[Lickorish and Martin][1] constructed, for each $r$, a triangulation of the $3$-ball whose $r$th barycentric subdivision collapses, but $(r-1)$th doesn't. The basic idea, going back to Bing, is to triangulate a cube with a knotted hole, where the missing knot has bridge index $2^r+1$, and then fill back a small part of the hole --- so that topologically, no hole remains, but the $3$-ball now contains a knot triangulated by a single edge.

  [1]: http://www.ams.org/journals/tran/1969-137-00/S0002-9947-1969-0238288-X/S0002-9947-1969-0238288-X.pdf