Design Guidelines and Troubles
for DYNAMICBELLOWSTM andCountermeasures
1. Introduction

catalog, but for Bellows that absorb displacement
perpendicular to the axis and in a bending direction,

The functions of welded metal DYNAMICBELLOWSTM

selection is often difficult.

（hereafter“Bellows”）are to absorb displacement

Also, when the Bellows’inner pressure is greater thanparallel with the axis, perpendicular to the axis, and in

the outside pressure to maintain airtightness, bucklinga bending direction as shown in Figure1. The Bellows

sometimes develops as shown in Figure2, resulting inabsorb displacement while maintaining airtightness.

the risk of deformation and breakage. Therefore, care

When selecting Bellows that absorb displacement

is required when designing.

parallel with the axis, it is easy to select from theSummer 2018
No.35
Displacement parallel with axis（Δx）

Before buckling

Figure2

After buckling

Bellows’buckling

In addition, when Bellows are attached in a horizontal
direction, the Bellows’own weight causes horizontal
deflection as shown in Figure3. Therefore, care isrequired like the case of buckling.
Displacement perpendicular to axis（Δ y）

Horizontal deflection

Displacement in bending direction（Δθ）

Figure19Types of displacement

Figure3

Bellows’horizontal deflection

Technical Papers

This article examines the case of our standard

In this case, Equation（1）becomes:

product V series（catalog No. PC08）, and describes theselection method for Bellows that absorb displacement…………（３）

perpendicular to the axis, and design considerations
for buckling and horizontal deflection.

Equation（2）becomes:

2. Design guidelines for Bellows…………（４）2-1）‌Concept of displacement perpendicular

However, Δ x is displacement parallel with the axis

to the axison the contraction side（mm）.

When investigating the Bellows’displacement
perpendicular to the axis, first the displacement

Regarding Sections 2-1）and 2-2）
, the shape of Bellows

perpendicular to the axis is converted into thewith displacement perpendicular to the axis is suchdisplacement parallel with the axis, using the equation:that expanded parts and contracted parts are combined
as shown in Figure4. Therefore, regarding the…………（１）
where, Δ x（y）：‌value converted into displacement

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parallel with axis（mm）
D：Bellows’outside diameter（mm）

Δ y ：‌displacement perpendicular toaxis（mm）
L：Bellows’length（mm）
When the converted allowable displacement parallelwith the axis is known, the allowable displacement

Before application of displacement

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perpendicular to the axis can then be calculated by
the equation:…………（２）where, Δ y（max）：‌allowable displacement
perpendicular to axis（mm）.
2-2）‌Combination of displacement parallel withaxis and displacement perpendicular to axis
In Section 2-1）
, only displacement perpendicular to theaxis was given. However, generally, Bellows are often

After application of displacement

used with both displacement perpendicular to the axis
and displacement parallel with the axis. According toEquation（1）, the smaller L is, the larger the converted

Expansion side

displacement parallel with the axis Δ x（y）. Therefore,

Contraction side

the calculation should be made to minimize L under
the influence of displacement parallel with the axis.

Figure4

Shape of displacement perpendicular to axis
10

Design Guidelines and Troubles for DYNAMICBELLOWSTM and Countermeasuresconverted displacement parallel with the axis Δ x（y）,

length = 15mm, Contraction length = 5mm

allowance ranges should be prepared for both the

In this case, the Bellows length L = 10mm, Δ x =±5mm.

expansion side and the contraction side.

According to Equation（2）,Therefore, a Bellows with a length of L is given thevalue of“＋Δ x（y）and‐Δ x（y）.”
In the case of 2 blocks:

3. Selection procedure for V series

Expansion and contraction length = 20mm, Expansion

3-1）‌When only displacement perpendicular

In this case, the Bellows length L＝20mm, Δ x = ±10

to the axis is present
Based on the design guidelines described above, we

length = 30mm, Contraction length = 10mmmm.
According to Equation（2）,consider the selection of V series（Catalog No. PC08）.
Refer to the catalog and follow the steps.
Figure5 shows the selection flow. Each step is described

In the same manner, increase the block number andbelow.

calculate for each block number. Table1 shows theresults.

【Step 1】Select Bellows’ size（nominal diameter）

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【Step 2】‌Calculate allowable displacement perpendicular
to axis for each block number【Step 3】‌Select the type which can accommodate thenecessary displacement perpendicular to axis

Figure5

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Selection procedure for V Series

Table1

V20’s block number and displacement amount

Block number

1

2

3

4

5

L（mm）

10

20

30

40

50

Δ x（mm）±

5

10

15

20

25

Δ y（max）
（mm）

0.4

1.6

3.7

6.6

10.4

【Step 3】 D etermine the type which canaccommodate the necessary displacement
perpendicular to the axis

Select the block number which can accommodate the【Step 1】Select the Bellows size (nominal diameter)

necessary displacement perpendicular to the axis andBased on the space in which the Bellows is attached,

select the Bellows type.

the Bellows size（nominal diameter）is selected. Here,For example, if the necessary displacement

we selected V20.

perpendicular to the axis is 5mm, the block number is 4

Catalog value of V20（1-block specification）

according to Table 1. So, the required type is V20–111–4.

Internal diameter = 20mm,

Regarding the Bellows length L, note that the length

Outside diameter = 40mm

excludes the length of fittings on both ends. In the

Expansion and contraction length = 10mm,case of V20–111–4, this length on both ends is 3mm, so

Expansion length = 15mm,

the overall length is 46mm.

Contraction length = 5mm【Step 2】Calculate‌the allowable displacement
perpendicular to the axis for each block number3-2）Displacement perpendicular to the axis
and displacement parallel with the axis

For the selected size
（nominal diameter）, the allowable

In the case where there is both displacement

displacement perpendicular to the axis is calculated

perpendicular to and parallel with the axis, as shownfor each block number.

in Section 2-2）, the calculation should be made withIn the case of 1 block:the length which becomes minimum under the effects

Expansion and contraction length = 10mm, Expansionof displacement parallel with the axis. The selection

11

Technical Papers

flow is the same as in Figure5.

applied, is 10 mm（15mm – 5mm = 10mm）.

【Step 1】Select the Bellows size（nominal diameter）

According to Equation（4）,Based on the space in which the Bellows is attached,
the Bellows size（nominal diameter）is selected. Here,
we selected V30.

In the same manner, increase the block number andCatalog value of V30（1-block specification）

calculate for each block number. Table2 shows theInternal diameter =30mm, Outside diameter =50mm

results.

Expansion and contraction length = 10mm, Expansion
length = 15mm, Contraction length = 5mm

Table2‌V30’s block number, displacement parallel with theaxis, displacement perpendicular to the axis

【Step 2】Calculate‌the allowable displacement

Block number

2

3

4

5

6

perpendicular to the axis for each block

L（mm）

20

30

40

50

60

Δ x（mm）±

5

5

5

5

5

For the selected size
（nominal diameter）, the allowable

Δ x（y）
（mm）±

5

10

15

20

25

displacement perpendicular to the axis against

Δ y（max）
（mm）

0.5

1.6

3.5

6.0

9.1

numberdisplacement parallel with the axis is calculated foreach block number. When the necessary displacement【Step3】‌Determine the type which canparallel with the axis is 10mm and the block number

accommodate the necessaryis 1, the converted displacement parallel with the axis

displacement perpendicular to the axis

In the case of 2 blocks:

Select the block number which can accommodate thenecessary displacement perpendicular to the axis andExpansion and contraction length = 20mm, Expansion

select the Bellows type.

length = 30 mm, Contraction length = 10mm

For example, if the necessary displacement

Bellows length L = 20mm; when there is only

perpendicular to the axis is 5mm, the block number isdisplacement parallel with the axis, then Δ x of up to

5 according to Table2. So, the required type is V30–±10mm is acceptable.

111–5. The Bellows length including the fitting length

When the necessary displacement parallel with theon both ends is 56mm.

axis of 10 mm is Δ x = ±5mm, Δ x（y）, the converted
value of allowable displacement parallel with the axis
when displacement perpendicular to the axis is

4. Design troubles and countermeasuresapplied, is 5mm（10mm – 5mm = 5mm）.

4-1）Investigation of Bellows’buckling

According to Equation（4）,One of the design problems of Bellows is buckling.
Bellows, like columns and springs, suffer buckling due
to compressive forces and lose straightness. Refer toIn the case of 3 blocks:

Figure2.

Expansion and contraction length = 30mm, Expansion

This buckling threshold value can be approximatedlength = 45mm, Contraction length = 15mm

by using the widely-used Euler’s formula:

Bellows length L = 30mm; when there is only…………（５）displacement parallel with the axis, Δ x of up to ±15mm
is acceptable.

where, Fcr：buckling load limit（N）

When the necessary displacement parallel with the

EI：Bellows’bending stiffness（N・mm２）

axis of 10mm is Δ x = ±5mm, Δ x（y）, the converted

L：Bellows’length

value of allowable displacement parallel with the axis
when displacement perpendicular to the axis is（mm）

Note : ‌Regarding EI, refer to Valqua Review Vol. 40,
1st issue. １）
12Summer 2018

Δ x（y）is 0. Therefore, at least 2 blocks are needed.

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Design Guidelines and Troubles for DYNAMICBELLOWSTM and Countermeasures4-2）Bellows’buckling pressure4-3）Investigation of V Series’buckling

When Bellows are subjected to internal pressure, theNext, we examine the buckling of our standard

limit pressure which triggers buckling, or buckling

product V Series（Catalog No. PC08）. Refer to thepressure, can be calculated assuming that thrustcatalog and follow the steps.

develops in response to the pressure applied to the

In the catalog, one column shows the block number

effective area of the Bellows, and that the buckling

limitation. This shows the case in which the V Series’

load is equivalent to the thrust.

allowable pressure of 0.1 MPa is outside pressure andthe case in which the allowable pressure is the…………（６）

internal pressure. Among these block numbers, when
the allowable pressure is internal pressure, the block…………（７）
where, Pcr：Bellows’buckling pressure（MPa）
2

A：Bellows’effective area（mm ）number limitation is determined based on thebuckling pressure explained in Section 4-2）.
An example of calculating the buckling pressure isshown below.

d ：Bellows’inner diameter（mm）
D：Bellows’outside diameter（mm）
In this case, the longer the Bellows length L becomes,

Example 1. According to Equation（6）when the blocknumber is 4 which is V20’s block number limit,

the smaller the buckling pressure becomes, reducing

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No.35

the strength against buckling. Therefore, whencalculating, the maximum service length（expansion

In this case, the Bellows length L is calculated using

length）should be used.

the expansion length.

When the pressure significantly exceeds this limit

The details of EI are omitted here.

pressure, the Bellows will buckle and not expand or

When the block number is 4, L is 60mm（expansion

contract, which may result in deformation, a shorter

length 15mm × 4 blocks = 60mm）.

lifespan or early breakage.
Figure6 shows a case of breakage due to buckling.

Regarding the Bellows’effective area A, according toEquation（7）,Example 2. According to Equation（6）, when theblock number is 6 which is V100’s block number limit,‌In this case, when the block number is 6, theBellows length L is 90mm（15mm × 6 blocks =90mm）.
Regarding the Bellows’effective area A, according toEquation（7）,Figure6

13

Breakage due to buckling of Bellows

Technical Papers

Based on the results of the above calculation, the【Example 2】‌When a shaft is absent on the inside of the
Bellows

inner pressure of 0.1 MPa is not greater than thebuckling pressure, so buckling does not develop.

The structure becomes more complex and accurate

When the block number which can be used for the

positioning of the guide rod is needed.

necessary displacement parallel with the axis isselected and the selected block number is greaterthan the block number limitation, the guide shown in

Bush
Block number limitation

Guide rodthe catalog should be attached. This guide is shown in
Section 4-4）below.
4-4）Countermeasures to prevent buckling
When the block number exceeds the block number
limitation, a guide should be attached. An example ofthe structure is as follows.
【Example 1】‌When a shaft is present on the inside of
the Bellows
Figure7 is a frequently used structure.
Block number limitation

Figure9

Guide rods are placed on the surroundings

Shaft

location where the block number is equal to or lessthan the block number limitation.
When displacement perpendicular to the axis ispresent, guides cannot be placed. Therefore,

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As shown above, guides should be placed at every

Guidedisplacement perpendicular to the axis which is
Lubricant coating

Figure7

A guide inner surface is coated with lubricant

greater than the block number limitation is not
acceptable.
4-5）Block number limitation using outside pressure

Block number limitation

The V Series’block number limitation is shown whenShaft

the outside pressure is 0.1 MPa. In this case, we do
not calculate the buckling pressure; we calculate the
Bellows’horizontal deflection（refer to Figure3）toprevent the horizontal deflection from exceeding
1mm, using the equation:

PTFE bush

Figure8 A PTFE bush is attached to the guide inner surface

where, y（max）：Bellows’horizontal deflection（mm）
W ：Bellows’weight（kg）
L：Bellows’length（mm）

EI ：Bellows’bending stiffness（N・mm２）
In this equation, the effects of outside pressure are
ignored. When outside pressure is applied, the
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Design Guidelines and Troubles for DYNAMICBELLOWSTM and Countermeasures

Bellows’deflection is considered to decrease.
Therefore, in this equation, we calculate values when

5. Conclusion

the Bellows inside/outside pressure is atmosphericThis article described the case of our standard

pressure.

DYNAMICBELLOWSTM V series（catalog No. PC08）
and examined the selection method when Bellows

Example 1）‌According to Equation（8）,when the blockabsorb displacement perpendicular to the axis, andnumber is 9 which is V20’s block number

design considerations regarding the block number

limit,

limitation.
As noted already, when Bellows with a length longer
than the block number limitation are used, there are‌In this case, the Bellows length L is calculated using

risks of deformation, a shorter lifespan, and breakage,

the expansion length.

so care is required.

The details of EI are omitted here.

We hope that this article will help prevent design

When the block number is 9, L is 135mm（15mm ×problems when selecting the V Series.

9 blocks = 135mm）. Also, the Bellows’weight W is
0.13kg.

6. References
１）Valqua Review Vol.39, 4th issue, Lecture on seals（23）

block number is 20 which is V210’s block

Vol.39, 5th issue, Lecture on seals（24）

number limit,

Vol.39, 6th issue, Lecture on seals（25）

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Example 2）‌According to Equation （8）, when theVol.39, 7th issue, Lecture on seals（26）
Vol.40, 1st issue, Lecture on seals（27）
Vol.40, 9th issue, Lecture on seals（28）
In this case, when the block number is 20, L is 300 mm

No.35（expansion length 15mm × 20 blocks = 300mm）andthe Bellows’weight W is 5.32kg.
4-6）‌Countermeasures to prevent horizontal
deflection
When the block number exceeds the block number
limitation, a guide should be attached. An example ofthe structure is similar to the example shown in
Section 4-4）.

Masafumi Ina
Sales Group
Technical Solution Division15