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Method of Finite Elements I
Chapter 6
2D Elements
*slides are prepared in collaboration with Dr. S. Triantafyllou, Assistant Professor at the University of No?ingham
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Method of Finite Elements I 30-Apr-10
Today’s Lecture Contents
• Continuum Elements – Plane Stress – Plane Strain
• Structural Elements – Plate Elements
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Method of Finite Elements I 30-Apr-10
FE Classification
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Method of Finite Elements I 30-Apr-10
2D vs. 3D Formulations
Three-‐‑dimensional elasticity problems are very difficult to solve. Thus we will first develop governing equations for two-‐‑dimensional problems, and will explore two basic theories:
-‐‑ Plane Strain -‐‑ Plane Stress
The basic theories of plane strain and plane stress represent the fundamental plane problem in elasticity. While these two theories apply to significantly different types of two-‐‑dimensional bodies, their formulations yield very similar field equations.
Since all real elastic structures are three-‐‑dimensional, theories set forth here will be approximate models. The nature and accuracy of the approximation will depend on problem and loading geometry
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Method of Finite Elements I 30-Apr-10
Plane Strain Consider an infinitely long cylindrical (prismatic) body as shown. If the body forces and tractions on lateral boundaries are independent of the z-‐‑coordinate and have no z-‐‑component, then the deformation field can be taken in the reduced form
x
y
z R
u = u(x, y) , v = v(x, y) , w = 0
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Method of Finite Elements I 30-Apr-10
Examples of Plane Strain
x
y
z
x
y
z
P
Long Cylinders Under Uniform Loading
Semi-‐‑Infinite Regions Under Uniform Loadings
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Plane Strain Equations
0,2
)()(
2)(,2)(
=τ=τµ=τ
σ+σν=+λ=σ
µ++λ=σµ++λ=σ
yzxzxyxy
yxyxz
yyxyxyxx
eee
eeeeee
Equilibrium Equations
0
0
=+∂σ∂
+∂τ∂
=+∂τ∂
+∂σ∂
yyxy
xxyx
Fyx
Fyx
Strain Compatibility
yxe
xe
ye xyyx
∂∂∂
=∂∂
+∂∂ 2
2
2
2
2
2
Strain Displacement Relations
ex =∂u∂x
, ey =∂v∂y
, exy =12
∂u∂y
+ ∂v∂x
⎛⎝⎜
⎞⎠⎟
ez = exz = eyz = 0
Strains vs. Stresses
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Method of Finite Elements I 30-Apr-10
Plane Stress Consider a where one dimension, eg. along z, is small in comparison to the other dimensions in the problem. Since the region is thin in the z-‐‑direction, there can be liMle variation in the stress components through the thickness, and thus they will be approximately zero throughout the entire domain. Finally, since the region is thin in the z-‐‑direction it can be argued that the other non-‐‑zero stresses will have liMle variation with z. Under these assumptions, the stress field can be taken as
0
),(
),(),(
=τ=τ=σ
τ=τ
σ=σσ=σ
yzxzz
xyxy
yy
xx
yxyxyx
x
y
z R
2h
yzxzz ττσ ,,
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Examples of Plane Stress Problems
Thin Plate With Central Hole
Circular Plate Under Edge Loadings
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Method of Finite Elements I 30-Apr-10
Plane Stress Equations
Strain Displacement Relations
Equilibrium Equations
0
0
=+∂σ∂
+∂τ∂
=+∂τ∂
+∂σ∂
yyxy
xxyx
Fyx
Fyx
Strain Compatibility
yxe
xe
ye xyyx
∂∂∂
=∂∂
+∂∂ 2
2
2
2
2
2
0,1
)(1
)(
)(1,)(1
==τν+=
+ν−
ν−=σ+σν−=
νσ−σ=νσ−σ=
yzxzxyxy
yxyxz
xyyyxx
eeE
e
eeE
e
Ee
Ee
021,0
21
21,,,
=⎟⎠⎞⎜
⎝⎛
∂∂+
∂∂==⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂+
∂∂=
⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂+
∂∂=
∂∂=
∂∂=
∂∂=
xw
zue
yw
zve
xv
yue
zwe
yve
xue
xzyz
xyzyx
Strains vs. Stresses
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Method of Finite Elements I 30-Apr-10
Plane Stress/ Strain Elasticity v Plane Strain
Premise 1:
Premise 2: Loads are applied only within the plane Premise 3: The applied loads are independent of z Premise 4: No load is applied on the boundary surfaces normal to the
Conclusion:
Premise 5: The edge surfaces are rigid
Stress Tensor
Strain Tensor
Constitutive Matrix
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Method of Finite Elements I 30-Apr-10
Plane Stress/ Strain Elasticity v Plane Stress
Premise 1:
Premise 2: Loads are applied only within the plane Premise 3: The applied loads are independent of Premise 4: No load is applied on the boundary surfaces normal to the
Conclusion:
Stress Tensor
Strain Tensor
Constitutive Matrix
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Method of Finite Elements I
What are the types of Finite Elements used in plane stress/strain formulations?
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Method of Finite Elements I 30-Apr-10
Constant Strain Triangle The Constant Strain Triangle element is historically the first finite element ever used in engineering practice (Argyris, 1960, Turner, 1956) for the evaluation of stress distribution in wing panels.
The CST Element
v Plane Element i. The displacement field varies within the x-‐‑y plane
v Constant strain field
i. The strain does not vary within the element v Elastic Material Behaviour
1 2
3
We number the nodes in counter-‐‑clockwise order. In this way, the normal vector to the 123 surface will point towards the positive z axis and the resulting area of the triangle will be a positive quantity.
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Method of Finite Elements I 30-Apr-10
The CST Element
Remember that the deformation is the first derivative of the displacement field Compatibility equations in
the 2D plane
Therefore if we ask for the strain field to be constant
The displacement field has to be a linear function of
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The CST Element
1 2
3
Therefore, the following “candidate” displacement field approximation is considered:
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The CST Element Therefore the shape functions of the CST element are readily derived as
where the shape function matrix assumes the following form
and
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CST Shape Functions and if we plot these shape functions over the surface of the element:
1
2
3 1
3
2
1
2
3
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v The CST element is fairly accurate within zones of small variation of stresses. However a fine mesh is required in every other case in order for FEA to converge to an accurate solution.
v As already discussed during the Galerkin lectures, the FEA solution is refined either by increasing the number of finite elements or by increasing the order of the interpolating (test) functions used in the FE formulation. Thus, there is a trade-‐‑off between required mesh-‐‑size and interpolation complexity for the same degree of accuracy.
v Based on that rationale, the Quadrilateral 4-‐‑node (Q4) finite element
was introduced in an effort to effectively model the stress variations of plane elasticity problems.
The Q4 Element
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The Q4 Element
1 2
3 4
v Plane Element i. The displacement field varies only within the x-‐‑y plane
v Elastic Material Behaviour
v Formulation Assumptions
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The Q4 Element
We number the nodes in counter-‐‑clockwise order. In this way, the normal vector to the 1234 surface will point towards the positive axis and the resulting area of the triangle will be a positive quantity.
1 2
3 4
v Plane Element i. The displacement field varies only within the x-‐‑y plane
v Elastic Material Behaviour
v Formulation Assumptions
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Method of Finite Elements I 30-Apr-10
The Q4 Element v Formulation Assumptions
The following bilinear “candidate” displacement field approximation is considered:
1 2
3 4
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Method of Finite Elements I 30-Apr-10
The Q4 Element
The following bilinear “candidate” displacement field approximation is considered: 1 2
3 4
The formulation is indifferent to the coordinate system. So why not make things easier?
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Method of Finite Elements I 30-Apr-10
The Q4 Element
The arbitrary nodal displacement values are introduced at the r.h.s. of the interpolation equation:
Exactly the same procedure as in the CST element!
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Method of Finite Elements I 30-Apr-10
The Q4 Element Following the standard procedure, the arbitrary nodal displacement values are introduced at the r.h.s. of the interpolation equation:
Therefore the Shape Function Matrix is derived form the following:
Shape Functions
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Method of Finite Elements I 30-Apr-10
The Q4 Element The compatibility relations are again expressed in matrix form
and therefore by substituting the interpolation equation into the r.h.s.
where now
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Method of Finite Elements I 30-Apr-10
Q4 Stiffness Matrix The stiffness matrix is derived as
and given that the element is rectangular
and if the thickness is constant
The evaluation of the stiffness matrix can be performed analytically. Integration involves only linear expressions of and
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Consistent nodal force vector Consider the case of distributed loading along the element’s 12 side
1 2
3 4
Equivalent nodal vector due to a traction load
In this case
For example, the component along the 12 side is
External Work due to surface tractions
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Method of Finite Elements I 30-Apr-10
Consistent nodal force vector Consider the case of distributed loading along the element’s 12 side
1 2
3 4
Equivalent nodal vector due to a traction load
In this case
External Work due to surface tractions
In matrix form
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Method of Finite Elements I 30-Apr-10
Consistent nodal force vector Consider the case of distributed loading along the element’s 12 side
1 2
3 4
Equivalent nodal vector due to a traction load
In this case
External Work due to surface tractions
Consistent Load Vector
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Method of Finite Elements I 30-Apr-10
Taxonomy of Finite Elements
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The Euler/Bernoulli beam theory
v Uniaxial Element i. The longitudinal direction is sufficiently larger than the
other two v Prismatic Element
i. The cross-‐‑section of the element does not change along the element’s length
v Euler/ Bernoulli assumption i. Upon deformation, plane sections remain plane AND
perpendicular to the beam axis
Assumptions
But what happens if two dimensions are sufficiently larger that the third one?
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The plate problem
• The slab thickness is sufficiently smaller that the two leading dimensions
• Consequently, the three
dimensional problem is reduced to a two-‐‑dimensional one and the plate problem is examined at the mid-‐‑surface.
• If the thickness of the slab is then the mid-‐‑surface is located at a distance from each lateral surface
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The plate problem Two possible loading states
Case 1: Loads applied within the plane Case 2: Loads applied perpendicular to the mid-‐‑surface
Plane stress (or membrane) problem Bending
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Method of Finite Elements I 30-Apr-10
The plate problem Two possible loading states
Case 1: Loads applied within the plane Case 2: Loads applied perpendicular to the mid-‐‑surface
Plane stress (or membrane) problem Bending
i. Plate theory is only concerned with the response of the body due to bending loads
ii. The combined response under membrane and bending conditions of plane (or curved) surfaces is treated under the framework of shell theory
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Method of Finite Elements I 30-Apr-10
Plate Theories
thick thin very thin
L/t 5-‐‑10 5-‐‑100 >100
characteristics transverese shear deformations
no transverse shear deformations
Geometrically nonlinear
Plate theory Reissner, Mindlin Kirchhoff Von Karman
Beam Theory Timoshenko Euler, Bernoulli
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Method of Finite Elements I 30-Apr-10
Kirchhoff-‐‑Love Plate Theory Assumptions • The plate is thin in the sense that the thickness
is small compared to the leading dimensions, but not so thin that the lateral deflection becomes comparable to .
• The plate thickness is either uniform or varies slowly so that three-‐‑dimensional stress effects are ignored.
• The plate is symmetric in fabrication about the mid-‐‑surface.
• Applied transverse loads are distributed over plate surface areas of dimension or greater.
• The support conditions are such that no significant extension of the mid-‐‑surface develops. Augustus Edward Hough Love
(1863-‐‑1940)
Gustav Robert Kirchhoff (1824-‐‑1887)
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Method of Finite Elements I 30-Apr-10
Kirchhoff-‐‑Love Plate Theory Assumptions
From (i), since is very small, the variation of with respect to z can be neglected. Therefore:
Additionally, the following kinematic assumption is introduced: “Planes perpendicular to the mid-‐‑surface will remain plane and perpendicular to the deformed
mid-‐‑surface”
This is the two-‐‑dimensional equivalent of the Euler-‐‑Bernoulli kinematic
assumption for beams!!
Augustus Edward Hough Love (1863-‐‑1940)
Gustav Robert Kirchhoff (1824-‐‑1887)
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Method of Finite Elements I 30-Apr-10
Kirchhoff-‐‑Love Plate Theory Remember in the one-‐‑dimensional beam problem...
Point A displacement (that’s because the section remains plane)
(that’s because the plane remains perpendicular to the neutral axis)
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Method of Finite Elements I 30-Apr-10
Kirchhoff-‐‑Love Plate Theory Similarly in the two-‐‑dimensional plate problem – Bending with respect to axis
Point A displacement (that’s because the section remains plane)
(that’s because the plane remains perpendicular to the neutral axis)
Positive rotation with respect to results in positive displacements
along
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Method of Finite Elements I 30-Apr-10
Kirchhoff-‐‑Love Plate Theory Similarly in the two-‐‑dimensional plate problem – Bending with respect to axis
Point A displacement (that’s because the section remains plane)
(that’s because the plane remains perpendicular to the neutral axis)
Positive rotation with respect to results in negative displacements
along
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Method of Finite Elements I 30-Apr-10
Kirchhoff-‐‑Love Plate Theory Compatibility Relations (Strain-‐‑Displacement)
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Method of Finite Elements I 30-Apr-10
Kirchhoff-‐‑Love Plate Theory Stress-‐‑strain relations (Considering elastic isotropic material)
and
1st Remark The constitutive relation results when we substitute the zero deformation terms (derived in the previous slide) to the three-‐‑dimensional elastic stress-‐‑strain relations. The derived relation is identical to the plane-‐‑strain case. Thus, according to the Kirchhoff-‐‑Love assumptions, every infinitesimal particle within the plate is in a plane-‐‑strain condition
2nd Remark An immediate consequence is that the shear stress components and vanish
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Method of Finite Elements I 30-Apr-10
Kirchhoff-‐‑Love Plate Theory Stress-‐‑strain relations (Considering elastic isotropic material)
and
Substituting for the strain components with respect to the displacement
As expected, all the stress components are linear functions of
(I)
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Method of Finite Elements I 30-Apr-10
Kirchhoff-‐‑Love Plate Theory Stress Resultants
Since the normal stress distribution is not uniform they give rise to a moment vector (created from the couple of tension and compression forces.
Since the shear stress distribution is not uniform they also give rise to a moment vector (created from the couple of tension and compression forces.
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Method of Finite Elements I 30-Apr-10
Kirchhoff-‐‑Love Plate Theory
Stress Resultants However, are not the only stress-‐‑resultants. For these moments to be in equilibrium, a pair of shear forces must exist. Considering an infinitesimal mid-‐‑surface element the positive moments are defined as:
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Method of Finite Elements I 30-Apr-10
Kirchhoff-‐‑Love Plate Theory Stress Resultants If a distributed load is applied onto the element, equilibrium with respect to the vertical axis z results in:
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Method of Finite Elements I 30-Apr-10
Kirchhoff-‐‑Love Plate Theory Stress Resultants Similarly, the following equations are derived, considering moment equilibrium with respect to axes
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Method of Finite Elements I 30-Apr-10
Kirchhoff-‐‑Love Plate Theory Differential Form with respect to Substituting the stress-‐‑displacement relations (I) into the definition of the moment components the following relations are derived
the bending rigidity of the plate
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Method of Finite Elements I 30-Apr-10
Kirchhoff-‐‑Love Plate Theory
Square simply supported plate subject to uniform distributed loading
bending in both directions
pay a?ention to the way the corner regions deform….
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Method of Finite Elements I 30-Apr-10
Kirchhoff-‐‑Love Plate Theory This is the effect of the twisting moment
bending in both directions
darker areas denote larger values of twisting moments
and this is why specifications make sure that an additional amount of reinforcement is provided for corner areas of slabs
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Method of Finite Elements I 30-Apr-10
Kirchhoff-‐‑Love Plate Theory The Kirchhoff-‐‑Love plate theory • extends the Euler/Bernoulli beam assumptions to the two-‐‑dimensional
case • Based on that, every significant measure of rotation, force, moment is
evaluated with respect to the vertical deflection Assumptions: The main kinematic assumption is that “Plane surfaces remain plane and perpendicular to the mid-‐‑surface of the plate”. General Remarks The Kirchhoff-‐‑Love theory predicts a zero distribution of shear stresses along the z direction. Thus, it can only be applied in problems where the variation of such stresses is expected to be small and their mean value does not deviate from 0. Such can be considered the case of thin plates.