Since $(\sin α- \frac {3}{5})+(\cos α- \frac {4}{5})i$ is a pure imaginary number, we have $$\begin{cases} \sin α- \frac {3}{5}=0, \\ \cos α- \frac {4}{5}\neq 0. \end{cases}$$

From the first equation, we find that $\sin α = \frac {3}{5}$. The second part of the condition, $\cos α- \frac {4}{5}\neq 0$, implies that $\cos α$ is not equal to $\frac {4}{5}$.

Now we need to determine the quadrant of the angle $α$. Since the sine of $α$ is positive and $\cos α \neq \frac {4}{5}$, we deduce that $α$ must be in the second quadrant where cosine is negative. Therefore, $\cos α=-\frac {4}{5}$.

Using the angle addition formula for sine, we get
$$\sin (α+ \frac {π}{4})= \sin α \cos \frac {π}{4} + \cos α \sin \frac {π}{4} = \frac {3}{5} \cdot \frac {\sqrt {2}}{2} - \frac {4}{5} \cdot \frac {\sqrt {2}}{2} = -\frac {\sqrt {2}}{10}.$$

Hence, the answer is $\boxed{-\frac {\sqrt {2}}{10}}$.