## Introduction

The definition of simple shear is shown in the following figure. The top surface of the specimen is displaced to the right a distance *dx*, while the bottom is held fixed and the distance between the top and bottom surfaces (*dy*) is held constant. The ratio *dx*/*dy* is often called \(\gamma\) and is equal to the shear strain at small deformations.

## Large Deformations

At large shear deformations we can still use gamma to quantify the amount of shear that is applied, but gamma in this case is no longer equal to the engineering shear strain. The following figure shows how the engineering shear strain and the true shear strain evolve with the applied value of gamma.

The left side of the figure also shows the deformed shape of a simple shear specimen with an initial height that is about equal to its width. Here are a few key observations:

- For gamma values that are less than about 0.8, the engineering shear strain and the true shear strain are basically equal to gamma.
- At large gamma values the engineering shear strain reaches a max value of 2.0.
- At large gamma values the true shear strain starts to
**decrease**. - If the shear angle gamma exceeds a value of 3, then the specimen will start to experience a true tensile strain that exceeds the shear strain. I therefore suggest that shear tests do not go to gamma values that are larger than 3.

To understand what is going on it is useful to consider some basic equations. Let’s start with the definition of simple shear expressed in terms of the deformation gradient \( \mathbf{F} \):

\[

\mathbf{F} = \begin{bmatrix}

1 & \gamma & 0\\

0 & 1 & 0\\

0 & 0 & 1

\end{bmatrix}

\]

It is easy to show that the left stretch tensor in this case is given by:

\[

\mathbf{v} = \frac{1}{\sqrt{4+\gamma^2}}\begin{bmatrix}

2+\gamma^2 & \gamma & 0\\

\gamma & 2 & 0\\

0 & 0 & 1

\end{bmatrix}

\]

The engineering shear strain is therefore given by:

\[

\varepsilon_{12} = \frac{\gamma}{\sqrt{1 + \gamma^2/4}} = \frac{1}{\sqrt{\frac{1}{\gamma^2} + \frac{1}{4}}}

\]

This shows that as \(\gamma \rightarrow \infty\), the engineering shear strain approaches 2.

It is more difficult to derive a closed formed expression for the true shear strain, but fortunately it is easy to perform the calculation numerically using the computer language Julia. If you have not tried Julia before, you should. It is like Matlab, but it is faster and it is free!

Here is the Julia code to calculate and plot the different shear strains:

` ````
```using LinearAlgebra
using PyPlot
# cd("d:")
# pwd()
# include("shear_strain_calculations.jl")
function getShearStrain(gamma, trueStrainFlag)
F = [1 gamma 0; 0 1 0; 0 0 1]
b = F * transpose(F)
v = sqrt(b)
if trueStrainFlag
strain = log(v) # true tensorial strain
else
strain = v - I # eng tensorial strain
end
return 2*strain[1,2]
end
N = 100
gamma = range(0.0, stop=10.0, length=N)
NE12 = zeros(N)
LE12 = zeros(N)
for i = 1 : length(gamma)
NE12[i] = getShearStrain(gamma[i], false)
LE12[i] = getShearStrain(gamma[i], true)
end
xlabel("Gamma")
ylabel("Shear Strain")
grid(true)
plot(gamma, NE12, color="blue", linewidth=1.5, linestyle="-", label="Engineering Shear Strain")
plot(gamma, LE12, color="red", linewidth=1.5, linestyle="--", label="True Shear Strain")
legend()
savefig("shear_strain_calculations.png", dpi=150)

## Influence of Specimen Size

The most common way to perform a simple shear experiment is to use an adhesive to attach one or two specimens in a single or double lap shear fixture. As quick FE study will show, however, that the deformation field in the specimens in these tests is not just simple shear, but it also contains a more complicated multiaxial deformation field at the edges. The following figures show that the error caused by these edge effects can be more than 15%, which is terrible! The accuracy of the experiments can be improved by using specimens with a height that is much smaller than it width. As an example, if the shear specimens have a width that is 5X larger then their height, then the experimental error is less than 3%.

## Summary

Do the following to get accurate simple shear data:

- Only go to a max shear angle gamma of 3.0
- Max sure the shear specimens have width that is 5X their height.

## 2 thoughts on “Simple Shear”

Yuhai XiangIt seems that the left stretch tensor is incorrect, I used your v to calculate v*v, it is not equal to b, pls check it, thanks.

JorgenCan you provide some more details? I tried the following quick test. It seems OK to me.

julia> gamma = 0.9

0.9

julia> F = [1 gamma 0; 0 1 0; 0 0 1]

3×3 Array{Float64,2}:

1.0 0.9 0.0

0.0 1.0 0.0

0.0 0.0 1.0

julia> b = F * transpose(F)

3×3 Array{Float64,2}:

1.81 0.9 0.0

0.9 1.0 0.0

0.0 0.0 1.0

julia> v = sqrt(b)

3×3 Array{Float64,2}:

1.28125 0.410365 0.0

0.410365 0.911922 0.0

0.0 0.0 1.0

julia> v*v

3×3 Array{Float64,2}:

1.81 0.9 0.0

0.9 1.0 0.0

0.0 0.0 1.0