If you want $S^{i0}$ to be zero, then the spin $S^{\mu\nu}$ is the angular momentum of the body about its energy centroid in the lab frame.
Let's see how this comes about.
Suppose that the we have a conserved and symmetric energy-momentum tensor:
$$
\partial_\mu T^{\mu\nu}=0, \quad T^{\mu\nu}=T^{\nu\mu}
$$
which is non-zero only within the body of interest.
Let $x_{\rm A}^\mu$ be a space-time event, $\Sigma$ a spacelike surface, and define the angular momentum about $x_{\rm A}$ by
$$
M^{\mu\nu}_{\rm A} = \int_\Sigma\left\{(x^\mu-x_{\rm A}^\mu)T^{\gamma\nu}- (x^\nu-x_{\rm A}^\nu)T^{\gamma\mu}\right\}d\Sigma_\gamma
$$
Then $M^{\mu\nu}_{\rm A} $ is a tensor and independent of the choice of the choice of $\Sigma$.
We now choose a lab frame and define the mass-centroid $X^i_{\rm L}$ in that frame by
$$
\left\{\int_{t=\rm const.} T^{00}d^3x \right\}\,X^i_{\rm L}= \int_{t=\rm const.} x^iT^{00}d^3x.
$$
Note that
$$
\partial_t \int T^{00}d^3x = \int \partial_0 T^{00}d^3x = - \int \partial_j T^{j0}d^3x=0,
$$
and
$$
\partial_t \int x^i T^{00}d^3x = \int x^i\,\partial_0 T^{00}d^3x = -\int x^i\,\partial_j T^{j0}d^3x= \int \delta^i_j \,T^{j0}d^3x= p^i.
$$
So, differentiating its definition with respect to $t$, we read off that the ordinary three-velocity of the centroid is
$$
\dot {\bf X}_{\rm L}= {\bf p}/E.
$$
Here
$$
E= \int T^{00}d^3x, \quad p^i = \int T^{0i}d^3x.
$$
Now take $\Sigma$ to be the lab-frame surface $t=$const with $x^\mu_{\rm A}$ a point in that surface. Then
$$
M^{i0}_{\rm A} = \int_\Sigma\left\{(x^i-x_{\rm A}^i)T^{00}- (x^0-x_{\rm A}^0)T^{0i}\right\}d^3x\\
= (X^i_{\rm L}-x^i_{\rm A})E.
$$
(The second term is zero because $x^0- x^0_{\rm A}$ is zero everywhere in the integral.)
Thus $M^{i0}_A$ is zero when ${\rm A}$ is the centroid in the lab frame. If we replace the lab frame with a frame having four-velocity $v^\mu$ we have that
$M^{\mu\nu}_{\rm A} v_\nu=0$ if and only if ${\rm A}$ is the mass centroid in that frame.
Define the centre of mass $X^i_{\rm CM}$ to be the mass-centroid in the frame where the three momentum $p^i=0$, then the spin or intrinsic angular momentum $S^{\mu\nu}$ is usually defined to be the angular momentum about the centre of mass. Thus a more natural condition is $S^{\mu\nu}p_\nu=0$.
The total angular momentum about an arbitrary point $x_A$ is then
$$
M^{\mu\nu}= (x_{\rm CM}-x_A)^\mu p^\nu- (x_{\rm CM}-x_a)^\nu p^\mu + S^{\mu\nu}\\
= L^{\mu\nu}_A+S^{\mu\nu}.
$$
A rather useful combination is the totally antisymmetric Pauli-Lubanski tensor (invented by Myron Matthisson)
$$
S^{\lambda\mu\nu}= p^\lambda S^{\mu\nu}+ p^\mu S^{\nu\lambda}+p^\nu S^{\lambda\mu}.
$$
This object turns out to be independendent of the choice of $x_A$ and is also sometimes called the intrinsic spin.
