Equations (2.26a,b.c) show that an infinitesimal displacement gradient is composed of the sum of strain and rotation tensors. In the case of finite straining, a similar decomposition can be made to the deformation gradient F. Let some intermediate position vector dx' lie between the mapping of line element vector dX in its initial position to its current position dx, in Fig. 2.6. The current position may be traced with two alternative local sequential mappinp: (i) a stretch dx' = U dX, followed by a rigid rotation dx = R dx', or (ii) a rigid rotation dx* = R dX, followed by a stretch dx = Vdx*. Thus, the final position dx is
dX (2.34a) Comparing eqs(2.21b) and (2.34a) it follows that F is
or F
i}= dxj dXj = R
mU^ = V
mR
v(2.34b) where U and V are the right and left stretch tensors, being positive, symmetric and possessing the same principal stretch values. Since det F = p
olp > 0, R is an orthogonal rotation matrix, obeying
RR
T= R
TR = I or R
ikR
ik= R
MR
v= 4, (2.35a) F, R, U and V will all depend upon position unless the deformation is homogenous, where x = FX and the components of F, R» U and V become constants. Combining eqs(2.34b) and
(2.35a) gives relationships between U and V;
U = R
T¥ R and ¥ = RUR
T(2.35b,c) Now, from eqs(2.34b) and (2.23c), G and its components G
8, may be written as
G = F
TF = (RU)
T(RU) G
v= F
&JF
ltBut as U
T= U and IU = U = (R^R^U^
G = U
2= SjJ^Uy = U^ (2.36)
Let U* contain the principal stretches a
ufifj and ^ of U. Substituting eq(2.36) into eq(2.30a), gives
A
im1= m
r(V'
l)m (2.37a)
STRAIN ANALYSIS
Taking the principal stretch directions of U* as co-ordinates, so that nij is a unit vector aligned with the major principal stretch direction, eq(2.37a) gives
4
0 0
0
4
0 0 0
4,
1 0 0
= ar, (2.37b)
Thus, from eq(2.37b), A^. = at. Similarly, when further unit vectors i% = {0 1 0}T and n% = {0 0 1 }T are aligned with their respective principal directions: A, , = a2 and
X{m j = af3. This means that the stretch ratios for the principal directions are the principal values of U«
From eqs (2.34b) and (2.28d), B and its components fl8, may each be written as
= F FT = (VR)(VR)T
= V(RRT)VT =
= But as
(2.38)
Let the principal directions of U be given by the three unit vectors 1%* (i = 1,2,3), with their components in the Xt frame. They will satisfy an equation similar to eq(1.23b):
(U-flfilJu^O Since RT Ru* = I u* = u*, eq(2.39a) can be re-written as
R (U - s-.IXRTR)Ui* = [ RURT - R (^I)RT ] (Ru,*) = 0 Substituting from eq(2.35c)
(2.39a)
(V - as ; ) = 0 (2.39b) When the principal directions of V are defined by the unit vectors v*s (i = 1,2, 3) they will satisfy the equation
(V-AD< = 0 (2.39c) Comparing eqs(2.39b and c) shows that: (i) the principal values of U and V are the same since af = $ and (ii) R connects the principal directions of U and V within: v,* = Ru* or u*
= RT v*. Equations (2.36) and (2.38) show that U and V provide an equivalent physical interpretation of G and B though G and B remain the more convenient measures of strain.
Principal stretch tensors U* and V* are calculated from the square roots of the eigen values of G and B respectively. The 3 * 3 transformation matrix M of eigen vector components (for G and B) will not, in general, equal the components of the rotation matrix R as the following example shows. M appears in tensor transformation laws for U and V, similar to eq( 1.21b);
and V*
Reversing eqs(2.40a,b) gives (see eq 1,20b)
U = MTU*M and V = MTV*M
(2.40a,b)
(2.40c,d) Alternatively, eqs(2.35b,c) allow the calculation of U from V and vice-versa, without first finding their principal values.
Example 2.6 Find U, V and R for a Lagrangian description of fee following motion; JK, = aXj + bXt ,Xj = ~ a%i + bX2 and x3 = cX% (where a>b>c). Show mat the principal values of U and V are equal.
Having found the components Fit = dxtl SX}, G is found from eq(2.23c) as
G = FTF = a b
0
- a b 0
0 0 c
a -a
0 b b
0
0 0 c
=
2 a2 0 0 0 2b2 0 0 0 c2
from which U is obtained:
U = VG =
0 0
0 Jib 0 0 0 c
= IT
This result reveals that the directions of the principal, right stretches are aligned with fee co- ordinates X,. The rotation R then follows from eq(2.34b) as
R = FU
a -a
0 b b 0
0 0 c
1
yfla
0
0 0
1 0
%/2£>
0 0 lie
4- 4-o -4- 4-o
0 0 1
These components are direction cosines corresponding to rotating Xx and Xz by 45° about fee fixed X3 axis. B is found from eq(2.28d) as
B a
-a0 b b 0
0 0 c
a b 0
-a b 0
0
0 c=
(a
2+ A
2) (-a
2+ b
2) 0 (-a
2+ b
2) (a
2+ b
2) 0
0 0
STRAIN ANALYSE
The stretch tensor V is more conveniently found from V* = V B* when B* is referred to its principal axes. The principal values of B* (i.e. eigen values) are the three real and positive roots Pi (i' = 1,2,3) of the characteristic equation for B. They follow from the expansion to the determinant (B - J3J) = 0:
0 0
c-fi
= 0
which gives fit = 2a*, fix = 2b2 and /?3 = c2. Hence, we have
B =
2 a2 0 0 0 2&2 0
0 0 c2
V = -/B =
-/2a 0 0 0 J2b 0 0 0 c
showing that the principal values of U and V (i.e. the components of U* and V*) are equal.
The eigen vectors v/ (i = 1,2,3), which define the orthogonal, principal axes of V*, are found from the solution to:
For i = l , let Vj* = 1^+ »%%+ «ii% where Ui, u2 and i% are unit vectors aligned with the co- ordinates Xj (see Fig. 2.6) and \, my and n1 are the intercepts (direction cosines) with Xf. In eq(i) we write vector components Vj* = {^ mt n,}T and substitute for matrix B to give:
(a1 + b% - A ) h + (- a2 + b2) m, = 0 (- a1 + b2) lx + (a1 + b1 - A) « i = 0
Substituting fl1 = 2a2"m eqs(ii)-(iv) we find ^=1/^2, mt=- 1/V5 and n% = 0 giving:
Similarly, for 1 = 2 and 3, the eigen vectors are:
u2 and v3* =
(ii) (iii) (iv)
in which vs • va = v2 • v3 = vs • Vj = 0 confirms orthogonality in these vectors. The components of these vectors define the transformation matrix M in eq(2.40a-d) so mat:
V =
1/V2 0 1A/2 1/-/2 0 0 0 1
0 0
0 0 0 0 c
1A/2
1//2
0-1/^2 1//2
0 0 0 1
=
0 ) 0 0 c
Finally, we may check this result from F = V R : 0 ) 0 0 c
1//2 0
1//2
1/V20 0 0 1
= a
- a0
b b 0
0 0 c
2.6 Strain Definitions
It is seen that there is no single fundamental definition of strain. Indeed, the expressions for the natural, Lagrangian, Eulerian strains and the extension ratio will all differ, even for a simple uniaxial, finite deformation. Thus, the flow behaviour of a material will depend upon the chosen definition of strain. The Lagrangian finite strain tensor, L in eq(2.27b,c), is more commonly used for referring finite deformation in a solid to its original shape.
For infinitesimal strains, the differential products in eqs(2.27b) and (2.28b) may be ignored. The Eulerian and Langrangian finite strain tensors E and L, will then reduce to eq{2.3), implying that there is negligible displacement between the material and spatial co- ordinates. Seth [4] related the three uniaxial strains to the extension ratio within a single formula. Given initial and final lengths /„ and I respectively, these strains are expressed from eqs{2.11b), (2.30c) and (2.31c) as
e« = (l/a)[l - (IJI)*] = (l/a)[l - ( l / i )B]
where a = - 1, a = - 2 and a = + 2 according to the engineering definition and those of Lagrange and Euler respectively. Other values of a may lead to simpler constitutive relations, e.g. when a is chosen to provide linear stress-strain behaviour, but the corresponding stress definition becomes uncertain and the tensorial nature of strain is lost.
The reader will find further discusion of strain tensors and their transformation properties in books on continuum mechanics [5-8].
References
1. Ludwik P. Elements der Technologischen Mechanik, 1909, Springer, Berlin.
2. Hencky H. "Uber die form des elastizitatsgesetzes bei ideal elastischen stoffen", Z.T.P.
1928,9,214-223.
3. Green A. E. and Adkins J. E. Large Elastic Deformations, 1960, Oxford University Press, London.
4. Seth B. R. "Generalised strain measure with applications to physical problems", Proc:
Second Order Effects, IUTAM, Pergamon Press, 1964,
5. Farrashkhalaut M. and Miles J. P. Tensor Methods for Engineers and Scientists, 1990, Ellis-Horwood.
6. Lai W. M., Rubin D. and Krempl E. Introduction to Continuum Mechanics, Pergamon Engineering Series, Vol.17,1978, Pergamon Press.
7. Mase G. Continuum Mechanics, 1970, Schaum Outline Series, McGraw-Hill, 8. Mase G. E. and Mase G. T. Continuum Mechanics for Engineers, 1992, CRC Press.
STRAIN ANALYSIS 63 Exercises
2.1 A displacement vector u is described by its three components «(in spacial co-ordinates x,, as
* a and % = (ij
Detennine the components of infinitesimal strain and rotation matrices, «j and a^, for a point P(0,1,2), whenfcj = J^= 1 and&3 = 2. What men is me change in the right angle between a pair of perpendicular lines passing through P with direction cosines: (1A/3,1/^3, - 1//3) and (W14,2/^14,3A/14) with respect to *,» % and JCJ?
2 2 The following infinitesimal, micro-strain components; eu = 600, cM = em = 0, £,2 = e2l = 200,
* » = sw - 400 m<* £» = £3i = 200 0*U multipied by 10 *) define a strain tensor. Determine the magnitude and direction of the principal strains, the maximum and octahedral shear strains. What is the state of strain for a plane with normal direction cosines a= 0.53, fl= 0.35 and y = 0.77?
[Answers: 800,200, - 400,600,9.1,622 ( x 10 *)]
2 3 The components of a micro-strain tensor are en = 100, en = eI3 = 0,em = - 100, sn = en = 500 and eli=eil = - 500 in x,, Xj and % co-ordinates. Transform these to Jt/, x[ and.%' co-ordinates given that the direction cosines for the x{- direction are: lu = 1/-/14, lu = 2/i/14, ln = 3/^/14 and those for the JC/- direction are: £„ = l ja = 1, IB = - 1.
[Answers: en' = - 128.6, ea' = 666.7, £B' = 538.1, en' = 138.9, eH' = 198 and £2S' = - 276.23 2.4 Under plane-sixain deformation, a line element increases in length by 17% in the Xt - direction and by 22% in the X2 - direction. What is the percentage increase in length and the rotation of a line element originally at 45° to the reference configuration XY and X2 ?
[Answers; 27%, 2.4* to X{\
2.5 Given the three components of a spacial co-ordinate function x = x(X, 0 are xx = A,(X, + BX3), j%=AJXJ and % = ASX3, where A and B are constants at time t, determine: (i) the components of the deformation gradients F and F \ (ii) the left and right Cauehy Green deformation tensors G, G'r, B and B " ' and (iii) the Lagrangian and Eulierian strain tensors, L and E respectively.
2.6 The spacial co-ordinates x,, in Exercise 2.5, describe homogenous deformation at a given time t.
Find the new direction and the stretch of a line element originally equally inclined to a reference configuration X,- (i = 1,2 and 3) at time t = 0.
2.7 Given that the spacial components of motion, x = x(X, I), are:
show that the material components of motion X = X(x, t), are:
Xi = I - xx + x2 (e' - *)] / [K(K - e' - e - ) ] [K(K- e1- e ' ) ]
2 J The spacial position vector x has components: xl=Xi,x2 = Xi + KX$ and xi = Xi + ECX2, where K is a constant. Determine the components of the displacement vector u = x - X (see Fig. 2.6) in both material and spacial co-ordinates.
[Answers: «, = 0, Uj = KX% = Mxs - Kx^il - K1) and M, = KXt = k{xt - J&3)/(1 - K1)]
2.9 Show that the components of G and L for x( in Exercise 2.8 are given as:
1 0 0
0 l*K2
2K 0 21?
1+KZ
0 0 0
0 K3
2K 0 2K K2
Taking K = 1, determine the magnitude and direction of the principal Lagrangian strains.
Z10 Given that the material components of the displacement vector U (see Fig. 2,6) are: Ux = Kl X32, U2=X^Xt and V3 = X^Xt, determine the components of the material displacement gradient dU,fdXj and the material deformation gradient Ftj = dxJdXj. Hence confirm the derivative Fff = dU,/dXj + 3ir
2.11 Calculate the stretch ratio for a line element vector aligned with the X, - axis and another with direction cosines (0, lMJ, 1A/2) w.r.t. Xt, following a shear deformation whose components x,, are given in Exercise 2 J . Hint, use eq(2.13c) with the corresponding right Cauchy-Green components Gs from Exercise 2.9. [Answer: 1, (1
2.12 Show from eq(2.28d) that the extension ratio 1 + K, found in Exercise 2,11, applies when inverting the components from xt in x = x{X) to X, in X = X(x) upon forming the inverse of the left
Cauchy-Green components By., Explain why this is.
2.13 Determine ttie extension ratios X^ and A^ from eqs(2.30a) and (2.3 la) for a line element originally aligned with the X2 - axis, following a motion with xt components given in Exercise 2.8. Explain why these differ.
2.14 Determine, for the unit square shown in Fig. 2.8a, the extension ratio for the diagonal PR and the change in the angle OPQ, when it is subjected to the same Lagrangian description x = x(X) of motion
(see Example 2.5).
2.15 Determine the deformation gradient for the Lagrangian motion x1=Xl + 2Xj, x, = Xt - 2X3 and XJ = - 2Xi + 2Xj + JKj. Find R, U and V by polar decomposition and show that the principal values of U and V are equal.