Given:

$M=2\mathrm{.}60Nm$

$=3\mathrm{.}10\frac{N}{m}$

$L_{}=2\mathrm{.}40m$

$L_{BC}=2\mathrm{.}60m$

$L_{CD}=2\mathrm{.}45m$

Find:

internal shear $(V)$ at position $x=0\mathrm{.}45m$

enter answer with $2$ decimal places, units of $Nm$