Analysis of Deformation Deformation: When Loads Or Forces Are Applied to a Solid Body This Causes the Body to Change Shape Or Geometry

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Chapter 5: Analysis of Deformation Deformation: When loads or forces are applied to a solid body this causes the body to change shape or geometry. Such a change is called “deformation” of that body. In fluids, the applied forces cause the fluids to “flow” instead of to “deform”.

Consider a piece of rubber in the form of a string of constant cross-section and length L0, call this the “initial length”. If a force is applied to the two ends of the string, it will stretch, i.e. change shape or deform, to a length L. Call this length the “current length” produced by the applied forces. How would one quantify such a deformation in an objective way? One might say that to measure or quantify the deformation, we can use the difference in length, ∆L = L – L0, for this purpose. This measure, however, is not informative as the absolute value of ∆L is not as important as the relative ratio of ∆L to the original length L0 say. For example, if ∆L is 1mm this can be seen as either too little if L0 = 1m or too much if L0 = 2mm say. This objective measure of deformation is what we call a “strain” measure. Hence, one measure of deformation or strain, call it ∈, is L − L L ∈= 0 = −1 = λ −1 L0 L0

where the ratio L/L0 is termed the “stretch” and denoted by λ here. Another measure of strain might be expressed as L − L L − ∈′= 0 =1− 0 = 1− λ 1 L L where it might be elected to normalize by the current length L instead. Notice that a prime was added to ∈ here to distinguish it from the different strain measure given above first.

If the deformation, i.e. change of length, is infinitesimal, then the two measures of strain are the same essentially. For example, take L = 1.00 and L0 = 1.01, then ∈ = ∈´ = 0.01. However, if L = 2 and L0 = 1, then ∈ = 1 and ∈´ = ½. and the two measures of strain give different results. It is to be noted at this juncture that other strain measures exists but will be left for future discussions.

There are two main modes of deformation. One of them is induced by normal stresses like pulling on or compressing a flat board of some thickness. See figure below. Here rectangles were sketched on the board (on its outside boundary) along its thickness before the force was applied. Such markers help us visualize the ensuing deformation after load application. Note in this particular example that the rectangles were stretched or elongated in one direction, the applied tensile stress direction, and compressed or shortened in the transverse or normal to that direction. Another mode of deformation is obtained by again marking a box of material, and looking at it from its frontal view, and then shearing the bottom and top faces of the box with shear forces. The ensuring deformation is shown on the right-hand side. In this type of deformation, the sides of each rectangle maintained original dimensions but changed interior angles. Any general deformation is composed of these two modes superposed.

F F

Example of deformation due to tension or compression

Example of deformation due to shearing

Strain in a material: Consider a small element of material of differential dimensions ABCD, originally a rectangle that is subjected to stresses and thus deforms to assume the shape A´B´C´D´. Such an element does not have to be stand-alone and can be considered as part of a larger body of material. Assume the analysis to be in 2D for simplicity. In 3D, the ideas are the same, and we thus can generalize the 2D results to 3D deformation. ∂u dy ∂ y C´ α´

∂v D´ dy ∂ β B´ y D´´ C´´ ∂v dy dx ∂x α A´, A´´ B´´ dx v D C ∂u dx dy ∂x

A B dx y u x

Point A is displaced u units in the x-direction and v units in the y-direction. In the above figure, everything is exaggerated, e.g. the size of the element and the displacements. Here we are only concerned about small displacements or deformations of the infinitesimal element. Notice that the deformed shape A´BĆ´D´ can be obtained from ABCD by three operations in sequence: 1- Rigid body translation to A´´B´Ć´´D´´ 2- Uniform stretching of A´´B´Ć´´D´´ in the x and y directions to A´´´B´´Ć´´´D´´´ which changes the lengths of element’s sides 3- Uniform shearing of the element to A´BĆ´D´. This shearing does not change the length of the sides of A´´´B´´Ć´´´D´´´ but changes the angles between two originally perpendicular sides of the element.

dy C´ A B dx

║ D´ C´´´ D´´ C´´ B´ D´´´

+ + A´´ B´´ A´´´ B´´´ A´

During step 1, AB = A′′B′′ = DC = D′′C′′ , and AD = A′′D′′ = BC = B′′C′′ which preserves the side lengths but distorts or changes the angle between any two originally perpendicular sides of the element. Notice that this step does not have to be purely rigid body translation. It can also be a rigid body rotation that also preserves the original side lengths. However, for simplicity of illustration only the translation is shown. Notice that during step 2, the side lengths of A´´B´Ć´´D´´ change. Therefore, A′′′B′′′ ≠ A′′B′′, D′′′C′′′ ≠ D′′C′′ , and A′′′D′′′ ≠ A′′D′′, B′′′C′′′ ≠ B′′C′′ . Finally during step 3, the side lengths stay the same (i.e. A′B′ = A′′′B′′′, D′C′ = D′′′C′′′, A′D′ = A′′′D′′′, B′C′ = B′′′C′′′ ). However, during this step, the initial right angles change. For example, the angle ∟ D´´Á´´´B´´´ becomes — DÁ´B´ which is less that 90o by an amount equal to α+α´. Similarly, for the angle ∟ B´´Ć´´´D´´´. However, the angle — A´BĆ´ is greater than 90o by an amount equal to α+α´. Similarly for — A´DĆ´. Notice that in step 1, the rectangle ABCD simply translates to A´´B´Ć´´D´´.

Before we carry on, notice a couple of things. First, the displacements of point A (the lower left corner of the element) in the x and y-directions are given by u and v (or equivalently by ux and uy or by u1 and u2), respectively. Since we randomly picked a

material element, hence the displacements of point A (the lower left corner of the material element) will depend on the spatial position of this element or point within the larger continuum or body that contains it, i.e. u = u(x,y) and v = v(x,y). In general, u = u(x,y,z), v = v(x,y,z) and there is a third component of displacement w perpendicular to the plane of u and v which is also a function of the three spatial coordinates, i.e. w = w(x,y,z).

For example, if the displacements of point A in a body, or in an element in a body, is known as u = u(x,y) and v = v(x,y), then the displacements of any nearby point P (like points C, B or D) can be found from u = u(x,y) and v = v(x,y) using Taylor series expansion as follows: ∂ ∂ ∂2 ∂2 ∂2 P = + u + u + u 2 + u 2 u u u dx dy (dx) (dy) + dxdy +L ∂x ∂y ∂x2 ∂y2 ∂x∂y Similarly, ∂ ∂ ∂2 ∂2 ∂2 P = + v + v + v 2 + v 2 v v v dx dy (dx) (dy) + dxdy +L ∂x ∂y ∂x2 ∂y2 ∂x∂y , where dx = xP - xA = xP - x, dy = yP - yA = yP - y, uP and vP represent the displacements of any point P in the material knowing the displacements u and v of another point A (here we are assuming continuous displacement distributions or function in the continuum, otherwise it is not be possible to take the derivatives).

Considering very small variations, i.e. gradients, of displacement and that points A and P are very close from each other to begin with, we can drop the higher order derivatives in the last two equations and write (or assume to first order accuracy): ∂u ∂u u P ≅ u + dx + dy ∂x ∂y ∂v ∂v vP ≅ v + dx + dy ∂x ∂y If point P happens to be point B, then we have ∂u u B ≅ u + dx (since dy = 0 here) ∂x ∂v vB ≅ v + dx (since dy = 0 here) ∂x Also, ∂u u D ≅ u + dy (since dx = 0 here) ∂y ∂v vD ≅ v + dy (since dx = 0 here) ∂y

From this last figure, we see that we can define two types of deformation measures or strain to which the element is subjected. One type of strain is associated with stretching of element ABCD (or alternatively A´´B´´C´´D´´). This is step 2. This type of strain which involves stretching or contracting of a material element is termed “extensional” or “normal strains”. If the strains are extensional then they are called “tensile” strains, if

they are contractional then they are called “compressive” strains. The other type of strain is associated with shearing of element A´´´B´´Ć´´´D´´´ by changing the angles between element sides while preserving their lengths (e.g. from A´´´B´´Ć´´´D´´´ to A´BĆ´D´). The shearing preserves the lengths of element sides but changes the angles between its sides. This deformation describes what is called “shearing” or “shear strain”.

The above two definitions of strain can be mathematically formulated or stated as follows. Normal, extensional or stretching strains (also called tensile or compressive stresses) in A′′′B′′′ − AB A′′′B′′′ − dx A′′′B′′′ the x-direction: e = = = −1 xx AB dx dx A′′′D′′′ − AD A′′′D′′′ − dy A′′′D′′′ The normal strain in the y-direction: e = = = −1 yy AD dy dy

Notice that the above definition of normal strains produces unitless or dimensionless strains since the changes in length are relative or normalized to the original unstretched lengths. Similar to stress definition, the first subscript indicates the direction of the face or the plane that is normal to the extended line (AB or AD in this case). The second subscript indicates the direction of the extension itself.

To define shear strain, we can write: π γ = − β = change in angle (in radians) = α +α′ xy 2 , where the first subscript indicates the face normal direction, and the second subscript indicates the shearing direction.

Notice in the above definitions of strain that if we are only assuming small changes in displacements and small changes of angles, we should then expect that the resulting strains be much less than unity (i.e. exx << 1, eyy << 1, and γxy << 1)

The shearing strain γxy is called the “engineering shear strain” and is equal to 2exy, where exy is called the tensorial shear strain. exy is given by exy = ½ (α+α´) which represents the average angle change between α (or angle B´´A´´B´) and α´ (or angle D´´A´´D´).

From the definitions of strain, we can write: ′ ′ 2 = ′′′ ′′′ 2 = + 2 (A B ) (A B ) (dx(1 exx )) But from the last sketch, ∂u ∂v (A′B′) 2 = (dx+ dx)2 + ( dx)2 ∂x ∂x Equate the last two equations, we get: ∂u ∂u ∂v (1+ 2e + e2 )(dx)2 = (dx)2 + 2( )(dx)2 + ( ) 2 (dx)2 + ( )2 (dx) 2 (*) xx xx ∂x ∂x ∂x

Assuming very small deformation (i.e. displacements and gradients of displacement <<1), we can then drop the squared or higher-order terms from the last equation to get ∂u e = (A) xx ∂x Similarly, we can find ∂v e = (B) yy ∂y Also, from the previous sketch ∂v ∂v dx ∂ ∂ α = x = x ∂u ∂u dx + dx 1+ ∂x ∂x ∂u ∂u Assuming <<1⇒1+ ≅ 1 ∂x ∂x ∂v Hence, α = ∂x ∂u Similarly, α′ = ∂y ∂u ∂v γ = + (C) xy ∂y ∂x , where the partial derivatives are positive if A´´´B´´´ and A´´´ D´´´ rotate inward as shown in the sketch.

Equations (A), (B) and (C) are called the strain-displacement relations for “small strain theory” (also called “infinitesimal strain theory”). In some texts, this is also called “small displacement theory”.

Notice that using (C), the tensorial shear strain (or simply shear strain) is given by γ ′ xy α +α 1 ∂u ∂v e = = = ( + ) (D) xy 2 2 2 ∂y ∂x which is simply the average of the two angles α and α´.

So far we have dealt with 2D or planar strains (equations (A)-(D)). In 3D, the strains become: ∂u e = xx ∂x ∂v e = yy ∂y ∂w e = zz ∂z ∂u ∂v γ = + xy ∂y ∂x

∂u ∂w γ = + xz ∂z ∂x ∂v ∂w γ = + yz ∂z ∂y The last equations for strain can be written more concisely using index (indicial) or tensor notation as: 1  ∂u ∂u  e =  i + j  ,i = 1 3, j =1 3 ij  ∂ ∂  L L 2  x j xi  From the last notation, we notice that 1  ∂u ∂u  e =  j + i  = e ji  ∂ ∂  ij 2  xi x j 

2 Now, lets go back to (*), if we only drop exx from that equation, we get: 2 2 ∂u 1  ∂u   ∂v   e = +  +  xx ∂  ∂   ∂   x 2  x   x   2 2 ∂v 1  ∂u   ∂v   e = +   +    yy ∂y 2  ∂y   ∂y        and ∂ ∂ ∂ ∂ ∂ ∂ = 1  u + v + u u + v v  exy   2  ∂y ∂x ∂x ∂y ∂x ∂y  The above strain equations are called the strain-displacement for “finite strains” or “large strains”

The 3-D version of the above strain equation is: 2 2 2 ∂u 1  ∂u   ∂v   ∂w   e = +  + +  xx ∂  ∂   ∂   ∂   x 2  x   x   x   2 2 2 ∂v 1  ∂u   ∂v   ∂w   e = +   +   +    yy ∂y 2  ∂y   ∂y   ∂y          2 2 2 ∂w 1  ∂u   ∂v   ∂w   e = +  + +  zz ∂  ∂   ∂   ∂   z 2  z   z   z   ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ = = 1  u + v + u u + v v + w w  exy eyx   2  ∂y ∂x ∂x ∂y ∂x ∂y ∂x ∂y  ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ = = 1  u + w + u u + v v + w w  exz ezx   2  ∂z ∂x ∂x ∂z ∂x ∂z ∂x ∂z  ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ = = 1  v + w + u u + v v + w w  eyz ezy   2  ∂z ∂y ∂y ∂z ∂y ∂z ∂y ∂z 

The last equations can be written concisely as: 1  ∂u ∂u ∂u ∂u  e =  i + j + k k  ij  ∂ ∂ ∂ ∂  2  x j xi xi x j 

It is important to note that in developing the strain-displacement relations above, no assumption was made regarding the constitution of the material (whether it is isotropic or anisotropic, homogeneous or inhomogeneous). The whole derivation was solely carried out on geometric grounds or arguments.

Notice also that the different strains can be considered to be the components of a two- dimensional array, or second-rank tensor in other words, e, such that

exx exy exz  =   e eyx eyy eyz  for 3-D   ezx ezy ezz  And

exx exy  e =   for 2-D eyx eyy 

And just as we solved for the “principal stresses” if given a stress tensor σ, we can solve for the “principal strains” if given a strain tensor e. This is done by solving for the roots of the characteristic equation that results from − exx e exy exz − = eyx eyy e eyz 0 − ezx ezy ezz e The resulting characteristic equation can be written as: 3 2 e − e I 1 − eI 2 − I 3 = 0 where I 1 , I 2 and I 3 are called the invariants of the strain tensor and are given by = + + = I 1 exx eyy ezz eii e e e e e e = − xx xy − xx xz − yy yz = 2 + 2 + 2 − − − I 2 exy exz eyz exxeyy exxezz eyyezz eyx eyy ezx ezz ezy ezz

exx exy exz = = I 3 eyx eyy eyz e

ezx ezy ezz We can also find the “principal strain directions” just as we’ve done for the stresses.

Now since strain e or eij is a second-rank tensor, it transforms accordingly: = β β e p′q′ p′i q′j eij Or = β β eij ip′ jq′e p′q′

which means that if one knows the strain tensor in one orthogonal coordinate system, it is possible to find the strain tensor with respect to another orthogonal coordinate system if the rotation or transformation matrix between the two is known or can be established at least.

The Strain in a material: Another Look The deformation in a material can be looked at differently than what was presented above. Consider a small material volume or particle that initially was located at point P with coordinates (a1, a2, a3), this is called the “original configuration”. After deformation, or at a given time of consideration during the deformation, the particle moved to point Q with coordinates (x1, x1, x3). This is called the “current” or “final configuration”. The → vector PQ is called the displacement vector. It is also denoted by the symbol u. Hence, = − ui xi ai If the displacement vector of every particle in the original configuration is known, then one can construct the deformed body, i.e. the current or final configuration, from such knowledge. Moreover, the coordinates, i.e. the xi’s, of every point in the deformed body must have depended on the location of that point in the body prior to deformation, i.e. on (a1, a2, a3). This is because there is a one-to-one correspondence between these two locations. Mathematically, this is stated as = xi xi (a1 ,a2 ,a3 ) Vice versa, it can be easily seen that the inverse relationship should also exist, i.e. that there is a reverse mapping that maps the current configuration onto the original one. Mathematically, this is stated as = ai ai (x1 , x2 , x3 ) Hence, the displacement vector can be written as = − ui (a1 ,a2 ,a3 ) xi (a1 ,a2 ,a3 ) ai or as = − ui (x1 , x2 , x3 ) xi ai (x1 , x2 , x3 )

(x1, x2, x3) u

P (a1, a2, a3)

x1 Consider an infinitesimal line element connecting the point P(a1, a2, a3) to a neighboring point P´(a1+da1, a2+da2, a3+da3). The square of the length ds0 of PP´ in the original undeformed configuration is given by 2 = 2 + 2 + 2 = ds0 da1 da2 da3 dai dai When P and P´ are deformed to point Q(x1, x1, x3) and Q´( x1+dx1, x2+dx2, x3+dx3), respectively, the square of the length ds of the element QQ´ in the current or final deformed configuration is given by 2 = 2 + 2 + 2 = ds dx1 dx2 dx3 dxi dxi But using the equations above, we know that ∂x ∂a dx = i da da = i dx i ∂ j, i ∂ j a j x j The last two equations can be re-written using the Kronecker delta as ∂ ∂a aβ ds 2 = δ da da = δ α dx dx 0 αβ α β αβ ∂ ∂ i j xi x j ∂ ∂x xβ ds 2 = δ dx dx = δ α da da αβ α β αβ ∂ ∂ i j ai a j Note now that one can now write

∂ ∂ ∂x xβ  ∂x xβ  ds 2 − ds 2 = δ α da da − δ da da = δ α − δ da da 0 αβ ∂ ∂ i j ij i j  αβ ∂ ∂ ij  i j ai a j  ai a j  Or ∂ ∂ ∂a aβ  ∂a aβ  ds 2 − ds 2 = δ dx dx − δ α dx dx = δ − δ α dx dx 0 ij i j αβ ∂ ∂ i j  ij αβ ∂ ∂  i j xi x j  xi x j  We define here two different “strain tensors” ∂ 1  ∂xα xβ  E = δ − δ  ij  αβ ∂ ∂ ij  2  ai a j  ∂ 1  ∂aα aβ  e = δ − δ  ij  ij αβ ∂ ∂  2  xi x j  So that 2 − 2 = ds ds0 2Eij dai da j 2 − 2 = ds ds0 2eij dxi dx j

The strain tensor Eij, was introduced by Green and St.-Venant and is called “Green’s strain tensor”. The strain tensor eij was introduced by Cauchy for infinitesimal strains (small-displacement theory) and by Almansi and Hamel for finite strains (i.e. finite-displacement theory). Eij is often referred to as the “Lagrangian strain tensor” and eij is often referred to as the “Eulerian strain tensor”, in analogy to terminology in hydrology or fluid mechanics, as the derivatives are taking with respect to the old or initial configurations in Eij and with respect to the current or final configuration in eij. As can be seen from the last equations, the tensors Eij and eij are symmetric, i.e. Eij = Eji, eij = eji Another observation about the above equations is that if the line length of QQ´ 2 = 2 remains unchanged from PP´ then we must have ds ds0 which implies that Eij = eij = 0. If the length of small lines ALL remain unchanged during the displacement then the body must have experienced rigid body motion and no deformation. Hence, a necessary indeed sufficient condition for rigid body motion definition is that the strain tensor Eij or eij must be zero when evaluated at every point in the body.

Q´ Q

(x , x , x ) u 1 2 3 P´

(a1, a2, a3)

x 1

Strain Components in terms of Displacement: From before, we know that we can write uα = xα − aα (α = 1,2,3) This last equation can be re-arranged to be

xα = uα + aα or a α = x α − u α Taking derivatives, we can write ∂xα ∂uα ∂aα ∂uα = + δ and = δ − ∂ ∂ αi ∂ αi ∂ ai ai xi xi Based on this, the strain tensors can be written as ∂ 1   ∂uα  uβ   E = δαβ  + δα  + δ β  −δ  ij   ∂ i  ∂ j  ij  2  ai  a j     ∂u ∂ ∂ ∂  = 1  j + ui + uα uα   ∂ ∂ ∂ ∂  2  ai a j ai a j 

∂ 1   ∂uα  uβ  e = δ −δαβ δα − δ β −  ij  ij  i ∂  j ∂  2  xi  x j     ∂u ∂ ∂ ∂  = 1  j + ui − uα uα   ∂ ∂ ∂ ∂  2  xi x j xi x j  Replacing x1, x1, x3 with x, y, z, and a1, a2, a3 with a, b, c, and u1, u2, u3 with u, v, w, we can write the above as 2 2 2 ∂u 1  ∂u   ∂v   ∂w   E = +   +   +    aa ∂a 2  ∂a ∂a ∂a         2 2 2 ∂v 1  ∂u   ∂v   ∂w   E = +   +   +    bb ∂b 2  ∂b ∂b ∂b         2 2 2 ∂w 1  ∂u   ∂v   ∂w   E = +   +   +    cc ∂c 2  ∂c ∂c ∂c         ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ = 1  u + v +  u u + v v + w w  Eab    2  ∂b ∂a  ∂a ∂b ∂a ∂b ∂a ∂b  ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ = 1  u + w +  u u + v v + w w  Eac    2  ∂c ∂a  ∂a ∂c ∂a ∂c ∂a ∂c  ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ = 1  v + w +  u u + v v + w w  Ebc    2  ∂c ∂b  ∂b ∂c ∂b ∂c ∂b ∂c  2 2 2 ∂u 1  ∂u   ∂v   ∂w   e = −   +   +    xx ∂x 2  ∂x ∂x ∂x         2 2 2 ∂v 1  ∂u   ∂v   ∂w   e = −   +   +    yy ∂y 2  ∂y   ∂y   ∂y          2 2 2 ∂w 1  ∂u   ∂v   ∂w   e = −   +   +    zz ∂z 2  ∂z ∂z ∂z         1  ∂u ∂v  ∂u ∂u ∂v ∂v ∂w ∂w  e =  + −  + +  xy  ∂ ∂  ∂ ∂ ∂ ∂ ∂ ∂  2  y x  x y x y x y  ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ = 1  u + w −  u u + v v + w w  exz    2  ∂z ∂x  ∂x ∂z ∂x ∂z ∂x ∂z  1  ∂v ∂w  ∂u ∂u ∂v ∂v ∂w ∂w  e =  + −  + +  yz  ∂ ∂  ∂ ∂ ∂ ∂ ∂ ∂  2  z y  y z y z y z 

If the relative difference between the displacements of one point in the material and another neighboring one, i.e. the gradient of displacement, is very small then the products of these gradients is much smaller compared to the gradients themselves and can be neglected. In this case eij reduces to the “Cauchy infinitesimal strain tensor” ∂ 1  u j ∂u  e = E =  + i  ij ij  ∂ ∂  2  xi x j  In the case of infinitesimal displacement, which means that (x1, x1, x3) is essentially (a1, a2, a3), the distinction between the Lagrangian and Eulerian strain tensors disappears since it then makes no difference if derivatives are carried out with respect to the original or deformed configuration.

Rigid Body Rotation during Displacement: Consider again points P and P´ from before. Those two points are separated by a small differential vector dx, where point P lies at x and point P´ lies at x+dx, and for this discussion lets assume infinitesimal displacements such that there is not much difference between the initial and final configurations. For the components of du, we can write ∂u du = i dx i ∂ j x j We can re-write the last equation as 1  ∂u ∂u  1  ∂u ∂u  du =  i + j dx +  i − j dx i  ∂ ∂  j  ∂ ∂  j 2  x j xi  2  x j xi  1424 434 1424 434 e ω ij ji

If the infinitesimal strain tensor, eij, tends to zero, i.e. there is no deformation in the body, then the difference in the displacement of these two points is due to rigid-body motion by ω definition. Hence, the tensor ji must be related to the rotation of that material element or particle associated with point P and its immediate neighborhood. This tensor is called the “rotation tensor” of the displacement field ui. Therefore, in this case, we can write the last equation as = ω dui ji dx j Note, however, that the rotation tensor is anti-symmetric, i.e. ω = −ω ij ji For any such anti-symmetric tensor we can always find what is called a “dual vector” associated with this tensor. We build this vector as 1 ω = ∈ ω k 2 kij ij If this is a dual vector, i.e. one that is derived from the tensor and associated with it, we should also be able to get the inverse relationship as well, i.e. express the tensor in terms ∈ of the dual vector. To show that, multiply the last equation by ijk 1 1 1 1 ∈ ω = ∈ ∈ ω = ∈ ∈ ω = (δ δ −δ δ )ω = (ω −ω ) = ω ijk k 2 ijk kmn mn 2 kij kmn mn 2 im jn in jm mn 2 ij ji ij

Hence ω =∈ ω ij ijk k

The vector ω above is called the “rotation vector” of the displacement field ui.

Based on all of the above, we can now write = ω = −ω = −∈ ω =∈ ω = × dui jidx j ijdx j ijk k dx j ikj k dx j (ω dx)i Therefore du = ω ×dx

Finite Strain Components: For large strains it is not as easy as for small strains to give a geometric interpretation of the deformation. We do so here for one component of strain E11. Consider rectangular coordinates and a line element of material that is da before deformation ensues. Let the components of da be da1 = ds0, da2 = 0, da3 = 0. Let the extension of the this element E1 be defined by − = ds ds0 E1 ds0 or = + 2 = + 2 2 = + 2 2 ds (1 E1)ds0 ⇒ ds (1 E1) ds0 (1 E1) (da1) From before we know that 2 − 2 = = 2 2 = 2 + 2 = 2 + 2 ds ds0 2Eij daida j 2E11(da1) ⇒ ds ds0 2E11(da1) (da1) 2E11(da1) From the last two equations, we can write = + 2 − 2E11 (1 E1) 1 This last equation gives the meaning of E11 in terms of E1. Alternately, we can write = + − E1 1 2E11 1 If E11 is small compared to 1, i.e. we have the case of small strains, the last equation reduces to = E1 E11 Which is the geometric interpretation of strain in the case of small strains.

Infinitesimal Strain Components in Polar Coordinates: The relationships between rectangular Cartesian coordinates and polar coordinates are given by − y x = r cosθ, θ = tan 1 , x y = r sinθ, r 2 = x2 + y2 , z = z

∂r x ∂r y = = cosθ , = = sinθ, ∂x r ∂y r

∂θ y sinθ ∂θ x cosθ = − = − , = = ∂x r 2 r ∂y r 2 r From the theory of partial differentiation, we can write ∂ ∂ ∂ ∂ ∂θ = r + ∂x ∂r ∂x ∂θ ∂x ∂ ∂ ∂ ∂ ∂θ = r + ∂y ∂r ∂y ∂θ ∂y The last derivatives can be transformed into derivatives with respect to x and y ∂ ∂ sinθ ∂ = cosθ − , ∂x ∂r r ∂θ

∂ ∂ cosθ ∂ = sinθ + ∂y ∂r r ∂θ

In polar coordinates, the displacement vector u has components ur, uθ, uz as shown in the figure.

θ-direction y y´-direction

r-direction u x´-direction uy uθ

r θ x

These components can be resolved into the x, y and z directions of the Cartesian coordinate system to give the ux, uy, and uz components. The relationship between the Cartesian and Polar displacement components is given by ′ = β ui iju j or = β ′ ui jiu j

In expanded form, the above equations can be written as = θ − θ u x ur cos uθ sin = θ + θ u y ur sin uθ cos = u z u z

The strain tensor in polar coordinates is given by

err erθ erz  e = e e e   θr θθ θz  ezr ezθ ezz  The transformation between strains in Cartesian coordinates and in polar coordinates is governed by = β β e p′q′ p′i q′j eij Which, in expanded form, can be written as = 2 θ + 2 θ + θ θ err exx cos eyy sin 2cos sin exy = 2 θ + 2 θ − θ θ eθθ exx sin eyy cos 2cos sin exy e = e zz zz = − θ θ + 2 θ − erθ (eyy exx )sin cos (2cos 1)exy = θ + θ erz exz cos eyz sin = − θ + θ eθz exz sin eyz cos

Using the definitions ∂u e = x xx ∂x ∂u e = y yy ∂y ∂u e = z zz ∂z 1  ∂u ∂u  =  x + y  exy   2  ∂y ∂x  ∂u ∂u = 1  x + z  exz   2  ∂z ∂x  1  ∂u ∂u  =  y + z  eyz   2  ∂z ∂y  one can now combine all of the previous equations to get ∂u e = r rr ∂r ∂ ur 1 uθ eθθ = + r r ∂θ

∂u e = z zz ∂z ∂u ∂u u = = 1  1 r + θ − θ  erθ eθr   2  r ∂θ ∂r r  ∂u ∂u = = 1  r + z  erz ezr   2  ∂z ∂r  ∂u ∂u = 1  1 z + θ  ezθ   2  r ∂θ ∂z 

The last equations are the expressions of the strain tensor components in a polar coordinate system (using polar coordinates and displacement components in polar coordinates).

Of course, an alternative to deriving the strain components in polar coordinates is to resort to graphical interpretation of the different strain components. This is done in the textbook and is not repeated here. The important thing is that both ways give exactly the same final results.

Strain-Compatibility Relations: If we are given the displacements u, v and w as a function of spatial coordinates, then we can derive from them the different strain components using the displacement- strain relationships we’ve learned before. However, if we are given the strain components (six of them in the 3-D case), we can not in general integrate the strains (using the displacement-strain relationships) to uniquely determine the 3 displacement components u, v and w. To uniquely determine the strains, we need to have more relations among the strain components. These relations are termed the “strain compatibility relations” and they are now illustrated for the 2-D case. Consider a case of “plane strain” where all non- vanishing strains lie in the xy-plane and any strains associated with the normal to the plane are all zeroes. This state of strain can happen if u = u(x,y), v = v(x,y) and w = constant. For here, we get (assuming infinitesimal strains) ∂ ∂ ∂ ∂ = u = v = 1  u + v  exx , eyy , exy   ∂x ∂y 2  ∂y ∂x  = = = ezz exz eyz 0 In this case, there is a relation (a “strain compatibility relation”) that exists between the planar non-vanishing strains. We can obtain this relation by differentiation as follows: ∂2 ∂3 ∂2 ∂3 ∂2 ∂3 ∂3 = u = v = u + v exx , eyy , 2 exy ∂y2 ∂x∂y2 ∂x2 ∂x2∂y ∂x∂y ∂x∂y2 ∂x2∂y Now, notice that the from the last equations, we can write 2 ∂2 ∂2 ∂ e eyy exy xx + = 2 ∂y2 ∂x2 ∂x∂y

This is a relation among three infinitesimal strain components. It is also given for 2D deformation and thus called “strain compatibility for plane strain”. In the general 3D case, six second-order equations or conditions exist among six components of strain.

To reiterate, if one is given the displacements (u,v,w) in terms of the spatial coordinates, then one can simply substitute them into the displacement-strain relationships, i.e. differentiate and find the strains using small strain theory. As long as the displacements represent continuous functions of the spatial coordinates, then the strain will also be continuous functions that satisfy the compatibility equations.

However, if one instead is given the strains as a function of the spatial coordinates, one has to make sure first that these functions satisfy the strain-compatibility equations, otherwise this is not a permissible strain state distribution of the material, and one can not uniquely determine the displacements out of the given strains.

The physical meaning of “incompatible” strains is that the given strain distributions prescribes a condition of either tearing or overlapping in the material (i.e. a condition of discontinuity in the displacements).

Example 1 on strains: Consider a square plate of unit size deformed as shown in the figure. Determine the strain components. Note: This state of deformation is termed “simple shear”

a 2 x2

1 1 30o

O a1 O x1 1 1

Solution: The mapping of the material point coordinates before deformation (a1,a2,a3) and after deformation (x1,x2,x3) is given by = + 1 = = x1 a1 a2 , x2 a2 , x3 a3 3 Where 1/√3 = tan30o Inversely, we can write = − 1 = = a1 x1 x2 , a2 x2 , a3 x3 3

Hence, the displacement components are = − = 1 = 1 = − = = − = u1 x1 a1 a2 x2 , u2 x2 a2 0, u3 x3 a3 0 3 3 The strain components are = = = E11 E33 E13 E23 = E22 1/ 6 E = 1 2 3 12 = = = e11 e33 e13 e23 = − e22 1/ 6 = e12 1 2 3

Example 2 on strains: ABCD is a unit square in the xy-plane (see figure) and is part of a large deformable body. If the strain in the body is small, uniform and given by 1 2 3   x   × −3 2 D C 2 1 0 10 3 0 2 1

A B 1

O x1

What is the change in length of the line AC due to deformation?

Solution: The original line length of AC before deformation is ds0 = √2=1.41421356 The new line length of AC is ds, the difference between the squares of these two lengths is given by: 2 − 2 = ds ds0 2eij dxi dx j 2 − 2 = + + ds ds0 2[e1 j dx1dx j e2 j dx2dx j e3 j dx/ 3dx j ] 2 − 2 = + + + ds ds0 2[e11dx1dx1 e12dx1dx2 e21dx2dx1 e22dx2 dx2 ] ds 2 − ds 2 = 2[e (dx )2 + 2e dx dx + e (dx )2 ] 0 11 1 12 1 2 22 2 = = dx1 dx2 1 ⇒ ds 2 − 2 = 2[1× (1)2 + 2× 2×1×1+1× (1)2 ]×10−3 ds 2 = 2 + 0.012 = 2.012 ⇒ ds =1.41845 − = = ds ds0 change in length 0.004236

Example 3 on strains: A displacement field is defined by = = − + = = − = = − + u u1 Cx2 Bx3 , v u2 Cx1 Ax3 , w u3 Bx1 Ax2 Show that this field represents pure rotation and no deformation.

Solution: For deformation to happen, we need to have non-zero strain eij component(s). Check: ∂u ∂u e = e = = 1 = 0 xx 11 ∂ ∂ x x1 ∂v ∂u e = e = = 2 = 0 yy 22 ∂ ∂ y x2 ∂w ∂u e = e = = 3 = 0 zz 33 ∂ ∂ z x3 1  ∂u ∂v  1  ∂u ∂u  1 e = e =  +  =  1 + 2  = ()− C + C = 0 xy 12  ∂ ∂   ∂ ∂  2  y x  2  x2 x1  2 1  ∂u ∂w  1  ∂u ∂u  1 e = e = + =  1 + 3  = ()B − B = 0 xz 13  ∂ ∂   ∂ ∂  2  z x  2  x3 x1  2 1  ∂v ∂w  1  ∂u ∂u  1 e = e =  +  =  2 + 3  = ()− A + A = 0 yz 23  ∂ ∂   ∂ ∂  2  z y  2  x3 x2  2 Since e = 0 from above, there is no deformation associated with this displacement field.

Check now w , the rotation tensor: ω = ω = ω = 11 22 33 0 1  ∂u ∂u  1 ω = −ω =  1 − 2  = ()− C − C = −C 21 12  ∂ ∂  2  x2 x1  2 1  ∂u ∂u  1 ω = −ω =  1 − 3  = ()B − (−B) = B 31 13  ∂ ∂  2  x3 x1  2 1  ∂u ∂u  1 ω = −ω =  2 − 3  = ()− A − A = −A 32 23  ∂ ∂  2  x3 x2  2 Since the components of the rotation tensor are not all zeroes, the displacements of points in the body can be due to rotation (i.e. rigid-body motion) but not deformation (i.e. strain).

Example 4 on strains: A 45o strain-rosette measures longitudinal strain along the axes shown in the figure. At a point P, the following measurements were made = × −4 ′ = = × −4 = × −4 x2 e11 5 10 , e11 e1′1′ 4 10 , e22 7 10 x´1 Determine the shear strain e12 at the point.

o 45

o 45 P x1 Solution: ′ Using strain tensor transformation, we can related e11 to e11, e22 , and e12 . Using such a relation, we can solve for the only unknown e12 as follows: = β β ep′q′ p′i q′j eij

= β β = β β + β β + β β ⇒ e1′1′ 1′i 1′ j eij 1′1 1′ j e1 j 1′2 1′ j e2 j 1′3 1′ j e3 j Where  1 1  0 o o  2 2   cos45 sin 45 0     1 1 β = (β ) = − sin 45o cos45o 0 = − 0 ij       2 2  0 0 1  0 0 1     = β β + β + β ⇒ e1′1′ 1′1 ( 1′1e11 1′2e12 / 1′3e13 ) + β β + β + β 1′2 ( 1′1e21 1′2e22 / 1′3e23 ) + β β + β + β / 1′3 ( 1′1e31 1′2e32 / 1′3e33 ) = β 2 + β β + β 2 e1′1′ ( 1′1 ) e11 2 1′1 1′2e12 ( 1′2 ) e22 = 1 + 1 1 + 1 e1′1′ e11 2( )( )e12 e22 2 2 2 2

1 −4 1 −4 −4 e = e ′ ′ − (e + e ) = 4×10 − (5 + 7)×10 = −2×10 12 1 1 2 11 22 2

Velocity Fields in a Fluid and the Compatibility Condition: For a fluid, we are more interested in the rate of displacement (i.e. velocity) of a fluid particle measured with respect to some coordinate system than simply with the displacement itself. This is because things are happening at a higher time rate in fluids undergoing deformation (typically called flow) than in solids. Hence, instead of describing a displacement field for the continuum fluid material, once usually talks about a continuous velocity field with three components of velocity u, v and

w which are functions of the spatial coordinates u = u(x1,x2,x3), v = v(x1,x2,x3), w = w(x1,x2,x3) In index notation we simply write the above functions as vi = vi (x1,x2,x3)

Now considering two fluid particles P and P´ which are located at some time next to one another at x and x+dx respectively. The difference in their velocities can be stated as ∂v dv = i dx i ∂ j x j ∂ ∂ Where the partial derivatives vi / x j are evaluated at the particle P. We can re-write the last equation as dv 1  ∂v ∂v  1  ∂v ∂v  i =  i + j  +  i − j   ∂ ∂   ∂ ∂  dx j 2  x j xi  2  x j xi  1424 434 1424 434 V Ω ij ji Ω , where Vij is called the “rate-of-deformation tensor” and ji is called the “spin tensor”. = Ω = −Ω Notice that Vij V ji , i.e. it is symmetric, and that ij ji , i.e. it is anti-symmetric. As we learned before, for any anti-symmetric tensor one can always find a dual vector Ω that is related to this tensor Ω =∈ Ω Ω k kij ij or = curl v where the vector Ω is called the “vorticity vector”

Finally, similar to strains, the components of the rate-of-deformation tensor are coupled via an equation of compatibility (also called an “equation of integrability”) that is given in 2D by ∂ 2V ∂ 2V ∂ 2V xx + yy = 2 xy ∂y 2 ∂x 2 ∂x∂y