It should be symmetrical except for particular problems such as multi-phase flow.

Symmetry of stiffness matrix of structure under conservative forces stems from the fact that the work of generalized inner forces does not depend on the path between two equilibrium states.

I feel symmetry of stiffness matrix is not depending on the forces and moment acting on the element. it is purely depends on the properties of the element.

To answer to Kahled question, the stiffness matrix in conventional Finite Element sense can also be defined through the defintion of the Elastic Potential (W, existent only for conservative forces):

W=1/2*u^T*K*u; (1)

where u in R^n is the vector of degree of freedom, K in R^(nxn) is the stiffness matrix and ^T is the transposition .

the Elastic Potential beeing a scalar:

W^T=W

so that:

(1/2*u^T*K*u)^T=1/2*u^T*K^T*u=1/2*u^T*K*u

in order to have this relationship to be true for any value of u in R^n,

K=K^T

so that K has to be symmetric.

Anyway often engineers refer to Stiffness matrix as the matrix of coefficient of the linear system of equation for static balance for small deformations hypothesis.

K*u=F (2)

where F in R^n is the vector of generalized forces.

Definition coming from equation (2) is not sufficient to ensure stiffness matrix symmetry. In fact simply changing the order of your balance equation the coefficient matrix can loose its symmetry. That is why, using Euler equations in conservative system, you can still get non symmetric matrix of coefficient.

With abuse of language some authors refer to the matrix of coefficient as stiffness matrix in any case. One should refer to stiffness matrix only when the coefficient matrix of equation (2) respect the definition of (1), so that the stiffness matrix has to be symmetric.

I think that symmetry of stiffness matrix can be proven as follows. Let the linearized equilibrium state of the system have a form K u=F, where K={kij} is square matrix NxN, u and F are generalized displacements and forces respectively. Let consider four states: 1) ui=0, i=1,...,N; 2) ui=0, i≠α, uα≠0; 3) ui=0, i≠ß, uß≠0; 4) ui=0, i≠α, i≠ß, uα≠0, uß≠0. Since the system is conservative, W12+W24=W13+W34, where Wij is work of linearly varying generalized forces on the generalized displacements between states i and j. Since W12=0.5 kααuα2; W24=kßαuαuß+0.5 kßßuß2; W13=0.5 kßßuß2; W34=kαßuαuß+0.5 kααuα2, we have kαß=kßα.