Introduction to 3D Elasticity

MANE 4240 & CIVL 4240Introduction to Finite Elements

Introduction to 3D Elasticity

Prof. Suvranu De

Reading assignment:

Appendix C+ 6.1+ 9.1 + Lecture notes

Summary:

• 3D elasticity problem•Governing differential equation + boundary conditions•Strain-displacement relationship•Stress-strain relationship

•Special cases2D (plane stress, plane strain)Axisymmetric body with axisymmetric loading

• Principle of minimum potential energy

1D Elasticity (axially loaded bar)

x

y

x=0 x=L

A(x) = cross section at xb(x) = body force distribution (force per unit length)E(x) = Young’s modulusu(x) = displacement of the bar at x

x

1. Strong formulation: Equilibrium equation + boundary conditions

Lxbdxd

0;0

LxatFdxduEA

xatu

00Boundary conditions

Equilibrium equation

F

3. Stress-strain (constitutive) relation : ε(x) E(x)

E: Elastic (Young’s) modulus of bar

2. Strain-displacement relationship:dxduε(x)

Problem definition3D Elasticity

xy

z

Surface (S)

Volume (V)

u v

w

x

V: Volume of bodyS: Total surface of the bodyThe deformation at point x =[x,y,z]T

is given by the 3 components of its displacement

wvu

u

NOTE: u= u(x,y,z), i.e., each displacement component is a function of position

3D Elasticity: EXTERNAL FORCES ACTING ON THE BODY

Two basic types of external forces act on a body1. Body force (force per unit volume) e.g., weight, inertia, etc2. Surface traction (force per unit surface area) e.g., friction

BODY FORCE

xy

z Surface (S)

Volume (V)

u v

w

x

Xa dVXb dV

Xc dVVolume element dV Body force: distributed

force per unit volume (e.g., weight, inertia, etc)

c

b

a

XXX

X

NOTE: If the body is accelerating, then the inertia force

may be considered as part of X

w

vu

u

u~ XX

xy

zST

Volume (V)

u v

w

x

Xa dVXb dV

Xc dVVolume element dV

py

pz

px

Traction: Distributed force per unit surface area

z

y

x

ppp

T S

SURFACE TRACTION

3D Elasticity: INTERNAL FORCES

If I take out a chunk of material from the body, I will see that, due to the external forces applied to it, there are reaction forces (e.g., due to the loads applied to a truss structure, internal forces develop in each truss member). For the cube in the figure, the internal reaction forces per unit area(red arrows) , on

each surface, may be decomposed into three orthogonal components.

xy

z

Volume (V)u v

w

x

Volume element dV

x

y

z

yz

yx

xyxz

zyzx

3D Elasticity

zx

yz

xy

z

y

x

xy

z

x

y

z

yz

yx

xy

xz

zyzx

x, y and z are normal stresses.The rest 6 are the shear stressesConventionxy is the stress on the face perpendicular to the x-axis and points in the +ve y directionTotal of 9 stress components of which only 6 are independent since

xzzx

zyyz

yxxy

The stress vector is therefore

Strains: 6 independent strain components

zx

yz

xy

z

y

x

Consider the equilibrium of a differential volume element to obtain the 3 equilibrium equations of elasticity

0

0

0

czyzxz

byzyxy

axzxyx

Xzyx

Xzyx

Xzyx

Compactly;

0 XT

xz

yz

xy

z

y

x

0

0

0

00

00

00where

EQUILIBRIUM EQUATIONS

(1)

3D elasticity problem is completely defined once we understand the following three concepts

Strong formulation (governing differential equation + boundary conditions)

Strain-displacement relationship

Stress-strain relationship

1. Strong formulation of the 3D elasticity problem: “Given the externally applied loads (on ST and in V) and the specified displacements (on Su) we want to solve for the resultant displacements, strains and stresses required to maintain equilibrium of the body.”

xy

z

Su

ST

Volume (V)

u v

w

x

Xa dVXb dV


py

pz

px

VinXT 0 Equilibrium equations (1)

Boundary conditions

1. Displacement boundary conditions: Displacements are specified on portion Su of the boundary

uspecified Sonuu

2. Traction (force) boundary conditions: Tractions are specified on portion ST of the boundaryNow, how do I express this mathematically?

xy

z

Su

ST

Volume (V)

u v

w

x

Xa dVXb dV


py

pz

px

Traction: Distributed force per unit area

z

y

x

ppp

T S

Traction: Distributed force per unit area

z

y

x

ppp

T S

n

nx

ny

nz

ST

TS

py

px

pz

If the unit outward normal to ST :

z

y

x

nnn

n

Then

zzyzyxxz

zyzyyxxy

zxzyxyxx

nnnnnnnnn

z

y

x

ppp

nx

ny

ST

In 2D

dy

dx

ds

x

y

n

x

y

ndsdy

ndsdx

cos

sin

TSpy

px

dy

dxdsx

y

xy

xy

Consider the equilibrium of the wedge in x-direction

yxyxxx

xyxx

xyxx

nnpdsdx

dsdyp

dxdydsp

Similarlyyyxxyy nnp

2. Strain-displacement relationships:

xw

zu

yw

zv

xv

yuzwyvxu

zx

yz

xy

z

y

x

Compactly; u

xz

yz

xy

z

y

x

0

0

0

00

00

00

wvu

u

zx

yz

xy

z

y

x

(2)

x

y

A B

CA’

B’

C’

vudy

dx

dxxv

xdxuu

dyyu

dyyvv

xu

xv

tanβtanβββ)B'A'(C'angle2π

yv

dy

dyvdyyvvdy

ACACC'A'

xu

dx

dxudxxuudx

ABABB'A'

2121

xy

y

x

1

2

In 2D

3. Stress-Strain relationship:

Linear elastic material (Hooke’s Law) D (3)

Linear elastic isotropic material

22100000

02210000

00221000

000100010001

)21)(1(

ED

Special cases:

1. 1D elastic bar (only 1 component of the stress (stress) is nonzero. All other stress (strain) components are zero)Recall the (1) equilibrium, (2) strain-displacement and (3) stress-strain laws2. 2D elastic problems: 2 situations

PLANE STRESSPLANE STRAIN

3. 3D elastic problem: special case-axisymmetric body with axisymmetric loading (we will skip this)

PLANE STRESS: Only the in-plane stress components are nonzero

x

y

Area element dA Nonzero stress components xyyx ,,

y

xxy

xy

h

DAssumptions:1. h<<D2. Top and bottom surfaces are free from traction3. Xc=0 and pz=0

PLANE STRESS Examples:1. Thin plate with a hole

2. Thin cantilever plate

y

xxy

xy

Nonzero strains: xyzyx ,,,

Isotropic linear elastic stress-strain law

xy

y

x

xy

y

x E

2100

0101

1 2

Hence, the D matrix for the plane stress case is

2100

0101

1 2

ED

D

yxz

1

Nonzero stresses: xyyx ,,PLANE STRESS

PLANE STRAIN: Only the in-plane strain components are nonzero

Area element dA

Nonzero strain components

y

x

xyxy

Assumptions:1. Displacement components u,v functions of (x,y) only and w=0 2. Top and bottom surfaces are fixed 3. Xc=0 4. px and py do not vary with z

xyyx ,,

x

y

z

PLANE STRAIN Examples:1. Dam

2. Long cylindrical pressure vessel subjected to internal/external pressure and constrained at the ends

1

Slice of unit thickness

x

y

z

y

xxy

xyz

Nonzero stress:

Isotropic linear elastic stress-strain law

xy

y

x

xy

y

x E

22100

0101

211

Hence, the D matrix for the plane strain case is

22100

0101

211

ED

D

z x y

xyzyx ,,,

Nonzero strain components: xyyx ,,

PLANE STRAIN

Example problem

x

y

3

2 1

4

2

2The square block is in plane strain and is subjected to the following strains

2

2 3

2

3x

y

xy

xy

xy

x y

Compute the displacement field (i.e., displacement components u(x,y) and v(x,y)) within the block

Solution

Recall from definition

)3(

)2(3

)1(2

32

2

yxxv

yu

xyyv

xyxu

xy

y

x

Integrating (1) and (2)

)5()(),(

)4()(),(

23

12

xCxyyxv

yCyxyxu

Arbitrary function of ‘x’

Arbitrary function of ‘y’

Plug expressions in (4) and (5) into equation (3)

0)()(

)()(

)()(

)3(

21

322312

3223

12

32

xxC

yyC

yxxxCy

yyCx

yxx

xCxyy

yCyx

yxxv

yu

Function of ‘y’ Function of ‘x’

)constanta()()( 21 CxxC

yyC

Hence

Integrate to obtain

22

11

)()(

DCxxCDCyyC

D1 and D2 are two constants of

integration

Plug these back into equations (4) and (5)

23

12

),()5(

),()4(

DCxxyyxv

DCyyxyxu

How to find C, D1 and D2?

Use the 3 boundary conditions

0)0,2(0)0,0(0)0,0(

vvu

To obtain

00

0

2

1

DDC

Hence the solution is

3

2

),(

),(

xyyxv

yxyxu

x

y

3

2 1

4

2

2

Principle of Minimum Potential Energy

Definition: For a linear elastic body subjected to body forces X=[Xa,Xb,Xc]T and surface tractions TS=[px,py,pz]T, causing displacements u=[u,v,w]T and strains and stresses , the potential energy is defined as the strain energy minus the potential energy of the loads involving X and TS

U-W

xy

z

Su

ST

Volume (V)

u v

w

x

Xa dVXb dV


py

pz

px

TSS

T

V

T

V

T

dSTudVXu

dV

W

21U

Strain energy of the elastic body

V

T

V

T dVDdV 21

21U

DUsing the stress-strain law

In 1D

L

xVVAdxEdVEdV

0

22

21

21

21U

In 2D plane stress and plane strain

V xyxyyyxx dV

21U

Why?

Principle of minimum potential energy: Among all admissible displacement fields the one that satisfies the equilibrium equations also render the potential energy a minimum.

“admissible displacement field”: 1. first derivative of the displacement components exist2. satisfies the boundary conditions on Su

Introduction to 3D Elasticity

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