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( 67 .)


σ ( x0 , t) = G{µ( x0 , „ ) ; „ ¤ t}, (13.5)
where F{}, as used in Eq. (13.1), has been replaced by G{} due to adaptations in
the argument, with
1 T
µ= ∇ 0u + ∇ 0u , (13.6)
∇ 0 · σ + ρ0 q = ρ0 u. (13.7)
Again these equations have to be satis¬ed for all x0 in V0 and for all times t.
The ¬rst equation (a formal functional expression) indicates that the current local
213 13.2 Solution strategies for deforming solids

Cauchy stress tensor σ is fully determined by the (history of the) local linear strain
tensor µ. The second equation actually implies that the balance of momentum
(equation of motion) is now related to the reference con¬guration instead of the
current con¬guration. The balance of mass is now redundant through replacing ρ
by ρ0 . With respect to the boundary conditions and initial conditions, the same
statements apply as in the previous section.

13.2.3 Linear elasticity theory, dynamic

In Section 13.2.2 geometrical linearity has been introduced to simplify the gen-
eral set of equations given in Section 13.2.1. If we add physical linearity to this
(meaning that the relation between stress and strain is described by Hooke™s law as
formulated in Chapter 12) linear elasticity theory results. The associated (linear)
equations, that have to be satis¬ed for all x0 in V0 and for all times t, read:
σ = K’ tr( µ) I + 2Gµ, (13.8)
1 T
µ= ∇ 0u + ∇ 0u ,
∇ 0 · σ + ρ0 q = ρ0 u. (13.9)
With respect to boundary conditions and initial conditions the same procedure
must be followed as in the previous section.

13.2.4 Linear elasticity theory, static

For slowly evolving (quasi-static) processes inertia effects can be neglected. In
that case it is suf¬cient to consider only the relevant current state and time is no
longer of interest. The current displacement ¬eld u( x0 ) and the current stress ¬eld
σ ( x0 ) have to be determined on the basis of the equations:
σ = K’ tr( µ) I + 2Gµ, (13.10)
1 T
µ= ∇ 0u + ∇ 0u , (13.11)
Solution strategies for solid and fluid mechanics problems

∇ 0 · σ + ρ0 q = 0. (13.12)
The equation of motion is reduced to the equilibrium equation. Only boundary
conditions are necessary to solve this set. Initial conditions are no longer appli-
cable. In every point of the outer surface of V0 three (scalar) relations have to be
speci¬ed. In this context, to ¬nd a unique solution for the displacement ¬eld, it is
necessary to suppress movement of the con¬guration as a rigid body. This will be
further elucidated in the following example.
Example 13.1 Consider a homogeneous body with reference volume V0 under a given hydro-
static pressure p. For the stress ¬eld in V0 satisfying the (dynamic) boundary
conditions and the equilibrium equations, it can be written:
σ ( x0 ) = ’pI for all x0 in V0 .
For the strain ¬eld, according to Hooke™s law it is found:
µ( x0 ) = ’ I for all x0 in V0 .
A matching displacement ¬eld is, for example,
u( x0 ) = ’ x0 .
It is easy to verify that the above given solution satis¬es all equations. However,
because in the given problem description the displacement as a rigid body is not
prescribed the solution is not unique. It can be proven that the general solution for
the components of the displacement vector reads:
ux = ’ x0 + c1 ’ c6 y0 + c5 z0
uy = ’ y0 + c2 + c6 x0 ’ c4 z0
uz = ’ z0 + c3 ’ c5 x0 + c4 y0 ,
with ci ( i = 1, 2, . . . 6) yet undetermined constants. The constants c1 , c2 and c3
represent translations in the coordinate directions and the constants c4 , c5 and c6
(small) rotations around the coordinate axes.

13.2.5 Linear plane stress theory, static

Consider a ¬‚at membrane with, in the reference con¬guration, constant thickness
h. The ˜midplane™ of the membrane coincides with the x0 y0 -plane, while in the
215 13.2 Solution strategies for deforming solids




ux x0

thickness h
mechanical load

Figure 13.1
Con¬guration with plane stress state.

direction perpendicular to that plane the domain for z0 is given by: ’h/2 ¤ z0 ¤
h/2. The thickness h is supposed to be small with respect to the dimensions ˜in
the plane™. The loading is parallel to the midplane of the membrane, see Fig. 13.1.
The midplane will continue to be a symmetry plane after deformation. It is
assumed that straight material line segments, initially perpendicular to the mid-
plane will remain straight (and perpendicular to the midplane) after deformation.
For the displacements this means

ux = ux ( x0 , y0 ) , uy = uy ( x0 , y0 ) . (13.13)

The relevant strain components of the linear strain matrix µ for the membrane are

µxx =
µyy = (13.14)
µxy = µyx = + .
‚y0 ‚x0

In general, from the other components µzz will certainly not be zero, while
µxz = µzx and µyz = µzy will vanish. Actually, these other strain components are
not important to set-up the theory.
With respect to the stress, it is assumed that only components ˜acting in the
x0 y0 -plane™ play a role and that for those components (see Fig. 13.2) it can be
Solution strategies for solid and fluid mechanics problems





Figure 13.2
Stress components in a plane stress state.

σxx = σxx ( x0 , y0 )
σyy = σyy ( x0 , y0 ) (13.15)
σxy = σyx = σxy ( x0 , y0 ) .

The stress components σzz , σxy = σyx and σxz = σzx are assumed to be zero
Based on Hooke™s law (see Section 12.2 for the fully elaborated expression in
components), using σzz = 0 it is found:
µzz = ’ ( µxx + µyy ) , (13.16)
and by exploiting this equation, the description of the (linearly elastic) material
behaviour for plane stress becomes
σxx = ( µxx + νµyy ) (13.17)
1 ’ ν2
σyy = ( νµxx + µyy ) (13.18)
1 ’ ν2
σxy = µxy , (13.19)
( 1 + ν)
thus coupling the relevant stresses and strains. It should be noted that in this case
for the material parameters the Young™s modulus E and the Poisson™s ratio ν have
been used, instead of the compression modulus K and the shear modulus G in
previous sections.
217 13.2 Solution strategies for deforming solids

The stresses have to satisfy the equilibrium equations. For plane stress this
means that only equilibrium ˜in the plane™ (in the case considered therefore in
the x0 - and y0 -direction) results in non-trivial equations:
+ + ρ0 qx = 0 (13.20)
‚x0 ‚y0
‚σxy ‚σyy
+ + ρ0 qy = 0. (13.21)
‚x0 ‚y0
Summarizing, it can be stated that in case of plane stress eight scalar functions
of the coordinates in the midplane have to be calculated: the displacements ux and
uy , the strains µxx , µyy and µxy , and the stresses σxx , σyy and σxy . For this objective
we have eight equations at our disposal: the strain/displacement relations (three),
the constitutive equations (three) and the equilibrium equations (two). In addition,
for each boundary point of the midplane two scalar boundary conditions have to
be speci¬ed and for uniqueness of the displacement ¬eld rigid body motion has to
be suppressed.

Example 13.2 In Fig. 13.3 a simple plane stress problem is de¬ned for a rectangular membrane
(length 2l, width 2b and thickness h) with linearly elastic material behaviour
(Young™s modulus E and Poisson™s ratio ν). The mathematical form for the
boundary conditions reads:
for x0 = ± l ’ b ¤ y0 ¤ b it holds: σxx = ± + β
σxy = 0
for y0 = ± b ’ l ¤ x0 ¤ l it holds: σyy = 0
σxy = 0.

σyy = σxy = 0
thickness h y0

x0 y0
σxx = ± + β
2b b
σxy = 0


Figure 13.3
A simple plane stress problem.
Solution strategies for solid and fluid mechanics problems

Herewith, the external load is speci¬ed using the constants ± and β; the constant
± is representative for the ˜normal™ force in the x0 direction and the constant β
for the ˜bending™. With the above given boundary conditions the displacement of
the membrane as a rigid body is not suppressed; uniqueness of the displacement
solution is obtained, if it is additionally required that the material point coinciding


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