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?
Best Answer
Your initial calculations are correct. The pin force is indeed $\tfrac{m g}{4} $ a fact that kind of surprised me the first time I encountered this problem.
Your idealization on the second part is where things were missed. I am using the sketch below, and I am counting positive directions as downwards (same as gravity) and positive angles as clock-wise. Notice each half-bar has mass $m/2$ and mass moment of inertia about its center of mass $ \tfrac{1}{12} \left( \tfrac{m}{2} \right) \left( \tfrac{\ell}{2} \right)^2 = \tfrac{m \ell^2}{96}$
Let's look at the equations of motion for the two half-bars as they are derived from the free body diagrams.
$$ \begin{aligned} \tfrac{m}{2} a_G & = \tfrac{m}{2} g - F_C - F_A \\ \tfrac{m \ell^2}{96} \alpha & = -\tau_C - \tfrac{\ell}{4} F_C + \tfrac{\ell}{4} F_A \\ \tfrac{m}{2} a_H & = \tfrac{m}{2} g + F_C \\ \tfrac{m \ell^2}{96} \alpha & = \tau_C - \tfrac{\ell}{4} F_C \\ \end{aligned} $$
And consider the kinematics, where it all acts like a rotating rigid bar, with point accelerations $a_G = \tfrac{\ell}{4} \alpha$ and $a_H = \tfrac{3 \ell}{4} \alpha$
The solution to the above 4×4 system of equations is
$$ \begin{aligned} F_A & = \tfrac{m g}{4} & F_C & = \tfrac{m g}{16} \\ \alpha & = \tfrac{3 g}{2 \ell} & \tau_C &= \tfrac{m g \ell}{32} \end{aligned} $$
I think because you did not account for the torque transfer $\tau_C$ between the half-bars, you got $F_C = \tfrac{m g}{8}$ which is incorrect.
Note, I used PowerPoint and IguanaTex plugin for the sketches.