Ebook The finite element method in engineering (4th edition) Part 2

APPLICATION TO SOLID
MECHANICS PROBLEMS


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8
BASIC EQUATIONS AND
SOLUTION PROCEDURE

8.1 INTRODUCTION
As stated in Chapter 1, the finite element method has been most extensively used in the
field of solid and structural mechanics. The various types of problems solved by the finite
element method in this field include the elastic, elastoplastic, and viscoelastic analysis of
trusses, frames, plates, shells, and solid bodies. Both static and dynamic analysis have
been conducted using the finite element method. We consider the finite element elastic
analysis of one-, two-, and three-dimensional problems as well as axisymmetric problems
in this book.
In this chapter, the general equations of solid and structural mechanics are presented.
The displacement method (or equivalently the principle of minimum potential energy) is
used in deriving the finite element equations. The application of these equations to several
specific cases is considered in subsequent chapters.
8.2 BASIC EQUATIONS OF SOLID MECHANICS
8.2.1 Introduction
The primary aim of any stress analysis or solid mechanics problem is to find the distribution of displacements and stresses under the stated loading and boundary conditions. If
an analytical solution of the problem is to be found, one has to satisfy the following basic
equations of solid mechanics:
Number of equations
Type of equations
Equilibrium equations

Stress–strain relations
Strain–displacement
relations
Total number of equations

In 3-dimensional
problems

In 2-dimensional
problems

In 1-dimensional
problems

3
6
6

2
3
3

1
1
1

15

8

3

279


280

BASIC EQUATIONS AND SOLUTION PROCEDURE

The unknown quantities, whose number is equal to the number of equations available, in
various problems are given below:
In 3-dimensional
problems

Unknowns
Displacements
Stresses

u, v, w
σxx , σyy , σzz ,
σxy , σyz , σzx
εxx , εyy , εzz , εxy
εyz , εzx

Strains
Total number of unknowns

15

In 2-dimensional

problems

In 1-dimensional
problems

u, v
σxx , σyy , σxy

u
σxx

εxx , εyy , εxy

εxx

8

3

Thus, we have as many equations as there are unknowns to find the solution of any stress
analysis problem. In practice, we will also have to satisfy some additional equations, such
as external equilibrium equations (which pertain to the overall equilibrium of the body
under external loads), compatibility equations (which pertain to the continuity of strains
and displacements), and boundary conditions (which pertain to the prescribed conditions
on displacements and/or forces at the boundary of the body).
Although any analytical (exact) solution has to satisfy all the equations stated previously, the numerical (approximate) solutions, like the ones obtained by using the finite
element method, generally do not satisfy all the equations. However, a sound understanding of all the basic equations of solid mechanics is essential in deriving the finite element
relations and also in estimating the order of error involved in the finite element solution
by knowing the extent to which the approximate solution violates the basic equations,
including the compatibility and boundary conditions. Hence, the basic equations of solid

mechanics are summarized in the following section for ready reference in the formulation
of finite element equations.
8.2.2 Equations
(i) External equilibrium equations
If a body is in equilibrium under specified static loads, the reactive forces and moments
developed at the support points must balance the externally applied forces and moments.
In other words, the force and moment equilibrium equations for the overall body (overall
or external equilibrium equations) have to be satisfied. If φx , φy , and φz are the body
forces, Φx , Φy , and Φz are the surface (distributed) forces, Px , Py , and Pz are the external
concentrated loads (including reactions at support points such as B, C, and D in Figure
8.1), and Qx , Qy , and Qz are the external concentrated moments (including reactions at
support points such as B, C, and D in Figure 8.1), the external equilibrium equations can
be stated as follows [8.1]:
⎫
Φx ds + φx dv +
Px = 0 ⎪
⎪
⎪
⎪
⎪
⎪
S
V
⎪
⎪
⎪
⎬
Φy ds + φy dv +
Py = 0
(8.1)

⎪
⎪
S
V
⎪
⎪
⎪
⎪
⎪
Φz ds + φz dv +
Px = 0 ⎪
⎪
⎭
S

V


BASIC EQUATIONS OF SOLID MECHANICS

281

Figure 8.1. Force System for Macroequilibrium for a Body.

For moment equilibrium:
(Φz y − Φy z) ds +
S

(φz y − φy z) dv +
V

(Φx z − Φz x) ds +
S

(φx z − φz x) dv +
V

(Φy x − Φx y) ds +
S

(φy x − φx y) dv +

⎫
Qx = 0⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎬
Qy = 0
⎪
⎪
⎪
⎪
⎪
⎪

⎪
Qz = 0⎪
⎪
⎭

(8.2)

V

where S is the surface and V is the volume of the solid body.
(ii) Equations of internal equilibrium:
Due to the application of loads, stresses will be developed inside the body. If we consider
an element of material inside the body, it must be in equilibrium due to the internal
stresses developed. This leads to equations known as internal equilibrium equations.
Theoretically, the state of stress at any point in a loaded body is completely defined
in terms of the nine components of stress σxx , σyy , σzz , σxy , σyx , σyz , σzy , σzx , and σxz ,
where the first three are the normal components and the latter six are the components
of shear stress. The equations of internal equilibrium relating the nine components of
stress can be derived by considering the equilibrium of moments and forces acting on the
elemental volume shown in Figure 8.2. The equilibrium of moments about the x, y, and z
axes, assuming that there are no body moments, leads to the relations
σyx = σxy ,

σzy = σyz ,

σxz = σzx

(8.3)

These equations show that the state of stress at any point can be completely defined by

the six components σxx , σyy , σzz , σxy , σyz , and σzx . The equilibrium of forces in x, y,


282

BASIC EQUATIONS AND SOLUTION PROCEDURE

Figure 8.2. Elemental Volume Considered for Internal Equilibrium (Only the Components of
Stress Acting on a Typical Pair of Faces Are Shown for Simplicity).

and z directions gives the following differential equilibrium equations:
⎫
∂σxx
∂σxy
∂σzx
+
+
+ φx = 0⎪
⎪
⎪
⎪
∂x
∂y
∂z
⎪
⎪
⎪
⎪
⎬
∂σxy

∂σyy
∂σyz
+
+
+ φy = 0
⎪
∂x
∂y
∂z
⎪
⎪
⎪
⎪
⎪
⎪
∂σzx
∂σyz
∂σzz
⎭
+
+
+ φz = 0⎪
∂x
∂y
∂z

(8.4)

where φx , φy , and φz are the body forces per unit volume acting along the directions x,
y, and z, respectively.

For a two-dimensional problem, there will be only three independent stress components
(σxx , σyy , σxy ) and the equilibrium equations, Eqs. (8.4), reduce to
⎫
∂σxx
∂σxy
⎪
+
+ φx = 0⎪
⎪
⎬
∂x
∂y
∂σxy
∂σyy
+
∂x
∂y

⎪
⎪
⎭
+ φy = 0⎪

(8.5)

In one-dimensional problems, only one component of stress, namely σxx , will be there and
hence Eqs. (8.4) reduce to
∂σxx
+ φx = 0
∂x

(8.6)


BASIC EQUATIONS OF SOLID MECHANICS

283

(iii) Stress–strain relations (constitutive relations) for isotropic materials
Three-dimensional case In the case of linearity elastic isotropic three-dimensional
solid, the stress–strain relations are given by Hooke’s law as
⎧ ⎫
⎫ ⎧
⎫
⎧
εxx0 ⎪
σxx ⎪
εxx ⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪

⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
ε
σ
ε
⎪
⎪
⎪
⎪
⎪
yy
yy
yy
0⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨ ⎬
⎬ ⎨
⎬

⎨
εzz
σzz
εzz0
ε=
= [C]σ + ε0 ≡ [C]
+
εxy ⎪
σxy ⎪
εxy0 ⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪

⎪
ε
σ
ε
⎪
⎪
⎪
⎪
⎪
⎪
yz
yz
yz
0
⎪ ⎭
⎪
⎪
⎪
⎪
⎩
⎩
⎭ ⎪
⎭
⎩
εzx
σzx
εzx0

(8.7)

where [C] is a matrix of elastic coefficients given by
⎡

1
⎢−v
⎢
1 ⎢
⎢−v
[C] =
E⎢
⎢ 0
⎣ 0
0

−v
1
−v
0
0
0

−v
−v
1
0
0
0

0
0

0
2(1 + v)
0
0

0
0
0
0
2(1 + v)
0

⎤
0
⎥
0
⎥
⎥
0
⎥
⎥
0
⎥
⎦
0
2(1 + v)

(8.8)

ε0 is the vector of initial strains, E is Young’s modules, and v is Poisson’s ratio of

the material. In the case of heating of an isotropic material, the initial strain vector is
given by
⎫
⎧
ε
⎪
xx
⎪
0
⎪
⎪
⎪
⎪
⎪
⎪
ε
⎪
⎪
yy
0
⎪
⎪
⎬
⎨
εzz0
ε0 =
= αT
εxy0 ⎪
⎪
⎪

⎪
⎪
⎪
⎪
εyz0 ⎪
⎪
⎪
⎪
⎪
⎩
⎭
εzx0

⎧ ⎫
1⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
1
⎪
⎪
⎪
⎬
⎨ ⎪
1
0⎪

⎪
⎪
⎪
⎪
⎪
⎪
⎪
0
⎪
⎪
⎪
⎭
⎩ ⎪
0

(8.9)

where α is the coefficient of thermal expansion, and T is the temperature charge.
Sometimes, the expressions for stresses in terms of strains will be needed. By including
thermal strains, Eqs. (8.7) can be inverted to obtain
⎧
⎧ ⎫
⎫
⎧ ⎫
σ
1⎪
ε
⎪
⎪
⎪

⎪
xx
xx
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
1
σ
ε
⎪
⎪
⎪
⎪
⎪

⎪
yy
yy
⎪
⎪
⎪
⎪
⎪
⎨
⎬
⎨ ⎪
⎬
⎨ ⎬
EαT
1
σzz
εzz
σ=
= [D](ε − ε0 ) ≡ [D]
−
0⎪
σxy ⎪
εxy ⎪
⎪
⎪
1 − 2v ⎪
⎪
⎪
⎪
⎪

⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
0
σ
ε
⎪
⎪
⎪
⎪
⎪
⎪
yz ⎪
yz ⎪
⎪
⎪
⎪
⎩
⎭

⎭
⎩ ⎭
⎩ ⎪
0
σzx
εzx

(8.10)


284

BASIC EQUATIONS AND SOLUTION PROCEDURE

where the matrix [D] is given by
⎡

1−v
⎢ v
⎢
⎢ v
⎢
⎢
⎢ 0
E
⎢
[D] =
(1 + v)(1 − 2v) ⎢
⎢
⎢

⎢ 0
⎢
⎣
0

v
1−v
v

v
v
1−v

0

0

0

0

0

0

0
0
0
1 − 2v
2

0
0
0

0
0
0

0

0

0

1 − 2v
2

0

0

0

1 − 2v
2

⎤
⎥
⎥

⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦

(8.11)

In the case of two-dimensional problems, two types of stress distributions, namely plane
stress and plane strain, are possible.
Two-dimensional case (plane stress) The assumption of plane stress is applicable
for bodies whose dimension is very small in one of the coordinate directions. Thus, the
analysis of thin plates loaded in the plane of the plate can be made using the assumption
of plane stress. In plane stress distribution, it is assumed that
σzz = σzx = σyz = 0

(8.12)

where z represents the direction perpendicular to the plane of the plate as shown in
Figure 8.3, and the stress components do not vary through the thickness of the plate (i.e.,
in z direction). Although these assumptions violate some of the compatibility conditions,
they are sufficiently accurate for all practical purposes provided the plate is thin. In this
case, the stress–strain relations, Eqs. (8.7) and (8.10), reduce to
ε = [C]σ + ε0

(8.13)

y

y

x

z

Figure 8.3. Example of a Plane Stress Problem; A Thin Plate under Inplane Loading.


BASIC EQUATIONS OF SOLID MECHANICS

285

where
⎧
⎫
⎧ ⎫
⎨σxx ⎬
⎨εxx ⎬
σ = σyy
ε = εyy ,
⎩
⎭
⎩ ⎭
εxy

σxy
⎤
⎡
1 −v
0
1 ⎣
⎦
−v
1
0
[C] =
E
0
0 2(1 + v)
⎧ ⎫
⎧
⎫
⎨1⎬
⎨εxx0 ⎬
in the case of thermal strains
ε0 = εyy0 = αT 1
⎩ ⎭
⎩
⎭
0
εxy0

(8.14)

(8.15)

and
⎧ ⎫
⎨1⎬
EαT
1
σ = [D](ε − ε0 ) = [D]ε −
1−v ⎩ ⎭
0

(8.16)

with
⎡

1
⎢
E ⎢v
[D] =
1 − v2 ⎣
0

v
1

0
0

0

1−v
2

⎤
⎥
⎥
⎦

(8.17)

In the case of plane stress, the component of strain in the z direction will be nonzero and
is given by (from Eq. 8.7)
εzz = −

v
−v
1+v
(σxx + σyy ) + αT =
(εxx + εyy ) +
αT
E
1−v
1−v

(8.18)

while
εyz = εzx = 0

(8.19)

Two-dimensional case (plane strain) The assumption of plane strain is applicable
for bodies that are long and whose geometry and loading do not vary significantly in
the longitudinal direction. Thus, the analysis of dams, cylinders, and retaining walls
shown in Figure 8.4 can be made using the assumption of plane strain. In plane strain
distribution, it is assumed that w = 0 and (∂w/∂z) = 0 at every cross section. Here,
the dependent variables are assumed to be functions of only the x and y coordinates
provided we consider a cross section of the body away from the ends. In this case, the
three-dimensional stress–strain relations given by Eqs. (8.7) and (8.10) reduce to
ε = [C]σ + ε0

(8.20)


286

BASIC EQUATIONS AND SOLUTION PROCEDURE
x
z

y
(a) Dam
z

x
x
z
y
y
(c) Retaining wall

(b) Long cylinder

Figure 8.4. Examples of Plane Strain Problems.

where
⎧
⎫
⎧ ⎫
⎨σxx ⎬
⎨εxx ⎬
σ = σyy ,
ε = εyy ,
⎩
⎩ ⎭
⎭
εxy
σxy
⎤
⎡
1−v
−v
0
1+y ⎣
−v
1 − v 0⎦ ,
[C] =
E
0
0
2

⎧ ⎫
⎧
⎫
⎨1⎬
⎨εxx0 ⎬
ε0 = εyy0 = (1 + v)αT 1
⎩ ⎭
⎩
⎭
0
εxy0

(8.21)

in the case of thermal strains

(8.22)

and
⎧ ⎫
⎨1⎬
EαT
1
σ = [D](ε − ε0 ) = [D]ε −
1 − 2v ⎩ ⎭
0

(8.23)

BASIC EQUATIONS OF SOLID MECHANICS

287

with
⎡
[D] =

E
(1 + v)(1 − 2v)

1−v
⎢ v
⎣
0

v
1−v
0

⎤
0
0 ⎥
⎦
1 − 2v
2

(8.24)

The component of stress in the z direction will be nonzero and is given by

σzz = v(σxx + σyy ) − EαT

(8.25)

σyz = σzx = 0

(8.26)

and

One-dimensional case In the case of one-dimensional problems, all stress components
except for one normal stress are zero and the stress–strain relations degenerate to
ε = [C]σ + ε0

(8.27)

where
ε = εxx ,
[C] =

σ = σxx ,

1
,
E

(8.28)

ε0 = εxx0 = αT in the case of thermal strain

(8.29)

σ = [D](ε − ε0 ) = [D]ε − EαT {1}

(8.30)

[D] = [E]

(8.31)

and

with

Axisymmetric case In the case of solids of revolution (axisymmetric solids), the stress–
strain relations are given by
ε = [C]σ + ε0
where
⎧ ⎫
εrr ⎪
⎪
⎪
⎬
⎨ ⎪
εθθ
,
ε=
εzz ⎪
⎪
⎪

⎪
⎩ ⎭
εrz

⎧ ⎫
σrr ⎪
⎪
⎪
⎨ ⎪
⎬
σθθ
σ=
,
σzz ⎪
⎪
⎪
⎪
⎩ ⎭
σrz

(8.32)


288

BASIC EQUATIONS AND SOLUTION PROCEDURE
⎡

1 −v
1 ⎢

1
⎢−v
[C] =
E ⎣−v −v
0
0
⎫
⎧
εrr0 ⎪
⎪
⎪
⎪
⎬
⎨
εθθ0
ε0 =
= αT
εzz0 ⎪
⎪
⎪
⎪
⎭
⎩
εrz0

⎤
−v
0
⎥
−v

0
⎥,
⎦
1
0
0 2(1 + v)
⎧ ⎫
1⎪
⎪
⎪
⎬
⎨ ⎪
1
in the case of theremal strains
1⎪
⎪
⎪
⎪
⎩ ⎭
0

(8.33)

(8.34)

and
⎧ ⎫
1⎪
⎪
⎪

⎬
⎨ ⎪
EαT
1
σ = [D](ε − ε0 ) = [D]ε −
1⎪
1 − 2v ⎪
⎪
⎭
⎩ ⎪
0

(8.35)

with
⎡

1−v
⎢ v
⎢
E
⎢ v
[D] =
⎢
(1 + v)(1 − 2v) ⎢
⎣
0

v
1−v

v

v
v
1−v

0

0

0
0
0

⎤

⎥
⎥
⎥
⎥
⎥
1 − 2v ⎦
2

(8.36)

In these equations, the subscripts r, θ, and z denote the radial, tangential, and axial
directions, respectively.
(iv) Stress–strain relations for anisotropic materials
The stress–strain relations given earlier are valid for isotropic elastic materials. The term

“isotropic” indicates that the material properties at a point in the body are not a function
of orientation. In other words, the material properties are constant in any plane passing
through a point in the material. There are certain materials (e.g., reinforced concrete,
fiber-reinforced composites, brick, and wood) for which the material properties at any
point depend on the orientation also. In general, such materials are called anisotropic
materials. The generalized Hooke’s law valid for anisotropic materials is given in this
section. The special cases of the Hooke’s law for orthotropic and isotropic materials will
also be indicated.
For a linearly elastic anisotropic material, the strain–stress relations are given by the
generalized Hooke’s law as [8.7, 8.8]
⎧ ⎫
ε1 ⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
ε
2
⎪
⎪
⎪
⎪
⎪
⎨ ε3 ⎪
⎬

⎡

C11
⎢C12
⎢
=⎢ .
⎪
⎣ ..
ε23 ⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
ε ⎪
C16
⎪
⎩ 13 ⎪
⎭
ε12

C12
C22

···
···

C26

···

⎧ ⎫
σ1 ⎪
⎪
⎪
⎤⎪
⎪
⎪
⎪
C16 ⎪
σ
2
⎪
⎪
⎪
⎪
⎪
⎨
⎬
⎥
C26 ⎥ σ3 ⎪
⎥
⎦⎪
σ23 ⎪
⎪
⎪
⎪

⎪
⎪
⎪
⎪
⎪
⎪
σ
C66 ⎪
13
⎪
⎩ ⎪
⎭
σ12

(8.37)


BASIC EQUATIONS OF SOLID MECHANICS

289

where the matrix [C] is symmetric and is called the compliance matrix. Thus, 21 independent elastic constants (equal to the number of independent components of [C]) are needed
to describe an anisotropic material. Note that subscripts 1, 2, and 3 are used instead of
x, y, and z in Eq. (8.37) for convenience.
Certain materials exhibit symmetry with respect to certain planes within the body.
In such cases, the number of elastic constants will be reduced from 21. For an orthotropic
material, which has three planes of material property symmetry, Eq. (8.37) reduces to
⎧ ⎫ ⎡
ε1 ⎪
C11

⎪
⎪
⎪
⎪
⎪
⎪
⎢
⎪
C12
ε
⎪
2⎪
⎪
⎨ ⎪
⎬ ⎢
⎢C13
ε3
=⎢
⎢ 0
ε23 ⎪
⎪
⎪
⎢
⎪
⎪
⎪
⎪
⎪
⎣ 0
ε

⎪
⎪
13
⎪
⎩ ⎪
⎭
0
ε12

C12
C22
C23
0
0
0

C13
C23
C33
0
0
0

0
0
0
C44
0
0

0
0
0
0
C55
0

⎤⎧ ⎫
σ1 ⎪
0 ⎪
⎪
⎪
⎪
⎪
⎪
⎪
0 ⎥
σ
⎪
⎪
2
⎪
⎥⎪
⎨
⎬
0 ⎥
σ
3
⎥
0 ⎥

σ23 ⎪
⎪
⎥⎪
⎪
⎪
⎪
⎪
⎪
⎦
0 ⎪
σ
⎪
13
⎪
⎩ ⎪
⎭
C66
σ12

(8.38)

where the elements Cij are given by
C11

1
=
,
E11

C12

v21
=−
,
E22

C13

v32
,
E33

C33

C22 =

1
,
E22

C23 = −

C44 =

1
,
G23

C55 =

1
,
G13

C66

⎫
v31 ⎪
=−
,⎪
⎪
E33 ⎪
⎪
⎪
⎪
⎪
⎬
1
=
,
E33 ⎪
⎪
⎪
⎪
⎪
⎪
⎪
1
⎪
⎭

=
G12

(8.39)

Here, E11 , E22 , and E33 denote the Young’s modulus in the planes defined by axes 1,
2, and 3, respectively; G12 , G23 , and G13 represent the shear modulus in the planes 12,
23, and 13, respectively; and v12 , v13 , and v23 indicate the major Poisson’s ratios. Thus,
nine independent elastic constants are needed to describe an orthotropic material under
three-dimensional state of stress. For the specially orthotropic material that is in a state
of plane stress, σ3 = σ23 = σ13 = 0 and Eq. (8.38) reduces to
⎧ ⎫ ⎡
C11
⎨ ε1 ⎬
⎣
ε2 = C12
⎩ ⎭
0
ε12

C12
C22
0

⎤⎧ ⎫
0 ⎨ σ1 ⎬
0 ⎦ σ2
⎩ ⎭
C66
σ12

(8.40)

which involves four independent elastic constants. The elements of the compliance matrix,
in this case, can be expressed as
1
E11
1
=
E22
v12
v21
=−
=−
E11
E22
1
=
G12

C11 =
C22
C12
C66

(8.41)


290

BASIC EQUATIONS AND SOLUTION PROCEDURE

The stress–strain relations can be obtained by inverting the relations given by
Eqs. (8.37), (8.38), and (8.40). Specifically, the stress–strain relations for a specially
orthotropic material (under plane stress) can be expressed as
⎧ ⎫ ⎡
Q11
⎨ σ1 ⎬
⎣
σ2 = Q12
⎩ ⎭
0
σ12

Q12
Q22
0

⎤⎧ ⎫
0 ⎨ ε1 ⎬
0 ⎦ ε2 ≡ [Q]ε
⎩ ⎭
Q66
ε12

(8.42)

where the elements of the matrix [Q] are given by
⎫
E22

⎪
⎪
1 − v12 v21 ⎪
⎪
⎪
⎪
⎬

Q11 =

E11
,
1 − v12 v21

Q12 =

v21 E11
v12 E22
=
1 − v12 v21
1 − v12 v21

Q66 = 2G12

Q22 =

⎪
⎪
⎪
⎪

⎪
⎪
⎭

(8.43)

If the material is linearly elastic and isotropic, only two elastic constants are needed to
describe the behavior and the stress–strain relations are given by Eq. (8.7) or Eq. (8.10).
(v) Strain–displacement relations
The deformed shape of an elastic body under any given system of loads and temperature
distribution conditions can be completely described by the three components of displacement u, v, and w parallel to the directions x, y, and z, respectively. In general, each of
these components u, v, and w is a function of the coordinates x, y, and z. The strains
induced in the body can be expressed in terms of the displacements u, v, and w. In this
section, we assume the deformations to be small so that the strain–displacement relations
remain linear.
To derive expressions for the normal strain components εxx and εyy and the shear strain
component εxy , consider a small rectangular element OACB whose sides (of lengths dx
and dy) lie parallel to the coordinate axes before deformation. When the body undergoes
deformation under the action of external load and temperature distributions, the element
OACB also deforms to the shape O A C B as shown in Figure 8.5. We can observe that
the element OACB has two basic types of deformation, one of change in size and the other
of angular distortion.
Since the normal strain is defined as change in length divided by original length, the
strain components εxx and εyy can be found as

εxx

change in length of the fiber OA that lies
in the x directon before deformation
=

original length of the fiber OA
dx + u +
=

∂u
· dx − u − dx
∂x
∂u
=
dx
∂x

(8.44)


BASIC EQUATIONS OF SOLID MECHANICS

u+

∂u ·dy
∂y
C′

B′
v+
B

291

∂u ·dy

∂y

C

q2
A′
dy

q1

O′

y

u+

v
A

O

∂u ·dx
∂x

v+

∂y ·dx
∂x

u

x

dx

Figure 8.5. Deformation of a Small Element OACB.
and

εyy

change in length of the fiber OB that lies
in the y directon before deformation
=
original length of the fiber OB
dy + v +
=

∂v
· dy
∂y
dy

− v − dy
=

∂v
∂y

(8.45)

The shear strain is defined as the decrease in the right angle between the fibers OA

and OB, which were at right angles to each other before deformation. Thus, the expression
for the shear strain εxy can be obtained as
∂v
∂u
u+
·dx −v
·dy −u
∂x
∂y
+
∂u
∂v
dx+ u+
dy + v +
·dx −u
·dy −v
∂x
∂y
v+

εxy = θ1 +θ2

tanθ1 +tan θ2

If the displacements are assumed to be small, εxy can be expressed as
εxy =

∂v
∂u
+

∂y
∂x

(8.46)

The expressions for the remaining normal strain component εzz and shear strain
components εyz and εzx can be derived in a similar manner as
∂w
,
∂z
∂w
∂v
=
+
,
∂y
∂z

εzz =

(8.47)

εyz

(8.48)


292

BASIC EQUATIONS AND SOLUTION PROCEDURE

and
εzx =

∂u
∂w
+
∂z
∂x

(8.49)

In the case of two-dimensional problems, Eqs. (8.44)–(8.46) are applicable, whereas
Eq. (8.44) is applicable in the case of one-dimensional problems.
In the case of an axisymmetric solid, the strain–displacement relations can be
derived as
∂u
∂r
u
=
r
∂w
=
∂z
∂w
∂u
+
=
∂z
∂r

εrr =
εθθ
εzz
εrz

(8.50)

where u and w are the radial and the axial displacements, respectively.
(vi) Boundary conditions
Boundary conditions can be either on displacements or on stresses. The boundary conditions on displacements require certain displacements to prevail at certain points on
the boundary of the body, whereas the boundary conditions on stresses require that the
stresses induced must be in equilibrium with the external forces applied at certain points
on the boundary of the body. As an example, consider the flat plate under inplane loading
shown in Figure 8.6.
In this case, the boundary conditions can be expressed as
u = v = 0 along the edge AB
(displacement boundary conditions)

y

B
C
p

b
D

A
a

Figure 8.6. A Flat Plate under Inplane Loading.

x


BASIC EQUATIONS OF SOLID MECHANICS

293

and
σyy = σxy = 0 along the edges BC and AD
σxx = −p,

σyy = σxy = 0 along the edge CD

(stress boundary conditions)
It can be observed that the displacements are unknown and are free to assume any values
dictated by the solution wherever stresses are prescribed and vice versa. This is true of all
solid mechanics problems.
For the equilibrium of induced stresses and applied surface forces at point A of
Figure 8.7, the following equations must be satisfied:
x σxx

+

y

σxy +

z

σxz = Φx

x σxy

+

y

σyy +

z

σyz = Φy

x σxz

+

y

σyz +

z

σzz = Φz

(8.51)

Φy
N (Normal)
A
y

Φx

Φz
x

z
(a) Components of the surface force

ᐉx ·dS
y

ᐉz ·dS
σxz

Φx

σxx
x
z

Normal

σxy
Surface area, dS
ᐉy ·dS

(b) Equillibrium of internal stresses and surface forces around point A

Figure 8.7. Forces Acting at the Surface of a Body.


294

BASIC EQUATIONS AND SOLUTION PROCEDURE

where x , y , and z are the direction cosines of the outward drawn normal (AN) at
point A; and Φx , Φy , and Φz are the components of surface forces (tractions) acting at
point A in the directions x, y, and z, respectively. The surface (distributed) forces Φx ,
Φy , and Φz have dimensions of force per unit area. Equation (8.51) can be specialized to
two- and one-dimensional problems without much difficulty.
(vii) Compatibility equations
When a body is continuous before deformation, it should remain continuous after deformation. In other words, no cracks or gaps should appear in the body and no part should
overlap another due to deformation. Thus, the displacement field must be continuous as
well as single-valued. This is known as the “condition of compatibility.” The condition of
compatibility can also be seen from another point of view. For example, we can see from
Eqs. (8.44)–(8.49) that the three strains εxx , εyy , and εxy can be derived from only two
displacements u and v. This implies that a definite relation must exist between εxx , εyy ,
and εxy if these strains correspond to a compatible deformation. This definite relation is
called the “compatibility equation.” Thus, in three-dimensional elasticity problems, there
are six compatibility equations [8.2]:
∂ 2 εyy
∂ 2 εxy
∂ 2 εxx
+
=

∂y 2
∂x2
∂x∂y

(8.52)

∂ 2 εyy
∂ 2 εzz
∂ 2 εyz
+
=
∂z 2
∂y 2
∂y∂z

(8.53)

∂ 2 εzz
∂ 2 εxx
∂ 2 εzx
+
=
∂x2
∂z 2
∂x∂z

(8.54)

1 ∂
2 ∂x

∂εxy
∂εyz
∂εzx
−
+
∂z
∂x
∂y

∂ 2 εxx
=
∂y∂z

(8.55)

1 ∂
2 ∂y

∂εxy
∂εyz
∂εzx
+
−
∂z
∂x
∂y

∂ 2 εyy
=

∂z∂x

(8.56)

1 ∂
2 ∂x

∂εxy
∂εyz
∂εzx
−
+
+
∂z
∂x
∂y

∂ 2 εzz
=
∂x∂y

(8.57)

In the case of two-dimensional plane strain problems, Eqs. (8.52)–(8.57) reduce to a single
equation as
∂ 2 εxx
∂ 2 εyy
∂ 2 εxy
+
=

∂y 2
∂x2
∂x∂y

(8.58)

For plane stress problems, Eqs. (8.52)–(8.57) reduce to the following equations:
∂ 2 εyy
∂ 2 εxy
∂ 2 εxx
,
+
=
∂y 2
∂x2
∂x∂y

∂ 2 εzz
∂ 2 εzz
∂ 2 εzz
=0
=
=
∂y 2
∂x2
∂x∂y

(8.59)

In the case of one-dimensional problems the conditions of compatibility will be automatically satisfied.

FORMULATIONS OF SOLID AND STRUCTURAL MECHANICS

295

8.3 FORMULATIONS OF SOLID AND STRUCTURAL MECHANICS
As stated in Section 5.3, most continuum problems, including solid and structural mechanics problems, can be formulated according to one of the two methods: differential equation
method and variational method. Hence, the finite element equations can also be derived
by using either a differential equation formulation method (e.g., Galerkin approach)
or variational formulation method (e.g., Rayleigh–Ritz approach). In the case of solid
and structural mechanics problems, each of the differential equation and variational
formulation methods can be classified into three categories as shown in Table 8.1.
The displacement, force, and displacement–force methods of differential equation formulation are closely related to the principles of minimum potential energy, minimum
complementary energy, and stationary Reissner energy formulations, respectively. We use
the displacement method or the principle of minimum potential energy for presenting the
various concepts of the finite element method because they have been extensively used in
the literature.
8.3.1 Differential Equation Formulation Methods
(i) Displacement method
As stated in Section 8.2.1, for a three-dimensional continuum or elasticity problem, there
are six stress–strain relations [Eq. (8.10)], six strain–displacement relations [Eqs. (8.44)–
(8.49)], and three equilibrium equations [Eqs. (8.4)], and the unknowns are six stresses
(σij ), six strains (εij ), and three displacements (u, v, and w). By substituting Eqs. (8.44)–
(8.49) into Eqs. (8.10), we obtain the stresses in terms of the displacements. By substituting these stress–displacement relations into Eqs. (8.4), we obtain three equilibrium
equations in terms of the three unknown displacement components u, v, and w. Now these
equilibrium equations can be solved for u, v, and w. Of course, the additional requirements
such as boundary and compatibility conditions also have to be satisfied while finding the
solution for u, v, and w. Since the displacements u, v, and w are made the final unknowns,
the method is known as the displacement method.

(ii) Force method
For a three-dimensional elasticity problem, there are three equilibrium equations, Eqs.
(8.4), in terms of six unknown stresses σij . At the same time, there are six compatibility
equations, Eqs. (8.52)–(8.57), in terms of the six strain components εij . Now we take
any three strain components, for example, εxy , εyz , and εzx , as independent strains and

Table 8.1. Methods of Formulating Solid and Structural Mechanics Problems

Differential equation formulation methods

Displacement
method

Force
method

Displacementforce method
(mixed method)

Variational formulation methods

Principle
of minimum
potential
energy

Principle
of minimum
complementary energy

Principle
of stationary Reissner
energy


296

BASIC EQUATIONS AND SOLUTION PROCEDURE

write the compatibility equations in terms of εxy , εyz , and εzx only. By substituting the
known stress–strain relations, Eq. (8.10), we express the three independent compatibility
equations in terms of the stresses σij . By using these three equations, three of the stresses
out of σxx , σyy , σzz , σxy , σyz , and σzx can be eliminated from the original equilibrium
equations. Thus, we get three equilibrium equations in terms of three stress components
only, and hence the problem can be solved. Since the final equations are in terms of stresses
(or forces), the method is known as the force method.
(iii) Displacement–force method
In this method, we use the strain–displacement relations to eliminate strains from the
stress–strain relations. These six equations, in addition to the three equilibrium equations,
will give us nine equations in the nine unknowns σxx , σyy , σzz , σxy , σyz , σzx , u, v, and w.
Thus, the solution of the problem can be found by using the additional conditions such as
compatibility and boundary conditions. Since both the displacements and the stresses (or
forces) are taken as the final unknowns, the method is known as the displacement–force
method.
8.3.2 Variational Formulation Methods
(i) Principle of minimum potential energy
The potential energy of an elastic body πp is defined as
πp = π − Wp

(8.60)

where π is the strain energy, and Wp is the work done on the body by the external
forces. The principle of minimum potential energy can be stated as follows: Of all possible
displacement states (u, v, and w) a body can assume that satisfy compatibility and given
kinematic or displacement boundary conditions, the state that satisfies the equilibrium
equations makes the potential energy assume a minimum value. If the potential energy,
πp , is expressed in terms of the displacements u, v, and w, the principle of minimum
potential energy gives, at the equilibrium state,
δπp (u, v, w) = δπ(u, v, w) − δWp (u, v, w) = 0

(8.61)

It is important to note that the variation is taken with respect to the displacements in
Eq. (8.61), whereas the forces and stresses are assumed constant. The strain energy of a
linear elastic body is defined as
π=

1
2

ε T σ dV

(8.62)

V

where V is the volume of the body. By using the stress–strain relations of Eq. (8.10), the
strain energy, in the presence of initial strains ε0 , can be expressed as
π=

1
2

ε T [D]ε dV −
V

ε T [D]ε0 dV

(8.63)

V

The work done by the external forces can be expressed as
φ¯ T U · dV +

Wp =
V

¯ T U · dS1
Φ
S1

(8.64)


FORMULATIONS OF SOLID AND STRUCTURAL MECHANICS

297

⎧ ⎫

⎧ ⎫
¯ x⎬
⎨Φ
⎨φ¯x ⎬
¯ y = vector of prescribed surface
¯ = Φ
where φ¯ = φ¯y = known body force vector, Φ
⎩¯ ⎭
⎩¯ ⎭
Φz
φz
⎧ ⎫
⎨u⎬
forces (tractions), U = v = vector of displacements, and S1 is the surface of the body
⎩ ⎭
w
on which surface forces are prescribed. Using Eqs. (8.63) and (8.64), the potential energy
of the body can be expressed as
πp (u, v, w) =

1
2

φ¯ T U · dV −

ε T [D](ε − 2ε0 ) dV −
V

V

¯ T U · dS1
Φ

(8.65)

S1

If we use the principle of minimum potential energy to derive the finite element equations, we assume a simple form of variation for the displacement field within each element
and derive conditions that will minimize the functional I (same as πp in this case). The
resulting equations are the approximate equilibrium equations, whereas the compatibility conditions are identically satisfied. This approach is called the “displacement” or
“stiffness” method of finite element analysis.
(ii) Principle of minimum complementary energy
The complementary energy of an elastic body (πc ) is defined as
π ) − work done by the applied
πc = complementary strain energy in terms of stresses (˜
˜
loads during stress changes (Wp )
The principle of minimum complementary energy can be stated as follows: Of all possible stress states that satisfy the equilibrium equations and the stress boundary conditions,
the state that satisfies the compatibility conditions will make the complementary energy
assume a minimum value.
If the complementary energy πc is expressed in terms of the stresses σij , the principle
of minimum complementary energy gives, for compatibility,
δπc (σxx , σyy , . . . , σzx ) = δ˜
π (σxx , σyy , . . . , σzx )
˜ p (σxx , σyy , . . . , σzx ) = 0
− δW

(8.66)

It is important to note that the variation is taken with respect to the stress components in

Eq. (8.66), whereas the displacements are assumed constant. The complementary strain
energy of a linear elastic body is defined as
π
˜=

1
2

σ T ε dV

(8.67)

V

By using the strain–stress relations of Eqs. (8.7), the complementary strain energy, in
the presence of known initial strain ε0 , can be expressed as∗
π
˜=

∗

1
2

σ T ([C]σ + 2ε0 ) dV

(8.68)

V

The correctness of this expression can be verified from the fact that the partial derivative of π
˜
with respect to the stresses should yield the strain–stress relations of Eq. (8.7).


298

BASIC EQUATIONS AND SOLUTION PROCEDURE

The work done by applied loads during stress change (also known as complementary work)
is given by
˜p =
W

¯ dS2
ΦT U

(Φx u
¯ + Φy v¯ + Φz w)
¯ dS2 =
S2

(8.69)

S2

where S2 is the part of the
of the body on which the values of the displacements
⎧ surface
⎫

¯⎬
⎨u
¯
v¯ . Equations (8.68) and (8.69) can be used to express the
are prescribed as U =
⎩ ⎭
w
¯
complementary energy of the body as
πc (σxx , σyy , . . . , σzx ) =

1
2

¯ · dS2
ΦT U

σ T ([C]σ + 2ε0 ) · dV −
V

(8.70)

S2

If we use the principle of minimum complementary energy in the finite element analysis,
we assume a simple form of variation for the stress field within each element and derive
conditions that will minimize the functional I (same as πc in this case). The resulting
equations are the approximate compatibility equations, whereas the equilibrium equations
are identically satisfied. This approach is called the “force” or “flexibility” method of finite
element analysis.

(iii) Principle of stationary Reissner energy
In the case of the principle of minimum potential energy, we expressed πp in terms of
displacements and permitted variations of u, v, and w. Similarly, in the case of the
principle of minimum complementary energy, we expressed πc in terms of stresses and
permitted variations of σxx , . . . , σzx . In the present case, the Reissner energy (πR ) is
expressed in terms of both displacements and stresses and variations are permitted in U
and σ. The Reissner energy for a linearly elastic material is defined as
[(internal stresses) × (strains expressed in terms of

πR =
V

displacements) − complementary energy in terms of stresses] · dV
− work done by applied forces
σxx ·

=
V

∂u
∂v
+ σyy ·
+ · · · + σzx
∂x
∂y

∂w
∂u
+
∂x

∂z

φ¯x · u + φ¯y · v + φ¯z · w · dV −

−
V

−

−π
˜ · dV

¯x · u + Φ
¯y · v + Φ
¯ z · w · dS1
Φ
S1

{(u − u
¯)Φx + (v − v¯)Φy + (w − w)Φ
¯ z } · dS2
S2

σT ε −

=
V

1 T
σ [C]σ − φ¯T U · dV

2
¯ )T Φ · dS2
(U − U

¯ dS1 −
UT Φ

−
S1

S2

(8.71)


FORMULATIONS OF SOLID AND STRUCTURAL MECHANICS

299

The variation of πR is set equal to zero by considering variations in both displacements
and stresses:

---

---

---

(8.72)

---

---

---

=0

---

∂πR
∂πR
∂πR
δu +
δv +
δw
∂u
∂v
∂w
------------------------------

∂πR
δσij +
∂σij
-----

---

δπR = Σ

gives equilibrium equations
and boundary conditions

gives stress–
displacement
equations

The principle of stationary Reissner energy can be stated as follows: Of all possible stress and displacement states the body can have, the particular set that makes the
Reissner energy stationary gives the correct stress–displacement and equilibrium equations along with the boundary conditions. To derive the finite element equations using the
principle of stationary Reisssner energy, we must assume the form of variation for both
displacement and stress fields within an element.
(iv) Hamilton’s principle
The variational principle that can be used for dynamic problems is called the Hamilton’s
principle. In this principle, the variation of the functional is taken with respect to time.
The functional (similar to πp , πc , and πR ) for this principle is the Lagrangian (L)
defined as
L = T − πp = kinetic energy − potential energy

(8.73)

The kinetic energy (T ) of a body is given by

T =

1
2

˙ ˙
ρU T U dV

(8.74)

V

⎧ ⎫
⎨ u˙ ⎬
˙
where ρ is the density of the material, and U = v˙ is the vector of velocity components
⎩ ⎭
w˙
at any point inside the body. Thus, the Lagrangian can be expressed as
L=

1
2

˙ ˙
ρU T U − ε T [D]ε + 2U T φ¯ dV +
V

¯ dS1
UT Φ

(8.75)

S1

Hamilton’s principle can be stated as follows: Of all possible time histories of displacement
states that satisfy the compatibility equations and the constraints or the kinematic boundary conditions and that also satisfy the conditions at initial and final times (t1 and t2 ), the
history corresponding to the actual solution makes the Lagrangian functional a minimum.

300

BASIC EQUATIONS AND SOLUTION PROCEDURE

Thus, Hamilton’s principle can be stated as

δ

t2
t1

L dt = 0

(8.76)

8.4 FORMULATION OF FINITE ELEMENT EQUATIONS (STATIC ANALYSIS)
We use the principle of minimum potential energy for deriving the equilibrium equations
for a three-dimensional problem in this section. Since the nodal degrees of freedom are
treated as unknowns in the present (displacement) formulation, the potential energy πp has
to be first expressed in terms of nodal degrees of freedom. Then the necessary equilibrium
equations can be obtained by setting the first partial derivatives of πp with respect to each
of the nodal degrees of freedom equal to zero. The various steps involved in the derivation
of equilibrium equations are given below.
Step 1:

The solid body is divided into E finite elements.

Step 2:

The displacement model within an element “e” is assumed as
⎫
⎧
⎨ u(x, y, z) ⎬
U = v(x, y, z) = [N ]Q(e)
⎭
⎩
w(x, y, z)

(8.77)

where Q(e) is the vector of nodal displacement degrees of freedom of the element, and
[N ] is the matrix of shape functions.
Step 3: The element characteristic (stiffness) matrices and characteristic (load) vectors
are to be derived from the principle of minimum potential energy. For this, the potential
energy functional of the body πp is written as (by considering only the body and surface
forces)
E

πp(e)

πp =
e=1

(e)

where πp

is the potential energy of element e given by (see Eq. 8.65)

πp(e) =

1
2

V (e)

U T φ¯ dV

¯ dS1 −
UT Φ

ε T [D](ε − 2ε0 ) dV −
(e)

S1

(e)

(8.78)

V (e)

where V (e) is the volume of the element, S1 is the portion of the surface of the element
¯ are prescribed, and φ¯ is the vector
over which distributed surface forces or tractions, Φ,
of body forces per unit volume.

FORMULATION OF FINITE ELEMENT EQUATIONS (STATIC ANALYSIS)

301

The strain vector ε appearing in Eq. (8.78) can be expressed in terms of the nodal
displacement vector Q(e) by differentiating Eq. (8.77) suitably as
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨

∂u

∂x
∂v
∂y
∂w
∂z

⎫
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎬

⎡

∂
⎢ ∂x
⎢
⎢ 0
⎧ ⎫
⎢
ε
⎢
⎪
xx
⎪
⎪
⎪
⎢
⎪
⎪
⎪
⎪
εyy ⎪
⎢ 0
⎪
⎪
⎪
⎢
⎨ ⎬
εzz
⎢
=⎢ ∂
=

ε=
εxy ⎪
⎢ ∂y
⎪
⎪
∂u
∂v ⎪
⎪
⎪
⎪
⎪
⎢
⎪
⎪
⎪
⎪
+
⎪
⎪
⎪
⎪
εyz ⎪
⎢
⎪
⎪
⎪
⎪
⎪
⎪
⎪

∂y
∂x
⎢
⎪
⎩ ⎭ ⎪
⎪
⎪
⎢ 0
⎪
⎪
εzx
⎪
⎪
⎢
⎪
⎪
∂v
∂w
⎪
⎪
⎢
⎪
⎪
+
⎪
⎪
⎪
⎪
⎣ ∂
∂z

∂y
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎭
⎩ ∂w + ∂u ⎪
∂z
∂x
∂z

⎤
0
∂
∂y

0

⎥
⎥
0 ⎥
⎥
⎥
∂ ⎥⎧ ⎫
⎥

⎨u⎬
∂z ⎥
⎥
v = [B]Q(e)
0 ⎥
⎥⎩ ⎭
⎥ w
⎥
∂ ⎥
⎥
∂y ⎥
⎥
∂ ⎦
∂x

0
∂
∂x
∂
∂z
0

(8.79)

where
⎡

∂
⎢ ∂x
⎢

⎢ 0
⎢
⎢
⎢
⎢ 0
⎢
⎢
[B] = ⎢ ∂
⎢ ∂y
⎢
⎢
⎢
⎢ 0
⎢
⎢
⎣ ∂
∂z

⎤
0
∂
∂y
0
∂
∂x
∂
∂z
0

0

⎥
⎥
0 ⎥
⎥
⎥
∂ ⎥
⎥
∂z ⎥
⎥
[N ]
0 ⎥
⎥
⎥
⎥
∂ ⎥
⎥
∂y ⎥
⎥
∂ ⎦
∂x

(8.80)

The stresses σ can be obtained from the strains ε using Eq. (8.10) as
σ = [D](ε − ε0 ) = [D][B]Q(e) − [D]ε0

(8.81)

Substitution of Eqs. (8.77) and (8.79) into Eq. (8.78) yields the potential energy of the

element as
πp(e) =

1
2

T

T

Q(e) [B]T [D][B]Q(e) dV −
V (e)

V (e)
T

T

Q(e) [N ]T φ¯ dV

¯ dS1 −
Q(e) [N ]T Φ

−
(e)

S1

Q(e) [B]T [D]ε0 dV

(8.82)

V (e)

In Eqs. (8.78) and (8.82), only the body and surface forces are considered. However,
generally some external concentrated forces will also be acting at various nodes. If P
∼c
denotes the vector of nodal forces (acting in the directions of the nodal displacement