# 38a-Solved problem 10-1-design of steel continuous beam-2-4.

Last Updated on June 7, 2024 by Maged kamel

## 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 considered the first span of the continuous beam, for which we estimated the plastic moment value. 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 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.

For the static method, we equate 2Mp with wn*L^2/8, Wn is 6.66 kips/ft, and L=30 Ft. Mp value is found to be 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 value for the plastic moment for the second span as obtained from the statical 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 get the value of Mp.
In the design of steel continuous beam-2-4, To create a mechanism, we consider the presence of two hinges to create collapse: one hinge at the edge and one hinge 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, the 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 in equation I, after simplifying, we get another equation for x^2+60x-900=0.

Solve for the value of X; we can get the x value equal to 12.426 ft and the Mp value equal to 3.33*x^2=514.204 Ft.kips. We could get the same value for Mp, when considering the sketch of Bm 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, Which is the same value as the statical 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 values of the plastic moment for each span for Mp. The design of the steel continuous beam-2-4 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 design value, Mp=666 ft. kips.

The corresponding Mult=600.0 ft. kips, which is=0.9*666=600 ft. kips. The Mp=Fy*Zx, the required 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. the lRfd value φbMp=0.9050*160/12=600.0 ft.kips=Mult. Our section for the design is W21x68, based on Lr=0.