CEN/TR 13121-5:2017

SLOVENSKI STANDARD
01-maj-2018
1DG]HPQLUH]HUYRDUMLLQSRVRGHL]XPHWQLKPDVRMDþDQLKVVWHNOHQLPLYODNQL
GHO3ULPHUL]UDþXQD
GRP tanks and vessels for use above ground - Part 5: Example calculation of a GRP-
vessel
Ta slovenski standard je istoveten z: CEN/TR 13121-5:2017
ICS:
23.020.10 1HSUHPLþQHSRVRGHLQ Stationary containers and
UH]HUYRDUML tanks
2003-01.Slovenski inštitut za standardizacijo. Razmnoževanje celote ali delov tega standarda ni dovoljeno.

CEN/TR 13121-5
TECHNICAL REPORT
RAPPORT TECHNIQUE
May 2017
TECHNISCHER BERICHT
ICS 23.020.10
English Version
GRP tanks and vessels for use above ground - Part 5:
Example calculation of a GRP-vessel

This Technical Report was approved by CEN on 18 April 2017. It has been drawn up by the Technical Committee CEN/TC 210.

CEN members are the national standards bodies of Austria, Belgium, Bulgaria, Croatia, Cyprus, Czech Republic, Denmark, Estonia,
Finland, Former Yugoslav Republic of Macedonia, France, Germany, Greece, Hungary, Iceland, Ireland, Italy, Latvia, Lithuania,
Luxembourg, Malta, Netherlands, Norway, Poland, Portugal, Romania, Serbia, Slovakia, Slovenia, Spain, Sweden, Switzerland,
Turkey and United Kingdom.
EUROPEAN COMMITTEE FOR STANDARDIZATION
COMITÉ EUROPÉEN DE NORMALISATION

EUROPÄISCHES KOMITEE FÜR NORMUNG

CEN-CENELEC Management Centre: Avenue Marnix 17, B-1000 Brussels
© 2017 CEN All rights of exploitation in any form and by any means reserved Ref. No. CEN/TR 13121-5:2017 E
worldwide for CEN national Members.

Contents Page
European foreword . 5
Introduction . 6
1 Scope . 7
2 General . 7
3 Dimensions of the tank . 7
4 Building materials . 9
5 Loadings (9) . 9
6 Limit strain for laminate (8.2.2) . 11
7 Influence factors (7.9.5.2) . 11
8 Partial safety factors (Table 12) . 12
9 Combination factors (Table 11) . 12
10 Analysis of the cylinder . 12
10.1 Influence factor A . 12
10.2 Characteristic strength values . 13
10.3 Moduli of elasticity . 13
10.4 Analysis of the cylinder in axial direction . 13
10.4.1 Proof of strength (Ultimate limit state) . 14
10.4.2 Proof of strain (Serviceability limit state) . 17
10.4.3 Stability proof (Ultimate limit state) . 19
10.5 Analysis of the cylinder in tangential direction . 21
10.5.1 Strength analysis (Ultimate limit state) . 21
10.5.2 Proof of strain (Serviceability limit state) . 23
10.5.3 Stability proof for the cylindrical shell tangential (Ultimate limit state) . 23
10.5.4 Critical buckling pressure for rings (Ultimate limit state) . 24
10.6 Earthquake design of the cylinder . 26
10.6.1 Analysis of the cylinder in axial direction . 26
10.6.2 Analysis of the cylinder in tangential direction . 29
11 Opening in the cylinder . 30
11.1 Analysis in circumferential direction . 31
11.1.1 Proof of strength . 31
11.1.2 Proof of strain . 31
11.2 Analysis in axial direction . 32
11.2.1 Proof of strength . 32
11.2.2 Proof of strain . 32
12 Analysis of the skirt . 33
12.1 Internal forces of the skirt . 33
12.2 Proof of strength (Ultimate limit state) . 34
12.2.1 Design value of actions . 34
12.2.2 Design value of corresponding resistance . 34
12.2.3 Verification . 35
12.3 Proof of strain (Serviceability limit state) . 35
12.3.1 Design value of actions . 35
12.3.2 Limit design value of serviceability criterion. 35
12.3.3 Verification . 35
12.4 Stability proof (Ultimate limit state) . 35
12.4.1 Design value of actions . 35
12.4.2 Design value of corresponding resistance . 36
12.4.3 Verification . 36
12.5 Earthquake design of the skirt . 36
12.5.1 Internal forces Earthquake . 36
12.5.2 Proof of strength (Ultimate limit state) . 37
12.5.3 Proof of strain (Serviceability limit state) . 37
12.5.4 Stability proof (Ultimate limit state) . 38
13 Overlay laminate connection skirt - vessel . 39
13.1 Proof of strength (Ultimate limit state) . 39
13.1.1 Design value of actions . 39
13.1.2 Design value of corresponding resistance . 40
13.1.3 Verification . 40
13.2 Proof of strain (Serviceability limit state) . 40
13.2.1 Design value of actions . 40
13.2.2 Limit design value of serviceability criterion. 40
13.2.3 Verification . 40
13.3 Seismic design of the skirt overlay . 41
13.3.1 Proof of strength (Ultimate limit state) . 41
13.3.2 Proof of strain (Serviceability limit state) . 41
14 Analysis of the bottom . 42
14.1 Influence factor A . 42
14.2 Characteristic strength values . 42
14.3 Moduli of elasticity . 42
14.4 Actions, which cause internal forces for the bottom . 42
14.5 Strength analysis (Ultimate limit state) . 42
14.5.1 Design value of actions . 42
14.5.2 Proof of strain (Serviceability limit state) . 44
14.5.3 Stability proof of the bottom (Ultimate limit state) . 45
15 Lower part of the cylinder (Region 1) . 46
15.1 Strength analysis (Ultimate limit state) . 46
15.1.1 Design value of corresponding resistance . 47
15.1.2 Verification . 47
15.2 Proof of strain (Serviceability limit state) . 47
15.2.1 Design value of actions . 47
15.2.2 Limit design value of serviceability criterion. 47
15.2.3 Verification . 47
15.3 Earthquake design of region 1 (Ultimate limit state) . 48
15.3.1 Strength analysis (Ultimate limit state) . 48
15.3.2 Proof of strain (Serviceability limit state) . 48
16 Upper part of the skirt (Region 2) . 49
16.1 Strength analysis (Ultimate limit state) . 49
16.1.1 Design value of corresponding resistance . 50
16.1.2 Verification . 50
16.2 Proof of strain (Serviceability limit state) . 50
16.2.1 Design value of actions . 50
16.2.2 Limit design value of serviceability criterion. 50
16.2.3 Verification . 50
16.3 Seismic design of region 2 (Ultimate limit state) . 51
16.3.1 Strength analysis (Ultimate limit state) . 51
16.3.2 Design value of corresponding resistance . 51
16.3.3 Verification . 51
16.4 Proof of strain (Serviceability limit state) . 51
16.4.1 Design value of actions . 51
16.4.2 Limit design value of serviceability criterion . 51
16.4.3 Verification . 52
17 Flange design . 52
18 Anchorage . 57
18.1 Anchorage for wind loads (Permanent / Transient situation) . 57
18.1.1 Uplifting anchor force . 57
18.1.2 Anchor shear force. 57
18.2 Anchorage for seismic loads (Seismic design situation) . 57
18.2.1 Uplifting anchor force . 57
18.2.2 Anchor shear force. 58

European foreword
This document (CEN/TR 13121-5:2017) has been prepared by Technical Committee CEN/TC 210 “GRP
tanks and vessels”, the secretariat of which is held by SFS.
Attention is drawn to the possibility that some of the elements of this document may be the subject of
patent rights. CEN [and/or CENELEC] shall not be held responsible for identifying any or all such patent
rights.
Introduction
EN 13121 consists of the following parts:
— EN 13121-1, GRP tanks and vessels for use above ground — Part 1: Raw materials — Specification
and acceptance conditions
— EN 13121-2, GRP tanks and vessels for use above ground — Part 2: Composite materials — Chemical
resistance
— EN 13121-3, GRP tanks and vessels for use above ground — Part 3: Design and workmanship
— EN 13121-4, GRP tanks and vessels for use above ground — Part 4: Delivery, installation and
maintenance
— CEN/TR 13121-5, GRP tanks and vessels for use above ground — Part 5: Example calculation of a
GRP-tank (this report)
These five parts together define the responsibilities of the tank or vessel manufacturer and the
materials to be used in their manufacture.
EN 13121-1 specifies the requirements and acceptance conditions for the raw materials - resins, curing
agents, thermoplastics linings, reinforcing materials and additives. These requirements are necessary in
order to establish the chemical resistance properties determined in EN 13121-2 and the mechanical,
thermal and design properties determined in EN 13121-3. Together with the workmanship principles
determined in Part 3, requirements and acceptance conditions for raw materials ensure that the tank or
vessel will be able to meet its design requirements. EN 13121-4 of this standard specifies
recommendations for delivery, handling, installation and maintenance of GRP tanks and vessels. This
part of EN 13121 gives guidance in use of the standard. CEN/TC 210 has found it necessary to publish
an example calculation of a vessel according to EN 13121-3 due to the standards complexity, and for the
understanding of how the standard complies with EN 1990:s principles and requirements for safety,
serviceability and durability of structures.
The design and manufacture of GRP tanks and vessels involve a number of different materials such as
resins, thermoplastics and reinforcing fibres and a number of different manufacturing methods. It is
implicit that vessels and tanks covered by this standard are made only by manufacturers who are
competent and suitably equipped to comply with all the requirements of this standard, using materials
manufactured by competent and experienced material manufacturers.
Metallic vessels, and those manufactured from other isotropic, homogeneous materials, are
conveniently designed by calculating permissible loads based on measured tensile and ductility
properties. GRP, on the other hand, is a laminar material, manufactured through the successive
application of individual layers of reinforcement. As a result there are many possible combinations of
reinforcement type that will meet the structural requirement of any one-design case. This allows the
designer to select the laminate construction best suited to the available manufacturing facilities and
hence be most cost effective.
1 Scope
This Technical Report gives guidance for the design of a vessel using the standard EN 13121-3 GRP
tanks and vessels for use above ground. The calculation is done according to the advanced design
method given in EN 13121-3:2016, 7.9.3 with approved laminates and laminate properties.
2 General
Vessels or vessel structures may contain such structural elements or solutions for which this standard
does not provide sufficient guidance. In that case, other methods shall be used in order to obtain a safe
structure.
This example calculation is based on a pressurized GRP vessel with an internal diameter of D 3000 mm.
The cylindrical parts of the vessel are filament wound. Its bottom and roof are torispherical dished ends
that are hand laid up using mixed laminates. Protection against medium attack is obtained by a
chemical resistance layer (CRL).
The tank is located outdoors in a seismic area.
IMPORTANT – This example doesn’t cover all necessary verifications for the calculation of the GRP tank.
Additional verifications have to be performed for the roof, the upper cylinder, etc.
3 Dimensions of the tank
Sketch of the tank dimensions:
General Dimensions:
Diameter: D = 3 000 mm
Total height: H = 8 000 mm
tot
Cylinder:
Thickness cylinder 1: t = t = 9,2 mm
Cyl,1 C1
Thickness cylinder 2: t = t = 11,7 mm
Cyl,2 C2
Thickness cylinder at roof: t = 30,0 mm
Z,R
Thickness cylinder at bottom: t = 46,1 mm
Z,B
Total cylinder length: l = 6 610 mm
Cyl.tot
Distance between stiffeners: l = 3700 mm l = 3303 mm
s.1 s.2
Thickness of the stiffener: t = 20 mm
S
Width of the stiffener: b = 260 mm
S
Skirt:
Thickness skirt: t = 17,0 mm
Sk
Thickness overlay laminate: t = 7,0 mm
Height of the skirt: H = 890 mm
Sk
Roof:
Thickness calotte: t = 13,0 mm
R
Radius calotte: R = 3000 mm
R
Thickness knuckle roof: tRk = 30,0 mm
Radius knuckle: r = 300 mm
Rk
Height of the roof: H = 590 mm
R
Bottom:
Thickness calotte: t = 16,5 mm
B
Radius calotte: R = 3 000 mm
B
Thickness knuckle: t = 45,0 mm
Bk
Radius knuckle: r = 300 mm
Bk
Height of the bottom: H = 590 mm
B
4 Building materials
Resin type: UP-resin, Resin group 4
5 Loadings (9)
LC 1: Dead load
The assumed dead loads for the separate tank parts are:
Roof: W = 4 kN Area load: w = 0,57 kN/m
R,k R,k
Cylinder + rings: W = 19 kN
C,k
Bottom: W = 4 kN Area load: w = 0,57 kN/m
B,k B,k
Skirt: W = 3 kN
Sk
Total dead load of the vessel: W = 30 kN
tot
LC 2: Liquid filling
Density of the medium ρ = 1,30 kg/dm
liquid
Filling height h = 7 000 mm
liquid
Volume V = 52,0 m
LC 3: Long time design overpressure
Design pressure PS = 2,000 bar ≡ 0,20 N/mm
op.l
LC 4: Short time design overpressure
Design pressure PS = 2,500 bar ≡ 0,25 N/mm
op.s
LC 5: Long time design negative pressure
Design pressure PS = 0,000 bar ≡ 0,00 N/mm
ep.l
LC 6: Short time design negative pressure
Design pressure PS = 0,050 bar ≡ 0,005 N/mm
ep.s
LC 7: Wind (9.2.2)
Peak velocity pressure q = 0,8 kN/m (EN 1991–1-4)
p
Force coefficient (cylindrical vessel) c = 0,8
f
External pressure arising from wind load:
p =0,6⋅=q 0,6⋅0,8=0,48 kN/m²
wind p
LC 8: Snow (9.2.1)
Characteristic snow load s = 0,85 kN/m (EN 1991–1-3)
k
Shape coefficient μ = 0,80
Snow load
p = s⋅=µ 0,85⋅0,8=0,68 kN/m²
snow k
LC 9: Personnel loading (9.2.8)
Live load on the roof p = 1,5 kN/m
access
LC 10: Temperature
Design temperature TS = 50°C
Difference in temperature ΔT = 20 K
LC 11: Earthquake (9.2.3.4)
Reference peak ground a = 1,00 m/s
gR
acceleration
Importance factor γ = 1,4
Design ground acceleration
a a ⋅ γ 1,00⋅ 1, 4 1, 40 ms/²
g gR 1
Ground type according to D
EN 1998–1
Viscous damping 5 %
Control periods of the T = 0,20 s T = 0,8 s T = 2,0 s
B C D
response spectrum
Soil factor S = 1,35
Behaviour factor q = 1,5
Bending modulus cylinder E = 19 000 N/mm
ϕ,b
tangential
Bending modulus cylinder E = 12 000 N/mm
x,b
axial
Modulus of elasticity for
E=1,5⋅ E ⋅=E 1,5 ⋅ 19 000⋅ 12 000=22 650N / mm²
e φ,,b xb
short time impact
Cylinder thickness lower t approximately t = 17 mm
1/2 Sk
= = =
third
Vibration period


ρ ⋅⋅h hh2

liquid liquid liquid liquid

T ⋅⋅D 0,,628⋅ + + 1 49


Et⋅ D D
e 12 



 
1,33 ⋅⋅7200 7200 2 7200

T ⋅⋅3,,0 0 628⋅++ 1, 49 0,15 s
 
3 
3000 3000
 
22650 ⋅⋅17,0 10


Design spectrum T ≤ T
C 25,,25
S T=a⋅ S⋅=1, 40⋅ 1,35⋅=3,15 ms/²
( )
D g
(plateau area):
q 1,5
Total mass of the vessel
W
tot
W +⋅V ρ + 52⋅ 1,30 70,66 Tonnen
(approximately)
G liquid
g 9, 81
Horizontal load (Base shear)
H ≅ S T⋅=W 3,15⋅ 70,66= 222,6 kN
( )
AE D G
Overturning moment

 
h
WH
liquid
tot tot

 
M ≅⋅ V ρ ⋅ + H − H + ⋅ ⋅ ST
( )
AE,tot liquid Sk B D
 
22g
 

 

7 000 8 000
−−33
 

M ≅⋅ 52 1,30 ⋅ + 800 − 590 ⋅ 10 + ⋅ ⋅ 10 ⋅ 3,15 =828,5 kNm
AE,tot
  
2 9, 81 2

 
6 Limit strain for laminate (8.2.2)
For the used UP resin is:
The roof is made of a mixed laminate ε = ε = 0,25 %
lim,R d,R
The bottom is made of a mixed laminate ε = ε = 0,25 %
lim,B d,B
The cylinder is made of a wound laminate 0° ε = ε = 0,20 % ε = ε = 0,27 %
lim,x,Cyl d,x,Cyl lim,ϕ,Cyl d,ϕ,Cyl
/90°
The skirt is made of a wound laminate 0° /90° ε = ε = 0,20 % ε = ε = 0,27 %
lim,x,Sk d,x,Sk lim,ϕ,Sk d,ϕ,Sk
7 Influence factors (7.9.5.2)
Influence factor A A = 1,0
1 1
Influence factor A A = 1,4 (Table A.4 of EN 13121–2)
2 2
Medium category 2, T = 50°C
d
HDT of the used resin HDT = 90 °C
Influence factor A
3   
TS −°20 C 50 − 20
A 1,,00+ 0 4⋅ 1,,00+ 0 4⋅ 1,20
  
HDT −°30 C 90 − 30
  
Influence factor A A = 1,0
4 4
Influence factor A The influence factor A depends on laminate type and is selected separately
5 5
for each kind of laminate.
= = =
= = =
= =
=
8 Partial safety factors (Table 12)
Situation
Action Symbol
P/T A/AE
Independent permanent actions (s.a): γ 1,35 1,00
G,sup
unfavourable γ 1,00 1,00
G,inf
favourable γ 1,35 1,00
G,sup
For liquid filling γ 0 0
G,inf
unfavourable γ 1,50 1,00
Q,sup
favourable γ 0 0
Q,inf
Independent variable actions: γ 1,00
A
unfavourable γ 1,00
AE
favourable
Accidental actions:
Seismic actions:
9 Combination factors (Table 11)
In the following table are shown the relevant Ψ-factors for this example.
Action ψ ψ ψ
0 1 2
Pressures: 1,0 1,0 1,0
- Long term pressures 0 0 0
- Short-term pressures
Imposed loads in buildings, category (see EN 1991–1-1) 0 0 0
- Category H: roofs
a)
Snow loads on buildings (see EN 1991–1-3) : 0,5 0,2 0
Remainder of CEN Member States,
- for sites located at altitude H ≤ 1000 m a.s.l.
Wind loads on buildings (see EN 1991–1-4) 0,6 0,2 0
Temperature (non-fire) in buildings (see EN 1991–1-5) 0,6 0,5 0
10 Analysis of the cylinder
The cylinder is made of a wound laminate 0° / 90°. For mechanical properties are used historic test
data. They are verified with tests in accordance to 7.9.3.
10.1 Influence factor A5
— For stress analysis
25 years: a) Axial A = 1,60 b) Tangential A = 1,20
5B.Cyl.25y.x 5B.Cyl.25y.ϕ
3 months:   A = 1,40  A = 1,15
5B.Cyl.3m.x 5B.Cyl.3m.ϕ
Shorttime:   A = 1,00  A = 1,00
5B.Cyl.sh.x 5B.Cyl.sh.ϕ
— For stability analysis
25 years: a) Axial A = 1,60 b) Tangential A = 1,20
5I.Cyl.25y.x 5I.Cyl.25y.ϕ
3 months:   A = 1,40  A = 1,15
5I.Cyl.3m.x 5I.Cyl.3m.ϕ
Short time:   A = 1,00  A = 1,00
5I.Cyl.sh.x 5I.Cyl.sh.ϕ
Check for minimum design factors K and F:
If the value of K does not reach a minimum of 4 (advanced design) only for longtime loads, the A
5B
values should be increased.
K 4
Minimum A 1,13
5B
10,,⋅ 1 4 ⋅ 12, 0 ⋅ 10,,⋅⋅1 4 1,5
AAA⋅ ⋅ ⋅ A ⋅ γγ⋅ ( )
( )
1 23 4 M Fi,
If the value of F does not reach a minimum of 2,7 (advanced design) only for longtime loads, the A
5I
values should be increased.
 
F 27,
 
Minimum A 10, 10,,10

5I
 AAA⋅ ⋅ ⋅ A ⋅ γγ⋅  10,,⋅ 1 4 ⋅ 12, 0 ⋅ 10,,⋅⋅1 4 1,5

1 23 4 M Fi,
 
All A values are greater than the minimum A values.
5 5
10.2 Characteristic strength values
— For tension
2 2
a) Axial f = 130 N/mm b) Tangential f = 400 N/mm
Cyl.x.t.k Cyl.ϕ.t.k
— For bending
2 2
a) Axial f = 150 N/mm b) Tangential f = 480 N/mm
Cyl.x.b.k Cyl.ϕ.b.k
10.3 Moduli of elasticity
— For tension
2 2
a) Axial E = 12500 N/mm b) Tangential E = 21000 N/mm
Cyl.x.t Cyl.ϕ.t
— For bending
2 2
a) Axial E = 12000 N/mm b) Tangential E = 19000 N/mm
Cyl.x.b Cyl.ϕ.b
10.4 Analysis of the cylinder in axial direction
1. Step) Calculate all characteristic internal forces from the actions, which may cause internal forces in
axial direction
WW+
R,,k Cyl k 4 + 19
LC 1 Dead load: n ⋅=10 2, 44 N/ mm
x.W
D ⋅⋅ππ3 000
LC 2 Liquid filling: n = 0
x,hp
3 000
D
LC 3 Long time design over pressure: n PS ⋅= 0,20⋅ 150,00 N/ mm
x.PS.op l op.l
= =
= =
≥= ≥==
===
3 000
D
LC 4 Short time design over pressure: n PS ⋅= 0,25⋅ 187,50 N/ mm
x.PS.op s op.s
3 000
D
LC 5 Long time design negative pressure: n PS ⋅= 0,000⋅ 0,00 N/ mm
x.PS.ep l ep.l
3 000
D
LC 6 Short time design negative pressure: n PS ⋅= 0,005⋅ 3,75 N/ mm
x.PS.ep s ep.s
LC 7 Wind: Wind causes internal forces due to pressure and moment.
LC 7 A Wind moment:
2 2
cq⋅⋅ l + H ⋅ D
0, 8 ⋅ 0, 8 ⋅ 6 610 + 590 ⋅ 3 000
( )
( )
f p cyl.tot R
−9
M 10 49,77 kNm
Cyl.wind
4 ⋅ M
Cyl.wind 4 ⋅ 49,77
n ⋅= 10 7,04 N/ mm
x.M.wind
ππ⋅⋅D 3 000
3 000
D
−3
LC 7 B Wind pressure n p ⋅= 0, 48⋅ ⋅ 10 0,36 N/ mm
x.pwind wind
3 000
D
−3
LC 8 Snow: n p ⋅= 0,68⋅ ⋅ 10 0,51 N/ mm
x.psnow snow
4 4
3 000
D
−3
LC 9 Personnel loading:
n = p ⋅= 1,5⋅ ⋅ 10 = 1,13 N/ mm
x.paccess access
LC 10 Temperature: The cylinder can expand freely. No axial forces occur.
10.4.1 Proof of strength (Ultimate limit state)
10.4.1.1 Design value of actions
2. Step) Find the decisive combination of actions
Fundamental combination for persistent or transient design situations
long term load time load time
E γ⋅ G⋅ A ⊕ γ⋅ Q⋅ A ⊕ γψ⋅ ⋅ QA⋅
∑∑
d G, j k, j 5 Q,1 k,1 5 Q,i oi,,k i 5
To find the decisive combination of the separate actions, all actions are written down in the E matrix.
As we need A ∙γ -fold loads for the strength analysis, the actions are multiplied with A ∙γ .
5B F 5B F
For γ factors refer to Table 12 of EN 13121-3.
F
The A ∙γ fold internal forces are determined as follows nn= ⋅⋅γ A
5B F
xd,,R xk, F 5B
Then the Ψ-Matrix with the for the separate load cases corresponding Ψ-factors has to be created.
For Ψ factors refer to Table 11 of EN 13121-3.
In this example 4 combinations for compression loads and 2 combinations for tension loads are created.
With this number of combinations, all possibilities to get the maximum n load are checked.
x,d,R
Each variable action is one time the predominant action with Ψ = 1,0.
=
= =
= =
= =
⋅= ==
= =
= =
= =
In the columns of the Ψ-matrix is determined, which load cases are combined with witch Ψ-factor in
each load combination.
Because the liquid column doesn’t create any cylinder axial forces for this kind of vessel, all Ψ-factors
for the load case LC 2 are 0. This would be different for example, if the vessel would be suspended with
a support ring.
The way of calculation is shown as an example with a spreadsheet analysis.
E-Matrix
LC Action n γ A n Ψ -Matrix
x,k F 5B x,d,R 0
[N/mm]   [N/mm] CO.1 CO.2 CO.3 CO.4 CO.5 CO.6

nx,W =
LC 1 2,44 1,35 1,60 5,27 1,0 1,0 1,0 1,0 0,0 0,0

LC 2 nx,hp= 61,31 1,35 1,60 132,4 1,0 1,0 1,0 1,0 0,0 0,0

LC 3 n = 150,0 1,50 1,60 360,0 0,0 0,0 0,0 0,0 1,0 0,0
x,op.l
LC 4 nx,op.s = 187,5 1,50 1,00 281,3 0,0 0,0 0,0 0,0 0,0 1,0

LC 5 n = 0,00 1,50 1,60 0,00 1,0 1,0 1,0 0,0 0,0 0,0
x,ep.l
LC 6 nx,ep.s = 3,75 1,50 1,00 5,63 0,0 0,0 0,0 1,0 0,0 0,0

LC 7A n = 7,04 1,50 1,00 10,56 1,0 0,6 0,6 0,0 1,0 0,0
x,M.wind
LC 7B nx,p.wind = 0,36 1,50 1,00 0,54 1,0 0,6 0,6 0,0 0,0 0,0

LC 8 n = 0,51 1,50 1,40 1,07 0,5 1,0 0,5 0,5 0,0 0,0
x,psnow
LC 9 nx,paccess = 1,13 1,50 1,00 1,69 0,0 0,0 1,0 0,0 0,0 0,0

LC 10 n = 0,00 1,00 1,60 0,00 1,0 1,0 1,0 1,0 1,0 1,0
x,∆T
nx,d,R * Ψ-Matrix
Compression Tension
CO.1 CO.2 CO.3 CO.4 CO.5 CO.6
[N/mm] [N/mm] [N/mm] [N/mm] [N/mm] [N/mm]

𝑛𝑛
𝑥𝑥,𝑑𝑑,𝑅𝑅∙𝜓𝜓
5,27 5,27 5,27 5,27 0,00 0,00
0,00 0,00 0,00 0,00 0,00 0,00
0,00 0,00 0,00 0,00 360,0 0,00

0,00 0,00 0,00 0,00 0,00 281,3

0,00 0,00 0,00 0,00 0,00 0,00
0,00 0,00 0,00 5,63 0,00 0,00
10,56 6,34 6,34 0,00 10,56 0,00

0,54 0,32 0,32 0,00 0,00 0,00
0,54 1,07 0,54 0,54 0,00 0,00
0,00 0,00 1,69 0,00 0,00 0,00
0,00 0,00 0,00 0,00 0,00 0,00
Σ = 16,9 13,0 14,2 11,4 370,6 281,3

Then the internal load n of a load case has to be multiplied with each corresponding Ψ-factor as it’s
x,d,R
shown in the n * Ψ-Matrix below. After that, you have to summarize the columns of the several
x.d,R
combinations.
The maximum value of the sums from combination 1 to 4 reflects the decisive compression load n
x.d,R
and is marked in red. The maximum value of the sums from combination 5-6 shows the decisive tension
load nx.d,R and is marked in yellow.
Decisive design axial force: n = 370,6 N/mm (from CO.5)
x,d,R
10.4.1.2 Design value of corresponding resistance
Characteristic limit unit load: U = f ⋅=t 130⋅ 11,7= 1 521 N/ mm
Lam,Cyl,x,k Cyl.x.t k Cyl
U 1 521
Lam,Cyl,,x k
Design value of limit unit load U 647 N/ mm
Lam,Cyl,,x Rd
AAA⋅ ⋅ ⋅ A ⋅ γ 10,,⋅ 1 4 ⋅ 12, 0 ⋅ 10,,⋅ 1 4
1 23 4 M
10.4.1.3 Verification
It shall be verified that:
ER≤
dd
E
d
η ≤=1 0,57≤ 1
So the utilization is given with
R 647
d
10.4.2 Proof of strain (Serviceability limit state)
10.4.2.1 Design value of actions
Characteristic combination (used to verify limit strain)
E = GQ⊕⊕⋅ψ Q
d,rare ∑∑k, j k,1 ∑ o,,1 k i
The characteristic combination is determined without taking into account any A or γ factor.
5 F
The design internal forces are determined as follows nn=
xd,,ε x,k
The way to determine the decisive load combination is similar as for the ultimate limit state.
= =
===
Combination of actions
E-Matrix
LC Action n n  Ψ -Matrix
x,k x,d,ε 0
[N/mm] [N/mm]  CO.1 CO.2 CO.3 CO.4 CO.5 CO.6
LC 1 n = 2,44 2,44  1,0 1,0 1,0 1,0 0,0 0,0
x,W
LC 2 n = 61,31 61,31  1,0 1,0 1,0 1,0 0,0 0,0
x,hp
LC 3 n = 150,00 150,0  0,0 0,0 0,0 0,0 1,0 0,0
x,op.l
LC 4 n = 187,50 187,5  0,0 0,0 0,0 0,0 0,0 1,0
x,op.s
LC 5 n = 0,00 0,00  1,0 1,0 1,0 0,0 0,0 0,0
x,ep.l
LC 6 n = 3,75 3,75  0,0 0,0 0,0 1,0 0,0 0,0
x,ep.s
LC 7A n = 7,04 7,04  1,0 0,6 0,6 0,0 1,0 0,0
x,M.wind
LC 7B n = 0,36 0,36  1,0 0,6 0,6 0,0 0,0 0,0
x,p.wind
LC 8 n = 0,51 0,51  0,5 1,0 0,5 0,5 0,0 0,0
x,psnow
LC 9 n = 1,13 1,13  0,0 0,0 1,0 0,0 0,0 0,0
x,paccess
LC 10 n = 0,00 0,00  1,0 1,0 1,0 1,0 1,0 1,0
x,ΔT
n * Ψ-Matrix
x,d,ε
Compression Tension
CO.1 CO.2 CO.3 CO.4 CO.5 CO.6
[N/mm] [N/mm] [N/mm] [N/mm] [N/mm] [N/mm]
2,44 2,44 2,44 2,44 0,00 0,00
0,00 0,00 0,00 0,00 0,00 0,00
0,00 0,00 0,00 0,00 150,0 0,00
0,00 0,00 0,00 0,00 0,00 187,5
0,00 0,00 0,00 0,00 0,00 0,00
0,00 0,00 0,00 3,8 0,00 0,00
7,04 4,22 4,22 0,00 7,04 0,00
0,36 0,22 0,22 0,00 0,00 0,00
0,26 0,51 0,26 0,26 0,00 0,00
0,00 0,00 1,13 0,00 0,00 0,00
0,00 0,00 0,00 0,00 0,00 0,00
Σ = 10,1 7,4 8,3 6,4 157,0 187,5
Decisive design axial force: n = 187,5 N/mm (from CO.6)
x,d,ε
10.4.2.2 Limit design value of serviceability criterion
E-Modulus axial: E = 12500 N/mm
Cyl.x.t
Limit unit load for strain design: X E ⋅=t 12 500⋅ 11,7 146 250 N/ mm
Lam,Cyl,,x t Cyl,,x t Cyl
10.4.2.3 Verification
It shall be verified that: EC≤
dd
Limit design strain axial: ε = 0,20 %
lim,x,Cyl
n
187,5
xd,,ε
≤ ε ⋅=10 0,13 %≤ 0,20 %
lim,,x Cyl
Verification:
X 146 250
Lam,Cyl,,x t
10.4.3 Stability proof (Ultimate limit state)
10.4.3.1 Design value of actions
Fundamental combination for persistent or transient design situations
long term load time load time
E γ⋅ G⋅ A ⊕ γ⋅ Q⋅ A ⊕ γψ⋅ ⋅ QA⋅
∑∑
d G, j k, j 5 Q,1 k,1 5 Q,i oi,,k i 5
A ⋅ γ
For the stability analysis we need fold loads. Because there are different A5I values for axial
5IF
and tangential direction, for axial stability A AA⋅ has to be used.
5I 5Ix,,5I φ
The A ⋅ γ fold internal forces are determined as follows n n ⋅⋅γ AA⋅
5IF x,,d cr x,k F 5I,x 5I,φ
=
=
=
=
= =
For the stability proof only compression loads have to be checked.
E-Matrix
LC Action nx,k γF A5I,x A5I,ϕ nx,d,cr  Ψ0-Matrix
[N/mm]    [N/ CO.1 CO.2 CO.3 CO.4
mm]
LC 1 n = 2,44 1,35 1,60 1,20 3,88  1,0 1,0 1,0 1,0
x,W
LC 2 n = 61,31 1,35 1,60 1,20 97,4 0,0 0,0 0,0 0,0
x,hp
LC 5 n = 0,00 1,50 1,60 1,20 0,00  1,0 1,0 1,0 0,0
x,ep.l
LC 6 n = 3,75 1,50 1,00 1,00 5,63  0,0 0,0 0,0 1,0
x,ep.s
LC n 7,04 1,50 1,00 1,00 10,5 1,0 0,6 0,6 0,0
x,M.wind
7A = 6
LC nx,p.wind 0,36 1,50 1,00 1,00 0,54  1,0 0,6 0,6 0,0
7B =
LC 8 n 0,51 1,50 1,40 1,15 0,86  0,5 1,0 0,5 0,5
x,psnow
=
LC 9 nx,paccess 1,13 1,50 1,00 1,00 1,69  0,0 0,0 1,0 0,0
=
LC nx,ΔT = 0,00 1,00 1,60 1,20 0,00  1,0 1,0 1,0 1,0
n * Ψ-Matrix
x,d,cr
Compression
CO.1 CO.2 CO.3 CO.4
[N/mm] [N/mm] [N/mm] [N/mm]
3,88 3,88 3,88 3,88
0,00 0,00 0,00 0,00
0,00 0,00 0,00 0,00
0,00 0,00 0,00 5,63
10,56 6,34 6,34 0,00
0,54 0,32 0,32 0,00
0,43 0,86 0,43 0,43
0,00 0,00 1,69 0,00
0,00 0,00 0,00 0,00
Σ = 15,4 11,4 12,7 9,9
Decisive design axial force: n = 15,4 N/mm (from CO.1)
x,d,cr
10.4.3.2 Design value of corresponding resistance
If we assume there is a cut out DN 600 in the lower part of the cylinder, the value for k is
d
co
with: 4,,5≥ 35
Dt⋅⋅3 000 11,7
= =
0,,54 0 54
Coefficient: k 0,357
D 3 000
1 +
1 +
200 ⋅ t
200 ⋅ 11,7
Factor for bending: k = 1,0
B
E-Modulus axial tension: E = 12500 N/mm
Cyl.x.t
E-Modulus tangential bending: E = 19000 N/mm
Cyl.ϕ.b
Characteristic limit buckling load:
2 2
t 11,7
n =k⋅ E ⋅ E ⋅=0,357⋅ 19 000⋅ 12 500⋅ =251,/4 N mm
cr ϕb x
D 3000
10.4.3.3 Verification
It shall be verified that: ER≤
dd
So the utilization is given with
n
x,,d cr 15, 4
η ≤=1 0,14≤ 1
x,cr
n / AAA⋅ ⋅ ⋅ A ⋅ γ 251 / 1,,0 ⋅ 1 4 ⋅ 1,20 ⋅ 1,,0 ⋅ 1 4
( ) ( )
cr 1 23 4 M
10.5 Analysis of the cylinder in tangential direction
Actions, which cause internal forces in tangential direction
−3
— LC 2: Hydrostatic pres.P ρ ⋅⋅gh −H 1,30⋅9,81⋅ 7,00− 0,59⋅10
( )
( )
hp liquid liquid B
0,0817 N/mm²
— LC 3: Overpressure long time PS = 0,200 N/mm
op.l
— LC 4: Overpressure short time PS = 0,250 N/mm
op.s
— LC 5: Negative pressure long time PS = 0,000 N/mm
ep.l
— LC 6: Negative pressure short time PS = 0,005 N/mm
ep.s
— LC 7: Wind pressure p = 0,000 48 N/mm
wind
10.5.1 Strength analysis (Ultimate limit state)
10.5.1.1 Design value of actions
The combination of actions is done similar to the calculation in axial direction but now an example in
Matrix notation is shown. The external pressures are marked in grey. For the strength and strain
analysis they are not decisive but they are listed to show all opportunities.
The most severe combination of n actions may be determined as follows:
load time
— Write all n loads line by line (with γ and A if necessary) in an nx1-matrix Ε.
F 5
— Write all m possible combinations column by column (in terms of combination values) in an nxm-
matrix Ψ. Each column of this matrix represents a possible combination of loads.
— Multiply the transpose of E with Ψ and find the maximum entry in the resulting matrix. The column
number of the maximum is the column number of the most severe combination in Ψ.
= = =
= =
===
For example:
long term

γ ⋅⋅GA

Fg,,1 c,1 5

111 1
load time

 
γ ⋅⋅Q A
1 ψ ψψ
Fq,,1 c,1 5
o,1 oo,,11
 
load time
T

ψ 1 ψψ
E , ψ ⇒= EEmax⋅ψ
γ ⋅⋅Q A   )
o,2 oo,,22 d (
Fq,,2 c,2

ψψ 1 ψ 
load time
oo,,33 o,3

γ ⋅ QA⋅
 
Fq,,33c, 5 ψψψ 1

ooo,,,444

load time

γ ⋅⋅Q A
Fq,,44c, 5

Matrix of actions for the example:

EA⋅⋅γ
FB5.
   
0,081 7 ⋅⋅1,,35 1 20 0,132 4

PA⋅⋅γ
hp F.g 5.B Cyl,25 y.φ
   

0,200 0 ⋅⋅1,,50 1 20 0,360 0
PS ⋅⋅γ A
   
op.l F.p 5.B Cyl,25 y.φ
0,250 0 ⋅⋅1,,50 1 00 0,375 0
   
PS ⋅⋅γ A
E =   E = E =
op.s F.p 5.B Cyl,sh.φ
0,000 ⋅⋅1,
...