SIST ISO/TS 13725:2002
(Main)Hydraulic fluid power -- Cylinders -- Method for determining the buckling load
Hydraulic fluid power -- Cylinders -- Method for determining the buckling load
Transmissions hydrauliques -- Vérins -- Méthode de détermination du flambage
La présente Spécification technique spécifie une méthode de détermination du flambage qui a) prend en considération la géométrie complète du vérin de transmissions hydrauliques et pneumatiques, cela signifie qu'elle ne traite pas le vérin comme une barre équivalente, b) peut être étendue pour être utilisée pour tous les types de fixation de vérin et de raccordement de l'extrémité de la tige, c) inclut un coefficient de sécurité, k, à déterminer par la personne effectuant les calculs et à noter avec les résultats de ces calculs, d) prend en compte une possible charge désaxée, e) prend en compte la masse du vérin hydraulique ou pneumatique, ce qui signifie qu'aucune charge transversale appliquée au vérin n'est négligée, f) peut prendre en compte un désalignement, mais uniquement s'il est situé dans le plan de gravité du vérin, g) est facile à transcrire sous la forme d'un programme simple pour ordinateur. Les résultats donnés par cette méthode ont été comparés favorablement à ceux donnés par plusieurs méthodes déjà utilisées dans l'industrie des vérins hydrauliques dans la plage de 25 mm à 200 mm avec des tiges de piston de 12 mm à 140 mm. En conséquence, il convient d'aborder avec précaution la conception de vérins de dimensions supérieures ou inférieures en utilisant cette méthode. NOTE Cette méthode est basée principalement sur des travaux originaux de Fred Hoblit, Critical buckling load for hydraulic actuating cylinders, Product Engineering - July 1950.
Fluidna tehnika - Hidravlika - Valji - Metoda določanja uklonske sile
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TECHNICAL ISO/TS
SPECIFICATION 13725
First edition
2001-08-15
Hydraulic fluid power — Cylinders —
Method for determining the buckling load
Transmissions hydrauliques — Vérins — Méthode de détermination du
flambage
Reference number
ISO/TS 13725:2001(E)
©
ISO 2001
---------------------- Page: 1 ----------------------
ISO/TS 13725:2001(E)
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ii © ISO 2001 – All rights reserved
---------------------- Page: 2 ----------------------
ISO/TS 13725:2001(E)
Contents Page
Foreword.iv
Introduction.v
1 Scope .1
2 Symbols and units.2
3 Method to be used .6
4 Method for pin- mounted hydraulic cylinders .7
5 Method for hydraulic cylinders fixed at the beginning of the cylinder tube and pin mounted at
the end of the piston rod.10
6 Method for hydraulic cylinders pin mounted at the beginning of the cylinder tube and fixed at
the end of the piston rod.13
7 Method for hydraulic cylinders fixed at their two ends .16
8 Method for hydraulic cylinders fixed at the beginning of the cylinder tube and free at the end
of the piston rod.19
9 Method for hydraulic cylinders fixed at their two ends with a free move allowed at the end of
the piston rod.22
10 Numerical results.25
10.1 Dimensions of the calculated cylinder and material characteristics .25
10.2 Results .25
11 Flow chart.26
© ISO 2001 – All rights reserved iii
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ISO/TS 13725:2001(E)
Foreword
ISO (the International Organization for Standardization) is a worldwide federation of national standards bodies (ISO
member bodies). The work of preparing International Standards is normally carried out through ISO technical
committees. Each member body interested in a subject for which a technical committee has been established has
the right to be represented on that committee. International organizations, governmental and non-governmental, in
liaison with ISO, also take part in the work. ISO collaborates closely with the International Electrotechnical
Commission (IEC) on all matters of electrotechnical standardization.
International Standards are drafted in accordance with the rules given in the ISO/IEC Directives, Part 3.
The main task of technical committees is to prepare International Standards. Draft International Standards adopted
by the technical committees are circulated to the member bodies for voting. Publication as an International
Standard requires approval by at least 75 % of the member bodies casting a vote.
In other circumstances, particularly when there is an urgent market requirement for such documents, a technical
committee may decide to publish other types of normative document:
� an ISO Publicly Available Specification (ISO/PAS) represents an agreement between technical experts in an
ISO working group and is accepted for publication if it is approved by more than 50 % of the members of the
parent committee casting a vote;
� an ISO Technical Specification (ISO/TS) represents an agreement between the members of a technical
committee and is accepted for publication if it is approved by 2/3 of the members of the committee casting a
vote.
An ISO/PAS or ISO/TS is reviewed every three years with a view to deciding whether it can be transformed into an
International Standard.
Attention is drawn to the possibility that some of the elements of this Technical Specification may be the subject of
patent rights. ISO shall not be held responsible for identifying any or all such patent rights.
ISO/TS 13725 was prepared by Technical Committee ISO/TC 131, Fluid power systems, Subcommittee SC 3,
Cylinders.
iv © ISO 2001 – All rights reserved
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ISO/TS 13725:2001(E)
Introduction
Historically, cylinder manufacturers in the fluid power industry have experienced very few rod buckling failures,
most likely due to the conservative factors of safety employed in the designs and factors of safety recommended to
their users. Many countries and some large companies have developed their own methods for determining the
buckling load.
The method presented in this Technical Specification has been developed to comply with the requirements
formulated by ISO working group TC 131/SC 3/WG 1 during their meeting of November 1995.
© ISO 2001 – All rights reserved v
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TECHNICAL SPECIFICATION ISO/TS 13725:2001(E)
Hydraulic fluid power — Cylinders — Method for determining the
buckling load
1 Scope
This Technical Specification specifies a method for the determination of buckling load which
a) takes into account the complete geometry of the fluid power cylinder, meaning it does not treat the fluid power
cylinders as an equivalent bar,
b) can be extended to be used for all types of cylinder mounting and rod end connection,
c) includes a factor of safety, k, to be determined by the person performing the calculations and reported with the
results of the calculations,
d) takes into account a possible off-axis loading,
e) takes into account the weight of the fluid power cylinder, meaning it does not neglect all transverse loads
applied on the fluid power cylinder,
f) can take into account a misalignment, but only if it is situated in the plane of cylinder selfweight, and
g) is easy to transcribe under the form of a simple computer program.
The results given by this method have been compared favourably to those given by several methods already used
in the industry for fluid power cylinders in the range 25 mm to 200 mm with 12 mm to 140 mm piston rods.
Accordingly, larger- or smaller-sized cylinder designs should be approached with caution when using this method.
NOTE This method is based mainly on original work by Fred Hoblit (Critical buckling load for hydraulic actuating cylinders,
Product Engineering, July 1950).
© ISO 2001 – All rights reserved 1
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ISO/TS 13725:2001(E)
2 Symbols and units
See Figures 1 and 2 and Table 1.
DD�
��
1e 1i
L �
p
��
��2
L �
3
2
Figure 1 — Cylinder
2 © ISO 2001 – All rights reserved
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ISO/TS 13725:2001(E)
Cylinder tube: 1 column
Rotational spring joining the two columns
Piston rod: 1 column
Figure 2 — Model of the hydraulic cylinder
© ISO 2001 – All rights reserved 3
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ISO/TS 13725:2001(E)
Table 1 — Symbols and units
Symbol Meaning Unit
C Stiffness of a spring N/mm
D
Diameter mm
D Outside diameter of the cylinder tube mm
1e
D
Inside diameter of the cylinder tube mm
1i
D Outside diameter of the piston rod mm
2
e Distance. The loading of an eccentrically loaded column is equivalent to a mm
a
concentric axial force F and end moment M = Fe
e
d
2
E Modulus of elasticity of cylinder tub material
N/mm
1
2
E Modulus of elasticity of piston rod material
N/mm
2
f Deflection of a slender bar mm
F Axial force N
F Euler buckling load N
euler
4
I
Moment of inertia
mm
4
I Moment of inertia of the cylinder tube
mm
1
4
I Moment of inertia of the piston rod
mm
2
k Factor of safety
L Cylinder tube length mm
1
L Piston rod length mm
2
L Length of the portion of rod situated inside the cylinder tube mm
3
Distance between centreline of piston and piston rod bearing
L Piston length mm
p
M Moment N/mm
M Fixed-end moment at the beginning of the cylinder tube of a fixed hydraulic N/mm
a
cylinder
M Moment at junction of cylinder tube and piston rod N/mm
bc
M Fixed-end moment at the end of the piston rod of a fixed hydraulic cylinder N/mm
d
M Maximal moment in the piston rod N/mm
max
q
1
kF�
EI�
11
q
2
kF�
EI�
22
r Radius of piston rod mm
R Reaction at the beginning of the cylinder tube N
a
R Reaction at the end of the piston rod N
d
4 © ISO 2001 – All rights reserved
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ISO/TS 13725:2001(E)
Symbol Meaning Unit
R Reaction between cylinder tube and piston rod N
bc
X
Distance from the end of a bar mm
Y Deflection of a slender bar at distance X mm
2
Acceleration to take into account inertial forces
� ;g mm/s
� Distance mm
Small value
�
Crookedness angle between the deflection curve of the cylinder tube and the rad
�
deflection curve of the piston rod
� Slenderness ratio: ratio of the column length to the radius of gyration
3
� Weight per unit volume of cylinder tube material
kg/mm
1
3
Weight per unit volume of piston rod material
� kg/mm
2
2
� Stress
N/mm
2
Yield point of a material
� N/mm
e
2
� Stress at the Euler buckling load
N/mm
euler
2
Maximum compressive stress
� N/mm
max
2
� Ultimate strength of a material
N/mm
s
Angle of the deflection curve at the beginning of the cylinder tube rad
�
a
� Angle of the deflection curve at the end of the cylinder tube rad
b
Angle of the deflection curve at the beginning of the piston rod rad
�
c
� Angle of the deflection curve at the end of the piston rod rad
d
Angle at the beginning of the cylinder tube rad
�
a
� Angle at the end of the piston rod rad
b
© ISO 2001 – All rights reserved 5
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ISO/TS 13725:2001(E)
3 Method to be used
See Table 2.
Tableau 2 — Method to be used
Mounting case Method
Method for pin-mounted hydraulic
cylinders.
See chapter 4.
Method for hydraulic cylinders
fixed at the beginning of the
cylinder tube and pin-mounted at
the end of the piston rod
See chapter 5.
Method for hydraulic cylinders pin
mounted at the beginning of the
cylinder tube and fixed at the end
of the piston rod.
See chapter 6.
Method for hydraulic cylinders
fixed at their two ends.
See chapter 7.
Method for hydraulic cylinders
fixed at the beginning of the
cylinder tube and free at the end of
the piston rod.
See chapter 8.
Method for hydraulic cylinders
fixed at their two ends with a free
move allowed at the end of the
piston rod.
See chapter 9.
6 © ISO 2001 – All rights reserved
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ISO/TS 13725:2001(E)
4 Method for pin-mounted hydraulic cylinders
k = [seeclause1, c)].
kF kF
2 2
q � q �
1 2
EI EI
11 22
s =sin (q L ) C =cos (q L ) s =sin (q L ) C =cos (q L )
1 1 1 1 1 1 2 2 2 2 2 2
Step 1
Find the critical buckling load (“force_det”) in solving the following equation by trial and error:
kFL s s – 3E I q C s – 3E I q C s =0
3 1 2 2 2 1 1 2 2 2 2 2 1
Step 2
Choose force_a = force_det*epsilon, force_b = force_det*(1 – epsilon)
Calculate the greatest stress � in the piston rod when the hydraulic cylinder is axially loaded by force_a (see
a
computation procedure hereafter).
Calculate the value f_a =� –� where� is the yield point of piston rod material.
a e e
Calculate the greatest stress � in the piston rod when the hydraulic cylinder is axially loaded by force_b.
b
Calculate the value f_b =� –�
b e
If the value f_a is greater than 0, the greatest allowable compressive load is 0.
If the value f_b is smaller than 0, the greatest allowable compressive load is “force_det”,
otherwise, choose force_c = force_a
until the interval (force_d, force_c) is small enough.
Write force_d = force_c
Calculate force_c
force_c = force_a – (force_b – force_a) * f_a / (f_a – f_b)
Calculate the greatest stress� in the piston rod when the hydraulic cylinder is axially loaded by force_c.
c
Calculate value f_c =� –�
c e
If the value (f_a times f_c) is greater than zero,
write force_a = force_c and f_a = f_c
Then, write force_b = force_c and f_b = f_c
The greatest allowable compressive load is “force_c”.
© ISO 2001 – All rights reserved 7
---------------------- Page: 12 ----------------------
ISO/TS 13725:2001(E)
Stress step 1
Solve the following set of equations using a numerical method:
00 0 0 0 L –L 00
1 2
00 0 0 1 –1 –11 –1
L
3
00 0 10 0 0 0
3EI
22
10 1 0 0 0 0 0 0
�
00 L 10 kFL 00 0
1 1
q L – sin (q L ) q [1 – cos (q L )] kF [q L – sin (q L )] kF sin (s L )
00 0 0 0
1 1 1 1 1 1 1 1 1 1 1 1 1
01 1 0 0 0 0 0 0
00 –L 10 0 kFL 00
2 2
q L – sin (q L ) q [1 – cos (q L )] –kF [q L – sin (q L )] kF sin (q L )
00 00 0
2 2 2 2 2 2 2 2 2 2 2 2 2
0
��R��
a
��
��
0
R
d
����
��
�� 0
R
bc
����
�
22
��
��
M
���LD��Dg
bc 11 1e 1i
����
4
��
��
�
22 2
����
����LD Dg�kFe
11��1e 1i a
��8
��
����
22
���DD gq 2
��
����1c� os qL
�� 11��e 1i 1 ��
L
��11
� ��
1
�
a ��
������
2
42
q
��
�� ��1
��
����
�
2
��
��
�
�LDg
d 22 2
����
4
��
��
�
22
����
�
���LDgkFe
b
22 2 d
��8
��
����
2 2
��
��1c� os qL
��
� �Dgq L
����22
�� ��
22 2 2
�
�
c ��
����
2
4 2
q
��
����2
������
����
8 © ISO 2001 – All rights reserved
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ISO/TS 13725:2001(E)
Stress step 2
Calculate the maximum bending moment M in the piston rod.
max
The equation of the deflection curve of the rod is (equ_def_rod):
2
yC��sinqx C cos (qx)�Cx�Cx�C
��
12 2 2 3 4 5
kF
2
q �
2
EI
22
222
� �
��
����DgEI D gL
1 � �
22 22 2 2 2
CR��L��M� cos (qL )� 1�kFL ��
��
��� �
1bc2bc 22 2d
kF sin (q L ) 4kF 8
22 ��
� �
� �
2
��
1 � �DgEI
22 22
CM�� �
��
2bc
kF 4kF
��
2
��� Dg
22
C �
3
8kF
Rk�F�
bc d
C �
4
kF
2
��
1 � �DgE I
22 22
CM�� �
��
5bc
kF 4kF
��
The bending moment at distance “x” of the junction between cylinder tube and piston rod is:
2
2
��
dy � �Dg
22 2
moment = EI �� x �()R �kF� x�M �kFy (equ_mom_rod)
22�� bc d bc
2
8
dx
��
This moment has a maximum at distance x , which satisfies one of the following conditions:
m_max
x =0
m_max
��
��
C
1
arctan ��n
��
��
3
C ��
��
�� d y
2
��
0< x < L and x = �� 0
m_max 2 m_max ��
3
q
dx
2 ��
x = L
m_max 2
At this distance, the value M of the bending moment can easily be evaluated using equations “equ_def_rod”
max
and “equ_mom_rod”.
Stress step 3
Calculate the greatest stress in the piston rod.
Neglecting transverse stress, one can write that the greatest stress occurs where the bending moment has a
maximum value M and that its value is equal to:
max
4kF 32M
max
� ��
max
23
��DD
22
© ISO 2001 – All rights reserved 9
---------------------- Page: 14 ----------------------
ISO/TS 13725:2001(E)
5 Method for hydraulic cylinders fixed at the beginning of the cylinder tube and pin
mounted at the end of the piston rod
k=[seeclause1,c)].
kF kF
2 2
q � q �
1 2
EI EI
11 22
s =sin (q L ) C =cos (q L ) s =sin (q L ) C =cos (q L )
1 1 1 1 1 1 2 2 2 2 2 2
Step 1
Find the critical buckling load (“force_det”) in solving the following equation by trial and error:
kFL s (L q C + L q C – s )+3E I (L + L )q (q s s – q C C )+3E I q C s +3E I q C s =0
3 2 1 1 1 2 1 1 1 2 2 1 2 1 1 1 2 2 1 2 2 2 1 1 2 2 2 2 2 1
Step 2
Choose force_a = force_det*epsilon, force_b = force_det*(1 – epsilon)
Calculate the greatest stress � in the piston rod when the hydraulic cylinder is axially loaded by force_a (see
a
computation procedure hereafter).
Calculate value f_a =� –� where� is the yield point of piston rod material.
a e e
Calculate the greatest stress� in the piston rod when the hydraulic cylinder is axially loaded by force_b.
b
Calculate value f_b =� –�
b e
If value f_a is greater than 0, the greatest allowable compressive load is 0.
If value f_b is smaller than 0, the greatest allowable compressive load is “force_det”,
otherwise, choose force_c = force_a
until the interval (force_d, force_c) is small enough.
Write force_d = force_c
Calculate force_c
force_c = force_a – (force_b – force_a) * f_a / (f_a – f_b)
Calculate the greatest stress� in the piston rod when the hydraulic cylinder is axially loaded by force_c.
c
Calculate value f_c =� –�
c e
If the value (f_a times f_c) is greater than zero,
write force_a = force_c and f_a = f_c
Then, write force_b = force_c and f_b = f_c
The greatest allowable compressive load is “force_c”.
10 © ISO 2001 – All rights reserved
---------------------- Page: 15 ----------------------
ISO/TS 13725:2001(E)
Stress step 1
Solve the following set of equations using a numerical method:
00 0 0 0 0 L 0 –L 00
1 2
00 0 0 0 1 –10 –11 –1
L
3
00 0 01 0 0 0 0 0
3EI
22
00 0 0 0 0 1 –100 0
10 1 0 0 0 0 0 0 0 0
00 L 1 –10 kFL 00 0 0
1 1
kF [q L – �
1 1
q L – sin (q L ) q [1 – cos (q L )] kF sin (s L )
00 00 00 0
1 1 1 1 1 1 1 1 1
sin (q L )]
1 1
q L cos (q L ) – kF [q L cos (q L ) –
1 1 1 1 1 1 1 1
– q [1 – cos (q L )] kF sin (q L )
00 00 0
1 1 1 1 1
sin (q L ) sin (q L )]
1 1 1 1
01 1 0 0 0 0 0 0 0 0
–L kFL
00 100 0 0 00
2 2
q L – sin (q L ) – q [1 – cos (q L )] – kF [q L – kF sin (q L )
00 00 0 0 0
2 2 2 2 2 2 2 2 2 2 2
sin (q L )]
2 2
0
��R � �
a
�� � �
R 0
d
� �
��
� �
�� 0
R
bc
� �
��
0
M
bc � �
��
�� � �
�
22
���LD D g
��M � �
a 11��1e 1i
4
� �
��
� �
�� �
22 2
� ���LD D g�kFe
� 11��1e 1i a �
��
8
�� � �
22
�� � �
���DD gq 2
����
1c� osqL
11��e 1i 1 ��
L
��11
��
�� � 1 �
�
��
a ��
� 2 �
�� �
42
q
��1
��
� �
��
� �
��
22
��� DD� g
��
11��e 1i 1c� os qL
�� � � � �
��L 11
1 ��
��qL cos qL��2 sin qL
� � � �
� ��
�� � 11 11 1 1 �
a ��
42 q
��
1
� �
��
� �
��
�
2
�LDg
� �
��� 22 2
d
4
�� � �
�� � � �
22
� ���LDgkFe
b 22 2 d
�� � �
8
� �
��
22
� �� �
�� ��
1c� os qL
� �
� �Dgq L
��22
��
22 2 2
� �
��� ��
��
c
2
42
�� � q �
��
�� 2
� �� �
�� � �
� �
© ISO 2001 – All rights reserved 11
---------------------- Page: 16 ----------------------
ISO/TS 13725:2001(E)
Stress step 2
Calculate the maximum bending moment M in the piston rod.
max
The equation of the deflection curve of the rod is (equ_def_rod):
2
yC��sinqx C cos (qx)�Cx�Cx�C
��
12 2 2 3 4 5
kF
2
q �
2
EI
22
222
� �
��
����DgEI DgL
1
22 22 2 2 2
� �
CR��L��M� cos (qL )�1�kFL ��
��
��
1bc2bc 222d
kF sin (q L ) 4kF 8
� �
22 ��
� �
2
��
1 � �DgEI
22 22
CM�� � �
��
2bc
kF 4kF
��
2
��� Dg
22
C �
3
8kF
Rk�F�
bc d
C �
4
kF
2
��
1 � �DgE I
22 22
CM�� �
��
5bc
kF 4kF
��
The bending moment at distance “x” of the junction between cylinder tube and piston rod is:
2
2
��
d y � �Dg
22 2
moment = E I �� x �()R � kF� x� M � kFy (equ_mom_rod)
22�� bc d bc
2
8
dx
��
This moment has a maximum at distance x , which satisfies one of the following conditions:
m_max
x =0
m_max
��
��
C
1
arctan ��n
��
��
3
C ��
��
�� d y
2
��
0< x < L and x = �� 0
m_max 2 m_max ��
3
q
dx
2 ��
x = L
m_max 2
At this distance, the value x of the bending moment can easily be evaluated using equations “equ_def_rod”
m_max
and “equ_mom_rod”.
Stress step 3
Calculate the greatest stress in the piston rod. Neglecting transverse stress, one can write that the greatest stress
occurs where the bending moment has a maximum value M and that its value is equal to:
max
4kF 32M
max
� ��
max
23
��DD
22
12 © ISO 2001 – All rights reserved
---------------------- Page: 17 ----------------------
ISO/TS 13725:2001(E)
6 Method for hydraulic cylinders pin mounted at the beginning of the cylinder tube and
fixed at the end of the piston rod
k =[see clause1,c)].
kF kF
2 2
q � q �
1 2
EI EI
11 22
s =sin (q L ) c =cos (q L ) s =sin (q L ) C =cos (q L )
1 1 1 1 1 1 2 2 2 2 2 2
Step 1
Find the critical buckling load (“force_det”) in solving the following equation by trial and error:
kFL s (–L q C – L q C + s )+3E I (L + L )q (q C C – q s s ) – 3E I q C s – 3E I q C s =0
3 1 1 2 2 2 2 2 2 2 2 1 2 2 1 1 2 2 1 2 2 2 1 1 2 2 2 2 2 1
Step 2
Choose force_a = force_det*epsilon, force_b = force_det*(1 – epsilon)
Calculate the greatest stress � in the piston rod when the hydraulic cylinder is axially loaded by force_a (see
a
computation procedure hereafter).
Calculate value f_a =� –� where� is the yield point of piston rod material.
a e e
Calculate the greatest stress� in the piston rod when the hydraulic cylinder is axially loaded by force_b.
b
Calculate value f_b =� –�
b e
If value f_a is greater than 0, the greatest allowable compressive load is 0.
If value f_b is smaller than 0, the greatest allowable compressive load is “force_det”,
otherwise, choose force_c = force_a
until the interval (force_d, force_c) is small enough.
Write force_d = force_c
Calculate force_c
force_c = force_a – (force_b – force_a) * f_a / (f_a – f_b)
Calculate the greatest stress� in the piston rod when the hydraulic cylinder is axially loaded by force_c
c
Calculate value f_c =� –�
c e
If the value (f_a times f_c) is greater than 0,
write force_a = force_c and f_a = f_c
Then, write force_b = force_c and f_b = f_c
The greatest allowable compressive load is “force_c”.
© ISO 2001 – All rights reserved 13
---------------------- Page: 18 ----------------------
ISO/TS 13725:2001(E)
Stress step 1
Solve the following set of equations using a numerical method:
L –L
00 0 0 0 0 00 0
1 2
00 0 0 0 1 –1 –10 1 –1
L
3
00 0 01000 0 0
3EI
22
10 1 0 00000 0 0
L kFL
00 1 100 1 00 0 0
kF [q L –
1 1
q L – sin (q L ) q [1 – cos (q L )] kF sin (s L )
00 1 1 1 1 1 1 1 00 00 1 1 0 �
sin (q L )]
1 1
00 0 0 00011 0 0
01 1 0 00000 0 0
–L kFL
00 110 0 00 0
2 2
q L – sin (q L ) –q [1 – cos (q L )] –kF [q L –sin (q L )] kF sin (q L )
00 2 2 2 2 2 2 2 00 0 2 2 2 2 00 2 2
q L cos (q L ) – –kF [q L cos (q L ) –
2 2 2 2 2 2 2 2
q [1 – cos (q L )] kF sin (q L )
00 2 2 2 00 0 2 2 00
sin (q L ) sin (q L )]
2 2 2 2
��R 0
��
a
����
R 0
d
��
��
����
R 0
bc
��
��
�
22
����
M
���LD D g
bc
11��1e 1i
��
��
4
����
�
22 2
��
��
M
���LD D g�kFe
d 11��1e 1i a
����8
��
��
22
���DD gq 2
��
������1c� os qL
11��e 1i 1 ��
L
��11
��
1
�
�� ��
����
2
42
q
����
��
1
��
��
��
�
�
a
0
����
��
��
�
2
�
�����LDg
d 22 2
4
��
��
����
�
2 2
�
��LD gk� Fe
��d
��22 2 d
8
����
��
��22
��
��1c� os qL
� �
� �Dgq L
��22
��
22 2 2
�����
��
b ��
2
�� 42
��
q
��
2
��
����
��
��22
��
��
1c� os �qL �
� �Dg��L 22
��
22 2
����
�
qL cos qL��2 sin q L
� � � �
c��22 22 2 2
��
��42 q
2
��
��
����
��
��
����
14 © ISO 2001 – All rights reserved
---------------------- Page: 19 ----------------------
ISO/TS 13725:2001(E)
Stress step 2
Calculate the maximum bending moment M in the piston rod.
max
The equation of the deflection curve of the rod is (equ_def_rod):
2
yC��sinqx C cos (qx)�Cx�Cx�C
��
12 2 2 3 4 5
kF
2
q �
2
EI
22
222
� �
��
1 ����DgEI D gL
� �
22 22 22 2
C��R L��M� cos (q L )�1� kFL ��
��
1b�c2��bc 222d �
kF sin (q L ) 4kF 8
22 ��
� �
� �
2
��
� �DgEI
1
22 22
CM�� � �
��
2bc
kF 4kF
��
2
��� Dg
22
C �
3
8kF
Rk�F�
bc d
C �
4
kF
2
��
1 � �DgE I
22 22
CM�� �
��
5bc
kF 4kF
��
The bending moment at distance “x” of the junction between cylinder tube and piston rod is:
2
2
��
d y � �Dg
22 2
moment = E I �� x � �R � kF� x� M � kFy (equ_mom_rod)
�
22�� bc d bc
2
8
dx
��
This moment has a maximum at distance x , which satisfies one of the following conditions:
m_max
x =0
m_max
��
��
C
1
arctan ��n
��
��
3
C ��
��
�� d y
2
��
0< x < L and x = �� 0
m_max 2 m_max ��
3
q
dx
2 ��
x = L
m_max 2
At this distance, the value M of the bending moment can easily be evaluated using equations “equ_def_rod”
max
and “equ_mom_rod”.
Stress step 3
Calculate the greatest stress in the piston rod.
Neglecting transverse stress, one can write that the greatest stress occurs where the bending moment has a
maximum value M and that its value is equal to:
max
4kF 32M
max
� ��
max
23
��DD�
22
© ISO 2001 – All rights reserved 15
---------------------- Page: 20 ----------------------
ISO/TS 13725:2001(E)
7 Method for hydraulic cylinders fixed at their two ends
k =[see clause1,c).
kF kF
2 2
q � q �
1 2
EI EI
11 22
s =sin (q L ) C =cos (q L ) s =sin (q L ) C =cos (q L )
1 1 1 1 1 1 2 2 2 2 2 2
Step 1
Find the critical buckling load (“force_det”), in solving the following equation by trial and error:
2
kFL s [(L + L )q q C C – q C s – q C s )] + 3E I (L + L )q q (q C s + q C s ) – 3E I q s s +
3 1 1 2 1 2 1 2 1 1 2 2 2 1 2 2 1 2 1 2 1 2 1 2 1 2 2 2 1 1 2
2
6E I q q (C C – 1) – 3E I q s s =0
2 2 1 2 1 2 2 2 2 1 2
Step 2
Choose force_a = force_det*epsilon, force_b = force_det*(1 – epsilon)
Calculate the greatest stress � in the piston rod when the hydraulic cylinder is axially loaded by force_a (see
a
computation procedure hereafter).
Calculate value f_a =� –� where� is the yield point of piston rod material.
a e e
Calculate the greatest stress� in the piston rod when the hydraulic cylinder is axially loaded by force_b.
b
Calculate value f_b =� –�
b e
If value f_a is greater than 0, the greatest allowable compressive load is 0.
If value f_b is smaller than 0, the greatest allowable compressive load is “force_det”,
otherwise, choose force_c = force_a
until the interval (force_d, force_c) is small enough.
Write force_d = force_c
Calculate force_c
force_c = force_a – (force_b – force_a)*f_a / (f_a – f_b)
Calculate the greatest stress� in the piston rod when the hydraulic cylinder is axially loaded by force_c.
c
Calculate the value f_c =� –�
c e
If the value (f_a times f_c) is greater than zero,
write force_a = force_c and f_a = f_c
Then, write force_b = force_c and f_b = f_c
The greatest allowable compressive load is “force_c”.
Stress step 1
Solve the following set of equations using a numerical method:
16 © ISO 2001 – All rights reserved
---------------------- Page: 21 ----------------------
ISO/TS 13725:2001(E)
L –L
0 0 0 0 000 0 00 0
1 2
0 0 0 0 001 –10 –10 1 –1
L
3
00 0 001 0 0 0 0 0 0
3EI
22
0 0 0 0 000 1 –10 0 0 0
1 0 1 0 000 0 0 0 0 0 0
L
00 1 –10 0 kFL 00 0 kF sin (q L ) 0
1
1 1 1
q L – – q [1 – kF [(q L –
1 1 1 1 1
00 000 00 0 0 0
sin (q L ) cos (q L )] sin (q L )]
1 1 1 1 1 1
q L kF [(q L
1 1 1 1
�
q [1 –
1
00 cos (q L ) – 000 cos (q L ) – kF sin (q L ) 00 0 0
1 1 1 1 1 1
cos (q L )]
1 1
sin (q L ) sin (q L )]
1 1 1 1
0 0 0 0 000 0 0 1 1 0 0
0 1 1 0 000 0 0 0 0 0 0
–L
00 1 010 0 0 kFL 0
2
2
q L – – q [1 –
2 2 2
– kF [q L – sin (q L )] kF sin (q L )
00 000 0 0 00
2 2 2 2 2 2
sin (q L ) cos (q L )]
2 2 2 2
q L
2 2
q [1 – – kF [q L cos (q L ) –
2 2 2 2 2
cos (q L ) – kF sin (q L )
00 000 0 0 00
2 2 2 2
cos (q L )] sin (q L )]
2 2 2 2
sin (q L )
2 2
��R � 0 �
a
�� � �
R 0
d
�� � �
�� � �
R 0
bc
�� � �
M 0
bc
�� � �
�� � �
�
22
M ���LD D g
�� � �
a
11��1e 1i
4
�� � �
�� � �
�
22 2
M ��� L D D g� kFe
��d � �
11��1e 1i a
8
�� � �
22
�� � �
2
���DD gq��
��
11��e 1i 1 1c� os �qL
�
��L 11
��
�� � 1 �
�
��
��
�� � 2 �
42
q
��
��1
�� � �
�� � �
22
� ��� D �Dg��
��
11��e1i 1c� os qL
��� �
�� � L 11 �
1 ��
�
a ����q L cos qL��2 sin qL
� � � �
�� � 11 11 11 �
��
42 q
��
1
��
�� � �
�� � �
� 0
a
�� � �
�
�� � 2 �
�LDg
�
22 2
d
�� � �
4
�� � �
�
22
�� � �
�
���LDgkFe
d
22 2 d
�� � �
8
�� � �
22
��
��
1c� os qL
� �
�� � � �Dgq��L 22 �
22 2 2��
�
��
��
b
�� � �
2
42
q
��
2
�� � �� �
�� � �
2
� �Dg��L 1
22 2
�� � �
�
qL cos qL��2 sin qL ��1� cos qL
� � � � � �
c
��22 22 22 2 2
��
�� � �
42 q
�� � 2 �
��
�� � �
�� � �
Stress step 2
Calculate the maximum bending moment M in the piston rod.
max
© ISO 2001 – All rights reserved 17
---------------------- Page: 22 ----------------------
ISO/TS 13725:2001(E)
The equation of the deflection curve of the rod is (equ_def_rod):
2
yC��sin��qx C cos (qx)�Cx�Cx�C
12 2 2 3 4 5
kF
2
q �
2
EI
22
222
� �
��
1 ����DgEI DgL
� �
22 22 22 2
CR��L��M� cos (qL )�1�kFL ��
� �� �
1bc2��bc 222d
kF sin (q L ) 4kF 8
22 ��
� �
� �
2
��
1 � �DgEI
22 22
CM�� � �
��
2bc
kF 4kF
��
2
��� Dg
22
C �
3
8kF
...
SLOVENSKI STANDARD
SIST ISO/TS 13725:2002
01-julij-2002
)OXLGQDWHKQLND+LGUDYOLND9DOML0HWRGDGRORþDQMDXNORQVNHVLOH
Hydraulic fluid power -- Cylinders -- Method for determining the buckling load
Transmissions hydrauliques -- Vérins -- Méthode de détermination du flambage
Ta slovenski standard je istoveten z: ISO/TS 13725:2001
ICS:
23.100.20 +LGUDYOLþQLYDOML Cylinders
SIST ISO/TS 13725:2002 en
2003-01.Slovenski inštitut za standardizacijo. Razmnoževanje celote ali delov tega standarda ni dovoljeno.
---------------------- Page: 1 ----------------------
SIST ISO/TS 13725:2002
---------------------- Page: 2 ----------------------
SIST ISO/TS 13725:2002
TECHNICAL ISO/TS
SPECIFICATION 13725
First edition
2001-08-15
Hydraulic fluid power — Cylinders —
Method for determining the buckling load
Transmissions hydrauliques — Vérins — Méthode de détermination du
flambage
Reference number
ISO/TS 13725:2001(E)
©
ISO 2001
---------------------- Page: 3 ----------------------
SIST ISO/TS 13725:2002
ISO/TS 13725:2001(E)
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© ISO 2001
All rights reserved. Unless otherwise specified, no part of this publication may be reproduced or utilized in any form or by any means, electronic
or mechanical, including photocopying and microfilm, without permission in writing from either ISO at the address below or ISO's member body
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Printed in Switzerland
ii © ISO 2001 – All rights reserved
---------------------- Page: 4 ----------------------
SIST ISO/TS 13725:2002
ISO/TS 13725:2001(E)
Contents Page
Foreword.iv
Introduction.v
1 Scope .1
2 Symbols and units.2
3 Method to be used .6
4 Method for pin- mounted hydraulic cylinders .7
5 Method for hydraulic cylinders fixed at the beginning of the cylinder tube and pin mounted at
the end of the piston rod.10
6 Method for hydraulic cylinders pin mounted at the beginning of the cylinder tube and fixed at
the end of the piston rod.13
7 Method for hydraulic cylinders fixed at their two ends .16
8 Method for hydraulic cylinders fixed at the beginning of the cylinder tube and free at the end
of the piston rod.19
9 Method for hydraulic cylinders fixed at their two ends with a free move allowed at the end of
the piston rod.22
10 Numerical results.25
10.1 Dimensions of the calculated cylinder and material characteristics .25
10.2 Results .25
11 Flow chart.26
© ISO 2001 – All rights reserved iii
---------------------- Page: 5 ----------------------
SIST ISO/TS 13725:2002
ISO/TS 13725:2001(E)
Foreword
ISO (the International Organization for Standardization) is a worldwide federation of national standards bodies (ISO
member bodies). The work of preparing International Standards is normally carried out through ISO technical
committees. Each member body interested in a subject for which a technical committee has been established has
the right to be represented on that committee. International organizations, governmental and non-governmental, in
liaison with ISO, also take part in the work. ISO collaborates closely with the International Electrotechnical
Commission (IEC) on all matters of electrotechnical standardization.
International Standards are drafted in accordance with the rules given in the ISO/IEC Directives, Part 3.
The main task of technical committees is to prepare International Standards. Draft International Standards adopted
by the technical committees are circulated to the member bodies for voting. Publication as an International
Standard requires approval by at least 75 % of the member bodies casting a vote.
In other circumstances, particularly when there is an urgent market requirement for such documents, a technical
committee may decide to publish other types of normative document:
� an ISO Publicly Available Specification (ISO/PAS) represents an agreement between technical experts in an
ISO working group and is accepted for publication if it is approved by more than 50 % of the members of the
parent committee casting a vote;
� an ISO Technical Specification (ISO/TS) represents an agreement between the members of a technical
committee and is accepted for publication if it is approved by 2/3 of the members of the committee casting a
vote.
An ISO/PAS or ISO/TS is reviewed every three years with a view to deciding whether it can be transformed into an
International Standard.
Attention is drawn to the possibility that some of the elements of this Technical Specification may be the subject of
patent rights. ISO shall not be held responsible for identifying any or all such patent rights.
ISO/TS 13725 was prepared by Technical Committee ISO/TC 131, Fluid power systems, Subcommittee SC 3,
Cylinders.
iv © ISO 2001 – All rights reserved
---------------------- Page: 6 ----------------------
SIST ISO/TS 13725:2002
ISO/TS 13725:2001(E)
Introduction
Historically, cylinder manufacturers in the fluid power industry have experienced very few rod buckling failures,
most likely due to the conservative factors of safety employed in the designs and factors of safety recommended to
their users. Many countries and some large companies have developed their own methods for determining the
buckling load.
The method presented in this Technical Specification has been developed to comply with the requirements
formulated by ISO working group TC 131/SC 3/WG 1 during their meeting of November 1995.
© ISO 2001 – All rights reserved v
---------------------- Page: 7 ----------------------
SIST ISO/TS 13725:2002
---------------------- Page: 8 ----------------------
SIST ISO/TS 13725:2002
TECHNICAL SPECIFICATION ISO/TS 13725:2001(E)
Hydraulic fluid power — Cylinders — Method for determining the
buckling load
1 Scope
This Technical Specification specifies a method for the determination of buckling load which
a) takes into account the complete geometry of the fluid power cylinder, meaning it does not treat the fluid power
cylinders as an equivalent bar,
b) can be extended to be used for all types of cylinder mounting and rod end connection,
c) includes a factor of safety, k, to be determined by the person performing the calculations and reported with the
results of the calculations,
d) takes into account a possible off-axis loading,
e) takes into account the weight of the fluid power cylinder, meaning it does not neglect all transverse loads
applied on the fluid power cylinder,
f) can take into account a misalignment, but only if it is situated in the plane of cylinder selfweight, and
g) is easy to transcribe under the form of a simple computer program.
The results given by this method have been compared favourably to those given by several methods already used
in the industry for fluid power cylinders in the range 25 mm to 200 mm with 12 mm to 140 mm piston rods.
Accordingly, larger- or smaller-sized cylinder designs should be approached with caution when using this method.
NOTE This method is based mainly on original work by Fred Hoblit (Critical buckling load for hydraulic actuating cylinders,
Product Engineering, July 1950).
© ISO 2001 – All rights reserved 1
---------------------- Page: 9 ----------------------
SIST ISO/TS 13725:2002
ISO/TS 13725:2001(E)
2 Symbols and units
See Figures 1 and 2 and Table 1.
DD�
��
1e 1i
L �
p
��
��2
L �
3
2
Figure 1 — Cylinder
2 © ISO 2001 – All rights reserved
---------------------- Page: 10 ----------------------
SIST ISO/TS 13725:2002
ISO/TS 13725:2001(E)
Cylinder tube: 1 column
Rotational spring joining the two columns
Piston rod: 1 column
Figure 2 — Model of the hydraulic cylinder
© ISO 2001 – All rights reserved 3
---------------------- Page: 11 ----------------------
SIST ISO/TS 13725:2002
ISO/TS 13725:2001(E)
Table 1 — Symbols and units
Symbol Meaning Unit
C Stiffness of a spring N/mm
D
Diameter mm
D Outside diameter of the cylinder tube mm
1e
D
Inside diameter of the cylinder tube mm
1i
D Outside diameter of the piston rod mm
2
e Distance. The loading of an eccentrically loaded column is equivalent to a mm
a
concentric axial force F and end moment M = Fe
e
d
2
E Modulus of elasticity of cylinder tub material
N/mm
1
2
E Modulus of elasticity of piston rod material
N/mm
2
f Deflection of a slender bar mm
F Axial force N
F Euler buckling load N
euler
4
I
Moment of inertia
mm
4
I Moment of inertia of the cylinder tube
mm
1
4
I Moment of inertia of the piston rod
mm
2
k Factor of safety
L Cylinder tube length mm
1
L Piston rod length mm
2
L Length of the portion of rod situated inside the cylinder tube mm
3
Distance between centreline of piston and piston rod bearing
L Piston length mm
p
M Moment N/mm
M Fixed-end moment at the beginning of the cylinder tube of a fixed hydraulic N/mm
a
cylinder
M Moment at junction of cylinder tube and piston rod N/mm
bc
M Fixed-end moment at the end of the piston rod of a fixed hydraulic cylinder N/mm
d
M Maximal moment in the piston rod N/mm
max
q
1
kF�
EI�
11
q
2
kF�
EI�
22
r Radius of piston rod mm
R Reaction at the beginning of the cylinder tube N
a
R Reaction at the end of the piston rod N
d
4 © ISO 2001 – All rights reserved
---------------------- Page: 12 ----------------------
SIST ISO/TS 13725:2002
ISO/TS 13725:2001(E)
Symbol Meaning Unit
R Reaction between cylinder tube and piston rod N
bc
X
Distance from the end of a bar mm
Y Deflection of a slender bar at distance X mm
2
Acceleration to take into account inertial forces
� ;g mm/s
� Distance mm
Small value
�
Crookedness angle between the deflection curve of the cylinder tube and the rad
�
deflection curve of the piston rod
� Slenderness ratio: ratio of the column length to the radius of gyration
3
� Weight per unit volume of cylinder tube material
kg/mm
1
3
Weight per unit volume of piston rod material
� kg/mm
2
2
� Stress
N/mm
2
Yield point of a material
� N/mm
e
2
� Stress at the Euler buckling load
N/mm
euler
2
Maximum compressive stress
� N/mm
max
2
� Ultimate strength of a material
N/mm
s
Angle of the deflection curve at the beginning of the cylinder tube rad
�
a
� Angle of the deflection curve at the end of the cylinder tube rad
b
Angle of the deflection curve at the beginning of the piston rod rad
�
c
� Angle of the deflection curve at the end of the piston rod rad
d
Angle at the beginning of the cylinder tube rad
�
a
� Angle at the end of the piston rod rad
b
© ISO 2001 – All rights reserved 5
---------------------- Page: 13 ----------------------
SIST ISO/TS 13725:2002
ISO/TS 13725:2001(E)
3 Method to be used
See Table 2.
Tableau 2 — Method to be used
Mounting case Method
Method for pin-mounted hydraulic
cylinders.
See chapter 4.
Method for hydraulic cylinders
fixed at the beginning of the
cylinder tube and pin-mounted at
the end of the piston rod
See chapter 5.
Method for hydraulic cylinders pin
mounted at the beginning of the
cylinder tube and fixed at the end
of the piston rod.
See chapter 6.
Method for hydraulic cylinders
fixed at their two ends.
See chapter 7.
Method for hydraulic cylinders
fixed at the beginning of the
cylinder tube and free at the end of
the piston rod.
See chapter 8.
Method for hydraulic cylinders
fixed at their two ends with a free
move allowed at the end of the
piston rod.
See chapter 9.
6 © ISO 2001 – All rights reserved
---------------------- Page: 14 ----------------------
SIST ISO/TS 13725:2002
ISO/TS 13725:2001(E)
4 Method for pin-mounted hydraulic cylinders
k = [seeclause1, c)].
kF kF
2 2
q � q �
1 2
EI EI
11 22
s =sin (q L ) C =cos (q L ) s =sin (q L ) C =cos (q L )
1 1 1 1 1 1 2 2 2 2 2 2
Step 1
Find the critical buckling load (“force_det”) in solving the following equation by trial and error:
kFL s s – 3E I q C s – 3E I q C s =0
3 1 2 2 2 1 1 2 2 2 2 2 1
Step 2
Choose force_a = force_det*epsilon, force_b = force_det*(1 – epsilon)
Calculate the greatest stress � in the piston rod when the hydraulic cylinder is axially loaded by force_a (see
a
computation procedure hereafter).
Calculate the value f_a =� –� where� is the yield point of piston rod material.
a e e
Calculate the greatest stress � in the piston rod when the hydraulic cylinder is axially loaded by force_b.
b
Calculate the value f_b =� –�
b e
If the value f_a is greater than 0, the greatest allowable compressive load is 0.
If the value f_b is smaller than 0, the greatest allowable compressive load is “force_det”,
otherwise, choose force_c = force_a
until the interval (force_d, force_c) is small enough.
Write force_d = force_c
Calculate force_c
force_c = force_a – (force_b – force_a) * f_a / (f_a – f_b)
Calculate the greatest stress� in the piston rod when the hydraulic cylinder is axially loaded by force_c.
c
Calculate value f_c =� –�
c e
If the value (f_a times f_c) is greater than zero,
write force_a = force_c and f_a = f_c
Then, write force_b = force_c and f_b = f_c
The greatest allowable compressive load is “force_c”.
© ISO 2001 – All rights reserved 7
---------------------- Page: 15 ----------------------
SIST ISO/TS 13725:2002
ISO/TS 13725:2001(E)
Stress step 1
Solve the following set of equations using a numerical method:
00 0 0 0 L –L 00
1 2
00 0 0 1 –1 –11 –1
L
3
00 0 10 0 0 0
3EI
22
10 1 0 0 0 0 0 0
�
00 L 10 kFL 00 0
1 1
q L – sin (q L ) q [1 – cos (q L )] kF [q L – sin (q L )] kF sin (s L )
00 0 0 0
1 1 1 1 1 1 1 1 1 1 1 1 1
01 1 0 0 0 0 0 0
00 –L 10 0 kFL 00
2 2
q L – sin (q L ) q [1 – cos (q L )] –kF [q L – sin (q L )] kF sin (q L )
00 00 0
2 2 2 2 2 2 2 2 2 2 2 2 2
0
��R��
a
��
��
0
R
d
����
��
�� 0
R
bc
����
�
22
��
��
M
���LD��Dg
bc 11 1e 1i
����
4
��
��
�
22 2
����
����LD Dg�kFe
11��1e 1i a
��8
��
����
22
���DD gq 2
��
����1c� os qL
�� 11��e 1i 1 ��
L
��11
� ��
1
�
a ��
������
2
42
q
��
�� ��1
��
����
�
2
��
��
�
�LDg
d 22 2
����
4
��
��
�
22
����
�
���LDgkFe
b
22 2 d
��8
��
����
2 2
��
��1c� os qL
��
� �Dgq L
����22
�� ��
22 2 2
�
�
c ��
����
2
4 2
q
��
����2
������
����
8 © ISO 2001 – All rights reserved
---------------------- Page: 16 ----------------------
SIST ISO/TS 13725:2002
ISO/TS 13725:2001(E)
Stress step 2
Calculate the maximum bending moment M in the piston rod.
max
The equation of the deflection curve of the rod is (equ_def_rod):
2
yC��sinqx C cos (qx)�Cx�Cx�C
��
12 2 2 3 4 5
kF
2
q �
2
EI
22
222
� �
��
����DgEI D gL
1 � �
22 22 2 2 2
CR��L��M� cos (qL )� 1�kFL ��
��
��� �
1bc2bc 22 2d
kF sin (q L ) 4kF 8
22 ��
� �
� �
2
��
1 � �DgEI
22 22
CM�� �
��
2bc
kF 4kF
��
2
��� Dg
22
C �
3
8kF
Rk�F�
bc d
C �
4
kF
2
��
1 � �DgE I
22 22
CM�� �
��
5bc
kF 4kF
��
The bending moment at distance “x” of the junction between cylinder tube and piston rod is:
2
2
��
dy � �Dg
22 2
moment = EI �� x �()R �kF� x�M �kFy (equ_mom_rod)
22�� bc d bc
2
8
dx
��
This moment has a maximum at distance x , which satisfies one of the following conditions:
m_max
x =0
m_max
��
��
C
1
arctan ��n
��
��
3
C ��
��
�� d y
2
��
0< x < L and x = �� 0
m_max 2 m_max ��
3
q
dx
2 ��
x = L
m_max 2
At this distance, the value M of the bending moment can easily be evaluated using equations “equ_def_rod”
max
and “equ_mom_rod”.
Stress step 3
Calculate the greatest stress in the piston rod.
Neglecting transverse stress, one can write that the greatest stress occurs where the bending moment has a
maximum value M and that its value is equal to:
max
4kF 32M
max
� ��
max
23
��DD
22
© ISO 2001 – All rights reserved 9
---------------------- Page: 17 ----------------------
SIST ISO/TS 13725:2002
ISO/TS 13725:2001(E)
5 Method for hydraulic cylinders fixed at the beginning of the cylinder tube and pin
mounted at the end of the piston rod
k=[seeclause1,c)].
kF kF
2 2
q � q �
1 2
EI EI
11 22
s =sin (q L ) C =cos (q L ) s =sin (q L ) C =cos (q L )
1 1 1 1 1 1 2 2 2 2 2 2
Step 1
Find the critical buckling load (“force_det”) in solving the following equation by trial and error:
kFL s (L q C + L q C – s )+3E I (L + L )q (q s s – q C C )+3E I q C s +3E I q C s =0
3 2 1 1 1 2 1 1 1 2 2 1 2 1 1 1 2 2 1 2 2 2 1 1 2 2 2 2 2 1
Step 2
Choose force_a = force_det*epsilon, force_b = force_det*(1 – epsilon)
Calculate the greatest stress � in the piston rod when the hydraulic cylinder is axially loaded by force_a (see
a
computation procedure hereafter).
Calculate value f_a =� –� where� is the yield point of piston rod material.
a e e
Calculate the greatest stress� in the piston rod when the hydraulic cylinder is axially loaded by force_b.
b
Calculate value f_b =� –�
b e
If value f_a is greater than 0, the greatest allowable compressive load is 0.
If value f_b is smaller than 0, the greatest allowable compressive load is “force_det”,
otherwise, choose force_c = force_a
until the interval (force_d, force_c) is small enough.
Write force_d = force_c
Calculate force_c
force_c = force_a – (force_b – force_a) * f_a / (f_a – f_b)
Calculate the greatest stress� in the piston rod when the hydraulic cylinder is axially loaded by force_c.
c
Calculate value f_c =� –�
c e
If the value (f_a times f_c) is greater than zero,
write force_a = force_c and f_a = f_c
Then, write force_b = force_c and f_b = f_c
The greatest allowable compressive load is “force_c”.
10 © ISO 2001 – All rights reserved
---------------------- Page: 18 ----------------------
SIST ISO/TS 13725:2002
ISO/TS 13725:2001(E)
Stress step 1
Solve the following set of equations using a numerical method:
00 0 0 0 0 L 0 –L 00
1 2
00 0 0 0 1 –10 –11 –1
L
3
00 0 01 0 0 0 0 0
3EI
22
00 0 0 0 0 1 –100 0
10 1 0 0 0 0 0 0 0 0
00 L 1 –10 kFL 00 0 0
1 1
kF [q L – �
1 1
q L – sin (q L ) q [1 – cos (q L )] kF sin (s L )
00 00 00 0
1 1 1 1 1 1 1 1 1
sin (q L )]
1 1
q L cos (q L ) – kF [q L cos (q L ) –
1 1 1 1 1 1 1 1
– q [1 – cos (q L )] kF sin (q L )
00 00 0
1 1 1 1 1
sin (q L ) sin (q L )]
1 1 1 1
01 1 0 0 0 0 0 0 0 0
–L kFL
00 100 0 0 00
2 2
q L – sin (q L ) – q [1 – cos (q L )] – kF [q L – kF sin (q L )
00 00 0 0 0
2 2 2 2 2 2 2 2 2 2 2
sin (q L )]
2 2
0
��R � �
a
�� � �
R 0
d
� �
��
� �
�� 0
R
bc
� �
��
0
M
bc � �
��
�� � �
�
22
���LD D g
��M � �
a 11��1e 1i
4
� �
��
� �
�� �
22 2
� ���LD D g�kFe
� 11��1e 1i a �
��
8
�� � �
22
�� � �
���DD gq 2
����
1c� osqL
11��e 1i 1 ��
L
��11
��
�� � 1 �
�
��
a ��
� 2 �
�� �
42
q
��1
��
� �
��
� �
��
22
��� DD� g
��
11��e 1i 1c� os qL
�� � � � �
��L 11
1 ��
��qL cos qL��2 sin qL
� � � �
� ��
�� � 11 11 1 1 �
a ��
42 q
��
1
� �
��
� �
��
�
2
�LDg
� �
��� 22 2
d
4
�� � �
�� � � �
22
� ���LDgkFe
b 22 2 d
�� � �
8
� �
��
22
� �� �
�� ��
1c� os qL
� �
� �Dgq L
��22
��
22 2 2
� �
��� ��
��
c
2
42
�� � q �
��
�� 2
� �� �
�� � �
� �
© ISO 2001 – All rights reserved 11
---------------------- Page: 19 ----------------------
SIST ISO/TS 13725:2002
ISO/TS 13725:2001(E)
Stress step 2
Calculate the maximum bending moment M in the piston rod.
max
The equation of the deflection curve of the rod is (equ_def_rod):
2
yC��sinqx C cos (qx)�Cx�Cx�C
��
12 2 2 3 4 5
kF
2
q �
2
EI
22
222
� �
��
����DgEI DgL
1
22 22 2 2 2
� �
CR��L��M� cos (qL )�1�kFL ��
��
��
1bc2bc 222d
kF sin (q L ) 4kF 8
� �
22 ��
� �
2
��
1 � �DgEI
22 22
CM�� � �
��
2bc
kF 4kF
��
2
��� Dg
22
C �
3
8kF
Rk�F�
bc d
C �
4
kF
2
��
1 � �DgE I
22 22
CM�� �
��
5bc
kF 4kF
��
The bending moment at distance “x” of the junction between cylinder tube and piston rod is:
2
2
��
d y � �Dg
22 2
moment = E I �� x �()R � kF� x� M � kFy (equ_mom_rod)
22�� bc d bc
2
8
dx
��
This moment has a maximum at distance x , which satisfies one of the following conditions:
m_max
x =0
m_max
��
��
C
1
arctan ��n
��
��
3
C ��
��
�� d y
2
��
0< x < L and x = �� 0
m_max 2 m_max ��
3
q
dx
2 ��
x = L
m_max 2
At this distance, the value x of the bending moment can easily be evaluated using equations “equ_def_rod”
m_max
and “equ_mom_rod”.
Stress step 3
Calculate the greatest stress in the piston rod. Neglecting transverse stress, one can write that the greatest stress
occurs where the bending moment has a maximum value M and that its value is equal to:
max
4kF 32M
max
� ��
max
23
��DD
22
12 © ISO 2001 – All rights reserved
---------------------- Page: 20 ----------------------
SIST ISO/TS 13725:2002
ISO/TS 13725:2001(E)
6 Method for hydraulic cylinders pin mounted at the beginning of the cylinder tube and
fixed at the end of the piston rod
k =[see clause1,c)].
kF kF
2 2
q � q �
1 2
EI EI
11 22
s =sin (q L ) c =cos (q L ) s =sin (q L ) C =cos (q L )
1 1 1 1 1 1 2 2 2 2 2 2
Step 1
Find the critical buckling load (“force_det”) in solving the following equation by trial and error:
kFL s (–L q C – L q C + s )+3E I (L + L )q (q C C – q s s ) – 3E I q C s – 3E I q C s =0
3 1 1 2 2 2 2 2 2 2 2 1 2 2 1 1 2 2 1 2 2 2 1 1 2 2 2 2 2 1
Step 2
Choose force_a = force_det*epsilon, force_b = force_det*(1 – epsilon)
Calculate the greatest stress � in the piston rod when the hydraulic cylinder is axially loaded by force_a (see
a
computation procedure hereafter).
Calculate value f_a =� –� where� is the yield point of piston rod material.
a e e
Calculate the greatest stress� in the piston rod when the hydraulic cylinder is axially loaded by force_b.
b
Calculate value f_b =� –�
b e
If value f_a is greater than 0, the greatest allowable compressive load is 0.
If value f_b is smaller than 0, the greatest allowable compressive load is “force_det”,
otherwise, choose force_c = force_a
until the interval (force_d, force_c) is small enough.
Write force_d = force_c
Calculate force_c
force_c = force_a – (force_b – force_a) * f_a / (f_a – f_b)
Calculate the greatest stress� in the piston rod when the hydraulic cylinder is axially loaded by force_c
c
Calculate value f_c =� –�
c e
If the value (f_a times f_c) is greater than 0,
write force_a = force_c and f_a = f_c
Then, write force_b = force_c and f_b = f_c
The greatest allowable compressive load is “force_c”.
© ISO 2001 – All rights reserved 13
---------------------- Page: 21 ----------------------
SIST ISO/TS 13725:2002
ISO/TS 13725:2001(E)
Stress step 1
Solve the following set of equations using a numerical method:
L –L
00 0 0 0 0 00 0
1 2
00 0 0 0 1 –1 –10 1 –1
L
3
00 0 01000 0 0
3EI
22
10 1 0 00000 0 0
L kFL
00 1 100 1 00 0 0
kF [q L –
1 1
q L – sin (q L ) q [1 – cos (q L )] kF sin (s L )
00 1 1 1 1 1 1 1 00 00 1 1 0 �
sin (q L )]
1 1
00 0 0 00011 0 0
01 1 0 00000 0 0
–L kFL
00 110 0 00 0
2 2
q L – sin (q L ) –q [1 – cos (q L )] –kF [q L –sin (q L )] kF sin (q L )
00 2 2 2 2 2 2 2 00 0 2 2 2 2 00 2 2
q L cos (q L ) – –kF [q L cos (q L ) –
2 2 2 2 2 2 2 2
q [1 – cos (q L )] kF sin (q L )
00 2 2 2 00 0 2 2 00
sin (q L ) sin (q L )]
2 2 2 2
��R 0
��
a
����
R 0
d
��
��
����
R 0
bc
��
��
�
22
����
M
���LD D g
bc
11��1e 1i
��
��
4
����
�
22 2
��
��
M
���LD D g�kFe
d 11��1e 1i a
����8
��
��
22
���DD gq 2
��
������1c� os qL
11��e 1i 1 ��
L
��11
��
1
�
�� ��
����
2
42
q
����
��
1
��
��
��
�
�
a
0
����
��
��
�
2
�
�����LDg
d 22 2
4
��
��
����
�
2 2
�
��LD gk� Fe
��d
��22 2 d
8
����
��
��22
��
��1c� os qL
� �
� �Dgq L
��22
��
22 2 2
�����
��
b ��
2
�� 42
��
q
��
2
��
����
��
��22
��
��
1c� os �qL �
� �Dg��L 22
��
22 2
����
�
qL cos qL��2 sin q L
� � � �
c��22 22 2 2
��
��42 q
2
��
��
����
��
��
����
14 © ISO 2001 – All rights reserved
---------------------- Page: 22 ----------------------
SIST ISO/TS 13725:2002
ISO/TS 13725:2001(E)
Stress step 2
Calculate the maximum bending moment M in the piston rod.
max
The equation of the deflection curve of the rod is (equ_def_rod):
2
yC��sinqx C cos (qx)�Cx�Cx�C
��
12 2 2 3 4 5
kF
2
q �
2
EI
22
222
� �
��
1 ����DgEI D gL
� �
22 22 22 2
C��R L��M� cos (q L )�1� kFL ��
��
1b�c2��bc 222d �
kF sin (q L ) 4kF 8
22 ��
� �
� �
2
��
� �DgEI
1
22 22
CM�� � �
��
2bc
kF 4kF
��
2
��� Dg
22
C �
3
8kF
Rk�F�
bc d
C �
4
kF
2
��
1 � �DgE I
22 22
CM�� �
��
5bc
kF 4kF
��
The bending moment at distance “x” of the junction between cylinder tube and piston rod is:
2
2
��
d y � �Dg
22 2
moment = E I �� x � �R � kF� x� M � kFy (equ_mom_rod)
�
22�� bc d bc
2
8
dx
��
This moment has a maximum at distance x , which satisfies one of the following conditions:
m_max
x =0
m_max
��
��
C
1
arctan ��n
��
��
3
C ��
��
�� d y
2
��
0< x < L and x = �� 0
m_max 2 m_max ��
3
q
dx
2 ��
x = L
m_max 2
At this distance, the value M of the bending moment can easily be evaluated using equations “equ_def_rod”
max
and “equ_mom_rod”.
Stress step 3
Calculate the greatest stress in the piston rod.
Neglecting transverse stress, one can write that the greatest stress occurs where the bending moment has a
maximum value M and that its value is equal to:
max
4kF 32M
max
� ��
max
23
��DD�
22
© ISO 2001 – All rights reserved 15
---------------------- Page: 23 ----------------------
SIST ISO/TS 13725:2002
ISO/TS 13725:2001(E)
7 Method for hydraulic cylinders fixed at their two ends
k =[see clause1,c).
kF kF
2 2
q � q �
1 2
EI EI
11 22
s =sin (q L ) C =cos (q L ) s =sin (q L ) C =cos (q L )
1 1 1 1 1 1 2 2 2 2 2 2
Step 1
Find the critical buckling load (“force_det”), in solving the following equation by trial and error:
2
kFL s [(L + L )q q C C – q C s – q C s )] + 3E I (L + L )q q (q C s + q C s ) – 3E I q s s +
3 1 1 2 1 2 1 2 1 1 2 2 2 1 2 2 1 2 1 2 1 2 1 2 1 2 2 2 1 1 2
2
6E I q q (C C – 1) – 3E I q s s =0
2 2 1 2 1 2 2 2 2 1 2
Step 2
Choose force_a = force_det*epsilon, force_b = force_det*(1 – epsilon)
Calculate the greatest stress � in the piston rod when the hydraulic cylinder is axially loaded by force_a (see
a
computation procedure hereafter).
Calculate value f_a =� –� where� is the yield point of piston rod material.
a e e
Calculate the greatest stress� in the piston rod when the hydraulic cylinder is axially loaded by force_b.
b
Calculate value f_b =� –�
b e
If value f_a is greater than 0, the greatest allowable compressive load is 0.
If value f_b is smaller than 0, the greatest allowable compressive load is “force_det”,
otherwise, choose force_c = force_a
until the interval (force_d, force_c) is small enough.
Write force_d = force_c
Calculate force_c
force_c = force_a – (force_b – force_a)*f_a / (f_a – f_b)
Calculate the greatest stress� in the piston rod when the hydraulic cylinder is axially loaded by force_c.
c
Calculate the value f_c =� –�
c e
If the value (f_a times f_c) is greater than zero,
write force_a = force_c and f_a = f_c
Then, write force_b = force_c and f_b = f_c
The greatest allowable compressive load is “force_c”.
Stress step 1
Solve the following set of equations using a numerical method:
16 © ISO 2001 – All rights reserved
---------------------- Page: 24 ----------------------
SIST ISO/TS 13725:2002
ISO/TS 13725:2001(E)
L –L
0 0 0 0 000 0 00 0
1 2
0 0 0 0 001 –10 –10 1 –1
L
3
00 0 001 0 0 0 0 0 0
3EI
22
0 0 0 0 000 1 –10 0 0 0
1 0 1 0 000 0 0 0 0 0 0
L
00 1 –10 0 kFL 00 0 kF sin (q L ) 0
1
1 1 1
q L – – q [1 – kF [(q L –
1 1 1 1 1
00 000 00 0 0 0
sin (q L ) cos (q L )] sin (q L )]
1 1 1 1 1 1
q L kF [(q L
1 1 1 1
�
q [1 –
1
00 cos (q L ) – 000 cos (q L ) – kF sin (q L ) 00 0 0
1 1 1 1 1 1
cos (q L )]
1 1
sin (q L ) sin (q L )]
1 1 1 1
0 0 0 0 000 0 0 1 1 0 0
0 1 1 0 000 0 0 0 0 0 0
–L
00 1 010 0 0 kFL 0
2
2
q L – – q [1 –
2 2 2
– kF [q L – sin (q L )] kF sin (q L )
00 000 0 0 00
2 2 2 2 2 2
sin (q L ) cos (q L )]
2 2 2 2
q L
2 2
q [1 – – kF [q L cos (q L ) –
2 2 2 2 2
cos (q L ) – kF sin (q L )
00 000 0 0 00
2 2 2 2
cos (q L )] sin (q L )]
2 2 2 2
sin
...
SPÉCIFICATION ISO/TS
TECHNIQUE 13725
Première édition
2001-08-15
Transmissions hydrauliques — Vérins —
Méthode de détermination du flambage
Hydraulic fluid power — Cylinders — Method for determining the buckling
load
Numéro de référence
ISO/TS 13725:2001(F)
©
ISO 2001
---------------------- Page: 1 ----------------------
ISO/TS 13725:2001(F)
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© ISO 2001
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Imprimé en Suisse
ii © ISO 2001 – Tous droits réservés
---------------------- Page: 2 ----------------------
ISO/TS 13725:2001(F)
Sommaire Page
Avant-propos.iv
Introduction.v
1 Domaine d'application.1
2 Symboles et unités .2
3Méthode à utiliser .6
4Méthode pour vérins hydrauliques montés sur axe .7
5Méthode pour vérins hydrauliques fixésau début du tube de vérin et axe montéà l’extrémité
de la tige de piston .10
6Méthode pour vérins hydrauliques montésau début du tube et fixés à l’extrémité de la tige de
piston .13
7Méthode pour vérins hydrauliques fixés à leurs deux extrémités .16
8Méthode pour vérins hydrauliques fixésau début du tube et libres à l’extrémité de la tige de
piston .19
9Méthode pour vérins hydrauliques fixés à leurs deux extrémités avec une liberté de
mouvement à l’extrémité de la tige de piston.22
10 Résultats numériques .25
10.1 Dimensions du vérin calculé et caractéristiques du matériau.25
10.2 Résultats.25
11 Organigramme .26
© ISO 2001 – Tous droits réservés iii
---------------------- Page: 3 ----------------------
ISO/TS 13725:2001(F)
Avant-propos
L'ISO (Organisation internationale de normalisation) est une fédération mondiale d'organismes nationaux de
normalisation (comité membres de l'ISO). L'élaboration des Normes internationales est en général confiée aux
comités techniques de l'ISO. Chaque comité membre intéressé par une étude aledroit de fairepartie ducomité
technique créé à cet effet. Les organisations internationales, gouvernementales et non gouvernementales, en
liaison avec l'ISO participent également aux travaux. L'ISO collabore étroitement avec la Commission
électrotechnique internationale (CEI) en ce qui concerne la normalisation électrotechnique.
Les Normes internationales sont rédigées conformément aux règles données dans les Directives ISO/CEI, Partie 3.
La tâche principale des comités techniques est d'élaborer les Normes internationales. Les projets de Normes
internationales adoptéspar lescomités techniques sont soumis aux comités membres pour vote. Leur publication
comme Normes internationales requiert l'approbation de 75 % au moins des comités membres votants.
Dans d'autres circonstances, en particulier lorsqu'il existe une demande urgente du marché,uncomité technique
peut décider de publier d'autres types de documents normatifs:
� une Spécification publiquement disponible ISO (ISO/PAS) représente un accord entre les experts dans un
groupe de travail ISO et est acceptée pour publication si elle est approuvée par plus de 50 % des membres
votants du comité dont relève le goupe de travail;
� une Spécification technique ISO (ISO/TS) représente un accord entre les membres d'un comité technique et
est acceptée pour publication si elle est approuvée par 2/3 des membres votants du comité.
Les ISO/PAS et ISO/TS font l'objet d'un nouvel examen tous les trois ans afin de décider éventuellement de leur
transformation en Normes internationales.
L'attention est appelée sur le fait que certains des élémentsdelaprésente Spécification technique peuvent faire
l'objet de droits de propriété intellectuelle ou de droits analogues. L'ISO ne saurait être tenue pour responsable de
ne pas avoir identifié de tels droits de propriété et averti de leur existence.
L'ISO/TS 13725 a étéélaborée par le comité technique ISO/TC 131, Transmissions hydrauliques et pneumatiques,
sous-comité SC 3, Vérins.
iv © ISO 2001 – Tous droits réservés
---------------------- Page: 4 ----------------------
ISO/TS 13725:2001(F)
Introduction
Historiquement, les fabricants de vérins dans l’industrie des transmissions hydrauliques et pneumatiques n’ont
expérimenté que très peu de défaillances de la tige par flambage, dues le plus souvent aux facteurs de sécurité
modérés utilisés pour la conception et aux facteurs de sécurité recommandés à leurs utilisateurs. Beaucoup de
pays et quelques très grandes sociétés ont développé leur propre méthode de détermination du flambage.
La méthode faisant l’objetdelaprésente Spécification technique a été développée pour répondre aux prescriptions
formulées par le groupe ISO/TC 131/SC 3/GT 1 lors de sa réunion de novembre 1995.
© ISO 2001 – Tous droits réservés v
---------------------- Page: 5 ----------------------
SPÉCIFICATION TECHNIQUE ISO/TS 13725:2001(F)
Transmissions hydrauliques — Vérins — Méthode de
détermination du flambage
1 Domaine d'application
La présente Spécification technique spécifie une méthode de détermination du flambage qui
a) prend en considération la géométrie complète du vérin de transmissions hydrauliques et pneumatiques, cela
signifie qu’elle ne traite pas le vérin comme une barre équivalente,
b) peut être étendue pour être utilisée pour tous les types de fixation de vérin et de raccordement de l’extrémité
de la tige,
c) inclut un coefficient de sécurité, k, à déterminer par la personne effectuant les calculs et à noter avec les
résultats de ces calculs,
d) prend en compte une possible charge désaxée,
e) prend en compte la masse du vérin hydraulique ou pneumatique, ce qui signifie qu’aucune charge
transversale appliquéeau vérin n’est négligée,
f) peut prendre en compte un désalignement, mais uniquement s’il est situé dans le plan de gravité du vérin,
g) est facile à transcrire sous la forme d’un programme simple pour ordinateur.
Les résultats donnés par cette méthode ont été comparés favorablement à ceux donnés par plusieurs méthodes
déjà utilisées dans l’industrie des vérins hydrauliques dans la plage de 25 mm à 200 mm avec des tiges de piston
de 12 mm à 140 mm. En conséquence, il convient d’aborder avec précaution la conception de vérins de
dimensions supérieures ou inférieures en utilisant cette méthode.
NOTE Cette méthode est basée principalement sur des travaux originaux de Fred Hoblit, Critical buckling load for
hydraulic actuating cylinders, Product Engineering – July 1950.
© ISO 2001 – Tous droits réservés 1
---------------------- Page: 6 ----------------------
ISO/TS 13725:2001(F)
2 Symboles et unités
Voir Figures 1 et 2 et Tableau 1.
��DD�
1e 1i
L �
p
��
��2
L �
3
2
Figure 1 — Vérin
2 © ISO 2001 – Tous droits réservés
---------------------- Page: 7 ----------------------
ISO/TS 13725:2001(F)
Tube du vérin: 1 colonne
Ressort rotatif joignant les deux colonnes
Tige de piston: 1 colonne
Figure 2 — Méthode de modélisation
© ISO 2001 – Tous droits réservés 3
---------------------- Page: 8 ----------------------
ISO/TS 13725:2001(F)
Tableau 1 — Symboles et unités
Symbole Signification Unité
C Raideur du ressort N/mm
D Diamètre mm
D Diamètre intérieur dutubede vérin mm
1i
D Diamètre extérieur du tubedevérin mm
1e
D Diamètre extérieurdelatigedepiston mm
2
e Distance. La charge d’une colonne chargée excentriquement est équivalente mm
a
à un effort axial concentrique F et au moment de l’extrémité M = Fe
e
d
2
E Module d’élasticité du matériau du tube du vérin
1 N/mm
2
E Module d’élasticité du matériau de la tige du vérin
N/mm
2
f
Déflexion d’une barre mince mm
F
Force axiale N
F
Effort de flambage d’Euler N
euler
4
I Moment d’inertie
mm
4
I Moment d’inertie du tube de vérin
mm
1
4
I Moment d’inertiedelatigedepiston
mm
2
k Coefficient de sécurité
L Longueur du tube de vérin mm
1
L Longueur de la tige de piston mm
2
L Longueur de la partie de la tige à l’intérieur du tube du vérin mm
3
Distance entre l’axedupiston et larotulede la tige depiston
L Longueur du piston mm
p
M Moment N/mm
M Moment d’extrémité fixe à l’extrémité du tube d’un vérin hydraulique fixe N/mm
a
M Moment à la jonction du tube de vérinet de la tigedepiston N/mm
bc
M Moment d’extrémitéà l’extrémité de la tige de piston d’un vérin hydraulique N/mm
d
fixe
M Moment maximal dans la tige de piston N/mm
max
4 © ISO 2001 – Tous droits réservés
---------------------- Page: 9 ----------------------
ISO/TS 13725:2001(F)
Symbole Signification Unité
kF�
q
1
EI�
11
kF�
q
2
EI�
22
r Rayondelatige de piston mm
R Réaction au début du tube du vérin N
a
R Réaction à l’extrémité de la tige de piston N
d
R Réaction entre le tube du vérinetlatigedepiston N
bc
X Distance depuis l’extrémité d’une barre mm
Y Déflexion d’une barre mince à une distance X mm
2
Accélération pour prendre en compte les forces d’inertie
� ;g
mm/s
�
Distance mm
�
Petite valeur
� Angle de courbure entre la courbe de déflexiondutube de vérin et la courbe rad
de déflexiondelatigedepiston
Élancement: rapport de la longueur d’une colonne par rapport au rayon de
�
giration
3
� Poids par unité de volume du matériau du tube de vérin
kg/mm
1
3
� Poids par unité de volume du matériau de la tige de piston
kg/mm
2
2
Contrainte
� N/mm
2
Limite élastique du matériau
� N/mm
e
2
Contrainte à la charge de flambage d’Euler
�
N/mm
euler
2
� Effort de compression maximal
N/mm
max
2
Résistance à la rupture d’un matériau
� N/mm
s
Angledelacourbededéflexion au début du tube de vérin rad
�
a
Angledelacourbededéflexion à l’extrémité du tube de vérin rad
�
b
� Angledelacourbededéflexion au début de la tige de piston rad
c
Angledelacourbededéflexion à l’extrémité de la tige de piston rad
�
d
© ISO 2001 – Tous droits réservés 5
---------------------- Page: 10 ----------------------
ISO/TS 13725:2001(F)
Tableau 1 (suite)
Symbole Signification Unité
Angleaudébutdutubedevérin rad
�
a
� Angle à l’extrémité de la tige de piston rad
b
3Méthode à utiliser
Voir Tableau 2.
Tableau 2 — Méthode à utiliser
Cas de montage Méthode
Méthode pour vérins hydrauliques
montés sur axe.
Voir article 4.
Méthode pour vérins hydrauliques
fixésaudébut du tube de vérin et
axe montéà l’extrémité de la tige
de piston.
Voir article 5.
Méthode pour vérins hydrauliques
montésaudébut du tube et fixés à
l’extrémité de la tige de piston.
Voir article 6.
Méthode pour vérins hydrauliques
fixés à leurs deux extrémités.
Voir article 7.
Méthode pour vérins hydrauliques
fixésaudébut du tube et libres à
l’extrémité de la tige de piston.
Voir article 8.
Méthode pour vérins hydrauliques
fixés à leurs deux extrémitésavec
une liberté de mouvement à
l’extrémité de la tige de piston.
Voir article 9.
6 © ISO 2001 – Tous droits réservés
---------------------- Page: 11 ----------------------
ISO/TS 13725:2001(F)
4Méthode pour vérins hydrauliques montés sur axe
k = [voir article 1 c)].
kF kF
2 2
q � q �
1 2
EI EI
11 22
s =sin (q L ) C =cos (q L ) s =sin (q L ) C =cos (q L )
1 1 1 1 1 1 2 2 2 2 2 2
Étape 1
Trouver l’effort de flambage critique («force_det»)enrésolvant l’équation suivante par essais et erreurs:
kFL s s – 3E I q C s – 3E I q C s =0
3 1 2 2 2 1 1 2 2 2 2 2 1
Étape 2
Choisir force_a = force_det*epsilon, force_b = force_det*(1 – epsilon)
Calculer la plus grande contrainte � dans la tige de piston lorsque le vérin hydraulique est chargé axialement par
a
force_a (voir procédure de calcul ci-après).
Calculer la valeur f_a =� –� où� est la limite élastique du matériau de la tige de piston.
a e e
Calculer la plus grande contrainte � dans la tige de piston lorsque le vérin hydraulique est chargé axialement par
b
force_b.
Calculer la valeur f_b =� –�
b e
Si la valeur f_a est supérieure à 0, la charge de compression la plus élevée permise est égale à 0.
Si la valeur f_b est inférieure à 0, la charge de compression la plus élevée permise est égaleà«force_det»
autrement, choisir force_c = force_a
jusqu’à ce que l’intervalle (force_d, force_c) soit assez petit.
Écrire force_d = force_c
Calculer force_c
force_c = force_a – (force_b – force_a) * f_a / (f_a – f_b)
Calculer la plus grande contrainte � dans la tige de piston lorsque le vérin hydraulique est chargé axialement par
c
force_c.
Calculer la valeur f_c =� –�
c e
Si la valeur (f_a fois f_c) est supérieure à 0,
écrire force_a = force_c et f_a = f_c
Alors, écrire force_b = force_c et f_b = f_c
Alors la charge de compression la plus élevée permise est égale à«force_c».
© ISO 2001 – Tous droits réservés 7
---------------------- Page: 12 ----------------------
ISO/TS 13725:2001(F)
Contrainte étape 1
Résoudre le jeu suivant d'équations en utilisant une méthode numérique:
00 0 0 0 L –L 00
1 2
00 0 0 1 –1 –11 –1
L
3
00 0 10 0 0 0
3EI
22
10 1 0 0 0 0 0 0
�
00 L 10 kFL 00 0
1 1
q L – sin (q L ) q [1 – cos (q L )] kF [q L – sin (q L )] kF sin (s L )
00 0 0 0
1 1 1 1 1 1 1 1 1 1 1 1 1
01 1 0 0 0 0 0 0
00 –L 10 0 kFL 00
2 2
q L – sin (q L ) q [1 – cos (q L )] –kF [q L – sin (q L )] kF sin (q L )
00 00 0
2 2 2 2 2 2 2 2 2 2 2 2 2
0
��R��
a
��
��
0
R
d
����
��
�� 0
R
bc
����
�
22
��
��
M
���LD��Dg
bc 11 1e 1i
����
4
��
��
�
22 2
����
����LD Dg�kFe
11��1e 1i a
��8
��
����
22
���DD gq 2
��
����1c� os qL
�� 11��e 1i 1 ��
L
��11
� ��
1
�
a ��
������
2
42
q
��
�� ��1
��
����
�
2
��
��
�
�LDg
d 22 2
����
4
��
��
�
22
����
�
���LDgkFe
b
22 2 d
��8
��
����
2 2
��
��1c� os qL
��
� �Dgq L
����22
�� ��
22 2 2
�
�
c ��
����
2
4 2
q
��
����2
������
����
8 © ISO 2001 – Tous droits réservés
---------------------- Page: 13 ----------------------
ISO/TS 13725:2001(F)
Contrainte étape 2
Calculer le moment de flexion maximal M dans la tige de piston.
max
L'équation de la courbe de déflexion de la tige est (equ_def_rod):
2
yC��sinqx C cos (qx)�Cx�Cx�C
��
12 2 2 3 4 5
kF
2
q �
2
EI
22
222
� �
��
����DgEI D gL
1 � �
22 22 2 2 2
CR��L��M� cos (qL )� 1�kFL ��
��
��� �
1bc2bc 22 2d
kF sin (q L ) 4kF 8
22 ��
� �
� �
2
��
1 � �DgEI
22 22
CM�� � �
��
2bc
kF 4kF
��
2
��� Dg
22
C �
3
8kF
Rk�F�
bc d
C �
4
kF
2
��
1 � �DgE I
22 22
CM�� �
��
5bc
kF 4kF
��
Le moment de flexion à une distance «x» de la jonction entre le tube du vérin et latigede pistonest:
2
2
��
d y � �Dg
22 2
moment = E I �� x �()R � kF� x� M � kFy (equ_mom_rod)
22�� bc d bc
2
8
dx
��
Ce moment a un maximum à une distance x qui satisfait l'une des conditions suivantes:
m_max
x =0
m_max
��
��
C
1
arctan ��n
��
��
3
C ��
��
�� d y
2
��
0< x < L et x = �� 0
m_max 2 m_max ��
3
q
dx
2 ��
x = L
m_max 2
À cette distance, la valeur M du moment de flexion peut être facilement évaluéeenutilisantles équations «equ-
max
def_rod» et «equ_mom_rod».
Contrainte étape 3
Calculer la contrainte maximale dans la tige de piston.
En négligeant les contraintes transversales, on peut écrire que la contrainte maximale se produit lorsque le
moment de flexion a une valeur maximale M et que cette valeur est égale à:
max
4kF 32M
max
� ��
max
23
��DD
22
© ISO 2001 – Tous droits réservés 9
---------------------- Page: 14 ----------------------
ISO/TS 13725:2001(F)
5Méthode pour vérins hydrauliques fixésaudébut du tube de vérin et axe montéà
l’extrémité de la tige de piston
k =[voir article1c)].
kF kF
2 2
q � q �
1 2
EI EI
11 22
s =sin (q L ) C =cos (q L ) s =sin (q L ) C =cos (q L )
1 1 1 1 1 1 2 2 2 2 2 2
Étape 1
Trouver la charge de flambage critique («force_det»)en résolvant l'équation suivante par essais et erreurs:
kFL s (L q C + L q C – s )+3E I (L + L )q (q s s – q C C )+3E I q C s +3E I q C s =0
3 2 1 1 1 2 1 1 1 2 2 1 2 1 1 1 2 2 1 2 2 2 1 1 2 2 2 2 2 1
Étape 2
Choisir force_a = force_det*epsilon, force_b = force_det*(1 – epsilon)
Calculer la plus grande contrainte � dans la tige de piston lorsque le vérin hydraulique est chargé axialement par
a
force_a (voir procédure de calcul ci-après).
Calculer la valeur f_a =� –� où� est la limite élastique du matériau de la tige de piston.
a e e
Calculer la plus grande contrainte � dans la tige de piston lorsque le vérin hydraulique est chargé axialement par
b
force_b.
Calculer la valeur f_b =� –�
b e
Si la valeur f_a est supérieure à 0, la charge de compression la plus élevée permise est égale à 0.
Si la valeur f_b est inférieure à 0, la charge de compression la plus élevée permise est égaleà«force_det»
autrement, choisir force_c = force_a
jusqu’à ce que l’intervalle (force_d, force_c) soit assez petit.
Écrire force_d = force_c
Calculer force_c
force_c = force_a – (force_b – force_a) * f_a / (f_a – f_b)
Calculer la plus grande contrainte � dans la tige de piston lorsque le vérin hydraulique est chargé axialement par
c
force_c.
Calculer la valeur f_c =� –�
c e
Si la valeur (f_a fois f_c) est supérieure à 0,
écrire force_a = force_c et f_a = f_c
Alors, écrire force_b = force_c et f_b = f_c
Alors la charge de compression la plus élevée permise est égale à«force_c».
10 © ISO 2001 – Tous droits réservés
---------------------- Page: 15 ----------------------
ISO/TS 13725:2001(F)
Contrainte étape 1
Résoudre le jeu suivant d'équations en utilisant une méthode numérique:
00 0 0 0 0 L 0 –L 00
1 2
00 0 0 0 1 –10 –11 –1
L
3
00 0 01 0 0 0 0 0
3EI
22
00 0 0 0 0 1 –100 0
10 1 0 0 0 0 0 0 0 0
00 L 1 –10 kFL 00 0 0
1 1
kF [q L – �
1 1
q L – sin (q L ) q [1 – cos (q L )] kF sin (s L )
00 00 00 0
1 1 1 1 1 1 1 1 1
sin (q L )]
1 1
q L cos (q L ) – kF [q L cos (q L ) –
1 1 1 1 1 1 1 1
– q [1 – cos (q L )] kF sin (q L )
00 00 0
1 1 1 1 1
sin (q L ) sin (q L )]
1 1 1 1
01 1 0 0 0 0 0 0 0 0
–L kFL
00 100 0 0 00
2 2
q L – sin (q L ) – q [1 – cos (q L )] – kF [q L – kF sin (q L )
00 00 0 0 0
2 2 2 2 2 2 2 2 2 2 2
sin (q L )]
2 2
0
��R � �
a
�� � �
R 0
d
� �
��
� �
�� 0
R
bc
� �
��
0
M
bc � �
��
�� � �
�
22
���LD D g
��M � �
a 11��1e 1i
4
� �
��
� �
�� �
22 2
� ���LD D g�kFe
� 11��1e 1i a �
��
8
�� � �
22
�� � �
���DD gq 2
����
1c� osqL
11��e 1i 1 ��
L
��11
��
�� � 1 �
�
��
a ��
� 2 �
�� �
42
q
��1
��
� �
��
� �
��
22
��� DD� g
��
11��e 1i 1c� os qL
�� � � � �
��L 11
1 ��
��qL cos qL��2 sin qL
� � � �
� ��
�� � 11 11 1 1 �
a ��
42 q
��
1
� �
��
� �
��
�
2
�LDg
� �
��� 22 2
d
4
�� � �
�� � � �
22
� ���LDgkFe
b 22 2 d
�� � �
8
� �
��
22
� �� �
�� ��
1c� os qL
� �
� �Dgq L
��22
��
22 2 2
� �
��� ��
��
c
2
42
�� � q �
��
�� 2
� �� �
�� � �
� �
© ISO 2001 – Tous droits réservés 11
---------------------- Page: 16 ----------------------
ISO/TS 13725:2001(F)
Contrainte étape 2
Calculer le moment maximal de courbure M dans la tige de piston.
max
L’équation de la courbe de déflexion de la tige est (equ_def_rod):
2
yC��sinqx C cos (qx)�Cx�Cx�C
��
12 2 2 3 4 5
kF
2
q �
2
EI
22
222
� �
��
����DgEI DgL
1 � �
22 22 22 2
C��R L��M� cos (q L )�1� kFL ��
��
��� �
1bc2bc 222d
kF sin (q L ) 4kF 8
22 ��
� �
� �
2
��
1 � �DgEI
22 22
CM�� � �
2b��c
kF 4kF
��
2
��� Dg
22
C �
3
8kF
Rk�F�
bc d
C �
4
kF
2
��
1 � �DgE I
22 22
CM�� �
��
5bc
kF 4kF
��
Le moment de courbure à une distance «x» de la jonction entre le tube du vérin et latigede pistonest:
2
2
��
d y � �Dg
22 2
moment = E I �� x �()R � kF� x� M � kFy (equ_mom_rod)
22�� bc d bc
2
8
dx
��
Ce moment est maximal à une distance x qui satisfait l’une des conditions suivantes:
m_max
x =0
m_max
��
��
C
1
arctan ��n
��
��
3
��C ��
��2 d y
��
0< x < L et x = �� 0
m_max 2 m_max ��
3
q
dx
2 ��
x = L
m_max 2
À cette distance, la valeur M du moment de flexion peut être facilement évaluéeenutilisantles équations
max
«equ_def_rod» et «equ_mom_rod».
Contrainte étape 3
Calculer la contrainte maximale dans la tige de piston. En négligeant les contraintes transversales, on peut écrire
que la contrainte maximale se produit lorsque le moment de flexion a une valeur M , et que sa valeur est égale
max
à:
4kF 32M
max
� ��
max
23
��DD
22
12 © ISO 2001 – Tous droits réservés
---------------------- Page: 17 ----------------------
ISO/TS 13725:2001(F)
6Méthode pour vérins hydrauliques montésaudébut du tube et fixés à l’extrémité de
la tige de piston
k =[voir article1c)].
kF kF
2 2
q � q �
1 2
EI EI
11 22
s =sin (q L ) C =cos (q L ) s =sin (q L ) C =cos (q L )
1 1 1 1 1 1 2 2 2 2 2 2
Étape 1
Trouver la charge critique de flambage («force_det»)en résolvant l'équation suivante par essais et erreurs:
kFL s (–L q C – L q C + s )+3E I (L + L )q (q C C – q s s ) – 3E I q C s – 3E I q C s =0
3 1 1 2 2 2 2 2 2 2 2 1 2 2 1 1 2 2 1 2 2 2 1 1 2 2 2 2 2 1
Étape 2
Choisir force_a = force_det*epsilon, force_b = force_det*(1 – epsilon)
Calculer la plus grande contrainte � dans la tige de piston lorsque le vérin hydraulique est chargé axialement par
a
force_a (voir procédure de calcul ci-après).
Calculer la valeur f_a =� –� où� est la limite élastique du matériau de la tige de piston.
a e e
Calculer la plus grande contrainte � dans la tige de piston lorsque le vérin hydraulique est chargé axialement par
b
force_b.
Calculer la valeur f_b =� –�
b e
Si la valeur f_a est supérieure à 0, la charge de compression la plus élevée permise est égale à 0.
Si la valeur f_b est inférieure à 0, la charge de compression la plus élevée permise est égaleà«force_det»
autrement, choisir force_c = force_a
jusqu’à ce que l’intervalle (force_d, force_c) soit assez petit.
Écrire force_d = force_c
Calculer force_c
force_c = force_a – (force_b – force_a) * f_a / (f_a – f_b)
Calculer la plus grande contrainte � dans la tige de piston lorsque le vérin hydraulique est chargé axialement par
c
force_c.
Calculer la valeur f_c =� –�
c e
Si la valeur (f_a fois f_c) est supérieure à 0,
écrire force_a = force_c et f_a = f_c
Alors, écrire force_b = force_c et f_b = f_c
Alors la charge de compression la plus élevée permise est égale à«force_c».
© ISO 2001 – Tous droits réservés 13
---------------------- Page: 18 ----------------------
ISO/TS 13725:2001(F)
Contrainte étape 1
Résoudre le jeu suivant d'équations en utilisant une méthode numérique:
L –L
00 0 0 0 0 00 0
1 2
00 0 0 0 1 –1 –10 1 –1
L
3
00 0 01000 0 0
3EI
22
10 1 0 00000 0 0
L kFL
00 1 100 1 00 0 0
kF [q L –
1 1
q L – sin (q L ) q [1 – cos (q L )] kF sin (s L )
00 1 1 1 1 1 1 1 00 00 1 1 0 �
sin (q L )]
1 1
00 0 0 00011 0 0
01 1 0 00000 0 0
–L kFL
00 110 0 00 0
2 2
q L – sin (q L ) –q [1 – cos (q L )] –kF [q L –sin (q L )] kF sin (q L )
00 2 2 2 2 2 2 2 00 0 2 2 2 2 00 2 2
q L cos (q L ) – –kF [q L cos (q L ) –
2 2 2 2 2 2 2 2
q [1 – cos (q L )] kF sin (q L )
00 2 2 2 00 0 2 2 00
sin (q L ) sin (q L )]
2 2 2 2
��R 0
��
a
����
R 0
d
��
��
����
R 0
bc
��
��
�
22
����
M
���LD D g
bc
11��1e 1i
��
��
4
����
�
22 2
��
��
M
���LD D g�kFe
d 11��1e 1i a
����8
��
��
22
���DD gq 2
��
������1c� os qL
11��e 1i 1 ��
L
��11
��
1
�
�� ��
����
2
42
q
����
��
1
��
��
��
�
�
a
0
����
��
��
�
2
�
�����LDg
d 22 2
4
��
��
����
�
2 2
�
��LD gk� Fe
��d
��22 2 d
8
����
��
��22
��
��1c� os qL
� �
� �Dgq L
��22
��
22 2 2
�����
��
b ��
2
�� 42
��
q
��
2
��
����
��
��22
��
��
1c� os �qL �
� �Dg��L 22
��
22 2
����
�
qL cos qL��2 sin q L
� � � �
c��22 22 2 2
��
��42 q
2
��
��
����
��
��
����
14 © ISO 2001 – Tous droits réservés
---------------------- Page: 19 ----------------------
ISO/TS 13725:2001(F)
Contrainte étape 2
Calculer le moment maximal de flexion M dans la tige de piston.
max
L'équation de la courbe de déflexion de la tige est (equ_def_rod):
2
yC��sinqx C cos (qx)�Cx�Cx�C
��
12 2 2 3 4 5
kF
2
q �
2
EI
22
222
� �
��
����DgEI D gL
1 � �
22 22 22 2
C��R L��M� cos (q L )�1� kFL ��
��
��� �
1bc2bc 222d
kF sin (q L ) 4kF 8
22 ��
� �
� �
2
��
1 � �DgEI
22 22
CM�� � �
��
2bc
kF 4kF
��
2
��� Dg
22
C �
3
8kF
Rk�F�
bc d
C �
4
kF
2
��
1 � �DgE I
22 22
CM�� �
��
5bc
kF 4kF
��
Le moment de flexion à une distance «x» de la jonction entre le tube de vérin et latigede pistonest:
2 2
��
d y � �Dg
22 2
moment = E I �� x � R � kF� x� M � kFy (equ_mom_rod)
� �
22�� bc d bc
2
8
��dx
Le moment est maximal à une distance x qui satisfait l'une des conditions suivantes:
m_max
x =0
m_max
��
��
C
1
arctan ��n
��
��
3
C ��
��
�� d y
2
��
0< x < L et x = �� 0
m_max 2 m_max ��
3
q
dx
2 ��
x = L
m_max 2
À cette distance, la valeur M du moment de flexion peut facilement être évaluéeenutilisantles équations
max
«equ_def_rod» et «equ_mom_rod».
Contrainte étape 3
Calculer la contrainte maximale dans la tige de piston.
En négligeant les contraintes transversales, on peut écrire que la contrainte maximale se produit lorsque le
moment de flexion a une valeur maximale M et que sa valeur est égale à:
max
4kF 32M
max
� ��
max
23
��DD�
22
© ISO 2001 – Tous droits réservés 15
---------------------- Page: 20 ----------------------
ISO/TS 13725:2001(F)
7Méthode pour vérins hydrauliques fixés à leurs deux extrémités
k =[voir article1c)].
kF kF
2 2
q � q �
1 2
EI EI
11 22
s =sin (q L ) C =cos (q L ) s =sin (q L ) C =cos (q L )
1 1 1 1 1 1 2 2 2 2 2 2
Étape 1
Trouver l'effort critique de flambage («force_det»)enrésolvant l'équation suivante par essais et erreurs:
2
kFL s [(L + L )q q C C – q C s – q C s )] + 3E I (L + L )q q (q C s + q C s ) – 3E I q s s +
3 1 1 2 1 2 1 2 1 1 2 2 2 1 2 2 1 2 1 2 1 2 1 2 1 2 2 2 1 1 2
2
6E I q q (C C – 1) – 3E I q s s =0
2 2 1 2 1 2 2 2 2 1 2
Étape 2
Choisir force_a = force_det*epsilon, force_b = force_det*(1 – epsilon)
Calculer la plus grande contrainte � dans la tige de piston lorsque le vérin hydraulique est chargé axialement par
a
force_a (voir procédure de calcul ci-après).
Calculer la valeur f_a =� –� où� est la limite élastique du matériau de la tige de piston.
a e e
Calculer la plus grande contrainte � dans la tige de piston lorsque le vérin hydraulique est chargé axialement par
b
force_b.
Calculer la valeur f_b =� –�
b e
Si la valeur f_a est supérieure à 0, la charge de compression la plus élevée permise est égale à 0.
Si la valeur f_b est inférieure à 0, la charge de compression la plus élevée permise est égaleà«force_det»
autrement, choisir force_c = force_a
jusqu’à ce que l’intervalle (force_d, force_c) soit assez petit.
Écrire force_d = force_c
Calculer force_c
force_c = force_a – (force_b – force_a) * f_a / (f_a – f_b)
Calculer la plus grande contrainte � dans la tige de piston lorsque le vérin hydraulique est chargé axialement par
c
force_c.
Calculer la valeur f_c =� –�
c e
Si la valeur (f_a fois f_c) est supérieure à 0,
écrire force_a = force_c et f_a = f_c
Alors, écrire force_b = force_c et f_b = f_c
Alors la charge de compression la plus élevée permise est égale à«force_c».
16 © ISO 2001 – Tous droits réservés
---------------------- Page: 21 ----------------------
ISO/TS 13725:2001(F)
Contrainte étape 1
Résoudre le jeu suivant d'équations en utilisant une méthode numérique:
L –L
0 0 0 0 000 0 00 0
1 2
0 0 0 0 001 –10 –10 1 –1
L
3
00 0 001 0 0 0 0 0 0
3EI
22
0 0 0 0 000 1 –10 0 0 0
1 0 1 0 000 0 0 0 0 0 0
L
00 1 –10 0 kFL 00 0 kF sin (q L ) 0
1
1 1 1
q L – – q [1 – kF [(q L –
1 1 1 1 1
00 000 00 0 0 0
sin (q L ) cos (q L )] sin (q L )]
1 1 1 1 1 1
q L kF [(q L
1 1 1 1
�
q [1 –
1
00 cos (q L ) – 000 cos (q L ) – kF sin (q L ) 00 0 0
1 1 1 1 1 1
cos (q L )]
1 1
sin (q L ) sin (q L )]
1 1 1 1
0 0 0 0 00
...
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