Module02 Kinematic Deformation

8/17/2019 Module02 Kinematic Deformation

http://slidepdf.com/reader/full/module02-kinematic-deformation 1/15

Module 2

Kinematics of deformation and Strain

Learning Objectives

• develop a mathematical description of the local state of deformation at a material point

• understand the tensorial character of the resulting strain tensor

• distinguish between a compatible and an incompatible strain field and understand themathematical requirements for strain compatibility

• describe the local state of strain from experimental strain-gage measurements

• understand the limitations of the linearized theory and discern situations where non-linear effects need to be considered.

2.1 Local state of deformation at a material point

Readings: BC 1.4.1

Deformation described by deformation mapping :

x = ϕ(x) (2.1)

We seek to characterize the local state of deformation of the material in a neighborhood of a point P . Consider two points P and Q in the undeformed:

P : x = x1e1 + x2e2 + x3e3 = xiei (2.2)Q : x + dx = (xi + dxi)ei (2.3)

and deformed

P : x = ϕ1(x)e1 + ϕ2(x)e2 + ϕ3(x)e3 = ϕi(x)ei (2.4)

Q : x + dx =

ϕi(x) + dϕi

ei (2.5)

29



30 MODULE 2. KINEMATICS OF DEFORMATION AND STRAIN

e1

e2

e3

dx

P

Q

x

dx

P

Q

x

u

Figure 2.1: Kinematics of deformable bodies

configurations. In this expression,

dx = dϕiei (2.6)

Expressing the differentials dϕi in terms of the partial derivatives of the functions ϕi(x je j):

dϕ1 = ∂ϕ1

∂x1dx1 +

∂ ϕ1

∂x2dx2 +

∂ϕ1

∂x3dx3, (2.7)

and similarly for dϕ2

, dϕ3, in index notation:

dϕi = ∂ϕi

∂x jdx j (2.8)

Replacing in equation (2.5):

Q : x + dx =

ϕi + ∂ϕi

∂x jdx j

ei (2.9)

dx

i = ∂ϕi

∂x jdx jei (2.10)

We now try to compute the change in length of the segment −→P Q which deformed into segment−−→P Q. Undeformed length (to the square):

ds2 = dx2 = dx · dx = dxidxi (2.11)

Deformed length (to the square):

(ds)2 = dx2 = dx · dx = ∂ϕi

∂x jdx j

∂ϕi

∂xk

dxk (2.12)



2.1. LOCAL STATE OF DEFORMATION AT A MATERIAL POINT 31

The change in length of segment−→P Q is then given by the difference between equations (2.12)

and (2.11):

(ds)2 − ds2 = ∂ϕi

∂x jdx j

∂ϕi

∂xk

dxk − dxidxi (2.13)

We want to extract as common factor the differentials. To this end we observe that:

dxidxi = dx jdxkδ jk (2.14)

Then:

(ds)2 − ds2 = ∂ϕi

∂x j

dx j∂ϕi

∂xk

dxk − dx jdxkδ jk

=∂ϕi

∂x j

∂ϕi

∂xk

− δ jk

dx jdxk

2 jk : Green-Lagrange strain tensor

(2.15)

Assume that the deformation mapping ϕ(x) has the form:

ϕ(x) = x + u (2.16)

where u is the displacement field . Then,

∂ϕi

∂x j=

∂xi

∂x j+

∂ui

∂x j

= δ ij + ∂ui

∂x j(2.17)

and the Green-Lagrange strain tensor becomes:

2ij = δ mi + ∂ um

∂xi δ mj + ∂ um

∂x j − δ ij

=δ ij + ∂ui

∂x j+

∂u j

∂xi

+ ∂ um

∂xi

∂um

∂x j

− δ ij

(2.18)

Green-Lagrange strain tensor : ij = 1

2

∂ui

∂x j+

∂ u j

∂xi

+ ∂ um

∂xi

∂um

∂x j

(2.19)

When the absolute values of the derivatives of the displacement field are much smaller than1, their products (nonlinear part of the strain) are even smaller and we’ll neglect them. Wewill make this assumption throughout this course (See accompanying Mathematica notebookevaluating the limits of this assumption). Mathematically:

∂ui

∂x j

1 ⇒ ∂um

∂xi

∂um

∂x j∼ 0 (2.20)

We will define the linear part of the Green-Lagrange strain tensor as the small strain tensor :

ij = 1

2

∂ui

∂x j

+ ∂ u j

∂xi

(2.21)




Concept Question 2.1.1. Strain fields from displacements.

The purpose of this exercice is to determine strain fields from given displacements.

1. Find the linear and nonlinear strain fields associated with the following displacements

u

a

1 = x1x2(2− x1)− c1x2 + c2x

3

2,

ua2 = −c3x2

2(1 − x1)− (3− x1)x2

1

3 − c1x1.

2. Find the linear strain fields associated with the following displacements

ub1 = x3

1x2 + 2c1c32x1 + 3c1c2

2x1x2 − c1x1x32,

ub2 = −2c3

2x2 − 3

2c2

2x22 +

1

4x4

2 − 3

2c1x2

1x22.

Solution: The expression to calculate the nonlinear (nl) strains in function of the

displacements isεnlij =

1

2

∂ui

∂x j+

∂ u j

∂xi

+ ∂um

∂xi

∂um

∂x j

. (2.22)

When the derivatives of the displacement components are small in comparison to one,

i.e. ∂um∂xi

, ∂um∂xj

1, the product

∂um∂xi

∂um∂xj

can be neglected, and the previous equation

simplifies to the following linear (l) expression

εlij = 1

2

∂ui

∂x j+

∂u j

∂xi

. (2.23)

When we apply the Equation 2.23 to the field (ua1, ua

2), we obtain the following linear(l) strain tensor

εla =

2x2(1− x1) −c1 + (3c2+c3)

2 x22

−c1 + (3c2+c3)2 x2

2 −2c3x2(1− x1)

.

On the other hand, the Equation 2.22 allows us to calculate the nonlinear (nl) straintensor for the field (ua

1, ua2)

εnla =

εnl11 εnl12

εnl12 εnl22

,

where

εnl11 = 2x2(1− x1) [1 + x2(1− x1)] + 1

2

−c1 + c3x22 − x1(2− x1)

2,

εnl22 = 2c3x2(1− x1) [−1 + c3x2(1− x1)] + 1

2

−c1 + 3c2x22 + x1(2− x1)

2,

εnl12 = −c1 + (3c2 + c3)

2 x2

2

+x2(1 − x1)

x1(2− x1)(1 + c3) + c1(−1 + c3) + (3c2 − c23)x2

2

.



2.2. TRANSFORMATION OF STRAIN COMPONENTS 33

The linear (l) strain tensor for the displacement field (ub1, ub

2) is

εlb =

3x2

1x2 + 2c1c32 + 3c1c2

2x2 − c1x32

12 x3

1 + 32 c1c2

2x1 − 3c1x1x22

12 x3

1 + 32 c1c2

2x1 − 3c1x1x22 −3c1x2

1x2 + x32 − 3c2

2x2 − 2c32

.

2.2 Transformation of strain components

Readings: BC 1.5.1, 1.6.2, 1.5.2, 1.6.3, 1.6.4

Given: ij, ei and a new basis ek, determine the components of strain in the new basis kl

ij = 1

2

∂ ui

∂ x j+

∂ u j

∂ xi

(2.24)

We want to express the quantities with tilde on the right-hand side in terms of their non-tildecounterparts. Start by applying the chain rule of differentiation:

∂ ui

∂ x j=

∂ ui

∂xk

∂xk

∂ x j(2.25)

Transform the displacement components:

u = umem = ulel (2.26)

um(em · ei) = ul(el · ei) (2.27)

umδ mi = ul(el · ei) (2.28)

ui = ul(el

·ei) (2.29)

take the derivative of ui with respect to xk, as required by equation (2.25):

∂ ui

∂xk

= ∂ul

∂xk

(el · ei) (2.30)

and take the derivative of the reverse transformation of the components of the position vectorx:

x = x je j = xkek (2.31)

x j(e j · ei) = xk(ek · ei) (2.32)

x jδ ji = xk(ek

·ei) (2.33)

xi = xk(ek · ei) (2.34)

∂xi

∂ x j=

∂ xk

∂ x j(ek · ei) = δ kj(ek · ei) = (e j · ei) (2.35)

Replacing equations (2.30) and (2.35) in (2.25):

∂ ui

∂ x j=

∂ ui

∂xk

∂xk

∂ x j=

∂ul

∂xk

(el · ei)(e j · ek) (2.36)




Replacing in equation (2.24):

ij = 1

2

∂ul

∂xk

(el · ei)(e j · ek) + ∂ul

∂xk

(el · e j)(ei · ek)

(2.37)

Exchange indices l and k in second term:

ij = 1

2

∂ul

∂xk

(el · ei)(e j · ek) + ∂ uk

∂xl

(ek · e j)(ei · el)

= 1

2

∂ul

∂xk

+ ∂uk

∂xl

(el · ei)(e j · ek)

(2.38)

Or, finally:

ij = lk(el · ei)(e j · ek) (2.39)

Concept Question 2.2.1. 2d relations for strain tensor rotation.

In two dimensions, let us consider two basis ei and ek such that e1 is oriented at an angleθ with respect to the axis e1. ij and ij are, respectively, the components of a strain tensor expressed in the ei and ek bases (i.e. they correspond to the same state of deformation.Using the following expression introduced in the class notes,

ij = lk(el · ei)(e j · ek)

derive the following relations:

11 = 11 cos2 θ + 22 sin2 θ + 12 sin 2θ

22 = 11 sin2 θ + 22 cos2 θ − 12 sin 2θ

12 = −11 − 22

2 sin2θ + 12 cos 2θ

Note: It is also usual to find the following expressions for 11 and 22 in textbooks:

11 = 11 + 22

2 +

11 − 22

2 cos2θ + 12 sin 2θ

22 = 11 + 22

2 + 22 − 11

2 cos2θ − 12 sin 2θ

Solution: First, let us recall the following trigonometric relations between thevectors of ei and ek:

e1 · e1 = cos θ e1 · e2 = − sin θ

e2 · e2 = cos θ e2 · e1 = sin θ




Using (2.39), it is possible to write the following:

11 = 11(e1 · e1)2 + 22(e2 · e2)2 + 212(e1 · e1)(e1 · e2)

= 11 cos2 θ + 22 sin2 θ + 12 sin 2θ

22 = 11(e1

·e2)2 + 22(e2

·e2)2 + 212(e1

·e2)(e2

·e2)

= 11 sin2 θ + 22 cos2 θ − 12 sin 2θ

22 = 11(e1 · e1)(e2 · e1) + 22(e2 · e1)(e2 · e2)

+ 12(e1 · e1)(e2 · e2) + 21(e2 · e1)(e2 · e1)

= −11

2 sin2θ +

22

2 sin2θ + 12(cos2 θ − sin2 θ)

= −11 − 22


The expresssions given in the remark can be derived from these using the following trigono-metric relations:

cos2

θ =

1 + cos 2θ

2 sin2

θ =

1

−cos2θ

2

Concept Question 2.2.2. Principal strains and maximum shear strain in 2d.

Using the relations introduced in Problem 2.2.1, show that given the components ij of a 2d strain tensor in a basis ei:

1. The principal strains can be computed as follows:

1,2 = 11 + 22

2 ± 11 − 22

2

2+ 2

12

and the principal directions of strain for angles with respect to e1 satisfy:

tan2θ p = 212

11 − 22

2. The maximum shear strain can be computed as follows:

max

12 = 11

−22

2 2

+ 2

12

and the normal of the planes of maximum shear form angles with respect to e1

tan2θs = −11 − 22

212.

Conclude that the direction of maximum shear is always oriented at an angle equal to45o with respect to the principal directions of strain.




Solution: Principal strains: The characteristic polynomial χ() corresponding tothe strain tensor components ij is:

χ() = det(ij − δ ij) = (11 − )(22 − )− 212

= 2 − (11 + 22) + (1122 − 212)

The roots of the characteristic polynomial are:

1,2 = 11 + 22

2 ±

11 − 22

2

2

+ 212

To find the angle θ p formed by the principal directions and the basis vecto e1, use the factthat the shear strains vanish in principal directions:

0 = −11 − 22

2 sin2θ p + 12 cos 2θ p ⇒ tan2θ p =

212

11 − 22

Maximum shear strain: The maximum shear strain can be found by simply finding the

value of the argument θ in the expression for transforming the shear strain component whichmakes the derivative of 12 with respect to θ vanish:

max12 = −11 − 22

2 sin2θs + 12 cos 2θs

∂12

∂θ = −2

11 − 22

2 cos2θs + 12 sin 2θs

= 0

By taking the square of the two previous equations and summing them, it is easy to showthat:

max 212 =

11 − 22

2

2

+ 212

The second equation leads directly to the angular relation:

tan2θs = 22 − 11

212

From the trigonometric relation: tan (α + π2 ) = − 1

tanα it is also easy to see that:

tan

2

θ p + π

4

= − 1

tan2θ p = −11 − 22

212= tan 2θs

Thus, proving that θs = θ p + π4

.

Concept Question 2.2.3. Strain tensor rotation.Consider the following problem of a square of unit area subject to the following strain

components in the basis given, Figure 2.3(a). :

11 = 3.4 × 10−4 22 = 1.1 × 10−4 12 = 9.0 × 10−5

Since the square has its edge of unit length, the changes in length in the directions e1 ande2 are directly equal to 11 and 22, respectively. The shear strain 12 is equal to half of thedecrease in angle in A (for infinitesimal angles).




1

1A e1

11

e2

22

212

(a) Deformed unit square

ea1

ea2

A e1

e2

30o

(b) different initial orientation of the unit square

Figure 2.2: Deformed unit square and oriented new initial configuration.

1

1A e1

11

e2

22

212

(a) Deformed unit square

ea1

ea2

A e1

e2

30o

(b) different initial orientation of the unit square

Figure 2.3: Deformed unit square and oriented new initial configuration.




1. Determine the strain components on a square initialy oriented at an angle equal to30o to the axis e1 as shown on Figure 2.3(b). Sketch in this case, the deformedconfiguration.

2. Determine the principal strains and sketch the deformed configuration.

3. Determine the maximum shear strain and sketch the deformed configuration.

Solution: For the solution of this problem, we are going to use extensively therelations introduced in Problem 2.2.1. Let us first compute the two following ratios:

11 + 22

2 = 2.25× 10−4 11 − 22

2 = 1.15 × 10−4

Orientation at an angle θ = 30o: The value of the strain tensor in the basis eai

are asfollows:

a11 = 11 + 22

2

+ 11 − 22

2

cos2θ + 12 sin 2θ

= 2.25× 10−4 + 1.15× 10−4 × 1

2 + 9.0 × 10−5 ×

√ 32

= 3.6 × 10−4

a22 = 11 + 22

2 +

22 − 11

2 cos2θ − 12 sin 2θ

= 2.25× 10−4 − 1.15 × 10−4 × 1

2 + 9.0 × 10−5 ×

√ 3

2 = 9.0 × 10−5

a12 = −11 − 22


= −1.15 × 10−4 ×√

3

2 + 9.0 × 10−5 × 1

2 = −5.5× 10−5

Figure 2.4(a) shows the deformed configuration corresponding to this case.Principal strains: Using the relation introduced in Problem 2.2.2, the principal strainsare:

1,2 = 11 + 22

2 ±

11 − 22

2

2

+ 212 =

= 3.7 × 10−4 : 1

= 8.0 × 10−5 : 2

and their respective direction can be computed as:

tan2θb = 212

11 − 22⇒

θb

1 ≈ 19o

θb2 ≈ 109o

In order to find which of the two angles solution of the equation above is associated withwhich value of principal strain, one can test these values of θb in the expression of 11 given inProblem 2.2.1. Figure 2.4(b) shows the deformed configuration corresponding to this case.Maximum shear strain: Following the relations introduced in Problem 2.2.2, we cancompute the absolute value of the maximal shear strain as:

max12 =

11 − 22

2

2

+ 212 =

(1.15 × 10−4)2 + (9.0 × 10−5)2 = 1.46 × 10−4.



2.3. COMPATIBILITY OF STRAINS 39

Using the fact that the maximum shear direction is oriented at an angle of 45o to one of theprincipal strain direction, let us consider the case of maximum shear obtained for an angleθc = 19o + 45o = 64o starting from e1. We obtain 12(θc) = −max

12 and contend that forthis angle the maximum negative shear strain is obtained. Figure 2.4(c) shows the deformedconfiguration corresponding to this case.

ea1

a11

ea2 a22

2a12

A e1

e2

30o

(a) Deformed configuration with initial orientation of 30o

eb1

b11

eb2

b22

A e1

e2

19o

(b) Deformed configuration of principal strain

ec1

c

11

ec2

c22

2c12

A e1

e2

65o

(c) Deformed configuration of maximum shear

Figure 2.4: Several deformed configuration of a unit square.

2.3 Compatibility of strains

Readings: BC 1.8

Given displacement field u, expression (2.21) allows to compute the strains componentsij. How does one answer the reverse question? Note analogy with potential-gradient field. Inthis section, we will restrain ourselves to small perturbation theory where the displacementsand the rotations of a deformable solid are infinitesimal. Let us first restrict the analysis to




two dimensions. The small strain tensor is defined as the symmetric part of the displacementgradient ∂ui

∂xj:

ij = 1

2 ∂ui

∂x j+

∂ u j

∂xi (2.40)

We define the skew-symmetric part of ∂ui∂xj

as:

ωij := 1

2

∂ui

∂x j− ∂ u j

∂xi

(2.41)

Concept Question 2.3.1. Properties of ωij

1. Verify that ω ji = −ωij , i.e. ωij is skew-symmetric Solution:

ω ji = 1

2

∂u j

∂xi

− ∂ui

∂x j

= −1

2

∂ui

∂x j− ∂u j

∂xi

= ωij

2. Verify that ij + ωij = ∂ui∂xj

Solution:

ij + ωij = 1

2

∂ui

∂x j+

∂ u j

∂xi

+

1

2

∂ui

∂x j

− ∂ u j

∂xi

=

∂ui

∂x j

For the two-dimensional setting, the components are as follows:

ω11 = ω22 = 0, ω12 = −ω21 = 1

2

∂u1

∂x2− ∂ u2

∂x1

(2.42)

We have seen in a previous section of this module, that ij describes the change of lengthof a vector dx due to deformation. We will now see that ωij represents the infinitesimalrotation of the vector dx from the initial to the deformed configuration. ωij is thus namedthe infinitesimal rotation tensor .

Consider an infinitesimal rotation of a vector −→

P Q in the neighborhood of a point P .

For this transformation, the strain tensor vanishes. Such a transformation can only be arotation of

−→P Q into

−−→P Q by an angle θ ( θ 1) as depicted in the following figure:

P

Q

dx

θQ

dx

Figure 2.5: infinitesimal rotation of a vector dx




From Figure 2.5, it is possible to express dx in terms of θ and dx:

dx =

cos θ sin θ

− sin θ cos θ

dx ≈

1 θ

−θ 1

dx (2.43)

Altenatively, from (2.17), it is possible to express dx in terms of ω12

and dx:

dx = (δ ij + ωij) dx j =

1 ω12

−ω12 1

dx (2.44)

By identification of the transformation matrix components, we conclude that ω12 = −ω21 ≈ θ

corresponds indeed to an infinitesimal rotation in the plane of normal e3. Similar conclusionscan be drawn on the remaining components: ω31 = −ω13 corresponds to an infinitesimalrotation in the plane of normal e2 and ω23 = −ω32 corresponds to an infinitesimal rotationin the plane of normal e1.

The compatibility of strain is intricately related to the continuity of infinitesimal rota-tions. In two dimensions, this can be readily expressed by requiring the equality of the mixedpartials of ω12: ∂ 2ω12

∂x1∂x2= ∂ 2ω12

∂x2∂x1. To this end, differentiate ω12 with respect to x1:

∂ω12

∂x1=

1

2

∂ 2u1

∂x2∂x1− ∂ 2u2

∂x21

(2.45)

= 1

2

∂ 2u1

∂x2∂x1+

∂ 2u1

∂x2∂x1−

∂ 2u2

∂x21

+ ∂ 2u1

∂x2∂x1

(2.46)

= ∂11

∂x2

− ∂ 12

∂x1(2.47)

and now with respect to x2:

∂ 2ω12

∂x1∂x2=

∂ 211

∂x22 − ∂ 212

∂x1∂x2(2.48)

Similarly, we can find that:∂ω12

∂x2=

∂12

∂x2

− ∂ 22

∂x1(2.49)

which differentiated with respect to x1 gives:

∂ 2ω12

∂x2∂x1=

∂ 212

∂x2∂x1− ∂ 222

∂x12

(2.50)

Equating the mixed partials in equations (2.48) and (2.50) we obtain:

2 ∂ 212

∂x1∂x2=

∂ 211

∂x22

+ ∂ 222

∂x21

(2.51)

The following concept question generalizes this result to obtain all of the equations of strain compatibility in three dimensions.




Concept Question 2.3.2. Strain compatibility equation in 3d.

The purpose of this exercise is to derive the strain compatibility equations in 3d usingthe approach followed in class for the 2d case.

1. Apply the equality of mixed partials to the small rotation tensor:

∂ 2

ωij

∂xk∂xl = ∂

2

ωij

∂xl∂xk

and show that the following relations hold:

∂ 2ik

∂x j∂xl

− ∂ 2 jk

∂xi∂xl

= ∂ 2il

∂x j∂xk

− ∂ 2 jl

∂xi∂xk

(2.52)

2. How many relations are defined by (2.52) and how many strain compatibility equationsare required in order to ensure that a unique displacement may be computed from agiven small strain tensor?

3. Notice that for i = j or l = k , (2.52) is automatically verified. How many non-trivialrelations can be derived from (2.52)? Are all these relation independant?

Solution: Let us remind first that the small rotation tensor is defined as:

ωij = 1

2 (ui,j − u j,i)

Thus, the gradient of small rotation reads:

ωij,k = 1

2 (ui,jk − u j,ik)

By adding and substracting uk,ij form the right-hand side of the previous relation, it is to

express the gradient of small rotation only in terms of the derivatives of the componenentof the small strain tensor:

ωij,k = 1

2

ui,jk + uk,ij

2ik,j

− (u j,ik − uk,ij) 2jk,i

=

1

2

ik,j − jk,i

Thus, the mixed derivatives: ωij,kl and ωij,lk of the small rotation tensor have the followingexpressions:

ωij,kl = 1

2

ik,jl − jk,il

ωij,lk =

1

2il,jk−

jl,ikThe equality of mixed partials implies:

ik,jl − jk,il = il,jk − jl,ik

Since i, j, k,l can take any value in {1, 2, 3} respectively, (2.52) comprises 34 = 81 relations.It is easy to verify that the only non-trivial relations from (2.52) can be obtained for i = j

and k = l.




i = j and k = l

1 2 and 1 22 3 and 2 33 1 and 2 11 2 and 1 3

2 3 and 2 13 1 and 3 2

Thus, obtaining the 6 following relations:

11,22 + 22,11 = 212,12

22,33 + 33,22 = 223,23

33,11 + 11,33 = 231,31

12,23 + 23,12 = 22,31 + 31,22

23,31 + 31,23 = 33,12 + 12,33

31,12 + 12,31 = 11,23 + 23,11

These six relations are linearly dependent and it is possible to show that if only three arethem are verifed then the remaining three are.

Documents

Module02 Kinematic Deformation