Assuming the axiom of choice there is a neat way defining inaccessible cardinals as uncountable, regular, strong limit cardinals.

Without the axiom of choice we have several notions of inaccessibility (cf. this paper by Blass, Dimitriou, and Lowe). For the purpose of the question, when we say that $\kappa$ is inaccessible in $ZF$ we mean that it is an uncountable, regular limit cardinal, and for $\alpha<\kappa$ we have no surjection from $V_\alpha$ onto $\kappa$.

Consider the following example:

Suppose now that $V\models ZF+DC_\kappa+\lnot DC_{\kappa^+}$ and $\kappa\in V$ is inaccessible. In particular it is clear that $V_\kappa\models ZFC$. The last is quickly proved since $|V_\alpha|<\kappa$ (since the cardinals are comparable, and $\kappa\nleq |V_\alpha|$ from its inaccessibility).

If we now consider $V$, it is an end extension of $V_\kappa$, however $V$ violates the axiom of choice.

Suppose that there are two inaccessibles, $\lambda<\kappa$, then $V_\lambda$ has two end extensions, one is $V_\kappa$ which is a model of $ZFC$, and the other is $V$ which is not a model of choice.

Question: Can we violate other axioms a well? Could we somehow arrange a model that for an inaccessible $\kappa$ we have $V_\kappa\models V=L$, while above it we may have something wider such as a measurable cardinal? Or the negation of the axiom of choice?

Is it possible to "gradually" break some axioms? That is a model with a class of large cardinals such that at each point we get slightly more complicated than the previous step?

The motivation comes from several points at which it seems that we "measure how far $V$ is from $L$" by using forcing axioms, or large cardinals, and so on.