Stress-strain Material Laws

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5Stress-StrainMaterial Laws

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Lecture 5: STRESS-STRAIN MATERIAL LAWS 5 2

TABLE OF CONTENTS

Page

5.1. Introduction 535.1.1. Material Behavior Assumptions . . . . . . . . . . . 535.1.2. The Tension Test Revisited . . . . . . . . . . . . 535.1.3. Elastic Modulus and Poissons Ratio . . . . . . . . . . 555.1.4. Shear Modulus . . . . . . . . . . . . . . . . 555.1.5. Thermal Strain . . . . . . . . . . . . . . . . . 55

5.2. Generalized Hooke s Law in 3D 575.2.1. Strain-Stress Relations . . . . . . . . . . . . . . 575.2.2. Stress-Strain Relations . . . . . . . . . . . . . . 585.2.3. Thermal Effect in 3D . . . . . . . . . . . . . . 59

5.3. Generalized Hooke s Law in 2D 595.3.1. Plane Stress . . . . . . . . . . . . . . . . . . 595.3.2. Plane Strain . . . . . . . . . . . . . . . . . 510

5.4. Example: An In ating Balloon 5115.4.1. Problem Description . . . . . . . . . . . . . . . 5115.4.2. When Will the Balloon Burst? . . . . . . . . . . . 512

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5 3 5.1 INTRODUCTION

5.1. Introduction

So far we have introduced mechanical stresses in Lecture 1 and strains in Lecture 4. How do theyconnect? Through the behavior of the structural material. This behavior is expressed in constitutiveequations , also called material laws .

5.1.1. Material Behavior Assumptions

There is a very wide range of materials used for structures, with drastically different behavior. Inaddition materials can go through different response regimes: elastic, plastic, viscoelastic, crackingand localization, fracture. In the present lecture we restrict attention to a very speci c class andresponse regime by making the following behavioral assumptions.

1. Macroscopic . The material is mathematically modeled as a continuum body. Constitutivefeatures at the meso, micro and nano levels: grains, molecules and atoms, are ignored.

2. Elasticity . That means the stress-strain response is reversible and consequently the materialhas a preferred natural state. This state is assumed to be taken in the absence of loads at areference temperature . By convention we will say that the material is then unstressed andundeformed . On applying loads, and possibly temperature changes, the material developsnonzero stresses and strains, and moves to occupy a deformed con guration.

3. Linearity . The relationship between strains and stresses is linear. Doubling stresses doublesstrains, and viceversa.

4. Isotropy . Thepropertiesof thematerial are independent of direction. This is a good assumptionfor materials such as metals, concrete, plastics, etc. It is notadequate for heterogenousmixturessuch as composites or reinforced concrete, which are anisotropic by nature. The substantialcomplication introduced by anisotropic behavior justify its exclusion from an introductorytreatment. This topic is developed further in the Aerospace Materials senior course.

5. Small Strains . Deformations are considered so small that changes of geometry are neglected asthe loads are applied. Violation of this assumption would require the introduction of nonlinearrelations between displacements and strains, again substantially complicating the equations.

5.1.2. The Tension Test Revisited

The rst acquaintance of an engineering student with lab-controlled material behavior is usuallythrough tension tests carried out during the rst Mechanics, Statics and Structures sophomorecourse. Test results are usually displayed as axial nominal strain versus axial nominal strain, asillustrated in Figure 5.1 for the test of a mild steel specimen up to failure. Several response regionsare indicated there: linearly elastic, yield, strain hardening, localization and failure. These arediscussed in the aforementioned course, and studied further in the senior course on AerospaceMaterials. It is suf cient to note here that we shall be concerned with the linearly elastic regionthat occurs before yield. In that region the 1D Hooke s law is assumed to hold.

Of course material behavior may depart signi cantly from that shown in Figure 5.1. Three distinctavors: brittle, moderately ductile and nonlinear-from-start, are shown schematically in Figure 5.2.Brittle materials such as glass, rock, ceramics, concrete-under-tension, etc., exhibit primarily linearbehavior up to near failure by fracture. Metallic alloys used in aerospace, such as Aluminum and

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Nominal stress = P/A 0

0

Yield

Linearelasticbehavior(Hooke's lawis valid)

Strainhardening

Maxnominalstress

Nominalfailurestress

Localization

Mild SteelTension Test

Nominal strain = L /L

Elasticlimit

Undeformed state

L0 PP

Figure 5.1. Typical tension test behavior of mild steel, which displays a well de ned yieldpoint and yield region.

Brittle(glass,ceramic,concrete)

Moderately ductile(Al alloy)

Nonlinear from start(rubber,polymers)

Figure 5.2. Three material response avors as displayed in a tension test.

Nominal stress = P/A 0

0

Mild steel(highly ductile)

High strength steel

Nominal strain = L /L

Tool steel

Figure 5.3. Different steel grades have approximately the same elastic modulus, but verydifferent post-elastic behavior.

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5.1.5. Thermal Strain

A temperature change of T with respect to a base or reference level produces a thermal strain

T = T , ( 5.5)

where is the coef cient of thermal dilatation, measured in 1 / F or 1 / C . This is typicallypositive: > 0 and very small: 0. The stress is zero when the bar is at the reference temperature T re f .Find which axial stress develops if the temperature changes to T = T re f + T . Since the bar length cannotchange, the combined axial strain must be zero:

xx = =

E + T = 0, ( 5.7)

Solving for gives = E T (5.8)

Since E and are positive, a rise in temperature, i.e., T > 0, will produce a negative axial stress, and the barwill be in compression . This is an example of the so-called thermally induced stress or simply thermal stress .It is the reason behind the use of expansion joints in pavements, rails and bridges. The effect is important inorbiting vehicles such as satellites, which undergo extreme (and cyclical) temperature changes from sun toEarth shade.

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5 7 5.2 GENERALIZED HOOKES LAW IN 3D

(a)

a b

c

xx xx

yy

yy

zz

zz

Initial shape

y

x z

(c)b a

c

Final shape

Final shape

(b)

a b

c

(d)

a b

c

Final shape

Figure 5.5. Derivation of 3D Generalized Hooke s Law for normal stresses and strain, byimagining carrying out three tension tests along { x , y, z}, respectievly.

5.2. Generalized Hooke s Law in 3D

5.2.1. Strain-Stress Relations

We now generalize the foregoing equations to the three-dimensional case, still assuming that thematerial is elastic and isotropic. Condider a cube of material aligned with the axes { x , y, z}, asshown in Figure 5.5, and imagine that three tension tests , labeled (1), (2) and (3) are conductedalong x , y and z, respectively. Pulling the material by applying xx along x will produce normalstrains

(1) xx =

xx E

, (1) yy = xx

E , (1) zz =

xx E

. (5.9)

Next, pull the material by yy along y to get the strains

(2) yy =

yy E

, (2) xx = yy

E , (2) zz =

yy E

. (5.10 )

Finally pull the material by zz along z to get

(3) zz =

zz E

, (3) xx = zz E

, (3) yy = zz E

. (5.11 )

In the general case the cube is subjected to combined normal stresses xx , yy and zz . Since weassumed that the material is linearly elastic, the combined strains can be obtained by superposition :

xx = (1) xx +(2)

xx +(3)

xx = xx E

yy

E

zz E

=1

E xx yy zz .

yy = (1) yy +(2)

yy +(3)

yy = x x

E +

yy E

zz E

=1

E xx + yy zz .

zz = (1) zz +(2)

zz +(3)

zz = x x

E

yy E

+ zz E

=1

E xx yy + zz .

(5.12 )

The shear strains and stresses are connected by the shear modulus as

xy = yx = xyG

= yxG

, yz = zy = yzG

= zyG

, zx = xz = zxG

= xzG

. (5.13 )

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The foregoing equations may be expressed in matrix form as

xx

yy

zz

xy yz zx

=

1 E

E

E 0 0 0

E 1

E

E 0 0 0

E

E 1

E 0 0 0

0 0 0 1G 0 0

0 0 0 0 1G 0

0 0 0 0 0 1G

xx yy zz xy yz zx

. (5.14 )

5.2.2. Stress-Strain Relations

To get stresses if the strains are given, an expedient method is to invert the matrix equation (5.14).This gives

xx yy zz xy yz zx

=

E (1 ) E E 0 0 0

E E (1 ) E 0 0 0 E E E (1 ) 0 0 00 0 0 G 0 00 0 0 0 G 00 0 0 0 0 G

xx

yy

zz

xy yz zx

. (5.15 )

in which E =

E (1 2)( 1 + )

(5.16 )

The six relations in (5.15) written out in long form are

xx =E

(1 2)( 1 + )(1 ) xx + yy + zz ,

yy =E

(1 2)( 1 + ) xx + (1 ) yy + zz ,

zz =E

(1 2)( 1 + ) xx + yy + (1 ) zz ,

xy = G xy , yz = G yz , zx = G zx .

(5.17 )

The combination a v = 13 ( xx + yy + zz ) is called the mean stress , or average stress . Thenegative of a v is called the pressure : p = a v . The combination v = xx + yy + zz is calledthe volumetric strain , or dilatation . The negative of v is known as the condensation . Both pressureand volumetric strain are invariants , that is, their value does not change if axes { x , y, z} are rotated.An important relation between pressure and volumetric strain can be obtained by adding the rstthree equations in (5.17), which upon simpli cation yields

p = E

3(1 2) v= K v . (5.18 )

This coef cient K is called the bulk modulus . If Poisson s ratio approaches 12 , which happens fornear incompressible materials, K .

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5 9 5.3 GENERALIZED HOOKES LAW IN 2D

Remark 5.1 . In the solid mechanics literature p is also de ned (depending on author s preferences) as p = a v = 13 ( xx + yy + zz ), which is the negative of the above one. If so p = + K v . The de nition p = a v is the most common one in uid mechanics.

5.2.3. Thermal Effect in 3D

To incorporate the effect of a temperature change T with respect to a baseor reference temperature,add T to the three normal strains in (5.12):

x x =1

E xx yy zz + T ,

yy =1

E xx + yy zz + T , .

zz =1

E xx yy + zz + T .

(5.19 )

No change in the shear strain-stress relation is needed because if the material is linearly elastic

and isotropic, a temperature change only produces normal strains. The strain-stress matrix relation(5.14) expands to

xx

yy

zz

xy yz zx

=

1 E

E

E 0 0 0

E 1

E

E 0 0 0

E

E 1

E 0 0 0

0 0 0 1G 0 0

0 0 0 0 1G 0

0 0 0 0 0 1G

xx yy zz xy yz zx

+ T

111000

. (5.20 )

Inverting this relation provides the stress-strain relations that account for a temperature change:

xx yy zz xy yz zx

=

E (1 ) E E 0 0 0 E E (1 ) E 0 0 0 E E E (1 ) 0 0 00 0 0 G 0 00 0 0 0 G 00 0 0 0 0 G

xx

yy

zz

xy yz zx

E T 1 2

111000

,

(5.21 )where E is de ned in (5.16). Note that if all mechanical normal strains xx , yy , and zz vanish, thenormal stresses in (5.21) are nonzero if T = 0. Those are called initial thermal stresses , and areimportant in engineering systems exposed to large temperature variations, such as turbine engines,satellites or reentry vehicles.

5.3. Generalized Hooke s Law in 2D

Two specializations of the foregoing 3D equations to two dimensions are of interest in the applica-tions: plane stress and plane strain . Plane stress is more important in Aerospace structures, whichtend to be thin, so in this course more attention is given to that case. Both are reviewed next.

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5.3.1. Plane Stress

In this case all stress components with a z component are assumed to vanish. For a linearly elasticisotropic material, the strain and stress matrices take on the form

xx xy 0 yx yy 00 0 zz

, xx xy 0 yx yy 00 0 0

(5.22 )

Note that the zz strain, often called the transverse strain or thickness strain in applications, ingeneral will be nonzero because of Poisson s ratio effect. The strain-stress equations are easilyobtained by going to (5.17) and setting zz = yz = zx = 0:

xx =1

E xx yy , yy =

1 E

xx + yy , zz =

E xx + yy ,

xy = xyG

, yz = zx = 0.(5.23 )

which in matrix form, omitting known zero components, is

xx

yy

zz

xy

=

1 E

E 0

E 1

E 0 E

E 0

0 0 1G

xx yy xy

. (5.24 )

Inverting the matrix composed by the rst, second and fourth rows of the above relation gives thestress-strain equations

xx yy xy

= E E 0

E E 00 0 G

xx

yy

xy. (5.25 )

in which E = E /( 1 2) . In long form

xx =E

1 2( xx + yy ), yy =

E 1 2

( yy + xx ), xy = G xy . (5.26 )

5.3.2. Plane Strain

In this case all strain components with a z component are assumed to vanish. For a linearly elasticisotropic material, the strain and stress matrices take on the form

xx xy 0 yx yy 00 0 0

, xx xy 0 yx yy 00 0 zz

(5.27 )

Note that the zz stress, which is called the transverse stress in applications, in general will notvanish. The strain-to-stress relations can be easily obtained by setting zz = yz = zx = 0 in

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5 11 5.4 EXAMPLE: AN INFLATING BALLOON

(5.23). This gives

xx =E

(1 2)( 1 + )(1 ) xx + yy ,

yy =E

(1 2)( 1 + ) xx + (1 ) yy ,

zz =E

(1 2)( 1 + ) xx + yy ,

x y = G xy , yz = 0, zx = 0.

(5.28 )

which in matrix form, with the zero components removed, is

xx yy zz xy

=

E (1 ) E 0 E E (1 ) 0 E E 00 0 G

xx

yy

xy. (5.29 )

Inverting the system provided by extracting the rst, second and fourth rows of (5.29) gives thestrain-stress relations, which are omitted for simplicity.

The effect of temperature changes can be incorporated in both plane stress and plane strain relationswithout any dif culty.

5.4. Example: An In ating Balloon

This is a generalization of Problem 3 of Recitation #2. The main change is that all data is expressedand kept in variable form until the problem is solved. Speci c numbers are plugged in at the end.

This will be the only problem in the course where some features of nonlinear mechanics appear.

These come in by writing the governing equations in both the initial and nal geometries, withoutlinearization.

5.4.1. Problem Description

The problem is depicted in Figure 5.6. A spherical rubber balloon has initial diameter D0 underination pressure p0 . This is called the initial conguration . The pressure is increased by p sothat the nal pressure is p f = p0 + p . The balloon assumes a spherical shape with nal diameter D f = D0 , inwhich > 1. This will be called the nal conguration . The initial wall thickness ist 0


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Initial (reference)diameter D , underinflation pressure p

0

0

0

0

Final (deformed)diameter D = D

inflationpressure

p = p + p

For simplicity assumea spherical ballon shapefor all calculations

f

f

D = D0f

0 D

Figure 5.6. In ating balloon example problem.

Since the balloon is assumed to remain spherical and its thickness is very small compared to itsdiameter, the above strains hold at all points of the balloon wall, and are the same in any directiontangent to the sphere. If we choose the sphere normal as local z axis, the wall is in a plane stressstate.

Next we introduce material laws. We will assume that rubber obeys the two-dimensional, planestress generalized Hooke s law (5.26) with respect to the Eulerian strain measure, with effectivemodulus of elasticity E and Poisson s ratio .1 Setting x x = yy = E f and x y = 0 therein andaccounting for the initial stress 0 , we obtain the inplane normal stress in the nal con guration:

x x = yy = f = 0 +

E

1 2 ( E

f + E

f ) = 0 +

E

1 E

f = 0 +

E

1 1

. (5.31 )

The normal inplane wall stress is the same in all directions, so it is called simply 0 and f , forinitial and nal con gurations, respectively. The inplane shear stress vanishes in all directions.

Assume D 0 , t 0 , E and are given as data. An interesting question: what is the relation between p(the excess or gage pressure) and the diameter D f = D0? And, is there a maximum pressure thatwill cause the balloon to burst?

5.4.2. When Will the Balloon Burst?

To relate p and it is necessary to express the wall stresses 0 and f in terms of geometry and

internal pressure. This is provided by equation (3.10) in Lecture 3, derived for a thin-wall sphericalvessel. In that equation replace p , R and t by quantities in the initial and nal con gurations:

0 = p0 R0

2t 0=

p0 D 04t 0

, f = p f R f

2 t f =

( p0 + p ) D f 4 t f

=( p0 + p ) D 0

4 t f . (5.32 )

1 This is a very rough approximation because the constitutive equations for rubber (and polymers in general) are highlynonlinear. But getting closer to reality would take us into the realm of nonlinear elasticity, which is a graduate-leveltopic.

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5 13 5.4 EXAMPLE: AN INFLATING BALLOON

1.5 21 2.5 3 3.5 4

10

2

3

4

5

6

7

= D /D

p (MPa)

=1/2

=0

f 0

Figure 5.7. In ating pressure (in MPa) versus diameter expansion ratio = D f / D 0 for a balloon with E = 1900 MPa, D 0 = 50 mm, p0 = 0 MPa,

t 0 = 0.18 mm, 1 4 and Poisson s ratios = 0 and = 12 .

All quantities in the above expressions are known in terms of the data, except t f . A kinematicanalysis beyond the scope of this course shows that

t f = 1 + 21

2 1 t 0 . (5.33 )

We can check (5.33) by inserting two limit values of Poisson s ratio:

= 0: t f = t 0 . This is correct since the thickness does not change. = 1/ 2: t f = t 0 /

2 . Is this correct? If = 1/ 2 the material is incompressible and does notchange volume. The initial and nal volume of the thin-wall spherical balloon areV 0 = D 20 t 0 and V f = D

2 f t f =

2 D 20 t f , respectively. On setting V 0 = V f andsolving for t f we get t f = t 0 / 2 .

To obtain p in terms of , replace (5.33) into (5.32), equate this to (5.31) and solve for p . Theresult provided by Mathematica is

p =4 Et 0(1 )( 2 + 2(1 2)) + D 0 p0( 1 )( 4 + 2 (2 4))

D 0 4 (1 )(5.34 )

This expression simpli es considerably in the two Poisson s ratio limits:

p | = 0 =4 E t 0 ( 1) + D 0 p0 (2 )

D 0 2(5.35 )

p | = 1/ 2 =8 E t 0 ( 1) + D 0 p0 (2 3)

D 0 4(5.36 )

Pressure versus diameter ratio curves given by (5.35) and (5.36) are plotted in Figure 5.7 forthe numerical values indicated there. Those values correspond to the data used in Problem 3 of Recitation 2, in which = 1/ 2 was speci ed from the start.Rubber (and, in general, polymer materials) are nearly incompressible; for example 0.4995 forrubber. Consequently, the response depicted in Figure 5.7 for = 1/ 2 is more physically relevantthan the other one.

Do the response plots in Figure 5.7 tell you when an in ating balloon is about to collapse? Yes.This is the matter of Homework Exercise 3.5.

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Stress-strain Material Laws