Simple Indeterminate Structure

An indeterminate structure is when there are more reactions or supports than are needed to keep the structure in static equilibrium. As an example, an objected supported by three rods, as shown on the left, is indeterminate to the first degree, i.e. there is one redundant support. If it is redundant, why are indeterminate structures used? They help distribute loads to more supports which can reduce individual loads and moments. A general explanation of indeterminate structures can be found in the Indeterminate Objects section of the Statics eBook.

    Stress and Deflection Superposition

Simple Application of the
Superposition Principle


A useful concept in solving indeterminate structures is the Principle of Superposition. Simply stated, a problem can be reduced to smaller problems and the solutions to the smaller problems can be added together to give the solution to the original problem.

This can be better understood by considering the deflection of a rod with two forces as shown in the diagram. This problem can be reduced to two separate problems, each with one force. The deflection of the rod due to each individual load is calculated. The the total deflection of the rod is simply the total of the two separate deflections. In equation form,

     F = F1 + F2

     δF = δF1 + δF2

This is a simple problem, but it serves as the basis for superposition. The solution of individual problems can be superimposed to give the combined solution. It also works for stress calculations. It can even be used for thermal expansion.

This concept does have limits. It can only be used when the material responds in a linear fashion (true for all problems in basic solid mechanics). This idea will be used many times in the future.

    Indeterminate Axial Structures

Simple Indeterminate
Structure Example

Force Equilibrium for Example


Structures that have too many supports are classified as indeterminate structures. This section examines only 1D or axial indeterminate structures where all displacements occur in the same direction. An example of this is a weight supported by three rods as shown on the left.

Equilibrium Equation(s)
Since there are too many supports, equilibrium equations, ΣF = 0 and ΣM = 0, will not be sufficient to solve for the force unknowns. Thus, additional equation(s) are needed.

For the given example, the equilibrium equations can relate the rod forces to the weight of the weight. This gives

     2Fal + Fst = W     (example equilibrium)

However, the exact force in either rod cannot be determined with just the equilibrium equations.


Deflections Without Constraint


Compatibility Equation(s)
The compatibility equations are based on member deflections and will vary for each problem. This makes it difficult to solve since there is no single equation that be used for all problems like the equilibrium equations (ΣF = 0 and ΣM = 0).

Continuing with the example with three rods, compatibility condition requires that all deflections are the same. If there was no constraint (the weight) then each rod would deflection differently as shown in the diagram on the left.

However, all the deflections are constrained by the weight, so they have the same deflection,


Deflections With Constraint


     δT = δal = δst     (example compatibility)

This condition is indicated in the diagram on the left. This relationship is only valid for this example and other problems will have different relationships.

Combining Equations
The final solution for the given example is obtained by combining both the equilibrium and compatibility equations,



Simplifying gives,


Substituting into the equilibrium equation gives,


The force in the steel rod can now be determined.

While other problems will use different compatibility equations the solution process is the same.

If the structure has more than one redundant support, then more compatibility equations are needed. For each redundant support, a separate compatibility equation is needed.

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