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TMR4205 Buckling and Ultimate Strength of Marine Structures

Chapter 1: Elastic-Plastic Analyses of Beams, Frames and Plates

by

Professor Jørgen Amdahl

MTS 2005-01-19

TMR4205 Buckling and Ultimate Strength of Marine Structures 1. Elastic-Plastic Analyses of Beams, Frames and Plates of 45

CONTENTS

1 ELASTIC-PLASTIC ANALYSIS OF BEAMS, FRAMES AND PLATES ............. 3

1.1 Kinematic - versus elastic-plastic analysis of beams............................................................................. 3 1.1.1 Kinematic analysis ............................................................................................................................ 3 1.1.2 Step-wise elastic-plastic analysis. ..................................................................................................... 4 1.1.3 Load - deformation relationship........................................................................................................ 7

1.2 Elastic-plastic analysis of frames.......................................................................................................... 12

1.3 Plastic analysis of beams and frames under cyclic loading ................................................................ 16 1.3.1 Elastic unloading............................................................................................................................. 16 1.3.2 Cyclic loading ................................................................................................................................. 17 1.3.3 Kinematic analysis of shakedown load ........................................................................................... 19 1.3.4 Example - Simple portal frame ....................................................................................................... 20

1.4 Elastic-plastic analysis of truss-works. ................................................................................................ 24 1.4.1 Behaviour in the elastic range: ........................................................................................................ 26 1.4.2 Behaviour in the post-buckling range ............................................................................................. 28

1.5 Plastic capacity of beams undergoing large lateral deflections. .................................................... 34 1.5.1 Plastic load-carrying capacity for uniform load and clamped ends ................................................ 39

1.6 Plastic load-carrying capacity of plates subjected to large lateral deformation .............................. 39

1.7 Use of plastic methods in design ........................................................................................................... 42


1 ELASTIC-PLASTIC ANALYSIS OF BEAMS, FRAMES AND PLATES

1.1 Kinematic - versus elastic-plastic analysis of beams.

A B C

A B C

P

l l

EI EI

Figure 1Two-span beam

The simple two-span beam shown in Figure 1 - discussed in Section 2.1 of Søreide's book- is reconsidered. First, the plastic collapse mechanism is calculated by means of the mechanism method. This calculation yields no information of the response history prior to collapse. However, by using well known principles from elastic analysis (unit load method) the sequence of plastic hinges as well as the total collapse load may be estimated in a two-step procedure where the calculation model is modified after the first hinge.

1.1.1 Kinematic analysis

A B C

lll2δθδθ

P

Figure 2 Assumed collapse mechanism

The collapse mechanism assumed is shown in Figure 2. It contains one plastic hinge under the concentrated load and one hinge at support B. The internal virtual work reads:

( )δ δ δ δW M Mi P P= + =2 3Θ Θ Θ (1)

The external virtual work is:


δ δW P w Pl

e = =2

Θδ (2)

Equating internal and external virtual work there is obtained

P PM

lcP= = 6

(3)

The corresponding bending moment diagram is easily drawn, since it is inherently assumed that the fully plastic moment is attained under the load and at support B. The corresponding shear force is derived from the inclination of the bending moment diagram. It is seen from Figure 3 that the sum of the support reactions equals the total vertical load.

A B C

Pc

Mp

Mp

4M

lM

lP P+2

Ml

PM

lP

Figure 3 Bending moment distribution at collapse

1.1.2 Step-wise elastic-plastic analysis. The above calculation of the plastic collapse load provides no information as to the sequence of plastic hinge formation. For this purpose elastic analysis has to be performed. It is carried out in two steps, i) In the elastic range ii) After the formation of first plastic hinge. i) Elastic range It is seen that the number of unknown reactions is three whereas the number of available equilibrium equations is two (disregarding the obvious condition of zero horizontal force). The beam is therefor one time statically indeterminate. Applying the unit load method, it is convenient to consider the bending moment at support B, X, as the unknown quantity.


A B

M0

M1

Pl1

4

1

XP1

C

Figure 4 Bending moment in the statically determinate system and for a unit moment

at support B.

The bending moment distribution in the statically determinate system, M0, and for a unit moment at support B, M1, is easily calculated as shown in Figure 4. The following relationships apply:

Θ100 1 1 1 1 1

224

12

13 4

12

12 4

12

16

696

= =−

+ +⎧⎨⎩

⎫⎬⎭= −∫

M MEI

dxl

EIP l P l P l P l

EIl

(4)

Θ111 1 2 1 1

13

23

= = ⋅ ⋅⎧⎨⎩

⎫⎬⎭=∫

M MEI

dxl

EIl

EIl

(5)

The unknown bending moment t at support B, X, is determined from the continuity requirement:

Θ Θ10 11 0+ =X (6)

This yields:

X P= − =ΘΘ

10

111

664

l (7)

The total bending moment is obtained by simple superposition

M M XM= +0 1 (8)


and is shown in Figure 5.

A B C1364 1Pl

664 1Pl

A B

12 2P l

P1

P2

Hinge C

Figure 5 Bending moment distribution in the elastic range, M1, and after formation of first hinge, M2

The maximum moment occurs under the concentrated load. Therefore, the first plastic hinge will be formed here for a load P1:

1364

64131 1P l M P

MlP

P= ⇒ = (9)

When the hinge has been formed the beam can take on no more moment under the load. This can be modelled by inserting a hinge at this section. The beam can, however, take some additional loading by cantilever action. Since the beam is statically determinate after the introduction of the hinge, the bending moment distribution is easily calculated as shown in Figure 5. The next section to become critical is support B. Further loading can here be accommodated until the sum of the moments from the elastic phase and the second phase becomes equal to the plastic bending moment, i.e.:

( )M M MB P1 2+ = (10)

664

12

613

12

14131 2 2 2Pl P l M P l M P

MlP P

P+ = + = ⇒ = (11)

After the second hinge the beam can take no more loading, it has turned into a mechanism. The total load is accordingly:


P P PM

lM

lcP P= + =

+=1 2

64 1413

6 (12)

that is the same result obtained with the mechanism method. However, this calculation has given more information as to the sequence of formation of the hinges. The load at first yield is given by

PMly

y=

6413

Thus, the extra capacity obtained with plastic analysis can conveniently be written

α⋅== 22.16478

y

P

y

c

MM

PP

(13)

where α denotes the shape factor. The reserve capacity contains two contributions: One is due to redistribution of stresses over the cross-section. It depends upon the shape of the cross-section as quantified by the shape factor. The other contribution stems from redistribution of forces (moments) from highly utilised sections/members to less utilised sections/members. In the example the left part of beam AB takes no more load after the first hinge, and its share of the additional loading in the second phase is taken by the right half and member BC. The amount of additional loading that can be taken after first hinge depends upon the structural configuration and the boundary conditions; Generally, the more statically indeterminate structure, the larger is the capability of force redistribution and the larger is the reserve strength.

1.1.3 Load - deformation relationship The vertical displacement under the load can easily be calculated using the unit load method. A unit load is applied at the load point and the corresponding bending moment distribution M2 (to be distinguished from the M1 distribution used previously) is determined. The vertical displacement is given by

δ100 2= ∫

M MEI

dxl

(14)

M2 in the elastic range can simply be determined from the elastic distribution by setting P1 =1. However, it is easier to use the distribution governing just after the formation of the first hinge (recall that the M2 distribution represents partial derivative of the bending moment with respect to the load, i.e. the incremental change in bending moment for an incremental change


of the load). Both methods yield the same result, but the latter approach yields simpler calculations. M2 just after the hinge is found by setting P2=1. With reference to Figure 6 the deformation at first hinge is given by

δ1

22 1913 2

16

613 2

12

613

113

2313 24

= − + +⎧⎨⎩

⎫⎬⎭=

lEI

Ml

Ml

MM l

EIP P pP

(15)

A B CM P

613

M P

A B

12

l1

=+

1913

M P

613

M P

C

M0

M2

l2

Figure 6 Moment distributions for calculation of deformation just after formation of first hinge

The deformation at the formation of the second hinge can be obtained in the same manner where M0 now represents the final plastic bending moment distribution. The same M2 distribution can be used, since we now have to calculate the deformation just prior to the second hinge (After the second hinge the beam has turned into a mechanism and it is impossible to calculate deformations)


A BM P

M P

A B

12 l1

=

+

2 M P

M P

C

M0

M2

l2

Figure 7Moment distributions for calculation of deformation just prior to second hinge

Using the moment distributions in Figure 7 the following result is obtained:

EIlMlMlMlM

EIl P

PPP 245

31

21

261

222 2

2 =⎭⎬⎫

⎩⎨⎧ ++−=δ

(16)

The resulting load -deformation relationship is plotted in Figure 9. It is seen that the stiffness is reduced significantly after the first hinge is formed. Alternatively the additional deformation after the first hinge may be computed using the additional moment in phase 2, only, as indicated in Figure 8

EIlM

EIlM

EIlM

MlMEI

dxEIMM

pPP

PPl

241323

245

527

31

137

31

21372

222

2

2012

−==

⎭⎬⎫

⎩⎨⎧ +==− ∫δδ

(17)


M0

A B C

PM137

A B

12

l1

C

M2

l2

A B

1

C

M3

l2

2

. Figure 8 Incremental moment-distributions for the second load step. It should also be noticed that phase the fist hinge has to sustain a finite rotation during the second phase. For slender cross-sections a potential failure mode for during severe rotations is local buckling on the compression side. Therefore, the codes specify compactness requirements to the slenderness (diameter to thickness ratios for tubes, flange width to thickness ratios for I-profiles etc.) of the cross-section. The requirements depend upon whether the cross-section has to undergo plastic rotations or whether it is the last to occur in the mechanism. The plastic hinge rotations can also be estimated by means of the unit load method. It is recalled that if a unit moment is applied the corresponding rotation at the given location can be calculated. Since we are not interested in the rotation of the beam as such, but the relative rotation of the two ends meeting at the hinge, we have to apply a couple of moments as indicated by the M3-diagram in Figure 9. It is realised that the relative rotation is zero up to the formation of the first hinge. It is therefor only necessary to use the additional moment produced in the second phase of the load history, given by M0. The plastic rotation in the hinge comes out to be

EIlMlMlM

EIdx

EIMM P

PPlp 127

312

137

31

21

21

1371

230 =⎟⎟

⎠

⎞⎜⎜⎝

⎛⋅+⎟

⎠⎞

⎜⎝⎛ +∫ ==θ

(18)


In Eurocode 3 for steel structures it is specified that the cross-section shall be able of sustaining plastic rotations determined by the above procedure. However, no quantitative information as to rotation capacity for different cross-sections is given.

Load versus vertical deformation for two-span beam

0

1

2

3

4

5

6

0 1 10/13 5 8Mpl

2/24EI

MP/

l

First hinge

Second hinge

Figure 9 Elasto-plastic load-deformation relationship for two-span beam

Remark: The above analysis provides useful insight in the physical basis for the mechanism method. Despite its simplicity the method captures the governing physical behaviour. As illustrated in Figure 10 the elastic deformations change continuously during loading until a complete mechanism is obtained. Once the mechanism is formed, the bending moments remain constant throughout the structure. Consequently, the elastic deformations also remain constant. Hence, the only internal work in the structure is that caused by the plastic hinges during rigid body rotations (where slope discontinuities are created in the hinges). The virtual displacement field assumed in the mechanism method is the additional displacements created after the mechanism is formed, confer Figure 11.

Elastic deformations prior to second hinge

Elastic deformations after second hingeremain constant

A C

P

A C

P

B

B

Figure 10 Elastic - and mechanism deformations.


Load versus vertical deformation for two-span beam

0

1

2

3

4

5

6

0 1 10/13 5 8Mpl

2/24EI

MP/l

Second hiFirst

Elastic def.Elasto-plastic def.

Mechanisme- "virtual deformation

Figure 11 Illustration of various contributions to vertical displacement under the load.

1.2 Elastic-plastic analysis of frames

The portal frame shown in Figure 12a is subjected to a horizontal load 0.5P and a vertical load P. All members have elastic bending stiffness EI. The plastic capacity in bending is MP for the columns and 1.2MP for the beam.

There are three potential collapse mechanisms for the frame: beam collapse of member DC, the side-sway mechanism and the combined mechanism. In the present case the combined mechanism is critical with plastic hinges in corner C and under the vertical load as illustrated in Figure 12b. By means of kinematic analysis the critical load is easily calculated to be cr PP 3.67 M L= .

Alternatively, the collapse load will be derived by means of incremental elastic-plastic analysis. In addition to the critical load this also yields information about deformations and plastic rotations in the frame.

0.7L 0.7L

0.5PP

L

C

A B

D

MP

1.2MP

MP

0.5PP

(a) (b)

0.7L 0.7L

0.5PP

L

C

A B

D

MP

1.2MP

MP

0.5PP

(a) (b) Figure 12 Portal frame: a) Geometry, b) Critical mechanism


Initially, the structure behaves fully elastically. In this phase it is one time statically indeterminate. The bending moment distribution in the frame has to be determined by an adequate method, e.g. the unit load method. The resulting bending moment diagram is shown in Figure 13.

A

D C

B(b)A

D C

B(a)

0.35MP

0.62MP

MP0.13P1L 0.37P1L

0.23P1L

Figure 13 Bending moment distribution: a) elastic range, b) M1 - at first hinge

The first plastic hinge will form at the point where the bending moment attains a maximum, i.e. in corner C. This occurs at a load P1 when:

p1 p 1

M0.37P L M P 2.7

L= ⇒ = (5.1)

The frame can now take no more moment in corner C. In the second phase of loading this is modelled by introducing a hinge in corner C. By this the frame becomes statically determinate. The bending moment distribution in phase 2 is shown in Figure 14a.

C

A

D

B

(a)

A

D C

B

(b)

0.5P2L0.83Mp

0.6P2L

1.2Mp

Mp

Figure 14 Bending moment distribution: a) phase 2 increment, b) M2 - total at second hinge

In phase 2 a new hinge can either occur in corner D or under the concentrated load. In the present case the latter event is critical. The second hinge forms when the load in the second phase, P2, attains a value given by:

( ) p2 p 2

M0.6P L 1.2 0.62 M P 0.97

L= − ⇒ = (5.2)

After the second hinge has formed the portal frame has turned into a mechanism. The bending moment distribution is shown in Figure 14b. The sum of the loads in phase 1 and phase 2 yields the collapse load of the frame:

pu 1 2

MP P P 3.67

L= + = (5.3)

This is identical with the results of the kinematic analysis. The horizontal displacement at the formation of the first and second hinges can also be determined. It is easiest to determine the displacement just after the first hinge. The displacement is then given by

1c

M MdxEI

δ = ∫ (5.4)


where M1 is the bending moment at formation of first hinge and M is the bending moment due to a unit load at corner C after the hinge has formed, refer Figure 15a.

A

D C

B

(a)

L 1

A

D

B

(b)

L 11

Figure 15 Bending moment distribution due to a) M - Unit load in C, b) M Unit moment couple in C

Thus, 2

1C1 P

22P

P P P

M M Ldx 0.35MEI EI 3

M L0.7L 5 1 10.35M 0.62M M 0.378EI 12 2 12 EI

1⎛ ⎞δ = = ⎜ ⎟⎝ ⎠

⎛ ⎞+ + − =⎜ ⎟⎝ ⎠

∫ (5.5)

Correspondingly, the displacement just prior to formation of the second hinge can be calculated using the M2 -distribution:

22

C2 P

22P

P P P

M M Ldx 0.83MEI EI 3

M L0.7L 5 1 10.83M 1.2M M 0.88EI 12 2 12 EI

1⎛ ⎞δ = = ⎜ ⎟⎝ ⎠

⎛ ⎞+ + − =⎜ ⎟⎝ ⎠

∫ (5.6)

Numerical analysis is carried out using the following data: L = 5m, moment of inertia I = 8.7E-4 m4, E= 2.1 MPa, σy = 300 MPa for the columns and σy 360 MPa for the beam. Plastic section modulus for the columns: WP = 4. 61E-3 m3. This yields P1 = 0.746 MN, P2 = 0.269 MN, Pu = 1.015 MN, δc1 = 0.072 m, and δc2 = 0.167 m.

The theoretical predictions are compared with results from simulation with USFOS in Figure 16. There appears to be a good correspondence. The major reason for the discrepancy is the p-δ effect: finite side-sway displacement induces additional bending moments in the frame.

0

0.2

0.4

0.6

0.8

1

1.2

0 0.05 0.1 0.15 0.2 0.25 0.3

Horizontal displacement [m]

Load

P [M

N]

USFOS

Calculations

Figure 16 Load versus horizontal displacement of portal frame


Codes that allow plastic or elastic-plastic methods of analysis (e.g. Eurocode 3) generally require that rotational evaluation of plastic hinges be performed. The plastic rotation, θpl, of the hinge in corner C up to the formation of the second hinge can be estimated by introducing a unit moment couple in corner C as shown in Figure 15b. This yields

2pl P P

PP

M M L 0.7L 1dx 0.83M 0.83M 1.2M MEI EI 3 EI 2 2

M LL 1 M 0.72EI 3 EI

1⎛ ⎞ ⎛P P

1 ⎞θ = = + + −⎜ ⎟ ⎜⎝ ⎠ ⎝ ⎠

⎛ ⎞+ − =⎜ ⎟⎝ ⎠

∫

(5.7)

With the given data there is obtained θpl = 0.027. This agrees fairly well with the value calculated by USFOS: 0.029.


1.3 Plastic analysis of beams and frames under cyclic loading

1.3.1 Elastic unloading What happens if the beam is unloaded from the fully plastic state? It is tentatively assumed that the unloading takes place elastically. The resulting bending moment distribution can be considered to be the sum of the fully plastic distribution and the elastic distribution corresponding to unloading as shown in Figure 17.

C

MP

MP

Pc

Pc

Loading

Unloading

1364

7864

P l Mc P=

664

3664

P l Mc P=

+Pc

Fully plastic

- Pc

Elastic unloading

+

=

2864

M P

2864

Ml

P 2864

Ml

P

2864

Ml

P

1464

M PP = 0

Resultantmoment

Deformed shape

Figure 17 Elasto-plastic loading followed by elastic-unloading

It is seen that there exists a residual bending moment distribution in the unloaded state. This distribution is caused by plastic rotation of the hinge under the load in stage 2 of the initial loading process. The residual moment filed is purely internal. The resultant external force and bending moment are zero although the reaction forces at each support are different from zero. It is readily seen that if the beam is reloaded up to P = Pc the response will be purely elastic. This can be expressed such that the beam has adapted to the load pattern in the sense that a


residual bending moment distribution has been created which is favourable with respect to the response for the given load pattern. If the beam is loaded in the reverse direction it is seen that some of the bending moment capacity under the load already is "utilised" by the residual moment field. The available

capacity for the reversed load is only 5064

Mp. Thus the reverse load, Pr, can only attain

a value of

lM3.85 =

lM

1350 = P

orM 64

50 = lP6413

ppr

pr

(19)

before a hinge is created. The range within which the load can be varied without creating a hinge is thus

lM

13128 = P - Pc and

> l

M 1378 ,

lM

1350- <

pr

pp

(20)

Before loading the elastic domain is

lM

13128 = P - P or

> l

M 1364 ,

lM

1364 - <

p11

+1

pp

(21)

Thus the size of the elastic domain is not changed by the residual moment distribution, but is only subjected to a shift in the direction of the dominant loading. If the beam is subjected to a load with even excursion in both directions, no beneficial moment distribution can be set up.

1.3.2 Cyclic loading Let us now consider the same beam subjected to two loads P1 and P2, which vary independently of each other such that

(0 < P < P)

(0 < P < P)

1

2


The static collapse load is the same as under a single load, namely: P Mc lP= 6 Three potential extreme bending moment distributions will occur for the following load combinations: (P1,P2) = (P,0), (0,P), (P,P), respectively. This is shown along with the envelope curve in Figure 18. The maximum value of the bending moment under the loads is 13 64 Pl and at point B 12 64 Pl . Obviously the beam will behave elastically if the loads are varied between 0 and

/lM 4.92 = lM pP1364 . However, can the elastic range be increased by creating a residual moment pattern?

P1 P2

2m m

1364

Pl

6 664+

Pl

1064

Pl

364

Pl

Figure 18 Beam under cyclic loading

The residual moment and the load-induced moment should nowhere exceed the plastic bending moment. It is necessary to consider both extremities of the bending moment at each section. From the bending moment distributions the following requirements can be formulated:

{ )P,P(action load ofPoint 21

M m + Pl 643 ii)

M m - Pl6413 i)

p

p

≤

≤

{ Bsupport At

M 2m + 0 iv)

M 2m + Pl 6412 iii)

p

p

≤

≤

Of the four requirements, conditions i) and ii) are the critical. Solving for equalities there is


obtained:

lM

381 = m

lM

5.05 = l

M 3864 3 = P = P

p

Ppc

⋅

(22)

Thus, if the maxima for loads P1, P2 are smaller than P Mc lP= 505. the beam will eventually shakedown to an elastic state. The shakedown limit is only slightly larger than the initial elastic state. ( 4 92. M lP ). This is because the bending moment under the load and at support B are almost equal and cannot be relieved very much by the residual moment. The shakedown load may also be calculated by means of the mechanism approach in a manner similar to the one used for static collapse.

1.3.3 Kinematic analysis of shakedown load Recalling that the load induced and residual bending moment everywhere should be smaller or equal to the plastic bending moment, this can be written:

i i pM + m M i≤ (23)

where the suffix i denotes a given section. It is obvious that this inequality needs to be checked for both extremities of the bending moment. Since these bending moments are supposed to be in equilibrium with the external load, a virtual work expression can be established. Assuming a kinematically admissible displacement field, (mechanism) the following expression is obtained

ii i i i i i p i M + m M iΣ Σ Σδ φ δ φ δ φ≤ (24)

where δϕi is the virtual rotation in the plastic hinges. By definition the residual bending moment is in equilibrium with zero external load. Hence, the second term vanishes. Accordingly:

ii i p i M P M iΣ Σ( )δ φ δ φ≤ (25)

Because the actual bending moment distribution is in equilibrium with - and is expressed as a function of the external loads, the left hand term can be interpreted as external virtual work and the right hand as internal virtual work The difference between a static approach and the shakedown calculation is that the first uses the bending moment distribution in the fully


plastic state, whereas the latter is based upon the elastic distribution. It is also observed that solving for the critical load an upper bound estimate of the dynamic collapse load is obtained, exactly as in the static case. In applying equation (25) the moment extremity in compliance with assumed mechanism shall be used. For the two-span beam the bending moment distributions shown in Figure 18 yield the following virtual work expression s

lM5.05 =

lM

3964 3 P

M3 = W

Pl6439 = Pl

6412 + Pl2

6413 = W

ppc

pi

e

⋅≤⇒

δθδ

δθδθδθδ

(26)

In the present case where the assumed mechanism is the correct one the true collapse load is obtained.

1.3.4 Example - Simple portal frame The portal frame is subjected to a vertical load, V, and a horizontal load, H.

V H

Figure 19 Portal frame a) Static collapse loads There are three possible mechanisms i) Sidesway

δ δθ

δθW Hl

dW M

HM

l

e

i P

P

==

=

2

2

δθ δθ


ii) Beam

δ δθ

δθW Vl

dW M

VM

l

e

i P

P

==

=

4

4

iii) Combined mechanism

δ δθ δθ

δθW Hl Vl

dW M

H VM

l

e

i P

P

= +=

+ =

4

4

These results are plotted in the diagram below:

0

1

2

3

4

5

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

H/(MP/ )

V/(M

P/)

Beam static

Combined static

Sidesway static

Figure 20 Failure lines for portal frame For H > V Sidesway mechanism governs V > H Combined mechanism governs Beam mechanism is not critical. b) Elastic bending momentsThe frame is one time statically indeterminate. It is made determinate by removing the horizontal restraint at right support and consider this reaction as the unknown, X.

δθ 2δθ

2δθ 2δθ


V H

H1

V1

X

V2 Consider first the horizontal load

H

H

H/2 H/2

1 1

H

2

38}122

31{1

34}

212

31{1

11

10

3

11

3

10

HXEIlllllll

EI

EIHHH

EI

−==

=⋅⋅⋅+⋅⋅=

=⋅⋅+⋅⋅=

δδ

δ

δ

Next, consider vertical load:

V

V/2 V/2

1 1V /2

2EI

lV = 21 l l

2Vl 2

EII =

3

10 ⋅δ

3EIl8 =

3

11δ

X = - = - 3

16V10

11

δδ


Moment distribution From the vertical load From the horizontal load

V/2 V/2

1 1

H /2

5V /16

3V /163V /16 H /2

c) Incremental collapse From the bending moment diagrams in pt. b) the following extreme values for the bending moment are found when 0 < V < V00 < H < H0

Table 1 Extreme values of bending moments

Section

Mmax

Mmin

A B C

A B C

3Vl/16 5Vl/16 3Vl/16 + Hl/2

- Hl/2 0 0

Beam mechanism: δ δθ

δθ δθ δθ

δ δ

W M

dWVl Vl Vl Hl

W W VH M

l

i P

e

e iP

=

= + + +

= ⇒ + =

4

316

516

316 2

24

( )

Sidesway mechanism: δ δθi pW = 2 M

δθδθδ )2

Hl + 163Vl( + Hl = W e 2

δ δe iPW = W H +

316

V = 2M

l⇒


Combined mechanism: δ δθi pW = 4 M

δθδθδ 2 )2

Hl + 163Vl( +

165Vl = W e 2

lM4

= H + V W = Wp

ie ⇒δδ

This criterion is identical to the static collapse load for the combined mechanism. It tells simply that the structure will adapt to the load pattern, so that it will behave elastically after some load repetitions in the range (0 < V < V0, 0 < H < H0). Thus there, is no reduction in the dynamic collapse load as compared to the static collapse load. The reason for this is that the bending moments a section B and C never change sign.

0

0,5

1

1,5

2

2,5

3

3,5

4

4,5

5

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

H/(MP/ )

Beam static

Combined static and cyclic

Sidesway static

Beam cyclic

Sidesway cyclic

Figure 21 Failure lines for static and incremental collapse mechanisms

The cyclic, sidesway load is sketched in Figure 21. It is seen that the domain for which the frame shakes down is somewhat reduced for the sidesway mechanism. This is due to the fact that the vertical load produces a negative bending moment in pt. A compared to the sidesway mechanism.

1.4 Elastic-plastic analysis of truss-works. Example: Jack-up platform Figure 22 shows a jack-up platform. This case is studied in the linear, elastic range in the course: marine Technology 2. Here, an elastic plastic analysis will be performed.


Figure 22 Jack-up platform subjected to ship collision

The ship is supposed drifting sideways onto the platform. Two legs are hit simultaneously., yielding symmetric conditions in the two opposing planes. Hence, a 2-dimensional analysis of one plane can be performed. The task is to calculate the relationship between the collision load- and horizontal deformation of the contact point for the truss-work up to collapse. It is assumed that the axial stiffness of the members is much larger than the bending stiffness. This means that a truss-work model can be applied. Further, it is assumed that the inertia forces caused by the acceleration of the platform (notably the topside) are so small that they can be disregarded. This means that the entire collision load is transferred down to the sea floor through the truss-work.


The forces in the truss-work in the first storey below the collision load are analysed. The model used is shown in Figure 23. It is supposed that the area of the legs is twice the area of the braces. The following data will be assumed: a = 10 m

Legs : d x t = 500 x 25 [mm] 2A = 3.730 10-2 m2 Braces : d x t = 250 x 26.6 [mm] A = 1.865 10-2 m2 Material: σY = 248 MPa Plastic yielding of tension member: AN YP σ= Critical buckling load for braces: Pcrcr NAN 75.0=σ=

.

A B

C D

2EA

EA

P

X1

a

a

2EA

EA

EA

Figure 23 Truss-work model of platform

1.4.1 Behaviour in the elastic range: The truss-work has five truss-forces and four support reactions, in all nine unknown reactions. There are eight equilibrium equations (two in each of the joints). This yields a one time statically indeterminate structure. The horizontal reaction in support B is considered unknown. Using the unit load method the forces in the statically determinate system (N0-system) and the unit load system ( 1N -system) are shown in Figure 24. Positive sign means tension, negative sign means compression. C D P0 -1

√2

-1 -1

√2

P0

- √2P


Figure 24Forces in N0- and 1N -system

The horizontal displacement of point B in the N0-system and the 1N -system can be determined by means of the formula

∑ =⎟⎠⎞

⎜⎝⎛

==

n

k

k

ikjkij ji

lEA

NNu

11,0,

(27)

This yields

( )∑ ⎟⎠⎞

⎜⎝⎛ +−=⎟

⎠⎞

⎜⎝⎛ −+−==

kk

k

kk

EAPaaPaP

EAL

EANN

u21221

21222110

10 (28)

( ) ( )∑ +=⎟⎠⎞

⎜⎝⎛ ⋅⋅+⋅+⋅==

kk

k

kk

EAaaaa

EAL

EANN

u 242212122221111

11 (29)

Compatibility yields the following equation

011110 =+ uXu (30)

where X is the unknown reaction. This yields

PPX 43.0284241

1 =++

= (31)

and the forces in each member can be calculated from

1101 NXNN += (32)

The result is shown in Figure 25


A B

C D

P1

-0.43P1

0.61P1 -0.81P1

0.57P1

0.57P1 0.43P

P1

P1-0.43P1

Figure 25 Axial forces in the elastic range - N1

The most heavily member is the compression brace AD. Since the buckling load is smaller than the plastic yield load, failure will first occur in the form of buckling of brace AD. The load level at which this takes place is given by:

PP NPNP 93.075.081.0 11 =⇒= (33)

The force in brace BC is

PBC NPN 57.061.0 1 == (34)

The horizontal deformation at buckling of brace AD is given by

( ) ( )EA

aP89.157.043.021261.081.043.0 1222221

11

1 =⎟⎠⎞

⎜⎝⎛ ++++=∑= aaa

EAPl

EANN

u ii i

iiD

(35)

1.4.2 Behaviour in the post-buckling range If a member yields in tension the axial force remains constant and equal to NP. By further loading the member can be removed from the model. This is completely analogous to inserting a plastic hinge in elastic-plastic analysis of beams and frames.


-NcrCompression

Np

u

Tension

Figure 26 Idealised force-end shortening relationship for braces.

For buckling members it is too simple to assume the member force remains constant after buckling. In practice the member force will drop in the post-buckling region. A more realistic model is to assume that the axial force reduces linearly with the end shortening as indicated in Figure 26. This may be modelled such that the stiffness is negative and a fraction c of the initial, elastic stiffness, i.e.:

lEAcckk elep −=−=

As far as the response after buckling is concerned the previous calculation carried out for the elastic system can be repeated, now with kel replaced by kep. This results in:

( )∑ ⎟⎠⎞

⎜⎝⎛ +−−=⎟

⎠⎞

⎜⎝⎛ −+==

kk

k

kk

cEAPaaPaP

cEAL

EANN

u212211

212221110

10 (36)

∑ ⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ −+=⎟

⎠⎞

⎜⎝⎛ ⋅⋅+−+⋅==

kk

k

kk

cEAaaa

caa

EAL

EANN

u 22112212122212221111

11 (37)

For the second phase of loading we therefor get

011210 =+ uXu (38)

where X2 is the unknown reaction. This yields

24114

24122

⎟⎠⎞

⎜⎝⎛ −+

−=

c

cPX

(39)

and the forces in each member can be calculated from

1202 NXSN += (40)


It is also noticed that if c > 0.586, P becomes negative. This means that the truss-work can take no more load, but has to unload after failure of brace AD. On the basis of analysis with USFOS it is found that a good representation of the post-buckling stiffness for the brace AD is obtained with c = 0.5 This yields X2= 6.22. The resulting forces are shown in Figure 27. It is observed that the drop in capacity of brace AD is substantial (note tensile force!). This drop can be transferred to the tension brace BC, which also can take some extra loading. The reserve capacity in brace BC after the elastic phase is (1-0.57)NP = 0.43NP. This means that it yields at a load of

Pp NPNP 05.043.08.8 22 =⇒= (41)

The total capacity becomes

( ) Pptot NNPPP 98.005.093.021 =+=+= (42)

A B

C D

P2

-6.22P2

8.80P2 +7.38P2

-5.22P2

5.22P2 6.22P

P2

P2-6.22P2

Figure 27 Axial forces in second stage of loading - N2.

The horizontal deformation in phase 2 is given by

( )EA

aP1.2722.622.521280.838.7122.6 222222222

2 =⎟⎟⎠

⎞⎜⎜⎝

⎛++⎟

⎠⎞

⎜⎝⎛ +−+=∑= aa

ca

EAPl

EANN

u ii i

iiD (43)

Substituting actual values there is obtained: P1 = 4.31 MN, uD1 = 0.021 m P2 = 0.23 MN, uD2 = 0.016 m Ptot= 4.54 MN, uDtot = 0.037 m

The force-deformation relationship for the truss-work and axial force versus collision load for three of the braces is plotted in Figure 28 and


-5

-3

-1

1

3

5

7

0 1 2 3 4 5

Load [MN]

Compression brace AD

Tension brace BC

Brace CD

Figure 29 along with USFOS results. The agreement is good. The actual force- end shortening behaviour for brace AD is shown in Figure 30. The calibration of the constant c was based upon the linearised response. It is observed that the drop in load is very pronounced just after buckling. The large negative stiffens (c > 0.586) explains the intermediate drop in load-carrying capacity experienced just after buckling of brace AD. After yielding brace BC can take no more load. Consequently, brace CD also takes no more load. Simple equilibrium of joint D gives

aEAcP22

13 −= (44)


0

1

2

3

4

5

0 0,01 0,02 0,03 0,04 0,05 0,06

Upper corner displacement [m]

USFOS

Hand calculations

where the constant c < 0.50. Consequently, truss-work has to unload; i.e. the ultimate load has been attained. Figure 28 Collision load versus horizontal displacement

-5

-3

-1

1

3

5

7

0 1 2 3 4 5

Load [MN]

Compression brace AD

Tension brace BC

Brace CD

Figure 29 Axial force versus collision load


-4

-3

-2

-1

0-4,00E-02-3,00E-02-2,00E-02-1,00E-020,00E+00

End displacement [m]

At maximum load (2.44E-02, 1.84)

Equivalent linear relationship

Figure 30 Axial force in brace AD versus end- shortening


1.5 Plastic capacity of beams undergoing large lateral deflections.

This section presents derivation of the capacity of beams under large lateral deflections. The derivation is an alternative to the approach adopted in Section 3.2 in Søreide’s book. Consider the beam in Figure 31.

l2

l2

P

A B

Figure 31 Simply supported - axially free beam.

According to rigid-plastic theory the beam undergoes collapse once the bending moment at midspan reaches the fully plastic bending moment, i.e. for

P lMc

P4=

(45)

or

PMlc

P=4

(46)

If the horizontal boundary condition at end B is changed from free to fixed, the collapse state shown in Figure 32 is obtained.

l2

l2

P

MMN N w

θ

A B

Figure 32 Simply supported- axially fixed beam Due to the fixity against inward displacement the beam elongates as soon as it undergoes finite displacement. This elongation gives rise to tensile membrane forces. Note that θ and w now represents finite (true) and not virtual rotation and displacements.


From simple equilibrium considerations the following relationship is obtained

PMl

Nw

lMl

Nwl

= + = +4

22

44

(47)

The first term is the conventional bending term, the second term represents the contribution to

the load-carrying from the membrane forces. The term kNl

= 4 can be considered as a

geometric stiffness. It depends upon the value off the axial force. Consequently, a beam which has an initial tensile axial force has a greater stiffness than the conventional bending stiffness. In the fully plastic state the cross-sectional forces have to obey the plastic interaction function for the cross-section. Interaction function for different cross-section s are derived Section 2.3. If we assume a rectangular cross-section, the interaction equation takes the from

FM

MN

NP P= +

⎛⎝⎜

⎞⎠⎟ − =

2

1 0

(48)

or

M MN

NPP

= −⎛⎝⎜

⎞⎠⎟{ }1

2

(49)

Combining equations (47) and (49) there is obtained

PM Pl

N

nP

N

M Pw= − +

41 2{( ( ) ) }

(50)

or PP

NN

NM

wc P= − +1 2( )

P

(51)

The problem is now to determine the axial force as a function of lateral displacement, i.e. N = N(w) We then introduce Drucker’s postulate: The incremental, plastic displacement vector, δvp is normal to the yield surface. The plastic displacement vector has two components: 1) plastic rotation in the yield hinge 2) plastic elongation for the member as it undergoes finite displacements The incremental, plastic displacement vector is sketched along with the yield surface in Figure 33. It corresponds to a given force state (combination of axial force and bending


moment). It will change continuously during the deformation process in accordance with the evolution of the force state.

M ,θ

N e,

δδθδ

vep =

⎡⎣⎢

⎤⎦⎥

F = 0

Figure 33 Illustration of the normality criterion

Mathematically the Druckers’ postulate, often denoted the normality criterion, can be expressed as

δδθδ

δλ∂ ∂∂ ∂

v eF MF N=

⎛⎝⎜

⎞⎠⎟ =

⎛⎝⎜

⎞⎠⎟

(52)

where θ and e represent plastic rotation and elongation respectively, λ is a proportionality constant, the symbol δ signifies a virtual - or incremental quantity and ∂ represents partial derivation. With the present yield condition there is obtained

δθ δλ=1

M P

(53)

δ δλeN

N P=

22

(54)

Remark: Physically λ represents plastic work in the hinge since δλ δθ= M P The proportionality constant λ can be eliminated yielding

NN

e MPP

2

2δ δ= θ

(55)

We also need to establish the geometric relationship between the plastic rotation and the plastic elongation. From Figure 34 it is seen that the total elongation can be expressed as


l2

wl

e2+

Figure 34 Elongation model

el

wl l w

l

l wl

wl

= + − = + −

≈ + =

( ) { }

{ }

2 2 21

41

21

2

2 22

2

2

2

2

(56)

The incremental change of the plastic elongation is found by taking the total differential

δδ

δθew wl

w= =2

(57)

where we have utilised the fact that

θ δθδ

= → =2 2wl

wl

(58)

It is observed that there is a nonlinear relation between rotation and elongation. The incremental change of elongation for an increment in rotation is proportional with the total deformation. This is also intuitively correct, the rotation affects the elongation more at large displacements. Combining equations (55) and (57) we get

NN

w MPP

2

2δθ δθ=

(59)

⇒ =N

N P

wN PM P2

(60)

We have now obtained the desired relationship between axial force and lateral deformation. The

expression can be simplified by introducing NM

bh

bh h

P

P

Y

Y

= =σ

σ2

4

4 so that

NN

wh

whP

= ≤212

(61)


The restriction set on the lateral displacement is obvious; the axial force cannot be larger than the plastic axial capacity. Combining equations (51) and (61) the load-carrying capacity can be expressed as

PP

wh

NM

wh

wh

wh

wh

wh

c

P

P= − + = − +

= + ≤

1 4 2 1 4 8

1 412

22

2

2

( ) ( ) ( )

( ) ,

2

(62)

The first two terms in the expression represents the bending contribution, the last term the membrane contribution. It is observed that the bending contribution dominates initially, but decreases rapidly when the displacement increases. Conversely, the membrane contribution increases with increasing displacement.

The above expression holds true for wh

≤12

when the axial force becomes fully plastic. At

this point the bending contribution vanishes. For further deformation the axial force remains constant and equal to the fully plastic force. The load-carrying capacity in this range is

P Nwl

Ml

NM

wPP P

P= =4

4

(63)

or PP

wh

whc

= ≥412

(64)

The solution is illustrated in the Figure 35. If the beam can develop membrane forces, there is

0

0,5

1

1,5

2

2,5

3

3,5

4

0 0,2 0,4 0,6 0,8 1

Displacement w/h

Late

ral l

oad

P/P

0 Pure membrane

Membrane

Membrane + bending

Pure bending

Bending

Figure 35 Load-carrying capacity for rectangular beam at large deformations


no unique collapse load. The capacity increases monotonously for increasing lateral displacement.

1.5.1 Plastic load-carrying capacity for uniform load and clamped ends In section 3.2.3 in Søreide’s book it is shown that the same relationships can be used for a beam subjected to uniformly distributed load provided that the collapse load is substituted by

PP

qq

where qMlc c

cP→ = 8 2

(65)

For a fully clamped plate the collapse mechanism consists also of hinges at the support. This increases the plastic bending work by a factor of two. The plastic elongation is distributed by one half to the center hinge and by one half to the support hinges. Apart from this the derivation of the plastic load-carrying capacity is identical This yields the following relationships:

PP

wh

wh

PP

wh

wh

P ql M

l

c

c

c cp

= + ≤

= ≥

= =

1 1

2 1

28

2( )

(66)

All these equations can be summarised in a single set of equations:

PP

z z

PP

z z

Pinned ends

zwh

P ql M

lClamped ends

zwh

P ql M

l

c

c

c cp

c cp

= + ≤

= ≥

= = =

= = =

1 1

2 1

22

4

28

2

:

,

:

,

(67)

1.6 Plastic load-carrying capacity of plates subjected to large lateral deformation The plastic collapse of plates under uniform pressure with boundaries free to displace axially is studied in Section 3.1 in Søreide’s book. If the plate edges are fixed against inward displacements, membrane forces will develop at finite deformation in the same manner as for


beams. The rooftop mechanism used for pure bending may be assumed valid also for large lateral deformations. The analysis is based upon the one used for beams with rectangular cross-section, where the cross-section now represents a plate strip with unit width. The plate mechanism is assumed to consist of a number of plate strips where the lateral deformation and consequently the force state vary in the oblique hinge lines. By integrating the contribution for all the plate-strips is possible to obtain closed-form solutions for the load-carrying capacity. They were first presented by Jones and Walker: «Large deflections of rectangular plates» Journal of Ship Research, Vol. 15, 1971, pp. 164-171. The solutions can be summarised in the following relationships:

qq

z z

qq

zz

z

Pinned ends

zwh

qh

a

Clampd ends

zwh

qh

a

c

c

cY

cY

= ++ −

−

⎛

⎝⎜

⎞

⎠⎟ ≤

= +−−

−⎛⎝⎜

⎞⎠⎟ ≥

= =

= =

13 2

9 31

2 12

31

31 1

26

12

22

2

2

2 2

2

2 2

α αα

α αα

σα

σα

( )

( )( )

:

:

(68)

h is plate thickness, a and b denote plate length and width, respectively. α is a plate aspect parameter defined as

α = + −⎛

⎝⎜

⎞

⎠⎟

ba

ba

ba

3 2( ) . The similarity with the equations for the rectangular beam is

noticed. The term in the large bracket may be interpreted as a plate factor. The solutions are depicted in Figure 36 and Figure 37. In Figure 36 the relative load-carrying capacity as a function of the lateral displacement. It is observed that long plates experience a larger relative increase in the capacity than quadratic plates. This is also shown in Figure 37 where the capacities are normalised versus the collapse load for a plate strip in bending. A quadratic plate is three times as strong as a plate-strip (a/b=100) in the bending phase, whereas it is only approximately 50% stronger for w/h=3


0

1

2

3

4

5

6

0 1 2 3Relative displacement [w/h]

Load

-car

ryin

g ca

paci

ty [q

/q0]

a/b = 100

5 3 2 1

Figure 36 Plastic load-carrying capacities of plates as a function of lateral displacement

0123456789

0 1 2 3

Relative displacement [w/h]

Load

-car

ryin

g ca

paci

ty [q

/qc p

late

strip

]

1 = a/b

3, 2 100,5

Figure 37 Plastic load-carrying capacity of plates relative to plate-strip in bending


1.7 Use of plastic methods in design

For building structures plastic methods of analysis have equal status as elastic methods provided that all plastic hinges satisfy compactness requirements and is adequately restrained against lateral displacement. In Eurocode 3 part 1.1 there is distinguished between three types of plastic analysis:

i) Rigid plastic analysis

All elastic deformations are neglected and plastic deformations are assumed to be concentrated at plastic hinge locations

i) Elastic-perfectly plastic

The cross-sections are assumed to be linearly elastic-perfectly plastic, i.e. gradual plastification in bending is not accounted for. Plastic deformations are assumed to be concentrated at plastic hinge locations

iii) Elasto-plastic analysis

Gradual plastification of the cross-section after yielding of extreme fibers are taken into account. Plastic deformations spread partially along the member, i.e. they are no longer assumed concentrated in the hinges. In this cases it is necessary to integrate stresses over cross-sections as well as along the member.

Figure 38 Illustration of different plastic methods of analysis in the case of bending

Rigid-plastic theory should not be used when second order effects are important (i.e. when deformations create additional bending moments- the so-called P-δ effect, see Figure 39 for illustration). Elastic-perfectly plastic and elasto-plastic methods may take this effect into account.

Elastic-plastic

M/MP Rigid-plastic 1

Curvature

Elasto-plastic First yield

MY/MP


H V

δ

δ

Figure 39 Examples of P-δ effects

For ship structures the use of plastic methods in design is very limited for several reasons:

i) Less use of fundamental methods»first principles»)

ii) Frames often consist of slender members which due not satisfy «compact section» requirements (plate girders)

iii) Loads are often alternating - in the case of which the benefit of plastic design is reduced

The design of offshore structures is generally based upon advanced method. However, the typical offshore structure- the jacket- carries the load primarily by truss-work action. Hence, second-order effect and buckling must be taken into account. Therefor the conventional plastic methods used for buildings are not very relevant. However, very advanced methods of analysis -like nonlinear finite element methods - are being taken more and more into account. Such methods contain all the particularities of elastic-plastic methods and include also the buckling phenomenon.

Literature

/1/ Heyman, J.: Plastic Design of Frames, Vol.1-2, Cambridge University Press, 1971

/2/ Massonet, C.F. and Save, M. A.: Plastic Analysis and Design of Beams and Frames, Blaidsdell, 1969

/3/ Jones, N. and Walker, R. M.: «Large deflections of rectangular plates» Journal of Ship Research, Vol. 15, 1971, pp. 164-171.

/4/ Søreide, T.H.: Ultimate Load Analysis of Marine Structures, Tapir 1981

/5/ Eurocode 3 - Design of steel structures. Part 1.1 General rules and rules for buildings DD ENV 1993-1-1


INDEX

C

clamped ends · 38 continuity requirement · 4

D

Drucker’s postulate · 34

E

elastic domain · 16 elastic range

behaviour · 12 elastic-plastic analysis · 11 elastic-plastic methods · 42 elasto-plastic methods · 41 elongation

of beam · 33, 34, 35, 36, 38 envelope curve · 17 Eurocode 3 · 14

F

first principles · 42 first yield

load at · 6

G

geometric stiffness · 34

I

interaction equation · 34

K

kinematic analysis · 11, 12 kinematically admissible

mechanism analysis · 18

L

local buckling compactness · 9

M

mechanism · 2, 5, 6, 7, 9, 10, 18, 19, 20, 22, 23, 38, 39

membrane forces effect of · 33, 34, 37, 38

N

normality criterion · 35

P

plastic displacement vector · 34 plastic hinge · 2, 3, 5, 9, 12, 27, 41 plastic work · 35

R

redistribution of forces reserve capacity · 6

reserve capacity · 6, 29 reserve strength · 6 residual moment · 15, 16, 17, 18 rigid body

rotaion · 10 Rigid-plastic analysis · 41 rotation

of plastic hinge · 9, 10, 15, 18, 33, 34, 35, 36

S

second order effects · 41 shape factor · 6 side-sway displacement · 13 statically determinate · 4, 5, 25 statically indeterminate · 3, 6, 20, 25

T

truss-work · 24, 25, 29, 31, 42

U

unit load method · 2, 3, 6, 9, 25 USFOS · 13, 14

V

virtual displacement mechanisme method · 10

Y

yield surface · 34

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