Vehicle Crash Dynamics P12 pot

Since the combined mass will give a greater deflection, this term is larger than *S and, hence, the maximum deflection, * ,will be smaller for the case with spring weight.. After impact,

(6.67)

Fig 6.41 Vertical Bar in

Drop Test Fig 6.42 EquivalentSpring in Drop Test

Equating the total energy at the start and end (potential and kinetic) yields:

which is a quadratic in * and can be rearranged as follows:

Given the values of W = 5 lbs, h = 3.0 feet and k = 20 lbs/ft, *S = 5/20 = 0.25 ft Substituting this

in the above equation gives for the maximum deformation * = 1.5 feet or 6 *S.

6.5.2.2 Drop Test Using a Spring Having Finite Weight

Let us repeat the drop test, but now assume that the bar has appreciable mass, Wb, as shown in Fig 6.41 The uniform bar-mass model is represented by a spring-mass model shown in Fig 6.42 The spring mass is uniformly distributed along the length of the spring and has a finite weight, Ws =

Wb

First, it should be noted that the spring mass will contribute to the kinetic energy during deflection

of the spring, since each differential element of mass along the length will move downward with some kinetic energy during the deflection This will decrease the potential energy of the system The uppermost element of mass will move the same amount as the block, since it is immediately below it The next element will move slightly less, etc The lowest element will not move at all, since it is contiguous with the base of the spring

Second, the block, upon falling through a height and gaining speed, now impacts another mass (the spring) and the change in speed upon impacting the spring needs to be ascertained using the principle of conservation of momentum [1]

(6.69)

(6.70)

(6.71)

(6.72)

(6.73)

The derivation of the equations involved in this process is covered in an article by Spotts [2] Spotts develops an equivalent mass of the spring, accounting for the different amounts of movement

of the mass elements in the spring during deflection This equivalent mass is 1/3 of the spring mass and it is then assumed that the spring has no mass and an “effective” mass is mounted on top of the spring as a lumped mass That is, the original spring is replaced by a weightless spring and on top of this spring is attached a block having 1/3 the original weight of the spring

First, the speed of the block at the point of contacting the spring, vh, must be determined This

is obtained in the same manner as above, equating the total energy at the start (at height h) and at the end (when contacting the spring):

Next, apply the conservation of momentum to the system of block and lumped mass at the time the two make contact The spring weight will be defined as WS It will be assumed that the impact is totally inelastic (e = 0) The two masses then move as one at a common velocity of vS:

The kinetic energy of the combined masses will now be:

Note that the kinetic energy is less than that just prior to collision (Wh) This is a necessary result whenever there is an inelastic collision, where a portion of the energy is lost to heat and deformation The maximum deflection of the spring, *, can now be determined by equating the total energy at the top of the spring and at the point where the masses come to a stop at a distance * down from the top:

Equating the energies:

This, again, is a quadratic in * and can be solved in the same manner as before The result is:

where *S is the static deflection of the spring produced by the block, W (*S = W'k), and *SS is the static deflection produced by the combined weights W and WS'3

(6.75)

(6.76)

(6.77)

Eq (6.73) differs from the case without spring mass only in that the term in the denominator, *ss, appears in place of the term *S Since the combined mass will give a greater deflection, this term is larger than *S and, hence, the maximum deflection, * ,will be smaller for the case with spring weight Using the data supplied (W = 5 lb, Ws = 9 lb, and K = 20 lb/in, h = 3 ft), we obtain

The deformation with a nonzero spring weight is seen to be less than that without any spring weight: 1.25 ft versus 1.50 ft This is expected since a portion of the block initial kinetic energy is converted to heat, energy that would have otherwise been used to deform the spring

The formula for the dynamic deflection, *, can be rearranged as follows:

The expression in the bracket is referred to as the dynamic amplification factor [DAF] The minimum value of DAF is two which occurs when h = 0 Although the drop height is zero, however, due to sudden dynamic loading, the dynamic deflection is equal to two times the static deflection

G-Force Loading Due to Impact

The maximum deceleration, a, due to the impact occurs at the maximum deflection, *

6.5.2.3 Horizontal Impact on a Bar/Spring

The initial horizontal impact velocity of the weight on the bar is va, as shown in Fig.6.43 After impact, the combined impactor weight and the effective bar (or spring) weight move at the common velocity of vS (see the spirng equivalent of the bar shown in Fig 6.44) The maximum deflection can then be computed:

Fig 6.43 A Bar Struck By a

Weight Fig 6.44 Spring Equivalent of aStruck Bar

(6.78)

Fig 6.45 Impact on a Horizontal

Beam Equivalent of a StruckFig 6.46 Spring

Horizontal Bar

6.5.2.4 Vertical Impact on a Beam/Spring

It can also be shown that in a vertical drop test where a block impacts on a horizontal free-supported beam, the effective spring weight is 17/35 of the weight of the bar (Figs 6.45 and 6.46) Compared to the impact on the vertical bar (or spring), a smaller portion of the initial kinetic energy

is converted to deform the horizontal beam

6.5.3 Rebound Criterion in a Two-Mass Impact

Two test setups, shown in Figs 6.47 and 6.48, are used to analyze the separation kinematics between the impactor and component In both cases, one vertical and the other horizontal, we will derive the relationship between the mass ratio and coefficient of restitution such that immediately after separation, both objects move in the same direction

Fig 6.47 A Drop Tower Test

Fig 6.48 Two Car Collision and

Separation

(6.79)

The drop height of an impactor, m1, is h, as shown in Fig 6.47 The horizontal impact velocity

is v1, as shown in Fig 6.48 e is defined as the coefficient of restitution (COR)

Find: the separation velocity, v1', of m1 after impact and with the condition of not rebounding Let v1: initial impact velocity

vc: common velocity of m1 and m2 after contact

v1N, v2N: rebound velocities of m1 and m2, respectively

The separation velocity ratio shown in (3) of Eq (6.79) is plotted and shown in Fig 6.49 The shaded section in the plot is bounded by both the mass ratio and coefficient of restitution ranging from

0 to 1 In this shaded section, the separation velocity ratio is negative

Case Study(exercise) In the example in Section 6.5.2.2 on the “Drop Test Using a Spring Having

Finite Weight,” we assumed that the striking mass, W, would stick with the struck mass, WS, until the time that both masses reach the maximum deformation, * To ensure that the striking mass does not rebound upward at the time of separation, what is the maximum e (coefficient of restitution) allowed between the two masses? [Ans.: e #3W/WS]

Fig 6.49 Two-Particle Impact Rebound Condition

Fig 6.50 Middle Car Rear-Ended Twice

6.5.4 Separation Kinematics in a Multi-Mass Impact

In laboratory component impact tests, such as those of windshield breakage or body mounts, it

is imperative to be able to reproduce the transient decelerations observed in actual crash tests However, unwanted signals often appear in the component tests As an example, repetitive loading due to multiple impacts between the striking object (impactor) and the struck object often produces

an additional signal following that due to the first impact

The problem of repetitive loading is often caused by a mismatch of the coefficients of restitution (COR) and the mass ratios among the impactor, test component, and supporting fixture To control the repetitive loading on an object, the separation kinematics of that object has to be understood first The relative values of the mass ratios and the COR values can then be adjusted to minimize any repetitive loading signals that may occur

In the case of a vehicle collision, repetitive loading due to multiple impacts can cause additional injury to an occupant in an accident As an example, let us assume that a car (Car #2), right behind another car (Car #3), which is stopped at an intersection, is rear-ended by a third car (Car #1) as shown

in Fig 6.50 Depending on the weights of the cars and the coefficients of restitution of the structures that are engaged, the driver in Car #2, the middle one of the three cars, may suffer a more severe neck injury due to double whiplashes Assume that the masses of the three vehicles are the same (M = 1,

1, 1), the COR’s between #1 and #2 and between #2 and #3 are the same, 0.15, and the initial impact speed of Car #1 is 35 mph

Before performing the computation of the separation kinematics, let us examine what happens to the middle car, Car #2 The first whiplash occurs in the first impact, when it is rear-ended by Car #1

at a speed of 35 mph In this impact, the velocity change of Car #2 is 20.1 mph The percentage of initial kinetic energy dissipated by the structure is 48.9 % (shown on the right side of plot) at the end

of the first impact In the second impact, Car #2 hits Car #3 at a speed of 20.1 mph Since Car #2 is slowed down to 8.6 mph, which is the separation velocity of Car #2 after the second impact, it gets rear-ended a second time by Car #1, which is traveling at 14.9 mph The second rear-ended velocity change of car #2 is 3.6 mph (= 12.2 ! 8.6) Following this third impact, Car #2 picks up speed and hits Car #3 again This fourth impact is the last one Afterwards, the car in the front (Car #3) is moving away from Car #2 and Car #2 moves away from Car #1 In this example, a total of four collisions occur and the drivers in Car #2 and #3 both receive double whiplashes

6.5.4.1 Separation Kinematics in a 3-Vehicle Collision

The separation kinematics of the cars involved can be solved by successively applying the equations for momentum and the coefficient of restitution at the times of impact and separation for each of the impacts Eq (6.80) shows the derivation and the relationship between the two coefficients

of restitution such that at least the middle car receives double whiplashes

Equation (5) shown in the Eq (6.80) specifies the relationship between the two coefficients of restitution such that second car is subjected to multiple rear impacts Note that this condition does not depend on the impact speed

For the data shown in Fig 6.50, the two mass ratios are equal to one, and the two COR’s are 0.15; substituting these numbers into (5) of Eq (6.80), one gets 0.15 > !0.48 Therefore, the condition for the middle car subjected to multiple rear impacts, as shown in Fig 6.50, is satisfied However, if the middle car is twice as heavy as each of the other two cars, then r1 becomes 0.5, and r2 becomes 2 Computing the values of both sides of (5) of Eq (6.80), it is found 0.15 Ý 0.78 Since the multiple impact condition is not met, car #2 receives only one rear impact as shown in Fig 6.51 The only one rear-ended velocity change of car #2 is 13.4 mph, smaller than 20.1 mph shown before in Fig 6.50 The percentage of initial kinetic energy loss (%E loss) by the structure is 65.2% (on the right column

of the plot) at the end of the second impact event This points out the advantage of driving a larger car, as the driver in such a car is subjected to one whiplash only in the case described above

Fig 6.51 Middle Car (with larger weight)

Rear-Ended Once

Fig 6.52 Middle Car (with High Rear COR)

Rear-Ended Once

The middle car does not have to be very heavy to receive only one rear impact If the section of the structure engaging in the impact between cars #1 and #2 is such that it has a COR of 55, the middle car would be subjected to only one rear impact as shown in Fig.6.52 This is because when r1 = r2 =1 and e2 = 15, the maximum value of e1 for Car #2 to have multiple rear impacts can be derived by using the condition shown in (5) of Eq (6.80) The condition is e1 < 0.4 Therefore, a value of e1 = 55, not satisfying the limit condition, would insure that Car #2 would have only one rear impact The velocity change of Car #2 is 27.1 mph, higher than the previous two cases This is because the percentage of initial kinetic energy dissipated by the structure is only 34.9% (on the right side of plot) at the end of second impact

6.5.5 COR, Times of Dynamic Crush, and Separation Time

A simple analytical relationship based on the relative separation velocity and relative approach velocity can be derived for the coefficient of restitution, e Given two timings obtained from a crash test, tm and tf, e can then be computed

Let define

a: Constant deceleration or ASW (Average Square Wave) v: Velocity change up to tm

tm: Time of (maximum) dynamic crush tf: Final separation time

e: Coefficient of restitution

Fig 6.53 Velocity Changes in the

Deformation and Rebound Phases

(6.81)

Either in the vehicle-to-(fixed rigid) barrier (VTB) test or in the vehicle-to-vehicle (VTV) test, the relative acceleration of the two objects in the deformation phase can be approximated by an average square wave as shown in Fig 6.53 The area under the square wave is the relative approach velocity The vehicle deceleration in the restitution phase is approximated by a triangle, and the area

is the rebound or relative separation velocity Eq (6.81) shows the derivation and relationship between the three parameters, tm, tf, and e

Using available crash test results for cars and trucks in rigid barrier and vehicle-to-vehicle crashes, a summary of the two timings and the coefficients of restitution are shown in Table 6.2

Table 6.2 Relationship Between tf, tm, and e

Crash Mode e tf/tm = 1+2e tf/tm from

Crash Tests

tm, ms, from Crash Tests

Vehicle-to-Vehicle

in line

6.5.6 Coefficient of Restitution and Stiffness in Vehicle Crashes

This section deals with the determination of e, the coefficient of restitution, for a two-car collision

using car-barrier data A study made by Prasad [3] investigated the coefficient of restitution for vehicle structures and its application in estimating velocity changes in vehicle collisions

The coefficient of restitution, e, for a two-car central collision can be obtained using data from

car-barrier collisions The value obtained will be an approximation since it assumes that the amount

of energy each car loses is the same in both the car-to-car and the barrier collisions The values of k,

the slope of the force-deflection curve for each barrier crash, must also be known

First, let us obtain a measure of the lost energy in a barrier crash as it relates to e The definition

of e is:

(6.83)

(6.84)

(6.85)

Here, a fixed reference system is assumed (relative to the earth) Then, the car to the left has an initial speed of v1; after the collision, its speed is v1N The other car, or barrier if one is used, has a speed of v2 and then v2N, as shown in Fig 6.48

Note that e is invariant to the reference system used, as long as the system is moving at a constant

speed If one uses a system moving at a speed w with respect to the fixed system, each velocity term

in the equation would be less by the amount w Since each relative speed will have two w’s, one

positive and one negative, these would cancel, leaving the relative speed unchanged Thus, e will stay

the same for any reference moving at a constant speed

As has been shown earlier, at some time during the collision, the two cars will have the same speed, called the common speed, vc At this point, the maximum deformation between two cars occurs and it is at the end of deformation phase (or loading phase) Beyond this point, the two cars will start

to unload and separate, releasing any elastic rebound (potential) energy until the two cars are totally apart from each other

For a car-fixed barrier collision, assuming a coordinate system at rest with respect to the barrier,

we see that v2 is zero and v2r will also be zero (since the mass of the barrier is assumed to be infinite) Then:

If e equals 1.0, the kinetic energy after collision equals that before; i.e., there is no loss If e

equals 0.0, the kinetic energy after collision is zero; i.e., the car is at rest, resting against the barrier Also, the initial kinetic energy is equal to the stored (i.e., available or elastic rebound) energy obtained during the collision

Now, consider a two-car central collision, with a moving reference system which moves at the common velocity, vC At the time the speeds of the two cars are equal, their speed will be the common speed In the common speed reference system used, their speeds will be zero After separation, their speeds will be v1N and v2N Applying the conservation of momentum in this reference system:

The kinetic energy after separation will be equal to the rebound energy, i.e., the energy released

by each car during the unloading (or restitution) phase This rebound energy is retrievable upon separation of the two cars As was noted earlier, this energy in the case of a car-barrier collision is

e2 × (Initial kinetic energy) Assuming the rebound energy in each car in the case of a car-to-car collision is equal to that obtained when the car collides with a fixed barrier, we can equate the rebound energy in the car-barrier crash and the final kinetic energy in the car-to-car crash Then:

The values of )E1 and )E2 can be obtained from Eqs (7.34) and (7.35) in Section 7.4 of Chapter

7 on central collisions: