I am trying to find some examples of Fraïssé classes that do not have the strong amalgamation property. Anyone?
In this edit, the relevant theorem is properly stated, and a previous "fake" example has been replaced by a genuine one (in light of corrective comments by Joel Hamkins and Emil Jeřábek). Goldstern and Hamkins have provided nice examples. I will add: Theorem A Fraïssé class has the strong amalgamation property if and only if the automorphism group of its Fraïssé limit $M$ has no algebraicity (i.e, the algebraic closure of any finite set X within $M$ coincides with X itself). The above theorem appears as Exercise 8 of sec. 7.1 of Hodges' text on Model Theory, as well as [(2.15), p. 37] of Peter Camerons's 1990monograph Oligomorphic permutation groups. According to the Hodges text, it is due to James Schmerl (Journal of Symbolic Logic, 45, pp.585611, 1980).



I am not sure what this is good for (and if I understand your terminology), but here is a trivial example: Take a language with one unary predicate $P$, and let $F$ be the class of all finite structures in which there is at most one element $e$ such that $P(e)$ holds. If I am not mistaken, the family of (finite) distributive lattices is a less trivial example. 


Consider the language with one unary predicate symbol $U$, and let $K$ be the class of all finite structures $\langle A, U\rangle$ for which $A$ is finite and there is at most one $a\in A$ with $U(a)$. This class is closed under isomorphic copies and substructures; it has the joint embedding property, since any two such structures can be mapped into a third, and it has the amalgamation property, since embedded copies of $A$ in two structures $B$ and $C$ can be amalgamated into a fourth structure $D$, just by mapping the copies of $A$ into $D$, and also mapping the single points satisfying $U$, if any, to such a point in $D$, and otherwise mapping the rest of $B$ and $C$ into $D$ to distinct points. So it is a Fraïssé class. But $K$ does not have the strong amalgamation property, since if $A$ has no point satisfying $U$, but $B$ and $C$ do, then the images of $B$ and $C$ in $D$ will necessarily both include the unique point in $D$ satisfying $U$ in common, but this is not in the image of $A$ in $D$ since $A$ had no such point. 


The following paper gives an example showing why distributive lattices have amalgamation but not strong amalgamation. E Fried, G Grätzer. Strong amalgamation of distributive lattices. Journal of Algebra, Volume 128, Issue 2, 1 February 1990, Pages 446–455. doi:10.1016/00218693(90)90033K Their example is the following. Let A and B be the 4element Boolean algebras A = {s0,s1,s2,a} and B = {s0,s1,s2,b} with s0 bottom and s1 top. Let S = {s0,s1,s2} be the intersection of A and B. S is a distributive lattice, indeed a sublattice of both A and B. Since both a and b are complements of s1, the 4 element Boolean algebra D = {s0,s1,s2,c} where c identifies a and b, provides an amalgamation of A and B wrt S. But S is not the intersection (formally: pullback) of A and B in D (as both A>D and B>D are isomorphisms), therefore D is not a strong amalgamation. Moreover, it follows from D being the pushout of (S>A,SB) that no other amalgamation of A and B wrt S is strong. 

