2 I &? + 3y

1 f 9K dw | 1 J dv dw I 9w

21, 8z dx j 2 { dz By j 3z

and the rotation matrix is

0 I ( du _ dv

^{ly &

1 | 9M _ dw

l["dz ~dx

U Bu 3v 2\dy Bx

0 1J dv 9w

l

du Bw 3z ~Bx 1 f dv _ Bw

| " !

All direct and shear strain components appear within a linear combination of displacement derivatives, i.e. the Eulerian infinitesimal strain tensor.

9M, du

dx. dx.

or

^(2.3)

The Eulerian rotation tensor is defined as 1 dui _ du,

dx. dx. or

^(2.4)

in whch xt are the spacial, or current, co-ordinates of the deformed body (see Fig. 1.19).

When the corresponding engineering notation is required, the subscripts i and j on the symbols u and x are to be read as: ut = u, % = v, J% = w, xt = x, x^ = y and x3 = z. Note that u and V in eqs(2.3) and (2.4) define column matrices of their components: u = {wj a2 «3}T

and V = { 3/9JCJ a/clxj 9 / &3}T. Hence uT and VT define their row matrices. Equation (2.2b) shows that the 3 x 3 strain matrix E is symmetric (i.e. etj = t0 and that the 3 x 3 rotation matrix Q is skew symmetric (i.e. &% = - 6$. Under applied principal stresses, when all the rotations ca^ = 0, the deformation is said to be irrotational.

2.2.2 Strain Transformations in Tensor and Matrix Notations

It may be deduced that the transformation properties of the symmetric 3 x 3 strain matrix ev

are identical to those of Cauchy stress, the latter being a symmetric 3 x 3 matrix (cff = c^().

Thus, the transformation of an Eulerian strain tensor, following a rotation in the orthogonal axes from xt to JC/, will obey a law equivalent in form to eq( 1,22a):

In full, eq(2.5a) becomes

or

^(2.5a)

*n °ia *i3

**2l C2t T3

^31 *32 *33

hi hi hi

hz hi

h2

^l

n

hi ?33

JP JP JP

e

n % %

% % ^3

% *S2 %

hz h,

hi hi

hi

hz

hi

(2.5b)

It follows that there is a correspondence between the stress and strain transformation expressions. For this to be achieved, rmust be interchanged with the tensor definition of shear strain (yf2). For example, to find the normal strain on an oblique plane in terms of engineering co-ordinate strains, eq( 1.12b) is converted from stress to strain as follows:

e = + 4 n2 + 2 (Ime^ + mnen + lnea) (2.6a) (2.6b)

STRAIN ANALYSIS 37

The principal strain cubic is similarly deduced from eq( 1.24b) as

where the strain invariants are

(2.7a)

e

y

+ e

t

= 3, = tr E

4 = h=

(2.7b) - tr E

2

] (2.7c)

= det E (2.7d) Because of Ifae identical nature of stress and strain transformations, the directions of the principal sfress and strain axes are coincident It is not possible to convert the general shear stress expression (in eq(1.12c)) directly to shear strain without firstly specifying the initial perpendicular directions to which the shear strain applies. Performing the matrix multiplication (2.5b) for any one tensor shear strain (i.e. with i * j) provides a method for finding the shear strain between a pair of perpendicular lines. For example, with i = 1 and j = 2, eq(2.5b) becomes e

u

' = l

lp

l^ s

m

. In the engineering notation this is equivalent to an

abbreviated matrix multiplication:

m,

^{£ £}_{xy y} «,*

(2.8a)

(2.8b) Alternatively, the generalised shear strain (eq(2.8b)) will appear within the tensor and matrix notations as

¥iy=l

li

e

ij

l

ls

= n

l

"En

1

' (2.8c)

where column matrices u / = { l

n

l

u

l

n

}

r

and %' = { 4i In l-a 1

T

express the direction

cosines of unit vectors aligned with a given pair of perpendicular directions: x / and JC/. It

is possible to identify the correspondence between r i n the stress eq( 1.12c) and ffl in

eq(2.8b). Referring to the plane ABC Fig. 1.7a, the perpendicular directions are those

defined by the direction cosines (1.13a»b,c) for the shear stress rand those (1.8a,b,c) for the

normal stress a. That is, in eq(2.8b) 4, m

t

and B

2

are identified with I,, m

s

and n

s

in

eqs(1.13a-c) and l

lt

m

1

and n

t

with (, m and n in eqs(Ua-c). Equation (2.8b) will then

supply the shear strain for original perpendicular directions: one parallel to the normal sixess

a and the other aligned with the direction of the resultant shear stress r in Fig. 1.7a.

Example 2.1 The state of strain at a point is specified, in the engineering notation, by the following components of its strain tensor (x 10"*): ex= 1, s^= 1, st = 4, sv= ~ 3, £^ = ^2 and Eyz = - V*2. Determine: (i) the normal strain component in a direction defined by the unit vector* u = (1/2)11, - (l/2)Uy + (1A/2)UJ, (ii) the shear strain between the normal and a perpendicular direction whose unit vector equation is u = - (1/2)^ + (l/2)u^ + {1/^2)1^, (iii) the invariants and (iv) the principal strains.

(i) Substituting I = 1/2, m = - 1/2 and n = 1/^2 together with the component strains into eq(2.6a) gives the normal strain

4 = I(lf2f x 1] + [(- 1/2)2x 1] + K1A/2)2 x 4 ]

+ 2 [(l/2)(- l/2)(- 3) + (- l/2)(l/v*2)(- y/2) + (l/2)(l//2)(\/2)] = 6

(ii) The engineering shear sixain f i s found from eq(2.8b). The direction cosines: Ij = 1/2, mt — - 1/2 and nt = 1/V2 apply to the normal direction and 4 = ~ 1/2,»% = 1/2 and % = apply to the perpendicular direction. (Note: lt 4 + tn^ w2 + nx n2 = 0.)

y/2 x 10'* = [ (l/2)(- 1/2) x 1] + [ (- l/2)(l/2) x l ] + [ (l/i/2)(l/%/2) x 4 ] - l/2)(- l/2)](- 3) + [(

+ (- 1/2X1/^)1(1/2) = 0

This zero value for y shows that there is no change in the right angle between the two perpendicular directions, i.e. they are principal directions and therefore the normal stxain found in (i) is one of the principal strains.

(iii) The invariants Jlf J2 and /3 are found from eqs(2.7b,c,d):

hi 10"8 = [(lXD + (1X4) + (4X1) - (- 3)2 - (- v^2)2 - (^2)2] = - 4 hi 10"12 = det (£y) = 1(4 - 2) + 3( - 12 + 2) + V2(3/2 - 42) = - 24 (iv) The invariants give a principal strain cubic, from eq(2.7a)

e% - ( 6 x m-*)e% - (4 x 10"%)e+ (24 x 10"12) = 0

for which one root is 6 x 10" *. The two remaining principal strains may be found from the coefficients a, b and c within the quadratic equation (as2 + be+c) as follows:

[s - (6 x 10"*)Ka£J +be+ c) = r* - (6 x 10"*)g2 - (4 x 10"n)s+ (24 x 1 0 "n) = 0 Equating coefficients gives a = 1, b = 0 and c = - 4 x 10"8. The quadratic equation simplifies to e2 - 4 x 10"8 = 0, for which the roots are: £ = ± 2 x 1 0 "4. Thus, the principal strains are identified as: et = 6 x 10"*, e% = 2 x 10"* and e3 = - 2 x 1 0 * .

2.2.3 Reduction to Plane Strain

Putting I = cose; m = cos (90°- «•) = sinar and n = cos 90°= 0 in eq(2.6b), leads to the normal strain in the direction of x' in Fig. 2.2a.

STRAIN ANALYSIS 39

r

(b)

Figure 2.2 (a) Plane co-ordinate and (b) displacement transformations

ej = e% l

= ex

m1 + ez n n2

sin2a+ yv simrcosar

ey) cos 2a + sin 2 a (2.9a) The normal strain in the / - direction is found from setting I = cos (90 + a) = - sino; m = cosfif and n = 0 in eq(2.9a). This gives

By c o s 2 smec cosa

) cos 2a - sin 2m (2.9b)

The shear strain between the primed directions x' and / is fovmd from eq(2.8b). For the x1- direction, we set: lt = cosa, m 1 = cos (90°- a) = sxaa and n x = cos 90° = 0 and for the / - direction, set J2 = cos (90 + a) = - sins, m^ = coses and % = 0:

,,(cos2 a - sin1 a) , sin 2a + Vkey sin 2a+%yxy cos 2a

c o s 2 a (2.9c)

The strains given in eqs(2.9a-c) may be confirmed from direct differentiation of the displacements according to the infinitesimal strain definitions (2.1a-c). Figure 2.2a provides the geometrical relationships between the co-ordinates:

x = x1 cosflf - / sine

y = x' sinar + / cosa

The displacements of a point P as it moves to P* (see Fig. 2.2b), are given by u = M cos^ + v sinar

V = - u sina+ v cosa

where M and v are the displacements of P* along x and y and M' and v' are the displacements of "P' along x1 and y', as shown. The normal strain and shear strains in the x'-y plane are found from eqs(2.1a-c):

ex' = du'/dx1 = (du'/dx)(dxldx!) + (

= [ (du/dx)(du'Bu) + (dvBx)(du'idv)](dxfdxf) + [(du/By)(di/Bu) + (dvfdy)(3dBv)](dyB£)

= [(du/dx)cosa+ (8v/&)sin<3J3cosflf+ [(du/dy)cosm+ (dvidy)sinm]sina

= EX cos2flf+ e^ %\n%m +

= ^ (ex + ey) + ¥i (ex - £y) cos 2 « + % %y sin 2 or

[(du/dx)(dv'fdu)

= [ - (du/dy) sin^+ (dv/dy) cos«r]cosff + [ - (du/dx) since + (dv/dx) cma](~ sina)

= V6 (e^ + £y) - i4 (e, - gy) cos 2 » -

= (dv'Bx)(Bx/B/) + (Bv'By^ByBx1) + (du'Bx)(dx/dy') + (du'fdy)(dy/dy')

= [{duldx)(dv'Bu)+(dvBx)(dv'fdv)](dxBx') +[(duBy)(dv'Bu) + (Bvfdy)(dv'Bv)](ByBx') + [(8M/&)(8M'/3«) + (dvBx)(du'/dv)](dx/dy) + [(Butdy){du'Bu) + (BviBy)(du'/Bv)](dy/By)

= [ - (du/dx) sinff + (BvBx) eosflf ]eos«r+ [ - (Bufdy) sinaf+ {dvBy) cosff ]sinar + [(du/dx) COS0+ (dv/dx) suw]( - sina) + [(dufdy) cosa+ (dvBy) sin»]cos^

= - 2(3M/3x)sinflrcosa+ 2(5v/8y)sinflfeosaf+ [(BvBx) + (Bufdyy\(coszei - sin2^)

¥iy^ = - ¥i (ex - ey) sin 2a + Viyxy cos 2ee

It is shown in Section 2.4 that infinitesimal strain transformation equations refer to line elements in the current configuration (the deformed solid). However, with the small elastic strains involved here, there is negligible error in choosing line elements in the original reference configuration (the undeformed solid). Thus, the strain components ex, s, and yv

may be identified with the nominal (engineering) strains in which displacements are more conveniently referred to the initial geometry.

23 Large Strain Definitions

When strains exceed 1%, we cannot employ the same approximations used previously for deriving the infinitesimal strain-displacement relations. A number of large (finite) sixain definitions are available and these are reviewed here. They can be applied wherever large deformations arise, say, in the stretching of rubbery materials and in deforming metallic materials at very rapid rates under impact loading.

2.3.1 Natural, True or Logarithmic Strain

The theory of metal plasticity employs an incremental strain definition. This was first employed by Ludwik [1] and Hencky [2], who referred the change in length <S of a given line element to its current length, I. This gives an incremental strain 3e= Mil, for which the true, or natural, strain is characterised by any one of the integrated quantities:

STRAIN ANALYSIS 41

(2.10a) i = l n ( l + e) (2.10b)

e-

where the nominal (engineering) strain e = dUl, refers <3? to its original length la. These two measures of strain, e and s, arc approximately equal for nominal strain values less than 2%. The logarithmic strain proYides the correct measure of the final strain when defommtion takes place in a series of increments. For example, let an initial length lB increase to a final length lf, under an incremental tensile loading for which there are six intermediate lengths:

4, 4» 4» If h and 4 • It follows from eq(2.10a) that the final, true strain s = In (lft la), will be given by the sum of the increments:

e= In ft/ IJ + In (y 4) + In (l3f 4) + In (l4i 4) + In (lsf Q + In (l6f ls) + In (lfl l6)

=in K/,/ 0(4/ my my hw my hw wi=i» (V i)

Here, as with all proportional loading paths, the final strain does not depend upon the intermediate strains, i.e. the same strain would be achieved if a single continuous loading were to produce lf. Applying the engineering strain definition to each stage of the defommtion shows that successive strains are not additive under uniaxial loading. That is,

(h - Qii

^a

+ (4 - y/4 + (i

³

- 4)/4 + (h - hVh + (h - Wh + (h - Wk * (t

^f

- i

^o

yi

^a

The logarithmic strain measure will therefore account for the influence of a strain path more reliably than engineering strain. This particularly applies when the load path changes direction under a non-proportional loading, giving a final strain that depends upon the history of strain. The essential tensorial character of normal strain is retained within this logarithmic definition but there is not a corresponding shear strain. Large shear strain is expressed as y = tan$=xll (see Fig. 1.2) but this will only approximate to the engineering shear strain i.e, y = tan$ «$rad, for distortion angles $ < 10°.

2.3.2 Extension Ratio

The extension ratio is a measure of large deformation, often used in rubber and polymer mechanics. It is, simply, the ratio between the final length I and the initial length la:

where the unit vector u subscript denotes the original direction of lB. Elastomeric fibres and polymers can stretch by multiples of their original lengths to give typical engineering strains of between 500% and 600%. The extension ratio bears a simple relationship to the engineering strain e. When lo is aligned with a unit vector u, in the x - co-ordinate direction:

If an incompressible (constant volume) deformation of a unit cube occurs withour shear, eq(2.1 lb) supplies a relation between the principal extension ratios:

(1 + e,)(l + e2)(l + <%} - i = 0 or J, 4 4 , - 1 = 0 (2.12a,b) Note that it is only when the strains are small can the left-hand side of eq(2.12a) be approximated to

«! + et + e3 = 0 giving Ay + Aj + Ay = 3

2,3.3 Finite Homogenous Strains (a) Direct Strain

Consider uni-direetional straining of a line element PQ, aligned with the x - direction, as shown in Fig. 2.3. Let end points P and Q, with co-ordinates P(x, 0,0) and Q(x + &c, 0,0), move to new positions P* and Q'.

y 4

P* Q' m a