1.3 Strain Measures

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Physically, we are interested in the change from dX to dx and not as much in the actual

size of dx. After all, assuming the body to be stress-free at the outset of the calculation,

it is the change of dX that generates the stress field in a mechanical problem. The vector

connecting the initial position of a material particle at X to its new position x at time t is

called the displacement U(X, t) of that particle at time t . One can write (see Figure 1.2)

u = U = o + x − X (1.67)

o is the vector connecting the spatial frame of reference {gk

} with the material frame {GK}.

Since the displacement connects a material vector with a spatial vector, it does not uniquely

belong to the material nor to the spatial frame, and both upper case notation U and lower

case notation u will be used. The component notation yields

U = UKGK (1.68)

and

u = ukgk. (1.69)

The difference between dS2 and ds2 can be written as (Equation (1.41))

ds2 − dS2 = (CKL − GKL) dXK dXL (1.70)

as well as (Equation (1.47))

ds2 − dS2 = (gkl − b

−1

kl ) dxk dxl . (1.71)

x1

x2

x3

X1

X2

X3

G1

G2

G3

I 1 I 2

I 3

g1

g2

g3

i1

i2

i3

X

U

dX

x

dx

U + dU

O

Figure 1.2 Displacement vectors

10 DISPLACEMENTS, STRAIN, STRESS AND ENERGY

Now, the Lagrangian strain tensor E (also sometimes called the Green–Lagrange strain

tensor) is defined by

EKL := 12

(CKL − GKL) (1.72)

and the Eulerian strain tensor e (also sometimes called the Euler–Almansi strain tensor)

by

ekl := 12

(gkl − b

−1

kl ). (1.73)

Accordingly,

ds2 − dS2 = 2EKL dXK dXL (1.74)

= 2ekl dxk dxl . (1.75)

E and e are second-order tensors and can be interpreted as measures for the change of

length in a body. Using Equation (1.67) one can write

ds2 − dS2 = dx · dx − dX · dX

= (dU + dX) · (dU + dX) − dX · dX

= dU · dX + dX · dU + dU · dU

= (U,K · X,L + X,K · U,L + U,K · U,L) dXK dXL. (1.76)

Since U = UMGM and dX = dXNGN, one finds

∂U

∂XK

= ∂

∂XK

(UMGM) = ∂UM

∂XK

GM + UM ∂GM

∂XK

= ∂UM

∂XK

GM + UM ∂2ZL

∂XK∂XM

IL

= ∂UM

∂XK

GM + UM ∂2ZL

∂XK∂XM

∂XN

∂ZL

GN

= _∂UM

∂XK

+ UN ∂2ZL

∂XK∂XN

∂XM

∂ZL

_GM

=: UM;KGM (1.77)

and

∂X

∂XL

= GL. (1.78)

UM;

K is the covariant derivative of UM and can also be written as

UM;

K

= UM

,K

+ UN _ M

KN

_ (1.79)

DISPLACEMENTS, STRAIN, STRESS AND ENERGY 11

where

_ M

KN

_ := ∂2ZL

∂XK∂XN

∂XM

∂ZL

(1.80)

are called the Christoffel symbols of the second kind. Hence,

ds2 − dS2 = (UM;

KGLM + UM;

LGKM + UM;

KUN;

LGMN) dXK dXL. (1.81)

Comparison of Equation (1.74) with Equation (1.81) finally yields

2EKL = UM;

KGLM + UM;

LGKM + UM;

KUN;

LGMN. (1.82)

Similarly, one finds

2ekl = um;

kglm − um;

lgkm + um;

kun

;lgmn. (1.83)

It is important to note that the extra term in Equation (1.77) derives from the fact that

GM is not necessarily constant in space. The expression UM;

K is also called the covariant

derivative covariant derivative of U (Eringen 1980). For rectangular coordinates, the unit

vectors do not vary in space and Equations (1.82) and (1.83) reduce to

2EKL = UM

,KGLM + UM

,LGKM + UM

,KUN

,LGMN (1.84)

and

2ekl = um

,kglm − um

,lgkm + um

,kun

,lgmn. (1.85)

Furthermore, the distinction between {GK} and {GK} fades since both bases are identical,

and GKL is the unit tensor. Consequently, Equations (1.84) and (1.85) can be further

simplified to

2EKL = UK,L + UL,K + UM,KUM,L (1.86)

and

2ekl = uk,l + ul,k − um,kum,l. (1.87)

The above equations establish a relationship between displacements and strains. This

relationship is nonlinear owing to the last terms in Equations (1.82) to (1.87). In problems

with small deformations, the nonlinear terms are frequently neglected, leading to the linear

strain ˜E KL, in rectangular coordinates:

˜E

KL := 12

(UK,L + UL,K). (1.88)

Defining the infinitesimal rotation as

˜R

KL := 12

(UK,L − UL,K) (1.89)

Equation (1.86) can be rewritten as

EKL = ˜EKL + 12

( ˜EMK + ˜RMK)( ˜EML + ˜RML) (1.90)

12 DISPLACEMENTS, STRAIN, STRESS AND ENERGY

X

Y

x

y

θ

θ0

x,X

y, Y

R

Figure 1.3 Finite rotation of a rod

showing that for the linear strain to be a good approximation for the actual strain, both the

linear strain and the linear rotation must be small. Accordingly, for a rod freely rotating

about one of its ends, linear strains are a poor approximation of the real strains. This is

easily shown. Consider a rod of length R rotating about the origin (Figure 1.3). The original

position is

X = R cos θ0

Y = R sin θ0. (1.91)

The final position is characterized by

x = R cos(θ 0 + θ)

y = R sin(θ 0 + θ). (1.92)

Consequently, the displacements amount to

UX = x − X = X(cos θ − 1) − Y sin θ

UY = y − Y = X sin θ + Y(cos θ − 1). (1.93)

The infinitesimal strains yield

˜E

XX = UX,X = cos θ − 1

˜E

YY = VY,Y = cos θ − 1

˜E

XY = (UX,Y + UY,X)/2 = 0 (1.94)

DISPLACEMENTS, STRAIN, STRESS AND ENERGY 13

which shows that ˜EXX and ˜EYY are generally not zero. Since a rigid body motion must

not generate strains, this clearly shows that the infinitesimal strains are not suited for finite

rotations. The Lagrangian strain tensor, on the other hand, vanishes. For instance,

EXX = UX,X + (U2

X,X

+ U2

Y,X)/2

= (cos θ − 1) +            (cos θ − 1)2 + sin2 θ_2 = 0. (1.95)

This is especially important for slender structures such as shells and beams in which strains

are usually small but rotations can be large.

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