3D gradient, X grad Δy/Δx, Z grad Δy/Δz what is Y grad?

This is very confusing.
If w = f(x, y, z), then ##∇w = <\frac{\partial w}{\partial x}, \frac{\partial w}{\partial y}, \frac{\partial w}{\partial z}>##. What function are you working with?

I don't know what this means.

 

What does endY - beginY (or endY/beginY, whatever) mean?

In the equation w = f(x, y, z), w is the dependent variable in what I'm showing as a function, written using normal notation. I'm trying to get you to write something meaningful.

 

Still not clear. Mathematically, a cylinder extends infinitely far along its central axis. For example, x² + z² = 1 is a right circular cylinder of radius 1, with its central axis lying along the y-axis. When you stretch the cylinder, do you mean lengthwise (along the axis) or laterally, so that its cross-sections deform to ellipses?

BTW, the word "gradient" has different meanings in American English vs. British English. Here in America we talk about the slope of a line, and reserve gradient to mean the vector of partial derivatives of a function of two or more variables.

 

You are trying to describe the deformation of a cylindrical surface. Does the surface stretch axially? Does the surface stretch radially. Is there angular shear? Is the deformation axially symmetric? Or, is the deformation totally arbitrary?

Chet

 

Is it just the axis of the cylinder that is being displaced (by shearing the cylinder like a deck of cards), or is there more being done, so that the circles don't remain circles, or the radius of the circles increases, or the vertical distance between the circles gets larger?

Chet

 

Is this a solid cylinder or a thin tube?
chet

 

You are trying to determine the deformation gradient tensor for the deformation you illustrated in post #9, correct?

Chet

 

Well, if I understand the problem correctly, I'll try to lead you through the solution.

Let x₀, y₀, and z₀ represent the coordinates of a material point on the cylinder before the cylinder has been deformed, and let x, y, and z be the coordinates of the same material point on the cylinder after the cylinder has been deformed. If θ represents the angle that the axis of the cylinder makes with the z (upward) direction in the deformed configuration of the cylinder, please express x, y, and z in terms of x₀, y₀, and z₀ and θ. (Assume that, in the deformed configuration), the axis is tilted in the x direction.

Chet

 

Well, this is not what I have in mind, but if you are happy with this solution, then I am.

Chet

 

[tex]x=x_0+tan\theta z_0[/tex]
[tex]y=y_0[/tex]
[tex]z=z_0[/tex]
So, if two neighboring material points in the original configuration of the body are located at x₀ and x₀+dx₀, y₀ and y₀+dy₀, and z₀ and z₀+dz₀, these same two neighboring material points in the deformed configuration of the body are located at x and x+dx, y and y+dy, and z and z+dz, with
[tex]dx=dx_0+\tanθ dz_0[/tex]
[tex]dy=dy_0[/tex]
[tex]dz=dz_0[/tex]

So,
[tex]\left( \begin{array}{c}dx\\dy\\dz\end{array} \right)=\left( \begin{array}{ccc}1&0&tanθ\\0&1&0\\0&0&1\end{array}\right)\left( \begin{array}{c}dx_0\\dy_0\\dz_0\end{array} \right)[/tex]

The 3x3 matrix is the deformation gradient tensor components.

Chet

 

I really don't understand what you are referring to here. Maybe you learned this stuff differently from me. I don't understand what your 4x4 matrix is supposed to represent. The transformation I gave maps an arbitrary differential position vector between two material points in the initial undeformed configuration of the cylinder into the corresponding differential position vector between the same two material points in the deformed configuration. From this, you can determine all the strains, which I assume, is your objective.

Chet