CM 2014_GG_Plates and shells

UNIVERSITA’ DEGLI STUDI DI ROMA “SAPIENZA”
Corso di
COSTRUZIONI METALLICHE
PLATE AND SHELLS:
BUCKLING AND NON – LINEAR
ANALYSIS
Prof. Ing. Franco Bontempi
Ing. Giordana Gai
Roma, 5 Dicembre 2014
gai.giordana@gmail.com
giordana.gai@uniroma1.it
Corso di Costruzioni Metalliche
Prof. Ing. Franco Bontempi, Ing. Giordana Gai

OUTLINE
Corso di Costruzioni Metalliche 2
PLATES
• 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)
1/92

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

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
Hipotesis
GEOMETRY
 No imperfections
 Thin plate: one dimension is very small with respect on the other two
MATERIAL (Steel)
 Linear Elastic
LOADS
 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 conditions
for each condition of loads

1.COMPRESSIVE FORCES
Plate simply supported along four edges
m = number of half – waves parallel to the
direction of N
n = number of half – waves perpendicular to the
direction of N
Deflection surface
Critical load
The minimum value of the critical load is
obtained by putting n = 1
The plate buckles in a shape that can have several half-waves in the direction of
compression but only one half wave in the perpendicular direction.

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 expression
where k is a numerical factor depending on a/b
6/92

• Short plates buckles in only one half – wave
• Long plates can have very large “m”
(comparable to the ratio a/b)
• The curve m = 1 has a minimum for square plate; before and after the ratio a/b = 1, the value k
increases.
• 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).
• Critical value of the compressive stress
7/92

Plate simply supported along two edges perpendicular to
the direction of N
Deflection surface
Solution depends on boundary conditions
along the edges y = 0 and y = b
• y = 0 simply supported; y = b free
• y = 0  zero displacement and zero bending
moment
• y = b  zero shear and zero bending moment
k Ncr
8/92

the direction of N
• y = 0 built in; y = b free
• y = 0  zero displacement and zero rotation
• y = b  zero shear and zero bending moment
k Ncr
The curve m = 1 has a minimum for
a/b = 1.635
9/92

the direction of N
• y = 0 built in; y = b built in
• y = 0  zero displacement and zero rotation
• y = b  zero displacement and zero rotation
k Ncr
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
10/92

Plate simply supported along two edges parallel to the
direction of N
• y = 0 and y = b  zero bending moment and shear equal to the
action force (N w/ y)
There are two possible forms of buckling
•The first buckling form is antisymmetrical up to a value of
b/a=1.316
• then, for 1.316 < b/a < 2.632 the buckling form is symmetrical
and then the value of k remains close to 2.31.
11/92

Plate simply supported along two edges parallel to the
direction of N
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.
13
Corso di Costruzioni Metalliche Prof. Ing. Franco Bontempi, Ing. Giordana Gai
12/92

2.BENDING MOMENT FORCES
•  = 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 pure
compression a/b = 1).
13/92

3.SHEARING FORCES
In this case, buckling is caused for the compressive stresses
acting along one of the two principal diagonals.
The critical value of shear stresses can be calculated by using
k, that depends on the ratio a/b.
14/92

4.SHEARING AND COMPRESSIVE FORCES
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.
15/92

5.SHEARING AND BENDING MOMENT FORCES
α is the ratio a/b
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 of
bending stress is small.
16/92

NUMERICAL EXAMPLE
a = b = L 50 cm
s 1 cm
E 2E+08 KPa
ν 0.30
Comparison between analytical and numerical code solutions
 Analytical solution: by Timoshenko’s theory
 Numerical codes: SAP 2000 and Straus 7
LOAD CASES ANALYZED:
1. Compressive forces
2. Bending moment forces
3. Shearing forces
4. Shearing and compressive forces
5. Shearing and bending moment forces
17/92

ANALYTICAL SOLUTION
Ncr = k Nl
Nl = π2Et3/12(1-ν2)b2 = 723KN/m
k  from the schedule
NUMERICAL SOLUTION
Buckling analysis
Mesh used = 30x30 finite elements
ISOP4
SAP 2000
Straus 7
18/92

CASE k Buckled
Plate
1 1
2 4
3 6.5
4 5
5 7.7
6 5.7
CASE k Buckled
Plate
7 1.44
8 1.7
9 10.2
10 2.05
11 2.3
12 2.3

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%
CASE 1 CASE 2 CASE 5
xy_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

ANALYTICAL SOLUTION
Ncr = k Nl
Nl = π2Et3/12(1-ν2)b2 = 723KN/m
k  from the schedules
NUMERICAL SOLUTION
Buckling analysis
ISOP4
SAP 2000
Straus 7
21/92

23
CASE k Pcr [KN/m] Buckled Plate
Differences Straus7-
Analytical Solution
[%]
Differences
SAP2000-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
COMPRESSI
ON
22/92

3.SHEARING FORCES
ANALYTICAL SOLUTION
τcr = k Nl
Nl = π2Et3/12(1-ν2)b2 = 723KN/m
k  from the schedule
NUMERICAL SOLUTION
Buckling analysis
ISOP4
SAP 2000
Straus 7
23/92

25
3.SHEARING FORCES
CASE k Pcr
[KN/m] Buckled Plate w[m]_
XY view
Differences
Straus7-
Analytical
Solution [%]
Differences
SAP2000-
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%
24/92

ANALYTICAL SOLUTION
• by using the approximate
expression, where N0cr and T0cr are
those calculated before (respectivly
compressive and shear forces only)
• by using the schedule
NUMERICAL SOLUTION
Buckling analysis
ISOP4
SAP 2000
Straus 7
25/92

27
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%
26/92

ANALYTICAL SOLUTION
• by using the approximate
expression, where N0cr and T0cr are
those calculated before (respectivly
compressive and shear forces only)
• by using the schedule
NUMERICAL SOLUTION
Buckling analysis
ISOP4
SAP 2000
Straus 7
27/92

29
Numerical Code Solution Analytical Expression Differences %
Tcr = Ncr [KN/m] Ncr = Ncr0[1-(Tcr/Tcr0)2]0.5 [KN/m] Numerical Code – Analytical
Solution
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
28/92

PLATES
Non – linear analysis

ELASTIC MATERIAL
We run displacement control - nonlinear analysis to investigate post – critical
behavior.
31
Numerical example
Numerical code : Straus 7
Mesh : 10x10 finite elements ISOP4
Step : 200 increments of 0.001 m
Boundary conditions: 2 edges simply supported
a = b = L 50 cm
s 1 cm
E 2E+08 KPa
ν 0.30
30/92

32
ELASTIC MATERIAL
 from Buckling analysis
 from NL analysis (plate straight) with large
displacement
 from NL analysis (plate not perfectly straight)
with large displacement
(only geometric nonlinearity)
POST CRITICAL BEHAVIOR = stable, with hardening
31/92

ELASTIC – PLASTIC MATERIAL
(only material
nonlinearity)
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 (ideal plate) NLM (real plate)
= 235*103*0.5*0.01=1175 KN
32/92

34
500
400
300
200
100
0
0 0.05 0.1 0.15 0.2
P [KN]
Dx [m]
NLM_not perfectly straight plate
NLG_not perfectly straight plate
450
400
350
300
250
200
150
100
50
0
0 0.05 0.1 0.15 0.2
P [KN]
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 (real plate)
The plate starts to reach the yelding
limit in the lateral zone, but rapidly
it is totally yelded (orange line)
33/92

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 plate
with s=1cm has an intermediate
behavior.
We compared the three “orange curve” by
adimensionalyzing respect on their
yielding maximum force.
35
34/92

36
65% of the yielding limit
Coupled behaviour of yielding and buckling
35/92

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.
It is unusual to find a sharp corner that separates the
elastic behavior from the plastic one.
37
Some permanent set usually takes place at a stress
lower than the yield point
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
36/92

• 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.
38
• 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
37/92

39
Approximate solution
1) 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
38/92

2) Determination of the components of the strain in the middle plane and the corresponding
strain energy of the plate
40
3) Determination of the energy of bending
39/92

4) By using the solution method of minimum of strain energy we determinate the constant f, C1, C2
41
By solving
the system
we obtain
We obtain a solution for “f” only if :
40/92

42
5) Determination of the stresses σx and σy
By plotting the diagram of the stresses for the ultimate
compressive force (Nr,d) we can notice:
• The ultimate compressive force is n times bigger than
the critical compressive force
• The compressive stress diagram is non-uniform
41/92

Why can the ultimate compressive force in a plate be n times bigger than the critical one?
For a compressive force greater than the critical one, the vertical fibres instead of collapse ,as
they would do if we analyzed a single strut, can absorb a further load because their deflection
out of the plane is hindered by the flexural rigidity of the horizontal fibres.
For the restraints of the plate that we are considering, we know that the part of the horizontal
fibres near of the edge is more rigid because is close to the restraints( that are considering having
an infinite rigidity). So the vertical fibres near to the vertical edge are able to absorb a greater
post-critical load because they can count on a greater flexural rigidity of the horizontal fibres.
43
Why is the compressive stress diagram non-uniform ?
42/92

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.
43/92

Enhanced solution
2) We determinate the ultimate load that the plate can carry by using the Tresca yield criterion.
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
44/92

46
Numerical example
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
45/92

3
2.5
2
1.5
1
0.5
0
σx /σcr
-1 -0.5 0 0.5 1
Analytical approximate
47
solution
asse y / a
Cross section x = a
0.2
0
-1 -0.5 0 0.5 1
-0.2
-0.4
-0.6
-0.8
-1
-1.2
-1.4
-1.6
σy /σcr
asse x /a
Cross section y = a
46/92

48
1) We run the non linear static analysis by considering the non linear material
Straus 7 approximate solution
σy 235 MPa
εy 1.175*10-3
εu 7.5*10-2
Analytical solution Numerical solution
σy 235 Mpa
Step 4.7
Fcr 558 kN
Fy 2350 kN Fy 2622.6 kN
errore 10%
47/92

2) We run the non linear static analysis by considering the non linear material and the non liner
geometry
49
Geometric non - linearity Material non - linearity
Numerical solution Analytical solution
Fu 1450.8 kN Fu 554.7 kN
48/92

49/92

-1 -0.5 0 0.5 1
51
Cross section x = a
4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
σx /σcr
asse y/a
-1 -0.5 0 0.5 1
Straus7 Analytical solution
1.5
1
0.5
0
-0.5
-1
-1.5
-2
σy /σcr
asse x/a
Cross section y = a
Analytical solution Straus7
Analytical enhanced solution
Straus7 approximate solution
50/92

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 = 0
ux = 0
51/92

-1 -0.5 0 0.5 1
53
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
σy /σcr
asse x/a
Sezione y = a
Analytical solution Strauss7
Sezione x = a
3.5
3
2.5
2
1.5
1
0.5
0
σx /σcr
asse y/a
-1 -0.5 0 0.5 1
Strauss7 Analytical solution
Analytical enhanced solution
Straus7 approximate solution
52/92

0.6
0.4
0.2
0
-1 -0.5 0 0.5 1
-0.2
-0.4
-0.6
-0.8
-1
-1.2
-1.4
-1.6
σy/σcr
asse x/a
C.S. y = a analytical solutions
P.A.S.10 A.S.10
C.S. x = a analytical solutions
3
2.5
2
1.5
1
0.5
0
σx /σcr
asse y/a
-1 -0.5 0 0.5 1
P.A.S.10 A.S.10
54
Comparison
We 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
53/92

Comparison
We compare the results provided by the approximate and the enhanced analysis
C.S. x = a Numerical solutions
4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
σx /σcr
asse y/a
-1 -0.5 0 0.5 1
P.S7.S.10 S7.S.10
C.S. y = a Numerical solutions
1.5
1
0.5
0
-1 -0.5 0 0.5 1
-0.5
-1
-1.5
σy /σcr
asse x/a
P.S7.S.10 S7.S.10
54/92
Legend:
P.A.S. Enhanced analytical solution
A.S.Analytical solution
P.S7.S. Enhanced Straus7 solution
S7.S. Straus7 solution

SHELLS

THIN SHELLS
The membrane theory
•We consider a shell where the thickness is small in comparison to its other dimensions
•We assume an axial-simmetric load.
The components of the stresses
the component of
the stresses
tangent to the
meridian
σ1
is Caused
by the resultant of
the external
vertical forces
the component of
stresses tangent to
the parallel
σ2
is
by the force Z
Caused
by the changing
of direction of
the stresses σ1
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
56/92

THIN SHELLS
The membrane theory
Hypothesis:
very small ratio thickness / minimal
radius
Thesis :
No flexural and torsional rigidity for
the membrane
Dimostration
1) We subdivided the shell in elementary strips of unitary width that follow the geometry of the
meridians and parallels
2) 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 curvature
4) 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
57/92

THIN SHELLS
The membrane theory
very small ratio thickness / minimal
No flexural and torsional rigidity for
Conclusion
Hypothesis:
radius
Thesis :
the membrane
For 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
58/92

THIN SHELLS
The membrane theory
Compatibiliy conditions
1) 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.
59/92

THIN SHELLS
The membrane theory
The membrane forces
We 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
60/92

THIN SHELLS
The membrane theory
The membrane forces
S2  integral of the stresses tangent to the parallels
61/92
We determinate the membrane forces due to the axial-simmetrical load

THIN SHELLS
Numerical example 1
Calculate the membrane forces of a hemisphere tank full of water
DATA
•R = 3 m
•s = 0.1 m
•γW = 10 kN/m3
62/92

THIN SHELLS
Numerical example 1
• Analytical solution
1) We calculate the volume of the hemisphere
2) We calculate the membrane forces
63/92

THIN SHELLS
In order to fulfil the requirements of
the membrane theory, we apply an
uniform distributed load that varies
linearly with the Z.
We define traslational restraints for the
edge of the structure in order to have a
boundary condition compatible with the
requirements of a pure membrane state.
• Sap2000 solution
Numerical example 1
64/92

S1
S2
THIN SHELLS
Numerical example 1
65/92

THIN SHELLS
Numerical example 1
SAP 2000 BELLUZZI
Step
z θ S1 S2 S1 S2
Error
S1
Error
S2
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%
66/92

50
40
30
20
10
0
-10
-20
-30
-40
THIN SHELLS
Numerical example 1
0 10 20 30 40 50 60 70 80 90
stress [KN/m]
θ [deg]
Comparison
S1 _ SAP S2 _ SAP S1 _analitica S2 _analitica
67/92

THIN SHELLS
1) The stresses of the element depend only on the Z variable.
2) The S1 stress is always positive  The meridians are always in traction
3) 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.
• Conclusion
Numerical example 1
68/92

THIN SHELLS
Numerical example 2
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
69/92

THIN SHELLS
Numerical example 2
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
70/92

THIN SHELLS
Numerical example 2
In order to fulfil the requirements of the
membrane theory, we apply a uniform
distributed load that varies linearly with the Z.
In addition we define traslational restraints
for the edge of the structure in order to have
boundary condition compatible with the
requirements of a pure membrane state.
71/92

THIN SHELLS
Numerical example 2
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%
The stress S2 varies between a maximum positive value that is reached in the internal
circumference and a minimum negative value in the external circumference .
That means that the external circumference is in traction while the internal is in compression.
• Conclusion
72/92

THICK SHELLS
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 in
addition 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 are
distributed in all element.
73/92

THICK SHELLS
Differential equation
Hypothesis
- We consider a pressure that act on a cylindrical tube from the internal to the external of the
volume.
- Pressure has to be uniformly distributed on the same parallel and can vary along the
meridian.
- External bending moment or radial forces can be present in one or both of the cylinder edges
Development
1) We subdivide the elements in strips of unitary width
2) We assume that the pressure is supported by the
longitudinal strips
74/92

THICK SHELLS
Development
3) 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)
75/92

THICK SHELLS
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 acting
on the tube surface.
-The solution is stationary because depends
on to the distributed load acting on the entire
element.
-Represent the effect of external forces or
bending moment acting on the edges of the
element.
-The solution dissipates by going away from
the loaded edge .
76/92

The general solution:
THICK SHELLS
By considering a tube that is very long in comparison with its other dimensions and assuming
that the variable x starts where the transverse shear forces and bending moments are applied, the
constant 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 transverse
shear (H) assume relevant value just in the
part of the element near the loaded edge and
can be ignored in the rest of the volume.
77/92

THICK SHELLS
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 thickness
of the element are small.
The general solution:
78/92
By considering a tube that is very long in comparison with its other dimensions and assuming
that the variable x starts where the transverse shear forces and bending moments are applied, the
constant 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.

THICK SHELLS
Numerical example
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
79/92

THICK SHELLS
Numerical example
The particular solution The general solution
We determinate the constant by using the boundary conditions:
We determinate the parameters
α 13.13 1/m
v 0.3
B 18.31 KN/m2
β 2.0E+06 kN/m3
a 12.85
C3 -2.46E-05 m
C4 -2.50E-05 m
80/92

THICK SHELLS
Numerical example
4.00
3.00
2.00
1.00
0.00
-1.00
-2.00
T [KN]
T = B w’’’
X [m]
Trasverse shear - T
0 1 2 3 4 5
5
4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
X [m]
Displacement - W
W [m]
0 0.00001 0.00002
0.20
0.15
0.10
0.05
0.00
-0.05
M [KNm]
0 1 2 3 4 5
X [m]
Bending moment- M
The bending moment (M) and the
transverse shear (H) can be considered
as local effects
M = B w’’
81/92

THICK SHELLS
Numerical example
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 results
provided by the numerical and the analytical analysis
82/92

SHELLS
Buckling analysis

CYLINDRICAL SHELLS WITH CIRCULAR SECTION
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
Corso di Costruzioni Metalliche Undeformed Shell
85/92

87
Numerical example
Boundary conditions: two edges simply supported or built in
L 1 m
Width 1 m
E 2E+08 KPa
ν 0.30
86/92

88
trendline : sixth-order polynomiale
• 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
87/92

SHELLS
Non – linear analysis

90
We chose one of the analyzed shells and we run
nonlinear analys.
Displacement control:
Step: 150 increments of 0.006 m
Force control:
Step: 150 increments of 50 KN
Numerical example
Boundary conditions: two edges simply supported
L 1 m
Width 1 m
h 0.4 m
s 0.02 m
89/92

Snap – through buckling
Looking at the red line, we see that when the curve reaches the maximum the plate buckles
laterally, 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 is
completely changed.
91
1 2 3
This is a typical non-eulerian
buckling: the behaviour of this
kind of structures is nonlinear
even before the buckling limit.
90/92

VERY FLAT SHELLS
In this case the axial stress, due to the action of the thrust H, becomes important: the buckled
form is influenced by the presence of high axial stresses and is not extensionless.
92
Numerical example
L 1 m
Width 1 m
h 0.05 m
Ends hinged
5000
4500
4000
3500
3000
2500
2000
1500
1000
500
0
Force control - Nonlinear analysis
0 0.02 0.04 0.06 0.08 0.1
P [KN]
Dy [m]
s = 1 cm
s = 2 cm
s = 3 cm
91/92

93
DIFFERENCES %
VERY FLAT SHELLS
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).
92/92

BIBLIOGRAFY
• Timoshenko, Gere, “Theory of elastic stability”
• Belluzzi, “Scienza delle costruzioni – Volume terzo”
• Bontempi, Sgarbi, Malerba, “Sulla robustezza strutturale dei ponti a arco molto ribassato”
94

CM 2014_GG_Plates and shells

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