Plastic Analysis Beams Frames

Ultimate/Accidental Limit State Analysis and Design

NUS July 10-12, 2006 Analysis and Design for Robustness of Offshore Structures NUS – Keppel Short Course

1

Outline Introduction Introduction Plastic hinge concept Pl ti th d f b d l t Plastic methods for beams and plates Brief on mechanism analysis Stiffness of beams including geometry

effect Tensile fracture


2

Introduction

wave,current

wind,

loadsoads

Bracing configuration

piles,f d ti


3

foundation

Ch t i ti l b l l dCharacteristic response - global load versus global displacement for an offshore structure

load level member fracture ?load level

limit load

member fracture ?

wave,current

wind,

loads

member instability

first plastic hinge

global post-collapse

load redistribution

first yield

ConnectionsMembers

piles,foundation


4

deck displacement

Can reserve systems effects be utilized?

For ULS design ductility For ULS design - ductility requirements must be complied with :

fracture, local buckling, cyclic effects etc.

A id l i Accidental actions - systems effects must be considered

Ship collision Explosions Fires Ship collision, Explosions, Fires, Dropped objects, Specified damages..


5

Definitions of ultimate resistance inDefinitions of ultimate resistance in intact and damaged conditions

capacity,intact structure global load

reservestrength

response,damaged

100-yearload level

residual strength

reservestrength

global displacement

structureresidual strength


6

Nonlinear analysis of offshore structuresNonlinear analysis of offshore structureschallenges

Effects:

• Nonlinear material, geometry, load,

Local and global failure modes:

• Yielding, buckling,fracture,

Modeling

• Beam columns, shell, solids, springs

St t i t ti t• Stress-strain representations, stress-resultant approach

Load control procedures


7

Design against accidental actionsDesign against accidental actions according to e.g. NORSOK

Step1Damage due to accidentalPlasticPlastic Damage due to accidental

actions

Step 2Elastic

Plastic

Elastic

Plastic

Step 2Resistance of damaged

structure to designPlasticPlastic

structure to design environmental loads

Partial safety factors = 1,0


8

f y f

Material modelling for plastic analysis

stress strain relationshipstress strain relationship

y

Rigid-perfectly plastic

yElastic-perfectly plastic

y ~0.001 p ~20 p u ~ 100 p


9

y p p u p

Plastic hinge conceptPlastic hinge concept

Plastic hinge


10

Plastic hinge

Plastic hinge conceptPlastic hinge concept

Elastic

Elastic-plastic

Fully plastic

Note distribution of plasticity


11

Note distribution of plasticity

Plastic hinge concept Plastic moment rectangular cross-sectiong


12

Plastic hinge conceptPlastic hinge conceptPlastic moment circular cross-section

2M f d 2p yM f d t Thin-walled tube: d >>t


13

Pl ti hi tPlastic hinge conceptNormalised moment

Plastic

Elastic


14

Normalised max. strain

Plastic hinge concept


15

Plastic collapse resistancePlastic collapse resistanceKinematic analysis

Pc

q q

w2q

e cW P w 2e pW M External virtual work Internal virtual work

Kinematics2

w

Kinematics

4 pMP Plastic collapse load


16

cPL

Plastic collapse load

Elastic-plastic collapse analysis of clamped beamTotal resistance

q 21Beam end:12p

q LM

l

M M =q1l 2

2 21 2

2 22 1

12

In the middle: 24 8

1

pq L q LM

q L q L

M = Mp = 112

M = q1l 224+

2 12 1

18 24 3

q L q L q q

Total resistance

M = q2l 2

8

+

=2121

164 Mqqqq p

c

ota es sta ce

M = Mp

M M

2121 3 Lqqqqc


17

M = Mp

Elastic-plastic collapse analysis of clamped beamComparison with usfosComparison with usfos

Fy 330D 0.5t 0.02Wp 0.004608 calculated mean diameterWp 0.004611 usfospArea 0.030159 calcuated mean diameterArea 0.03016 usfosI 0.000869 calculated mean diameterI 8.70E-04 usfosL 10qc 0.243461 calculatedqc 240000 usfos referencew load 0.1E+6disp 1 yield 2.60E-02 first yeild hinge calculateddisp2 6.94E-02 3 hinges calculated

plastic rotation 1.39E-02 calcuated

Plastic rotation vs

displacement

Load factor vs displacement


18

Elastic-plastic collapse analysis of clamped beamLoad versus mid span deformation

q1qLMLqw p1 1

241

q2 qLMLq p55 224

2q1

w1 EILMw

qq

qEIEIw

pc

c

24/

24384

21

1

w2 EILMw

qq

qq

EIEIqw

pc

c

p

24/

24384

21

222

Deformation in step 1Deformation in step 2.

Deformation in step 1.

q / qc

1

k = 1k = 0.2

Hi t id

0.75

Hinge at mid span

Hinges at ends

02 w

Mpl 2/ 24 EI0 0.75 1

Load mid span deformation


19

Load - mid span deformation

Elastic-plastic collapse analysis of clamped beamPlastic rotation analysis

M M0M1 1/2Mp0

1

Plastic rotation at ends

For rectangular cross-section

0 11 2

1 2 1 112 3 2 6

pp p

MM M dx MEI EI EI

Plastic rotation at ends

For rectangular cross section2

3

/ 46 /12 2

yp

fy hE h h

A il bl f li l ti f tl l ti t i lAvailable for linear elastic-perfectly plastic material

/ 20.75

4 / 2 / 3p y

ep y

hh


20

Strains in elasto-plastic regionp gCantilever beam

2220

00

4 12 1 13 3

h

yy p p

yM dy M Mh

11pxM M

0 1 0

31

y x

1/3 1/30 2 412 1 1y y yd d

0

max1 1

1 13

31

y y ydx dxh h hx

41 yep


21

3ep h

Compactness requirements for various cross-sectionscross sections

b

Flytning ytterste fiber

TVERRSNITTELLER

TVERR-SNITTSDEL

TRYKKKRAFTOG / ELLERMOMENT

TVERRSNITTSKLASSE

Full plastisk

Siste flyteledd

Lokal knekning

bt £ 1.0

Ef

bt £ 1.2

Ef

bt £ 1.3

Ef

TVERRSNITTLIVPLATE

TRYKKKRAFTOG

TVERRSNITTSKLASSE

F llSi tFl t iL k l

t

t

b

t

a· b

TRYKK

MOMENT

t 1.0 fy t 1.2 fy t 1.3 fy

bt £ 2.0

Efy

bt £ 2.6

Efy

bt £ 3.3

Efy

KT 5

.6

b E£ 1 b E£1 2 b E£1 3MOMENT

LIVPLATE OGMOMENT Full

plastiskSiste

flyteleddFlytning

ytterste fiberLokal

knekning

dt £ 2.5

Efy

dt £ 3.8

Efy

dt £ 4.2

Efyd t

1/2d

bt

b

t

MOMENT OG TRYKKRAFT

TRYKKE

TEN

KA

N B

ES

TEM

ME

S E

TTE

R P

Kbt

Efy

£ 1a

bt

Efy

£ 1.2a

bt

Efy

£1.3a

bt £ 0.30

Efy

b E0 3

bt £ 0.33

Efy

bt £ 0.43

Efy

b E0 33 b E0 43 ETTE

R P

KT

5.6

£ 0

.15,

N

p = f d

·d·t

£ 1.

0N N

p

£ 0

.10,

N

= s

·d·t

£ 0

.10,

N

= s

·d·t

d1

d2

t

t

b

b

t

a· b

MOMENT OG TRYKKRAFT

KA

PA

SITb

tEfy

£ 0.3a

bt

Efy

£ 0.33a

bt

Efy

£ 0.43a

bt £ 1.1

Efy

bt £ 1.25

Efy

bt £ 1.5

Efy

PAS

ITE

TEN

KA

N B

ES

TEM

ME

S E

3

)N Np

E f y0

£

N Np

0.15

£

0

)N Np

E f y

5

)N Np

E f y0

£

N Np

9

)N Np

E f y0

£

N Np

TVERRSNITTS-KLASSE 1 OG 2

dt

b t1t1

t1

dt £ 0.056

Efy

dt £ 0.078

Efy

dt £ 0.112

Efy

b E

KA

d t£

2.50

(1 -

0.93

d t£

2.20

(1 -

0.2

d t£

3.80

(1 -

0.55

d t£

4.20

(1 -

0.59

TVERRSNITTS-KLASSE 3


22

b2

b1t1

t1 bt £ 0.4

Efy

Bending moment axial–forceBending moment axial force interaction

Mechanism analysis works well for beam and frames where the resistance is

d b b digoverned by bending

In many structures the resistance contribution from axial force importantcontribution from axial force important, either initially (truss-works) or during force redistribution (beams under finite deformations)


23

Pl ti hi tPlastic hinge conceptBending moment- axial force interaction

Generalised yield criteria

2 M N

Tube

12 sin 1 0p p

M NFM N

TubeTube2

1 0p p

M NFM N

Compression

p p Rectangular cross.Bending


24

Plastic resistance for beam with concentrated l d t id (1)load at midspan (1)

P

w D,t

Pipe section

E q u i l ib r iu m

8 2M wR NN

Pipe section

8 2

/ 2R = N

B e n d in g m o m e n t – a x ia l f o r c e in te ra c t io n

p

NN

cos 02p p

M NF M N

( ) ?N N w U n k n o w n


25

U n k n o w n

p

MM

Plastic resistance for beam with concentrated load at midspan (2)concentrated load at midspan (2)

P

w

Kinematics

Plastic elongation in each hinge Plastic elongation in each hinge

2 221 1~

2 2 2 2wu w

wu w

Plastic rotation in each hinge

/ 2w / 2

w


26

Plastic resistance for beam with concentrated load at midspan (2)concentrated load at midspan (2)

P

w

Kinematics Plastic elongation in each hinge

2 21 1N w N

21 1~2 2 2 2 2 2

N w Nu wk k

u

Plastic rotation in each hinge w w

/ 2w / 2

w


27

Plastic resistance for beam with concentrated load at midspan (3)

P

concentrated load at midspan (3)

w Plastic flow - normality criterion

FM

v

criterion

p

NN p

NN

1

p FuN

v

M

pv

M1

21 i

p

wM

w N

0M NF

pM pM


28

1 sin2 2p p

NwN N

cos 02p p

NFM N

Plastic resistance for beam with concentrated l d id (4)load at midspan (4)

P

w

R esu lts o f an a lysis

w N w

12 p

w N wD N D

W h /D > 1 W h en w /D > 1

0 1pwN N MD


29

D

Plastic resistance curve for beam withPlastic resistance curve for beam with concentrated load at midspan (5)

w

P

w

Collapse model for beam with fixed ends

wwwwR 1 < Dw

Dw

Dw+)

Dw(-1 =

RR 2

o

u arcsin

1>ww=Ru


30

1 > D

D2

Ro

Plastic resistance curve for beam withPlastic resistance curve for beam with concentrated load at midspan (6)

P

w

P

8

6

8

Transition from bending & mebrane to pure tension at

The displacment at this transition is denoted

2

4

R/R

0 w/D =1 R/R0 = /2 transition is denoted characteristic displacment wc

0

2

0 1 2 3 4

Bending only


31

0 1 2 3 4Deformation w/D

Plastic resistance curve for beam with d l d id (7)concentrated load at midspan (7)

P KK

w

Kinematics- Plastic elongation in each hinge

2 221 1~N w Nu w

w Nu w

In real structures beam ends ends are not fully fixed. The axial flexibility of

2 2 2 2 2 2u w

k K

2

u wK

the adjacent structure may be represented by a linear spring with stiffness K. This affects the kinematic relationship for plastic axial elongation. Closed form solution is no longer possible, but simple incremental equation may be

l d i ll


32

solved numerically

Elastic-plastic resistance curve for tubular pbeam with conc. Load at midspan

Factor c includes the effect of elastic flexibility at ends

Bending & membrane5,5

6

6,5

Rigid plastic Bending & membraneMembrane only

k kw

F - R

3,5

4

4,5

5

5,5

0 0.1

0.2

0,3

0 5

Rigid-plastic

1

1,5

2

2,5

3R/ 1

0.5

0.05cKw4c

c2

c1

0

0,5

1

0 0,5 1 1,5 2 2,5 3 3,5 4

Deformation w

Afc

y


33

Tensile Fracture

According to plastic theory no limitation to resistance and energy dissipation in beams gy pwith axial restraintUltimately the member will undergoUltimately the member will undergo

fracture due to excessive strainingIn order to predict fracture a strain modelIn order to predict fracture a strain model

for the plastic hinges must be developed


34

Strain hardening paradox

In plastic analysis the stress-strain curve is assumed rigid-plastic or linear-elastic perfectly plastic

If the material behavior is really like this, the b b h b i l i l b l dmember behaves brittle in a global sense and

plastic theory cannot be appliedSt i h d i i i l i di t ib ti l tiStrain hardening is crucial in distributing plastic

strains axially in the member, so that significant energy dissipation can be achieved


35

energy dissipation can be achieved

M

Y max hY hY

M

Strain Stressdistribution

Approximate stressdistribution

S d b b l l

40

45

50

Stress-strain distribution - bilinear material

25

30

35

40

Stra

in

Hardening parameter H = 0.005

Maximum strain cr/Y

= 5040

P

x

5

10

15

20

S = 40 = 20

No hardening

00 0.05 0.1 0.15 0.2 0.25 0.3 0.35

x/

Axial variation of maximum strain for a cantilever beam


36

Axial variation of maximum strain for a cantilever beam with circular cross-section

Assumption: Bilinear stress-strain relationship

Tensile Fract reTensile Fracture• The critical strain in parent material dependsThe critical strain in parent material depends

upon: stress gradients dimensions of the cross section presence of strain concentrations material yield to tensile strength ratiomaterial yield to tensile strength ratio material ductility

• Critical strain (NLFEM or plastic analysis)

zoneplasticoflength:5,t65.00.02 tcr


37

pg,cr

Critical deformation for tensile fracture in yield hinges

1/εc4c12cc

dw

1crfwf

1

c c

2displacement factor2

crcrPlplp

1w d

κεε

WW14c

321c

c1c

Y

l i l h fH

WW1

εε

P

cr

plastic zone length factor 1H

WW1

εε

Wεc

Py

cr

Pylp

axial flexibility factor2

cc

axial flexibility factor f c1c

non-dim. plastic stiffness

ycr

ycrp

εεff

E1

EE

H

c1 = 2 for clamped ends

= 1 for pinned ends

c = non-dimensional spring stiffness

l 0.5l the smaller distance from location of collision load

cr = critical strain for rupture

Ef

ε yy yield strain

fy = yield strength


38

to adjacent joint

W = elastic section modulus

WP = plastic section modulus

fcr = strength corresponding to crdc = D diameter of tubular beams

= 2hw twice the web height for stiffened plates= h height of cross-section for symmetric I-profiles

Deformation at rupture for a fully clamped beamDeformation at rupture for a fully clamped beam as a function of the axial flexibility factor c

5

3 5

4

4.5

2.5

3

3.5

w/D

/D = 30

c= 0

/D = 20

c = 0

1

1.5

2 = 0.05 = 0.5 = 1000

= 0.05 = 0.5 = 1000

0

0.5

0 20 40 60 80 100 120


39cry

Tensile fracture in yield hingesD t i ti f HDetermination of H

f fA2 A2

A1= A2

E

H E

fcr

E

H Efcr

A1 A1

A2

cr cr

Determination of plastic stiffness

H E

fEven if the stress strain curve lies below the true relationship such that the energy dissipation for the fiber is smaller, the hardening exaggeration Use true yield

, g ggmay give too large energy dissipation in the member as a whole

ystress


40

Erroneous determination of plastic stiffness

Tensile fracture in yield hinges

• Recommended values for cr and H for different steel gradesg

Steel grade cr HS 235 20 % 0 0022S 235 20 % 0.0022S 355 15 % 0.0034S 460 10 % 0.0034


41

Plastic hinge conceptBending moment –axial force history

M,P


42

Stiffness matrix for beam with axial force

w

B

N

QA

MA

QBMB

Nw(x)

wAwB

XX = x

0EIw X N wQ wM

Differential equilibrium equation

, 0A Axx AEIw X N wQ wM


43


A 3 5 2 2 3 5 2 2 AQ 12EI

6EI

12EI

6EIw

1

tan 1

tanh

A 3 2 2 4 AM

Q

=

4EI 6EI 2EI

12EI 6EI

w

2

2

1

3 1 2

13

14

34

1 2

2

1

3 1 2

13

14

34

1

B

B

3 5 2 2

3

B

B

Q

M 4EI

w

4 1 2

5 1 2

4 412

32

4 1 2

5 1 2

4 412

32

symmetry

E

N2 N

2

E 2

EIN

l


44

EN l


5

2

3

4

N-1

0

1

-4 -3 -2 -1 0 1 2 3 4- val

ue

E

N2 N

-4

-3

-2

-6

-5

Axial force E


45

Buckling of column E l 1Buckling of column – Example 1

N NA B

l

K

2EI 22

3 4

4 3

K 0 0123 4, ;4 3

242

N N ECritical force


46

Buckling of column – Example 2

N A B

l

K4EI

0 2N N KK 3 0 2 EN N K

1Buckling length k

12

0 7.


47

Buckling of column – Example 3

NA B

l

K

2EI 6 3- 3 23

5 2

22

3

2 2 23 5 20 12 9 0l l K

Critical force 23 5 212 9 0.25 EN N


48

The stiffness matrix for beam with axial force contains ll i f i d d di h b kliall information needed to predict the exact buckling

load for beams subjected to end forces

A 3 5 2 2 3 5 2 2 AQ 12EI

6EI

12EI

6EIw

1

tan 1

tanh

A 3 2 2 4 AM

Q

=

4EI 6EI 2EI

12EI 6EI

w

2

2

1

3 1 2

13

14

34

1 2

2

1

3 1 2

13

14

34

1

B

B

3 5 2 2

3

B

B

Q

M 4EI

w

4 1 2

5 1 2

4 412

32

4 1 2

5 1 2

4 412

32

symmetry

E

N2 N

2

E 2

EIN

l


49

EN l

Documents

Plastic Analysis Beams Frames