1

Chapter 3

Basic Elements

2

Stress-Strain Review

Textbook: 3.1

3

Review of Solid Mechanics

• The analysis of any solid elastic body has to define and develop the following quantities and/or relations

– Stress

– Strain (strain-displacement relations)

– Constitutive Properties (Stress-Strain relations)

– Compatibility

– Equilibrium Equations

– Boundary Conditions

4

SURFACE TRACTION

• Surface traction (Stress)

– The entire body is in equilibrium with external forces (f1 ~ f6)

– The imaginary cut body is in equilibrium due to external forces (f1, f2, f3) and internal forces

– Internal force acting at a point Pon a plane whose unit normal is n:

– The surface traction depends on the unit normal direction n.

– Surface traction will change when n changes.

– unit = force per unit area (pressure)

f1

f2

f3

f4

f6

f5

y

z

F

n

f1

f2

f3

P A

x

( )

A 0lim

An F

T

T(n)x y zT T Ti j k T(n) 2 2 2

x y zT T T T

5

NORMAL AND SHEAR STRESSES

• Normal and shear stresses– Decompose T(n) into normal and tangential components

n normal stress stress component parallel to n

n shear stress stress component perpendicular to n

n

n

T(n)

n

P

2( ) 2 2n n

nT

( )n

nT n

What if T(n) and n are in the same direction?

2( ) 2

n nnT

6

CARTESIAN STRESS COMPONENTS

– Surface traction changes according to the direction of the surface.

– Impossible to store stress information for all directions.

– Let’s store surface traction parallel to the three coordinate directions.

– Surface traction in other directions can be calculated from them.

– Consider the x-face of an infinitesimal cube

xx xy

xz

x y

z

x

y

zF

(x)xx xy xzT i j k

x

x

x

xxx A 0

x

yxy A 0

x

zxz A 0

x

Flim

A

Flim

A

Flim

A

T i j + k(x ) (x ) (x ) (x )x y zT T T

x y zF F FF i j k

7

CARTESIAN COMPONENTS cont.

– First index is the face and the second index is its direction

– When two indices are the same, normal stress, otherwise shear stress.

– Continuation for other surfaces.

– Total nine components

– Same stress components are defined for the negative planes.

xx yy

zz

xy

xz

yx

yz

zx zy

x y

z

x

y

z F

Comp. Description

xxNormal stress on the x face in the x dir.

yyNormal stress on the y face in the y dir.

zzNormal stress on the z face in the z dir.

xyShear stress on the x face in the y dir.

yxShear stress on the y face in the x dir.

yzShear stress on the y face in the z dir.

zyShear stress on the z face in the y dir.

xzShear stress on the x face in the z dir.

zxShear stress on the z face in the x dir.

8

CARTESIAN COMPONENTS cont.

• Sign convention– Positive when tension and negative when compression.

– Shear stress acting on the positive face is positive when it is acting in the positive coordinate direction.

xx x

xy y

sgn( ) sgn( ) sgn( F )

sgn( ) sgn( ) sgn( F )

n

n

9

STRESS TRANSFORMATION

– If stress components in xyz-planes are known, it is possible to determine the surface traction acting on any plane.

– Consider a plane whose normal is n.

– Surface area (ABC = A)

– The surface traction

– Force balance (h → 0)

x

y

z

B

A

C

zz zx

zy

yy

yz yx

xx

xz

xy

T(n) n

P

( ) ( ) ( ) ( )x y zT T Tn n n nT i j k

( )x x xx x yx y zx zF T A An An An 0n

( )x xx x yx y zx zT n n nn

z x yPAB An ; PBC An ; PAC An

x

x y z y

z

n

n n n n

n

n i j k

10

STRESS TRANSFORMATION cont.

• All three-directions

• Matrix notation

– []: stress matrix; completely characterize the state of stress at a point

• Normal and shear components

( ) [ ]nT n

( )n

2( ) 2n n

[ ]n

n

T n n n

T

( )x xx x yx y zx z

( )y xy x yy y zy z

( )z xz x yz y zz z

T n n n

T n n n

T n n n

n

n

n

T{ } [ ]{ }n n

xx yx zx

xy yy zy

xz yz zz

[ ]

11

SYMMETRY OF STRESS TENSOR

– Stress tensor should be symmetric9 components 6 components

– Equilibrium of the angular moment

– Similarly for all three directions:

– Let’s use vector notation:

xy

yx

x

y

xy

yx

O

l

l

A B

C D

xy yx

xy yx

M l( ) 0

xy yx

yz zy

xz zx

xx

yy

zz

yz

zx

xy

{ }

xx yx zx

xy yy zy

xz yz zz

[ ]

12

COORDINATE TRANSFORMATION

– When [σ]xyz is given, what would be the components in a different coordinate system x′y′z′ (i.e., [σ]x′y′z′)?

– Unit vectors in x′y′z′-coordinates:

– b1 = {1, 0, 0}T in x′y′z′ coordinates, while in xyz coordinates

– the rotational transformation matrix

– Stress does not rotate. The coordinates rotate

x

y

z

x y

z

b1 b2

b3

1 1 1 11 2 3{b , b , b }b

1 2 31 1 1

1 1 2 2 3 32 2 21 2 33 3 3

b b b

b , b , b

b b b

b b b

1 2 31 1 1

1 2 3 1 2 32 2 21 2 33 3 3

b b b

[ ] [ ] b b b

b b b

N b b b Direction cosine matrix

13

COORDINATE TRANSFORMATION cont.

– [N] transforms a vector in the x′y′z′ coordinates into the xyz coordinates, while [N]T transforms a vector in the xyz coordinates into the x′y′z′ coordinates.

– Consider bx′y′z′ = {1, 0, 0}T:

– Stress transformation: Using stress vectors,

– By multiplying [N]T the stress vectors can be represented in the x′y′z′ coordinates

– The first [N] transforms the plane, while the second transforms the force.

Tx y z xyz[ ] [ ] [ ] [ ]N N

1 2 3( ) ( ) ( ) 1 2 3

xyz xyz xyz[ ] [ ] [ ] [ ] [ ]b b bT T T b b b N

x

y

z

x y

z

b1 b2

b3

11

1 1 1xyz x ' y ' z ' 2

13

b

[ ] b

b

b N b

14

Equilibrium Equations (2-D)

Equilibrium of a cube in xy-plane with length dxand dy in X-and Y-directions and unit thickness in the Z-direction

Equilibrium of forces in X-direction

x x x xy xy xyd dy dy d dx dx F 0 X

x

y

2

dxxx x

2

dxxx x

2

dyyy y

2

dyyy y

2

dyyx y

2

dyyx y

2

dxxy x

2

dxxy x

15

Equilibrium of forces in X-direction

0 dxdyFdxdxddydyd xyxyxyxxx X

dxx

d xx

0 dxdyFdxdxddydyd Xxyxyxyxxx 1 1 2 2

dyy

d xyxy

0 dxdyFdxdydy

ddxdy

dx

dX

xyx

xyxXF dxdy 0

x y

16

2-D Equilibrium Equations

xyxx

xy yy

F 0x y

F 0x y

The force equilibrium provide the relations shown below referred to as differential equation of equilibrium

3D casexyxx xz

X

xy yy yzY

yzxz zzZ

F 0x y z

F 0x y z

F 0x y z

17

Strain

• Why do we need the strain measures? Will displacement not suffice?

• Strain better quantifies the deformation of the body and eliminates rigid body motion/rotation

• Strain in very general terms is a measure of relative deformation

– Relative to what?

• Undeformed body : Lagrangian strain

• Deformed body: Eulerian strain

18

STRAIN

• Strain: a quantitative measure of deformation– Normal strain: change in length of a line segment

– Shear strain: change in angle between two perpendicular line segments

• Displacement of P = (u, v, w)

• Displacement of Q & R

P(x,y,z) Q

R P'(x+u,y+v,z+w)

Q'

R'

x

y

z

xy

Q

Q

Q

uu u x

xv

v v xxw

w w xx

R

R

R

uu u y

y

vv v y

y

ww w y

y

19

STRAIN

– Strain is defined as the elongation per unit length

– Tensile (normal) strains in x- and y-directions

– Strain is a dimensionless quantity. Positive for elongation and negative for compression

P Px ux

y

uy

x xxx x 0

y yyy y 0

u ulim

x xu u

limy y

Textbook has different, but more rigorous derivations

20

SHEAR STRAIN

– Shear strain is the tangent of the change in angle between two originally perpendicular axes

– Shear strain (change of angle)

– Positive when the angle between two positive (or two negative) faces is reduced and negative when the angle is increased.

– Valid for small deformation

P

ux

uy

2

1

/2 – 12

y1 1

x2 2

utan

xu

tany

y yx x

xy 1 2 x 0 y 0

u uu ulim lim

x y x y

y xxy xy

u u1 1

2 2 x y

x

y

Engineering shear strain

Tensor shear strain

21

STRAIN MATRIX

– Strain matrix and strain vector

– Normal component:

– Coordinate transformation:

xx xy xz

yx yy yz

zx zy zz

[ ]

nn [ ]n n

Tx y z xyz[ ] [ ] [ ] [ ]N N

xx

yy

zz

yz

zx

xy

{ }

22

Compatibility

• Deformation must be such that the pieces fit together without any gaps or overlap.

• Why is this an issue?

• In 2-D we require only 2 displacements u, and v to describe deformation, but have three strain quantities x,y, and xy. This implies only two of the three strain terms are independent.

x

ux

y

v

y

y

u

x

vxy

2

2

2

2

2

2

2

22

xyy

v

xx

u

yyxyx

xy

23

Stress-Strain Relations

• The stress-strain relations in solid mechanics is often referred to as the Hooke’s Law

• Hooke’s law of proportionality stated as “extension is proportional to the force” refers to the axial extension of a bar under an axial force

• This can be extended to 3-D stress/strain state referred to as the Generalized Hooke’s Law relates the components of the 3-D stress state to 3-D strains as follows.

zx

yz

xy

z

y

x

zx

yz

xy

z

y

x

cccccc

cccccc

cccccc

cccccc

cccccc

cccccc

666564636261

565554535251

464544434241

363534333231

262524232221

161514131211

24

Generalized Hooke’s Law

• In the most general form the generalized Hooke’s Lawrequires 36 constants to relate the terms of a 3-D Stress state to its corresponding 3-D strain state for an elastic material

• However, from symmetry of the strain energy terms, it can be shown that cij = cji

• This reduces the number of unknown constants to 21

zx

yz

xy

z

y

x

zx

yz

xy

z

y

x

cccccc

cccccc

cccccc

cccccc

cccccc

cccccc

665646362616

565545352515

464544342414

363534332313

262524232212

161514131211

25

Hooke’s Law: Orthotropy and Isotropy

• If we assume the x, y and x coordinates provide the planes of symmetry we can further reduce the number of constants to 9.

zx

yz

xy

z

y

x

zx

yz

xy

z

y

x

c

c

c

ccc

ccc

ccc

66

55

44

332313

232212

131211

00000

00000

00000

000

000

000

Orthotropic material

Isotropy assumes that there is no directional variation on property. Using this argument we can obtain

665544

231312

332211

ccc

ccc

ccc

26

Hooke’s law : Engineering Elastic Constants

The two engineering elastic constants used to relate stress to strain for isotropic materials are the Elastic modulus, Eand the Poisson’s ratio .

For uniaxial loading, strain in the loading direction obtained from Hooke’s law, states

Transverse to loading direction

The relation between the shear stress component and its corresponding shear-strain component is called the modulus of rigidity or modulus of elasticity in shear and is denoted by the letter G.

Ex

x

Ex

xzy

G

12

EG

27

LINEAR ELASTICITY (HOOKE’S LAW) cont.

• Isotropic Material:– Stress in terms of strain:

– Strain in terms of stress

Shear Modulus

{ } [ ] { }C

xx xx

yy yy

zz zz

xy xy yz yz zx zx

1E

11 1 2

1

G , G , G

xx xx

yy yy

zz zz

xy yz zxxy yz zx

11

1E

1

, ,G G G

E

G2(1 )

28

TYPES OF 2D PROBLEMS

• Governing D.E.

• Definition of strain

• Stress-Strain Relation

– Since stress involves first-order derivative of displacements, the governing differential equation is the second-order

x

y

2

dxxx x

2

dxxx x

2

dyyy y

2

dyyy y

2

dyyx y

2

dyyx y

2

dxxy x

2

dxxy x

bx

by

xyxxx

xy yyy

b 0x y

b 0x y

xx yy xy

u v u v, ,

x y y x

xx 11 12 13 xx

yy 21 22 23 yy

31 32 33xy xy

C C C

C C C { } [ ]{ }

C C C

C

29

TYPES OF 2D PROBLEMS cont.

• Boundary Conditions– All differential equations must be accompanied by boundary conditions

– Sg is the essential boundary and ST is the natural boundary

– g: prescribed (specified) displacement (usually zero for linear problem)

– T: prescribed (specified) surface traction force

• Objective: to determine the displacement fields u(x, y) and v(x, y) that satisfy the D.E. and the B.C.

, on

, on

g

T

S

S

=

=

u g

n Ts

30

PLANE STRESS PROBLEM

• Plane Stress Problem:– Thickness is much smaller than the length and width dimensions

– Thin plate or disk with applied in-plane forces

– z-directional stresses are zero at the top and bottom surfaces

– Thus, it is safe to assume that they are also zero along the thickness

– Non-zero stress components: σxx, σyy, τxy

– Non-zero strain components: εxx, εyy, εxy, εzz

x

y fx

fy zz xz yz 0

31

PLANE STRESS PROBLEM cont.

• Stress-strain relation

– Even if εzz is not zero, it is not included in the stress-strain relation because it can be calculated from the following relation:

• How to derive plane stress relation?– Solve for zz in terms of xx and yy from the relation of zz = 0 and Eq.

(1.57)

– Write xx and yy in terms of xx and yy

xx xx

yy yy2

12xy xy

1 0E

1 0 { } [ ]{ }1

0 0 (1 )

C

zz xx yy( )E

32

PLANE STRAIN PROBLEM

• Plane Strain Problem– Thickness dimension is much larger than other two dimensions.

– Deformation in the thickness direction is constrained.

– Strain in z-dir is zero

– Non-zero stress components: σxx, σyy, τxy, σzz.

– Non-zero strain components: εxx, εyy, εxy.

0, 0, 0zz xz yz

Plane strain model

33

PLANE STRAIN PROBLEM cont.

• Plan Strain Problem– Stress-strain relation

– Even if σzz is not zero, it is not included in the stress-strain relation because it can be calculated from the following relation:

xx xx

yy yy

12xy xy

1 0E

1 0 { } [ ]{ }(1 )(1 2 )

0 0

C

zz xx yy

E( )

(1 )(1 2 )

34

EQUIVALENCE

• A single program can be used to solve both the plane stress and plane strain problems by converting material properties.

From To E

Plane strain Plane stress

Plane stress Plane strain

2

11

E

2

11

E

1

1

35

STRAIN ENERGY

• Force Deformation Stress Stored Energy

• Strain Energy Density:

• 3-D situation:

• Use principal stress-strain relation

• Strain Energy Density:

U Strain energy

E

i

i

in terms of principal stresses

0

1U

2

0 1 1 2 2 3 3

1U ( )

2

1 1 2 3

2 2 1 3

3 3 1 2

1( )

E1

( )E1

( )E

2 2 2

0 1 2 3 1 2 2 3 1 3

1U 2 ( )

2E

36

Homework

• For a plane stress problem, the strain components in the x–y plane at a point P are computed as

(a) Compute the state of stress at this point if Young’s modulus E = 2×1011

Pa and Poisson’s ratio = 0.3. (b) What is the normal strain in the z–direction? (c) Compute the normal strain in the direction of n = (1, 1, 1).

• The state of stress at a point is given by

(a) Determine the strains using Young’s modulus of 100 GPa and Poisson’s ratio of 0.25. (b) Compute the strain energy density using these stresses and strains. (c) Calculate the principal stresses. (d) Calculate the principal strains from the strains calculated in (a). (e) Show that the principal stresses and principal strains satisfy the constitutive relations. (f) Calculate the strain energy density using the principal stresses and strains.

2 2xx yy xy0.125 10 , 0.25 10

80 20 40

[ ] 20 60 10 MPa

40 10 20

37

Interpolation and Element Matrix Formation

Textbook: 3.2, 3.3

38

Interpolation

• For given nodal values ui, how to calculate u(x) within the element?

• A continuous function that satisfies prescribed conditions (nodal values, ui) at a finite number of points (nodes).

• Polynomials are the usual choice for FEM

i iu(x ) u Property of interpolation

x1 x2 x3

u(x)

x

u1u2

u3

39

Polynomial interpolation (1D)

• For (n+1) node element, solution u(x) is approximated by n+1 monomials

• Impose interpolation condition u(xi) = ui, i = 1, …, n+1

• Calculate unknown coefficients

ni

ii 0

u(x) a x or u(x) { }

X a

2 n

T

0 1 2 n

1 x x ... x

a a a ... a

X

a

1{ } [ ] { }a A d

n1 01 1

n2 12 2

nn 1 nn n

u a1 x x

u a1 x x{ } [ ]{ }

u a1 x x

d A a

40

Polynomial Interpolation (1D)

• Solution in terms of nodal values

• Coefficients ai do not have any physical meaning

• {d} are nodal values of the solution

• Property of interpolation

1u(x) (x) [ ] { } (x) { } X A d N d

Interpolation or shape function

i j

1 if i jN (x )

0 if i j

n 1

ii 1

N (x) 1

41

Degree of Continuity

• In FEM, field quantities u(x) are interpolated in piecewise fashion over each element

• This implies that u(x) is continuous and smooth within the element

• However, u(x) may not be smooth between elements (but continuous)

• An interpolation function with Cm-continuity provides a continuous variation of the function and up to the m-derivatives at the nodes

– For 1D bar element, if the displacement u(x) is C0 then displacements are continuous between elements, but the strains are not

42

Degree of Continuity

Function u1 is C0 continuous while u2 is C1 continuous

C0 C1

xx0 x1 x2

du

dx

2du

dx

1du

dx

Elem1 Elem2

C0C1

u(x)

xx0 x1 x2

Elem1 Elem2

u1(x)u2(x)

43

Example: 1D linear interpolation function

• Linear interpolation between x1 and x2

• Two points: (x1, u1) and (x2, u2)

• Assume linear polynomial (why?)u(x) = a0 + a1x

• Apply interpolation property

x1 x2

u(x)

x

u1u2

u(x)

L

1 1 0

2 2 1

u 1 x a

u 1 x a

1u(x) (x) [ ] { } (x) { } X A d N d

x1 2

x1 x2

u2u1

1

1

21

2 1

x xN

x x

1

2

uu

u

N

12

2 1

x xN

x x

2 1

2 1 2 1

x x x x

x x x x

N

44

Example: Quadratic Interpolation

• Three points(x1, u1), (x2, u2), (x3, u3)

• Quadratic approx.u(x) = a0 + a1x + a2x2

1 2 3x

x1 x2 x3

x1 x2 x3

x1 x2 x3

x1 x2 x3

1

1

1

2 31

2 1 3 1

x x x xN

x x x x

1

2

3

u

u(x) (x) u

u

N

1 32

1 2 3 2

x x x xN

x x x x

1 23

1 3 2 3

x x x xN

x x x x

u1

u2 u3

45

Lagrange Interpolation Formula

• Shape functions shown for the C0 interpolations are special forms of the Lagrangian interpolation functions

n

k kk 1

f(x) N (x)f

1 k 1 k 1 nk

1 k k 1 k k 1 k n k

x x ... x x x x ... x xN

x x ... x x x x ... x x

46

C1 Interpolation

• Also called Hermite interpolation (Hermite polynomials)

• Use the ordinate and slope information at the nodes to interpolate

u

x

C0 interpolation curve

C1 interpolation curvex

u

ux2

ux1

u

u

47

Beam Element Interpolation

• Start with v(x) = a0 + a1x + a2x2 + a3x3

• Impose 4 conditions:

1 2

,x 1 ,x 2

v(0) v v(L) v

v (0) v (L)

48

2D and 3D Interpolation

• The 2D and 3D shape functions follow the same procedure as for 1D

• We now have to start with shape functions that have two or more independent terms.

• For example, a linear interpolation in 2D from 3 nodes will require an interpolation function

• If there are two or more components (e.g., u, v and w displacements) then the same interpolation function is used for all components

1

2

3

a

f(x,y) 1 x y a

a

49

Principle of Virtual Work

• The principle of virtual work states that at equilibrium the strain energy change due to a small virtual displacement is equal to the work done by the forces in moving through the virtual displacement.

• A virtual displacement is a small imaginary change in configuration that is also a admissible displacement

• An admissible displacement satisfies kinematic boundary conditions

• Note: Neither loads nor stresses are altered by the virtual displacement.

50

Principle of Virtual Work

• The principle of virtual work can be written as follows

– {u} = {u, v, w}T: virtual displacement

– {}: virtual strain due to {u}

• Interpolation:

• Strain-displacement:

• Constitutive relation:

• Altogether

– Need to satisfy all kinematically admissible displacements {d}

T T T{ } { }dV { u} {F}dV { u} { }dS

{ } [ ]{ } { } [ ]{ } u N d u N d

{ } [ ]{ } [ ] [ ][ ] { } [ ]{ } ε B d B N ε B d

{ } [ ]{ }σ E ε

T T T T{ } [ ] [ ][ ]dV{ } [ ] { }dV [ ] { }dS 0 d B C B d N F N

51

Stiffness Matrix and Load Vector

• Equations of equilibrium

• Element stiffness matrix

• Element load vector

• Refer textbook for loads due to initial strain and initial stress

e[ ]{ } { }k d r

T T T[ ] [ ][ ]dV { } [ ] { }dV [ ] { }dS B E B d N F N

T[ ] [ ] [ ][ ]dV k B E B

T Te{ } [ ] { }dV [ ] { }dS r N F N

52

Homework

• Solve Problem 3.2-3

• Solve Problem 3.2-5

53

Plane Elements (CST, LST)

Textbook: 3.4, 3.5

54

CST ELEMENT• Constant Strain Triangular Element

– Decompose two-dimensional domain by a set of triangles.

– Each triangular element is composed by three corner nodes.

– Each element shares its edge and two corner nodes with an adjacent element

– Counter-clockwise or clockwise node numbering

– Each node has two DOFs: u and v

– displacements interpolation using the shape functions and nodal displacements.

– Displacement is linear because three nodal data are available.

– Stress & strain are constant.

u1

v1

u2

v2

u3

v33

1

2 x

y

55

CST ELEMENT cont.

• Displacement Interpolation– Since two-coordinates are perpendicular, u(x,y) and v(x,y) are

separated.

– u(x,y) needs to be interpolated in terms of u1, u2, and u3, and v(x,y) in terms of v1, v2, and v3.

– interpolation function must be a three term polynomial in x and y.

– Since we must have rigid body displacements and constant strain terms in the interpolation function, the displacement interpolation must be of the form

– The goal is how to calculate unknown coefficients αi and βi, i = 1, 2, 3, in terms of nodal displacements.

1 2 3

1 2 3

u(x,y) x y

v(x,y) x y

1 1 2 2 3 3u(x,y) N (x,y)u N (x,y)u N (x,y)u

56

CST ELEMENT cont.

• Displacement Interpolation– x-displacement: Evaluate displacement at each node

– In matrix notation

– Is the coefficient matrix singular?

1 1 1 1 2 1 3 1

2 2 2 1 2 2 3 2

3 3 3 1 2 3 3 3

u(x ,y ) u x y

u(x ,y ) u x y

u(x ,y ) u x y

1 1 1 1

2 2 2 2

3 3 3 3

u 1 x y

u 1 x y

u 1 x y

u1

v1

u2

v2

u3

v33

1

2

57

CST ELEMENT cont.

• Displacement Interpolation

– where

– Area:

1

1 1 1 1 1 2 3 1

2 2 2 2 23 31 12 2

3 3 3 3 32 13 21 3

1 x y u f f f u1

1 x y u y y y u2A

1 x y u x x x u

1 2 3 3 2 23 2 3 32 3 2

2 3 1 1 3 31 3 1 13 1 3

3 1 2 2 1 12 1 2 21 2 1

f x y x y , y y y , x x x

f x y x y , y y y , x x x

f x y x y , y y y , x x x

1 1

2 2

3 3

1 x y1

A det 1 x y2

1 x y

58

CST ELEMENT cont.

• Substitute into the interpolation equation

1 1 1 2 2 3 3

2 23 1 31 2 12 3

3 32 1 13 2 21 3

1(f u f u f u )

2A1

(y u y u y u )2A1

(x u x u x u )2A

1 2 3

1 1 2 2 3 3 23 1 31 2 12 3 32 1 13 2 21 3

1 23 32 1

2 31 13 2

3 12 21 3

u(x,y) x y

1(f u f u f u ) (y u y u y u )x (x u x u x u )y

2A1

(f y x x y) u2A1

(f y x x y) u2A1

(f y x x y) u2A

N1(x,y)

N2(x,y)

N3(x,y)

59

CST ELEMENT cont.

• Displacement Interpolation– A similar procedure can be applied for y-displacement v(x, y).

– N1, N2, and N3 are linear functions of x- and y-coordinates.

– Interpolated displacement changes linearly along the each coordinate direction.

Shape Function

1

1 2 3 2

3

1

1 2 3 2

3

u

u(x,y) [N N N ] u

u

v

v(x,y) [N N N ] v

v

1 1 23 32

2 2 31 13

3 3 12 21

1N (x,y) (f y x x y)

2A1

N (x,y) (f y x x y)2A1

N (x,y) (f y x x y)2A

60

CST ELEMENT cont.

• Displacement Interpolation– Matrix Notation

– [N]: 2×6 matrix, {q}: 6×1 vector.

– For a given point (x,y) within element, calculate [N] and multiply it with {q} to evaluate displacement at the point (x,y).

1

1

1 2 3 2

1 2 3 2

3

3

u

v

N 0 N 0 N 0 uu{ }

0 N 0 N 0 N vv

u

v

u

{ (x,y)} [ (x,y)]{ }u N q

61

CST ELEMENT cont.

• Strain Interpolation– differentiating the displacement in x- and y-directions.

– differentiating shape function [N] because {q} is constant.

3 3i

xx i i ii 1 i 1

NuN (x,y)u u

x x x

3 3i

yy i i ii 1 i 1

NvN (x,y)v v

y y y

3 3i i

xy i ii 1 i 1

N Nu vu v

y x y x

1 1 23 32

2 2 31 13

3 3 12 21

1N (x,y) (f y x x y)

2A1

N (x,y) (f y x x y)2A1

N (x,y) (f y x x y)2A

62

CST ELEMENT cont.

• Strain Interpolation

– [B] matrix is a constant matrix and depends only on the coordinates of the three nodes of the triangular element.

– the strains will be constant over a given element

• Element stiffness matrix

1

123 31 12

232 13 21

232 23 13 31 21 12

3

3

uuvx y 0 y 0 y 0uv 1

{ } 0 x 0 x 0 x [ ]{ }vy 2A

x y x y x yuu v

y x v

B q

T T

V[ ] [ ] [ ][ ]dV [ ] [ ][ ]At k B E B B E B

63

CST ELEMENT cont.

• Property of CST Element– Since displacement is linear in x and y, the triangular element deforms

into another triangle when forces are applied.

– an imaginary straight line drawn within an element before deformation becomes another straight line after deformation.

– Consider a local coordinate such that = 0 at Node 1 and = a at Node 2.

– Displacement on the edge 1-2:

– Since the variation of displacement is linear, the displacements should depend only on u1 and u2, and not on u3.

3

1

2

a

1 2

3 4

u( )

v( )

64

CST ELEMENT cont.

• Property of CST Element

• Inter-element Displacement Compatibility– Displacements at any point in an element can be computed from nodal

displacements of that particular element and the shape functions.

– Consider a point on a common edge of two adjacent elements, which can be considered as belonging to either of the elements.

– Then the nodes of either triangle can be used in interpolating the displacements of this point.

– However, one must obtain a unique set of displacements independent of the choice of the element.

– This can be true only if the displacements of the points depend only on the nodes common to both elements.

1 2 1 1 2 2

1 2 1 1 2 2

u( ) 1 u u H u H ua a

v( ) 1 v v H v H va a

3

1

2x

a

Element 1

Element 2

3

1

2

65

Performance of CST Element

• CST element is STIFF– Bending problem: linearly varying stress & strain

– CST can only represent constant stress & strain (Lack of capability)

– Spurious shear stress-F

1 m

5 m F

1

23

45

67

89

10

xx

x

Exact solution:

Disp = 7.5mm

Stress = 70MPa

66

Performance of CST Element

• Y-normal and shear stress are supposed to be zero

• What will happen if an element shows stiff behavior?– For given loads, displacements and stresses are lower

– Solution converges slowly with increasing No. of elements

yy Plot xy Plot

67

CST ELEMENT cont.

• Discussions– CST element performs well when strain gradient is small.

– In pure bending problem, xx in the neutral axis should be zero. Instead, CST elements show oscillating pattern of stress.

– CST elements predict stress and deflection about ¼ of the exact values.

• Strain along y-axis is supposed to be linear. But, CST elements can only have constant strain in y-direction.

• CST elements also have spurious shear strain.

0 1 2

0 1 2

( , )

( , )

u x y a a x a y

v x y b b x b y

1

2

2 1

xx

yy

xy

ua

xv

by

u va b

y x

How can we improve accuracy?

What direction?u2

v21 2

3

68

Linear Strain Triangle (LST)

• Six nodes with 12 DOFs– 3 corner nodes (1, 2, 3)

– 3 mid-side nodes (4, 5, 6)

• Displacements are quadratic polynomial

• Strains are linear polynomials

69

Polynomial Pyramid

• When an element has N nodes, how to choose bases for interpolation?

• The order of polynomial determines the level of accuracy

2 2

3 2 2 3

1

x y

x xy y

x x y xy y

2 20 1 2 3 4 5

xx 1 3 4

u(x,y) a a x a y a x a xy a y

u(x,y) a 2a x a y

x

70

Linear Strain Triangle (LST)

• The quadratic displacement field in terms of generalized coordinates:

• The linear strain field:

• When would this element be exact? And when would be approximate?

2 21 2 3 4 5 6

2 27 8 9 10 11 12

u x y x xy y

v x y x xy y

x 2 4 5

y 9 11 12

xy 3 8 5 10 6 11

2 x y

x 2 y

( ) ( 2 )x (2 )y

71

Performance of LST

72

Homework

• Using two CST elements, solve the simple shear problem described in the figure and determine whether the CST elements can represent the simple shear condition accurately or not. Material properties are given as E = 10 GPa, = 0.25, and thickness is h = 0.1 m. The distributed force f = 100 kN/m2 is applied at the top edge.

f

1 2

34

x

y

1 m

1 m

73

Plane Elements (Q4, Q8, Q9)

Textbook: 3.6, 3.7

74

Bilinear Quadrilateral (Q4)

• 4 nodes with 8 DOFs

• Displacement field:

• Artificial anisotropy: sides are stiffer than diagonals

1 2 3 4

5 6 7 8

u x y xy

v x y xy

75

Q4: Strain Fields

• Strain fields:

• Observation 1: x f(x) Q4 cannot exactly model the beam where x x

x 2 4

y 7 8

xy 3 6 4 8

y

x

( ) x y

76

Q4: Behavior in Pure Bending of a Beam

• Observation 2: When 4 0, x varies linearly in y - desirable characteristic for a beam in pure bending because normal strain varies linearly along the depth coordinate. But xy0 is undesirable because there is no shear strain.

• Fig. (a) is the correct deformation in pure bending while (b) is the deformation of Q4 (sides remain straight).

• Physical interpretation: applied moment is resisted by a spurious shear stress as well as flexural (normal) stresses.

77

Q4: Interpolation functions

• Easy to obtain interpolation functions

where matrix N is 2x8 and the shape functions are

78

Q4: Shape (Interpolation) Functions

• N1=1, N2=N3=N4=0 at node 1, x=-a, y=-b, so u= N1 u1= u1 at that node.

• Similarly Ni=1 while all other Ns are zero at node i.

• Fig 3.6-2

• See Eqn. 3.6-6 for strain-displacement matrix (= Nd=Bd).• Q4 converges properly with mesh refinement and works

better than CST in most problems.

79

Coarse mesh results

• Q4 element is over-stiff in bending. For the following problem, deflections and flexural stresses are smaller than the exact values and the shear stresses are greatly in error:

80

• Sxx Plot

• Stress is constant along the x-axis (pure bending)

• linear through the height of the beam

• Deflection is much higher than CST element. In fact, CST element is too stiff. However, stress is inaccurate.

BEAM BENDING PROBLEM cont.

Max v = 5.1mm

81

RECTANGULAR ELEMENT

• y-normal stress and shear stress are supposed to be zero.

yy Plot xy Plot

1

2

3

4

/ ( 1)

/ ( 1)

/

/

N x y

N x y

N x y

N x y

1

2

3

4

/ ( 1)

/

/

/ ( 1)

N y x

N y x

N y x

N y x

xx is a linear function of y alone

yy is a linear function of x alone

xy is a linear function of x and y

4I

xx II 1

Nu

x

4

Iyy I

I 1

Nv

y

82

Quadratic Quadrilateral (Q8)

• 4 corner nodes and 4 side nodes and 16 nodal dof.

3.7-1

83

Quadratic Quadrilateral (Q8): Displacement

• The displacement field, which is quadratic in x and y:

• Two types of shape functions (=x/a, =y/b) :

• The edges x=a deform into a parabola (i.e., quadratic displacement in y) (same for y=b)

84

Quadratic Quadrilateral (Q8): Strains

• The strain field:

• Strains have linear and quadratic terms. Hence, Q8 can represent many strain states exactly.

• For example, states of constant strain, bending strain, etc.

85

BEAM BENDING PROBLEM cont.

• 8-Node Rectangular Elements

• Tip Displacement = 7.5 mm, Exact!

86

BEAM BENDING PROBLEM cont.

• If the stress at the bottom surface is calculated, it will be the exact stress value.

Sxx Syy

87

Homework

• For a rectangular element shown in the figure, displacements at the four nodes are given by {u1,v1,u2,v2,u3,v3,u4,v4} = {0.0, 0.0, 1.0, 0.0, 2.0, 1.0, 0.0, 2.0}. Calculate displacement (u, v) and strain εxx at (x, y) = (2, 1)

x

y

1 (0,0) 2 (3,0)

3 (3,2)4 (0,2)

88

Choice of interpolation, improved elements

Textbook: 3.8, 3.9, 3.10

89

Polynomial Pyramid

• When an element has N nodes, how to choose bases for interpolation?

• The order of polynomial determines the level of accuracy

2 2

3 2 2 3

1

x y

x xy y

x x y xy y

2 20 1 2 3 4 5

xx 1 3 4

u(x,y) a a x a y a x a xy a y

u(x,y) a 2a x a y

x

90

Choice of Interpolation

• For three-node triangular element, Can we choose different interpolation functions?

• Elements must be able to display rigid body motion and constant strain field

• Then, mesh refinement is to produce convergence toward correct results

• Choice of interpolation must consider “balance”

– No favor to a particular direction (x or y)

1 2 3( )u x a a x a y 2 2

1 2 3( )u x a x a xy a y

91

Choice of Interpolation

• Q4 elements use xy rather than x2 or y2

• Interpolation must provide geometric isotropy or frame invariance

• Can Q4 element use x2 + y2 instead of xy?

• Complete polynomial – include all lower-order polynomials

Terms CST LST Q4 Q8(Q9)

Constant 1 1 1 1

Linear x y x y x y x y

Quadratic x2 xy y2 xy x2 xy y2

Cubic x2y xy2

Quartic (x2y2)

92

Improved Triangles and Quadrilaterals

• Without adding more nodes, add additional DOFs or change numerical integration to improve element performance

• Drilling DOF: rotational DOF about axis normal to the plane.

• A CST with these added to each node has 9 DOF.

– For three-node triangle element, each node has u, v, and rz DOFs

– Much better than CST, but not as good as LST

93

Q6: Additional Bending Shape Functions

• Q4: Artificial shear deformation under pure bending

• Additional shape functions to solve the issue

• Strain xx can vary linear along x-dir.

• Shear strain xy can vanish for pure bending

• Nodeless DOFs, a1, a2, a3, and a4, are condensed in the element level (total 12 DOFs)

=

=

= + - + -

= + - + -

å

å

42 2

1 21

42 2

3 41

( , ) (1 ) (1 )

( , ) (1 ) (1 )

i ii

i ii

u N s t u s a t a

v N s t v s a t a

Bubble modes

94

Modeling Bending with the Q6 Element

•Modeling the previous bending problem with Q6 elements gives the following stresses:

é ù ì üì ü ï ïï ïï ï ï ïê ú =í ý í ýï ï ï ïê ú ï ï ï ïî þë û î þ

dd da d

ad aa a

K K fd

K K fa

-= -1{ } { }aa a ada K f K d

- -- = -1 1[ ]{ } { }dd da aa ad d da aa aK K K K d f K K f

95

Homework

• Solve Problem 3.9-3 and 3.9-4

96

Equivalent nodal forces, stress calculation

Textbook: 3.11, 3.12, 3.13

97

Nodal loads

• Consistent (work-equivalent) loads

– FEM only takes nodal forces (discrete matrix equation)

– Distributed loads must be converted to equivalent nodal forces

– Both must produce the same amount of work (force * displacement)

• Potential of applied load (negative)

• How to calculate it?

eW { } { } dV dS T T Td r u F u Φ

Equivalentnodal forces

Distributedloads

T T T TdV { } [ ] dV { } [ ] dV Tu F d N F d N F

Integrateshape function

98

Example: Beam under uniform loads

L L 2 3 22

1 1 2 1 20 0

2x x 1 2 1 qLM qN dx q(x )dx qL ( )

L L 2 3 4 12

• Normal forces are obvious. For moments

99

Work equivalent (consistent) normal loads

• Normal surface traction on a side of a plane element whose sides remain straight (q is force/length):

• Shape function along the edge 4-3

• Equivalent nodal forces

4

3

q1[ ] [a x a x] F [ ]

q2a

N N

a Te a

a 4T

a3

4

3

{ } [ ] Fdx

q[ ] [ ]dx

q

q2 1aq3 1 2

r N

N N

100

Loads on quadratic sides

• Quadratic side has different shape function

• But, the procedure is the same

2 22

1[ ] [x(x a) 2(a x ) x(x a)]

2a N

4 4a T

7 7a

3 3

4

7

3

F q

F [ ] [ ]dx q

F q

4 2 1 qa

2 16 2 q15

1 2 4 q

N N

101

Multiple elements

• Calculate equivalent nodal forces per each element

• Use the same assembly process

q(x)

1 2 3 Separate

Equivalent nodal forces

(1)1f

(1)2f

(2)1f

(2)2f

Assembly

F1 F2 F3

102

Distributed Shear Traction

• Shear traction on a side of a plane element whose sides remain straight (q is force/length)

• Q4 element and two LSTs share the top midnode so that the nodal loads from Q4 and the right LST are combined.

103

Uniform Body Force• Work-equivalent nodal forces corresponding to weight as a

body force

• LST has no vertex loads and vertex loads of Q8 are upwards!

• The resultant in all cases is W, the weight of the element.

104

Stress calculation

• Hooke’s law with strain-displacement relation (Plane stress)

• Stress & strain have greater error than displacement

• Stresses are most accurate inside elements, but not on the element boundary (discontinuous stress on the boundary)

0 2

1 0

1 01

0 0 (1 ) / 2 0

x

y

xy

TE

Tσ E Bd ε Bd

105

Stress averaging

• Stress averaging– Contour-plotting algorithms are based on nodal values

– Stress is discontinuous at nodes

– Extrapolated stresses are averaged at nodes -> Cause error

– Difference b/w actual and averages stress values are often used as criterion of accuracy

Stress

Elem 1 Elem 2 Elem 3

Averaged nodal stress

Stress at integration

point

106

Estimating errors

• Error estimation– Check accuracy of current analysis

– Criterion for mesh refinement

– Gauss point stress , averaged nodal stress *

– Difference in stresses

– Strain energies

– Error estimation

– the current mesh size is considered to be appropriate, if ≈ 0.05

*E

( e )

2NE

Ve 1

U dV2E

( e )

2NEE

E Ve 1

U dV2E

E

E

U

U U

107

Improving stresses at nodes and boundaries

• Stress should not be averaged when material properties or geometries are discontinuous

• Invariants: Stresses that are independent of coordinate sys.– Von Mises stress or “effective” stress

– Used for yield criterion

1/22 2 2 2 2 21

( ) ( ) ( ) 62

e x y y z z x xy yz xz

108

Examples of poor meshing

• Do not create unnecessary discontinuities!

109

Nature of Finite Element Solution

• FEA solution is not exact, but approximate

• Adjacent elements have identical displacement at nodes

• Displacements may or may not be continuous across element boundaries

• Displacement, stress and strains are continuous within an element

• Equilibrium of nodal forces and moments is satisfied

• Equilibrium is usually not satisfied at or across element boundaries

• Equilibrium is usually not satisfied within elements (only averaged sense)

110

Example of discontinuities in un-averaged stresses

111

Homework

• Solve Problem 3.11-2

• Solve Problem 3.11-4