Solved problem-8-34-for a nominal uniform load.
Another solved problem-8-34, we have a continuous beam with three spans, the first span length=24′ and the second span length=36′, while the last span length=24′, the beam has a distributed uniform loading, the section of the beam is W27x84, Fy=50 KSI.
If we consider the first span, which we want to convert to a statically determinate system, then estimate the number of plastic hinges for an unstable system. We need to have 3 Hinges. The first hinge will be at the left support.
The second hinge is at the point where there is a maximum positive moment. The last hinge will be at the right support.
In our solved problem-8-34, for the first span, a question will be asked, whether it is an end span or a continuous span?
For the case of the intermediate span, the parabola value is W*L^2/8 for the maximum positive moment.
We have two negative Plastic moment ordinates, from which we drop the value of W*L^2/8. We look for the maximum value for Mp value at the hatched region.
Due to symmetry, the location of the maximum positive value will be in the middle.If we consider the first span, which we want to convert to a statically determinate system, then estimate the number of plastic hinges for an unstable system. We need to have 3 Hinges. The first hinge will be at the left support.
The second hinge is at the point where there is a maximum positive moment.
The last hinge will be at the right support. In our example, for the first span, a question will be asked, whether it is an end span or a continuous span.
For the case of the intermediate span, the parabola value is W*L^2/8 for the maximum positive moment.
We have two negative Plastic moment ordinates, from which we drop the value of W*L^2/8.
We look for the maximum value for Mp value at the hatched region.
Due to symmetry, the location of the maximum positive value will be in the middle.
If we consider the case of the end span, where the left span has a zero value for moment Mp, the place of the Maximum moment is located at a distance <0.50*L.
Solved problem-8-34-Plastic nominal uniform load Wn for the first span by the upper bound theorem.
So the first span, in this case, is a continuous beam from both sides, for which if we wish to solve by the statical method,2Mp=W*L^2/8, then Mp=W*L^2/16. The same result can be obtained by using the upper bound theorem.
The place of the plastic hinge is at 0.50*L=12′.
The slope of each side due to the deflection delta Δ, θ= tan θ=Δ/12. The external work= internal work, the external work=Wn*(0.50*Δ*24)=12*Wn*Δ. We have three Mp due to the three hinges. The internal work=Mp*θ+Mp*θ+Mp*(θ+θ).
The internal work=Mp*θ*412*Wn*Δ=Mp*θ*4=Mp*4*Δ/12. Δ goes with Δ.Wn=4*Mp/144.
In the solved problem-8-34 the first span the Mn=Mp=FyZx, we have fy=50 ksi, Zx=244 inch3, Mp=50244/12=1016.66 Ft.kips. We will substitute, to get the Wn value.
Wn=4Mp/144.=41016.66/144=28.24 kips/ft.
Wn for the second span by the upper bound theorem-solved problem-8-34.
Due to symmetry, the deflection is Δ at the mid-span.The slope at each angle=Δ/18. The external work= internal work. Wn*36*(0.50*Δ)=18*Wn*Δ.
4Mp*θ, but our theta θ =tan θ=Δ/18, the internal work=4*Mp*Δ/18.
That was the value for Wn, nominal uniform load, for the first span, while for the second span, we have the maximum positive moment =W*L^2/8, the span length=36′. The number of plastic hinges= 3 hinges.
In some textbooks, the system is returned to a simple beam, then the number of redundant is estimated.
We have two negative moments, so we need two hinges to cancel the moment effect and return to a stable statically determinate. For an unstable system, an additional hinge is to be placed in the middle span.
Wn=(1/18)*(4/18)*(1016.66)=12.55 kips/ft, we will select the smaller value between(28.24,12.55), which is Wn=12.55 kips/ft.
This is a pdf file used in the illustration of this post and the previous post.
Have more information about the structural analysis –III.
For the next post, 38-solved problem 10-1-Design of steel section for continuous beam part-1/3.
