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satisfy,

σm± = σ±± m± (4.9)

In the next chapter we shall show that for isotropic materials the principal
directions m± of the Cauchy stress coincide with the principal Eulerian triad
n± introduced in the previous chapter.
Note that σ is a spatial tensor; equivalent material stress measures asso-
ciated with the initial con¬guration of the body will be discussed later. Note
also that the well-known symmetry of σ has not yet been established. In
fact this results from the rotational equilibrium equation, which is discussed
in the following section.



4.2.2 STRESS OBJECTIVITY
Because the Cauchy stress tensor is a key feature of any equilibrium or ma-
terial equation, it is important to inquire whether σ is objective as de¬ned
in Section 3.15. For this purpose consider the transformations of the nor-
mal and traction vectors implied by the superimposed rigid body motion Q
shown in Figure 4.4 as,

˜n
t(˜ ) = Qt(n) (4.10a)
n = Qn
˜ (4.10b)

Using the relationship between the traction vector and stress tensor given
by Equation (4.7a) in conjunction with the above equation gives,

σ = QσQT
˜ (4.11)

The rotation of σ given by the above equation conforms with the de¬nition
of objectivity given by Equation (3.135), and hence σ is objective and a
valid candidate for inclusion in a material description. It will be shown later
that the material rate of change of stress is not an objective tensor.
7
4.3 EQUILIBRIUM



Q
x3


˜
t t
n
˜
n


p ˜
p

x1
x2




FIGURE 4.4 Superimposed rigid body motion.


4.3 EQUILIBRIUM

4.3.1 TRANSLATIONAL EQUILIBRIUM
In order to derive the di¬erential static equilibrium equations, consider the
spatial con¬guration of a general deformable body de¬ned by a volume v
with boundary area ‚v as shown in Figure 4.5. We can assume that the body
is under the action of body forces f per unit volume and traction forces t
per unit area acting on the boundary. For simplicity, however, inertia forces
will be ignored, and therefore translational equilibrium implies that the sum
of all forces acting on the body vanishes. This gives,

t da + f dv = 0 (4.12)
‚v v


Using Equation (4.7a) for the traction vector enables Equation (4.12) to
be expressed in terms of the Cauchy stresses as,

σn da + f dv = 0 (4.13)
‚v v
The ¬rst term in this equation can be transformed into a volume integral
by using the Gauss theorem given in Equation (2.139) to give,

(·σ + f ) dv = 0 (4.14)
v
where the vector ·σ is de¬ned in Section 2.4.1. The fact that the above
8 STRESS AND EQUILIBRIUM



t
X 3, x 3
n


`
v



‚v

f


time = t
V
X1 , x1
X 2, x 2


time = 0

FIGURE 4.5 Equilibrium.



equation can be equally applied to any enclosed region of the body implies
that the integrand function must vanish, that is,

·σ + f = 0 (4.15)

This equation is known as the local (that is, pointwise) spatial equilibrium
equation for a deformable body. In anticipation of situations during a solu-
tion procedure in which equilibrium is not yet satis¬ed, the above equation
de¬nes the pointwise out-of-balance or residual force per unit volume r as,

r = ·σ + f (4.16)


EXAMPLE 4.2: Rectangular block under self-weight (ii)
It is easy to show that the stress tensor given in Example 4.1 satis-
¬es the equilibrium equation. For this purpose, note ¬rst that in this
particular case the forces f per unit volume are f = ’ρge2 , or in

(continued)
9
4.3 EQUILIBRIUM




EXAMPLE 4.2 (cont.)
component form,
0
[f ] =
’ρg
Additionally, using De¬nition (2.134), the two-dimensional components
of the divergence of σ are,
‚σ11 ‚σ12
+ 0
‚x1 ‚x2
[·σ] = =
ρ0 g dX2
‚σ21 ‚σ22 2
+ dx
‚x1 ‚x2
which combined with the mass conservation equation ρdx1 dx2 = ρ0 dX1 dX2
and the lack of lateral deformation implies that Equation (4.14) is sat-
is¬ed.



4.3.2 ROTATIONAL EQUILIBRIUM
Thus far the well-known symmetry of the Cauchy stresses has not been es-
tablished. This is achieved by considering the rotational equilibrium of a
general body, again under the action of traction and body forces. This im-
plies that the total moment of body and traction forces about any arbitrary
point, such as the origin, must vanish, that is,

x—t da + x—f dv = 0 (4.17)
‚v v
where it should be recalled that the cross product of a force with a position
vector x yields the moment of that force about the origin. Equation (4.7a)
for the traction vector in terms of the Cauchy stress tensor enables the above
equation to be rewritten as,

x—(σn) da + x—f dv = 0 (4.18)
‚v v

Using the Gauss theorem and after some algebra, the equation becomes*

* To show this it is convenient to use indicial notation and the summation convention whereby
repeated indices imply addition. Equation (2.136) then gives,

Z Z
E ijk xj σkl nl da = (E ijk xj σkl ) dv
v ‚xl
‚v

‚σkl
Z Z
= +
E ijk xj E ijk σkj dv
‚xl
v v
Z Z
(x— · σ)i dv + (E : σT )i dv
=
v v
10 STRESS AND EQUILIBRIUM



t
X3, x 3
n


`
v
δv


‚v


f

V time = t
X1, x 1
X2, x 2


time = 0

FIGURE 4.6 Principle of virtual work.




E : σ T dv +
x—(·σ) dv + x—f dv = 0 (4.19)
v v v

where E is the third-order alternating tensor, de¬ned in Section 2.2.4 (E ijk =
1 if the permutation {i, j, k} is even, “1 if it is odd, and zero if any indices
are repeated.), so that the vector E : σ T is,
® 
σ32 ’ σ23
E : σ T = ° σ13 ’ σ31 » (4.20)
σ21 ’ σ12
Rearranging terms in Equation (4.19) to take into account the translational
equilibrium Equation (4.15) and noting that the resulting equation is valid
for any enclosed region of the body gives,
E : σT = 0 (4.21)
which, in view of Equation (4.20), clearly implies the symmetry of the
Cauchy stress tensor σ.
11
4.4 PRINCIPLE OF VIRTUAL WORK



4.4 PRINCIPLE OF VIRTUAL WORK

Generally, the ¬nite element formulation is established in terms of a weak
form of the di¬erential equations under consideration. In the context of
solid mechanics this implies the use of the virtual work equation. For this
purpose, let δv denote an arbitrary virtual velocity from the current position
of the body as shown in Figure 4.6. The virtual work, δw, per unit volume
and time done by the residual force r during this virtual motion is r · δv,
and equilibrium implies,

δw = r · δv = 0 (4.22)

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