M. Vable Notes for finite element method: Intro to FEMIntroduction to Finite Element Method

There are two version of FEM1. Flexibility Method or Force Method: 2. Stiffness Method or Displacement Method.

The set of equations in the stiffness method are the equilibrium equa-tions relating displacements of points.

Rayleigh-Ritz is an approximate method based on energy principle by which we can obtain equilibrium equations in matrix form.

The learning objectives in this chapter is:

1. Understand Rayleigh-Ritz method and its application to axial mem-bers, torsion of circular shafts, and symmetric bending of beams.

2. Understand the perspective, the key issues, the terminology, and steps used in solving problems by finite element method.8-1

M. Vable Notes for finite element method: Intro to FEMRayleigh Ritz

Rayleigh Ritz method is an approximate method of finding displace-ments that is based on the theorem of minimum potential energy.

The method is restricted to conservative systems that may be linear or non-linear.

Potential energy function (): U W=U is the strain energy W is the work potential of a force.

Internal virtual work: Wint U=External virtual work: Wext W=Virtual work principle: Wint Wext U W 0= = =

The zero virtual variation in the potential energy function occurs where the slope of the potential energy function with respect to the parameters defining the potential function is zero.

The parameters defining the potential function are the generalized dis-placements.

The zero slope condition can occur where the potential function is:

maximumUnstable equilibriumsaddle pointNeutral equilibriumminimumStable equilibrium8-2

M. Vable Notes for finite element method: Intro to FEMMinimum Potential Energy

The theorem of minimum potential energy can be stated as:

Of all the kinematically admissible displacement functions the actual displacement function is the one that minimizes the potential energy function at stable equilibrium.

Corollary

The better approximation of displacement function is the one that yields a lower potential energy.

The greater the degrees of freedom, lower will be the potential energy for a given set of kinematically admissible functions.

ProcedureLet the displacement u(x) be represented by a series of kinematically admissible functions fi(x), i.e.,

u x( ) Cifi x( )i 1=

n

=Substitute in Potential Energy: C1 C2 C3Cn, ,( )=Minimize Potential energy:

Ci 0= i 1 n,=

Set of n-algebraic equations in the unknown Ci.8-3

M. Vable Notes for finite element method: Intro to FEMRayleigh Ritz for Axial Members

Axial strain energy density:

Ua12---EA

xddu

2=

Displacement Approximation: u x( ) Cifi x( )i 1=

n

=

xddu Cj xd

dfj

j 1=

n

=

Ua12---EA

xddu xd

du

12---EA Cj xd

dfj

j 1=

n

Ck xd

dfk

k 1=

n

= =

Ua12---EA CjCk xd

dfj

xddfk

k 1=

n

j 1=

n

=Axial strain energy

UA Ua xd

0

L

12--- CjCk EA xddfj

xddfk

xd

0

L

k 1=

n

j 1=

n

= =

Define: Kjk EA xddfj

xddfk

xd

0

L

= ----Stiffness matrix EA can be function of x. Kjk Kkj= Stiffness Matrix is symmetric8-4

M. Vable Notes for finite element method: Intro to FEMAxial Strain Energy: UA12--- CjCkKjk

k 1=

n

j 1=

n

=Work potentialpx(x) axial distributed force Fq concentrated axial forces applied at m xq pointsThe work potential of these forces can be written as:

WA px x( )u x( ) xd0

L

Fqu xq( )q 1=

m

+=Substituting the displacement approximation.

WA Cj px x( )fj x( ) xd0

L

j 1=

n

FqCjfj xq( )j 1=

n

q 1=

m

+=

Define: Rj px x( )fj x( ) xd0

L

Fqfj xq( )q 1=

m

+= ---Load Vector

WA CjRjj 1=

n

=Axial potential energy function:

A UA WA 12--- CjCkKjkk 1=

n

j 1=

n

CjRjj 1=

n

= =8-5

M. Vable Notes for finite element method: Intro to FEMAxial potential energy function:

A UA WA 12--- CjCkKjkk 1=

n

j 1=

n

CjRjj 1=

n

= =

Minimize the potential energy function: CiA 0=

12--- Ci

Cj Ck Cj CiCk+ Kjk

k 1=

n

j 1=

n

CiCj Rj

j 1=

n

0= i 1 n,=

Note: CiCj 1 i j=

0 i j

=

12--- CkKik

k 1=

n

CjKjij 1=

n

+ Ri 0= i 1 n,=

12--- Cj Kij Kji+( )

j 1=

n

Ri 0= i 1 n,=

Using symmetry of stiffness matrix: Kji Kij=

KijCjj 1=

n

Ri= i 1 n,= or K[ ] C{ } R{ }=At Equilibrium only

A 12--- Cj KjkCkk 1=

n

j 1=

n

CjRjj 1=

n

=

A 12--- CjRjj 1=

n

CjRjj 1=

n

12--- CjRjj 1=

n

= =8-6

M. Vable Notes for finite element method: Intro to FEMStrain energy at equilibrium

UAWA

2--------- A( )12--- CjRj

j 1=

n

= = = at equilibrium More the degree of freedom, lower is the potential energy (more nega-

tive), higher is the strain energy stored in the structure. With greater degree of freedom the structure becomes less stiff and

hence can store more energy. In FEM mesh refinement schemes try to get a more uniform distribu-

tion of strain energy across the structure.8-7

M. Vable Notes for finite element method: Intro to FEMSymmetric Bending of Beams

Bending strain energy density:Ub12---EIzz x2

2

d

d v 2

=

Displacement Approximation:v x( ) Cifi x( )i 1=

n

=Work Potential:py(x) distributed load Fq concentrated transverse forces applied at m1 xq pointsMq concentrated transverse forces applied at m2 xq points

WA py x( ) x( ) xd0

L

Fqv xq( )q 1=

m1

Mq xddv xq( )q 1=

m2

+ +=

Stiffness Matrix: Kjk EIzz( )x2

2

d

d fj

x2

2

d

d fk

xd

0

L

=Load Vector:

Rj py x( )fj x( ) xd0

L

Fqfj xq( )q 1=

m1

Mq xddfj xq( )

q 1=

m1

+ +=8-8

M. Vable Notes for finite element method: Intro to FEM8.6 For the beam shown in Fig. P8.6 determine one and two parameter Rayleigh-Ritz displacement solution using the approximation for the bending displacement given below. Calculate the potential energy func-tion for the one and two parameters.

w kips / in

L inL in

x

y

A B C

Fig. P8.6

v x( ) Ci ix2L--------sini 1=

n

=8-9

M. Vable Notes for finite element method: Intro to FEMFinite Element Method (FEM)

In Rayleigh-Ritz the kinematically admissible function is over the entire structure. In FEM the kinematically admissible function is piecewise continuous over small (finite) domains called the elements. Potential energy is a scaler quantity and can be written as:

i( )i 1=

n

=where, n is the number of elements (structural members). This perspective of reducing the complexity of analysis of large struc-

tures to analysis of simple individual members (elements) is what makes the finite element method such a versatile and popular analysis tool in structural (and engineering application) analysis.

Definition 1 Nodes are points on the structure at which displacements and rotations are to be found or prescribed.

Definition 2 Element is a small domain on which we can solve the boundary value problem in terms of the displacements and forces of the nodes on the element.

Definition 3 The discrete representation of the structure geometry by elements and nodes is called a mesh.

Definition 4 The process of creating a mesh (discrete entities) is called discretiza-tion.

Definition 5 Interpolation function is a kinematically admissible displacement func-tion defined on an element that can be used for interpolating displace-ment values between the nodes.

Definition 6 The mesh, boundary conditions, loads, and material properties repre-senting the actual structure is called a model.

Definition 7 Element stiffness matrix relate the displacements to the forces at the element nodes.

Definition 8 Global stiffness matrix is an assembly of element stiffness matrix that relates the displacements of the nodes on the mesh to applied external forces.8-10

M. Vable Notes for finite element method: Intro to FEMLagrange polynomials

Polynomial functions that yield function value at finite number of points.

Linear Approximation: u x( ) C1 C2x+=x

x1 x2

u1 u2

Node 1 Node 2

Conditions: u x1( ) u1= u x2( ) u2=

u x( ) u1x x2x1 x2-----------------

u2x x1x2 x1-----------------

+=

Define: L1 x( )x x2x1 x2-----------------

= L2 x( )x x1x2 x1-----------------

=

u x( ) u1L1 x( ) u2L2 x( )+ uiLi x( )i 1=

2

= =

L1(x) x

xL2(x)

1

18-11

M. Vable Notes for finite element method: Intro to FEMQuadratic displacement function: u x( ) C1 C2x C3x2+ +=x

x1x2

u1 u3

Node 1 Node 3

u2

Node 2

x3

Alternative Approach: u x( ) uiLi x( )i 1=

n

= [

u xj( ) uiLi xj( )i 1=

n

uj= = Li xj( ) 1 i j=0 i j

=

L1 x( ) a1 x x2( ) x x3( )= L1 x( )x x2x1 x2----------------- x x3

x1 x3-----------------

=

L2 x( ) a2 x x3( ) x x1( )= L2 x( )x x1x2 x1----------------- x x3

x2 x3-----------------

=

L3 x( ) a3 x x1( ) x x2( )= L3 x( )x x1x3 x1----------------- x x2

x3 x2-----------------

=8-12

M. Vable Notes for finite element method: Intro to FEML1(x) x

x

L2(x) 1

1

1

x

Node 1 Node 2 Node 3

L3(x)

nth order polynomial:Li x( )x xj( )xi xj( )

--------------------

j 1=i j

n

= i 1 n 1+( ),=

1 2 3

u1u3 u5

4 5

u2 u4element 1 element 2

1 2 3

u1u2 u3

element 1 element 2

Lagrange polynomials ensure that the function these are used to repre-sent will be continuous at the element ends.

The continuity of the derivative of the function cannot be ensured at the element end irrespective of the order of polynomials when Lagrange polynomials are used for representing the function.8-13

M. Vable Notes for finite element method: Intro to FEMAxial elementsNotation: superscript to designate quantities at element level and no

superscript at the global level. u31( )and u32( ) l refer to displacement of

node 3 for element 1 and 2. u3 refers to displacement of node 3 on the actual structure.

Displacements: u e( ) uie( )Li x( )

i 1=

n

=

Axial strains: xxe( ) xddu e( ) ui

e( )xd

dLi

i 1=

n

= =

Axial Stress: xxe( ) E e( )xxe( ) E e( ) ui e( ) xddLi

i 1=

n

= =

Internal axial force: N e( ) A e( )xxe( ) E e( )A e( ) uie( ) xddLi

i 1=

n

= =In Equations for Rayleigh-Ritz substitute:Ci ui

e( )= and fi x( ) Li x( )=

Element stiffness matrix: Kjke( ) E e( )A e( )

xddLj

xddLk

xd

0

L

=Element load vector:

Rje( ) px x( )Lj x( ) xd

0

L

F1e( )Lj x1( ) Fne( )Lj xn( )+ +=For px 0= R1

e( ) F1e( )= Rn

e( ) Fne( )= Rj

e( ) 0= j 2 n 1( ),=8-14

M. Vable Notes for finite element method: Intro to FEM Fje( ) is positive in the positive direction uj

e( ) .

Assembly of global matrix and global load vector The governing criteria is that the displacement function at the node

where two elements meet must be continuous i.e., be the same.

Assume there is no distributed force i.e., px 0=Use two quadratic axial elements

P/2

P/2

1 2 3 4 5

1 2 3 1 2 3

Element 1 Element 2

xA B C

Virtual variation in potential energy in an element:

m( )ui

m( ) m( )uim( )

i 1=

n

=

m( ) uim( ) Kijm( )ujm( ) Rim( )( )j 1=

n

i 1=

n

=

1( )u11( )

u21( )

u31( ) T

K111( ) K12

1( ) K131( )

K211( ) K22

1( ) K231( )

K311( ) K32

1( ) K331( )

u11( )

u21( )

u31( )

F11( )

0

F31( )

=8-15

M. Vable Notes for finite element method: Intro to FEM 2( )u12( )

u22( )

u32( ) T

K112( ) K12

2( ) K132( )

K212( ) K22

2( ) K232( )

K312( ) K32

2( ) K332( )

u12( )

u22( )

u32( )

F12( )

0

F32( )

=

P/2

P/2

1 2 3 4 5

1 2 3 1 2 3

Element 1 Element 2

xA B C

u11( ) u1= u2

1( ) u2= u31( ) u3=

u12( ) u3= u2

2( ) u4= u32( ) u5=

1( )u11( )

u21( )

u31( ) T

K111( ) K12

1( ) K131( )

K211( ) K22

1( ) K231( )

K311( ) K32

1( ) K331( )

u11( )

u21( )

u31( )

F11( )

0

F31( )

=

1( )u1u2u3u4u5

T

K111( ) K12

1( ) K131( ) 0 0

K211( ) K22

1( ) K231( ) 0 0

K311( ) K32

1( ) K331( ) 0 0

0 0 0 0 00 0 0 0 0

u1u2u3u4u5

F11( )

0

F31( )

00

=8-16

M. Vable Notes for finite element method: Intro to FEM 2( )u1u2u3u4u5

T 0 0 0 0 0

0 0 0 0 0

0 0 K112( ) K12

2( ) K132( )

0 0 K212( ) K22

2( ) K232( )

0 0 K312( ) K32

2( ) K332( )

u1u2u3u4u5

00

F12( )

0

F32( )

=

Total potential energy 1( ) 2( )+=

u1u2u3u4u5

T K111( ) K121( ) K131( ) 0 0

K211( ) K22

1( ) K231( ) 0 0

K311( ) K32

1( ) K331( ) K11

2( )+( ) K122( ) K132( )

0 0 K212( ) K22

2( ) K232( )

0 0 K312( ) K32

2( ) K332( )

u1u2u3u4u5

F11( )

0

F31( ) F1

2( )+0

F32( )

=

The element stiffness matrix and element load components that add corresponds to the degree of freedom associated with the shared node of the elements.

Incorporating the external concentrated forces

P/2

P/2

1 2 3 4 5

1 2 3 1 2 3

Element 1 Element 2

xA B C

The force F31( ) is in the direction of u3

1( )

The force F12( ) is in the direction of u1

2( )8-17

M. Vable Notes for finite element method: Intro to FEMThe applied force is in the direction of u3 .

F11( ) RA= F3

1( ) F12( )+ P= F3

2( ) RC=

u1u2u3u4u5

T K111( ) K121( ) K131( ) 0 0

K211( ) K22

1( ) K231( ) 0 0

K311( ) K32

1( ) K331( ) K11

2( )+( ) K122( ) K132( )

0 0 K212( ) K22

2( ) K232( )

0 0 K312( ) K32

2( ) K332( )

u1u2u3u4u5

RA0P0

RC

=

Incorporating the boundary conditions on displacements The stiffness matrix in is singular, which reflects the fact that the two

element structure can move as a rigid body. u1 0= u1 0= u5 0= u5 0=

0u2u3u40

T K111( ) K121( ) K131( ) 0 0

K211( ) K22

1( ) K231( ) 0 0

K311( ) K32

1( ) K331( ) K11

2( )+( ) K122( ) K132( )

0 0 K212( ) K22

2( ) K232( )

0 0 K312( ) K32

2( ) K332( )

0u2u3u40

RA

0P0

RC

=

or

u2u3u4

T K221( ) K231( ) 0

K321( ) K33

1( ) K112( )+( ) K122( )

0 K212( ) K22

2( )

u2u3u4

0P0

=

By principle of minimum potential energy: 0= 8-18

M. Vable Notes for finite element method: Intro to FEMK221( ) K23

1( ) 0

K321( ) K33

1( ) K112( )+ K12

2( )

0 K212( ) K22

2( )

u2u3u4

0P0

=

Reaction Forces:

RA K121( )u2 K13

1( )u3+= RC K312( )u3 K32

2( )u4+=8-19

M. Vable Notes for finite element method: Intro to FEMElement strain energyIn Rayleigh-Ritz method we showed that the strain energy is half the work potential of the forces at equilibrium.

UAe( ) 1

2--- uje( )Rj

e( )

j 1=

n

=For px 0= R1

e( ) F1e( )= Rn

e( ) Fne( )= Rj

e( ) 0= j 2 n 1( ),=

UAe( ) 1

2--- u1

e( )F1e( ) un

e( )Fne( )+( )=

u1e( )

F1e( )

une( )

Fne( )

u1e( )

N1e( )

une( )

Nne( )

F1e( ) N1

e( )= Fne( ) Nn

e( )=

For px 0= N1e( ) Nn

e( ) N e( )= =

UAe( ) 1

2--- une( )Nn

e( ) u1e( )N1

e( )( ) 12--- une( ) u1

e( )( )N e( )= =

Mesh Refinement Elements with high strain energy identify the region of the body

where mesh should be refined.The h-method of mesh refinement reduces the size of element. The p-method of mesh refinement increases the order of polynomial in an element. The r-method of mesh refinement relocates the position of a node. Combinations: hr-method, hp-method, hpr-method8-20

M. Vable Notes for finite element method: Intro to FEMCoordinate transformation

xG

yG

uG31( )

vG31( )

uG11( )

uG21( )

vG11( )

vG21( ) u3

1( )

u11( )

u21( )

The displacement vector in global orientation: D1e( )

uG1e( )i vG1

e( ) j+=The unit vector: eL icos jsin+=Axial displacement: u1

1( ) D1e( )

eL uG1e( ) cos vG1e( ) sin+= =

u1e( )

u2e( )

u3e( )

cos sin 0 0 0 00 0 cos sin 0 00 0 0 0 cos sin

uG1e( )

vG1e( )

uG2e( )

vG2e( )

uG3e( )

vG3e( )

T[ ]

uG1e( )

vG1e( )

uG2e( )

vG2e( )

uG3e( )

vG3e( )

= =

The [T] matrix in the above equation is 3 x 6 matrix relating the local and the global coordinate system.

For n nodes on the element the matrix [T] would be n x 2n size. 8-21

M. Vable Notes for finite element method: Intro to FEMTransformation equation in matrix form: u e( ){ } T[ ] uGe( )

=

Potential energy in matrix form:

e( ) u e( ){ }T K e( )[ ] u e( ){ } R e( ){ }( )= or

e( ) uGe( ) T

T[ ]T K e( )[ ] T[ ] uGe( )

R e( ){ }

= or

e( ) uGe( ) T

T[ ]T K e( )[ ] T[ ] uGe( )

T[ ]T R e( ){ }

= or

Define: KGe( )[ ] T[ ]T K e( )[ ] T[ ]= RGe( )

T[ ]T R e( ){ }=

e( ) uGe( ) T

KGe( )[ ] uGe( )

RG

e( )

=

Linear and quadratic axial elementsWe assume the following:

---the distributed load px is evaluated at the mid point of the element and has a uniform value po

---the cross-sectional area (A) is evaluated at the mid point of the element.---the modulus of elasticity (E) is constant over the element. 8-22

M. Vable Notes for finite element method: Intro to FEMLinear: u e( ) x( ) u1e( )L1 x( ) u2e( )L2 x( )+=

L1 x( )x x2x1 x2-----------------

=

L2 x( )x x1x2 x1-----------------

=

u e( ){ }u1

e( )

u2e( )

=

K e( )[ ] EAL--------1 11 1

= R e( ){ } poL2---------11

F1e( )

F2e( )

+=

Quadratic: u e( ) x( ) u1e( )L1 x( ) u2e( )L2 x( ) u3e( )L3 x( )+ +=

L1 x( )x x2x1 x2----------------- x x3

x1 x3-----------------

=

L2 x( )x x1x2 x1----------------- x x3

x2 x3-----------------

=

L3 x( )x x1x3 x1----------------- x x2

x3 x2-----------------

=

u e( ){ }u1

e( )

u2e( )

u3e( )

=

K e( )[ ] EA3L--------

7 8 18 16 8

1 8 7

= R{ } poL6

---------141

F1

e( )

0

F3e( )

+=

e( ) u e( ){ }T K e( )[ ] u e( ){ } R e( ){ }( )=8-23

M. Vable Notes for finite element method: Intro to FEMSteps in FEM procedure1. Obtain element stiffness and element load vector.2. Transform from local orientation to global orientation.3. Assemble the global stiffness matrix and load vector.4. Incorporate the external loads5. Incorporate the boundary conditions.6. Solve the algebraic equations for nodal displacements.7. Obtain reaction force, stress, internal forces, strain energy.8. Interpret and check the results.9. Refine mesh if necessary, and repeat the above steps.8-24

M. Vable Notes for finite element method: Intro to FEM8.8 A rectangular tapered aluminum bar (Eal = 10,000 ksi, = 0.25) is shown in Figure 8.8. The depth of the tapered section varies as: h x( ) 4 0.04x= .Find the stress at point B, displacement at point C and the strain energy in each element using the following FEM model: one linear element in AB and one quadratic element in BC.

P = 10 kips

50 in

4 in

1 in

A B h(x)x

10 in

C

Fig. P8.88-25

M. Vable Notes for finite element method: Intro to FEM8.10 A force F= 20 kN is applied to the roller that slides inside a slot as shown in Fig. P8.10. Both bars have an area of cross-section of A = 100 mm2 and a Modulus of Elasticity E = 200 GPa. Bar AP and BP have lengths of LAP= 200 mm and LBP= 250 mm respectively. Deter-mine the displacement of the roller and the reaction force on the roller using linear elements to represent each bar.

P

110oF

B

A

Fig. P8.108-26

M. Vable Notes for finite element method: Intro to FEMSymmetric beam elements

In beam bending problems deflection v and its derivative xddv must be

continuous at all points on the beam including the element ends. Lagrange polynomials cannot be used for representing v because it

would result in slope being discontinuous at the element end irrespec-tive of the order of polynomial.

Deflection Approximation: v x( ) C1 C2x C3x2 C4x3+ + +=

Conditions: v x1( ) v1e( )= xd

dv x1( ) 1e( )=

v x2( ) v2e( )= xddv x2( ) 2e( )=

v x( ) f1 x( )v1e( ) f2 x( )1e( ) f3 x( )v2e( ) f4 x( )2e( )+ + +=

f1 x( ) 1 3x x1

L-------------- 2 2 x x1L--------------

3+=

f2 x( ) Lx x1

L-------------- 2 x x1L--------------

2 x x1L-------------- 3+=

f3 x( ) 3x x1

L-------------- 2 2 x x1L--------------

3=

f4 x( ) Lx x1

L--------------

2 x x1L--------------

3+=8-27

M. Vable Notes for finite element method: Intro to FEMConditions on interpolating functions

v x1( ) v1e( ) f1 x1( )v1e( ) f2 x1( )1e( ) f3 x1( )v2e( ) f4 x1( )2e( )+ + += =f1 x1( ) 1= f2 x1( ) 0= f3 x1( ) 0= f4 x1( ) 0=

xddv x1( ) xd

df1 x1( )v1e( ) xddf2 x1( )1e( ) xd

df3 x1( )v2e( ) xddf4 x1( )2e( )+ + +=

1e( )=

xddf1 x1( ) 0= xd

df2 x1( ) 1= xddf3 x1( ) 0= xd

df3 x1( ) 0=.

f1(x) x

xf2(x)

1

1

zero slope

zero slope

Node 1 Node 2

f3(x) x

xf4(x)

1

1

zero slope

zero slope

Node 1 Node 2

Hermite Polynomials for Beam Bending8-28

M. Vable Notes for finite element method: Intro to FEMElement Stiffness Matrix:From Rayleigh-Ritz the stiffness matrix is:

Kjk EIzz( )x2

2

d

d fj

x2

2

d

d fk

xd

0

L

=Assume: the area moment of inertia (I) is evaluated for the cross-section at the

mid point of the element. the modulus of elasticity (E) is constant over the element.

v e( ){ }

v1e( )

1e( )

v2e( )

2e( )

= K e( )[ ] 2EIL3---------

6 3L 6 3L

3L 2L2 3 L L2

6 3 L 6 3 L

3L L2 3 L 2L2

=

Element Load VectorFrom Rayleigh-Ritz the load vector is:

Rj py x( )fj x( ) xd0

L

Fqfj xq( )q 1=

m1

Mq xddfj xq( )

q 1=

m1

+ += Assume point forces and moments can only be applied at element end

nodes.

Rje( ) py x( )fj x( ) xd

0

L

F1e( )fj x1( ) F2e( )fj x2( ) M1e( ) xddfj x1( ) M2e( ) xddfj x2( )+ + + +=8-29

M. Vable Notes for finite element method: Intro to FEM the distributed load py is evaluated at the mid point of the element and has a uniform value po.

R e( ){ } p0L12---------6L6L

F1e( )

M1e( )

F2e( )

M2e( )

+=

Positive directions for displacements, slopes, nodal forces and moments.

v1e( ) v2

e( )

1e( ) 2

e( )F1

e( ) F2e( )

M1e( ) M2

e( )

Be( ) v e( ){ }T

K e( )[ ] v e( ){ } R e( ){ }( )=8-30

M. Vable Notes for finite element method: Intro to FEM8.23 Using a single beam element for AB and a single beam element for BC in Fig. P8.23, determine (a) the slope at B and C (b) reaction force and moment at A.

L 2L

w

xA B C

Fig. P8.238-31

Introduction to Finite Element MethodRayleigh RitzMinimum Potential EnergyRayleigh Ritz for Axial MembersSymmetric Bending of Beams8.6 For the beam shown in Fig. P8.6 determine one and two parameter Rayleigh-Ritz displacement solution using the approximation for the bending displacement given below. Calculate the potential energy function for the one and two parameters.Fig. P8.6

Finite Element Method (FEM)Definition 1 Nodes are points on the structure at which displacements and rotations are to be found or prescribed.Definition 2 Element is a small domain on which we can solve the boundary value problem in terms of the displacements and forces of the nodes on the element.Definition 3 The discrete representation of the structure geometry by elements and nodes is called a mesh.Definition 4 The process of creating a mesh (discrete entities) is called discretization.Definition 5 Interpolation function is a kinematically admissible displacement function defined on an element that can be used for interpolating displacement values between the nodes.Definition 6 The mesh, boundary conditions, loads, and material properties representing the actual structure is called a model.Definition 7 Element stiffness matrix relate the displacements to the forces at the element nodes.Definition 8 Global stiffness matrix is an assembly of element stiffness matrix that relates the displacements of the nodes on the mesh to applied external forces.

Lagrange polynomialsAxial elements8.8 A rectangular tapered aluminum bar (Eal = 10,000 ksi, n = 0.25) is shown in Figure 8.8. The depth of the tapered section var...Fig. P8.8

8.10 A force F= 20 kN is applied to the roller that slides inside a slot as shown in Fig. P8.10. Both bars have an area of cross...Fig. P8.10

Symmetric beam elements8.23 Using a single beam element for AB and a single beam element for BC in Fig. P8.23, determine (a) the slope at B and C (b) reaction force and moment at A.Fig. P8.23