Fundamental-Finite-Element-Analysis-and-Applications

UNITS 83

Using the Lagrange multiplier method, the solution is obtained as follows:
Augmented system of equations:

142.693 -'-36.8215 -46.0268 36.8215 -96.6667 O. 0 0 0 0 0 0 dl O.
-36.8215 150.29 36.8215 -29.4572 O. O. O.
-46.0268 36.8215 142.693 -36.8215 O. O. 00 0 0 0 0~ O.
-29.4572 150.29 O. -120.833 0 0 -4s 0 0 0 d3 O.
36.8215 -36.8215 O.
0 0 -1 0 -1 1 d4 -20.
O.
-96.6667 O. O. O. 142.693 36.8215 0 0 0 1 0 0 ds -20.
0
O. O. O. -120.833 36.8215 150.29 0 0 0 0 1 0 d6 0
0 0 0 0 00 0 0 s -1 0 0 d7 0
4 0
0 0 0 1 0 0 0 dB
000 00 s
0 0 -4s -1 0 4
0 1 0 0 0 0 it!

000 01 0 -1 0 0 0 0 0 A2

00 0 -1 0 1 0 0 0 0 0 0 1\3

000 10 0 0 0 0 0 0 0 A4

Solution:

(dl = 0.172849, dz = 0.0764461, d3 = -'-0.139174, d4 = -4.68418 X lO-IS,
ds = 0.292296, d6 = -1.62088 X 10-17, d7 = 0.292296, ds = -0.539337,
AI = -20., Az = -25., A3 = -30.7628, A4 = -60.)

1.7 UNITS

It is important to use a consistent system of units during finite element calculations. When
using the U.S. customary system of units, it is convenient to use the pound-inch (PI) units.
Thus, the nodal coordinates will be entered in inches, the modulus of elasticity in pounds
per square inch (psi), mass density in pounds per inch cubed (lb,/in3) , thermal conductivity
in British thermal units per hour-inch-degrees Fahrenheit [Btu/(hr . in . OF)], and coefficient
of thermal expansion in inch per inch per degrees Fahrenheit (in/in/°F). When using the
International System of Units, it is recommended to use the newton-millimeter (N . mm)
units. Thus the nodal coordinates will be entered in millimeters, the modulus of elasticity in
megapascals (MPa), mass density in metric tonnes per millimeter cubed (mt/mm"), thermal
conductivity in watts per millimeter-degrees Celsius [W/(mm· °C)),"and coefficient of
thermal expansion in millimeters per millimeter per degrees Celsius (mm/mm/°C). Typical
numerical values for these quantities for concrete, steel, and aluminum are as follows:

Concrete (medium strength)

Quantity PI System N -rnmSystem

Modulus of elasticity 4.64 X 106 psi . 0.32 X lOs MPa
Mass density 2.24 x 1O-4 1b,/ in3 '2.4 x 10-9 mt/rnm''
Thermal conductivity 0.048 Btu/(hr . in . OF) 0.001 W/(mm· 0c)
Coefficient of 6.1 x 10-6 in/in/T'
i 1 x 1O-6mm/mm/' C
thermal expansion

84 FINITE ELEMENT METHOD: THE BIG PICTURE

Steel (carbon and alloys):

Quantity PI System N· mm system

Modulus of elasticity 30'x psi 2.1 x MPa
Mass density 7.32 x 1O-4 1b,/ in3 7.83 x 10-9 mt/mm''
Thermal conductivity 2.07 Btu/(hr· in- OF) 0.043 W/(mm· 0C)
Coefficient of 12 X 10-6 mm!mm!°C
6.5 x 10- 6 in/in/T'
thermal expansion

Aluminum alloys:

Quantity PI System N -rnm System

Modulus of elasticity 10.4 x psi 0.72 x MPa
Mass density 2.62 x 1O-4 1b / in3 2.8 x 10-9 mt/mnr'
Thermal conductivity 0.21 W/(mm' °C)
Coefficient of ll 23 X 10- 6 mm!mm!°C

thermal expansion 10.11 Btu/(hr· in . OF)
12.8 x 10- 6 in/in/OF

Commonly used conversion factors:

Length Given in Multiply by To get
Force or weight
Weight to mass in 25.4 mm
Ib! 4.4484
Weight to mass. Ib! 1/386.4 N
Mass 1.02 x 10- 4
Mass density /N 0.17524 Ibm
Power 1.0694 x 10- 5 mt
Temperature Ibm 0.2931 mt
Coefficient of thermal Ib,/in3 ("F - 32)/1.8 mt/mm"
Btu/hr W
expansion 1.8 °C
Thermal conductivity OF mm!mm!°C
Convection coefficient in/in/OF 0.02077
Modulus of elasticity 6.411 x 10-3 W/(mm·°C)
Btu/(hr . in . ~F) 6.895 x 10-3 W/(mm2 • 0C)
or stress Btu/(hr . in2 • OF)
MPa
psi

PROBLEMS

Element Equations
1.1 A triangular finite element for a 2D heat flow problem is shown in Figure 1.28.
Write the finite element equations for this element. Assume that the heat conduction

PROBLEMS 85

coefficient is 1.4 W/m· °C. The convection heat loss takes place along side 2-3 of
the element. The average convection heat transfer coefficient is 20W/m' °C and the
surrounding air temperature is 2YC. There is no heat generation inside the element.

y(m) 3

0.2

0.15

0.1

o- - - - - - - x(m)
0.05 0.1 0.15 0.2
Figure 1.28. Triangular element

1.2 A triangular finite element for a plane stress problem is shown in Figure 1.29. Write

= =the finite element equations for this element. Assume E 20 X 106 Nzcm", v 0.3,

thickness = 2 em, and a uniform pressure of 300 N/cm 2 normal to side 1-3 of the

element.

y(cm) 3 y(m) 3
0.1
0.1

o o
o
-0.05
Figure 1.29. Triangular element 0.01

Figure 1.30. Triangular element

1.3 A triangular finite element for a 2D heat flow problem is shown in Figure 1.30.
Write the finite element equations for this element. Assume that the heat conduction
coefficient is 1.17 W/m' "C. The convection heat loss takes place along side 3-1 of
the element. The average convection heat transfer coefficient is 17 W/m' °C and the
surrounding air temperature is 32°C. There is no heat generation inside the element.

1.4 Consider two-dimensional steady-state heat flow in a V-grooved solid body with
a cross section as shown in Figure 1.31. A hot liquid flowing through the groove
maintains the surface at a temperature of 170°F. The thermal conductivity of the
material is k = 0.04 Btu/hr- in- "F, The bottom surface is insulated. The sides have

=a convection coefficient of h '0.03 Btu/hr . in2 • OF with 'the outside temperature

of 70°F. Determine the temperature distribution in the solid. Taking advantage of
symmetry and using only two triangular elements, the model is as shown in the
figure. Develop finite element equations for element 1.

86 FINITE ELEMENT METHOD:THE BIG PICTURE

4

Insulated

Figure 1.31. V-grooved solid

1.5 Develop finite element equations for element 2 of the model shown in Figure 1.31.

1.6 The cross section of a long concrete dam (thermal conductivity = 0.6 W/m . DC)

with base = 3 m and height = 4 m is shown in Figure 1.32. The face of the dam

is exposed to heat flux qo = 800 W/m2 from sun. The vertical face is subjected to

convection by water at 15DC with convection heat transfer coefficient 150 W/m2.DC.
It is proposed to determine the temperature distribution in the dam using only one
triangular element. Set up the system of finite element equations.

Figure 1.32. Concrete dam

Assembly of Element Equations
1.7 A three-bar truss is shown in Figure 1.33. Write the element equations and carry out
the assembly of the element matrices to form the global system of equations. The
area of cross section of each element is 500 mm2 and E = 210 GPa. Use N . mm
units in your calculations.

PROBLEMS 87

10 leN
0.5

0.4

0.3
0.2

0.1

0
(m)

0 0.1 0.2 0.3 0.4 0.5 0.6
Figure 1.33. Plane truss

1.8 A three-bar truss is shown in Figure 1.34. Write the element equations and cany out
the assembly of the element matrices to form the global J( matrix and the R vector.

=The area of cross section of each element is 500 mm2 and E 210 GPa. Use N . mm

units in your calculations.

0.6
0.5
0.4

45° 20 leN
0.3
0.2
0.1

o2

o 0.1 0.2 0.3 0.4 0.5

Figure 1.34. Plane truss

1.9 A finite element model employs triangular plane stress elements and consists of only
six nodes. Each node has two degrees of freedom. The first element is connected to
the model through nodes 1,4, and 6 and has the following coefficient matrix and the
right-hand-side vector:

246 198 221 179 188 194 10
6
198 273 182 242 221 186 9

k= 221 182 215 167 198 185 r = 10
179 242 167 243 225 183
8
188 221 198 225 266 180 7

194 186 185 183 180 201

88 FINITE ELEMENTMETHOD:THE BIG PICTURE

Carry out the assembly of the element matrices to form the global K matrix and the
R vector. Note that this is the first element that is being assembled, and hence the
global matrices are all zeros prior to assembly of this element.

1.10 Form the global system of equations for the two-element finite element model
shown in Figure 1.35. The equations for the individual elements are as follows:

][T0.04 =1

-0.04 T3 ]

[o
Element I: -00..0543 00.23 [500.4]
Element 2: 0.23 0.49
T4 50.4

0.02 -0.06][T [0]0.04 1 ] = 0

0.02 0.02 -0.04 T4
[ -0.06 -0.04 0.1
T2 0

4

2

Figure 1.35: Two-element model

Handling Essential Boundary Conditions and Solution for Nodal Unknowns

I

1.11 In Problem 1.7 you were asked to obtain a global system of equations for a plane

truss. Continue further with the analysis and obtain the reduced system of equations

after adjusting for specified displacement boundary conditions. Solve the resulting

system for the unknown nodal displacements. Compute reactions at the supports

and verify that your solution satisfies overall equilibrium. .

1.12 The global K matrix and the R vector for a finite element system are as follows:

63993.2 79991.4 0 0 0 0 -63993.2 -79991.4
99989.3 0 0 0 0 -79991.4 -99989.3
79991.4 O. O. 0 0
D O. 210000. 0 0 O. O.
0 0 0 0 239682. -59920.6 O. -210000.
0 0 0 -59920.6 14980.1 -239682;
K= 0 0 O. O. -239682. 59920.6 59920.6 59920.6
0 -79991.4 O. -210000. 59920.6 -14980.1 303675. -14980.1
-99989.3 20070.9
0 20070.9
324969.
-63993.2

-79991.4

RT =(0. O. O. o. O. O. 14142.1 -24142.1)

PROBLEMS 89

Obtain the reduced system of equations after imposing the following essential
boundary conditions:

dl = -0.1; dz = 0.2; d3 =d4 = ds =d6 = 0

Solve the final system of equations for the remaining two nodal unknowns.

1.13 After assembly a finite element model yields the following global system of equa-
tions:

358815 -447 -2060 -60 -1053 365 dl 2
-447 553547 -1916 -85 -313 3
-2060 -1916 182342 1479 931 dz -8
-60 -85 -1 -1321 239 5
-1053 . -313 -1 435603 277735 1501 = d3 -7
365 931 1479 -1321 -1056 d4 -4
239
1501 -1056 ds
33134 d6

Get the reduced system of equations after imposing essential boundary conditions

dl = -1, dz = 0, and d4 = 4. Solve the reduced system for remaining nodal un-

knowns.

1.14 A typical truss to support a highway is as shown in Figure 1.36. Because of a con-
struction error, the support a~de 5 rom too tall. The contractor is con-
sidering using jack hammers to force the truss to fit it into place. You are asked to
conduct a finite element analysis to make sure that the stresses in the truss elements
will remain within allowable limits under the given loading and the forced fit. A
colleague of yours has started the analysis already and has obtained the following
global stiffness matrix after assembly of the first eight elements (N . rom units are
used):

7560. -1920. -5000. 0 00 0 0 -2560. 1920. 0 0 III

-1920. 1440. 0 o· 00 0 0 1920. -1440. 0 0 VI

-5000. 0 12560. -1920. -5000. 0 0 0 0 0 -2560. 1920. u2

0 0 -1920. 8106.67 0 0 00 0 -6666.67 1920. -1440. v2

0 0 -5000. _. 0 10000. 0 -5000. 0 0 0 0 0 u3

0000 0 6666.67 00 0 0 0 -6666.67 v3

0 0 0 0 -5000. 0 5000. 0 0 0 0 0 114

0000 00 00 0 0 0 0 v4

-2560. 1920. 0 0 00 0 0 7560. -1920. -5000. 0 liS

1920. -1440. 0 -6666.67 0 0 0 0 -1920. 8106.67 0 0 Vs

0 0 -2560. 1920. 00 0 0 -5000. 0 7560. -1920. 116

0 0 1920. -1440. 0 -6666.67 0 0 0 0 -1920. 8106.67 v6

(a) Form the equations for element 9 and assemble them to form the global system
of equations. Use the following numerical values:

E = 200000N/romz; A = 100rom2; P =20000N

(b) Taking boundary conditions into consideration, determine the final system of
equations to solve for nodal displacements.

90 FINITEELEMENTMETHOD:THE BIG PICTURE
pp

o4

-3 48 (m)
Figure 1.36. Bridge truss 12
o

Complete Solution over Each Element

1.15 The following nodal temperatures are obtained for the heat flow problem in the V-
grooved solid shown in Figure 1.31. Compute the temperature distribution and its x
and y derivatives over each element. By comparing the temperature derivatives over
the two elements, comment on the quality of the solution.

Node Temperature

1 99.4118
2 170
3
/4 28.8235
170

1.16 The correct nodal displacements, in mm, for the three-bar in Problem 1.7 are as
follows:

uv

10 o
20 o

30 o

4 0.0506837 -0.0736988

Compute axial strains, axial stresses, and axial forces in each element. Using the
computed axial forces, draw a free-body diagram of all forces acting at node 4.
By summing forces in the horizontal and vertical directions, show that the joint
equilibrium is satisfied.

1.17 Determine joint deflections, stresses in each member, and support reactions for the
plane truss shown in Figure 1.37. The members have a cross-sectional area of 8 cm2
and are made of steel (E == 200 GPa). Show all calculations.

PROBLEMS 91

1.5
10kN

0.5

o

o 0.5 1.5 (m)

Figure 1.37. Plane truss

1.18 Determine the temperature at the centroid of the concrete dam cross section shown
in Figure 1.32, assuming the temperatures at the nodes are 10, 20, and 30°C, respec-

tively.

Solution of Linear Equations
1.19 Solve the following system of equations using Choleski decomposition:

-; -~ ~ -1oo.5 0 ][Xxx32l] = -2
0 8
0 [2 ]04· 9 1.25
0.5
[ 00 .. -1.5
0.5 x4

000 0.5 3.25 Xs -3.25

1.20 Solve the following system of equations using Choleski decomposition:

1.21 Factor the following matrix using Choleski decomposition:

2 -1 0 000

-1 2-1 o0 0

o -1 3 -1 O' 0

o 0-1 3 -1 0
o0 0
-1 2-1
000
o -1 2

92 FINITE ELEMENT METHOD: THE BIG PICTURE

Using the factored matrix, obtain solutions for the following right-hand sides:

b(l) = ~l b(2) = ~l b(3) = ~l

0 2 2
0 0 0
0 0 0
0 0 0
0 0
~2

1.22 Factor the following matrix using Choleski decomposition:

1 1 1 1 0 0 1 0 -1 0
1 12 3 1 2 0 1 2 1 2
1 33 111 22 12
1 11522 12 32
0 2 1263 20 54
0 0 1234 2 1 53
1 1 2 1 2 2 7 2 -1 4
0 22201 29 25
-1 1 1 3 5 5 -1 2 24 4
0 22243 45 47

Using the factored matrix, obtain solutions for the following right-hand sides:

0 15 6

-8 43 10

-3 43 8
14
b(l) = 39 r 27 b(3) = 12
32 12 7
1 ,/ b(2) = 14 0
-45 29
56 59 19
1 -24
79 21

-2

63

1.23 Solve the system of equations in Problem 1.19 using the basic conjugate gradient
method.

1.24 Solve the system of equations in Problem 1.20 using the basic conjugate gradient
method.

1.25 -Solve the system of equations with the given coefficient matrix and the first right-
hand side in Problem 1.21 using the basic conjugate gradient method.

1.26 Solve the system of equations with the given coefficient matrix and the first right-
hand side in Problem 1.22 using the basic conjugate gradient method.

1.27 Solve the system of equations in Problem 1.19 using the conjugate gradient method
with Jacobi preconditioning.

PROBLEMS 93

1.28 Solve the system of equations in Problem 1.20 using the conjugate gradient method
with Jacobi preconditioning.

1.29 Solve the system of equations with the given coefficient matrix and the first right-
hand side in Problem 1.21 using the conjugate gradient method with Jacobi precon-
ditioning.

1.30 Solve the system of equations with the given coefficient matrix and the first right-
hand side in Problem 1.22 using the conjugate gradient method with Jacobi precon-
ditioning. .

1.31 Solve the system of equations with the given coefficient matrix and the first right-
hand side in Problem 1.21 using the conjugate gradient method with incomplete
Choleski preconditioning with 112 == 1. The solution should converge in one iteration.
Why?

1.32 Solve the system of equations with the given coefficient matrix and the first right-
hand side in Problem 1.22 using the conjugate gradient method with incomplete
Choleski preconditioning with 112 == 1.

Multipoint Constraints

1.33 A finite element system results in the following system of global equations:

2 -2 0 o -2

-2 6 4 o 28 ]
0.5
049 -1.5 -3.25
1.25
[ o 0-1.5 0.5
o0 0

Using the Lagrange multiplier approach, solve for the nodal unknowns if the model

requires the following multipoint constraint: .

dl+dz == 1

1.34 A finite element system results in the following system of global equations:

2· -1 0 0 0 0 dl -1
2
-1 2 -i 0 0 0 dz 0
0
0 -1 3 -1 0 0 d3 0
0
0 0 -1 3 -1 0 d4
0 0 0 -1 2 -1 ds
0 0 0 0 -1 2 d6

Using the Lagrange multiplier approach, solve for the 'nodal unknowns if the model
requires the following multipoint constraints:

94 FINITE ELEMENTMETHOD:THE BIG PICTURE

1.35 Solve Problem 1.33 using the penalty function approach.
1.36 Solve Problem 1.34 using the penalty function approach.

Computational Projects

1.37 Using a finer mesh and the available finite element software, determine the temper-
ature distribution through a solid with a V-groove as shown in Figure 1.31. Take
advantage of symmetry and model only half of the domain.

1.38 Determine the temperature distribution through a solid with a circular groove as
shown in Figure 1.38. ,A hot fluid flowing through the groove keeps the temperature
of the circular surface at 150°C. The bottom surface of the solid is in contact with
a coolant that maintains it at DoC. The top and side surfaces are well insulated. The
solid is made of two different materials, with thermal conductivities k1 =: lOW/rn- "C
and k2 =: 20 Wlm . DC. Take advantage of symmetry and model only half of the
domain.

f

10 m

~I

Figure 1.38. Solid with a circular groove

1.39 Heating wires are embedded in a concrete slab, as shown in cross section in Figure
1.39, to keep snow from accumulating on the slab. In a proposed design wires are
placed at 2 em from the top and at 4-cm intervals. The heat generated by each wire
is 10 Wlm length. The bottom of the slab is insulated. T):J.e top is exposed to a
convection heat loss with a convection coefficient of h =: 30 W1m2 •DC. The thermal
conductivity of concrete is 1.2 Wlm . DC. Determine the slab surface temperature
when the air temperature is -SOC. Note that, because of symmetry, we need to model

T

6cm

1

Figure 1.39. Heating wires embedded in concrete

PROBLEMS 95

only the 2 ern x 6 em dark shaded area. There will be no heat flow across the sides
of this area. The heating wire represents a concentrated point heat source.

1.40 Steps of a staircase are supported by two identical trusses made of 2-in-nominal-

diameter pipes (outside diameter = 2.375 in, wall thickness =. 0.154 in). One side E::o 'lqi e.G("l 5;

truss is shown in Figure 1.40. There are 12 steps each with 7.5 in rise and 11.5 in
run. The load from the steps is assumed to be applied as concentrated downward
loads of 100 lb at each nodal point. Determine nodal deflections, stresses in each
member, and support reactions.

Verify computer results by performing the following calculations by hand:

(a) Use computed reaction forces and applied external forces to check over all
equilibrium of forces.

(b) Isolate a typical node in the truss and draw a free-body diagram of the forces
acting at that node. Verify that the equilibrium is satisfied at the node.

(c) Pass a vertical section through the truss. Draw the free body diagram of the
truss on the right-hand section. Verify that the section is in equilibrium.

Figure 1.40. Staircase truss

1.41 The lower portion of an aluminum step bracket, shown in Figure 1.41, is subjected
to a uniform pressure 'of 20 Nzrnrrr'. The left end is fixed and the right end is free.

The dimensions of the bracket are thickness = 3 rum, L = 150 rum, b = 10 rum,

Figure 1.41. Step bracket

96 FINITE ELEMENTMETHOD: THE BIG PICTURE

'0~~l(JX..,t:J~I ~and h = 24 mm. Fillets of 3 mm radius are provided at the two reentrant comers.
'.<. l.:J" The material properties are E = 70,000 N/mm2 and v = 0.3. Determine stresses

M~ ,.J in the bracket using a planar finite element model (using say Plane42 element of
ANSYS). Compare solutions based on plane stress and plane strain assumptions.
R ':f':- (The situation can also be modeled using bearn/frame elements, three-dimensional
(.0

solid elements, and plate elements.)

1.42 The top surface of an S-shaped aluminum block, shown in Figure 1.42, is subjected
to a uniform pressure of 20 N/mm2 . The bottom is fixed. The dimensions of the

block are t = 3 mm, L = 15 mm, b = 10 mm, and h = 24 mm. The material
= =properties are E 70,000N/mm2 and v. 0.3. Determine stresses in the block using

a planar finite element model (using say Plane42 element of ANSYS). Compare

solutions based on plane stress and plane strain assumptions. (The situation can

also be modeled using bearn/frame elements, three-dimensional solid elements, and

plate elements.)

Figure 1.42. S-shaped aluminum block
1.43 Design optimization. Aluminum fins are to be used to cool an integrated circuit (IC)

chip as shown in Figure 1.43. On~ side of the chip is insulated. The chip dissipates
electrical energy at a rate of 150 W1m. The thermal conductivity of aluminum is
170 W1m . K. The convection coefiicient of the fin surface is 55 W1m2 • K, and the
ambient air temperature is 30·C.

ooling fms

C Chi
sulatiEn
Figure 1.43. Aluminum fins for cooling IC chip

PROBLEMS 97

Since the length of the fins is long, we can take a cross section and model as a
plane problem. Furthermore, taking advantage of symmetry, we need to model half
of the section, as shown in Figure 1.44. The bottom of the fin receives heat flux from
the IC chip. Because of symmetry, no heat can flow across the left side. Convection
heat loss takes place along the remaining exposed sides.

Figure'1.44. Planar model for aluminum fins for cooling Ie chip

The dimensions d, w, and t are fixed as follows:
d= 3mm; w=3mm; t=lmm

The design goal is to determine the optimum height h of the fins so that the temper-
ature ofthe IC chip does not exceed 70QC. Assume a starting value of h = 5 mm.

CHAPTER TWO

MATHEMATICAL FOUNDATION OF THE
FINITE ELEMENT METHOD

From a mathematical point of view the finite element method is a special form of the
well-known Galerkin and Rayleigh-Ritz methods for finding approximate solution of dif-
ferential equations. In both methods the governing differential equation first is converted
into an equivalent integral form. The Rayleigh-Ritz method employs calculus of variations
to define an equivalent variational or energy functional. A function that minimizes this en-
ergy functional represents a solution of the governing differential equation. The Galerkin
method uses a more direct approach. An approximate solution, with one or more unknown
parameters, is chosen. In general, this lassumed solution will not satisfy the differential
equation. The integral form represents the residual obtained by integrating the error over
the solution domain. Employing a criteria adopted to minimize the residual gives equations
for finding the unknown parameters.

For most practical problems solutions of differential equations are required to satisfy
not only the differential equation but also the specified boundary conditions at one or more
points along the boundary of the solution domain. In both methods some of the boundary
conditions must be satisfied explicitly by the assumed solutioris while others are satisfied
implicitly through the minimization process. The boundary conditions are thus divided
into two categories, essential and natural. The essential boundary conditions are those that
must explicitly be satisfied while the natural boundary conditions are incorporated into the
integral formulation. In general,therefore, the approximate solutions will not satisfy the
natural boundary conditions exactly.

The basic concepts will be explained with reference to the problem of axial deformation
of bars. The derivation of the governing differential equation is considered in the next sec-
tion. Approximate solutions using the classical form of Galerkin and Rayleigh-Ritz meth-
ods are presented. Finally the methods are cast into the form that is suitable for developing
finite element equations.
98

AXIAL DEFORMATION OF BARS 99

2.·1 AXIAL DEFORMATION Of BARS

2.1.1 Differential Equation for Axial Deformations

Consider a bar of any arbitrary cross section subjected to loads in the-axial direction only,
as shown in Figure 2.1. The area of cross section is denoted by A and it could vary over
the length of the bar. The modulus of elasticity is denoted by E. The bar may be subjected
to a distributed axial load q(x, t) along its length. The load can vary over the bar length and

may also be a function of time. The axial displacement is denoted by u(x, t).

The governing differential equation can be written by considering equilibrium of a dif-
ferential element as shown in Figure 2.2. Note the sign convention adopted in drawing the
free-body diagram assumes that the tension in the bar is positive. The axial force at x is
denoted by F. The axial force at x + dx is F + (aFlax) dx based on the Taylor series expan-

asion. The acceleration is indicated by ii == 2ul at 2. From Newton's second law of motion

a force, called the inertia force, is produced that is proportional to the mass of the element
and acts in the direction opposite to the direction of motion. Denoting the mass density

= =by p, the mass per unit length is ni Ap dx and thus the inertia force is mii Ap dx ii.

The only other force acting on the element is the applied axial load q(x, t). Considering
summation of forces in the x direction, we have

ax= =..
. aF .. aF
A(x)p dx u(x, t) + F q dx + F + ax dx =? Apu q +

The axial force is related to the axial stress 0:.: as follows:

F =ACT.'r

Axial load, q(x,t)

x, u(x,t)

EA(x)
Figure 2.1. Axially loaded bar

F

m=Apdx
I--dx ---l
Figure 2.2. Forces acting on a differential element in an axially loaded bar

100 MATHEMATICAL FOUNDATION OF THE FINITE ELEMENTMETHOD

Assuming linear elastic material, axial stress is related to the axial strain Ex through the
modulus of elasticity. Thus

Assuming small displacements, the axial strain is related to the first derivative of the axial
displacement. Thus the axial force is related to displacement as follows:

au
F =AE ax

Substituting this into the equilibrium equation, the governing differential equation for axial
deformation of bars is as follows:

aax (AE aaxu)+q =Apu..

This is a second-order partial differential equation. Since the equation involves second-
order derivatives with respect to both x and t, we need two boundary conditions and two
initial conditions for a proper solution. The initial conditions are specified displacement

=and velocity along the bar at time t 0:

it(x,O) = Vo

where Uo and Vo are the specified values. The boundary conditions involve specification of

displacement or its first derivative at the ends. Since the first derivative of displacement is

related to the axial force, the derivative boundary condition is expressed in terms of the

applied force. The possible boundary conditions at the right end of the bar x = XI are as

follows: /

or XI -? right end of the bar

where u; is a specified displacement and Pxf is a specified force. Both these quantities could
'~l
be functions of time. Care must be exercised in assigning proper signs to force boundary

condition terms. Applied forces are considered positive when they act in the positive co-

ordinate direction. From the free-body diagram in Figure 2.2, it should be clear that, if a

force is specified at the left end of a bar, then the appropriate force boundary condition

must include a negative sign as follows:

oX -? left end of the bar

If the forces and displacements do not vary with time, we have a static analysis situation
and the equilibrium equation is an ordinary second-order differential equation as follows:

d ( dU)dx AE dx + q(x) = 0; Xo < x < XI

AXIAL DEFORMATION OF BARS 101

with the boundary conditions of the form

or

or

An integration of the second-order differential equation, if possible, would yield two arbi-
traryconstants, and we would need two conditions to evaluate these constants. Thus at least
two boundary conditions must be specified for a unique solution. On physical grounds, it is
easy to see that we must specify displacement at least at one point along the bar to prevent
the whole bar from moving as a rigid body. The second bouridary condition could be either
of the displacement or the force type. Note that at the same point both a displacement and
a force cannot be specified independently. The reason for this is that, if a displacement is
specified, then the corresponding force represents a reaction at the point and ingeneral is

2.1.2 Exact Solutions of Some Axial Deformation Problems

It is possible to directly integrate the differential equation to obtain exact solutions for the
axial deformation problems in which the loading and the area of cross section are simple
functions of x. Solutions for a uniform bar and a tapered bar subjected to linearly varying
load are presented in this section. In later sections these exact solutions will be used to
check the quality of the approximate solutions that are obtained using the Galerkin, the
Rayleigh-Ritz, and the finite element methods.

Axial Deformation of a Uniform Bar Consider a uniform bar fixed at one end and
subjected to a static point load at the other end, as shown in Figure 2.3. The bar is also
subjected to a linearly varying axial load q(x) = ex, where e is a given constant. The
problem is described in terms of the following boundary value problem:

u(O) = 0; O<x<L

EA du(L) =P

dx .

'}---I»-P

L £!>/
Figure 2.3. Uniform axially loaded bar

102 MATHEMATICAL FOUNDATION OF THE FINITE ELEMENTMETHOD

An exact solution of the problem can easily be obtained by integrating the differential
equation twice and then using the boundary conditions to evaluate the resulting integration
constants. Integrating both sides of the differential equation once, we get

2.

EA du + ex =C
dx 2
I

where C1 is an integration constant. Integrating once again, we get

EAu(x) + c;2 =C1x + C2

(5

where C2 is another integration constant. Rearranging terms,

Using the boundary conditions £1(0) = 0 ~ C2 = 0
and

Thus the exact solution of the problem is

_ x(6P + 3eL2 - ex2)

u(x) - , 6EA

!

Axial Deformation of a Tapered Bar Consider a tapered bar fixed at one end and
subjected to a static point load at the other end, as shown in Figure 2.4. The bar is also

subjected to a linearly varying axial load q(x) = ex, where e is a given constant. The

problem is described in terms of the following boundary value problem:

=ddx (EA(x) ddUx ) + ex 0; O<x<L

=A(x) Ao - A_O_- A_LX
L

£1(0) = 0; dueL) _

EAL t;lx -P

where Ao is the area of cross section at x =0 and AL is that at x =L.

Integrating both sides of the differential equation once, we get

E (Ao - -Ao-L--A Lx)d-dUx+e2-x?-_ C

1

AXIAL DEFORMATION OF BARS 103

p

Figure 2.4. Tapered axially loaded bar
where C1 is an integration constant. Straightforward integration once again is not possible
because of the presence of the x du/dx term. However, using more advanced techniques
for solution of ordinary differential equations, it is possible to obtain the following general
solution in terms of two integration constants C1 and C2 :

Using the boundary conditions, the integration constants are evaluated as follows:

The exact solution of the problem can now be written as follows:

1.

u(x)=- 3 (L(c(-1+r)x(-2L+(-1+r)x)
4(-1 + r) EAo

+ 2(2P('-1 +:)2 + cL2(-2 + r)r) log[L]

- 2(2P(-1 + r)2 + cL2(-2 + r)r) log[L + (-1 + r)xJ))

=where r AJAo' the ratio of areas at two ends of the bar:

The only difference between this example and the previous one is that here the area

of cross section of the bar varies linearly. The solution is much more difficult to obtain

and the final solution expression is quite complicated as well. This example demonstrates

how quickly analytical solutions become difficult to obtain and shows the importance of

numerical methods for practical problems.

As should be expected, by taking the limit as Ao -) AL == A, i.e., r -) 1, it can be shown

that this solution reduces to the one for a uniform bar presented in the previous section.

Also note that, when c = 0 (no distributed axial load), the axial force at any point in the

bar is equal to P, which obviously should be the case from statics.

104 MATHEMATICAL FOUNDATION OF THE FINITEELEMENTMETHOD
u(x)

2 r=0.2

1.5 r=O.4

0.5 r=0.6
0.2 0.4 0.6 0.8 r=0.8
r=1.2
r=2

r=6
r=10

x

Figure 2.5. Solutions of tapered axially loaded bar subjected to end load

Solutions for several values of r are plotted in Figure 2.5 with the following numerical
values:

c = 0; p= 1; Ao = 1; L= 1; E=1

2.2 AXIAL DEFORMATION OF BARS USING GALER KIN METHOD

The governing differential equation for axial deformation of bars is the following second-

order differential equation: ./

d ( dU) + q =:= 0;

dx AE dx

where Xoand XI are the coordinates of the ends of the bar. The primary unknown is the axial

displacement u(x). Once the displacement is known, axial strain, stress, and force can be

computed from the following relationships: .

du F =AD;:
E =d-x''
x

At the ends either a displacement or an axial force can be specified. Thus the boundary
conditions for the problem are of the following form:

or

or

. AXIAL DEFORMATION OF BARS USING GALERKIN METHOD 105

WhBre uxQ' P.tO' ... are appropriate specified values. Mathematically spealcing,for a second-

order differential equation either u is specified or its first derivative du/dx is specified. Both
u and du/dx cannot have specified values at the same point.

2.2.1 Weak Form for Axial Deformations

In the Galerkin method we assume a general form of the solution. This assumed solution
must contain some unknown parameters whose values are determined so that the error
between the assumed solution and the exact solution is as small as possible. The assumed
solution can be of any form. As an example, we will consider a solution in the form of a
polynomial:

where ao'al , ..• are the unknown parameters. In general, this assumed solution will not sat-
isfy the differential equation for all values of x. When this assumed solution is substituted
into the differential equation, the error in satisfying the governing differential equation is

'*d( dft)e(x) ;= dx AE dx + q(x) 0

The tilde (~) over u is used to emphasize that it is an assumed solution and not necessarily
the same as the exact solution of the differential equation. In the following development we
. will have to perform several mathematical operations, such as differentiation and integra-
tion, on the assumed solution. Carrying the tilde symbol through all these manipulations
becomes tedious. Since we are dealing primarily with the approximate solutions, for nota-
tional convenience, we will drop the tilde and use only u instead to indicate an approximate
solution. An occasional reference to the exact solution will be indicated by using the word
exact explicitly..

The total error, called the residual, for the entire solution domain can be obtained by
integrating e(x) over the domain. However, in a straight integration of e, the negative and
positive errors at different points may cancel each other. To avoid error cancellation, e(x) is
multiplied by a suitable weighting function and then integrated over the solution domain.
Since there are n unknown parameters, we need n weighting functions to establish weighted
residual equations as follows:

i =0, 1, ... , n

where wi(x) are suitable weighting functions. This form is known as the weak form. The
differential equation is the strong form because it requires error term e(x) to vanish at every
x. The integral form makes the total error go to zero but does not necessarily satisfy the
governing differential equation for all values of x.

In a method known as the least-squared weighted residual method the error term is
squared to define the total squared error as follows:

106 MATHEMATICAL FOUNDATION OF THE FINITEELEMENTMETHOD

The necessary conditions for the minimum of the total squared error give n equations that
can be solved for the unknown parameters:

.i = 0, 1,...

Thus in the least-squares method the weighting functions are the partial derivatives of the
error term e(x).

Least-squares weighting functions:

i =0, l, ... .n

A more popular method in the finite element applications is the Galerkin method. In this
method, instead of taking partial derivatives of the error function, the weighting functions
are defined as the partial derivatives of the assumed solution.

Galerkin weighting functions:

i = 0,1, ... , n

Thus the Galerkin weighted residual method defines the following n equations to solve for
the unknown parameters:

i = 0,1, ...

It turns out that for a large number of engineering applications the Galerkin method gives
the same solution as another popular method, the Rayleigh-Ritz method, presented in a
later section. Furthermore, since the least-squares method has no particular advantage over
the Galerkin method for the kinds of problems discussed in this book, only the Galerkin
method is presented in detail.

So far in developing the residual we have considered error in satisfying the differential
equation alone. A solution must also satisfy boundary conditions. In order to be able to
introduce the boundary conditions into the weighted residual, we use mathematical manip-
ulations involving integration by parts.

AXIAL DEFORMATION OF BARS USINGGALERKINMETHOD 107

'I'he integration-by-parts formula is used to rewrite an integral of a product of a deriva-
tive of a function, say f(x), and another function, say g(x), as follows:

Note that the integrand must involve the product of a function and the derivative of
another function. The application of the formula produces two terms that are evaluated
at the ends of integration domain and another integral in which the derivative shifts from
one function to the other.

For a second-order differential equation, we know that the boundary conditions specify

either u or du/dx at the ends. Thus we are looking for a way of introducing these terms into

the weighted residual. Since the integration-by-parts formula gives rise to boundary terms,

it is precisely the tool that we need. Note that the highest derivative term in the residual

e(x) involves a second derivative on u. Using integration by parts on this term will result

in two terms evaluated at the integration limits that are nsed to incorporate the boundary

conditions into the residual. .

For the axial deformation problem, the weighted residual is

L:I (~ (AE~~)e(x)w/(x)dx = LXI + q(X))W/(X) dx = 0; i=O,l, ...

Note that, in general, A and E could be functions of x, and therefore, care must be taken
when carrying out differentiation and integration. Also q(x) and w/(x) are arbitrary func-
tions of x at this stage and cannot be taken out of the integral. Writing the two terms in the
weighted residual as separate integrals, we have

XI d (dU) LXI . =. dx AE dx wi(x)dx + q(x)wi(x) dx 0; i = 0,1, ...
LXo Xo

The first integral contains the second-order derivative on u and is written exactly in the

form to which integration by parts is applicable with f(x) = AE(du/dx) and g(x) = w/(x).

Thus the application of integration by parts to the first integral gives

Jx(XoI J(XI °A(x/)E(x/)
du(x) wi(x/)-A(xo)E(dxuo(x)o~) w/(xo)- du dw.
d/ AE dx dxl dx +
q(x)w/(x)dx=
Xo

The first two terms in the weak form give us a way to incorporate the specified force or
derivative boundary conditions into the weak form. If a force PxO is applied at end xo' then

-A(xo)E(xo)d~u(xo) = P.to:=;. Second term in the weak form: P.tow;Cxo)

If a force px/ is applied at end xI' then

A(x/)E(x/) dL~~/) = PXI :=;. First term in the weak form: P.t Wi(x/)
l

108 MATHEMATICAL FOUNDATION OF THE FINITE ELEMENTMETHOD

Thus, when a force is specified, the boundary condition can be naturally incorporated
into the weak form. Hence this type of boundary condition is called a natural boundary
condition (NBC).

The situation is very different if the u is specified at one or both ends. The derivative
du/dx at the corresponding point is unknown. (Physically AE du/dx at the point represents
the unknown reaction.) The specified displacement boundary condition therefore cannot
be incorporated into the weak form directly. The assumed solutions must satisfy this type
of boundary condition explicitly, and thus such a boundary condition is called the essential
boundary condition (EBC). An assumed solution that satisfies the EBC is known as an
admissible solution. Since the weighting functions are partial derivatives of the assumed
solution, with an admissible assumed solution, all weighting functions corresponding to
the location of an EBC are zero. Therefore, the boundary term in the weak form vanishes
at the point where anEBC is specified. Thus we must deal with the boundary terms as
follows:

Boundary Specified Boundary Term in Requirement on the

Condition Term Value the Weak Form Assumed Solution 'IYpe

-A(xo)E(xo)C!Ld;o) P.to P.towj(xo) None NBC

A(x/)E(x/) dL~~/) Px, PX,wj(x/) None NBC
uXo None Must satisfy EBC
u(xo) None Must satisfy EBC
!t(x/)

For structural problems, the weak form can be interpreted as the well-known principle of
virtual displacements. To see this, assume we have specified force boundary conditions at
the ends. Then the weak form can be Written as follows:

Rearranging terms, we have

If we interpret wj(x) as a virtual displacement, then the right-hand side is the virtual work
done by the applied forces. The left-hand side is the total internal virtual work, since
E du/dx == 0:;, is the axial stress and dw.rdx is axial virtual strain. Thus the weak form
implies that when a bar is given a virtual displacement, then the external virtual work is
equal to the total internal virtual work, which is a statement of the principle of virtual
displacements. Since this principle is widely used in structural mechanics, it is one of the
main reasons why the Galerkin weighted residual. method is more popular in developing
finite element equations. Also recall that the virtual displacements are required to satisfy
the displacement (i.e., essential) boundary conditions.

AXIAL DEFORMATION OF BARS USING GALERKIN METHOD 109

From the final weak form we note that another advantage of the integration by parts is
that the order of the derivative on the terms remaining inside the integral sign is reduced
by 1. Thus for a second-order problem the weak form involves only first-order derivatives.

aIt may not appear to be a big deal here, but it has important consequences in develop-

ing simple finite elements for practical problems. It will be seen in later example that,
when dealing with a fourth-order problem, integration by parts must be carried out twice
to reduce the highest order derivative to 2.

2.2.2 Uniiorm Bar SUbjected to linearly Varying Axial load

We now use the Galerkin method to find approximate solutions for a uniform bar (EA
constant) fixed at one end and subjected to a static point load at the other end, as shown
in Figure 2.3. Thebar is also subjected to a linearly varying axial load q(x) = ex, where
e is a given constant. The problem is described in terms of the following boundary value
problem:

O<x<L

u(O) = 0; EA dueL) =P

dx

The following exact solution of the problem was obtained in Section 2.1:

u(x) _ x(6P + 3eL2 - ex2)
- .
6EA

=With the solution domain fromIf), L), the EA constant, q(x) ex, the essential boundary

condition u(O) = 0 => w(O) = 0, and the natural boundary condition EAu' (L) =P, the weak

form specific to this problem is as follows:

lLPw;(L) + (~AE~: ~:i + eXWi(X))dX=0

EBC: u(O) =0

This weak form can now be used to find a variety of approximate solutions to the problem.

Linear Solution The simplest possible solution that we can assume is a linear poly-
nomial. An approximate solution of the problem is obtained using the following starting
solution:

u(X) =ao +xa l

To satisfy the essential boundary condition, we must have

u(O) = 0 ==> ao + Oal = 0 ==> ao = 0

Thus the admissible assumed solution is

u(x) = xa l giving

110 MATHEMATICAL FOUNDATION OF THE FINITE ELEMENTMETHOD

Substituting into the weak form, we have

Jo(L(Pw;CL) + =-AE(a1) ddwxi + cxw j ) dx 0

There is only one unknown parameter left in the solution and therefore we need only one
equation to find it. The Galerkin weighting function is

axaWl --1',

Substituting this into the weak:form, we have

PL + lL(-AEa l + c2)dx = 0,

Carrying out integration and simplifying, we get

Solving this equation for a p we get
-CL2 - 3P
3EA

Thus a linear approximate solution for the problem is as follows:

(cL2 + 3P)x

u(x) =~r1: = 3EA

Quadratic Solution A better solution can be obtained if we start with a quadratic poly-
nomial:

To satisfy the essential boundary condition, we must have

Thus the admissible assumed solution is

Substituting into the weak form; we have

'AXIAL DEFORMATION OF BARS USINGGALERKIN METHOD 111

There are two unknown parameters left in the solution and therefore we need two equations
to find them. We get these equations by using the two Galerkin weighting functions

axaWl = I:' WI(O) =0;

ow =2.:c' W2(0) = 0;
_ax2

Substituting WI into the weak form, we have

Substituting w2 into the weak form, we have

Solving the two equations, we have

-7cL2 -12P
l2EA

Thus a quadratic approximate solution is as follows:

1)- 2+ x - - cL 2_ - 7 c Ll 22E-Al 2 P x _- (12P+cL(7L-3x))x
I (x - a l - 4EA x l2EA
a2A

Cubic Solution The exact solution of the problem is a cubic polynomial. As demon-
strated below, the Galerkin method finds an exact solution if we start with a cubic polyno-

mial:

To satisfy the essential boundary condition, we must have

Thus the admissible assumed solution is

Substituting into the weak form, we have

112 MATHEMATICAL FOUNDATION OF THE FINITEELEMENTMETHOD

There are three unknown parameters left, which we find by using the three Galerkin weight-
ing functions into the weak form, giving the following equations:

au aWl -1' WI(O) = 0; wI(L) = L
wI =aa-l =x; ax - , wz(L) = LZ
Wz(O) =0;
Wz= -au =xz. awz _ 2x' =w3(L) L3
aaz ' ax - , W3(0) = 0;

W3 = -au =x3 . aw =3xz;
aa3
' _3
ax

Substitute admissible solution and weights into the weak form and perform integrations to

get the following: .

Weight Equation

t(e - 3EAa3)L3 - EAazL2 + PL-EAaIL= 0

~(e - 6EAa3)L4 -1EAa2L3 + PL2 - EAaIL2 =0

!(e - 9EAa3)L5 - ~EAa2L4 + PL3 - EAa IL3 = 0

Solving these equations,

-eL2 -2P

a l =- 2EA

Thus a cubic solution is as follows:

Since this is the exact solution, trying any higher order polynomial will not make any

difference.
Carefully note the distinction between the two types of boundary conditions. To make

the assumed solution admissible, we required it to satisfy only the displacement boundary

condition [u(O) = 0]. We never explicitly require the assumed solution to satisfy the force

boundary condition. We can easily see that for this example the linear and quadratic solu-
tions in fact do not satisfy this boundary condition. Only the cubic solution satisfies this

condition exactly.

u(x) au/ax au(L)/ax EA au(L)/ax
(eL2 + 3P)x
Linear eLz + 3P eL2+ 3P eL2
Quadratic 3EA 3EA -+P
Cubic (12P + eL(7L - 3x))x 3EA 3
4eL2 + 12P eL2
12EA 12P + eL(7L - 3x) eLx 12EA 4EA eL2 P
x(3eL2 - ex2 + 6P) 12+
12EA 4EA 2eL2 + 6P eL2
6EA 6EA 3EA
3eL2 - ex2 + 6P ex2 P

6EA 3EA

AXIAL DEFORMATION OF BARS USINGGALERKIN METHOD 113

The logic in using the terminology essential and natural boundary conditions should now
be clear. Essential boundary conditions are those that must explicitly be satisfied by the as-
sumed solution while the natural boundary conditions are only implicitly satisfied through
the weak form. A solution that satisfies the differential equation and all boundary condi-
tions is obviously the exact solution.

2.2.3 Tapered Bar SUbjected to linearly Varying Axial Load

Consider now the tapered bar fixed at one end and subjected to a static point load at the
other end, as shown in Figure 2.4. The bar is also subjected to a linearly varying axial load
q(x) : ex, where e is a given constant. The problem is described in terms of the following
boundary value problem:

d( dU)dx EA dx + ex : 0; 0 < x < L

A(x) : A _ Ao - ALx: (L + (-1 + r)x)Ao
oL
L

u(O): 0; ErA dueL) : P
o dx

where Ao is the area of cross section at x : 0, A L is that at x: L, and r : ALIAo. The weak

form specific to this problem is the same as that for the uniform bar in the previous section,
except that A is now a function of x:

(L(Jo _.PwJL) + du dw. + ) dx: 0
-AE dx d; exwj(x)

EBC: u(O): 0

Linear Solution Starting~ssumed solution: u(x) : ao+ xa 1

The admissible solution must satisfy EBC:

EBC Equation
u(O) ~ 0 ao : 0

Thus the admissible assumed solution is u(x) : xa 1•
Weighting function -7 {x}
Substitute into the wealeform and perform integrations to get:

Weight Equation
x
LP + eL4/3 - ( l/2 ) rE a 1A 2 - (l12)Ea 1A 2 : 0

oL oL

L

114 MATHEMATICAL FOUNDATION OF THE FINITEELEMENTMETHOD

Solving this equation,

a- _(CL3/3) - PL
1-
---(:-l:/-:2::-):L-=E-':A:::o-:----''(-:1-:/-2-:)-L:-.r-:E:-A=-:o-

Substituting into the admissible solution, we get the following solution of the problem:

u (x )32r=EcxAL-o2+-+3-6EP-Ax-o
.

Quadratic Solution Starting assumed solution: u(x) = a2x2 + a1x + ao

The admissible solution must satisfy the EBC:

EBC Equation

u(O) = 0

Thus the admissible assumed solution is u(x) = a2x2 + a1x.
Weighting functions -} (x,x2)

Substitute into the weak form and perform integrations to get:

Weight Equation

x

/

Solving these equations,

-cL2 - 6crL2 - 12Pr -CL2 + 7crL2 - 12P + 12Pr
2(r2 + 4r + I)EAo ' 4L(r2 + 4r + I)EA o

Substituting into the admissible solution, we get the following solution of the problem:

x(c(2L(6r + 1) - Trx + x)L2 + 12P(2Lr - xr + x))
u(x)= - ' - - ' - - ' - -.- -4' -L-(-?-+; 4;r- -+' -I)-E-A-o- ' - - - - - - ' - ' -

.The exact solution of the problem was derived earlier. The linear and quadratic solutions
are compared with the exact solution in Figure 2.6. The following numerical values are
used:

= =c = 0; P= 1; Ao 1; L= 1; E 1; r -- 12

The quadratic solution is reasonably close to the exact solution. Of course the solution can
be improved further by starting with a cubic or a higher-order polynomial.

ONE-DIMENSIONAL BVP USINGGALERKIN METHOP 115

u(x)

1.75 - - Quadratic
1.5
1.25 - - Exact

1
0.75

0.5
0.25

x

0.2 0.4 0.6 0.8

Figure 2.6. Solutions of tapered axially loaded bar subjected to end load

2.3 ONE-DIMENSIONAL BVP USING GALERKIN METHOD

So far only the axial deformation problem has been used to illustrate the Galerkin method.
The method in fact is applicable to any differential equation. It is particularly well suited
to boundary value problems (BVPs) in which a solution must satisfy differential equations
and several boundary conditions over the domain. This section first summarizes the overall
solution procedure and then presents several examples of finding approximate solutions of
one-dimensional boundary value problems.

2.3.1 Overall Solution Procedure Using Galerldn Method

Given a boundary value problem, the overall procedure for obtaining an approximate so-
lution using the Galerkin method is summarized here. Several examples in this section
further clarify the details.

(i) Construct the wealcform.
1. Move all terms in the differential equation to the left-hand side. With an as-
sumed solution the left-hand side now represents the residual e(x).
2. Define the total weighted residual. It consists of integral of the left-hand side of
the differential equation multiplied by a weighting function Wi'
3. Use integration by parts to incorporate boundary conditions. For a second-order
problem, all terms involving second-order derivatives are integrated by parts
once. The highest order of remaining terms in the residual should therefore
be 1. For a fourth-order problem integration by parts is used twice on all terms
involving fourth-order derivatives and once' on the terms involving third-order
derivatives. The highest order of remaining terms in the residual should there-
fore be 2.
4. Identify essential and natural boundary conditions.For second-order problems
boundary conditions involving first-order derivatives are natural because they

116 MATHEMATICAL FOUNDATION OF THE FINITEELEMENTMETHOD

can be incorporated into the weak form. For fourth-order problems boundary
conditions involving second- and third-order derivatives are natural. All other
boundary conditions are essential. Incorporate natural boundary conditions into
the weak form.

(ii) Construct an admissible assumed solution.

1. Start with an assumed solution with some unknown parameters and enough
terms in it so that it is possible to evaluate the residual. Any suitable function
can be used for the assumed solution. However, since polynomials are easy to
differentiate and integrate, it is convenient to start with a polynomial of desired
order.

2. Set some of the parameter values such that the essential boundary conditions
are satisfied regardless of the values of the remaining unknown parameters.
If there are no essential boundary conditions, then the solution already is an
admissible solution and we can go to the next step. Otherwise set up equations
to satisfy essential boundary conditions. Solve these equations for some of the
parameters. Substitute these parameter values back into the starting solution to
get an admissible assumed solution.

(iii) Set up equations and solve for the unlmown parameters.

1. Determine a set of weighting functions by differentiating the admissible solu-
tion with respect to the unknown parameters.

2. Set up a system of equations by substituting each weighting function in tum
into the weak form.

3. Solve the system of equations for the unlmown parameters.

(iv) Approximate solution.

1. Substitute computed values of the parameters into the admissible solution to
get the approximate solution,

2. Check the quality of the solution by substituting it into the differential equation
and the natural boundary conditions to see how well it satisfies them.

For convenience the prime and superscript notations for derivatives will be used fre-

quently in the following examples. Thus the first and second derivatives of a function
u(x) will be denoted as u'(x) == du(x)/dx and ul/(x) == d2u/dr. When third- or higher

order derivatives are written, the prime notation becomes cumbersome and therefore a
superscript notation will be adopted. For example, the third and fourth derivatives of a
function u(x) will be denoted as u(3)(x) == d3u/d~ and u(4)(x) == d4u/dx4.

Example 2.1 Second-Order Equation Find an approximate solution of the following
boundary value problem:

rul/ + 2xu' +x = 1; 1<x<2

u(l) = 2; u' (2) + 2u(2) = 5

ONE-DIMENSIONAL BVP USING GALER KIN METHOD 117

The weak form is derived first followed by detailed calculations of quadratic and cubic
approximate solutions.

With u(x) as an assumed solution, the residual is

=e(x) u"(x)x'l + 2u' (x)x + x-I

Multiplying by w;(x) and writing integral over the given limits, the Galerkin weighted
residual is

Using integration by parts, the order of derivative in x'lwju" can be reduced to 1 as follows:

Combining all terms, the weighted residual now is as follows:

Consider the boundary terms
4wP)u'(2) - wj(l)u'(l)

Given the NBC for the problem:

2u(2) + u'(2) - 5 = 0

Rearranging: (u'(2) -7 5 - 2u(2))
Given the EBC for the problem:

u(l) - 2 =0

Therefore with admissible solutions (those satisfying the EBC): (wj(l) -7 0)
Thus the boundary terms in the weak form reduce to

4(5 - 2u(2))w i(2)
and assuming admissible solutions, the final weak form is as follows:

1 =2

4(5 - 2u(2))wi(2) + «x - l)wi - x'lu'w;}dx 0
EBC: u(l) - 2 = 0

118 MATHEMATICAL FOUNDATION OF THE FINITEELEMENTMETHOD

=Quadratic Solution Starting assumed solution: u(x) a2x2 + alx + ao

The admissible solution must satisfy the EBC:

EBC Equation

u(l) - 2 =0

Solving this equation, ao = -al - a2 + 2
Thus the admissible assumed solution is u(x) = a2x2 + alx - a l - a2 + 2.

Weighting functions .... [x - 1, x2 - l}

Substitute into the weak form and perform integrations to get:

Weight Equation

x-I -~ - j~, + 4(5 - 2(a[ + 3a2 + 2)) + ~ = 0

x2 - 1 H-~ - l2~a, + 12(5 - 2(al + 3a2 + 2)) + = 0

Solving these equations, at = [i6~; a2 = -lstf3

Substituting into the admissible solution, we get the following solution of the problem:

u(x) = -1090x2 + 4533x + 2329

2886

Cubic Solution Starting assumed solution: u(x) = a3x3 + a2x2 + alx + ao

The adrriissible solution must satisfy the EBC:

,/

EBC Equation

Solving this equation, ao = -a j - a2 - a3 + 2
=Thus the admissible assumed solution is u(x) a3x3 + a2x2 + ajx - a l - a2 - a3 + 2.

Weighting functions .... {x - 1,x2 - 1, x3 - I}

Substitute into the weak form and perform integrations to 'g~t:

Weight Equation
x- 1
-~ - ~ - ~ + 4(5 - 2(a j + 3a2 + 7a 3 + 2)) + ~ = 0
x2 - 1
x3 - 1 H-~ - l2~a2 - 63a 3 + 12(5 - 2(a l + 3a2 + 7a 3 + 2)) + = 0

¥o.:-~ - 63a 2 - ll~al + 28(5 - 2(a j + 3a2 + 7a 3 + 2)) + = 0

8J?;Solving these equations, a l = 2g69 a2 = - n~6~; a3 = is4g554

ONE-DIMENSIONAL BVP USINGGALERKIN METHOD 119

e(x) - - Quadratic
0.5

1.2
-0.5

-1

-1.5

Figure 2.7. Error in satisfying the differential equation

Substituting into the admissible solution, we get the following solution of the problem:

=u(x) 22365i3 - 144070x2 + 322764x + 13765

107412

Substituting these approximate solutions into the differential equation, we get the error in
satisfying the differential equation. This error is plotted in Figure 2.7.

u(x) Error, e(x)

Quadratic _ 545.t'! + 1511.< + 2329 _ 1090..' + 1992x _ 1
Cubic
1443 962 2886 481 481

7455x3 _ 72035i' + 26897x + 13765 22365x3 _ 72035..' + 62745x _ 1
8951 8951 8951
35804 53706 8951 107412

To demonstrate convergence, solutions up to fifth order are computed. All solutions and
their derivatives are compared in Figure 2.8. The higher order solutions are almost on top
of each other, indicating the solution convergence. The first derivatives of solutions show
greater discrepancy, but they also converge as more terms are included in the assumed
solution .

• MathematicafMATLAB Implementation 2.1 on the Book Web Site:
Second-order BV? using the Galerkin method

2.3.2 Higher Order Boundary Value Problems

The example problems considered so far in this chapter have all been second-order dif-
ferential equations. They are called second order because the highest derivative present
in the differential equatiori is 2. For the second-order problems, a boundary condition in
which only u is specified is essential, and the one involving first derivative of u is called
natural. The assumed solutions are not required to satisfy the natural boundary conditions.
They are actually incorporated into the weak form and are satisfied approximately with
the approximation getting better as the number of parameters in the assumed solution is
increased.

120 MATHEMATICAL FOUNDATION OF THE FINITE ELEMENT METHOD

u(x) ------ U2
2.4 - - U3

2.3

- - U5

2.2
2.1

1.2 . 1.4 1.6 1.8 x
2

du zdx - - U3
1 --U5

~.

0 . 8 -..'.......

0.6

0.4

0.2

1.2 1.4 1.6 1.8 x
2

Figure 2.8. Comparison of solutions and their first derivatives

!

The Galerkin method applies to higher order differential equations as well. For example,
consider a fourth-order differential equation of the following form:

4u

d + 1= Q.
dx4 '
or

The superscript (4) over u(x) indicates fourth derivative of u(x) with respect to x.
With u(x) an assumed solution, the residual is

e(x) = u(4)CX) + 1

Multiplying by wi(x) and writing the integral over the given limits, the Galerkin weighted
residual is

ONE·DIMENSIONAL BVP USINGGALERKIN METHOD 121

Using integration by parts, the order of derivative in Wiu(4) can be reduced to 2 as follows:

Combining all terms, the weighted residual now is as follows:

Consider the boundary terms

Each one of these terms gives rise to two possibilities:

- Wi ( XO)U(3)(XO) Either -u(3)(xO) is known or wi(xO) = 0

wi (xo)u" (xo) Either u"(xo) is known or wi (xo) = 0
W/X,)U(3) (x,) Either u(3)(x,) is known or wi(x,) = 0

-Wi(x,)u" (x,) Either -u"(x,) is known or wi (x,) = 0

From these requirements the possible boundary conditions are as follows:

NBC EBe

1 -u(3)(xO) is given or w/xo) = 0 ===> Must satisfy u(xo) boundary condition

2 £I" (xo) is given =or wi(xo) 0 ===> Must satisfy u'(xo) boundary condition

3 u(3)(x,) is given Of- ...w/x,) = 0 ===> Must satisfy u(x,) boundary condition

4 -u"(x,) is given or wj(x,) = 0 ===> Must satisfy u'(x,) boundary condition

Thus for a fourth-order problem boundary conditions involving £I and £I' both are essential
while those involving £I" and u(3) are natural. A practical problem governed by a fourth-
order differential equation is that of beam bending. As will be seen in Chapter 4, for beams,
the bending moment is proportional to £I" while the shear force is proportional to £1(3). Thus
specified bending moments and shear forces represent natural boundary conditions and
displacements and slopes represent essential boundary conditions.

The generalization to further higher order differential equations should now be obvious.
For a general boundary value problem in which the highest derivative present is of the
order 2p (an even order), we should be able to carry out integration by parts p times.
Each integration by parts introduces boundary terms that give rise to essential and natural
boundary conditions. For the general case the classification of boundary conditions is as
follows:

122 MATHEMATICAL FOUNDATION OF THE FINITEELEMENT METHOD

. Those with the order from 0 to p - 1 are the essential boundary conditions .
. Those with the order from p to 2p - 1 are the natural boundary conditions.

The assumed solutions must satisfy all essential boundary conditions for any value of the
unknown parameters. Solutions that satisfy the essential boundary conditions and have the
necessary continuity for required derivatives are called admissible solutions.

Example 2.2 Fourth-Order Equation Find an approximate solution of the following
fourth-order boundary value problem:

=U(4)(X) + 8u"(x) + 4u(x) 10; O<x<S

= = = =with the boundary conditions u(O) 0; u'(O) 1; u(S) 2; u'(S) O. The superscript

(4) on u indicates its fourth derivative. The weak form is derived first followed by detailed
calculations of fourth- and fifth-order solutions.

With u(x) as an assumed solution, the residual is

e(x) = 4u(x) + 8u"(x) + u(4)(x) - 10

Multiplying by w;(x) and writing the integral over the given limits, the Galerkin weighted
residual is

Using integration by parts, the order of derivative in w;u(4) can be reduced to 2 as follows:

i i./ 5
5 (-w;u(3)) dx
=(w;u(4»)dx w;(S)u(3)(S) - w;(0)u(3)(0) +

i i5 5
=(-W;U(3») dx w;(O)u"(O) - w;(S)u"(S) + (u"w;') dx

Combining all terms, the weighted residual now is as follows:

i5

w;(O)u" (0) - w;(S)u" (S) - w;(O)zP)(O) +w;(S)u(3)(S) + (2w;(2u +4u" - S) +u"w;') dx = 0

Consider the boundary terms

Given the EBC for the problem:

=u(O) 0; u'(O) -1 =0; u(S) - 2 = 0; u'(S) =0

ONE-DIMENSIONAL BVP USINGGALERKINMETHOD 123

Therefore, with admissible solutions (those satisfying EBe)
w;(O) -7 0;

Assuming admissible solutions, the final weak form is as follows:

is (2wpu + 4u" - 5) + u"W;/) dx = 0

Fourth-Order Solution Starting assumed solution: u(x) = a4x4+a3x3 + azxz + a l x + ao

The admissible solution must-satisfy the EBC:

EBC. Equation

u(O) =0 aO = 0
u'(O) -1 = 0
u(5) - 2 =0 al -1 = 0 =0
u' (5) =0 + 51a0al z++275a5az3++l2550a03a4+=6205a4
ao + - 2
al

Solving these equations, ao =0; a] = 1; az =25a4 - ~; a3 = ]~5 -10a4

Thus the admissible assumed solution is

.x3 . 4xz

u(x) = a4x4 - 10a4.x3 + 125 + 25a4~ - 25 + x

Weighting function -7 {x4 - 10x3 + 25xz}

Substitute into the weak form and perform integrations to get:

Weight Equation
]88750a4 _ 23875 - 0

63 4Z-

=So1vm· g th'IS equati.on, a4 _}5Q7Z3O

Substituting into the admissible solution, we get the following solution of the problem:

u(x) = x(14325x3 - 142646x2 + 346045x + 75500)

75500

Fifth-Order Solution Starting assumed solution: u(x) = a5x5 + a4x4 + a3x3 + a2 2 +
x

alx+ ao

The admissible solution must satisfy the EBC:

EBC Equation

u(O) =0 aO = 0

u'(O) -1 = 0 a] -1 = 0

u(5) - 2 =0 ao + 5a] + 25a z + 125a3 + 625a4 + 3l25a5 - 2 =0
u' (5) = 0 a] + lOa2 + 75a3 + 500a4 + 3125a5
=0

124 MATHEMATICAL FOUNDATION OF THE FINITE ELEMENT METHOD

Solving these equations,

Thus the admissible assumed solution is

Weighting functions -7 {X4 - 1Ox3 + 25x2,.x5 - 75x3 + 250x2}

Substitute into the weak form and perform integrations to get:

Weight Equation

x4 - 1Ox3 + 25x2 188750a. + 2359375a, _ 23875 - 0

.x5 - 75x3 + 250x2 63 63 4 2 -

2359375"4 + 319843750a, _ 873625 - 0

63 693 126-

SoIvm· g these equati.ons, Q4 - - 818230355801'. Q5 - 1348612750
-
-

Substituting into the admissible solution, we get the following solution of the problem:

=u(x) x(576367x4 - 3014525x3 - 12905605x2 + 65195225x + 22083750)

22083750

Substituting these approximate solutions into the differential equation, we get the error in
satisfying the differential equation. A plot of the error shown in Figure 2.9 indicates that
both solutions have significant error. Since the differential equation is quite complicated,
we need a fairly high-order polynomial to get reasonable solution. To demonstrate conver-
gence, solutions up to 12th order are computed. These solutions are compared in Figure

e(x) _._- 4th order
- - 5th order
100
80 x

60 \.

40 \

\

20\"

Figure 2.9. Error in satisfying differential equation

ONE-DIMENSIONAL BVP USINGGALERKINMETHOD 125
'-'-'-' u6
ujx) - - Us
6
5 --'UlO
4 - - Ul2
3
2

x
2345

du jdx - - Us

--UlO

\ - - U12
-2 f\\\r\--j"i'Il,fr x
\,~ /\ I
-4 .~/'

Figure 2.10. Comparison of solutions and theirfirstderivatives

2.10. The LOth- and 12th-order solutions are fairly close to each other, indicating that they
represent good solutions to the problem.

Fourth-order . Error, e(x)
Fifth-order
14325x4-l42646x'+689845i'-1636252x+ 1281380
18875

2(576367i'-3014525x4+10149075.t'l-7153375i'-115492500x+187484375)
11041875

• MathematicafMATLAB Implementation 2.2 on the Book Web Site:
Fourth-order BVP using the Galerkin method

Example 2.3 Third-Order Equation The even-order equations, such as the second

and the fourth order, are common in engineering applications. However, the Galerki~

method can be applied to odd-order equations as well. To demonstrate this, we consider

the following third-order boundary value problem:

=U(3)(X) + 2u(x) ~; O<x<l

126 MATHEMATICAL FOUNDATION OF THE FINITE ELEMENTMETHOD

= = =with the boundary conditions u(O) 1; u(I) 2; u'(I) 3. The superscript (3) on u

indicates its third derivative.
The weak form is derived first followed by detailed calculations of a quadratic solution

in Mathematica.
With u(x) as an assumed solution the residual is

= +e(x) -:t? 2u(x) + u(3)(x)

Multiplying by wi(x) and writing the integral over the given limits, the Galerkin weighted

residual is

11( -wi 2 + 2uwj + wju(3))dx = 0
x

Using integration by parts twice on the W ju(3) term, we have

1 11 1
=(Wiu(3))dx wi(I)ul/(I) -wj(O)ul/(O) + (-w;'ul/)dx

1 = 11 1
(-W;'Ul/) dx u'(O)w;'(O) - u'(I)w;'(I) + (u'w/,)dx

Combining all terms, the weighted residual now is as follows:

11

=u'(O)w;'(O) - u'(I)w;'(l) - wj(O)ul/(O) + wi(I)ul/(l) + (u'w/, - (:t? - 2u)wj) dx 0

=The boundary condition u'(1) 3 can be directly incorporated into the weak form, giving

1. . 1 •

u'(O)w;(O) - 3w;(l) - wj(O)ul/(O) +)'vi(l)ul/(l) + (u'w;' - (:t? - 2u)w)dx = 0

The other two boundary conditions are considered essential:

WeakForm[w_, u_] ;=
Simplify[«D[u, x] D[w, x])/.x-?O)- «3D[w, x])/.x-? 1)

- «D [u , {x , 2}] v) I. x -? 0) +( (D Iu, {x , 2}] w)I. x -? 1)

+ Integrate [D[u , x] D[w, {x , 2}] - (x-2 - 2u)w, {x , 0, 1}]]

We start with a quadratic polynomial ux with three parameters aD' aI' and a2 as follows:
ux = aO + a1 x + a2 x-2;

The first task is to make it an admissible solution. We must satisfy the two essential bound-
ary conditions:

ebc1 = (uxz . x-« 0) == 1

aO == 1

ebc2= (ux/vx-» 1) ==2

aO + al + a2 == 2

ONE-DIMENSIONAL BVP USING GALERKIN METHOD 127

Solving these equations, we get
sol = Solve [{ebc1, ebc2} , {aO, a1}J
({aO ~ 1, a1 ~ 1 - a2)}
Substituting the solution into ux gives an admissible solution as follows:

u=ux/. so l I [1J J
a2x2 + (l - a2)x + 1

Now we can set up the equation needed to find the remaining coefficient:

-w=D[u, a2J; eq1=WeakForm[w, uJ ==0

1.

60 (64 a2 - 147) == 0

Solving this equation,
sol = Solve [{eqf.} , {a2}J

{{a2~ l;:n

Substituting the solution into the assumed admissible solution, we have a quadratic ap-
proximate solution of the problem as follows:
ux = Simplify [ul . sol [[1J J J

614(147x2 - 83x + 64)

It is possible to obtain an exact analytical solution of the problem as follows:
Clear[uJ; soi=DSolve[uJ' [xJ +2u[xJ ==x-2, u Ixl , x] [[1, 1, 2JJ;
exactSol = Simplify [soli . Solve [{ (5011. x ~ 0) == 1,

(soli .x~ 1) ==2, D[sol, xJI .x~ 1) ==3}J [[1]JJ

128 MATHEMATICAL FOUNDATION OF THE FINITEELEMENTMETHOD

u(x) - - - - Approximate ;/
2 - - - Exact
1.8
1.6

/1.4

1.2

%'

0.6 0.8

Figure 2.11. Comparison of the approximate and the exact solutions

{D[exactSol, {x, 3}] + 2 exactSol, exactS 011 . x -70,

exactSol/ . x -7 1, D[exactSol, xl 1.x -7 1}1ISimplify

{x2, 1,2, 3}

As can be seen from Figure 2.11, the quadratic approximate solution compares very well
with the exact solution. A cubic solution will almost be indistinguishable from the exact
solution:

Needs ["Graphics 'Legend' "] ; __
Plot [{ux, exactSol}, {x , 0, 1}, PlotStyle -7 {Hue [0.6] , Hue [0. 9]},

AxesLabel-7 {"x", "u Cx) "}, PlotLegend -7 {"Approx", "Exact"},
LegendPosition -7 {-1/2, O}, Legendshadow -7 None,
PlotRange -7 All] j

2.4 RAYLEIGH-RITZ METHOD

As is the case with the Galerkin method, the Rayleigh-Ritz method also requires an equiv-
alent integral formulation. However, the derivation of this equivalent integral form is based
on fundamentally different concepts. For many practical problems,the integral form is
based on energy considerations. For structural problems, the potential energy is an example
of such an integral form that has been used successfully in developing a large number of
finite elements. Similar energy-based integral forms are available for other applications.
If the energy function is not known from physical considerations, it may still be possible
to derive an equivalent integral formulation using mathematical manipulations. However,
such manipulations involve a branch of calculus known as the calculus of variations. A
review of basic calculus of variations and mathematical manipulations to derive integral
formulations is presented in Appendix B. In the main body of this text the use of the
Rayleigh-Ritz method will be restricted to structural problems for which the potential en-
ergy function is well known.

RAYLEIGH·RITZ METHOD 129

"Once the equivalent integral form is known, the steps in the Rayleigh-Ritz method are
siillilar to those in the Galerkin 'method. A solution is assumed with some unknown pa-
rameters. The assumed solution is made admissible by requiring that it explicitly satisfy
the essential boundary conditions. The energy form reduces to a function of unknown pa-
rameters once the assumed solution is substiiuted into the energy function. The necessary
conditions for the minimum of this function then give the required equations for finding
suitable values for the unknown parameters.

2.4.1, Potential Energy for Axial Deformation of Bars

Consider a linear elastic bar of any arbitrary cross section subjected to loads in the axial
direction only, as shown in Figure 2.1. The area of cross section is denoted by A(x) and
could vary over the Iength of the bar. The modulus of elasticity is denoted by E. The bar
may be subjected to distributed axial load q(x) along its length. In addition, there may be
concentrated axial loads Pi applied at some locations Xi on the bar. The axial displacement

'is denoted by u(x). Denoting the axial stress by 0:, and corresponding strain by Ex, the axial

strain energy is defined as follows:

where the integration is over V, the volume of the bar. For axial deformations stress and
strain are assumed to be constant over a cross section, and thus the volume integral can be
expressed as a one-dimensional integral

U=l2Lx'a:Ex Axl:.1d. X

Xo

where Xo and x, are end coordinates of the bar. Using Hooke's law and the strain-

displacement relationship 0:, =EE., =E du/dx yields

The potential of the applied loads is equal to the negative of the work done by all external '
applied forces:

Lx,

W = qu dx + Ipiu(x)

Xo

where u(x) is the axial displacement at the location of the concentrated applied load Pi'

Using these expressions, the potential energy for axial deformation of bars is defined as

follows: .

(X, (du)2 (X, :

EA(x) dx dx - },. q(x)u(x) dx -
1 ILII =U - W = Piu(x)

Xo Xo

130 MATHEMATICAL FOUNDATION OF THE FINITEELEMENTMETHOD

Using calculus of variations, it is possible to show that a function u(x) that minimizes the
potential energy is a solution of the differential equation governing the axial deformation
of bars.

2.4.2 Overall Solution Procedure Using the Rayleigh-Ritz Method

Given a boundary value problem, the overall procedure for obtaining an approximate solu-
tion using the Rayleigh-Ritz method is as follows:

(i) Obtain the equivalent energy form. For structural problems use the potential en-
ergy. For general boundary value problems use calculus of variations to construct
an equivalent energy form (see Appendix B). Clearly identify the essential and
natural boundary conditions.

(ii) Construct an admissible assumed solution.

I. Start with an assumed solution with some unknown parameters and enough
terms in it so that it is possible to evaluate the terms in the energy form. Any
suitable function can be used for the assumed solution. However, since poly-
nomials are easy to differentiate and integrate, it is convenient to start with a
polynomial of desired order.

2. Set some of the parameter values such that the essential boundary conditions are

satisfied regardless of the values of the remaining unknown parameters. If there

are no essential boundary conditions, then the solution already is admissible

and we can go to the next step. Otherwise set up equations to satisfy essential

boundary conditions. Solve these equations for some of the parameters. Substi-

tute these parameter values ~ack into the starting solution to get an admissible

assumed solution. '

(iii) Set up equations and solve for the unknown parameters.

1. Substitute the admissible assumed solution into the energy form and carry out
the integration. The resulting expression should be a function only of the un-
known parameters.

2. Set up a system of equations by using necessary conditions for the minimum of
the energy form. Since the minimum of the energy function corresponds to the
solution of the boundary value problem, we get a system of equations by setting
to zero the partial derivatives of the energy function with respect to unknown
parameters.

3. Solve the system of equations for the unknown parameters.

(iv) Approximate solution.

I. Substitute computed values of parameters into the admissible solution to get the
approximate solution.

2. Check the quality of the solution by substituting it into the differential equation
and the natural boundary conditions to see,how well it satisfies them.

RAYLEIGH-RITZ METHOD 131

2.-4.3 Uniform Bar Subjected to Linearly Varying Axial Load

We now use the Rayleigh-Ritz method to find approximate solutions of a uniform bar (EA

constant) fixed at one end and subjected to a static point load at the other end, as shown in

Figure 2.3. The bar is also subjected to a linearly varying axial load qEx) = ex, where e is a

given constant. The potential energy for the problem is as follows:

I r (du)2 rL
2"EA
=Jo Jon
dx dx - e xu dx - Pu(L)

EEC: u(O) = O.

The simplest possible solution that we can assume is a linear polynomial. Using this as the
starting solution, an" approximate solution of the problem is obtained as follows:

=Linear Solution Starting assumed solution: u(x) ao + xa 1

The admissible solution must satisfy the EEC:

EEC Equation

u(O) =0 ao =0

=Thus the admissible assumed solution is u(x) xa 1•

Substituting the admissible solution into IT and carrying out integration,

For the miriimum of IT, set its derivatives with respect to parameters to 0:

Solving this equation,

Substituting into the admissible solution, we get the following solution of the problem:

(eL2 + 3P)x
u=

3EA
A better solution can be obtained if we start with a quadratic polynomial:

132 MATHEMATICAL FOUNDATION OF THE FINITE ELEMENTMETHOD

Quadratic Solution Starting assumed solution: u(x) == a2x2 + alx + ao

The admissible solution must satisfy the EEC:

EEC Equation
u(O) == 0

Thus the admissible assumed solution is u(x) == a2x2 + alx.
Substituting the admissible solution into Il and carrying out integration,

For the minimum of Il, set its derivatives with respect to parameters to 0:

an == 0: -C3L3 + EAa2L2 - PL + EAalL == 0

Ba,

aaan == O: - 4cL4 PL2 + EAa lL2 == 0

. + 1EAa2L3 -

2

Solving these equations,

-7cL2 - 12P
12EA

Substituting into the admissible solution, we get the following solution of the problem:

u == (l:?? + cL(7L - 3x))x

12EA

The exact solution is a cubic polynomial. As demonstrated below, if we start with a cubic
polynomial, the Galerkin method finds this exact solution:

Cubic Solution Starting assumed solution: u(x) == a3x3 + a2x2 + alx + ao

The admissible solution must satisfy the EEC:

EEC Equation
u(O) == 0

Thus the admissible assumed solution is u(x) == a3x3 + a2x2 + alx. \
\
Substituting the admissible solution into n and carrying out integration,

n == foa3(9EAa3 - 2c)L5 + ia2(6EAa3 - c)L4 + ~(2EAa~ - cal + 3EAa la 3
3)L

+ EAa 2 + ~EAaiL - P(a3L3 + a2L2 + alL)
la2L