CA LC U L AT I O N S · D ES I G N · A PPL I CAT I O N S B . 3 . 1
Design calculations for
snap fit joints in plastic parts
Contents
1.
Introduction
3
2.
Requirements for snap-fit joints
3
3.
Basic types
3.1
3.2
of snap-fit joint
leg snap-fit
Barbed leg snap-fit supported
on
3.3
3.4
4.
both sides
4
4
Cylindrical snap-fit
Ball and socket snap-fit
Critical dimensions for
5
snap-fit joint
permissible undercut depth Hmax.
and maximum permissible
elongation s,^
a
6
Maximum
4.2
Elastic modulus E
10
4.3
Coefficient of friction M
Assembly angle a\ and retaining
10
Design calculations for snap-fit joints
Barbed leg snap-fit
5.1
5.2
Cylindrical snap-fit
5.3
6.
4
4.1
4.4
5.
4
Barbed
Ball and socket
snap-fit
Calculation
6.1
6.2
6.3
6.4
examples
leg snap-fit
Cylindrical snap-fit
Ball and socket snap-fit
Barbed leg snap-fit supported
Barbed
angle 2
6
11
12
12
13
14
16
16
16
18
both sides
18
7.
Demoulding ofsnap-fit joints
20
8.
Applications
21
Barbed
21
on
8.1
8.2
8.3
9.
leg snap-fit
Cylindrical snap-fit
Ball and socket snap-fit
Explanation of symbols
10. Literature
23
24
24
25
Introduction
1.
2.
Snap-fits are formfitting joints which permit great design
flexibility. All these joints basically involve a projecting
lip, thicker section, lugs or barbed legs moulded on one
part which engage in a corresponding hole, recess or
undercut in the other. During assembly, the parts are
elastically deformed. Joints may be non-detachable or
detachable, depending on design (figs. 4 and 5). Nondetachable joints can withstand permanent loading even
at high temperatures. With detachable joints, it is neces
sary
to test
in each individual
deformation which
can
case the permanent load
be permitted in the joint. In the
unloaded state, snap-fit joints are under little or no stress
and are therefore not usually leaktight. By incorporating
sealing elements, e.g. O-rings, or by using
leaktight joints can also be obtained.
an
adhesive,
Snap-fits are one of the cheapest methods of joining
plastic parts because they are easy to assemble and no
additional fastening elements are required.
Hostaform
Acetal
copolymer (POM)
Hostacom
Reinforced
polypropylene (PP)
Celanex
Polybutylene terephthalate (PBT)
@Vandar
Impact-modified
polybutylene terephthalate (PBT-HI)
lmpet
Polyethylene terephthalate (PET)
=
registered
trademark
Requirements
for snap-fit joints
Snap-fits
position.
used
fix
two
cases,
it is
parts together in a certain
important to exclude play
between the assembled parts (e. g. rattle-free joints
for automotive applications). The axial forces to be
transmitted are relatively small. In the majority of appli
cations, the joints are not subject to permanent loads
are
In
some
to
(e. g. from internal pressure).
Special fasteners such as rivets and clips also work on the
snap-fit principle. They should be easy to insert, suitable
for blind fastening, require low assembly force and be
able to bridge the tolerances of the mounting hole.
3.
Basic types
of snap-fit joint
The undercut
outside
depth H
is the difference between the
of the barb and the inside
edge
edge
of the hole
(% 1):
The parts with an undercut can be cylindrical, spherical
or barbed. There are three corresponding types of snapfit
joint:
1
Barbed
undercut
The
.
2.
3.
leg snap-fit
Cylindrical snap-fit
Ball and socket snap-fit
In
leg
is deflected
designing a
vent
by
barbed
overstressing
at
depth H
this
leg,
amount
care
therefore be
Barbed
as
large
LI
(1)
L2
during assembly.
should be taken
to
pre
point of support
radius r (fig. 1) should
the vulnerable
because of the notch effect. The
3.1
=
as
possible.
leg snap-fit
Hg.l
Barbed
3.2
Fig.
leg snap-fit supported on
both sides
3
1, Pj
T
-\
_vu
i
S
t
HJ1
T t
-1 >~
(/}
1
/R
l
spring elements supported on one or
both sides and usually pressed through holes in the
mating part (fig. 1). The hole can be rectangular, circular
or a slot. The cross-section of the barbed leg is usually
rectangular, but shapes based on round cross-sections are
also used. Here, the originally cylindrical snap-fit is
divided by one or several slots to reduce dimensional
rigidity and hence assembly force (fig. 2).
Barbed
legs
are
This
on
joint employs
a
barbed
both sides. The undercut
spring element supported
depth H is the difference
edge of the barb and the width of
(fig. 3). Hence as in formula (1) we
between the outside
the
receiving
hole
obtain:
undercut
depth
H
Lt
L2
(la)
snap-joint may be detachable or non-detachable
depending on the design of the retaining angle.
This
Fig.2
3.3
Cylindrical snap-fit
Cylindrical snap-fits consist of cylindrical parts with a
lip or thick section which engage in a corre
sponding groove, or sometimes just a simple hole in the
mating part.
moulded
Fig.
4: Non-detachable
compression ( ) of the shaft
joint
^1DG
100%
(4)
+ ~^-WO%
(5)
_==*.
^
UG
elongation (+)
of the hub
e2
,
=
AV*
i-TC
As it is
known how the undercut
depth H is appor
mating parts, it is assumed for sim
plicity that only one part undergoes a deformation e
corresponding to the whole undercut depth H.
not
tioned between the
H
-100%
s
e=^^-100%
DK
or
Dr,
Ball and socket
3.4
Fig.
5: Detachable
snap-fit
joint
Fig.
The difference between the
largest
6
diameter of the
shaft DG and the smallest diameter of the hub DK is the
undercut
depth
Ball and socket
H.
snap-fits (fig. 6)
are
mainly
used
as
motion
A ball
undercut
depth H
=
(2)
DK
DG
transmitting joints.
corresponding socket;
ence
DG largest diameter of the shaft [mm]
DK smallest diameter of the hub [mm]
between the ball diameter DG and the socket open
ing diameter DK.
undercut
The parts
are
deformed
by
the
amount
depth
H
As
are
a
can
also be described
ADC
result of these diameter
deformed
as
as
follows:
=
DG
DK
(7)
Because the shaft is solid and therefore very rigid, the
hole undercut depth H must be overcome by expanding
deformed
JDK
(3)
changes,
the shaft and hub
+
H
DG ball diameter
[mm]
DK socket opening diameter [mm]
the hub. As
=
depth
of this undercut
depth during assembly. The diameter of the shaft is
reduced by ADC, and the diameter of the hub increased
by +ZlDK.
So the undercut
or ball section engages in a
the undercut depth H is the differ
,
as
a
result of this diameter
-
elongation
change,
the hub is
follows:
e
=
DG-DK
^
jL>K
H
100%
100%
=
DK
(8)
the deformation is lower. So barbed
Critical dimensions
4.
legs
are
stressed
much less than
cylindrical snap-fits. As a result of this,
higher elongation is permissible and in many cases is
necessary for design reasons.
for a snap-fit joint
of the type of
snap-fit there is a linear relation
depth H and elongation e. The
maximum permissible undercut depth Hmax. is limited by
the specified maximum permissible elongation e^^
Irrespective
between the undercut
.
The
load-carrying capacity
of
snap-fits depends
on
non-rectangular barbed leg cross-sections, the follow
ing relationships apply between undercut depth H and
deformation e in the outer fibre region (outer fibre elon
gation):
For
the
elastic modulus E and coefficient of friction //. It can be
matched to the requirements of the joint by adjusting
undercut
angle
K2
depth
H and
assembly angle
i
or
semicircular
I2
(10)
100
r
retaining
(see section 4.4).
third of
a
circle
I2
Hmax.
cross-section
Maximum
4.1
en
Hmax.= 0.578
cross-section
maximum
permissible undercut depth Hm
permissible elongation &max.
=
en
0.580
(H)
100
r
and
quarter of a circle
cross-section
Hmax.=
0.555-^-^-
(12)
In barbed
legs (fig. 7), the following relation applies
between undercut depth H (= deflection) as a result of
deflection force FB and elongation or compression in the
outer fibre region of the barbed leg cross-section
(rectangular section):
These
A
undercut
depth Hmax.
=
-|- - ^
-
also
apply approximately to leg
comparison
(9)
depth Hmax.
sectors
of
an
annulus.
between formula 9 and formulae 10
12 shows that the maximum
for barbed
legs
permissible
to
undercut
with cross-sections in the
form of segments of a circle is 15% lower than that of
a rectangular barbed leg cross-section (assumption:
barbed
leg length
[mm]
barbed leg height
[mm]
permissible elongation [/o]
x.
relationships
cross-sections in the form of
h
-I).
The maximum
Fig.
permissible undercut depth Hmax. for
barbed legs of different length and height with a
rectangular cross-section can be read off figs. 10 to 13.
7
Kg-
9
\
'fi
Fig.
8:
Elongation
in cross-section A- A
s
'T
a
\
n t
(fig. 7)
1
The maximum
permissible undercut depth Hmax. for
leg snap-fits supported on both sides can be
calculated with the aid of fig. 14, irrespective of the
material. Fig. 14 applies for emax
6% (see calculation
example 6.4).
barbed
The maximum deformation
critical
region
A
-
(fig. 8) only applies
in the
A, fig. 7, while in other cross-sections
=
Fig.
10: Maximum
undercut
permissible
depth Hn
Fig.
12: Maximum
permissible
undercut de
for Hostacom M 4 N01 and G 3 N01
for Hostaform and Hostalen PP
a
3
o^
<^i
~13^
E
1
X
4^
n *iln*L
mm
30
M
^
20
max.
=
t
\
-%
\
S Is
A
10
Sl=15
s
s
sl
=
10
\
s
mmN.
V
S
sv
\
mm
X
4
s
2
Si
=
5
\
Maximu
\
\
\\
mi n
permis ble
s,
Vl
\
V
x=
SJ
Xj
fö
=
10
20
mrr [
X
mn
s
rn
V
15 mm\
=
\
m\
\.
Sk
s
\
1
s
x
\
pbo
30
^s
s
s
=
s
>
N
op
|5
\
:
KJ
p^
i
H
f
oj
N
S.
v
\
\
N
\
s\
3
undercut
mm
\
s
6
20
=
mm
\.
s
8
Hmax.
depth
N
s_
i"
8%
y&^
s 1=50 mm
2%
=
^|
LU u=L
~
max.
X 1= 5
4*
5
Ui
s
mm
v
X
V
\
pfo
\
S
\
o l^
I
0 5
0.8
2345
1.0
Height
Fig.
11: Maximum
of barbed
permissible
leg
0.5
6mm8
undercut
345
1.0
0.8
h
Height
depth Hn
Fig.
13: Maximum
of barbed
permissible
undercut
depth Hmax.
for Hostacom G 2 N01 and M 2 N01
for Hostaform C 9021 CV 1/30
mm
t
L
max.
=
6%
P
^\
y
s
rr
10
8
6
""
X
N
s.
^V 1
X
X
on
\
N
JV
S
=
20
\
mm
\
Sj=15mmS
X
s
X
4
30r
x\r
s
\
S
s
5
\
\
Si
=
10
X
mm
3
I
I
2
X
0.8
1.0
345
Height of barbed leg h
6mm8
5
s
X
1.5
0.8
V
S
X
mr n
s
\
N^
1
0.1
0.5
6mm 8
leg h
1.0
345
Height
of barbed
leg h
X
6mm8
The undercut
depth H is
r
,
calculated
maximum
1
s
thickness of
Fig.
maximum
14: Barbed
undercut
J~lmax.
m
^ ()
depth
Dt
(14)
DC
100
outside diameter of the shaft [mm] in cylindrical
snap-fits or ball diameter [mm] in ball and socket
snap-fits
The maximum
leg [mm]
permissible elongation (table 1) [%]
leg snap-fit supported
both
on
sides; relative undercut depth
j
spring leg thickness for
function of barb
as a
emax.
=
permissible elongation of materials with
a
yield point (e. g. Hostaform) should be about
a third of the elongation at yield stress es (fig. 15a).
For materials without a definite yield point (e. g. glass
fibre reinforced Hostacom, fig. 15b), the maximum per
missible elongation (see table 1) should be about a third
of the elongation at break SR.
definite
TT
width and
permissible
en
(-1)
barb width [mm]
length of hole [mm]
Smax.
follows:
(-!)' (>+4)
12
b
as
6%
Fig. 15a: For materials
(e. g. Hostaform)
with
a
definite
yield point
1.0
0.8
0.6
0.4
relative
spring leg thickness
y
=
0.01
0.2
es
3
El-
0.1
0.08
Fig. 15b: For materials without
point os (e. g. Hostacom)
0.06
a
0.04
|
TB
0.02
eü
0.01
0.008
0.006
IE.
0.004
Fig.
0.002
0.4
0.5
Relative barb width
With
cylindrical snap-fits
the maximum
permissible
-r
and ball and socket
snap-fits,
depth can be
permissible elongation
undercut
calculated from the maximum
emax.
16
F
^o
0.001
0.3
SR
3
(%) using the formula:
0.6
ES
'
/
definite
yield
os
Table 1 :
Maximum
permissible elongation
Material
emax.
for determination of the maximum
Maximum
Barbed
leg
permissible
permissible elongation
undercut
emax.
depth Hn
(%)
Cylindrical snap-fits,
snap-fits
ball and socket
Hostaform C 52021
Hostaform C 27021
Hostaform C 13021
Hostaform C 13031
Hostaform C 9021
Hostaform C 2521
Hostaform C 9021 K
Hostaform C 9021 M
Hostaform C 9021 TF
Hostaform T 1020
Hostaform S 9063/S 27063
Hostaform C 9021 GV 1/30
1.5
Hostaform S 9064/S 27064
10
0.8
Hostacom M2 N02
Hostacom M2 N01
Hostacom G2 N01
Hostacom M4 N01
Hostacom G2 N02
Hostacom Ml U01
Hostacom G3 N01
1.5
1.0
S 1.0
S 0.5
Vandar 4602 2
^ 3.0
^2.0
Celanex 2500
S 2.0
1.0
S 1.0
S 0.5
Hostacom M4 U01
Impet
2600 GV 1/30
Celanex 2300 GV 1/30
Elastic modulus E
4.2
The elastic modulus E0 is defined in DIN 53 457 as the
slope of the tangent to the stress-strain curve at the
origin (fig. 16,
page
Fig. 17: Secant modulus Es as a function of outer fibre
elongation (based on 3-point flexural test)
(el%/min)
8).
a
E0
at
=
the
point
e
=
0
(15)
Celanex 2300 GV 1/30
b Hostaform C 9021 GV 1/30
c
Hostacom G 3 N01
d Hostacom M 4 N01
With greater elongation, e. g. Si (fig. 16), the elastic modu
lus is smaller because of the deviation from linearity
between
The elastic modulus then
and
a
e.
corresponds
slope of a secant which is drawn from the origin
through the e\ point of the stress strain curve. This is
known as secant modulus Es and is dependent on the
magnitude of elongation e
to
the
e
Hostaform C 9021
f
Celanex 2500
g Hostacom M 2 N01
h Hostacom G 2 N01
i
Vandar4602Z
.
The
following applies:
N/mm2
,a
7500
Es
=
f(8)
S,
(16)
7000
This
snap-fits.
modulus ES is used in design calculations for
Fig. 17 plots the secant modulus against elonga
tion
to
V
\
secant
e
up
barbed
the maximum
N
6500
\\
permissible elongation for
legs.
6000
r
450(f
v
4.3
Coefficient offriction fj.
assembling snap-fits, friction has to be overcome. The
degree of friction depends on the materials used for the
mating elements, surface roughness and surface loading.
Table 2 gives coefficient of friction ranges for various
combinations of mating element materials. The friction
values quoted are guide values only.
In
w
4000
|
3500
\
3000
Table 2
1500
element materials
Coefficient of friction //
V
\
2500
2000
Mating
\
\\
vV
S3
"V
Vs sf__ .ü!^
_\ sjX
^ ^1\
^S
1000
[*-_
1
'
^^^
---^
-
^s
l"**
Hostaform/Hostaform
0.2
to
0.3
Hostaform/other
0.2
to
0.3
Hostaform/steel
0.1
to
0.2
Hostacom/Hostacom
0.4
Hostacom/other
plastics
plastics
to
0.4
0.2
to
0.3
Impet/Impet
Impet/other plastics
Impet/steel
0.2
to
0.3
0.2
to
0.3
0.1
to
0.2
Vandar/Vandar
0.3
to
0.4
Vandar/other
0.2
to
0.3
Vandar/steel
0.2
to
0.3
Celanex/Celanex
0.2
to
0.3
Celanex/other
0.2
to
0.3
0.1
to
0.2
Celanex/steel
10
plastics
-
500
1
1
calculation
1
example
3456
0.3
plastics
*-
T
0
Hostacom/steel
=
25*^.
Elongation
E
6.2
4.4
The
Assembly angle
assembly angle
a\
at
and
(figs.
retaining angle
18 and
a2
Fig.
18: Detachable
Fig.
19: Non-detachable
Fig.
20
F]
assembly force required
joint
19), along with the
barb dimensions and coefficient of friction fj, between
the mating elements (table 2), determines the required
force F,
(fig. 20). The greater a\ the higher the
required. With a large assembly angle
and
(! ä 45)
high coefficient of friction //, it may no
longer be possible for parts to be assembled. The barb
then shears off rather than being deflected. The recom
mended assembly angle for barbed legs and cylindrical
15 to 30.
snap-fits is i
assembly
assembly
force
joint for
2
=
90
=
With ball and socket
snap-fits, the assembly angle cannot
freely chosen. It depends on the maximum permissible
socket opening diameter DK (fig. 27).
be
The
retaining angle 2 (figs. 18 and 19) decides how much
loading the joint can stand. The maximum load-bearing
capacity is reached when the retaining angle is a2 90
(fig. 19). During long-term loading and/or in the event
of elevated ambient temperatures, the retaining angle 2
should always be 90. The joint is then permanent. For
detachable joints, a retaining angle 2
45 should be
provided, preferably a2 30 to 45.
=
=
=
=
11
Barbed
5.1
Design calculations
for snap-fit joints
5.
Fig.
leg snap-fit
22
The
load-bearing capacity of snap-fits under steady
(short-term) stress depends primarily on:
1
.
the mechanical
particularly
properties of the plastics concerned,
as expressed by the elastic
stiffness
modulus ES,
2. the
design of the snap-fit, i. e. wall thickness,
depth H, retaining angle 2-
undercut
Load-bearing capacity is
which the
assembly
joint
can
defined
pull-out force F2
opposite direction to
separating.
as
the
stand in the
without the parts
possible to design the direction of
snap-fit assembly right angles to the actual loading
direction F during service (fig. 21). Then the load-bearing
capacity of the joint is not determined by pull-out
force F2 but by the break resistance or shear strength of
the vulnerable cross-section. This design technique is
most often used with ball and socket snap-fits.
In many cases, it is
at
The
assembly force FI and pull-out force F2 (fig. 22)
legs can be calculated from the formula:
3H
Fl,2
secant
ft
i
2
The factor
fig.
tr
// +
1
1
Ii2
[N]
(17)
jM-tan!^
depth
modulus
-^
^
can
be taken
l-w-tanai.2
from
directly
'
23.
Fig.
V- +
c
i-i
23:
Factor -r^
1
\JL
tan
gl,2
tan
1,2
(from formulae 17, 22 and 25)
assembly/retaining angle i, 2
j
tan
[mm]
[N/mm2] (Fig. 17)
moment of inertia [mm4] (table 3)
barbed leg length [mm]
coefficient of friction (table 2)
assembly angle
[]
retaining angle
[]
Es
1
r
ES -J
=
undercut
H
J
Fig. 21
for
barbed
as a
function of
Table 3
Barbed
leg
Moment of inertia
cross-
[mm4]
section
^\1
xNSN
rectangle
semicircle
b'h3
b
i,p.
wnere
|2
^
0.110 r4
|>_
0.0522 r4
circle
quarter
of a circle
^_
0.0508 r4
_j
h
r
leg width [mm]
leg height [mm]
radius [mm]
third
of
12
a
15
30
45
90
60
Assembly/retaining angle
t
,
2
retaining angle a2 90, the pull-out force F2
is determined by the shear-stressed area and the
shear strength TB of the plastic used.
With the
=
Fig.
24
Table 4
Ultimate tensile
Material
strength OR and
strength OB
tensile
The shear
stress
TS is
[N/mm2]*
Hostaform C 52021
65
Hostaform C 27021
64
Hostaform C 13021
65
Hostaform C 13031
71
Hostaform C 9021
64
Hostaform C 2521
62
Hostaform C 9021 K
62
Hostaform C 9021 M
64
Hostaform C 9021 TF
49
Hostaform T 1020
=
A
[N/mm2]
or
F2max.
110
Hostaform S 27063
50
Hostaform S 9063
53
Hostaform S 27064
42
Hostaform S 9064
42
5.2
Hostacom M2 N02
19
Hostacom M2 N01
33
Hostacom M4 N01
33
Hostacom G2 N01
32
Hostacom G2 N02
70
Hostacom G3 N01
80
Hostacom Ml U01
36
Hostacom M4 U01
33
Celanex 2500
(18)
Taking into account ultimate tensile strength OR or
tensile strength 0B (table 4), the following holds true
shear strength
Fig.
=
for
TB
=
0.6
CTR
(19)
TB
=
0.6
OB
(20)
=
b
64
Hostaform C 9021 GV 1/30
A TB
c
rB
[N]
(21)
Cylindrical snap-fit
25
65
Celanex 2300 GV 1/30
150
Celanex 2300 GV 3/30
50
Vandar 4602 Z
40
Impet
Ts
2600 GV 1/30
165
assembly force FI and pull-out force F2 for cylin
snap-fits unlike for barbed legs can only be
roughly estimated. This is because the length a (fig. 25)
which is deformed during assembly of the parts with
consequent increase in assembly force FI is unknown.
The length a depends on both the wall thickness of the
hub and the undercut depth H. A useful guide to a has
proved to be twice the width b of the moulded lip.
The
drical
I Test
specimen injection
moulded
according to
DIN 16770 part 2.
-
-
13
The
assembly
force FI and
pull-out
force F2
can
be calcu
Fig.
lated from the formula:
ratio
M+
->u
r>
c
Fu-p.rt.IV2b
1,2
rxn
[N]
'
p
joint
DG
b
outside diameter of the hub [mm]
width of the moulded lip [mm]
fj.
coefficient of friction
pressure
2
y
J
H
pressure p, the
s
o
1
[N/mm2]
^-Es-^
smallest diameter of the hub
DK
DK
(table 2)
depth H and joint
following relationship applies:
=
function of diameter
[N/mm2]
Between undercut
p
as a
(22)
assembly angle []
retaining angle []
ai
K
or
DG
tan
f_
r
Geometry factor
DG
Ji
26:
(23)
[mm]
1.2
The geometry factor K
depends
on
1.5
the dimensions of the
Diameter ratio
a
_
L>G
snap-fit:
mv +
K=
VDGj
1
Fig.
+1
or -pj
UK
27
(24)
fuy-i
loj
outside diameter of the hub
Da
DG
[mm]
outside diameter of the shaft [mm]
Here it is assumed that the whole undercut
accommodated
by expansion
depth
H is
of the hub. With thin-
walled shafts, the shaft deforms as well but this can be
ignored in the case described here. Fig. 26 shows the geo
metry factor K
as a
function of the diameter ratio Da/Dc.
Ball and socket
5.3
snap-fit
In this
design (fig. 27), the assembly angle j and retain
ing angle 2 and hence assembly force FI and pull-out
force F2
The
and
14
are
the
Table 5
same.
assembly/retaining angle is between 8 (e
16 (e
4%), depending on elongation.
=
1%)
=
=
-^100%
UK
Assembly angle a\
Retaining angle 2
a
DG
1
8
0.07
2
11.4
0.10
3
13.9
0.12
4
15.9
0.14
assembly or pull-out force,
cylindrical snap-fits are used:
To estimate
Fi
=
T-.
F2
T^2
=
p
n
Dé
a
fs~
DG
'
ß + tan
T
i
l jM-tana
the formulae for
joint
pressure
relationship
pressure p
r-Nn
[N]
[N/mm2]
DG ball diameter [mm]
f deformation length divided by the
a
DG l ball diameter (table 5)
coefficient of friction (table 2)
H
a.
assembly or retaining angle [] (table 5)
p
The
can
(25)
P
=
by
the
rJ'Es'T
H
undercut
DK
ES
socket
K
geometry factor
secant
depth H and joint
following formula (23):
between undercut
be described
tN/mm2]
depth [mm]
opening diameter [mm]
modulus [N/mm2] (fig. 17)
my
bJ
(26)
+ i
K=
+1
fAiiDj 1
15
Calculation
6.
b) Assembly force FI
examples
For the
Barbed
6.1
leg snap-fit
P
assembly
3H
force FI formula
ES J
_
(17) applies:
// + tani
I3
\-fjL-
tan 0.1
The top and bottom
plates of a time switch are to be
detachably joined by two diagonally opposite spacers
and two barbed legs. The hole diameter in the top plate
is DK
8 mm. The pull-out force F2 required per barbed
leg is 50 N. The barbed legs are to be injection moulded
For the Hostaform/steel
from Hostaform C 9021 and will have
that the friction coefficient fi
H
=
0.3
ES
=
2800 N/mm2
mm
(fig. 17).
=
cross-section
a
slotted circular
=
(fig. 28).
Using table
Fig.
mating elements, it is assumed
0.2 (table 2).
28
3
we
obtain for the semicircular
cross
section:
J
So
assembly
Each
=
to
=
28.2 mm4
out as
3-0.3-2800-28.2
0.2 + 0.577
153
1
0.2
-
0.577
18.5 N
securing
each have
44
0.11
=
force FI works
F,=
FI
0.110 r4
=
element
comprises
be deflected
element is therefore 2
by
FI
=
barbed
legs which
assembly force per
two
H. The
37 N.
c) Pull-out force F2
The
pull-out force F2 is calculated in the same way as
45 is substituted for
assembly force except that 2
The pull-out force is thus
=
a)
What should the dimensions of the barbed
What assembly force FI is required ?
b)
c)
What
pull-out
force F2 is obtained?
a) The maximum permissible
chosen
leg be?
outer
F2
fibre
elongation
1 % For the semicircular crossto be emax.
section, the following applies using formula (10):
=
H
=
is
2
31.6 N
pull-out
63
a
force of
pull-out
N, which is greater than the required
force of 50 N.
0.578-^-smax.
Cylindrical snap-fit
6.2
The
1 is chosen
=
31.6N
Each element withstands
r--^-
H
=
0.578
to
be 15
body
of
a
rubber-tyred
roller is
to
be made in
-0.01
that the roller bears
is used
as
directly
onto a
steel axle, Hostaform
the construction material.
H= 0.3mm
a) What should the dimensions of the snap-fit be
(undercut depth H) ?
The diameter of the undercut is calculated from
+
2H
=
1 mm, the
angle
16
8.6
mm.
The slot width is chosen
assembly angle
a2 45.
a\ 30
and the
two
parts which are permanently joined together (fig. 29).
Because of the relatively high stress involved and the fact
mm
4
DK
a\.
to
be
retaining
b) What assembly
force FI is
required?
c) What is the pull-out force F2?
a) Maximum permissible undercut depth Hmax.
To determine the maximum
permissible
undercut
b
H
=
2
tan
2
-tan
depth
H
Hmax., it is assumed that only the hub is deformed.
The greatest elongation takes place at the diameter DK
16 mm.
which is expanded during assembly to DG
is 6max.
Fig.
=
permissible elongation
4%, according
to
30
0.64
=
The maximum
i
2-0.577
for Hostaform
table 1.
b
=
0.55
mm
29
The
joint
pressure p is calculated from formula
H
<f^
K
7r*
Qil
^:
0
^
r
c
1
E*-i
P"W
F
(23).
1
With
1Q?
The
^^r
I"
L>G
=
secant
=
fig.
1.5
modulus for
(fig. 17)
3S^
=
=
emax.
a
is Es
26 shows
a
value for K of 3.6.
4% for Hostaform
1800 N/mm2.
\
~B
So the
joint
pressure works
out as
p-0.04.Jff
p
The
=
assembly force FI
20 N/mm2
is
0.2 + 0.577
Fi
permissible undercut depth
according to Formula (14):
So the maximum
calculated
max.
TT
can
=
20-yt-16-2 -0.55
1-0.2-0.577
be
FI
=
970.8 N
p\
_
100
c) Pull-out force F2
16
~
100
Hmax.
=
0.64
DK
=
DG-H
=
16
=
15.36
DK
The diameter DK is chosen
assembly
F!
=
mm
-
According
mm
be 15.4
mm.
JT
(22) applies:
fj, +
2b
DG
1
fj.
to
formula
TB
OB
force FI, formula
p
=
0.64
force FI
b) Required assembly
For the
to
90, the joint is perma
retaining angle 2
nent. The force required to separate the mating elements
can be calculated from the shear strength rB and the
shear-stressed area A (shear surface).
Because the
tan !
=
(20) the shear strength
=
assembly angle
a\
is 30.
for Hostaform/Hostaform
OB
62 N/mm2 e.g. for Hostaform C 2521
TB
=
0.6
TB
=
37.2 N/mm2
(table 4)
62
tan !
The shear surface in this
The
0.6
is
case
is
The coefficient of friction
mating
elements is asumed
to
A
be /A
0.2 (table 2). The width b of the undercut can be
determined from the assembly angle a\ and the undercut
=
it
DG b
=
n
16
=
27.6 mm2
=
depth
H.
A
0.55
17
So
using
formula
(21), the pull-put force
F2
=
max.
=
A
ist:
b) Assembly
For
TB
=
=
1
27.6-37.2
=
=
pull-out
%, table 5 gives
The deformation
~-
F2max.
e
force FI
0.07
1027N
to
8.
retaining angle of 2
ball
the
diameter
is
by
=
divided
length
according
a
force F2
table 5.
For Hostacom/Hostacom the coefficient of friction is
Ball and socket
6.3
In
a
car,
the
mitted via
snap-fit
movement
of the accelerator
pedal
is
linkage
joint connecting the pedal
the
to
=
0.4
(table 2).
D
trans
the carburettor. A ball and socket
to
a
(JL
For
14
=
Y^r"
JLG
~5~~
1-75 for K
=
linkage (fig. 30) and
required to have a
made from Hostacom G 3 N 01 is
=
(26).
my + 1
force F2 of at least 100 N. The ball diameter
8 mm, the outside diameter Da
14 mm.
pull-out
DG
formula
using
0
VDj
K=
+1
AY.
=
loj
Fig.
30
14
(
V
+ 1
\7.92J
-+1
P1_Y_
\7.92j
K=2.94
17 the
According to fig.
G3N01fore
=
joint
a)
How
large
should the socket
opening
4400 N/mm2.
=
can
pressure
from formula
modulus of Hostacom
l%is
Es
The
secant
be calculated with H
What
assembly
force F]
or
pull-out
DG
DK
(23):
diameter DK be?
[N/mm2]
P=D~'Es'"K
b)
=
force F2 is
0.1
obtained?
1
4400-
'
7.92
=
p
a) Socket opening diameter DK
According to table 1 the maximum permissible
gation for Hostacom G3 N01 is emax. 1%.
The
elon
assembly
or
2.94
18.89 N/mm2
pull-out force
is then
(formula 25):
=
T
U +
fJ.
a_
A.
T^-f
tan
L
^p-^Dë-g:-^fc'
//tan
1
Thus
using
formula
(8)
DG
e
~
DK
_
=
-
=
18.89 -;r-82- 0.07-
a
0.4 + 0.14
1-0.4-0.14
100%
DK
Fi.2
DK
=
152 N
DG
=
r>
+ 1
100
Barbed
6.4
DK
leg snap-fit supported on
=
0.01 + 1
The
housing
halves of
box-shaped moulding made
be non-detachably joined
barbed leg snap-fits supported on both sides (fig. 31).
two
from Hostacom M2 N01
DK
18
both sides
=
7.92
mm
by
2
a
are to
For
an
assumed
31
Fig.
a
spring
element thickness of
3
s
/
"T"
*-b-
1
f
=
3 mm,
=
"örf
0-15 is obtained.
El
1
\\
V)
KP
s
element thickness ratio of
1
f
spring
\ \
k
\
\, l
J
ir
With the aid of
,
fig. 14,
an
undercut ratio of
p_r
=
-p
0.019
is determined.
1
1
1
What should the dimensions of the
The
receiving
holes in the
The maximum
to
1
u
1
The undercut H of the barb is then calculated from
H
snap-fit joints
moulding
are
permissible elongation
1
=
emax.
20
be?
H
=
0.019
1
=
0.019
20
0.4
mm
mm.
according
table 1 is
6max.
=
6%
Note:
The width of the barb is assumed
This
gives
a
to
barb width ratio of
--04'4
~
1
~
20
be b
=
8
mm.
possible flow line in the region of the spring element
provide a weak point. By increasing wall thickness
at this point, design strength can be improved (see also
C.3.4 Guidelines for the design of mouldings in engineer
ing plastics, p. 25, no. 18).
A
could
19
With
Demoulding of
snap-fit joints
7.
cylindrical snap-fits,
a tubular part under com
than
under
is
tension. The hub of a
pression greater
easier
is
to demould than the
snap-fit (fig. 32a) generally
parting line of the mould can
run through an undercut edge, for example with a
through hole and inwardly projecting lip (fig. 32a) or
with an outwardly projecting lip (fig. 32b).
shaft. In
The undercut
has
to
on
which the effect of the
snap-fit depends
injection moulding. The im
whether the parts can be directly
be demoulded after
portant question here is
demoulded or whether it is necessary to bed the under
cut in slides, followers or collapsible cores.
In the
some
more
inner and
There is
this. The maximum per
missible deformation values quoted in table 1 can of
course
ing.
no
be
general
answer to
well
applied equally
usually arise
Problems
demould-
parts during
from the introduction of
to
deformation forces into the component. These can result
in local stretching of the part or cause the ejector to press
into the part, among other undesirable consequences.
A disadvantage here is that the demoulding temperature
considerably above room temperature
correspondingly low.
is
it should be remembered that
the dimensional stiffness of
and hence
cases,
frequent
outer
the
case
of
a
blind hole
faces of the undercut
(fig. 33),
must
be demoulded
in succession. When the mould has
cylinder
It takes
1 is
core
opened (A), the
mould cavity by ejector
pressed out of the
2 along with it until
stop
4 is
reached
32
mate
Fig.
_
A
\
split core
^*
\\\\\\\\\\
plastic
20
33
plastic part
<
part
(B).
Through further movement of the ejector, the cylinder
is stripped from the core. Expansion of the hub by an
amount corresponding to undercut depth is not pre
vented (C).
rial stiffness is
Fig.
the
3.
Photo 2 shows Hostaform fasteners which
Applications
8.
facilitate
assembly, particularly
Nos. 1, 2 and 3
Barbed
8.1
leg snap-fit
No. 4 is
a
are
used
cable holder
to
as
in
mass
considerably
production.
fix interior trim in
used in
washing
cars.
machines and
dishwashers. No. 5 is
Photo 1 shows
of the
mability
examples of snap-fits in which the deforcylindrical snap-fit has been increased by
of slots. In the top half of the picture there are
rollers with Hostaform bearings for dishwashers.
means
two
In the left
f
roller, each barbed leg is deflected by
a clip with a similar function. Here
snap-fit is secured by driving a pin into the hollow
shank (expanding rivet). The clips for fixing car exterior
trim (no. 6) work on the same principle. No. 7 shows
the hinge fixing for a detergent dispenser tray flap on a
washing machine.
the
during assembly. With a barbed leg length
2.5 mm, the
a barbed leg height of h
maximum elongation at the vulnerable cross-section of
the leg support point is:
=
of 1
0.75
=
7
mm
mm
and
=1
The lower half of the
=
f-h
=
picture
0.058
=
5.f
shows how
a
Hostaform
bush is fixed. The bush is secured
bearing
end by a
axially at one
leg and at the other by a flange.
bush is prevented by flattening off the
barbed
Rotation of the
flange.
In all the
i
=
45
examples shown, the assembly angle
the retaining angle 2
90 and the joints
=
,
are
non-detachable.
Photo 1
Photo 2
21
photo 3 another application from the automotive
industry is shown. This is a Hostaform plug box which
snap-fits into the fascia panel. The part is made in two
symmetrical halves which are inserted into each other.
In
Photo 4 shows
lid, which
Photo 3
22
a
Hostaform release lever for
is secured
by
two
pairs
of barbed
a car
Photo 5 shows that
non-cylindrical housing parts can
legs. This air filter intake is
made from Hostacom G2 N01 In assembling the two
halves, the barbed legs are not deflected but the mount
ing holes are elastically deformed.
also be
joined by
barbed
.
boot
legs.
Photo 5
Cylindrical snap-fit
8.2
Photo 7 shows
design
Photo 6 shows
pneumatic positioning device for con
trolling the flaps in air conditioning systems. The two
Hostaform halves are snap-fitted together, thereby at the
same time forming a seal by means of an O-ring. The
operating pressure is 0.2 to 0.8 bar. The undercut depth
is H
86.5
84
2.5 mm. Owing to the different wall
thickness of the shaft and hub, the hub is extended more
than the shaft during assembly. The diameter difference
is apportioned between 1.56 mm expansion of the hub
and 0.94 mm compression of the shaft.
=
a
=
to
the
carburettor
secured
by
adjuster for a
previous example.
an
vacuum.
the
Here,
too,
car,
which is similar in
It is controlled
a
snap-fit joint connecting
The
by the
diaphragm is
rubber
the
two
halves.
assembly diameter is DG 60.8 mm and the under
1.6 mm. Assuming that during assembly
cut depth H
only the hub is expanded, the maximum permissible
elongation is
=
=
1.6
e
The
100%
=
assembly angle
45 .
angle 2
is i
=
45
=
2.6%.
and the
retaining
=
Photo 6
Photo 7
23
8.3
Ball and socket
snap-fit
9.
Photo 8 shows parts of a carburettor linkage made from
Hostaform. The ball, with a diameter of DG
7.8 mm,
Explanation of symbols
=
bears in
a
socket with
a
diameter of 7.85
mm.
feature of this
circular but
corresponds
The
special
design is the socket opening which is
elliptical. The major axis of the ellipse
to
Symbol
Unit
Explanation
A
mm
area
a
mm
deformation
not
the ball diameter DG
7.8 mm, the
in
In
this
mm
direction, the dia
length.
=
socket
minor axis is 7.5
meter
difference is
b
mm
length (ball
snap-fit
barb width
supported
H
=
7.8
-
7.5
mm
=
0.3
(barbed leg snap-fit
both sides)
on
mm
that this diameter difference is
spread evenly
around the circumference, during assembly the parts will
be expanded by
Assuming
and
Da
mm
Dr
mm
mm
outside diameter of hub
largest diameter of the
(cylindrical snap-fit)
ball diameter
(ball
shaft
and socket
snap-fit)
0.3
e
100
=
=
2%.
DK
mm
2-7.5
smallest diameter of the hub
(cylindrical snap-fit)
mm
socket diameter
(ball
and socket
snap-fit)
modulus
Es
N/mm2
secant
F,
N
assembly
F2
N
pull-out
h
mm
barbed
H
mm
undercut
J~Mnax.
mm
maximum
Photo 8
(fig. 17)
force
force
leg height
depth
permissible
undercut
depth
mnr
moment
of inertia
geometry factor
mm
mm
(fig. 26)
difference between outside
of
mm
(table 3)
and inside
leg
barbed
edge
leg length
length of receiving hole
(barbed leg snap-fit supported
on both sides)
H
N/mm2
mm
joint pressure p
=
Es
^=
J_
K
wall thickness
o
assembly angle
o
retaining angle
%
elongation
%
maximum
%/min
rate
of
permissible elongation
elongation
coefficient of friction
24
edge
of hole
N/mm2
tensile
N/mm2
ultimate tensile
N/mm2
shear
(table 2)
strength (table 4)
strength
strength (table 4)
10. Literature
Engineering plastics
Design Calculations Applications
[1]
H. Schmidt:
[2]
K.
VDI-Z, No. 5, 1972
Oberbach, D. Schauf: Schnappverbindungen
aus
[3]
Fügen durch Schnappverbindungen,
Kunststoff, Verbindungstechnik,
and 8, 1977
W. W. Chow: Snap-fit
Modern Plastics
design
Nos. 6, 7
Publications
A.
concepts.
International, August
1977
so
far in this series:
Engineering plastics
A. 1.1 Grades and properties
A. 1.2 Grades and properties
A. 1.4 Grades and properties
A. 1.5 Grades and properties
Vandar, Impet
A.2.1 Calculation
A.2.2
B.
-
-
-
Hostaform
Hostacom
Hostalen GUR
Celanex,
Characteristic values and
examples
Hostacom
calculation
-
principles
Hostaform
calculation
A.2.3
-
-
Characteristic values and
examples
Design of technical mouldings
B.I.I Spur gears with gearwheels made from
Hostaform, Celanex and Hostalen GUR
B.2.2 Worm gears with
worm
wheels made from
Hostaform
B.3.1
B.3.2
B.3.3
B.3.4
B.3.5
B.3.7
Design calculations for snap-fit joints in
plastic parts
Fastening with metal screws
Plastic parts with integrally moulded threads
Design calculations for press-fit joints
Integral hinges in engineering plastics
Ultrasonic welding and assembly of
engineering plastics
of technical mouldings
Indirectly heated,
system
conductive
torpedo
thermally
Hot runner system
Indirectly heated,
thermally conductive torpedo
Design principles and examples of moulds
for processing Hostaform
Machining Hostaform
Design of mouldings made from
engineering plastics
Guidelines for the design of mouldings
in engineering plastics
Outsert moulding with Hostaform
C. Production
C.2.1 Hot
C.2.2
C.3.1
C.3.3
C.3.4
C.3.5
runner
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-
25
World-Class Engineering Polymers
■ Celanex® thermoplastic polyester (PBT)
■ Celcon® and Hostaform® acetal copolymer (POM)
■ Celstran® and Compel® long fiber
reinforced thermoplastics (LFRT)
■ Fortron® polyphenylene sulfide (PPS)
■ GUR® ultra-high molecular
weight polyethylene (UHMW-PE)
■ Impet® thermoplastic polyester (PET)
■ Riteflex® thermoplastic polyester elastomer (TPC-ET)
■ Vandar® thermoplastic polyester alloy (PBT)
■ Vectra® liquid crystal polymer (LCP)
NOTICE TO USERS: To the best of our knowledge, the information
contained in this publication is accurate; however, we do not assume any
liability whatsoever for the accuracy and completeness of such information.
Any values shown are based on testing of laboratory test specimens and
represent data that fall within the standard range of properties for natural
material. Colorants or other additives may cause significant variations
in data values. Any determination of the suitability of this material for
any use contemplated by the users and the manner of such use is the sole
responsibility of the users, who must assure themselves that the material
subsequently processed meets the needs of their particular product or
use, and part design for any use contemplated by the user is the sole
responsibility of the user. The user must verify that the material, as
subsequently processed, meets the requirements of the particular product
or use. It is the sole responsibility of the users to investigate whether any
existing patents are infringed by the use of the materials mentioned in this
publication.
Please consult the nearest Ticona Sales Office, or call the numbers listed
above for additional technical information. Call Customer Services for
the appropriate Materials Safety Data Sheets (MSDS) before attempting to
process our products. Ticona engineering polymers are not intended for
use in medical or dental implants.
Except as otherwise noted, all of the trademarks referenced herein are
owned by Ticona or its affiliates. Fortron is a registered trademark of
Fortron Industries LLC.
© 2009
Ticona
Contact Information
Americas
Ticona Engineering Polymers
Product Information Service
8040 Dixie Highway
Florence, KY 41042
USA
Tel.: +1-800-833-4882
Tel.: +1-859-372-3244
Customer Service
Tel.: +1-800-526-4960
Tel.: +1-859-372-3214
Fax: +1-859-372-3125
email: prodinfo@ticona.com
Europe
Ticona GmbH
Information Service
Professor-Staudinger-Straße
65451 Kelsterbach
Germany
Tel.: +49 (0)180-584 2662 (Germany)*
+49 (0)69-305 16299 (Europe)**
Fax: +49 (0)180-202 1202
See example below for rate information:
* 0.14 €/min + local landline rates
**0.06 €/call + local landline rates
email: infoservice@ticona.de
Ticona on the web: www.ticona.com
Asia
Celanese (China) Holding Co., Ltd.
3F, China Development Bank Tower
500 South Pu Dong Road
Shanghai, 200120
P.R. China
Customer Service
Tel.: +86-21-3861 9266
Fax: +86-21-3861 9599
email: infohelp@ticona.com
www.ticona.cn