Is an Eulerian lattice shellable? The notion of Eulerian lattice generalizes the notion of face lattice of a convex polytope.  
(Bruggesse-Mani): The boundary complex of a convex polytope is shellable.
(Björner-Wachs): A poset is said to be shellable if its order complex is shellable. A bounded poset is shellable if and only if its proper part is shellable. 
(Wachs): The order complex of the face poset of a simplicial complex is its barycentric subdivision.
Pre-question: Is the shellability kept by taking the barycentric subdivision?
If so, the face lattice of a convex polytope is shellable. This leads to:
Question: Is an Eulerian lattice shellable?
 A: The simplest example of a nonshellable (even non-Cohen Macaulay) Eulerian lattice is the disjoint union of two boolean algebras of rank three with their bottom elements identified and their top elements identified. The total number of elements is 14.
A: I think that the answer to your question ought to be "not always".  
There are examples of non-shellable balls and spheres.  A very readable account of these, together with some of the history, can be found at this blog entry.
Now if you take the face lattice of a sphere, that is an Eulerian lattice.  Passing from a complex $\Delta$ to the order complex of the face lattice of $\Delta$ yields the barycentric subdivision of $\Delta$.
So to find a negative answer to your question, find a non-shellable sphere whose barycentric subdivision also fails to be shellable.  The construction of such spheres is briefly discussed in this paper.  There is likely a source somewhere that constructs such a sphere more explicitly.
Update:  Also, the answer to your pre-question is "yes".  Indeed, the barycentric subdivision of any shellable complex is vertex-decomposable, a somewhat stronger property.
