Here's a few examples
POVMs with two components
These are POVMs involving only two matrices $M_1,M_2\ge0$ with $M_1+M_2=I$.
This implies that they are mutually diagonalisable, as discussed e.g. in this question. Using their (mutual) eigenvectors as basis, we can therefore always write them as
$$\newcommand{\on}[1]{{\operatorname{#1}}} M_1 = \on{diag}(s_1,..., s_N),
\qquad M_2 = \on{diag}(1-s_1,...,1-s_N),$$
where $N$ is the dimension of the underlying space and $s_i\in[0,1]$.
Note that $M_i$ are not, in general, projections.
For a concrete example, consider
$M_1 = M_2 = \on{diag}(1/2,1/2). %\begin{pmatrix}1/2 & 0 \\ 0 & 1/2\end{pmatrix}.$
Of course, this represents a completely useless measurement, as $\on{Tr}(\rho M_i)=1/2$ for all $\rho$.
POVMs that are sum of projectors
Suppose $N=3$, $\newcommand{\bs}[1]{\boldsymbol{#1}}\newcommand{\PP}{\mathbb P}\newcommand{\ketbra}[1]{\lvert #1\rangle\!\langle #1\rvert}$ and consider $M_1=\ketbra1+\ketbra2$ and $M_2=\ketbra3$. This is a POVM that's also not sum of trace-1 projectors.
You can also show that, if $N=2$, the only POVMs that are sum of projectors have the form $\{\PP(\bs u),\PP(-\bs u)\}$, where $\PP(\bs u)\equiv \frac12 (I+\bs u\cdot\bs \sigma)$ projects onto the $+1$ eigenspace of $\bs u\cdot\bs \sigma\equiv\sum_k u_k \sigma_k$, with $\sigma_k$ the Pauli matrices and $u\in\mathbb R^3$ with $\|\bs u\|=1$.
This can be seen from the above observation about diagonalisability. Another proof is found around page 19 of these notes by Nicholas Wheeler (Link to pdf).
Still considering the qubit case $N=2$, we can more generally take a set of $n$ vectors $\mathbf a_i$ such that $\sum_i \mathbf a_i=0$ and $\|\mathbf a_i\|\le1$, and then
$$\sum_i \frac{2}{n}\PP(\mathbf a_i) = I,$$
so that $\big\{\frac{2}{n}\PP(\mathbf a_i)\big\}_{i=1}^n$ is an $n$-element qubit POVM. This is also shown in Wheeler's notes linked above.
A question related to this is Are three POVM measurements on a single qubit physically realizable?.
State discrimination
Another common scenario is unambiguous state discrimination. Given two pure states $\newcommand{\ket}[1]{\lvert #1\rangle}\ket\psi,\ket\varphi$, the optimal POVM to distinguish between them is
$$M_1 = \frac{1}{1+\lvert\langle\varphi|\psi\rangle\rvert}\ketbra{\varphi^\perp},
\quad
M_2 = \frac{1}{1+\lvert\langle\varphi|\psi\rangle\rvert}\ketbra{\psi^\perp},
\quad
M_3 = I - M_1 - M_2.
$$
On the other hand, the POVM that optimally distinguishes between two states, uses $M_1=\sum_{i: \lambda_i>0}\mathbb{P}(|\lambda_i\rangle)$, $M_2=\sum_{i: \lambda_i<0}\mathbb{P}(|\lambda_i\rangle)$, where $|\lambda_i\rangle$ is the eigenstate with eigenvalue $\lambda_i$ of $\mathbb{P}(|\psi\rangle)-\mathbb{P}(|\phi\rangle)$, where $|\psi\rangle,|\phi\rangle$ are the states we are trying to discriminate.
This is also discussed in What is the Helstrom measurement?.
For example, to optimally discriminate between $|0\rangle$ and $|+\rangle$, we get
$$M_1 = \frac1{2(2-\sqrt2)}\mathbb{P}((-1+\sqrt2)|0\rangle+|1\rangle)
= \frac1{2(2-\sqrt2)} \begin{pmatrix}3-2\sqrt2 & \sqrt2-1\\\sqrt2-1&1\end{pmatrix}, \\
M_2 = \frac1{2(2+\sqrt2)}\mathbb{P}((-1-\sqrt2)|0\rangle+|1\rangle)
= \frac1{2(2+\sqrt2)} \begin{pmatrix}3+2\sqrt2 & -\sqrt2-1\\-\sqrt2-1&1\end{pmatrix}.
$$
Example without any specific structure
A generic example with $N=3$, given in this answer, is
$$
M_1 =\frac{1}{9}
\begin{pmatrix}
3 & 2 & -1\\
2 & 3 & -1\\
-1 & -1 & 3\\
\end{pmatrix},
\quad
M_2 =\frac{1}{9}
\begin{pmatrix}
3 & -1 & 2\\
-1 & 3 & -1\\
2 & -1 & 3\\
\end{pmatrix},
\quad
M_3 =\frac{1}{9}
\begin{pmatrix}
3 & -1 & -1\\
-1 & 3 & 2\\
-1 & 2 & 3\\
\end{pmatrix}.
$$
