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radou
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Dec17-06, 03:00 PM
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1.3. A FORMAL OUTLINE (II).

If we consider a statical system consisting of m uniform bars, every of which is straight, it can be shown that the potential energy of a single bar j equals:

[tex]U_{j} = \int_{0}^{L_{j}}\left(\frac{N^2}{2EA}+\frac{M^2}{2EI}\right)dx \ \text{.}[/tex]

(Note that the notation is inconsistent due to practical reasons - the j-th bar has a local coordinate system associated with it, so it is not correct for the variables of indegration to be dx for every bar of the system.)

The upper expression for deformational potential energy is derived by defining the work of internal forces of the system on differential elements, since that work must equal the potential energy of deformation stored in that element. The original expression consists of six terms, since the resulting force and torque can, in space, be decomposed to three components each. Since we assume all the loads act in the xz plane, we are left with the normal force N and the bending torque M only (the bars bend in the xz plane). The influence of the shear force (perpendicular to N) on the potential energy of deformation is neglible, so the expression is left with two terms only. It is also assumed that the module of elasticity E, the area of the cross section A, and the moment of inertia of the cross section I are all constant for every bar of the system.

Further on, the potential energy of the whole system consisting of m bars, is:

[tex]U=\sum_{j=1}^m \int_{0}^{L_{j}}\left(\frac{N^2}{2EA}+\frac{M^2}{2EI}\right)dx \ \ \text{(1).}[/tex]

Since the principle of superposition holds (the material behaviour is linearly elastic), the internal forces N and M can be expressed in the form:

[tex]N(x) = N^0(x)+\sum_{k=1}^n X_{k}n_{k}(x)[/tex]

[tex]M(x) = M^0(x)+\sum_{k=1}^n X_{k}m_{k}(x) \ \ \text{(2).}[/tex]

In words, the forces at every point are represented as the sum of forces due to the original load applied to the bar structure ([itex]N^0(x)[/itex] and [itex]M^0(x)[/itex]) and the forces due to the n generalized forces [itex]X_{k}[/itex] ([itex]\sum_{k=1}^n X_{k}n_{k}(x)[/itex] and [itex]\sum_{k=1}^n X_{k}m_{k}(x)[/itex]). The functions [itex]n_{k}(x)[/itex] and [itex]m_{k}(x)[/itex] represent the internal forces at some point due to a unit load applied to the structure at the point and in the direction of the generalized force [itex]X_{k}[/itex] (which arises from the principle of superposition, too).

As considered in the previous section, we first apply Castigliano's theorem, which states, once again:

[tex]\delta_{i} = \frac{\partial U}{\partial X_{i}} \ \text{.}[/tex]

After applying this to expression (1), we have:

[tex]\delta_{i} = \frac{\partial U}{\partial X_{i}} =\sum_{j=1}^m \int_{0}^{L_{j}}\left(\frac{N}{EA}\frac{\partial N}{\partial X_{i}} +\frac{M}{EI} \frac{\partial N}{\partial X_{i}} \right)dx \ \text{.}[/tex]

After plugging equations (2) into the expression above and rearranging, we obtain:

[tex]\delta_{i} = \sum_{k=1}^n X_{k} \sum_{j=1}^m\int_{0}^{L_{j}}\left( \frac{n_{k}(x)n_{i}(x)}{EA} + \frac{m_{k}(x)m_{i}(x)}{EI} \right) dx + \sum_{j=1}^m \int_{0}^{L_{j}} \left( \frac{N^0(x)n_{i}(x)}{EA}+\frac{M^0(x)m_{i}(x)}{EI} \right) dx \ \text{.}[/tex]

The expression above represents the displacement at the point i where the generalized force [itex]X_{i}[/itex] acts. The displacement can be a translation, rotation, or a relative translation or rotation (depends on how we choose the primary system). Further on, we set:

[tex]\delta_{ik}=\sum_{j=1}^m\int_{0}^{L_{j}} \left( \frac{n_{k}(x)n_{i}(x)}{EA}+\frac{m_{k}(x)m_{i}(x)}{EI} \right) dx \ \text{, and}[/tex]

[tex]\delta_{i0} = \sum_{j=1}^m \int_{0}^{L_{j}} \left( \frac{N^0(x)n_{i}(x)}{EA} + \frac{M^0(x)m_{i}(x)}{EI} \right) dx \ \ \text{(3).}[/tex]

Since the displacement at every point must vanish (see previous section), we obtain the system of equations:

[tex]\delta_{i} = 0, i = 1, \cdots, n \ \text{, i.e.}[/tex]

[tex]\sum_{k=1}^n X_{k} \delta_{ik} + \delta_{i0} = 0, i = 1, \cdots, n \ \ \text{(4),}[/tex]

which represents the system of equations of compatibility of the force method. We shall once again repeat what has already been said in the previous section; [itex]X_{k}[/itex] represents the generalized force acting at point k, [itex]\delta_{ik}[/itex] represents the generalized displacement at the point i, caused by the force k, and [itex]\delta_{i0}[/itex] represents the generalized displacement at point i, caused by the original loads applied to the system. By solving the system of equations (4) we obtain the values of the unknown forces [itex]X_{1}, \cdots, X_{n}[/itex] which represent the reaction forces in the supports we have removed (or the forces acting on the added hinges which are preventing relative displacement). After doing so, one merely has to use equations (2) to obtain the functions of the internal forces of the structure.

Since the system of equations (4) is, as stated before, equivalent to the system of equations (3') in the previous section, there is another important conclusion which arises. It is easily seen that the system of equations represents a necessary condition for the function U to have an extremum. Further on, it can be seen that this extremum is a minimum, hence we can conclude that, in a statically indeterminate system, the unknowns, i.e. the n generalized forces [itex]X_{k}[/itex] must have such a value for which the potential energy of deformation U of the system is minimized. This is referred to as the principle of minimum potential energy.

The theory behind the force method is now slightly brightened up, but every reader has the right to ask himself how to actually calculate the values of the integrals (3), which are needed to solve the system of equations (4). The answer to that question is given in the next section.