Beams and frames

• Beams are slender members used for supporting transverse loading.

• Beams with cross sections symmetric with respect to loading are considered.

EIMdxvdE

yI

M

///

22 =

∈=

−=

σ

σ

Potential energy approach

Strain energy in an element of length dx is

∫

∫

∫

⎟⎟⎠

⎞⎜⎜⎝

⎛=

=

A

A

A

IinertiaofmomenttheisdAy

dxdAyEIM

dAdxdU

2

22

2

21

21 σε

The total strain energy for the beam is given by-

( )∫=L

dxdxvdEIU0

22 /21

( ) ∑∑∫∫ −−−=Πk

kkm

mm

LL

vMvppvdxdxdxvdEI '

00

22 /21

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 Approachp

dx

V

V+dV

M M+dM

•Here we start from equilibriumof an elemental length.

dV/dx = pdM/dx =V

EIMdxvd // 22 =

( ) 022

2

2

=− pdxvdEI

dxd

( )∫ =Φ⎥⎦⎤

⎢⎣⎡ −

Ldxpdx

vdEIdxd

0 22

2 0

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

02

2

02

2

2

2

02

2

002

2

2

2

=Φ

−Φ

−Φ⎟⎟⎠

⎞⎜⎜⎝

⎛+

Φ⎟⎟⎠

⎞⎜⎜⎝

⎛+Φ−

Φ∫∫

L

x

xL

x

xlL

k

k

m

m

dxd

dxvdEI

dxd

dxvdEI

dxvdEI

dxd

dxvdEI

dxddxpdx

dxd

dxvdEI

∑∑∫∫ =Φ−Φ−Φ−Φ

kkk

mmm

LL

Mpdxpdxdxd

dxvdEI 0'

002

2

2

2

Further simplifying-

Φ and M are zero at support..at xm shear force is pm and at xkBending moment is -Mk

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

Q1 Q3 Q5 Q7 Q9

Q2 Q8Q6Q4 Q10

e1 e3e2 e4

[ ]TQQQQQ 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 definedin terms of ζ from –1 to 1.•The shape functions for beam elements differ from those definedearlier. Therefore, we define ‘Hermite Shape Functions’

H1

H4

H3

H21

1

Slope=0

Slope=0

Slope=0

Slope=0

Slope=0

Slope=1

Slope=1

Slope=0

4,3,2,132 =+++= idcbaH iiiii Kζζζ

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

2141

1141

2141

24

23

22

21

−+=

++=

+−=

+−=

ζζ

ζζ

ζζ

ζζ

H

H

H

H

Finding out values of coefficients and simplifying,

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

2433

1211)( ⎟⎟

⎠

⎞⎜⎜⎝

⎛++⎟⎟

⎠

⎞⎜⎜⎝

⎛+=

ζζζ

ddvHvH

ddvHvHv

The coordinates transform by relationship-

ζ

ζ

ζζ

dldx

xxxx

xxx

e

2

22

21

21

1221

21

=

−+

+=

++

−=

(x2-x1) is the length of element le

⎥⎦⎤

⎢⎣⎡=

=

+++=

=

4321

44332211

2,,

2,

22)(

,2

HlHHlHH

whereHqv

qHlqHqHlqHv

Thereforedxdvl

ddv

ee

ee

e

ζ

ζ

( )

qd

Hdd

Hdl

qdx

vd

equationaboveinngsubstitutid

vdldx

vdandddv

ldxdv

dxdxvdEIU

T

e

T

ee

ee

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

==

= ∫

2

2

2

2

42

2

2

2

2

2

22

16

42

/21

ζζ

ζζ

⎥⎦⎤

⎢⎣⎡ +−+−

=⎟⎟⎠

⎞⎜⎜⎝

⎛22

31,23,

2231,

23

2

2ee ll

dHd ζζζζζ

Where-

∫−

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

⎟⎠⎞

⎜⎝⎛ +

+−

+−+−

−⎟⎠⎞

⎜⎝⎛ +−

+−+−

=1

1

22

2

22

22

22

3

431

)31(834916

91)31(83

431

)31(8349)31(8349

821 qd

l

lsymmetric

lll

ll

lEIqU

e

e

eee

ee

e

Te ζ

ζ

ζζζ

ζζζζ

ζζζζζζ

qkqU

ddd

eT

e 21

2032 1

1

1

1

1

1

2

=

=== ∫∫∫−−−

ζζζζζ

Note that-

This result can be written as-

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−

−−−−

−

=

22

22

3

4626

6126122646

612612

eeee

ee

eeee

ee

ee

llll

llllll

ll

lEIk

Where Ke 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.

qHdplpvdx e

le⎟⎟⎠

⎞⎜⎜⎝

⎛= ∫∫

−

1

12ζ

Substituting the value of H we get-

T

eeeee

e

l

plplplplf

where

qfpvdxT

e

⎥⎦

⎤⎢⎣

⎡−=

=∫

12,

2,

12,

2

22

This is equivalent to the element shown below-

1 2le

p

21 e

Ple/2 Ple/2Ple

2/12 -Ple2/12

The point loads Pm and Mk are readily taken care of byintroducing the nodes at the point of application.

Introducing local-global correspondence from potential energy approach we get-

.

0

21

vectorntdisplacemevirtualadmissiblewhere

FKQ

QFKQQ

TT

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 Krr 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-

HqvanddxdMV

dxvdEIM === 2

2

We have,

⎪⎪

⎭

⎪⎪

⎬

⎫

⎪⎪

⎩

⎪⎪

⎨

⎧

−−

−

+

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−

−−−−

−

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

12

212

2

4626

6126122646

612612

2

2

4

3

2

1

22

22

3

4

3

2

1

e

e

e

e

eeee

ee

eeee

ee

e

pl

plpl

pl

qqqq

llll

llllll

ll

lEI

RRRR

V1 = R1 V2 = -R3 M1 = -R2 M2 = R4

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

TT

l

qdxHHsq

Hqv

dxsv

21

21

0

2

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−−−

−

−

=

= ∑

22

22

422313

221561354313422

135422156

420

21

eeee

ee

eeee

ee

ese

se

e

se

T

llll

llllll

ll

slk

foundationelasticformatrixstiffnessiskwhere

kq

PLANE FRAMES•Plane structure with rigidly connected members.

•Similar to beams except that axial loads and deformationsare 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-

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−

=

1000000000

000000010000000000

lmml

lmml

L

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

Combining two stiffness and rearranging at proper locations weget element stiffness matrix as-

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−−−

−

−

−

−

=

eeee

eeee

ee

eeee

eeee

ee

e

lEI

lEI

lEI

lEI

lEI

lEI

lEI

lEI

lEA

lEA

lEI

lEI

lEI

lEI

lEI

lEI

lEI

lEI

lEA

lEA

k

460260

61206120

0000

260460

61206120

0000

22

2323

22

2323

'

LkLK

LqkL

qkWapproachsGalerkinby

LqkLq

qkqU

eTe

eTT

eTe

eTT

eTe

'

'

'

'

'

'','

21

''21

=

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 onmember, we have-

FKQformglobalIn

fLf

plplplplf

wherefLqfq

T

T

eeee

TTT

=

=

⎥⎦

⎤⎢⎣

⎡−=

=

,'

12,

2,0,

12,

2,0'

'''

22