How to compute the Picard-Lefschetz monodromy matrix of a non-semistable fiber?

Let $f:X\to B$ be a family of curves of genus $g$ over a smooth curve $B$. Let $F_0$ be a singular fiber.

If $F_0$ is a semistable fiber, the monodromy matrix can be gotten by the classical Picard-Lefschetz formula.

If $F_0$ is non-semistable, I don't know how to compute its monodromy matrix. For example, in Namikawa and Ueno's paper[1], they can compute the Picard-Lefschetz monodromy matrix for each type of singular fiber of genus 2. It's not clear to me how they did that.

[1] Namikawa, Y. and Ueno, K., The complete classification of fibres in pencils of curves of genus two, Manuscripta math., Vol. 9 (1973), 143-186.

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To achieve semi-stable reduction, you have to alternately blow-up singular points in the special fibre, and then make ramified base-changes. The latter operation just extracts a root of the monodromy operator (i.e. if $\gamma$ is a generator of $\pi_1$ of the punctured $t$-disk, and we set $t = s^n$, then $\gamma = \tau^n,$ where $\tau$ is a generator of $\pi_1$ of the punctured $s$-disk), so it is easy to see how the monodromy matrix changes. And blowing up a point in the special fibre doesn't change the monodromy action around the puncture at all.
This response is over $\mathbb{C}$, because I'm not sure what the monodromy matrix means in other settings: Shrink $B$ to a disc around $b_0$. Let $b_0$ be the point of $B$ under $F_0$ and let $B' \to B$ be your branched cover, of degree $k$. The monodromy of $B'$ is the $k$-th power of the monodromy of $B$. So you can compute the monodromy of $B$ up to any ambiguity about taking the $k$-th root of a matrix. – David Speyer Apr 20 '10 at 13:26