58)+(Δ/12.42)). Wi=Mp*(42.42*Δ/(17.58*12.42)).

We have 99.90*Δ=Mp*(42.42*Δ/(17.58*12.42)), Mp =514.203 ft. kips, Which is the same value from the statical method which is Mp=514.203 Ft.kips.

#### The final step for the Solved problem 10-1 design of steel continuous beam-2-4.

In the next slide, there is a sketch that shows the different values for each span for Mp. Design of steel continuous beam-2-4, which is the second part 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 biggest value for design which is Mp=666 ft. kips.

The corresponding Mult=600.0 ft.kips, which is=0.9*666=600 ft.kips. The Mp=Fy*Zx, Zx=Mp/Fy=666*12/50, (inch kips/kip/inch^2)=159.84 inch3.

From table 3-2 sorted by Zx, the lightest weight for the W section is W21x68, which has Zx=160.0 inch3.φb*Mp=0.90*50*160/12=600.0 ft.kips=Mult. Our section for the design is W21x68, the design is based on Lr=0.

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

Have more information about the structural analysis –III.

The next post will be –Moment values for continuous beam by three-moment equations-3/4.