1) The document discusses the theory of elasticity and its application to finite element analysis. It uses the example of determining stresses in a cantilever beam under a end load to illustrate the process.
2) Historically, stresses were determined by assuming functional forms and checking equations for body and surface equilibrium. Finite element methods provide a way to model complex shapes and materials.
3) The example shows determining the stresses and displacements in the beam step-by-step using the theory of elasticity, including deriving the equations and applying boundary conditions.
Stress and Strains, large deformations, Nonlinear Elastic analysis,critical load analysis, hyper elastic materials, FE formulations for Non-linear Elasticity, Nonlinear Elastic Analysis Using Commercial Finite Element Programs, Fitting Hyper elastic Material Parameters from Test Data
Stress and Strains, large deformations, Nonlinear Elastic analysis,critical load analysis, hyper elastic materials, FE formulations for Non-linear Elasticity, Nonlinear Elastic Analysis Using Commercial Finite Element Programs, Fitting Hyper elastic Material Parameters from Test Data
All of material inside is un-licence, kindly use it for educational only but please do not to commercialize it.
Based on 'ilman nafi'an, hopefully this file beneficially for you.
Thank you.
MA 243 Calculus III Fall 2015 Dr. E. JacobsAssignmentsTh.docxinfantsuk
MA 243 Calculus III Fall 2015 Dr. E. Jacobs
Assignments
These assignments are keyed to Edition 7E of James Stewart’s “Calculus” (Early Transcendentals)
Assignment 1. Spheres and Other Surfaces
Read 12.1 - 12.2 and 12.6
You should be able to do the following problems:
Section 12.1/Problems 11 - 18, 20 - 22 Section 12.6/Problems 1 - 48
Hand in the following problems:
1. The following equation describes a sphere. Find the radius and the coordinates of the center.
x2 + y2 + z2 = 2(x + y + z) + 1
2. A particular sphere with center (−3, 2, 2) is tangent to both the xy-plane and the xz-plane.
It intersects the xy-plane at the point (−3, 2, 0). Find the equation of this sphere.
3. Suppose (0, 0, 0) and (0, 0, −4) are the endpoints of the diameter of a sphere. Find the
equation of this sphere.
4. Find the equation of the sphere centered around (0, 0, 4) if the sphere passes through the
origin.
5. Describe the graph of the given equation in geometric terms, using plain, clear language:
z =
√
1 − x2 − y2
Sketch each of the following surfaces
6. z = 2 − 2
√
x2 + y2
7. z = 1 − y2
8. z = 4 − x − y
9. z = 4 − x2 − y2
10. x2 + z2 = 16
Assignment 2. Dot and Cross Products
Read 12.3 and 12.4
You should be able to do the following problems:
Section 12.3/Problems 1 - 28 Section 12.4/Problems 1 - 32
Hand in the following problems:
1. Let u⃗ =
⟨
0, 1
2
,
√
3
2
⟩
and v⃗ =
⟨√
2,
√
3
2
, 1
2
⟩
a) Find the dot product b) Find the cross product
2. Let u⃗ = j⃗ + k⃗ and v⃗ = i⃗ +
√
2 j⃗.
a) Calculate the length of the projection of v⃗ in the u⃗ direction.
b) Calculate the cosine of the angle between u⃗ and v⃗
3. Consider the parallelogram with the following vertices:
(0, 0, 0) (0, 1, 1) (1, 0, 2) (1, 1, 3)
a) Find a vector perpendicular to this parallelogram.
b) Use vector methods to find the area of this parallelogram.
4. Use the dot product to find the cosine of the angle between the diagonal of a cube and one of
its edges.
5. Let L be the line that passes through the points (0, −
√
3 , −1) and (0,
√
3 , 1). Let θ be the
angle between L and the vector u⃗ = 1√
2
⟨0, 1, 1⟩. Calculate θ (to the nearest degree).
Assignment 3. Lines and Planes
Read 12.5
You should be able to do the following problems:
Section 12.5/Problems 1 - 58
Hand in the following problems:
1a. Find the equation of the line that passes through (0, 0, 1) and (1, 0, 2).
b. Find the equation of the plane that passes through (1, 0, 0) and is perpendicular to the line in
part (a).
2. The following equation describes a straight line:
r⃗(t) = ⟨1, 1, 0⟩ + t⟨0, 2, 1⟩
a. Find the angle between the given line and the vector u⃗ = ⟨1, −1, 2⟩.
b. Find the equation of the plane that passes through the point (0, 0, 4) and is perpendicular to
the given line.
3. The following two lines intersect at the point (1, 4, 4)
r⃗ = ⟨1, 4, 4⟩ + t⟨0, 1, 0⟩ r⃗ = ⟨1, 4, 4⟩ + t⟨3, 5, 4⟩
a. Find the angle between the two lines.
b. Find the equation of the plane that contains every point o ...
11 - 3
Experiment 11
Simple Harmonic Motion
Questions
How are swinging pendulums and masses on springs related? Why are these types of
problems so important in Physics? What is a spring’s force constant and how can you measure
it? What is linear regression? How do you use graphs to ascertain physical meaning from
equations? Again, how do you compare two numbers, which have errors?
Note: This week all students must write a very brief lab report during the lab period. It is
due at the end of the period. The explanation of the equations used, the introduction and the
conclusion are not necessary this week. The discussion section can be as little as three sentences
commenting on whether the two measurements of the spring constant are equivalent given the
propagated errors. This mini-lab report will be graded out of 50 points
Concept
When an object (of mass m) is suspended from the end of a spring, the spring will stretch
a distance x and the mass will come to equilibrium when the tension F in the spring balances the
weight of the body, when F = - kx = mg. This is known as Hooke's Law. k is the force constant
of the spring, and its units are Newtons / meter. This is the basis for Part 1.
In Part 2 the object hanging from the spring is allowed to oscillate after being displaced
down from its equilibrium position a distance -x. In this situation, Newton's Second Law gives
for the acceleration of the mass:
Fnet = m a or
The force of gravity can be omitted from this analysis because it only serves to move the
equilibrium position and doesn’t affect the oscillations. Acceleration is the second time-
derivative of x, so this last equation is a differential equation.
To solve: we make an educated guess:
Here A and w are constants yet to be determined. At t = 0 this solution gives x(t=0) = A,
which indicates that A is the initial distance the spring stretches before it oscillates. If friction is
negligible, the mass will continue to oscillate with amplitude A. Now, does this guess actually
solve the (differential) equation? A second time-derivative gives:
Comparing this equation to the original differential equation, the correct solution was
chosen if w2 = k / m. To understand w, consider the first derivative of the solution:
−kx = ma
a = −
k
m
⎛
⎝
⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟
x
d 2x
dt 2
= −
k
m
x x(t) = A cos(ωt)
d 2x(t)
dt 2
= −Aω2 cos(ωt) = −ω2x(t)
James Gering
Florida Institute of Technology
11 - 4
Integrating gives
We assume the object completes one oscillation in a certain period of time, T. This helps
set the limits of integration. Initially, we pull the object a distance A from equilibrium and
release it. So at t = 0 and x = A. (one.
All of material inside is un-licence, kindly use it for educational only but please do not to commercialize it.
Based on 'ilman nafi'an, hopefully this file beneficially for you.
Thank you.
Forklift Classes Overview by Intella PartsIntella Parts
Discover the different forklift classes and their specific applications. Learn how to choose the right forklift for your needs to ensure safety, efficiency, and compliance in your operations.
For more technical information, visit our website https://intellaparts.com
All of material inside is un-licence, kindly use it for educational only but please do not to commercialize it.
Based on 'ilman nafi'an, hopefully this file beneficially for you.
Thank you.
MA 243 Calculus III Fall 2015 Dr. E. JacobsAssignmentsTh.docxinfantsuk
MA 243 Calculus III Fall 2015 Dr. E. Jacobs
Assignments
These assignments are keyed to Edition 7E of James Stewart’s “Calculus” (Early Transcendentals)
Assignment 1. Spheres and Other Surfaces
Read 12.1 - 12.2 and 12.6
You should be able to do the following problems:
Section 12.1/Problems 11 - 18, 20 - 22 Section 12.6/Problems 1 - 48
Hand in the following problems:
1. The following equation describes a sphere. Find the radius and the coordinates of the center.
x2 + y2 + z2 = 2(x + y + z) + 1
2. A particular sphere with center (−3, 2, 2) is tangent to both the xy-plane and the xz-plane.
It intersects the xy-plane at the point (−3, 2, 0). Find the equation of this sphere.
3. Suppose (0, 0, 0) and (0, 0, −4) are the endpoints of the diameter of a sphere. Find the
equation of this sphere.
4. Find the equation of the sphere centered around (0, 0, 4) if the sphere passes through the
origin.
5. Describe the graph of the given equation in geometric terms, using plain, clear language:
z =
√
1 − x2 − y2
Sketch each of the following surfaces
6. z = 2 − 2
√
x2 + y2
7. z = 1 − y2
8. z = 4 − x − y
9. z = 4 − x2 − y2
10. x2 + z2 = 16
Assignment 2. Dot and Cross Products
Read 12.3 and 12.4
You should be able to do the following problems:
Section 12.3/Problems 1 - 28 Section 12.4/Problems 1 - 32
Hand in the following problems:
1. Let u⃗ =
⟨
0, 1
2
,
√
3
2
⟩
and v⃗ =
⟨√
2,
√
3
2
, 1
2
⟩
a) Find the dot product b) Find the cross product
2. Let u⃗ = j⃗ + k⃗ and v⃗ = i⃗ +
√
2 j⃗.
a) Calculate the length of the projection of v⃗ in the u⃗ direction.
b) Calculate the cosine of the angle between u⃗ and v⃗
3. Consider the parallelogram with the following vertices:
(0, 0, 0) (0, 1, 1) (1, 0, 2) (1, 1, 3)
a) Find a vector perpendicular to this parallelogram.
b) Use vector methods to find the area of this parallelogram.
4. Use the dot product to find the cosine of the angle between the diagonal of a cube and one of
its edges.
5. Let L be the line that passes through the points (0, −
√
3 , −1) and (0,
√
3 , 1). Let θ be the
angle between L and the vector u⃗ = 1√
2
⟨0, 1, 1⟩. Calculate θ (to the nearest degree).
Assignment 3. Lines and Planes
Read 12.5
You should be able to do the following problems:
Section 12.5/Problems 1 - 58
Hand in the following problems:
1a. Find the equation of the line that passes through (0, 0, 1) and (1, 0, 2).
b. Find the equation of the plane that passes through (1, 0, 0) and is perpendicular to the line in
part (a).
2. The following equation describes a straight line:
r⃗(t) = ⟨1, 1, 0⟩ + t⟨0, 2, 1⟩
a. Find the angle between the given line and the vector u⃗ = ⟨1, −1, 2⟩.
b. Find the equation of the plane that passes through the point (0, 0, 4) and is perpendicular to
the given line.
3. The following two lines intersect at the point (1, 4, 4)
r⃗ = ⟨1, 4, 4⟩ + t⟨0, 1, 0⟩ r⃗ = ⟨1, 4, 4⟩ + t⟨3, 5, 4⟩
a. Find the angle between the two lines.
b. Find the equation of the plane that contains every point o ...
11 - 3
Experiment 11
Simple Harmonic Motion
Questions
How are swinging pendulums and masses on springs related? Why are these types of
problems so important in Physics? What is a spring’s force constant and how can you measure
it? What is linear regression? How do you use graphs to ascertain physical meaning from
equations? Again, how do you compare two numbers, which have errors?
Note: This week all students must write a very brief lab report during the lab period. It is
due at the end of the period. The explanation of the equations used, the introduction and the
conclusion are not necessary this week. The discussion section can be as little as three sentences
commenting on whether the two measurements of the spring constant are equivalent given the
propagated errors. This mini-lab report will be graded out of 50 points
Concept
When an object (of mass m) is suspended from the end of a spring, the spring will stretch
a distance x and the mass will come to equilibrium when the tension F in the spring balances the
weight of the body, when F = - kx = mg. This is known as Hooke's Law. k is the force constant
of the spring, and its units are Newtons / meter. This is the basis for Part 1.
In Part 2 the object hanging from the spring is allowed to oscillate after being displaced
down from its equilibrium position a distance -x. In this situation, Newton's Second Law gives
for the acceleration of the mass:
Fnet = m a or
The force of gravity can be omitted from this analysis because it only serves to move the
equilibrium position and doesn’t affect the oscillations. Acceleration is the second time-
derivative of x, so this last equation is a differential equation.
To solve: we make an educated guess:
Here A and w are constants yet to be determined. At t = 0 this solution gives x(t=0) = A,
which indicates that A is the initial distance the spring stretches before it oscillates. If friction is
negligible, the mass will continue to oscillate with amplitude A. Now, does this guess actually
solve the (differential) equation? A second time-derivative gives:
Comparing this equation to the original differential equation, the correct solution was
chosen if w2 = k / m. To understand w, consider the first derivative of the solution:
−kx = ma
a = −
k
m
⎛
⎝
⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟
x
d 2x
dt 2
= −
k
m
x x(t) = A cos(ωt)
d 2x(t)
dt 2
= −Aω2 cos(ωt) = −ω2x(t)
James Gering
Florida Institute of Technology
11 - 4
Integrating gives
We assume the object completes one oscillation in a certain period of time, T. This helps
set the limits of integration. Initially, we pull the object a distance A from equilibrium and
release it. So at t = 0 and x = A. (one.
All of material inside is un-licence, kindly use it for educational only but please do not to commercialize it.
Based on 'ilman nafi'an, hopefully this file beneficially for you.
Thank you.
Forklift Classes Overview by Intella PartsIntella Parts
Discover the different forklift classes and their specific applications. Learn how to choose the right forklift for your needs to ensure safety, efficiency, and compliance in your operations.
For more technical information, visit our website https://intellaparts.com
Sachpazis:Terzaghi Bearing Capacity Estimation in simple terms with Calculati...Dr.Costas Sachpazis
Terzaghi's soil bearing capacity theory, developed by Karl Terzaghi, is a fundamental principle in geotechnical engineering used to determine the bearing capacity of shallow foundations. This theory provides a method to calculate the ultimate bearing capacity of soil, which is the maximum load per unit area that the soil can support without undergoing shear failure. The Calculation HTML Code included.
We have compiled the most important slides from each speaker's presentation. This year’s compilation, available for free, captures the key insights and contributions shared during the DfMAy 2024 conference.
Understanding Inductive Bias in Machine LearningSUTEJAS
This presentation explores the concept of inductive bias in machine learning. It explains how algorithms come with built-in assumptions and preferences that guide the learning process. You'll learn about the different types of inductive bias and how they can impact the performance and generalizability of machine learning models.
The presentation also covers the positive and negative aspects of inductive bias, along with strategies for mitigating potential drawbacks. We'll explore examples of how bias manifests in algorithms like neural networks and decision trees.
By understanding inductive bias, you can gain valuable insights into how machine learning models work and make informed decisions when building and deploying them.
NO1 Uk best vashikaran specialist in delhi vashikaran baba near me online vas...Amil Baba Dawood bangali
Contact with Dawood Bhai Just call on +92322-6382012 and we'll help you. We'll solve all your problems within 12 to 24 hours and with 101% guarantee and with astrology systematic. If you want to take any personal or professional advice then also you can call us on +92322-6382012 , ONLINE LOVE PROBLEM & Other all types of Daily Life Problem's.Then CALL or WHATSAPP us on +92322-6382012 and Get all these problems solutions here by Amil Baba DAWOOD BANGALI
#vashikaranspecialist #astrologer #palmistry #amliyaat #taweez #manpasandshadi #horoscope #spiritual #lovelife #lovespell #marriagespell#aamilbabainpakistan #amilbabainkarachi #powerfullblackmagicspell #kalajadumantarspecialist #realamilbaba #AmilbabainPakistan #astrologerincanada #astrologerindubai #lovespellsmaster #kalajaduspecialist #lovespellsthatwork #aamilbabainlahore#blackmagicformarriage #aamilbaba #kalajadu #kalailam #taweez #wazifaexpert #jadumantar #vashikaranspecialist #astrologer #palmistry #amliyaat #taweez #manpasandshadi #horoscope #spiritual #lovelife #lovespell #marriagespell#aamilbabainpakistan #amilbabainkarachi #powerfullblackmagicspell #kalajadumantarspecialist #realamilbaba #AmilbabainPakistan #astrologerincanada #astrologerindubai #lovespellsmaster #kalajaduspecialist #lovespellsthatwork #aamilbabainlahore #blackmagicforlove #blackmagicformarriage #aamilbaba #kalajadu #kalailam #taweez #wazifaexpert #jadumantar #vashikaranspecialist #astrologer #palmistry #amliyaat #taweez #manpasandshadi #horoscope #spiritual #lovelife #lovespell #marriagespell#aamilbabainpakistan #amilbabainkarachi #powerfullblackmagicspell #kalajadumantarspecialist #realamilbaba #AmilbabainPakistan #astrologerincanada #astrologerindubai #lovespellsmaster #kalajaduspecialist #lovespellsthatwork #aamilbabainlahore #Amilbabainuk #amilbabainspain #amilbabaindubai #Amilbabainnorway #amilbabainkrachi #amilbabainlahore #amilbabaingujranwalan #amilbabainislamabad
Harnessing WebAssembly for Real-time Stateless Streaming PipelinesChristina Lin
Traditionally, dealing with real-time data pipelines has involved significant overhead, even for straightforward tasks like data transformation or masking. However, in this talk, we’ll venture into the dynamic realm of WebAssembly (WASM) and discover how it can revolutionize the creation of stateless streaming pipelines within a Kafka (Redpanda) broker. These pipelines are adept at managing low-latency, high-data-volume scenarios.
Industrial Training at Shahjalal Fertilizer Company Limited (SFCL)MdTanvirMahtab2
This presentation is about the working procedure of Shahjalal Fertilizer Company Limited (SFCL). A Govt. owned Company of Bangladesh Chemical Industries Corporation under Ministry of Industries.
DESIGN AND ANALYSIS OF A CAR SHOWROOM USING E TABS
LECT_01.pptx
1. MAE 5563 Finite Element Methods
Professor J.K. Good
Fall 2019
SR71 1966
Apollo Saturn V
1966
Is the topic of
finite
elements
relevant to
these
amazing
machines?
Computers
UNIVAC 1 1951
IBM 360 1964
Theory ~1940
Hrennikoff
Courant
Software
NASTRAN 1968
ANSYS 1969
Any finite element
calculations on these
marvels were
performed by hand….
2. • Why are we here?
• Most BS level engineers have limited
stress analysis skills.
• B.S. coursework focuses on:
– Beams in bending and shear
– Thin & thick wall vessels
– Truss structures
– Students may apply FE software but cannot
attest to the validity of the results.
• Reality - Today’s structures are often
complex in shape and in material
properties. What are the stresses in the
red plate? If you model these stresses,
when do you know you have correct
results?
3. These problems were solved historically using
the Theory of Elasticity. The Stress Method
requires these steps:
1. Assume functional forms of sx, sy, sz,
txy, tyz, and tzx(x,y,z).
2. Check Body Equilibrium
3. If the functional forms of sx, sy, sy, txy,
tyz, and tzx do not satisfy these equations,
start over!
sx
x
txy
y
txz
z
fx 0
txy
x
sy
y
tyz
z
fy 0
txz
x
tyz
y
sz
z
fz 0
P1
P2
4. 4. Check surface equilibrium
The Tx tractions in the boss regions
integrated over the area should yield
the applied loads P1 and P2.
4. If the assumed forms of stress
don’t satisfy these equations, start
over, assume new forms.
n
x
y
x y z
n cos n,x n cos n,y n cos n,z
sxnx txyny txznz Tx
txynx syny tyznz Ty
txznx tyzny sznz Tz
P1
P2
0
on free boundaries
x y n
T T T
5. 6. Convert stresses to strains using
constitutive equations:
x x y z
1
E
s s s
y y x z
1
E
s s s
z z x y
1
E
s s s
xy
E
2 1
txy yz
E
2 1
tyz
zx
E
2 1
tzx
6. 7. Integrate strains to yield
displacements u, v, w(x,y,z):
8. Are these displacements (u, v,
and w) compatible with the
external constraints? Yes: You
are done! No: Start over!
x
u
x
y
v
y
z
w
z
xy
u
y
v
x
yz
v
z
w
y
zx
w
x
u
z
u,v,wwelded boundary
?
0
7. Theory of Elasticity – A
“Simple” Example
• Let’s consider an end loaded
cantilever beam that we were
familiar with as undergraduates.
• Based on our experience:
• 2D Body Equilibrium is satisfied by
these assumed forms:
x 4 y
2
4
xy 2
d xy (Euler) 0
d
b y (Parabolic Shear Stress)
2
s s
t
xy
x
x
f 0
x y
t
s
xy y
y
f 0
x y
t s
c
c
L
x
y
P
1
u
v
Assume the
body intensity
forces fx and fy
are negligible
8. • On the upper and lower boundaries
where y=+c, surface equilibrium
dictates that the shear stress must
vanish:
• On the loaded end (x=0) the shear
stress txy integrated over the area must
be equal to the applied load if surface
equilibrium is to be satisfied:
• Substituting back we have:
• These results appear to be in exact
agreement with what we learned in
strength of materials course work.
t
2
4 2
xy 2 4 2
y c
d 2b
b c =0 d
2 c
c c
2
2
xy 2 2
2
c c
b 3 P
dy b y dy P b
4 c
c
t
s s
t
3
x y
3
2
2 2
xy 2
3 P Pxy 2
xy (I= c ) 0
2 I 3
c
3P y P VQ
1 c y
4c 2I It
c
9. • If in fact we look closer this may or
may not be the case. To be exact the
load P would have to be applied
parabolically to the tip of the beam
(x=0) and be reacted parabolically at
the beam root (x=L). If they are not
our stress solution may be in error at
the tip and root but by St.Venant’s
principle we can state that they
become accurate away from these
surfaces.
• St. Venant’s Principle:
If the loading distribution on a small section
of an elastic body is replaced by another
loading which has the same resultant force
and moment as the original loading, then no
appreciable changes will occur in the stresses
in the body except in the region near the
surface where the loading is altered. Rivello
10. • Now we must determine the
deformations which result from
the stress functions that have
satisfied body and surface
equilibrium. First we convert
the stress functions to strain
functions.
• Constitutive Relationships
x
x x y
u 1 Pxy
x E E EI
s
s s
x
y y x
v 1 Pxy
y E E EI
s
s s
xy 2 2
xy
v u P
c y
x y G 2IG
t
11. • Integrating the x and y strains yields:
where f(y) and f1(x) are as yet
unknown functions of y and x only.
• If we substitute u and v into the shear
strain equation we have:
• Let’s now collect the terms which are
functions of x only, y only, and
constants:
2
Px y
u f y
2EI
2
1
Pxy
v f x
2EI
2 2
1
2 2
df y df x
Px Py
2EI dy 2EI dx
P
c y
2IG
2
1
df x
Px
F x
2EI dx
2 2
df y Py Py
G y
dy 2EI 2IG
2
Pc
2IG
a constant
12. • The shear strain equation can now be
rewritten as:
which means that F(x) must be some
constant we will call d and G(y) must
be some constant e and thus:
from F(x):
from G(y):
Now we can determine f(y) and f1(x):
F x G y
2
Pc
d e
2IG
2
1
d f x Px
d
dx 2EI
2 2
d f y Py Py
e
dy 2EI 2IG
3 3
Py Py
f y ey g
6EI 6IG
3
1
Px
f x dx h
6EI
13. • If we substitute these expressions back into
our expression for u and v we have:
• Now we must determine the constants e, g,
d, and h using kinematic (displacement
related) boundary conditions. First we know
at the root of the cantilever (x=L, y=0) that
u and v are zero. From our equations above:
• The deflection curve for the elastic axis of
the beam can be found by setting y=0 in
v(x):
2 3 3
Px y Py Py
u ey g
2EI 6EI 6IG
2 3
Pxy Px
v dx h
2EI 6EI
3
PL
g 0 h= - dL
6EI
3 3
y 0
Px PL
v d L x
6EI 6EI
14. • We now have more than one route by
which we can proceed. Our first route
will be to assume the slope of the
deflection of the root of the cantilever
will be zero:
• Substituting into the derivative of our
deflection equation (v)y=0 yields:
• Earlier we found:
• We have found all the constants now
and can write u and v:
x L,y 0
v
0
x
2
PL
d
2EI
2 2 2
Pc PL Pc
e d so e=
2IG 2EI 2IG
2 3 3 2 2
Px y Py Py PL Pc
u y
2EI 6EI 6IG 2EI 2IG
15. • The deflection equation for the elastic
axis is:
• At x=L this yields the deflection
PL3/3EI, the value we found before in
strength of materials.
• The 2nd route we could have chosen
would be to set the slope at the root to
zero using the expression:
• Our equation for u was:
2 3 2 3
Pxy Px PL x PL
v
2EI 6EI 2EI 3EI
3 2 3
y 0
Px PL x PL
v
6EI 2EI 3EI
x L
y 0
u
0
y
2 3 3
Px y Py Py
u ey g
2EI 6EI 6IG
0
x L,y 0
u 0
16. • If we set x=L we have:
and taking the derivative wrt y yields:
• Earlier we found:
and substituting into our earlier
equation for the deflection curve:
we can find the deflection curve for
the elastic axis:
2 3 3
PL y Py Py
u ey
2EI 6EI 6IG
2 2
x L
y 0
u PL PL
e 0 e
y 2EI 2EI
2 2 2
Pc PL Pc
e d so d=-
2IG 2EI 2IG
3 3
y 0
Px PL
v d L x
6EI 6EI
3 2 3 2
y 0
Px PL x PL Pc
v L x
6EI 2EI 3EI 2IG
17. • The slope of the deformation wrt x is:
and will be non zero at x=L:
• At the beam tip where x=0 the
deflection is now:
• The 1st term is the deformation due to
bending strain. The 2nd term is the
deformation due to shear strain. For
large L (L>20c) you will find the 1st
term is dominant. For small L (L<20c)
you find the 2nd term becomes
important.
2 2 2
y 0
dv Px PL Pc
dx 2EI 2EI 2IG
2
y 0
x L
dv Pc
dx 2IG
3 2
y 0
x 0
PL Pc L
v
3EI 2IG
18. • Theory of Elasticity
– Benefits: Closed form solutions for
stresses, strains, and deformations
as functions of x,y and z are
obtained. This can be the best
format for providing solutions to
problems to other users.
– Disadvantages: A good deal of
mathematical prowess is required to
solve non-trivial problems. It may
require much iteration and time to
come to a solution which satisfies all
the stress (kinetic) and deformation
(kinematic) boundary conditions.
• A Finite Element Solution
10000 lb
10000 lb
E = 30*106 psi
= 0.29
t = 0.5
19. u (in)
v (in)
What functions of x and y would produce these
displacements?
21. • This is why we are here. The
theory of elasticity is very
difficult to apply for the complex
problems we must solve.
• We must resort to other
methods which may not give us
closed form functions for the
displacements, strains and
stresses. Discrete results are
better than none!
• Anyone can execute a finite
element code. My objective is to
explain the science of the
method to you and how you
should know when your output
results are correct.
22. Potential Energy &
Equilibrium
• Definition: The total potential
energy of an elastic body is:
= Strain Energy+Work Potential
(U) (WP)
• Definition: U is the strain energy
due to stresses and strains within
the elastic body. For linear elastic
problems at a point in space the
strain energy is:
T
T
Vol
1
dU dV and
2
1
U= dV
2
s
s
dU
sx
x for the body.
23. • The Work Potential (WP) is the ability
of external and internal forces to
perform work on the body as these
forces move through deformations
which resulted from the stresses and
strains. The WP is always defined to
be negative in expressions although
calculation may result in a positive
contribution to the total potential:
T
V
T T
i i
i
S
WP u f dV
u T dS u P
WP due to
inertial forces
WP due to
surface tractions
WP due to
concentrated loads
24. • The total potential is thus:
• Principle of Minimum Potential
Energy: “Among all compatible
states of deformation of an elastic
body in stable equilibrium, the true
deformations are those for which
the total potential is a minimum.” -
Rivello, Theory and Analysis of
Aircraft Structure, McGraw-Hill,
1969.
• Refer to Example 1.1
=
1
2
s
T
V
dV u
T f
V
dV
u
T T
S
dS ui
T
Pi
i
25.
26. The Rayleigh-Ritz Method
• We have seen how the PMPE applies
to linearly elastic discrete problems.
Now we will apply it to continuums.
• A set of deformations are assumed.
The form of deformation may be
polynomial functions but need not be:
where a1, a2, and a3 are constant
coefficients for which we shall solve.
1, 2, and 3 may be polynomials,
transcendentals, etc.
• The deformation function (u) must
satisfy kinematic boundary
conditions.
• We then develop the total potential as
a function of the constant coefficients.
u a11 a22 a33
27. So: =(a1, a2, a3 )
where these constant coefficients are
three independent unknowns for
which we must solve.
• The variation in the total potential is:
• Since the forms of variation are
arbitrary:
which generates 3 equations by
which we can solve for our 3
unknowns.
• Refer to Example 1.2.
a1
a1
a2
a2
a3
a3 0
1 2 3
0 0 0
a a a
28.
29.
30. u 1
a x 0 x 1
du
dx
1
a
u 1
a 2 x
1 x 2
du
dx
1
a
1
2
EA
2
du
dx
dx
0
1
1
2
EA
2
du
dx
dx 2 1
a
1
2
Piecewise Continuous Functions:
1
2
a 2 1
a
1
a
2 1
a 2 0
1
a 1
u x s = E
du
dx
1 0 x 1
u 2 x s = E
du
dx
1 1 x 2
This yields the exact solution:
31. The expression for the Total
Potential can be applied to
different types of structure other
than the rod in Example 1.2.
Beam structures are examples.
Classic beams are defined as
structures whose length dimension
exceeds any cross sectional
dimension by a factor of ten.
In such cases the only substantial
stresses and strains are calculated
from Euler’s expressions:
sx
My
I
and x
My
EI
32. Also recall that the curvature of a
beam (2nd derivative of the
displacement) at some location x is
equal to the bending moment at that
location divided by the bending
stiffness:
The strain energy is:
M and EI are at best functions of x
and thus:
and:
d2
v
dx2
M
EI
U
1
2
M
EI
2
Ey2
A
x
dAdx but y2
dA I
A
2
T
2
Vol Vol
1 1 My
U dVol dVol
2 2 EI
s
U
1
2
d2
v
dx2
2
EI
x
dx
33. Example: Use of Rayleigh-Ritz
Method on a Beam
Pick displacement functions which
satisfy or can be made to satisfy
the kinematic boundary conditions
v(0) = v’(0) = 0. I selected:
Our strain energy expression
yields:
4 2 2
x
v a 1 Cos b x 6L x
2L
W
x
y,v
L
2 5 4 2 8
3
5a 15360abL 12288 b L
U EI
320 L
34. The Work Potential results from
the surface traction:
The Total Potential is the sum of
the strain energy and the work
potential. Now we minimize the
Total Potential with respect to the
unknown coefficients a and b.
Now we have two equations with
which we can solve for a and b.
WP uT
s
Tds W aL
9bL5
5
2aL
a
10a5
15360bL4
320L3 EI WL 1
2
0
b
24576bL8
15360aL4
320L3 EI
9WL5
5
0
35. Solving yields:
Now a and b can be substituted
back into the assumed
displacement function to yield the
solution:
a
4WL4
8 25
EI 6 960
6
6
W 5120 2560 3
b
128EI 960
4
6
6
4 2 2
6
4 WL 8 25 x
v 1 Cos
2L
EI 960
W 5120 2560 3
x 6L x
128EI 960
36. How accurate is the solution?
Without comparison you cannot be
sure. You can expect that the
values of a and b are the best
possible solutions for the
displacement functions chosen
because the Principle of Minimum
Potential Energy has been
employed.
We can substitute numerical values
for the inputs and compare:
W=10 lb/in
E = 10,000,000 psi
I = 1/12 in4
L = 20 in
38. Although the % errors are not
appealing, these are small numbers
where % error can be meaningless.
A chart shows our solution to be
quite acceptable.
.00
.05
.10
.15
.20
.25
0 5 10 15 20
X position (in)
Deflection
(in)
Rayleigh-Ritz
Exact
v
(in)
39. In general we wish to solve
problems where exact answers are
not known. The trial functions we
choose for the deformation may be
good or bad choices for
representing the true deformations.
How do we decide we have
obtained a good solution?
We must select several trial
functions, use the theory of minimum
total potential to evaluate unknown
coefficients, and compare the total
potentials that resulted from all trial
functions.
Whichever trial function yields the
lowest total potential will be the best
solution by our theory.
If several of our trial functions yield
comparable total potentials, we gain
confidence that our solution(s) are
good.
40. In our example of the uniformly loaded
cantilever we might propose several
solutions.
The 1st solution might be the trial function
proposed in the example:
a and b are known and we can solve for the
total potential:
Let’s pretend we do not know the exact
solution is exact and that it is just another
trial solution candidate. It’s total potential
is:
4 2 2
x
v a 1 Cos b x 6L x
2L
2 5
W L
TP1 0.0242647
EI
2 5
W L
TP2 0.025
EI
2
2
T
2
x S
1 d v
= EI dx u T dS
2 dx
41. A 3rd trial function might be the first
half of the function I chose for the
example:
We must now resolve the problem
for a new constant coefficient aa:
and the total potential is:
Finally a 4th trial function might be
the second half of my original trial
function:
We must now resolve the problem
for a new constant coefficient bb:
x
v aa 1 Cos
2L
4
5
32 2 WL
aa
EI
2 5
W L
TP3 0.0216892
EI
4 2 2
v bb x 6L x
3W
bb
128EI
42. and the total potential is:
If we produce a chart of our
solutions we can see how much
they differ. We see solutions 1 and
2 are similar. We see solutions 3
and 4 are similar.
Remember all our solutions satisfy
the kinematic boundary conditions
for the cantilever: v(0)=v’(0)=0.
Which is really “best”?
2 5
W L
TP4 0.0210938
EI
0
0.1
0.2
0.3
0 5 10 15 20
v
(in)
x (in)
sol1 sol2
sol3 sol4
43. How do we evaluate what the best
solution is? Our 2nd solution had
the minimum total potential. By the
Theory of Minimum Total Potential
it is the best solution.
-0.026
-0.025
-0.024
-0.023
-0.022
-0.021
-0.02
-0.019
TP1 TP2 TP3 TP4
Total
Potential
(W
2
L
5
/EI)
Trial Solutions
44. Read Chapter 1 in
Chandrupatla and Belegundu.
This chapter discusses
principles we will use to
develop and understand finite
element theory.
We will ensure we understand
these theories by first applying
them to continuum problems.
I will assign homework next
lecture. Have you ever done
symbolic mathematics using
codes like Mathematica,
Mathcad or Matlab? These are
all available to you through
OSU. I suggest you download
one and be prepared to use it.