CH.7. PLANE LINEAR ELASTICITY Continuum Mechanics Course (MMC) - - - PowerPoint PPT Presentation
CH.7. PLANE LINEAR ELASTICITY Continuum Mechanics Course (MMC) - ETSECCPB - UPC Overview Plane Linear Elasticity Theory Plane Stress Simplifying Hypothesis Strain Field Constitutive Equation Displacement Field The
Continuum Mechanics Course (MMC) - ETSECCPB - UPC
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
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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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A lineal elastic solid is subjected to body forces and prescribed
The Linear Elastic problem is the set of equations that allow
t
,0 ,0 b x t x
, , t t b x t x
Initial actions: Actions through time:
,t u x
,t x
,t x
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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.
2 2
, , , t t t t u x x b x
, 2 t Tr x 1
1 , , 2
S
t t x u x u u
This is a PDE system of 15 eqns -15 unknowns: Which must be solved in the space.
,t u x
,t x
,t x
3 unknowns 6 unknowns 6 unknowns
3
R R
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For some problems, one of the principal stress directions is known
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 ELASTICITY
REMARK
The isothermal case will be studied here for the sake of simplicity. Generalization of the results
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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.
x xy xy y xyz
, , , , , ,
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:
e L
,t b x
*
,t u x
*
,t t x
*
,t t x
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The strain field is obtained from the inverse Hooke’s Law: As And the strain tensor for plane stress is:
, ,
x x x y t
, ,
y y x y t
, , x y t
1 2 1 , , 2
x xy xy y z
x y t
with
1
z x y
1
z xz yz
Tr E E
1
1 1 2(1 ) 2 1 1 2 1 2
x x y xy xy xy y y x xz xz z x y yz yx
E E E E
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Operating on the result yields:
plane stress
C
2
1 1 1 1 2
x x y y xy xy
E
plane stress
C
Constitutive equation in plane stress (Voigt’s notation)
1 2(1 ) 2 1 2 2
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
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The displacement field is obtained from the geometric equations,
. These are split into:
Those which do not affect the displacement : Those in which appears:
, ,
S
t t x u x
z
u
z
u
, ,
x x
u x y t x
, ,
y y
u x y t y
, , 2
y x xy xy
u u x y t y x
, , , ,
x x y y
u u x y t u u x y t
Integration in .
, , ( , , ) ( , , , ) 1 ( , ) , , 2 ( , ) ( , ) , , 2
z z x y z x z z xz xz z y z z yz yz
u x y t x y t u x y z t z u x y u u x y t z x x u z t u x y u u x y t z y y
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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,
1
z x y
z xz xz xz yz
, , ,
z
u x y z t does not intervene in the problem
, ,
x y
u x 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.
, ,
y
u x y t
, ,
x
u x y t
, , ,
z
u x y z t
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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.
Slab loaded on the mean plane Deep beam
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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.
x y
u u u
, , , ,
x x y y
u u x y t u u x y t
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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
In the central section, considered as the “analysis section” the
,t b x
*
,t u x
*
,t t x
z
u
x
u z
y
u z
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The strain field is obtained from the geometric equations: And the strain tensor for plane strain is:
1 2 1 , , 2
x xy xy y
x y t
REMARK
The name “plane strain” arises from the fact that all strain is contained in the x-y plane.
, , , , , , , , , , , , , , , , , ,
x z x z y x z y xz y y x z xy yz
u x y t u x y t x z u x y t u x y t u x y t y z x u x y t u x y t u x y t u x y t y x z y
z
u
x
u z
y
u z 19
Introducing the strain tensor into Hooke’s Law and
As And the stress tensor
for plane strain is:
Tr G 1
2 2 ( )
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 x
G
, , x y t
, ,
x x x y t
, ,
y y x y t
, ,
z z x y t
, ,
xy xy x y t
, ,
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
plane strain
C
1 1 1 1 1 1 2 1 1 2 2 1
x x y y xy xy
E
plane strain
C
Constitutive equation in plane strain (Voigt’s notation)
1 2 1 1 2 1
x x y x y
E G
2 1
xy xy xy
E G
1 2 1 1 2 1
y y x y x
E G
2 Tr 1
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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
z xz yz xz yz
z
u
z x y
, ,
x y
u x y t u u
, ,
x y xy
x y t
, ,
x y xy
x y t
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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
Any section not close to the extremes can be considered a symmetry plane and satisfies:
z xz yz
z
u
x
u z
y
u z
x y
u u u
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3D problems which are typically assimilated to a plane strain state are:
Pressure pipe Tunnel Continuous brake shoe Solid with blocked z displacements at the ends
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A lineal elastic solid is subjected to body forces and prescribed
The Plane Linear Elastic problem is the set of equations that
Actions:
, , x y t u
, , 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 t b x y t b
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The Plane Linear Elastic Problem is governed by the equations:
1. Cauchy’s Equation of Motion.
Linear Momentum Balance Equation.
2 2
, , , t t t t u x x b x 2D
2 2 2 2 2 2 xy x xz x x xy y yz y y yz xz z z z
u b x y z t u b x y z t u b x y z t
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The Plane Linear Elastic Problem is governed by the equations:
2. Constitutive Equation (Voigt’s notation).
Isotropic Linear Elastic Constitutive Equation.
, : t x C
2D
C
x y xy
x y xy
With , and
2
1 1 1 1 2 E C E E
PLANE STRESS
2
1 1 E E
PLANE STRAIN
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The Plane Linear Elastic Problem is governed by the equations:
3. Geometrical Equation.
Kinematic Compatibility.
1 , , 2
S
t t x u x u u
This is a PDE system of 8 eqns -8 unknowns: Which must be solved in the space.
,t u x
,t x
,t x
2 unknowns 3 unknowns 3 unknowns
2
2D
x x y y y x xy
u x u y u u y 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 :
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
n n n
x xy xy y
with
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INTIAL CONDITIONS (boundary conditions in time) Affect the time argument of the unknowns. Generally, they are the known values at :
Initial displacements: Initial velocity:
t
, ,0
x y
u x y u u
v , , , ,0 , v
x x y y t
u x y t x y x y u t
u u v
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The 8 unknowns to be solved in the problem are: Once these are obtained, the following are calculated explicitly:
, ,
x xy xy y
x y t
1 2 , , 1 2
x xy xy y
x y t ( , , )
x y
u x y t u u
PLANE STRESS PLANE STRAIN
1
z x y
z x y
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Traditionally, plane stress states where graphically represented
Isostatics or stress trajectories Isoclines Isobars Maximum shear lines Others: isochromatics, isopatchs, etc.
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System of curves which are tangent to the principal axes of stress
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.
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
Mohr’s Circle of a neutral point Mohr’s Circle of a singular point
x y xy
x 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.
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Consider the general equation of an isostatic curve: Solving the 2nd order eq.:
y f x
2
2 2 1
xy x y
y y
2
1
x y xy
y y
2
2 2tg tg 2 1 tg tg
xy x y
dy y d x
2
' 1 2 2
x y x y xy xy
y Differential equation
, 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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Locus of the points along which the principal stresses are in the
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.
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To obtain the general equation of an isocline with angle , the
1
2 tg 2
xy x y
Algebraic equation
, 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.
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Locus of the points along which the principal stress ( or ) is
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 .
1
2
1
2
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Take the equation used to find the principal stresses and principal
Algebraic equation
2 2 1 1 1 2 2 2 2 2
, 2 2 , 2 2
x y x y xy x y x y xy
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.
min
max
REMARK The two planes form a 45º angle with the principal stress directions and, thus, are orthogonal to each
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Consider the general equation of a slip line , the relation
Then,
4
1 tan 2 tan 2 2 tan 2
2 tan 2
xy x y
y f x
2
1 2 tan tan 2 tan 2 2 1 tan tan
x y xy not
dy y d x
2
2 2 1
x y xy
y y
2
4 1
xy x y
y y
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Solving the 2nd order eq.:
Differential equation of the slip lines
2
2 2 ' 1
xy 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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For some problems the elastic problem can be solved independently for
Two problem types:
PLANE ELASTICITY
x xy xy y xyz
, , , , , ,
x x y y xy xy
x y t x y t x y t
e L
1 2 1 , , 2
x xy xy y z
x y t
with
1
z x y
The displacement field is obtained from
, ,
S
t t x u x
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x y
u u u
, , , ,
x x y y
u u x y t u u x y t
In the central section,
z
u
x
u z
y
u z
1 2 1 , , 2
x xy xy y
x y t
, ,
x xy xy y z
x y t
with
z x y
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For both problem types the unknowns are:
The additional unknowns (w.r.t. the general problem) are either null,
independently obtained or irrelevant:
1
z x y
z xz xz xz yz
, , ,
z
u x y z t does not intervene in the problem
, ,
x y
u x y t u u
, ,
x y xy
x y t
, ,
x y xy
x y t
z x y
z xz yz xz yz
z
u
PLANE STRAIN PLANE STRESS
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The PLANE Linear Elastic Problem:
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 t b x y t b
Responses:
,t u x
,t x
,t x
2 unknowns 3 unknowns 3 unknowns
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The PLANE Isotropic Linear Elastic Problem:
Cauchy’s Equation of Motion Constitutive Equation
2 2 2 2 xy x x x xy y y y
u b x y t u b x y t
Geometric Equation
C
2
1 1 1 1 2 E C
; E E
PLANE STRESS:
2 ;
1 1 E E
PLANE STRAIN:
with
; ;
y y x x x y xy
u u u u x y y x
This is a PDE system of 8 eqns -8 unknowns: Which must be solved in the space.
,t u x
,t x
,t x
2 unknowns 3 unknowns 3 unknowns
2
R R
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The PLANE Isotropic Linear Elastic Problem (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 n
with
, ,0 x y u
, , , ,0 ,
not t
x y t x y x y t
u u v
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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
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.
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