I was afraid my cursory response wouldn't be enough . . .
What is shear locking?
This is not intended to be an exhaustive discussion, so my apologies to the hardcore solid mechanicians; this is intended to explain in easiest terms this phenomenon.
Take the example which I earlier gave of the cantilever beam. Solid elements cannot "bend", per se.
So for a single solid element in cantilever bending,
(where the x's are the nodes)
Let's presume I hold the left vertical face fixed and apply a bending moment to the right side:
The original and deformed configurations would look like:
ORIGINAL DEFORMED
x x x x
becomes
x x x x
If this is a fully-integrated 1st order element, the "bending" displacement plays itself out as a contraction of the bottom portion and an extension of the top portion of the element. Hence, the bending mode effectively can only be reacted by a shear mode in the element.
One could calculate the strain at the top and bottom of a cantilever beam. If a solid element in bending has a strain of +e on the top face and -e on the bottom face, the energy due to this is much less than the energy due to shearing of that same amount. However, the solid element can ONLY react this load condition via shear. These parasitic shear stresses result in an overly-stiff element. For first-order fully-integrated elements, this can introduce significant error for a coarse mesh. This overly-stiff behavior is referred to as shear locking.
Rahulsrp suggests using reduced-integration elements to avoid this; however, they present their own unique problems in that they have no hourglass stiffness. They have no way of resisting this shear behavior, therefore they have spurious zero-energy shear modes, which can propogate through the mesh.
The way to get around these issues:
1) Use 2nd order elements
2) Use a finer mesh--a fine mesh will eventually converge to the "correct" solution
3) Some codes have special fully-integrated elements which account for this behavior. They are sometimes referred to as "incompatible modes" elements.
And despite rahulsrp's statement, you CAN in fact use solid elements for "thin" structures; you just need a fine enough mesh and need to be able to maintain reasonable element ratios (after all, "thin" is a relative term). As previously stated, once one has addressed these two issues, it is generally much more computationally expensive to use "solid" elements instead of "shells".
Just so there is no confusion: one CAN use solids for these thin structures, but I would recommend using shell elements instead for most cases.
Brad
.