San Jose State University
Department of Mechanical Engineering
ME 160 Introduction to Finite Element Method
Chapter 4
Finite Element Analysis in Stress
Analysis of Elastic Solid Structures
Instructor
Tai-Ran Hsu, Professor
Part 1
Introduction to
Fundamentals of Theory of Linear Elasticity
What defines elastic solids?
A solid deforms in response to external actions (e.g., forces, heat, etc.). The deformation is completely reversible – meaning the
solid returns to its original shape after the removal of the external actions.
Two Types of Elastic Solids
Type 1: Linear elasticity of materials:
This type of elasticity occurs to solids undergoing small deformations, such as springs that exhibit linear relationships
between the applied force (F) and the induced elongation (x) that can be represented by a mathematical formula as: F = kx
where k is a constant known as the rate or spring constant. Many metallic materials fall into the category of linear elastic solids
It can also be stated as a linear relationship between stress (σ) and strain (ε) in stretching or compressing a thin rod by
The expression: σ = Eε where E Is known as the elastic modulus or Young's modulus
Type 2: Nonlinear elasticity of materials:
This type of solids behaves as elastic materials as described above, but can exhibit large deformations, such as rubbers and polymers
The FE formulation presented in this course will be based on linear elasticity theory
Linear Elastic Behavior of Solids
Fundamental assumptions
(1) The material is treated as a continuous medium (or a continuum). In the words, the material is homogeneous with no
internal defects or voids of significant sizes
(2) The material is isotropic – meaning its properties are uniform in all directions
(3) The material has no memory
(4) The material exhibits the same properties in tension and compression
Definitions of stress and strain
Uniaxial elongation of a thin rod:
(1) Engineering strain (e):
It is defined with reference to the original shape of the solid. Mathematically it is expressed as:
= ΔL/L0 = L/L0 - 1
(2) True strain (ε):
It is measured on the basis of the immediate proceeding length of the rod sample.
Mathematically it is expressed as:
A0
L0
L
 
L L L L
LL 
  1 L 0  2 L 1  .........  L n  
0
1
n


(3) Relationship between ε and e:
ΔL
F
(4) Engineering stress (S):
S
 L
dL
 n 
L0 L
 L0 

L
ε = ℓn(1+e)
Insta tan eous load
P

Original cross  sec tional area A0
(5) True stress (σ): σ = P/A
(6) Relationship between S and σ: σ = (1 + e) S
Stress-Strain Curves of Materials
Stretching of thin rods of most engineering materials will exhibit the stress vs. strain relations illustrated in the figure below:
Stresses
True
0
Engineering
Designations:
A’ = proportional limit
A = elastic limit
B = yield point
m =necking point
f = rupture point
S0 = Yield strength of material
Su = ultimate tensile strength of material
Strains
Typical elastic deformation of engineering materials:
(1)
(2)
(3)
(4)
(5)
Very small deformation with strain up to 0.1%
Straight linear relationship between the stress and strain, resulting in a constant stiffness of the .
The slope of line OA’ is called the Young’s modulus (E), representing the stiffness of the material
Completely recoverable strain (or deformation) after the applied load is removed
The yield stress (So) or σy is defined to be the interception of ϵoffset= 0.2% of the σ-ϵ curve. It is a measure of materials
exceeding the elastic limit, and undergoing plastic deformation (an irreversible deformation)
Physics of Deformable Solids subjected to External Forces
Original State:
2 Physical Consequences:
RESPONDES
To Allied
Foreces:
2 things will
happen:
after applied forces:
The Many Components of Displacements and Stresses:
(Element) Displacements:
{U}T: {Ux,Uy,Uz}
in a deformed solid
A small element
located at x,y,z:
See detailed definitions
on the next slide:
Induced Stress Components in Deformable Solids subjected to External Forces
Solid subjected to external forces:
Result in:
Because the forces applied to a general 3-D solid, the induced stresses are MULTI-directional designated by σαβ:
σ = magnitude, subscript α = the axis normal to the plane of action, subscript β = the direction the stress component points to
So, stress component σyy = stress component acting on the plane normal to the y-axis and pointing to the y-direction, whereas
stress component σxy = stress components acting on the face normal to the x-axis but points to the y-direction.
In theory, there are nine (9)
stress components
 xx
everywhere

inside the solid:     yx
 zx

 xy  xz 

 yy  yz 
 zy  zz 
But in reality there
are only six (6)
independent stress
components with:   
σxy =σyx , σxz =σzx ,
σyz = σzx
  xx


 SYM

 xy  xz 

 yy  yz 
 zz 
 1 
 
For FE formulation:
 2
 3 
σ1=σxx,σ2=σyy,,
     (4.1)
σ3=σzz, σ4=σxy,
 4 
σ5=σyz, σ6=σxz
 5 
 
 6 
Induced Stresses in Deformable Solids subjected to External Forces – Cont’d
z
Stress components with same subscripts such as:
σxx, σyy, σzz are called NORMAL stress components,
with unit: psi, or Pascal (Pa) = N/m2 .
σzz
D
σzy
σzx
A
σxz
C
B
σyx
σxy
σyz
G
σxx
E
Stress components with different subscripts such
as: σxy, σxz, σyx, σyz, σzx, σzy are SHEARING stress
components. Shear stress has the unit of change of
y angle from the original right angle to the angle with
the stress, i.e (π/2) – θ. The unit is thus “angle
change” in radians (rad)
σyy
F
x
Effects of Normal and Shear Stresses to Solid Deformation
y
Normal stresses change size:
contraction
-σxx
σxx
Exist NORMAL
to surfaces
Shear stresses change shape:
y
Exist on the SURFACES
-σxy
σxy
elongation
x
x
Relationships between Primary unknown and Secondary Unknowns in FE analyses
Primary Unknown:
Element displacements: {U(x,y,z)} and
Nodal displacements: [N(x,y,z)]{u}T
in FE formulation: {U} = [N]{u}
where {N(x,y,z)} = Interpolation function
Strain-displacement relations
Linkage to the 1st secondary unknown:
Element strains: {ε(x,y,z)}=[B(x,y,z)]{U(x,y,z)} →
corresponding to element stresses: {σ(x,y,z)}
Generalized Hook’s Law
Element strains and stresses, and nodal displacement
are the solutions of FE analyses.
2nd Secondary Unknown:
Element stresses {σ(x,y,z)}=[C]{ε}
Multiaxial Deformations of Elastic Solids
Uniaxial stress with uniaxial deformation:
F
Stress: σxx
Strain: εxx
x
x
L
εxx = ΔL/L
L
ΔL
Uniaxial stress with biaxial deformation with Poisson’s ratio:
y
For solids with significant cross-section, a uniaxial stress may
produce biaxial deformation such as illustrate in the figure in the left.
Lateral contraction:
- εyy
F
The lateral contraction deformation is represented by: - εyy
0
Longitudinal stretching:
+ εxx
x
F
Poisson’s ratio defines as:
 yy
 

Longitudinal stretching strain  xx
Lateral contraction strain
● Poisson ratio exists in most multi-axially loaded elastic solids, and this effect needs to be included in all
FE formulations. This effect also leads to the use of Generalize Hooke’s Law for the formulations
Generalized Hooke’s Law for Solids with Multi-axial Stresses:
Stress vs. strain relationship for deformed solids with 3-D deformation
Elongation in one-direction causes contractions
in other directions, and vise versa.
Total strains in three directions induced by the three normal stresses
in x-direction
in y-direction
in z-direction
in which E = Young’s modulus, and γ = Poisson’s ratio of the material
The following expression express the stresses in terms of strains – the Hooke’s law:
1   x  y  z 
 xx 
E


 


 yy 
 x  1   y  z 
  1  1  2     1   
y
z
 zz 
 x
One may derive the uniaxial stress situation as a special case from the above expression to obtain:
σxx = Eεx
by substituting: σyy = σzz = 0, and εy = -γεx and εz = -γεx in the generalized Hooke’s Law
Element Strain-Displacement Relations for FE Formulation
There are six (6) strain components corresponding to the stress components in interior of the deformed solid.
These strain components are related to the displacements of the solid induced by the external forces.
Because deformation of the solid CONTINUOUSLY varying throughout the solid, the following relationship exits:

 x

 xx   0
  
 yy  
 zz   0
 
 xy  
 yz   y
  
 xz   0


Element Strains
 z
Corresponding to
By theory of elasticity:
0

y
0

x

z
0

Element
0
Displacements

U(x,y,z)
0

  U  x, y, z 
x


z  U  x, y, z  (4.2)
 y


0 U  x, y, z 

 z


y 

x 
Element stresses
Example: For uni-axial elongation or contraction of a rod:
Element displacement {U} = {Ux(x)}, The corresponding strain in element is:  xx
or: {ε(x,y,z)} = [D]{U(x,y,z)}

 x

0


0
D   

 y

0


 z
U x  x  dU x  x  


x
dx
0

y
0

x

z
0

0

0


z 

0



y 

x 
(4.3)
(4.4)
Element Stress-Strain Relations – the Generalized Hooke’s Law
According to generalized Hooke’s law for MULTI-Axial stress state,
the following relationship between the element stress and strain exists:
or:
{σ} = [C]{ε}
(4.6)
where [C] is the elasticity matrix with:

0
0
0 
1  


1



0
0
0
 xx 

  xx 
 

1 
0
0
0   
 yy 

  yy 
1  2
 zz 
0
0   zz 

E
2
 

  




1


1

2

1

2

xy
 xy 

0  
 yz 

  yz 
2
 

1  2   
 xz 
SYM

  xz 
2 



Example: For uni-axially loaded rod:
 xx  E xx  E
dU x  x 
dx
(4.5)

0
0
0 
1  


1



0
0
0



1 
0
0
0 


1  2
0
0 

E
C  
2

1  1  2  
1  2
SYM
0 


2

1  2 


2




where E = Young’s modulus, and γ = Poisson’s ratio
where Ux(x) is the displacement in the rod
(4.7)
Physics of Solid Deformation by
External Forces
Displacements:
{Φ(x,y,z)}T: {Φx(x,y,z,Φy(x,y,z,Φz(x,y,z}
Small element located
at x,y,z:
Total 9 stress components at (x,y,z)
 xx  xy  xz 
    yx  yy  yz 
 zx  zy  zz 


Part 2
FE Formulation of
Deformable Elastic Solids
Derivation of Element Equations
In Step 4, Chapter 3, we derived the element equation using the Rayleigh-Ritz method to take a form:
We will select tetrahedron elements as the basis for FE formulation for general 3-D solid structures
[Ke]{q}= {Q}
(1.27) in the textbook
where [Ke] = Element matrix
{q} = Vector of primary unknown quantities at the nodes of the element
{Q} = Vector of element nodal actions (e.g., forces)
Element equations for each tetrahedron element in the FE model for a
structure are then assembled to establish “overall stiffness equation”
for determining nodal displacements of all nodes in the structure.
[K]{Φ} = {R}
K    K em 
m
FE Model for s Structure made of
Tetrahedron Elements
where
1
m=total number of elements in the FE model
Derivation of Element Equations-Cont’d Principle of deriving element equation using Rayleigh-Ritz variational principle
From Chapter 2: Let us determine a suitable “functional” to derive the element equation.
A general form of functional:
   
 
 








f

,
,
..........
dv

g

,
,.........
.



 ds
v  r

s
r



v = volume, s = surface
and then apply the Variational principle on:
  
  
 1
 

   

  2   0
    





  
from which equations of each element are derived:
 
 
 
 0,
 0,
 0,..............
1
2
3
The “functional” for a deformed solid subjected to external forces is “POTENTIAL ENERGY” (P) in the situation.
The potential energy associated with a deformed solid can be defined as:
P=U-W
(4.8)
where U = the strain energy in a deformed solid, and W = the work done to the deformed solid by external forces acting on
the solid body in the volume and surface of the solid
Derivation of Element Equations-Cont’d
(a)
Strain energy in a deformed solid:
(C)
As we mentioned early in this Chapter that a solid in (a) deforms
into a new shape in (c) – but not indefinitely.
● It stopped further deformation after deformed by certain
amount.
● It reaches a new state of equilibrium. WHY???
Imagine the following phenomenon:
Stretch a free-hung spring by a weight W.
The spring will elongate, but only by a finite amount.
(b)
Ask yourself: WHY?
Answer: the elongation of the spring develops a “resistance,”
which increases as the spring elongates. The spring ceases
further elongation when the “resistance” in the spring balances
the applied weight (force). We say the spring – and the applied
weight reaches a new state of equilibrium, which stop the spring
(d)
From further elongation.
Next: what will happen to the spring after the weight is removed?
You will say that the spring returns to its original length, but WHY??
Answer: because a form of ENERGY was stored in the stretched
spring when it is elongated. This ENBERGY is released to restore
the spring to its original shape after the external force (the weight)
Now, you know why the solid in (a) ceases to deform further after
the application of the system of external forces {p} has been applied was removed.
to the solid in (b). And you would know that there is such ENERGY associated with the solid deformation developed in the solid called
“STRAIN ENERGY.” which is responsible for restoring the solid to its original shape after the applied forces are removed in “ELASTIC” solids.
T
1
Mathematical expression of strain energy in State (c) is:
(3.9) textbook
U      dv
2 v
Derivation of Element Equations-Cont’d
The potential energy in a deformed solid is:
Potential energy in a deformed solid subjected to external forces:
P=U-W
(4.8)
Strain energy:

 xx  

  

 yy  

 zz  
T
1
1 
U      dv    xx  yy  zz  xy  yz  xz    dv
2 v
2 v
 xy  

 yz  





 xz  

1
   xx xx   yy yy   zz zz   xy xy   yz yz   xz xz dv
2 v
(4.9)
Both the strain and stress components are function of (x,y,z), and dv = (dx)(dy)(dz) = the volume of given points
in the deformed solid.
Strain energy is a scalar quantity.
Derivation of Element Equations – cont’d
Work done to deform the solid:
Potential energy in a deformed solid subjected to external forces:
Definition of “work”: Work (W) = Force x Displacement (deformation)
Two kinds of forces: (1) body forces (uniformly distributed throughout the volume of the solid (v)), e.g., the weight,
(2) surface tractions, e.g., the pressure or concentrated forces acting on the boundary surface (s)
Mathematical expression of work: W 
v ( x, y, z )  f dv  s x, y, z  tds
T
   x  x, y, z   y  x, y, z 
v
T
 fx 
 
 z x, y, z  f y dv
f 
 z
   x  x, y, z   y  x, y, z 
s
t x 
 
 z x, y, z t y ds
t 
 z
(4.10)
where {Φ(x,y,z)} = the displacement of the solid at (x,y,z), {f} = body forces, and {t} = the surface tractions , and
ds = the part of the surface boundary on which the surface tractions apply
Derivation of Element Equations – cont’d
Potential energy in a deformed solid subjected to external forces:
So, the potential energy stored in a deformed solid is: P = U – W, or:
T
1
P  U  W     x, y, z    x, y, z dv 
2 v
  x, y, z   f dv    x, y, z  tds 
T
v
T
s
(4.11)
Following the Rayleigh-Ritz Variational principle, the equilibrium condition for the deformed solid should
satisfy the following conditions:
P 
 0
 
From which, equations for each element may be derived from:
P 
P 
P 
 0,
 0,
 0,..............
1
2
3
Derivation of Element Equations – cont’d
for FE mesh of discretized solids
What we had formulated was for continuum solids. We will now derive the ELEMENT EQUATION for discretized solids in FE mesh:
We need to make distinction between the ELEMENT quantities and the NODAL quantities.
The primary quantity in FE analysis is DISPLACEMENTS. We need teo make distinction between the Element displacements and
the Nodal displacements.
The element displacement is: {Φ(x,y,z)} with three components:
Φx(x,y,z) = the element displacement component along the x-direction
Φy (x,y,z) = the element displacement component along the y-direction, and
Φz(x,y,z) = the element displacement component along the z-direction
Derivation of Element Equations – cont’d
for FE mesh of discretized 3-D solids with tetrahedron elements
We realize that TETRAHEDRON and HEXAHEDRN elements are used in FE models for general 3-D solid structures.
The tetrahedron elements are the “basic elements” for this type of structures, because hexahedron elements
are made up by 4 or more tetrahedron elements.
Our FE formulation for general 3-D solid structures will thus be based on TETRAHEDRON elements
We notice that tetrahedron elements has four (4) associate nodes: Φ1, Φ2, Φ3, and Φ4 with FIXED (specified) COORDINATES.
Each node has three (3) displacement components too. These nodal displacement components are:
 T  1x
1 y 1z 2 x 2 y 2 z 3 x 3 y 3 z 4 x 4 y 4 z T
where Φ1x, Φ1y, Φ1z = displacements in Node 1 in 3 directions; Φ2x, Φ2y, Φ2z = displacements in Node 2 in 3 directions;
Φ3x, Φ3y, Φ3z = displacements in Node 3 in 3 directions; Φ4x, Φ4y, Φ4z = displacements in Node 4 in 3 directions
Derivation of Element Equations – cont’d for FE mesh of discretized 3-D solids with tetrahedron elements
We mentioned previously that the functional that we will use to derive the element equations in FE formulation of solid structures
Is the potential function in the solid as show below:
P  U W 
T
1




 x, y, z dv 

x
,
y
,
z

v
2
  x, y, z   f dv    x, y, z  tds 
T
v
T
s
(4.11)
Now because the ELEMENTS in discretized solid) Also, these elements are interconnected by the NODES to simulate the original
solid structures. This “link” requires the FE formulation involves the functional with both the ELEMENT and NODAL quantities in
the formulation.
This “link” is established using the INTERPOLATION FUNCTION that relates the
element quantities with the corresponding nodal quantities such s:
The 12 nodal displacements:
The 3 element displacements:  x  x, y, z 
 x, y, z    y x, y, z   N x, y, z  
  x, y, z 
 z

The INTEERPOLATION function
 T  1x
1 y 1z 2 x 2 y 2 z 3 x 3 y 3 z 4 x 4 y 4 z T
Derivation of Element Equations – cont’d
Key equations to construct the functional - the potential energy
1. The element displacements vs. nodal displacements via Interpolation function:
{Φ(x,y,z)} = [N(x,y,z)] {Φ}
2. Element strain vs. nodal displacements:
{ε(x,y,z)} = [D]{Φ(x,y,z)}
in which [D] in Equation (4.4)
(4.3)
Hence
{ε} = [D][N(x,y,z)]{Φ}= [B(x,y,z)]{Φ}
with [B(x,y,z)] = [D][N(x,y,z)]
(4.12)
(4.13)
3. Element stresses vs. nodal displacements:
{σ} = [C]{ε}
in which the elasticity matrix [C] in Equation (4.7)
Hence
{σ} = [C] [B(x,y,z)]{Φ}
Typical tetrahedron element
for 3-D FE models
4. Strain energy with nodal displacements:
T
1
U      dv
v
(4.6)
(4.14)
(4.9)
2
Hence
or
U
U
1
Bx, y, z  T C Bx, y, z  dv

v
2
T
1
T




C B( x, y, z ) dv

B
(
x
,
y
,
z
)
2 v
(4.15)
(4.16)
Derivation of Element Equations – cont’d
The functional for Variational process
We mentioned that the functional for deriving the element equations for
discretized solid structure is the POTENTIAL ENERGY (P) as shown below:
P  U W 
T
1




 x, y, z dv 

x
,
y
,
z
2 v
  x, y, z   f dv    x, y, z  tds 
T
v
T
s
(4.11)
By substituting the Strain energy expressed in Equation (4.16) into the above equation, we get:
P( ) 
1
T
T




C B( x, y, z ) dv

B
(
x
,
y
,
z
)
2 v
    N ( x, y, z )  f dv     N ( x, y, z ) tds
T
T
T
v
T
(4.17)
s
Due to the fact that nodal displacement {Φ} have “fixed value” but not a function og (x,y,z), so they can be factore
Out of the integration with respect to (x,y,z). We thus have the following:

T1
T
T
P( )        B( x, y, z ) C B( x, y, z )dv  
2 v

T
T
T
T
    N ( x, y, z )  f dv      N ( x, y, z ) tds 
v
 s



(4.18)
Derivation of Element Equations – cont’d
Element equation by Variational process
All elements in discretized solids subjected to external force require to satisfy
the following condition that:
 P 
  
 1
P 
P  

  2   0
    





  
By substituting the potential energy P in Equation (4.18) into the above equation:
 T1


T
T
       B( x, y, z ) C B( x, y, z )dv  

v
P 
 
2


0



 
T
T
T
T

    N ( x, y, z )  f dv      N ( x, y, z ) tds  

v
 s




(4.19)
Derivation of Element Equations – cont’d
Element equation by Variational process
The above variation results in:
 B( x, y, z ) C B( x, y, z )dv  
  N ( x, y , z )  f dv    

T
v
N ( x, y, z )T tds 
T
v

s
 0
Upon moving the last two items to the right-hand side:
 B( x, y, z ) C B( x, y, z )dv  
  N ( x, y , z )  f dv   
T
v
T
v
s
N ( x, y, z )T tds 
(4.20)
We may represent Equation by the following element equation:
(4.21)
[Ke] {Φ} = {q}
where
K e   vB( x, y, z )T C Bx, y, z dv
  
= Element stiffness matrix
(4.22)
Nodal displacement copmponents
q  vN x, y, z T  f dv  sN x, y, z  tds 
T
Nodal force matrix
{f} = Body forces
{t} = Surface tractions
[N(x,y,z)] in Step 3, Chapter 3, [B(x,y,z)] in Equation (4.13), [C] in Equation (4.7)
(4.23)
Examples of FE Stress Analysis of Solid Structures
NOTE: In FE stress analysis of solid structures, it is customer to represent the element displacements by:
and nodal displacements by: {u}.
U x  x, y, z 
U ( x, y, z )  U y x, y, z 
U  x, y, z 
 z

The relationship between the element displacements and the nodal displacements are:
{U} = [N(x,y,z)] {u}
where [N(x,y,z)] = the Interpolation function. It is a row matrix for 1-D bar elements,
rectangular matrices for 2- or 3-D elements.
● Interpolation function enables the determination of the primary quantities in the element with specified coordinate (x,y,z)
with the same primary quantities of the associate nodes
Part 3
Finite Element Formulation for
One-dimensional Bar elements
1. Bar elements subjected to unidirectional load
2. Truss bar elements
3. Beam bending elements
One-Dimensional Stress Analysis of Bar Elements
Derive the interpolation function using general formulation:
We will need first to derive the [h] matrix in Equation (1.13) by the following computations:
  x   1   2 x
Φ1
We have:
●
L
x1 = 0
x2 = L
Figure 1.6 Linear Interpolation Function
of a Bar Element
and
 2   1   2 x2
From which we will have
Φ2
●
1  1   2 x1
x
1  1 x1  1 

    Aa

1
x
2 2
 2 
   
with the matrix [A] to be:
1 x
A   1  and
1 x2 
A 
1 x1
 x2  x1
1 x2
1  x2  x1 
 h
x2  x1  1 1 
By following Equation (1.15), we have the interpolation function of a bar element to be:
The inverse matrix of [A] is:
A1

 1  x2 x1   1
T
N  x   R h  1 x
 1 1    L  x2  x   x1  x 
x

x

 2 1
For the present case with x1 = 0, and x2 = L, we have the interpolation function to be:
 x x
N  x   1 

 L L
(1.7)
Derive the element equation for a typical 1-D bar element: a bar element made of one material with Young’s modulus E.
The bar is subjected two forces F1 and F2 as shown in the figure below. Establish the element equation
u1 Node 1
●
F1
Node 2
L
x=x1
●
u2
F2
x
Because the bar is made of one material, and the applied forces are along
the length of the bar. So the bar will only deform along the length of the bar
e.g., the element displacement U is a function of coordinate x only:
x=x2
We thus have the element displacement to be U = U(x). The corresponding displacements at the two nodes are:
{u}T = {u1
u2}.
The relationship between the element displacement and the corresponding nodal displacements are:
{U(x)} = {N(x)} {u}
(4.24)
in which {N(x)} is the interpolation function for one-D bar element. It is a row matrix with two columns
We had derived this interpolation function before with assumption that the element displacement {U(x)} follows
a linear polynomial function: U(x)= α1 + α2x, and found the interpolation function using this liner function to be
In the same form as in Equation (1.7) we derived using general formulation method:
N x   1  x x 
(4.25)
 L L
with L = x2 – x1
Because the bar element is subjected to force along the x-coordinate only, and both the induced stress and strain are
in the direction of x-coordinate only. We thus obtain from Equation (4.4) and obtain:
D 

d

x dx
From Equation (4.13), we thus have the [B] matrix to be:
B( x)  DN ( x)T  d 1  x x    1
dx  L L   L
1 1
   1 1
L L
For the same reason of being uniaxial stress distribution, i.e., σxx = Eεxx by uniaxial Hooke’s law, we have the
elasticity matrix [C] = E – the Young’s modulus from Equation (4.7) with Poisson’s ratio γ = 0. We thus has the
element equation of the deformed bar expressed in the form:
(4.26)
[Ke] {Φ} = {q}
with
K e   vB( x, y, z ) C Bx, y, z dv  x Bx  EBx ( Adx)
T
x2
T
1
x2  1
EA x2  1
1

T 
    1 1 E   1 1 Adx  2    1 1dx
x1
L x1  1 
L
 L

EA x2  1  1
EA   x2  x1    x2  x1  EA  1  1
 2  
dx  2 

L x1  1 1 
L   x2  x1   x2  x1   L  1 1 
(4.27)
Example 4-1 Show the element equation for the bar element in the figure, and determine displacement at Node 2
u1
F1
●
The nodal forces {q} for the bar element are expressed as: {q}T = {-F1
Node 1
x=x1
Node 2
L
●
u2
F2
x
F2}
We may thus express the element equation for the bar element to be:
x=x2
EA  1  1 u1   F1 
 



L  1 1  u2   F2 
(a)
Because there is only one bar element in the structure, the overall stiffness equation in Equation (1.28)
is identical to that of element equation in Equation (4.24).
However, adjustments of the now overall stiffness equation with the three matrices in Equation (4.24) would be required,
as expressed:
 K aa
K
 ba
K ab  qa   Ra 
  
K bb  qb   Rb 
where {qa} = specified (known) nodal quantities; {Rb} = specified (known) applied resulting actions, from which we may obtain:
{qb} = [Kbb]-1 ({Rb} – [Kba]{qa})
(6.12) (PRINCIPAL REF)
The specified nodal unknown {qa} in this case is u1 = 0 and F1=0, the required unknown {qb} is u2, We thus need no adjustment of the
Matrices in the overall stiffness equation in (4.24). The required unknown quantity u2 is obtained from Equation (6.12, REF) to be:
u2 
EA 1
F2   10  EA F2
L 1
L
Example 4-2 Deformation and stress in a compound bar made of two different materials
Use the FEM to determine the displacements at the joint of a compound road made of copper and aluminum induced by a
uniaxial force P = 30000 N at of the end of the rod as shown in the Figure A below:
Figure A Compound Rod subjected to a Uniaxial Force
The compound rod has a cross-sectional area A = 650 mm2 and the Young’s moduli Ecu = 10300 MPa and Eal =69000 MPa.
Solution:
The situation shown in Figure A indicates that the rod is expected to elongate along the same direction in the x-axis, as shown in Figure B.
Figure B FE model for a Compound Bar
Example 4-2 – Cont’d
The FE model in Figure B indicates the following:
(1) There are total 3 nodes in the structure
(2) Nodal coordinates:
Node 1 at x1 = 0; Node 2 at x2 = 915 mm, and Node 3 at x3 = 1220 mm
(3) The length of Element 1 = L1 = 915 mm; the length of Element 2 = L2 = 305 mm
(4) Both elements have a cross-sectional area: A1 =A2 = A = 650 mm2
(5) Displacements at the 3 nodes are: {u}T = {u1 u2 u3} with u1 = 0
Our solution begins with developing the “element equations” for both elements in the FE model:
Element 1 made of copper:
Coefficient matrix for Element 1:
K  
1
e
E1 A1  1  1 10300 x 650  1  1  7.317  7.317 6


x10 N / m
L1  1 1 
915 x103  1 1   7.317 7.317 
Element equation for Element 1:
 7.317 x106

6
 7.317 x10
 7.317 x106  u1   p1 
 
6 
u
7.317 x10   2   p2 
where p1 and p2 are forces art Node 1 and 2 respectively
(a)
Example 4-2 – Cont’d
Element equation for Element 1:
 7.317 x106

6
 7.317 x10
 7.317 x106  u1   p1 
 
6 
7.317 x10  u2   p2 
Element 2 made of aluminum:
Coefficient matrix for element 2:
K   ELA
2
e
2
2
2
 1  1 69000 x650  1  1  147.05  147.05 6
 1 1   305 x103  1 1    147.05 147.05  x10 N / m



 

(b)
Element equation for Element 2:
 147.05 x106

6
 147.05 x10
 147.05 x106  u2   p2 
 
6 
u
147.05 x10   3   p3 
Due to the fact that the present case involves TWO elements with Node 2 being common to both these element,
we need to assemble the coefficient matrices by following the established rule by summing up the values of Node 2
from both element coefficient matrices.
Example 4-2 – Cont’d
 
We thus assemble the overall stiffness matrix by adding K e1 In Equation (a) and K e2  in Equation (b) in the following way:
7.317
K   106  0
 0
 7.317
0 
7.317  147.05  147.05 106
0
147.05 
0 
7.317  7.317
 0

154
.
367

147
.
05


 0
0
147.05 
(c)
The numbers in boldface in Equation (c) are those associated with Node 2.
The overall stiffness equation for the bar structure with specified loading/boundary conditions is thus expressed in the
following partitioned matrices as:
0  u1  0  p1  0 
7.317  7.317

 

106  0
154.367  147.05  u2    p2  0 
 0
0
147.05   u3   p3  30000
(d)
The two unknown nodal displacements u2 and u3 may be obtained by the following equations using the partitions
in Equation (d):
154.367  147.05 u2   0 
106 
 


u
0
147
.
05
30000

 3  

resulting in the displacement of the compound bar at the joint (Node 2) to be u2 = 1.94 mm and the displacement
at the free end u3 = 2.04 mm. The total elongation of the rod is 3.98 mm.
Example 4-2 – Cont’d
Strain in Elements
Now that we have solved the displacements at the 3 nodes, we may use the train-displacement relations to
determine the induced strains in both these elements:
 T

  1xx
 xx2 
1
2
where  xx and  xx are the strains in Element 1 and 2 respectively
The strain-displacement relationship is available from the expression: {ε} = [B]{u} in Equation (4.3), in which the matrix {B} is:
Bx  

d
N x   d  x  x2
dx
dx  x1  x2

x  x1   1

x1  x2   x1  x2

1 

x1  x2 
We have: Node 1 at x1 = 0; Node 2 at x2 = 915 mm, and Node 3 at x3 = 1220 mm, leading to: the length of
Element 1 = L1 = 915 mm; the length of Element 2 = L2 = 305 mm. We may thus express the [B(x)] for both
elements to be:
B1   
1
 915
1 
 1
 , and B2   
915 
 305
1 

305 
Example 4-2 – Cont’d
From the [B] matrices for both elements, we may compute the strains in Element 1 and 2 as follows:

1
xx
 1
 
 915
 1
 305
 xx2  
1 u1  0
1
1.94
u2 
 0.21%


915  u2  915
915
for element 1, and
1 u2 
u2
u3
1.94 2.04
0.1







 0.033%
 
305 u3 
305 305
305 305
305
for element 2
Stresses in elements
We may use the Hooke’s Law to determine the stresses in each of these two elements from their corresponding strains
For element 1 with Ecu = 10300 MPa:
 1xx  C1  1xx  Ecu 1xx  10300 x0.0021  1.03MPa
For element 2 with Eal = 69000 MPa:
 xx2  C2  xx2  Eal xx2  69000 x0.00033  22.77 MPa
Finite Element Formulation of Elastic Solid Structures
One-Dimensional Bar Elements
for Truss Structures
FE Formulation of Truss Elements Using 1-D Bar Elements
Common Trusses
FE Formulation of Truss Elements Using 1-D Bar Elements - Cont’d
The characteristics of a truss element can be summarized as follows (a quote from Dr. Agarwal’s
lecture notes):
1. Truss is a slender member (length is much larger than the cross-section).
2. It is a two-force member i.e. it can only support an axial load and cannot support a
bending load. Members are joined by pins (no translation at the constrained node,
but free to rotate in any direction). Being a two-force member, the force acting in the member is in the
length-direction only.
3. The cross-sectional dimensions and elastic properties of each member are constant along its length.
4. The element may interconnect in a 2-D or 3-D configuration in space.
We will formulate 2-D configuration only. Meaning the truss member deforms in a plane
defined by x-y coordinates.
5. The bar elements for truss structures is mechanically equivalent to a spring, since it has
no stiffness against applied loads except those acting along the axis (the length direction) of the member.
6. However, unlike a spring element, discussed in previous chapters, a truss element can be oriented in any
direction in a plane, and the same element is capable of tension as well as compression along its longitudinal
directions.
FE Formulation of Truss Elements Using 1-D Bar Elements – Cont’d
Simple Bridge Structure:
FE model with 1-D bar elements with designated element and node numbers. All elements are
interconnected by “frictionless hinges”:
3
ο
Element (1)
(3)
Node 1
ο
(2)
ο2
P1
5
(6)
(5)
(4)
ο
(7)
ο
4
P2
(8)
(11)
(9)
ο6 (10)
P3
ο7
ο ο οο
FE Formulation of Truss Elements Using 1-D Bar Elements – Cont’d
A fundamental phenomenon in Statics is that “hinged” joint of any member in
A frame or truss cannot resist moment. Consequently, any member in s truss
As shown in the figure at the left with hinge joints (as illustrated in the lower
Figure will have forces acting along the length of the member.
Consequently, the derivation of the interpolation function in the FE formulation of
these members may begin with the bar element in Example 4-1
y
Truss members along x-direction of y-direction only:
x
The element equations for these elements will be in similar forms
as for the 1-D bar elements in Examples 4-1 and 4-2.
Inclined Truss Members:
The element equation for those truss members that are inclined with the x-coordinate with angle θ,
however, will be in different forms, as will be derived in the following procedures.
FE Formulation of Inclined Truss Elements Using 1-D Bar Elements
Inclined truss bar element in x-y plane: Element (1), (5), (7) and (11):
y
vj
Fj
j
(xj,yj)
vi
The inclined truss bar elements such as shown in the figure at right
may have TWO displacement components, but these elements can only
elongate or contract in the LONGITUDINAL direction only.
i
Fi
The same applies to the induced strain and stress.
So, These are regarded as “a special bar elements.”
uj
ui
(xi,yi)
0
These bar elements remain having two nodes: Node i and Node j
Located at specified coordinates (xi, yi) and (xj, yj) respectively.
Each node has two displacement components:
ui and vi = displacement of Node i in respective x- and y- directions
uj and vj = displacement of Node j in respective x- and y-directions
Fi and Fj are the forces acting at Node I and Node j respectively
x
FE Formulation of Truss Elements Using 1-D Bar Elements – Cont’d
Derivation of element equation
[Ke]{Φ} = {q}
(4.21)
The situation in the bar element in truss structures is unique in the way that these bar elements are solid element of “two-force members”
– meaning that the force is acting along the length of the bar only. Consequently, the induced displacements that ae effective to the
elongation or contraction of the bar (truss) member elements are the ones in the length-direction only. The strain and stress in the truss
members are along the length of the bar too.
The displacement components in an inclined bar element in the plane defined by the (x,y) coordinate system in the figure below do not
contribute in producing the displacement, strain and stress in the bar element for this reason . However their “equivalence” to the ones
along the length direction of the bar element do.
Consequently, we need to convert the current situation in the inclines bar element in the left to the equivalent situation in the right
of the figure through a coordinate transformation process of the nodal displacement components and the applied nodal forces
Equivalent Typical Bar Element
FE Formulation of Truss Elements Using 1-D Bar Elements – Cont’d
Derivation of element equation – Cont’d
Transformation of Nodal displacement components: From the “global coordinates” to “Local coordinates”- The latter
is used for the FE formulation as with 1-D bar elements
Truss bar element:
Typical bar element:
Converting
“truss bar element”
to equivalent “typical
bar element”
By referring to the above figure, we have the following relationship in transforming the nodal displacement components from
The global coordinates (x,y) in the left figure to the local coordinate (x) in the right of the figure:
at Node i:
 i  ui cos   vi sin   cui  svi
(4.28a)
at Node j:
 j  u j cos   v j sin   cu j  sv j
(4.28b)
 ui 
 
 i  c s 0 0  vi 
Equations (4.28a,b) may be expressed in matrix form:       
 u 

0
0
c
s
j
 j 
  
or in a shorthand version of: {δ} = [T]{u}
v j 
with the matrix [T] being the transformation matrix
where c = cosθ and s = sinθ
(4.29a)
(4.29b)
FE Formulation of Truss Elements Using 1-D Bar Elements – Cont’d
Derivation of element equation – Cont’d
Transformation of Nodal force components:
From “global coordinates” to “Local coordinates”
Transformation of nodal forces can be done in a similar way as we did with nodal displacements. However, it would be easier to do it using
the work done to the element by these forces, as work done is a scalar quantity, which is easier to transform in space than the vector
quantities such as displacements.
Since the work done to the element by nodal forces may be expressed as: W   T  f   uT F  for the same element in both local and global
coordinate systems, or in a long-hand form:
 Fix 
F 
 f1 
 iy 
T
T
W  1  2    ui vi u j v j   or    f   u F 
 f2 
 F jx 
 F jy 
Substituting {δ} = [T]{u} in Equation (26b) into the above expression, we get: ([T]{u}])T{f} = {u}T{F}, leading to:{u}T[T]T{f} = {u}T{F}
We will thus have the nodal force transformation by the following relationship:
[T]T {f} = {F}
where {f} = nodal forces in “typical bar element”, {F} = nodal forces in “inclined truss element”
(4.30)
FE Formulation of Truss Elements Using 1-D Bar Elements – Cont’d
Derivation of element equation – Cont’d
Element equation in Local Coordinate System
Now that we have made the real situation in the “Global” coordinates (x,y) to be equal to the situation in 1-D “Local” coordinate (x) situation
through a “transformation” process:
Inclined truss bar element:
Transformation
Matrix:
T   
Equivalent bar element:
=
c s 0 0

0 0 c s 
(4.26a)
Previously derived interpolation function:
N x 
Real situation in
“Global” coordinates
Coordinate Transform:
 x
 1 
 L
x

L
Previously derived “element equation”
with modified notation of nodal quantities:
EA  1  1  i   f i 
(4.24)
 




f

1
1
L 
 j   j 
or K e     f 
 AE 
 AE
L
L
with K e   

AE
AE

L
L 
(4.27)
FE Formulation of Truss Elements Using 1-D Bar Elements – Cont’d
Derivation of element equation – Cont’d
Element equation in Local Coordinate System
EA  1  1  i   f i 
 

L  1 1   j   f j 
(4.24)
 AE 
 AE
L
L  (4.27)
K e     f  with K e    AE
AE

L
L 
Relationship of nodal forces between the two coordinate systems: {F} in global coordinate system = [T]T{f} in Local coordinate system .
The element equation in global coordinate system thus have the form:
K e     f 
(4.31)
The nodal forces in the Local coordinate system {f} can be viewed as a transformed forces {F} rom the global coordinate systems with
the relationship: [T]T {f} = {F} as shown in Equation (4.27), or {f} = ([T]T)-1, leading to:
[Ke]{δ} = ([T]T)-1 {F}
But we also have already derived the following relationships: {δ} = [T]{u} in Equation (4/26b), and [T]T = {F} in Equation (4.27).
By substituting these relationships into the above expression, we will get the following “reversed transformation” of element equation
from the Local coordinate system to Global coordinate systems:
(4.32)
REAL inclined truss members: [K ][T]{u} = ([T]T)-1 {F} Equivalent 1-D bar elements
e
FE Formulation of Truss Elements Using 1-D Bar Elements – Cont’d
Derivation of element equation – Cont’d
Element equation in Local Coordinate System
The Element Equation of inclined truss elements is: [Ke][T]{u} = ([T]T)-1 {F}
(4.33)
We realize the fact that the transformation matrix: [T]T = [T]-1 (Reference No. 2). Thus Equation (4.33) has a new form of:
[Ke][T]{u} = ([T]-1)-1 {F} leads to: [Ke][T]{u} = [T]{F} with ([T]-1)-1 = [T] in the last expression.
Now, if we multiply both sides by [T]-1, and use the relationship of [T]-1 = [T]T, we have the element equation of
the inclined truss element (i.e., the bar element in the global coordinate system) to be:
The element Equations for
inclined truss members:
 AE
 AE 
L
AE 
L
L
where K e    AEL

[T]T[Ke][T]{u} = {F}
(4.34)
c s 0 0 with A, L= cross-sectional area and the length of the truss element

0 0 c s  respectively, c=cosθ and s=sinθ
and [T ]  
FE Formulation of Truss Elements Using 1-D Bar Elements – Cont’d
Element equation for Inclined Truss Members
where
The stiffness matrix:
K e u F 
 ui 
v 
u   i 
u j 
v j 
and
 Fix 
F 
 iy 
F    
 F jx 
 F jy 
 c2
cs  c 2  cs 


s 2  cs  s 2 
 AE   cs
K e      2
2
cs 
 L   c  cs c

2
2 
cs
s 
  cs  s
(4.35)
(4.36a)
(4.36b)
(4.37)
1) This stiffness matrix represents the stiffness of a single element with incline angle θ
2) It is symmetric about the diagonal line of the square matrix
3) Since there 4 unknown nodal displacements - meaning 4 degree-of-freedom (dof), the matrix is of the size
of (4x4)
4) The terms c and s represent the cosine and sine values of the orientation of the element with the
horizontal plane, rotated in a counter clockwise direction (+ve direction)
Example 4-3 FE Stress Analysis of a Truss Structure:
A truss with 3 members joined by frictionless hinges as shown in the figure.
Element 1 and 2 are made of aluminum, and element 3 is made of steel.
Materials
Cross-sectional
Area (A) (10-6m2)
Young’s Modulus (E)
(106 N/m2)
Yield Strength (σy)
(106 N/m2)
Member (1)
Aluminum
200
70,000
170
Member (2)
Aluminum
200
70,000
170
Member (3)
Steel
100
200,000
21,000
Solution:
We realize the fact that there are 3 members, each has its own dimension, and material with properties listed in the above table.
So, we may conveniently construct the FE model of the truss structure by using 3 elements 1 with node pair 1-3, element 2
with node pair 2-1, and element 3 with node pair 2-3, as shown in the figure.
We will derive the element equations for all these 3 elements by using the formulations for truss elements with: Equation (4-24) for
the horizontal Element 1, and Equations (4.35) and (4.37) for the inclined elements 2 and 3.
Example 4-3 FE Stress Analysis of a Truss Structure - cont’d
Derivation of element equations
Element 1:
L1 = 0.26 m, E1 = 70000 MPa, A1 =200x10-6 m2:
(1)
u1x
Being horizontal, we have the incline angle θ = 0, which leads to:
c = cosθ=1, c2 = 1, s=sinθ = 0, s2 = 0 and cs = 0
v3y
v1y
Node 3
Node 1
u3x
We also calculate the coefficient of the stiffness matrix of element 1 as follows:



E1 A1 70000 106 200 10 6

 53.84 106 N m
L1
0.26
L1 = 0.026 m
Using Equation (4.37), the stiffness matrix for Element 1 is thus:
 c2
cs  c 2  cs 
1

0
2
2
cs
s

cs

s
E
A
  53.84 106 
K e1  1 1   2
 1
L1  c  cs c 2
cs 


2
2 

cs

s
cs
s
0


 
0 1
0 0
0 1
0 0
0
0
0

0
(a)
Example 4-3 FE Stress Analysis of a Truss Structure - cont’d
Derivation of element equations
Element 2:
L2 = 0.15 m, E2 = 70000 MPa, A2 =200x10-6 m2:
v1y
u1x
L2=150 mm
Node 1
c = cos 90o = 0, c2 = 0; s = sin 90o = cos0o = 1, s2 = 1 and cs = 0 ,
and

u2x
v2y


E2 A2 70000 106 200 10 6

 31.06 106 N m
L2
0.15
(2)
Node 2
In this Element 2, we have the inclined angle θ=90, which lead to:
Hence the stiffness matrix for Element 2 is:
0 0
0 1
K e2  31.06 106 
0 0

0  1
 
0 0
0  1
0 0

0 1
(b)
Example 4-3 FE Stress Analysis of a Truss Structure - cont’d
Derivation of element equations
Element 3:
L3 = 0.3 m, E3 = 200,000 MPa, A3 =100x10-6 m2:
v3y
y
Node 3
Element 3 has an inclined angle θ = 30o leading to:
u3x
v2y
u2x
Node 2
X
c= cos 30o = 0.866, c2 = 0.75, s = sin 30o = 0.5, s2 = 0.25 and cs = 0.433;
Also the coefficient of the stiffness matrix:



E3 A3 200000 106 100 10 6

 66.67 106 N m
L3
0.3
The stiffness matrix for element 3 is:
0.433
 0.75  0.433
 c2
cs  c 2  cs 
 0.75

 0.433

2
2
0
.
25

0
.
433

0
.
25
cs
s

cs

s
E
A
  66.67 106 

K e3  3 3   2
2
  0.75  0.433 0.75
0.433 
L3  c  cs c
cs 



2
2 
0.433
0.25 
cs
s 
  cs  s
 0.433  0.25
 
(c)
Example 4-3 FE Stress Analysis of a Truss Structure - cont’d
Derivation of element equations
From Equation (4.21):
[Ke] {Φ} = {q} with a general expression for element equations, we are now in the position to express the
same equations for the 3 elements in the current truss structure as follows.
The 3 Element equations for the truss structure:
Element (1) with Node 1 and 3
 53.84
 0
106 
 53.84

 0
0  53.84
0
0
0 53.84
0
0
0 u1x   f1x 
   
0  v1 y   f1 y 
  
0 u3 x   f 3 x 

0 v3 y   f 3 y 
Element (2) with Node 1 and 2
0
0

31.6
6 0
10
0
0

0  31.6
0
0  u1x   f1x 
   
0  31.6  v1 y   f1 y 
  
0
0  u2 x   f 2 x 

0 31.6  v2 y   f 2 y 
Element (3) with Node 2 and 3
28.87
 50
 28.86 u2 x   f 2 x  in which u1x, u2x and u3x = displacement components in x-direction
 50
of Node 1, 2 and 3 respectively
 28.87
 v   f 
16
.
67

28
.
87

16
.
67
v1y, vy2 and v3y = displacement components in y-direction
  2 y    2 y 
106 
of Node 1, 2 and 3 respectively
  50
 28.87
50
28.87  u3 x   f 3 x 


f1x, f2x and f3x = Nodal force at Node 1, 2 and 3 respectively
16.67  v3 y   f 3 y 
 28.87  16.67 28.87
f1y, f2y and f3y = Nodal force at Node 1, 2 and 3 respectively
Example 4-3 FE Stress Analysis of a Truss Structure - cont’d
Assembly of element equations for Overall Stiffness Equation
By following the description on “assembly of element stiffness matrices” in Step 5 with diagram of Chapter 3 Steps in Finite
Element Analysis, we may assemble the three (3) truss element stiffness matrices shown above in the following form:
or in a neat form as shown in the next slide:
Example 4-3 FE Stress Analysis of a Truss Structure - cont’d
Element stiffness matrix and Overall Stiffness Equation of the Truss Structure
The overall stiffness matrix of the truss structure:
(d)
With this overall stiffness matrix, we may establish the overall stiffness equation for the truss structure as shown below:
(e)
Example 4-3 FE Stress Analysis of a Truss Structure - cont’d
Boundary and Loading conditions of the Truss Structure
We recognize the following specified conditions for the discretized truss structures:
The boundary conditions (displacement at nodes):
1) With Node 1 being completely fixed: u1x = v1y = 0
2) Node 2 is allowed to move in vertical (y) direction: u2x = 0
The loading conditions:
All except Node 3 has one force acting in the y-direction:
f1x = f1y = f2x = f2y = f3x = 0 and f3y = 0.4 kN = 400 N
The overall stiffness equation in Equation (e) after the substitution of the above specified boundary and loading conditions has the form:
0
0
0
 53.84
0  u1x  0   f1x  0 
 53.84
 0
v  0   f  0 
31
.
6
0

31
.
6
0
0
  1y


 1y
 0
0
50
28.87
 50
 28.87  u2 x  0  f 2 x  0 
6
10 



v
f

0
0

31
.
6
28
.
87
48
.
27

28
.
87

16
.
67

 2 y   2 y

 53.8
0
 50
 28.87 103.84 28.87   u3 x   f 3 x  0 


 

0
 28.87  16.67 28.87
16.67   v3 y   f 3 y  400
 0
Example 4-3 FE Stress Analysis of a Truss Structure - cont’d
Overall Structure Stiffness Equation with given Boundary and Loading Conditions for the Truss Structure
0
0
0
 53.84
0  0   0 
 53.84
 0
 0   0 
31
.
6
0

31
.
6
0
0

  



0
 0
0
50
28.87
 50
 28.87     0 
106 
   

v
0

31
.
6
28
.
87
48
.
27

28
.
87

16
.
67
0

 2 y  

 53.8
0
 50
 28.87 103.84 28.87  u3 x   0 

  

0
 28.87  16.67 28.87
16.67  v3 y   400
 0
(f)
Now, if we follow Step 6 in Chapter 3 on Steps in FE Analysis, we may partition Equation (f) in the following way:
where {qa} = specified (known) nodal
quantities;
{Rb} = specified (known) applied resulting
actions, from which we may obtain:
{qb} = [Kbb]-1 ({Rb} – [Kba]{qa})
qa   0
v2 y 
qb   u3 x 
v 
 3y 
Ra   0
 0  31.6 28.87 
 48.27  28.87  16.67 


6
 0  K   10  53.8
0
 50  K bb   106  28.87 103.84 28.87 

 ba



Rb    0 
  16.67 28.87
 0
16.67 
0  28.87
 400


Example 4-3 FE Stress Analysis of a Truss Structure - cont’d
Solution of 3 unknown nodal displacements in truss structure
We will get the 3 unknown nodal displacements from the portioned over
stiffness equation by the expression: {qb} = [Kbb]-1 {Rb}:
v2 y 
 48.27  28.87  16.67 
 

6 
u3 x   10  28.87 103.84 28.87 
v 
  16.67 28.87
16.67 
 3y 
1
 3.1646
867.36 10 19
 0 


8 
19
0

10
1.86


867.36 10
 400
 3.1646
 3,2166



3.1646   0 


 3.2166  0 
14.734   400
Solve for the unknown displacement components at Node 2 and 3:
u2y = -12.65x10-6 m, u3x = 12.86x10-6 m and v3y = -58.94x10-6 m with negative signs meaning downward direction
Example 4-3 FE Stress Analysis of a Truss Structure - cont’d
Solution of secondary unknowns of induced strains and stresses in all three elements in the truss structure
We will determine the strain components and then stress components in each of the 3 elements
in the truss structure.
We should bear in mind that truss members are “two force members,” in which only the in-line
displacement components will produce strains and stresses.
Element 1:
u1x
ο
X=0
L1
ο
X=L1
u3x
with Node 1 and Node 3
We have the relationship between the element displacement U(x) and the
x corresponding nodal displacement components {u1x u3x}T along the line of
the bar element in the figure I the left:
 x
x u1x  
x
x
U  x   N  x u  1 
   1  u1x  u3 x
L1
 L1 L1 u3 x   L1 
where the interpolation function of this simplex bar element is given in Equation (1.5) textbook
We will use Equation (4.3): the element strain {ε} = [B(x)]{u} to compute the only strain, εxx in the element by using Equation (4.12):
 x    xx  Bx u

d
in which the matrix [B(x)] is given in Equation (4.13) to be: [B(x)] = [D][N(x)], with D  x  dx for the
present one-dimensional case. We thus have: Bx   d 1  x
dx 
L1
x

L1 
We thus have the strain εxx in element 1 to be:
Example 4-3 FE Stress Analysis of a Truss Structure - cont’d
Solution of secondary unknowns of induced strains and stresses in all three elements in the truss structure
We will determine the strain components and then stress components in each of the 3
elements in the truss structure.
We should bear in mind that truss members are “two force members,” in which only the
in-line displacement components will produce strains and stresses.
Element 1:
u1x
ο
X=0
with Node 1 and Node 3
in global coordinates
L1=0.26 m
ο
u3x
x
[T]
with θ=0o
X=L1
1 0 0 0

0 0 1 0 
The transformation matrix: T   
 ui 
 
 i  1 0 0 0  vi 
The relationship of nodal displacements in Local coordinate system       
 u 

0
0
1
0
 j 
and the corresponding global coordinate systems is: {δ} = [T]{u}, or:
 j 
v j 
But we obtained the nodal displacements from the previous calculations to be:
ui = u1x =0, vi = v1x =0, uj = u3x = 12.86x10—6m, vj = v3y = -58.94x10-6m
= Element 1
in local coordinate
Example 4-3 FE Stress Analysis of a Truss Structure - cont’d
Solution of secondary unknowns of induced strains and stresses in all three elements in the truss structure
0



 i  1 0 0 0 
0
0

 
      




6 
6 
 j  0 0 1 0  12.86 10  12.86 10 
 58.94 10 6 
We have the strain in the element is obtained by using Equation (4.12): {ϵ} = [B(x)]{Φ}, and [B(x)] = [D]{N(x)] (3.13)

 

We thus have:
B  DN x   d 1  x x    1 1 
dx 
The strain in Element 1 is thus:

 xx    1
 L1
L1
L1   L1
L1 
1   i   1
   
L1  j   0.26
0

1 
 49.46 10 6 m / m

6 
0.26 12.86 10 
The corresponding stress in Element 1 can be obtained from Equation (4.6): {σ}=[C]{ϵ}=(70000x106)(49.46x10-6)=3.43 MPa
Example 4-3 FE Stress Analysis of a Truss Structure - cont’d
Solution of secondary unknowns of induced strains and stresses in all three elements in the truss structure - Cont’d
Element 2:
with Node 1 and Node 2
We have computed that u1 = v1 = 0, u2 = 0 and v2 = u2y = -12.65x10-6 m
Let us transform the coordinate system in the lower left figure to one-dimensional bar
element using Equation (4.29) to the following situation:
 ui 
 
Node i = Node 2
Node j = Node 1
 i  c s 0 0  vi 
      
 
δi
δj
 j  0 0 c s  u j 
v j 
We have : c = cos 90o =0, and s = sin 90o = 1.0, leading to:
y
v1
u1
Node 1
u2
Node 2
v2
ui  u 2  0



6 
 i  0 1 0 0 vi  v2  u2 y  12.65 10 
      



u

u

0
0
0
0
1
j
j
1


 




v j  v1  0
 i   12 10 6 
The above expression leads to: δi = -12x10-6 m and δj = 0 or:       

x

0
j

  
We have the strain in the element is obtained by using Equation (4.12): {ϵ} = [B(x)]{Φ}, and [B(x)] = [D]{N(x)] (3.13)
d 
x
x  1
1
We thus have: B   DN x   1 
  

dx 
The strain in Element 2 is thus:
L2
L2   L2

 xx    1
 L2
L2 
1   i   1
   
L2  j   0.15
1  12 10 6 
6

  80 10 m / m
0.15 
0

The corresponding stress in Element 2 can be obtained from Equation (4.6): {σ}=[C]{ϵ}=(70000x106)(80x10-6)=5.6 MPa
Example 4-3 FE Stress Analysis of a Truss Structure - cont’d
Solution of secondary unknowns of induced strains and stresses in all three elements in the truss structure - Cont’d
Element 3:
with Node 2 and Node 3
We obtained the nodal
Displacements from
The previous calculation
to be as shown
v3j = -58.94x10-6
y
Node 3
v2i = -12.65x10-6
Node 2
Transformation matrix
[T]
x
u2i = 0
δi=u2x
ο
L3=0.3 m
The transformation matrix [T] has the following form with θ = 30o:
cos 30o
T   
 0
sin 30
0
0
cos 30o
u3j=12.86x10-6
0
0
0  0.866 0.5

 
0 0.866 0.5
sin 30o   0
ο
δj=u3x
x
Example 4-3 FE Stress Analysis of a Truss Structure - cont’d
Solution of secondary unknowns of induced strains and stresses in all three elements in the truss structure - Cont’d
Element 3:
with Node 2 and Node 3
The nodal displacements in the equivalent Element 3 in the Local coordinates can be related to the real Element 3 in the global
coordinates by the transformation matrix to be:
ui  u 2 i  0



6 
 i  0.866 0.5
0
0   vi  v2i  12.65 10  
  6.325  6
6.325 10 6
      


 x10
  u  u  12.86 10 6  
6 

0
0
0
.
866
0
.
5

18
.
3332


11
.
1368

29
.
47
10
3j
 j

 
 j 
 
6
v j  v3 j  58.94 10 
But since the interpolation function of the equivalent Element 3 in the local coordinate is:
And the matrix [B(x)] = [D][N(x,y,z)] from Equation (4.13) in the present case to be:
Bx  

N x   1 
d 
x
1


dx  L3

x
L3
x  1
  
L3   L3
x

L3 
1

L3 
(1.7)
Example 4-3 FE Stress Analysis of a Truss Structure - cont’d
Solution of secondary unknowns of induced strains and stresses in all three elements in the truss structure - Cont’d
Element 3:
with Node 2 and Node 3
The strain {ε} in terms of nodal displacements in Element 3 may be obtained by using Equation (4.12): {ε}=[D]{N(x)}:
 1
    xx  
 L3
1   i   1
   
L3  j   0.3
1  6.325 
6

 10
0.3  18.3332
 (21.0833  61.1107) 10 6  82.194 10 6 m
The induced stress in Element 3 is {σ} can be computed by using Equation (4.6), or {σ} = [C]{ε} with [C] = E3 = 200,000x106 Pa (N/m2)
We thus have the stress in element 3 to be:
{σ} = σxx = E3εxx = (200000x106)(-82.194x10-6) = - 16438800.33 N/m2 or -16.44 MPa (a compressive stress)
Finite Element Formulation of Elastic Solid Structures
One-Dimensional Bar Elements
for Bending of Beams
Quick Review of Simple Beam Theory
1) Like truss members, beams are slender members (length is much larger than the cross-section) in structures.
2) However, unlike the truss members, beam members can resist forces applied laterally or transversely to their axes.
y
Original straight beam:
Induced
deflection by
applied forces
P – concentrated force
W(x) – distributed load
y
per unit length
x
x
x
Deflected state
Induced deflection: y(x)
3) Beams can also resist the applied moments that cause the beams to bend, i.e. beams can have deflection
induced by applied moments (e.g., torque in the x-y plane)
4) Beams may be straight or curved. We will focus on straight beams only .
5) The application of unilateral forces or moments will not only cause the beam to bend in shapes, but also
having deflections, y(x) perpendicular to the beam axes, as shown in the above figure.
6) So, the actions to beam can be: concentrated forces, distributed load and moments
The induced reactions are: Deflection of the beam y(x), bending moments M(x) and shear forces V(x),
and the bending stresses (normal stress σn(x) – normal to the cross- section of the beam, and the shear
stress σs(x) - acting on the cross-section of the beam)
Quick Review of Simple Beam Theory – Cont’d
Construction of induced bending moment and shear force distributions (diagrams)
y
y
Uniform distribution load: w N/m
P
B
A
Ra 
Pb
L
a
b
L
x
Rb 
Pa
L
B
A
Ra 
wL
2
x
L
Rb 
x
wL
2
Bending Moment Diagram:
M(x)
M x  
Pb
x
L
M x  
M(x)
Pa
L  x 
L
Pab
L
wL2
8
x
V(x)
Shear Force Diagram:
pb
L
pa
L
x
x
V(x)
wL
2
wL
2
x
Quick Review of Simple Beam Theory – Cont’d
Induced Shear Stresses in Bent Beams by Applied Force
There are two types of stresses induced to the beam subjected to external forces:
(1) The normal stress along the x-coordinate. It exists in perpendicular to the
cross-section of the beam (σxx), and
(2) The shear stress (σxz) that exists on the surfaces of the beam cross-section.
It also exists on the face along the x-coordinate (σyx) with the same magnitude
Expression for normal stress (σxx):  xx x    M x  y
I
(4.38)
where M(x) = bending moment at location x, y = distance from center of the beam
cross-section to the depth of the cross-section in y direction, and I = section
y
moment of inertia
b
3
bh
For beams with rectangular cross-section:
I
z
H
12
For beams with circular cross-section:
I
d 4
y
64
Section moment of inertia for beams of other cross-sections, including I-cross-sections
are available from mechanical engineering handbooks
z
d
Quick Review of Simple Beam Theory – Cont’d
Induced Shear Stresses in Bent Beams by Applied Force
Expression for shearing (σxz) or (σyx):
These stresses are induced by the shear force V(x) at the various cross-sections along the beam
y
c1
z
dA
y
y1 H
 yx
b
V x 

Ib

c1
y1
ydA 
V x 
Q
Ib
(4.39)
where Q=shear moment
For beams with rectangular cross-sections, dA = bdy, resulting in:
by 2
2
Q   bydy 
y1
2
h
We have

V x   h 2
  y12  with
 yx x  
2I  4

h
2
y1

b  h2
   y12 
2 4

 
yx max

3V
bh
at the center line of the beam where y1=0
Quick Review of Simple Beam Theory – Cont’d
Euler-Bernoulli Theory of Beam Bending
This theory relates the induced deflection (deformation of beams) and the applied forces
y
Original straight beam:
Induced
deflection by
applied forces
P – concentrated force
w(x) – distributed load
y
per unit length
x
x
x
Deflection and Bending Moment M(x)
and Shear force V(x) relations:
d 2 v x  M  x 

2
dx
EI
Applied
Forces
Deflected state
Induced deflection: v(x)
d 2 v x 
d 3v  x 
M  x   EI
and V  x   EI
2
dx
dx 3
Induced Deflection v(x) may e obtained by solving the 4th order differential equation:
d 4 v x 
EI
0
4
dx
where E = Young’s modulus of the beam material
I = Section moment of inertia
Induced
Deflection v(x)
& M(x), and V(x)
(4.40a,b)
FE Formulation of Beam Elements
Principal reference: “A First Course in the FEM” 6th Edition, Cengage Learning by Daryl Logan, 2017
y
Uniform distribution load: w N/m
y
B
A
x
Beam
Element
B
A
v(x)
x
L
ρ
Node i
ρ = Radius of curvature
deflection curve at x
Θ = slope of deflection
curve at x
V(x) = Deflection at x
Node j
x
x
v(x)
L
Primary Quantities
x=0
 vi 
 
In the element: The deflection v(x)
 i
At the nodes: The deflections v and slope θ, or as expressed as: d     at Node i and Node j
v j 
 j 
x
Datum
line
L
x=L
FE Formulation of Beam Elements
Principal reference: “A First Course in the FEM” 6th Edition, Cengage Learning by Daryl Logan, 2017
Beam elements
Contrary to the “truss elements,” beam elements deforms from its original straight shape into curved bent shape due to lateral
(or transverse) forces or applied couples (moments).
Actions and Induced Reactions in Beam Elements
Mi
Deformed shape
Node i
vi
Vi
X=0
Original shape
x
L
Node j
θ(x)
X
vj
Vj
X=L
Mj
in the element
Induced
Reactions
at Node i
at Node j
Linear
displacement
v(x)
Linear
displacement
vi
Linear
displacement
vj
Rotation
θ(x)
Rotation
θi
Rotation
θj
Actions
Lateral forces Vi
and Vj
Moments Mi and
Mj
FE Formulation of Beam Elements – Cont’d
Principal reference: “A First Course in the FEM” 6th Edition, Cengage Learning by Daryl Logan, 2017
Derive Interpolation Function
We assume the traverse displacement of the beam element follows a linear
polynomial function o the form:
vx   a1 x 3  a2 x 2  a3 x  a4
(4.41)
in which a1, a2, a3 and a4 are constant coefficients
By substituting the coordinates of Node I and Node j into the assumed
element deflection in Equation (4.38), we get:
x=0
dv x 
 a3
dx x 0
At Node i:
vi  v x  x 0  a4
At Node j:
v j  v x  x  L  a1 L3  a2 L2  a3 L  a4
x=L
i 
2
j 
dv x 
 3a1 L2  2a2 L  a3
dx x  L
FE Formulation of Beam Elements – Cont’d
Principal reference: “A First Course in the FEM” 6th Edition, Cengage Learning by Daryl Logan, 2017
Derive Interpolation Function
We may thus express the element deflection v(x) in terms of the nodal deflections by
substituting the 4 constant coefficients into Equation (3.38) and after re-arranged terms
to yield:
1
2

v x    3 vi  v j   2  i   j  x 3
L
L

1
 3

  2 vi  v j   2 i   j  x 2   i x  vi
L
 L


The above expression can also be expressed as: v x   N iv
where
N iv 

1
3
2
3
2
x

3
x
L

L
L3

1
N jv  3  2 x 3  3 x 2 L
L
We thus have:
v(x) = [N(x)]{d}
The element
deflections
The nodal
deflections


N i 
N j
N i
N jv

1 3
2 2
3
x
L

2
x
L

xL
L3

1
 3 x 3 L  x 2 L2
L

with Interpolation function: N  x   N iv

 vi 
 
 i
N j  
v j 
 j 
(4.43)

N i
(4.42)
N jv
N j 
(4.44)
FE Formulation of Beam Elements – Cont’d
Principal reference: “A First Course in the FEM” 6th Edition, Cengage Learning by Daryl Logan, 2017
Derive Element Strain εxx – Displacement v(x) Relation
We realize the fact that the rotation (or the slope) of the deflected beam is:   dvx  for 0  x  L
dx
An expanded view of a deformed beam in x-direction:
Contracted top edge
-u
Undeformed neutral axis
Rotation of the beam section at x
Stretched bottom edge
X
dx
We may find from the above diagram of expanded beam that the stretch u can be obtained by: u ( x)   y   y
But from theory of elasticity:
du
d  dv x  
d 2 v x 
 xx 
or  xx  x, y    y 
  y
dx
dx  dx 
dx 2
(4.45)
dv( x)
dx
FE Formulation of Beam Elements – Cont’d
Principal reference: “A First Course in the FEM” 6th Edition, Cengage Learning by Daryl Logan, 2017
Derive Element Stress σxx – strain εxx Relation
We will have the normal stress σxx= Eεxx  
and the shear stress:
 yx x  
M x  y
I
(4.38)
V x 
Q (4.39) with Q to be the shear moment
Ib
The relationships between the bending moment M(x) and Shear force V(x) are expressed in Equation (4.40a) and (4.40b) respectively:
d 2 v x 
d 3v  x 
M  x   EI
and V  x   EI
dx 2
dx 3
FE Formulation of Beam Elements – Cont’d
Principal reference: “A First Course in the FEM” 6th Edition, Cengage Learning by Daryl Logan, 2017
Derive the Element Stiffness Equations
We may express the applied actions to the beam element in applied forces fiy, fjy and moments mi, mj in terms of the
element deflections V(x) represented by the assumed polynomial function in Equation (4. 41) to be:
The beam element:
mi
Deformed shape
Node i
vi
Original shape
vj mj
L
X=0
d 3v  x 
EI
12vi  6Li  12v j  6L j 
f iy  V  EI

3
3
dx x 0 L
d 3v  x 
EI
f jy  V   EI
 3  12vi  6 L i  12v j  6 L j 
3
dx x  L L
Nodal moments:
X
θ(x)
x
fiy
Nodal forces:
Node j
-fjy
X=L
d 2 v x 
EI
2
2
mi   M   EI

6
Lv

4
L


6
Lv

2
L
j
i
i
j
dx 2 x 0 L3


d 2 v x 
EI
m j  M  x   EI

6 Lvi  2 L2 i  6 Lv j  4 L2 j
2
3
dx x  L L


FE Formulation of Beam Elements – Cont’d
Principal reference: “A First Course in the FEM” 6th Edition, Cengage Learning by Daryl Logan, 2017
Derive the Element Stiffness Equations
The above expressions may e express in the following matrix form for the element stiffness equations:
6 L  12 6 L   vi   f iy 
 12
  

2
2 
EI  6 L 4 L  6 L 2 L   i   mi 
  
3


L  12  6 L 12  6 L v j   f jy 

2
2 
6
L
2
L

6
L
4
L

  j   m j 
Stiffness matrix [Ke] x
Nodal unknown
quantities(transverse
displacements & rotations
=
(4.46)
Applied nodal transverse forces
& moments
We thus have the stiffness equation of a beam element to be:
6 L  12 6 L 
 12

2
2 
EI  6 L 4 L  6 L 2 L 
ke   3
L  12  6 L 12  6 L 

2
2 
6
L
2
L

6
L
4
L


(4.47)
FE Formulation of Beam Elements – Cont’d
Principal reference: “A First Course in the FEM” 6th Edition, Cengage Learning by Daryl Logan, 2017
Determine the deflection of a cantilever beam at the half span and at the point under the load. Dimensions and
Example 4
applied lateral force are shown in the figure. The beam is made of a material with Young’s modulus E = 10000 MPa
ℓ/2=0.5 m
Node 1
•
•
x
Node 2
P = 20 N
•
Node 3
We realize the fact that we are seeking solutions at the mid-span and at the point
under the applied force. It id reasonable to discretize the beam into to two (2) elements,
as shown below:
ℓ=1m
Beam Cross-section:
b = 1 cm
h = 2 cm
Section moment of inertia (I) is:


bh 3 10  2 2 10  2
I

12
12
Solution:

3
 0.667 10 8 m 4
EI = (1010)(0.667x10-8) = 66.7
Element 2:
Element 1:
Node 1
v1,, θ1
Node 2
• L = 0.5 m •
v2,θ2
Node 2
v2,, θ2
Node 3
• L = 0.5 m •
v3,θ3
We will first derive the element equations given in Equation (4.46) with element stiffness
matrices shown in Equation (4.47) for both Element 1 and 2.
FE Formulation of Beam Elements – Cont’d
Example 4
For Element 1 with L1 = 0.5 m:
6 L1
 12

2
EI  6 L1 4 L1
1
ke  3
L1  12  6 L1

2
6
L
2
L
1
 1
 
 12
 6 L1
12
 6 L1
6 L1 
3  12 3 
 12
 3

2 L12 
1

3
0
.
5

 533.6 
 12  3 12  3
 6 L1 



4 L12 
3
0
.
5

3
1


(a)
Element equation for Element 1 is:
3  12 3   v1   f1 
 12
 3
    m 
1

3
0
.
5
  1    1 
533.6 
 12  3 12  3 v2   f 2 

   
3
0
.
5

3
1

  2  m2 
in which v1,θ1, v2, θ2 are the respective deflections and rotations in Node 1 and Node 2 respectively, whereas
f1, m1 and f2, m2 are the applied lateral forces and moments at Node 1 and Node 2 respectively
(b)
FE Formulation of Beam Elements – Cont’d
Example 4
Due to the fact that Element 2 has the same length and is made of the same material as Element 1, with the only difference
of the associated nodes, we can express the same element equation for Element 1 for element 2 as shown below:
3  12 3  v2   f 2 
 12
 3
   m 
1

3
0
.
5
  2    2 
533.6 
 12  3 12  3  v3   f 3 

   
3
0
.
5

3
1

  3  m3 
(c)
Assembly of element equations in (b) and (c) for Overall stiffness equations:
Because Node 2 happens to be common node shared by Element 1 and 2, we need to
assemble the element equation following the “MAP” stipulated in Step 5 in Chapter 3:
0
K   K e1  K e2  
Sum
Node 2
0
Applied loads at Node 2 in Element 1 and 2 should be summed up too in the load matrix
FE Formulation of Beam Elements – Cont’d
Example 4
We thus have the assembled overall stiffness equations as:
3
 12
 3
1

 12  3
533.6 
 3 0.5
 0
0

0
 0
 12
3
12  12
 3  3
 12
3
3
0
0   v1   f1 
0.5
0
0  1   m1 
 3  3  12 3  v2   f 2 
    
1  1  3 0.5  2  m2 
3
12  3  v3   f 3 
   
0.5
 3 1   3  m3 
The assembled overall stiffness equation for the beam structure is:
3  12 3
0
0   v1   f1 
 12
 3
    m 
1

3
0
.
5
0
0

 1   1 
 12  3 24
0  12 3  v2   f 2 
533.6 
    
2
 3 0.5  2  m2 
 3 0.5 0
 0
0  12  3 12  3  v3   f 3 

   
0
3 0.5  3 1   3  m3 
 0
We may solve the six unknown responses at the three nodes in the beam structure from Equation (d)
(d)
FE Formulation of Beam Elements – Cont’d
Example 4
We realize that the following boundary and applied loading conditions apply:
v1 and θ1 = 0 for having Node 1 being fixed at the built-in end, and
f1, m1, f2,m2 and m3 = 0. the only non-zero load is the applied force f3 = -20 N at Node 3
We thus have the overall stiffness equation of the beam structure with s[ecified boundary
and loading conditions take the form:
3  12 3
0
0  0   0 
 12
 3
 0   0 
1

3
0
.
5
0
0

  

 12  3 24
0  12 3  v2   0 
533.6 
   


3
0
.
5
0
2

3
0
.
5
0

 2  

 0
0  12  3 12  3  v3   20

  

0
3 0.5  3 1   3   0 
 0
(e)
The four unknown at Node 2 and 3 can be solved from Equation (e) by partitioning the above matrix equations following
Step 6 in Chapter 3.
FE Formulation of Beam Elements – Cont’d
Example 4
3  12 3
0
0  0   0 
 12
 3
 0   0 
1

3
0
.
5
0
0

  

 12  3 24
0  12 3  v2   0 
533.6
   


3
0
.
5
0
2

3
0
.
5
0

 2  

 0
0  12  3 12  3  v3   20

  

0
3 0.5  3 1   3   0 
 0
In the form of:
 K aa
K
 ba
K ab  qa   Ra 
  

K bb  qb   Rb 
where {qa} = specified (known) nodal quantities; {Rb} = specified (known) applied resulting actions, from which
we may obtain:
{qb} = [Kbb]-1 ({Rb} – [Kba]{qa})
(6.12, REF)
In the present case, we have {qa} = {0
0}T
We thus have the 4 unknowns obtained by:
{qb} = [Kbb]-1 {Rb}
FE Formulation of Beam Elements – Cont’d
3  12 3
0
0  0   0 
 12
 3
 0   0 
1

3
0
.
5
0
0

  

 12  3 24
0  12 3  v2   0 
533.6
   


3
0
.
5
0
2

3
0
.
5
0

 2  

 0
0  12  3 12  3  v3   20

  

0
3 0.5  3 1   3   0 
 0
Example 4
{qb} = [Kbb]-1 {Rb}:
 v2 
 
qb    2 
 v3 
 3 
0  12 3 
 24
 0
2
 3 3 

K bb  
 12  3 12  3


 3 0.5  3 1 
 0 
 0 
Rb    
 20
 0 
One will find that:
0.3333
 1
K bb 1  
0.8333

 1
1 0.8333
4
3
3 2.6667
4
4
1
4
4

8
FE Formulation of Beam Elements – Cont’d
Example 4
We may thus solve for the 4 primary unknown quantities in {qb} from the following equations:
Given conditions:
v1 = 0, θ1 = 0
 v2 
0.3333
 

1  1
 2
 
 v3  533.6 0.8333

 3 
 1
1 0.8333
4
3
3 2.6667
4
4
1  0 
0.8333   20 
 0.03123 


 0.1124 
4  0 
1  3   20  





  

4  20 533.6 2.6667   20 
0
.
099995



 4   20  
 0.1499 
8   0 
From which we obtain the following solutions:
Deflections at Node 2, v2 = -0.03123 m = -3.123 cm, at Node 3, v3=-0.09995 m = -9.9995 cm
Slope at Node 2, θ2 = -0.1124 rad, at Node 3 θ3=-0.1499 rad.
Check with solutions from classical beam theory:
P
ℓ=1 m
A
X
•
ℓ/2
x
X=0
y(x)
The induced deflection in the cantilever beam by the applied force P = 20 N is
1 P 3

yx   
x  3 2 x  2 3 
6 EI
One may find the deflection at Point A at x=0 (equivalent to Node 3) = -0.0999 m
and the deflection at Point B at x = ℓ/2 (equivalent to Node 2) =-0.03123m. Both
these values fully agree with the solutions we obtained from the FE analysis.
Part 4
Finite Element Formulation for
Two-Dimensional Stress Analysis of Solids
with Plate Elements
The Need for Stress Analysis of Solids of Plane Geometry
There are many machine components that involve the geometry that prevent engineers using traditional methods to conduct stress
Analysis in their design process. For example, perforated tapered plates are commonly use, with the holes for bolted joints with
Other parts of the machine. The figure on the left below shows a tapered perforated plate, and the perforated plate in the right
also involve the plate with two curve notches.
Conventional wisdom shows that serious “stress concentration” can occur in the vicinity of a structure with drastic geometry changes.
The parts near the holes and the tapered areas and the notches in both plates are vulnerable for stress concentrations.
We have mentioned that the problem of plate with a small hole may be solved by the theory of “advanced strength of materials.”
However, plates with tapered and notched edges as shown above can not be handled by any existing method from classical
theories of elasticity, and the FEM appears the only available method of the solution.
Examples of FE Stress Analysis of Solid Structures
Two-Dimensional Stress Analysis of Solids with Plate Elements – Cont’d
This type of structures are typically thin in their thickness in comparison to the bulk volume of the overall structure.
As such only three (3) out of total six (6) independent stress components need to be considered in the analysis,
as illustrated below:
Plane structure subject
to in-plane forces:
Induced stresses:
ο
x
Note: Shear stresses exist on the thickness or “edges.”
y

The in-plane displacement components in the solid: U ( x, y )  
u ( x, y ) 

 v  x, y  
The stress components in the solid: {σ(x,y)}T = {σxx(x,y
σyy(x,y) σxy(x,y)}T
The strain components in the solid: {ε(x,y)}T = {εxx(x,y) εyy(x,y) εxy(x,y)}T
u(x,y) = the component along the x-coordinate
V(x,y) = the component along the y-coordinate
Examples of FE Stress Analysis of Solid Structures
Two-Dimensional Stress Analysis of Solids with Plate Elements – Cont’d
The strain – displacement relation in Equation (4.3) is modified to the new form:

 
u  x, y 

 

x

  x
v x, y 
 
 x, y   
0

y

 





u
x
,
y

v
x
,
y

 

 y
x   y


0

  u ( x, y )


y   v x, y 


x 
(4.48)
and the stress-strain relation in Equation (4.5) becomes:

 
1

0
 xx


 ( x, y )  E 2  1 0   yy 
1  
1    
0
0

  xy 
2 

(4.49)
Formulation FE analysis for Two-Dimensional Stress Analysis of Solids with
Plate Elements
Let us now formulating the FE for plate structures such as with the discretization in a tapered plate as illustrated
In the figure below:
We begin our FE formulation of the plate structure with the expression of the “element displacement” components
{U(x,y)} in terms of the corresponding “nodal displacements” using an interpolation function {N(x,y)} as follows:
u  x, y 
  N  x, y u


v
x
,
y


U x, y   
 ui 
u 
Element displacements
Interpolation
Nodal displacements
 j
function

u m 



u

 
where the nodal displacement components,
 vi 
v j 
 

 vm 

(4.)
(4.)
Formulation FE analysis for Two-Dimensional Stress Analysis of Solids with Plate Elements – Cont’d
Derivation of interpolation function {N(x,y)}
We assume the elements used in this FE analysis are the “simplex” elements, meaning that the
Element displacements follow linear polynomial functions in relating their nodal displacements:
and
We will thus have:
u x, y   1   2 x   3 y
along the x-coordinate
vx, y    4   5 x   6 y
along the y-coordinate
u x, y 

vx, y 
U x, y   
1 x y 0 0 0 
1  2  3  4  5  6 T


0 0 0 1 x y 
or in an alternative matrix form:
(4.28)
U x, y   Rx, y 
where α1, α2, α3, α4, α5, α6 in the matrix {α}T are constants to be determined with specified nodal coordinates later.
The matrix [R(x,y)] in Equation (4.28) has the form:
Rx, y   
1 x y 0 0 0

0 0 0 1 x y 
(4.29)
Formulation FE analysis for Two-Dimensional Stress Analysis of Solids with Plate Elements – Cont’d
Derivation of interpolation function {N(x,y)} - cont’d
With the specified nodal coordinates:
ui = α1 + α2xi + α3yi,
uj = α1 + α2xj + α3yj, um = α1 + α2xm + α3ym, and
vi = α4 + α5xi + α6yi,
vj = α4 + α5xj + α6yj, vm = α4 + α5xm + α6ym
We are able to expand Equation (4.28) into the following form for the nodal displacements:
 ui  1 xi
 u  1 x
j
 j 
um  1 xm
u     
 vi  0 0
 v j  0 0
  
vm  0 0
Displacement components
of 3 nodes in the element
yi
0
0
yj
0
0
ym
0
0
0
1
xi
0
1
xj
0
1 xm
0  1 
0   2 
0   3 
 
yi   4 
y j   5 
 
ym   6 
(4.30)
Constant coefficients
Specified coordinates of
the 3 nodes in the element
Formulation FE analysis for Two- Dimensional Stress Analysis of Solids with Plate Elements – Cont’d
Derivation of interpolation function {N(x,y)} - cont’d
Equation (4.30) may be expressed in a different form of:
{u} = [A]{α}
(4.)
in which the matrix [A] has the form:
1 xi
1 x
j

1 x
A   m
0 0
0 0

0 0
yi
0
0
yj
0
0
ym
0
0
0
1
xi
0
1
xj
0
1 xm
0
0 
0

yi 
yj 

ym 
(4.)
We may obtain the solution of the unknown coefficient matrix in Equation (4.31) as:
   A1u  hu
where the matrix:
[h] = [A]-1
with [A]-1 = the inverse of the nodal coordinate matrix [A] in Equation (4.32)
(4.)
(4.)
Formulation FE analysis for Two- Dimensional Stress Analysis of Solids with Plate Elements – Cont’d
Derivation of interpolation function {N(x,y)} - cont’d
We recall Equation (4.28) with:
u x, y 
U x, y   



v
x
,
y


1 x y 0 0 0 
1  2  3  4  5  6 T


0 0 0 1 x y 
Also with Equation (4.29) with:
Rx, y   
1 x y 0 0 0

0 0 0 1 x y 
(4.28)
(4.29)
We will obtain the following expression after substituting the matric {α} in Equation (4.33) into Equation (4.28), yielding:
U x, y   Rx, y hu
Displacement components
in Element ijm
(4.35)
Displacement components
of 3 nodes in the element
By comparing Equation (4.35) with Equation (4.26), we have the interpolation function of this simplex element to be:
[N(x,y)] = [R(x,y)][h]
with Matrices [R(x,y)] in Equation (4.29) and [h] in Equation (4.34)
(4.36)
Formulation FE analysis for Two- Dimensional Stress Analysis of Solids with Plate Elements – Cont’d
The interpolation function {N(x,y)} - cont’d
We may refer to the interpolation function {N(x,y)} that we derived for the 3-node
 ui  plate elements in
Chapter 3 for a comparable case with the current analysis with the form of:  
u  x, y 
  N  x, y u  N i  x, y  N j  x, y 


v
x
,
y


U x, y   
An alternative
expression:
where
u  x, y 
 Ni






N
x
,
y
u


0


v
x
,
y



U x, y   

1
x j ym  xm y j  y j  ym x  xm  x j y
A
N j  x, y  
1
xm yi  xi ym   ym  yi x  xi  xm y 
A

1
j xi y j  x j yi   yi  y j x  x j  xi y
A
0
Ni
Nj
0
Nj
0
Nm
0
0
Nj
0
Ni
Nm
0
0
Nj
 ui 
u 
 j
0  um 
  (4. b)
N m   vi 
v j 
 
vm 

N i  x, y  
N m  x, y  
 vi 
 u   N i
N m  x, y  j   
v j   0
um 
 
vm 
 ui 
v 
 i
0   u j 
 
N m   v j 
um 
 
vm 

(3.2)
A  xi y j  x j yi   x j ym  xm y j   xm yi  xi ym   the area of the element made of triangle (ijm)
(3.2a)
(4. a)
Formulation FE analysis for Two- Dimensional Stress Analysis of Solids with Plate Elements – Cont’d
Derivation of Element equation
Because the element equation is derived by minimizing the Potential energy in gthe deformed solid, we need
to derive the expression of “strain energy” in terms of nodal displacements (the primary quantities in the analysis).
We will first express the element strain vs. nodal displacements by Equation (4.12) as:
(4.12)
{ϵ(x.y)} = [B]{u}
(4.13)
[B(x,y,z)] = [D][N(x,y,z)]
where
By using Equation (4.4) for [D] and Equation (4.36b) for [N(x,y)}, we get the matrix [B(x,y)] in the following:
 y j  y m 
B  1  0
2A
 xm  x j 

 ym  yi  yi  y j 
0
xi  xm 
0
x j  xi 
0
( xm  x j )
 y j  ym 
0
xi  xm 
 ym  yi 
0 
x j  xi 
yi  y j 
(4.37)
The potential energy in the deformed element , according to Equation (4.18) is:
Pu 
1
T
T




C Budv
u
B
2 v
  u N  x, y   f dv   u N  x, y  tds
T
v
T
T
s
T
(4.38)
Formulation FE analysis for Two- Dimensional Stress Analysis of Solids with Plate Elements – Cont’d
Derivation of Element equation
The element equation is obtain by minimizing the potential energy in Equation (4.38) as:
Pu
0
u
Leading to the following element equation:
Ke u  p
where
K e   Element stiffness matrix   BT C Bdv
(4.39)
(4.40)
v
and the nodal force matrix
p 
Nodal forcwe matrix 
T






tds




N
x
,
y
f
dv

N
x
,
y
v
s
T
(4.41)
The integration in Equation (4.40) with respect to the volume of the element may turn out to be tedious. However, if the size of
the element is not too large, this integration may be approximated by the following expression without significant error:
[Ke] ≈ [B]T[C][B] (wA)
(4.42)
in which w is the thickness of the plane element, and A is the plane area that can be computed by Equation (3.2a).
A  xi y j  x j yi   x j ym  xm y j   xm yi  xi ym   the area of the element made of triangle (ijm)
Numerical example on FE analysis for Two-Dimensional Stress Analysis of Solids with a Plate Element
Example (Example 7.2 from the cited reference):
A structure made of a triangular plate defined by three corners at A,B and C. A force P is applied at corner C as shownin
the figure.
Find the following:
(a)
(b)
(c)
(d)
The displacement of the plate at corner C
The displacement in the plate
The stresses and strains in the plate, and
The reactions at the two fixed corners
Solution:
We assume that only ONE element is used for the analysis.
This triangular plate element has three nodes I, j and m located
as shown at the right:
The coordinates of the 3 nodes are:
{xi = 6, yi=0), (xj=6, yj=0) and (xm=3, ym=4)
Numerical example on FE analysis for Two-Dimensional Stress Analysis of Solids with a Plate Element – Cont’d
(a) To determine the nodal displacements:
We will first obtain the [A] matrix using Equation (4.32) with the specified nodal coordinates:
and with Equation (4.34) to obtain the [h] matrix:
And then using Equation (4.36) to obtain the interpolation function in the form of Equation (4.36b) as:
N x, y   

1  0.167 x  0.125 y  0.167 x  0.125 y 
0
0
0.25 y
0
0
1  0.167 x  0.125 y 
0
0 
0.167 x  0.125 y  0.25 y 
Numerical example on FE analysis for Two-Dimensional Stress Analysis of Solids with a Plate Element – Cont’d
We are now ready to derive the element matrix for the structure with the newly derived interpolation
function. We will obtain first the [B] matrix from using Equation (4.37) as:
0 0
 4 4 0 0
B  1  0 0 0  3  3 6
24
  3  3 6  4 4 0
and the [C] matrix FROM Equation (4.7) and the coefficient matrix of Equation (4.25):
0 
 1 0.3
C   10.99 106 0.3 1 0 
 0
0 0.35
We are now ready to determine the element stiffness matrix in Equation (4.39) from Equations (4.40) and (4.42) as:
19.15  12.85  6.3 7.8  0.6  7.2

19.15  6.3 0.6  7.8 7.2 


12.6  8.4 8.4
0 
K e   228958.32

SYM
14
.
6
3
.
4

18



14.6  18 


36 

Numerical example on FE analysis for Two-Dimensional Stress Analysis of Solids with a Plate Element – Cont’d
Constructing the “element equation:”
We realize the following boundary and applied force conditions:
ui, vi, vj = 0 (fixed ends), and the applied nodal forces:
pix = piy = pjx = pjy = 0, and pmx = p cos 30o = 866 lb, and pmy = p sin 30o = -500 lb
The “element equation” according to Equation (4.39) becomes:
19.15  12.85  6.3 7.8  0.6  7.2 ui  0   pix  0 

 u   p  0 
19
.
15

6
.
3
0
.
6

7
.
8
7
.
2
jx


 j  

12.6  8.4 8.4
0   um   pmx  866 
228958.32 



p

0
v

0
SYM
14
.
6
3
.
4

18
iy

 i
 


14.6  18  v j  0  p jy  0 


 

p


500


v
36


  m   my

Because there is only one element in the structure, the above element equation is also the “overall stiffness
equation” of the structure, from which we may solve the “displacement components of ALL nodes after
Making necessary interchange of rows and columns in the above equation (according to the rule stipulatedin
Step 5 in Chapter 3.
Numerical example on FE analysis for Two-Dimensional Stress Analysis of Solids with a Plate Element – Cont’d
Thus, after the necessary interchanges of rows and columns, we reached the partition of the element equation
In the form of Equation (6.12, Ref) as:
7.8  0.6  12.85
 19.15
 7.8
14.6 3.4
0.6

  0.6
3.4 14.6
 7.8
228958.32 
 12.85 0.6  7.8 19.15
  6.3  8.4 8.4
 6.3

 18  18
7.2
  7.2
 6.3  7.2  0   0 
 8.4  18   0   0 
8.4  18   0   0 
   

 6.3 7.2   u j   0 
12.6
0  um   866 
  




v
0
36   m   500
The 3 nonzero nodal displacements can be solve from the above portioned overall stiffness equation by the
following simultaneous equations:
19.15  6.3 7.2  u j   0 
  

228958.32   6.3 12.6 0  um    866 
 7.2
0
36  vm   500
The above simultaneous equations may be solved by either matrix inversion method or Gaussian elimination method, with:
uj = 0.16x10-3 inch, um = 0.38x10-3 inch and vm = -0.093x10-3 inch
Numerical example on FE analysis for Two-Dimensional Stress Analysis of Solids with a Plate Element – Cont’d
(b) The displacements in the element:
We may use the interpolation function {N(x,y)} to determine the displacement components everywhere in the element:
For the displacement in the x-direction:
●P
u(x,y) = (1 –- 0.167x – 0.125y)ui + (0.167x -0.125y)uj + 0.25 um = 0 +(0.167x-0.125y)x0.16x10-3+0.25x0.38x10-3y
V(x,y) = (1 –- 0.167x – 0.125y)vi + (0.167x -0.125y)vj + 0.25 vm = -0.09264x0.25x10-3y
For instance the displacements at Point P(3,2) have the values of:
u(3,2) = 0 +(0.167x3-0.125x2)x0.16x10-3+0.25x0.38x10-3 x2 = 0.23x10-3 inch
v(3,2) = -0.09264x0.25x10-3x2 = -.04632x10-3 inch
(c) The strain components in the element by using Equation (4.12) with {ε} = [B]{u}:
 ui  0 
 u  0.16 

 xx  x, y 
0 0  j
 4 4 0 0
 26.65 


u  0.38 


3
6
 x, y    yy x, y   1  0 0 0  3  3 6  m
 10   23.16 10
  x, y  24   3  3 6  4 4 0  vi  0 
 75 

 v  0 


 xy

j


vm  0.09
Numerical example on FE analysis for Two-Dimensional Stress Analysis of Solids with a Plate Element – Cont’d
The stresses in the element may be obtained by using the generalized Hooke’s Law in Equation (4.25):

 
1

0
 xx
0   26.65 
 1 0.3
 216.5 


 ( x, y )  E 2  1 0   yy   10.99 106 0.3 1 0   23.16 106   166.67 psi
1  
1    
 288.66 
 0
0 0.35  75 


0 0
  xy 
2 

(d) The reactions at all nodes:
One may derive the following expression for the nodal forces: R   B   dv  B T  WA with W  thickness of the plte, and A  plane area
T
v
 Rix 
  4 0  3
  866 
R 
4

 0 
0

3
jx
 

  216.5 


 Rmx  1  0
 866 
0
6 

R     
  166.67  12 1  

R
0

3

4

327
.
3
24
 iy 

  288.66 




 R jy 
 0 3 4 
 827.3 
 




  500 
6
0 
 Rmy 
 0
The reactions at Node i therefore have numerical values at: Rix = 866 lb towards left, and Riy=327.3 lb in the downward direction
Numerical example on FE analysis for Two-Dimensional Stress Analysis of Solids with a Plate Element – Cont’d
Important lesson learned from this numerical example:
We noticed that the stresses and strains in the element (and thus the triangular
plate structure re CONSTANT:
 xx  x, y   26.65 
 x, y    yy x, y    23.16 106
  x, y   75 

 xy
 
and
 xx   216.5 
 ( x, y )   yy    166.67 psi
   288.66 

 xy  
– meaning there is no variation of stresses and strains throughout the entire structure. This is obviously not realistic!!
The reason for what has happened in this (and the other)numerical example is because we used “linear polynomial” in deriving
the interpolation function – resulting in using “SIMPLEX” element in the FE analysis. “Simplex elements” offers “simple
mathematical formulation in FEA, but results in constant stresses and strains in elements.
That was the reason why engineers need to place many more (smaller) elements in the area with conceivable high gradients of
primary unknown quantities, such as in the following cases:
This is good lesson for any intelligent FE user to learn and exercise
Part 5
Finite Element Formulation of
Stress Analysis of Axisymmetric Solids Structures
References: “A First Course in the Finite Element Method,” 5th and 6th edition, Daryl L. Logan,
Cengage Learning, 2012
Structures with axisymmetric geometry are common in practice ; Any solid with its geometry symmetric to an axis
- Called the “axis of geometry-usually designated by the z-axis” is classified as “axisymmetric solid.”
Examples: Wheels, Cylinders of constant or variable radius along the z-axis, such as pressure vessels,
Coordinate systems (r, ϴ,z) :
Z-Axis of symmetry
r
z
Classifications of Axisymmetric solids :
Wheels and thin circular plates:
2-D axisymmetric solids with 3 components of stress and strains
as defined in the case of thin plates
Cylinders with constant or variable radius: 3-D axisymmetric solids with four (4) stress components:
σrr (r,z) = Stress in the radial direction,
σθθ(r,z) = Stress in the tangential θ-direction (or “hoop direction)
σrz(r,z) = Stress in z-direction, and
σrz (r,z) = Shearing stress on the surface of
the solid element that is
σzr=σrz
perpendicular to the r-coordinate
but in the z-direction
σrz
Formulas Derived from Theory of Elasticity
Displacements:
u(r,θ,z) = displacement component in r-direction
w(r,θ,z) = displacement component in z-direction
Strains:
 rr r , z  
   r , z  
u r , z 
r
u r , z 
r
Normal strain in along r-coordinate
(4.43a)
Normal strain along θ-coordinate
(4.43b)
wr , z 
Normal strain along z-coordinate
z
u r , z  wr , z 
Shearing strain
 rz r , z  

z
r
 zz r , z  
(4.43c)
(4.43d)
Formulas Derived from Theory of Elasticity - Cont’d
Stress-Strain Relation – derived from Generalized Hooke’s law


1 
 rr 
 
 
1 

E
 zz 

 
 

1 




1


1

2

  

0
0
 rz 
 0
0 
0 
0 
1  2 
2 
(4.44)
Finite Element Formulation of Solid Structures of Axisymmetric Geometry
Step 1 Select Element Type:
Our FE formulation will be based on a typical triangular element.
The element has three nodes with two degree-of-freedom per node,
e.g., ui and wi at node i, etc. The stresses and strains in the element are
shown in Equations (4.43) and (4.44)
Axis of symmetry
Derivation of Stiffness Matrix – cont’d
Step 2 Select Element Displacement Functions:
We will stick to simplex elements. So, the displacements in the element will be of the forms
of simple linear polynomial functions as:
Displacement in r-direction:
u r , z   1   2 r   3 z
(4.45a)
Displacement in z-direction:
wr , z    4   5 r   6 z
(4.45b)
 ui 
The six nodal displacements are expressed as follow:
w 
i
 d i   

  u j 
Axis of symmetry
Nodal displacements: d    d j    
d   w j 
 m  u 
m
Displacements in 3 nodes
 
wm 
For example: ui  u ri , zi   1   2 ri   3 zi
wi  wri , zi    4   5 ri   6 zi
at Node i
(4.46)
Displacement components in 3 nodes
Derivation of Stiffness Matrix – cont’d
Step 3 Derive Interpolation Function:
The general displacement in the triangular torus element can thus be expressed as:
1 
 
 2
u
   r   3 z  1 r z 0 0 0  3 
       1 2

  



r


z
w
0
0
0
1
r
z
   4
5
6 

 4 
 5 
 
Substituting the nodal coordinates into Equation (4.47) will result in:
 6 
1
 i  j  m   ui 
1  1 ri zi   ui 
  
   1     u 
 2   1 rj z j   u j  
j
m  j 
 i
2
A
  1 r z  u 
  i  j  m  um 
m
m  m
 3 


1
 i  j  m   wi 
 4  1 ri zi   wi 
  
   1      w 
 5   1 rj z j   w j  
j
m  j 
 i
2
A
  1 r z  w 
  i  j  m  wm 
m
m  m
 6 


where
 i  rj z m  z j rm  j  rm zi  z m ri  m  ri z j  zi rj
 i  z j  zm
 j  z m  zi
 m  zi  z j
 i  rm  rj
 j  ri  rm
 m  rj  ri
(4.49)
(4.47)
(4.48a)
(4.48b)
Derivation of Stiffness Matrix – cont’d
Step 3 Derive Interpolation Function – cont’d:
The interpolation function is:
 Ni
N r , z   
0
where
Ni 
1
 i   i r   i z 
2A
Nj 
0
Ni
Nj
0
0
Nj
1
 j   j r   j z 
2A
with the cross-sectional area A obtained from:
0 
N m 
Nm
0
Nm 
1 ri
2 A  1 rj
1 rm
1
 m   m r   m z 
2A
z1
zj
zm
Element and Nodal displacements relation:
 u r , z   N i
   



w
r
,
z

 0
0
Ni
Nj
0
0
Nj
(4.50)
Nm
0
(4.51)
(4.52)
 ui 
w 
 i
0   u j 
 
N m   w j 
 um 
 
wm 
(4.53)
Derivation of Stiffness Matrix – cont’d
Step 4 Define the Strain-displacement and Stress-Strain Relationship:
Equations (4.13) and (4.43) are used to derive the strain-nodal displacements relation.
The strain components:
2

 rr  

  

6

    zz   1
3z 




2
    r

r
  rz     

3
5


Using the relationships derived in Equations (4.48a,b), we may express the element strains and nodal displacements
in the following expression:
 ui 
i
0
j
0
m
0  w 

 rr 
 i


 
0
i
0
j
0
 m  u 
 zz  1 
 j
       i


z

 

z


z
j
j
i
m
m
(4.54)

2
A



0



0



0
i
j
m
  

 w j 
r
r
r
r
r
  rz 
r 
 u m 





i
i
j
j
m
m 



w
 m 
 i  rj z m  z j rm  j  rm zi  z m ri  m  ri z j  zi rj
with
 i  z j  zm
 j  z m  zi
 m  zi  z j
(4.49)
 i  rm  rj
 j  ri  rm
 m  rj  ri
Derivation of Stiffness Matrix – cont’d
Step 4 Define the Strain-displacement and Stress-Strain Relationship:
i

 rr 

 
0
1 
 zz 
       i
iz
   2 A    i 
r

r
  rz 


i

0
j
0
m
i
0
j
0
0
i
j
r
j 
j
 jz
r
0
j
m
r
 m 
m
 mz
r
 ui 
0  w 
 i
 m  u 
 j
 
0  w j 
 m  um 

wm 

(4.54)
Equation (4.54) will be expressed in another way to conform Equation (4.37) for the derivation of the element stiffness equation:
 rr 
 
    zz   Bi 
  

  rz 

B 
j
 ui 
w 
 i
u 
Bm  j 
wj 
 um 
 

wm 

(4.55)
i


0
1 
where Bi  

z
2 A  i  i  i
r
r

i
0
 i 
0

 i 
(4.56)
The other components of {B} matrix in Equation (4.55) may be obtained by
Substituting j for i in Equation (4.56) for [Bj] and m in Equation (4.56) for [Bm]
We may thus express the element strains in terms of nodal displacements in the following expression:
(4.57)
{ε} = [B]{d}
in which nodal displacement matrix {d} is expressed in Equation (4.46). One needs to notice that the [B] matrices
are function of (r,z), which makes the non-constant strains in torus element with linear element displacement functions
Derivation of Stiffness Matrix – cont’d
Step 4 Define the Stress-Strain and Stress-nodal displacement Relationships:
We have relate the element stresses and element strains by the generalized Hooke’s law for multi-axially loaded solids:
{σ} = [C]{ε}
(4.60)
Yet the element strains and nodal displacements are related in Equation (4.57) for torus elements:
{ε} = [B]{d}
(4.57)
We can thus relate the element stresses and nodal displacements to be:
{σ} = [B][C] {d}
where
B  Bi  B j  Bm 
with
i


0
1 
Bi    i
z
2A
 i  i
r
r

i
(4.58)
0
 i 
0

 i 
etc. shown in Equation (4.56)
Derivation of Stiffness Matrix – cont’d
Step 4 Define the Stress-Strain and Stress-nodal displacement Relationships – Cont’d:
The elasticity matrix [C] for torus element can be derived from Generalized Hooke’s Law in Equation (4.7):


1 
 
1 

E

C  

1 
1  1  2   
0
0
 0
0 
0 
0 
1  2 
2 
(4.59)
Step 5 The Element stiffness matrix
Following Equation (4.39), we have the element equation for torus element to be:
[Ke]{d} = {q}
(4.60)
vB C Bdv  2 A B C Brdrdz 
T
where
[Ke] = element stiffness matrix =
T
(4.61)
The following approximate method may be used to compute the [Ke] matrix by letting:
rr 
ri  rj  rm
3
and
zz
zi  z j  z m
3
, and thus use
B  B   Br , z 
 
We may thus compute the [Ke] matrix by the expression: ke   2r A B C B 
in computing [Ke} in Equation (4.61)
(4.62)
Derivation of Element Stiffness Equation
By following the general formulations of element stiffness equation:
[Ke]{φ} = {q}
(4.21)
with [Ke] expressed in Equation (4.22), {φ} = nodal displacements and {q} = applied nodal forces expressed in Equation (4.23)
For triangular torus elements, the element stiffness equation takes the form:
[Ke]{d} = {q}
(4.63)
where {d} =nodal displacements and {q} = applied nodal forces with mathematical expressions derived
from Equation (4.23) to be:
q  vN r , z T  f b dv  sN r , z   f s ds
T
(4.64)
in which the interpolation function [N(r,z)] is expressed in Equations (4.50) and (4.51). The two types of forces in
Equation (4.64) with {fb} = body forces, and {fs} = surface tractions appear frequently in axisymmetric solid
structures, as will be elaborated with mathematical expressions in the following.
Derivation of Element Stiffness Equation
Body forces {fb} in axisymmetric structures
Body forces {f} in finite element formulations means the forces that are distributed in the solid structures. Weight is one form of
body force. The centrifugal forces generated by the spinning mass, such as spinning wheels (i.e., flywheels), disks or cylinder (rotors)
are common place in structures of axisymmetric geometry. We will formulate {f} of the latter types of body forces as follows: Rb
The mathematical expression of the body forces {fb} in Equation (4.64) takes the form:
 Rb 
rdrdz
Z b 
 f b   2 A N r , z T 
(4.65)
where Rb = ω2ρr with ω = angular velocity of the spinning mass and ρ = mass density of the material.
Zb = body force per unit volume of the material along the z-direction
Body force at Node i:
 f bi   2 A N i 
T
z
N i T
where
Zb
Rb
N
 i
0
 Rb 
 rdrdz
Z b 
0
N i 
(4.67)
Alternately, with approximation if the triangular
torus element is not too large, we may use:
r
with
R   2 r
b
(4.66)
 Rb 
 
Z b 
2 r A  Rb 
 fb  
 
3 Z b 
 Rb 
 
Z b 
(4.67)
Derivation of Element Stiffness Equation
Surface traction forces {fs} in axisymmetric structures
Surface traction means the distributed forces acing on the surface of the solid structure, such as pressure loadings.
For structures of axisymmetric geometry, such as pressure vessels of cylindrical geometry there could be pressure loadings
Acting on the inner or outside surface of the pressure vessels. So, we will take a close look at this kind of load and also
Derive mathematical expression for the surface tractions acting at nodes of torus elements.
The surface traction (or forces) can be expressed as:
 f s   sN s T T ds
(4.68)
where [Ns]=the interpolation function evacuated along the surface where the surface traction acts
The radial and axial pressure pr and pz may thus expressed as:
 f s   sN s 
T
 pr 
 ds
 pz 
(4.69)
Surface forces along the surface j-m with r = rj:
z
pz
For Node j, substituting Nj from Equation (4.51) into Equation (4.69), leads to:
   
0
f sj  zzjm 1  j 0j j        ppr 2 rj dz
2A
j
j
j  z 
m
pr
j
i
r
0
0
 
evaluated at r=rj, z=z
2 rj z m  z j   pr 


f

  (4.70)
s
The total distribution of surface force to node I,j and m is:
2
 pz 
 pr 
 
 p z 
Finite Element Formulation of Solid Structures of Axisymmetric Geometry
Step 6 Assemble the element equations to Overall Stiffness Equations
Step 7 Solution of primary unknown (nodal displacements) from overall stiffness equations
Step 8 Solve secondary unknown quantities of strains and stresses in all elements of the discretized model
Example
Stress analysis of a pressure ring
Material: Aluminum with following properties:
1.5”
Pi = 5 psi
2” dia
4” dia
3”
Young’s modulus E =70,000 x 106 psi
Poisson’s ratio ν = 0.3
Finite element model with one element:
Nodal coordinates:
ri = 1
rj = 2
zi = 0
zj = 1.5
Z
rm = 1
zm = 3
Cross-sectional area of the element from Equation (4.52):
1 ri
2 A  1 rj
1 rm
zi 1 1 0
z j  1 2 1.5  3
zm 1 1 3
from which the cross-sectional area A = 1.5 in2
m (1,3)
Pi = 5 psi
i
1”
2”
J (2,1.5)
(1,0)
r
Example
Stress analysis of a pressure ring – cont’d
Derive the Interpolation function:
Components of the interpolation function from Equation (4.49):
 i  rj z m  z j rm  2  3  1.5 1  4.5
 i  z j  z m  1`.5  3  1.5
 i  rm  rj  1  2  1
 j  rm zi  z m ri  1 0  3 1  3
 j  z m  zi  3  0  3
 j  ri  rm  1  1  0
 m  zi  z j  0  1.5  1.5
 m  ri z j  zi rj  11.5  0  2  1.5
 m  rj  ri  2  1  1
The interpolation function:
 Ni
N r , z   
0
0
Ni
Nj
0
0
Nj
Nm
0
0 
N m 
(4.50)
with the components Ni, Nj and Nm expressed in Equation (4.51) to be:
Ni 
1
4.5  1.5r  z 
3
Nj 
1
 3  3r 
3
Nm 
1
1.5  1.5r  z ) 
3
(a)
Example
Stress analysis of a pressure ring – cont’d
Element strain and nodal displacements:
i

 rr 

 
0
 zz  1 
       i
iz

2
A



i
  

r
r
  rz 

i

 1.5


0
1
 4.5
z
 1.5 
3
r
 r
1

Equation (4.54):
0
j
0
m
i
0
j
0
0
i
0
1
j
r
j 
j
3
0
 jz
r
0
j
0
0
m
r
 m 
m
 1.5
0
3
1.5
z
 3 0 0
 1.5 
r
r
r
 1.5
0
3
1
0
 mz
r
 ui 
0  w 
 i

 m  u 
 j
 
0  w j 
 m  um 
wm 
 ui 
0   wi 
 

1   u j 
 
0  w j 

 1.5  um 
 
wm 
Example
Stress analysis of a pressure ring – cont’d
Elasticity matrix:
1 
 
E

C  
1  1  2   
 0
2.33

10  1
 4.03846 10
 1

 0



0 
1 
0  700000 106

1 
0   1.3  0.4
1  2 
0
0
2 
1
1
0 
2.33 1
0 
1
2.33
0 

0
0
0.67 
0.7 0.3 0.3 0 
0.3 0.7 0.3 0 

.
0.3 0.3 0.7 0 


0
0 0.2 
0
(4.59)
Example
Stress analysis of a pressure ring – cont’d
Element stiffness matrix [Ke]:
K e   vB C Bdv  2 A B C Brdrdz 
T
T
(4.61)
We approximate [Ke] in Equation (4.61) by using the average r and z in lieu of integration.
These average coordinates
r
ri  rj  rm
3
The [B] matrix, with {[Bi]
r and z are defined as:

[Bj]
zi  z j  z m 0  1.5  3
1 2 1
 1.3333 inches, and z 

 1.5 inches
3
3
3
[Bm]} with [Bi], [Bj] and [Bm] expressed in Equation (4.56) will now be replace by:
B r , z   B r , z  B r , z  B r , z 
i
where
0 
 1.5

 1 
1 0
Bi r , z   3 1.25 0 


  1  1.5
j
 3

1 0
B j r , z  
3 0.75

 1


m
0
0
0

3
(b)
0 
 1.5

1 
1 0
Bm r , z   3  0.75 0 


1

1
.
5


Example
Stress analysis of a pressure ring – cont’d
Element stiffness matrix [Ke]-cont’d:
B r , z  matrix in the form:
We may express the
The transpose of the
B r , z 
T
B r , z 
 1.5
0 

 1 
1  0
B r , z   3 1.25 0 

  1  1.5
 3
 0

0.75

 1
0
0
0

3
0 
 1.5
 0

1

 
 0.75
0 


 1.5 
 1
(c)
Is:
1 
1  1.5 0 1.25  1  3 0 0.75  1 1.5 0  0.75
 

1 0
 1.5 0 0
0
3   0 1
0
 1.5 
3  0
(d)
By substituting Equation (d), (4.59) and (c) into (4.61), we will have the element stiffness matrix in the following form:
K e   2 1.33331.5B r , z  C B r , z 
T
(e)
Example
Element stiffness matrix [Ke]-cont’d:
[Bi]T
[Bj]T
 1.5 0 1.25  1  3 0

 0 0
0

1
0

1
.
5

 

1
0   1.5
0   3
2.33 1
 1
  0
  0
2
.
33
1
0

1


 
 
 1
1
2.33 0  1.25
0  0.75

 
 
0
0
0
.
67

1

1
.
5

 
  1
50.72 1010
K e  
3
[C]
[Bi]
[Bm]T
0.75  1 1.5 0  0.75
1 

0
3   0 1
0
 1.5 
0
0
0

3
[Bj]
0 
 1.5
 0

1

 
 0.75
0 


1

1
.
5


[Bm]
from which we may express the stiffness matrices relating to the three nodes i, j and m:
 
 
 
   
50 1010
Bi T C  Bi for Node i
K 
3
50 1010
T
j
Ke 
B j C  B j for Node j
3
50 1010
m
Bm T C Bm  for Node m
Ke 
3
i
e
 
(f)
Example
Element stiffness matrix [Ke]-cont’d:
We thus have the following stiffness matrices relating to the 3 nodes for the present example to be:
 
1.255 
50 1010
T
10 5.0803
Bi  C  Bi  16.9110 
K 

1
.
255
3
.
8375
3


 
i
e
 
 27.45  2.01
C  B j  16.9110 
for Node j

 2.01 6.03 
 0.255
50 1010
T
10  4.9731
Bm  C Bm   16.9110 

for Node m

3
 0.255 3.8375 
   
50 1010
K 
Bj
3
j
e
K 
m
e
for Node i
T
(g1)
10
(g2)
(g3)
Example
Element stiffness equation
 
 K ei
We may express the element stiffness equation in the form:  0

 0

0
K ej
0
 
By substituting the 3 sub-stiffness matrices in Equations (g1), (g2) and (g3)
into Equation (h), we will have the element equation in the following form:
 ui   qri 
w  q 
0   i   zi 
  u j   qrj 
0     
w
q zj 
m  j 

Ke 
 um  qrm 
   
wm  q zm 
 
0
0
0
0   ui   qri 
5.0803 1.255
 1.255 3.8375
 w   q 
0
0
0
0

  i   zi 
 0
0
27.45  2.01
0
0   u j   qrj 
10
16.9110 
    
0
 2.01 6.03
0
0   w j   q zj 
 0
 0
0
0
0
4.9731  0.255  um  qrm 

   
0
0
0
 0.255 3.8375  wm  q zm 
 0
(h)
(j)
The nodal forces qri, qzi, qrj, qzj, qrm and qzm in Equations (h) and (j)are the APPLIED FORCES at Node i, j and m respectively.
Example
Overall stiffness equation
The Element stiffness equation in Equation (h) is used as the Overall stiffness equation for the structure because the structure has
only one element. We may thus use this equation to compute the six displacements at all 3 nodes.
We observe from the loading to the structure to be the pressure Pi applied to the inner surface in the radial direction. The uniform
pressure applied to the inner surface may be converted to the concentrated forces acting at Nodes i and m by the following formula:
m
z
z
qrm
m
Pi=5 psi
0
i
r
0
qri
i
r
The equivalent nodal forces are:
qri  qrm 
2 ri z m  zi 
Pi   ri z m  zi Pi  3.1413  0 5  47.1 lb f
2
Example
Solution of nodal displacements from overall stiffness equations
We are now ready to solve the displacement components at each of the three nodes in the element by inputting the
applied forces at the nodes, as shown in the following equations [K]{d} ={q} with numerical values:
0
0
0
0   ui   qri  47.1 
5.0803 1.255
 1.255 3.8375
 w   q  0 
0
0
0
0


  i   zi
 0
0
27.45  2.01
0
0   u j   qrj  0 
10
16.9110 
   

0
 2.01 6.03
0
0   w j   q zj  0 
 0
 0
0
0
0
4.9731  0.255  um  qrm  47.1

  

0
0
0
 0.255 3.8375  wm   q zm  0 
 0
[K]
{d}
{q}
The nodal displacement {d} in Equation (k) may be solved by matrix inversion:
d  
1
1
q


K
16.911010
(m)
(k)
Example
Solution of nodal displacements from overall stiffness equations
The inverse of [K] matrix in Equation (k) is:
0
0
0
0 
0.2145  0.07
  0.07 0.2835

0
0
0
0


 0
0
0.03734 0.01245
0
0 
1
K   

0
0
0
.
01245
0
.
17
0
0


 0
0
0
0
0.2018 0.01341


0
0
0
0
0
.
01341
0
.
2615


We thus have the 6 nodal displacements from Equation (m) to be:
 ui 
w 
 i
 u j 
1
d     
10
w
16
.
91

10
 j
 um 
 
wm 
10.0892 
 59.66 
  3.297 
 19.5




 0 
 0 
12 

  10 
 inches
0
0




 9.5048 
 56.33 




0.06316
 3.74 
Example
Strains in the structure
The element, and thus structure strains may be obtained from Equation (4.54) with r  r  1.3333", z  z  1.5" and 2 A  3 in 2
 1.5

 rr 

 
0
 zz  1 
       4.5
z

5
.
1


3
  
r
 r
  rz 
1

0
1
3
0
0
0
 1.5
0
z
1.5
3
 1.5 
 3 0 0
r
r
r
1
3
0
 1.5
0
 ui 
0   wi 
 

1   u j 
 
0  w j 

 1.5  um 
 
wm 
 59.66 
0   19.5  48
0  1.5
3
0
 1.5

1   0  7.75 12
0
0
0
1
10 12  0

10




0  0   29 
0.75 0 0.75
0
3 0.75


5
.
1

1
3
0
5
.
1

1

 56.33  6.77 



 3.74 
Example
Stresses in the structure
Stresses in the structure {σ} = [C]{ε} from Equation (4.60):
1
0    58 
 rr 
2.33 1
 
 1
 7.75
2
.
33
1
0
 zz 
 12
 
      4.03846 1010  
 10
 1
1
2.33
0   29 
  


 rz 
0
0
0
0
.
67

 6.77 
 295.2   11.92 

  1.3265


 2   32.85 
 4.03846 10 

 psi
 51.95   2.097 
 13.6   0.55 




We thus have the induced stresses in the wall of this pressure ring to be:
σrr = -11.92 psi, σzz = -1.3265 psi, σθθ= 2.097 psi and σrz = 0.55 psi


SUMMARY
1. This chapter begins with a brief overview of key formulations of linear theory of elasticity relevant to the
finite element (FE) formulation for stress analysis of deformable elastic solid structures subjected to
applied forces, either in the form of body force or as surface tractions.
2. The deformation of the solids is supposed to be small so that the induced stresses are within the elastic
limit of the material
3. The FE formulation is based on the theories of linear elasticity
4. General FE formulation is presented for “simplex elements” only. (simplex elements is defined with
element displacements follow linear polynomial functions)
5. FE formulations on 1-D bar elements, 2-D triangular plate elements and 3-D triangular torus elements are
presented for the following specific types of structures:
(a) Solid bars subject to forces and deformation along the length of the bar
(b) Trusses in 2-D plane with planar displacements at nodes but force and stresses along the length of the bar
(c) Beam bending in planes
(d) Thin plates with in-plane loading and deformation
(e) Axisymmetric structures
SUMMARY – cont’d
6. Key FE equations are:
a) Element displacements and nodal displacement:
Element displacements   N r   with   Nodal displacements
The interpolation function N r  are available in :
Equation (1.7) for 1-D bar and truss elements; Equations (4.42) to (4.44) for beam elements;
Equation (4.36) for triangular plate elements; Equations (4.50) and (4.51) for triangular torus elements
b) Element strains and element displacements:    D  with D  in Equation (4.4)
c) Element strains and nodal displacements:    D N r    B  
d) Element stresses and element strains and nodal displacements:
e) Strain energy in element:
U
   C    C B 
T
1
1
1 T
T
T








 







dv


B
C
B

dv

2 v
2 v
2
 B C Bdv
T
v
f) Work done to the element by applied body forces {f} and surface tractions {t}:
W   N r   f dv   N r  tds
T
v
g) Potential energy in elements:
T
s
    U   W   U  W
(4.41)
(4.38)
SUMMARY – cont’d
h) Element equation with minimization of potential energy:
  
 0 leading to :
 
j) Element equations:
[Ke]{φ} = {q}
Element stiffness matrix: [Ke] = element stiffness matrix =
Applied nodal forces:
q  vN r T  f dv  sN r  tds
T
k) Overall Stiffness equations: [K]{φ} = {Q}
with Overall stiffness matrix: K    K eM 
n
M 1
M
and overall applied loading matrix: Q   q
1
vB C Bdv
T