<<

. 22
( 54 .)



>>

t t2 1
® 
’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±±

<<

. 22
( 54 .)



>>