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radou
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Dec18-06, 01:44 PM
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An introduction to the analysis of statically indeterminate systems

1.3. A FEW COMMENTS.

As mentioned, calculation of the intergals (3) from the previous section is the most important operative tool which one has to develop in order to use the force method.

A formal proof shall not be presented here, but it can be seen that the summands in the expressions (3) are of the form

[tex]k \int_{0}^{L_{j}} f(x)\cdot g(x) dx \ \text{,}[/tex]

where k stands for [itex]\frac{1}{EI}[/itex] or [itex]\frac{1}{EA}[/itex], and it is assumed that the module of elasticy E, the moment of inertia of the cross section I, and the area of the cross section A are all constant. Further on, since at least one of the functions f or g presents the internal force (bending moment or normal force) caused by a generalized unit force acting at a point where a support has been removed (see previous section), at least one of the functions shall be a linear function (i.e. the internal force diagram shall be linear). This is the fact from which Vereshchagin derived the method of integration of the expressions (3) - one merely has to find the area of the diagram (between 0 and Lj, of course) which is the diagram of the internal force caused by the original loads acting on the primary system (not the diagram of the generalized unit loads), and multiply it (that area) with the ordinate of the unit load diagram at the point of the centroid of area of the non unit load diagram. This will be shown in an example in the next section, and the reader shouldn't be worried if the description of the method of integration was confusing.

Another useful fact considering the equations (3) is that [itex]\delta_{ik} = \delta_{ki}[/itex], which follows directly from Maxwell's theorem, which states that the displacement at the point and in the direction of the first unit load caused by the second unit load equals the displacement at the point and in the direction of the second unit load, caused by the first unit load. This theorem holds for a linearly elastic body.

After determining the values of (3), i.e. of [itex]\delta_{ik}[/itex] and [itex]\delta_{i0}[/itex], we plug them back into the system of equations of compatibility (4). The system can be represented in matrix form:

d X + do = 0.

(Note that the matrix d is symmetric.)
Solving the system is a standard linear algebra procedure, and it is left to the reader which method to choose.

As mentioned before, after solving the system, we can, using the equations of equilibrium, find all the reactions in the system - we merely have to solve a statically determinate system with the original loads acting on it, along with the loads [itex]X_{1}, \cdots, X_{n}[/itex]. Further on, by using equations (2) from the previous section, we can find the values of the internal forces N and M at any point of the system, and sketch normal force and bending moment diagrams, which is one of the standard assignments in structural statics.

In the next (last) section the reader will be provided with an example of a simple statically indeterminate system treated with the force method.