®

’1 t 0

1

F ’1 (t) = 3 ° 0 ’1 t»

(t ’ 1)

t 0 ’1

Find the velocity of the particle, (a) initially at X = (1, 1, 1) at time

t = 0 and (b) currently at x = (1, 1, 1) at time t = 2. Using J = dv/dV

show that at time t = 1 the motion is not realistic.

ˆ

7. Show that at the initial con¬guration (F = I) the linearization of C in

the direction of a displacement u is:

ˆ

DC[u] = 2µ = 2 µ ’ 1 (trµ)I

3

CHAPTER FOUR

STRESS AND EQUILIBRIUM

4.1 INTRODUCTION

This chapter will introduce the stress and equilibrium concepts for a de-

formable body undergoing a ¬nite motion. Stress is ¬rst de¬ned in the

current con¬guration in the standard way as force per unit area. This leads

to the well-known Cauchy stress tensor as used in linear analysis. We will

then derive the di¬erential equations enforcing translational and rotational

equilibrium and the equivalent principle of virtual work.

In contrast to linear small displacement analysis, stress quantities that

refer back to the initial body con¬guration can also be de¬ned. This will be

achieved using work conjugacy concepts that will lead to the Piola“Kirchho¬

stress tensors and alternative equilibrium equations. Finally, the objectivity

of several stress rate tensors is considered.

4.2 CAUCHY STRESS TENSOR

4.2.1 DEFINITION

Consider a general deformable body at its current position as shown in

Figure 4.1. In order to develop the concept of stress it is necessary to study

the action of the forces applied by one region R1 of the body on the remaining

part R2 of the body with which it is in contact. For this purpose consider

the element of area ∆a to normal n in the neighborhood of spatial point p

shown in Figure 4.1. If the resultant force on this area is ∆p, the traction

1

2 STRESS AND EQUILIBRIUM

x3

R2

R2

t

n

∆p

n

∆a

p

’n

x1

’t

R1

R1 x2

FIGURE 4.1 Traction vector.

vector t corresponding to the normal n at p is de¬ned as,

∆p

t(n) = lim (4.1)

∆a’0 ∆a

where the relationship between t and n must be such that satis¬es Newton™s

third law of action and reaction, which is expressed as (see Figure 4.1),

t(’n) = ’t(n) (4.2)

To develop the idea of a stress tensor, let the three traction vectors

associated with the three Cartesian directions e1 , e2 , and e3 be expressed

in a component form as (see Figure 4.2),

t(e1 ) = σ11 e1 + σ21 e2 + σ31 e3 (4.3a)

t(e2 ) = σ12 e1 + σ22 e2 + σ32 e3 (4.3b)

t(e3 ) = σ13 e1 + σ23 e2 + σ33 e3 (4.3c)

Although the general equilibrium of a deformable body will be discussed

in detail in the next section, a relationship between the traction vector

t corresponding to a general direction n and the components σij can be

obtained only by studying the translational equilibrium of the elemental

tetrahedron shown in Figure 4.3. Letting f be the force per unit volume

acting on the body at point p (which in general could also include inertia

3

4.2 CAUCHY STRESS TENSOR

e3

t (e2)

32

22

e2

12

e1

FIGURE 4.2 Stress components.

e3

t (n)

t(’e2 ) e2

n

da1

da

da2

e1

da 3

FIGURE 4.3 Elemental tetrahedron.

terms), the equilibrium of the tetrahedron is given as,

3

t(n) da + t(’ei ) dai + f dv = 0 (4.4)

i=1

where dai = (n · ei ) da is the projection of the area da onto the plane or-

thogonal to the Cartesian direction i (see Figure 4.3) and dv is the volume

of the tetrahedron. Dividing Equation (4.4) by da, recalling Newton™s third

4 STRESS AND EQUILIBRIUM

law, using Equations (4.3a“c), and noting that dv/da ’ 0 gives,

3

daj dv

t(n) = ’ t(’ej ) ’f

da da

j=1

3

= t(ej ) (n · ej )

j=1

3

= σij (ej · n) ei (4.5)

i,j=1

Observing that (ej · n)ei can be rewritten in terms of the tensor product as

(ei — ej )n gives,

3

t(n) = σij (ej · n) ei

i,j=1

3

= σij (ei — ej )n

i,j=1

3

= σij (ei — ej ) n (4.6)

i,j=1

which clearly identi¬es a tensor σ, known as the Cauchy stress tensor, that

relates the normal vector n to the traction vector t as,

3

t(n) = σn; σ= σij ei — ej (4.7a,b)

i,j=1

5

4.2 CAUCHY STRESS TENSOR

EXAMPLE 4.1: Rectangular block under self-weight (i)

X2 x2

e2

H

dx 1 h

X 1 x1

A simple example of a two-dimensional stress tensor results from the

self-weight of a block of uniform initial density ρ0 resting on a fric-

tionless surface as shown in the ¬gure above. For simplicity we will

assume that there is no lateral deformation (in linear elasticity this

would imply that the Poisson ratio ν=0).

Using De¬nition (4.1), the traction vector t associated with the unit

vertical vector e2 at an arbitrary point at height x2 , initially at height

X2 , is equal to the weight of material above an in¬nitesimal section

divided by the area of this section. This gives,

h

(’ y ρg dx2 ) e2 dx1

t(e2 ) =

dx1

where g is the acceleration of gravity and h is the height of the block

after deformation. The mass conservation Equation (3.57) implies that

ρdx1 dx2 = ρ0 dX1 dX2 , which in conjunction with the lack of lateral

deformation gives,

t(e2 ) = ρ0 g(H ’ X2 ) e2

Combining this equation with the fact that the stress components σ12

and σ22 are de¬ned in Equation (4.3) by the expression t(e2 ) = σ12 e1 +

σ22 e2 gives σ12 = 0 and σ22 = ’ρ0 g(H ’ X2 ). Using a similar process

and given the absence of horizontal forces, it is easy to show that the

traction vector associated with the horizontal unit vector is zero and

consequently σ11 = σ21 = 0. The complete stress tensor in Cartesian

components is therefore,

0 0

[σ] =

0 ρ0 g(X2 ’ H)

The Cauchy stress tensor can alternatively be expressed in terms of its

6 STRESS AND EQUILIBRIUM

principal directions m1 , m2 , m3 and principal stresses σ±± for ± = 1, 2, 3 as,

3

σ= σ±± m± — m± (4.8)

±=1

where from Equations (2.57a“b), the eigenvectors m± and eigenvalues σ±±