MATLAB Assignment

CEE 213—Deformable Solids

© Keith D. Hjelmstad 2014

The Mechanics Project School for Sustainable Engineering and the Built Environment

Ira A. Fulton Schools of Engineering Arizona State University

CEE 213—Deformable Solids

Multiaxial States of Strain Motion, deformation, and strain

CEE 213—Deformable Solids

© Keith D. Hjelmstad 2014

Multiaxial States of Strain Motion, deformation, and strain

The Mechanics Project Arizona State University

CEE 213—Deformable Solids Contents. 1. What is strain? 2. Axial bar revisited. 3. Torsion bar revisited. 4. Beam shearing and bending. 5. Biaxial deformation. 6. Uniform and non-uniform deformation.. 7. Rotation and translation. 8. Shearing and extension. 9. Uniform motion. 10. The motion map. 11. What does the motion map do? 12. The lengths of lines. 13. Definition of the strain tensor. 14. Definition of normal strain. 15. Definition of shear strain. 16. What is εab? 17. Plane strain—strain transformations (1,2,3,4). 18. Maximum normal strain 19. Maximum shear strain (1,2) 20. Strain invariants. 21. Mohr’s circle. 22. Mohr’s circle construction. 23. Principal strains (1.2). 24. Summary.

CEE 213—Deformable Solids

© Keith D. Hjelmstad 2014

Multiaxial Strain What is strain?

What is strain? In the study of deformable body mechanics we seek to describe how the material in a body deforms and how that deformation induces stress. To help us think about this problem we will consider two states of the body: the undeformed state (before loads are applied) and the deformed state (the position of the body after the loads have been applied). The motion of the body is the path it takes to get from the undeformed state to the deformed state. The deformation map is simply the function that describes that motion. Strain is an internal quantity that helps us to describe the nature of the deformation. In essence, strain characterizes how a line etched in the undeformed body looks in the deformed body and how angles between lines in the undeformed body look in the deformed configuration (i.e., change in angle). As an example, consider the line AB in the undeformed configuration (red). After deformation that same line has translated, rotated, and stretched to become the line A’B’ in the deformed configuration (blue). The line has changed length; it has stretched; it has strained. In this set of notes we will try to put that idea into a context that will allow us to faithfully describe the concept of strain as a means of completely describing what happens to a body when it deforms. Strain is “change in length divided by original length” but we need to figure out what that means for 2D and 3D states of deformation.

Deformation is a mapping from the undeformed configuration to

the deformed configuration.

B A

A’

B’

Undeformed Deformed

Deformation map

CEE 213—Deformable Solids

© Keith D. Hjelmstad 2014

Multiaxial Strain Axial bar revisited

Axial bar revisited. The first state of strain that we encountered in our study of deformable solids was that associated with the axial bar (shown at left). The axial bar changes length due to the application of an axial force. If the force is tensile then the bar gets longer. If the force is compressive then the bar gets shorter. We define the strain to be the change in length divided by the original length

o

o o

L L L L L

ε ∆ − = =

That is the average strain over the finite length of the bar. For the small (gray) piece we can write

Strain is a quantity that characterizes the internal state of deformation of the body. It is change in length per original length.

oL

x∆x ( )u x x+ ∆

( )u x

1e

1e

Undeformed

Deformed

L

0

( ) ( )( ) lim x

u x x u x dux x dx

ε ∆ →

+ ∆ − = =

∆

This limiting process yields a relationship between the strain at a point to the displacement. The displacement function u(x) is the deformation mapping. A mapping is simply a function that takes the material point x and produces its displacement (i.e., where it goes). Axial displacement and strain is the simplest case of strain because change happens only in one direction.

CEE 213—Deformable Solids

© Keith D. Hjelmstad 2014

Undeformed γ

γ

Deformed

L

x∆x

1e

1e

Undeformed

Deformedγ

Lϕ

Multiaxial Strain Torsion bar revisited

Torsion bar revisited. The second state of strain that we encountered in our study of deformable solids was that associated with the bar in torsion (shown at left). If we etch a little square on the surface of the circular bar when it is undeformed then we can see what happens to it when the bar is twisted. The square gets mapped into a rhombus (sides remain the same length, but the angles do not remain right angles). This situation is depicted below:

The bar twisted by end couples puts each element into a state of pure shear.

This form of deformation is called shearing and it suggests that strain must be more than just change in length because there is no change in length of lines in going from square to rhombus. This second form of deformation concerns the change in angle of lines that start out perpendicular. The measure of deviation in the example above is the angle γ. We can relate the shear strain to the rotation angle of the bar as (ϕ is the mapping function)

dr dx ϕγ =

CEE 213—Deformable Solids

© Keith D. Hjelmstad 2014

σ

τ

τ

σ

Multiaxial Strain Beam shearing and bending

z

x

V

M

N

Beam shearing and bending. The third state of deformation that we encountered in our study of deformable solids was that associated with a beam in flexure (shown at left). The beam in flexure gives rise to both shear stress and normal stress associated with N, V, and M. In particular, they come from the stress formulas

( ), ( )

N Mz VQ z A I I b z

σ τ= + =

These stresses are associated with strains (both change in length and change in angle), but there is something a little odd about beam theory. We make the assumption in beam theory that plane sections remain plane and that the angle of orientation of the cross section θ is equal the the slope of the deflection curve w'. This assumption causes the shear strain to be zero (an approximation, it is just very small for most beams). Because of this oddity, we will generally compute the beam stresses and then compute the strains from those stresses using the constitutive model (i.e., Hooke’s Law). We can, of course, compute the normal strain for fibers in the axial direction as:

( , ) ( )x z x zε κ=

where κ is the curvature.

CEE 213—Deformable Solids

© Keith D. Hjelmstad 2014

What happens to lines and angles under the deformation map depends upon the initial orientation of the piece.

Biaxial deformation. To get an idea of what the motion implicit in the deformation map implies about the straining of the material, consider the example illustrated at left. Imagine that we have a flexible fabric on which we lay out a grid in the undeformed state. We have a grid of 12 by 5 squares on the grid. On the undeformed grid we draw three squares oriented at different angles. As the fabric deforms the drawn squares do, too. Observe the undeformed and deformed grids. The deformation map is uniform—i.e., every square in the undeformed grid becomes a parallelogram in the deformed grid. Every square gets mapped the same way. Now look at what happens to the squares that were etched on the undeformed grid. Note that the corners of the deformed squares are located at the same grid intersections. The first square deforms into a rectangle; the other two deform into parallelograms. If you think about the motion in terms of what happens to different lines and the angles between lines, you can see that the outcome depends upon the initial orientation of the etched square. In this observation is the basis for our study of motion, deformation, and strain. Our aim is to characterize the deformation in terms of normal and shear strains. Normal strains are about change in length; shear strains are about changes in angles.

Multiaxial Strain Biaxial deformation

Deformation Map

Undeformed

1e

2e

Deformed 1e

2e

CEE 213—Deformable Solids

© Keith D. Hjelmstad 2014

Multiaxial Strain Uniform and non-uniform deformation

Uniform and non-uniform deformation. The motion of a body is a change of location of every particle in the body. The deformation map is the function that describes the relationship between the original location and the deformed location. A uniform motion is a deformation in which every point experiences the same deformation. In the sketch at top left, every little square in the undeformed configuration deforms in exactly the same way. Every point in the body feels the same amount of strain. A circular bar subjected to an end force and torque has the same state of strain at every point on the surface. A non-uniform motion is a deformation in which the deformation changes from place to place. In the sketch at bottom left each little square in the undeformed configuration deforms differently, with the points in the lower left straining less than the points in the upper right parts of the body. This state is the more general case. In a beam in planar bending the normal strain changes differently from the shear strain. Hence, each point through the depth of a beam is in a different state of strain (which brings up the question: “Where is the maximum strain?” If we focus on a local region of a body we can assume that the deformation is uniform over that small region. We will examine the strain associated with uniform motion.

Uniform motion

Non-uniform motion

CEE 213—Deformable Solids

© Keith D. Hjelmstad 2014

Multiaxial Strain Rotation and translation

The motion of a body includes rigid body motion—translation and rotation.

Rotation and translation. The motion of a body is generally composed of a rigid motion and a deformational motion. A rigid body motion includes a translation and a rotation but not line in the body changes length and no two lines in the body change their relative angle with respect to each other. For rigid body motion there is no straining (normal or shear) of the material associated with the motion. A case of rigid body motion for motion in a plane is illustrated at left. We can characterize the final position of each particle by keeping track of the translation of the lower left corner, the angle of rotation of the entire body, and the position of the particle relative to the left corner (in the case we note that the gray square is four units right and four units up from the lower left corner). Rigid body mechanics is the main subject of the course Dynamics in which we focus on the motion of bodies that do not deform. The assumption of rigidity is built into the kinematics through the position vector. In deformable body mechanics we will focus on quasi-static deformations (i.e., acceleration is negligible). We will need to formulate the concept of strain in such a way that it is not affected by rigid body motions.

CEE 213—Deformable Solids

© Keith D. Hjelmstad 2014

Multiaxial Strain Shearing and extension

Motion including translation

Motion with translation removed

Shearing and extension. For our analysis of deformation we will eliminate the translational part of the motion by moving the deformed piece to have a common origin with the undeformed piece. At left the motion is shown with and without the translational component of the motion. It is easy to see that moving the pieces to have a common origin does not change the deformation of the body. The extension and shearing do have an effect on the lengths of lines and relative angle between lines. We call that effect strain. We will show that, in a very fundamental way, strain is all about the change in length and angle of lines. Even in the presence of shearing and extension, rotation and translation have no additional effect on the lengths of lines in a body. It should also be evident that we have the freedom to rotate the deformed piece, too. The sketches below show the deformed piece rotated up to vertical or rotated down to horizontal. We will see how to deal with this ambiguity in the analysis of strain.

CEE 213—Deformable Solids

© Keith D. Hjelmstad 2014

Multiaxial Strain Uniform motion

1

1

d

c

b

a

a c+

b d+

Uniform motion. In order to have uniform deformation there are some features that must be included in the motion. To see what the requirements are, consider the sketch shown at left. The red square is the undeformed configuration and the blue parallelogram is the deformed configuration. We have eliminated the influence of translation by moving the lower left corners to be coincident. The lower right corner displaces in such a way that it moves a distance a in the e1 direction and a distance b in the e2 direction. The upper left corner displaces in such a way that it moves a distance c in the e1 direction and a distance d in the e2 direction. Both of these displacements are emphasized with colored rectangles that are sized to span the motion. Now consider the upper right corner. It is not free to move independently because to do so would result in non-uniform motion. If we put the two rectangles end to end, we can see that the location of the deformed point that defines the upper right corner must be located in such a way that it moves moves a distance a+c in the e1 direction and a distance b+d in the e2 direction. The upshot of this discussion is that there are only four possible variables needed to describe uniform motion.

A uniform deformation can be described uniquely by specifying the four parameters a, b, c, and d. That is equivalent to specifying the location of the two corners originally at (1,0) and (0,1).

CEE 213—Deformable Solids

© Keith D. Hjelmstad 2014

Multiaxial Strain The motion map

1 1

a c b d + 

=  +  F

=z F x

, x y

ξ η

    = =       

x z

The motion map. If we can track the positions of every point in the deformed body then we can also find the change in the lengths of lines that join them to the origin. Hence, we can find the strains in the body from the motion map. Consider, for example, the point P, which is located at the position x in the undeformed configuration. This point becomes P’ in the deformed configuration. Let us describe these two points with their coordinates:

How do you know what the map does? Just apply it to certain points and see where they end up. For example, the point x=(1,0) becomes the point z=(1+a, b). It turns out that the tensor F contains all of the information needed to describe uniform deformation.

x z

,x ξ

,y η

P P’

O

1

1

d

c

b

a

a c+

b d+

The motion map can be described with the simple relationship

Where the deformation tensor (or matrix) has the components

CEE 213—Deformable Solids

© Keith D. Hjelmstad 2014

Multiaxial Strain What does the motion map do?

Point A

Point B

Point C

1

1

d

c

b

a

a c+

b d+C B

A

What does the motion map do? Let’s see what happens when we map the three points at the vertices of the original unit square. The motion map is

1 0 1 , ,

0 1 1A B C      

= = =           

x x x

1 1 1 1 0

a c a b d b + +    

=    +     

=z F x

The position vectors from the origin to the three points A, B, and C are

The deformed position vectors from the origin to the three points A, B, and C can be computed from the map as

The deformation map works for those three points and for any point inside the domain of the body. Note that the only thing that changes is the vector components of the original points. The deformation map is the same for all points.

1 0 1 1 1

a c c b d d +    

=    + +    

1 1 1 1 1 1

a c a c b d b d + + +    

=    + + +    

C C=z F xCx

Ax Az

Cx Cz

Bx Bz

CEE 213—Deformable Solids

© Keith D. Hjelmstad 2014

Multiaxial Strain The lengths of lines

The lengths of lines. Because we are describing the lines connecting the origin O to the point P we can use the fact that the square of the length of a vector is the vector dotted with itself. The square of the length of the line OP is

2 T o = ⋅ =x x x x

The square of the length of the line O’P’ can be computed as

x =z Fx

,x ξ

,y η

P P’

O

( ) ( )2 TT f = ⋅ = =z z z z Fx Fx

2 T T f  =  x F F x

Let us carry out the matrix multiply

From linear algebra we know that the transpose of the product of two matrices is the product of the transposes of the matrices in reverse order (remember that?). Thus,

( ) ( ) ( ) ( ) ( ) ( )

2 2

2 2

2 2

2 2

1 1 1 1

1 1 1

1 1 1

1 2 1 2

T a b a c c d b d

a b a c d b

a c d b d c

a a b c b ac bd c b ac bd d d c

+ +    =    + +     + + + + +  =  + + + + +   + + + + + +

=  + + + + + + 

F F

1 2 1 2

T a b c b c d + + 

≈  + +  F F

If the deformation is small, i.e.,

, , , and 1a b c d 

Then we can neglect squares and products of these quantities. Hence, we can approximate

CEE 213—Deformable Solids

© Keith D. Hjelmstad 2014

Multiaxial Strain The definition of the strain tensor

The definition of the strain tensor. If the deformations are small then

1 0 2 2

0 1 xx xyT

xy yy

ε ε ε ε   

≈ + = +      

F F I E

xx xy

xy yy

ε ε ε ε  

=    

E

( )1 2, ,xx xy yya b c dε ε ε= = + =

1 2 1 2

T a b c b c d + + 

≈  + +  F F

Let’s make the following notational definitions:

Then we can write

where

We call E the strain tensor and the three values εxx, εxy, and εyy the components of the strain tensor. The quantity E will help to describe the changes in lengths of lines and, hence, we can use it to connect with the idea of strain as being change in length divided by original length.

x =z Fx

,x ξ

,y η

P P’

O

It is interesting to note that the off- diagonal elements of E were both b+c. It is a fact of linear algebra that the matrix FTF must be symmetric. What this means is that b and c always appear together in summed form. This result is because a rotation of the deformed piece does not cause additional strain. Although b and c are independent in the motion, they combine to cause the shear strain.

CEE 213—Deformable Solids

© Keith D. Hjelmstad 2014

Multiaxial Strain The definition of normal strain

The definition of normal strain. If the deformations are small then

Now, if the deformations are small the sum ℓo+ℓf is approximately equal to 2ℓo. Hence,

Which is exactly the definition of the strain (change in length divided by original length). Therefore, we can compute the strain associated with any direction (defined by the vector x) as

o f

,x ξ

,y η

P P’

O

Note that the quantity

θ

2 T o = x x

( )( )2 2

2 22 2 f o f of o

n o o

ε − +−

= =    

 

 

( )2 1 T

n o

ε = x Ex 

Strain is change in length divided by original length. Take a look at this derivation:

( )( ) 2

2 2

f o o f o n

o o

ε − −

≈ =   

 

 

[ ]

2

2

2

2 2

T f

T

T T

T o

=

= +

= +

= +

z z

x I E x

x x x Ex x Ex





2 2

22 f o

n o

ε −

=  



is called the Lagrangian strain and is useful in cases where the deformations are not small. There are several definitions of strain that are used in different contexts. Here we will stick with engineering strain.

CEE 213—Deformable Solids

© Keith D. Hjelmstad 2014

Multiaxial Strain The definition of shear strain

The definition of shear strain. Let us compute the cosine of the angle between the two deformed vectors xa and xb. Recall that the cosine can be computed from the dot product as

The strains εaa and εbb are small compared with one so we can neglect them to give a final expression

The lengths of the deformed vectors:

( )2 2cos T

a b o

β = x Ex 

( ) ( ) ( )

( )( ) [ ]

( )( )

( )( )

( )( )

2

2

2

2

cos

1 1

1 1 1

21 1 1

1 2 1 1

1 2 1 1

a b

a b

a b

o aa o bb

T T a b

o aa bb

T a b

o aa bb

T T a b a b

o aa bb

T a b

o aa bb

β

ε ε

ε ε

ε ε

ε ε

ε ε

⋅ =

⋅ =

+ +

= + +

+ =

+ +

+ =

+ +

= + +

z z z z

Fx Fx

x F F x

x I E x

x x x Ex

x Ex

 







( ) ( )1 , 1a o aa b o bbε ε= + = +z z 

ax

az

x

y

θ

bx

bz

θ

β

aγ bγ

0a b⋅ =x x

2T = +F F I E

From earlier…

Orthogonality of the original vectors:

Key mathematical truths…

CEE 213—Deformable Solids

© Keith D. Hjelmstad 2014

Multiaxial Strain Plane strain—strain transformations

cos sina o

θ θ

  =  

  x 

( )

{ }

{ }

2 1

cos cos sin

sin

cos sin cos sin

cos sin

T aa a a

o

xx xy aa

xy yy

xx xy aa

xy yy

ε

ε ε θ ε θ θ

ε ε θ

ε θ ε θ ε θ θ

ε θ ε θ

=

   =   

   + 

=  + 

x Ex 

2 2cos 2 cos sin sinaa xx xy yyε ε θ ε θ θ ε θ= + +

Plane strain—strain transformations. If the motion is purely in the x-y plane then we call the condition plane strain. In this state there is no deformation in the z direction. Let us examine how strain transforms in a plane strain situation. First take an undeformed vector at angle θ from the horizontal el axis). The vector has components

Let us compute the strain εaa associated with this original line from

What we get is a formula for transforming the original strain components into an expression for the component εaa.

ax

az

x

y

coso θ

sino θ o

f

θ

The strain tensor E provides a way to compute the strain of a line that is initially oriented along xa. Note that

f o aa

o

ε −

=  



for this particular line.

CEE 213—Deformable Solids

© Keith D. Hjelmstad 2014

Multiaxial Strain Plane strain—strain transformation

sin cosb o

θ θ

−  =  

  x 

( )

{ }

{ }

2 1

sin sin cos

cos

sin cos sin cos

sin cos

T bb b b

o

xx xy bb

xy yy

xx xy bb

xy yy

ε

ε ε θ ε θ θ

ε ε θ

ε θ ε θ ε θ θ

ε θ ε θ

=

−   = −   

   − + 

= −  − + 

x Ex 

2 2sin 2 cos sin cosbb xx xy yyε ε θ ε θ θ ε θ= − +

Plane strain—strain transformations. Next, let us take an undeformed vector at angle θ from the vertical e2 axis). This vector is perpendicular to xa. The vector has components

Let us compute the strain εbb associated with this original line from

What we get is a formula for transforming the original strain components into an expression for the component εbb.

ax

az

x

y

sino θ

coso θ

θ

bx

bz

o

f

θ

If these equations get you thinking about the stress transformation equations you are not off track. We will get exactly the same relationships. But notice how different the derivations are!

CEE 213—Deformable Solids

© Keith D. Hjelmstad 2014

Multiaxial Strain Plane strain—strain transformation

cos sin ,

sin cosa o b o

θ θ θ θ

−    = =   

    x x 

[ ]( )

2

2 2

1cos

1 22

a b a b o

T T a b a b

o o

β = ⋅ = ⋅

= − =

z z Fx Fx

x I E x x Ex



 

( ) ( )2 22 2 sin cos 2 cos sinab yy xx xyε ε ε θ θ ε θ θ= − + −

Plane strain—strain transformations. To complete the analysis of strain transformations let’s look at the change in angles between xa and xb (which are initially 90 degrees apart). First take an undeformed vector at angle θ from the horizontal el axis). The vector has components

Let us compute the cosine of the angle β between the two deformed lines using the dot product definition:

ax

az

x

y

θ

bx

bz

θ

β

aγ bγ

( ) ( )1 2cos cos sina b a bβ π γ γ γ γ= − − = +

( )sin a b a bγ γ γ γ+ ≈ +

From trigonometry:

If the angles are very small then:

From our original interpretation of motion (see the meaning of b+c) we can define the shear strain

2 cosab a bε β γ γ= = +

{ }

{ }

sin cos 2 cos sin

cos

sin cos cos 2 cos sin

sin cos

xx xy

xy yy

xx xy

xy yy

ε ε θ β θ θ

ε ε θ

ε θ ε θ β θ θ

ε θ ε θ

−   =   

   − + 

=  − + 

For the case at hand:

CEE 213—Deformable Solids

© Keith D. Hjelmstad 2014

Multiaxial Strain What is εab?

What is εab? We can get a better understanding for the nature of the component εab that we defined to represent the cosine of the angle between the two deformed vectors za and zb by looking at a specific case. The state of strain shown in the figures at left has

0.4 0.2 0.3xx yy xyε ε ε= = − =

(These are very large strains, but they help to see the straining in the pictures). The first picture shows the undeformed and deformed body in the original x-y coordinate system. You can see that the edge along the x axis has gotten longer, the edge along the y axis has gotten shorter, and the shearing is such that the angle β is less that 90 degrees. If we rotate the undeformed (red) block to exactly the principal angle (0.3374 radians in this case) then we can observe that there is no shear. Hence, the angle β is exactly 90 degrees. Finally, if we rotate the undeformed (red) block by twice the principal angle, the angle β is greater than 90 degrees. In our convention we put half of the shearing angle on each side—that is what εab is. It is the amount that β differs from 90 degrees, split in two. You can see that for any angle θ we can compute the angle εab. Sometimes the deformation is such that it is positive (as in the third picture) and sometimes negative (as in the first picture).

0.0000

0.3374

0.6747

θ =

θ =

θ =

CEE 213—Deformable Solids

© Keith D. Hjelmstad 2014

Multiaxial Strain Plane strain—strain transformation

( ) ( )

2 2

2 2

2 2

cos 2 sin cos sin

sin 2 cos sin cos

2 2 cos sin 2 cos sin

aa xx xy yy

bb xx xy yy

ab yy xx xy

ε ε θ ε θ θ ε θ

ε ε θ ε θ θ ε θ

ε ε ε θ θ ε θ θ

= + +

= − +

= − + −

2 2

sin 2 2sin cos cos2 cos sin

θ θ θ

θ θ θ

=

= −

( )

2 2

2 2

1 2

cos sin 2 sin

sin sin 2 cos

sin 2 cos2

aa xx xy yy

bb xx xy yy

ab yy xx xy

ε ε θ ε θ ε θ

ε ε θ ε θ ε θ

ε ε ε θ ε θ

= + +

= − +

= − +

Plane strain—strain transformations. The analysis of the last few pages has led to the strain transformation equations

We can put these equations into a slightly different form by introducing the trigonometric identities

Substituting these in we get

Note the direct similarity between these equations and the equations for stress transformations. All of the analyses we did for stress apply similarly to strain (including Mohr’s circle).

θ

1e

2e

be ae

One way to think about the strain transformation equations is that they represent a different choice of coordinate system for describing the strain tensor. Physically, it amounts to a different choice of the initial square and associated deformation.

CEE 213—Deformable Solids

© Keith D. Hjelmstad 2014

Multiaxial Strain Maximum normal strain

Maximum normal strain. The strain transformation equations for normal strain (previous page) suggest that these values depend upon the orientation θ that defines how we etch the little square on the undeformed body. From an engineering standpoint, then, we might wonder if there is a worst case direction that gives the maximum normal strain. Let’s compute the maximum with respect to θ.

2 cos sin 2 cos2 2 sin cos 0

2 sin cos 2 cos2 2 cos sin 0

aa xx xy yy

bb xx xy yy

d d

d d

ε ε θ θ ε θ ε θ θ θ ε ε θ θ ε θ ε θ θ θ

= − + + =

= − − =

Note that both give the same equation to determine θp :

( ) ( )1

2

cos sin cos2 0

sin 2 cos2 0

yy xx p p xy p

yy xx p xy p

ε ε θ θ ε θ

ε ε θ ε θ

− + =

− + =

We can solve for the angle θp (box at left). This equation actually yields two angles: One gives the angle for the maximum normal strain, one for the minimum. These max/min normal stresses are called principal strains. One very important observation is that the equation for maximum normal strain is exactly the same as the equation for zero shear strain εab. Hence, the principal planes are free of shear strain.

2 tan 2 xy

p xx yy

ε θ

ε ε =

−

The angle θp that gives the maximum and minimum normal strain (for plane strain) is given by

Use this angle in the strain transformation relationships to find the values of the principal strains.

θ

1e

2e

be ae

CEE 213—Deformable Solids

© Keith D. Hjelmstad 2014

Multiaxial Strain Maximum shear strain

Maximum shear strain. We can also ask if there is an orientation θ that gives the maximum shear strain. Let’s compute the maximum with respect to θ.

( )cos2 2 sin 2 0ab yy xx xy

d d ε ε ε θ ε θ θ

= − − =

Note that this gives an equation to determine the angle θm at which the maximum shear happens:

It is hard not to notice that the expression for the angle of maximum shear is very similar to the one for maximum normal strain. In fact, the tangents of the two angles are negative reciprocals of each other. Note carefully the appearance of each term in each expression! You can see how these values relate in the example on the following page. Note that at the orientation angle θm the normal strains do not vanish, but they are equal to each other at that angle. We could actually use εaa = εbb to find the angle θm.

The angle θm describes the orientation that gives the maximum shear strain (for plane strain). It is not the same angle as θp (which gives the orientation of the principal planes— i.e., the angle of orientation of the square associated with the principal strains).

( )1 2 sin 2 cos2ab yy xx xyε ε ε θ ε θ= − +

tan 2 2

yy xx m

xy

ε ε θ

ε −

=

θ

1e

2e

be ae

CEE 213—Deformable Solids

© Keith D. Hjelmstad 2014

tan 2 2

yy xx m

xy

ε ε θ

ε −

=

Multiaxial Strain Maximum shear strains

( )

( )

( )

1 2

1 2

2 21 2

2 2

sin 2 cos2

2

2

( ) 4

m yy xx m xy m

yy xx xy yy xx xy

yy xx xy

yy xx xy

h h

γ ε ε θ ε θ

ε ε ε ε ε ε

ε ε ε

ε ε ε

= − +

−    = − +   

   

− + =

− +

2 mθ yy xxε ε−

2 xyε

h

Maximum shear strains. The angle θm associated with maximum shear strain, is given by

The angle, then, must be the one shown in the right triangle at left. It should be evident that the hypotenuse of the triangle is

We can use the relationship given in the box at left in the formula for εab to compute the maximum shear strain γm.

Consider the triangle construction above. We can compute the sine and cosine of the angle 2θm as

We can use these in the formula for shear stress to compute maximum shear stress (as shown at right).

2 21 4 ( )m yy xx xyγ ε ε ε= − +

Simplifying this expression we get an expression for the maximum shear strain

2 2( ) 4yy xx xyh ε ε ε= − +

2 2

2 2

sin 2 ( ) 4

2 cos2

( ) 4

yy xx m

yy xx xy

xy m

yy xx xy

ε ε θ

ε ε ε

ε θ

ε ε ε

− =

− +

= − +

CEE 213—Deformable Solids

© Keith D. Hjelmstad 2014

Multiaxial Strain Strain invariants

Strain invariants. The strain transformation equations are

( )

2 2

2 2

1 2

cos sin 2 sin

sin sin 2 cos

sin 2 cos2

aa xx xy yy

bb xx xy yy

ab yy xx xy

ε ε θ ε θ ε θ

ε ε θ ε θ ε θ

ε ε ε θ ε θ

= + +

= − +

= − +

If we simply add the first two equations together we get

aa bb xx yyε ε ε ε+ = +

This relationship is called a strain invariant because it implies that if we add the two normal strains together for the square oriented in any direction we get the same result. The variation of the strain components as a function of θ are shown in the following example:

aaε bbε abε

aa bbε ε+

6

6

6

5 10 3 10

12 10

xx

yy

xy

ε ε

ε

−

−

−

= ×

= − ×

= ×

Example of stress transformations

(red) (blue) (green)

aa

bb

ab

ε ε ε θ

1ε

2ε maxγ

π

aa bbε ε+

xxε

xyε

yyε

10

15

5−

10−

0

5

610−×

θ

1e

2e

be ae

CEE 213—Deformable Solids

© Keith D. Hjelmstad 2014

Multiaxial Strain Mohr’s circle

The interpretation of the “sign” of the shear strain is similar to the case for stress. Shearing “to the right” and shearing “to the left” are possible. So we use those as “plus” and “minus” in the plot..

Mohr’s circle. Christian Otto Mohr was the first person to realize that if you plot the pairs (εaa, εab) and (εbb, εab) for all values of the angle θ the result is a circle. We call this construction Mohr’s circle. As such, the strain transformation equations represent a parametric form of the equation of a circle. A little algebra can prove this assertion. Note that the principal strains like on the shear free axis and represent the maximum and minimum normal strains. The maximum shear strain is the radius of the circle. You can also see the orientation angles θ on the plot. Mohr’s circle provides a great visualization of the state of strain.

Christian Otto Mohr (1835-1918)

γ

( , )xx xyε ε

( , )yy xyε ε

1ε

( , )aa abε ε

( , )bb abε ε

mγ

2 pθ

2 mθ

mγ

ε2ε

2θ

CEE 213—Deformable Solids

© Keith D. Hjelmstad 2014

Multiaxial Strain Mohr’s circle construction

xyε

1 2 ( )xx yyε ε−

2 21 4 ( )xx yy xyR ε ε ε= − + The radius of Mohr’s

Circle can be computed from the Pythagorean theorem.

The center of Mohr’s Circle is the average of the normal stresses (an invariant).

Mohr’s circle construction. The Mohr circle construction can only be done for states of plane strain. Beyond being a very clever way to illustrate the state of strain, it provides an organization for doing the calculations of principal strains and maximum shear strain. In fact, because the sum of the normal strains is invariant, we can compute the center of the circle as the average of the normal strains in any orientation. Hence,

( )1 2center xx yyε ε= +

The radius of the circle can be computed by the Pythagorean theorem noting that the height of the triangle is the shear stress and the base is half the difference between the normal stresses.

2 21 4 ( )xx yy xyR ε ε ε= − +

γ

ε

( , )xx xyε ε

( , )yy xyε ε

1ε2ε R

1 2 ( )xx yyε ε+ 1 2 1 2( )ε ε+

1 2 ( )aa bbε ε+

( , )aa abε ε

( , )bb abε ε

CEE 213—Deformable Solids

© Keith D. Hjelmstad 2014

Multiaxial Strain Principal strains

λ= =z Fn n

where λ is an (as yet) unknown scalar magnitude of the stretch. We can note, since rotation of the deformed piece is arbitrary, we can assume b=c. Hence, F = I +E. Thus, the following are equivalent

where I is the identity matrix. This equation is called an eigenvalue problem. The solution is to find values of ε and n that satisfy the equation. This equation has a solution only if

[ ]det 0ε− =E I

Principal strains. At an angle with no shear strain the mapping causes only extension in the normal direction. Hence,

which is called the characteristic equation. This equation can be used to determine ε. The unit vector n can then be found from the original eigenvalue problem equation.

The principal plane is the etching (i.e., the angle θp) on which there is no shear strain. The equation for shear strain on the oblique plane at angle θ is

2 tan 2 xy

p xx yy

ε θ

ε ε =

−

( )1 2 sin 2 cos2ab yy xx xyε ε ε θ ε θ= − +

If the shear strain is zero then we can find the angle of the plane

θ

1 nλ ε= +

n 1

[ ]ε− =E I n 0

[ ] ( )

( )1

λ

λ

λ

− =

+ − =   − − =  

F I n 0

I E I n 0

E I n 0 1ε λ= −

CEE 213—Deformable Solids

© Keith D. Hjelmstad 2014

It appears that there are always two principal strains. Is it possible for there to be only one value? Yes, if

, 0xx yy xyε ε ε ε= = =

Then ε1=ε2.

This state has equal strain in all directions in the plane.

Multiaxial Strain Principal strains

( )( ) ( ) ( )

2

2 2

det 0

0

0

xx xy

xy yy

xx yy xy

xx yy xx yy xy

ε µ ε ε ε µ

ε µ ε µ ε

µ ε ε µ ε ε ε

−  = − 

− − − =

− + + − =

You can see that the characteristic equation for plane strain is simply a quadratic equation. Solving by the quadratic formula gives

( ) ( ) ( ) ( ) ( ) ( ) ( )

2 21 1 1,2 2 4

2 2 21 1 2 4

2 21 1 2 4

2

xx yy xx yy xx yy xy

xx yy xx xx yy yy xx yy xy

xx yy xx yy xy

ε ε ε ε ε ε ε ε

ε ε ε ε ε ε ε ε ε

ε ε ε ε ε

= + ± + − −

= + ± + + − +

= + ± − +

( ) ( )

( ) ( )

2 21 1 1 2 4

2 21 1 2 2 4

xx yy xx yy xy

xx yy xx yy xy

ε ε ε ε ε ε

ε ε ε ε ε ε

= + − − +

= + + − +

The two roots of the characteristic equation are called the principal strains and have the specific form

Principal strains. So, let’s compute the principal values of strain using the characteristic equation

θ

1 nλ ε= +

n 1

CEE 213—Deformable Solids

© Keith D. Hjelmstad 2014

Multiaxial Strain Summary

Summary. The concept of plane strain occurs in situations where the strain perpendicular to the plane is zero. This might occur, for example, in a thick body that is restrained from motion on the two face (e.g., a dam). The analysis of strain revolves around the description of the deformation map. In these notes we have considered a uniform deformation in which all points in the body feel the same state of strain. While this is generally not the case for solid bodies, it is approximately true in a small neighborhood of a point in the body. Strain is a measure of the distortion o)f the material (usually caused by stress). Strain is the change in length divided by the original length (normal strain) or the change in the relative angle of two lines (shear strain).

This set of notes has presented the analysis needed to explore the state of strain at a point in a body in a condition of plane strain. We encountered the concept of strain transformation which showed how the strain tensor components change as we etch a square at the point in question at different angles of orientation. The derivations all come from considerations of the motion map that result from reorientation of the unit material square. From the equations of strain transformation we have explored the conditions of maximum normal and shear strain and we have discovered how these can be computed from the strain components given in the original coordinate system. We found that the principal strains (maximum/minimum normal strains) occur at orientations where the shear strain is zero. The orientation of maximum shear strain has equal normal strains. We noted the similarity with stress transformations (including Mohr’s circle).

B A

A’

B’

Undeformed Deformed

Deformation map

9 Deform--Multiaxial Stress.pdf

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