Here's one. Let $\mathbb{D}$ be the monoidal category of finite ordinals. Thus, the objects are the natural numbers (including 0), a map $m \to n$ is an order-preserving function $\{1, \ldots, m\} \to \{1, \ldots, n\}$, and the monoidal structure is addition. The object $1$ is a monoid in $\mathbb{D}$, in a unique way. This makes $T = 1 + (-)$ into a monad on $\mathbb{D}$.

I claim that $T$ admits no strength. A strength on $T$ would consist of a map
$$
t_{m, n}\colon m + 1 + n \to 1 + m + n
$$
for each $m$ and $n$, satisfying some axioms. Readers might wish to stop reading here, because perhaps it's clear that no sensible such $t$ can exist (bearing in mind that maps have to be order-preserving). But ploughing on:

$t_{0, 0}$ must be the identity map on $1$, and the naturality square for the unique maps $0 \to m$ and $0 \to n$ then tells us that $t_{m, n}\colon m + 1 + n \to 1 + m + n$ must send the copy of $1$ in the domain to the copy of $1$ in the codomain.

On the other hand, the unit axiom (i.e. the second triangle on the Wikipedia page) tells us that $t_{m, n}\colon m + 1 + n \to 1 + m + n$ must send each element of $m$ in the domain to the corresponding element of $m$ in the codomain. So, for instance, $t_{1, 0}\colon 1 + 1 \to 1 + 1$ is the non-identity bijection. This is not order-preserving — contradiction.

(Why did I think of this example? Because I wanted to find the most generic possible example of a category equipped with a monad. Well, the initial category equipped with a monad is the empty category, which clearly isn't going to answer your question, so I wanted the free category equipped with a monad and an object. This is exactly $\mathbb{D}$, equipped with the monad $T$ and the object $0$.)