herimite shape function for beam

7/27/2019 Herimite Shape Function for Beam

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Beams and frames

• Beams are slender members used forsupporting transverse loading.

• Beams with cross sections symmetric with

respect to loading are considered.

EI dxvd

E

y

I

M

//

/

22 =

∈=

−=

σ

σ



Potential energy approach

Strain energy in an element of length dx is

∫∫

∫

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ =

=

A

A

A

I inertiaof moment theisdA y

dxdA y

EI

M

dAdxdU

2

2

2

2

2

1

2

1 σε

The total strain energy for the beam is given by-

( )∫= L

dxdxvd EI U 0

22 /2

1



( ) ∑∑∫∫

−−−=Πk

k k

m

mm

L L

v M v p pvdxdxdxvd EI '

00

22 /2

1

Potential energy of the beam is then given by-

Where-

-p is the distributed load per unit length-pm is the point load at point m.

-Mk is the moment of couple applied at point k

-vm

is the deflection at point m

-v’k is the slope at point k.



Galerkin’s Approach p

dx

V

V+dV

MM+dM

•Here we start from equilibrium

of an elemental length.

dV/dx = pdM/dx =V

EI M dxvd // 22 =

( ) 02

2

2

2=− p

dxvd EI

dx

d

( )∫ =Φ⎥⎦⎤

⎢⎣⎡ −

L

dx pdx

vd EI dx

d 0

2

2

20

For approximate solution by Galerkin’s approach-

Φ is an arbitrary function using same basic functions as v



Integrating the first term by parts and splitting the interval 0 to L

to (0 to xm), (xm to xk ) and (xk to L) we get-

02

2

0

2

2

2

2

0

2

2

00

2

2

2

2

=Φ

−Φ

−Φ⎟⎟

⎠

⎞⎜⎜

⎝

⎛ +

Φ⎟⎟ ⎠

⎞

⎜⎜⎝

⎛

+Φ−

Φ

∫∫ L

x

x L

x

xl L

k

k

m

m

dx

d

dx

vd EI

dx

d

dx

vd EI

dx

vd EI

dx

d

dx

vd

EI dx

d

dx pdxdx

d

dx

vd

EI

∑∑∫∫ =Φ−Φ−Φ−

Φ

k k k

mmm

L L

M pdx pdxdx

d

dx

vd

EI 0

'

002

2

2

2

Further simplifying-

Φ and M are zero at support..at xm shear force is pm and at xk

Bending moment is -Mk



•Beam is divided in to elements…each node has two degrees of

freedom.

•Degree of freedom of node j are Q2j-1 and Q2j

• Q2j-1 is transverse displacement and Q2j is slope or rotation.

FINITE ELEMENT FORMULATION

Q1Q3

Q5 Q7 Q9

Q2Q8Q6Q4 Q10

e1 e3e2 e4

[ ]T

QQQQQ 10321 ,, K

=

Q is the global displacement vector.



e

q1

q4q2

q3

Local coordinates-

],,,[

],,,[

'

22

'

11

4321

vvvv

qqqqq T

=

=

•The shape functions for interpolating v on an element are defined

in terms of ζ from –1 to 1.

•The shape functions for beam elements differ from those defined

earlier. Therefore, we define ‘Hermite Shape Functions’



H1

H4

H3

H2

1

1

Slope=0

Slope=0

Slope=0

Slope=0

Slope=0

Slope=1

Slope=1

Slope=0



4,3,2,132 =+++= id cba H iiiii Kζ ζ ζ

Each Hermite shape function is of cubic order represented by-

The condition given in following table must be satisfied.

10010000ζ=1

00001001ζ=-1

H’4H4H’3H3H’2H2H’1H1



( ) ( )

( ) ( )

( ) ( )

( ) ( )1141

214

1

114

1

214

1

24

2

3

2

2

2

1

−+=

++=

+−=

+−=

ζ ζ

ζ ζ

ζ ζ

ζ ζ

H

H

H

H

Finding out values of coefficients and simplifying,



Hermite functions can be used to write v in the form-

2

433

1

211)( ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ++⎟⎟

⎠

⎞⎜⎜⎝

⎛ +=

ζ ζ ζ

d

dv H v H

d

dv H v H v

The coordinates transform by relationship-

ζ

ζ

ζ ζ

d l

dx

x x x x

x x x

e

2

22

21

21

1221

21

=

−+

+=

++−=

(x2-x1) is the length of

element le



⎥

⎦

⎤⎢

⎣

⎡=

=

+++=

=

4321

44332211

2

,,

2

,

22)(

,

2

H l

H H l

H H

where

Hqv

q H l

q H q H l

q H v

Therefore

dx

dvl

d

dv

ee

ee

e

ζ

ζ



( )

qd

H d

d

H d

lq

dx

vd

equationaboveinngsubstituti

d

vd

ldx

vd

and d

dv

ldx

dv

dxdxvd EI U

T

e

T

ee

e

e

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ =

==

= ∫

2

2

2

2

42

2

2

2

2

2

22

16

42

/2

1

ζ ζ

ζ ζ

⎥⎦

⎤⎢⎣

⎡ +−+−=⎟⎟

⎠

⎞⎜⎜⎝

⎛

22

31,

2

3,

22

31,

2

32

2

ee ll

d

H d ζ ζ

ζ ζ

ζ

Where-



∫−

⎥⎥⎥⎥

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎢⎢⎢⎢

⎣

⎡

⎟ ⎠

⎞⎜⎝

⎛ +

+−

+−+−

−⎟ ⎠

⎞⎜⎝

⎛ +−

+−+−

=1

1

2

2

2

22

2

2

22

3

4

31

)31(8349

16

91)31(

8

3

4

31

)31(8349)31(8349

8

2

1qd

l

lsymmetric

lll

ll

l

EI qU

e

e

eee

ee

e

T

e ζ

ζ

ζ ζ ζ

ζ ζ ζ

ζ

ζ ζ ζ ζ ζ ζ

qk qU

d d d

e

T

e2

1

203

21

1

1

1

1

1

2

=

=== ∫∫∫−−−

ζ ζ ζ ζ ζ

Note that-

This result can be written as-



⎥⎥⎥⎥

⎥

⎦

⎤

⎢⎢⎢⎢

⎢

⎣

⎡

−

−−−

−

−

=

22

22

3

4626

612612

2646

612612

eeee

ee

eeee

ee

e

e

llll

ll

llll

ll

l EI k

Where K e is element stiffness matrix given by

It can be seen that it is a symmetric matrix.



Load vector-

•We assume the uniformly distributed load p over the element.

q Hd

pl

pvdx

e

le ⎟⎟ ⎠

⎞

⎜⎜⎝

⎛ =

∫∫ −

1

12 ζ

Substituting the value of H we get-

T

eeeee

e

l

pl pl pl pl f

where

q f pvdxT

e

⎥⎦

⎤⎢⎣

⎡−=

=∫

12,

2,

12,

2

22



This is equivalent to the element shown below-

1 2le

p

21e

Ple/2 Ple/2

Ple2/12 -Ple2/12

The point loads Pm and Mk are readily taken care of by

introducing the nodes at the point of application.



Introducing local-global correspondence from potential energy

approach we get-

.

0

2

1

vector

nt displacemevirtualadmissiblewhere

F KQ

QF KQQ

T T

T

=Ψ

=Ψ−Ψ

−=Π

And from Galerkin’s approach we get-



BOUNDARY CONSIDERATIONS

•Let Qr = a…….single point BC

•Following Penalty approach, add 1/2C(Qr -a)2 to Π

•C represents stiffness which is large in comparison

with beam stiffness terms.

•C is added to K rr and Ca is added to Fr to get-

KQ = F

•These equations are solved to get nodal displacements.

CCa

Dof =(2i-1)

CCa

Dof = 2i



Shear Force and Bending Moment-

Hqvand

dx

dM V

dx

vd EI M ===

2

2

We have,

⎪⎪

⎭

⎪⎪

⎬

⎫

⎪⎪

⎩

⎪⎪

⎨

⎧

−

−

−

+⎪⎪

⎭

⎪⎪

⎬

⎫

⎪⎪

⎩

⎪⎪

⎨

⎧

⎥⎥⎥

⎥⎥

⎦

⎤

⎢⎢⎢

⎢⎢

⎣

⎡

−

−−−

−

−

=⎪⎪

⎭

⎪⎪

⎬

⎫

⎪⎪

⎩

⎪⎪

⎨

⎧

12

2

12

2

4626

612612

2646

612612

2

2

4

3

2

1

22

22

3

4

3

2

1

e

e

e

e

eeee

ee

eeee

ee

e

pl

pl

pl

pl

q

q

q

q

llll

ll

llll

ll

l

EI

R

R

R

R

V1 = R 1 V2 = -R 3 M1 = -R 2 M2 = R 4



Beams on elastic support

•Shafts supported on ball, roller, journal bearings

•Large beams supported on elastic walls.

•Beam supported on soil (Winkler foundation).

•Stiffness of support contributes towards PE.

•Let ‘s’ be the stiffness of support per unit length.

additional term =

∑ ∫

∫

=

=

e e

T T

l

qdx H H sq

Hqv

dxsv

2

1

2

10

2



⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−−−

−−

=

= ∑

22

22

422313221561354

313422135422156

420

2

1

eeee

ee

eeee

ee

es

e

se

e

s

e

T

llllll

llll

ll

slk

foundationelastic

for matrixstiffnessisk where

k q



PLANE FRAMES

•Plane structure with rigidly connected members.

•Similar to beams except that axial loads and deformations

are present.

•We have 2 displacements and 1 rotation at each node.

•3 dof at each node.

q =[q1, q2, q3, q4, q5, q6]T



X

Yq1

q2

q4

q5

q3 (q’3)

X ’ Y ’

1

2

Global and local coordinate systems

θ

q’4q’5

q’1q’2

q6 (q’6)



q’ =[q’1, q’2, q’3, q’4, q’5, q’6]

l,m are the direction cosines of local coordinate system. X’Y’

• l = cos(θ)

• m =sin(θ)

We can see from the figure that-

•q3 = q’3

•q3

= q’6which are rotations with respect to body.



q’ = Lq

Where-

⎥⎥⎥⎥

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎢⎢⎢⎢

⎣

⎡

−

−

=

100000

00000000

000100

0000

0000

lmml

lm

ml

L

q’2, q’3,q’5 and q’6 are beam element dof while q’1andq’4

are like rod element dof.



Combining two stiffness and rearranging at proper locations we

get element stiffness matrix as-

⎥⎥⎥⎥

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎥

⎦

⎤

⎢⎢⎢⎢

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎢

⎣

⎡

−

−−−

−

−

−

−

=

eeee

eeee

ee

eeee

eeee

ee

e

l

EI

l

EI

l

EI

l

EI

l

EI

l

EI

l

EI

l

EI

l

EA

l

EAl

EI

l

EI

l

EI

l

EI

l

EI

l

EI

l

EI

l

EI

l

EA

l

EA

k

460

260

612

0

612

0

0000

260

460

6120

6120

0000

22

2323

22

2323

'



Lk LK

Lqk L

qk W

approachsGalerkinby

Lqk Lq

qk qU

eT e

eT T

eT

e

eT T

eT

e

'

'

'

'

'

''

,'2

1

''2

1

=

Ψ=

Ψ=

=

=

Strain energy of an element is given by-

Element stiffness matrix in global form can be written as-



X

Y

p

Y’

X’

If there is distributed load on

member, we have-

F KQ

formglobal In f L f

pl pl pl pl f

where

f Lq f q

T

T

eeee

T T T

=

=

⎥⎦

⎤⎢⎣

⎡−=

=

,'

12,

2,0,

12,

2,0'

'''

22

herimite shape function for beam

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