What Does Simple Shear Deformation Tell Us About Material Behavior?

In summary: \cdot x\centerdot x_2\cdot x_3=x_1i_1+γi_1i_2+i_2i_2+i_3i_3\cdot x_2\cdot x_3; c=i_1i_1+i_2i_2+i_3i_3\cdot x_1=−i_1i_1+i_2i_2+i_3i_3\cdot x_1i_2
  • #1
Bertbos
2
0

Homework Statement


Let {ii}, i = 1, 2, 3 be an orthonormal basis as shown in Figure 2 and consider a
simple shear deformation from R to R′ defined as
[itex]\hat{y}[/itex](x) = x + γ(x · i2)i1 , where the scalar γ corresponds to amount of shear.

Homework Equations


For this homogeneous deformation, compute the deformation gradient F
and the translation vector c. Compute the change in length of fibers of
unit length in R aligned with the basis vectors ii (i = 1, 2, 3). What can
you say about fibers aligned with i1 and i3? Compute the change in angle
of pairs of fibers aligned with ii, ij (for i, j = 1, 2, 3, i ≠ j).

The Attempt at a Solution


If I could get [itex]\hat{y}[/itex](x) in a form of [itex]\hat{y}[/itex](x) = Fx + c I could compute the different variables, but I don't know how to get the equation in the right format. Computing F can be done by F = ∇y, with ∇=∇x ; however, I don't know how to compute the translation vector c. Could anyone help me?
 
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  • #2
Let x = x1i1+x2i2+x3i3

y=y1i1+y2i2+y3i3

So, [tex]x\centerdot i_2=x_2[/tex]

and γ(x · i2)i1=γx2i1

so
y1i1+y2i2+y3i3
=x1i1+x2i2+x3i3
+γx2i1
So now, resolve this equation into its 3 components (i.e., into the coefficients of the three unit vectors). Also, write out y = Fx + c in component form, and compare the expressions.
 
  • #3
Chestermiller said:
Let x = x1i1+x2i2+x3i3

y=y1i1+y2i2+y3i3

So, [tex]x\centerdot i_2=x_2[/tex]

and γ(x · i2)i1=γx2i1

so
y1i1+y2i2+y3i3
=x1i1+x2i2+x3i3
+γx2i1
So now, resolve this equation into its 3 components (i.e., into the coefficients of the three unit vectors). Also, write out y = Fx + c in component form, and compare the expressions.

Thank you very much. Sometimes I've trouble to write the equations is the right form. Your explanation was very clear and now it's easy to me to solve the problem!
 
  • #4
Bertbos said:
Thank you very much. Sometimes I've trouble to write the equations is the right form. Your explanation was very clear and now it's easy to me to solve the problem!

There's an even simpler way of doing it that might appeal to you even more.

Write [tex]x=(i_1i_1+i_2i_2+i_3i_3)\centerdot x[/tex]
and [tex](x \centerdot i_2)i_1=i_1i_2\centerdot x[/tex]

So [tex]y=x+γ(x \centerdot i_2)i_1=(i_1i_1+i_2i_2+i_3i_3)\centerdot x+γi_1i_2\centerdot x=(i_1i_1+γi_1i_2+i_2i_2+i_3i_3)\centerdot x[/tex]

So, [itex]F=i_1i_1+γi_1i_2+i_2i_2+i_3i_3[/itex]
 
  • #5


Hello,

Thank you for your question. I would be happy to assist you with finding a solution for your homework problem.

To begin, let us first define the deformation gradient F. It is a matrix that relates the change in position of a point in the deformed configuration to its original position in the undeformed configuration. In other words, it describes how the shape of an object changes under deformation. In this case, we have a simple shear deformation, which is a type of homogeneous deformation, meaning that the deformation gradient F is constant throughout the entire body.

To compute the deformation gradient F, we can use the following equation:

F = ∇y = [∇y1, ∇y2, ∇y3]

Where ∇y is the gradient of the transformation function y, and y is defined as:

y(x) = x + γ(x · i2)i1

To simplify the calculation, we can expand the equation for y to obtain:

y(x) = x + γ(x1i1 + x2i2 + x3i3)

Now, taking the gradient of y, we get:

∇y = [1 + γi1, γi2, γi3]

Therefore, the deformation gradient F is:

F = [1 + γi1, γi2, γi3]

Next, to compute the translation vector c, we can use the following equation:

c = y(0) - 0

Where y(0) represents the position of the origin in the deformed configuration. In this case, since the origin remains unchanged in the deformed configuration, c = 0.

Now, to compute the change in length of fibers of unit length in R aligned with the basis vectors ii (i = 1, 2, 3), we can use the following equation:

Δl = ||Fii|| - 1

Where ||Fii|| represents the norm of the i-th column of the deformation gradient F. In this case, since the deformation gradient F is constant and equal to [1 + γi1, γi2, γi3], we can simply compute the norm of each column to obtain:

||Fii|| = √(1 + γ²)

Therefore, the change in length of fibers aligned with i1 and i3 is:

Δl1 = √(1 + γ²) - 1
Δ
 

Related to What Does Simple Shear Deformation Tell Us About Material Behavior?

1. What is simple shear?

Simple shear is a type of deformation where the applied force is parallel to one of the object's surfaces, causing the object to deform in a way that preserves the original shape but changes its orientation. This type of shear is commonly seen in materials that are being cut or stretched.

2. How is simple shear analyzed?

Simple shear is often analyzed using mathematical models and equations, such as the shear strain equation and the stress-strain relationship. These models help scientists and engineers understand how different materials will respond to shear forces and how to predict their behavior.

3. What factors affect the analysis of simple shear?

The analysis of simple shear is affected by several factors, including the type of material being sheared, the magnitude and direction of the applied force, and the geometry of the object. Other factors such as temperature, time, and external stresses can also play a role in the analysis.

4. What are the applications of simple shear analysis?

Simple shear analysis has several applications in various fields, including engineering, material science, and geology. It is used to understand the behavior of materials under stress, design structures that can withstand shear forces, and predict how geological formations may change due to shear deformation.

5. What are the limitations of simple shear analysis?

While simple shear analysis is a useful tool, it has some limitations. It assumes that the material being sheared is homogeneous, isotropic, and follows linear elastic behavior. In reality, many materials do not meet these assumptions, which can lead to errors in the analysis. Additionally, simple shear analysis does not take into account other types of deformation, such as bending and torsion, which may also occur simultaneously.

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