Contradiction between Force and Torque equations A thin uniform rod of mass $M$ and length $L$ and cross-sectional area $A$, is free to rotate about a horizontal axis passing through one of its ends (see figure).

What is the value of shear stress developed at the centre of the rod, immediately after its release from the horizontal position shown in the figure?
Firstly, we can find the angular acceleration $\alpha$ of the rod by applying the Torque equation about hinge point A as follows:
$$ \frac{MgL}{2} = \frac{ML^2}{3} \alpha$$
$$\alpha = \frac{3g}{2L}$$
So the acceleration of the centre of the rod equals $\alpha \cdot \frac{L}{2} = \frac{3g}{4}$, Hence the hinge force $F_H = \frac{Mg}{4}$
Now consider an imaginary cut at the centre of the rod, dividing it into two halves. To account for the effect of one half on the other, we can add a shear force $F$ acting tangentially to the cross-section area on each half.

Now focusing on the left-most half, the acceleration of its centre of mass should be equal to $\alpha$ times its distance from hinge point A
So, $a_{cm} = {\frac{L}{4}} \cdot {\frac{3g}{2L}} = \frac{3g}{8}$
Keeping in mind that the mass of the left half is $\frac{M}{2}$ and applying force equation, we get the following:
$$\frac{Mg}{2} + F - F_H = \frac{3Mg}{16}$$
$$F - F_H = \frac{-5Mg}{16}$$
$$F = \frac{Mg}{4} - \frac{5Mg}{16} = -\frac{Mg}{16}$$
But if we apply the Torque equation about hinge point A, for the left half:
$$\frac{Mg}{2} \cdot \frac{L}{4} + F \cdot \frac{L}{2} = \frac{\left(\frac{M}{2}\right)\left(\frac{L}{2}\right)^2}{3} \cdot \alpha $$
$$ \frac{MgL}{8} + \frac{FL}{2} = \frac{ML^2}{24} \cdot \frac{3g}{2L}$$
$$ \frac{FL}{2} = \frac{MgL}{16} - \frac{MgL}{8} = -\frac{MgL}{16}$$
$$F = -\frac{Mg}{8}$$
Why is there a contradiction?

I think you can obtain the results like this (your approach)

I )   obtain    $~F_H~,a_{CM}~,\alpha~$ with
sum of the forces at the center of mass
$$M\,a_{CM}=M\,g-F_H$$
sum of the torques at the center of mass
$$I_{CM}\,\alpha=\frac{F_H\,L}{2}$$
and the kinematic equation
$$\tan(\alpha)\approx\alpha=\frac{2\,a_{CM}}{L}$$
with $~I_{CM}=\frac{M\,L^2}{12}$
$\Rightarrow$
$$F_H= \frac{M\,g}{4}~,a_{CM}=\frac{3\,g}{4}~,\alpha=\frac{3\,g}{2\,L}$$
II) for the left side where you „cut“ the rod

sum of the forces at the center of mass G
\begin{align*}
 &\frac{M\,a_{G}}{2}=\frac{M\,g}{2}-F_H-F_C\tag 1
\end{align*}
sum of the torques at the center of mass
\begin{align*}
I_{G}\,\alpha=\frac{F_H\,L}{4}-\frac{F_C\,L}{4}-\tau_C \tag 2
\end{align*}
and the kinematic equation
$$\frac{a_{CM}}{\frac L2}=\frac{a_{G}}{\frac L4}\quad\Rightarrow a_G=\frac 38\,g$$
with $~I_G~=\frac{1}{12}\frac{M}{2}\left(\frac{L}{2}\right)^2~$ and equation (1), (2) you obtain
\begin{align*}
 &F_C=\frac{1}{16}\,M\,g\quad,\tau_C=\frac{1}{32}\,M\,g\,L
\end{align*}
