Structural Mechanics - MIT 2002

7/27/2019 Structural Mechanics - MIT 2002

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MIT - 16.20 Fall, 2002

Unit 1

Introduction and Design Overview

Paul A. Lagace, Ph.D.Professor of Aeronautics & Astronautics

and Engineering Systems

Paul A. Lagace 2001


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Need to study structural mechanics to design properly toprevent failure

There is no doubt that any of the disciplines of Aeronautics andAstronautics can contribute to an accident

-engine failure-etc.

But, the vast majority of non-human induced accidents is due to structural(material) failure (ultimately).

Purpose of 16.20: Provide you with the tools to properlyDesign Aerospace Structures to assure structural integrity

(i.e., it doesnt fail)Note, 16.20 mainly oriented in past to aircraft structures because that iswhere the main experience lies. We will try to generalize and showexamples for space structures.

Paul A. Lagace 2001 Unit 1 - p. 2


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Aeronautics and Astronautics deal with three major categories of

structures:1. Aircraft (atmospheric vehicles)2. Launch vehicles3. Space structures (partially a civil engineering task?)

(Note: Transatmospheric vehicles can becombinations of 1 and 2the Shuttle is!)

IMPORTANT: Many of the design considerations for these

three categories are different, but the same techniques andconcepts are used to analyze the structures (basically)

In fact, except for special design considerations, the techniques used for

all structures are basically the same:



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Possible considerationsStructure type

BuildingsShipsCarsSpace stationsAirplanes

The difference is often in the degree of refinement of the structuralanalysis (generally more refined in A & A!)

We will teach basic techniques and concepts and usespecific examples. But, the technique may apply to anotherstructural type as well.

Example: (aircraft to space station)Fuselage --> space station living habitat (pressurized cylinders)



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Overview of Structural Design Process(Review from U.E.)

Purpose: Assure structural integrity while minimizing cost

In aerospace structures, cost often means weight.Why?

Saving a pound of weight means more- payload (extra passengers, more satellites)

- fuel (longer distance, longer duration via extendedstation keeping)

Amount industries (civilian) are willing to pay to save a pound of weight:Satellites $10,000 - $20,000 (w/o servicing)Transport Aircraft $100 - $200General Aircraft $25

Automobile

$0.00Paul A. Lagace 2001 Unit 1 - p. 5


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Sometimes willing to pay for performance

(military, FWCpolar orbit)Factors in determining cost

Material cost Waste amount Manufacturing Subassembly/assembly Durability and maintenance Useful life

An engineer must consider all these. In 16.20 we will focus onstructural integrity and methods to assess such.

Definition of structural integrity :

Capability of a structure to carry out the operation for which it wasdesigned.



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Many aspects

Required loads Required deformations Corrosion resistance (e.g., no penetration on

pressure vessels)

Many aspects to failure (we will discuss later)



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Design Process

5. Determine capability of structure to carry loads(of box 2) subject to design restrictions (of box 1)

4.

3.

size components

2. environment to which structure is subjected

1.

2

1

Manufacturing & Maintenanceconsidered throughout

Determine internal stresses and deflections

Layout structural arrangement, select materials,

Determine applied loads and operating

Design restrictions/specifications



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Iterative loops

Big loop 2 : May find there is no way to attain what was asked for

Loop 1 : Go through this all the time. Get more and more

specific on design each time- Use more refined techniques each time- Iterate to get most efficient structure

#1 usually given from above

#2 tells us what we need to considerin 16.20

Learn to do #4 and #5 Use knowledge attained in 16.20 and by doing

#4 a number of times to think about #3

Current use of IPTscustomers involved



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Unit 2

Loads and Design ConsiderationsReadings:

Rivello (Ch. 1)Cutler book (at leisure)G 7.1



Paul A. Lagace 2001


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Sources of Stresses and Strains

Depends on type of structure

Aircraft Launch Vehicles Space Structures

General Other Considerations



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Can generally divide these into: Normal operational effects (regular use) Environmental effects (internal stresses, material property

degradation) Isolated effects (lightning, impact)

In a (large company) Design group does general management Loads group determines operating conditions This is passed on to stress group that analyzes stresses and

deformations Materials group provides material ultimates, etc. Need to understand each part

NOTE:

New approach in companies: IPT (IntegratedProduct Teams)

DBT (Design Build Teams) - people from eachbranch including manufacturing and marketing

even more important to understandvarious factors



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Factors, Margins, etc.

Two important definitions for static considerations

Limit Load/Stress/Condition: Maximum load/stress/condition wherestructure shows no permanent deformation.

Ultimate Load/Stress/Condition: Maximum load/stress/conditionwhere structure does not fail.

Definition is key; often defined asbreak (i.e., carry no more load)

Operationally, the limit load is the maximum load the structure is expectedto see

The ultimate load provides a factor of safety for unknownsUltimate Factor of Safety (U.F.S.) = Ultimate Load

Limit LoadThis is a design value



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F.O.S. is also a Factor of Ignorance

This accounts for

probability & statistics

(also in material allowances)U.F.S. = 1.5 for Aircraft

1.25 for Spacecraft (unmanned)

Design is usually conservative and an additional Margin of Safety(M.O.S.) is used/results

Limit MOS = Tested Limit X - Limit XLimit X

Ultimate MOS = Tested Ultimate X - Ultimate XUltimate X

An MOS is an experimental reality.Paul A. Lagace 2001 Unit 2 - p. 5


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Try to minimize M.O.S.

- Too conservative

too heavy- Not conservative enough plane falls out of sky (thingshave flown with negative M.O.S.)

So, begin with operational envelope, the way the structure will be used Aircraft --> v-n diagram Spacecraft, etc.

Then add special conditions (gusts, etc.)

Also need to account for Environmental effects

- change in material properties- causes stresses and strains

Special conditions- Lightning

- Impact- etc.



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Fatigue (cyclic loading)

- Effect on material properties- Damage growth

Example

A fighter aircraft has a gross take-off weight of 30,000 lbs. In a test of onewing, the wing fails at a total loading of 243,000 lbs. What is the marginof safety?

Definition of M.O.S. = Tested Ultimate - UltimateUltimate

We know: Tested Ultimate = 243,000 lbs.

(for one wing)How do we get the Design Ultimate?

Design Ultimate = Design Limit x Factor of Safety

For aircraft, F.O.S. = 1.5



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How do we get the Design Limit?Use the v-n diagramFrom U.E., max n for fighter = 9 = limit n

(Note: n for level flight = 1 loading for level flight = weight)

Design Limit = nlimit

x weight= 9 x 30,000 lbs. = 270,000 lbs.

But , each wing carries 1/2 of this

Design limit load for one wing = 135,000 lbs.

Design ultimate load for one wing = 202,500 lbs.

Finally, M.O.S. = 243,000 lbs. - 202,500 lbs.202,500 lbs.

= 40,500 lbs. = 0.2 = M.O.S202,500 lbs.+ 20% margin



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What is failure?

It depends on use..may be deflection based (CTE example).may be fracture.may be buckling

.

.

.

Overall, you must consider:Static Stresses

- Fracture- Yielding

- BucklingDeflections- Clearances- Flutter- Vibration- Tolerances (e.g., C. T. E.)



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Life- Damage accumulation- Fatigue

For aircraft, design guidelines provided by FAA F.A.R.s (Federal AviationRegulations)

Part 23 Part 25 Part 27 Part 29

Aircraft Aircraft HelicopterUnder Above

14,000 lbs. 14,000 lbs.

USAF guidelines (e.g., Damage Tolerance Regulations)



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Unit 3(Review of) Language of

Stress/Strain AnalysisReadings:

B, M, P A.2, A.3, A.6Rivello 2.1, 2.2T & G Ch. 1 (especially 1.7)



Paul A. Lagace 2001


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Recall the definition of stress: = stress = intensity of internal force at a point

Figure 3.1 Representation of cross-section of a general body

Fn

F s

F Stress = lim A

A 0

There are two types of stress: n (F n) 1. Normal (or extensional): act normal to the plane of the

element s (F s) 2. Shear: act in-plane of element

Sometimes delineated as



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And recall the definition of strain: = strain = percentage deformation of an infinitesimal element

Figure 3.2 Representation of 1-Dimensional Extension of a body

L = lim L

L 0

Again, there are two types of strain:n 1. Normal (or extensional): elongation of elements 2. Shear: angular change of element

Sometimes delineated as

Figure 3.3 Illustration of Shear Deformation

shear

deformation!Paul A. Lagace 2001 Unit 3 - p. 3


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Since stress and strain have components in several directions, we need anotation to represent these (as you learnt initially in Unified)

Several possible Tensor (indicial) notation Contracted notation will review here Engineering notation and give examples Matrix notation in recitation

IMPORTANT: Regardless of thenotation, the equations and conceptshave the same meaning learn, be comfortable with, be able to

use all notations

Tensor (or Summation) Notation Easy to write complicated formulae Easy to mathematically manipulate Elegant, rigorous Use for derivations or to succinctly express a set of equations or a long

equationPaul A. Lagace 2001 Unit 3 - p. 4


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Example: xi = fij y j

Rules for subscripts NOTE: index subscript Latin subscripts (m, n, p, q, ) take on the values 1, 2, 3 (3-D)

Greek subscripts ( , , ) take on the values 1, 2 (2-D)

When subscripts are repeated on one side of the equationwithin one term, they are called dummy indices and are to besummed on

Thus:3

fij y j = fij y j j=1

But fij y j + g i ... do not sum on i !

Subscripts which appear only once on the left side of the equationwithin one term are called free indices and represent a separateequation



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Thus:

xi = .. x1 = ..x2 = ..x3 = ..

Key Concept: The letters used forindices have no inherent meaning inand of themselves

Thus: x i = fij y j

is the same as: xr = frs y s or x j=

f y i ji

Now apply these concepts for stress/strain analysis:

1. Coordinate System

Generally deal with right-handed rectangular Cartesian: y m



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Figure 3.4 Right-handed rectangular Cartesian coordinate systemy

3 , z

y2, y

Compare notations y1 , x

zy3

yy2

xy1

EngineeringTensor

Note: Normally this is so, but always checkdefinitions in any article, book, report, etc. Keyissue is self-consistency, not consistency with aworldwide standard (an official one does not

exist!)Paul A. Lagace 2001 Unit 3 - p. 7


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2. Deformations/Displacements (3)

Figure 3.5 p(y 1, y 2, y 3),small p

(deformed position)P(Y 1, Y2, Y3)

Capital P(original position)

um = p(y m) - P(y m)

--> Compare notations

Tensor Engineering Direction inEngineering

u 1 u xu 2 v y

u 3 w z



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3. Components of Stress (6)

mn Stress Tensor 2 subscripts 2nd order tensor6 independent components

Extensional

11 22 33

Note: stress tensor is symmetricmn = nm

Shear

12 = 21 23 = 32 13 = 31

due to equilibrium (moment) considerations

Meaning of subscripts: mnstress acts in n-direction

stress acts on facewith normal vector inthe m-direction



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Figure 3.6 Differential element in rectangular system

NOTE: If face has a negative normal,positive stress is in negative direction


= yz sometimes= xz used for= xy shear stresses

Tensor Engineering 11 x 22 y

33 z 23 yz 13 xz 12 xy


MIT 16 20 ll


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4. Components of Strain (6)

mn Strain Tensor2 subscripts 2nd order tensor

6 independent components

Extensional Shear112233

12 = 2123 = 3213 = 31

NOTE (again): strain tensor is symmetricmn = nm

due to geometrical considerations(from Unified)


MIT 16 20 F ll 2002


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Meaning of subscripts not like stress

mn m = n extension along mm n rotation in m-n plane

BIG DIFFERENCE for strain tensor:

There is a difference in the shear components ofstrain between tensor and engineering (unlikefor stress).

Figure 3.7 Representation of shearing of a 2-D element

angularchange


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--> total angular change = 12 = 12 + 21 = 2 12(recall that 12 and 21 are the same due to

geometrical considerations)But , engineering shear strain is the total

angle: 12 = xy = xy--> Compare notations

= yz sometimes= xz used for= xy shear strains

Thus, factor of 2 will pop upWhen we consider the equations of elasticity, the2 comes out naturally.

(But, remember this physical explanation)

Tensor Engineering11 x22 y

33 z223 = yz213 = xz2

12=

xy


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CAUTION When dealing with shear strains, must know if they aretensorial or engineeringDO NOT ASSUME!

5. Body Forces (3)fi internal forces act along axes

(resolve them in this manner -- can always dothat)


fzf3

fyf2

fxf1

EngineeringTensor


MIT - 16.20 Fall, 2002


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,

6. Elasticity Tensor (? will go over later)

Emnpq relates stress and strain(we will go over in detail, recall introduction in Unified)

Other Notations

Engineering Notation One of two most commonly used Requires writing out all equations (no shorthand) Easier to see all components when written out fully

Contracted Notation Other of two most commonly used Requires less writing

Often used with composites (reduces four subscripts onelasticity term to two) Meaning of subscripts not as physical Requires writing out all equations generally (there is contracted

shorthand)


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1

4

3

2

5

6

--> subscript changesTensor Engineering Contracted

11 x22 y

33 z

23, 32 yz13, 31 xz

12, 21 xy

--> Meaning of 4, 5, 6 in contracted notation Shear component Represents axis (x n) about which

shear rotation takes place via:

m = 3 + nmor xn m

Figure 3.8 Example:Rotation about y 3


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Matrix notation

Super shorthand Easy way to represent system of equations Especially adaptable with indicial notation Very useful in manipulating equations (derivations, etc.)

Example: xi

= Aij

y j

x = A y~ ~ ~ ~ matrix (as underscore)

tildex1 A11 A12 A13 y1 x2 = A21 A22 A23 y2 x3 A31 A32 A33 y3

(will see a little of this mainly in 16.21)

KEY: Must be able to use various notations. Dont rely on

notation, understand concept that is represented.Paul A. Lagace 2001 Unit 3 - p. 17

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Unit 4Equations of ElasticityReadings:R 2.3, 2.6, 2.8T & G 84, 85B, M, P 5.1-5.5, 5.8, 5.9

7.1-7.96.1-6.3, 6.5-6.7

Jones (as background on composites)



Paul A. Lagace 2001

MIT - 16.20 Fall, 2002


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Lets first review a bit from Unified, saw that there are 3 basic considerations in elasticity:

1. Equilibrium2. Strain - Displacement3. Stress - Strain Relations (Constitutive Relations)

Consider each:1. Equilibrium (3)

F i = 0, Mi = 0

Free body diagrams Applying these to an infinitesimal element

yields 3 equilibrium equationsFigure 4.1 Representation of general infinitesimal

element


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11 + 21 + 31 + f1 = 0 (4-1)y1 y2 y312 + 22 + 32 + f2 = 0 (4-2)y1 y2 y313 + 23 + 33 + f3 = 0 (4-3)y1 y2 y3

+ mnm

nyf 0=

2. Strain - Displacement (6)

Based on geometric considerations

Linear considerations ( I.e., small strains only -- we will talk aboutlarge strains later )(and infinitesimal displacements only)


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11 = u1 (4-4)y1

22 = u2 (4-5)y2

33 = u3

(4-6)y3

21 = 12 = 1 u1 + u2 2 y2 y1

31 = 13 = 1 u1 + u3 2 y3 y1

32 = 23 = 1 u

2 + u3

2 y3 y2

(4-7)

(4-8)

(4-9)

1 um + unmn = 2 yn ym


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3. Stress - Strain (6)

mn = Emnpq pqwell come back to this

Lets review the 4th important concept:

Static DeterminanceThere are there possibilities (as noted in U.E.)

a. A structure is not sufficiently restrained(fewer reactions than d.o.f.)

degrees offreedom

DYNAMICSb. Structure is exactly (or simply) restrained

(# of reactions = # of d.o.f.) STATICS (statically

determinate)Implication: can calculate stresses via

equilibrium ( as done in Unified )Paul A. Lagace 2001 Unit 4 - p. 5

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c. Structure is overrestrained(# reactions > # of d.o.f.)

STATICALLYINDETERMINATE

must solve for reactionssimultaneously with stresses, strains, etc.

in this case, you must employ the stress-strain equations

--> Overall, this yields for elasticity:

15 unknowns and 15 equations6 strains = mn 3 equilibrium ( )6 stresses = mn 6 strain-displacements ( )3 displacements = u m 6 stress-strain ( - )

IMPORTANT POINT:The first two sets of equations are universal (independent of thematerial) as they depend on geometry (strain-displacement) andequilibrium (equilibrium). Only the stress-strain equations are

dependent on the material.Paul A. Lagace 2001 Unit 4 - p. 6

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One other point: Are all these equations/unknowns independent? NO

Why? --> Relations between the strains and displacements (due togeometrical considerations result in the Strain Compatibility Equations(as you saw in Unified)

General form is:

2nk + 2m l 2n l

2mk = 0y y l yk y yk y lm yn m yn

This results in 6 strain-compatibility (in 3-D).

What a mess!!!What do these really tell us???

The strains must be compatible, they cannot be prescribed inan arbitrary fashion.

Lets consider an example:

Step 1: consider how shear strain ( 12) is related to displacement:

1 u1 u2

+ 12 = 2 y2 y1


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Note that deformations (u m) must be continuoussingle-valued functions for continuity. (or itdoesnt make physical sense!)

Step 2: Now consider the case where there are gradients in the strainfield

12 0, 12 0y1 y2This is the most general case and most likely in a generalstructure

Take derivatives on both sides:212 1

3u1 3u2

= 2 + 2y y2 2

y y2 y y2

1 1 1

Step 3: rearrange slightly and recall other strain-displacementequations

u1 = 1 ,u2 2= y1 y2


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2 2 2


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212 1

211 + 222

= 2y y2 2 y2 y1

21

So, the gradients in strain are related in certain ways sincethey are all related to the 3 displacements.

Same for other 5 cases

Lets now go back and spend time with the

Stress-Strain Relations and the ElasticityTensor

In Unified, you saw particular examples of this, but we now want togeneralize it to encompass all cases.

The basic relation between force and displacement (recall 8.01) is HookesLaw:

F = kxspring constant (linear case)


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If thi i t d d t th th di i l d li d


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If this is extended to the three-dimensional case and applied overinfinitesimal areas and lengths, we get the relation between stress and

strain known as:Generalized Hookes law:

mn = Emnpq pq

where E mnpq is the elasticity tensor

How many components does this appear to have?m, n, p, q = 1, 2, 3

3 x 3 x 3 x 3 = 81 componentsBut there are several symmetries:

1. Since mn = nm (energy considerations)

E mnpq = E nmpq(symmetry in switching first two

indices)2. Since pq = qp (geometrical considerations)

E mnpq = E mnqpPaul A. Lagace 2001 Unit 4 - p. 10


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R l i 21 i d d f h l i i


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Results in 21 independent components of the elasticity tensor Along diagonal (6) Upper right half of matrix (15)

[dont worry about 2s]

Also note : 2s come out automaticallydont put them in ~For example: 12 = E 1212 12 + E 1221 21

= 2E 1212 12

These E mnpq can be placed into 3 groups: Extensional strains to extensional stresses

E1111

E1122

E 2222 E1133E 3333 E 2233

e.g., 11 = E 1122 22 Shear strains to shear stresses

E1212 E1213E1313 E1323E 2323 E 2312


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e g = 2E


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e.g., 12 2E 1223 23 Coupling term: extensional strains to shear stress or

shear strains to extensional stressesE1112 E2212 E3312E1113 E2213 E 3313E

1123E

2223E

3323e.g., 12 = E 1211 11

11 = 2E 1123 23

A material which behaves in this

manner is fully anisotropicHowever, there are currently no useful engineering materials whichhave 21 different and independent components of E mnpq

The type of material (with regard to elastic behavior) dictates the numberof independent components of E mnpq :


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2Isotropic

3Cubic

5Transversely Isotropic*

6Tetragonal

9Orthotropic

13Monoclinic

21Anisotropic

# of IndependentComponents of E mnpq

Material Type

UsefulEngineering

Materials

CompositeLaminates

Basic

CompositePly

Metals(on average)

Good Reference: BMP, Ch. 7*not in BMP

For orthotropic materials (which is as complicated as we usually get),there are no coupling terms in the principal axes of the material


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When you apply an extensional stress no shear strains arise


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When you apply an extensional stress, no shear strains arisee.g., E 1112 = 0

(total of 9 terms are now zero) When you apply a shear stress, no extensional strains arise

(some terms become zero as forprevious condition)

Shear strains (stresses) in one plane do not cause shear strains(stresses) in another plane

( E 1223 , E 1213 , E 1323 = 0)

With these additional terms zero, we end up with 9 independentcomponents:

(21 - 9 - 3 = 9)

and the equations are:


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11 E1111 E1122 E1133 0 0 0 11


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13

11 E1111 E1122 E1133 0 0 0 11 E1122 E 2222 E 2233 0 0 0

22

22

33 E1133 E 2233 E 3333 0 0 0 33 = 23

0 0 0 2E 2323 0 0 23 13 0 0 0 0 2E 1313 0 13

12

0 0 0 0 0 2E 1212 12 For other cases, no more terms become zero, but the terms are notIndependent.

For example, for isotropic materials: E1111 = E 2222 = E 3333 E1122 = E 1133 = E 2233 E2323 = E 1313 = E 1212 And there is one other equation relating E 1111 , E 1122 and E 2323

2 independent components of E mnpq(well see this more when we do engineering constants)


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Why then do we bother with anisotropy?


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Why, then, do we bother with anisotropy?Two reasons:

1. Someday, we may have useful fully anisotropic materials(certain crystals now behave that way) Also, 40-50 years ago,people only worried about isotropy

2. It may not always be convenient to describe a structure (i.e.,write the governing equations) along the principal material axes.

How else?Loading axes

ExamplesFigure 4-2

wing

rocket case fuselage


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In these other axis systems, the material may have more elastic


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In these other axis systems, the material may have more elasticcomponents. But it really doesnt.

(you cant create elastic components just by describing a material ina different axis system, the inherent properties of the material stay thesame).

Figure 4-3 Example: Unidirectional composite (transversely isotropic)

No shear / extension coupling Shears with regard to loadingaxis but still no inherentshear/extension coupling

In order to describe full behavior, need to doTRANSFORMATIONS

(well review this/expand on it later)


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--> It is often useful to consider the relationship between


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pstress and strain (opposite way). For this we use

COMPLIANCE

mn = S mnpq pqwhere: S mnpq = compliance tensor



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Meaning of each:


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Elasticity term E mnpq : amount of stress ( mn ) related to thedeformation/strain ( pq )

Compliance term S mnpq : amount of strain ( mn ) the stress ( pq )causes

These are useful in defining/ determining the engineering constants

All of this presentation on elasticity (and what you had inUnified ) is based on assumptions which limit their

applicability:

which we will review / introduce / expand on in the next lecture.

CAUTION

Small strain Small displacement / infinitesimal (linear) strain

Fortunately, most engineering structures are such that these assumptionscause negligible error.


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However, there are cases where this is not true:


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Manufacturing (important to be able to convince)

Compliant materials Structural examples: dirigibles,

So lets explore:

Large strain and the formal definition of strainWhat we defined before are the physical manifestation of strain /

deformation Relative elongation Angular rotation

Strain is formally defined by considering the diagonal length of a cube:Figure 4-4 undeformed x3

(small letters)

x2

Paul A. Lagace 2001 x1 Unit 4 - p. 22

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and looking at the change in length under general (and possibly large)d f


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deformation:

Figure 4-5 deformed(capital letters) x3

x2

x1

The formal definition of the strain tensor is:

2 22 mn dxm dxn = (dS ) - (ds )

2 11 dx1 dx1 + 2 22 dx 2 dx 2 + 2d 33 dx 3 dx 3+ 2( 12 + 21) dx1 dx 2 + 2( 13 + 31) dx1 dx 3

2 2+ 2( 23 + 32 ) dx 2 dx 3 = (dS ) (ds )


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where mn = formal strain tensor.


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This is a definition . The physical interpretation is related tothis but not directly in the general case.

One can show (see BMP 5.1 - 5.4) that the formal strain tensor is relatedto relative elongation (the familiar l ) via:

l

relative elongation in m-direction:

Em = m1 m+ 2 1 (no summation on m)

and is related to angular change via:2 mnsin = mn (1 + Em ) (1 + En )

Thus, it also involves the relative elongations!

Most structural cases deal with relatively small strain. If the relativeelongation is small (


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Em = 1 + 2 mm 1

2 (Em + 1) = 1 + 2 mm

E 2 + 2E m = 2 mmm

but if E m


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2 mnsin

= mn (1 + Em ) (1 + En )

for small elongations (E m


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consider

Large Displacement and Non-Infinitesimal(Non-linear) Strain

See BMP 5.8 and 5.9The general strain-displacement relation is:

1 u u u u mn = m + n + r s rs2

xn

xm

xm

xn

Where:rs = Kronecker delta

The latter terms are important for larger displacements but are higher

order for small displacements and can then be ignored to arrive backat:

1 um + unmn = 2 yn ym


MIT - 16.20 Fall, 2002

How to assess?L k


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Look at

um ur u svs. rsxn xm xnand compare magnitudes

Small vs. large and linear vs. nonlinear will depend on: material(s) structural configuration mode of behavior the loading

Examples Rubber in inflated structures

Large strain (Note : generally means larger displacement) Diving board of plastic or wood

Small strain but possibly large displacement (will look at thismore when we deal with beams)


MIT - 16.20 Fall, 2002

Floor beam of steelS ll t i d li t i (N t li t i t l b


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Small strain and linear strain (Note: linear strain must also be small)

Nextback to constitutive constantsnow their physicalreality


MIT - 16.20 Fall, 2002


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Unit 5Engineering Constants

Readings:Rivello 3.1 - 3.5, 3.9, 3.11



Paul A. Lagace 2001

MIT - 16.20 Fall, 2002

We do not characterize materials by their E mnpq . The E mnpq areuseful in doing transformations manipulations etc


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useful in doing transformations, manipulations, etc.

We characterize materials by theirENGINEERING CONSTANTS

(or , Elastic Constants)

(what we can physically measure)

There are 5 types

1. Longitudinal (Youngs) (Extensional) Modulus: relates

extensional strain in the direction of loading to stress in thedirection of loading.

(3 of these)

2. Poissons Ratio: relates extensional strain in the loadingdirection to extensional strain in another direction.

(6 of theseonly 3 are independent)


MIT - 16.20 Fall, 2002

3. Shear Modulus: relates shear strain in the plane of shearloading to that shear stress.


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g

(3 of these)

4. Coefficient of Mutual Influence: relates shear strain due to shearstress in that plane to extensional strain or , relates extensional

strain due to extensional stress to shear strain.(up to 18 of these)

5. Chentsov Coefficient: relates shear strain due to shear stress in

that plane to shear strain in another plane.(6 of these)

Lets be more specific:

1. Longitudinal Modulus1) E 11 or E xx or E 1 or E x: contribution of 11 to 112) E 22 or E yy or E 2 or E y: contribution of 22 to 223) E 33 or E zz or E 3 or E z: contribution of 33 to 33


MIT - 16.20 Fall, 2002

In general: Emm = mm due to mm applied onlymm


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mm

(no summation on m)

2. Poissons Ratios ( negative ratios)1) 12 or xy: (negative of) ratio of 22 to 11 due to 11

2) 13 or xz: (negative of) ratio of 33 to 11 due to 113) 23 or yz: (negative of) ratio of 33 to 22 due to 224) 21 or yx: (negative of) ratio of 11 to 22 due to 225) 31 or zx: (negative of) ratio of 11 to 33 due to 336) 32 or zy: (negative of) ratio of 22 to 33 due to 33

In general: nm = mm due to nn applied onlynn

(for n m)

Important: nm mn


MIT - 16.20 Fall, 2002

However, these are not all independent. There are relationsknown as reciprocity relations (3 of them)


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p y ( )

21 E11 = 12 E 22

31 E11 = 13 E 33

32 E 22 = 23 E33

3. Shear Moduli1) G 12 or G xy or G 6: contribution of (2) 12 to 122) G 13 or G xz or G 5: contribution of (2) 13 to 133) G 23 or G yz or G 4: contribution of (2) 23 to 23

In general: Gmn = mn due to mn applied only 2mn

factor of 2 here since it relates physical quantitiesshear stress mn

Gmn = shear deformation (angular charge) mn


MIT - 16.20 Fall, 2002

4. Coefficients of Mutual Influence ( negative ratios)(also known as coupling coefficients)


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Note: need to use contracted notation here:1) 16 : (negative of) ratio of (2) 12 to 11 due to 11

2) 61 : (negative of) ratio of 11 to (2) 12 due to 12

3) 26 (5) 36 (7) 14 (9) 24

4) 62 (6) 63 (8) 41 (10) 42

11) 34 (13) 15 (15) 25 (17) 35

12) 43 (14) 51 (16) 52 (18) 53

5. Chentsov Coefficients ( negative ratios)1) 46 : (negative of) ratio of (2) 12 to (2) 23 due to 232) 64 : (negative of) ratio of (2) 23 to (2) 12 due to 12

3) 45 : (negative of) ratio of (2) 13 to (2) 23 due to 234) 54 : (negative of) ratio of (2) 23 to (2) 13 due to 135) 56 : (negative of) ratio of (2) 12 to (2) 13 due to 136) 65 : (negative of) ratio of (2) 13 to (2) 12 due to 12



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MIT - 16.20 Fall, 2002

in general:

E G


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nm Em = mn Gn(m = 1, 2, 3) no sum

(n = 4, 5, 6)

and

46 G6 = 64 G445 G5 = 54 G456 G6 = 65 G5

in general:

nm Gm = mn Gn(m = 4, 5, 6) no sum

(m n)


MIT - 16.20 Fall, 2002

This gives 21 independent (at most)engineering constants:


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Total 3 E n 6 nm 3 G m 18 nm 6 nm

Indpt.: 3 3 3 9 3 = 21

--> Now that we have defined the terms, we wish to write theengineering stress-strain equations

Recall compliances:

mn = S mnpq pq

and consider only the first equation:

11 = S 1111 11+ 2S 1123

1 = S 11 1 +

Paul A. Lagace 2001

+ S 1122 22 + S 1133 33

23 + 2S 1113 13 + 2S 1112 12(well have to use contracted notation, so)

S 12 2 + S 13 3 + S 14 4 + S 15 5 + S 16 6(Note: 2s disappear!)

Unit 5 - p. 9

MIT - 16.20 Fall, 2002

Consider each of the compliance terms separately:

Case 1: Only 11 applied


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1 = S 11 1and we know

E1 = 1 due to 1 only11 S 11 = E1

Case 2: Only 22 applied1 = S 12 2

We need two steps here.The direct relation to 2 is from 2:

E 2 = 2 due to 2 only2

and we know21 =

1 due to 2 only2

2 = E 2 due to 2 only

1 21Paul A. Lagace 2001 Unit 5 - p. 10

MIT - 16.20 Fall, 2002

Thus:S 12 =

21 (due to 2 only)E 2


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E 2--> to make this a bit simpler (for later purposes), we recall the reciprocity

relation:12 E 2 = 21 E1

21

12 = E2 E1

Thus:

S 12 = 12E1

Case 3: Only 3 appliedIn a similar manner we get:

S 13 = 13

E1Case 4: Only 4 ( 23) applied

1 = S 14 4Again, two steps are needed.



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MIT - 16.20 Fall, 2002

get:S 15 =

15E1


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E1and

S 16 = 16E1

With all this we finally get:1 14 4 5 6 ]1 = E1

[1 12 2 13 3 15 16

we used the reciprocity relations so

we could pull out this commonfactor.

We can do this for all the other cases.In general we write:

1 6n m = 1

Note: nn = -1

If we let s be sPaul A. Lagace 2001 Unit 5 - p. 13

n = E nm m

MIT - 16.20 Fall, 2002

Engineering Stress-Strain Equations


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General Form11 = E1

[1 12 2 13 3 15 1614 4 5 6 ]

1

2=

E 2 [ 21 1+

2 23 3 4 5 6 ]

24 25 26

13 = E 3[31 1 2 + 3 34 4 35 5 36 6 ] 32

1 4 = 4 = G4 [41 1 42 2 43 3 + 4 45 5 46 6 ]

1 5 = 5 = G5[51 1 52 2 53 3 54 4 + 5 56 6 ]

1 6 = 6 = G6 [ 1 62 2 63 3 64 4 65 5 + 6 ] 61


MIT - 16.20 Fall, 2002

In general:


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1 6n = E nm mn m = 1Note: nn = -1 s --> s

We have developed these for a fully anisotropic material. Again,there are currently no useful engineering materials of this nature.Thus, these would need to be reduced accordingly.


MIT - 16.20 Fall, 2002

Orthotropic Case:In material principal axes, there is no coupling between extension and


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shear and no coupling between planes of shear, so:all mn = 0

Thus, only the following constants remain:E

1

12,

21G

12E2 13 , 31 G13E3 23 , 32 G23

3 + 3 + 3 = 9(same as E mnpq, better be! )


MIT - 16.20 Fall, 2002

So the six equations become:

11 = [1 12 2 13 3 ]


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1E1

[ 1 12 2 13 3 ]

12 = E2[ 21 1 + 2 23 3 ]

13 = E 3[31 1 32 2 + 3 ]

14 = 4G23

15 = 5G13

16 = 6G12


MIT - 16.20 Fall, 2002

matrix form:

1 1 12 13 0 0 0 1


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0

1 E1 E1 E1 2

21E1

2 23 0 0 0 2

E 2 E 2 3 31 32 1 0 0 0

3

= E 3 E 3 E 3 4 0 0 0 1 0 0 4 G23 1 5 0 0 0 0

G130 5

1 0 0 0 0 0 G12 66

= S ~ ~ ~this is, in fact, the compliance matrix


MIT - 16.20 Fall, 2002

Thus: If we know the engineering constants (through tests -- this is

upcoming)


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Relate engineering constants to S mnpq

Get Emnpq by inversion of S mnpq matrix (combine steps to directlyget relationships between E mnpq and the engineering constants)

Isotropic CaseAs we noted in the last unit, as we get to materials with less elasticconstants (< 9) than an orthotropic material, we no longer have any morezero terms in the elasticity or compliance matrix, but more nonzero terms

are related.For the isotropic case:

All extensional moduli are the same:E1 = E 2 = E 3 = E

All Poissons ratios are the same: 12 = 21 = 13 = 31 = 23 = 32 =

All shear moduli are the same:G

4 = G

5 = G

6 = G


MIT - 16.20 Fall, 2002

And, there is a relationship between E, and G (from Unified):E

G = +


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2 1 )(Thus, there are only 2 independentconstants.

We now have all the relationships to do the manipulations,but we need to measure the basic properties. We musttherefore talk about

TestingTesting is used for a variety of purposes.

Depending on the purpose, the technique and care will vary. The

fidelity needed in the testing depends on the use.Basically, apply a load (stress) condition and measure appropriateresponses:


MIT - 16.20 Fall, 2002

Strain

Displacements


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Failure(or maybe vice versa)

Concerns in test specimens Boundary conditions and introduction of load

Stress concentrations

Achievement of desired stress state

Cost and ease of use also important (again, depends onuse of test)

Many of these concerns will depend on the material / configuration andload condition. This generally involves issues of scale.

Properties of a material / structure depends on the scale at which youlook at it.

Scale level of homogenization (average behavior over a certainsize)


MIT - 16.20 Fall, 2002

Example 1: Atoms make up a materialFigure 5.1 Representation of atomic bonding as springs


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The behavior of the materials is some combination of theatoms and their bonds.

Example 2: CompositeFigure 5.2 Representation of unidirectional composite

fibers in a matrix

Fibers and matrix respond differently. Average theirresponse to get composite properties



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MIT - 16.20 Fall, 2002

Figure 5.4 Tension specimens


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Tapered Dogbone Straight-edged couponBar Specimen (composites) stressconcentration problem

2. CompressionSimilar specimens can be used, but must beware of buckling(global and local instabilities)

Possibility of local reinforcement to prevent buckling.


MIT - 16.20 Fall, 2002

3. ShearVery hard to apply pure shear.

Figure 5 5 P ibl h i


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Figure 5.5 Possible shear specimens

Tube

Iosipescu

(beam theory shows area ofpure shear -- test section)

Refer to: ASTM (American Society for Testingand Materials) Annual Book of Standards

Voluntary test standards are contained there.Paul A. Lagace 2001 Unit 5 - p. 26

MIT - 16.20 Fall, 2002

what to do with test dataTest data must be reduced to give the engineeringconstants.


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Figure 5.6 Typical stress-strain data (for ductile material)

The engineering constants are defined (somewhat

arbitrary) as various parts of this curve. Generally withinthe initial linear region.

Often use linear regression



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MIT - 16.20 Fall, 2002

We have now developed the general 3-D stress-strain relations. But weoften deal with a problem where we can simplify (model as) to a 2-Dsystem. Two important cases to next consider:


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Plane Stress

Plane Strain


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Unit 6 Plane Stress and Plane Strain

Readings:T & G 8, 9, 10, 11, 12, 14, 15, 16



Paul A. Lagace 2001

MIT - 16.20 Fall, 2002

There are many structural configurations where we do nothave to deal with the full 3-D case.

First lets consider the models


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First let s consider the models Lets then see under what conditions we can

apply them

A. Plane Stress

This deals with stretching and shearing of thin slabs.Figure 6.1

Representation of Generic Thin Slab


MIT - 16.20 Fall, 2002

The body has dimensions such thath


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consider later)

Thus, the plate is thin enough such that there is no variation ofdisplacement (and temperature) with respect to y 3 (z).

Furthermore, stresses in the z-direction are zero (small order ofmagnitude).

Figure 6.2 Representation of Cross-Section of Thin Slab


MIT - 16.20 Fall, 2002

Thus, we assume: zz = 0

yz = 0


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0 xz = 0

= 0z

So the equations of elasticity reduce to:

Equilibrium

11 + 21 + f1

= 0 (1)y1 y2

12 + 22 + f2 = 0 (2)y1 y2

(3rd equation is an identity) 0 = 0(f3 = 0)In general: + f = 0y


MIT - 16.20 Fall, 2002

Stress-Strain (fully anisotropic)Primary (in-plane) strains

11 = [ 1 12 2 16 6 ] (3)


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1 E 1[ 1 12 2 16 6 ] (3)

1 2 = E

2

[ 21 1 + 2 26 6 ] (4)

16 = G6

[61 1 62 2 + 6 ] (5)

Invert to get:*

= E

Secondary (out-of-plane) strains

(they exist, but they are not a primary part of the problem)13 = E 3

[ 311 32 2 36 6 ]


MIT - 16.20 Fall, 2002

14 = G4[ 41 1 42 2 46 6 ]

1 = [ ]


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5 = G5[51 1 52 2 56 6 ]

Note: can reduce these for orthotropic, isotropic(etc.) as before.

Strain - DisplacementPrimary

11 = u1 (6)y1

22 = u2 (7)y2

12 = 1

u1

+ u2

(8)2 y2 y1



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MIT - 16.20 Fall, 2002

This further implies from above(since = 0 )

y3No in-plane variation


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pu3 = 0y

but this is not exactly true

INCONSISTENCY

Why? This is an idealized model and thus an approximation. Thereare, in actuality, triaxial ( zz , etc.) stresses that we ignore here as

being small relative to the in-plane stresses!(we will return to try to define small)

Final note: for an orthotropic material, write the tensorialstress-strain equation as:

2-D plane stress = ( , , , , = 1 2)

E



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MIT - 16.20 Fall, 2002

Figure 6.3 Representation of Long Prismatic Body


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Dimension in z - direction is much, much larger than inthe x and y directions

L >> x, y


MIT - 16.20 Fall, 2002

(Key again: where are limits to >>??? wellconsider later)

Since the body is basically infinite along z, the important loads are in thex - y plane (none in z) and do not change with z:


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x y plane (none in z) and do not change with z:

= = 0

y3 z

This implies there is no gradient in displacement along z, so (excludingrigid body movement):

u3 = w = 0

Equations of elasticity become:

Equilibrium:Primary

11 + 21 + f1 = 0 (1)y1 y2

12 + 22 + f2 = 0 (2)y1 y2Paul A. Lagace 2001 Unit 6 - p. 11

MIT - 16.20 Fall, 2002

Secondary13 + 23 + f3 = 0y

1y

2 13 and 23 exist but do not enter into primary


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13 23 p yconsideration

Strain - Displacement

11 = u1 (3)y1

22 = u2 (4)y2

12 = 1 u1 + u2

(5)

2 y2 y1

Assumptions

= 0, w = 0

give:y3

13 = 23 = 33 = 0(Plane strain)


MIT - 16.20 Fall, 2002

Stress - Strain(Do a similar procedure as in plane stress)

3 Primary11 = (6)


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11 ... (6) 22 = ... (7)12 = ... (8)

Secondary13 = 0 23 = 0

orthotropic ( 0 for anisotropic)

33 0INCONSISTENCY: No load along z,

yet 33 ( zz) is non zero.

Why? Once again, this is an idealization. Triaxial strains ( 33 )actually arise.

You eliminate 33 from the equation set by expressing it in terms of via ( 33 ) stress-strain equation.


MIT - 16.20 Fall, 2002

SUMMARY Plane Stress Plane Strain

length (y 3) >> in-planedimensions (y 1 y2)

thickness (y 3)


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Eliminate 33

from eq.Set by using 33 - eq. and expressing 33in terms of

Eliminate 33

from eq. setby using 33 = 0 - eq.and expressing 33 interms of

Note:

3333, u 3SecondaryVariable(s):

, , u , , u PrimaryVariables:

i3 = 0 i3 = 0ResultingAssumptions:

only

/ y3 = 0

33


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Figure 6.4 Pressure vessel (fuselage, space habitat) Skin

in order of 70 MPa(10 ksi)

p o 70 kPa (~ 10 psi for living environment) zz


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MIT - 16.20 Fall, 2002

Problem: get expressions for (whatever) in one axis system in terms of(whatever) in another axis system

(Review from Unified)

Recall: nothing is inherentlychanging, we just describe a body


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from a different reference.

Use ~ (tilde) to indicate transformed axis system.Figure 7.2 General rotation of 3-D rectangular axis system

(still rectangular cartesian)


MIT - 16.20 Fall, 2002

Define this transformation via direction cosines

~~l mn = cosine of angle from y m to y n

Notes: by convention, angle is measured positivecounterclockwise (+ CCW)


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counterclockwise (+ CCW)(not needed for cosine)

~ ~ since cos is an even functionl mn = l nmcos ( ) = cos (- )(reverse direction)

~ ~But

lmn

lmn

angle differs by 2 !

The order of a tensor governs the transformation needed. An n th ordertensor requires an n th order transformation (can prove by showing link of

order of tensor to axis system via governing equations).



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MIT - 16.20 Fall, 2002

Not only do we sometimes want to change the orientation ofthe axes we use to describe a body, but we find it more

convenient to describe a body in a coordinate system otherthan rectangular cartesian. Thus, consider


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Other Coordinate Systems

The easiest case isCylindrical (or Polar in 2-D) coordinates

Figure 7.3 Loaded disk

Figure 7.4 Stress around a hole


MIT - 16.20 Fall, 2002

Figure 7.5 Shaft


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Define the point p by a different set of coordinates other than y 1, y2, y 3

Figure 7.6 Polar coordinate representation

Volume = rd drdz


MIT - 16.20 Fall, 2002

Use , r, z where:y1 = r cos

y2 = rsin

y3 = zare the mapping functions


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are the mapping functionsNow the way we describe stresses, etc. change

--> Differential element is now differentRectangular cartesian

Figure 7.7 Differential element in rectangular cartesian system

Volume = dy 1 dy 2 dy 3


MIT - 16.20 Fall, 2002

CylindricalFigure 7.8 Differential element in cylindrical system


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Volume = rd drdz

Generally:dy1 --> drdy2 --> rd dy3 --> dz (get from mapping functions)



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MIT - 16.20 Fall, 2002

(engineering shear strains)

r = u + 1 ur u r r r

z = 1 u3 + u r z


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r z

zr

= ur + u3z r

Stress - Strain Equations become, however, more complicated and noteasy to map into another coordinate systems.

Why? Unless the material is isotropic, the properties change withdirection (if the material principal axes are rectangular orthogonal).

Emnpq = l l qu E rstumr ns l pt l So, for cylindrical coordinates:

Emnpq ( )

a function of

Paul A. Lagace 2001Unit 7 - p. 14


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MIT - 16.20 Fall, 2002

For cylindrical case: = r =

= zactual mapping functions:


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actual mapping functions:

F y 11 ( , y 2 , y 3 ) = y22y1

2+

2 ( ,F y y 2 , y 3 ) = tan-1 (y2 / y1)1

F y y 2 , y 3 ) = y33 ( ,1Other cases

Lets next consider some general solution approaches

Paul A. Lagace 2001Unit 7 - p. 17

MIT - 16.20 Fall, 2002

Unit 8Solution Procedures


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Readings:R Ch. 4T & G 17, Ch. 3 (18-26)

Ch. 4 (27-46)Ch. 6 (54-73)



Paul A. Lagace 2001


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MIT - 16.20 Fall, 2002

Lets consider exact techniques. A common, and classic, one is:

Stress Functions

Relate six stresses to (fewer) functions defined in such a mannerh h id i ll i f h ilib i di


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that they identically satisfy the equilibrium conditon Can be done for 3-D case Can be done for anisotropic (most often orthotropic) case

--> See: Lekhnitskii, Anisotropic Plates,Gordan & Breach, 1968.

--> Lets consider plane stress (eventually) isotropic

8 equations in 8 unknowns

- 2 equilibrium - 3 strains- 3 strain-displacement - 2 displacements- 3 stress-strain - 3 stresses

Paul A. Lagace 2001

Unit 8 - p. 3


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MIT - 16.20 Fall, 2002

Step 2: Use these in the plane stress compatibility equation:

2xx 2yy

2xy+ =

x yy2

x2

(E6)

we get quite a mess! After some rearranging andmanipulation, this results in:


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p ,

V V

+

+

=

( )

+ ( )

( )

+

4

4

4

2

2

4

2

2

2

2

2

2

2

21

x y y

ET

xT

y y

2

2

x

x

(*)

temperature term wehavent yet considered

= coefficient of thermal expansionT = temperature differential

This is the basic equation for isotropic planestress in Stress Function form

Recall : is a scalarPaul A. Lagace 2001

Unit 8 - p. 8


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MIT - 16.20 Fall, 2002

this function, and accompanying governing equation, couldbe defined in any curvilinear system (well see one suchexample later) and in plane strain as well.

Butwhats this all useful for???This may all seem like magic. Why were the s assumed as theywere?


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were?

This is not a direct solution to a posed problem, per se , but is knownas

The Inverse Method

In general, for cases of plane stress without body force or temp ( 4 = 0):1. A stress function (x, y) is assumed that satisfies the

biharmonic equation2. The stresses are determined from the stress function as

defined in equations (8-1) - (8-3)3. Satisfy the boundary conditions (of applied tractions)4. Find the ( structural ) problem that this satisfies

Paul A. Lagace 2001

Unit 8 - p. 10


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MIT - 16.20 Fall, 2002

(Rule of Thumb) Generally, the area where specificsare important extends into the body for a distance equalto about the greatest linear dimension of the portion of

the surface on which the loading / B.C. occurs. This allows us to get solutions for most parts of a

structure via such a method.


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But failure often originates/occurs in a region of load

introduction/boundary condition(example: where do nailed/screwed boards

break?)

Examples (for stress functions)

Example 1 (assume T = 0, V = 0 [no body forces])Pick = C 1 y2

this satisfies 4 = 0C1 is a constantwill be determined by satisfying the

B.C.sPaul A. Lagace 2001

Unit 8 - p. 13

MIT - 16.20 Fall, 2002

Using the definitions:

2 xx = 2 = 2C 1y

2 = = 0yy

x2


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= 0xy

gives the state of stress

What problem does this solve???

Uniaxial loadingFigure 8.1 Representation of Uniaxial Plane Stress Loading

applied stress (uniformstress on end)

Paul A. Lagace 2001

Unit 8 - p. 14

MIT - 16.20 Fall, 2002

Check the B.C.s:@ x = l , 0

xx =

o =

2C 1

C1 = o

2 = 0xy


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@ y = w2

= 0yy = 0xy

Thus: =

o y22

Paul A. Lagace 2001

Unit 8 - p. 15


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MIT - 16.20 Fall, 2002

Figure 8.2 Representation of uniaxial test specimen and resultantstress state

x

y


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These grips allow no displacement in the y-direction

--> v = 0 both @ x = 0, lThe solution gives:

v = y 0 in general @ x = 0, l

E

But , far from the effects of load introduction, the solution holds.Near the grips, biaxial stresses arise. Often failure occurs here.(Note : not in a pure uniaxial field. This is a common problem with testspecimens.)


MIT - 16.20 Fall, 2002

Example 2

Lets now get a bit more involved and consider the

Stress Distribution Around a HoleFigure 8.3 Configuration of uniaxially loaded plate with a hole


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Large (infinite) plate subjected to uniform far - field tension

Since the local specific of interest is a circle, it makes sense to usepolar coordinates.

By using the transformations to polar coordinates of Unit #7, we find:


MIT - 16.20 Fall, 2002

1 1 2 rr = r r+

r2 2+ V

2

+ V = r2

1 2 1 + r = 2


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r r r r2

These are, again, defined such that equilibrium equations areautomatically satisfied.

where: fr = V

f = 1 V

r r

f + f 1 frand V exists if: =r r r

The governing equation is again:

4 = E 2 (T) (1 )2 V for plane stress, isotropic



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MIT - 16.20 Fall, 2002

Step 2: Determine stressesPerforming the derivatives from the - relations in polar coordinatesresults in:

rr =B

20 + 2C 0 + 2A 2

6B4

2 4D 2 cos2 r r r 2

B0 2C 2A6B 2 12C 2 2


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= 20 + 2C 0 + 2A 2 + 4

2 + 12C 2r2

cos2

r r

2 in r = 2A 2 6B

42 + 6C 2r

2 2D 2 s 2 r rNote that we have a term involving r 2. As r gets larger, the stresseswould become infinite. This is not possible. Thus, the coefficient C 2must be zero:

C2 = 0

So we have five constants remaining:A2, B 0, B2, C 0, D 2

we find these by



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MIT - 16.20 Fall, 2002

Figure 8.5 Stress condition at boundary of hole


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So we (summarizing) appear to have 4 B. C.s: rr = o + o cos2 @ r = 2 2

r = o sin2 @ r = 2

rr = 0 @ r = a

r = 0 @ r = a



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MIT - 16.20 Fall, 2002

Figure 8.6 Polar coordinate configuration for uniaxially loaded platewith center hole


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So we have the solution to find the stress field around a hole. Letsconsider one important point. Wheres the largest stress?

At the edge of the hole. Think of flow around the hole:

Figure 8.7 Representation of stress flow around a hole


MIT - 16.20 Fall, 2002

@ = 90, r = a

a 2 a 4o = xx = 2

o 1 + a

2

2

1 + 3a

4 (1)

= 3 o

Define the:


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Stress Concentration Factor (SCF) = local stressfar - field stress

SCF = 3 at hole in isotropic plate

The SCF is a more general concept. Generally the sharper thediscontinuity, the higher the SCF.

The SCF will also depend on the material. For orthotropic materials, itdepends on E x and E y getting higher as E x /E y increases. In a uni-directionalcomposite, can have SCF = 7.

Can do stress functions for orthotropic materials, but need to go tocomplex variable mapping

--> (See Lekhnitskii as noted earlier)Paul A. Lagace 2001 Unit 8 - p. 28


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If these thermal expansions / contractions are resisted by some means,then thermal stresses can arise. However , thermal stresses is amisnomer, they are really

stresses due to thermal effects -- stresses are alwaysmechanical(well see this via an example)

--> Consider a 3-D generic material.


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Then we can write:T ij = ij T (9 1)

i, j = 1, 2, 3 (as before)

ij = 2nd order tensorThe total strain of a material is the sum of the mechanical strain and thethermal strain .

mechanical

thermalM T ij = ij + ij (9 2)

totalPaul A. Lagace 2001 Unit 9 - p. 3

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(actual) total strain ( ij): that which you actually measure; thephysical deformation of the part

T thermal strain ( ij ): directly caused by temperature differencesM

mechanical strain ( ij ): that part of the strain which is directlyrelated to the stress

Relation of mechanical strain to stress is:M = S


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ij

Sijkl

kl

compliance

Substituting this in the expression for total strain (equation 9-2) and usingthe expression for thermal strain (equation 9-1), we get:

ij = S ijkl kl + ij T

S ijkl kl = ij ij TWe can multiply both sides by the inverse of the compliancethat is

merely the elasticities:1

Sijkl

= E ijkl



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Stresses will arise due to the mechanical strain and these are theso-called thermal stresses.

Due to equilibrium there must be a reaction at the boundaries.

(must always have dA = Force for equilibrium)Think of this as a two-step process

Figure 9.3 Representation of stresses due to thermal expansion astwo-step process


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p p

expands due to T, T

Reactionforce of boundaries related to mechanical strain, M

M = -T

total = 0 (no physical deformation)


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Values of C.T.E.s

Note: ij = ji

Anisotropic Materials6 possibilities: 11 , 22 , 33 , 12 , 13 , 23

T can cause shear strains

not true in engineering materials


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Orthotropic Materials3 possibilities: 11 , 22 , 33

T only causes extensional strainsNotes: 1. Generally we deal with planar structures and

are interested only in 11 and 222. If we deal with the material in other than the

principal material axes, we can have an 12

Transformation obeys same law as strain (its a tensor).


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2-D form:

= l l

11 , 22 (in- plane values) 12 = 0 (in material axes)

3-D form:

ij = l l kl


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j j li k

So, in describing deformation in some axis system at an angle tothe principal material axes..

Figure 9.4 Representation of 2-D axis transformation

~y2

y1~

y2

+ CCW

y1Paul A. Lagace 2001 Unit 9 - p. 9

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11 = cos2 11 + sin

2 22 22 = sin

2 11 + cos2 22

12 = cos sin ( 22 11) only exists if 11 22

[isotropic no shear]


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Isotropic Materials1 value: is the same in all directions

Typical Values for Materials:

Units: x 10 -6 /Fin/in/Fstrain/F

strain/F


Material C.T.E.

Uni Gr/Ep (perpendicular to fibers)

Uni Gr/Ep (along fibers)

TitaniumAluminum

Steel

16

-0.2

512.5

6


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= (T) C.T.E. is a function of temperature (see MIL HDBK 5for metals). Can be large difference.

Implication: a zero C.T.E. structure may not truly be

attainable since it may be C.T.E. at T 1 but not at T 2 !--> Sources of temperature differential (heating)

ambient environment (engine, polar environment, earthshadow, tropics, etc.)


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aerodynamic heating radiation (black-body)

--> Constant T (with respect to spatial locations)

In many cases, we are interested in a case where T (from some referencetemperature) is constant through-the-thickness, etc. thin structures structures in ambient environment for long periods of time

Relatively easy problem to solve. Use: equations of elasticity equilibrium stress-strain



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Convection most important in aircraft:

Aerodynamic Heatinglook at adiabatic wall temperature

2TAW = 1 + 1 r M T2

where: = specific heat ratio (1.4 for air)r = "recovery factor" (0 8 0 9)


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r = "recovery factor" (0.8 - 0.9)M = Mach numberT = ambient temperature ( K)

TAW is maximum temperature obtained on surface (for zero heat flux)

Note: @40,000 ft. M = 2 TAW = 230FM = 3 TAW = 600F

(much above M = 2, cannot use aluminum

since properties are too degraded)worse in reentry



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q = - Ts 4

surface temperatureheatflux Stefan-Boltzmann constant

emissivity(a material property)

2. Absorptivity


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heatq = I s angle factor

flux intensity of sourceabsorptivity(a material property)

Figure 9.5 Representation of heat flux impinging on structure

angle ofstructurelike the sun

Is (intensity of source)


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Heat conductionThe general equation for heat conduction is:

Tqi

T = kij

T

x jwhere:

T = temperature [K]T wattsq i = heat flux per unit area in i direction

m2


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mT wattsk ij = thermal conductivity m K

(material properties)TThe k ijare second order tensors


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consider:Figure 9.6 Representation of structure exposed to two environments

Environment 1 Environment 2


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look at a strip of width dz:Figure 9.7 Representation of heat flow through infinitesimal strip of

materialTq z

TTq z +

qz dz

z



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Bottom line : use these equations to solve for temperature distribution instructure subject to B.C.s

T (x, y, z) gives T (x, y, z)

Note : These variations can be significant

Example:Figure 9.8 Representation of plate in space


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4q in

sun side black space side

q in

= I s - Ts4 qout = Ts


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could get other cases where T peaks in the center, etc.Result :

Internal stresses (generally) arise if T varies spatially. (unless it isa linear variation which is unlikely given the governing equations).

Why?consider an isotropic plate with T varying only in the y-direction


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Figure 9.9 Representation of isotropic plate with symmetric y-variationof T about x-axis


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(for the time being, limit T to be symmetric withrespect to any of the axes)


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At first it would seem we get a deformation of a typical cross-sectionA-B as:

x

y


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This basic shape would not vary in x.Note, however, that this deformation in the x-direction (u) varies in y.

u 0

y shear strain exists!

But , T only causes extensional strains. Thus, this deformationcannot occur.

(in some sense, we have planesections must remain plane)

Thus, the deformation must be:Paul A. Lagace 2001 Unit 9 - p. 26

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locally

In order to attain this deformation, stresses must arise. Consider

two elements side by sideU d f d D f d


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Undeformed Deformed

T

greater

These two must deform the same longitudinally, so there must bestresses present to compress the top piece and elongate the bottompiece


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Thus:x = x (x)y = y (y)

This physical argument shows we have thermal strains, mechanicalstrains and stresses.

Causes


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self-equilibrating

a a ydx = 0

b b xdy = 0


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Solution Technique

No different than any other elasticity problem. Use equations of elasticity

subject to B. C.s. exact solutions stress functions

recall:

4 = E 2 (T) (1 )2 V


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etc.(see Timoshenko)

--> Does this change for orthotropic materials?NO (stress-strain equations change)

Weve considered Thermal Strains and Stresses , now lets

look at the other effect:


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Degradation of Material Properties (due tothermal effects)

Here there are two major categories

1. Static Properties

Modulus, yield stress, ultimate stress, etc. change withtemperature (generally T property )

(see Rivello)


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temperature (generally, T property ) Fracture behavior (fracture toughness) goes through a

transition at glass transition temperatureductile brittle

Tg


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Figure 9.10 Representation of variation of ultimate stress withtemperature


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Figure 9.11 Representation of change in stress-strain behavior withtemperature

ductile as T increases)(generally, behavior is more

--> Thus, must use properties at appropriate temperature in analysisMIL-HDBK-5 has much data


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2. Time-Dependent Properties

There is a phenomenon (time-dependent) in materials known as creep .This becomes especially important at elevated temperature.

Figure 9.12 Representation of creep behavior

Hang a load P and

monitor strain with time


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resulting strain-timebehavior

This keeps aluminum from being used in supersonic aircraft in criticalareas for aerodynamic heating.


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Other Environmental Effects

Temperature tends to be the dominating concern, but others

may be important in both areas atomic oxygen degrades properties UV degrades properties etc.

Same effects may cause environmental strains like thermalstrains:


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Example - moistureMaterials can absorb moisture. Characterized by aswelling coefficient = ij

Same operator as ij (C. T. E.) except it operates on moistureconcentration, c:

sij = ij c

swelling moisture concentrationstrain swelling

coefficient


MIT - 16.20 Fall, 2002

and then we have:ij = ij

M+ ij

T+ ij

S

totalThis can be generalized such that the strain due to an environmentaleffect is:

environmental

strain Eij = ij X


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environmentalenvironmentaloperator scalar

and the total strain is the sum of the mechanical strain(s) and theenvironmental strains

A strain of this type has become important in recent work.This deals with the field of


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Piezoelectricity

A certain class of materials, known as piezoelectronics, have a coupling

between electric field and strain such that:electric field causes deformation/strainstrain results in electric field

This can be looked at conceptually the same way as environmentalstrains except electric field is a vector (not a scalar). Thus, the basicrelationship is:


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piezoelectric

ijp

= d E kijkwhere:

Ek = electric fieldd ijk = piezoelectric constant

units = [strain/field]a key difference here is that the operator (d ijk) is a third-order tensor

(how transform? 3 direction cosines)


MIT - 16.20 Fall, 2002

And we add this strain to the others to get the total strain

(consider the case with only mechanical and piezoelectric strain)

ij=

ijM

+ ij

p

Again, only the mechanical strain is related directly to the stress:ij = S ijmn mn + d ijk Ek

inverting gives:

ij = E ijmn mn E ijmn dmnk Ek( t h th it hi f i di !)


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(watch the switching of indices!)thus we have piezoelectric-induced stresses of:

E ijmn d mnk E kif the piezoelectric expansion is physically resisted.Again, equilibrium ( = F) must be satisfied.

But , unlike environmental cases, the electric field is not just an externalparameter from some uncoupled equation of state but there is a coupledequation:


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Di = e ik Ek + d inm mnnote switch in indices since this is transpose ofdielectric constant from previous equation

where:e ik = dielectric constantDi = electrical charge

--> Normally, when piezoelectric materials are utilized, E-field control

is assumed. That is, E k is the independent variable and the electricalcharge is allowed to float and take on whatever value results. But,


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g ,when charge constraints are imposed the simultaneous set of equations:

mn = Emnij ij Emnij d ijk Ek

Di = e ik Ek + d inm mnmust be solved. This is coupled with any other sources of strain(mechanical, etc.)


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Piezoelectrics useful for

sensors control of structures (particularly dynamic effects)

Note: electrical folk use a very differentnotation (e.g., S = strain)

Now that weve looked at the general causes of stress andstrain and how to manipulate, etc., consider generalstructures and stress and strain in that context


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structures and stress and strain in that context.



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III Torsion


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III. Torsion


MIT - 16.20 Fall, 200

Structural Mechanics - MIT 2002

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