DE8427945U1 - Bumpers for all types of vehicles - Google Patents

Description

Jüttemann, Friealielm, 5030 Hürth.

1 spring design for bumpers of motor vehicles of any kind.

2.1 Motor vehicle engineering

2.2 Previous design of the bumpers:

f 5 Bumpers are designed to protect the bodywork from damage

protect against minor impacts and prevent two vehicles from getting caught when parking. They should also enable a vehicle to be pushed in an emergency.

10 They also have a certain protective function in pedestrian accidents.

This definition shows that they have a function and are not a stylistic element. -s Unfortunately, the forms used today are

15 not very suitable because they are mounted too close to the body panels and are too weak,

j to absorb shocks.

I Bumper mounting (height):

I A uniform altitude is a prerequisite for

f 20 the protective function.

This requirement of existing standards is often disregarded,

I because the vehicles are too different in size and

• Spring detection.

[ According to current status and ISO standard

) 25 2958/73 a reference height ( level ) of 445mm applies

above ground for operating weight empty and with

&igr;: &igr; ) 3 persons payload of 750 N each for 4- to 5-seater

vehicles (after agreement with the USA probably 330/350 mm) 30 From 1980, compliance with this standard will be

. ECE - draft required. The problem lies in the

; Compliance with the reference levels for both load

f cases. It is only possible through very high (wide)

Bumpers and / or vertically arranged over-

I 35 covering elements (horns), whose function

\l is unsafe, especially in the case of diagonal impact.

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Design and strength

In the event of an impact (pendulum impact) at the front, rear and corners, no permanent damage should occur, nor should the adjustment of the headlights or the functions of the hood and trunk locks be impaired. In addition, the contact surfaces should be provided with rubber or plastic protection.

These conditions can only be met if the impact energy is absorbed not only by the bumper itself but also by its fastening (plastically deformable supports). For this reason, the bumper must have sufficient clearance (minimum clearance 70 mm) from, or through appropriate design of the adjoining body parts to allow for displacement. Hydraulic dampers can be provided as fastening elements, as used for USA &tgr; vehicles.

find a way; easier and cheaper, however, are ver- j

malleable sheets or foam-filled plastic parts that are easily replaceable (see 2.6). Figure 1 shows a schematic of such a bumper including the fastening element.

(Design proposal for bumpers with "limited energy absorption" (ISO 2958 or FM VSS 215).) The question arises whether it is right at all to stick to the existing concept of a bumper ^- or to modify the bodywork instead.

especially at the front so that it better meets the requirements of a small collision as well as pedestrian protection. Especially in |

In the USA, such deformable plastic front parts have already been implemented. (Figure 2: Plastically deformable front section on a steel support).

I chose the solution shown in 2.6.

in

2.3 When two motor vehicles collide, the deformation should be "transformed" from a plastic impact into an elastic impact.

2.4 The above-mentioned subject matter of the invention can be used commercially in every branch of the automobile industry.

2.5 The invention serves to increase the safety of the vehicle and its passengers, as well as to reduce accident costs and damage expenses for insurance companies.

With conventional bumper attachment, rear-end collisions and head-on collisions result in significant deformation of the bodywork*. This leads to serious personal injuries.

2.6 The following applies in principle to energy absorption by deformation:

When hitting a wall, the kinetic energy of the vehicle

Δε = lmv ² _Q

m = vehicle mass ( U = 1 Nm )

V ₀ = impact speed

practically only absorbed by the vehicle, namely through kinetic and elastic deformations as well as through plastic deformations (strain work).
Expressed mathematically:

Eq.1 Delta E = Imv* = Σ( E _{elast> + Eplaste} ) ( J ) During an accident impact, the proportion of elastic energy absorption is small. It is mainly plastic deformation that occurs, which represents the main energy absorption.
Note: The right term of Eq.1 should be reduced by the springs to Delta E , .

the, i.e. elimination of Delta E

4 · Λ i *

IV

The following should be noted;

Def.: Collision processes are characterized by the fact that/ compared to the movement process before and after the collision, very large forces act on the bodies involved in the collision during a very short time and significantly change the state of motion of the bodies. S- and S_ denote the centers of gravity of the bodies.

The collision is called central if the collision normal passes through the centers of gravity of the two bodies. If this is not the case, it is called an eccentric collision. Furthermore, a distinction is made between straight and oblique collisions, depending on whether V/ the velocity vectors of the contact points

the body will coincide with the collision normal or not.

When impacted, bodies are deformed. If the deformations disappear completely after the impact so that the bodies return to their original shape, the impact is called elastic. If the deformations do not disappear, the deformations are called plastic deformations. Straight central impact:

From what has been said it follows;

A straight central impact cannot initiate a rotational, ·. movement.

Furthermore, the following applies to elastic collision:

The two colliding bodies form a closed mechanical system, and the momentum and energy conservation laws apply. It is:

PA ^=P E ^=P 0 ^=COnSt · '" ^E kA ^=E kE ^=E k0 ^=COnSt · In elastic collisions, two collision sections are distinguished:

First shock section

Due to the forces occurring at the point of contact, which are equal and opposite according to the reaction axiom, body 2 is accelerated and body 1 is decelerated. Both bodies are deformed by the contact and mass forces. At the end of the first impact section, the deformations are greatest and the two bodies have the same weight.

150 speed V ₁ =v _n =v

1p 2p &rgr;

Second shock section

The deformations disappear partially or completely. Body 2 is further accelerated and body 1 is further decelerated. The second collision is completed when the bodies no longer touch each other.

Their speed after the collision is denoted by V _1E and v _2E for elastic collision. If the deformations do not decrease after the first collision phase (plastic collision), the second collision phase is omitted and the two bodies move after the

first collision section with the common velocity &ngr;.

Application of the law of conservation of momentum:

It be
Pp=P α ^=η0 =const . and P O m.. &ngr; 1 ^{+m2 V2 V2A} =Φ P ^=rn 1 ^V p ^{+ m} 2 V ^(m _{1+ m2} ) V
P ⁼ P; \ ^=P O and thus
Eq. 2 ^V P Butt ^{m1 V1A} + ITIp ^m1 +in., m + m

Furthermore, the elastic collision is: ^V 1A- ^V p ^=V p" ^V 1E ^{; V} 2A- ^V p ⁼ V ^V 2E

and subtraction

^V1A - ^{V2A =} V2F - ^V1E

i.e. the relative velocities to each other are inversely equal before and after the elastic collision.

At the end of the first collision section, the kinetic energy of the body system is

^v2

^E kp ^{= (m} 1 ^+m 2 ⁾ 2

According to the law of conservation of energy, the C 180 difference from the kinetic energy E, . = E,-

before the collision and the kinetic energy E, at

end of the first collision section is equal to the maximum elastic energy E',

"Max

- 1 ^(m 1 ^+m 2 ^)v p

^with \o - \ »I*? * \ Vl «nd v _p

₃ . l( _m ν ² + m ν ² - (m n. ) ~lIl ^A i- ² ! ^2A i-Pmax 2 ^{( 1 1A 2 2A C 1 2>} <"&Iacgr;^+&Igr;&eegr; 2)

as you can easily see

^{1m 1m} 2

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Plastic impact

In the case of a plastic collision between two bodies, the law of conservation of momentum applies as in the case of an elastic collision, but not the law of conservation of energy. The changes in shape do not decrease in the case of a plastic collision, i.e. compared to the elastic collision, the second collision phase is eliminated in the case of a plastic collision. After the plastic collision, the two bodies move at the same speed v. This can be calculated using equation 2, which follows from the law of conservation of momentum.

^m 1 ^v 1A
· ν = 1-1−

^m2

The maximum energy stored during elastic impact according to equation 3 corresponds to the deformation work performed during plastic impact

2 *

) ²

The right side of equation 1 includes a plastic deformation of the vehicle after an impact, due to the plastic energy.

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My design is characterized by the fact that the right-hand term of equation 1 is reduced by Delta ^E pj _ast , ^ie

Delta E = Delta E ·, _t (see note)

This is achieved as follows: instead of the usual bumper attachment, springs are mounted between the bumper and the body frame. These springs are attached to the front and rear. With them I have carried out a so-called 'shock transformation', i.e.

Delta E = W&rdquor; (W= spring energy). r F

The space that performs the so-called 'transformation ¹ ' I call JÜTTEMANN SPACE (~&Pgr;).

Given the large number of spring types, I decided on the cylindrical helical compression spring with a round cross-section. You could just as well use cylindrical helical compression springs with a rectangular cross-section, or non-cylindrical helical compression springs with a round or rectangular cross-section, as well as disc springs or leaf springs. These springs are subjected to compression. It is important to ensure that the slenderness factor A does not significantly exceed a certain value so that the spring does not buckle.

The following applies: ( &mdash;.-&mdash; ) ² <--!&mdash; Av 6.935

For values of &ngr; A < 2.635 there is no danger of buckling, since in this case no real solution exists.
With v=1 it follows for A '·

A < 2.635

This condition is met because in the calculations A = 2/5.

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Spring (general)

Areas of responsibility: 1. Energy storage

2. Energy transfer

3. Energy conversion or energy destruction The requirement I have for the springs is energy conversion.

Prerequisite: The linear force law applies. Then F = es ; where c is the spring constant (spring rate)

W&rdquor; = spring tension work

- JfIs) ds = J ¹ C

it ²

csds = &mdash;

^S0

When the spring is tensioned, the spring force is opposite to the displacement, i.e.

it ²

w = - . &mdash; F= load * 2

s,f = spring travel

Spring stiffness c = f spring characteristic

(see illustration 3 spring characteristics) Helical compression springs are rotationally symmetrically wound torsion bars with a circular or rectangular cross-section. The deformations are achieved by forces acting in the direction of the spring axis (axis of rotation). The spring material is subjected to torsion. (see illustration 4) Stress distribution in the bar cross-section of a helical compression spring (schematic).

Forces and moments that occur in the bar cross-section of a helical compression spring are illustrated in the cross-section of a helical compression spring coil. (see Figure 5, Forces and moments on a helical compression spring coil).

= ora/s ; c_ = 44.44 m/s &ngr; 9-m = 40000kg; v_&rdquor; = 44.44 m/s

^{1m 1m} 2
It is m ¹ = ^ """m" ^m " ^{will be} 9 ^bigger ' Je .

bigger iru and now is

Since the bumper and therefore also the springs are exposed to less than 10,000 impacts in practice, one cannot speak of dynamic loading. This is referred to as static loading.

Calculation;

To calculate the spring, the force F that the spring is subjected to pressure must be known or calculated. The maximum force F that occurs must be taken into account. This means that |

^max3 (·

the spring is designed for the theoretical force ( F _{max th} ).

The actual shock sequence takes a fraction of a second.

Assumption: t = 0.2s
It is: F = ma or F = cs_,&rdquor; where c = const.

Discussion of this formula: F = ma

F increases if m increases ; if a = const* or

F increases when a increases; with m = const.

Since Delta ν = V ₁ - V ₂ is not as large as Delta

m = m ₁ - m _o , it follows that F_ ₌ &ldquor; at m and

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for example: m.. = wall
m ₁ » m_; m ₀ = 40000 kg ( truck ) _n

Max

Therefore: F ^u = 8888.88 kN = 9000 kN max tn.

There are also 50t or 60t mobile cranes, but these have a speed of 11 >. 1T m/s or

16.66 m/s . They do not reach F and fall

Max

thus out of consideration.

Discussion of equation 3: \

_ 1 ^m i ^m 2 , _. &ldquor; _l2

&Ggr;&Pgr;3.&KHgr;

E - 1 es ²

Pmax ^{2 max}

XI
m^ = m^ ,' m ₁ , m ₀ arbitrary, but fixed

V ₁ = 70 m/s and m ₁ = 2500kg (car)

V ₂ = 44/44 m/s and m ₂ = 40000 kg (truck)

Furthermore, Delta v ² = ( ^V _1A "V _2A ) ² , for this the following applies: The larger the difference between the two speeds, the larger the expression in brackets becomes, ie with 1* V ₁ = 70 m/s and ν _{�Ogr;�Lgr;} = γ m/s
or v _1A = 0 m/s and v _2A =70 m/s

and 2. V ₁ = 44.44 m/s and v_ = 0 m/s or V _1A = 0 m/s and v _2A =44.44 m/s

The expression in brackets is always positive, since this

is squared.

This results in the following E

As you can see, the items listed under 2. |

Values Extreme values for the bracket expression. ' " |

^{With m} max ^{and v} max ^{we get}

( 44.44 ) Nm

= 19749136 Nm = 2-10 ¹⁰ Nmm

If the velocities V ₁ and v ₂ at the collision j*

oppositely directed, i.e. that the forces |

, are also directed in opposite directions, then

I subtract the spring path of the collision partners from each other. The same applies if V ₁ and v ₂ are in the same direction, but v_ is larger than v^. If the collision partners meet at an angle « then E is smaller than E

P P

^rmax

The basic principle is: uniform power transmission

to the entire bumper width It is Delta E = E _k ; E _k = W _p (linear identifier)

^{And E} k ⁼ \ ^cs max ^{= W} F

711

Since there are 40 (20 per impact partner) equally stressed springs, "is

^E ' " * ^{40 OS}

. * ²⁰ " max

The springs are arranged in two rows with 10 springs per row.

TJI

The suspension travel is: &rgr;

2 max -vr«« , _&ngr;&pgr;}2}_ &ngr;

max 2Ö~c ^K N7mn~ '

- 2 - 10 ¹⁰

( _y J 2Ö ^T 72638,88

^{13766.73mm2 =} 117.33171mm 117.3mm

The maximum spring force is then c = 73000 N/mm

This c = 73000 N/mm can be reduced by adding a third row in addition to two rows.

^F max ^{= cs} max ^{( 1} ^ ^{N/mm }}

= 73000-117.3
= 8562900 N = 8/6-10 ⁶ N is less than 9-10 ⁶ N

^{(see F} max th. ^}

Since s _max only occurs when one of the collision partners is at rest ( ν = 0 ), the spring deflection ^s max k ^e '^ collision is reduced by s, with the corresponding mass of the collision partners and corresponding energy. It is

¹ V-. ^{= s} m= _v ^{+ L} n ' ^L n ⁼ block length of the spring

= (117.3 + 164 )mm L _Q = total length of

Spring = 281.3 mm

Selected spring length: 300 mm

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I ' i - i^. + 2*0.75 ; i _Q = total number of turns

=3+1.5 i^ = number of spring

Windings

ω 300mm

⁼ D" ⁼ 120mm ^{= 2} ' ⁵ ' ^d \ ^h * ^no risk of buckling, since 370 ^m vA ^ 2.635

A =

Slimming factor

m 12 0rpm ^ · ,■,»■,-< ^»-ij _w = -^ = ~4Öinm ⁼ ■ w = winding ratio

(see Figure 6)

Since the springs are to occupy a given arbitrary but fixed space, they must fulfill the following conditions:

( see calculation and

Illustration 6 ) 380

^L o = 310 mm D
a = 160 mm ^D m = 120 mm ^D i = 80 mm d = , 40 mm if = · 3 G = 78450 N/ ^mm2

i .&ngr; To explain, let us give an example:

i 385 Prerequisite: Central impact

! Let m ₁ = 2500 kg and V ₁ = 70 m/s

i m_ = 2500 kg and v ₂ = 0 m/s

I ' both vehicles collide. li It is then

I 390 E _n = \ -1-2- (v -v) ² = 3062500 Nm

&rgr; 2 In ₁ +m ₂ 12

Furthermore, let m ₁ = 40000 kg and V ₁ = 44.44 m/s m&ldquor; = 40000 kg and v&ldquor; = 0 m/s Then

&bull; · m &igr;

f, < · ■

XIV

1 ^m i' ^m 2 2

E = I -L-£- (v^-v^) = 19749136 Nm

^ ^m i ^+m

= 2-10 ⁷ Nm

Since there are 40 equally stressed springs, 20 per vehicle

? ₉ E

E _n = W- * 20 cs ^z -^s* = ^p

&rgr;&Ggr;

s ² = 3062500000 Nmm 1452777.77N/mm

^s2 = ^{2108.030604mm2}

s = 45.91329441mm = 46mm

2 £inax.

max 2Ö c

_2· 1o2° Nmm

T452777,77K7mm'

_smax = 117.33171mm = 117.3mm

Due to the collision between the two vehicles, s must be subtracted from s _max .

So

^{s> = s} _ma « ** ^{s =} 117,3 mm - 46 mm = 71,3 mm

&Pgr;&Igr;&agr;.&KHgr;

s ¹ is the spring deflection of the car's springs subjected to impact.

In contrast, the spring travel of the stressed springs on a truck is 46 mm. If a truck drives into a Malier/, s s is 117.3 mm, i.e. there is enough spring travel (156 mffi)

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XV

The springs in the car are designed for a speed ( v _m= ^ ) of 250 km/h.

^v max ^{for the truck}

^160km /h.

The spring should be made of .51 CrMoV4 spring steel.

I # * I 4

it*

XVI

Arrangement and design of the springs on the motor vehicle

The prerequisite for arranging the springs on the vehicle is that the bumpers are attached at the same height.

Arrangement of the springs on the vehicle:

1. The springs must be installed across the entire width of the vehicle so that they touch each other. {Minimum width of the vehicle 1600 mm).

2. The springs are mounted in two rows (or possibly three or four rows for trucks) one above the other.

3. The springs should, if possible, take up the space between the bumper and the body.

4. Spring guides must be fitted to the body and bumpers. This is useful to prevent the springs from buckling and becoming damaged.

5. In addition to the springs, gas pressure dampers can also be used, which are located inside the spring. This relieves additional load on the springs, which results in a shortening of the spring travel.

6. The remaining space is to be filled with foam. Everything together forms a unit.

5 The vehicles equipped with them therefore make a major contribution to the safety of their occupants and that of other road users.

For the extra safety of the vehicle and its occupants, the springs can also be attached to the sides in addition to the front and rear as impact protection.

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&bull; IIIIII

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Claims (1)

ft t · Claims ^Bumper for vehicles of any kind^J> for protecting the bodywork against damage, characterized in that the bumper &rgr; 5 is designed by means of springs in such a way that the impacts which occur are converted into elastic impacts by a transformation, whereby no permanent deformation of the bodywork occurs even at high speed. P 1o 2» Bumper according to claim 1, characterized in that ten contacting Q springs of a certain alloy in a Rows are arranged next to each other, each two or more rows on top of each other, with the P 15 gaps filled with plastic foam &ggr;,-earth j . 3« Bumper according to claim 1, characterized in that spring guides are attached to the body and trim of the bumper so that the springs do not buckle under load and thus the protection for the vehicle and its occupants is no longer guaranteed. hV, 4t* * t · I