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. rod_2

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?

  • 1
    $\begingroup$ Did you forget the bending moment between the two halves? $\endgroup$ Commented Jul 15, 2022 at 17:22

2 Answers 2


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.

  • 2
    $\begingroup$ The perennial lesson: make sure to draw a complete free-body diagram instead of just labeling loads on the original geometry. $\endgroup$ Commented Jul 15, 2022 at 18:12
  • $\begingroup$ Definitely, especially in dynamics, it is quite easy to miss things or to flip a sign somewhere. Draw your FBDs kids, and label which directions are positive. $\endgroup$ Commented Jul 15, 2022 at 18:14
  • $\begingroup$ Thank you, I've understood my error and the right approach. But one thing that still troubles me, why is it right to take the point of action of $F_C$ to be the centre of the rod? I concluded it would be some point below the centre, but I couldn't confirm as that was purely based on my visualization. $\endgroup$
    – Sankalp
    Commented Jul 16, 2022 at 3:30
  • $\begingroup$ Or is it that the point of action does not matter here, as the bar is thin (and would matter for a bar with significant dimensions)? $\endgroup$
    – Sankalp
    Commented Jul 16, 2022 at 4:27
  • $\begingroup$ @Sankalp, it does not matter how you split it up. Taking the middle point makes it easier as you have the same mass and mmoi on both sides. $\endgroup$ Commented Jul 16, 2022 at 14:32

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

enter image description here I ) obtain $~F_H~,a_{CM}~,\alpha~$ with

sum of the forces at the center of mass


sum of the torques at the center of mass


and the kinematic equation


with $~I_{CM}=\frac{M\,L^2}{12}$


$$F_H= \frac{M\,g}{4}~,a_{CM}=\frac{3\,g}{4}~,\alpha=\frac{3\,g}{2\,L}$$

II) for the left side where you „cut“ the rod

enter image description here

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*}


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