CM 2014_GG_Plates and shells

PLATE AND SHELLS:BUCKLING AND NON – LINEAR

ANALYSIS

Corso diCOSTRUZIONI METALLICHE

Prof. Ing. Franco Bontempi

Corso di Costruzioni MetallicheProf. Ing. Franco Bontempi, Ing. Giordana Gai

UNIVERSITA’ DEGLI STUDI DI ROMA “SAPIENZA”

[email protected]

Ing. Giordana Gai

Roma, 5 Dicembre 2014

[email protected]

2Corso di Costruzioni MetallicheProf. Ing. Franco Bontempi, Ing. Giordana Gai


OUTLINEPLATES

• Buckling analysis Compressive forces Bending moment forces Shearing forces Shearing and compressive forces Shearing and bending moment forces

• Non – linear analysis (elastic material)• Non – linear analysis (elastic – plastic material)

SHELLS• Thin shells (membrane theory)• Thick shells• Buckling analysis• Non – linear analysis (elastic material)

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PLATES

Buckling analysis


UNIVERSITA’ DEGLI STUDI DI ROMA “SAPIENZA” 2/92

How can we calculate the critical load for a plate?• Analytical method:by solving equations of equilibrium it can be difficult to find the solution (the

equations are fourth order - partial differential equations)• Energy methods:by equaling the work done by acting forces with strain energy it is useful if we are

interested in approximate solutions.

4

HipotesisHipotesisGEOMETRY No imperfections Thin plate: one dimension is very small with respect on the other twoMATERIAL (Steel) Linear ElasticLOADS Loads are applied in the middle plane of the plate



The critical value of the forces acting on a plate depends on: Ratio a/b (length / width) Thickness s Material properties (E and ν) Boundary conditionsfor each condition of loads



1.COMPRESSIVE FORCES

The plate buckles in a shape that can have several half-waves in the direction ofcompression but only one half wave in the perpendicular direction.

m = number of half – waves parallel to thedirection of Nn = number of half – waves perpendicular to thedirection of N

Deflection surface

Critical load

The minimum value of the critical load isobtained by putting n = 1

PlatePlate simplysimply supportedsupported alongalong fourfour edgesedges



Euler load for a strip of length “a” and unit width

the stability of the continuous plate is greater than the stability of an isolated strip

(depends on the ratio a/b and on the number m)

• If a < b the minimum is obtained by putting m = 1(only one half – wave )

• If a > b it is used a simplified expressionwhere k is a numerical factor depending on a/b




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• Short plates buckles in only one half – wave

• Long plates can have very large “m” (comparable to the ratio a/b)

• If m > 1 , the plate buckles in more half – waves and each half is in the condition of a simply supported plate of length a/(number of half – waves).

• The curve m = 1 has a minimum for square plate; before and after the ratio a/b = 1, the value k increases.

• Critical value of the compressive stress




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PlatePlate simplysimply supportedsupported alongalong twotwo edgesedges perpendicularperpendicular totothe direction the direction ofof NN

Solution depends on boundary conditions along the edges y = 0 and y = b

Deflection surface

•• y = 0 y = 0 simplysimply supportedsupported; y = b free; y = b free

• y = 0 zero displacement and zero bending moment • y = b zero shear and zero bending moment

k Ncr




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•• y = 0 y = 0 builtbuilt in; y = b freein; y = b free

• y = 0 zero displacement and zero rotation

• y = b zero shear and zero bending moment

The curve m = 1 has a minimum fora/b = 1.635



1.COMPRESSIVE FORCESPlatePlate simplysimply supportedsupported alongalong twotwo edgesedges perpendicularperpendicular toto

the direction the direction ofof NN

k Ncr

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•• y = 0 y = 0 builtbuilt in; y = b in; y = b builtbuilt inin

• y = 0 zero displacement and zero rotation

• y = b zero displacement and zero rotation

The smallest value of k is for 0.6 < a/b < 0.7.

In this case, a long compressed plate buckles in comparatively short waves



1.COMPRESSIVE FORCESPlatePlate simplysimply supportedsupported alongalong twotwo edgesedges perpendicularperpendicular toto

the direction the direction ofof NN

k Ncr

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There are two possible forms of buckling

• y = 0 and y = b zero bending moment and shear equal to the action force (N w/ y)

•The first buckling form is antisymmetrical up to a value ofb/a=1.316

• then, for 1.316 < b/a < 2.632 the buckling form is symmetricaland then the value of k remains close to 2.31.



1.COMPRESSIVE FORCESPlatePlate simplysimply supportedsupported alongalong twotwo edgesedges parallelparallel toto the the

direction direction ofof NN

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13

The curve of antysimmetrical buckling can be used also for the case of a plate simply supported along 3 sides (this plate is in the same conditions as the half of the antisymmetrical plate described before).

To calculate k, that curve can be used , using 2b/a instead of b/a.



1.COMPRESSIVE FORCESPlatePlate simplysimply supportedsupported alongalong twotwo edgesedges parallelparallel toto the the

direction direction ofof NN

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• = 0 pure compression• = 2 pure bending moment• = 1 bending moment and compression

For each case k is calculated and then the critical value of N0.

η = 2

Considering m=1, it has a minimum when a / b = 2/3.

For cases with η > 2, the value of the ratio at which k presents a minimum increases (for purecompression a/b = 1).



2.BENDING MOMENT FORCES

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In this case, buckling is caused for the compressive stressesacting along one of the two principal diagonals.

The critical value of shear stresses can be calculated by usingk, that depends on the ratio a/b.



3.SHEARING FORCES

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The axial stresses reduces the stability of the plate submitted at the action of shear stresses.

The interaction between shear and compression is highly negative.



4.SHEARING AND COMPRESSIVE FORCES

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In this case we have both compressive that tension stresses.

The diagram shows that for τ / τ cr < 0.4 the effect of shearing stress on the critical value ofbending stress is small.

α is the ratio a/b



5.SHEARING AND BENDING MOMENT FORCES16/92

Comparison between analytical and numerical code solutionsAnalytical solution: by Timoshenko’s theory Numerical codes: SAP 2000 and Straus 7

a = b = L 50 cms 1 cmE 2E+08 KPaν 0.30

LOAD CASES ANALYZED:1. Compressive forces2. Bending moment forces3. Shearing forces4. Shearing and compressive forces5. Shearing and bending moment forces



NUMERICAL EXAMPLENUMERICAL EXAMPLE

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ANALYTICAL SOLUTIONNcr = k Nl

Nl = π2Et3/12(1-ν2)b2 = 723KN/mk from the schedule

NUMERICAL SOLUTIONBuckling analysisMesh used = 30x30 finite elementsISOP4

SAP 2000

Straus 7



1.COMPRESSIVE FORCES18/92

CASE k BuckledPlate

1 1

2 4

3 6.5

4 5

5 7.7

6 5.7

CASE k BuckledPlate

7 1.44

8 1.7

9 10.2

10 2.05

11 2.3

12 2.3



CASE 1 CASE 2 CASE 5xy_view xz_view xy_view xz_view xy_view xz_view

Pcr = 723 KN/m Pcr = 2892.2 KN/m Pcr = 5560.2 KN/m

CASE Analytical Pcr[KN/m]

Differences Straus7-Analytical Solution [%]

Differences SAP2000-Analytical Solution [%]

1 723.0 -5.0% 1.8%7 1041.2 -2.8% 3.8%

8 1229.2 -2.9% 3.7%10 1482.2 -1.0% 3.1%11 1663.0 2.7% 1.7%

12 1663.0 3.8% 2.7%2 2892.2 -0.2% 5.0%4 3615.2 5.6% 5.6%

6 4121.4 0.5% 4.5%3 4699.8 3.4% 3.4%5 5560.2 -0.4% 3.0%

9 7375.1 -1.6% -1.7%



ANALYTICAL SOLUTIONNcr = k Nl

Nl = π2Et3/12(1-ν2)b2 = 723KN/mk from the schedules


SAP 2000

Straus 7



2.BENDING MOMENT FORCES21/92

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CASE k Pcr [KN/m] Buckled PlateDifferences Straus7-Analytical Solution

[%]

DifferencesSAP2000-Analytical

Solution [%]

25.6 18510 1.15% 0.15%

39.5 28530.4 -9.85% 1.73%

7.8 5639.8 -0.17% 0.27%

15.0 10845.7 -2.41% -1.72%

PURE BENDING

BENDING AND

COMPRESSION



2.BENDING MOMENT FORCES

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ANALYTICAL SOLUTIONτcr = k Nl

Nl = π2Et3/12(1-ν2)b2 = 723KN/mk from the schedule


SAP 2000

Straus 7



3.SHEARING FORCES23/92

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CASE k Pcr[KN/m] Buckled Plate w[m]_

XY view

DifferencesStraus7-

AnalyticalSolution [%]

DifferencesSAP2000-Analytical

Solution [%]

9.7 7013.6 4.3% 6.2%

15.0 10845.7 3.1% 3.1%

12.3 8893.5 -1.5% -0.3%



3.SHEARING FORCES

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ANALYTICAL SOLUTION• by using the approximateexpression, where N0cr and T0cr are those calculated before (respectivlycompressive and shear forces only)

• by using the schedule


SAP 2000

Straus 7



4.SHEARING AND COMPRESSIVE FORCES25/92

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Numerical Code Solution Analytical Expression Differences %

Tcr = Ncr [KN/m] Ncr =Ncr0[1-(Tcr/Tcr0)2] [KN/m]

Numerical Code –Analytical Solution

SAP 2000 2315.5 2666.1 13%

STRAUS 7 2452.4 2887.2 2%



4.SHEARING AND COMPRESSIVE FORCES

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ANALYTICAL SOLUTION• by using the approximateexpression, where N0cr and T0cr are those calculated before (respectivlycompressive and shear forces only)

• by using the schedule


SAP 2000

Straus 7




29

Numerical Code Solution Analytical Expression Differences %

Tcr = Ncr [KN/m] Ncr = Ncr0[1-(Tcr/Tcr0)2]0.5 [KN/m] Numerical Code – AnalyticalSolution

SAP 2000 6103.1 6906.5 12%

STRAUS 7 6241.3 6709.6 7%

SHEAR SHEAR+BENDING SHEAR+COMPRESSION

λcr = 6708.5 λcr = 6241.3 λcr = 2452.4




PLATES

Non – linear analysis



We run displacement control - nonlinear analysis to investigate post – criticalbehavior.

31

NumericalNumerical exampleexampleNumerical code : Straus 7Mesh : 10x10 finite elements ISOP4Step : 200 increments of 0.001 mBoundary conditions: 2 edges simply supported

a = b = L 50 cm

s 1 cm

E 2E+08 KPa

ν 0.30



ELASTIC MATERIAL

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from Buckling analysis

from NL analysis (plate straight) with largedisplacement

from NL analysis (plate not perfectly straight) with large displacement

POST CRITICAL BEHAVIOR = stable, with hardening

(only geometric nonlinearity)



ELASTIC MATERIAL

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We insert elastic-plastic diagram into material properties and we run again NL analysis.

33

σy 235 MPa

εy 1.175*10-3

εu 7.5*10-2

NLM (NLM (idealideal plateplate) ) NLM (NLM (realreal plateplate) )

(only material nonlinearity)

= 235*103*0.5*0.01=1175 KN



ELASTIC – PLASTIC MATERIAL

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0

100

200

300

400

500

0 0.05 0.1 0.15 0.2

P [K

N]

Dx [m]

NLM_not perfectly straight plate

NLG_not perfectly straight plate

050

100150200250300350400450

0 0.05 0.1 0.15 0.2

P [K

N]

Dx [m]

Buckling Yielding_not perfectly straight plate

If we consider the plate with an initial imperfection, the elastic-plastic limit is really near to the buckling curve

NLM+NLG (realNLM+NLG (real plate) plate) The plate starts to reach the yelding

limit in the lateral zone, but rapidlyit is totally yelded (orange line)




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35

We studied two other plates: one thicker(s=3 cm) and one thiner (s=2mm).

The thicker reaches the yielding limit and then collapse, the thiner follows the buckling curve: the previous platewith s=1cm has an intermediate behavior.

We compared the three “orange curve” byadimensionalyzing respect on theiryielding maximum force.



ELASTIC – PLASTIC MATERIAL34/92

36

65% of the yielding limit



Coupled behaviour of yielding and buckling



ELASTIC – PLASTIC MATERIAL35/92

37

Experiments show that due to the interaction between the buckling and the yielding behaviour, when the compressive stress reaches the yield point of the material, the plate buckles.

Zone 1 buckling of plate when the compressive stress reach the yield point

Zone 2 buckling of plate when the stresses remain within the elastic limit

Some permanent set usually takes place at a stress lower than the yield point

It is unusual to find a sharp corner that separates the elastic behavior from the plastic one.




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• Different behaviour of plates and struts when the stresses are beyond the proportional limit

• The critical load for a strut is considered as the ultimate load, instead a thin buckled plate can carry a bigger load than the critical load at which buckling begins.

• In order to explain the post-critical behavior of plates we analyze a compressive square plate simply supported by using two different analytical solutions

Approximate solution Enhanced solution




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39

Approximate solutionApproximate solution1) Definition of the boundary condition

Simply supported plate with the lateral expansion in the x direction prevented by a rigid frame

This approximate expression of the components of the displacements satisfy the boundary conditions




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40

2) Determination of the components of the strain in the middle plane and the corresponding strain energy of the plate

3) Determination of the energy of bending



ELASTIC – PLASTIC MATERIALApproximate solutionApproximate solution

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41

4) By using the solution method of minimum of strain energy we determinate the constant f, C1, C2

By solving the system we obtain

We obtain a solution for “f” only if :




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42

5) Determination of the stresses σx and σy

By plotting the diagram of the stresses for the ultimatecompressive force (Nr,d) we can notice:

• The ultimate compressive force is n times bigger thanthe critical compressive force

• The compressive stress diagram is non-uniform




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43

For the restraints of the plate that we are considering, we know that the part of the horizontalfibres near of the edge is more rigid because is close to the restraints( that are considering havingan infinite rigidity). So the vertical fibres near to the vertical edge are able to absorb a greaterpost-critical load because they can count on a greater flexural rigidity of the horizontal fibres.

Why can the ultimate compressive force in a plate be n times bigger than the critical one?

Why is the compressive stress diagram non-uniform ?

For a compressive force greater than the critical one, the vertical fibres instead of collapse ,asthey would do if we analyzed a single strut, can absorb a further load because their deflectionout of the plane is hindered by the flexural rigidity of the horizontal fibres.




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44

The total compressive force acting on the plate at a unit compression necr is

The factor c gives the relative diminishing of the resistance of compression of the plate due to buckling

A1 = A2

We can state that this resistance is equivalent to that of a flat plate having a width equal to c.




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45

For thin plate the approximate solution that we calculated is not sufficiently accurate.

1)The solution can be improved by changing the boundary conditions.

we assume that the later bars of the frame keep the lateral edges of the plate straight but moving freely laterally.

We can resolve the problem by following the steps shown in the approximate solution

2) We determinate the ultimate load that the plate can carry by using the Tresca yield criterion.



ELASTIC – PLASTIC MATERIALEnhanced solutionEnhanced solution

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A square plate simply supported

DATA• a = b = 500mm• t = 10 mm• E = 2E5 MPa• Perfect elastic-plastic constitutive law • σy = 235 MPa• εy = 1.175E-3



ELASTIC – PLASTIC MATERIALNumericalNumerical exampleexample

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47

Analytical approximate solution

0

0.5

1

1.5

2

2.5

3

-1 -0.5 0 0.5 1

σx /σcr

asse y / a

Cross section x = a

-1.6

-1.4

-1.2

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

-1 -0.5 0 0.5 1

σy /σcr

asse x /a

Cross section y = a




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48

1) We run the non linear static analysis by considering the non linear material

σy 235 MPa

εy 1.175*10-3

εu 7.5*10-2

Analytical solution Numerical solution

σy 235 MpaStep 4.7

Fcr 558 kN

Fy 2350 kN Fy 2622.6 kN

errore 10%




Straus 7 approximate solution

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2) We run the non linear static analysis by considering the non linear material and the non liner geometry

Geometric non - linearity Material non - linearity

Numerical solution Analytical solutionFu 1450.8 kN Fu 554.7 kN




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51

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

-1 -0.5 0 0.5 1

σx /σcr

asse y/a

Cross section x = a

Straus7 Analytical solution-2

-1.5

-1

-0.5

0

0.5

1

1.5

-1 -0.5 0 0.5 1

σy /σcr

asse x/a

Cross section y = a

Analytical solution Straus7

Analytical enhanced solution

Straus7 approximate solution




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52

In order to obtain an enhanced solution, we repeat the analysis by considering different boundary conditions

Boundary condition:• Horizontal edges: uz=0• Right vertical edge: uz = 0• Left vertical edge: uz = 0ux = 0




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53

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

-1 -0.5 0 0.5 1

σy /σcr

asse x/a

Sezione y = a

Analytical solution Strauss7

0

0.5

1

1.5

2

2.5

3

3.5

-1 -0.5 0 0.5 1

σx /σcr

asse y/a

Sezione x = a

Strauss7 Analytical solution




Analytical enhanced solution

Straus7 approximate solution

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-1.6-1.4-1.2

-1-0.8-0.6-0.4-0.2

00.20.40.6

-1 -0.5 0 0.5 1

σy/σcr

asse x/a

C.S. y = a analytical solutions

P.A.S.10 A.S.10

0

0.5

1

1.5

2

2.5

3

-1 -0.5 0 0.5 1

σx /σcr

asse y/a

C.S. x = a analytical solutions

P.A.S.10 A.S.10

54

ComparisonComparisonWe compare the results provided by the approximate and the enhanced analysis

Legend:P.A.S. Enhanced analytical solution A.S.Analytical solution P.S7.S. Enhanced Straus7 solution S7.S. Straus7 solution




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00.5

11.5

22.5

33.5

44.5

-1 -0.5 0 0.5 1

σx /σcr

asse y/a

C.S. x = a Numerical solutions

P.S7.S.10 S7.S.10

-1.5

-1

-0.5

0

0.5

1

1.5

-1 -0.5 0 0.5 1

σy /σcr

asse x/a

C.S. y = a Numerical solutions

P.S7.S.10 S7.S.10



ELASTIC – PLASTIC MATERIALComparisonComparison

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Legend:P.A.S. Enhanced analytical solution A.S.Analytical solution P.S7.S. Enhanced Straus7 solution S7.S. Straus7 solution

We compare the results provided by the approximate and the enhanced analysis

SHELLS



• We consider a shell where the thickness is small in comparison to its other dimensions

The components of the stresses

the component of the stresses

tangent to the meridian

σ1is Caused

by the resultant ofthe externalvertical forces

the component ofstresses tangent tothe parallel

σ2is

Caused

by the force Z

by the changingof direction ofthe stresses σ1



THIN SHELLSThe membrane theoryThe membrane theory

• We assume an axial-simmetric load.

Components of the external forces:X Component that is tangent to the meridian of the shell Z Component that is normal to the surface of the shell

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Hypothesis:very small ratio thickness / minimal

radius

Thesis :No flexural and torsional rigidity for

the membrane

Dimostration1) We subdivided the shell in elementary strips of unitary width that follow the geometry of the

meridians and parallels2) Because the strips of the parallels are compressed or in traction they modify their radius but

they remain circulars.3) Due to the deformation of the parallels, the strips of the meridians can deform and

consequently change their curvature4) The change of the meridian curvature causes not uniformly distributed stresses along the

thickness (σ’1).5) The total stress tangent to the meridians σ1tot = σ1+σ1’6) σ1’<< σ1 due to the small thickness (small ymax) and the small change of meridians curvature




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ConclusionFor shell characterized by a very small ratio thickness / minimal-radius

the stresses can be considered uniformly distributed along the thickness

Consequence

1) In each point of the membrane the stress state is plane.2) No bending or shear transverse stresses in any point of the shell




Hypothesis:very small ratio thickness / minimal

radius

Thesis :No flexural and torsional rigidity for

the membrane

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Compatibiliy conditions1) Boundary condition and deformation constraints must be compatible with the requirements of a pure membrane field.

2) There must not be concentrated loads.

3) Shell geometry must not change.




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The membrane forcesWe determinate the membrane forces due to the axial-simmetrical load

S1 integral of the stresses tangent to the meridians

Q is the resultant of the vertical loads




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S2 integral of the stresses tangent to the parallels




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We determinate the membrane forces due to the axial-simmetrical load The membrane forces

DATA•R = 3 m•s = 0.1 m•γW = 10 kN/m3

Calculate the membrane forces of a hemisphere tank full of water



THIN SHELLSNumerical example 1Numerical example 1

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• Analytical solution 1) We calculate the volume of the hemisphere

2) We calculate the membrane forces




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In order to fulfil the requirements ofthe membrane theory, we apply anuniform distributed load that varieslinearly with the Z.

We define traslational restraints for theedge of the structure in order to have aboundary condition compatible with therequirements of a pure membrane state.



THIN SHELLS

• Sap2000 solution

Numerical example 1Numerical example 1

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S1

S2



THIN SHELLS



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SAP 2000 BELLUZZI

Stepz θ S1 S2 S1 S2

ErrorS1

ErrorS2

m deg KN/m KN/m KN/m KN/m % %

1 0.000 0 45.53 44.65 45.00 45.00 -1.2% 0.8%

2 0.333 10 44.45 42.61 44.66 43.97 0.5% 3.1%

3 0.667 20 43.04 38.94 43.66 40.92 1.4% 4.8%

4 1.000 30 41.32 34.10 42.06 35.88 1.8% 5.0%

5 1.333 40 39.33 27.94 39.97 28.98 1.6% 3.6%

6 1.667 50 37.16 20.40 37.55 20.31 1.0% -0.5%

7 2.000 60 34.91 11.52 35.00 10.00 0.3% -15.1%

8 2.333 70 32.77 1.40 32.61 -1.83 -0.5% 176.1%9 2.667 80 30.99 -9.84 30.77 -15.14 -0.7% 35.0%

10 3.000 90 29.90 -22.16 30.00 -30.00 0.3% 26.2%




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-40

-30

-20

-10

0

10

20

30

40

50

0 10 20 30 40 50 60 70 80 90stre

ss [K

N/m

]

θ [deg]

Comparison

S1 _ SAP S2 _ SAP S1 _analitica S2 _analitica




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1) The stresses of the element depend only on the Z variable.2) The S1 stress is always positive The meridians are always in traction3) The stress S2 is positive near the bottom of the element and negative near the top of it The parallels are in compression near the top of the element and in traction near the bottom of it.



THIN SHELLS

• Conclusion


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Calculate the membrane forces of a toroidal tank full of water

S1e

S1i

DATA•Re = 4.5 m•Ri = 3.5 m•Rt = 0.5 m•S = 0.05 m•γW = 10 kN/m3




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1) By using the Guldino’s formulations we calculate the volume of the parts E and I.

2) By writing the equation equilibrium we determinate the stresses S1e and S1i.

3) By using the formula we calculate S2e and S2i




• Analytical solution

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In order to fulfil the requirements of themembrane theory, we apply a uniformdistributed load that varies linearly with the Z.

In addition we define traslational restraintsfor the edge of the structure in order to haveboundary condition compatible with therequirements of a pure membrane state.





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The stress S2 varies between a maximum positive value that is reached in the internalcircumference and a minimum negative value in the external circumference .That means that the external circumference is in traction while the internal is in compression.

SAP 2000 BELLUZZI

S1est S1int S2est S2int S1est S1int S2est S2int

KN/m KN/m KN/m KN/m KN/m KN/m KN/m KN/m

1.73 2.15 -10.72 13.28 1.84 2.12 -16.54 14.87

Error

S1est S1int S2est S2int

% % % %

6% 0% 35% 11%




• Conclusion

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These type of elements have a thickness that we can’t ignore during the analytical analysis

By considering the thickness of the shell, we must consider bending and shear stresses inaddition to the stresses that we regarded for the thin shells

The deformations generated from the bending and the shear stresses are caused by two factors:

1) External forces or bending moment applied at the edge of the shell.In this case the stresses quickly dampen and we can notice the presence of the bending and shear

stresses just as a local effect around the edge

2) External forces distributed on the surface of the element.In this case, the distributed forces deform the shells, so the bending and shear stresses aredistributed in all element.



THICK SHELLS

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Hypothesis- We consider a pressure that act on a cylindrical tube from the internal to the external of thevolume.- Pressure has to be uniformly distributed on the same parallel and can vary along themeridian.- External bending moment or radial forces can be present in one or both of the cylinder edges

Development1) We subdivide the elements in strips of unitary width2) We assume that the pressure is supported by the

longitudinal strips



THICK SHELLSDifferential equationDifferential equation

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Development3) The transversal strips contrast the deformation of longitudinal strips by reacting with radial forces (ρ) that act on the longitudinal strips.Aim:Determinate the relation between the radial force (ρ) and the deformation of the parallels

- We considerate a meridian segment

- We determinate the deformation of a generic parallel- We determinate the relation ρ = f(w)




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The differential equation of the elastic line

We know the relations between the component of displacement w and the other factors:

The stationary solution The general solution

-Represent the effect of the pressure P actingon the tube surface.-The solution is stationary because dependson to the distributed load acting on the entireelement.

-Represent the effect of external forces orbending moment acting on the edges of theelement.-The solution dissipates by going away fromthe loaded edge .




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The general solution:

By considering a tube that is very long in comparison with its other dimensions and assumingthat the variable x starts where the transverse shear forces and bending moments are applied, theconstant C1 and C2 can be considered equal to zero.

The dissipation of the displacements is regulated by the term e-αx that rapidly decrease with the increasing of the variable x.

The bending moment (M) and the transverseshear (H) assume relevant value just in thepart of the element near the loaded edge andcan be ignored in the rest of the volume.




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The dissipation of these local effects is as fast as the wavelength is small.The wavelength depends on the radius and the thickness of the element by the expression:

The dissipation of these local effects is as fast as the radius and the thicknessof the element are small.




The general solution:

78/92

By considering a tube that is very long in comparison with its other dimensions and assumingthat the variable x starts where the transverse shear forces and bending moments are applied, theconstant C1 and C2 can be considered equal to zero.

The dissipation of the displacements is regulated by the term e-αx that rapidly decrease with the increasing of the variable x.

Calculate the displacement w, the bending moment M and the transverse shear H associated to one longitudinal strip

DATA:•R = 1 m•L = 5 m •h = 5 m•s = 0.01•Material : Steel•Constitutive law : linear elastic •E = 2*10^8 kPa•γt = 10 kN/m3



THICK SHELLSNumerical exampleNumerical example

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The particular solution The general solution

We determinate the constant by using the boundary conditions:

We determinate the parametersα 13.13 1/mv 0.3B 18.31 KN/m2

β 2.0E+06 kN/m3

a 12.85C3 -2.46E-05 mC4 -2.50E-05 m





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-2.00

-1.00

0.00

1.00

2.00

3.00

4.00

0 1 2 3 4 5

T [KN]

X [m]

Trasverse shear - T

00.5

11.5

22.5

33.5

44.5

5

0 0.00001 0.00002

X [m]

W [m]

Displacement - W

-0.05

0.00

0.05

0.10

0.15

0.20

0 1 2 3 4 5

M [KNm]

X [m]

Bending moment- M

The bending moment (M) and thetransverse shear (H) can be consideredas local effects

T = B w’’’

M = B w’’





81/92

We run the analysis and then we export the displacements(w), the transverse shear (H) and the bending moment (M)of a single meridian in order to compare the resultsprovided by the numerical and the analytical analysis





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SHELLS

Buckling analysis



86

In the case of shells, the most interesting behaviour is that out of plane.• LOADS: perpendicular to the middle plane, along the revolution axis.• MATERIAL: Linear Elastic• No imperfection

Buckled Shell



CYLINDRICAL SHELLS WITH CIRCULAR SECTION

Undeformed Shell

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87

Numerical code : Straus 7Mesh : 10x10 finite elements ISOP4Boundary conditions: two edges simply supported or built in

L 1 m

Width 1 m

E 2E+08 KPa

ν 0.30




NumericalNumerical exampleexample

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88

• The shells with built in - ends presents higher buckling load• The shells with larger thickness presents higher buckling load• The curves presents a maximum for 0.3 < h/L < 0.4; then Pcr decreaeses

trendline : sixth-order polynomiale




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SHELLS

Non – linear analysis



90

We chose one of the analyzed shells and we runnonlinear analys.

Displacement control:Step: 150 increments of 0.006 mForce control:Step: 150 increments of 50 KN

Numerical code : Straus 7Mesh : 10x10 finite elements ISOP4Boundary conditions: two edges simply supported

L 1 m

Width 1 m

h 0.4 m

s 0.02 m





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91

Snap – through buckling

Looking at the red line, we see that when the curve reaches the maximum the plate buckleslaterally, and then flips over : the structure goes from a “arch behavior” to a “cable behavior”.When the structure is in the third position, the load can increase, but the resistence mechanism iscompletely changed.

1 2 3

This is a typical non-eulerianbuckling: the behaviour of thiskind of structures is nonlineareven before the buckling limit.




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92

In this case the axial stress, due to the action of the thrust H, becomes important: the buckledform is influenced by the presence of high axial stresses and is not extensionless.

Numerical code : Straus 7Mesh : 10x10 finite elements ISOP4

L 1 m

Width 1 m

h 0.05 m

Ends hinged

0

500

1000

1500

2000

2500

3000

3500

4000

4500

5000

0 0.02 0.04 0.06 0.08 0.1

P [K

N]

Dy [m]

Force control - Nonlinear analysis

s = 1 cm

s = 2 cm

s = 3 cm



VERY FLAT SHELLS


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93

DIFFERENCES %

Buckling –force control NL analysis

s = 1 cm s = 2 cm s = 3 cm

6.4 % 12.9 % 26.5 %

We find always the same way of buckling (snap – through) but, by increasing the thickness, we goes away from ideal buckling curve (green line).



VERY FLAT SHELLS92/92

94

• Timoshenko, Gere, “Theory of elastic stability”• Belluzzi, “Scienza delle costruzioni – Volume terzo”• Bontempi, Sgarbi, Malerba, “Sulla robustezza strutturale dei ponti a arco molto ribassato”



BIBLIOGRAFY

Date post:	08-Jul-2015
Category:	Engineering
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CM 2014_GG_Plates and shells

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