me 160 introduction to finite element method chapter 4 finite … · 2018-10-29 · induced stress...

Chapter 4 Finite Element Analysis in Stress

Analysis of Elastic Solid Structures

Instructor Tai-Ran Hsu, Professor

San Jose State University Department of Mechanical Engineering

ME 160 Introduction to Finite Element Method

Introduction to Fundamentals of Theory of Linear Elasticity

Part 1

This type of elasticity occurs to solids undergoing small deformations, such as springs that exhibit linear relationships between the applied force (F) and the induced elongation (x) that can be represented by a mathematical formula as: F = kx where k is a constant known as the rate or spring constant. Many metallic materials fall into the category of linear elastic solids It can also be stated as a linear relationship between stress (σ) and strain (ε) in stretching or compressing a thin rod by The expression: σ = Eε where E Is known as the elastic modulus or Young's modulus

Type 1: Linear elasticity of materials:

What defines elastic solids? A solid deforms in response to external actions (e.g., forces, heat, etc.). The deformation is completely reversible – meaning the solid returns to its original shape after the removal of the external actions.

Type 2: Nonlinear elasticity of materials: This type of solids behaves as elastic materials as described above, but can exhibit large deformations, such as rubbers and polymers

The FE formulation presented in this course will be based on linear elasticity theory

Two Types of Elastic Solids

https://en.wikipedia.org/wiki/Tensile_stress

https://en.wikipedia.org/wiki/Strain_(materials_science)

https://en.wikipedia.org/wiki/Elastic_modulus

https://en.wikipedia.org/wiki/Young's_modulus

Linear Elastic Behavior of Solids Fundamental assumptions

(1) The material is treated as a continuous medium (or a continuum). In the words, the material is homogeneous with no internal defects or voids of significant sizes (2) The material is isotropic – meaning its properties are uniform in all directions (3) The material has no memory (4) The material exhibits the same properties in tension and compression

Definitions of stress and strain

Uniaxial elongation of a thin rod:

L0 L

F

A0

(1) Engineering strain (e): It is defined with reference to the original shape of the solid. Mathematically it is expressed as: = ΔL/L0 = L/L0 - 1 (2) True strain (ε): It is measured on the basis of the immediate proceeding length of the rod sample. Mathematically it is expressed as:

ΔL

==

−++

−+

−= ∫∑

01

12

0

01

0

.........LLn

LdL

LLL

LLL

LLL L

Ln

n ε

(3) Relationship between ε and e: ε = ℓn(1+e)

(4) Engineering stress (S): 0sec

tanAP

areationalcrossOriginalloadeousInstaS =

−=

(5) True stress (σ): σ = P/A

(6) Relationship between S and σ: σ = (1 + e) S

Stress-Strain Curves of Materials Stretching of thin rods of most engineering materials will exhibit the stress vs. strain relations illustrated in the figure below:

Designations: A’ = proportional limit A = elastic limit B = yield point m =necking point f = rupture point S0 = Yield strength of material Su = ultimate tensile strength of material

Typical elastic deformation of engineering materials: (1) Very small deformation with strain up to 0.1% (2) Straight linear relationship between the stress and strain, resulting in a constant stiffness of the . (3) The slope of line OA’ is called the Young’s modulus (E), representing the stiffness of the material (4) Completely recoverable strain (or deformation) after the applied load is removed (5) The yield stress (So) or σy is defined to be the interception of ϵoffset= 0.2% of the σ-ϵ curve. It is a measure of materials exceeding the elastic limit, and undergoing plastic deformation (an irreversible deformation)

0

True

Engineering Stre

sses

Strains

Physics of Deformable Solids subjected to External Forces

Original State:

after applied forces:

RESPONDES To Allied Foreces:

2 things will happen:

2 Physical Consequences:

The Many Components of Displacements and Stresses:

(Element) Displacements:

{U}T: {Ux,Uy,Uz} in a deformed solid

A small element located at x,y,z:

See detailed definitions on the next slide:

Induced Stress Components in Deformable Solids subjected to External Forces

Because the forces applied to a general 3-D solid, the induced stresses are MULTI-directional designated by σαβ: σ = magnitude, subscript α = the axis normal to the plane of action, subscript β = the direction the stress component points to So, stress component σyy = stress component acting on the plane normal to the y-axis and pointing to the y-direction, whereas stress component σxy = stress components acting on the face normal to the x-axis but points to the y-direction.

[ ]

=

zzzyzx

yzyyyx

xzxyxx

σσσσσσσσσ

σ

In theory, there are nine (9) stress components everywhere inside the solid:

But in reality there are only six (6) independent stress components with: σxy =σyx , σxz =σzx , σyz = σzx

[ ]

=

zz

yzyy

xzxyxx

SYM σσσσσσ

σ{ }

=

6

5

4

3

2

1

σσσσσσ

σ

For FE formulation: σ1=σxx,σ2=σyy,, σ3=σzz, σ4=σxy, σ5=σyz, σ6=σxz

(4.1)

Solid subjected to external forces:

Result in:

Stress components with same subscripts such as: σxx, σyy, σzz are called NORMAL stress components, with unit: psi, or Pascal (Pa) = N/m2 . Stress components with different subscripts such as: σxy, σxz, σyx, σyz, σzx, σzy are SHEARING stress components. Shear stress has the unit of change of angle from the original right angle to the angle with the stress, i.e (π/2) – θ. The unit is thus “angle change” in radians (rad)

Induced Stresses in Deformable Solids subjected to External Forces – Cont’d

A B

C D

E F

G

x

y

z

σyy σyz

σyx

σzz

σzy σzx

σxx

σxz σxy

Effects of Normal and Shear Stresses to Solid Deformation

x

y

x

y Normal stresses change size:

σxx -σxx

Shear stresses change shape:

σxy -σxy

contraction

elongation

Exist on the SURFACES Exist NORMAL to surfaces

Relationships between Primary unknown and Secondary Unknowns in FE analyses

Primary Unknown: Element displacements: {U(x,y,z)} and

Nodal displacements: [N(x,y,z)]{u}T in FE formulation: {U} = [N]{u}

where {N(x,y,z)} = Interpolation function

2nd Secondary Unknown: Element stresses {σ(x,y,z)}=[C]{ε}

Linkage to the 1st secondary unknown: Element strains: {ε(x,y,z)}=[B(x,y,z)]{U(x,y,z)} → corresponding to element stresses: {σ(x,y,z)}

Strain-displacement relations

Generalized Hook’s Law

Element strains and stresses, and nodal displacement are the solutions of FE analyses.

x F

L x

Stress: σxx Strain: εxx

L ΔL

Uniaxial stress with uniaxial deformation:

Uniaxial stress with biaxial deformation with Poisson’s ratio:

εxx = ΔL/L

F

y

F x 0

Lateral contraction: - εyy

Longitudinal stretching: + εxx

For solids with significant cross-section, a uniaxial stress may produce biaxial deformation such as illustrate in the figure in the left. The lateral contraction deformation is represented by: - εyy

Poisson’s ratio defines as:

xx

yy

strainstretchingalLongitudinstrainncontractioLateral

εε

ν ==

● Poisson ratio exists in most multi-axially loaded elastic solids, and this effect needs to be included in all FE formulations. This effect also leads to the use of Generalize Hooke’s Law for the formulations

Multiaxial Deformations of Elastic Solids

Generalized Hooke’s Law for Solids with Multi-axial Stresses:

Stress vs. strain relationship for deformed solids with 3-D deformation

Total strains in three directions induced by the three normal stresses

in x-direction

in y-direction

in z-direction

in which E = Young’s modulus, and γ = Poisson’s ratio of the material

The following expression express the stresses in terms of strains – the Hooke’s law:

( )( )

( )( )

( )

−+++−+++−

−+=

zyx

zyx

zyx

zz

yy

xx E

εννενενεεννενενεεν

ννσσσ

11

1

211

One may derive the uniaxial stress situation as a special case from the above expression to obtain:

σxx = Eεx

by substituting: σyy = σzz = 0, and εy = -γεx and εz = -γεx in the generalized Hooke’s Law

Elongation in one-direction causes contractions in other directions, and vise versa.

Element Strain-Displacement Relations for FE Formulation

There are six (6) strain components corresponding to the stress components in interior of the deformed solid. These strain components are related to the displacements of the solid induced by the external forces. Because deformation of the solid CONTINUOUSLY varying throughout the solid, the following relationship exits:

( )( )( )

∂∂

∂∂

∂∂

∂∂∂∂

∂∂

∂∂

∂∂

∂∂

=

zyxUzyxUzyxU

xz

yz

xy

z

y

x

z

y

x

xz

yz

xy

zz

yy

xx

,,,,,,

0

0

0

00

00

00

εεεεεε

or: {ε(x,y,z)} = [D]{U(x,y,z)} (4.3)

(4.2)

[ ]

∂∂

∂∂

∂∂

∂∂∂∂

∂∂

∂∂

∂∂

∂∂

=

xz

yz

xy

z

y

x

D

0

0

0

00

00

00

(4.4)

Element Strains Corresponding to Element stresses

Element Displacements

U(x,y,z)

By theory of elasticity:

Example: For uni-axial elongation or contraction of a rod: Element displacement {U} = {Ux(x)}, The corresponding strain in element is: ( ) ( )

dxxdU

xxU xx

xx =∂

∂=ε

Element Stress-Strain Relations – the Generalized Hooke’s Law

According to generalized Hooke’s law for MULTI-Axial stress state, the following relationship between the element stress and strain exists:

( )( )

−

−

−−

−−

−+=

xz

yz

xy

zz

yy

xx

xz

yz

xy

zz

yy

xx

SYM

E

εεεεεε

ν

ν

νν

ννννν

νν

σσσσσσ

221

0221

00221

000100010001

211(4.5)

where E = Young’s modulus, and γ = Poisson’s ratio

or: {σ} = [C]{ε} (4.6) where [C] is the elasticity matrix with:

[ ] ( )( )

−

−

−−

−−

−+=

221

0221

00221

000100010001

211

ν

ν

νν

ννννν

ννSYM

EC (4.7)

Example: For uni-axially loaded rod: ( )dx

xdUEE xxxxx == εσ where Ux(x) is the displacement in the rod

Displacements: {Φ(x,y,z)}T: {Φx(x,y,z,Φy(x,y,z,Φz(x,y,z}

Small element located at x,y,z:

Physics of Solid Deformation by External Forces

[ ]

=

zzzyzx

yzyyyx

xzxyxx

σσσσσσσσσ

σ

Total 9 stress components at (x,y,z)

FE Formulation of Deformable Elastic Solids

Part 2

Derivation of Element Equations

In Step 4, Chapter 3, we derived the element equation using the Rayleigh-Ritz method to take a form:

[Ke]{q}= {Q}

where [Ke] = Element matrix {q} = Vector of primary unknown quantities at the nodes of the element {Q} = Vector of element nodal actions (e.g., forces)

(1.27) in the textbook

Element equations for each tetrahedron element in the FE model for a structure are then assembled to establish “overall stiffness equation” for determining nodal displacements of all nodes in the structure.

FE Model for s Structure made of Tetrahedron Elements

[K]{Φ} = {R}

where [ ] [ ]∑=m

meKK

1m=total number of elements in the FE model

We will select tetrahedron elements as the basis for FE formulation for general 3-D solid structures

Derivation of Element Equations-Cont’d Principle of deriving element equation using Rayleigh-Ritz variational principle

From Chapter 2: Let us determine a suitable “functional” to derive the element equation.

( ) { } { } { } { }∫∫

∂∂

+

∂∂

=sv

dsgdvf .,.........,..........,,rrφφφφφχ

( ){ } 02

1

=

•••∂∂∂∂

=∂∂ φ

χφχ

φφχ

and then apply the Variational principle on:

from which equations of each element are derived:

The “functional” for a deformed solid subjected to external forces is “POTENTIAL ENERGY” (P) in the situation.

The potential energy associated with a deformed solid can be defined as:

P = U - W (4.8)

where U = the strain energy in a deformed solid, and W = the work done to the deformed solid by external forces acting on the solid body in the volume and surface of the solid

A general form of functional: v = volume, s = surface

{ } { } { } .....,.........0,0,0321

=∂∂

=∂∂

=∂∂

φχ

φχ

φχ

Derivation of Element Equations-Cont’d

(b)

(a) (C)

(d)

Strain energy in a deformed solid: As we mentioned early in this Chapter that a solid in (a) deforms into a new shape in (c) – but not indefinitely. ● It stopped further deformation after deformed by certain amount. ● It reaches a new state of equilibrium. WHY???

Imagine the following phenomenon: Stretch a free-hung spring by a weight W. The spring will elongate, but only by a finite amount. Ask yourself: WHY? Answer: the elongation of the spring develops a “resistance,” which increases as the spring elongates. The spring ceases further elongation when the “resistance” in the spring balances the applied weight (force). We say the spring – and the applied weight reaches a new state of equilibrium, which stop the spring From further elongation. Next: what will happen to the spring after the weight is removed? You will say that the spring returns to its original length, but WHY?? Answer: because a form of ENERGY was stored in the stretched spring when it is elongated. This ENBERGY is released to restore the spring to its original shape after the external force (the weight) was removed.

Now, you know why the solid in (a) ceases to deform further after the application of the system of external forces {p} has been applied to the solid in (b). And you would know that there is such ENERGY associated with the solid deformation developed in the solid called “STRAIN ENERGY.” which is responsible for restoring the solid to its original shape after the applied forces are removed in “ELASTIC” solids. Mathematical expression of strain energy in State (c) is: { } { }dvU

T

vσε∫=

21 (3.9) textbook

Derivation of Element Equations-Cont’d Potential energy in a deformed solid subjected to external forces:

P = U - W The potential energy in a deformed solid is:

{ } { } { }

( )dv

dvdvU

v xzxzyzyzxyxyzzzzyyyyxxxx

v

xz

yz

xy

zz

yy

xx

xzyzxyzzyyxx

T

v

∫

∫∫

+++++=

==

σεσεσεσεσεσε

σσσσσσ

εεεεεεσε

21

21

21

Strain energy:

Both the strain and stress components are function of (x,y,z), and dv = (dx)(dy)(dz) = the volume of given points in the deformed solid. Strain energy is a scalar quantity.

(4.8)

(4.9)

Work done to deform the solid: Definition of “work”: Work (W) = Force x Displacement (deformation)

Two kinds of forces: (1) body forces (uniformly distributed throughout the volume of the solid (v)), e.g., the weight, (2) surface tractions, e.g., the pressure or concentrated forces acting on the boundary surface (s)

Mathematical expression of work:

{ } { } ( ){ } { }

( ) ( ) ( ){ }

( ) ( ) ( ){ } dsttt

zyxzyxzyx

dvfff

zyxzyxzyx

dstzyxdvfzyxW

z

y

x

s zyx

z

y

x

v zyx

T

s

T

v

+

=

+=

∫

∫

∫∫

,,,,,,

,,,,,,

,,),,(

φφφ

φφφ

φφ

Derivation of Element Equations – cont’d Potential energy in a deformed solid subjected to external forces:

where {Φ(x,y,z)} = the displacement of the solid at (x,y,z), {f} = body forces, and {t} = the surface tractions , and ds = the part of the surface boundary on which the surface tractions apply

(4.10)

Derivation of Element Equations – cont’d Potential energy in a deformed solid subjected to external forces:

So, the potential energy stored in a deformed solid is: P = U – W, or:

( ){ } ( ){ } ( ){ } { } ( ){ } { }( )∫ ∫∫ +−=−=v s

TTT

vdstzyxdvfzyxdvzyxzyxWUP ,,,,,,,,

21 φφσε (4.11)

( ){ } 0=∂∂φφP

Following the Rayleigh-Ritz Variational principle, the equilibrium condition for the deformed solid should satisfy the following conditions:

( ) ( ) ( ) ..............,0,0,0321

=∂∂

=∂∂

=∂∂

φφ

φφ

φφ PPP

From which, equations for each element may be derived from:

The element displacement is: {Φ(x,y,z)} with three components: Φx(x,y,z) = the element displacement component along the x-direction Φy (x,y,z) = the element displacement component along the y-direction, and Φz(x,y,z) = the element displacement component along the z-direction

Derivation of Element Equations – cont’d for FE mesh of discretized solids

What we had formulated was for continuum solids. We will now derive the ELEMENT EQUATION for discretized solids in FE mesh:

We need to make distinction between the ELEMENT quantities and the NODAL quantities. The primary quantity in FE analysis is DISPLACEMENTS. We need teo make distinction between the Element displacements and the Nodal displacements.

{ } { }TzyxzyxzyxzyxT

444333222111 φφφφφφφφφφφφφ =

We notice that tetrahedron elements has four (4) associate nodes: Φ1, Φ2, Φ3, and Φ4 with FIXED (specified) COORDINATES. Each node has three (3) displacement components too. These nodal displacement components are:

Derivation of Element Equations – cont’d for FE mesh of discretized 3-D solids with tetrahedron elements

We realize that TETRAHEDRON and HEXAHEDRN elements are used in FE models for general 3-D solid structures. The tetrahedron elements are the “basic elements” for this type of structures, because hexahedron elements are made up by 4 or more tetrahedron elements. Our FE formulation for general 3-D solid structures will thus be based on TETRAHEDRON elements

where Φ1x, Φ1y, Φ1z = displacements in Node 1 in 3 directions; Φ2x, Φ2y, Φ2z = displacements in Node 2 in 3 directions; Φ3x, Φ3y, Φ3z = displacements in Node 3 in 3 directions; Φ4x, Φ4y, Φ4z = displacements in Node 4 in 3 directions

Derivation of Element Equations – cont’d

( ){ } ( ){ } ( ){ } { } ( ){ } { }( )∫ ∫∫ +−=−=v s

TTT


21 φφσε

We mentioned previously that the functional that we will use to derive the element equations in FE formulation of solid structures Is the potential function in the solid as show below:

Now because the ELEMENTS in discretized solid) Also, these elements are interconnected by the NODES to simulate the original solid structures. This “link” requires the FE formulation involves the functional with both the ELEMENT and NODAL quantities in the formulation.

(4.11)

This “link” is established using the INTERPOLATION FUNCTION that relates the element quantities with the corresponding nodal quantities such s:

( ){ }( )( )( )

( )[ ]{ }φφφφ

φ zyxNzyxzyxzyx

zyx

z

y

x

,,,,,,,,

,, =

=

The 3 element displacements: The 12 nodal displacements:

{ } { }TzyxzyxzyxzyxT

444333222111 φφφφφφφφφφφφφ =

The INTEERPOLATION function

for FE mesh of discretized 3-D solids with tetrahedron elements

Derivation of Element Equations – cont’d Key equations to construct the functional - the potential energy

{ε(x,y,z)} = [D]{Φ(x,y,z)} (4.3) in which [D] in Equation (4.4) Hence {ε} = [D][N(x,y,z)]{Φ}= [B(x,y,z)]{Φ} (4.12) with [B(x,y,z)] = [D][N(x,y,z)] (4.13)

1. The element displacements vs. nodal displacements via Interpolation function: {Φ(x,y,z)} = [N(x,y,z)] {Φ}

2. Element strain vs. nodal displacements:

3. Element stresses vs. nodal displacements: {σ} = [C]{ε} (4.6) in which the elasticity matrix [C] in Equation (4.7) Hence {σ} = [C] [B(x,y,z)]{Φ} (4.14)

4. Strain energy with nodal displacements: { } { }dvU

T

vσε∫=

21 (4.9)

Hence ( )[ ]{ }( ) [ ] ( ){ }[ ]( )dvzyxBCzyxBUv

T φφ ,,,,21∫= (4.15)

{ } [ ] [ ][ ]{ }dvzyxBCzyxBU TT

vφφ ),,(),,(

21∫=or (4.16)

Typical tetrahedron element for 3-D FE models

Derivation of Element Equations – cont’d The functional for Variational process

( ){ } ( ){ } ( ){ } { } ( ){ } { }( )∫ ∫∫ +−=−=v s

TTT


21 φφσε

We mentioned that the functional for deriving the element equations for discretized solid structure is the POTENTIAL ENERGY (P) as shown below:

(4.11)

By substituting the Strain energy expressed in Equation (4.16) into the above equation, we get:

{ } [ ] [ ][ ]{ }

{ } [ ] { } { } [ ] { }dstzyxNdvfzyxN

dvzyxBCzyxBP

TT

s

TT

v

v

TT

),,(),,(

),,(),,(21)(

∫∫

∫−−

=

φφ

φφφ

(4.17)

{ } { } [ ] [ ][ ] { }

{ } [ ] { }( ) { } [ ] { }

−−

=

∫∫

∫

dstzyxNdvfzyxN

dvzyxBCzyxBP

T

s

TT

v

T

v

TTT

),,(),,(

),,(),,(21)(

φφ

φφφφ

Due to the fact that nodal displacement {Φ} have “fixed value” but not a function og (x,y,z), so they can be factore Out of the integration with respect to (x,y,z). We thus have the following:

(4.18)

Derivation of Element Equations – cont’d Element equation by Variational process

All elements in discretized solids subjected to external force require to satisfy the following condition that:

( ){ } 02

1

=

•••∂∂∂∂

=∂∂ φ

φ

φφ

P

P

P(4.19)

By substituting the potential energy P in Equation (4.18) into the above equation:

( ) { } { } [ ] [ ][ ] { }

{ } [ ] { }( ) { } [ ] { }0

),,(),,(

),,(),,(21

=

−−

∂∂

=∂

∂

∫∫

∫

dstzyxNdvfzyxN

dvzyxBCzyxBPT

s

TT

v

T

v

TTT

φφ

φφφ

φφφ

Derivation of Element Equations – cont’d Element equation by Variational process

[ ] [ ][ ]( ){ }

[ ] { }( ) [ ] { } 0),,(),,(

),,(),,(

=

−− ∫∫

∫dstzyxNdvfzyxN

dvzyxBCzyxB

T

s

T

v

v

T φ

The above variation results in:

Upon moving the last two items to the right-hand side:

[ ] [ ][ ]( ){ }

[ ] { }( ) [ ] { }( )dstzyxNdvfzyxN

dvzyxBCzyxBT

s

T

v

v

T

),,(),,(

),,(),,(

∫∫∫

+=

φ

(4.20)

We may represent Equation by the following element equation:

[Ke] {Φ} = {q} (4.21)

where [ ] [ ] [ ] ( )[ ]dvzyxBCzyxBKv

Te ,,),,(∫= = Element stiffness matrix (4.22)

{ } scopmponentntdisplacemeNodal=φ

{ } ( )[ ] { } ( )[ ] { } matrixforceNodaldstzyxNdvfzyxNqT

sv

T =+= ∫∫ ,,,, (4.23)

{f} = Body forces

{t} = Surface tractions

[N(x,y,z)] in Step 3, Chapter 3, [B(x,y,z)] in Equation (4.13), [C] in Equation (4.7)

Examples of FE Stress Analysis of Solid Structures

NOTE: In FE stress analysis of solid structures, it is customer to represent the element displacements by:

{ }( )( )( )

=

zyxUzyxUzyxU

zyxU

z

y

x

,,,,,,

),,(

and nodal displacements by: {u}.

The relationship between the element displacements and the nodal displacements are:

{U} = [N(x,y,z)] {u}

where [N(x,y,z)] = the Interpolation function. It is a row matrix for 1-D bar elements, rectangular matrices for 2- or 3-D elements.

● Interpolation function enables the determination of the primary quantities in the element with specified coordinate (x,y,z) with the same primary quantities of the associate nodes

Part 3

Finite Element Formulation for One-dimensional Bar elements

1. Bar elements subjected to unidirectional load

2. Truss bar elements

3. Beam bending elements

We will need first to derive the [h] matrix in Equation (1.13) by the following computations:

x1 = 0 x2 = L

L ● ●

Φ1 Φ2

( ) xx 21 ααφ +=

x

Figure 1.6 Linear Interpolation Function of a Bar Element

We have: 1211 xααφ += and 2212 xααφ +=From which we will have

{ } [ ]{ }aAxx

=

=

=2

1

2

1

2

1

11

αα

φφ

φ

with the matrix [A] to be:

[ ]

=

2

1

11

xx

A and 122

1

11

xxxx

A −==

The inverse matrix of [A] is: [ ] [ ]hxx

xxA =

−

−−

=−

111 12

12

1

By following Equation (1.15), we have the interpolation function of a bar element to be:

( ) { } [ ] { } ( ) ( ){ }xxxxL

xxxx

xhRxN T +−=

−−

== 1212

12

111

11

For the present case with x1 = 0, and x2 = L, we have the interpolation function to be:

( )

−=

Lx

LxxN 1 (1.7)

One-Dimensional Stress Analysis of Bar Elements

Derive the interpolation function using general formulation:

● ● u1 u2 Node 1 Node 2

x=x1 x=x2

L x

Derive the element equation for a typical 1-D bar element: a bar element made of one material with Young’s modulus E. The bar is subjected two forces F1 and F2 as shown in the figure below. Establish the element equation

F2 F1

Because the bar is made of one material, and the applied forces are along the length of the bar. So the bar will only deform along the length of the bar e.g., the element displacement U is a function of coordinate x only:

We thus have the element displacement to be U = U(x). The corresponding displacements at the two nodes are: {u}T = {u1 u2}.

The relationship between the element displacement and the corresponding nodal displacements are:

{U(x)} = {N(x)} {u} (4.24)

in which {N(x)} is the interpolation function for one-D bar element. It is a row matrix with two columns We had derived this interpolation function before with assumption that the element displacement {U(x)} follows a linear polynomial function: U(x)= α1 + α2x, and found the interpolation function using this liner function to be In the same form as in Equation (1.7) we derived using general formulation method:

( ){ }

−=

Lx

LxxN 1

with L = x2 – x1

(4.25)

Because the bar element is subjected to force along the x-coordinate only, and both the induced stress and strain are in the direction of x-coordinate only. We thus obtain from Equation (4.4) and obtain:

[ ]dxd

xD =

∂∂

=

From Equation (4.13), we thus have the [B] matrix to be:

[ ] [ ]{ } { }111111)()( −=

−=

−==

LLLLx

Lx

dxdxNDxB T

For the same reason of being uniaxial stress distribution, i.e., σxx = Eεxx by uniaxial Hooke’s law, we have the elasticity matrix [C] = E – the Young’s modulus from Equation (4.7) with Poisson’s ratio γ = 0. We thus has the element equation of the deformed bar expressed in the form:

[Ke] {Φ} = {q} (4.26)

with [ ] [ ] [ ] ( )[ ] ( )[ ] ( )[ ]

{ } ( ) { } { }

( ) ( )( ) ( )

−

−=

−−−−−−

=

−

−=

−

−

=

−

−=

==

∫

∫∫

∫∫

1111

1111

1111

111111

)(,,),,(

1212

121222

2

2

1

2

1

2

1

2

1

LEA

xxxxxxxx

LEAdx

LEA

dxLEAAdx

LE

L

AdxxBExBdvzyxBCzyxBK

x

x

x

x

x

x

T

Tx

xv

Te

(4.27)

Example 4-1 Show the element equation for the bar element in the figure, and determine displacement at Node 2

The nodal forces {q} for the bar element are expressed as: {q}T = {-F1 F2}

We may thus express the element equation for the bar element to be:

−

=

−

−

2

1

2

1

1111

FF

uu

LEA (a)

Because there is only one bar element in the structure, the overall stiffness equation in Equation (1.28) is identical to that of element equation in Equation (4.24). However, adjustments of the now overall stiffness equation with the three matrices in Equation (4.24) would be required, as expressed:

=

b

a

b

a

bbba

abaa

RR

qq

KKKK

where {qa} = specified (known) nodal quantities; {Rb} = specified (known) applied resulting actions, from which we may obtain: {qb} = [Kbb]-1 ({Rb} – [Kba]{qa}) (6.12) (PRINCIPAL REF)

The specified nodal unknown {qa} in this case is u1 = 0 and F1=0, the required unknown {qb} is u2, We thus need no adjustment of the Matrices in the overall stiffness equation in (4.24). The required unknown quantity u2 is obtained from Equation (6.12, REF) to be:

( ) ( )( )( ) 222 0111 F

LEAF

LEAu =−−=

● ● u1 u2 Node 1 Node 2

x=x1 x=x2

L x

F2 F1

Example 4-2 Deformation and stress in a compound bar made of two different materials Use the FEM to determine the displacements at the joint of a compound road made of copper and aluminum induced by a uniaxial force P = 30000 N at of the end of the rod as shown in the Figure A below:

The compound rod has a cross-sectional area A = 650 mm2 and the Young’s moduli Ecu = 10300 MPa and Eal =69000 MPa.

Solution:

The situation shown in Figure A indicates that the rod is expected to elongate along the same direction in the x-axis, as shown in Figure B.

Figure A Compound Rod subjected to a Uniaxial Force

Figure B FE model for a Compound Bar

Example 4-2 – Cont’d

The FE model in Figure B indicates the following: (1) There are total 3 nodes in the structure (2) Nodal coordinates: Node 1 at x1 = 0; Node 2 at x2 = 915 mm, and Node 3 at x3 = 1220 mm (3) The length of Element 1 = L1 = 915 mm; the length of Element 2 = L2 = 305 mm (4) Both elements have a cross-sectional area: A1 =A2 = A = 650 mm2

(5) Displacements at the 3 nodes are: {u}T = {u1 u2 u3} with u1 = 0

Our solution begins with developing the “element equations” for both elements in the FE model:

Element 1 made of copper:

[ ] mNxx

xLAEKe /10

317.7317.7317.7317.7

1111

1091565010300

1111 6

31

111

−

−=

−

−=

−

−= −

=

−−

2

1

2

166

66

10317.710317.710317.710317.7

pp

uu

xxxx

Coefficient matrix for Element 1:

Element equation for Element 1:

where p1 and p2 are forces art Node 1 and 2 respectively

(a)

Element 2 made of aluminum:

[ ] mNxx

xL

AEKe /1005.14705.14705.14705.147

1111

1030565069000

1111 6

32

222

−

−=

−

−=

−

−= −


Coefficient matrix for element 2:


=

−−

3

2

3

266

66

1005.1471005.1471005.1471005.147

pp

uu

xxxx

=

−−

2

1

2

166

66

10317.710317.710317.710317.7

pp

uu

xxxx


Due to the fact that the present case involves TWO elements with Node 2 being common to both these element, we need to assemble the coefficient matrices by following the established rule by summing up the values of Node 2 from both element coefficient matrices.

(b)


We thus assemble the overall stiffness matrix by adding [ ]1eK In Equation (a) and [ ]2

eK in Equation (b) in the following way:

[ ] ( )

−

−=

−+

−=

05.1470005.147367.1540

0317.7317.710

05.1470005.1470

0317.7317.710 66 147.057.317K (c)

The numbers in boldface in Equation (c) are those associated with Node 2.

The overall stiffness equation for the bar structure with specified loading/boundary conditions is thus expressed in the following partitioned matrices as:

===

=

=

−

−

30000000

05.1470005.147367.1540

0317.7317.710

3

2

1

3

2

16

ppp

uu

u(d)

The two unknown nodal displacements u2 and u3 may be obtained by the following equations using the partitions in Equation (d):

=

−30000

005.147005.147367.154

103

26

uu

resulting in the displacement of the compound bar at the joint (Node 2) to be u2 = 1.94 mm and the displacement at the free end u3 = 2.04 mm. The total elongation of the rod is 3.98 mm.


Strain in Elements

Now that we have solved the displacements at the 3 nodes, we may use the train-displacement relations to determine the induced strains in both these elements:

{ } { }21xxxx

T εεε =

where lyrespectiveandElementinstrainstheareand xxxx 2121 εε

The strain-displacement relationship is available from the expression: {ε} = [B]{u} in Equation (4.3), in which the matrix {B} is:

( )[ ] ( ){ }

−−

−=

−−

−−−

==212121

1

21

2 11xxxxxx

xxxxxx

dxdxN

dxdxB

We have: Node 1 at x1 = 0; Node 2 at x2 = 915 mm, and Node 3 at x3 = 1220 mm, leading to: the length of Element 1 = L1 = 915 mm; the length of Element 2 = L2 = 305 mm. We may thus express the [B(x)] for both elements to be:

[ ] [ ]

−=

−=

3051

3051,

9151

9151

21 BandB


From the [B] matrices for both elements, we may compute the strains in Element 1 and 2 as follows:

%21.0915

94.191510

9151

9151

22

11 ===

=

−= u

uu

xxε

%033.0305

1.0305

04.2305

94.1305305305

13051 32

3

22 ==+−=+−=

−=

uuuu

xxε

for element 1, and

for element 2

Stresses in elements

We may use the Hooke’s Law to determine the stresses in each of these two elements from their corresponding strains

For element 1 with Ecu = 10300 MPa:

[ ] MPaxEC xxcuxxxx 03.10021.010300111

1 ==== εεσ

For element 2 with Eal = 69000 MPa:

[ ] MPaxEC xxalxxxx 77.2200033.069000222

2 ==== εεσ

Finite Element Formulation of Elastic Solid Structures

One-Dimensional Bar Elements for Truss Structures

FE Formulation of Truss Elements Using 1-D Bar Elements

Common Trusses

The characteristics of a truss element can be summarized as follows (a quote from Dr. Agarwal’s lecture notes): 1. Truss is a slender member (length is much larger than the cross-section). 2. It is a two-force member i.e. it can only support an axial load and cannot support a bending load. Members are joined by pins (no translation at the constrained node, but free to rotate in any direction). Being a two-force member, the force acting in the member is in the length-direction only. 3. The cross-sectional dimensions and elastic properties of each member are constant along its length. 4. The element may interconnect in a 2-D or 3-D configuration in space. We will formulate 2-D configuration only. Meaning the truss member deforms in a plane defined by x-y coordinates. 5. The bar elements for truss structures is mechanically equivalent to a spring, since it has no stiffness against applied loads except those acting along the axis (the length direction) of the member. 6. However, unlike a spring element, discussed in previous chapters, a truss element can be oriented in any direction in a plane, and the same element is capable of tension as well as compression along its longitudinal directions.

FE Formulation of Truss Elements Using 1-D Bar Elements - Cont’d

Simple Bridge Structure:

FE Formulation of Truss Elements Using 1-D Bar Elements – Cont’d

ο

ο ο ο

ο ο ο ο ο ο

ο

Node 1

2

3

4

5

6 7

Element (1)

(2)

(3)

(4)

(5)

(6)

(7)

(8) (9)

(10)

(11)

FE model with 1-D bar elements with designated element and node numbers. All elements are interconnected by “frictionless hinges”:

P1 P2 P3


A fundamental phenomenon in Statics is that “hinged” joint of any member in A frame or truss cannot resist moment. Consequently, any member in s truss As shown in the figure at the left with hinge joints (as illustrated in the lower Figure will have forces acting along the length of the member.

Consequently, the derivation of the interpolation function in the FE formulation of these members may begin with the bar element in Example 4-1

x

y Truss members along x-direction of y-direction only: The or these element equations for these elements will be in similar forms as for the 1-D bar elements in Examples 4-1 and 4-2.

The element equation for those truss members that are inclined with the x-coordinate with angle θ, However, will be in different forms, as will be derived in the following procedures.

Inclined truss bar element in x-y plane: Element (1), (5), (7) and (11):

The inclined truss bar elements such as shown in the figure at right may have TWO displacement components, but these elements can only elongate or contract in the LONGITUDINAL direction only. The same applies to the induced strain and stress. So, These are regarded as “a special bar elements.”

i

j

vj

uj

Fj

ui

vi

Fi (xi,yi)

(xj,yj)

x

y

0

These bar elements remain having two nodes: Node i and Node j Located at specified coordinates (xi, yi) and (xj, yj) respectively. Each node has two displacement components: ui and vi = displacement of Node i in respective x- and y- directions uj and vj = displacement of Node j in respective x- and y-directions Fi and Fj are the forces acting at Node I and Node j respectively

FE Formulation of Inclined Truss Elements Using 1-D Bar Elements


Derivation of element equation

The situation in the bar element in truss structures is unique in the way that these bar elements are solid element of “two-force members” – meaning that the force is acting along the length of the bar only. Consequently, the induced displacements that ae effective to the elongation or contraction of the bar (truss) member elements are the ones in the length-direction only. The strain and stress in the truss members are along the length of the bar too. The displacement components in an inclined bar element in the plane defined by the (x,y) coordinate system in the figure below do not contribute in producing the displacement, strain and stress in the bar element for this reason . However their “equivalence” to the ones along the length direction of the bar element do. Consequently, we need to convert the current situation in the inclines bar element in the left to the equivalent situation in the right of the figure through a coordinate transformation process of the nodal displacement components and the applied nodal forces

Equivalent Typical Bar Element

Derivation of element equation – Cont’d


Transformation of Nodal displacement components: From the “global coordinates” to “Local coordinates”- The latter is used for the FE formulation as with 1-D bar elements

By referring to the above figure, we have the following relationship in transforming the nodal displacement components from The global coordinates (x,y) in the left figure to the local coordinate (x) in the right of the figure:

iiiii svcuvu +=+= θθδ sincosat Node i: (4.28a)

at Node j: jjjjj svcuvu +=+= θθδ sincos (4.28b)

where c = cosθ and s = sinθ

Equations (4.28a,b) may be expressed in matrix form: { }

=

=

j

j

i

i

j

i

vuvu

scsc00

00δδ

δ (4.29a)

or in a shorthand version of: {δ} = [T]{u} (4.29b) with the matrix [T] being the transformation matrix

Converting “truss bar element”

to equivalent “typical bar element”

Truss bar element: Typical bar element:



From “global coordinates” to “Local coordinates”

Transformation of nodal forces can be done in a similar way as we did with nodal displacements. However, it would be easier to do it using the work done to the element by these forces, as work done is a scalar quantity, which is easier to transform in space than the vector quantities such as displacements. Since the work done to the element by nodal forces may be expressed as: for the same element in both local and global coordinate systems, or in a long-hand form:

{ } { } { } { }FufW TT == δ

{ } { } { } { } { } { }Fufor

FFFF

vuvuff

W TT

jy

jx

iy

ix

jjii =

=

= δδδ2

121

Substituting {δ} = [T]{u} in Equation (26b) into the above expression, we get: ([T]{u}])T{f} = {u}T{F}, leading to:{u}T[T]T{f} = {u}T{F} We will thus have the nodal force transformation by the following relationship:

[T]T {f} = {F} (4.30) where {f} = nodal forces in “typical bar element”, {F} = nodal forces in “inclined truss element”

Transformation of Nodal force components:



Element equation in Local Coordinate System

Now that we have made the real situation in the “Global” coordinates (x,y) to be equal to the situation in 1-D “Local” coordinate (x) situation through a “transformation” process:

= [ ]

=

scsc

T00

00

Transformation Matrix:

(4.26a)

( ){ }

−=

Lx

LxxN 1

−

=

−

−

j

i

j

i

ff

LEA

δδ

1111

(4.24)

Real situation in “Global” coordinates

Previously derived interpolation function:

Previously derived “element equation” with modified notation of nodal quantities:

[ ]{ } { }fKe =δ [ ]

−

−=

LAE

LAE

LAE

LAE

Keor with

Coordinate Transform:

(4.27)

Inclined truss bar element: Equivalent bar element:




−

=

−

−

j

i

j

i

ff

LEA

δδ

1111

[ ]{ } { }fKe =δ [ ]

−

−=

LAE

LAE

LAE

LAE

Kewith

Relationship of nodal forces between the two coordinate systems: {F} in global coordinate system = [T]T{f} in Local coordinate system . The element equation in global coordinate system thus have the form:

[ ]{ } { }fKe =δ

The nodal forces in the Local coordinate system {f} can be viewed as a transformed forces {F} rom the global coordinate systems with the relationship: [T]T {f} = {F} as shown in Equation (4.27), or {f} = ([T]T)-1, leading to:

(4.31)

[Ke]{δ} = ([T]T)-1 {F}

But we also have already derived the following relationships: {δ} = [T]{u} in Equation (4/26b), and [T]T = {F} in Equation (4.27). By substituting these relationships into the above expression, we will get the following “reversed transformation” of element equation from the Local coordinate system to Global coordinate systems:

[Ke][T]{u} = ([T]T)-1 {F} (4.32)

(4.24)

(4.27)




[Ke][T]{u} = ([T]T)-1 {F} The Element Equation of inclined truss elements is: (4.33)

We realize the fact that the transformation matrix: [T]T = [T]-1 (Reference No. 2). Thus Equation (4.33) has a new form of:

[Ke][T]{u} = ([T]-1)-1 {F} leads to: [Ke][T]{u} = [T]{F} with ([T]-1)-1 = [T] in the last expression. Now, if we multiply both sides by [T]-1, and use the relationship of [T]-1 = [T]T, we have the element equation of the inclined truss element (i.e., the bar element in the global coordinate system) to be:

[T]T[Ke][T]{u} = {F} (4.34)

[ ]

−

−=

LAE

LAE

LAE

LAE

Kewhere and

=

scsc

T00

00][

with A, L= cross-sectional area and the length of the truss element respectively, c=cosθ and s=sinθ

Element equation for Inclined Truss Members


[ ]{ } { }FuKe =

{ }

=

j

j

i

i

vuvu

u { }

=

jy

jx

iy

ix

FFFF

F

[ ]

−−−−

−−−−

=

22

22

22

22

scsscscsccscscsscscsccsc

LAEKe

where and

The stiffness matrix:

1) This stiffness matrix represents the stiffness of a single element with incline angle θ 2) It is symmetric about the diagonal line of the square matrix 3) Since there 4 unknown nodal displacements - meaning 4 degree-of-freedom (dof), the matrix is of the size

of (4x4) 4) The terms c and s represent the cosine and sine values of the orientation of the element with the

horizontal plane, rotated in a counter clockwise direction (+ve direction)

(4.35)

(4.36a) (4.36b)

(4.37)

Example 4-3 FE Stress Analysis of a Truss Structure:

A truss with 3 members joined by frictionless hinges as shown in the figure. Element 1 and 2

are made of aluminum, and element is made of steel. 3

Materials Cross-sectional Area (A) (10-6m2)

Young’s Modulus (E) (106 N/m2)

Yield Strength (σy) (106 N/m2)

Member (1) Aluminum 200 70,000 170

Member (2) Aluminum 200 70,000 170

Member (3) Steel 100 200,000 21,000

We realize the fact that there are 3 members, each has its own dimension, and material with properties listed in the above table. So, we may conveniently construct the FE model of the truss structure by using 3 elements with node pair 1-3, element with node pair 2-1, and element with node pair 2-3, as shown in the figure. We will derive the element equations for all these 3 elements by using the formulations for truss elements with: Equation (4-24) for the horizontal Element 1, and Equations (4.35) and (4.37) for the inclined elements 2 and 3.

Solution:

1 2 3

Derivation of element equations

Example 4-3 FE Stress Analysis of a Truss Structure - cont’d

Element 1:

Node 1 Node 3

L1 = 0.026 m

v1y v3y

u1x u3x (1)

L1 = 0.26 m, E1 = 70000 MPa, A1 =200x10-6 m2:

Being horizontal, we have the incline angle θ = 0, which leads to: c = cosθ=1, c2 = 1, s=sinθ = 0, s2 = 0 and cs = 0 We also calculate the coefficient of the stiffness matrix of element 1 as follows:

( )( ) mNLAE 6

66

1

11 1084.5326.0

102001070000×=

××=

−

Using Equation (4.37), the stiffness matrix for Element 1 is thus:

[ ]

−

−

×=

−−−−

−−−−

×=

0000010100000101

1084.53 6

22

22

22

22

1

111


LAEKe (a)



Element 2: L2 = 0.15 m, E2 = 70000 MPa, A2 =200x10-6 m2: L 2=

150

mm

Node 1

Node 2 u2x

u1x

v1y

v2y

(2)

In this Element 2, we have the inclined angle θ=90, which lead to: c = cos 90o = 0, c2 = 0; s = sin 90o = cos0o = 1, s2 = 1 and cs = 0 , and

( )( ) mNL

AE 666

2

22 1006.3115.0

102001070000×=

××=

−

Hence the stiffness matrix for Element 2 is:

[ ]

−

−×=

101000001010

0000

1006.31 62eK (b)



Element 3: L3 = 0.3 m, E3 = 200,000 MPa, A3 =100x10-6 m2:

Node 2

Node 3

X

y u3x

v3y

u2x

v2y

Element 3 has an inclined angle θ = 30o leading to:

c= cos 30o = 0.866, c2 = 0.75, s = sin 30o = 0.5, s2 = 0.25 and cs = 0.433; Also the coefficient of the stiffness matrix:

( )( ) mNLAE 6

66

3

33 1067.663.0

1010010200000×=

××=

−

The stiffness matrix for element 3 is:

[ ]

−−−−

−−−−

×=

−−−−

−−−−

×=

25.0433.025.0433.0433.075.0433.075.0

25.0433.025.0433.0433.075.0433.075.0

1067.66 6

22

22

22

22

3

333


LAEKe (c)



The 3 Element equations for the truss structure:

[Ke] {Φ} = {q} From Equation (4.21): with a general expression for element equations, we are now in the position to express the

same equations for the 3 elements in the current truss structure as follows.

Element (1) with Node 1 and 3

=

−

−

y

x

y

x

y

x

y

x

ffff

vuvu

3

3

1

1

3

3

1

1

6

0000084.53084.530000084.53084.53

10


=

−

−

y

x

y

x

y

x

y

x

ffff

vuvu

2

2

1

1

2

2

1

1

6

6.3106.3100000

6.3106.3100000

10


=

−−−−

−−−−

y

x

y

x

y

x

y

x

ffff

vuvu

3

3

2

2

3

3

2

2

6

67.1687.2867.1687.2887.285087.285067.1687.2867.1687.2886.285087.2850

10

in which u1x, u2x and u3x = displacement components in x-direction of Node 1, 2 and 3 respectively v1y, vy2 and v3y = displacement components in y-direction of Node 1, 2 and 3 respectively f1x, f2x and f3x = Nodal force at Node 1, 2 and 3 respectively f1y, f2y and f3y = Nodal force at Node 1, 2 and 3 respectively

Assembly of element equations for Overall Stiffness Equation


By following the description on “assembly of element stiffness matrices” in Step 5 with diagram of Chapter 3 Steps in Finite Element Analysis, we may assemble the three (3) truss element stiffness matrices shown above in the following form:

or in a neat form as shown in the next slide:

With this overall stiffness matrix, we may establish the overall stiffness equation for the truss structure as shown below:

Element stiffness matrix and Overall Stiffness Equation of the Truss Structure


The overall stiffness matrix of the truss structure:

(d)

(e)

Boundary and Loading conditions of the Truss Structure


We recognize the following specified conditions for the discretized truss structures:

The boundary conditions (displacement at nodes):

1) With Node 1 being completely fixed: u1x = v1y = 0 2) Node 2 is allowed to move in vertical (y) direction: u2x = 0

The loading conditions: All except Node 3 has one force acting in the y-direction: f1x = f1y = f2x = f2y = f3x = 0 and f3y = 0.4 kN = 400 N

The overall stiffness equation in Equation (e) after the substitution of the above specified boundary and loading conditions has the form:

−======

=

===

−−−−−

−−−−−

−−

40000000

000

67.1687.2867.1687.280087.2884.10387.285008.5367.1687.2827.4887.286.31087.285087.285000

006.3106.310084.5300084.53

10

3

3

2

2

1

1

3

3

2

2

1

1

6

y

x

y

x

y

x

y

x

y

x

y

x

ffffff

vuv

uvu

Overall Structure Stiffness Equation with given Boundary and Loading Conditions for the Truss Structure


−

=

−−−−−

−−−−−

−−

40000000

000

67.1687.2867.1687.280087.2884.10387.285008.5367.1687.2827.4887.286.31087.285087.285000

006.3106.310084.5300084.53

10

3

3

2

6

y

x

y

vuv

Now, if we follow Step 6 in Chapter 3 on Steps in FE Analysis, we may partition Equation (f) in the following way:

(f)

where {qa} = specified (known) nodal quantities; {Rb} = specified (known) applied resulting actions, from which we may obtain:

{qb} = [Kbb]-1 ({Rb} – [Kba]{qa}) { } { }0=aq

{ }

=

y

x

y

b

vuv

q

3

3

2

{ } { }0=aR

{ }

−=

40000

bR[ ]

−−−

−=

87.28005008.5387.286.310

106baK [ ]

−−

−−=

67.1687.2867.1687.2884.10387.2867.1687.2827.48

106bbK

Solution of 3 unknown nodal displacements in truss structure


We will get the 3 unknown nodal displacements from the portioned over stiffness equation by the expression: {qb} = [Kbb]-1 {Rb}:

−

−−×

×=

−

−−

−−=

−

−

−

−

−

40000

734.142166,31646.32166.386.11036.867

1646.31036.8671646.310

40000

67.1687.2867.1687.2884.10387.2867.1687.2827.48

10 19

19

8

1

6

3

3

2

y

x

y

vuv

Solve for the unknown displacement components at Node 2 and 3: u2y = -12.65x10-6 m, u3x = 12.86x10-6 m and v3y = -58.94x10-6 m with negative signs meaning downward direction

Solution of secondary unknowns of induced strains and stresses in all three elements in the truss structure


We will determine the strain components and then stress components in each of the 3 elements in the truss structure. We should bear in mind that truss members are “two force members,” in which only the in-line displacement components will produce strains and stresses.

Element 1:

ο ο x u3x

u1x L1

X=0 X=L1

with Node 1 and Node 3

We have the relationship between the element displacement U(x) and the corresponding nodal displacement components {u1x u3x}T along the line of the bar element in the figure I the left:

( ) ( )[ ]{ } xxx

x uLxu

Lx

uu

Lx

LxuxNxU 3

11

13

1

11

11 +

−=

−==

where the interpolation function of this simplex bar element is given in Equation (1.5) textbook

We will use Equation (4.3): the element strain {ε} = [B(x)]{u} to compute the only strain, εxx in the element by using Equation (4.12):

( ){ } ( )[ ]{ }uxBx xx == εε in which the matrix [B(x)] is given in Equation (4.13) to be: [B(x)] = [D][N(x)], with [ ]dxd

xD =

∂∂

= for the

present one-dimensional case. We thus have: ( )[ ]

−=

11

1Lx

Lx

dxdxB We thus have the strain εxx in element 1 to be:



We will determine the strain components and then stress components in each of the 3 elements in the truss structure. We should bear in mind that truss members are “two force members,” in which only the in-line displacement components will produce strains and stresses.

Element 1: with Node 1 and Node 3 in global coordinates

ο ο x u3x

u1x L1=0.26 m

X=0 X=L1

= Element 1 in local coordinate

[T]

with θ=0o

The transformation matrix: [ ]

=

01000001

T

The relationship of nodal displacements in Local coordinate system and the corresponding global coordinate systems is: {δ} = [T]{u}, or:

{ }

=

=

j

j

i

i

j

i

vuvu

01000001

δδ

δ

But we obtained the nodal displacements from the previous calculations to be: ui = u1x =0, vi = v1x =0, uj = u3x = 12.86x10—6m, vj = v3y = -58.94x10-6m



{ }

×=

×−×

=

= −

−

− 6

6

6 1086.120

1094.581086.12

00

01000001

j

i

δδ

δ

We have the strain in the element is obtained by using Equation (4.12): {ϵ} = [B(x)]{Φ}, and [B(x)] = [D]{N(x)] (3.13) We thus have: [ ] [ ] ( ){ }

−=

−==

1111

111LLL

xLx

dxdxNDB

The strain in Element 1 is thus:

{ } mmLL j

ixx /1046.49

1086.120

26.01

26.0111 6

611

−− ×=

×

−=

−=

δδ

ε

The corresponding stress in Element 1 can be obtained from Equation (4.6): {σ}=[C]{ϵ}=(70000x106)(49.46x10-6)=3.43 MPa

Solution of secondary unknowns of induced strains and stresses in all three elements in the truss structure - Cont’d Example 4-3 FE Stress Analysis of a Truss Structure - cont’d

Element 2: with Node 1 and Node 2

x

y

Node 1

Node 2

v1

u1

u2

v2

We have computed that u1 = v1 = 0, u2 = 0 and v2 = u2y = -12.65x10-6 m

Let us transform the coordinate system in the lower left figure to one-dimensional bar element using Equation (4.29) to the following situation:

Node j = Node 1 Node i = Node 2

δi δj { }

=

=

j

j

i

i

j

i

vuvu

scsc00

00δδ

δ

We have : c = cos 90o =0, and s = sin 90o = 1.0, leading to:

{ }

====

×−=====

=

=−

00

1065.120

10000010

1

1

622

2

vvuu

uvvuu

j

j

yi

i

j

i

δδ

δ

The above expression leads to: δi = -12x10-6 m and δj = 0 or: { }

×−

=

=−

01012 6

j

i

δδ

δ

We have the strain in the element is obtained by using Equation (4.12): {ϵ} = [B(x)]{Φ}, and [B(x)] = [D]{N(x)] (3.13) We thus have: [ ] [ ] ( ){ }

−=

−==

2222

111LLL

xLx

dxdxNDB

The strain in Element 2 is thus: { } mmLL j

ixx /1080

01012

15.01

15.0111 6

6

22

−−

×=

×−

−=

−=

δδ

ε

The corresponding stress in Element 2 can be obtained from Equation (4.6): {σ}=[C]{ϵ}=(70000x106)(80x10-6)=5.6 MPa



u3j=12.86x10-6

v3j = -58.94x10-6

u2i = 0

v2i = -12.65x10-6

Node 2

Node 3

x

y We obtained the nodal Displacements from The previous calculation to be as shown

ο ο x δj=u3x

δi=u2x

L3=0.3 m

Transformation matrix

[T]

The transformation matrix [T] has the following form with θ = 30o:

[ ]

=

=

5.0866.000005.0866.0

30sin30cos000030sin30cos

oo

o

T



The nodal displacements in the equivalent Element 3 in the Local coordinates can be related to the real Element 3 in the global coordinates by the transformation matrix to be:

{ }( )

66

6

63

63

62

2

103332.18

325.61047.291368.11

10325.6

1094.581086.121065.12

0

5.0866.000005.0866.0 −

−

−

−

−

−

−

=

−×

=

×−==×==×−==

==

=

= x

vvuu

vvuu

jj

jj

ii

ii

j

i

δδ

δ

But since the interpolation function of the equivalent Element 3 in the local coordinate is: ( ){ }

−=

33

1Lx

LxxN (1.7)

And the matrix [B(x)] = [D][N(x,y,z)] from Equation (4.13) in the present case to be: ( )[ ]

−=

−=

3333

111LLL

xLx

dxdxB



The strain {ε} in terms of nodal displacements in Element 3 may be obtained by using Equation (4.12): {ε}=[D]{N(x)}:

{ }

m

LL j

ixx

66

6

33

10194.8210)1107.610833.21(

103332.18

325.63.0

13.0

111

−−

−

×−=×−−=

×

−

−=

−==

δδ

εε

The induced stress in Element 3 is {σ} can be computed by using Equation (4.6), or {σ} = [C]{ε} with [C] = E3 = 200,000x106 Pa (N/m2)

We thus have the stress in element 3 to be: {σ} = σxx = E3εxx = (200000x106)(-82.194x10-6) = - 16438800.33 N/m2 or -16.44 MPa (a compressive stress)

Finite Element Formulation of Elastic Solid Structures

One-Dimensional Bar Elements for Bending of Beams

Quick Review of Simple Beam Theory

1) Like truss members, beams are slender members (length is much larger than the cross-section) in structures. 2) However, unlike the truss members, beam members can resist forces applied laterally or transversely to their axes.

3) Beams can also resist the applied moments that cause the beams to bend, i.e. beams can have deflection induced by applied moments (e.g., torque in the x-y plane) 4) Beams may be straight or curved. We will focus on straight beams only . 5) The application of unilateral forces or moments will not only cause the beam to bend in shapes, but also having deflections, y(x) perpendicular to the beam axes, as shown in the above figure. 6) So, the actions to beam can be: concentrated forces, distributed load and moments The induced reactions are: Deflection of the beam y(x), bending moments M(x) and shear forces V(x), and the bending stresses (normal stress σn(x) – normal to the cross- section of the beam, and the shear stress σs(x) - acting on the cross-section of the beam)

x

y

x

y P – concentrated force

W(x) – distributed load per unit length

Deflected state x

Induced deflection: y(x)

Original straight beam: Induced

deflection by applied forces

Quick Review of Simple Beam Theory – Cont’d

x

y P

a

Construction of induced bending moment and shear force distributions (diagrams)

b L

A B

LPaRb =L

PbRa =

M(x)

x

( ) xL

PbxM =( ) ( )xL

LPaxM −=

Bending Moment Diagram:

x

y

x L

A B

2wLRb =2

wLRa =

V(x)

x Lpb

Lpa

LPab

Uniform distribution load: w N/m

M(x)

x 8

2wL

V(x)

x 2wL

2wL

Shear Force Diagram:


Induced Shear Stresses in Bent Beams by Applied Force

There are two types of stresses induced to the beam subjected to external forces: (1) The normal stress along the x-coordinate. It exists in perpendicular to the cross-section of the beam (σxx), and (2) The shear stress (σxz) that exists on the surfaces of the beam cross-section. It also exists on the face along the x-coordinate (σyx) with the same magnitude

Expression for normal stress (σxx): ( ) ( )I

yxMxxx −=σ

where M(x) = bending moment at location x, y = distance from center of the beam cross-section to the depth of the cross-section in y direction, and I = section moment of inertia

12

3bhI =For beams with rectangular cross-section: b

H

y

z

For beams with circular cross-section: y

z d

64

4dI π=

Section moment of inertia for beams of other cross-sections, including I-cross-sections are available from mechanical engineering handbooks

(4.38)


Induced Shear Stresses in Bent Beams by Applied Force

Expression for shearing (σxz) or (σyx):

These stresses are induced by the shear force V(x) at the various cross-sections along the beam

H

y

z

b

y y1

dA

c1

( ) ( )QIb

xVydAIb

xV c

yyx ∫ == 1

1

σ

For beams with rectangular cross-sections, dA = bdy, resulting in:

−=== ∫ 2

1

2222

4221

1

yhbbybydyQh

y

h

y

where Q=shear moment

We have ( ) ( )

−= 2

1

2

42yh

IxVxyxσ with ( )

bhV

yx3

max=σ at the center line of the beam where y1=0

(4.39)


Euler-Bernoulli Theory of Beam Bending

This theory relates the induced deflection (deformation of beams) and the applied forces

x

y

x

y P – concentrated force w(x) – distributed load per unit length

Deflected state x

Induced deflection: v(x)

Original straight beam: Induced

deflection by applied forces Applied

Forces

Induced Deflection v(x)

& M(x), and V(x)

( ) ( )EI

xMdx

xvd=2

2

Deflection and Bending Moment M(x) and Shear force V(x) relations:

Induced Deflection v(x) may e obtained by solving the 4th order differential equation:

( ) 04

4

=dx

xvdEI

where E = Young’s modulus of the beam material I = Section moment of inertia

( ) ( ) ( ) ( )3

3

2

2

dxxvdEIxVand

dxxvdEIxM == (4.40a,b)

FE Formulation of Beam Elements Principal reference: “A First Course in the FEM” 6th Edition, Cengage Learning by Daryl Logan, 2017

x

y

x L

A B

Uniform distribution load: w N/m

x

y

x L

A B v(x)

ρ

v(x) Datum line

x

ρ = Radius of curvature deflection curve at x Θ = slope of deflection curve at x V(x) = Deflection at x

Beam Element

Node i Node j

L Primary Quantities

In the element: The deflection v(x) At the nodes: The deflections v and slope θ, or as expressed as: { }

=

j

j

i

i

v

v

d

θ

θat Node i and Node j

x=0 x=L

FE Formulation of Beam Elements Principal reference: “A First Course in the FEM” 6th Edition, Cengage Learning by Daryl Logan, 2017

Beam elements

Contrary to the “truss elements,” beam elements deforms from its original straight shape into curved bent shape due to lateral (or transverse) forces or applied couples (moments).

x L

Node i Node j Mi

vi

Vi

θ(x) Mj vj

Vj

X=0 X=L

X

Deformed shape

Original shape

Actions and Induced Reactions in Beam Elements

Actions

Induced Reactions

in the element at Node i at Node j

Lateral forces Vi and Vj

Linear displacement

Linear displacement

Linear displacement

v(x) vi vj

Moments Mi and Mj Rotation Rotation Rotation θ(x) θi θj

FE Formulation of Beam Elements – Cont’d Principal reference: “A First Course in the FEM” 6th Edition, Cengage Learning by Daryl Logan, 2017

Derive Interpolation Function

We assume the traverse displacement of the beam element follows a linear polynomial function o the form:

( ) 432

23

1 axaxaxaxv +++= (4.41)

in which a1, a2, a3 and a4 are constant coefficients

x=0 x=L

By substituting the coordinates of Node I and Node j into the assumed element deflection in Equation (4.38), we get:

( ) 40axvv

xi ===

( )3

0

adx

xdv

xi ==

=

θAt Node i:

At Node j: ( ) 4322

23

1 aLaLaLaxvvLxj +++==

=

( )32

21 23 aLaLa

dxxdv

Lxj ++==

=

θ


Derive Interpolation Function We may thus express the element deflection v(x) in terms of the nodal deflections by substituting the 4 constant coefficients into Equation (3.38) and after re-arranged terms to yield:

( ) ( ) ( )

( ) ( ) iijiji

jiji

vxxL

vvL

xL

vvL

xv

++

+−−−+

++−=

θθθ

θθ

22

323

213

12

The above expression can also be expressed as: ( ) { }

=

j

j

i

i

jjviiv v

v

NNNNxv

θ

θθθ (4.42)

where ( )3233 321 LLxx

LNiv +−= ( )3223

3 21 xLLxLxL

Ni +−=θ

( )LxxL

N jv23

3 321+−= ( )223

3

1 LxLxL

N j −=θ

(4.43)

We thus have: v(x) = [N(x)]{d} with Interpolation function: ( )[ ] { }θθ jjviiv NNNNxN = (4.44)

The element deflections

The nodal deflections


Derive Element Strain εxx – Displacement v(x) Relation

We realize the fact that the rotation (or the slope) of the deflected beam is: ( ) Lxfordx

xdv≤≤= 0θ

-u

X dx

An expanded view of a deformed beam in x-direction:

Undeformed neutral axis

Contracted top edge

Stretched bottom edge

Rotation of the beam section at x

We may find from the above diagram of expanded beam that the stretch u can be obtained by: dx

xdvyyxu )()( −=−= θ

But from theory of elasticity: ( ) ( ) ( )2

2

,dx

xvdydx

xdvdxdyyxor

dxdu

xxxx −=

−== εε (4.45)


Derive Element Stress σxx – strain εxx Relation

We will have the normal stress σxx= Eεxx ( )I

yxM−= (4.38)

( ) ( )QIb

xVxyx =σand the shear stress: (4.39) with Q to be the shear moment

The relationships between the bending moment M(x) and Shear force V(x) are expressed in Equation (4.40a) and (4.40b) respectively:

( ) ( ) ( ) ( )3

3

2

2

dxxvdEIxVand

dxxvdEIxM ==

Derive the Element Stiffness Equations

We may express the applied actions to the beam element in applied forces fiy, fjy and moments mi, mj in terms of the element deflections V(x) represented by the assumed polynomial function in Equation (4. 41) to be:


x L

Node i Node j mi

vi

fiy

θ(x) mj vj

-fjy

X=0 X=L

X

Deformed shape

Original shape

( ) ( )jjiix

iy LvLvLEI

dxxvdEIVf θθ 6126123

03

3

+−+====

( ) ( )jjiix

i LLvLLvLEI

dxxvdEIMm θθ 22

30

2

2

2646 +−+=−=−==

( ) ( )jjiiLx

jy LvLvLEI

dxxvdEIVf θθ 61261233

3

−+−−=−=−==

( ) ( ) ( )jjiiLx

j LLvLLvLEI

dxxvdEIxMm θθ 22

32

2

4626 +−+====

The beam element:

Nodal forces:

Nodal moments:

The above expressions may e express in the following matrix form for the element stiffness equations:

−−−−

−−

=

j

jy

i

iy

j

j

i

i

mfmf

LLLLLL

LLLLLL

LEI

v

v

22

22

3

4626612612

2646612612

θ

θ

Derive the Element Stiffness Equations


(4.46)

We thus have the stiffness equation of a beam element to be:

[ ]

−−−−

−−

=

22

22

3

4626612612

2646612612

LLLLLL

LLLLLL

LEIke

(4.47)

Nodal unknown quantities(transverse

displacements & rotations = Stiffness matrix [Ke] x Applied nodal transverse forces

& moments


Example 4 Determine the deflection of a cantilever beam at the half span and at the point under the load. Dimensions and applied lateral force are shown in the figure. The beam is made of a material with Young’s modulus E = 10000 MPa

ℓ/2=0.5 m

x

ℓ = 1 m

P = 20 N

b = 1 cm

h = 2 cm

Beam Cross-section:

• • Solution: We realize the fact that we are seeking solutions at the mid-span and at the point under the applied force. It id reasonable to discretize the beam into to two (2) elements, as shown below:

• • Node 1 Node 2

Element 1:

v1,, θ1 v2,θ2 L = 0.5 m

• • Node 2 Node 3

Element 2:

v2,, θ2 v3,θ3 L = 0.5 m

Section moment of inertia (I) is: ( )( ) 48

3223

10667.012

1021012

mbhI −−−

×=×

==

EI = (1010)(0.667x10-8) = 66.7

We will first derive the element equations given in Equation (4.46) with element stiffness matrices shown in Equation (4.47) for both Element 1 and 2.

Node 1 • Node 2

Node 3

FE Formulation of Beam Elements – Cont’d

Example 4

[ ]

−−−−

−−

=

−−−−

−−

=

135.033123125.0313

312312

6.533

4626612612

2646612612

211

211

11

211

211

11

31

1

LLLLLL

LLLLLL

LEIke

For Element 1 with L1 = 0.5 m:

Element equation for Element 1 is:

=

−−−−

−−

2

2

1

1

2

2

1

1

135.033123125.0313

312312

6.533

mfmf

v

v

θ

θ

in which v1,θ1, v2, θ2 are the respective deflections and rotations in Node 1 and Node 2 respectively, whereas f1, m1 and f2, m2 are the applied lateral forces and moments at Node 1 and Node 2 respectively

(a)

(b)


Example 4

Due to the fact that Element 2 has the same length and is made of the same material as Element 1, with the only difference of the associated nodes, we can express the same element equation for Element 1 for element 2 as shown below:

=

−−−−

−−

3

3

2

2

3

3

2

2

135.033123125.0313

312312

6.533

mf

mf

v

v

θ

θ(c)

Assembly of element equations in (b) and (c) for Overall stiffness equations:

Because Node 2 happens to be common node shared by Element 1 and 2, we need to assemble the element equation following the “MAP” stipulated in Step 5 in Chapter 3:

0

0

[ ] [ ] [ ]=+= 21ee KKK Sum

Node 2

Applied loads at Node 2 in Element 1 and 2 should be summed up too in the load matrix


Example 4 We thus have the assembled overall stiffness equations as:

( ) ( )( ) ( )

=

−−−−

−++−−+−+−−

−−

3

3

2

2

1

1

3

3

2

2

1

1

135.0300312312005.0311335.03

312331212312005.031300312312

6.533

mf

mfmf

v

v

v

θ

θ

θ

=

−−−−

−−−−

−−

3

3

2

2

1

1

3

3

2

2

1

1

135.0300312312005.03205.03

312024312005.031300312312

6.533

mf

mfmf

v

v

v

θ

θ

θ

The assembled overall stiffness equation for the beam structure is:

(d)

We may solve the six unknown responses at the three nodes in the beam structure from Equation (d)


Example 4 We realize that the following boundary and applied loading conditions apply: v1 and θ1 = 0 for having Node 1 being fixed at the built-in end, and f1, m1, f2,m2 and m3 = 0. the only non-zero load is the applied force f3 = -20 N at Node 3

We thus have the overall stiffness equation of the beam structure with s[ecified boundary and loading conditions take the form:

−

=

−−−−

−−−−

−−

0200000

00

135.0300312312005.03205.03

312024312005.031300312312

6.533

3

3

2

2

θ

θv

v(e)

The four unknown at Node 2 and 3 can be solved from Equation (e) by partitioning the above matrix equations following Step 6 in Chapter 3.


Example 4

−

=

−−−−

−−−−

−−

0200000

00

135.0300312312005.03205.03

312024312005.031300312312

6.533

3

3

2

2

θ

θv

v

=

b

a

b

a

bbba

abaa

RR

qq

KKKK

In the form of:

where {qa} = specified (known) nodal quantities; {Rb} = specified (known) applied resulting actions, from which we may obtain: {qb} = [Kbb]-1 ({Rb} – [Kba]{qa}) (6.12, REF)

In the present case, we have {qa} = {0 0}T We thus have the 4 unknowns obtained by: {qb} = [Kbb]-1 {Rb}


Example 4

−

=

−−−−

−−−−

−−

0200000

00

135.0300312312005.03205.03

312024312005.031300312312

6.533

3

3

2

2

θ

θv

v

{qb} = [Kbb]-1 {Rb}:

{ }

=

3

3

2

2

θ

θv

v

qb[ ]

−−−−

−−

=

135.03312312

3320312024

bbK { }

−=

02000

bR One will find that: [ ]

=−

844146667.238333.0434118333.013333.0

1bbK


Example 4 We may thus solve for the 4 primary unknown quantities in {qb} from the following equations:

( )( )

( )( )

−=

−×−×

−×−×

=

−

=

1499.0099995.01124.003123.0

204206667.2

203208333.0

6.5331

02000

844146667.238333.0434118333.013333.0

6.5331

3

3

2

2

θ

θv

v

From which we obtain the following solutions: Deflections at Node 2, v2 = -0.03123 m = -3.123 cm, at Node 3, v3=-0.09995 m = -9.9995 cm Slope at Node 2, θ2 = -0.1124 rad, at Node 3 θ3=-0.1499 rad.

Given conditions: v1 = 0, θ1 = 0

Check with solutions from classical beam theory:

x y(x)

X

ℓ=1 m

ℓ/2 ( ) ( )323 23

61

+−−= xxEIPxy

The induced deflection in the cantilever beam by the applied force P = 20 N is

One may find the deflection at Point A at x=0 (equivalent to Node 3) = -0.0999 m and the deflection at Point B at x = ℓ/2 (equivalent to Node 2) =-0.03123m. Both these values fully agree with the solutions we obtained from the FE analysis.

A

P

•

X=0

Finite element Formulation for Two-Dimensional Stress Analysis of Solids

with Plate Elements

Part 4

There are many machine components that involve the geometry that prevent engineers using traditional methods to conduct stress Analysis in their design process. For example, perforated tapered plates are commonly use, with the holes for bolted joints with Other parts of the machine. The figure on the left below shows a tapered perforated plate, and the perforated plate in the right also involve the plate with two curve notches.

Conventional wisdom shows that serious “stress concentration” can occur in the vicinity of a structure with drastic geometry changes. The parts near the holes and the tapered areas and the notches in both plates are vulnerable for stress concentrations.

We have mentioned that the problem of plate with a small hole may be solved by the theory of “advanced strength of materials.” However, plates with tapered and notched edges as shown above can not be handled by any existing method from classical theories of elasticity, and the FEM appears the only available method of the solution.

The Need for Stress Analysis of Solids of Plane Geometry

This type of structures are typically thin in their thickness in comparison to the bulk volume of the overall structure. As such only three (3) out of total six (6) independent stress components need to be considered in the analysis, as illustrated below:


Two-Dimensional Stress Analysis of Solids with Plate Elements – Cont’d

ο x

y

Plane structure subject to in-plane forces: Induced stresses:

The in-plane displacement components in the solid: { } ( )

=yxvyxu

yxU,

),(),( u(x,y) = the component along the x-coordinate

V(x,y) = the component along the y-coordinate

The stress components in the solid: {σ(x,y)}T = {σxx(x,y σyy(x,y) σxy(x,y)}T

The strain components in the solid: {ε(x,y)}T = {εxx(x,y) εyy(x,y) εxy(x,y)}T

Note: Shear stresses exist on the thickness or “edges.”


The strain – displacement relation in Equation (4.3) is modified to the new form:

Two-Dimensional Stress Analysis of Solids with Plate Elements – Cont’d

( ){ }

( )

( )

( ) ( )( )

∂∂

∂∂

∂∂

∂∂

=

∂∂

+∂

∂∂

∂∂

∂

=yxvyxu

xy

y

x

xyxv

yyxu

yyxv

xyxu

yx,

),(0

0

,,

,

,

,ε (4.48)

and the stress-strain relation in Equation (4.5) becomes:

{ }

−−=

xy

yy

xxEyxεεε

νν

ν

νσ

2100

0101

1),( 2

(4.49)

Let us now formulating the FE for plate structures such as with the discretization in a tapered plate as illustrated In the figure below:

Formulation FE analysis for Two-Dimensional Stress Analysis of Solids with Plate Elements

We begin our FE formulation of the plate structure with the expression of the “element displacement” components {U(x,y)} in terms of the corresponding “nodal displacements” using an interpolation function {N(x,y)} as follows:

( ){ } ( )( ) ( ){ }{ }uyxN

yxvyxu

yxU ,,,

, =

= (4.)

where the nodal displacement components, { }

=

m

j

i

m

j

i

vvvuuu

u (4.)

Element displacements Nodal displacements Interpolation function

Formulation FE analysis for Two-Dimensional Stress Analysis of Solids with Plate Elements – Cont’d

Derivation of interpolation function {N(x,y)}

We assume the elements used in this FE analysis are the “simplex” elements, meaning that the Element displacements follow linear polynomial functions in relating their nodal displacements:

( ) yxyxu 321, ααα ++= along the x-coordinate

( ) yxyxv 654, ααα ++=and along the y-coordinate

We will thus have: ( ){ } ( )( )

{ }T

yxyx

yxvyxu

yxU

65432110000001

,,

,

αααααα

=

=

(4.28)

or in an alternative matrix form: ( ){ } ( )[ ]{ }αyxRyxU ,, =

The matrix [R(x,y)] in Equation (4.28) has the form: ( )[ ]

=

yxyx

yxR1000

0001,

where α1, α2, α3, α4, α5, α6 in the matrix {α}T are constants to be determined with specified nodal coordinates later.

(4.29)

Formulation FE analysis for Two-Dimensional Stress Analysis of Solids with Plate Elements – Cont’d

With the specified nodal coordinates:

ui = α1 + α2xi + α3yi, uj = α1 + α2xj + α3yj, um = α1 + α2xm + α3ym, and

vi = α4 + α5xi + α6yi, vj = α4 + α5xj + α6yj, vm = α4 + α5xm + α6ym

We are able to expand Equation (4.28) into the following form for the nodal displacements:

{ }

=

=

6

5

4

3

2

1

100010001000

000100010001

αααααα

mm

jj

ii

mm

jj

ii

m

j

i

m

j

i

yxyxyx

yxyxyx

vvvuuu

u (4.30)

Displacement components of 3 nodes in the element

Specified coordinates of the 3 nodes in the element

Constant coefficients

Derivation of interpolation function {N(x,y)} - cont’d

Formulation FE analysis for Two- Dimensional Stress Analysis of Solids with Plate Elements – Cont’d

Equation (4.30) may be expressed in a different form of:

{u} = [A]{α} (4.)

in which the matrix [A] has the form:

[ ]

=

mm

jj

ii

mm

jj

ii

yxyxyx

yxyxyx

A

100010001000

000100010001

(4.)

We may obtain the solution of the unknown coefficient matrix in Equation (4.31) as:

{ } [ ] { } [ ]{ }uhuA == −1α (4.)


with [A]-1 = the inverse of the nodal coordinate matrix [A] in Equation (4.32)

where the matrix: [h] = [A]-1 (4.)



We recall Equation (4.28) with:

( ){ } ( )( )

{ }T

yxyx

yxvyxu

yxU

65432110000001

,,

,

αααααα

=

=

(4.28)

Also with Equation (4.29) with: ( )[ ]

=

yxyx

yxR1000

0001, (4.29)

We will obtain the following expression after substituting the matric {α} in Equation (4.33) into Equation (4.28), yielding:

( ){ } ( )[ ][ ]{ }uhyxRyxU ,, = (4.35)

Displacement components of 3 nodes in the element

Displacement components in Element ijm

By comparing Equation (4.35) with Equation (4.26), we have the interpolation function of this simplex element to be:

[N(x,y)] = [R(x,y)][h] (4.36)

with Matrices [R(x,y)] in Equation (4.29) and [h] in Equation (4.34)


The interpolation function {N(x,y)} - cont’d

( ) ( ) ( ) ( )[ ]yxxxyyyxyxA

yxN jmmjjmmji −+−+−=1,

( ) ( ) ( ) ( )[ ]yxxxyyyxyxA

yxN miimmiimj −+−+−=1,

( ) ( ) ( ) ( )[ ]yxxxyyyxyxjA

yxN ijjiijjim −+−+−=1,

( ) ( ) ( ) )(ijmtriangleofmadeelementtheofareatheyxyxyxyxyxyxA miimjmmjijji =−+−+−=

where

(3.2)

We may refer to the interpolation function {N(x,y)} that we derived for the 3-node plate elements in Chapter 3 for a comparable case with the current analysis with the form of:

( ){ } ( )( ) ( ){ }{ } ( ) ( ) ( ){ }

=

==

=

m

m

j

j

i

i

mji

mji

m

m

j

j

i

i

mji

vuvuvu

NNNNNN

vuvuvu

yxNyxNyxNuyxNyxvyxu

yxU000

000,,,,

,,

,(4. a)

(3.2a)

( ){ } ( )( ) ( ){ }{ }

==

=

m

j

i

m

j

i

mji

mji

vvvuuu

NNNNNN

uyxNyxvyxu

yxU000

000,

,,

,An alternative expression: (4. b)

Formulation FE analysis for Two- Dimensional Stress Analysis of Solids with Plate Elements – Cont’d Derivation of Element equation

Because the element equation is derived by minimizing the Potential energy in gthe deformed solid, we need to derive the expression of “strain energy” in terms of nodal displacements (the primary quantities in the analysis).

We will first express the element strain vs. nodal displacements by Equation (4.12) as:

{ϵ(x.y)} = [B]{u} (4.12)

[B(x,y,z)] = [D][N(x,y,z)] (4.13) where

By using Equation (4.4) for [D] and Equation (4.36b) for [N(x,y)}, we get the matrix [B(x,y)] in the following:

[ ]( ) ( ) ( )

( ) ( )( ) ( ) ( ) ( ) ( ) ( )

−−−−−−−−−

−−−=

jiimmjijmijm

ijmijm

jiimmj

yyyyyyxxxxxxxxxxxx

yyyyyy

AB )(000

000

21

(4.)

The potential energy in the deformed element , according to Equation (4.18) is:

{ }( ) { } [ ] [ ][ ]{ }

{ } ( )[ ] { } { } ( )[ ] { }dstyxNudvfyxNu

dvuBCBuuP

TT

s

TT

v

v

TT

,,

21

∫∫

∫

−−

=

(4.)

Formulation FE analysis for Two- Dimensional Stress Analysis of Solids with Plate Elements – Cont’d Derivation of Element equation

The element equation is obtain by minimizing the potential energy in Equation (4.38) as:

{ }( ){ } 0=∂

∂uuP

Leading to the following element equation:

[ ]{ } { }puKe = (4.)

[ ] [ ] [ ][ ]dvBCBmatrixstiffnessElementKv

Te ∫== (4.)

and the nodal force matrix

{ } ( )[ ] { } ( )[ ] { }dstyxNdvfyxNmatrixforcweNodalps

TT

v ∫∫ +== ,, (4.)

The integration in Equation (4.40) with respect to the volume of the element may turn out to be tedious. However, if the size of the element is not too large, this integration may be approximated by the following expression without significant error:

[Ke] ≈ [B]T[C][B] (wA) (4.)

in which w is the thickness of the plane element, and A is the plane area that can be computed by Equation (3.2a). ( ) ( ) ( ) )(ijmtriangleofmadeelementtheofareatheyxyxyxyxyxyxA miimjmmjijji =−+−+−=

where

Numerical example on FE analysis for Two-Dimensional Stress Analysis of Solids with a Plate Element

Example (Example 7.2 from the cited reference): A structure made of a triangular plate defined by three corners at A,B and C. A force P is applied at corner C as shownin the figure.

Find the following: (a) The displacement of the plate at corner C (b) The displacement in the plate (c) The stresses and strains in the plate, and (d) The reactions at the two fixed corners

Solution:

We assume that only ONE element is used for the analysis. This triangular plate element has three nodes I, j and m located as shown at the right:

The coordinates of the 3 nodes are: {xi = 6, yi=0), (xj=6, yj=0) and (xm=3, ym=4)

Numerical example on FE analysis for Two-Dimensional Stress Analysis of Solids with a Plate Element – Cont’d

We will first obtain the [A] matrix using Equation (4.32) with the specified nodal coordinates:

and with Equation (4.34) to obtain the [h] matrix:

And then using Equation (4.36) to obtain the interpolation function in the form of Equation (4.36b) as:

( )[ ] ( ) ( )( ) ( )

−−−

−−−=

yyxyxyyxyx

yxN25.0125.0167.0125.0167.0100000025.0125.0167.0125.0167.01

,

(a) To determine the nodal displacements:


We are now ready to derive the element matrix for the structure with the newly derived interpolation function. We will obtain first the [B] matrix from using Equation (4.37) as:

[ ]

−−−−−

−=

044633633000000044

241B

and the [C] matrix FROM Equation (4.7) and the coefficient matrix of Equation (4.25):

[ ]

×=

35.000013.003.01

1099.10 6C

We are now ready to determine the element stiffness matrix in Equation (4.39) from Equations (4.40) and (4.42) as:

[ ]

−−

−−−

−−−−

=

36186.14184.36.1404.84.86.122.78.76.03.615.192.76.08.73.685.1215.19

32.228958SYM

Ke


Constructing the “element equation:”

We realize the following boundary and applied force conditions: ui, vi, vj = 0 (fixed ends), and the applied nodal forces: pix = piy = pjx = pjy = 0, and pmx = p cos 30o = 866 lb, and pmy = p sin 30o = -500 lb

The “element equation” according to Equation (4.39) becomes:

−======

=

==

=

−−

−−−

−−−−

50000

86600

00

0

36186.14184.36.1404.84.86.122.78.76.03.615.192.76.08.73.685.1215.19

32.228958

my

jy

iy

mx

jx

ix

m

j

i

m

j

i

ppp

ppp

vvv

uu

u

SYM

Because there is only one element in the structure, the above element equation is also the “overall stiffness equation” of the structure, from which we may solve the “displacement components of ALL nodes after Making necessary interchange of rows and columns in the above equation (according to the rule stipulatedin Step 5 in Chapter 3.

−

=

−−−−−−

−−−−−−−−−−−−

500866

0000

000

3602.718182.706.123.64.84.83.62.73.615.198.76.085.12

184.88.76.144.36.0184.86.04.36.148.7

2.73.685.126.08.715.19

32.228958

m

m

j

vuu


The 3 nonzero nodal displacements can be solve from the above portioned overall stiffness equation by the following simultaneous equations:

−=

−

−

500866

0

3602.706.123.62.73.615.19

32.228958

m

m

j

vuu

The above simultaneous equations may be solved by either matrix inversion method or Gaussian elimination method, with: uj = 0.16x10-3 inch, um = 0.38x10-3 inch and vm = -0.093x10-3 inch

Thus, after the necessary interchanges of rows and columns, we reached the partition of the element equation In the form of Equation (6.12, Ref) as:


(b) The displacements in the element:

We may use the interpolation function {N(x,y)} to determine the displacement components everywhere in the element:

For the displacement in the x-direction:

u(x,y) = (1 –- 0.167x – 0.125y)ui + (0.167x -0.125y)uj + 0.25 um = 0 +(0.167x-0.125y)x0.16x10-3+0.25x0.38x10-3y V(x,y) = (1 –- 0.167x – 0.125y)vi + (0.167x -0.125y)vj + 0.25 vm = -0.09264x0.25x10-3y

● P

For instance the displacements at Point P(3,2) have the values of: u(3,2) = 0 +(0.167x3-0.125x2)x0.16x10-3+0.25x0.38x10-3

x2 = 0.23x10-3 inch v(3,2) = -0.09264x0.25x10-3x2 = -.04632x10-3 inch

(c) The strain components in the element by using Equation (4.12) with {ε} = [B]{u}:

( ){ }( )( )( )

63 1075

16.2365.26

10

09.00038.016.00

044633633000000044

241

,,,

, −− ×

−=×

−======

−−−−−

−=

=

m

j

i

m

j

i

xy

yy

xx

vvv

uu

u

yxyxyx

yxεεε

ε


The stresses in the element may be obtained by using the generalized Hooke’s Law in Equation (4.25):

{ } psiEyx

xy

yy

xx

−=×

−

×=

−−= −

66.28867.1665.216

1075

16.2365.26

35.000013.003.01

1099.10

2100

0101

1),( 66

2

εεε

νν

ν

νσ

(d) The reactions at all nodes: One may derive the following expression for the nodal forces: { } [ ] { } [ ] { }( ) areaplaneAandpltetheofthicknessWwithWABdvBR TT

v==≈= ∫ ,σσ

{ }

−

−

−

=××

−

−−−

−−−

=

=

5003.8273.327

8660866

11266.28867.1665.216

060430430

600304304

241

my

jy

iy

mx

jx

ix

RRRRRR

R

The reactions at Node i therefore have numerical values at: Rix = 866 lb towards left, and Riy=327.3 lb in the downward direction


Important lesson learned from this numerical example:

We noticed that the stresses and strains in the element (and thus the triangular plate structure re CONSTANT:

( ){ }( )( )( )

61075

16.2365.26

,,,

, −×

−=

=yxyxyx

yx

xy

yy

xx

εεε

ε { } psiyx

xy

yy

xx

−=

=66.28867.1665.216

),(σσσ

σ

– meaning there is no variation of stresses and strains throughout the entire structure. This is obviously not realistic!!

and

The reason for what has happened in this (and the other)numerical example is because we used “linear polynomial” in deriving the interpolation function – resulting in using “SIMPLEX” element in the FE analysis. “Simplex elements” offers “simple mathematical formulation in FEA, but results in constant stresses and strains in elements. That was the reason why engineers need to place many more (smaller) elements in the area with conceivable high gradients of primary unknown quantities, such as in the following cases:

This is good lesson for any intelligent FE user to learn and exercise

SUMMARY

1. This chapter presents FE formulation for solids deformed by applied forces, either in the form of body force or as surface tractions. 2. The deformation of the solids is supposed to be small so that the induced stresses are within the elastic limit of the material 3. The FE formulation is based on the theories of Linear Elasticity

4. General FE formulation for “simplex element” (with interpolation functions follow linear polynomial functions) were derived. 5. FE formulations for special cases for: (a) one-dimensional bar elements, (b) 2-dimensional plate elements, and 3-dimensional torus elements are derived. 6. FE formulations of stress analysis of deformable solids uses minimization of potential energy in the deformed state to derive the element equations. 7. Because simplex elements are used for all these cases for simplicity in deriving element equations. However, these simplex element result in “constant” stresses and strains in each individual elements. Intelligent users will thus need to assign many more smaller elements in the regions where hgi gradients of primary quantities are expected to achieve better result.


U(x,y)

V(x,y) ( ) yxyxu 321, ααα ++= along the x-coordinate

( ) yxyxv 654, ααα ++= along the y-coordinate

We assume the element displacement components follow linear polynomial functions as shown below:

( ){ } ( )( )

{ }T

yxyx

yxvyxu

yxU

65432110000001

,,

,

αααααα

=

=

or to show the above relations in a matrix form:

(4.25)

or in an alternative matrix form: ( ){ } ( )[ ]{ }αyxRyxU ,, =

The matrix [R(x,y)] in Equation (4.26) has the form: ( )[ ]

=

yxyx

yxR1000

0001,

where α1, α2, α3, α4, α5, α6 in the matrix {α}T are constants to be determined with specified nodal coordinates later.

(4.27)

(4.26)

me 160 introduction to finite element method chapter 4 finite … · 2018-10-29 · induced stress...

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