Stress tensor vanishes on cylinder edge

A few comments:1. I agree with Chet that the students probably haven't seen the stress equilibrium equations. So it is nice to derive them from the tensor.2. I disagree with Chet on one point: let's not get into the habit of doing problems in one coordinate system. Physics is coordinate-free, and problems should be solved in the most convenient coordinate system. In this case, cylindrical is probably easier because the boundary conditions are stated in that coordinate system.3. I have no idea why the problem statement gives a tensor in terms of x2 and x3. In cylindrical coordinates, x2 and x3 are really angles, and the tensor should not depend on angles. The tensor should only
  • #1
jasonmcc
10
0

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
 
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  • #2
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 :)
 
  • #3
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.
 
  • #4
Chestermiller said:
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?
 
  • #5
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).
 
  • #6
BruceW said:
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.
 
  • #7
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:



[itex]0[/itex][tex]\frac{x_2}{\sqrt{x_2^2+x_3^2}}[/tex]
[tex]\frac{x_3}{\sqrt{x_2^2+x_3^2}}[/tex]

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.
 
  • #8
Chestermiller said:
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:



[itex]0[/itex][tex]\frac{x_2}{\sqrt{x_2^2+x_3^2}}[/tex]
[tex]\frac{x_3}{\sqrt{x_2^2+x_3^2}}[/tex]

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*}
 
  • #10
Thanks for all the help and action! I was traveling and away from my computer; great to see the discussion.
 

1. What is a stress tensor?

A stress tensor is a mathematical representation of the internal forces and stresses acting within a material or system. It is a second-order tensor that describes the distribution of forces and stresses at a point in a material or system.

2. What does it mean for the stress tensor to vanish on a cylinder edge?

When the stress tensor vanishes on a cylinder edge, it means that there are no internal forces or stresses acting on that point of the cylinder. This can occur in certain situations where the geometry or loading conditions of the cylinder result in a balanced distribution of forces and stresses.

3. Why is it important for the stress tensor to vanish on a cylinder edge?

This condition is important because it ensures that the cylinder is in a state of equilibrium and there are no unbalanced forces or stresses acting on it. It also allows for easier analysis and calculations of the stresses and forces within the cylinder.

4. Are there any real-world applications of the stress tensor vanishing on a cylinder edge?

Yes, this concept is commonly used in engineering and physics, particularly in the design and analysis of structures such as pipes and pressure vessels. It is also relevant in fluid dynamics, where the stress tensor vanishing on the surface of a flow domain can indicate a state of no shear stress.

5. Can the stress tensor vanish on other shapes besides a cylinder edge?

Yes, the stress tensor can also vanish on other curved surfaces, such as the edge of a sphere or a torus. It can also occur on flat surfaces in certain scenarios, such as a plate under certain loading conditions.

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