MECHANICS  THEORY

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 = F_{1} + F_{2}
δ_{F} = δ_{F1} + δ_{F2}
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
2F_{al} + F_{st} = 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


δ_{T} = δ_{al} = δ_{st} (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.



