Suppose $F:\mathcal C\to\mathcal D$ admits a left adjoint $F^L:\mathcal D\to\mathcal C$. Then
$$ T=FF^L:\mathcal D\to\mathcal D $$
is a monad, and we have a canonical comparison functor
$$ \mathcal C\longrightarrow \operatorname{Alg}_T(\mathcal D). $$
The Barr–Beck monadicity theorem gives sufficient conditions for this comparison to be an equivalence:
For ordinary categories, a common formulation of the second condition is that $\mathcal C$ admits coequalizers of $F$-split pairs, and $F$ preserves these coequalizers.
We give an explicit example where $F$ is conservative and admits a left adjoint, but fails the second condition.
Let
$$ \mathcal C=\mathrm{Ab}_{\mathrm{tf}}, \qquad \mathcal D=\mathrm{Set}, $$
where $\mathrm{Ab}_{\mathrm{tf}}$ denotes the category of torsion-free abelian groups.
Consider the forgetful functor
$$ F:\mathrm{Ab}_{\mathrm{tf}}\longrightarrow\mathrm{Set}. $$
Claim 1. The functor $F$ is conservative and admits a left adjoint.
Indeed:
The associated monad
$$ T=FF^L $$
is the usual free abelian group monad on $\mathrm{Set}$. Consequently,
$$ \operatorname{Alg}_T(\mathrm{Set})\simeq\mathrm{Ab}. $$
In particular,
$$ \mathrm{Ab}_{\mathrm{tf}} \not\simeq \operatorname{Alg}_T(\mathrm{Set}) $$
via the canonical comparison functor: its essential image consists only of torsion-free groups.
Thus $F$ is not monadic. Let us see explicitly how the second Barr–Beck condition fails.
Consider the following pair of morphisms in $\mathrm{Ab}_{\mathrm{tf}}$:
$$ \mathbb Z^2 \underset{g}{\overset{f}{\rightrightarrows}} \mathbb Z, $$
where
$$ f(a,b)=a+2b, \qquad g(a,b)=a. $$
Coequalizer in $\mathrm{Set}$.
The coequalizer identifies all integers differing by an even number:
$$ a+2b\sim a. $$
Therefore,
$$ \operatorname{coeq}_{\mathrm{Set}}(f,g) = \mathbb Z/2\mathbb Z $$
as a set with two elements.
Coequalizer in $\mathrm{Ab}_{\mathrm{tf}}$.
In the category of all abelian groups, the coequalizer would be
$$ \operatorname{coeq}_{\mathrm{Ab}}(f,g) = \mathbb Z/2\mathbb Z. $$
But $\mathbb Z/2\mathbb Z$ is not torsion-free.
In fact, the coequalizer in $\mathrm{Ab}_{\mathrm{tf}}$ is the zero group:
$$ \operatorname{coeq}_{\mathrm{Ab}_{\mathrm{tf}}}(f,g)=0. $$
To verify this, suppose $A$ is torsion-free and $h:\mathbb Z\to A$ satisfies $hf=hg$. Then
$$ 2h(1)=0. $$
Since $A$ is torsion-free, this forces $h(1)=0$, hence $h=0$.
Thus the zero group has the required universal property.
We obtain
$$ F\left(\operatorname{coeq}_{\mathcal C}(f,g)\right) = \{*\}, $$
whereas
$$ \operatorname{coeq}_{\mathcal D}(Ff,Fg) = \{0,1\}. $$
Therefore $F$ does not preserve this coequalizer.
However, Barr–Beck does not require preservation of all coequalizers, only those which become split coequalizers after applying $F$.
We now verify that this pair is indeed $F$-split.
Recall that a split coequalizer consists of
$$ A \underset{g}{\overset{f}{\rightrightarrows}} B \xrightarrow{q} Q $$
together with maps
$$ s:Q\to B, \qquad t:B\to A, $$
satisfying
$$ qs=\mathrm{id}_Q, \qquad ft=\mathrm{id}_B, \qquad gt=sq. $$
These maps are not required to preserve any additional algebraic structure if we work in $\mathrm{Set}$.
In our example, let
$$ q:\mathbb Z\to\{0,1\} $$
be reduction modulo $2$.
Define a section
$$ s:\{0,1\}\to\mathbb Z, \qquad s(0)=0,\quad s(1)=1. $$
Set $r(n)=s(q(n))$, so that $r(n)\in\{0,1\}$ is the parity representative of $n$.
Now define
$$ t:\mathbb Z\to\mathbb Z^2, \qquad t(n)= \left( r(n), \frac{n-r(n)}{2} \right). $$
We check the identities:
$$ \begin{aligned} q(s(i))&=i,\cr f(t(n))&=r(n)+2\frac{n-r(n)}2=n,\cr g(t(n))&=r(n)=s(q(n)). \end{aligned} $$
Consequently,
$$ qs=\mathrm{id}, \qquad ft=\mathrm{id}, \qquad gt=sq. $$
This proves that the coequalizer is split in $\mathrm{Set}$.
Conclusion. We have found an $F$-split pair in $\mathcal C$ whose coequalizer is not preserved by $F$.
Thus the second condition of the Barr–Beck theorem genuinely fails.
The essential point is that the splitting maps $s,t$ are set maps, not homomorphisms of abelian groups.
We can express the same obstruction using geometric realizations of simplicial objects, as in the higher-categorical formulation of Barr–Beck.
Consider the simplicial object $X_\bullet$ in $\mathrm{Ab}_{\mathrm{tf}}$ defined by
$$ X_n= \left\{ (a_0,\ldots,a_n)\in\mathbb Z^{n+1} \ \middle| a_i\equiv a_j\pmod 2 \text{ for all }i,j \right\}. $$
Each $X_n$ is a subgroup of $\mathbb Z^{n+1}$ and hence is torsion-free.
The face maps delete coordinates, and the degeneracy maps repeat coordinates.
In low degrees,
$$ X_0=\mathbb Z, $$
$$ X_1= \{(a,b)\in\mathbb Z^2:a\equiv b\pmod 2\}. $$
The maps $d_0,d_1:X_1\rightrightarrows X_0$ are the two coordinate projections (in either order).
After applying $F$.
The simplicial set $F(X_\bullet)$ is the nerve of the equivalence relation
$$ a\sim b \quad\Longleftrightarrow\quad a\equiv b\pmod 2. $$
Equivalently, it is the nerve of a groupoid with two connected components, consisting of even and odd integers, with a unique arrow between any two objects in the same component.
Its geometric realization in $\mathrm{Set}$ is
$$ {|}F(X_\bullet){|}_{\mathrm{Set}} = \{0,1\}. $$
Moreover, the augmentation
$$ F(X_\bullet)\longrightarrow\{0,1\} $$
is split as an augmented simplicial object.
Indeed, we can choose the representatives $0$ and $1$ in the two components. An extra degeneracy is given by
$$ s_{-1}(a_0,\ldots,a_n) = (r(a_0),a_0,\ldots,a_n), $$
where $r(a_0)$ is the parity representative.
This gives an explicit simplicial contraction over the two-element set.
Before applying $F$.
The geometric realization in $\mathrm{Ab}_{\mathrm{tf}}$ is the coequalizer of the two face maps
$$ X_1\rightrightarrows X_0. $$
Their difference has image $2\mathbb Z$, so the coequalizer in ordinary abelian groups is $\mathbb Z/2\mathbb Z$.
Passing to the torsion-free category kills this quotient. Therefore,
$$ {|}X_\bullet{|}_{\mathrm{Ab}_{\mathrm{tf}}}=0. $$
It follows that
$$ F({|}X_\bullet{|}) \not\cong {|}F(X_\bullet){|}. $$
We have thus exhibited an explicit $F$-split simplicial object whose geometric realization is not preserved by $F$.
For an adjunction
$$ F^L:\mathcal D \rightleftarrows \mathcal C:F, $$
the monad $T=FF^L$ describes how free objects in $\mathcal C$ interact.
Monadic reconstruction attempts to recover an arbitrary object $A\in\mathcal C$ from a simplicial resolution by free objects, of the form
$$ \cdots \rightrightarrows F^L T^2 F(A) \rightrightarrows F^L T F(A) \rightrightarrows F^L F(A). $$
After applying $F$, the augmented resolution is split, and its realization recovers $F(A)$.
To recover $A$ itself, one needs $F$ to preserve the relevant geometric realization. Conservativity can then detect that the resulting map is an isomorphism.
Thus the two Barr–Beck conditions have different roles:
In the torsion-free abelian group example, a quotient of free objects may create torsion. The category $\mathrm{Ab}_{\mathrm{tf}}$ cannot retain that torsion and instead replaces the quotient by its torsion-free reflection.
However, the monad $FF^L$ on $\mathrm{Set}$ is the ordinary free abelian group monad, which remembers no such torsion-free restriction.
Consequently,
$$ \operatorname{Alg}_{FF^L}(\mathrm{Set}) \simeq\mathrm{Ab} $$
rather than $\mathrm{Ab}_{\mathrm{tf}}$.
This is a concrete example showing why conservativity and the existence of a left adjoint do not suffice for monadic reconstruction.