 # 37-Solved problem 8-33 for Plastic nominal Uniform Load .

## Solved problem 8-33 for a plastic nominal uniform load.

### Brief content of the video.

It is required to evaluate the Plastic nominal uniform load for continuous beams with different uniform load values and end conditions, two problems will be solved, the first problem is 8-33 and the second problem is 8-34.
For the first problem 8-33.

We have continuous beams with three spans, for which span length is 16′ for each. The first span carries a uniform distributed load of 2*Wn, while the second span’s uniform load is Wn.
For the last span, it carries a uniform distributed load of 2*Wn. The section of the beams is W16x26. It is required to get the value of Wn. The video has a subtitle and closed caption in English.

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### Solved problem 8-33 estimate the nominal load Wn for the end span.

In solved problem 8-33. The given section of the beams is W16x26. It is required to get the value of Wn, the plastic nominal uniform load,  from table 1-1, we get the value of the plastic section modulus Zx, which is 44.20 inch3.

The given yield stress is 50 ksi for steel. The nominal, or we can estimate the plastic moment value Mn as equal to Mp=Fy*Zx= =50*44.20/12=154.166 Ft.kips.

The solution for the first span nominal Wn, to be done by using the upper bound theorem. We have proved, earlier, that for the end span, the Plastic hinge is present at a distance =0.414 *L from the left hinge, where L is the span length.

The proof of the distance value for the plastic hinge is shown in a previous video by using both the lower bound and upper bound theorem. We have a span distance L=16′, then the x distance to the plastic hinge=0.414*(16)=6.624′, due to the mechanism of failure, a deflection will occur by a value=Δ.

For the first span, the distance to the maximum deflection of 6.624′ from the left support.
The remaining distance is (16-6.624)= 9.376′ to the right support, we two angles that are θ1 and θ2. θ1= tan θ1=Δ/6.624, for θ2= tan θ2=Δ/9.376. We add θ1+θ2, we will have (A)*((1/6.624)+(1/9.376))=0.25762*Δ.

The external work=Load*span*average of the deflection value, We=2*Wn*16*(Δ/2)=16*Wn, that is the external work value.
The internal work, for the same span, Mp*(θ1+θ2) due to the plastic hinge at 0.414L.

There will be a negative moment due to continuity at the right support plastic hinge will be created, for that joint, the internal work=Mp*θ2.

The overall internal work for the first span will be =Mp*(θ1+θ2)+Mp*θ2=Mp*(θ1+θ2)+Mp*θ2=Mp(0.25762*Δ)+Mp*θ2,
for the second term, it is equal to Mp*(Δ/9.376), which is  Mp*0.1076 *Δ.
The total internal work=Mp*Δ((0.2576)+0.1076 )) =Mp*0.3643*Δ.

We will equate to the external work, which is =(32/2)*(Wn)*Δ, Δ delta goes with delta Δ. Wn*16= Mp*0.1067.
We have estimated Mp as 184.1666 ft. kips,

We can get the value of the plastic nominal uniform load Wn from the relation, Wn=0.02277*Mp, this is not the final value, since we will estimate from the second span, another value of Wn, we will select the lesser value of Wn.
Wn=0.02277*Mp.

This is the value of Wn as estimated from the first span plastic joint.

### Solved problem 8-33 estimate the plastic nominal load Wn for the 2nd span.

For the second span, it can be considered as a continuous beam from both sides, we have 2*Mp at the two edges and one Mp at the mid-span due to symmetry. We will check the internal work. The second span =16′. We have three Mp acting as shown in the sketch, due to deflection, we will have an angle θ at both sides. For θ =tan θ =Δ/8.The value of 2*Δ=Δ/4.

The external work is =Load* span * average deflection=Wn*16*Δ*0.50. The internal work= Mp(θ) +Mp(θ)+Mp(2*θ).

The total value of the internal work =4*Mp*θ. The external work=internal work, then Wn*16*Δ*0.50=4*Mp*θ, θ=Δ/8. Wn*16*Δ*0.50=4*Mp*(Δ/8). delta Δ goes with delta Δ.

Wn=Mp*(1/16). The nominal Wn as estimated from the second span=Mp/16=0.0625 *MP. Mp is the M-nominal of the given section W16*26.
We have two values for Wn, the first one for Wn=0.02277*Mp, while the second value for Wn=0.0625*Mp.

We cannot select the bigger value for Wn, since Mp is reached under a lighter value of load Wn, the lesser value for the plastic nominal uniform load Wn will be selected, it will as Wn=0.02277*Mp, Since Mn=Mp=184.166 Ft.kips, then Wn=0.02277*184.166, Wn=4.20 Kips/ft.

That is the value for the nominal uniform load for the three spans. But for the first span and the third span will carry a nominal uniform load=2*4.20=8.40 Kips/Ft.

This is the pdf file used in the illustration of this post and the next post.