ch.7. plane linear elasticity

CH.7. PLANE LINEAR ELASTICITYContinuum Mechanics Course (MMC) - ETSECCPB - UPC

Overview

Plane Linear Elasticity Theory

Plane Stress Simplifying Hypothesis Strain Field Constitutive Equation Displacement Field The Linear Elastic Problem in Plane Stress Examples

Plane Strain Simplifying Hypothesis Strain Field Constitutive Equation Stress Field The Linear Elastic Problem in Plane Stress Examples

2

Overview (cont’d)

The Plane Linear Elastic Problem Governing Equations

Representative Curves Isostatics or stress trajectories

Isoclines

Isobars

Shear lines

Others: isochromatics and isopachs

Photoelasticity

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4

Ch.7. Plane Linear Elasticity

7.1 Plane Linear Elasticity Theory

A lineal elastic solid is subjected to body forces and prescribed traction:

The Linear Elastic problem is the set of equations that allow obtaining the evolution through time of the corresponding displacements , strains and stresses .

Introduction

0t

,0

,0

b x

t x

,

,

t

t

b x

t x

Initial actions:

Actions through time:

, tu x , tx , tx

5

The Linear Elastic Problem is governed by the equations:1. Cauchy’s Equation of Motion.

Linear Momentum Balance Equation.

2. Constitutive Equation. Isotropic Linear Elastic Constitutive Equation.

3. Geometrical Equation. Kinematic Compatibility.

Governing Equations

2

0 0 2

,, ,

tt t

t

u x

x b x

, 2t Tr x 1

1, ,2

St t x u x u u

This is a PDE system of 15 eqns -15 unknowns:

Which must be solved in the space.

, tu x , tx

, tx

3 unknowns

6 unknowns

6 unknowns

3R R

6

For some problems, one of the principal stress directions is known a priori: Due to particular geometries, loading and boundary conditions

involved. The elastic problem can be solved independently for this direction. Setting the known direction as z, the elastic problem analysis is

reduced to the x-y plane

There are two main classes of plane linear elastic problems: Plane stress Plane strain

Plane Linear Elasticity

PLANE ELASTICITY

REMARK The isothermal case will be studied here for the sake of simplicity. Generalization of the results obtained to thermo-elasticity is straight-forward.

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8


7.2 Plane Stress

Simplifying hypothesis of a plane stress linear elastic problem:

1. Only stresses “contained in the x-y plane” are not null

2. The stress are independent of the z direction.

Hypothesis on the Stress Tensor

00

0 0 0

x xy

xy yxyz

, ,

, ,

, ,

x x

y y

xy xy

x y t

x y t

x y t

REMARK The name “plane stress” arises from the fact that all (not null) stress are contained in the x-y plane.

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These hypothesis are valid when:

The thickness is much smaller than the typical dimension associated to the plane of analysis:

The actions , and are contained in the plane of analysis (in-plane actions) and independent of the third dimension, z.

is only non-zero on the contour of the body’s thickness:

Geometry and Actions in Plane Stress

e L

, tb x * , tu x * , tt x

* , tt x

10

The strain field is obtained from the inverse Hooke’s Law:

As

And the strain tensor for plane stress is:

Strain Field in Plane Stress

, ,x x x y t

, ,y y x y t , ,x y t

1 02

1, , 02

0 0

x xy

xy y

z

x y t

with 1z x y

000

1

zxzyz

TrE E

1

1 1 2(1 )2

1 1 021 02

x x y xy xy xy

y y x xz xz

z x y yz yx

E E

E

E

11

Operating on the result yields:

Constitutive equation in Plane Stress

planestress C

2

1 01 0

110 0

2

x x

y y

xy xy

E

planestress C

Constitutive equation in plane stress

(Voigt’s notation)

1 2(1 )2

1 2 0

2 0

x x y xy xy xy

y y x xz xz

z x y yz yx

E E

E

E

2

2

1

1

2 1

x x y

y y x

xy xy

E

E

E

12

The displacement field is obtained from the geometric equations,. These are split into:

Those which do not affect the displacement :

Those in which appears:

Displacement Field in Plane Stress

, ,St tx u x

zu

zu , , x

xux y tx

, , yy

ux y t

y

, , 2 yxxy xy

uux y ty x

, ,, ,

x x

y y

u u x y tu u x y t

Integration in .

Contradiction !!!

0

0

, , ( , , ) ( , , , )1

( , ), , 2 0

( , )( , ), , 2 0

zz x y z

x z zxz xz

zy z z

yz yz

ux y t x y t u x y z tz

u x y u ux y tz x x

u z tu x y u ux y tz y y

13

The problem can be reduced to the two dimensions of the plane of analysis. The unknowns are:

The additional unknowns (with respect to the general problem) are either null, or independently obtained, or irrelevant:

The Lineal Elastic Problem in Plane Stress

1z x y

0z xz xz xz yz

, , ,zu x y z t does not intervene in the problem

, , x

y

ux y t

u

u , ,x

y

xy

x y t

, ,x

y

xy

x y t

REMARK This is an ideal elastic problem because it cannot be exactly reproduced as a particular case of the 3D elastic problem. There is no guarantee that the solution to and

will allow obtaining the solution to for the additional geometric eqns.

, ,yu x y t , ,xu x y t

, , ,zu x y z t

14

3D problems which are typically assimilated to a plane stress state are characterized by: One of the body’s dimensions is significantly smaller than the other two. The actions are contained in the plane formed by the two “large” dimensions.

Examples of Plane Stress Analysis

Slab loaded on the mean plane Deep beam

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7.3 Plane Strain

Simplifying hypothesis of a plane strain linear elastic problem:

1. The displacement field is

2. The displacement variables associated to the x-y plane are independent of the z direction.

Hypothesis on the Displacement Field

0

x

y

uu

u

, ,

, ,x x

y y

u u x y t

u u x y t

17

These hypothesis are valid when:

The body being studied is generated by moving the plane of analysis along a generational line.

The actions , and are contained in the plane of analysis and independent of the third dimension, z.

In the central section, considered as the “analysis section” the following holds (approximately) true:

Geometry and Actions in Plane Strain

, tb x * , tu x * , tt x

0zu

0xuz

0yuz

18

The strain field is obtained from the geometric equations:

And the strain tensor for plane strain is:

Strain Field in Plane Strain

1 02

1, , 02

0 0 0

x xy

xy yx y t

REMARK The name “plane strain” arises from the fact that all strain is contained in the x-y plane.

, ,, , 0

, , , ,, , 0

, , , ,, ,, , 0

x zx z

y x zy xz

y yx zxy yz

u x y t ux y tx z

u x y t u x y t ux y ty z x

u x y t u x y tu x y t ux y ty x z y

0zu

0xuz

0yuz

19

Introducing the strain tensor into Hooke’s Law and operating on the result yields:

As

And the stress tensor for plane strain is:

Stress Field in Plane Strain

Tr G 1

2

2 0

( ) 0

x x y x xy xy

y x y y xz xz

z x y x y yz yz

G G

G G

v G

2G x y

2 y xG

, ,x y t

, ,x x x y t

, ,y y x y t

, ,z z x y t

, ,xy xy x y t

0

, , 00 0

x xy

xy y

z

x y t

with z x y

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Introducing the strain tensor into the constitutive equation and operating on the result yields:

Constitutive equation in Plane Strain

planestrain C

1 01

11 0

1 1 2 11 20 0

2 1

x x

y y

xy xy

E

planestrain C

Constitutive equation in plane strain

(Voigt’s notation)

12

1 1 2 1x x y x y

EG

2 1xy xy xyEG

12

1 1 2 1y y x y x

EG

2Tr 1

21

The problem can be reduced to the two dimensions of the plane of analysis. The unknowns are:

The additional unknowns (with respect to the general problem) are either null or obtained from the unknowns of the problem:

The Lineal Elastic Problem in Plane Strain (summary)

0z xz yz xz yz

0zu

z x y

, , x

y

ux y t

u

u , ,x

y

xy

x y t

, ,x

y

xy

x y t

22

3D problems which are typically assimilated to a plane strain state are characterized by: The body is generated by translating a generational section with actions

contained in its plane along a line perpendicular to this plane. The plane stress hypothesis must be justifiable. This typically

occurs when:1. One of the body’s dimensions is significantly larger than the other two.

Any section not close to the extremes can be considered a symmetry plane and satisfies:

2. The displacement in z is blocked at the extreme sections.

Examples of Plane Strain Analysis

0z xz yz

0zu

0xuz

0yuz

0

x

y

uu

u

23

3D problems which are typically assimilated to a plane strain state are:

Examples of Plane Strain Analysis

Pressure pipe

Tunnel

Continuous brake shoe

Solid with blocked zdisplacements at the ends

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7.4 The Plane Linear Elastic Problem

A lineal elastic solid is subjected to body forces and prescribed traction and displacement

The Plane Linear Elastic problem is the set of equations that allow obtaining the evolution through time of the corresponding displacements , strains and stresses .

Plane problem

Actions:

, ,x y tu , ,x y t , ,x y t

*

*

, ,

, ,x

y

t x y t

t x y t

*t

*

*

, ,

, ,x

y

u x y t

u x y t

*u

On :

On :u

On :

, ,, ,

x

y

b x y tb x y t

b

26

The Plane Linear Elastic Problem is governed by the equations:1. Cauchy’s Equation of Motion.

Linear Momentum Balance Equation.

Governing Equations

2

0 0 2

,, ,

tt t

t

u x

x b x

2D

2

2

2

2

2

2

xyx xz xx

xy y yz yy

yzxz z zz

ubx y z t

ub

x y z tub

x y z t

27

The Plane Linear Elastic Problem is governed by the equations:2. Constitutive Equation (Voigt’s notation).

Isotropic Linear Elastic Constitutive Equation.

Governing Equations

, :t x C

2D

C

x

y

xy

x

y

xy

With , and

2

1 01 0

10 0 1 2

E

C

E E

PLANE STRESS

21

1

EE

PLANE STRAIN

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The Plane Linear Elastic Problem is governed by the equations:3. Geometrical Equation.

Kinematic Compatibility.

Governing Equations

1, ,2

St t x u x u u



, tu x , tx

, tx

2 unknowns

3 unknowns

3 unknowns

2

2D

xx

yy

yxxy

uxuy

uuy x

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Boundary conditions in space Affect the spatial arguments of the unknowns Are applied on the contour of the solid, which is divided into:

Prescribed displacements on :

Prescribed stresses on :

Boundary Conditions

u

* **

* *

, ,

, ,x x

y y

u u x y t

u u x y t

u

* **

* *

, ,

, ,x x

y y

t t x y t

t t x y t

t * t n

x

y

nn

n

x xy

xy y

with

30

INTIAL CONDITIONS (boundary conditions in time) Affect the time argument of the unknowns. Generally, they are the known values at :

Initial displacements:

Initial velocity:

Boundary Conditions

0t

, ,0 x

y

ux y

u

u 0

00

v, ,, ,0 ,

vx x

y yt

ux y tx y x y

ut

uu v

31

The 8 unknowns to be solved in the problem are:

Once these are obtained, the following are calculated explicitly:

Unknowns

, , x xy

xy y

x y t

12, ,

12

x xy

xy y

x y t

( , , ) x

y

ux y t

u

u

PLANE STRESS

PLANE STRAIN

1z x y

z x y

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7.5 Representative Curves

Traditionally, plane stress states where graphically represented with the aid of the following contour lines: Isostatics or stress trajectories Isoclines Isobars Maximum shear lines Others: isochromatics, isopatchs, etc.

Introduction

34

System of curves which are tangent to the principal axes of stress at each material point . They are the envelopes of the principal stress vector fields. There will exist two families of curves at each point:

Isostatics , tangents to the largest principal stress. Isostatics , tangents to the smallest principal stress.

Isostatics or Stress Trajectories

1

2

REMARK The principal stresses are orthogonal to each other, therefore, so will the two families of isostatics orthogonal to each other.

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Singular point: characterized by the stress state

Neutral point: characterized by the stress state

Singular and Neutral Points

Mohr’s Circle of a neutral point

Mohr’s Circle of a singular point

0x y xy

0x y

xy

REMARK In a singular point, all directions are principal directions. Thus, in singular points isostatics tend to loose their regularity and can abruptly change direction.

36

Consider the general equation of an isostatic curve:

Solving the 2nd order eq.:

Differential Equation of the Isostatics

y f x

2

2 21

xy

x y

yy

2 1 0x y

xy

y y

2

2 2 tgtg 21 tg

tg

xy

x y

dy yd x

2

' 12 2x y x y

xy xy

y

Differential equation of the isostatics

,x y Known this function, the eq. can be integrated to obtain a family of curves of the type:

y f x C

37

Locus of the points along which the principal stresses are in the same direction. The principal stress vectors in all points of an isocline are parallel to

each other, forming a constant angle with the x-axis.

These curves can be directly found using photoelasticity methods.

Isoclines

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To obtain the general equation of an isocline with angle , the principal stress must form an angle with the x-axis:

Equation of the Isoclines

1

2

tg 2 xy

x y

Algebraic equation

of the isoclines

,x y

For each value of , the equation of the family of isoclines parameterized in function of is obtained:

,y f x

REMARK Once the family of isoclines is known, the principal stress directions in any point of the medium can be obtained and, thus, the isostatics calculated.

39

Locus of the points along which the principal stress ( or ) is constant. The isobars depend on the value of the principal stress but not on

their direction. There will exist two families of isobars at each point, and .

Isobars

1 2

1 2

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Take the equation used to find the principal stresses and principal stress directions given in a certain set of axes:

Equation of the Isobars

Algebraic equation of the isobars

22

1 1 1

22

2 2 2

,2 2

,2 2

x y x yxy

x y x yxy

x y cnt c

x y cnt c

1 1 1

2 2 2

,

,

y f x c

y f x c

This eq. implicitly defines the family of curves of the isobars:

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Envelopes of the maximum shear stress (in modulus) vector fields. They are the curves on which the shear stress modulus is a maximum. Two planes of maximum shear stress correspond to each material

point, and . These planes are easily determined using Mohr’s Circle.

Maximum shear lines

minmax

REMARK The two planes form a 45º angle with the principal stress directions and, thus, are orthogonal to each other. They form an angle of 45º with the isostatics.

42

Consider the general equation of a slip line , the relation and

Then,

Equation of the maximum shear lines

4 1tan 2 tan 2

2 tan 2

2tan 2 xy

x y

y f x

2

1 2 tantan 2tan 2 2 1 tan

tan

x y

xy

notdy yd x

22

2 1x y

xy

yy

2 41 0xy

x y

y y

43

Solving the 2nd order eq.:

Equation of the maximum shear lines

Differential equation of the

slip lines

22 2

' 1xy xy

x y x y

y

,x y Known this function, the eq. can be integrated to obtain a family of curves of the type:

y f x C

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Summary

For some problems the elastic problem can be solved independently for one of the directions. The analysis is reduced to a plane.

Two problem types:1. Plane stress

Summary

PLANE ELASTICITY

00

0 0 0

x xy

xy yxyz

, ,

, ,

, ,

x x

y y

xy xy

x y t

x y t

x y t

e L

1 02

1, , 02

0 0

x xy

xy y

z

x y t

with

1z x y

The displacement field is obtained from

, ,St tx u x

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2. Plane strain

Summary (cont’d)

0

x

y

uu

u

, ,

, ,x x

y y

u u x y t

u u x y t

In the central section,

0zu

0xuz

0yuz

1 02

1, , 02

0 0 0

x xy

xy yx y t

0

, , 00 0

x xy

xy y

z

x y t

with z x y

52

For both problem types the unknowns are:

The additional unknowns (w.r.t. the general problem) are either null, independently obtained or irrelevant:

Summary (cont’d)

1z x y

0z xz xz xz yz

, , ,zu x y z t does not intervene in the problem

, , x

y

ux y t

u

u , ,x

y

xy

x y t

, ,x

y

xy

x y t

z x y

0z xz yz xz yz

0zu

PLANE STRAINPLANE STRESS

53

The PLANE Linear Elastic Problem:

Summary (cont’d)

Actions:

*

*

, ,

, ,x

y

t x y t

t x y t

*t

*

*

, ,

, ,x

y

u x y t

u x y t

*u

On :

On :u

On :

, ,, ,

x

y

b x y tb x y t

b

Responses:

, tu x , tx

, tx

2 unknowns

3 unknowns

3 unknowns

54

The PLANE Isotropic Linear Elastic Problem:

Summary (cont’d)

Cauchy’s Equation of Motion

Constitutive Equation

2

2

2

2

xyx xx

xy y yy

ubx y t

ub

x y t

Geometric Equation

C

2

1 01 0

10 0 1 2

E

C

;E E PLANE STRESS: 2 ;

1 1EE

PLANE STRAIN:

with

; ;y yx xx y xy

u uu ux y y x



, tu x , tx

, tx

2 unknowns

3 unknowns

3 unknowns

2R R

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The PLANE Isotropic Linear Elastic Problem (cont’d):

Summary (cont’d)

Boundary Conditions in Space

Initial Conditions

:u

* **

* *

, ,

, ,x x

y y

u u x y t

u u x y t

u

:

* **

* *

, ,

, ,x x

y y

t t x y t

t t x y t

t* t nwith

, ,0x y u 0

00

, ,, ,0 ,

not

t

x y tx y x y

t

u

u v

56

Plane stress states can be graphically represented with the curves:

Isostatics or stress trajectories: curves which are at each material point tangent to the principal axes of stress.

Isoclines: locus of the points along which the principal stresses are in the same direction. (Determined using photoelasticity methods.)

Isobars: locus of the points along which the principal stress is constant.

Maximum shear lines: envelopes of the maximum (in modulus) shear stress vector fields.

Isochromatics: curves along which the maximum shear stress is constant. (Determined using photoelasticity methods.)

Isopachs: curves along which the mean normal stress is constant.

Summary (cont’d)

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ch.7. plane linear elasticity

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