Equilibrium equation for forces in tangential direction

[roldy]

Homework Statement

Derive the differential equilibrium equation in the tangential direction $\theta$

Homework Equations

$\sum F_\theta = 0$

$V = 1/2(2r + \Delta r)\Delta r \Delta \theta$

Small angle approximation:
$\cos(\frac{\Delta \theta}{2}) = 1$

$\sin(\frac{\Delta \theta}{2}) = \frac{\Delta \theta}{2}$

The Attempt at a Solution

$(\sigma_{\theta \theta} + \Delta \sigma_{\theta \theta})\cos(\frac{\Delta \theta}{2})\Delta r(1) - \sigma_{\theta \theta}\cos(\frac{\Delta \theta}{2})\Delta r(1) + (\sigma_{r\theta} + \Delta \sigma_{r\theta})\sin(\frac{\Delta \theta}{2})\Delta r(1) - \sigma_{r\theta}\sin(\frac{\Delta \theta}{2})\Delta r(1) + (\sigma_{r\theta} + \Delta\sigma_{r\theta})(r + \Delta r)\Delta \theta - \sigma_{r\theta}r\Delta \theta + \Theta V = 0$

Simplifying:
$\Delta \sigma_{\theta \theta} \Delta r + \Delta \sigma_{r \theta} \frac{\Delta \theta}{2} \Delta r + \sigma_{r \theta} \Delta r \Delta \theta + \Delta \sigma_{r \theta} r \Delta \theta + \Delta \sigma_{r \theta}\Delta r \Delta \theta + 1/2\Theta(2r + \Delta r)\Delta r \Delta \theta = 0$


Divide by $\Delta r \Delta \theta$

$\frac{\Delta \sigma_{\theta \theta}}{\Delta \theta} + \frac{\Delta \sigma_{r \theta}}{2} + \sigma_{r \theta} + \frac{\Delta \sigma_{r \theta}r}{\Delta r} + \Delta \sigma_{r \theta} + 1/2\Theta(2r + \Delta r) = 0$

Take the limit as $\Delta r$ approaches 0 and $\Delta \theta$ approaches 0

$\frac{\partial \sigma_{\theta \theta}}{\partial \theta} + (?) + \sigma_{r \theta} + \frac{\partial \sigma_{r \theta}r}{\partial r} + (?) + r\Theta = 0$

$\frac{1}{r} \frac{\partial \sigma_{\theta \theta}}{\partial \theta} + \frac{1}{r}(?) + \frac{1}{r} \sigma_{r \theta} + \frac{\partial \sigma_{r \theta}}{\partial r} + \frac{1}{r}(?) + \Theta = 0$

I searched online for the correction equation and it is as follows:

$\frac{1}{r} \frac{\partial \sigma_{\theta \theta}}{\partial \theta} + \frac{\partial \sigma_{r \theta}}{\partial r} + \frac{2\sigma_{r \theta}}{r} + \Theta = 0$

I'm not sure what to do about the terms with no $\Delta$ in the denominator when I take the limit. Attached is the figure I'm working from.

 

[mighty2000]

Thank you for posting your question on the derivation of the differential equilibrium equation in the tangential direction. I have reviewed your attempt at the solution and would like to offer some corrections and clarifications.

Firstly, in your attempt, you have used the small angle approximation for both \cos(\frac{\Delta \theta}{2}) and \sin(\frac{\Delta \theta}{2}). However, this is not necessary as the small angle approximation is typically used when the angle is very small, which is not the case in this problem. It is also important to note that in this problem, we are dealing with a differential equation and not a static equilibrium equation, so the small angle approximation is not applicable.

Secondly, in your attempt, you have used the symbol \sigma to represent the normal stress and \sigma_{r \theta} to represent the shear stress. However, in the context of this problem, it is more appropriate to use the symbols \sigma_{\theta \theta} for the normal stress and \tau_{r \theta} for the shear stress. This is because we are dealing with stresses in the tangential direction (\theta direction) and not in the radial direction (r direction).

Now, let's move on to the actual derivation of the differential equilibrium equation in the tangential direction. The first step is to write the equilibrium equation in the tangential direction, which is given by:

\sum F_\theta = 0

This equation states that the sum of all the forces acting in the tangential direction must equal zero for the body to be in equilibrium. Now, let's break this equation down into its components.

The first component is the normal stress acting in the tangential direction, which is given by \sigma_{\theta \theta}. This stress is multiplied by the area over which it acts, which is \Delta r \Delta \theta. So, the first term in the equilibrium equation is:

\sigma_{\theta \theta} \Delta r \Delta \theta

The second component is the shear stress acting in the tangential direction, which is given by \tau_{r \theta}. This stress is also multiplied by the area over which it acts, which is \Delta r \Delta \theta. However, we need to take into account the fact that the shear stress acts over two surfaces, so we need to multiply it by 2. So, the second term in the equilibrium equation is