Finite element method for structural dynamic

and stability analyses

1

Prof C S Manohar Department of Civil Engineering

IISc, Bangalore 560 012 India

Module-9

Structural stability analysis

Lecture-31 3D beam element; plate element; imperfection sensitive

structures; beams on elastic foundation; rings and arches.

2

2 22

2

0 0

Euler-Bernoulli beam element

1

2 2

l ld v P dv

U EI dx dxdx dx

P P,EI l

4u

3u

2u

1u

x

y

4

1

i i

i

v x u x

2 2

3

2 2

2 2

2 2

12 6 12 6

6 4 6 2

12 6 12 6

6 2 6 4

36 3 36 3

3 4 3

36 3 36 330

3 3 4

G

l l

l l l lEIK

l ll

l l l l

l l

l l l lPK

l ll

l l l l

3

2 2 2

0 0

0 0 0

0 0 0

1

2

0 01

0 0 with =2

0 0

xx zx

V

e e e

xx xy xz

t

xy yy yz

V

u v w u u v v w wU dV

x x x y z y z y z

u

u

v N u N u G u

w

s

U s dV s

s

0 0 0

0 01

0 02

0 0

0 01

with 0 0 Geometric stiffness matrix2

0 0

xz yz zz

tt t

e e e e

V

t

V

s

U u G s G u dV u K u

s

s

K G s G dV

s

4

Axially deforming rod

l l

,x u ,x u

,z w ,z w

2u

2w

2

121

1 1 2 2

1 1 2 2

1 2;

u x N u N u

w x N w N w

l x xN N

x l

0

1 0 1 01

0 1 0 1

1 0 1 0

0 1 0 1

1 0 1 0

0 1 0 1

with xx

Gl

PK

L

P A

5

0 0 0 0

0 0

0 0

1

2 2

1

2 2

1

2 2

t t

t t t t

t t

t

t

PV t u t Ku t u Ju

Pu T KT u u T JT u

Pu Ku u Ju

K T KT

J T JT

Element stiffness matrix in the transformed coordinate system

Element stability matrix in the transformed coordinate system

K

J

Local to global

coordinate

transformation

6

How to analyse this type of problems?

7

1

Step-1 Assemble the reduced structure elastic stiffness matrix and

equivalent nodal force vector .

Step-2 Find response and determine the ax

K

F

X K F

Recipe for finding critical load for planar frames

ial thrust diagram

for the structure.

Step-3 Using the axial thrust information in the above step, formulate the

reduced structural geometric stiffness matrix

Step-4 Solve the eigenvalue problem

GK

K K

min

min

and estimate the lowest

eigenvalue, .

factor by which the prevailing loading pattern needs to be multiplied

so that the structure reaches a critical state (load factor).

The associated eige

G

nvector leads to the definition of the lowest buckling

mode shape.

8

0.2m

0.3m

3

210 GPa

=7800 kg/m

E

3m

040

4m

P

2

P

Determine critical value of P.

Plot the axial thrust diagram at the critical loading condition

Plot the corresponding buckling mode shape

9

Remark: beam column analysis using FEM

1 1,u P

2 2,u P4 4,u P

3 3,u P

P P

2 22 4

210 0

4

1

4 4 4 4 4

1 2 3 4

1 1 1 1 1

1

2 2

1, , ,

2 2

Equilibrium 0, 1,2,3,4

After imposing BCs one gets

l l

i i

i

i i

i

i j ij i j ij i i

i j i j i

k

v P vU EI dx EI dx u P

x x

v x u x

PU u u u u u u K u u J u P

Uk K PJ u Q

u

u K PJ

1

Q

10

P P

Q

B

Q l cR

l

A

QcR

l

BA

cl c

0 03

0 02

0 0

Summary

3 tan=

2 1 cos0

cos

tan

2

u uu

u

uu

u u

l uM M M u

u

Single

concentrated

load

at midspan

Exercise: revisit beam column problem and compare

FE solutions with the exact solutions

11

0 00 ; 0

We will consider this later.

MX CX K PJ X F t

X X X X

Generalization

12

Combined axial force and twisting

x

z

P xM, , , , ,xx zzE l A I I

Recall

Lecture-11

2 2 2

Strain of a fibre along longitudinal direction based on nonlinear

strain displacement relations is

1

2

x

x

v z

w y

du du dv dwe

dx dx dx dx

Symmetric

Cross section

13

2 2 2

2 22 2 2

2 2 2 2

;

1

2

Focussing on geometric effects

1 1 1

2 2 2

x x

x x xg

v z w y

du du dv dwe

dx dx dx dx

d d ddu dv dwe z y z y

dx dx dx dx dx dx

2

2 2

2 2

0

0

2 2

0

1 1;

2 2

&

Considering a two-noded element and using linear interpolations, we can obtain

1 1

1 1

l

x xPxx

V

xx P

A

Pg

d dPIU z y dAdx dx

dx A dx

PI z y dA

A

PIk

Al

1

11

10

9

8

7

6

5

43

2

12, , , , , , , ,x y zE G I I I A J l

, , , , , , ,u x t v x t w x t x t

Effects included: coupling between axial forces with

bending along two axes

torsion about longitudinal axis

3D beam element : a simplified model

Refer to

Lecture-12 Is it that computational effort increases and we need to

handle larger sized matrices? OR

Are there any new phenomena that we need to be

concerned about?

3D beam element

15

1 2 3 4 5 6 7 8 9 10 11 12

1

2

3

4

5

6

7

8

9

10

11

12

Z Z Z Z

y y y y

y y y y

Z Z Z z

Z Z Z z

y y y y

y y y y

Z Z Z Z

A A

B B B B

B B B B

T T

B B B B

B B B BK

A A

B B B B

B B B B

T T

B B B B

B B B B

16

2 2

2 2

1 0 0 0 0 1 0 0 0 0 0 0

6 60 0 0 0 0 0 0 0

5 10 5 10

6 60 0 0 0 0 0 0 0

5 10 5 10

0 0 0 0 0 0 0 0 0 0

20 0 0 0 0 0 0 0

10 15 10 30

20 0 0 0 0 0 0 0

10 15 10 30

1 0 0 0 0 0 1 0 0 0 0 0

6 60 0 0 0 0 0 0 0

5 10 5 10

6 60 0 0 0 0 0 0 0

5 10 5 10

0 0 0 0 0 0 0 0 0 0

0 01

P P

G

P P

l l

l l

I I

A A

l l l l

l l l l

PK

l

l l

l l

I I

A A

l

2 2

2 2

20 0 0 0 0 0

0 30 10 15

20 0 0 0 0 0 0 0

10 30 10 15

l l l

l l l l

3 2 3 2

2 2

3 2

3

3 2

2 2

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

0 0 0 0

12 6 12 6

6 4 6

12

0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

0 0 0 0

6 12 6

6 4 6

2

2

12

z z z z

y y y y

y y y y

z z z z

z

AE AE

l

G

E

EI EI EI EI

l l l l

EI EI EI EI

l l

l

KAE AE

l

I EI EI EI

l l l l

E

l l

J GJ

I EI EI EI

L l

l

l

l l l

EI

l

2 3 2

2 2

3 2 3 2

2 2

0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0

6 12 6

0

0

12 6 12

0 0 0 0 0 0

6

6 4

6 6

2

2

6

04

z z z

z z z z

y y y y

y y y y

EI EI EI

l l l

EI EI EI EI

l l l l

EI EI EI EI

l l l l

EI EI EI E

G

I

l l l l

J GJ

L L

18

The initial state of stress in the beam could include presence of

longitudinal force

torque along longitudinal axis

two bending moments

A more general formulation (within the framework

3D beam element

of Euler Bernoulli theory)

need to include

interactions of torque and BM (in addition to axial force),

spatial variation of existing torque and bending moments

asymmetric cross sections

0 0

2 2 2

0 0

1

1 0 0

1

2

1

2

1=

2

xx zx

V

xx zx

V

xx zx

u v wU dV

x z y

u v w u u v v w wdV

x x x y z y z y z

U U

u vU

x

0

2 2 2

0 0

leads to equivalent nodal forces due to .

1

2

contributes to the analysis o

V

xx zx

V

wdV

z y

u v w u u v v w wU dV

x x x y z y z y z

f stability of the structure.

, , , ,

1 1

2 2ij i j j i km k i m ju u u u

20

1 22 2

2 21 1

1 2 1 1 21 2 1 1 2

2 21 2

0 0 0 0 0 0 0 0 0 0

6 60 0 0

5 10 5 10

6 60 0

5 10 5 10

220

6 6 6 6

20 0

15 10 6 30 2

2

x x

y yx x x x x x

x x x x x xz z

y y y y yx P x Pz z z z z

x x x x xz z

x

G

F F

l l

M MF M F F M F

l l l l l l

F F M F F MM M

l l l l l l

M M M M MF I F IM M M M M

Al l l Al

F l M F F l MM M

l

F l

K

1 22 2

2 2

22

1 21 2

015 10 6 30

0 0 0 0 0

60

5 10

6

5 10

22

6 6

20

15

20

15

y yx x x x

x

yx x x

x x xz

y yx P z z

x

x

M MF M M F l

l l

F

l

MF M F

l l l

F F MM

l l l

M MF I M M

Al

F l

F l

sym

W McGuire, R H Gallagher, R D Zieman, 2000, Matrix structural

analysis, 2nd Edition, John Wiley, New York.

21

Geometric stiffness matrix for a thin plate bending element

Kirchoff-Loeve plate

x

,z w

y

yN

xyN

xyN

xN

/2 /2 /2

/2 /2 /2

Membrane forces

; ;

t t t

x xx y yy xy xy

t t t

N dz N dz N dz

22

Idea: examine the work done by the constant membrane forces on the

displacements associated with transverse deflections.

Membrane strains associated with small rotations & are given by

1

2xx

w w

x y

w

x

22

22

1; ; 2

2

Assume: , , are independent of ,

2

1 1

2 2

yy xy

x y xy

x xx y yy xy xy

A

x y xy

A

w w w

y x y

N N N w x y

U N N N dA

w w w wN N N dA

x y x y

w N

e e

w

xu G u

w

y

23

1

Isoparametric formulation

&

: matrix of derivatives of shape fun

e e

t x xy

xy y

I e

I

w

xw N u G u

w

y

N Nk G G dxdy

N N

w ww

xG u J

ww w

y

G

1 1

1 1

1 1

ctions wrt &

tt x xy

I I

xy y

N Nk G J J G d d

N N

24

1 1

1 1

1 1

Remarks

Membrane forces may vary in space and these variations may be accounted

for while evaluating the above integral

do not depend upon elastic prop

tt x xy

I I

xy y

N Nk G J J G d d

N N

k

erties of the plate (except to the extent

that these properties influence the initial membrane stresses that prevail).

The formulation can therefore be used for anisotropic material as well.

2 22

0

22 2

2 2

3D Timoshenko beam

1{

2

+ }

1

2

I

ly z

I y z

xy z z y x

yzx y

U U U

uU AE EI EI

x x x

v wGF GF GI dx

x x x

v wU N M

x x x

0

0

0

1

2

l

z

l

y x z x

l

y z z y x

dxx

v wM M dx

x x x x

M M

A V Perelmuter, CVslivker, 2013, Handbook of mechanical stability in engineering,

Vol.2, Stability of elastically deformable mechanical systems, World Scientific, New Jersey.

26

Imperfection sensitive structures

P PP

1k

2k

3k

A AA

B BB

I II III

l

27

1 2 3

12 3

A thought experiment

Design I,II and III in such a way that .

That is, select the spring constants , ,& such that .

It can be verified that , ,& cos

cI cII cIII

cI cII cIII

cI cII cIII

P P P

k k k P P P

kP P k l P k l

l

2

211 2 3 2 3

.

Thus, we are selecting , ,& such that cos .

If we were to fabricate the three models and perform experiments

it would be obeserved that no matter how carefully

the thre

cI cII cIII

kk k k k l k l

l

P P P

e models are fabricated and no matter how deligently the

experiments are performed.

It would also be observed that the scatter in the experimentally determined

value of critical loads would be such that

III II I

Scatter Scatter Scatter .

Indeed such behavior has been observed for the case when

I=beam, II=plate, and III=shell.

28

P

k A

B

I

P

2

2

2

2

11 cos

2

Equillibrium: 0 sin 0sin

Stability: 0 cos 0

Let the spring remain unstrained till > .

1cos cos

2

Equillibrium: 0 sin 0

V k Pl

V kk Pl P

l

Vk Pl

V k Pl

Vk Pl

Ideal

Imperfect

2

2

sin

Stability: 0 cos 0

kP

l

Vk Pl

29 -1.5 -1 -0.5 0 0.5 1 1.5

0

1000

2000

3000

4000

5000

6000

7000

theta

load P

Stable if

cos 0k Pl

Black: stableRed: Unstable

0

50

0

50

30

Remarks

Load-displacement curve for imperfect system rises

along a stable path without encountering any instability.

The deflection grows more rapidly as the load approaches

the critical value correspo

nding to a perfect system.

This is an example of a symmetric stable bifurcation

Beam structures display this type of behavior.

P

k

A

B

II

2

2

2

22 2

2

22

1sin 1 cos

2

Equillibrium: 0 sin cos sin 0

sin cos 0

Stability: 0 cos cos sin cos 0

Let the spring remain unstrained till > .

1sin sin

2

V k l Pl

Vkl Pl

kl Pl

Vkl Pl kl

V kl

Ideal

Imperfect

2

22

2

2

cos cos

Equillibrium: 0 sin sin cos sin 0

sin sin cos

sin

Stability: 0 cos sin sin sin cos 0

Pl

Vkl Pl

klP

Vkl Pl

32 -1.5 -1 -0.5 0 0.5 1 1.5

0

1000

2000

3000

4000

5000

6000

7000

theta

Load,P

Black: stableRed: Unstable

0 / 50

/ 20

33

Remarks

Load-displacement curve for imperfect system is not

always rising and stable; instead, it rises along a stable branch

and exhibits a limit point beyond which the path drops and becomes

unstable.

Greater the imperfection, lesser is the ability of structure to

carry the load.

This is an example of a symmetric unstable bifurcation

Plate structures display this type of behavior.

34

P

k

A

BP

B

O

cosL

cotL sinL

L

OB

OB

sinsin

s

s

L Ls

s

Structure is asymmetric

35

2

22 2 2

22 2 2

2 2 2

12 2 2 2

11 cos

2

sin

cot sin cos

1cot sin cos 1 cos

2 sin

Equillibrium: 0

cot sin cossin

1cot sin cos 2 cot sin

2

V k s s PL

Ls

s L L L

LV k L L L PL

V

V Lk L L L

L L L L L L

Ideal

2

2

2

cos 2 cos sin

sin

stability: 0

L

PL

V

36

P P

S

S

S

SUU

US

37

Remarks

The stable equillibrium path is asymetric with respect to the

undeflected configuration state wrt which the bifurcation takes place.

The stable load-displacement curve for imperfect system is

not

always rising and stable; instead, it rises along a stable branch

and exhibits a limit point beyond which the path drops and becomes

unstable.

Greater the imperfection, lesser is the ability of stru cture to

carry the load.

This is an example of an asymmetric unstable bifurcation

Shells (and frame) structures display this type of behavior.

38

Imperfection sensitive structures

P PP

1k

2k

3k

A AA

B BB

I II III

l

mP

mP

39

Illustrates the possibility of diverse type of behavior which could

be displayed by different structures.

The function encapuslates different types of structures.

can be expanded in Taylor's ser

U

U

ies about the equillibrium point

and different bifurcation patterns can be discerned by examining the

signs of various terms (Koiter's theory)

Nonlinear analysis helps us to examine the likely importanc e of geometric

imperfections in a given context.

Consider a simply supported, axially loaded beam on Winkler's foundation.

Reaction proportional to deflection

Same in tension and compression

Behaves like a series of springs

Beam on elastic foundation

(linear, elastic) without mutual interactions.

,EI lP P

k

2 22

2

2

0 0 0 0

4 2

4 2

1 1

2 2 2

0

BCs: Either 0 or 0 & Either 0 or 0

l l l ld y P dy

U EI dx dx ky dx p x y x dxdx dx

d y d yEI P ky

dx dx

y EIy Py y EIy

41

4 2

4 2

1

4

Consider a simply supported, axially loaded beam on Winkler's foundation.

0

0 0, 0, 0 0, 0

sinN

n

m

d y d yEI P ky

dx dx

y y L y y L

m xy x a

L

m mEI P

L

Simply supported beam on elastic foundation

2

2

2

0

Note: Notice the presence of in the denominator.

1 need not lead to the lowest critical load.

cr

kL

m kP EI

L m

L

m

m

42

2 2

2 2

3

,min

Smallest value of need not lead to the smallest value of .

is the smallest when 0

2 2 0

2 th

cr

cr

cr cr

cr

m P

m mP m EI k

L L

dP m P m

dm

k m kEI m

L L L EIm

L

k EIP EI k kEI

EI k

is value is independent of m

1P1P

2P2P

2 1

Special featureIt is possible that P P

43 101

102

103

104

105

106

108

109

1010

1011

1012

(l/r)2

Pcr

m=1

m=2

m=3

m=4

2 kEI

44

Exercise

Consider an Euler-Bernoulli beam element resting on a Winkler's

foundation. Develop the elastic and geometric stiffness matrices

for the element. Using the element thus developed determine

the critical load for a simply supported beam and illustrate the

results shown in the figure on the previous slide.

45

O

R

A

B

m

n1n

1m

1mm w

d

1 1

Buckling of a ring

=radius of the ring

AB=deformed configuration

mn=section of length in the undeformed

configuration

1

Upon deformation the section mn

becomes m n

Let =radius of the def

R

ds

dds Rd

ds R

1

ormed configuration

1

angle between tangent to the centre

at m and the perpendicular to the radius mO

d d

ds ds

dw

ds

46

1

2

1 2

2

2

1 1

angle between tangent to the centre at m and the perpendicular to the

radius mO

The corresponding angle at n is

Length m n :

1

dw

ds

dw d wds

ds ds

d wd ds

ds

dsds Rd R w d R w d ds wd w

R

d d

2 2

12 2

2 2 2

2 2 2 2

2 2 2

2 2 2

11

11 is ignored

1 1

d w d wd ds d ds

wds dsdsds ds ds R

ds wR

d d w w w d w w d w

ds ds R R R ds R ds

M w d w d w MRw

R EI R ds d EI

47

Ring under

Uniform

External

pressure

0 0

2 2 2

0 02

0, ,

hoop force=

w w M M

qR

d w MR Rw M qR w w

d EI EI

q

Ring with radius and thickness

Rh

Find critical value of pressure at which a neghbouring equillibrium (red) becomes possible

q

48

2 2

0 02

2 3 2

0 02

2 2 32 2

0 02

2

0 02

1

with 1

cos sin

sin cos

d w Rw M qR w w

d EI

d w qR Rw M qRw

d EI EI

d w R qRk w M qRw k

d EI EI

Rw A k B k M qRw

k EI

dwAk k Bk k

d

49

3

3

sin cos

For the mode shape we are looking for,

0 at =0, by symmetry2

0 at =0 0

0 at = sin 0 2,4,6...2 2

31 4c

c

dwAk k Bk k

d

dw

d

dwB

d

dw kk

d

q R EIq

EI R

Exercise: Using Euler-Bernoulli beam elements, verify this result.