CH.7. PLANE LINEAR ELASTICITY Continuum Mechanics Course (MMC)

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

Lecture 1

Lecture 2

Lecture 4

Lecture 3

Overview (cont’d)

The Plane Linear Elastic Problem Governing Equations

Representative Curves Isostatics or stress trajectories

Isoclines

Isobars

Maximum shear lines

3

Lecture 5

Lecture 6

4

Ch.7. Plane Linear Elasticity

7.1 Plane Linear Elasticity Theory

For some problems, one of the principal 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 not be studied here for the sake of simplicity. Generalization of the results obtained to thermo-elasticity is straight-forward.

7

8

Ch.7. Plane Linear Elasticity

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

σ ττ σ

≡

σ

( )( )( )

, ,

, ,

, ,

x x

y y

xy xy

x y t

x y t

x y t

σ σ

σ σ

τ τ

=

=

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

9

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 tσ σ=

( ), ,y y x y tσ σ= ( ), ,x y t=ε ε

( )

1 02

1, , 02

0 0

x xy

xy y

z

x y t

ε γ

γ ε

ε

≡

ε with ( )1z x yνε ε εν

= − +−

( )000

1

zxzyz

TrE E σ

ττ

ν ν

===

+= − +ε σ σ1

( )

( )

( )

1 2(1 )2

1 2 0

2 0

x x y xy xy xy

y y x xz xz

z x y yz yz

E E

E

E

νε σ νσ γ ε τ

ε σ νσ γ ε

νε σ σ γ ε

+= − = =

= − = =

= − + = =

11

Operating on the result yields:

Constitutive equation in Plane Stress

planestress= ⋅σ εC

2

1 01 0

110 0

2

x x

y y

xy xy

Eσ ν εσ ν ε

ντ ν γ

= − − = σ = ε

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 yz

E E

E

E

νε σ νσ γ ε τ

ε σ νσ γ ε

νε σ σ γ ε

+= − = =

= − = =

= − + = =

( )

( )

( )

2

2

1

1

2 1

x x y

y y x

xy xz

E

E

E

σ ε νεν

σ ε νεν

τ γν

= + −

= + −

=+

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 t=x u xε ∇

zu

zu( ), , x

xux y tx

ε ∂=∂

( ), , yy

ux y t

yε

∂=∂

( ), , 2 yxxy xy

uux y ty x

γ ε∂∂

= = +∂ ∂

( )( )

, ,, ,

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

νε ε εν

γ ε

γ ε

=

=

∂= − + = ⇒

− ∂∂ ∂ ∂ = = + = = ∂ ∂ ∂

⇒∂ ∂ ∂ = = + = =∂ ∂ ∂

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 Linear Elastic Problem in Plane Stress

( )1z x yνε ε εν

= − +−

0z xz yz xz yzσ τ τ γ γ= = = = =

( ), , ,zu x y z t does not appear in the problem

( ), , x

y

ux y t

u

≡

u ( ), ,x

y

xy

x y tεεγ

≡

ε ( ), ,x

y

xy

x y tσστ

≡

σ

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

15

16

Ch.7. Plane Linear Elasticity

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

ε γ

γ ε

≡

εREMARK 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

ε ε

ε γ

γ γ

∂ ∂= = =

∂ ∂∂ ∂ ∂

= = + =∂ ∂ ∂

∂ ∂∂ ∂= + = + =

∂ ∂ ∂ ∂

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 Gλ= +σ ε 2 ε1

( )( )( )

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

σ λ ε ε ε τ γ

σ λ ε ε ε τ γ

σ λ ε ε σ σ τ γ

= + + =

= + + = =

= + = + = =( )2 y xGλ ε λε= + +

( ), ,x y t=σ σ

( ), ,x x x y tε ε=

( ), ,y y x y tε ε=

( ), ,z z x y tε ε=

( ), ,xy xy x y tγ γ=

( )0

, , 00 0

x xy

xy y

z

x y tσ ττ σ

σ

≡

σ with ( )z x yσ ν σ σ= +

20

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

Constitutive equation in Plane Strain

planestrain= ⋅Cσ ε

( )( )( )

( )

1 01

11 0

1 1 2 11 20 0

2 1

x x

y y

xy xy

E

ννσ ε

ν νσ εν ν ν

τ γνν

−

− = + − − − − = σ = ε

planestrain= C

Constitutive equation in plane strain

(Voigt’s notation)

( ) ( )( )( )

12

1 1 2 1x x y x y

EG

ν νσ λ ε λε ε εν ν ν

− = + + = + + − −

( )2 1xy xy xyEGτ γ γν

= =+

( ) ( )( )( )

12

1 1 2 1y y x y x

EG

ν νσ λ ε λε ε εν ν ν

− = + + = + + − − ( ) 2Trλ µ= +σ ε ε1

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 yzε γ γ τ τ= = = = =

0zu =

( )z x yσ ν σ σ= +

( ), , x

y

ux y t

u

≡

u ( ), ,x

y

xy

x y tεεγ

≡

ε ( ), ,x

y

xy

x y tσστ

≡

σ

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 strain 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 yzε γ γ= = =

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 z displacements at the ends

24

25

Ch.7. Plane Linear Elasticity

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 tε ( ), ,x y tσ

( )( )

*

*

, ,

, ,x

y

t x y t

t x y t

=

*t

( )( )

*

*

, ,

, ,x

y

u x y t

u x y t

=

*u

On : σΓ

On : uΓ

On : Ω( )( )

, ,, ,

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

tρ ρ

∂⋅ + =

∂u x

x b x∇ σ

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

τσ τ ρ ρ

τ σ τρ ρ

ττ σ ρ ρ

∂∂ ∂ ∂+ + + =

∂ ∂ ∂ ∂

∂ ∂ ∂ ∂+ + + =

∂ ∂ ∂ ∂∂∂ ∂ ∂

+ + + =∂ ∂ ∂ ∂

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 Cσ ε

2D

= ⋅Cσ ε

x

y

xy

εεγ

=

ε x

y

xy

σστ

≡

σWith , and

( )2

1 01 0

10 0 1 2

Eν

νν

ν

= − −

C

E Eν ν==

PLANE STRESS

( )

21

1

EEνννν

=−

=−

PLANE STRAIN

28

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 uε ∇ ∇ ∇

This is a PDE system of 8 eqns -8 unknowns: Which must be solved in the space.

( ), tu x( ), txε

( ), txσ

2 unknowns

3 unknowns

3 unknowns

2+×

2D

xx

yy

yxxy

uxuy

uuy x

ε

ε

γ

∂=∂∂

=∂

∂∂= +∂ ∂

29

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

Γ

uΓ

σΓ

( )( )

* **

* *

, ,

, ,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 nσ

x

y

nn =

n

x xy

xy y

σ ττ σ

≡

σ

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 yx y t

σ ττ σ

≡

σ( )12, ,

12

x xy

xy y

x y tε γ

γ ε

≡

ε( , , ) x

y

ux y t

u

=

u

PLANE STRESS

PLANE STRAIN

( )1z x yνε ε εν

= +−

( )z x yσ ν σ σ= +

32

33

Ch.7. Plane Linear Elasticity

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 (orthogonal) families of curves at each point:

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

Isostatics or Stress Trajectories

1σ

2σ

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

35

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 xyσ σ τ= = =

0x y

xy

σ σ

τ

=

=

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

τσ σ

′=

− ′−( )2 1 0x y

xy

y yσ στ−

′ ′+ − =

( ) 2

2 2 tgtg 21 tg

tg

xy

x y

dy yd x

τ αασ σ α

α

= =− −

′= =

( ) 2

' 12 2x y x y

xy xy

yσ σ σ σ

τ τ

− −= − ± +

Differential equation of the isostatics

( ),x yφ 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

θ

38

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

θ1σ α θ=

( )2

tg 2 xy

x y

τθ

σ σ=

−Algebraic equation

of the isoclines

( ),x yϕ

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

θ

θ( ),y f x θ=

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

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

minτmaxτ

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

4πβ α= + ( ) 1tan 2 tan 2

2 tan 2πβ α

α = − = −

2tan 2 xy

x y

τα

σ σ=

−

( )y f x=

( ) ( )

( )

2

1 2 tantan 2tan 2 2 1 tan

tan

x y

xy

notdy yd x

σ σ ββα τ β

β

−= − = − =

−

′= = ( )22

2 1x y

xy

yy

σ στ− ′

− =′−

( )2 41 0xy

x y

y yτ

σ σ′ ′− − =

−

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

yτ τ

σ σ σ σ

= ± + − −

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

( )y f x C= +

44