Stress tensor vanishes on cylinder edge

[jasonmcc]

Homework Statement

Given a cylinder in the Ox1x2x3 coordinate system, such that x1 is in the Length direction and x2 and x3 are in the radial directions. The stress components are given by the tensor
$$
[T_{ij}] = \begin{bmatrix}Ax_2 + Bx_3 & Cx_3 & -Cx_2 \\ Cx_3 & 0 & 0 \\ -C_2 & 0 & 0\end{bmatrix}
$$
(a) Verify that in the absence of body forces the equilibrium equations are satisfied.
(b) Show that the stress vector vanishes at all points on the curved surface of the cylinder

(by the way I'm pretty sure that the T_31 is a typo and should be -Cx_2)

Homework Equations

The Attempt at a Solution

(a) is simple enough, I think:
$$
\sum \vec{F} = 0 = \nabla \cdot \mathbb{T} = \begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}
$$

However I have no idea what they mean or how to go about answering (b). If you did a diagram of an element on the edge, I think you could show there is no shear, but there would still be the compression represented by T_11

Thanks

 

[BruceW]

For part b, the question is asking about the stress vector, not the stress tensor. So you need to use the definition of the stress vector.

p.s. welcome to physicsforums :)

 

[Chestermiller]

The equation that you wrote down is equivalent to the three stress-equilibrium equations. Do you know how to write the stress-equilibrium equations in component form?

For part B, do you know how to write down the components of a normal to the cylinder surface in terms of x₂ and x₃. To get the stress vector on the cylinder surface, you need to take the dot product of this normal vector with the stress tensor. I'm sure you have been learning how to do this. Right? Incidentally, T_11 will not be involved in the components of this stress vector.

 

[Dustinsfl]

The equation that you wrote down is equivalent to the three stress-equilibrium equations. Do you know how to write the stress-equilibrium equations in component form?

For part B, do you know how to write down the components of a normal to the cylinder surface in terms of x₂ and x₃. To get the stress vector on the cylinder surface, you need to take the dot product of this normal vector with the stress tensor. I'm sure you have been learning how to do this. Right? Incidentally, T_11 will not be involved in the components of this stress vector.

I am working on this same problem and obtained the normal vector to be
$$
\begin{bmatrix}
0\\
r\cos\theta\\
-r\sin\theta
\end{bmatrix}
$$
Then when I take T.n, I get
$$
\begin{bmatrix}
Crx_3\cos\theta + Crx_2\sin\theta\\
0\\
0
\end{bmatrix}
$$
How do I get this identically equal to 0 now?

 

[BruceW]

you're close now. Just plug in the values of x2 and x3. you have define the normal vector to be (0,rcos(theta),-rsin(theta)) So this should tell you what x2 and x3 should be. (Think about how the normal vector depends on the position it is evaluated at, on the curved surface of the cylinder).

 

[Dustinsfl]

you're close now. Just plug in the values of x2 and x3. you have define the normal vector to be (0,rcos(theta),-rsin(theta)) So this should tell you what x2 and x3 should be. (Think about how the normal vector depends on the position it is evaluated at, on the curved surface of the cylinder).

If we want zero, then $x_2 = \cos\theta$ and $x_3 = -\sin\theta$, but I just picked that by inspection and not really knowing why.

 

[Chestermiller]

You can't mix cartesian and cylindrical coordinates. You must work in terms of one coordinate system or the other. In cartesian coordinates, the unit normal vector to the cylinder has components:



$0$$$\frac{x_2}{\sqrt{x_2^2+x_3^2}}$$
$$\frac{x_3}{\sqrt{x_2^2+x_3^2}}$$

See if you can figure out why these are the components. Then carry out the dot product with the stress tensor, and see what you get.

 

[Dustinsfl]

You can't mix cartesian and cylindrical coordinates. You must work in terms of one coordinate system or the other. In cartesian coordinates, the unit normal vector to the cylinder has components:



$0$$$\frac{x_2}{\sqrt{x_2^2+x_3^2}}$$
$$\frac{x_3}{\sqrt{x_2^2+x_3^2}}$$

See if you can figure out why these are the components. Then carry out the dot product with the stress tensor, and see what you get.

I got there but I had to take the scenic route through polar.

In the $(r,\theta)$ system, we have
\begin{alignat*}{3}
x_2(r,\theta) & = & r\sin\theta\\
x_3(r,\theta) & = & r\cos\theta\\
x_1(r,\theta) & = & x\\
\end{alignat*}
Then a normal vector on the cylinder in $(r,\theta)$ can be found:
$$
\mathbf{v} = \begin{vmatrix}
\hat{\mathbf{x}}_1 & \hat{\mathbf{x}}_2 & \hat{\mathbf{x}}_3\\
1 & 0 & 0\\
0 & r\sin\theta & r\cos\theta
\end{vmatrix} = \langle 0,r\cos\theta,r\sin\theta\rangle
$$
Now, let's normalize $\mathbf{v}$ and write it back in the $x_1, x_2$, and $x_3$ coordinate system.
$$
\hat{\mathbf{n}} = \begin{bmatrix}
0\\
\frac{x_2}{\sqrt{x_2^2 + x_3^2}}\\
\frac{x_3}{\sqrt{x_2^2 + x_3^2}}
\end{bmatrix}
$$
Let's now take $\hat{\mathbf{n}}^T\cdot\mathbb{T}$.
\begin{alignat*}{3}
\hat{\mathbf{n}}^T\cdot\mathbb{T} & = & \begin{bmatrix}
0 & \frac{x_2}{\sqrt{x_2^2 + x_3^2}} & \frac{x_3}{\sqrt{x_2^2 + x_3^2}}
\end{bmatrix}\begin{bmatrix}
Ax_2 + Bx_3 & Cx_3 & -Cx_2\\
Cx_3 & 0 & 0\\
-Cx_2 & 0 & 0
\end{bmatrix}\\
& = & \begin{bmatrix}
\frac{Cx_3x_2}{\sqrt{x_2^2 + x_3^2}} - \frac{Cx_2x_3}{\sqrt{x_2^2 + x_3^2}}\\
0\\
0
\end{bmatrix}\\
& = & \begin{bmatrix}
0\\
0\\
0
\end{bmatrix}
\end{alignat*}

 

[Chestermiller]

Excellent job.

Chet

 

[jasonmcc]

Thanks for all the help and action! I was traveling and away from my computer; great to see the discussion.