Engineering Mechanics-MIT NOTES
-
Upload
damini-thakur -
Category
Documents
-
view
143 -
download
4
Transcript of Engineering Mechanics-MIT NOTES
-
1.050 Engineering Mechanics I
Fall 2007
-
Notes and remarks Lecture Summary Slide
Content Survey
Lecture notes
Homework assignments (weekly)
Exams: 2 in-class quizzes, 1 finalAll exams are open-book
Grading:Two quizzes (25%)Final (25%)Homework assignment (50%)
-
Assignments
Homework / Problem Sets (50%) Assigned weekly on Wednesday, evaluated and returned to you
(ASAP) Build homework teams of three students:
Engineering is team work. We expect a true team work, in which everybody contributes equally to the result. This is testified by the team members signing a declaration that the signature confirms that all have equally contributed to the homework.
Typical teamwork: Each student works individually through the homework set. The team meets and discusses questions, difficulties and
solutions. Possibly, meet with TA or instructor.
You must reference your sources and collaborators, whether otherstudents, sources on the web, archived solutions from previous years etc
-
A few things wed like you to remember
We teach the class for you! At any time please let us know if youhave concerns or suggestions, or if you have difficulties. Well do the best to cater to your needs!
The goal is that you will have an excellent basis for engineering science in many other applications aside from the mechanics topic covered here
Our goal: Discover Engineering Mechanics with you starting at fundamental concepts (Newtons laws) to be able to apply theknowledge to complex engineering problems.
-
1.050: Engineering Mechanics Why are there no monsters on Earth?
Images removed due to copyright restrictions.
Normandy Bridge 900m (1990ies)
Can we build bridges Jack and the giantJack and the giantCopyright , The British LibraryCopyright , The British Library Between continents?
-
Hurricane Katrina
What caused major flooding in the city? Why did the levees break?
Geotechnical Design - Load < strength capacity - Failure (plasticity or fracture) - Mechanism
Photograph of floodwaters removed due to copyright restrictions.
Impact - 2 million people - Nationwide Life Line interruption
What caused this to happen? - Global warming? - Policy: Role of the federal government?
-
Minnesota bridge collapse Aging infrastructure -What caused the bridge to collapse? -Are our bridges safe? -Can we detect failure before tragedy happens?
Photographs of collapsed bridge removed due to copyright restrictions.
Fixing the problems -Retrofitting? -Rebuilding new bridges? -Funding? -- Policy change to allocate more funding to fix unfit infrastructure
-
Earthquake disasters
Earthquake in Peru (August 2007)
Map of Peru showing epicenter location removed due to copyright restrictions.
Structural Design - Service State (Elasticity)
Photographs of collapsed roads removed due to copyright restrictions. - Failure (Plasticity or Fracture) - Mechanism
Impact - Millions of people - Nationwide Life Line interruption - Economy
-
9-11: The Fall of the Towers North Tower: 8:46 am above 96th floor, failed at 10:28 am South Tower: 9:03 am above 80th floor, failed at 9:59 am
Immediate Question: How did the towers fail? - Mechanism Lecture 4
Three sequential photographs of tower collapse removed due to copyright restrictions.
-
Engineering science paradigm: Multi-scale view of materials
Buehler and Ackbarow, Materials Today, 2007 Courtesy Elsevier, Inc., http://www.sciencedirect.com. Used with permission.
-
Atomistic mechanisms of fracture
Simulations of atomistic fracture mechanisms
Reveals new fracture mechanism: Supersonic fracture
View the complete movie at: http://web.mit.edu/mbuehler/www/research/supersonic_fracture.mpeg.
Buehler et al., Nature, 2003; Nature, 2006
-
Fracture is linked to the mechanics of chemical bond breaking
Fracture mechanics
Mesoscale
Mechanics of chemical interactions
Buehler et al., Nature, 2003; Nature, 2006
-
Impact of cement on worldwide CO2 production Worldwide Cement Consumption
Worldwide Cement Consumption equates to 10% of worldwide CO2 Emission YEAR
Met
ric T
ons
(mill
ions
) 2001: 1.7 x 109 t/yr ~ 1 m3/capita/yr
~ 350 kg CO2 /capita/yr
2050: 3.2-7.5 x 109 t/yr
312
3
m109: m000,2500,1:
Total Car
Can drive 200 million times Around the world
Chaturvedi, S. and Ochsendorf, J., Global Environmental Impacts Due to Concrete and Steel, Structural Engineering International, 14/3, Zurich, Intl. Assoc. of Bridge and Structural Engineers, August 2004, 198-200.
Courtesy of John Ochsendorf. Used with permission.
-
Concrete: A complex multi-scale material New materials for construction industry? Ti, Mg based cement? New production pathways? Mortar
few mm
Cement paste Concrete 1 cm< 0.1 mm Molecular mechanics
Images of concrete from the nanometer to centimeter scale
C-S-H removed due to copyright restrictions.
< mPlatelets few 10 nm
Enables structures
Chemistry Kilometers Angstrom-nm
Image of suspension bridge removed due to copyright restrictions.
-
silica Dreierketten
Ca octahedra
Water layer
Ca octahedra
silica Dreierketten
Ca octahedra
Water layer
Ca octahedra
Opening molecular-nanoscale for engineering design
Production of green concrete
Reduce CO2 emission during production
Understand diffusion of radioactive waste through concrete
Long-term stability/durability avoid disasters
Environmental effects (chemicals, moisture,..)
Mechanical stability
-
Mechanics in life sciences
Elasticity of environment directs stem cell differentiation
Brain tissue
Muscle
Bone
D. Discher, Cell, 2006 Courtesy Elsevier, Inc., http://www.sciencedirect.com. Used with permission.
-
Mechanics in life sciences
Change of mechanics in diseases?
How can we use self-assembly to synthesize new materials?
Courtesy Elsevier, Inc., http://www.sciencedirect.com. Used with permission.
Buehler and Ackbarow, Materials Today, 2007Courtesy Elsevier, Inc., http://www.sciencedirect.com. Used with permission.
-
Mechanics in life sciences
Single point mutations in IF structure causes severe diseases such as rapid aging disease progeria HGPS (Nature, 2003; Nature, 2006, PNAS, 2006)
Cell nucleus loses stability under cyclic loading Failure occurs at heart (fatigue)
Substitution of a single DNA base: Amino acid guanine is switched to adenine
Experiment suggests that mechanical properties of nucleus change (Dahl et al., PNAS, 2006)
Images from the organismal to cell to molecular scales removed due to copyright restrictions.
-
1.050 Content overview I. Dimensional analysis
1. On monsters, mice and mushrooms 2. Similarity relations: Important engineering tools
II. Stresses and strength 2. Stresses and equilibrium 3. Strength models (how to design structures,
foundations.. against mechanical failure)
III. Deformation and strain 4. How strain gages work? 5. How to measure deformation in a 3D
structure/material?
IV. Elasticity 5. Elasticity model link stresses and deformation 6. Variational methods in elasticity
V. How things fail and how to avoid it 7. Elastic instabilities 8. Plasticity (permanent deformation) 9. Fracture mechanics
Lectures 1-3 Sept.
Lectures 4-15 Sept./Oct.
Lectures 16-19 Oct.
Lectures 20-31 Nov.
Lectures 32-37 Dec.
-
1.050 Content The contents of 1.050 will be important in several subjects
Spring: 1.060 Engineering Mechanics II Fluid Mechanics
Hydrostatics Hydrodynamics Open Channel Flow
Application in many engineering applications and in engineeringscience Biomechanics Molecular mechanics & molecular dynamics Microfluidics Environmental science and application Earthquake engineering Structural engineering Materials science
-
1.050 Content overview
I. Dimensional analysis Lecture 1: Introduction & Galileo's problem Lecture 2: Dimensional Analysis and Atomic Explosion Lecture 3: Dimension analysis and application to engineering
structures
II. Stresses and strength
III. Deformation and strain
IV. Elasticity
V. How things fail and how to avoid it
-
1.050 Engineering Mechanics
Lecture 2: Dimensional Analysis and Atomic Explosion
-
1.050 Content overviewI. Dimensional analysis
1. On monsters, mice and mushrooms Lectures 1-32. Similarity relations: Important engineering tools Sept.
II. Stresses and strength2. Stresses and equilibrium Lectures 4-153. Strength models (how to design structures,
foundations.. against mechanical failure) Sept./Oct.
III. Deformation and strain4. How strain gages work?5. How to measure deformation in a 3D Lectures 16-19
structure/material? Oct.
IV. Elasticity5. Elasticity model link stresses and deformation Lectures 20-316. Variational methods in elasticity Nov.
V. How things fail and how to avoid it7. Elastic instabilities8. Plasticity (permanent deformation) Lectures 32-379. Fracture mechanics Dec.
-
1.050 Content overviewI. Dimensional analysis
Lecture 1: Introduction & Galileo's problem Lecture 2: Dimensional Analysis and Atomic ExplosionLecture 3: Dimension analysis and application to engineering
structures
II. Stresses and strength
III. Deformation and strain
IV. Elasticity
V. How things fail and how to avoid it
-
Discorsi e Dimonstrazioni Matematiche intorno a Due Nuove Scienze (1638)
We clearly see, by what has been demonstrated, that it is impossible to magnify structures to large dimensions, whether in art or nature; [...] it would be equally impossible to create huge bone structures for humans, horses or other animals that would function normally, unless the material employed was much harder and more resistant than usual [...]. Obviously, if we wish to maintain the same proportions of a normal man in a giant, it would be necessary to find a harder and more resistant material to build his bones, or yet admit that his robustness would be proportionally smaller; as he grew up immeasurably we would see him collapse under his own weight.
-
Discorsi e Dimonstrazioni Matematiche intorno a Due Nuove Scienze (1638)
another way [...] of making giants or other large animals to live and move like the smaller ones:
this would be possible only by increasing the strength of the bones and also the strength of parts that support the weight and additional loads; but also keeping the same proportions the bone structure would resist only if its specific weight were reduced, as well as the specific weight of the flesh and all other parts supported by the bones.
Galileo Number
-
Exercise: Atomic Explosion
r(t)E
Trinity Test Nuclear Explosion, New Mexico, July 16, 1945 Library of Congress
-
Steps of Dimensional AnalysisRecipe Exercise1. Problem Formulation2. Dimensional Analysis
1. Build the exponent matrix of dimensions of N+1
2. Rank of matrix = k = number of dimensionally independent variables (see next slide)
3. Choose k independent variables, express N+1-k dimensionless variables
4. Determine exponents by solving linear system (see next+1 slide)
3. Dimensionless expression
Here: k = 3 N +1 k = 1
N +1= 4
-
Technique 1: Number of dimensionally independent variables Method 1: Look for the maximum number
of linearly independent rows or columns
Method 2: Rank of a matrix Manually: Identify the dimension of the
biggest sub-square matrix that has a non-zero determinant (math: non-singular)
Software: Matlab, Maple, Excel, etc
-
Technique 2:Determination of the exponents
Find a1, a2, a3 In a log-representation
Linear system: Ax=y
a 2 /5 1
a2 = 1/5 a3 1/5
-
G.I. Taylors Analysis* Top Secret: What is the D-Analysis energy E released by a
nuclear explosion? But: High speed
photographs were available, giving r and t
Air density = 2.5 kg/m3
const ~ O(1)
known known
mr 100~
mst 30=
(*) G.I. Taylor (1950)
-
G.I. Taylors Analysis (contd)
Back Analysis:
52 log r(t = 104 s)~ 8
1 E 2 log = 8 log( )104 = 12
E ~ 1021 erg = 100,000GJ
Comparison: ~ hourly energy production of 20 nuclear power plants
10.5
9.5
8.5
-3.0 -2.0 -1.0log t
52 log r
52
12
Elog r = log ( ) + log t
~8
-4
Figure by MIT OpenCourseWare, adapted from Taylor, G. I. "Formation of a Blast .Wave by a Very Intense Explosion. II. The Atomic Explosion of 1945." Proceedings of the Royal Society A 201 (1950): 175-186.
-
Summary Pi-Theorem
Most critical step: problem formulation if you forget one parameter on which the problems depends, the problem is ill-posed!
By means of dimensional analysis reduce the complexity of a problem from N+1 parameters to N+1-k parameters:
Some technique: Exponent matrix linear system Of critical importance for lab testing: instead of (N)a tests, you only
need to carry out (N-k)a tests Critical for model scaling: Model (e.g. human) and Prototype (e.g.
monster) must have the same invariants. Best invariants: not unique, some try and error you can always
recombine invariants as power functions of others. If N = k, jackpot you have the solution (close to a multiplying
constant). In the next lecture and recitation: Applications
-
1.050 Engineering Mechanics
Lecture 3: Dimension analysis and application to engineering
structures
-
1.050 Content overviewI. Dimensional analysis
1. On monsters, mice and mushrooms Lectures 1-32. Similarity relations: Important engineering tools Sept.
II. Stresses and strength2. Stresses and equilibrium Lectures 4-153. Strength models (how to design structures,
foundations.. against mechanical failure) Sept./Oct.
III. Deformation and strain4. How strain gages work?5. How to measure deformation in a 3D Lectures 16-19
structure/material? Oct.
IV. Elasticity5. Elasticity model link stresses and deformation Lectures 20-316. Variational methods in elasticity Nov.
V. How things fail and how to avoid it7. Elastic instabilities8. Plasticity (permanent deformation) Lectures 32-379. Fracture mechanics Dec.
-
1.050 Content overviewI. Dimensional analysis
Lecture 1: Introduction & Galileo's problem Lecture 2: Dimensional Analysis and Atomic Explosion Lecture 3: Dimension analysis and application to engineering
structures
II. Stresses and strength
III. Deformation and strain
IV. Elasticity
V. How things fail and how to avoid it
-
D-Analysis of Tall Buildings
Graphic of tall buildings removed due to copyright restrictions.
http://www.joelertola.com/grfx/grfx_update_feb_05/tall_buildings.jpg
-
Hurricane Katrina
Wind speeds 200 km/h
http://www.nasa.gov/images/content/126301main_Katrina_082805_516.jpghttp://www.asiatraveltips.com/newspics/074/BurjDubai.jpg
Photograph of skyscraper removed due to copyright restrictions.
-
Lab Results: Drag Coefficienton smooth objects
400200100604020106421
0.60.40.20.1
0.0610-1 100 101 102 103 104 105 106 107
Smooth Cylinder
Smooth Sphere
D
E
CB
A
Reynolds Number Re = 1 =1 UD
v
CD
= 2
0 =
2FD
a(U
D)2
aU2D2
CD =24Re
0 =FD
UDv
= F ( 1 = )
Figure by MIT OpenCourseWare.
-
1.050 Engineering Mechanics
Lecture 4: Stresses and StrengthStresses and Equilibrium
Discrete Model
-
1.050 Content overviewI. Dimensional analysis
1. On monsters, mice and mushrooms 2. Similarity relations: Important engineering tools
II. Stresses and strength 2. Stresses and equilibrium 3. Strength models (how to design structures,
foundations.. against mechanical failure)
III. Deformation and strain 4. How strain gages work? 5. How to measure deformation in a 3D
structure/material?
IV. Elasticity 5. Elasticity model link stresses and deformation 6. Variational methods in elasticity
V. How things fail and how to avoid it 7. Elastic instabilities 8. Plasticity (permanent deformation) 9. Fracture mechanics
Lectures 1-3 Sept.
Lectures 4-15 Sept./Oct.
Lectures 16-19 Oct.
Lectures 20-31 Nov.
Lectures 32-37 Dec.
-
1.050 Content overviewI. Dimensional analysis
II. Stresses and strength Lecture 4: Newtons laws, fall of the WTC towers Lecture 5: Stress vector and stress tensor Lecture 6: Hydrostatic problem Lecture 7: Soil mechanics / geostatics problem Lecture 8: Beam stress model Lecture 9: Beam model II and summary Lecture 10: Strength models
III. Deformation and strain
IV. Elasticity
V. How things fail and how to avoid it
-
Content lecture 41. Review: Newtons Laws of Motion
2. Application: Discrete Model Linear Momentum & Dynamic Resultant Theorem Angular Momentum & Dynamic Moment Theorem
3. Exercise: The Fall of the WTC Towers 1. Free Fall Assumption 2. Discrete Model 3. From Discrete to Continuum
Goal: Put Newtons Laws to work.
-
911
9-11-2001: The Fall of the Towers
North Tower: 8:46 am above 96th floor, failed at 10:28 am South Tower: 9:03 am above 80th floor, failed at 9:59 am
Immediate question: How did the towers fail?
Three sequential photographs of tower collapse removed due to copyright restrictions.
-
Physics BackgroundThe Three Laws of Motion of Isaac Newton (1642 1727):
1. Every body continues in its state of rest, or of uniform motion in a right line, unless it is compelled to change that state by forces impressed upon it.
2. The change of motion is proportional to the motive force impresses, and is made in the direction of the right line in which that force is impressed.
3. To every action there is always opposed an equal reaction: or, the mutual action of two bodies upon each other are alwaysequal, and directed to contrary parts.
Our Aim: Translate these Laws into powerful tools of Engineering Mechanics
-
Dynamic Resultant Theorem:Discrete Mass System
Linear motion of a mass is quantified by the linear momentum vector:
-
Dynamic Moment Theorem:Discrete System
The angular motion of a mass point i is quantified by the angular momentum vect:
cross product
-
9-11: engineering questions
Free Fall? Dynamic Resultant Theorem:
defdp dt
= m0a ez = m0 g ez
Integrate twice + Initial velocity
North Tower: South Tower:
0V
110=N
80=M
96=M
-
9-11: engineering questions (contd)
Return to Problem Formulation Dimensional Analysis
Exponent Matrix (k=3)
Pi-Theorem
0m0m
TmTm
,maxV
110=N
96=M
-
9-11: engineering questions (contd)
Kausels Discrete Mass Formulation
h
0 ,Vm 0
1 =V ? 1 = mm 0 + m
2 =V ?
2 = mm 20 + m ?= iV
0 += i immm ( )
?max
0
== +
=
MN
MN
VV mMN
mm
Sequence of 1-story free-falls: when mass collides with floor below, they continue together the free fall until next floor level. There is no resistance to this fall (neither Strength, drag force, etc)
-
Application of Dynamic Resultant Theorem
11, ii Vm
0=V
Linear Momentum before collision
i
ii
V mmm += 1
i
ii
V mmm += 1
0=V
pi1 = mi1Vi1 ez Linear Momentum after
collision
pi1 = miVi ezh
0=V
Instantaneous Conservation of Linear Momentum
pi = 0 Vi = mi1 Vi1mi
Time of free fall over inter-story Before After Collision Collision
height V Vti = i i1 g
-
Results of the Discrete Model
110
100
90
80
70
60
50
40
30
20
10
00 1 2 3 4 5 6 7 8 9 10
V0
Free Fall
er a
bove
Gro
und
After BeforeCollision Collision
oor N
umb
lF
V /V0
110
100
90
80
70
60
50
40
30
20
10
00 1 2 3 4 5 6 7 8 9 10
V0
Free Fall
er a
bove
Gro
und
After BeforeCollision Collision
oor N
umb
lF
V /V0
(M = 80) = 9.0s ( )M = 96 = 10.8sM = 96
M = 80
-
From the Discrete Model to the Continuum Model
Discrete Model Continuum Model Discrete mass system Continuous mass
mi = m0 + i m h / H = 1/110
-
Continuum Approach
0V0V
Differential Equation NN ==110110
for collapse front MM == 9696
MM == 8080
Boundary Conditions
Solution* yields Evaluate for
(*) with MATLAB
-
9-11: engineering questions (last)
Why did the towers not tilt? Think: Dynamic Moment Theorem
And ask yourself, whether the resulting Photograph of airplane about to strike the south moment would have been large enough
tower removed due to copyright restrictions. to reach the strength limit of a building designed to withstand the moment generated by forces of a hurricane (weight equivalence of 1000 elephants)
World Trade Center Towers (1973 2001) Engineer: Leslie E. Robertson
Boeing 767 aircraft approaching the South Tower (www) Max Fuel: 90 m3 - Total max weight ~ 500 tons Approaching Speed V ~ 691 km/h (NT) / 810 km/h (ST)
-
1.050 Engineering Mechanics
Lecture 4: Stresses and StrengthStresses and Equilibrium
Discrete Model
-
1.050 Content overviewI. Dimensional analysis
1. On monsters, mice and mushrooms 2. Similarity relations: Important engineering tools
II. Stresses and strength 2. Stresses and equilibrium 3. Strength models (how to design structures,
foundations.. against mechanical failure)
III. Deformation and strain 4. How strain gages work? 5. How to measure deformation in a 3D
structure/material?
IV. Elasticity 5. Elasticity model link stresses and deformation 6. Variational methods in elasticity
V. How things fail and how to avoid it 7. Elastic instabilities 8. Plasticity (permanent deformation) 9. Fracture mechanics
Lectures 1-3 Sept.
Lectures 4-15 Sept./Oct.
Lectures 16-19 Oct.
Lectures 20-31 Nov.
Lectures 32-37 Dec.
-
1.050 Content overviewI. Dimensional analysis
II. Stresses and strength Lecture 4: Newtons laws, fall of the WTC towers Lecture 5: Stress vector and stress tensor Lecture 6: Hydrostatic problem ApplicationsLecture 7: Soil mechanics / geostatics problem Lecture 8: Beam stress model Lecture 9: Beam model II and summary Lecture 10: Strength models
III. Deformation and strain
IV. Elasticity
V. How things fail and how to avoid it
-
Content lecture 5 1. 3-scale continuum model: Molecular scale,
representative volume element (REV), macro-scale
2. Stress vector, stress matrix and stress tensor Definition of stress vector Generalized expression as stress matrix Definition of stress tensor
3. Implement dynamic resultant theorem for REV Use Gauss theorem (divergence theorem) Develop differential equilibrium: Partial differential equation
-
1.050 Engineering Mechanics
Lecture 6: Stresses and Equilibrium
Application: Hoover Dam
-
1.050 Content overviewI. Dimensional analysis
1. On monsters, mice and mushrooms 2. Similarity relations: Important engineering tools
II. Stresses and strength 2. Stresses and equilibrium 3. Strength models (how to design structures,
foundations.. against mechanical failure)
III. Deformation and strain 4. How strain gages work? 5. How to measure deformation in a 3D
structure/material?
IV. Elasticity 5. Elasticity model link stresses and deformation 6. Variational methods in elasticity
V. How things fail and how to avoid it 7. Elastic instabilities 8. Plasticity (permanent deformation) 9. Fracture mechanics
Lectures 1-3 Sept.
Lectures 4-15 Sept./Oct.
Lectures 16-19 Oct.
Lectures 20-31 Nov.
Lectures 32-37 Dec.
-
1.050 Content overviewI. Dimensional analysis
II. Stresses and strength Lecture 4: Newtons laws, fall of the WTC towersLecture 5: Stress vector and stress tensorLecture 6: Hydrostatic problem ApplicationsLecture 7: Soil mechanics / geostatics problem Lecture 8: Beam stress model Lecture 9: Beam model II and summary Lecture 10: Strength models
III. Deformation and strain
IV. Elasticity
V. How things fail and how to avoid it
-
Content lecture 6
1. Review: 3-scale continuum model: Molecular scale, representative volume element (REV), macro-scale; stress vector and stress tensor
2. Implement dynamic resultant theorem for REV Use Gauss theorem (divergence theorem) Develop differential equilibrium: Partial differential
equation
3. Application: Hoover Dam (hydrostatic problem)
-
Photographs of Hoover Dam removed due to copyright restrictions.
http://www.concreteresources.net/images/graphics/clip_image004.jpg http://www.sdsuniverse.info/Upload/hoover_dam.jpg
-
Energy Production ~ 4 billion kilowatt-hours a year ~ 1.3 million people
-
Forces that act on Hoover Dam
( )nzpT =
)(nT
-
Forces that act on Hoover DamForce reduction formula
Forces from stress vector
Force Equivalence
( )nzpT = ye
xe xF
yF
yF
dydzdSx =
Surface on which stress Hoover Dam: F ~ 16 billion Newtonx Vector acts (weight equivalence of 20 million people,
or of the entire population of Australia)
-
1.050 Engineering Mechanics
Lecture 7: Application: Hoover Dam and Soil
Mechanics
-
1.050 Content overviewI. Dimensional analysis
1. On monsters, mice and mushrooms 2. Similarity relations: Important engineering tools
II. Stresses and strength 2. Stresses and equilibrium 3. Strength models (how to design structures,
foundations.. against mechanical failure)
III. Deformation and strain 4. How strain gages work? 5. How to measure deformation in a 3D
structure/material?
IV. Elasticity 5. Elasticity model link stresses and deformation 6. Variational methods in elasticity
V. How things fail and how to avoid it 7. Elastic instabilities 8. Plasticity (permanent deformation) 9. Fracture mechanics
Lectures 1-3 Sept.
Lectures 4-15 Sept./Oct.
Lectures 16-19 Oct.
Lectures 20-31 Nov.
Lectures 32-37 Dec.
-
1.050 Content overviewI. Dimensional analysis
II. Stresses and strength Lecture 4: Newtons laws, fall of the WTC towersLecture 5: Stress vector and stress tensorLecture 6: Hydrostatic problem ApplicationsLecture 7: Soil mechanics / geostatics problem Lecture 8: Beam stress modelLecture 9: Beam model II and summaryLecture 10: Strength models
III. Deformation and strain
IV. Elasticity
V. How things fail and how to avoid it
-
Content lecture 7
1. Application I: Hoover Dam (hydrostatic problem) finishing
2. Application II: Soil mechanics / foundation
-
1.050 Engineering Mechanics
II. Stresses and StrengthApplication in Structural
Mechanics
-
Program 8th Lecture1-050 CONTENT
I. Dimensional Analysis: II. Stresses & Strength
2. Stresses and Equilibrium 1. Discrete Model 2. Continuum Model 3. Beam Model
3. Strength Models III. Deformation and Strain
4. How Strain Gages work? IV. Elasticity
5. Elastic Model 6. Variational Methods in
Elasticity V. How Things Fail? And How
to avoid it.
TODAY: 1. Scales of Structural mechanics:
Section vs. Beam structure 2. Link between stresses and
forces and moments 3. Beam Equilibrium Conditions 4. Example
Goal: Construct a Force-Moment Beam Model
Appreciate the link between Continuum Model and Beam Model
-
Three Scale Approach
Beam Scale defined by beam length
Cross-section scale (height, width)
(h,b)
-
From the Continuum Scale to the Cross Section Scale
Continuum Quantity: Stress vector
T (n = ex ) = ex
Section Quantities: Forces
F S = e dS x S
Moments
M S = x ( ex )dS S
xex,
yey,
zez,
|
|h
b
dSnd
-
From the Cross Section Scale to the Beam Length Scale
Differential Force equilibrium
eezz M zM z +
dM+ z
dM z dxdx eezz dx dx e e dF S extyy
MM y +y + dxdxeeyy dMdM y y + f = 0 dx dx dx MM eexx xx ffextext FFxexexx
ee ee dx dx
xx xx
FFyeyeyyMM x +x +
dMdM x dxx dxeexx dx dx
FF eezz zz Differential Moment MM eeyy yy e ze z zzyy
MM ee equilibrium eezz
dM S +ex F S = 0dx
-
Formulation of a Beam Boundary Value Problem
Example Force and Moment Boundary Conditions
Sum of all forces and Moments along x is zero
x Differential Equilibrium of Section forces Section moments
z ext
egSf = R
yz
-
1.050 Engineering Mechanics
II. Stresses and StrengthExamples: Beam Statistics
-
Program 9th Lecture1-050 CONTENT
I. Dimensional Analysis: II. Stresses & Strength
2. Stresses and Equilibrium 1. Discrete Model 2. Continuum Model 3. Beam Model
3. Strength Models III. Deformation and Strain
4. How Strain Gages work? IV. Elasticity
5. Elastic Model 6. Variational Methods in
Elasticity V. How Things Fail? And How
to avoid it.
TODAY: 1. Review: Beam Stress Model 2. Formulation of a Beam
Boundary Value Problem 3. Statically Determined vs.
Statically Indetermined Beam Structures
4. Closure: Stresses & Equilibrium
Goal: Appreciate Force-Moment Beam Model for solving beam problems
-
Review: Beam Model
1. Scales in Structural Mechanics
2. Reduction Formulas: (from stresses to sectionforces and section moments)
3. Equilibrium: alongbeam axis, differential equilibrium of forcesand moments
1/ 3O(d )
-
Formulation of a Beam Boundary Value Problem
Example Force and Moment Boundary Conditions
Sum of all forces and Moments along x is zero
x Differential Equilibrium of Section forces Section moments
z ext
egSf = R
yz
-
Stresses & EquilibriumDiscrete System Continuum System Beam System
Elementary System Internal Stresses
Boundary Condition
Continuity Condition
Diff. Force Equilibrium
Diff. Moment Equilibrium
-
1.050 Engineering Mechanics
Lecture 10: Strength models
1D examples truss structures
-
1.050 Content overviewI. Dimensional analysis
1. On monsters, mice and mushrooms 2. Similarity relations: Important engineering tools
II. Stresses and strength 2. Stresses and equilibrium 3. Strength models (how to design structures,
foundations.. against mechanical failure)
III. Deformation and strain 4. How strain gages work? 5. How to measure deformation in a 3D
structure/material?
IV. Elasticity 5. Elasticity model link stresses and deformation 6. Variational methods in elasticity
V. How things fail and how to avoid it 7. Elastic instabilities 8. Plasticity (permanent deformation) 9. Fracture mechanics
Lectures 1-3 Sept.
Lectures 4-15 Sept./Oct.
Lectures 16-19 Oct.
Lectures 20-31 Nov.
Lectures 32-37 Dec.
-
1.050 Content overviewI. Dimensional analysis
II. Stresses and strength Lecture 8: Beam stress model Lecture 9: Beam model II and summary Lecture 10: Strength models: Introduction (1D) Lecture 11: Mohr circle strength criteria 3D Lecture 12: Application foundations
III. Deformation and strain
IV. Elasticity
V. How things fail and how to avoid it
-
Quiz I Covers first 15 lectures
QUIZ I: Dimensional analysis, stresses and strength
Monday October 15 in class
Start to prepare early!
-
PP
Surface roughness
Image of crack propagation removed due to copyright restrictions.See Figure 3 in: Buehler, Markus, et al. Threshold Crack Speed ControlsDynamical Fracture of Silicon Single Crystals. Physical Review Letters 99(2007): 165502.
-
A1
A3
A2
A1/A3=2 same strength 0
0
1
2
3
4
5
6
7
8
Rob
ustn
ess
ratio
gam
ma_
1/ga
mm
a_3
0 20 40 60 80 100
Angle phi
-
1.050 Engineering Mechanics
Lecture 11: Strength models
3D model Mohr Circle
-
1.050 Content overviewI. Dimensional analysis
1. On monsters, mice and mushrooms 2. Similarity relations: Important engineering tools
II. Stresses and strength 2. Stresses and equilibrium 3. Strength models (how to design structures,
foundations.. against mechanical failure)
III. Deformation and strain 4. How strain gages work? 5. How to measure deformation in a 3D
structure/material?
IV. Elasticity 5. Elasticity model link stresses and deformation 6. Variational methods in elasticity
V. How things fail and how to avoid it 7. Elastic instabilities 8. Plasticity (permanent deformation) 9. Fracture mechanics
Lectures 1-3 Sept.
Lectures 4-15 Sept./Oct.
Lectures 16-19 Oct.
Lectures 20-31 Nov.
Lectures 32-37 Dec.
-
1.050 Content overviewI. Dimensional analysis
II. Stresses and strength Lecture 8: Beam stress modelLecture 9: Beam model II and summaryLecture 10: Strength models: Introduction (1D)Lecture 11: Mohr circle strength criteria 3D Lecture 12: Application soil mechanics: How to build sandcastles Lecture 13: Strength criterion in beams (I/II) Lecture 14: Strength criterion in beams (II/II) convexity of strength domain Lecture 15: Closure strength models & review for quiz
III. Deformation and strain
IV. Elasticity
V. How things fail and how to avoid it
-
Christian Otto Mohr (1835-1918)
German civil engineer, one of the most celebrated of the 19th century
Important contributions in strength of materials, design of steel trusses, bridges
Professor of Mechanics at Stuttgart PolytechnicPhotograph of Mohr removed due to copyright restrictions. and Dresden Polytechnic
Student of Mohr: Foeppl, the doctoral advisor of Ludwig Prandtl, who was the advisor of Theodore von Krmn (Caltech)
-
1.050 Engineering Mechanics
Lecture 12: Strength models
3D model Mohr-Coulomb modelApplication to sand piles
-
1.050 Content overviewI. Dimensional analysis
1. On monsters, mice and mushrooms 2. Similarity relations: Important engineering tools
II. Stresses and strength 2. Stresses and equilibrium 3. Strength models (how to design structures,
foundations.. against mechanical failure)
III. Deformation and strain 4. How strain gages work? 5. How to measure deformation in a 3D
structure/material?
IV. Elasticity 5. Elasticity model link stresses and deformation 6. Variational methods in elasticity
V. How things fail and how to avoid it 7. Elastic instabilities 8. Plasticity (permanent deformation) 9. Fracture mechanics
Lectures 1-3 Sept.
Lectures 4-15 Sept./Oct.
Lectures 16-19 Oct.
Lectures 20-31 Nov.
Lectures 32-37 Dec.
-
1.050 Content overviewI. Dimensional analysis
II. Stresses and strength Lecture 8: Beam stress modelLecture 9: Beam model II and summaryLecture 10: Strength models: Introduction (1D)Lecture 11: Mohr circle strength criteria 3DLecture 12: Application soil mechanics: How to build sandcastles Lecture 13: Strength criterion in beams (I/II) Lecture 14: Strength criterion in beams (II/II) convexity of strength domain Lecture 15: Closure strength models & review for quiz
III. Deformation and strain
IV. Elasticity
V. How things fail and how to avoid it
-
Sand piles
Photograph of sand dunes removed due to copyright restrictions.
http://cic.nist.gov/lipman/sciviz/scan/jun24_ptC1a.jpg http://www.maths.bris.ac.uk/~majhs/desert.bmp
-
Photograph of ornate sand castle removed due to copyright restrictions.
http://fixiefoo.typepad.com/photos/uncategorized/sand_castle.jpg
-
Another photograph of sand castle removed due to copyright restrictions.
http://uzar.files.wordpress.com/2007/04/sandcastle.jpg
-
1.050 Engineering Mechanics
Lecture 13: Strength models
Strength models for beams (I/II)
-
1.050 Content overviewI. Dimensional analysis
1. On monsters, mice and mushrooms 2. Similarity relations: Important engineering tools
II. Stresses and strength 2. Stresses and equilibrium 3. Strength models (how to design structures,
foundations.. against mechanical failure)
III. Deformation and strain 4. How strain gages work? 5. How to measure deformation in a 3D
structure/material?
IV. Elasticity 5. Elasticity model link stresses and deformation 6. Variational methods in elasticity
V. How things fail and how to avoid it 7. Elastic instabilities 8. Plasticity (permanent deformation) 9. Fracture mechanics
Lectures 1-3 Sept.
Lectures 4-15 Sept./Oct.
Lectures 16-19 Oct.
Lectures 20-31 Nov.
Lectures 32-37 Dec.
-
1.050 Content overviewI. Dimensional analysis
II. Stresses and strength Lecture 8: Beam stress model Lecture 9: Beam model II and summary Lecture 10: Strength models: Introduction (1D) Lecture 11: Mohr circle strength criteria 3D Lecture 12: Application soil mechanics: How to build sandcastles Lecture 13: Strength criterion in beams (I/II) Lecture 14: Strength criterion in beams (II/II) convexity of strength domain Lecture 15: Closure strength models & review for quiz
III. Deformation and strain
IV. Elasticity
V. How things fail and how to avoid it
-
Photographs removed due to copyright restrictions.
http://www.civeng.unsw.edu.au/research/images/semi-continuous.jpg http://members.tripod.com/str_n_tips/eq/eq_rcc2/g23.jpg
-
Photograph of beam failure removed due to copyright restrictions.
http://www.emeraldinsight.com/fig/1100210504010.png
-
1.050 Engineering Mechanics
Lecture 14: Strength models for beams (II/II)
M-N couplingConvexity of strength domain
-
1.050 Content overviewI. Dimensional analysis
1. On monsters, mice and mushrooms 2. Similarity relations: Important engineering tools
II. Stresses and strength 2. Stresses and equilibrium 3. Strength models (how to design structures,
foundations.. against mechanical failure)
III. Deformation and strain 4. How strain gages work? 5. How to measure deformation in a 3D
structure/material?
IV. Elasticity 5. Elasticity model link stresses and deformation 6. Variational methods in elasticity
V. How things fail and how to avoid it 7. Elastic instabilities 8. Plasticity (permanent deformation) 9. Fracture mechanics
Lectures 1-3 Sept.
Lectures 4-15 Sept./Oct.
Lectures 16-19 Oct.
Lectures 20-31 Nov.
Lectures 32-37 Dec.
-
1.050 Content overviewI. Dimensional analysis
II. Stresses and strength Lecture 8: Beam stress model Lecture 9: Beam model II and summary Lecture 10: Strength models: Introduction (1D) Lecture 11: Mohr circle strength criteria 3D Lecture 12: Application soil mechanics: How to build sandcastles Lecture 13: Strength criterion in beams (I/II) Lecture 14: Strength criterion in beams (II/II) convexity of strength domain Lecture 15: Closure strength models & review for quiz
III. Deformation and strain
IV. Elasticity
V. How things fail and how to avoid it
-
Quiz I
Wednesday, October 17 in class Please be on time Covers first 15 lectures Open book
Preparation: Lecture material, PSs, recitation Old quizzes (posted) instead of PS this week Alberto will work through one example
(nanoindentation) in recitation Study old quizzes before recitation this week
-
Strength models Equilibrium conditions only specify statically admissible
stress field, without worrying about if the stresses canactually be sustained by the material S.A. From EQ condition for a REV we can integrate up(upscale) to the structural scale
Examples: Many integrations in homework and in class;Hoover dam etc.
Strength compatibility adds the condition that in addition to S.A., the stress field must be compatible with the strengthcapacity of the material S.C. In other words, at no point in the domain can the stressvector exceed the strength capacity of the material
Examples: Sand pile, foundation etc. Mohr circle
-
Strength models
Max. shear stress c
Tresca criterion
Max. tensile stress
c
0= c
Tension cutoff criterion
n v : f (T v )
-
Strength models
=
Max. shear stress
c cohesion
function of
Mohr-Coulomb
c=0 dry sand
-
Review: Beam models
Beam model: Special case of the general continuum model
Special geometry highly distorted system (much longer than wide)
Special form of stress tensor:
h,b
-
Link between section quantities and section stress field
y
Section force and moment distribution is due to a particular stress tensor distribution in the section
z
Reduction formulas dS=dzdy
y,z: C.S. in section
Torsion
Bending
-
Example
Stress distribution in section Equivalent normal force
Nx
-
Review: Beam modelsBeam modelContinuum model
dxdDifferential element
Equilibrium condition 0
Line force density
Hydrostatics (fluid): 2D:Simplification
+BCs z
+BCs x
-
Beam models: Moments
-
Cantilever beam Dead weight (gravity)
2D Example
EQ and solution
Qz
-
Example: Coupled M-N strength domain
l (1)(2)
(I) Moment distribution Normal force distribution
(II) Pl
(3) NMy=Pl N=P
1
(3)(2) in (I)
(1) in (I) 1
-
1.050 Engineering Mechanics
Lecture 15: Closure strength models & review
for quiz
-
1.050 Content overviewI. Dimensional analysis
1. On monsters, mice and mushrooms 2. Similarity relations: Important engineering tools
II. Stresses and strength 2. Stresses and equilibrium 3. Strength models (how to design structures,
foundations.. against mechanical failure)
III. Deformation and strain 4. How strain gages work? 5. How to measure deformation in a 3D
structure/material?
IV. Elasticity 5. Elasticity model link stresses and deformation 6. Variational methods in elasticity
V. How things fail and how to avoid it 7. Elastic instabilities 8. Plasticity (permanent deformation) 9. Fracture mechanics
Lectures 1-3 Sept.
Lectures 4-15 Sept./Oct.
Lectures 16-19 Oct.
Lectures 20-31 Nov.
Lectures 32-37 Dec.
-
1.050 Content overviewI. Dimensional analysis
II. Stresses and strength Lecture 8: Beam stress model Lecture 9: Beam model II and summary Lecture 10: Strength models: Introduction (1D) Lecture 11: Mohr circle strength criteria 3D Lecture 12: Application soil mechanics: How to build sandcastles Lecture 13: Strength criterion in beams (I/II) Lecture 14: Strength criterion in beams (II/II) convexity of strength domain Lecture 15: Closure strength models & review for quiz
III. Deformation and strain
IV. Elasticity
V. How things fail and how to avoid it
-
1.050 Engineering Mechanics
Lecture 16: Introduction: Deformation and
strain
-
1.050 Content overviewI. Dimensional analysis
1. On monsters, mice and mushrooms 2. Similarity relations: Important engineering tools
II. Stresses and strength 2. Stresses and equilibrium 3. Strength models (how to design structures,
foundations.. against mechanical failure)
III. Deformation and strain 4. How strain gages work? 5. How to measure deformation in a 3D
structure/material?
IV. Elasticity 5. Elasticity model link stresses and deformation 6. Variational methods in elasticity
V. How things fail and how to avoid it 7. Elastic instabilities 8. Plasticity (permanent deformation) 9. Fracture mechanics
Lectures 1-3 Sept.
Lectures 4-15 Sept./Oct.
Lectures 16-19 Oct.
Lectures 20-31 Nov.
Lectures 32-37 Dec.
-
1.050 Content overviewI. Dimensional analysis
II. Stresses and strength
III. Deformation and strain Lecture 16: Introduction: Deformation and strain Lecture 17: Strain tensor and small deformation Lecture 18: Mohr circle in strain space Lecture 19: Beam deformation
IV. Elasticity
V. How things fail and how to avoid it
-
Stresses Thermodynamics
Energy balance Momentum Deformation Geometrical analysisconservation
Lectures 1-15 Lectures 20-.. Lectures 16-19
-
0 0
1
2
3
4
1 2 3
Ferry alloy
Constantan alloy
40% Gold / Palladium
Nickel
% Strain
10% Rhodium / Platinum
% R R
Figure by MIT OpenCourseWare.
-
1.050 Engineering Mechanics
Lecture 17: Deformation and strain (contd)
-
1.050 Content overviewI. Dimensional analysis
1. On monsters, mice and mushrooms 2. Similarity relations: Important engineering tools
II. Stresses and strength 2. Stresses and equilibrium 3. Strength models (how to design structures,
foundations.. against mechanical failure)
III. Deformation and strain 4. How strain gages work? 5. How to measure deformation in a 3D
structure/material?
IV. Elasticity 5. Elasticity model link stresses and deformation 6. Variational methods in elasticity
V. How things fail and how to avoid it 7. Elastic instabilities 8. Plasticity (permanent deformation) 9. Fracture mechanics
Lectures 1-3 Sept.
Lectures 4-15 Sept./Oct.
Lectures 16-19 Oct.
Lectures 20-31 Nov.
Lectures 32-37 Dec.
-
1.050 Content overviewI. Dimensional analysis
II. Stresses and strength
III. Deformation and strain Lecture 16: Introduction: Deformation and strain Lecture 17: Strain tensor Lecture 18: Simplification for small deformation; Mohr circle in strain space Lecture 19: Beam deformation
IV. Elasticity
V. How things fail and how to avoid it
-
Goals of this lecture
Review: The main tool deformation gradient tensor
Applications to calculation of Volume change Surface normal and surface area change Length change Angle change
-
General solution procedure
Elasticity condition (no dissipation): d = W reflecting that dD = 0 (this is the result from analyzing the TD as done in class)
Step 1: Express d (x , x ,..) = dx + dx + ... = dx1 2 1 2 ix1 x2 xi
F F Step 2: Express W (1,2 ,..) = d1 + d2 + ... = d j1 2 j
Step 3: Solve equations
xi dxi = j
d j dxi ,d j
Collect all terms dxi and d j and set the entire expression to zero.
In EQ, the expression must be satisfied for all displacement changes dxi ,d j
-
Example II: Truss structure (1)
Problem statement: Structure of three trusses with applied force F d :
Forces in each truss
1 1
0
F d 1, N1 2 , N2 Distance L=1 between the trusses N = k1 1
N = k
3, N3 Trusses behave
2 2 like springs
Goal: Calculate displacements i ,0
for given
F d N3 = k3
-
Example II: Truss structure (2)Rigid bar: If two displacements 1,2 are specified can calculate the other
displacements (kinematic constraint):
Deformation 1 1 2 1 3
0
Therefore:
3 13 = 1 + 2 2 1 = 22 1 0 = 2 11 2 2
-
Example II: Truss structure (3) Solution procedure:
Elasticity condition (no dissipation): d = W
Step 1: d (1,2 ) = 1 k[(41 42 )d1 + ( 41 + 102 )d2 ]2
Step 2: W (1) = Fd
12
d1 + 23 d2
Step 3: Solve equations d = W dxi ,d j 12
k[(41 42 )d1 + ( 41 +102 )d2 ]= F d
12
d1 + 23 d2
2k 2k +
1 F d d + 2k + 5k 3 F d d
= 0 1 2 1 1 2 2
2 2 ! !=0 =0 for elastic EQ
-
Example II: Truss structure (4) This results in linear system of equations:
1
2
d
1ad bc
=
1
/ 2/ 2
d
d
F3F
=
b
2k 2k
M Solve for the unknown variables 1,2 ,..
Note that (forming the inverse of a 2x2 matrix):
2k 5k
a bc d c a
This can be used to calculate M 1
to solve for
1,2
-
Example II: Truss structure (5) This results in:
d2
M 1
Solve for the other unknown variables (utilize kinematic relationships and the
2
spring equations):
1 2
d5 =
F / 2 F1=
1/12 1/ 3
dk k26 3F / 2
3 = 7 /12F d
k1/12F dN =
iN k= 1i1/ 3F dN =
=
2F d 0 = 11/ 24 k N3 7 /12F
d
-
1.050 Engineering Mechanics
Lecture 22: Isotropic elasticity
-
1.050 Content overviewI. Dimensional analysis
1. On monsters, mice and mushrooms 2. Similarity relations: Important engineering tools
II. Stresses and strength 3. Stresses and equilibrium 4. Strength models (how to design structures,
foundations.. against mechanical failure)
III. Deformation and strain 5. How strain gages work? 6. How to measure deformation in a 3D
structure/material?
IV. Elasticity 7. Elasticity model link stresses and deformation 8. Variational methods in elasticity
V. How things fail and how to avoid it 9. Elastic instabilities 10. Plasticity (permanent deformation) 11. Fracture mechanics
Lectures 1-3 Sept.
Lectures 4-15 Sept./Oct.
Lectures 16-19 Oct.
Lectures 20-31 Oct./Nov.
Lectures 32-37 Dec.
-
1.050 Content overviewI. Dimensional analysis
II. Stresses and strength
III. Deformation and strain
IV. Elasticity Lecture 20: Introduction to elasticity (thermodynamics) Lecture 21: Generalization to 3D continuum elasticity Lecture 22: Special case: isotropic elasticity Lecture 23: Applications and examples
V. How things fail and how to avoid it
-
Important concepts: Isotropic elasticity
Isotropic elasticity = elastic properties do not depend on direction
In terms of the free energy change, this means that the change ofthe free energy does not depend on the direction of deformation
Rather, it depends on quantities that are independent on thedirection of deformation (i.e., independent of coordinate system)
Idea: Use invariants of strain tensor to calculate free energy change
Volume change Shape change (shear deformation)
Note: Invariants are defined as properties of strain tensor that areindependent of coordinate system (C.S.)
-
Important mathematical toolsTrace of a tensor
( ) = :1 = 11 + 22 + 33 = d d Relates to the chain of
tr d 0 volume of REV d Independent of C.S. 0 trace of a tensor is an
invariant
= 1 ( : T ) = 1 ij 2 Magnitude of a tensor (2nd order norm)2 2 i j
Note: Analogy to the magnitude of a tensor is the norm of a first order tensor (=vector), that is, its length
-
Overview: ApproachStep 1: Calculate change in volume v = tr( )= :1 Step 2: Calculate magnitude of angle change
Define strain deviator tensor = tensor that describes deformation without the volume change (trace of strain deviator tensor is zero!)
= 1 tr( ) tr e = 11 + 22 + 33 1 (3( 11 + 22 + 33 =e 1 ( ) )) 0 3 3
d = 2 e = 21 (e : eT ) = 2 1 eij 2 2 2 i j
Step 3: Define two coefficients to link energy change with deformation (spring model):
= 2 1 2
vK + 2 1 2
dG
Bulk modulus Shear modulus
-
NoteThe approach that the free energy under deformation depends only on volume change and overall angle change is not derived from physical principles
Rather, it is an assumption, which is made to model the behavior of a solid (modeling is finding a mathematical representation of a physical phenomenon)
Generally, models must be validated, for instance through experiments
Alternative approach: Calculation of from first principles by explicitly considering the atomistic scale of atomic, molecular etc. interactions
Spring 2008: 1.021J Introduction to Modeling and Simulation (Buehler, Radovitzky, Marzari) continuum methods, particle methods, quantum mechanics
-
Stress-strain relationTotal stress tensor = sum of contribution from volume change and contribution from shape change:
= + v d
v = v
= d d
Next step: Carry out differentiations
-
Stress-strain relation Total stress tensor = sum of contribution from volume change and contribution from shape change:
= + = v = d v d v d
1. Calculation of v
v v vv = 1 Kv
2 v =
= : = Kv12 v
v (tr( )) ( :1)v = K = = = 1 v v
-
Stress-strain relation2. Calculation of d d =
1 Gd 2
2
d d =
= e
: e = 2Ge :
1
13
11 = 2Ge
13 (e :1)1
d
=0
since:
1 (2e : eT ) e 1 tr(e) = 0 d = G = 2Ge = 1 11e 2 e 3
Note (definition of d ): tensor productNote (definition of e ):d = 2 e = 2
12 (e : eT ) e = 1
3v1 =
13
( :1) 1
-
Complete stress-strain relation3. Putting it all together: = +
v d
= Kv1+ 2Ge = Kv1+ 2G
1 v1
3
e = 1 tr( ) 1
3 Deviatoric part of the strain tensor
= K 2 G v1+ 2G Reorganized
3
-
Complete stress-strain relation =
K 2
3 G v1+ 2G =
K 2
3 G (11 + 22 + 33 )1+ 2G
Writing it out in coefficient form:
11 = K 2 G (11 + 22 + 33 )+ 2G11 3
22 = K 2 G (11 + 22 + 33 )+ 2G22 3
33 = K 2 G (11 + 22 + 33 )+ 2G33 3
12 = 2G12 23 = 2G2313 = 2G13
-
Complete stress-strain relationRewrite by collecting terms multiplying ii
11 = K + 4 G 11 +
K 2 G 22 + K 2 G 33 (1)
3 3 3 22 =
K 2 G 11 + K + 4 G 22 +
K 2 G 33 (2) 3 3 3
33 = K 2 G 11 +
K 2 G 22 + K + 4 G 33 (3)
3 3 3
collecting terms multiplying 12 ,23,13 12 = 2G12
23 = 2G23
13 = 2G13
-
Complete stress-strain relation
4 2 2 11 = K +
3 G 11 +
K
3 G 22 +
K
3 G 33 (1)
4 c1111 = K + G32c1122 = K G = c11333
-
Complete stress-strain relation
2 4 2 22 = K
3 G 11 +
K +
3 G 22 +
K
3 G 33 (2)
2c2211 = K G = c223334c2222 = K + G3
-
Complete stress-strain relation
33 = K 2 G 11 +
K 2 G 22 + K + 4 G 33 (3)
3 3 3
2c3311 = K G = c332234c3333 = K + G3
-
Complete stress-strain relation
12 = 2G12 23 = 2G23 13 = 2G13
c1212 = 2G c2323 = 2G c1313 = 2G
All other cijkl are zero
-
Summary: Expression of elasticity tensor
4 c1111 = c2222 = c3333 = K + G3
2 c1122 = c1133 = c2233 = K G3
c1212 = c2323 = c1313 = 2G
-
Examples numerical valuesConcrete
K = 14 GPa G = 10 GPa
Quartz (sand, stone..)
K = 27 GPa G = 26 GPa
Steel
K = 200 GPa G = 140 GPa
-
1.050 Engineering Mechanics
Lecture 23: Example detailed steps
1
Problem statement p
Note: p is applied pressure at the top of the soil layer K ,G given r
Goal: Determine (x r), (x r), (x r)
On the next few slides we will go through steps 1, 2, 3 and 4 to solve this problem.
g rH
x
z
Isotropic solid (soil) on a rigid substrate (infinitely large in x-y-directions)
rigid substrate (no displacement)
2
1
-
Reminder: 4-step procedure to solve elasticity problems
Step 1: Write down BCs (stress BCs and displacement BCs), analyze the problem to be solved (read carefully!)
Step 2: Write governing equations for stress tensor, strain tensor, and constitutive equations that link stress and strain, simplifyexpressions
Step 3: Solve governing equations (e.g. by integration), typically results in expression with unknown integration constants
Step 4: Apply BCs (determine integration constants)
3
Step 1: Boundary conditions
Write out all BCs in mathematical equations
Displacement BCs: At z=H: Displacement specified
r d (z = H ) = (0,0,0) or x
d = 0, yd = 0,z
d = 0
(no displacement at the interface between the soil layer and the rigid substrate)
Stress BCs: At z=0: Stress vector provided T r
d (n r = e r , z = 0) = pe r z z
Note: Orientation of surface and C.S. 4
2
-
Step 2: Governing equations
Write out all governing equations and simplify
Due to the symmetry of the problem (infinite in x- and y-directions), the solution will depend on z only, and there are no displacements in the r x- and y-directions (anywhere in the solution domain): = e r z z
Governing eqn. for strain tensor: ij =
j
i
x
2 1
+
i
j
x
Calculation of strain tensor simplifies (symmetry): zz
= z z
(*) Note : only 1 nonzero coefficient of strain tensor
Governing eqn. for stress tensor: div + g r 0= (contd next slide)
5
Step 2: Governing equations (contd)
Gravity only in z-direction
xx
xy
y xzGoverning eqn. for stress tensor: + + + g = 0
xy + + yz + g = 0 x y z y
z x
yy
x
xz + + + gz = 0
yz
y zz
x z
Due to symmetry, only dependence on z-direction
xz = 0 yz = 0
z z Final set of governing eqns. for stress tensor
zz(1)
+ g = 0 (note: gz = g ) 6z
3
-
Step 2: Governing equations (contd)
Link between stress and strain
Linear isotropic elasticity (considering that there is only one nonzero coefficient in the strain tensor, zz ):
11 = K 2 G 33 3
22 = K
2 G 33 3
33 = K + 4 G 33 (2) 3
7
Step 2: Governing equations (contd)
Now combine eqns. (*), (1) and (2):
Substitute (2) in (1): zzz
K +
43
G + g = 0 (4)
Substitute (*) in (4):
2
z 2 z K + 4
3 G + g = 0
2z = g xz yzz2 K + 4 G
(5) z
= 0 z
= 0 (6)
3
Step 2 results in a set of differential eqns. 8
4
-
Step 3: Solve governing eqns. by integration
gFrom (5):
zz = 4 z + C1 = zz (first integration) K + G
3
zz = K + 4 G
g z + C1
(knowledge of strain enables 3 K +
4 G to calculate stress via eq. (2)) 3
z = 21 g
4 z2 + C1z + C2 (second integration)
K + G3
From (6):
xz yz z
= 0 z
= 0 xz = const . = C3 yz = const . = C4
9 Solution expressed in terms of integration constants Ci
Step 4: Apply BCs
Stress boundary conditions: Integration provided that
xz = const . = C3 yz = const . = C4
Stress vector at the boundary of the domain:
T r
d (n r = e r z , z = 0) = pe r
z = ! xze
r x yze
r y zze
r z
BC Stress vector due to stress tensor at z = 0:
T r (n r = e r z , z = 0) = (z = 0) (e
r z )
Left and right side xz = C3 = 0, yz = C4 = 0 Note: Orientation of must be equal, surface and C.S. therefore: = pzz
10
5
-
Step 4: Apply BCs (contd)
Further,
4 g zz = K + G 4 z + C1 (general solution) 3 K + G
3
zz (z = 0) = C1 K + 4 G =
! p (at z=0, see previous slide)
3
This enables us to determine the constant C1
C1 = p 4K + G3 11
Step 4: Apply BCs (contd)
Displacement boundary conditions:
z = 1 g z2 p z + C2 (general solution, with C12 K + 4 G K + 4 G included)
3 3
Displacement is known at z = H:
(z = H ) = 1 g H 2 p H + C =! 0z 2 K + 4 G K + 4 G 2
3 3 This enables us to determine the constant C2
C2 = 1
g H 2 + pH
K + 4 G 2 12 3
6
-
Final solution (summary): Displacement field, strain field, stress field
z (z) = 14
g (H 2 z2 ) p(z H ) K + G 2
3
zz (z) = gz + p
4K + G3
zz (z) = gz + p
13
Solution sketch
Displacement profile Stress profile:
g 2 p H H 2 K +
4 G K + 4 G
3 3
z = 0 p
z = 0z zz z z
z = H z = H ( p + gH )
14
7
-
1.050 Engineering Mechanics
Lecture 24: Beam elasticity derivation of governing
equation
1
1.050 Content overviewI. Dimensional analysis
1. On monsters, mice and mushrooms 2. Similarity relations: Important engineering tools
II. Stresses and strength 3. Stresses and equilibrium 4. Strength models (how to design structures,
foundations.. against mechanical failure)
III. Deformation and strain 5. How strain gages work? 6. How to measure deformation in a 3D
structure/material?
IV. Elasticity 7. Elasticity model link stresses and deformation 8. Variational methods in elasticity
V. How things fail and how to avoid it 9. Elastic instabilities 10. Plasticity (permanent deformation) 11. Fracture mechanics
Lectures 1-3 Sept.
Lectures 4-15 Sept./Oct.
Lectures 16-19 Oct.
Lectures 20-31 Oct./Nov.
Lectures 32-37 Dec. 2
1
-
1.050 Content overviewI. Dimensional analysis
II. Stresses and strength
III. Deformation and strain
IV. Elasticity Lecture 20: Introduction to elasticity (thermodynamics)Lecture 21: Generalization to 3D continuum elasticityLecture 22: Special case: isotropic elasticityLecture 23: Applications and examplesLecture 24: Beam elasticity Lecture 25: Applications and examples (beam elasticity)Lecture 26: contd and closure
V. How things fail and how to avoid it 3
Goal of this lecture
Derive differential equations that can be solved to determine stress, strain and displacement fields in beam
Consider 2D beam geometry:
z
x
+ boundary conditions (force, clamped, moments)
Approach: Utilize beam stress model, strain model for beams and combine with isotropic elasticity
4
2
-
Stress
( ) =
0 xx
00 xz
ij 0 0 0 xz
Shape of stress tensor for 2D beam problem
N = xxdS Qz = xzdS S S
M y = z xxdS S
dM y =Qz d 2 M y = f zdx dx2
dN = f
dx x
Isotropic elasticity: = K G 1+ 2G 5 3 v
Strain Navier-Bernouilli beam model
= 0 + 0 zxx xx y 0 d 2z
0
y = 2 Curvaturedx d 0 0 = x Axial strain xx dx
Thus: d 0 d 2 0 x z xx = 2 zdx dx
Strain completely determined from displacement of beam reference axis
2
Derivation of beam constitutive equation in 3-step approach
Section number below corresponds to section numbering used in class
Step 1: Consider continuum scale alone (derive a relation between stress and strain for the particular shape of the stress tensor in beam geometry) 2.1)
Step 2: Link continuum scale with section scale (use reduction formulas) 2.2)
Step 3: Link section scale to structural scale (beam EQ equations) 2.3)
6
3
-
Overview
Continuum scale
Section scale
Structural scale
, yz MQN ,, )(),(),( ),(),(),(
xxx xMxQxN
zxy
yz
Reminder:
Rotation (slope) 0 z
slope z
x
Curvature (=first derivative of rotation)
0 y = dz
0
y 0 =
d 2 2 z 0
= dy
0
7dx dx dx
x F
000
2.1) Step 1 (continuum scale) xx 0 0
z Consider a beam in uniaxial tension: ( ) ij =
0 0 0
xx = K 2 G ( xx + yy + zz )+ 2G xx (1) 3
3 unknowns, 2 (2) yy = K 2 G ( xx + yy + zz )+ 2G yy = ! 0equations; can 3 eliminate one
variable and obtain zz =
K 2
3 G ( xx + yy + zz )+ 2G zz = ! 0 (3)relation between 2
remaining ones
Eqns. (2) and (3) provide relation between xx and yy , zz : 1 3K 2G = = = yy zz xx xx2 3K + G
8 =: Poissons ratio
4
-
10
Physical meaning Poissons effect
The Poisson effect refers to the fact that beams contract in the lateral directions when subjected to tensile strain
= = yy zz xx
d1
d2 d2 = (1+ yy )d1 = (1 xx ) 9
9KGFrom eq. (1) (with Poisson relation): = xx xx3K +G
=: E Youngs modulus
xxxx E =
This result can be generalized: In bending, the shape of the stress tensor is identical, for any point in the cross-section (albeit the component zz typically varies with the coordinate z)
Thus, the same conditions for the lateral strains applies xx (z) 0 0
( ) = 0 0 Therefore: We can use the same formulas! ij 0 0 0 0
5
-
11
2.2) Step 2 (link to section scale)
Now: Plug in relation xx = E xx into reduction formulas
Consider that xx = ddx x
0
ddx
2 2 z 0
z and thus xx = E
ddx x
0
ddx
2 2 z 0
z
Results in: Assume: E constant over S =0
S
ddx x
0 ddx
2 2 z 0
N = E ddx x
0
S
dS E dx d 2
2 z 0
S
zdSN = E z dS
=0
M y = S
E
ddx x
0
z ddx
2 2 z 0
z2
dS M y = E dx
dx 0
S
zdS E ddx
2 2 z 0
S
z2dS
= I Finally: N = ES dx
0
M y = EI d 2
2 z 0
Area
dx dx moment of inertia
2.3) Step 4 (link to structural scale)
Beam EQ equations: Beam constitutive equations:
d 4 0 zM = EI = fy 4 zdxd 2 M y = f zdx2 d 4 0 fz z= with: dx4 EI
z 0dN = f x
M y = EI ddx
2 2
0
d 2x = f x 2dx d 0 dx ES xN = ES dx
12
6
-
Beam bending elasticity Governed by this differential equation:
d 4 0 fz z= dx4 EI
Integration provides solution for displacement
Solve integration constants by applying BCs
Note:
E = material parameter (Youngs modulus)
I = geometry parameter (property of cross-section)
f = distributed shear force z
How to solve? Lecture 25 13
7
-
1.050 Engineering Mechanics I
Lecture 25: Beam elasticity problem solving technique
and examples
Handout 1
1.050 Content overviewI. Dimensional analysis
1. On monsters, mice and mushrooms 2. Similarity relations: Important engineering tools
II. Stresses and strength 3. Stresses and equilibrium 4. Strength models (how to design structures,
foundations.. against mechanical failure)
III. Deformation and strain 5. How strain gages work? 6. How to measure deformation in a 3D
structure/material?
IV. Elasticity 7. Elasticity model link stresses and deformation 8. Variational methods in elasticity
V. How things fail and how to avoid it 9. Elastic instabilities 10. Plasticity (permanent deformation) 11. Fracture mechanics
Lectures 1-3 Sept.
Lectures 4-15 Sept./Oct.
Lectures 16-19 Oct.
Lectures 20-31 Oct./Nov.
Lectures 32-37 Dec. 2
1
-
1.050 Content overviewI. Dimensional analysis
II. Stresses and strength
III. Deformation and strain
IV. Elasticity Lecture 20: Introduction to elasticity (thermodynamics)Lecture 21: Generalization to 3D continuum elasticityLecture 22: Special case: isotropic elasticityLecture 23: Applications and examplesLecture 24: Beam elasticityLecture 25: Applications and examples (beam elasticity) Lecture 26: contd and closure
V. How things fail and how to avoid it 3
Beam bending elasticity Governed by this differential equation:
d 4 fz z= dx4 EI
Integration provides solution for displacement
Solve integration constants by applying BCs
Note:
E = material parameter (Youngs modulus)
I = geometry parameter (property of cross-section)
fz = distributed shear force (force per unit length)
f z = pb0 where p0=pressure, b=thickness of beam in y-direction 4
2
-
4-step procedure to solve beam elasticity problems
Step 1: Write down BCs (stress BCs and displacement BCs), analyze the problem to be solved (read carefully!)
Step 2: Write governing equations for z , x ...
Step 3: Solve governing equations (e.g. by integration), results in expression with unknown integration constants
Step 4: Apply BCs (determine integration constants)
Note: Very similar procedure as for 3D isotropic elasticity problems 5
Difference in governing equations (simpler for beams)
Physical meaning of derivatives of z
d 4z = f z d 4z EI = f Shear force density
dx4 EI dx4 z
ddx
3 3 z =
QEI
z d 3
3 z EI = Qz Shear force dx
d 2z = M y d
2z EI = M Bending moment dx2 EI dx2 y
dz = y dz = y
Rotation (angle)
dx dx Displacement z z
6
3
-
8
Step-by-step example
z p = force/length
xl length
Step 1: BCs z (0) = 0 EI
x = 0 (0) = 0y z (l) = 0
x = l M y (0) = 0
7
Step 2: Governing equation
d 4z f z d 4z p= = dx4 EI dx4 EI
p applied in negative z-direction
Step 3: Integration ''' = p x + Cz 1EI 2
z '' =
p x + C1x + C2EI 2
'''' = p 3 2z EI z ' =
p x + C1
x + C2 x + C3EI 6 2
4 3 2
z = p x
+ C1 x
+ C2 x
+ C3x + C4EI 24 6 2
4
-
Step 4: Apply BCs
z ''' =
p x + C1 = Qz
EI EI
z
2
z '' =
p x + C1x + C2EI 2
3 2
z ' =
p x + C1
x + C2 x + C3EI 6 2
EI M
y
y
=
=
4 3 2
= p x
+ C1 x
+ C2 x
+ C3x + C4EI 24 6 2
Known quantities are marked 9
Step 4: Apply BCs (contd)
z (0) = 0 C4 = 0y (0) = 0 C3 = 0
z (l) = 0 p l4
+ C1 l3 + C2
l 2 = 0
EI 24 6 2
M y (0) = 0 p l 2
+ C1l +C 2 = 0EI 2
l3 l 2
l 4 C1 =
p 5 l C1
=
p 24 EI 8 6 l 1
2 C2 EI l
2 p 1 2 C = l 2 2 EI 8 10
5
-
Solution:
Qz (x) = p x 5 l 8
M y (x) = p 1 l2 + x
2
5 lx
8 2 8
y(x) = EIp
18
l 2 x + x 6
3
165 lx2
z (x) = p 1 l 2 x2 + x
4
5 lx3
EI 16 24 48
11
6
-
1.050 Engineering Mechanics I
Lecture 26 Beam elasticity how to sketch the solution
Another exampleTransversal shear in beams
Handout
1
1.050 Content overview I. Dimensional analysis
1. On monsters, mice and mushrooms Lectures 1-3 2. Similarity relations: Important engineering tools Sept.
II. Stresses and strength 3. Stresses and equilibrium Lectures 4-15 4. Strength models (how to design structures,
foundations.. against mechanical failure) Sept./Oct.
III. Deformation and strain 5. How strain gages work? 6. How to measure deformation in a 3D Lectures 16-19
structure/material? Oct.
IV. Elasticity 7. Elasticity model link stresses and deformation Lectures 20-31 8. Variational methods in elasticity Oct./Nov.
V. How things fail and how to avoid it 9. Elastic instabilities 10. Plasticity (permanent deformation) Lectures 32-37 11. Fracture mechanics Dec. 2
1.050 Content overview I. Dimensional analysis
II. Stresses and strength
III. Deformation and strain
IV. Elasticity Lecture 20: Introduction to elasticity (thermodynamics)Lecture 21: Generalization to 3D continuum elasticityLecture 22: Special case: isotropic elasticityLecture 23: Applications and examplesLecture 24: Beam elasticity Lecture 25: Applications and examples (beam elasticity)Lecture 26: contd and closure
V. How things fail and how to avoid it 3
Drawing approach Start from fz = EIz
'''' , then work your way up
Note sign changes: '''' ~ fz z +
''' ~ Qz z '' ~ Mz y ' ~ z y
+ ~ z z
At each level of derivative, first plot extreme cases at ends of beam
Then consider zeros of higher derivatives; determine points of local min/max
z represents physical shape of the beam (beam line) 4
1
-
Review: Finding min/max of functions
Example
f (x) = x2f (x) function of x
f ' (x) = 0 necessary condition for min/max
f ' (x) = 2x f ' ' (x) < 0 local maximum f ' ' (x) > 0 local minimum
f ' ' (x) = 0 inflection point
f ' ' (x) = 2
Example solved in lecture 25:
z p = force/length
x l length
EI
Qz (x) = p x
85 l y(x) = EI
p
81 l 2 x + x
6
3
165 lx2
1 2 x2 5 p 1 2 2 x4 5 3 M y (x) = p 8
l + 2
8 lx z (x) = EI
16
l x + 24
48
lx
65
p
fz (x) = p ~ z ''''
5 pl
8 3 8
pl Qz (x) = p
x
5 l ~ z '''
8
8
+= 3
4 22
48 5
2416 1)( lxx xl
EI pxz
48 13
EI pl
8
2plmin
max
M y (x) = p 1 l2 + x
2
5 lx
~ z
''
8 2 8
y(x) = p 1 l 2 x + x
3
5 lx2
~ z
'
EI 8 6 16
z (x) = p 1 l 2 x2 + x
4
5 lx3
EI 16 24 48
7
min
2
-
Illustration of various BCs Example with point load z
Free end r l PF = 0 = 0z
r xM = 0M = 0 y
Concentrated force = 0 z (0) = 0 Qz (l) = P
x
= 0 Step 1: BCs x = 0
(0) = 0 x = l
M (l) = 0Qz = P y y y
P Hinge (bending)
r = 0 Step 2: Governing equation d
4z = 0M = 0y = 0 dx
4 y
9 10
Example with point load (contd) Step 3: Integrate z
'''' = 0,z ''' = C1 =
Qz EI
Mz
'' = C1x + C2 = y
2 EI z
' = C1 x
+ C2 x + C3 = y23 2
z = C1 x
+ C2 x
+ C3 x + C46 2
Step 4: Determine integration constants by applying BCs
z (0) = 0 C4 = 0 y = z ' (0) = 0 C3 = 0
P PlM y (l) = EI ( l + C2 ) = 0 C2 = EI EI P
11Qz (l) = C1EI = P C1 = EI
Example with point load (contd)
f z = 0
Qz = P
M y = P(l x) P x2 Pl = xy EI 2 EI
P x3 Pl x2 = z EI 6 EI 2
12
3
-
f z (x) = 0 ~ z ''''
P
Qz (x) = P ~ z '''
Pl ''M y (x) = P(l x) ~ z
max 2
(x) =
P x
Pl x ~
' 1 Pl 2 y z EI 2 EI 2 EI
P x3 Pl x2 z (x) = ~ zmax EI 6 EI 2 1 Pl 3 13
3 EI
P
P x3 Pl x2 z (x) = 14EI 6 EI 2
Plotting stress distribution in beams Example: Plotting stress distribution in cross-section beams cross-section
Given: Section quantities known as a function of position x z Fixed x:Want: Calculate stress distribution in the section
0 xx = E( xx +y z)N = ES 0 (z)xx xxwith: M = EIy y
N (x) M y (x) N (x) M y (x) N M y xx (z; x) = E ES
+ EI
z = S
+ I
z 15
N > 0, M y > 0 xx (z) = S +
Iz
16
4
-
1.050 Engineering Mechanics I
Lecture 27
Introduction: Energy bounds in linear elasticity
1
-
1.050 Content overviewI. Dimensional analysis
1. On monsters, mice and mushrooms 2. Similarity relations: Important engineering tools
II. Stresses and strength 3. Stresses and equilibrium 4. Strength models (how to design structures, foundations..
against mechanical failure)
III. Deformation and strain 5. How strain gages work? 6. How to measure deformation in a 3D structure/material?
IV. Elasticity 7. Elasticity model link stresses and deformation 8. Variational methods in elasticity
V. How things fail and how to avoid it 9. Elastic instabilities 10. Plasticity (permanent deformation) 11. Fracture mechanics
Lectures 1-3 Sept.
Lectures 4-15 Sept./Oct.
Lectures 16-19 Oct.
Lectures 20-31 Oct./Nov.
Lectures 32-37 Dec.
2
-
1.050 Content overviewI. Dimensional analysis
II. Stresses and strength
III. Deformation and strain
IV. Elasticity Lecture 20: Introduction to elasticity (thermodynamics) Lecture 21: Generalization to 3D continuum elasticity Lecture 22: Special case: isotropic elasticity Lecture 23: Applications and examples Lecture 24: Beam elasticity Lecture 25: Applications and examples (beam elasticity) Lecture 26: contd and closure Lecture 27: Introduction: Energy bounds in linear elasticity (1D system) Lecture 28: Introduction: Energy bounds in linear elasticity (1D system),
contd
V. How things fail and how to avoid it Lectures 32..37 3
-
Convexity of a function
f (x) f | (b a) f (b) f (a)x x=a
secant
f (x)
tangent
x a b 4
-
Example system: 1D truss structure
We will use this example to illustrate all key concepts
5
-
Total external work
* & * & W d = F d + d R
Work done by Work done by prescribed prescribed forces displacements, Displacements force unknown unknown
6
lec1lec2lec3lec4lec5lec6lec7lec8lec9lec10lec11lec12lec13lec14lec15lec16lec17lec19lec22lec23lec24lec25lec26