Statically indeterminate

In statics, a structure is statically indeterminate when the static equilibrium equations are not sufficient for determining the internal forces and reactions on that structure.

Based on Newton's laws of motion, the equilibrium equations available for a two-dimensional body are
* $sum\; vec\; F\; =\; 0$: the vectorial sum of the forces acting on the body equals zero. This translates to ::Σ "H" = 0: the sum of the horizontal components of the forces equals zero;::Σ "V" = 0: the sum of the vertical components of forces equals zero;
* $sum\; vec\; M\; =\; 0$: the sum of the moments (about an arbitrary point) of all forces equals zero.

In the beam construction on the right, the four unknown reactions are "V"_"A", "V"_"B", "V"_"C" and "H"_"A". The equilibrium equations are:

Σ "V" = 0::"V"_"A" − "F"_"v" + "V"_"B" + "V"_"C" = 0

Σ "H" = 0::"H"_"A" − "F"_"h" = 0

Σ "M"_"A" = 0::"F"_"v" · "a" − "V"_"B" · ("a" + "b") - "V"_"C" · ("a" + "b" + "c") = 0.

Since there are four unknown forces (or "variables") ("V"_"A", "V"_"B", "V"_"C" and "H"_"A") but only three equilibrium equations, this system of simultaneous equations cannot be solved. The structure is therefore classified "statically indeterminate". Considerations in the material properties and compatibility in deformations are taken to solve statically indeterminate systems or structures.

tatically determinate

If the support at "B" is removed, the reaction "V"_"B" cannot occur, and the system becomes statically determinate. Note that the system is "completely constrained" here. The solution to the problem is

:$H\_A=F\_h$ ,:$V\_C=frac\{F\_vcdot\; a\}\{a+b+c\}$ ,:$V\_A=F\_v-V\_C$ .

If, in addition, the support at "A" is changed to a roller support, the number of reactions are reduced to three (without "H"_"A"), but the beam can now be moved horizontally; the system becomes "unstable" or "partially constrained". In order to distinguish between this and the situation when a system under equilibrium is perturbed and becomes unstable, it's preferable to use the phrase "partially constrained" here. In this case, the 2 unknowns "V"_"A" and "V"_"C" can be determined by resolving the vertical force equation and the moment equation simultaneously. The solution yields the same results as previously obtained. However, it's not possible to satisfy the horizontal force equation unless $F\_h=0$.

tatic indeterminacy

A system can be statically indeterminate even though its reactions are determinate as shown in Fig.(a) on the right. On the other hand, the system in Fig.(b) has indeterminate reactions, and yet, the system is determinate because its member forces, and subsequently the reactions, can be found by statics. Thus, in general, the static indeterminacy of structural systems depends on the internal structure as well as on the external supports.

The degree of static indeterminacy of a system is "M-N" where
* "M" is the number of unknown member forces, and optionally, reactions in the system;
* "N" is the number of independent, non-trivial equilibrium equations available for determining these "M" unknown forces.

If "M" includes reaction components, then "N" must include equilibrium equations along these reaction components, one for one. Thus, we may, in fact, choose to exclude reactions from the above relation.

ee also

*flexibility method
*Hardy Cross
*Christian Otto Mohr
*structural engineering