Let $X$ be a $k$-scheme, and let $A$ and $B$ be algebraic cycles on $X$ (i.e., irreducible subvarieties). Recall that $A$ is rationally equivalent to $B$ if there is a finite sequence of cycles $A_1, \dots, A_n$ on $X$ with $A_1 = A$ and $A_n = B$ such that for each $i$, there exists a flat family over $\mathbb{P}_k^1$ of cycles on $X$ interpolating between $A_i$ and $A_{i+1}$.
It follows immediately from the above definition that rational equivalence is transitive. However, I'm looking for something a bit stronger: if $A$ and $B$ are rationally equivalent cycles on $X$, does there exist a flat family over $\mathbb{P}_k^1$ of cycles on $X$ interpolating between $A$ and $B$? This is equivalent to asking whether the existence of such an interpolating family is a transitive condition.
It's false. In $\mathbb{P}^3$, a twisted cubic $C$ is rationally equivalent to an elliptic curve $E$ contained in a plane. But there is no flat family interpolating between $C$ and $E$.
Reason: the corresponding Hilbert scheme $H$ has two irreducible components, and $C$, $E$ lie in their interiors, so no single $\mathbb{P}^1$ links them. To be precise, the first component $H_1$ parametrizes twisted cubics, and the second $H_2$ parametrizes "plane cubics union a point". The intersection $H^1 \cap H^2$ corresponds to singular plane cubics with an embedded point at the node (the nonreduced structure pokes outside of the plane into the ambient $\mathbb{P}^3$).
See: http://www.uio.no/studier/emner/matnat/math/MAT4230/h10/undervisningsmateriale/Hilbertscheme.pdf