Editing Statically indeterminate (section)

==Statically determinate==
If the support at {{mvar|B}} is removed, the reaction {{math|'''V'''{{sub|''B''}}}} cannot occur, and the system becomes '''statically determinate''' (or '''isostatic''').<ref>{{Cite book|title=Structural mechanics: a unified approach|last=Carpinteri|first=Alberto|date=1997|publisher=E & FN Spon|isbn=0419191607|edition=1st|location=London|oclc=36416368|author-link=Alberto Carpinteri}}</ref> Note that the system is ''completely constrained'' here.
The system becomes an [[exact constraint]] [[kinematic coupling]].
The solution to the problem is:<ref name=":0" />

:<math>\begin{align}
  \mathbf H_A &= \mathbf F_h \\
  \mathbf V_C &= \frac{\mathbf F_v \cdot a}{a + b + c} \\
  \mathbf V_A &= \mathbf F_v - \mathbf V_C
\end{align}</math>

If, in addition, the support at {{mvar|A}} is changed to a roller support, the number of reactions are reduced to three (without {{math|'''H'''{{sub|''A''}}}}), but the beam can now be moved horizontally; the system becomes ''unstable'' or ''partly constrained''—a [[mechanism (engineering)|mechanism]] rather than a structure. In order to distinguish between this and the situation when a system under equilibrium is perturbed and becomes unstable, it is preferable to use the phrase ''partly constrained'' here. In this case, the two unknowns {{math|'''V'''{{sub|''A''}}}} and {{math|'''V'''{{sub|''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 is not possible to satisfy the horizontal force equation unless {{math|1='''F'''{{sub|''h''}} = 0}}.<ref name=":0" />