Introduction
In this chapter, we will describe some modeling guidelines, including generally recommended mesh size, natural subdivisions modeling around concentrated loads, and
more on use of symmetry and associated boundary conditions. This is followed by dis-cussion of equilibrium, compatibility, and convergence of solution. We will then consider interpretation of stress results.
Next; we introduce the concept of static condensation, which enables us to apply
the concept of the basic constant-strain triangle stiffness matrix to a quadrilateral element. Thus, both three-sided and four-sided two-dimensional elements can be used in
the finite element models of actual bodies.
We then show some computer program results. A computer program facilitates
the solution of complex, large-number-of-degrees-of-freedom plane stress/plane strain
problems that generally cannot be solved longhand because of the larger number of
equations involved. Also, problems for which longhand solutions do not exist (such
as those involving complex geometries and complex loads or where unrealistic, often
gross, assumptions were previously made to simplify the problem to allow it to be
described via a classical differential equation approach) can now be solved with a
higher degree of confidence in the results by using the finite element approach {with
its resulting system of algebraic equations}.

.A.

7.1 Finite Element Modeling
We will now discuss various concepts that should be considered when modeling any
problem for solution by the finite element method.

7.1 Finite Element Modeling

...

351

General Considerations

Finite element modeling is partly an art guided by visualizing physical interactions
taking place within the body. One appears (0 acquire good modeHng techniques
through experience and by working with experienced people. General-purpose programs provide some guidelines for specific types of problems f12, I5}. In subsequent
parts of this section, some significant concepts that should be considered are described.
In modeling, the user is first confronted with the sometimes difficult task of
understanding the physical behavior taking place and understanding the physical behavior of the various elements available for use. Choosing the proper type of element
or elements to match as closely as possible the physical behavior of the problem is
one of the numerous decisions that must be made by the user. Understanding the
boundary conditions imposed on the problem can, at times, be a difficult task. Also,
it is often difficult to determine the kinds of loads that must be applied to a body
and their magnitudes and locations. Again, working with more experienced users
and searching the literature can belp overcome these difficulties.

Aspect Ratio and Element Shapes
The aspect ratio is defined as the rglio of the longest dimension to the shortest dimension
of a quadrilateral element. In many cases) as the aspect ratio increases, the inaccuracy
of the solution increases. To illustrate this point, Figure 7-1(a) shows five different finite element models used to analyze a beam subjected to bending. The element used
here is the rectangular one described "in Section lO.2. Figure 7-1 (b) is a plot of the
resulting error in the displacement at point A of the beam versus the aspect ratio.
Table 7-1 reports a comparison of results for the displacements at points A and B
for the five models, and the exact solution [2}.
There are exceptions for which asPect ratios approaching 50 still produce satisfactory results; for example, if the stress gradient is close to zero at some location of the
actual problem, then large aspect ratios at that location still produce reasonable results.
In general, an element yields best results if its shape is compact and regular. Although different elements have different sensitivities to shape distortions, try to maintain (1) aspect ratios low as in Figure 7-1, cases 1 and 2, and (2) corner angles
of quadrilaterals near 90 Figure 7-2 shows elements with poor shapes that tend
to promote poor results. If few of these poor element shapes exist in a model, then
usually only results near these elements are poor. ]n the Algor program [12J, when
IX ~ 170° in Figure 7-2(c), the program automatically divides the quadrilateral into
two triangles.
D

•

Use of Symmetry

The appropriate use of syrnmetry* will often expedite the modeling of a problem. Use
of symmetry allows us to consider a reduced problem instead of the actual problem.
'" Again, reflective s}mmetry means correspondence in size, shape, and position of loads: material propenies;
and boundary conditions that are on opposite sides of a dividing line or plane.

352

~

7 Practical Considerations in Modeling; Interpreting Results
Y,v

/l---------------....l

4O,OOO-lb

A

T

8 in. ' A - - - -.. X.

~:: shear

U

l

Parabolic

1·. . .- - - - 4 8 i n · - - - - - - l - I

load distribution

E = 30 x Ilf psi
v =0.3
t

12

.

In.

= 1.0 in.
I.

x '2 In.

6 in. x I in.
elements (typical)

elements (typical)

(5) AR=24

48'
• In.

n

X

(4) AR=6

II.
'3 In.-

3 in. x 2 in .

elements (typical)

h

(3) AR =3.6

elements (typical)

(2) AR = 1.5

2.4 in. x

2~ in.

elements (tYPical)
(1) AR= 1.1

Figure 7-1 (a) Beam with loading; effects of the aspect ratio (AR) ilhJstrated by five
cases with different aspect ratios

Thus, we can use a finer subdivision of elements with less labor and computer costs.
For another discussion on the use of symmetry, see Reference [3].
Figures 7-3-7-:-5 illustrate the use of symmetry in modeling (1) a soil mass subjected to foundation loading, (2) a uniaxially loaded member with a fillet, and (3) a
plate with a.hole subjected to internal pressure. Note that at the plane of symmetry
the displacement in the direction perpendicular to the plane must be equal to zero.
This is modeled by the rollers at nodes 2-6 in Figure 7-3, where the plane of symmetry is the vertical plane passing through nodes 1-6, perpendicular to the plane of the
modeL In Figures 7-4(a) and 7-5(a), there are two planes of symmetry. Thus, we need
model only one-fourth of the actual members, as shown in Figure!:! 7-4(b) and 7-5(b).

7.1 Finite Element Modeling

Exact solution
----------

0

- 5

~

(I)

Q,

353

---

:Ie

(2)

-10

-=
'0

.

(3)

-15

~
::::>

-20

C
It)

e

-25

8eo

(4)

~ -30
::0
-35

.5

g
u

-40

E -4$

~

-SO

l.

(5)

-55
2

6'

4

8

10

Asped ratio

12

14

16

18

20

22

24

lonsest dimension
shortest dimension

Figure 7-1 (b) Inaccuracy of solution as a function of the aspect ratio (numbers in
parentheses correspond to the cases listed in Table 7-1)

Table 7-1

Case
1

2
3
4
5

Comparison of results for various aspect ratios

Aspect
Ratio
t.l
1.5
3.6
6.0
24.0

Exact solution [21

Number of Number of
Elements
Nodes
84

85
77
81
85

60
64
60
64
64

Vertical
DispJacement,
v (in.)
Point A Point B

Percent
Error in
Displacement
atA

-1.093
-1.078
-1.014
-0.886
-0.500

-0.346
-0.339

5.2
6.4

-0.328
-0.280
-0.158

23.0
56.0

-1.152

-0.360

11.9

Therefore, rollers are used at nodes along both the vertical·and horizontal planes of '
symmetIy.
As previously indicated in Chapter 3, in vibration and buckling problems, symmetry must be used with caution since symmetry in geometry does not imply symmetry in an vibration or buckling modes.

354

"

7 Practical Considerations in Modeling; Interpreting Results

'---__~I

h

,b»

h

b
(b) Approaching a triangular shape

p~

a C / a»{3

large and very small
comer angles

(c) Very

Figure 7-2

(d) Triangular quadrilateral

Elements with poor shapes

. -112 in. b---IO Ib/in.
6

'

7

24

36

42

4g

54

60

66

19'

31

37

43

'49

55

'61

I.~.-~---------%iD'----------~11

~--- Axis of sylll.lIletry

Figure 7-3 Use of symmetry applied to a soil mass subjected to foundation loading
(number of nodes 66, number of elements = SO) (254 cm 1 in.. 4.445 N = lib)

=

=

Natural Subdivisions at Discontinuities
Figure 7-6 illustrates various natural subdivisions for finite element discretization.
For instance, nodes are required at locations of concentrated loads or discontinuity
in loads, as shown in Figure 7-6(a) and (b). Nodal lines are defined by abrupt changes

of plate thickness, as in Figure 7-6(c). and by abrupt changes of material properties,
as in Figure 7-6(d) and (e). Other natural subdivisions occur at re~trant comers, as
in Figure 7-6(0. and along holes in members, as in Figure 7-5.

7.1 Finite Element Modeling

...

355

lA=m~
4 in.

EL~L

---"'i1. . .

- - 4 in.

I

..I.

4 in.

__---",;.I~.S~~~~~~~~~f3
1--3in.-J

200 Ib/in.

(a) Plane stress uniaxially loaded member wich fille'

(b) Enlarged finite element IllOdeI of the cross-batched quarter of the: member

(number of nodes = 78. number of elements

60) (2.54 em '" 1 in.)

Figure 7-4 Use of symmetry applied to a uniaxially loaded mer,nber with a fillet

Sizing of Efements and the hand

p Methods of Refinement

For structural problems, to obtain displacements, rotations, stresses, 'and strains,
many computer programs include two basic solution methods. (These same methods
apply to nonstructural problems as well.) These are called the h method and the
p method. These methods are then used to revise or refine a finite element mesh to improve the results in the next refined analysis. The goal of the analyst is to refine the
mesh to obtain the necessary accuracy by using only as many degrees of freedom as
final objective of this so called adaptive refinement is to obtain equal
necessary.
distribution of an error indicator over all elements.
The discretization depends on the geometry of the structure, the loading pattern,~ .
and the boundary conditions. For instance, regions of stress concentration or high
stress gradient due to fillets, holes, or re-entrant corners require a finer mesh near
those regions, as indicated in Figures 7-4, 7-5, and 7-6{f).
We will briefly describe the hand p methods of refinement and provide references for those interested in more in-depth understanding of these methods.

The

h Method of Refinement In the h method of refinement, we use the particular element
based on the shape functions for that element (for example, linear functions for the
bar, quadratic for the beam, bilinear for the CST). We then start with a baseline
mesh to provide a baseline solution for error estimation and to provide guidance for

356

...

7 Practical Considerations in Modeling; Interpreting Results

(a) Plate with hole under plane stress

y

_ _ _ Axis of symmetry

Irlc+-+....~r---.. x
(b) Finite element model of one-quart.er of the plate

Figure 7-5 Problem reduction using. axes of symmetry applied to a plate with a hole
subjected to tensile force

mesh revision. We then add elements of the same kind to refine or make smaller elements
in the model. Sometimes a uniform refinement is done where the original element size
(Figure 7-7a) is perhaps divided in two in both directions as shown in Figure 7-7b.
More often, the refinement is a nonuniform h refinement as shown in Figure 7-7c
(perhaps even a local refinement used to capture some physical phenomenon, such as
. a shock wave or a thin boundary layer in fluids) [19]. The mesh refinement is continued until the results from one mesh compare closely to those of the previously refined
mesh. It is also possible that part of the mesh can be enlarged instead of refined.
F or in~tance) in regions where the stresses do not change or change slowly1 larger

7.1 Finite Element Modeling

(a) Concentrated load

A.

(b) Abrupt change of

distributed load

.......... Nodal
line

III f.

't2

Material

CD

Material

(3)

Nodal
line

1'"

\ Node

(c) Abrupt change of
pJate thickness

(d) Abrupt change of

material properties

'P2

P

A

(e) Basic model of an implant (cross-hatched) in bone, located
at various depths X beneath the bony surface., using
rectangular elements,

(t)' Re-entrant comer, B
P

w

(g) Structure with a distributed load

Figure 7-6

Natural subdivisions at discontinuities

(h) Using dements to distribute the loading
and spread the concentrated load

357

358

...

7 Practical Considerations in Modeling; Interpreting Results

elements may be quite acceptable. The h-type mesh refinement strategy had its beginnings in !20-23). Many commercial computer codes, such 'as [12], are based on the h
refinement.

p Method of Refinement In the p method of refinement [24-28}, the polynomial pis
increased from perhaps quadratic to a higher-order polynomial based on the degree
of accuracy specified by the user. In the p method of refinement, the p method adjusts
the order of the polynomial or the p level to better fit the conditions of the problem,
such as the boundary conditions, the loading, and the geometry changes. A problem
is solved at a given p level, and then the order of the polynomial is nonnaI1y increased
while the element geometry remains the same and the problem is solved again. The
results of the iterations are compared to some set of convergence criteria specified by
the user. Higher-order polynomials nonnally yield better solutions. This iteration process is done automatically within the computer program. Therefore, the user does not
p

F
~

~

~
~
~

r

F

~

F

(a) Original mesh

(b) A uniformly refined mesh
F

F

%
%
%
~~

~
F

p

%
/

(c) A possible nonuniform h refinement

Figure 7-7

Examples of hand p refinement

(d) A possible uniform p refinement

7.1 Finite Element Modeling

Figure 7-7

A

359

(Continued)

need to manually change the size of elements by creating a finer mesh, as must 1;>e
done in the h method. (Tbe h refinement can be automated using a remeshing algorithm within the finite element software.) Depending on the problem, a coarSe mesh
will often yield acceptable results. An extensive discussion of error indicators and esti·
mates is given in the literature [l9}.
The p. refinement may consist of adding degrees of freedom to existing nodes,
adding nodes on existing boundaries between elements} andlor adding internal degrees
of freedom. A uniform p refinement (same refinement performed on all elements) is
shown in ·Figure 7-7d. One of the more common commercial computer programs,
Pro/MECHANICA 129], uses the p method exclusively. A typicar discretized finite element model ofa pulley using Pro/MECHANICA is shown in Figure 7-7e.

Transition Triangles
Figure 7-4 illustrates the use of triangular elements for transitions from smaller quadrilaterals to larger quadrilaterals. This transition is necessary because for simple CST
elements, intermediate nodes along element edges are inconsistent with the energy

360

..

7 Practical Considerations in Modeling; Interpreting Results

fonnulation of the CST equations. If intennediate nodes were used, no assurance of
compatibility would be possible~ and resulting, holes could occur in the deformed
model. Using higher-order elements, such as the linear-strain triangle described in
Chapter 8, allows us to use intennediate nodes along element edges and maintain
compatibility.

Concentrated or Point Loads and Infinite Stress
Concentrated or point loads can be applied to nodes of an element provided the ete·
ment supports the degree of freedom associated with the load. For instance, truss
elements and two- and three-dimensional elements support only translational degrees
of freedom, and therefore concentrated nodal moments cannot be applied to these
elements; only concentrated forces can be applied. However, we should realize that
physically concentrated forces are usually an idealization' and mathematical conve·
nience that represent a distributed load of high intensity acting over a small area.
According to classical linear theories of elasticity for beams, plates, and solid
bodies [2, 16, 17}, at a point loaded by a concentrated normal force there is finite displacement and stress in a beam, 'finite displacement but infinite stress in a plate, and
both infinite displacement and stress in a two- or three-dimensional solid body.
the consequences of the differing assumptions about the stress fields
These results
in standard linear theories of beams, plates, and solid elastic bodie~. A truly concen~
trated force would cause material under the load to yield, and linear elastic theories
do not predict yielding.
In a finite element' analysis, when a concentrated force is applied to a node of a
finite element model, infinite displacement and stress are never computed. A concen·
trated force on a plane stress or strain model has a number of equivalent distributed
loadings, which would not be expected to produce infinite displacements or infinite
stresses. Infinite displacements and stresses can be approached only as the mesh
around the load is highly refined. The best we can hope for is that we can highly refine
the mesh in the vicinity of 'the concentrated load as shown in Figure 7-6(a), with the
understanding that the deformations and stresses will be approximate around the
load) or that these stresses near the concentrated force are not the object of study,
while stresses near another point away from the force) such as B in Figure 7-6(f),
are of concern. The preceding remarks about concentrated forces apply to concentrated reactions as well.
Finally, another way to model with a concentrated forte is to use additional ele'ments and a single concentrated load as shown in Figures 7-6(h). The ~hape of the
distribution used to simulate a distributed load can be controlled by the relative stiffness of the elements above the loading plane to the actual structure by changing the
modulus of elasticity of these elements. This method spreads the concentrated load
over a number of elements of the actual structure.
Infinite stress based on elasticity solutions may also exist for special geometries
and loadings, such as the re...entrant comer shown in Figure 7-6(f). The stress is predicted to be infinite at the re...entranf corner. Hence, the finite element method based
on linear elastic material models will never yield convergence (no matter how many
times you refine the mesh) to a correct stress level at the re-entrant corner 118J.

are

7.1 Finite Element Modeling

...

361

We must either change the sharp re-entrant corner to one with a radius or use a theory
that accounts for plastic or yielding behavior in the material.
Infinite Medium

Figure 7-3 shows a typical model used to represent an infinite medium (a soil mass
subjected to a foundation load). The guideline for the finite element model is that
enough material must be included such that the displacements at nodes and stresses
within the elements become negl~gibly small at ioeations far from the foundation
load. Just how much of the mediUm should be modeled can be determined by a trialand-error procedure in which the horizontal and vertical distances from the load are
varied and the resulting effects on the displacements and stresses are observed. Alternatively, the experiences of other investigators working on similar problems may
prove helpful. For a homogeneous soil mass) experience has shown that the iniluence
of the footing becomes insignificant if the horizontal distance of the model is taken
as approximately four to six times the width of the footing and the vertical distance
is taken as approximately four to ten times the width of the footing [4-6}. Also, the
use of infinite elem~ts is described in Reference [13].
After choosing the ~orizontal and vertical dimensions of the model, we must
idealize the boundary conditions. "Usually, the horizontal displacement becomes negligible far from the load, and we restrain the horizontal movement of all the nodal
Points on that boundary (the right-side boundary in Figure 7-3): Hence, rollers are
used to restrain the horizontal motion along the right side. The bottoIl! boundary
can be completely fixed, as is modeled in Figure 7-3 by using pin suppo\:ts at each
nodal point along the bottom edge. Alternatively, the bottom can be cbnstrained
only against vertical movement. The choice depends on the soil condltions ~t the bot~
tom of the model. Usually, complete fixity is assumed if the lower boundary is taken
as bedrock.
In Figure 7-3, the left-side vertical boundary is taken to be directly under the
center of the load because symmetry has been assumed. As-we said before when discussing symmetry) all nodal points along the line of symmetry are restrained against
horizontal displacement.
Finally} Reference [111 is recommended for additional discussion regarding guidelines in modeling with different element types, such as beams, plane stress/plane strain,
and three-dimensional solids.
Connecting (Mixing) Different Kinds of Elements

Sometimes it becomes necessary in a model to mix different kinds of elements, Sl,lch as
beams and plane elements, such as CSTs. The problem with mixing these elements is
that they have different degrees of freedom at each node. The beam allows for transverse displacement and rotation at each -node, while the plane element only has"inplane displacements at each node. The beam can resist a concentrated moment at a
node, whereas a plane element (CST) cannot. Therefore~ if a beam element is CODnected to a plane element at a single node as shown in Figure 7-8(a), the result will
be a hinge connection at A. This means only a force can be transmitted through the

362

J..

7 PraCtical Considerations in Modeling; Interpreting Results

Plane elements

Plane elements

(a)

(b)

Figure 7-8 Connecting beam element to plane elements (a) No moment is
transferred, (b) moment is transferr,ed

node between the two kinds of elements. This also creates a mechanism, as shown by
the stiffness matrix being singular. This problem.can be corrected by extending the
beam into the plane element by adding one or more beam elements, .shown as AB,
for one beam element in Figure 7-8(b}. Moment can now be transferred through the
beam to the plane element. This extension assures that translational degrees of freedom of beam and plane element are connected at nodes A and B. Nodal rotations
are associated with only the beam element, AB. The calculated stresses in the plane element will not nonnally be accurate near node A.
For more examples of connecting different kinds of elements see Figures 1-5,
11-10, 12-10 and 16-31. These figures show examples of beam. ~nd plate elements
connected together (Figures 1-5, 12-10, and '16-31) and solid (brick) elements connected to plates (Figure 11-10).
Checking the Model

The discretized finite element model should be checked carefully before results are
computed. Ideally, a model should be checked by an analyst not involved in the preparation of the model, who is then more likely to be objective.
Preprocessors with their detailed graphical display capabilities (Figure 7-9) now
make it comparatively easy to find errors, particularly the more obvious ones involved
with a misplaced node or missing element or a misplaced load or boundary support.
Preprocessors include such niceties as color, shrink plots, rotated views, sectioning,
exploded views, and removal of hidden lines to aid in error detection.
Most commercial codes also include warnings regarding overly distorted element shapes and checking for sufficient supports. However, the user must still select
the proper element types, place supports and forces in proper locations, use oonsistent
units, etc., to obtain a successful analysis.
Checking the Results and Typical Postprocessor Results

The results should be checked for consistency by making sure that intended support
nodes have zero displacement, as required. If symmetry exists) then stresses and displacements should exhibit this symmetry. Computed results from the finite element
.. "'1111 !=\hould be compared with results from other available techniques, even if

7.2 Equilibrium and Compatibility of Finite Element Results

'"

363

Figure 7-9 Plate of steel (20 in. long, 20 in. wide, 1 in. thick, and with a Hn.'i'adius
hole) discretized using a preprocessor prowam [15] with automatic mesh generation

these techniques may be cruder than the finite element results. For instance, approximate mechanics ofmaterial formulas, experimental data, and numerical analysis of
simpler but similar problems may be used for comparison, particularly if you have
no real idea of the magmtude of the answers. Remember to use all results with some
degree of caution) as errors can crop up in such sources as textbook or handbook
comparison solutions and experimental results.
In the end, the analyst should probably spend as much time processing, checking, and analyzing results as is spent in data preparation.
Finally. we present some typical postprocessor results for the plane stress problem of Figure 7-9 (Figures 7-10 and 7-11). Other examples with results are shown
in Section 7.7.

:I

7.2 Equilibrium and Compatibility of Finite
Element Results
An approximate solution for a stress analysis problem using the finite element method
based on assumed displacement fields does not generally satisfy all the requirements
for equilibrium and compatibility that an exact theory-of-elasticity solution satisfies.
However. remember that relatively few exact solutions exist. Hence, the finite element
method is a vr;ry practical one for obtaining reasonable, but approximate~ numerical
solutions. Recall the advantages of the finite element method as described in Chapter 1
and as illustrated numerous times throughout this text.

364

...

7 Practical Considerations in Modeling; Interpreting Results

1000 psi

20 in.

Figure 7-10 Plate with a hole showing the deformed shape of a"plate superimposed
over an undeformed shape. Plate is fixed on the left edge and subjected to 10DO-psi
tensile stress along the right edge. Maximum horizontal displacement is 7.046 x
10-4 in. at the center of the right edge

We now describe some of the approximations generally inherent in finite element solutions.
'

1. 'Equilibrium of nodal forces and moments is satisfied. This is true
because the global equation E = K 51 is a nodal equilibrium equation
whose solution for 51 is such that the sums of all forces and moments
applied to each node are zero. Equilibrium of the whole structure is
also satisfied because the structUre'reactions are included in the global
forces and hence in the nodal equilibrium equations. Numerous
example problems, particUlarly involving truss and frame analysis in
Chapter 3 and 5, respectively, have illUstrated the equilibrium of
nodes and of total structures.
2. Equilibrium within an element is not always satisfied. However, for
the constant-strain bar of Chapter 3 and the constant-strain triangle of
'Chapter 6, element equilibrimn is satisfied. Also the cubic displacement function is shown to satisfy the basic beam equilibrium differential equation in Chapter 4 and hence to satisfy element force and

7.2 Equilibrium and Compatibility of finite Element Results

.&

365

2_

-..:t2:t

ZT'IIl.7

1
,--

',_1'"

1UO"'I%l
11S42.1>11

m.llOOC

6"'_
-=

1.:D!B:!..o12

Figure 7-11 Maximum principal stress contour (shrink fit plot) for a plate with hole.
Largest principal stresses of 3085 psi occur at the top and bottom of the hole, which
indicates a stress concentration of 3.08. Stresses were obtained by using an average of
the nodal values (called smoothing)

moment equilibrium. However, elements such as the linear-strain
triangle of Chapter 8, the axisymmetric element of Chapter 9, and the
rectangular element of Chapter 10 usually only approximately satisfy
the element equilibrium equations.
3. Equilibrium is not usually satisfied between elements. A differential
element including parts of two adjacent finite elements is usually not
in equilibrium (Figure 7-12). For line elements, such as used for truss
and frame analysis, interelement equilibrium is satisfied, as shown in
example problems in Chapters 3-5. However, for two- and threedimensional elements, interelement equilibrium is not usually satisfied.
For instance, the results of Example 6.2 indicate that the nonnal stress'
along the diagonal edge between the two elements is different in the
two elements. Also, the coarseness of the me.§h causes this lack of
interelement equilibrium to be even more pronounced. The normal
and shear stresses at a free edge usually are not zero even though
theory predicts them to be. Again, Example 6.2 illustrates this, with

366

£.

7 Practical Considerations in Modeling; Interpreting Results

~----------------7r--_5~lb

CD

,...--

lOin.

t _ _ .J

~------------------~~S~lb

20 in.

Ex.ample6.2

a,
0',. ::;

301 psi

r----'-----::~ • or..,..

=

301 psi

.!\=

ax

= 995

~

2.4 psi

~ t. r = 2.80 ~si

= - 2.4 psi

:.--a~

psi
0;

= 'OOSpa
'1:..,

440 psi

=::

~ '~PS;99Spa

= 397 psi

'Ij,,, =

0'., ::

Stresses on a differential
element common to both finite
elements, iIIus£rating violation
of equilibrium

- 2.4

-1.2 psi

Srress along the diagonal between elemenlS,
showing normal and shear suesses,
,
(I,.

and'tnl' Note: aft

and t.t are not

equa.l in magnitude but are opposite in
sign for the two elements, and so
interelement equilibrium is not satisfied

Figure 7-12 Example 6.2, illustrating violation of equilibrium of a differential
element and along the diagonal edge between two elements (the coarseness of the
mesh amplifies the violation of equilibrium)

free-edge stresses O'y and '1:xy not equal to zero. However, .as more
elements are used (refined mesh) the O'y and'txy stresses on the stressfree edges will approach zero.
4. Compatibility is satisfied within an element as lOrlg as the element
displacement field is continuous. Hence, individual elements do not
tear apart.
5. In the formulation of the el~ent equations, compatibility is invoked
at the nodes. Henc:e) elements remain connected at their common
nodes. Similarly, the structure remains connected to its support nodes
because boundary conditions are invoked at these nodes.
6. Compatibility mayor may not be satisfied along interelement
boundaries. For line elements such as bars and beams, interelement
boundaries are merely nodes. Therefore, the prec:eding statement 5
applies for these line elements: The constant-strain triangle of Chapter
6 and the rectangular element of Ghapter 10 remain straight-sided
when deformed.. Therefore, interelement compatibility existS for these
elements; that is, these plane elements deform along common lines

psi

7.3 Convergence of Solution

A

367

without openings, overlaps, or discontinuities. Incompatible elements,
those that allow gaps or overlaps between elements, ~an be acceptable
and even desirable. Incompatible element formulations, in some cases,
have been shown to converge more rapidly to the exact solution [I}.
(For more on this special topic, consult References [7J and 18].)

1

7.3 Convergence of Solution
In Section 3.2, we presented guidelines for the selection of so--called compatible and
complete displacement functions as they "related to the bar element. Those four guide.
lines are generally applicable, and satisfaction of them has been shown to ensure monotonic convergence of the solution of a particular problem [9J. Furthermore, it has
been shown {I 0] that these compatible and complete displacement functions used in
the displacement formulation of the finite element method yield an upper bound on
the true stiffness, and hence a lower bound on the displacement the problem, as
shown in Figure 1-13.
Hence, as the mesh size is reduced-that is, as the number of elements is
increased-we are ensured of monotonic convergence of the solution when compatible
and complete displacement functions are used. :txamples of this convergence are
given in References [1J and [11], and in Table 7-2 for the beam with loading shown

of

Exact solution

Number of elements
i~ ~----------------~----~
'\

~
is

"c

ompall" Ie
bdlsplatement
'

fonnulation

Figure :'J -13 Convergence of a finite element solution based on the compatible
displacement formulation
Table 7-2 Comparison of results for different numbers of elements

Case

Number of
Nodes

Number of
Elements

21
39

12

2

24

I

45
85
105

32
6480

3
1.5
1.2

2
3
4
5

Exact solution [2J

Aspect
Ratio

Vertical
Displacement, v (in.)
Point A
-0.740
-0.980
-0.875
-1.078
-1.100
-1.152

368

£

7 Practical Consiclerations in Modeling; Interpreting Results

in Figure 7-1(a). All elements in the table are rectangular. The results in Table 7-2 indicate the influence of the numbe~ of elements (or the number of degrees of freedom as
measured by the number of nodes) on the convergence toward a common solution,
in this case the exact one. We again observe the influence of the aspect ratio. The
higher the aspect ratio, even with a larger num~r of degrees of freedom, the worse
the answer, as indicated by comparing cases 2 and 3.

:l

7.4 Interpretation of Stresses~
In th~ stiffness or displacement formulation of the finite element method used
throughout this text, the primary quantities determined are the interelement nodal displacements of the assemblage. The secondary quantities, such as strain and stress in an
element, are then obtained through use of {t} = [B]{d} and {u} = [D][B]{d}. For elements using linear..rusplacement models, such as the bar and the constant-strain triangle, [BJ is constant, and since we assume [DJ to be constant, the stresses are constant
over the element. In this easel it is common practice to assign the stress to the centroid
of the element with acceptable results.
However. as il1ustrated in Section 3.11 for the axial member, stresses are not
predicted as accurately as the displacements (see Figures 3-32 and 3-33). For example, remember the constant-strain or constant-stress element has been used in mode1·
ing the beam in Figure 7-1. Therefore, the stress in each element is assumed constant.
Figure 7-14 compares the exact beam theory solution for bending stress through the
beam depth at the centroidallocation of the elements next to the wall with the finite
element solution of case 4 in 'Table 7-2. This finite element model consists of four
elements through the beam depth. Therefore) only four stress values are o.btained

y(in.)

4 t - - - - - - i i > 174.4

t2h 130.8
2

Finite element solution::;

e

Exact solution

39

43.6
--------.1'--'--100
........1-'5-0-200....1.-- (T'~ (ksi)

-39

-2
e-122

-3

"'------t-

-4

-174.4

Figure 7-14 Com parison of the finite element solution and the exact solution of

bending stress through a beam cross section

7.5 Static Condensation

A..

369

through the depth. Again, the best approximation of the stress appears to occur at the
midpoint of each element, since the derivative of displacement is better predicted between the nodes than at the nodes.
.
For higher-order elements, such as the linear-strain triangle of Chapter 8, [BI}
and hence the stresses, are functions of the coordinates. The common practice is then
to evaluate directly the stresses at the centroid of the element.
An alternative procedure sometimes is to use an average (possibly weighted)
value of the stresses evaluated at each node of the element. This averaging method is
often based on evaluating the stresses at the· Gauss poInts located within the element
(described in Chapter 10) and then interpolating to the e~ment nodes using the
shape functions of the specific element. Then these stresses in all elements at a common
node afe averaged to represent the stress at the node. This averaging process is called
smoothing. Figure 7-11 shows a maximum. principal stress "fringe carpet" (dithered)
contour plot obtained by smoothing.
Smoothing results in a pleasing, continuous plot which may not indicate some
serious problems with the model and the results. You should always view the unsmoothed contour plots as well. Highly discontinuous contours between elements in
a region of an unsmoothed plot indicate modeling probl~ms and typically require additional refinement of the element mesh in the suspect region. . 'If the discontinuities in an unsmoothed contour plot are small or are in regions
of little consequence) a smoothed contour plot can normally be used with a high
degree of confidence in the results. There are, however, exceptions when smoothing
leads to erroneous results. For instance, if the thickness or material stiffness changes
significantly between adjacent elements, the stresses will no~ally be different from
one element to the next. Smoothing will likely hide the actual results. Also, for shrinkfit problems involving one cylinder being expanded enough by heating to slip over the
.smaller one, the circumferential stress between the mating cylinders is normally quite
different [16].
The computer program examples in Section 7.7 show additional results, such as
displaced models, along with line contour stress plots and smoothed stress plots. The
stresses to be plotted can be von Mises (used in the maximum distortion energy theory
to predict failure of ductile materials subjected to static loading as described in
Sec.tion 6.5); Tresca (used in the Tresca or maximum shear stress theory also to predict
failUI;e of ductile materials subjected to static loading) [14, 16], and maximum and
minimum principal stresses.

"" 7.5 Static Condensation
We will now consider the concept of static condensation because this concept is used
in developing the stiffness matrix of a quadrilateral element in many computer
programs.
Consider the basic quadrilateral eiement with external nodes 1-4 shown in
Figure 7-15. An imaginary node 5 is temporarily introduced at the intersection of the
diagonals of the quadrilateral to create four triangles. We then superimpose the stiffness matrices of the four triangles to create the stiffness matrix of the quadrilateral

370

..

Pract~cal

7

Considerations in Modeling; Interpreting Results

y,"

I

3

CD //
" / ®
0)<,
/5,
.1/ CD "

4 "

Figure 7-15 Quadrilateral element
with an internal node

J

2

I

' - - - - - - - - x. u

element, where the internal imaginary node 5 degrees of freedom are said to be condensed out so as never to enter the final equations. Hence, only the degrees of freedom
associated with the four actual external comer nodes enter the equations.
'We begin the static condensation procedure by partitioning the equilibrium
equations as
{7.5.l )
where di is the vector of internal displacements corresponding to the imaginary' internal node (node 5 in Figure 7-15), fi is the vector of loads at the internal node, and
do and Fa are the actual nodal degrees of freedom and loads, respectively, at the
actual nodes. Rewriting Eq. (7.5.1), we have

[kll}{du } + [k!2l{di } = {Fa}

+ [k22J{dr} =

[k21 ]{da }

{F;}

(7.5.2)

(7.5.3)

Solving for {di } in Eq. (7.5.3), we obtain

{di } = -[k22

r

I

[k21 l{da }

+ [k22r l {Fi}

(7.5.4)

Substituting Eq. (7.5.4) into Eq_ (7.5.2), we obtain the condensed equilibriwn equation
[kd{d:J}
where

= {Pc}

(7.5.5)

[krJ = [kill - [k I2 J[k:rd- 1[k2d

(7.5.6)

[kI211k22rl {Fi}

(7.5.7)

{f;} = {F

lI }

and [k/.J and {F;.} are called the condensed stiffness matrix and the condensed load vec
101', respectively. Equation (7.5.5) can now be solved for the actual comer node displacements in the usual manner of solving simultaneous linear equations.
Both constant-strain triangular (CST) and constant-strain quadrilateral elements
are used to analyze plane stress/plane strain problems. The quadrilateral element has the
stiffness of four CST elements~ An advantage of the fOUI..csr quadrilateral is that the
solution becomes less dependent on the skew of the subdivision mesh, as shown in
u

7.5 Static Condensation

...

171

Figure 7-16. Here skew means the directional stiffness bias that can be built into a
model through certain discretization patterns, since the stiffness matrix of an element
is a function of its nodal coordinates) as indicted by Eq. (6.2.52). The four-CST
mesh of FigiJre 7-16(c) represents a reduction in the skew effect over the meshes of
Figure 7-16(a) and (b). Figure 7-16(b) is generally worse than Figure 7-16(a) because
the use of long, narrow triangles results in an element stiffness matrix that is stiffer
along the narrow direction of the triangle..
The resulting stiffness matrix of the quadrilateral element will be an 8 x 8 matrix
consisting of the stiffnesses of four triangles, as was shown in Figure 7-1 S. The stiff~
ness matrix is first assembled according to the usual direct stiffness method. Then we
apply static condensatiori as outlined in Eqs. (7.5.1)-(7.5.7) to remove the internal
node 5 degrees of freedom .
.The stiffness matrix of a typical triangular element (labeled dement 1 in Figure
7-15) with nodes 1,2, and 5 is given in general form by
[k(1)j

k(J) -12
k(l) -15
k(l)

-II

=

k(l)

[

-21

k(l}

-22

1

k(l)

(7.5.8)

-25

k~~) k;~ k;~)

where the superscript in parentheses again refers to the element number, and each submatrix [kV)] is of order 2 x 2. The stiffness matrix of the quadrilateral, assembled
using Eq. (7.5.8) along with similar stiffness matrices for elements 2-4 of Figure 7-15,
is given by the following (before static condensation is used):
(UI. VI)

V4)

(us, vs)

(U2' V2)

(U3) V3)

[k~~]

[0]

[ki~l

[ki~)] + rki~)J

[k~;)J

[0]

[k~~)] + [k~~l

rk~!)]

[k~~)] + [k~!)]

(U4)

(kii)]

+
[ki~)l

fk~;)]
[k;~)]

+
[J4~J

[OJ

[k~;)l

[k~~)]

[k]=

+
[k~~)I

[k~)]

rkl~)l

[OJ

[k~~)J

+

[ki~}] + [ki~)]

[k~)l

[kg)l

[kWl

[k~!ll

+

+

+

+

([ki~)l + [k~~)])
+

[k1i1j

[kg) 1

[kg)]

[k~)l

([k~~)] + [k;~)J)

']

(7.5.9)

372:

.A.., 7 Practical Considerations in Modeling; Interpreting Results

.

(a)

(c) _

(b)

Figure 7-16 Skew effects in finite element modeling

where the orders of the degrees of freedom are shown above the columns of the
stiffness matrix and the partitioning scheme used in static condensation is indicated
by the dotted lines. Before- static condensation is applied. the stiffness matrix is of
order 10 x 10.

Example 7.1

Consider the quadrilateral with internal node 5 and dimensions as shown in Figure 7-17
to illustraJ:e the application of-static condensation.

3

4

®
CD

I

~

5

<D
4 in.

0

I

2 in.

~

Figure 7-17 Quadrilateral with an
internal node

./2

Recall that the original stiffness matrix of the quadrilateral is lOx 10, but static
condensation will result in an 8 x 8 stiffness matrix after removal of the degrees of
freedom (us, tis) at node 5.
Using the CST stiffness matrix of Eq. (6.4.3) for plane strain, we have

-4

3

2

5
5

0.1
0.2 -1.6 -1.2
-0.2
2.6 -0.8 -5.6
1.2
1.5
-1.0
-1.6
[k(J}] = [k(3)] = ~
3.() 0.8 --5.6
4.16
0.0
3.2
Symmetry
11.2

1-5

1.0
3.0

(7.5.10)

7.5 Static Condensation

..

373

Similarly, from Figure 7-17, we can show that

2

3

4

1

5
5

1.5 - 1.0 -0.1
0.2 -1.4
0.8
3.0 -0.2 -2.6
1.2 -0.4
1.S
1.0 -1.4 -0.8
E
4.16
3.0 -1.2 -0.4
2.8
0.0
Symmetry
0.8

(7.S.!I)

where the numbers above the columns in Eqs. (7.SJ 0) and (7.5.11) indicate the orders
of the degrees of freedom associated with each stiffness matrix. Here the quantity in
the denominator ofEq. (6.4.3), 4A(1 + v){l - 2v), is equal to 4.16 in Eqs. (7.5.10)
and (7.5.11) because A = 2 in 2 and v is taken to be 0.3. Also, the thickness t of the element has been taken as I in. Now we can superimpose the stiffness terms as indicated
by Eq. (7.5.9) to obtain the general expression for a four-CST element. The resulting
assembled total stiffness matrix before static condensation is applied is given by

3.0

[k]

2.0
6.0

0.1
0.2
0.0
0.0 -0.1 -0.2: -3.0 -2.0
-0.2
2.6
0.0
0.0
0.2 -2.6: -2.0 -6.0
3.0 -2.0 -0.1
0.2
0.0
0.0: -3.0
2.0
1
6.0 -0.2 -2.6
0.0 . 0.0: 2.0 -6.0
3.0
2.0
0.1
0.2: -3.0 -2.0
6.0 -0.2
2.6: -2.0 -6.0
3.0 -2.0: -3.0
2.0

E

= 4.16

I

-~.~-:--~.Q_-~~~
:·12.0

0.0
24.0
(7.5.12)
After we partition Eq. (7.5.12) and use Eq. (7.5.6), the condensed stiffness matrix is
Symmetry

given by

ur

E

VI

U2

V2

0.20
2.08 1.00 -0.48
1.43
4.17 -0.20
2.08 -1.00
4.17

[kc] = 4.16

Symmetry

U3

V3

U4

V4

-0.92 -LOO -0.68 -0.20
-l.00 -1.83
0.20 -3.77
-0.68
0.20 -0.92
1.00
-0.20 -3.77
1.00 -1.83
2.08
0.20
1.00 -0.48
4.17 -0.20
1.43
2.08 -1.00
4.17
(7.5.13)

•

374

A

A

7 Practical Considerations in Modeling; Interpreting Results

7.6 Flowchart for the Solution
of Plane Stress/Strain Problems
In Figure 7-18, we present a flowchart ofa typical 'finite element process used for the
analysis of pJane stress and plane strain, problems on the basis of the theory presented
in Chapter 6.

.6..

7.7 Computer Program Assisted Step-by-Step
Solution, Other Models and Results for Plane
Stress/Strain Problems
In this section, we present.a computer-assisted step-by-step solution of a plane stress
problem, along with results of some plane stress/strain problems solved using a computer program (12]. These results illustrate the various kinds of difficult problems
that can be solved using a general-purpose computer program.

Draw Ihe geometry and apply forces
and boundary conditions

Define the element type and mechanical
properties (here the 2-D element is used)

Compute the element stiffuc:ss matrix

!, and the load veclor[in global coordinates
Perform static condensation if the element has internal degrees of
freedom (that is, if quadrilaterals are used)

Use the direct stiffness procedure to add If. and distributed loads[
to tbe proper locations in assemblage stiffness K and loads E.

Output results

Figure 7-18

Flowchart of plane stress/strain finite element process

7.7 Computer Program Assisted ,Step-by-Step Solution, Other Models and Results

A

375

The computer-assisted step-by-step problem is the bicycle wrench shown in
Figure "i-19(a). The following steps have been used to solve for the stresses in the
wrench~~ .
1. The first step is to draw the outline of the wrench using a standard
drawing program as shown in Figure 7-i9(a). The exact dimensions
of the wrench are obtained from Figure P7-35, where the overall
depth of the wrench is 2.0 em, the length is 14 em) and the sides of the
hexagons are 9 mm long for the middle one and 7 mm long for the
side ones. The radius of the enclosed ends is 1_50 em.
2. The second step is to use a two-dimensional mesh generator to create
the model mesh as shown in Figure 7-19(b).
3. The third step is to apply the boundary conditions to the proper
nodes using the proper boundary condition command. This is
shown in Figure 7-19(c) as indicated by the small @ signs at the
nodes on the inside of the left hexagonal shaped hole. The @ sign
indicates complete fixity for a node. This means these nodes' are
constrained from translating in the y and z directions in the plane of
the wrench.
4. The fourth step- requires us to select the surface where the distributed
loading is to be applied and then the magnitude of the surface
traction. This is the upper surface between the middle and right
hexagonal holes where the surface traction of 100 N/cm 2 is applied as
shown in Figure 7-19{d). In the computer program this surface
changes to the color red as selected by the user (Figure 7-19{c)).
S. In step five we choose the material properties. Here ASTM A·514
steel has been selected, as this is quenched apd tempered steel
with high yield strength and win allow for the thickness to be
minimized.
6. In step six we select the element type for the kind of analysis to be
performed. Here we select the plane stress element, as this is a good
approximation to the kind of behavior that is produced in a plane
stress analysis. For the plane stress element a thickness is required. An
initia:l guess of one em is made. This thickness appears to be
compatible with the other dimensions of the wrench.
7. The seventh step is an optional check of the model. If you choose to
perform this step you will see the boundary conditions now appear as •
triangles at the left nodes corresponding to the @ signs forful1 fixity
and the surface traction arrows, indicating the l~tion and direction
of the sUlface traction shown also in Figure 7-19(d).
8. In step eight we perform the stress analysis of the model.
. 9. In step nine we select the results, such as the displacement plot, the
principal stress plot, and the von Mises stress plot. The von Mises
stress plot is used to determine the failure of the wrench based on the
maximum distortion energy theory as described in Section 6.5. The
von Mises stress plot is shown in Figure 7-19(e). The maximum
von Mises stress indicated in Figure 7-19(e) is 502 MPa, and the yield

376

....

7 Practicaf Considerations in Modeling; Interpreting Results.

(a)

GilllillOtllilitG
(b)

(e)

(d)

--4_s....
1Ii-."....

~

..

. 3-..00&

>0'_

....-24_
20'_

looe-",..-

~

'7St.~

(e)

Figure 7--19 Bicycle wrench (a) Outline drawing of wrench, (b) meshed model of wrench,
(c) boundary conditions and selecting surface where surface traction will be applied, (d) checked
model showing the boundary conditions. and surface traction, and (e) von Mises stress plot
(compliments. of Angela Moe)

7.7 Computer Program Assisted Step-by-Step Solution, Other Models and Results

..

317

--~

"

",

........... \IlIIIJI< ·1 41l'J13e.Cl1 lbVC....2)

(b)

Figure 7-20 (a) Conneaing rod subjected to tensile loading and (b) resulting
principal stress throughout the rod

strength of the ASTM A-S14 steel is 690 MPa. Therefore, the wrench
is safe from yielding. Additional trials can be made if the factor of
safety is satisfied and if the maximum deflection appears to be
satisfactory.
Figure 7-20{a) shows a finite element model of a steel connecting rod that is
fixed on its left edge and loading around· tlie right inner edge of the hole with a total
force of 3000 lb. For more details, including the geometry of this rod, see Figure
P7-11 at the end of this chapter. Figure 7-20{b) shows the resulting maximum

378

.... 7 Practical Considerations in Modeling; Interpreting Results
-:;;....;;.:.;..;.......:-.;;...;.....:..;;.;.;;...-....'li..::;:......~..;...;;.::.;..--.;..=..O';'...:...::;...:....::.:..-~~:-;..:.;..;..(...~----.. ..;-'-_\...:-.---- -:~'

~,

~

Figure 7-21

von Mises stress. plot of overload protection device

principal stress plot. The largest principal stress of 12051 psi occurs at the top and bottom inside edge of the hole.
Figure 7-21 shows a finite element model along with the von Mises stress plot of
an overload protection device (see Problem 7-30 for details of this problem). The
upper member of the device was modeled. Node S at the shear pin location was constrained from vertical motion and a node at the roller E was constrained in the horizontal direction. An equilibrium load was applied at B along line BD. The magnitude
of this load was calculated as one that just makes the shear stress reach 40 MPa in the
pin at S. The largest von Mises stress of 178 MPa occurs at the inner edge of the cutout section.
Figure 7-22 shows the shrink plot of a finite element analysis of a tapered plate
with a hole in it, subjected to tensile loading along the right edge. The left edge was
fixed. For details of this problem see Problem 7-23. The shrink. plot separates the elements for a clear look at the model. The lugest principal stress of 19.'0 e6' Pa
(19.0 MPa) occurs at the edge of the hole, whereas the second largest principal stress
of 17.95 e6 Pa (17.95 MPa) occurs at the elbow between the smallest cross section
.
and where the taper begins.
Figure 7-23 shows the shrink fit plot of the maximum principal stresses in an
overpass subjected to vertiCal loading on the top edge. The largest principal stress of
56162 Iblft2 (390 psi) occUrs at the top inside edge. For more details of this problem
see Problem 7.20.

7.7 Computer Program Assisted Step-by-Step Solution, Other Models and Results

;

..

379

'~
1.1121DtkoOO1
1~_

-,~

---

'.'H~

1151_

7111_
5:I'OGn>4
~
~

.3.t121mM1• .oog

l[i:liii!ii!iili~l~

=!~

Figure 7-22 Shrink fit plot of principal stresses in a tapered plate with hole

Finally, Figure 7-24(a) shows a finite element discretized model of a steel spur
gear for stress analysis. The auto meshing feature resulted in very small elements at
the base of the tooth. The applied load of 164.8 Ib and the fixed nodes around the
inner hole of the gear
shown. Figure 7-24(b) shows an enlarged von Mises stress

are

Su.s:s
von Mists
Ib\l(ft"2)

56161.S6
50571.37
44lSO.SS
39390.4
33799.91
28209.42
22618.93
17028.44
11437.05
5847.4136
258.9776

Figure 7-23

Shrink fit plot of principal stresses in overpass (Compliments of David Walgrave)

380

...

7 Practical Considerations in Modeling; Interpreting Results

(a)

(b)

Figure 7-24 (a) Finite element model of a spur gear and (b) von Mises stress plot
(Compliments of Bruce Figi)

plot near the root of the tooth with the applied load acting on it. Notice that the largest stress of 4315 psi occurs at the left root of the tooth. The gear model has 27761
plane stress elements.

References

...

381

;1 References
[I] Desai. C. S., and Abel, J. F., 'Introdu.ction to the Finite Element Method, Van Nostrand
Reinhold, New York, J972.
[2] Timoshenko, S., and Goodier) 1., Theory of Elasticity, 3rd ed., McGraw-Hili, New York,
1970.
[3] Glockner, P. G., "Symmetry in Structural Mechanics," Journal of rhe Structural Division,
American Society of Civil Engineers, Vol. 99, No. STI, pp. 71-89, 1973.
[4J Yamada, Y., "Dynamic Analysis of Civil Engineering Structures," Reeent Advances in
Malrix MethQdsofStructural Analysis and Design, R. H. Gallagher, Y. Yamada, and J. T.
Oden, eds., University of Alabama Press, Tuscaloosa, AL, pp. 487-512, 1970.
[5] Koswara, H., A Finite Element Analysis of Underground Shelter Subjecled to Ground Shock
Load, M. S. Thesis, Rose·Hulman Institute of Technology, Terre Haute, IN, 1983.
16] Duruop, P., Duncan, J. M., and Seed, H. B., "Finite Element Analyses of Slopes in Soil,"
Journal of the Soil Meclumics and Fozmdations Division, Proceedings of the American
Society of Civil Engineers, Vol. 96, No. SM2, March 1970.
[7] Cook, R. D., Malkus, D. S., Plesha, M. E., and Witt, R. J., Concepts and Applications of
Finite Element Analysis, 4th ed., Wiley, New York, 2002.
[8) Taylor, R. L., Beresford, P. J., and Wilson, E. L., "A Nonconforming Element for Stress
Analysis," InterM tionar Journal for Numerical Methods in Engineering, Vol. to, No.6,
pp. 1211-1219, 1976.
!9J Melosh, R. J., "Basis for Derivation of Matrices for the Direct Stiffness Method," Journal
of the American Institute of Aeronautics and Astronautics, Vol. I, No.7, pp. 1631-1637.
July 1963.
PO] Fraeijes de Veubeke, B., "Upper and Lower Bounds in Matrix Structural Analysis,"
Matrix Methods of Structural Amziysis, AGARDograpb 72, B. Fraeijes de Veubeke, ed.,
MacmiI1an, New York, 1964.
fIll Dunder, V., and Ridlon, S.,' "Practical Applications of Finite Element Method," Journal of
the Structural Division, American Socic;;ty of Civil Engineers, No. STl, pp. 9-21, 1978.
[IlJ Linear Stress and Dynamics Reference Division, Docutech On-line DOCUD1entation, Algor,
Inc., Pittsburgh, PA t 5238.
[13J Bettess, P., "More on Infinite Elements," Internatiomzl Journal for Numerical Method.,: in
Engineering, Vol. 15, pp. 1613-1626~ 1980.
[14] Gere,1. M., Mechanics of Materials, 5th ed., Brooks/Cole Publishers, Pacific Grove, CA,
2001.
[IS} Superdraw Reference Division, Docutech On·1ine Documentation, Algor, Inc., Pittsburgh,
PA ]5238.
[16] Cook, R. D., and Young, W. c., Advanced Mechanics of Materials, Macmillan, New
York,1985.
(17] Cook, R. D., Finite Element Modeling/or Stress Analysis> Wiley, New York, 1995.
[18] Kuro\h-ski, P., "Easily Made Errors Mar FEA Results/' Machine Design, Sept. 13,2001.
[19} Huebner, K. H., Dewirst, D. L., Sr:n.ith, D. E., and Byrom, T. G., The Finite Element
Methodfor Engineers, Wiley, New York, 2001.
[20] Demkowicz, L., Devloo, P., and Oden, J. T., "On an h·Type Mesh-Refinement Strategy
Based on Minimization of Interpolation Errors," Comput: Methods Appl. Meck Eng.,
Vol. 53, 1985, pp. 67-89.
[21] Lohner, R., Morgan, K., and Zienkiewicz, O. C., "An Adaptive Finite Element Procedure
for Compressible High Speed Flows," Comput. Methods Appl. Meeh. Eng., Vol. 51, 1985,
pp. 441-465.

382

.&

7 Practical Considerations in Modeling; Interpreting Results

[22}
[23]
[24]

t2S}
[26]

[27}

[281
[29J

4.

Lohner~

R., "An Adaptive Finite Element Scheme for Transient Problems in CFD,"
Comput. Methods Appl. Meeh. End, Vol. 61, 1987, pp. 323-338.
Ramakrisbnan, R., Bey, K. S., and Thornton, E. A., "Adaptive Quadrilateral and Triangular Finite Element Scheme for Compressible Flows," AIAA J.; Vol. 28, No.1, 1990,
pp.51-59.
Peano, A. G., "Hierarchies of Conforming Finite Elements for Plane Elasticity and Plate
Bending," Comput. Match. Appl, Vol. 2, 1976, pp. 211-224.
Szabo, B. A., ':Some Recent Developments in Finite Element Analysis," Comput. Match.
Appl., Vol. 5, 1979, pp. 99-115.
Peano, A. G.~ Pasin~ A., Ric:cioni., R. and Sardena, L., "Adaptive Approximation in Finite
Element Structural Analysis," Comput. Struct., Vol. 10, 1979, pp. 332-342.
Zienkiewicz, O. c., Gago, J. P. de S. R, and Kelly, D. W., "The Hierarchical Concept in
Finite Element Analysis," Comput. StrucL, Vol. 16, No. 1-4, 1983, pp. 53-65.
Szaoo, B., A., "Mesh Design for the p-Version of the Finite Element Method," Comput.
Meth.ods Appl Meek Eng.) Vol. 55, 1986, pp. 181-197.
Toogood, Roger, ProlMECHANICA, Structural Tutorial, SDC Publications, 2001.

Problems
7.1 For the finite element mesh shown in Figure P7-I, comment on the goodness of the
mesh. Indicate the mistakes in the model. Explain and show how to correct them.

~IIIII

III
Figure P7-1
/:

C

............
../"
~

.

A

./
D

B

Figure P7-2

7.2 Comment on the mesh sizing in Figure P7-2. Is it reasonable? lfnot, explain why not.
7.3 What happens if the material property v = 0.5 in the plane strain caSe? Is this possible? Explain.

j

.

7.4 Under what conditions is the structure in Figure P7-4 a plane strain problem? Under
what conditions is the structure a pJane stress problem?

Problems

!--IOin.-t
%

T

1000 Ib/in.

lOin.

/

=--r

...

383

Figure P7-4

10 in.

'~_~_----J -.L

I---- in.---1
20

7.5 When do problems occur using the smoothing (averaging of stress at the nodes from
elements connected to the node) method for obtaining stress results?
7.6 What thickness do you think is used in computer programs for plane strain problems?
7.7 Which one of the CST models shown below is expected to give the best results for a
cantilever beam subjected to an end shear load? Why?

~]
I_

I I I I 14@."=4bL
6 @ 2" = 12"

~ 1111111111 nJ

m

12 @ 1" = 12"

- I Ifo/It.. - - - - - ' " " I

(a)

(b)

~-I-------li ::
!(c)

6"

-I-

6"

-I

(d)

Figure P7-7

7.8 Show that Eq. (7.5.13) is obtajned by static condensation of Eq. (7.5.12).

Solve the following problems using a computer program.. In some of these problems, we
suggest that students be assigned separate parts (or models) to facilitate parametric
studies.
'

•

7.9 Determine the free-end displacements and the element stresses for the plate discretized
into four triangular elements and subjected to the tensile forces ~hown in Figure P7-9.
Compare your results to the soIutio~ given in Section 6.5 Why are these results dif.;.
ferent? Let E = 30 x 106 psi, V = 0.30, and I = 1 in.

384

•

7 Practical Considerations in Modeling; Interpreting Results

C><J 1.:~1'
"*

10m.

~-20in.

S

1

Figure P7-9

5000 Ib

..I

7.10 Determine the stresses in the plate with the hole subjected to the tensile stress shown in
Figure P7-1O. Graph the stress variation ax versus the distance y from the hole. Let
E = 200 GPa, v = 0.25, and t 25 rnm. (Use approximately 25, 50, 75~ 100, and then
120 nodes in your finite element model.) Use symmetry as appropriate.

2S mm radius

IOkPa

WkPa

500mm

~-500mm

L.

.1

Figure P7-10.

S
S

.s

7.11 Solve the following problem of a steel tensile plate with a concentrated load applied at
the top, as shown in Figure P7-11. Determine at what depth the effect of the load dies
out. Plot stress Cly versus distance from the load. At distances of 1 in.~ 2 in., 4 in., 6 in.,
10 in., 15 in., 20 in., and 30 in. from the load~ list a y versus these distances. Let the
width of the plate be b = 4 in., thickness of the plate be t = 0.25 in., and length be
L == 40 in. Look up the concept of St. Venant's principle to'see how it explains the
stress behavior in this problem.
7.12 For the connecting rod shown in Figure P7-12, determine the maximum principal
stresses and their location. Let E = 30 X 106 psi, v = 0.25, t = 1 in., and P = 1000 lb .

7.13 Determine the maximum pIincipal stresses and their locations for the member with
fillet' subjected to tensile forces shown in Figure P7-13. Let E = 200 'GPa and
v = 0.25. Then let E 73 GPa and v 0.30. Let t = 25 mm for both cases. Compare
your answers for the two cases.

Problems
P

1000 Ib

I

'-_

•

I]U::=!...
I

1

_oJ

Figure P7-11

tl-in. radiUS]
O.5-in. radius ~

I Lin.

~

2 in.

·1 0.75 in.

1.
f - - - - - - - - - - 7Sm.

Figure P7-12

Figure P7-13

br

....

385

386

A

7 Practical Considerations in Modeling;

Interpr~ing

Results

7.14 Determine the stresses in the member with a re-entrant comer as shown in Figure P7-14.
At what location are the principal stresses largest? Let E = 30 X 10 6 psi and v = 0.25.
Use plane strain conditions.

--II ftLr::- 100 Iblln.
-,....-

!-IOio'-1
%

T
10 in.

1

1000 Ib/in.

D

/
I .

=T
JO in.
'/'--____--' --L
!--20in.--j
Figure P7-14

I

I

I"""

1
2D---t·1

Axis of symmetry

Figure P7-15

7.15 Determine the stresses in the soil mass subjected to the strip footing load shown in
Figure P7-15. Use a width of2D and depth of D. where D is 3, 4.6,8, and 10 ft. Plot
the maximum stress contours on your finite element model for each case. Compare
your results. Comment regarding your observations on modeling infinite media. Let
E = 30,000 psi and v = 0.30. Use plane strain conditions.

~

7.16 For the tooth implant sUbjected to loads shown in Figure P7-16) determine the maximum principal stresses. Let E = 1.6 X 106 psi arid v = 0.3 for the denta1 restorative
10 lb

o

I

1 7 . ' 35. 37.
32 lD.
32 m. 32 10.

r-'- - - - - - -

Figure P7-16

151b

55.

32 Ill.

36.

16 1tI.

2~ in.-------t·1

Problems

J;.

387

implant material (cross-hatched), and let E = I X 106 psi and v = 0.35 for the bony
material. Let X = 0.05 in., OJ in.) 0.2 in., 0.3 in., and 0.5 in., where X represents the
various depths of the implant beneath the bony surface. Rectangular elements are
used in the finite element" model shown in Figure P7-16. Assume the thickness of each
element to be t = 0.25 in.

S

7.17 Detennine the middepth deflection at the free end and the maximum principal stresses
and their location for the beam subjected to the shear load v~riation shown in Figure
P7-17. Do this using 64 rectangular elements all_of size 12 in. x in.; then all of size
6 in. x 1 in.; then all of size 3 in. x 2 in. Then use 60 rectangular elements all of size
2.4 in. x 2~ in.; then all of size 4.8 in. x l~ in. Compare the free-end deflections and the
maximum .principal stresses in each case to the exact solution. Let E = 30 X 106 psi,
v = 0.3, and t = 1 in. Comment on the accuracy of both displacements and stresses.

!

rI___~_-----,l~

T

4O.000-lb total shear
load parabolically

d;,"buI<d

t--1·~-----48in.--------II·

Figure P7-17

S

7.18 Detennine the stresses in the shear wall shown in Figure P7-18. At what location are
the principal stresses largest? Let E = 21 GPa, v = 0.25, twall = 0.10 m, and tbeam =
0.20m.
50 kN/m

1
8m

________ J __
1m

Beam

~::~

--r-

2m
~~~,--L

f=4m

~ IO'm - - - - - 0 0 1
Figure P7-18

J

388

..

7 Practical Considerations in Modeling; Interpreting Results

7.19 Determine the stresses in the plates with the round and square holes subjected to the
tensile stresses shown in Figure P7-19. Compare the largest principal stresses for each
plate. Let E = 210 GPa, v = 0.25, and t = 5 mm.

1 mmrod

~

XJ~rrvn

2Smm

2S-mm radius

Figure P7-19

. . 7.20 For the concrete overpass structUre shown in Figure P7-20, detennine the maximum
principal stresses and their locations. Assume plane strain conditions. Let E = 3.0 X
106 psi and v = 0.30.
2 k/ft

~~~l
18ft
ll.5-ft
radius

Tn

I

~\:~~

'\:

~lOft-t--l0ft-+-lOft-+-IOft~
Figure P7-20

•

in

7.21 For the steel culvert shown
Figure P7-21, detennine the maximum principal
stresses and their locations and the largest displacement and its location. Let
Esteel = 210 GPa and let v = 0.30.
.

. . 7.22 For the tensile member shown in Figure P7-22 with two holes, detemline the maxi·
mum principal stresses and their locations. Let E = 210 GPa, v 0.25, and t 10 mm.
Then let E :: 70 GPa and v == 0.30. Compare your results.

=

~

7.23 For the plate shown iIi Figure P7-23~ determine the maximUm principal stresses and
their locations. Let E = 210 GPa and v = 0.25.

Problems

t--

1.5 m

.&

389

20kN

3m

FigureP7-21

)'

0.3 m+-OA m-l-O.3 m

T

0.75 m

20 k:N

F:igure P7-22

t=lOmm

75-mm radius

~------Im-------·~I

T ,'0
25-mm radius

I

O.15m

,0.15 m+015

»

m~015

1=

tOmm

m-

'Figure P7-23

7.24 For the concrete dam shown subjected to water pressure in Figure P7-24. determine
the principal stresses. Let E = 3.5 X io 6 psi and v = 0.30. Assume plane strain conditions. Perform the analysis for'self-weight and then for hydrostatic (water) pressure
against the dam vertical face as shown.

S7.25

Determine the stresses in the wrench shown in Figure P7-25. Let E = 200 GPa and
v = 0.25, and assume uniform thickness t = 10 mm.

390

.&

7 Practical Considerations in Modeling; Interpreting Results

25ft

260ft:

T
1(lO it

14-----180 it

-----001·1

Figure P7-24

II

7.26 Determine the principal stresses in the bJade implant and the bony material shown in
Figure P7-26. Let Ebladc 20 GPa, Vblade = 0.30, Ebone = 12 GPa, and "bone. = 0.35.
Assume plane stress conditions with"t = 5 mm .

•

7.27 Determine the stresses in the plate shown in Figure P7-27. Let E
v = 0.25. The element thickness is 10 mm.

•

7.28 For the O.S in. thick canopy hook shown in Figure P7-28~ used to hold down an aircraft canopy, determine the maximum von Mises stress and maximum deflection. The
hook is subjected to a concentrated upward load of 22,400 Ib as shown. Asswne
boundary conditions of fixed supports over the lower half of the inside hole diameter.
The hook is made from AISI 4130 steel, quenched and tempered at 400 OF. (This
problem is compliments of Mr. Steven Miner.)

210 GPa and

.7.29 For the! in. thlc~ L-sbaped steel bracket shown in Figure P7-29, show that the stress
at the 90 degree re-entrant comer never converges. Try models with increasing numbers of elements to show this while plotting the maximum principat stresS in the
bracket. That is, start with one model, then refine the
around the re-entrant
comer and see what happens, say, after two refinements. Why? Then add a fillet, say,
of radius i in. and see what happens as you refine the mesh. A~ plot the maximum
principal stress for each refinement.

mesh

Problems

....

391

Figure P7-25

Use a computer program to belp solve the design-type problems, 7.30-7.36•
•

7.30 The machine shown in Figure P7-30 is an overload protection device that releases the
load when the shear pin S fails. Determine the maximum von Mises stress in the upper
part ABE if the pin shears when its shear stress is 40 MPa. Assume the upper part to
have a unifonn thickness of 6mm. Assume plane stress conditions for the upper part.
The part is made of 6061 aluminum alloy. Is the thickness sufficient to prevent failure
based on the maximum distortion energy theory? If not, suggest a better thickness.
(Scale aU dimensions as needed.)

•

7.31 The steel triangular plate in. thick shown in Figure P7-31 is bolted to a steel column
with }in.-diameter bolts in the pattern shown. Assuming the column and bolts are
very rigid relative to the plate and neglecting friction forces between the column and
pJate) detennine the highest load exerted on any bolt. The bolts should not be included

1

392

...

7 Practical Considerations in Modeling; Interpreting Results
66N
66N
6mm

T

Blade: implant ...............

Figure P7-26

3k.N

3kN

Figure P7-27

in,the·model. Just fix the nodes around the bolt circles and consider the reactions at
these nodes as the bolt loads. If i-in.-diameter bolts are not sufficient, recoIJJ.IIieild
another standard diameter. Assume a standard material for the bolts. Compare the
reactions from the finite element results to those .found by classical methods.
•

7.32 A l in. thick ~chine part supports an end load of 1000 lb as shown in Figure P7-32.
Determine the stress concentration factors for the two changes in geometry l09ated at
the radii shown on the lower side of the part. Compare the stresses you get to classical
. beam theory results with and without the change in geometry, that is, with 'a unifonn
depth of 1 in. instea4 of the additional matetial depth of 1.5 in. Assume standard
mild steel is used for the part. Recommend any changes you might make in the
geometry.

Problems

22,4001b

Figure P7-28

lin.

12m.

P=SOOlb

2 in.
12m.

Figure P7-29

•

393

394

...

7 Rractical Considerations in Modeling; Interpreting Results
120

30

60

f

..1.
T

Dimensions
in millimeters

figure P7-30 Overload protection device

o

0

Figure P7-31
connection

24

L

Steel triangular plate

Dimensions in inches

Figure P7-32 Machine part

Dimensions in inches

~.7.33

A plate with a hole off-centered is shown in Figure P7-33. Determine how close to the
top edge the hole can be placed before yielding of the A36 steel occurs (based on the
maximum distortion energy theory). The applied tensile stress is 10)000 psi, and
the plate thickness is i in. Now if the plate is made of 6061-T6 aluminum alloy with a
yield strength of 37 ksi, does this change your answer? If the plate thickness is changed
to ! in., how does this change the results? Use same total load as when the plate is ~ in.
thick.

Problems

:=T
:= I'-t==2.5
-__

A

395

----------------

0.5 i.n..dia.

______ lO,OOOpsi

25

fL-_ _+--_ _ _ _

---I

[---1-------5.0-------1

Dimensions in inches

Figure

~

P7~33

Plate with off-centered hole

7.34 One arm ofa crimper tool shown in Figure P7-34 is to be designed of 1080 as-rolted
steeL The loads and boundary conditions are shown in the figure. Select a thickness
for the arm based on the material not yielding with a factor of safety of 1.5. Recommend any other changes in the design. (Scale any other dimensions that you need.)

1.-.
.... ---8.oooo------....!,1
RO.l409

O.6000;}

!
(a) Crimper ann with dimensions (inches)

y

E=60lb

1oot----S.8 in.

RO.l409

.1

x--------~--------_.

B=S411b

*"
(b) Crimper arm loads and boundary condtions

Figure P7-34 Crimper arm

•

7.35 Design tJ:ie-bicycle wrench with the approximate dimerisions shown in Figure P1-35.
If you need to change dimensions explain why. The wrench should be made of steel or

396

...

7 Practical Considerations in Modeling; Interpreting Results

aluminum alloy. Determine the thickness needed based on the maximum distortion
energy theory. Plot the deformed shape of the Wrench and the principal stress and von
Mises stress. The boundary conditions are shown in the figure, and the loading is
shown as a distributed load acting over the right part of the wrench. Use a factor of
safety of 1.S against yielding.
R= J.50cm

The sides of the middle
hexagon are 9 rnm long.

The sides of the comer
.hexagons are 7 mm long.

Fixed all the way around this hexagon.

Figure P7 -35

SJ

Bicycle wrench

7.36 For the various parts shown in Figure P7-36 determine the best one to relieve stress.
Make the original part have a small radius of 0.1 in. at the inside re-entrant comers.
Place a uniform pressure load of 1000 psi on the right end of each part and fix the left
end. All units shoYlIl are taken in inches. Let the material be A 36 steeL

I
L30-.....;..'.-3.0---l.1
T
1.0

T
1.0
.1

+

3.0

f
1.0

J.-

Original design

Figure P7-36

Problems

... ' 397

T

TI

1.0

T

f~).o~

1

IL

3.0

-1..L-'- - ).0 - - - - - I .__
.-='-='-3-.0_------j----'----.J

3.0---1

L-

1

Radius

Taper

1--- 3.0

·1·

3.0------1
1-3/8

TT

R=O.5

..........

°~

in.rad

R=O.l

3.0

1

m. .
'4l.rad

1.0

1

3.0------j
Undercut

1.0

f

3

1~

Relief Holes
Figure P7-36

, ' 'I
'

..
: ;

(Continued)

1. rad

0i6ID.
1-318

I
3.0

1