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An introduction to the analysis of statically indeterminate systems

  1. Dec 16, 2006 #1

    radou

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    It is assumed that the reader is familiar with the basic concepts of structural analysis. Further on, I am aware of the possible inconsistencies and errors in this writing, and hereby invite all who have comments and objections to contribute and help to make this small tutorial useful for anybody interested in the analysis of statically indeterminate systems.


    1. THE FORCE METHOD.

    1.1. INTRODUCTION THROUGH AN EXAMPLE.

    Perhaps the most standard example of the force method application is the system in Figure 1. It is obvious that the system's degree of statical indeterminacy equals one, since the number of constraints equals 4, and the number of equations of equilibrium we can use equals 3. Note that, if we removed the roller support at point B, we would be left with a statically determinate console with a support at point A, which (the support) represents three kinematic constraints. Hence, we can refer to the support at point B as a statical constraint, since the system, after having this support removed, still remains geometrically stable (i.e. it does NOT become a mechanism).

    [​IMG]
    Figure 1.​

    The geometrical and material characteristics are all displayed in Figure 1. The module of elasticity equals E, and it is constant along the whole length L, as is the moment of inertia of the cross section, denoted by I (with respect to the y-axis, which is pointing 'towards us').

    Let us remove the roller support at point B, and 'replace it' with its reactive force [itex]R_{B}[/itex]. We shall explicitely use the equations of equilibrium for the z-direction (1), and we shall set the sum of the torques with respect to point A to equal zero (2):

    (1)
    [tex]\sum F_{z} = 0 \rightarrow R_{A} + R_{B} = qL \ \text{,}[/tex]
    (2)
    [tex]\sum M_{A} = 0 \rightarrow M_{A} + R_{B}L = \frac{qL^2}{2} \ \text{,}[/tex]

    where [itex]M_{A}[/itex] is the reaction torque at point A, [itex]R_{A}[/itex] the reaction force in the z-direction at the same point, and [itex]R_{B}[/itex] the reaction force in the same direction at point B. (It is assumed that the positive direction of the torques is counter-clockwise, and the positive direction of the vertical forces is in the positive direction of the z-axis).

    It is obvious that the system of two equations (1) and (2), which represents the conditions of equilibrium of the system, has no unique solution, since we are left with two equations and three unknowns, [itex]M_{A}[/itex], [itex]R_{A}[/itex] and [itex]R_{B}[/itex]. Further on, we can write the system of equations in the form:

    (1')
    [tex]R_{A} = qL - R_{B}[/tex]
    (2')
    [tex]M_{A} = \frac{qL^2}{2} - R_{B}L \ \text{.}[/tex]

    Now we can see the the equations of equilibrium will be satisfied for any value of the reaction [itex]R_{B}[/itex], i.e. the system has a one-parameter solution, where the parameter is [itex]R_{B}[/itex]. This means that there is an infinite number of equilibrium states of the system. But, the one we are trying to find is the one which satisfies the displacement conditions of the system - i.e. the vertical displacement at point B equals zero, since, in the original system there exists a support at point B which prevents a vertical displacement at that point.

    The vertical displacement of the point B has two contributions: one comes from the load q, and the other from the reactive force [itex]R_{B}[/itex] acting at that point. So, it can be expressed as the superposition of these two contributions, respectively:

    [tex]w_{B} = \frac{qL^4}{8EI} - R_{B}\frac{L^3}{3EI} \ \text{.}[/tex]

    (Note that the displacement of the reactive force [itex]R_{B}[/itex] has a negative sign, since the force is assumed to point in the negative direction of the z-axis.) As mentioned, the displacement must equal zero, which represents the additional condition we need to describe the desired 'real' state of equilibrium of the system, so [itex]w_{B} = 0[/itex] (3) implies:

    [tex]R_{B} = \frac{3}{8}qL \ \text{.}[/tex]

    Now we can finally solve equations (1') and (2'), i.e. calculate the reactive forces [itex]R_{A}[/itex] and [itex]M_{A}[/itex], and, of course, solve standard statics problems as before - find bending moment and shear force diagrams, displacements, etc.

    The provided example was supposed to serve as a brief introduction to the force method, whose summary can easily be distinguished from the very same example. First, a support, which represents a statical constraint is removed and replaced with a reactive force in the 'position and direction' of the support. Further on, the equations of equilibrium are set up. Finally,a kinematic condition (a condition of displacement, i.e. a geometric condition) is set up to determine which state of equilibrium is the real one. As a final illustraion, one can find deflection curves for some values of [itex]R_{B}[/itex] which do not satisfy the condition (3) in Figure 2, while Figure 3 represents the actual deflection curve of the bar, which is in consistence with the condition (3). (The console is placed on the horizontal axis, and the diagrams are plotted for L = 5. The vertical axis represents the vertical displacement w.)

    [​IMG]
    Figure 2.​

    In the end, it is of great importance to mention that this rough illustration of the force method concept applies to systems with more than one degree of statical indeterminacy - in general, for a statically indeterminate system with the degree of statical indeterminacy n, one merely has to pick n different supports (i.e. statical constraints) and replace them with forces acting in the points and the directions of the supports. Of course, this system, which is referred to as the primary system, is not allowed to be a mechanism, which means we can not remove just any support as we wish. In our example, we could as well add a hinge to point A, which physically means we have removed the part of the rigid support which transfers torques. After that, we would have to add a torque M to that point, and that torque should satisfy condition equivalent to (3), which would state that the rotation at point A should equal zero. In general, when dealing with statically indeterminate systems with a degree of statical indeterminacy equal to n, we have n displacement conditions (in our case we have only one - equation (3) ).

    [​IMG]
    Figure 3.​

    A more detailed and formal outline of the force method shall be introduced in section 1.2.
     
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  3. Dec 17, 2006 #2

    radou

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    1.2. A FORMAL OUTLINE (I).

    Under the assumption that the idea of the force method was successfully illustrated in Section 1.1., we shall now try to generalize and formalize that idea.

    It first has to be mentioned that the statical constraints that are to be removed from the system do not have to be external constraints, such as supports - one can add internal hinges or roller connections to the system too, which also minimizes the degree of statical indeterminacy. This is shown in Figure 4. However, one has to add a pair of genenralized forces (forces or moments) which are to prevent a relative generalized displacement (translation or rotation) at that point. The pair of generalized forces consists of two forces with an equal magnitude and direction, but of opposite orientation.

    [​IMG]
    Figure 4.​

    As mentioned before, for a system with a degree of statical indeterminacy which equals n, one has to remove (and/or add) a total number of n supports (and/or internal hinges). Of course, the resulting primary system must not be a mechanism, i.e. it has to have zero degrees of freedom. We can accomplish this in a lot of ways - it will be seen later that there are smarter ways to do that, since we shall need to sketch different internal force diagrams, which can become fairly complicated, depending on the supports we choose to remove (or the hinges we choose to add).

    After creating the primary system, we place the set of generalized forces [itex]\left\{X_{1}, \cdots, X_{n}\right\}[/itex] on the system in such a manner that each force acts at the point and in the direction of the removed support. In the example in Section 1.1., the generalized force was the reaction [itex]R_{B}[/itex], which was placed at the point of the removed support. (Note that the generalized forces may represent pairs of forces too, as shown in Figure 4.)

    After doing so, it is intuitively clear that one has to solve a system of n equations, each of which expresses a displacement condition at one of the points where the supports were removed (or where internal hinges were added - we shall, for practical reasons, from now on refer to this procedure as support removal only). These equations are often referred to as compatibility equations, since they guarantee that the primary system with the generalized forces acting on it must have the same displacements as the original system.

    We shall now analyse the structure of the i-th equation, which has the form:

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

    where [itex]X_{k}[/itex] represents the 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 (not the added generalized forces). In words, the total displacement at every point i has two contributions: one is from the generalized forces [itex]\left\{X_{1}, \cdots, X_{n}\right\}[/itex], and the other one comes from the loads applied to the original system. This total displacement must vanish at every point, which is expressed by (1). (It is important to understand that every generalized force has a contribution to the displacement at point i, i.e. at every point of the system.)

    We shall now present some theorems and principles which justify and explain the formulation of the n equations of compatibility, each of which has the form (1).

    Perhaps the most important theorem is Castigliano's (second) theorem, which states that the partial derivative of the potential energy of deformation of a linearly elastic body with respect to the i-th force acting on the body equals the displacement at the point and direction of the force, i.e.

    [tex]\delta_{i} = \frac{\partial U}{\partial F_{i}} \ \ \text{(2).}[/tex]

    The proof of this theorem shall not be presented here. We shall only mention that it relies on Betti's theorem (which can be found here), which is derived under the assumption that the principle of superposition holds.

    Without going more in-depth, one can intuitively assume that expression (2) will be useful in the force method, since the most important issue is to find displacements at certain points, while the generalized forces acting at those points remain the unknowns - which is why the method is called the force method.

    Indeed, if we set [itex]F_{i} = X_{i}[/itex], it can easily be seen that, for a system with a degree of statical determinacy which equals n, we shall have (in consistence with the idea of the force method laid out at the beginning of this section) n equations of the form

    [tex]\delta_{i} = 0, i = 1, \cdots, n \ \ \text{(3),}[/tex]

    which, after applying Castigliano's theorem, become:

    [tex]\frac{\partial U}{\partial X_{i}}=0, i = 1, \cdots, n \ \ \text{(3').}[/tex]

    The solution of the system is the ordered n-touple [itex](X_{1}, \cdots, X_{n})[/itex], which represents the reaction forces in all the removed supports. The primary system is now in the same state of equilibrium as the original system - the equations of equilibrium can easily be solved to retrieve the values of all the other reactive forces (and internal forces, automatically), and it can easily be seen that the displacement of the system equals the displacement of the original system at every point, which almost directly follows from condition (3).

    The only thing left to do now is to show how the deformational potential energy of the system, referred to as U, is constructed, and how the system of equations (3') is equivalent to the system (1).
     
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  4. Dec 17, 2006 #3

    radou

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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.
     
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  5. Dec 18, 2006 #4

    radou

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    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.
     
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  6. Dec 19, 2006 #5

    radou

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    1.4. 'OUTRODUCTION' THROUGH AN EXAMPLE.

    As a final example, we shall analyze the system shown in Figure 5. Let:
    F = 50 [kN],
    q = 25 [kN/m], and
    L = 5 [m].​
    The material and cross-sectional characteristics are:
    E = 3*107 [kN/m2],
    b/h = 0.5/0.5,
    I = 0.54/12 [m4], and
    A = 0.52 = 0.25 [m2]. ​

    [​IMG]
    (Missing figure)
    Figure 5​

    The total number of reactions is 5, and the number of independent equations of equilibrium we can use is 3, which implies that the system's degree of statical indeterminacy equals 2. The chosen primary system is displayed in Figure 6.

    [​IMG]
    (Missing figure)
    Figure 6​

    As it can be seen in the figure above, the primary system is chosen so that a a hinge is added at the support at point C, and the hinge support at point A is turned into a roller hinge support. Further on, at point C there is now a moment [itex]X_{2}[/itex] added, and at point A there is a horizontal force [itex]X_{1}[/itex]. Note that the (generalized) forces [itex]X_{1}[/itex] and [itex]X_{2}[/itex] act in the direction of the released support parts (or added hinges), i.e. the forces act in the direction of the new added degree of freedom. The system is now statically determinate - the number of reactions equals the number of equations
    of equilibrium we can use.

    The equations of compatibility (eq. (4) in Section 1.3.) are now:

    [tex]X_{1}\delta_{11}+X_{2}\delta_{12}+\delta_{10} = 0[/tex]
    [tex]X_{1}\delta_{21}+X_{2}\delta_{22}+\delta_{20} = 0 \ \text{(1).}[/tex]

    In order to solve the equations for [itex]X_{1}[/itex] and [itex]X_{2}[/itex] we have to calculate the coefficients [itex]\delta_{ik}[/itex] and [itex]\delta_{i0}[/itex], where i, k = 1, 2. In order to do that, we shall use internal force diagrams referred to as [itex]M^0[/itex], [itex]N^0[/itex], [itex]m_{1}[/itex], [itex]m_{2}[/itex], [itex]n_{1}[/itex] and [itex]n_{2}[/itex]. The 'capital letter' diagrams [itex]M^0[/itex] and [itex]N^0[/itex] represent the bending moment and normal force in the system due to the original loads applied to the system (the uniform load q and the concentrated force F). The other diagrams represent bending moments and normal forces due to the unit generalized displacement at a point i - for example, the diagram [itex]m_{2}[/itex] represents the bending moment in the system caused by the unit moment acting at point C in the same direction as the moment [itex]X_{2}[/itex]; the diagram [itex]n_{1}[/itex] represents the normal force in the system due to a unit concentrated force at point A acting in the same direction as the force [itex]X_{1}[/itex], etc.

    The diagrams are nothing but sketches of the functions [itex]M^0(x)[/itex], [itex]N^0(x)[/itex], [itex]m_{1}(x)[/itex], etc., which are calculated on the local coordinate systems of each member of the system, AB, and BC. The functions can, of course, easily be set up after calculating the reactions. The diagrams are shown in Figure 7 (Note that every diagram has its own scale, due to practical reasons.)

    [​IMG]
    (Missing figure)
    Figure 7​

    We shall only calculate the value of the coefficient [itex]\delta_{10}[/itex], which represents the displacement at the point where the force [itex]X_{1}[/itex] is acting, due to the loads F and q. By definition, we have:

    [tex]\delta_{10} = \sum_{j=1}^2 \int_{0}^{L_{j}} \left( \frac{N^0(x)n_{1}(x)}{EA} + \frac{M^0(x)m_{1}(x)}{EI} \right) dx[/tex]
    [tex] = \int_{0}^{L_{1}} \left( \frac{N^0(x)n_{1}(x)}{EA} + \frac{M^0(x)m_{1}(x)}{EI} \right) dx + \int_{0}^{L_{2}}\left( \frac{N^0(x)n_{1}(x)}{EA} + \frac{M^0(x)m_{1}(x)}{EI} \right) dx[/tex]
    [tex] = \frac{1}{EI} \int_{0}^{L_{1}} M^0(x)m_{1}(x) dx + \frac{1}{EI} \int_{0}^{L_{2}} M^0(x)m_{1}(x) dx + \frac{1}{EA} \int_{0}^{L_{1}} N^0(x)n_{1}(x) dx + \frac{1}{EA} \int_{0}^{L_{2}} N^0(x)n_{1}(x) dx \ \ \text{(2).}[/tex]

    In our case [itex]L_{1} = L_{2} = L[/itex]. We shall now use Vereshchagin's theorem to integrate the expressions above. We have:

    [tex]\delta_{10}
    = \frac{1}{EI} \left[ 2.5\cdot 125\cdot\frac{1}{2}\cdot(2.5+\frac{2}{3}\cdot 2.5)\right]
    + \frac{1}{EI} \left[ 125\cdot 5\cdot\frac{1}{2} \cdot ( \frac{2}{3}\cdot 5)
    + \frac{2}{3} \cdot \frac{25\cdot 5^2}{8} \cdot 5 \cdot (-\frac{1}{2}\cdot 5) \right][/tex]
    [tex] \ \ \ + \frac{1}{EA} \left[ 87.5\cdot 5 \cdot (1) \right]
    + \frac{1}{EA} \left[ 50 \cdot 5 \cdot (1) \right] [/tex]
    [tex]= 6.7583333\cdot 10^{-3} \ \text{.}[/tex]

    It is useful to compare the calculated term with the expression (2) to see how Vereshchagin's integration method is used. Every term multiplied by [itex]\frac{1}{EI}[/itex] or [itex]\frac{1}{EA}[/itex] consists of two factors. The first one represents the area of the diagram [itex]M^0[/itex] or [itex]N^0[/itex], and the second one (the bracketed one) represents the value of the functions [itex]m_{1}[/itex] or [itex]n_{1}[/itex] at the abscissa of the centroid of area of the [itex]M^0[/itex] or [itex]N^0[/itex] diagram. For example, the first term in the expression above consists of two factors: the first one, [itex]2.5\cdot 125\cdot\frac{1}{2}[/itex] is the area of the diagram [itex]M^0[/itex] along the length [itex]L_{1}[/itex], i.e. on the element AB. The second factor, [itex]2.5+\frac{2}{3}\cdot 2.5[/itex] equals the value of the [itex]m_{1}[/itex] diagram at the (local) abscissa of the centroid of area of the [itex]M^0[/itex] diagram. (This can be calculated easily - it is suggested for the reader to make a sketch and 'follow' the calculations laid out in order to avoid unnecessary confusion.)

    The value of the coefficient [itex]\delta_{11}[/itex] will be calculated,

    [tex]\delta_{11} = \frac{1}{EI} \left[ 5\cdot 5 \cdot \frac{1}{2} \cdot(\frac{2}{3}\cdot 5)\cdot 2 \right] + \frac{1}{EA} \left[ 1\cdot 5 \cdot (1) \cdot 2 \right] = 5.346666667 \cdot 10^{-4} \ \text{,}[/tex]

    and the other coefficients are left for the reader to calculate as an exercise. The values one should obtain after doing that are:

    [tex]\delta_{20} = -1.78333355\cdot 10^{-4} \ \text{,}[/tex]
    [tex]\delta_{22} = 1.069333333\cdot 10^{-5} \ \text{,}[/tex]
    [tex]\delta_{12} = \delta_{21} = 2.653333333\cdot 10^{-5} \ \text{.}[/tex]

    After plugging the coefficients in the system of equations (1), and solving, we have:

    [tex]X_{1} = -15.3986[/tex \ \text{, and}[/tex]
    [tex]X_{2} = 54.8855 \ \text{.}[/tex]

    All we have to do now is use the equations (2) from Section 1.3, i.e.

    [tex]N(x) = N^0(x)+X_{1}n_{1}(x) + X_{2}n_{2}(x)[/tex]
    [tex]M(x) = M^0(x)+X_{1}m_{1}(x) + X_{2}m_{2}(x)[/tex]

    to obtain the internal forces N and M at any point of the system. This is left as an exercise too. The final diagrams of the internal force are shown in Figure 8. (The reactions can be calculated too, of course.)

    [​IMG]
    (Missing figure)
    Figure 8​

    As a final exercise, the reader can solve the example from Section 1.1. with the same primary system as suggested in that section, and then try out a different primary system.
     
    Last edited by a moderator: Jun 25, 2011
  7. Jul 16, 2010 #6

    radou

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    Noticed figures are missing, will re-upload them in a day or two.
     
  8. May 2, 2011 #7

    radou

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    [​IMG]

    Figure 1.

    [​IMG]

    Figure 2.

    [​IMG]

    Figure 3.

    [​IMG]

    Figure 4.
     
  9. May 2, 2011 #8

    radou

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    Figures 1-4 uploaded, having trouble finding Figures 5-8, though. Will look into it.

    Edit: had to upload them separately, since I can't edit the old posts, I hope this isn't a problem.
     
  10. Jun 25, 2011 #9

    Redbelly98

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    If you can upload figures 5-8, I can edit the old posts to include them.
     
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