Formulation of Solid and Structural Mechanics

By

S. Ziaei Rad

Isfahan University of Technology

Methods of Formulating SSM Problems

• Differential Equation Formulation Methods

•Displacement Method•Force Method•Displacement-Force Method (Mixed Method)

•Variational Formulation Methods

•Principle of Minimum Potential Energy•Principle of Minimum Complementary Energy•Principle of Stationary Reissner Energy•Hamilton’s Principle (Lagrange’s Principle)

Basic Equation of Solid Mechanics

Type of Equation Number of Equations------------------------------------------

3D 2D 1D-----------------------------------------------------------------------Equilibrium Eqs. 3 2 1Stress-Strain Relations 6 3 1Strain-Displacement 6 3 1Relations -----------------------------------------------------------------------Total Number 15 8 3of Equations

Basic Equation of Solid Mechanics

wvu ,, vu,

Unknowns 3D 2D 1D-----------------------------------------------------------------------Displacements

Stresses

Strains

-----------------------------------------------------------------------Total Number 15 8 3of Unknowns

xyyyxx σσσ ,, xxσ

zxyzxy

zzyyxx

εεε

εεε

,,

,,

zxyzxy

zzyyxx

σσσ

σσσ

,,

,,

xyyyxx εεε ,,xxε

u

Equilibrium Equations

0

0

0

=+∂

∂+

∂

∂+

∂

∂

=+∂

∂+

∂

∂+

∂

∂

=+∂

∂+

∂

∂+

∂

∂

zzzyzzx

y

yzyyxy

xzxxyxx

zyx

zyx

zyx

φσσσ

φσσσ

φσσσ

Body Force

Stress-Strain Relations

{ } { }( )

−−

=−=

0

0

0

1

1

1

21][][ 0

ν

α

ε

ε

ε

ε

ε

ε

εε

σ

σ

σ

σ

σ

σ

TEDD

zx

yz

xy

zz

yy

xx

zx

yz

xy

zz

yy

xx

−

−

−−

−

−

−+=

2

2100000

02

210000

002

21000

0001

0001

0001

)21)(1(][

ν

ν

νννν

ννν

ννν

νν

ED

Strain-Displacement Relations

z

u

x

w

y

w

z

v

x

v

y

u

z

w

y

v

x

u

zx

yz

xy

zzyyxx

∂

∂+

∂

∂=

∂

∂+

∂

∂=

∂

∂+

∂

∂=

∂

∂=

∂

∂=

∂

∂=

ε

ε

ε

εεε

Boundary Conditions

Two Types of BC:1- on Displacements2-on Stresses

Compatibility Equations

yxyxzz

xzyxzy

zyyxzx

zxzx

zyyz

yxxy

zzzxyzxy

yyzxyzxy

xxzxyzxy

zxxxzz

yzzzyy

xyyyxx

∂∂

∂=

∂

∂+

∂

∂+

∂

∂−

∂

∂

∂∂

∂=

∂

∂−

∂

∂+

∂

∂

∂

∂

∂∂

∂=

∂

∂+

∂

∂−

∂

∂

∂

∂

∂∂

∂=

∂

∂+

∂

∂

∂∂

∂=

∂

∂+

∂

∂

∂∂

∂=

∂

∂+

∂

∂

εεεε

εεεε

εεεε

εεε

εεε

εεε

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

1

2

1

2

1

Compatibility EquationsWhen a body is continuous before deformation, it should remain continuous after deformation.

In other words, no cracks or gap should appear and no part should overlap another due to deformation.

Thus, the displacement field must be continuous as well as single valued. This is known as condition of compatibility

From another point of view, three strains are derived from u and v only.

xyyyxx εεε ,,

Principle of minimum potential energy

0),,(),,(),,( =−= wvuWwvuwvu pp δδπδπ

∫∫∫=v

TdV

2

1σεπ��

The operator is acting on displacement while stresses and forcesAre assumed constant.

∫∫∫∫∫∫ −=V

T

V

TdVDdVD ][ ][

2

10εεεεπ����

or

Principle of minimum potential energy

pp W−= ππ

Strain Energy Work done on the bodyby external forces

Of all possible displacement states a body can assumed whichSatisfy compatibility and given kinematic or displacement BCs,The state which satisfies the equilibrium equations makes thePotential energy assume a minimum value.

Principle of minimum potential energy

Principle of minimum potential energy

∫∫∫ ∫∫Φ+=V S

TT

p dSUdVUW ����

φ

=

z

y

x

φ

φ

φ

φ�

Φ

Φ

Φ

=Φ

z

y

x�

=

w

v

u

U�

Body forceSurface forces(Tractions)

DisplacementVector

Principle of minimum potential energy

∫∫∫ ∫∫∫ ∫∫Φ−−−=V V S

TTT

p dSUdVUdVDwvu ) 2]([2

1),,( 0

�������φεεεπ

If we use the principle of minimum potential energy:

1- We assume a simple form for displacement within each element2- The displacement field satisfy the compatibility relations3-Derive the condition that minimize the functional4- The resulting equation approximately satisfy the equilibrium equations.

The approach is called “displacement or stiffness method” ofFinite element analysis.

Principle of minimum complementary energy

pc W−= ππComplementary energy

Complementary strainenergy in term of stresses

Work done by theapplied loads duringStress changes

Of all possible stress states which satisfy the equilibrium equationsand the stress boundary conditions, the state that satisfy thecompatibility conditions will make the complementary energy assume a minimum value

Principle of minimum complementary energy

Principle of minimum complementary energy

If we use the principle of minimum complementary energy:

1- We assume a simple form for stress within each element2- The stress field satisfy the equilibrium equations3-Derive the condition that minimize the functional4- The resulting equation approximately satisfy the compatibility equations.

The approach is called “force or flexibility method” ofFinite element analysis.

Principle of stationary Reissner energy

forcesappliedbydonework

dVstressesoftermsinenergyarycomplement

ntdisplacemeoftermsinstrainsstressesInternal

Vp

−

∫∫∫

−×=π

Principle of stationary Reissner energy

Of all possible stress and displacement states the body can have, the particular set that makes the Reissner energy stationary gives the correct stress-displacement and equilibrium equations along with the boundary conditions.

1- assume the form of variation for both displacement and stress fields within elements.2- This leads to mixed formulation of finite element

Hamilton’s principle

energypotentialenergykineticTL p −=−= π

∫∫∫=V

TdVUUT����

ρ2

1

The variational principle that can be used for dynamic problems isCalled the Hamilton’s principle. The functional is define as:

Lagrangian

=

w

v

u

U

�

�

���

Hamilton’s principle

[ ]

∫∫

∫∫∫

Φ+

+−=

S

T

V

TTT

dSU

dVUDUUT

��

��������φεερ 2][

2

1

Of all possible time histories of displacement states which satisfyThe compatibility equations and the constraints or the kinematicBoundary conditions and which also satisfy the conditions at initialAnd final times (t1 and t2), the history corresponding to the actualSolution makes the Lagrangian functional a minimum.

0

2

1

=∫t

t

Ldtδ

Formulation of FE Equations(Static Analysis)

-Nodal DOFs are treated as unknowns (displacement formulation)-Principle of minimum potential energy is used.-Potential energy is expressed in terms of unknowns.-The first derivatives of the functional wrt each nodal dof, set equalto zero.

Step 1 solid body is divided into E finite elements

Step 2 the displacement model within an element e is assumed

eQN

zyxw

zyxv

zyxu

U��

][

),,(

),,(

),,(

=

=

Nodal displacement vector

Shape function matrix

Formulation of FE Equations(Static Analysis)

∑=

=E

e

e

pp

1

ππ

Step 3 The element stiffness matrix and load vector derivedFrom the principle of minimum potential energy.

∫∫∫ ∫∫ ∫∫∫−Φ−=e e e

V S V

TTTe

p dVUdSUdV φσεπ������

2

1

Element VolumePortion of elementSurface where distributedSurface force applied

Vector of bodyForce per unitvolume

Formulation of FE Equations(Static Analysis)

e

zx

yz

xy

zz

yy

xx

QB

w

v

u

xz

yz

xy

z

y

x

z

u

x

w

y

w

z

v

x

v

y

u

z

w

y

v

x

u

��][

0

0

0

00

00

00

=

∂

∂

∂

∂∂

∂

∂

∂∂

∂

∂

∂∂

∂∂

∂∂

∂

=

∂

∂+

∂

∂

∂

∂+

∂

∂

∂

∂+

∂

∂

∂

∂

∂

∂

∂

∂

=

=

ε

ε

ε

ε

ε

ε

ε

Formulation of FE Equations(Static Analysis)

][

0

0

0

00

00

00

][ N

xz

yz

xy

z

y

x

B

∂

∂

∂

∂∂

∂

∂

∂∂

∂

∂

∂∂

∂∂

∂∂

∂

=

( )00 ][]][[][ εεεσ�����

DQBDDe −=−=

Formulation of FE Equations(Static Analysis)

∫∫∫ ∫∫ ∫∫∫∫∫∫ −Φ−−=e e eeV S V

TTeTTe

V

TeeTTee

p dVNQdSNQdVDBQdVQBDBQ φεπ��������

][][]][[]][[][2

10

∑=

−=E

e

c

Te

pp PQ1

��ππ

=

MQ

Q

Q

Q�

�2

1

M=total number of displacementsOr total number of DOFs

Formulation of FE Equations(Static Analysis)

c

T

E

e V S V

TTT

E

e V

TT

p

PQ

dVNdSNdVDBQ

QdVBDBQ

e e e

e

��

����

��

−

+Φ+−

=

∑ ∫∫∫ ∫∫ ∫∫∫

∑∫∫∫

=

=

1

0

1

][][]][[

]][[][2

1

φε

π

The static equilibrium of the structure can be found by minimizingthis functional.

0 021

=∂

∂==

∂

∂=

∂

∂=

∂

∂

M

pppp

QQQor

Q

ππππ��

Formulation of FE Equations(Static Analysis)

∑ ∫∫∫ ∫∫ ∫∫∫

∑∫∫∫

=

=

+Φ++

=

E

e V S V

TT

c

E

e V

T

e e e

e

dVNdSNdVDBP

QdVBDB

1

0

1

][][]][[

]][[][

φε����

�

Global stiffness matrix

Global vector of nodaldisplacements

Vector of concentrated loads

Vector of nodal forceProduced by initial strains

Vector of nodal forceProduced by surface forces

Vector of nodal forceProduced by body forces

Formulation of FE Equations(Static Analysis)

∫∫∫

∫∫

∫∫∫

∫∫∫

=

Φ=

=

=

e

e

e

e

V

Te

b

S

Te

S

V

e

i

V

Te

dVNP

dSNP

dVDBP

dVBDBK

φ

ε

��

��

��

][

][

]][[

]][[][][

0

( )∑∑==

+++=

E

e

e

b

e

S

e

ic

E

e

ePPPPQK

11

][�����

Element stiffness matrix

Element load by initial strains

Element load by surface loads

Element load by body forces

Formulation of FE Equations(Static Analysis)

The load vectors are called kinematically consistent because they satisfy the virtual work equation.

Step 4 The equilibrium equations of the overall structure is:

PQK��

=][

Step 5 & 6 The nodal displacements will be obtained from the Solution of the above equation. The stresses will be calculatedFrom the nodal displacements.

Formulation of FE Equations(Static Analysis)

1- The calculation of stiffness matrix and load vectors neednumerical integration. In simple cases the evaluation of integrals is simple.

2-The obtained formulae remain the same for all types of elements.However, the order of stiffness matrix and load vectors will changefor different types of elements.

3-The stiffness matrix is symmetric as the matrix [D] is symmetric.

4-In some cases, it is simpler to calculate the stiffness matrix and load Vectors in local coordinates and then transfer them to the global one.

5- The stiffness matrix is singular and in order to solve the finalEquation, we need to impose some BCs.

Exercise

ωq1

q2

Node 1

Node 2

L Consider a uniform bar with length l andCross section A rotating with a constantSpeed. Using the centrifugal force as bodyForce, determine the stiffness matrix and Load vector of element using a linear Displacement model:

21 )/()/1()( qlxqlxxu +−=

Assume a linear stress-strain relation.

Formulation of FE Equations(Dynamic Analysis)

e

e

QBDD

QB���

��

]][[][

][

==

=

εσ

ε

Step 1: Idealize the body into E finite elements.

Step 2: Assume the displacement model of element e as

Step 3: Assume the displacement model of element e as

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

)()],,([

),,,(

),,,(

),,,(

),,,(

tQzyxNtzyxU

tQzyxN

tzyxw

tzyxv

tzyxu

tzyxU

e

e

��

��

��

=

=

=

Formulation of FE Equations(Dynamic Analysis)

pTL

Q

R

Q

L

Q

L

dt

d

π−=

=

∂

∂+

∂

∂−

∂

∂}0{�

���

�

∫∫∫ ∫∫ ∫∫∫−Φ−=e e e

V S V

TTTe

p dVUdSUdV φσεπ������

2

1

Use Lagrange equation

∫∫∫

∫∫∫

=

=

e

e

V

Te

V

Te

dVUUR

dVUUT

��

��

��

��

µ

ρ

2

1

2

1

Potential energy

Kinetic energy

Dissipative functionFor viscous damping

Formulation of FE Equations(Dynamic Analysis)

)()(][)(][

]][[][2

1

1

1

tPQdVtNdStNQ

QdVBDBQ

c

TE

e S V

TTT

E

e V

TT

p

e e

e

�����

��

−

+Φ−

=

∑ ∫∫ ∫∫∫

∑∫∫∫

=

=

φ

π

QdVNNQTE

e V

TT

e

��

��

= ∑∫∫∫

=1

][][2

1ρ

QdVNNQRE

e V

TT

e

��

��

= ∑∫∫∫

=1

][][2

1µ

Element mass matrix

Element damping matrix

Element stiffness matrix

Formulation of FE Equations(Dynamic Analysis)

QCQT

QMQT

PQQKQ

T

T

TT

p

��

��

��

��

����

][2

1

][2

1

][2

1

=

=

−=π

( ) e

c

E

e

e

b

e

s

E

e

e

E

e

e

E

e

e

PtPtPtP

CC

MM

KK

����++=

=

=

=

∑

∑

∑

∑

=

=

=

=

1

1

1

1

)()()(

][][

][][

][][Global stiffness matrix

Global mass matrix

Global damping matrix

Global load vector

)()(][)(][)(][ tPtQKtQCtQM���

���� =++

Finally:

Step 4: assemble the element matrices and load vectors

Formulation of FE Equations(Dynamic Analysis)

Step 5 & 6: Solve the equation by using boundary and initial conditions.Different methods are available for solving the obtained governingequation.

Once the nodal displacement time history calculated, the stresses andStrains time history can be found similar to the static case.

Mass matrix of a 1D bar

]/ /1[ lxlxN −=

== ∫∫∫ 21

12][][][ ALdVNNM

eV

Te ρρ

For 1D bar we had the following shape function

L

x

==

10

01

2][

ALM

e ρ

Consistent mass matrix

Lumped mass matrix

Consistent mass matrix in global coordinates

]][[][][

] ][[][2

1

][

][

][2

1

λλ

λλ

λ

λ

eT

eeTTe

ee

ee

eeTe

mM

QmQT

Qq

Qq

qmqT

=

=

=

=

=

��

��

��

��

��

��

�� The kinetic energy of an element

The transformation between the localAnd global coordinates

Consistent mass matrix in global coordinates

Note: For element with translational DOFs only the consistent mass matrixIs invariant wrt the orientation and position of axis.

[M]=[m] Truss, membrane elements and 3D elements like tetrahedralhaving only translational DOFs. However, this is not true forBeam, frame and plate elements.

Consistent mass matrix of 3D bar

166313 ][

)(

)(

)(

)( ××× =

= eQN

xw

xv

xu

xU��

−

−

−

=

lxlx

lxlx

lxlx

N

/00/100

0/00/10

00/00/1

][

=

== ∫∫∫

200100

020010

002001

100200

010020

001002

6

][][][][

Al

dVNNMme

V

Tee

ρ

ρ

Shape function matrix

Consistent mass matrix of 3D bar

Consistent mass matrix of 3D beam (frame)

=

00000000

00000000

0000000000

][

6543

6543

21

NNNN

NNNN

NN

N

2

23

6

3

23

5

2

223

4

3

323

3

2

1

32

2

32

/

/1

l

lxxN

l

lxxN

l

xllxxN

l

llxxN

lxN

lxN

−=

−−=

+−=

+−=

=

−=

Note: The terms corresponding toTorsion of beam are set to zero.The mass matrix for torsion willCalculate separately and will add To this one.

Consistent mass matrix of 3D beam (frame)

−−−

−

−

−

=

105000

210

110

140000

420

130

1050

210

11000

1400

420

1300

300000

6000

35

13000

420

130

70

900

35

130

420

13000

70

90

3

100000

6

1105

000210

110

1050

210

1100

3000

35

1300

35

130

3

1

][

22

22

2

2

llll

llllA

J

A

J

l

l

ll

llA

J

Alme ρ

Consistent mass matrix of 2D beam (frame)

−−−

=

105210

110

140420

130

35

130

420

13

70

90

3

100

6

1105210

110

35

130

3

1

][

22

2

llll

l

ll

Alme ρ

Consistent mass matrix of 2D beam

−−−

−

−

−

=

22

22

422313

221561354

313422

135422156

420][

llll

ll

llll

ll

Alm

e ρ

Transformation needed for derivation the element mass matrix inGlobal coordinates.

If the beam cross section is not small, the effect of rotary inertiaAnd shear deformation become important in the dynamic analysis.