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.


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.


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.


Lecture 1: Introduction & Galileo's problem Lecture 2: Dimensional Analysis and Atomic ExplosionLecture 3: Dimension analysis and application to engineering

structures



IV. Elasticity


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


Lecture 3: Dimension analysis and application to engineering

structures


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,


III. Deformation and strain4. How strain gages work?5. How to measure deformation in a 3D Lectures 16-19


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.


Lecture 1: Introduction & Galileo's problem Lecture 2: Dimensional Analysis and Atomic Explosion Lecture 3: Dimension analysis and application to engineering

structures



IV. Elasticity


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.


Lecture 4: Stresses and StrengthStresses and Equilibrium

Discrete Model






structure/material?



Lectures 1-3 Sept.


Lectures 16-19 Oct.

Lectures 20-31 Nov.

Lectures 32-37 Dec.


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


IV. Elasticity


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)


Lecture 4: Stresses and StrengthStresses and Equilibrium

Discrete Model






structure/material?



Lectures 1-3 Sept.


Lectures 16-19 Oct.

Lectures 20-31 Nov.

Lectures 32-37 Dec.


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


IV. Elasticity


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


Lecture 6: Stresses and Equilibrium

Application: Hoover Dam






structure/material?



Lectures 1-3 Sept.


Lectures 16-19 Oct.

Lectures 20-31 Nov.

Lectures 32-37 Dec.


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


IV. Elasticity


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)


Lecture 7: Application: Hoover Dam and Soil

Mechanics






structure/material?



Lectures 1-3 Sept.


Lectures 16-19 Oct.

Lectures 20-31 Nov.

Lectures 32-37 Dec.


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


IV. Elasticity


Content lecture 7

1. Application I: Hoover Dam (hydrostatic problem) finishing

2. Application II: Soil mechanics / foundation


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


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


Lecture 10: Strength models

1D examples truss structures






structure/material?



Lectures 1-3 Sept.


Lectures 16-19 Oct.

Lectures 20-31 Nov.

Lectures 32-37 Dec.


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


IV. Elasticity


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



3D model Mohr Circle






structure/material?



Lectures 1-3 Sept.


Lectures 16-19 Oct.

Lectures 20-31 Nov.

Lectures 32-37 Dec.


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


IV. Elasticity


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)



3D model Mohr-Coulomb modelApplication to sand piles






structure/material?



Lectures 1-3 Sept.


Lectures 16-19 Oct.

Lectures 20-31 Nov.

Lectures 32-37 Dec.


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


IV. Elasticity


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



Strength models for beams (I/II)






structure/material?



Lectures 1-3 Sept.


Lectures 16-19 Oct.

Lectures 20-31 Nov.

Lectures 32-37 Dec.


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


IV. Elasticity


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


Lecture 14: Strength models for beams (II/II)

M-N couplingConvexity of strength domain






structure/material?



Lectures 1-3 Sept.


Lectures 16-19 Oct.

Lectures 20-31 Nov.

Lectures 32-37 Dec.




IV. Elasticity


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


Lecture 15: Closure strength models & review

for quiz






structure/material?



Lectures 1-3 Sept.


Lectures 16-19 Oct.

Lectures 20-31 Nov.

Lectures 32-37 Dec.




IV. Elasticity



Lecture 16: Introduction: Deformation and

strain






structure/material?



Lectures 1-3 Sept.


Lectures 16-19 Oct.

Lectures 20-31 Nov.

Lectures 32-37 Dec.



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


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.


Lecture 17: Deformation and strain (contd)






structure/material?



Lectures 1-3 Sept.


Lectures 16-19 Oct.

Lectures 20-31 Nov.

Lectures 32-37 Dec.



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


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


Lecture 22: Isotropic elasticity






structure/material?



Lectures 1-3 Sept.


Lectures 16-19 Oct.

Lectures 20-31 Oct./Nov.

Lectures 32-37 Dec.




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


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


2 4 2 22 = K

3 G 11 +

K +

3 G 22 +

K

3 G 33 (2)

2c2211 = K G = c223334c2222 = K + G3


33 = K 2 G 11 +

K 2 G 22 + K + 4 G 33 (3)

3 3 3

2c3311 = K G = c332234c3333 = K + G3


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


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


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


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


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


Lecture 24: Beam elasticity derivation of governing

equation

1






structure/material?



Lectures 1-3 Sept.


Lectures 16-19 Oct.


Lectures 32-37 Dec. 2

1




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


Lecture 25: Beam elasticity problem solving technique

and examples

Handout 1






structure/material?



Lectures 1-3 Sept.


Lectures 16-19 Oct.


Lectures 32-37 Dec. 2

1




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


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


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


Lecture 26 Beam elasticity how to sketch the solution

Another exampleTransversal shear in beams

Handout

1


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,


III. Deformation and strain 5. How strain gages work? 6. How to measure deformation in a 3D Lectures 16-19


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




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


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


Lecture 27

Introduction: Energy bounds in linear elasticity

1



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?



Lectures 1-3 Sept.


Lectures 16-19 Oct.


Lectures 32-37 Dec.

2




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
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Engineering Mechanics-MIT NOTES