1. Our task is to estimate the axial displacement u at any section x
1-D ELASTICITY
Our first practical application of FE will be an elastic rod subjected to
a distributed axial load F(x) along its length. The rod may also have a
variable stiffness.
x dx
EA(x)
F(x)
Our task is to estimate the axial displacement u at any section x
We need a differential equation to describe this system, together with
boundary conditions.
Once we have these, we can then attempt to solve the problem by FE.

2. Consider a thin strip of the rod of length
Area A
1) EQUILIBRIUM
2) CONSTITUTIVE
3) STRAIN/
DISPLACEMENT
“small strain”

3. DIFFERENTIAL EQUATION
These three equation can be simplified as follows:
GOVERNING
EQUATION
Now we will attempt to solve the equation by FE

4. with one spatial dimension.
The 1-d Rod Element
This is the simplest 1-d finite element suitable for solving differential equations
with one spatial dimension.
Original problem 1 2 3 4 5
Element number
Node number
Finite element
discretization x L u1 u2
A typical 1-d element
Let the nodal displacements be u1 and u2.
Assume a linear variation of
from one end to the other.

5. Consider the following TRIAL SOLUTION across the element:1
where c1 and c2 are the “undetermined parameters”, and
y1 (x) = 1 and y2 (x) = x are the “trial functions”
The Trial Solution must satisfy the boundary conditions corresponding to
the displacement at each node of the element, thus: 2 3
Eliminate c1 and c2 from equations and to give: 2 1 3 or
where
“Shape Functions” and
“Undetermined parameters”

6. FINITE ELEMENT SOLUTION
Governing differential Equation
Trial solution
Matrix version
Substitute the Trial Solution into the governing equation
Residual

7. Galerkin weighting functions
Weight the Residual using Galerkin’s method
Assume EA and F are constant over each element. They can vary from one
element to the next of course!
Galerkin weighting functions
After integration by parts and the elimination of “boundary terms”, this can be written as,
or in matrix form as

8. This is the “ELEMENT STIFFNESS RELATIONSHIP”
where
Element stiffness matrix
Element nodal displacements
Element nodal forces
The Galerkin Finite Element Method has turned the
DIFFERENTIAL EQUATION
into the MATRIX EQUATION

9. Note how the distributed force on the element has been interpreted
as two forces “lumped” at the nodes: L F
Uniform distributed load
FL/2
Equivalent nodal loads
This always happens in FE analysis
Another example:
F(x)=Px/L
Linear distributed load L P
PL/6
PL/3
Equivalent nodal loads
So far, everything has been performed at the “element level”.
Now we must discuss attaching elements together and ASSEMBLY

10. ANOTHER TYPE OF ELEMENT MATRIX: THE MASS MATRIX
Consider a rod vibrating along its axis
Area A
1) EQUILIBRIUM
2) CONSTITUTIVE
3) STRAIN/
DISPLACEMENT
“small strain”

11. DIFFERENTIAL EQUATION
These three equation can be simplified as follows:
WAVE
EQUATION
Now we will attempt to solve the equation by FE

12. FINITE ELEMENT SOLUTION
Governing differential equation
Trial solution
Substitute the Trial Solution into the governing equation
Residual

13. Weight the Residual using Galerkin’s method
Assume EA and rA are constant over each element.
Galerkin weighting functions
This term leads to the stiffness matrix
as we saw previously
This term can be written as:
or in matrix form as

14. This is the “ELEMENT DYNAMIC EQUATION OF MOTION”
where
Element stiffness matrix
Element mass matrix
Element nodal accelerations
Element nodal displacements
The Galerkin Finite Element Method has turned the
DIFFERENTIAL EQUATION
into the MATRIX EQUATION

15. DIFFERENT FORMS OF THE MASS MATRIX
“Consistent Mass Matrix”
an alternative form is:
“Lumped Mass Matrix”
Lumped mass is sometimes preferred on account of its simplicity.
When assembled into [M], this form is easily inverted and can be
stored very efficiently.

16. ASSEMBLY
Once the element stiffness matrices
have been found, we can assemble
them into a “global stiffness matrix”
The element force vectors
must also be assembled in a “global
force vector”
“Assembly” collects the contribution from each ELEMENT matrix to form the
GLOBAL matrix for the whole finite element mesh.
“Assembly” involves mapping the freedom numbering at the element level
onto the freedom numbering at the global level.
“Assembly” is easily automated in computer codes provided the “connectivity”
information is available.
Connectivity information involves knowledge of:
Which node belongs to which element
(a typical node will be attached to several elements)
2) Which freedom belongs to which node
(a typical node will have more than one “degree of freedom”
e.g. u, v, w, p, etc….)

17. degrees-of-freedom at each node.
Assembly of a string of 1-d rod element with one degee-of-freedom per node
is very simple and can be performed by hand, however the logic of assembly
is exactly the same as it would be for a 3-d analysis involving multiple
degrees-of-freedom at each node.
Example:
Element numbering
Global freedom numbering
Node numbering 1 1 2 2 3 3 4 4 1 2 3 4 5
Which node belongs to which element?
This information is held in the g_num array
g_num=
Element 1
Element 2
etc.
Which freedom belongs to which node?
This information is held in the nf array
nf=
Node 1
etc.
Node 2

18. Local element freedom numbering1 2 3 4 x L 1 2
ith element
Local element freedom numbering x EA
assumed EA distribution (step)
actual EA distribution
Each element has a different stiffness matrix:
e.g. for the ith element:

19. The assembled global stiffness relationship will be of the form:
Store half the “band”
In general:
Note the symmetry and banding of
This has major implications for storage-saving strategies.
This system of linear equations can be solved using any appropriate method
e.g. Gaussian elimination, preconditioned conjugate gradient.

20. Once the global displacements have been found, strains and stresses
POSTPROCESSING
The method described so far is sometimes referred to as the “displacement finite element method” since the fundamental dependent variable is the nodal displacement.
Once the global displacements have been found, strains and stresses
can be retrieved by performing a final scan through all the elements.
For example in the 1-d rod problem, 1 2 3 4
in element 3,
hence

21. It is a “do nothing” boundary condition.
BOUNDARY CONDITIONS
A solution to a differential equations needs boundary conditions. For example
in the rod analysis: L x
The
boundary condition was handled explicitly via the data, however
the
boundary condition was obtained by default at the mesh boundary.
The derivative boundary condition
where
is the dependent variable
and
is normal to the mesh boundary, is obtained automatically in finite element
analysis.
It is a “do nothing” boundary condition.

22. FIXED VALUES OF THE DEPENDENT VARIABLE
A nodal value can be fixed using the “stiff spring” or “penalty” method.
First the global stiffness matrix [K] is assembled in the conventional way
treating all nodal variables as unknowns.
Assume we have 5 equations:
We wish to fix the following displacements:

23. STIFF SPRING METHOD
Add a “big” number to the diagonal(s) corresponding to the freedoms to be
fixed (in this example K22 and K55).
2) Modify the RHS vector (in this example F2 and F5) such that the forces
corresponding to the freedoms to be fixed are set equal to the augmented
diagonal term multiplied by the prescribed value.
modified stiffness
modified rhs
modified stiffness
modified rhs
Large numbers “swamp” the smaller ones giving the desired result:

24. holds the locations of the diagonals
EFFICIENT STORAGE OF SYMMETRIC BANDED MATRICES
In nearly all finite element analyses, the global matrices are banded,
And symmetrical.
The most efficient storage strategy is called the “skyline” method.
Consider a system of 6 equations:
Extra vector kdiag=[ ]
holds the locations of the diagonals

25. EA(x)=300 F(x)=5 WORKED EXAMPLE 2.01 2 3 4
Each element has the same EA=300 and L=0.5, hence all elements have the
same stiffness matrix given by:
Loading is uniform, so each element has the same force vector given by:

26. After assembly of [K] and {F} we have:
Solution by Gaussian elimination gives:

27. Alternative approach in which all freedoms are numbered (whether we know them or not).
2.0
EA(x)=300
F(x)=5 2 3 4 5 1
Assemble all five equations to give:

28. Computer programs would use the “stiff spring” approach to fix U1=0
For hand calculation purposes, its easier to enter the known value(s) and remove the row(s)
in [K] and {F} corresponding to the fixed freedoms, thus:
Initial
assembled
System:
Same equations and
solution as before
with different
freedom numbering.
Modified
system:
Rearrange the
remaining 4
equations:

29. We rarely have the luxury of an analytical option in FE analysis……..
Analytical check
We rarely have the luxury of an analytical option in FE analysis……..
…..but in this case we do.
Governing equation:
Integrate once:
hence
Integrate again:
hence x
u (analytical)
u(FE)
0.0
0.5
0.0146
1.0
0.025
1.5
0.0313
2.0
0.033
FE solution is
exact at the nodes
but not between
the nodes

30. Postprocessing Element 1: Element 2: Element 3: Element 4: 8.75 6.25
3.75
Element 3:
1.25
Element 4:

31. Next, if the applied loading involved the use of “equivalent nodal loads”
they must be subtracted from the “actions” for each element, e.g.
8.75
“actions”
Element 1: -
1.25
“equivalent nodal loads” =
10.0
7.50
“actual end forces”
Similarly for elements 2,3 and 4
Element 2:
7.50
5.00
“actual end forces”
Element 3:
5.00
2.50
“actual end forces”
Element 4:
2.50
0.00
“actual end forces”

32. Note internal and external equilibrium satisfied.
….FINALLY…
Reaction forces:
10.0
7.5 5
Element 1
0.5 5
5.0
Element 2
2.5 5
Element 3
0.0 5
Element 4
Note internal and external equilibrium satisfied.

33. Program 4.1 One dimensional analysis of axially loaded elastic rods
using 2-node rod elements.
Structure Chart for Program 4.1

34. PROGRAM p41 !
! Program 4.1 One dimensional analysis of axially loaded elastic rods
! using 2-node rod elements.
USE main; IMPLICIT NONE
INTEGER,PARAMETER::iwp=SELECTED_REAL_KIND(15)
INTEGER::fixed_freedoms,i,iel,k,loaded_nodes,ndof=2,nels,neq,nod=2, &
nodof=1,nn,nprops=1,np_types,nr
REAL(iwp)::penalty=1.0e20_iwp,zero=0.0_iwp
! dynamic arrays
INTEGER,ALLOCATABLE::etype(:),g(:),g_g(:,:),kdiag(:),nf(:,:),no(:), &
node(:),num(:)
REAL(iwp),ALLOCATABLE::action(:),eld(:),ell(:),km(:,:),kv(:),loads(:), &
prop(:,:),value(:)
! input and initialisation
OPEN(10,FILE='fe95.dat'); OPEN(11,FILE='fe95.res')
READ(10,*)nels,np_types; nn=nels+1
ALLOCATE(g(ndof),num(nod),nf(nodof,nn),etype(nels),ell(nels),eld(ndof), &
km(ndof,ndof),action(ndof),g_g(ndof,nels),prop(nprops,np_types))
READ(10,*)prop; etype=1; IF(np_types>1)READ(10,*)etype; READ(10,*)ell
nf=1; READ(10,*)nr,(k,nf(:,k),i=1,nr); CALL formnf(nf); neq=MAXVAL(nf)
ALLOCATE(kdiag(neq),loads(0:neq)); kdiag=0
! loop the elements to find global arrays sizes-----
elements_1: DO iel=1,nels
num=(/iel,iel+1/); CALL num_to_g(num,nf,g); g_g(:,iel)=g;
CALL fkdiag(kdiag,g)
END DO elements_1
DO i=2,neq; kdiag(i)=kdiag(i)+kdiag(i-1); END DO; ALLOCATE(kv(kdiag(neq)))
WRITE(11,'(2(A,I5))') &
" There are",neq," equations and the skyline storage is",kdiag(neq)

35. !-----------------------global stiffness matrix assembly------------------
kv=zero
elements_2: DO iel=1,nels
CALL rod_km(km,prop(1,etype(iel)),ell(iel)); g=g_g(:,iel)
CALL fsparv(kv,km,g,kdiag)
END DO elements_2
! read loads and/or displacements
loads=zero; READ(10,*)loaded_nodes,(k,loads(nf(:,k)),i=1,loaded_nodes)
READ(10,*)fixed_freedoms
IF(fixed_freedoms/=0)THEN
ALLOCATE(node(fixed_freedoms),no(fixed_freedoms),value(fixed_freedoms))
READ(10,*)(node(i),value(i),i=1,fixed_freedoms)
DO i=1,fixed_freedoms; no(i)=nf(1,node(i)); END DO
kv(kdiag(no))=kv(kdiag(no))+penalty; loads(no)=kv(kdiag(no))*value
END IF
! equation solution
CALL sparin(kv,kdiag); CALL spabac(kv,loads,kdiag); loads(0)=zero
WRITE(11,'(/A)')" Node Disp"
DO k=1,nn; WRITE(11,'(I5,2E12.4)')k,loads(nf(:,k)); END DO
! retrieve element end actions
WRITE(11,'(/A)')" Element Actions"
elements_3: DO iel=1,nels
CALL rod_km(km,prop(1,etype(iel)),ell(iel)); g=g_g(:,iel); eld=loads(g)
action=MATMUL(km,eld); WRITE(11,'(I5,2E12.4)')iel,action
END DO elements_3
STOP
END PROGRAM p41

36. 4 1 1
5 0 5
Data file

37. There are 4 equations and the skyline storage is 7
Node Disp
E-04
E-04
E-04
E-04
E+00
Element Actions
E E+00
E E+01
E E+01
E E+01

38. 4 2 1
1 0
Data file

39. There are 4 equations and the skyline storage is 7
Node Disp
E+00
E-01
E-01
E-01
E-01
Element Actions
E E+02
E E+02
E E+02
E E+02