Lower-Bound:
The area immediately under the notch, above the neutral axis is in tension σ = 2k. The area below the neutral axis is in compression σ = 2k.
where:
h = thickness of slab beneath the notch.
= magnitude of forces in tensile and compressive regions.
= distance between the two.
Equating the couples, 
Upper-Bound:
Assume failure occurs by sliding around a ‘plastic hinge’ along a circular arc of length l and radius r.
If the rotation is δθ, the internal work done 
along one arc.
External work = Mδθ by one moment.
where no assumptions have been made regarding l and r.
The upper bound theorem states that whatever values are taken for l and r will lead to an upper bound. Clearly we wish to find the lowest possible value.
From the above geometry,
and 
and so to find the lowest possible value of M, we minimise the function 
Let 
= 0 when 
Taking the lower bound and the upper bound as limits, we therefore find
This forms a good example of constraining the value of the external force between lower bound and upper bound. It is also a good example of how to produce a lower limit on an upper bound calculation.
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