unit m3 - web.mit.eduweb.mit.edu/.../fall/materials/lectures/m3.2-unified06.pdf · d l such a load...

Paul A. Lagace © 2006

MIT - 16.001/16.002 Fall, 2006

16.001/002 -- “Unified Engineering”Department of Aeronautics and Astronautics

Massachusetts Institute of Technology

Unit M3.2General Stress-Strain Behavior

Readings:A & J 3

CDL 5.1, 5.2, 5.4, 5.10

Unit M3.2 - p. 2Paul A. Lagace © 2006

MIT - 16.001/16.002 Fall, 2006

LEARNING OBJECTIVES FOR UNIT M3.2Through participation in the lectures, recitations, and workassociated with Unit M3.2, it is intended that you will beable to………

• ….explain the meaning of the elasticity and compliancetensors and analyze their mathematical details

• ….describe the behavior of a material in terms ofconstitutive response

• ….discuss engineering/elastic constants, theirmeasurement, and their relationship to tensors

• ….employ a continuum version of the constitutive lawof elasticity


MIT - 16.001/16.002 Fall, 2006

Will look at the model to relate stress and strain and considerhow we manipulate this mathematically and determine theproperties/characterization experimentally.

To orient ourselves, let’s first look at how actual materialsbehave and examine…

Uniaxial Stress-Strain

Consider a…Figure M3.2-1 Bar pulled by a force F

F F

Area = A

L dsuch that there is only uniaxial (s11) stress:

s1 1 = FA


MIT - 16.001/16.002 Fall, 2006

and the relative elongation is:

e1 1 = d

lSuch a load is generally applied via a testing machine to obtain sversus e experimentally (recall truss experiment).

Different types of material exhibit different stress-strain behavior.There are two general categories:Figure M3.2-2a Illustration of brittle behavior (e.g., glass, ceramics)

s

e

x


MIT - 16.001/16.002 Fall, 2006

Figure M3.2-2b Illustration of ductile behavior (e.g., metals)s

e

x

yield unload path

linear elasticregion

plasticregion

--> In elastic region, return path is to origin--> In plastic region, return path is parallel to original linear portion fi permanent strain remains

--> Let’s concentrate on the linear portion and model that behavior.

We look to work done in 1676 proposed by Hooke and consider:


MIT - 16.001/16.002 Fall, 2006

(Generalized) Hooke’s LawHooke said that force and displacement and also stress and strainare linearly related:

s = Ee --Hooke’s Law

(also think of F = kx)

Thus, the slope of the uniaxial stress-strain response in the linearregion is:

(as we’ve seen before)

se

= E Modulus of Elasticityforce / length2[ ]psi[ ] Pa[ ]ˆ M 106( ) ˆ G 109( )

Units:

Note:

force / length2[ ]psi[ ] Pa[ ]ˆ M 106( ) ˆ G 109( )


s = F

A, e = d

l



fi F / A

d / l = E fi d Fl

AE fi F / A

d / l = E fi d Fl

AE=


MIT - 16.001/16.002 Fall, 2006

We need to generalize this concept in order to relate general stress (asecond-order tensor) to general strain (a second-order tensor). We arriveat…..

--> Generalized Hooke’s Law

the elasticity tensorThis is a fourth-order tensor which is needed torelated two second-order tensors

smn = Emnpq e p q

Write out for a sample case (m = 1, n = 1)s1 1 = E1111 e1 1 + E1112 e1 2 + E1113 e1 3

+ E1121 e2 1 + E1122 e2 2 + E1123 e2 3

+ E1131 e3 1 + E1132 e3 2 + E1133 e3 3

(p = 1, sum on q)(p = 2, sum on q)(p = 3, sum on q)(sum on p)

Collect like terms (e.g. emn = enm) to get:s1 1 = E1111 e1 1 + E1122 e2 2 + E1133 e3 3

+ 2E1112 e1 2 + 2E1113 e1 3 + 2E1123 e2 3

(Note: recall 2’s for tensorial strain)


MIT - 16.001/16.002 Fall, 2006

But in order to consider this 3-D stress-strain relation in its entirety, weneed to consider the…

Elasticity and Compliance TensorsEmnpq is the “Elasticity Tensor”

How many components does this appear to have?m, n, p, q = 1, 2, 3fi 3 x 3 x 3 x 3 = 81 components

But there are several symmetries:1. Since smn = snm (equilibrium/energy considerations)

fi Emnpq = Enmpq

2. Since epq = eqp (geometrical considerations)

(symmetry in switching first two indices)

fi Emnpq = Emnqp (symmetry in switching last two indices)

3. From thermodynamic considerations fi Emnpq = Epqmn

(1st law of thermo)(symmetry in switching pairs of indices)


MIT - 16.001/16.002 Fall, 2006

Note that:• since smn = snm , the apparent 9 equations for stress are only 6

With these symmetrics, the resulting full 3-D stress-strain equations are (in matrix form):s11s22s33s23

s13

s12

Ï

Ì

Ô Ô Ô Ô

Ó

Ô Ô Ô Ô

¸

˝

Ô Ô Ô Ô

˛

Ô Ô Ô Ô

=

E1111 E1122 E1133 2E1123 2E1113 2E1112E1122 E2222 E2233 2E2223 2E2213 2E2212E1133 E2233 E3333 2E3323 2E3313 2E3312E1123 E2223 E3323 2E2323 2E1323 2E1223E1113 E2213 E3313 2E1323 2E1313 2E1213E1112 E2212 E3312 2E1223 2E1213 2E1212

È

Î

Í Í Í Í Í Í Í Í

˘

˚

˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙

=

e11e22e33e23e13e12

Ï

Ì

Ô Ô Ô Ô

Ó

Ô Ô Ô Ô

¸

˝

Ô Ô Ô Ô

˛

Ô Ô Ô Ô

s11s22s33s23

s13

s12

Ï

Ì

Ô Ô Ô Ô

Ó

Ô Ô Ô Ô

¸

˝

Ô Ô Ô Ô

˛

Ô Ô Ô Ô

=

E1111 E1122 E1133 2E1123 2E1113 2E1112E1122 E2222 E2233 2E2223 2E2213 2E2212E1133 E2233 E3333 2E3323 2E3313 2E3312E1123 E2223 E3323 2E2323 2E1323 2E1223E1113 E2213 E3313 2E1323 2E1313 2E1213E1112 E2212 E3312 2E1223 2E1213 2E1212

È

Î


˘

˚

˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙

=

e11e22e33e23e13e12

Ï

Ì

Ô Ô Ô Ô

Ó

Ô Ô Ô Ô

¸

˝

Ô Ô Ô Ô

˛

Ô Ô Ô Ô

• 2’s come out automatically since epq = eqp and Emnpq = Emnqp terms like Emnpq epq + Emnqp eqp = 2Emnpq epq

(factor of 2!)

Note: 2’s come naturally here

--> as we saw for case of m = 1, n = 1--> don’t put them in e~


MIT - 16.001/16.002 Fall, 2006

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

[don’t worry about 2’s]The components of the Emnpq can be placed into 3 groups:

• Extensional strains to extensional stresses

e.g., s11 = … E1122 e22 …

• Shear strains to shear stressesE1212 E1213E1313 E1323E2323 E2312

E1111 E1122E2222 E1133E3333 E2233

or:s = E e; smn = Emnpq e p q~ ~ ~

e.g., s12 = … 2E1223 e23 …

s = E e; smn = Emnpq e p q


MIT - 16.001/16.002 Fall, 2006

E1112 E2212 E3312E1113 E2213 E3313E1123 E2223 E3323

• Coupling terms: extensional strains to shear stresses or shear strains to extensional stresses

e.g., s12 = …E1211 e11…s11 = …2E1123 e23…

A material which behaves in this manner is “fully” anisotropic

The Compliance TensorJust as there is a general relationship between stress and strain:

smn = Emnpq e p q

Need to consider a “companion” to the elasticity tensor…


MIT - 16.001/16.002 Fall, 2006

there is an inverse relationship between strain and stress:

where: is the compliance tensorUsing matrix notation:

s = E efi

with: e = S s

~ ~ ~

~ ~ ~

E-1 s = e~ ~ ~inverse

comparing the two gives:

E-1 = S~ ~

fi E S = I~ ~ ~fiThe compliance matrix is the inverse of the elasticity matrix

emn = Smnpq s p q

Smnpq


MIT - 16.001/16.002 Fall, 2006

• Compliance term Smnpq: amount of strain (emn) caused by the stress (spq)

Meaning of the tensors and their components:

• Elasticity term Emnpq: amount of stress (smn) caused by/related to the deformation/strain (epq)

--> Final note…TransformationsThese are fourth order tensors and thus require 4 direction cosines totransform:

~ ~ ~ ~

~ ~ ~ ~

~

~

Emnpq = lmr lns lpt lq u Erstu

Smnpq = lmr lns lpt lq u Srstu

Note: the same symmetries apply to Smnpq as to Emnpq


MIT - 16.001/16.002 Fall, 2006

Classes of Stress-Strain Behavior(e.g., anisotropy, orthotropy, isotropy)

--> Good reference for this:Bisplinghoff, Mar, and Pian, “Statics

of Deformable Solids”, Addison,Wesley, 1965, Ch. 7.

Start out by making a table of the classes of material stress-strainbehavior and the associated number of independent components ofEmnpq:

Not all materials require all 21 components to describe their behavior. Wetherefore consider…..


MIT - 16.001/16.002 Fall, 2006

2Isotropic

3Cubic

5“Transversely Isotropic”*

6Tetragonal

9Orthotropic

13Monoclinic

21Anisotropic

# of IndependentComponents of Emnpq

Class of Stress-StrainBehavior

UsefulEngineering

Materials

CompositeLaminates

BasicComposite

Ply

Metals(on average)


MIT - 16.001/16.002 Fall, 2006

Consider some key cases:--> Anisotropic

• 21 independent components• Nonorthogonal crystals• Currently no useful engineering materials

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,but may use loading axes.

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

In these other axis systems, the material may appear to have “more”elastic components. But it really doesn’t.

(you can’t “create” elastic components just by describing a material ina different axis system, the inherent properties of the material stay thesame).


MIT - 16.001/16.002 Fall, 2006

Example of unidirectional composite (transverselyisotropic) in two different axis systems

Figure M3.2-3

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

In order to describe full behavior, need to do…TRANSFORMATIONS


MIT - 16.001/16.002 Fall, 2006

--> Orthotropic• Limit of current useful engineering materials• Needed for composite analysis• No coupling terms in the principal axes of the material

e.g., E1112 = 0

- No shear strains arise when extensional stress is applied and vice versa)

(total of 9 terms are now zero)- No extensional strains arise when shear stress is applied (and vice versa)

(some terms become zero as for previous condition)

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

E1223 , E1213 , E1323 = 0

E1112 , E1113 , E1123, E2212 , E2213 , E2223 , E3312 , E3313 , E3323 = 0


MIT - 16.001/16.002 Fall, 2006

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

(21 - 9 - 3 = 9)and the resulting equations are:

For other cases, no more terms become zero, but the terms are notIndependent.

s11s22s33s23

s13

s12

Ï

Ì

Ô Ô Ô Ô

Ó

Ô Ô Ô Ô

¸

˝

Ô Ô Ô Ô

˛

Ô Ô Ô Ô

=

E1111 E1122 E1133 0 0 0E1122 E2222 E2233 0 0 0E1133 E2233 E3333 0 0 0

0 0 0 2E2323 0 00 0 0 0 2E1313 00 0 0 0 0 2E1212

È

Î


˘

˚

˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙

=

e11e22e33e23e13e12

Ï

Ì

Ô Ô Ô Ô

Ó

Ô Ô Ô Ô

¸

˝

Ô Ô Ô Ô

˛

Ô Ô Ô Ô

s11s22s33s23

s13

s12

Ï

Ì

Ô Ô Ô Ô

Ó

Ô Ô Ô Ô

¸

˝

Ô Ô Ô Ô

˛

Ô Ô Ô Ô

=

E1111 E1122 E1133 0 0 0E1122 E2222 E2233 0 0 0E1133 E2233 E3333 0 0 0

0 0 0 2E2323 0 00 0 0 0 2E1313 00 0 0 0 0 2E1212

È

Î


˘

˚

˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙

=

e11e22e33e23e13e12

Ï

Ì

Ô Ô Ô Ô

Ó

Ô Ô Ô Ô

¸

˝

Ô Ô Ô Ô

˛

Ô Ô Ô Ô

--> 9 independent components(3 orthogonal axes with different responsesalong each)(e.g., orthogonal crystals, woods, composites)


MIT - 16.001/16.002 Fall, 2006

For example, consider….--> Isotropic

• No further zero terms (after orthotropic)• Components of elasticity tensor are related• Only 2 independent constants

• E1111 = E2222 = E3333 • E1122 = E1133 = E2233

• E2323 = E1313 = E1212 • And there is one other equation relating E1111 , E1122 and E2323

• Behavior of most metals, polymers--> elastic response the same in all directions

To better consider these cases we need to discuss:


MIT - 16.001/16.002 Fall, 2006

It is important to remember that generalized Hooke’s Law is a model ofthe stress-strain response. So the components must be measuredexperimentally.

The components of the tensors cannot be directly measured, butwe characterize materials by their…

“Engineering Constants”(or, Elastic Constants)

What we can physically measure for orthotropic materials (inclusive), there are 3 types:

1. Longitudinal (Young’s) (Extensional) Modulus: relates

(3 of these)

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

(Measurement of) Engineering Constants


MIT - 16.001/16.002 Fall, 2006

Examples:1. Longitudinal Modulus

Figure M3.3a Apply s11 only to a bar, measure e11 and e22:

s11

e11

slope = E1s11

s11

e11

e22

E1 = s1 1

e1 1 = Ex

2. Poisson’s Ratio: relates extensional strain in the loading

(6 of these…only 3 are independent)direction to extensional strain in another direction.

3. Shear Modulus: relates shear strain in the plane of shear(3 of these)


MIT - 16.001/16.002 Fall, 2006

1) E11 or Exx or E1 or Ex 2) E22 or Eyy or E2 or Ey

3) E33 or Ezz or E3 or Ez

In general: due to smm applied only

(no summation on m)

Emm = smmemm

Figure M3.3b Consider ratio of two longitudinal strains

2. Poisson’s Ratios (negative ratios)

e22

e11

slope = - n12

generally:


MIT - 16.001/16.002 Fall, 2006

negative ratio of e22 to e11 with s11 applied only!In general: nmn

strain in n - directionloading inm - direction

generally:1) n12 or nxy: (negative of) ratio of e22 to e11 due to s11 2) n13 or nxz: (negative of) ratio of e33 to e11 due to s11 3) n23 or nyz: (negative of) ratio of e33 to e22 due to s22 4) n21 or nyx: (negative of) ratio of e11 to e22 due to s22 5) n31 or nzx: (negative of) ratio of e11 to e33 due to s33 6) n32 or nzy: (negative of) ratio of e22 to e33 due to s33

In general: due to snn applied only

(for n ≠ m)Important: nnm ≠ nmn

nnm = -emmenn

n1 2 = -e2 2

e1 1 = nxy


MIT - 16.001/16.002 Fall, 2006

However, these are not all independent. There are relationsknown as “reciprocity relations” (3 of them) between Poisson’sratios and extensional moduli:

n21 E11 = n12 E22

n31 E11 = n13 E33

n32 E22 = n23 E33

fi only 3 Poisson’s ratios are independent!

3. Shear Moduli--> Apply s12 only and measure e12

factor of 2 because it is anengineering constant andthus use engineering strain

G1 2 = -s1 2

2e1 2 = Gxy


MIT - 16.001/16.002 Fall, 2006

1) G12 or Gxy or G6: contribution of (2)e12 to s12 2) G13 or Gxz or G5: contribution of (2)e13 to s13 3) G23 or Gyz or G4: contribution of (2)e23 to s23

In general: due to smn applied onlyGmn = smn2emn

factor of 2 here since it relates physical quantities

--> one can think about doing each case separately and measuringthe effects. Since this is linear, one can use superposition andthereby get the overall effect. This gives the stress-strain relationswith the engineering constant (compliance format)

generally:

shear stressshear deformation (angular change)

fi Gmn = t mn

g mn


MIT - 16.001/16.002 Fall, 2006

+3

E3

E2

E1

3

G23

G13

G12

n23, n32

n13, n31

= 93+

n12, n21

(same as Emnpq, better be!)

OrthotropicIn material principal axes, there is no coupling between extension andshear and no coupling between planes of shear, so the followingconstants remain:


MIT - 16.001/16.002 Fall, 2006

matrix form:

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

Note: 2’s now incorporated in engineering strain terms

e1

e2

e3

g 2 3

g 1 3

g 1 2

Ï

Ì

Ô Ô Ô Ô Ô Ô Ô Ô

Ó


¸

˝


˛


=

1E1

-n1 2

E1-

n1 3

E10 0 0

-n1 2

E1

1E2

-n3 2

E30 0 0

-n1 3

E1-

n2 3

E2

1E3

0 0 0

0 0 0 1G2 3

0 0

0 0 0 0 1G1 3

0

0 0 0 0 0 1G1 2

È

Î

Í Í Í Í Í Í Í Í Í Í Í Í Í Í Í Í

˘

˚

˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙

s1

s2

s3

s 2 3

s1 3

s1 2

Ï

Ì

Ô Ô Ô Ô Ô Ô Ô

Ó


¸

˝


˛


e = S s


MIT - 16.001/16.002 Fall, 2006

Note: the reciprocity relations can be used to get the equations in the following form:

e1 = 1E1

s1 - n1 2s 2 - n1 3s3[ ]

e2 = 1E2

- n2 1 s1 + s 2 - n2 3s3[ ]

e3 = 1E3

-n3 1 s1 - n3 2s2 + s 3[ ]

g 2 3 = 1G2 3

s23

g 1 3 = 1G1 3

s13

g 1 2 = 1G1 2

s12


MIT - 16.001/16.002 Fall, 2006

IsotropicAs we get to materials with less elastic constants (< 9) than anorthotropic material, we no longer have any more zero terms in theelasticity or compliance matrix, but more nonzero terms are related.

For the isotropic case:• All extensional moduli are the same:

E1 = E2 = E3 = E• All Poisson’s ratios are the same:

n12 = n21 = n13 = n31 = n23 = n32 = n• All shear moduli are the same:

G4 = G5 = G6 = G• And, there is a relationship between E, n and G:

G = E2 1 + n( )

Thus, there are only 2 independentconstants.


MIT - 16.001/16.002 Fall, 2006

e1

e2

e3

g 2 3

g 1 3

g 1 2

Ï

Ì


Ó


¸

˝


˛


=

1E

-nE

-nE

0 0 0

-nE

1E

-nE

0 0 0

-nE

-nE

1E

0 0 0

0 0 0 2 1 +n( )E

0 0

0 0 0 0 2 1 +n( )E

0

0 0 0 0 0 2 1+ n( )E

È

Î

Í Í Í Í Í Í Í Í Í Í Í Í Í Í Í

˘

˚

˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙

s1

s2

s3

s 2 3

s1 3

s1 2

Ï

Ì


Ó


¸

˝


˛


Can also write this out in full form:

This gives:


MIT - 16.001/16.002 Fall, 2006

e1 = 1E

s1 - ns2 - ns3( )

e2 = 1E

-ns1 + s2 - ns3( )

e3 = 1E

-ns1 - ns2 + s3( )

g 2 3 = s 2 3

G

g 1 3 = s1 3

G

g 1 2 = s1 2

GAs noted, there are only 2 independent elastic constants: E and n

Sometimes express as/use Lamé’s constants: m and l


MIT - 16.001/16.002 Fall, 2006

m = E2 1 + n( )

= G

l = nE1 + n( ) 1 - 2n( )

(can be derived by considering relationship between shear stress and principal stresses)

There is also another derived modulus known as the “bulk modulus”, k:

k = 3l + 2m3

= E3 1 - 2n( )

The bulk modulus characterizes the compressibility of a materialunder hydrostatic stress/pressure

hydrostatic = same on all sides(think of submerging cube in water)


MIT - 16.001/16.002 Fall, 2006

Figure M3.4 Illustration of hydrostatic stress/pressure

p

pp

x2

x3

x1

The volume of the block changes from V to V¢

And the bulk modulus, k, relates stress to volumetric strain:

Use this physical situation in the isotropic stress-strain equations:

fi D = volumetric strain = DVV

= V - VV

¢

s1 = s 2 = s3 = - p

p = k D


MIT - 16.001/16.002 Fall, 2006

gives:

each side of the cube changes length related to strain from L to:

So the new volume is:

since e is small (order of 0.01 - 0.02), we neglect higher order terms:

Now:

¢ L = L 1 - e( )

¢ V = ¢ L ( )3 = L3 1 - e( )3 = L3 1 - 3e + 3e 2 - e3( )

fi ¢ V = L3 1 - 3e( )

e1 = e2 = e3 = e = -pE

1 - 2n( )

DV = ¢ V - V = L3 1 - 3e - 1( ) = L3 -3e( )

D = DVV

= -L3 3e( )

L3 = - 3e

= 3 pE

1 - 2n( )


MIT - 16.001/16.002 Fall, 2006

In general, for all cases once we test and characterize the behavior, thisgives us the compliance form of the equations. If we want to get thecomponents of the tensors:

- convert to compliance tensor format- invert compliance matrix to get elasticity matrix and thus components of elasticity tensor

Finally recalling that:

k = pD

gives:

Q.E.D.k = E3 1 - 2n( )


MIT - 16.001/16.002 Fall, 2006

This unit has been devoted to establishing the model forstress-strain response of a material on a macroscopic basis.But there are reasons based on the microstructure thatcertain materials behave certain ways. We thus need to lookat the structure of materials to look at the physical basis forelastic properties.

unit m3 - web.mit.eduweb.mit.edu/.../fall/materials/lectures/m3.2-unified06.pdf · d l such a load...

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