If gravity bends space time, could gravity be manipulated to freeze time? You age at a different rate depending on the force of gravity. Astronauts age fractions of fractions of fractions of a second less than earthlings.
If you took a sphere of equal masses, separated by space, then found the exact center of the gravitational pulls of all masses. How would time react?
 A: Time does run more slowly inside a massive spherical shell than outside it, however you could not stop time this way because if you made the shell massive enough it would collapse into a black hole.
You need to be very careful talking about gravitational time dilation as it's easy to misunderstand what is happening. No observer will ever see their own clock running at a different speed, that is every observer still experiences time passing at the usual one second per second. However if two observers in different places compare their clocks they may find that their clocks are running at different rates.
The best know example of this is the static black hole, which is described by the Schwarzschild metric. If an observer at infinity and an observer at a distance $r$ from the black hole compare their clocks they will find the clock near the black hole is running more slowly. The ratio of the speeds of the clocks is:
$$ \frac{\Delta t_r}{\Delta t_\infty} = \sqrt{1 - \frac{2GM}{c^2r}} \tag{1} $$
If we graph this ratio as a function of $r/r_s$, where $r_s = 2GM/c^2$, we get:

and you can see that the ratio goes to zero, i.e. time freezes when $r/r_s = 1$. This value of $r$ is actually the position of the black hole event horizon, and what we've discovered is the well known phenomenon that time stops at the event horizon of a black hole. But note that when I say time stops I mean it stops relative to the observer at infinity. If you were falling into a black hole you wouldn't notice anything odd happening to your clock as you crossed the event horizon.
You specifically asked about a spherical shell. The easy way to calculate the time dilation is to use the weak field expression:
$$ \frac{\Delta t_r}{\Delta t_\infty} = \sqrt{1 - \frac{2\Delta\Phi}{c^2}} $$
where $\Delta\Phi$ is the difference in gravitational potential relative to infinity. If we have a spherical shell of mass $M$ and radius $r$ then the time dilation at the outer surface is just:
$$ \frac{\Delta t_{outer}}{\Delta t_\infty} = \sqrt{1 - \frac{2GM}{c^2r_{shell}}} \tag{2} $$
so it's the same as the expression for the black hole given in (1). If we can assume the thickness of the shell is negligable the potential remains constant as we cross the shell and go inside it, so everywhere inside the shell the time dilation remains constant and is given by equation (2). Now you can see why you can't stop time inside a spherical shell. To make the ratio of times zero you need to decrease the shell radius to $r_{shell} = 2GM/c^2$. But this is the black hole radius so our shell has now turned into a black hole.
