I think what is meant here is that the part of the strain energy responsible for the locking phenomenon is eliminated although the value of strain is averaged at each Guass point.

In the FE method the various integrations are performed numerically using Gauss Quadrature and the number of Gauss points used dictates the degree of the integrand that can be integrated exactly.  If a lesser number of points are used the method leads to an approximation.  The standard conforming elements used in most if not all commercial FE packages are described as over-stiff in the sense that in a coarse mesh the displacements resulting from a force driven problem will be less than the true values.  They will, however, converge to the correct values with mesh refinement.  To provide some counter to this over-stiff behaviour FE software offers reduced integration schemes which use less points than required for exact integration.  This 'fiddle' leads to better numerical results for some types of behaviour as the stiffness is reduced.  Unfortunately, however, it also induces other unwanted behaviours that can sometimes upset the results of an analysis - the hourglassing issue.  An hour glass mode might be considered as a spurious (in that it would not occur in reality and with exact integration) kinematic mode or mechanism.  A mechanism is a mode of deformation that requires no forces to drive it.  Hour glass modes occur at the single element level and may or maynot propogate to the entire mesh depending on the type and degree of the element you use.

So returning to your question, the stiffness matrix is formed by integration at the Gauss points but, as you point out, the displacements are based on nodal values.  So if in your FE model you load it such that the nodal displacements cause no strain at the Gauss points then that mode of displacement will induce no strain energy in the element and therefore the element will have no stiffness to resist this load.  With full integration there are no such hourglass modes.

To develop your understanding you might look at, for example, the plate-membrane (plane-stress) elements both four and eight noded forms and with reduced and full integration.  Try single element tests with applied nodal displacements and forces to excite or drive the element in one of it's hour glass modes.

In hourglass mode infinite deformations form at nodal points.These deformations are not a function of the applied forces or resultant stresses. Therefore, a very likely condition for such deformations is a zero-load condition. In such a case, the energy, the multiplication of stresses and strains, becomes zero. Actually, the formula expressing energy in form of pure strain is no longer valid because the Hook relation between stresses and strains has no validity as was mentioned at the beginning of the answer.

Dr. Angus Ramsay, in your reply, you have written that :

"if in your FE model you load it such that the nodal displacements cause no strain at the Gauss points then that mode of displacement will induce no strain energy in the element and therefore the element will have no stiffness to resist this load. "

- why are we looking at the strain values at the gauss point ?

As I understood, the use of guass points is only upto the point we arrive at stiffness matrix and rest is calculated from nodal values which we get after solving the matrix equation. Can you please help me on this point?

By using reduced integration which is inexact you will obtain a different stiffness matrix to that which would have been obtained if you had requested exact or 'full' integration.  The stiffness matrix resulting from full integration will be rank deficient by the number of rigid-body modes for the element but for reduced integration it will be rank deficient by the number of rigid-body modes PLUS any hourglass modes.  This means that there is a mechanism in the element which is the hourglass mode.  So if you drive the element (nodal forces or displacements) such as to excite the hourglass mode it will offer no resistance.  

You will recall that in Gauss Quadrature the integration is converted into a weighted summation of terms at the various Gauss Points.  So one can consider the stiffness matrix to be a sum of stiffness matrices calculated at each Gauss Point.  With full integration the sum is capable of resisting all deformation modes (apart from rigid-body modes).  But for reduced integration resistance to some modes of deformation are missing in addition to the rigid-body modes and the reason that they are missing is that these modes of deformation (hourglass modes) cause no strain energy at the Gauss Points and, therefore, add no stiffness to the summation when the stiffness matrix was formed.  The hourglass modes WOULD develop strain energy at other Gauss Points and therefore stiffness contributions would arise but as these points were not included with reduced integration then these terms are simply missing.

Consider the case of a plane stress/strain element modelling a beam in bending.  For the four-noded element with reduced integration the stiffness matrix is based on the stiffness of a single central Gauss Point and in pure bending this will lie on the neutral axis of the beam and so will not be subject to any strain.  This is then an hourglass mode.

Let me know whether this answer helps.