# Compatibility equations in elastcity

Strain field in deformable solid (continuum) has to follow compatibility equations so as to ensure single valued continuous displacement field.
There are in all 81 such equations and most of them are repeated, finally we are left with 6 equations.
It is quoted in text books that only 3 out of these 6 equations are independant.
If it is so why we need to use six equations to ensure possibility of strain field?
Only three equations should be enough.

Mech_Engineer
Gold Member
If I'm understanding your question correctly I think the difference lies in compression/tension vs. shear. There are three tension directions and three shear planes (for a total of 6 values), but only three of the six values are needed to find all other values.

Just to clarify there are 3 normal and 6 shear values,

σxx, σyy, σzz,

τxy, τyx
τzy, τyz
τxz, τzx

and the corresponding strain ε values to go with them

but symmetries reduce these.

AlephZero
Homework Helper
There are 9 terms in the strain tensor, but compatibility means the tensor must be siymmetric, so there are only 6 independent terms.

The off-diagonal terms are the shear strains, so there are 3 equal pairs of shear strains, ##\epsilon_{xy} = \epsilon_{yx}##, ##\epsilon_{yz} = \epsilon_{zy}##, and ##\epsilon_{zx} = \epsilon_{xz}##, plus the three direct strains.

The reason for that is just Euclidean geometry. If the strain tensor was not symmetric, a continuous region of space could deform into something containing holes or overlapping regions, whcih doesn't make sense for doing continuum mechanics.

I don't understand what you mean by "81 equations". There are 81 terms in the compliance (stress-strain) tensor, of which 21 are independent for a general anisotropic material (and only 2 are independent for isotropic materials) but that is a different concept from compatibiltiy, and "81 terms in a tensor" isn't te same as "81 equations".