Last Updated on February 6, 2026 by Maged kamel
Upper bound and Lower bounds.
A new subject to be discussed is the upper and lower bounds and the uniqueness theorem.
In our previous post, we had a beam fixed at one end at support A and a roller support at point B, with a load acting at point C at mid-span. The load value is 32λ, where λ is the collapse load divided by the working load. We want to estimate λ, to get the value of Pp for a beam with a length of 1.0 m.
We can get the value of λ by using statics.
We have two hinges at the collapse, one at joint A and the other at joint C. If we consider the beam in the sketch, with a load of 32*λ. We can estimate the reactions at A and C by taking moments about A and B. By = 16λ- 9, while Ay = 16λ + 9. Please refer to the following slide.
We verify the lambda value by taking the moment about point C on the right side; we have M(x) = Mp = (16λ- 9) * 0.50 = Mp. After simplifying, we get 8λ = 13.50, which implies that λ = 1.6875.
We verify the lambda value by taking the moment about point C on the left side: M(x) = Mp = – (16λ + 9) * 0.50 + 9 = Mp. After simplifying, we get 8λ = 13.50, which implies that λ = 1.6875.
The uniqueness theorem & lower and Upper bound theorems.
There is the Uniqueness theorem, which has three requirements and serves as a guide for controlling the development of the true hinge locations and the nominal load estimate. The first point is the equilibrium, for which we get the same value of Mp at the same point from the left side or from the right side.
The second point is the mechanism by which we make the system unstable by increasing the number of indeterminacies by 1. The yield is the bending moment, which must be for Mp, the biggest value in the diagram.
It has three conditions of Equilibrium, the mechanism conditions.
In the virtual work method, we assume a point at which the external and internal work are equal. At that point, we assume it is our chosen point, but someone else may select a different point and then obtain a different value of λ for this method under the upper bound theorem.
I quote: if a bending moment diagram is found that satisfies equilibrium and the mechanism (but not necessarily yields), what is meant by “yield” is that we could obtain a value higher than the plastic moment.
The corresponding load factor is greater than or equal to the true load factor at collapse. Will a question be asked: which point is correct and gives the exact value of Mp?
This method is called the unsafe theorem because, for an arbitrarily assumed mechanism, the load factor is either exactly right or exactly wrong; hence, it is called the upper-bound method. If the collapse loads are determined for all the possible mechanisms, then the actual collapse load will be the lowest of these (upper bound theorem). For the static load method, select the point that yields the highest Mp value.
In the Lower Bound Theorem, it satisfies two conditions: the equilibrium and yield, but not necessarily the mechanism.
The collapse load factor is related to the three theorems.
A practice problem for lower-bound points on the graph.
I will introduce a practice problem for the lower-bound discussion on the graph on the last slide. If we have a propped cantilever of length 10m and the plastic moment is 266.667 kN·m, to find a point on the lower portion of the graph, we assume a nominal load of 80 kN. We use the static (equilibrium) method by superimposing the two bending-moment graphs. The MA value for the case of P=80 KN is -150 KN · m, while Pl/4=200 KN · m. We will find that both Ma and Mc are lower than the required Mp, so no collapse will occur.
We assume the nominal load is 120 kN. We use the static (equilibrium) method by superimposing the two bending-moment graphs. The MA value for the case of P=120 KN is -225 KN · m, while PL/4=300 KN · m. We will find that both Ma and Mc have a lower value than Mp; this is not the correct solution.
We assume the nominal load is 160 kN. We use the static (equilibrium) method by superimposing the two bending-moment graphs. The MA value for the case of P=160 KN is -266.67 KN · m, while PL/4=400 KN · m. We will find that both Ma and Mc have the same value as Mp; this is the correct solution.
The graph shows the relation between the lower and upper bounds.
We will proceed to the graph illustrating the differences between the lower and upper bounds when the load or moment is Pp or Mp. This horizontal line is the bound or the actual load represented by F at the collapse.
Using the kinematic method will yield values that are either equal to or greater than the actual plastic load.
We can represent the three points B, C, and D obtained from the practice problem. The bound is the actual value, the kinematic theorem is the upper bound of collapse, and the static theorem gives a lower bound at points B and C, or matches the graph at point D.
Thanks a lot, I will attach the PDf document for the content of this post.
For useful information on structural analysis III, most of the data used in this post is quoted from the paper on structural analysis III.
The next post continues the introduction to the lower bound and uniqueness.