Chapter 11 - Stress, Strain and Deformation in...

2001, W. E. Haisler Chapter 8: Stress, Strain and Deformation in Solids 1

Chapter 8 - Stress, Strain and Deformation in SolidsFor a beam in bending, we are often interested in determining the transverse displacement along the beam as shown below.

To do that, we would start with Conservation of Linear Momentum. Unfortunately, COLM is in terms of stress, . Consequently, we must have some way to relate stress to deformation. We will need additional equations as follows:

Constitutive relations - relate stress to strain

2001, W. E. Haisler Chapter 8: Stress, Strain and Deformation in Solids 2Kinematic relations - relate strain to displacement (gradients)

2001, W. E. Haisler Chapter 8: Stress, Strain and Deformation in Solids 3In the study of the motion of a solid or fluid, we will find it necessary to describe the kinematic behavior of a continuum body by defining expressions called strains in terms of the gradients of displacement components. In the example below, we consider an elastic bar of length L. If the bar is subjected by an axial force F, it will stretch an amount as shown in figure b). The quantity /L is a measure of the change in length relative to the original length and is defined to be the axial strain for the bar.

2001, W. E. Haisler Chapter 8: Stress, Strain and Deformation in Solids 4In figure d), a shear load is applied that is parallel to the top surface as shown. The angle measures the amount the original right angle in figure c) has changed from a right angle, and the angle is related to the shear strain.

What causes strain?

Mechanical loads (forces, pressures, etc.)

Temperature change (thermal expansion)

Moisture absorption ** Stress (relationship between stress and strain is the

constitutive relationship)

2001, W. E. Haisler Chapter 8: Stress, Strain and Deformation in Solids 5In this chapter we will mathematically formalize these simple ideas to develop expressions for strains in terms of displacement components. We will consider two approaches:

1) mathematically precise approach and 2) a simpler geometrical approach.


Deformations in solids are characterized by displacements of points and by elongations and rotations of line segments in the solid.


Note the following geometry quantities:P moves to P*, Q moves to Q*, line segment P-Q deforms (stretches/rotates) to P*-Q*:

= position vector of point P (initial state)= position vector of point P (in deformed state)= displacement vector of point P (from initial to

deformed state)= vector line segment between P and Q (initial state)= vector line segment between P and Q (deformed

state) = displacement vector of point Q (from initial to

deformed state)


From geometry of the vectors, we can write two equations:

displacement vector of point P: and

or

The last two terms represent the change (gradient) in displacement with respect to position , i.e.,

displacement gradient matrix (3x3). Note: not

symmetric!

uux x uy x uz xux y uy y uz yux z uy z uz z


Definition of strain. Strain is a measure of the deformation and rotation of line segments. Consider two material elements & , which undergo deformations that will

bring them into new locations & ; respectively.


Previously, we obtained for a line segment :

We will obtain identical expressions for the two line segments and defined in the figure above:

and

Consider the dot product of these two vectors (a scaler result!)


Recall that is a 3x3 matrix. The “T” in means that the matrix is transposed. The underlined term is defined as so that and

Finite Strain TensorNote that the expression for the finite strain tensor


contains two distinctive terms:

linear in displacement gradients

quadratic in displacement gradients


Infinitesimal Strain Tensor

If the higher order terms are neglected from the finite strain tensor [E] (i.e., keep only linear displacement gradient terms), we obtain the infinitesimal strain tensor, []:

where [] is a 3x3 matrix given by:


Note: Both [E] and [] are symmetric matrices.


A 2-D geometrical look at (defined to be the change in length of a line segment dx which is originally oriented in the x direction) and undergoes displacements ux and uy:


= (change in length)/(original length) .

Note: (for small a). Thus, the last result is approximately:


The strain now becomes


After canceling the dx terms, the finite strain term becomes

finite strain

For small gradients, we assume that the squared terms are small compared to the un-squared terms and we obtain the linear, or small strain expression:

small strain


Shear strain (rotation of line segments):Note that

xy = . Geometrically, each of the two terms is

an angle as shown at the left. xy is called a shear strain and geometrically is 1/2 (average) of the angular rotation of line segments dx & dy which originally form a right angle. In


contrast, the engineering shear strain xy is defined as the sum

of these two angles, ie, xy = 2 xy = .

The definition of the engineering shear strain xy from a graphical viewpoint is an approximation (similar to the square root approximation made in xx ). From the geometry above,

. Define xy to be the sum of the

angular rotations, ie, xy = . (engineering shear strain)


As in the square root approximation made for xx (for the geometrical interpretation of strain), an assumption of “small” rotations has been made in defining the shear strain xy .


Some examplesConsider a bar of lenght L, fixed at the left end, and with a force P applied at the right end. The bar stretches an amount uL at the right end. If the “engineering” strain in the x direction is defined to be (change in length)/(original length), then xx = uL/L.

We could also find the expresssion for u(x) and apply the mathematical definition of xx . We have two boundary conditions on u(x): u(0)=0 and u(L)= uL. Assume

where C1 and C2 are constants. Applying B.C.s gives C1 = 0 and C2 = uL/L so that u(x) = (uL/L) x. Thus the

strain is given by = (uL/L).

2001, W. E. Haisler Chapter 8: Stress, Strain and Deformation in Solids 24Example: At the point "A" shown on the wheel, the displacement field has been determined to be as follows (using the finite element method):

Determine the strains at point "A" which is located at x=4in, y=5in.

Point A


, Note:


Now, do again, but consider finite strain.

Comparing the two results, is it OK to assume small strains for this problem?

undeformed

deformed

2001, W. E. Haisler Chapter 8: Stress, Strain and Deformation in Solids 27Some Thought Exercises

Suppose we have an 8x8 square area outlined on a larger chunk of planar material (like a plate). Suppose the plate is loaded in the x-y plane so that the square is displaced and deformed to a rectangle as shown below [new x-y coordinates starting at lower left corner and going CCW are: (8,4), (20,4), (20,12) and (8,12)]. Picture shows “deformed” and “undeformed” area. What are your guesses for the strains? There is stretching in the x direction, so . There is no stretching in the y direction, so . All right angles remain right angles, hence there is no shear strain and .

2001, W. E. Haisler Chapter 8: Stress, Strain and Deformation in Solids 28We can actually solve for the displacements and strains since we know the initial and final positions of the 4 corner points. Assume that the displacements are given by (a 2-D curve fit):

Four constants are chosen because we know information at 4 points. Now apply know conditions for the four corner points:

2001, W. E. Haisler Chapter 8: Stress, Strain and Deformation in Solids 29In each of the above, the displacement at a point is set equal to the “final position” – “initial position” of that point. Solve for the constants and substitute back into and to obtain:

Last equation says that displacement of all points in y direction () is a constant, which is consistent with the “picture” of

deformed and undeformed area. Now calculate the strains:

(a positive value indicates stretching)

(no stretching in y direction)

(no shear strain)

undeformed deformed

2001, W. E. Haisler Chapter 8: Stress, Strain and Deformation in Solids 30Suppose we have an 8x8 square area outlined on a larger chunk of planar material (like a plate). Suppose the plate is loaded in the x-y plane so that the square is displaced and deformed to a parallelogram as shown below [new x-y coordinates starting at lower left corner and going CCW are: (8,4), (16,4), (19,12) and (11,12)]. Picture shows “deformed” and “undeformed” area. What are your guesses for the strains? There is no stretching in the x direction, so

. There is no stretching in the y direction, so . However, original right angles are no longer right angles, hence there is some shear strain and

.

2001, W. E. Haisler Chapter 8: Stress, Strain and Deformation in Solids 31Suppose we have an 8x8 square area outlined on a larger chunk of planar material (like a plate). Suppose the plate is loaded in the x-y plane so that the square is displaced and deformed to a quadrilateral as shown below [new x-y coordinates starting at lower left corner and going CCW are: (9,4), (16.8,6), (17.5, 14.5) and (10.8,12.5)]. Picture shows “deformed” and “undeformed” area. What are your guesses for the strains? It appears that there is shortening or stretching in the x and y direction; hence, and

. Original right angles are no longer right angles, hence there is some shear strain and .


Note on Tensor Stress & Strain Transformation

For 2-D, Cauchy's formula provided the following relation:

(3.31)

The component of in the direction of the unit outward normal of a surface (or in x' direction is:

Substitute Cauchy's formula into the above, and write in both vector and matrix notation:


The expression gives the component of stress, , in the direction of the unit normal (or in the direction of

the x'-axis which makes an angle CCW from the x-axis).

Lets do both the vector and matrix operations to show that they are the same. First, Cauchy's formula is


Note . Now do the remaining vector operation for to obtain:

If we do this in matrix notion we obtain:

(5.3)


The above is called the stress transformation equation (see eq. 5.3 in the notes). The stress transformation transforms stresses from an (x,y) coordinate system to an (x',y') system where x' is rotated by an angle CCW from the x-axis.

We can similarly show that the strain transformation is given by:

The quantity is the component of strain in the direction of a unit normal . is often called the unit elongation in the direction (just as is the unit elongation in the or x-coordinate direction).


Notes: 1. Both and are second order tensors.

2. All second order tensors follow the same transformation form in transforming from (x,y,z) to another orthogonal coordinate system (x',y',z'), i.e.,

and


3. The same transformation applies to moments of inertia of a cross-section A (which is also a second order tensor):

, ,

With respect to the x'-y' coordinate system at some angle , we have:

Can also get by applying the coordinate transformation to the x-y moments of inertia written as a matrix: ( is unit vector in x' direction):

, then

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