The answer is known for $B_3$ and $B_4$:

$B_3=\langle \sigma_1,\sigma_2\mid \sigma_1\sigma_2\sigma_1=\sigma_2\sigma_1\sigma_2\rangle$ decomposes as $B_3=F_2\rtimes \mathbb Z$, where $F_2=B_3'=\langle\sigma_2\sigma_1^{-1},\sigma_1\sigma_2\sigma_1^{-2}\rangle$, and $\mathbb Z=\langle \sigma_1\rangle$.

Similarly,

$B_4=\langle \sigma_1,\sigma_2,\sigma_3\mid \sigma_1\sigma_2\sigma_1=\sigma_2\sigma_1\sigma_2, \sigma_1\sigma_3=\sigma_3\sigma_1,\sigma_2\sigma_3\sigma_2=\sigma_3\sigma_2\sigma_3 \rangle$ decomposes as $B_4=F_2\rtimes B_3=F_2\rtimes (F_2\rtimes \mathbb Z)$, where the outer $F_2$ is generated by $\sigma_3\sigma_1^{-1}$ and $\sigma_2\sigma_3\sigma_1^{-1}\sigma_2^{-1}$, and $B_3$ has generators as above.

See
Leonid Bokut, Andrei Vesnin, Grobner–Shirshov bases for some braid groups,
Journal of Symbolic Computation 41 (2006) 357–371
http://math.nsc.ru/~vesnin/papers/bokut-vesnin2006.pdf

or Theorem 2.1 (presentation for $B_n'$ for all $n$) here:

E. A. Gorin, V. Ya. Lin, “Algebraic equations with continuous coefficients and some problems of the algebraic theory of braids”, Mat. Sb. (N.S.), 78(120):4 (1969), 579–610
http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=sm&paperid=3572&option_lang=eng